4.6: Pretérito con cambios de raíz
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Objetivos
Identify the correct conjugations of stem-changing verbs in the preterit tense
Do you remember the three groups of verbs that change their stem vowel when it’s stressed? They were presented in section 5.4 of the Español 1 book. If you don’t remember, you may want to review how these verbs work in the present tense before moving on with this chapter’s topic: stem-changers in the preterit tense.
Stem-changing -AR and -ER verbs in the preterit
-AR and -ER verbs that have a stem change in the present tense, don’t have a stem change in the preterit tense. For example:
Stem-changing -IR verbs in the preterit
However, the -IR stem-changing verbs * do * have a change in the preterit: but it’s only in the third person (singular and plural), and -IR stem-changers will change their stem vowel to just the first letter of the stem-change they had in the present tense.
So, -IR verbs that changed from “e” to “ie” in the present will change just from “e” to “i” in the preterit. -IR verbs that changed from “o” to “ue” in the present will change just from “o” to “u” in the preterit. And -IR verbs that changed from “e” to “i” in the present will do just that in the preterit.
Conjugation of stem-changing -IR verbs in the preterit
- ¿Pediste unas prendas de ropa de Zara.com? ( Did you order/request some clothing from Zara.com? )
- No, pero Luisa pidió mucha ropa de ese sitio. ( No, but Luisa ordered lots of clothes from that site. )
- Ella se vistió de pies a cabeza en ropa de Zara.com. ( She dressed from head to toe in clothes from Zara.com. )
The best way to learn these conjugations is to say them out loud until they sound right to you. Practice, and train your ear! Once you hear the endings and changes right in your head, you’ll be able to spell them right on the page.


5: Air Temperature
The temperature of our atmosphere is controlled by a complex set of interactions between the biosphere, lithosphere, hydrosphere, and atmosphere. Energy is constantly exchanged between the surface and the air above, as well as circulating around the globe. Here we'll look at what controls the air temperature at a particular place by examining radiation and energy exchanges between the earth and air above. Then we'll examine how the global circulation of air and water affects air temperature.
By the end of the chapter you should be able to:
- Define temperature and describe how air temperature is measured.
- Explain what controls daily and seasonal temperature variations.
- Determine the temperature gradient using isotherms.
- Calculate average daily and annual temperature, daily and annual temperature range, determine seasonal temperature lag.
- Explain the global pattern of air temperature.
- Summarize the effects of global warming on temperature patterns.
See if you are ready for this chapter by Getting Ready for Chapter 5: Air Temperature.


6.3: Proof by mathematical induction II
Even where there is no confusion between mathematical induction and ‘scientific induction', students often fail to appreciate the essence of “proof by induction”. Before illustrating this, we repeat the basic structure of such a proof.
A mathematical result, or formula, often involves a parameter n , where n can be any positive integer. In such cases, what is presented as a single result, or formula, is a short way of writing an infinite family of results. The proof that such a result is correct therefore requires us to prove infinitely many results at once. We repeat that our only hope of achieving such a mind–boggling feat is
- to formulate the stated result for each value of n separately: that is, as a statement P ( n ) which depends on n ;
- then to use bare hands to check the “beginning” – namely that the simplest case (usually P (1)) is true;
- finally to find some way of demonstrating that,
– as soon as we know that P ( k ) is true, for some (unknown) ,
– we can prove that the next result is then automatically true
We can then conclude that
“ every statement P ( n ) is true”,
or that
“ P ( n ) is true, for all ”
Problem 229 Prove the statement:
“5 2n+2 – 24n – 25 is divisible by 576, for all ”.
When trying to construct proofs in private, one is free to write anything that helps as ‘rough work'. But the intended thrust of Problem 229 is two–fold:
-
to introduce the habit of distinguishing clearly between
(i) the statement P ( n ) for a particular n, and
(ii) the statement to be proved – namely “P( n ) is true, for all ”; and
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to draw attention to the “induction step” (i.e. the third bullet point above), where
(i) we assume that some unspecified P ( k ) is known to be true, and
(ii) seek to prove that the next statement must then be true.
The central lesson in completing the “induction step” is to recognize that:
to prove that is true, one has to start by looking at what says.
In Problem 229 says that
“ is divisible by 576”.
Hence one has to start the induction step with the relevant expression
and look for some way to rearrange this into a form where one can use P ( k ) (which we assume is already known to be true, and so are free to use).
It is in general a false strategy to work the other way round – by “starting with P ( k ), and then fiddling with it to try to get ”. (This strategy can be made to work in the simplest cases; but it does not work in general, and so is a bad habit to get into.) So the induction step should always start with the hypothesis of .
The next problem invites you to prove the formula for the sum of the angles in any polygon. The result is well–known; yet we are fairly sure that the reader will never have seen a correct proof. So the intention here is for you to recognise the basic character of the result, to identify the flaws in what you may until now have accepted as a proof, and to try to find some way of producing a general proof.
Problem 230 Prove by induction the statement:
“for each , the angles of any n –gon in the plane have sum equal to radians.”
The formulation certainly involves a parameter ; so you clearly need to begin by formulating the statement P ( n ). For the proof to have a chance of working, finding the right formulation involves a modest twist! So if you get stuck, it may be worth checking the first couple of lines of the solution.
No matter how P ( n ) is formulated, you should certainly know how to prove the statement P (3) (essentially the formula for the sum of the angles in a triangle). But it is far from obvious how to prove the “induction step”:
“if we know that P ( k ) is true for some particular , then must also be true”.
When tackling the induction step, we certainly cannot start with P ( k )! The statement says something about polygons with sides: and there is no way to obtain a typical –gon by fiddling with some statement about polygons with k sides. (If you start with a k–gon, you can of course draw a triangle on one side to get a –gon; but this is a very special construction, and there is no easy way of knowing whether all –gons can be obtained in this way from some k –gon.) The whole thrust of mathematical induction is that we must always start the induction step by thinking about the hypothesis of – that is in this case, by considering an arbitrary –gon and then finding some guaranteed way of “reducing” it in order to make use of P ( k ).
The next two problems invite you to prove some classical algebraic identities. Most of these may be familiar. The challenge here is to think carefully about the way you lay out your induction proof, to learn from the examples above, and (later) to learn from the detailed solutions provided.
Problem 231 Prove by induction the statement:
“ holds, for all ”
The summation in Problem 231 was known to the ancient Greeks. The mystical Pythagorean tradition (which flourished in the centuries after Pythagoras) explored the character of integers through the ‘spatial figures’ which they formed. For example, if we arrange each successive integer as a new line of dots in the plane, then the sum “ ” can be seen to represent a triangular number. Similarly, if we arrange each odd number in the sum “ ” as a “ k -by- k reverse L-shape”, or gnomon (a word which we still use to refer to the L-shaped piece that casts the shadow on a sundial), then the accumulated L-shapes build up an n by n square of dots - the “1” being the dot in the top left hand corner, the “3” being the reverse L-shape of 3 dots which make this initial “1” into a 2 by 2 square, the “5” being the reverse L-shape of 5 dots which makes this 2 by 2 square into a 3 by 3 square, and so on. Hence the sum “ ” can be seen to represent a square number.
There is much to be said for such geometrical illustrations; but there is no escape from the fact that they hide behind an ellipsis (the three dots which we inserted in the sum between “5” and “ ”, which were then summarised when arranging the reverse L-shapes by ending with the words “and so on”). Proof by mathematical induction, and its application in Problem 231 , constitute a formal way of avoiding both the appeal to pictures, and the hidden ellipsis .
Problem 232 The sequence
is defined by its first two terms , and by the recurrence relation:
(a) Guess a closed formula for the n th term u n .
(b) Prove by induction that your guess in (a) is correct for all
Problem 233 The sequence of Fibonacci numbers
is defined by its first two terms , and by the recurrence relation:
when
Prove by induction that, for all
Problem 234 Prove by induction that
is divisible by 19, for all
Problem 235
Use mathematical induction to prove that each of these identities holds, for all
(a)
(b)
(c)
(d)
(e)
Problem 236 Prove by induction the statement:
“ , for all ”.
We now know that, for all
n terms ) = n .
And if we sum these “outputs” (that is, the first n natural numbers), we get the n th triangular number:
.
The next problem invites you to find the sum of these “outputs”: that is, to find the sum of the first n triangular numbers.
Problem 237
(a) Experiment and guess a formula for the sum of the first n triangular numbers:
(b) Prove by induction that your guessed formula is correct for all
A We now know closed formulae for
“ ”
and for
“ ”.
The next problem hints firstly that these identities are part of something more general, and secondly that these results allow us to find identities for the sum of the first n squares:
for the first n cubes:
and so on.
Problem 238
(a) Note that
.
Use this and the result of Problem 237 to derive a formula for the sum:
.
(b) Guess and prove a formula for the sum
.
Use this to derive a closed formula for the sum:
.
It may take a bit of effort to digest the statement in the next problem. It extends the idea behind the “method of undetermined coefficients” that is discussed in Note 2 to the solution of Problem 237 (a).
Problem 239
(a) Given distinct real numbers
find all possible polynomials of degree n which satisfy
for some specified number b .
(b) For each
Given two labelled sets of real numbers
,
and
,
where the a i are all distinct (but the b need not be), there exists a polynomial f n of degree n , such that
We end this subsection with a mixed bag of three rather different induction problems. In the first problem the induction step involves a simple construction of a kind we will meet later.
Problem 240 A country has only 3 cent and 4 cent coins.
(a) Experiment to decide what seems to be the largest amount, N cents, that cannot be paid directly (without receiving change).
(b) Prove by induction that “ n cents can be paid directly for each ”.
Problem 241
(a) Solve the equation
. Calculate
, and check that
(b) Solve the equation
. Calculate
(c) Solve the equation
. Calculate
(d) Solve the equation
, where k is an integer. Calculate z
2
, and check that
(e) Prove that if a number
z
has the property that
is an integer, then
Problem 242 Let p be any prime number. Use induction to prove:
“ is divisible by p for all ”.


4.3: YouTube
Beginning of YouTube
YouTube was created in 2005 and originally was intended to be a dating website . Today YouTube is the biggest video sharing website known and used around the globe. YouTube’s mission is to give everyone a chance to make their voice heard across the world. This mission also includes allowing the content creators to use their freedom of speech and expression actively. To create the most personalized environment for its users, YouTube collects data over time to recommend the most relevant videos for them. YouTube collects data from its users by putting their information into an algorithm to figure out what is most relevant for users to view next. YouTube was only around for one year before Google decided to buy them. Google created Google AdSense , a program designed to put ads before, during, and after the videos based on the interests of the specific viewer. These advertisements are a significant part in funding for both YouTube and their content creators, which allows for consistent uploading of a seemingly infinite amount of videos on the site.
YouTube and the Media
YouTube has become a huge resource for people to access all kinds of content, including the news. The site has verified channels that a user can turn to get their news or see coverage of news that was seen on a different platform. This way of having access to video or news coverage from not only yesterday, or a week ago, but even multiple years ago, provides the user with access to information possibly not found elsewhere. The site also has a “ Trending ” page with videos that YouTube determines as the videos that are garnering views at the fastest rate. The videos on this page tend to gather the most attention because they are easily accessible. YouTube, while still trying to allow users to freely express themselves, moderates videos that are posted as well as comments on videos. The company has a ‘ Community Guidelines ’ page that discusses what kind of videos should not be uploaded.  Despite having these guidelines, the company relies heavily on users to report or flag content that they feel is not appropriate.
YouTube: A Valid Source?
There are a few concerns and critiques of YouTube that should be addressed. One main concern of YouTube is that the anonymity on the site can make it easier for people to blindly hate or bully someone while keeping their real identity a secret. Another issue is that the recommendations provided by the algorithm tend to push viewers toward more extreme content. This can amplify extremist political material and has led to children’s programming that is groteque and disturbing .
A common critique when it comes to YouTube is the copyright claims for certain videos, clips, and songs used in videos that restrict the amount of money YouTubers make. YouTube has made it so these companies or corporations can easily claim copyright and this can cause certain videos to be removed or blocked. The YouTuber may even have to pay the corporation or company for copyright claims when the YouTuber didn’t really do anything wrong. One way in which YouTube could solve this major issue would be to make it so that YouTube requires proof that copyright infringement is actually happening before a video gets shut down or blocked.
Recommended Reading
“ The Battle Between YouTube and the Mainstream Media ” by Joe Bish, Vice , May 3, 2017.
“ YouTube’s Copyright Strikes Have Become a Tool for Extortion ” by Shoshana Wodinsky, The Verge, Feb. 11, 2019.


12: Concepts and Tools
We introduce here the concepts and tools that are useful in studying nonlinear dynamical systems.
Fixed points and linear stability analysis
Consider the one-dimensional differential equation for \(x=x(t)\) given by
\[\dot{x}=f(x) \nonumber \]
We say that \(x_{*}\) is a fixed point, or equilibrium point, of \((12.1)\) if \(f\left(x_{*}\right)=0\) , so that at a fixed point, \(\dot{x}=0 .\) The name fixed point is apt since the solution to \((12.1)\) with initial condition \(x(0)=x_{*}\) is fixed at \(x(t)=x_{*}\) for all time \(t\) .
A fixed point, however, can be stable or unstable. A fixed point is said to be stable if a small perturbation decays in time; it is said to be unstable if a small perturbation grows in time.
We can determine stability by a linear analysis. Let \(x=x_{*}+\epsilon(t)\) , where \(\epsilon\) represents a small perturbation of the solution from the fixed point \(x_{*} .\) Because \(x_{*}\) is a constant, \(\dot{x}=\dot{\epsilon} ;\) and because \(x_{*}\) is a fixed point, \(f\left(x_{*}\right)=0\) . Taylor series expanding about \(\epsilon=0\) , we have
\[\begin{aligned} \dot{\epsilon} &=f\left(x_{*}+\epsilon\right) \\ &=f\left(x_{*}\right)+\epsilon f^{\prime}\left(x_{*}\right)+\ldots \\ &=\epsilon f^{\prime}\left(x_{*}\right)+\ldots \end{aligned} \nonumber \]
The omitted terms in the Taylor series expansion are proportional to \(\epsilon^{2}\) , and can be made negligible-at least over a short time interval-by taking \(\epsilon(0)\) small. The differential equation to be considered, \(\dot{\epsilon}=f^{\prime}\left(x_{*}\right) \epsilon\) , is therefore linear, and has the solution
\[\epsilon(t)=\epsilon(0) e^{f^{\prime}\left(x_{*}\right) t} \nonumber \]
The perturbation of the fixed point solution \(x(t)=x_{*}\) thus decays or grows exponentially depending on the sign of \(f^{\prime}\left(x_{*}\right)\) . The stability condition on \(x_{*}\) is therefore
\[x_{*} \text { is } \begin{cases}\text { a stable fixed point if } & f^{\prime}\left(x_{*}\right)<0, \\ \text { an unstable fixed point if } & f^{\prime}\left(x_{*}\right)>0 .\end{cases} \nonumber \]
For the special case \(f^{\prime}\left(x_{*}\right)=0\) , we say the fixed point is marginally stable. We will see that bifurcations usually occur at parameter values where fixed points become marginally stable.
Difference equations, or maps, may be similarly analyzed. Consider the one-dimensional map given by
\[x_{n+1}=f\left(x_{n}\right) \text {. } \nonumber \]
We say that \(x_{n}=x_{*}\) is a fixed point of the map if \(x_{*}=f\left(x_{*}\right) .\) The stability of this fixed point can then be determined by writing \(x_{n}=x_{*}+\epsilon_{n}\) so that \((12.4)\) becomes
\[\begin{aligned} x_{*}+\epsilon_{n+1} &=f\left(x_{*}+\epsilon_{n}\right) \\ &=f\left(x_{*}\right)+\epsilon_{n} f^{\prime}\left(x_{*}\right)+\ldots \\ &=x_{*}+\epsilon_{n} f^{\prime}\left(x_{*}\right)+\ldots . \end{aligned} \nonumber \]
Therefore, to leading-order in \(\epsilon\) ,
\[\left|\frac{\epsilon_{n+1}}{\epsilon_{n}}\right|=\left|f^{\prime}\left(x_{*}\right)\right| \text {, } \nonumber \]
and the stability condition on \(x_{*}\) for a one-dimensional map is
\[x_{*} \quad \text { is } \quad \begin{cases}\text { a stable fixed point if } & \left|f^{\prime}\left(x_{*}\right)\right|<1 \\ \text { an unstable fixed point if } & \left|f^{\prime}\left(x_{*}\right)\right|>1\end{cases} \nonumber \]
Here, marginal stability occurs when \(\left|f^{\prime}\left(x_{*}\right)\right|=1\) , and bifurcations usually occur at parameter values where the fixed point becomes marginally stable. If \(f^{\prime}\left(x_{*}\right)=0\) , then the fixed point is called superstable. Perturbations to a superstable fixed point decay especially fast, making numerical calculations at superstable fixed points more rapidly convergent.
The tools of fixed point and linear stability analysis are also applicable to higher-order systems of equations. Consider the two-dimensional system of differential equations given by
\[\dot{x}=f(x, y), \quad \dot{y}=g(x, y) . \nonumber \]
The point \(\left(x_{*}, y_{*}\right)\) is said to be a fixed point of \((12.7)\) if \(f\left(x_{*}, y_{*}\right)=0\) and \(g\left(x_{*}, y_{*}\right)=0 .\) Again, the local stability of a fixed point can be determined by a linear analysis. We let \(x(t)=x_{*}+\epsilon(t)\) and \(y(t)=y_{*}+\delta(t)\) , where \(\epsilon\) and \(\delta\) are small independent perturbations from the fixed point. Making use of the two dimensional Taylor series of \(f(x, y)\) and \(g(x, y)\) about the fixed point, or equivalently about \((\epsilon, \delta)=(0,0)\) , we have
\[\begin{aligned} \dot{\epsilon} &=f\left(x_{*}+\epsilon, y_{*}+\delta\right) \\ &=f_{*}+\epsilon \frac{\partial f_{*}}{\partial x}+\delta \frac{\partial f_{*}}{\partial y}+\ldots \\ &=\epsilon \frac{\partial f_{*}}{\partial x}+\delta \frac{\partial f_{*}}{\partial y}+\ldots, \\ \dot{\delta} &=g\left(x_{*}+\epsilon, y_{*}+\delta\right) \\ &=g_{*}+\epsilon \frac{\partial g_{*}}{\partial x}+\delta \frac{\partial g_{*}}{\partial y}+\ldots \\ &=\epsilon \frac{\partial g_{*}}{\partial x}+\delta \frac{\partial g_{*}}{\partial y}+\ldots, \end{aligned} \nonumber \]
where in the Taylor series \(f_{*}, g_{*}\) and the similarly marked partial derivatives all denote functions evaluated at the fixed point. Neglecting higher-order terms in the Taylor series, we thus have a system of odes for the perturbation, given in matrix form as
\[\frac{d}{d t}\left(\begin{array}{c} \epsilon \\ \delta \end{array}\right)=\left(\begin{array}{cc} \partial f_{*} / \partial x & \partial f_{*} / \partial y \\ \partial g_{*} / \partial x & \partial g_{*} / \partial y \end{array}\right)\left(\begin{array}{c} \epsilon \\ \delta \end{array}\right) \nonumber \]
The two-by-two matrix in (12.8) is called the Jacobian matrix at the fixed point. An eigenvalue analysis of the Jacobian matrix will typically yield two eigenvalues \(\lambda_{1}\) and \(\lambda_{2}\) . These eigenvalues may be real and distinct, complex conjugate pairs, or repeated. The fixed point is stable (all perturbations decay exponentially) if both eigenvalues have negative real parts. The fixed point is unstable (some perturbations grow exponentially) if at least one eigenvalue has a positive real part. Fixed points can be further classified as stable or unstable nodes, unstable saddle points, stable or unstable spiral points, or stable or unstable improper nodes.
Bifurcations
For nonlinear systems, small changes in the parameters of the system can result in qualitative changes in the dynamics. These qualitative changes are called bifurcations. Here we consider four classic bifurcations of one-dimensional nonlinear differential equations: the saddle-node bifurcation, the transcritical bifurcation, and the supercritical and subcritical pitchfork bifurcations. The differential equation that we will consider is in general written as
\[\dot{x}=f_{r}(x), \nonumber \]
where the subscript \(r\) represents a parameter that results in a bifurcation when varied across zero. The simplest differential equations that exhibit these bifurcations are called the normal forms, and correspond to a local analysis (i.e., Taylor series expansion) of more general differential equations around the fixed point, together with a possible rescaling of \(x\) .
Saddle-node bifurcation
The saddle-node bifurcation results in fixed points being created or destroyed. The normal form for a saddle-node bifurcation is given by
\[\dot{x}=r+x^{2} . \nonumber \]
The fixed points are \(x_{*}=\pm \sqrt{-r}\) . Clearly, two real fixed points exist when \(r<0\) and no real fixed points exist when \(r>0\) . The stability of the fixed points when \(r<0\) are determined by the
derivative of \(f(x)=r+x^{2}\) , given by \(f^{\prime}(x)=2 x .\) Therefore, the negative fixed point is stable and the positive fixed point is unstable.
We can illustrate this bifurcation. First, in Fig. 12.1(a), we plot \(\dot{x}\) versus \(x\) for the three parameter values corresponding to \(r<0, r=0\) and \(r>0\) . The values at which \(\dot{x}=0\) correspond to the fixed points, and arrows are drawn indicating how the solution \(x(t)\) evolves (to the right if \(\dot{x}>0\) and to the left if \(\dot{x}<0\) ). The stable fixed point is indicated by a filled circle and the unstable fixed point by an open circle. Note that when \(r=0\) , solutions converge to the origin from the left, but diverge from the origin on the right.
Second, we plot the standard bifurcation diagram in Fig. 12.1(b), where the fixed point \(x_{*}\) is plotted versus the bifurcation parameter \(r\) . As is the custom, the stable fixed point is denoted by a solid line and the unstable fixed point by a dashed line. Note that the two fixed points collide and annihilate at \(r=0\) , and there are no fixed points for \(r>0\) .
Transcritical bifurcation
A transcritical bifurcation occurs when there is an exchange of stabilities between two fixed points. The normal form for a transcritical bifurcation is given by
\[\dot{x}=r x-x^{2} . \nonumber \]
The fixed points are \(x_{*}=0\) and \(x_{*}=r\) . The derivative of the right-hand-side is \(f^{\prime}(x)=r-2 x\) , so that \(f^{\prime}(0)=r\) and \(f^{\prime}(r)=-r\) . Therefore, for \(r<0, x_{*}=0\) is stable and \(x_{*}=r\) is unstable, while for \(r>0, x_{*}=r\) is stable and \(x_{*}=0\) is unstable. The two fixed points thus exchange stability as \(r\) passes through zero. The transcritical bifurcation is illustrated in Fig. 12.2.
Pitchfork bifurcations
The pitchfork bifurcations occur when one fixed point becomes three at the bifurcation point Pitchfork bifurcations are usually associated with the physical phenomena called symmetry breaking. Pitchfork bifurcations come in two types. In the supercritical pitchfork bifurcation, the stability of the original fixed point changes from stable to unstable and a new pair of stable fixed points are created above (super-) the bifurcation point. In the subcritical bifurcation, the stability of the original fixed point again changes from stable to unstable but a new pair of now unstable fixed points are created below (sub-) the bifurcation point.
Supercritical pitchfork bifurcation
The normal form for the supercritical pitchfork bifurcation is given by
\[\dot{x}=r x-x^{3} \nonumber \]
Note that the linear term results in exponential growth when \(r>0\) and the nonlinear term stabilizes this growth. The fixed points are \(x_{*}=0\) and \(x_{*}=\pm \sqrt{r}\) , the latter fixed points existing only when \(r>0\) . The derivative of \(f\) is \(f^{\prime}(x)=r-3 x^{2}\) so that \(f^{\prime}(0)=r\) and \(f^{\prime}(\pm \sqrt{r})=-2 r\) . Therefore, the fixed point \(x_{*}=0\) is stable for \(r<0\) and unstable for \(r>0\) while the fixed points \(x=\pm \sqrt{r}\) exist and are stable for \(r>0 .\) Notice that the fixed point \(x_{*}=0\) becomes unstable as \(r\) crosses zero and two new stable fixed points \(x_{*}=\pm \sqrt{r}\) are born. The supercritical pitchfork bifurcation is illustrated in Fig. 12.3.
The pitchfork bifurcation illustrates the physics of symmetry breaking. The differential equation (12.12) is invariant under the transformation \(x \rightarrow-x\) . Fixed point solutions of this equation that obey this same symmetry are called symmetric; fixed points that do not are called asymmetric. Here, \(x_{*}=0\) is the symmetric fixed point and \(x=\pm \sqrt{r}\) are asymmetric. Asymmetric fixed points always occur in pairs, and mirror each others stability characteristics. Only the initial conditions determine which asymmetric fixed point is asymptotically attained.
Subcritical pitchfork bifurcation
In the subcritical case, the cubic term is destabilizing. The normal form (to order \(x^{3}\) ) is
\[\dot{x}=r x+x^{3} \nonumber \]
The fixed points are \(x_{*}=0\) and \(x_{*}=\pm \sqrt{-r}\) , the latter fixed points existing only when \(r \leq 0\) . The derivative of the right-hand-side is \(f^{\prime}(x)=r+3 x^{2}\) so that \(f^{\prime}(0)=r\) and \(f^{\prime}(\pm \sqrt{-r})=-2 r .\) Therefore, the fixed point \(x_{*}=0\) is stable for \(r<0\) and unstable for \(r>0\) while the fixed points \(x=\pm \sqrt{-r}\) exist and are unstable for \(r<0\) . There are no stable fixed points when \(r>0\) .
The absence of stable fixed points for \(r>0\) indicates that the neglect of terms of higher-order in \(x\) than \(x^{3}\) in the normal form may be unwarranted. Keeping to the intrinsic symmetry of the equations (only odd powers of \(x\) so that the equation is invariant when \(x \rightarrow-x\) ) we can add a stabilizing nonlinear term proportional to \(x^{5}\) . The extended normal form (to order \(x^{5}\) ) is
\[\dot{x}=r x+x^{3}-x^{5}, \nonumber \]
and is somewhat more difficult to analyze. The fixed points are solutions of
\[x\left(r+x^{2}-x^{4}\right)=0 . \nonumber \]
The fixed point \(x_{*}=0\) exists for all \(r\) , and four additional fixed points can be found from the solutions of the quadratic equation in \(x^{2}\) :
\[x_{*}=\pm \sqrt{\frac{1}{2}(1 \pm \sqrt{1+4 r})} . \nonumber \]
These fixed points exist only if \(x_{*}\) is real. Clearly, for the inner square-root to be real, \(r \geq-1 / 4\) . Also observe that \(1-\sqrt{1+4 r}\) becomes negative for \(r>0\) . We thus have three intervals in \(r\) to consider, and these regions and their fixed points are
\[\begin{aligned} r<-\frac{1}{4}: & x_{*}=0 \quad \text { (one fixed point); } \\ -\frac{1}{4}<r<0: & x_{*}=0, \quad x_{*}=\pm \sqrt{\frac{1}{2}(1 \pm \sqrt{1+4 r})} \quad \text { (five fixed points); } \\ r>0: \quad x_{*}=0, \quad x_{*}=\pm \sqrt{\frac{1}{2}(1+\sqrt{1+4 r})} \quad \text { (three fixed points). } \end{aligned} \nonumber \]
Stability is determined from \(f^{\prime}(x)=r+3 x^{2}-5 x^{4} .\) Now, \(f^{\prime}(0)=r\) so \(x_{*}=0\) is stable for \(r<0\) and unstable for \(r>0\) . The calculation for the other four roots can be simplified by noting that \(x_{*}\) satisfies \(r+x_{*}^{2}-x_{*}^{4}=0\) , or \(x_{*}^{4}=r+x_{*}^{2} .\) Therefore,
\[\begin{aligned} f^{\prime}\left(x_{*}\right) &=r+3 x_{*}^{2}-5 x_{*}^{4} \\ &=r+3 x_{*}^{2}-5\left(r+x_{*}^{2}\right) \\ &=-4 r-2 x_{*}^{2} \\ &=-2\left(2 r+x_{*}^{2}\right) \end{aligned} \nonumber \]
With \(x_{*}^{2}=\frac{1}{2}(1 \pm \sqrt{1+4 r})\) , we have
\[\begin{aligned} f^{\prime}\left(x_{*}\right) &=-2\left(2 r+\frac{1}{2}(1 \pm \sqrt{1+4 r})\right) \\ &=-((1+4 r) \pm \sqrt{1+4 r}) \\ &=-\sqrt{1+4 r}(\sqrt{1+4 r} \pm 1) \end{aligned} \nonumber \]
Clearly, the plus root is always stable since \(f^{\prime}\left(x_{*}\right)<0 .\) The minus root exists only for \(-\frac{1}{4}<r<0\) and is unstable since \(f^{\prime}\left(x_{*}\right)>0 .\) We summarize the stability of the various fixed points:
\[\begin{aligned} r<-\frac{1}{4}: \quad x_{*} &=0 \quad(\text { stable }) ; \\ -\frac{1}{4}<r<0: \quad x_{*} &=0, \quad(\text { stable }) \\ x_{*} &=\pm \sqrt{\frac{1}{2}(1+\sqrt{1+4 r})} \quad \text { (stable); } \\ x>0: \quad x_{*} &=\pm \sqrt{\frac{1}{2}(1-\sqrt{1+4 r})} \quad \text { (unstable); } \\ x_{*} &=0 \quad(\text { unstable }) \\ x_{*} &=\pm \sqrt{\frac{1}{2}(1+\sqrt{1+4 r})} \quad \text { (stable). } \end{aligned} \nonumber \]
The bifurcation diagram is shown in Fig. 12.4. Notice that there in addition to a subcritical pitchfork bifurcation at the origin, there are two symmetric saddlenode bifurcations that occur when \(r=-1 / 4\) .
We can imagine what happens to the solution to the ode as \(r\) increases from negative values, supposing there is some noise in the system so that \(x(t)\) fluctuates around a stable fixed point. For \(r<-1 / 4\) , the solution \(x(t)\) fluctuates around the stable fixed point \(x_{*}=0\) . As \(r\) increases into the range \(-1 / 4<r<0\) , the solution will remain close to the stable fixed point \(x_{*}=0 .\) However, a so-called catastrophe occurs as soon as \(r>0 .\) The \(x_{*}=0\) fixed point is lost and the solution will jump up (or down) to the only remaining fixed point. A similar catastrophe can happen as \(r\) decreases from positive values. In this case, the jump occurs as soon as \(r<-1 / 4\) . Since the behavior of \(x(t)\) is different depending on whether we increase or decrease \(r\) , we say that the system exhibits hysteresis. The existence of a subcritical pitchfork bifurcation can be very dangerous in engineering applications since a small change in the physical parameters of a problem can result in a large change in the equilibrium state. Physically, this can result in the collapse of a structure.
Hopf bifurcations
A new type of bifurcation can occur in two dimensions. Suppose there is some control parameter \(\mu\) . Furthermore, suppose that for \(\mu<0\) , a two-dimensional system approaches a fixed point by exponentially-damped oscillations. We know that the Jacobian matrix at the fixed point with \(\mu<0\) will have complex conjugate eigenvalues with negative real parts. Now suppose that when \(\mu>0\) the real parts of the eigenvalues become positive so that the fixed point becomes unstable. This change in stability of the fixed point is called a Hopf bifurcation. The Hopf bifurcations also come in two types: a supercritical Hopf bifurcation and a subcritical Hopf bifurcation. For the supercritical Hopf bifurcation, as \(\mu\) increases slightly above zero, the resulting oscillation around the now unstable fixed point is quickly stabilized at small amplitude, and one obtains a limit cycle. For the subcritical Hopf bifurcation, as \(\mu\) increases slightly above zero, the limit cycle immediately jumps to large amplitude.
Phase portraits
The phase space of a dynamical system consists of the independent dynamical variables. For example, the phase space of the damped, driven pendulum equations given by (11.15) is three dimensional and consists of \(\theta, u\) and \(\phi\) . The unforced equations have only a two-dimensional phase space, consisting of \(\theta\) and \(u\) .
An important feature of paths in phase space is that they can never cross except at a fixed point, which may be stable if all paths go into the point, or unstable if some paths go out. This is a consequence of the uniqueness of the solution to a differential equation. Every point on the phase-space diagram represents a possible initial condition for the equations, and must have a unique trajectory associated with that initial condition.
The simple pendulum is a conservative system, exhibiting a conservation law for energy, and this implies a conservation of phase space area (or volume). The damped pendulum is nonconservative, however, and this implies a shrinking of phase space area (or volume).
Indeed, it is possible to derive a general condition to determine whether phase space volume is conserved or shrinks. We consider here, for convenience, equations in a three-dimensional phase space given by
\[\dot{x}=F(x, y, z), \quad \dot{y}=G(x, y, z), \quad \dot{z}=H(x, y, z) . \nonumber \]
We can consider a small volume of phase space \(\Gamma=\Gamma(t)\) , given by
\[\Gamma(t)=\Delta x \Delta y \Delta z, \nonumber \]
where
\[\Delta x=x_{1}(t)-x_{0}(t), \quad \Delta y=y_{1}(t)-y_{0}(t), \quad \Delta z=z_{1}(t)-z_{0}(t) \nonumber \]
The initial phase-space volume at time \(t\) represents a box with one corner at the point \(\left(x_{0}(t), y_{0}(t), z_{0}(t)\right)\) and the opposing corner at \(\left(x_{1}(t), y_{1}(t), z_{1}(t)\right)\) . This initial phase-space box then evolves over time. To determine how an edge emanating from \(\left(x_{0}, y_{0}, z_{0}\right)\) with length \(\Delta x\) evolves, we write using a first-order Taylor series expansion in \(\Delta t\)
\[\begin{aligned} &x_{0}(t+\Delta t)=x_{0}(t)+\Delta t F\left(x_{0}, y_{0}, z_{0}\right) \\ &x_{1}(t+\Delta t)=x_{1}(t)+\Delta t F\left(x_{1}, y_{0}, z_{0}\right) \end{aligned} \nonumber \]
Therefore,
\[\Delta x(t+\Delta t)=\Delta x(t)+\Delta t\left(F\left(x_{1}, y_{0}, z_{0}\right)-F\left(x_{0}, y_{0}, z_{0}\right)\right) \nonumber \]
Subtracting \(\Delta x(t)\) from both sides and dividing by \(\Delta t\) , we have as \(\Delta t \rightarrow 0\) ,
\[\frac{d}{d t}(\Delta x)=\left(F\left(x_{1}, y_{0}, z_{0}\right)-F\left(x_{0}, y_{0}, z_{0}\right)\right) . \nonumber \]
With \(\Delta x\) small, we therefore obtain to first order in \(\Delta x\) ,
\[\frac{d}{d t}(\Delta x)=\Delta x \frac{\partial F}{\partial x}, \nonumber \]
where the partial derivative is evaluated at the point \(\left(x_{0}(t), y_{0}(t), z_{0}(t)\right)\) .
Similarly, we also have
\[\frac{d}{d t}(\Delta y)=\Delta y \frac{\partial G}{\partial y}, \quad \frac{d}{d t}(\Delta z)=\Delta z \frac{\partial H}{\partial z} \nonumber \]
Since
\[\frac{d \Gamma}{d t}=\frac{d(\Delta x)}{d t} \Delta y \Delta z+\Delta x \frac{d(\Delta y)}{d t} \Delta z+\Delta x \Delta y \frac{d(\Delta z)}{d t} \nonumber \]
we have
\[\frac{d \Gamma}{d t}=\left(\frac{\partial F}{\partial x}+\frac{\partial G}{\partial y}+\frac{\partial H}{\partial z}\right) \Gamma \nonumber \]
valid for the evolution of an infinitesimal box.
We can conclude from (12.25) that phase-space volume is conserved if the divergence on the right-hand-side vanishes, or that phase-space volume contracts exponentially if the divergence is negative.
For example, we can consider the equations for the damped, driven pendulum given by (11.15). The divergence in this case is derived from
\[\frac{\partial}{\partial \theta}(u)+\frac{\partial}{\partial u}\left(-\frac{1}{q} u-\sin \theta+f \cos \psi\right)+\frac{\partial}{\partial \psi}(\omega)=-\frac{1}{q} \nonumber \]
and provided \(q>0\) (i.e., the pendulum is damped), phase-space volume contracts. For the undamped pendulum (where \(q \rightarrow \infty\) ), phase-space volume is conserved.
Limit cycles
The asymptotic state of a nonlinear system may be a limit cycle instead of a fixed point. A stable limit cycle is a periodic solution on which all nearby solutions converge. Limit cycles can also be unstable. Determining the existence of a limit cycle from a nonlinear system of equations through analytical means is usually impossible, but limit cycles are easily found numerically. The damped, driven pendulum equations has no fixed points, but does have limit cycles.
Attractors and basins of attraction
An attractor is a stable configuration in phase space. For example, an attractor can be a stable fixed point or a limit cycle. The damped, non-driven pendulum converges to a stable fixed point corresponding to the pendulum at rest at the bottom. The damped, driven pendulum for certain values of the parameters converges to a limit cycle. We will see that the chaotic pendulum also has an attractor of a different sort, called a strange attractor.
The domain of attraction of a stable fixed point, or of a limit cycle, is called the attractor’s basin of attraction. The basin of attraction consists of all initial conditions for which solutions asymptotically converge to the attractor. For nonlinear systems, basins of attraction can have complicated geometries.
Poincaré sections
The Poincaré section is a method to reduce by one or more dimensions the phase space diagrams of higher dimensional systems. Most commonly, a three-dimensional phase space can be reduced to two-dimensions, for which a plot may be easier to interpret. For systems with a periodic driving force such as the damped, driven pendulum system given by (11.15), the two-dimensional phase space \((\theta, u)\) can be viewed stroboscopically, with the strobe period taken to be the period of forcing, that is \(2 \pi / \omega\) . For the damped, driven pendulum, the third dynamical variable \(\psi\) satisfies \(\psi=\psi_{0}+\omega t\) , and the plotting of the phase-space variables \((\theta, u)\) at the times \(t_{0}, t_{0}+2 \pi / \omega\) , \(t_{0}+4 \pi / \omega\) , etc., is equivalent to plotting the values of \((\theta, u)\) every time the phase-space variable \(\psi\) in incremented by \(2 \pi\) .
For systems without periodic forcing, a plane can be placed in the three-dimensional phase space, and a point can be plotted on this plane every time the orbit passes through it. For example, if the dynamical variables are \(x, y\) and \(z\) , a plane might be placed at \(z=0\) and the values of \((x, y)\) plotted every time z passes through zero. Usually, one also specifies the direction of passage, so that a point is plotted only when \(\dot{z}>0\) , say.
Fractal dimensions
The attractor of the chaotic pendulum is called strange because it occupies a fractional dimension of phase space. To understand what is meant by fractional dimensions, we first review some classical fractals.
Classical fractals
Cantor set
The Cantor set has some unusual properties. First, we compute its length. Let \(l_{n}\) denote the length of the set \(S_{n} .\) Clearly, \(l_{0}=1 .\) Since \(S_{1}\) is constructed by removing the middle third of \(S_{0}\) , we have \(l_{1}=2 / 3\) . To construct \(S_{2}\) , we again remove the middle third of the two line segments of \(S_{1}\) , reducing the length of \(S_{1}\) by another factor of \(2 / 3\) , so that \(l_{2}=(2 / 3)^{2}\) , and so on. We obtain \(l_{n}=(2 / 3)^{n}\) , and the \(\lim _{n \rightarrow \infty} l_{n}=0 .\) Therefore, the Cantor set has zero length.
The Cantor set, then, does not consist of line segments. Yet, the Cantor set is certainly not an empty set. We can see that at stage 1 , there are two interior endpoints \(1 / 3\) and \(2 / 3\) , and these will never be removed; at stage 2 , there are the additional interior endpoints \(1 / 9,2 / 9,7 / 9\) and \(8 / 9\) ; and at stage 3 , we add eight more interior endpoints. We see, then, that at stage \(k\) we add \(2^{k}\) more interior endpoints. An infinite but countable number of endpoints are therefore included in the Cantor set.
But the Cantor set consists of much more than just these endpoints, and we will in fact show that the Cantor set is an uncountable set. Recall from analysis, that an infinite set of points can be either countable or uncountable. A countable set of points is a set that can be put in a oneto-one correspondence with the set of natural numbers. As is well known, the infinite set of all rational numbers is countable whereas the infinite set of real numbers is uncountable. By listing the endpoints of the Cantor set above, each stage adding \(2^{k}\) more endpoints, we have shown that the set of all endpoints is a countable set.
In order to prove that the Cantor set is uncountable, it is helpful to make use of the base 3 representation of numbers. Recall that any number \(N\) , given by
\[N=\ldots \mathrm{a} * B^{3}+\mathrm{b} * B^{2}+\mathrm{c} * B+\mathrm{d} * B^{0}+\mathrm{e} * B^{-1}+\mathrm{f} * B^{-2}+\mathrm{g} * B^{-3}+\ldots, \nonumber \]
can be written in base \(B\) as
\[N=\ldots \text { abcd.efg } \ldots \text { (base } B \text { ) } \nonumber \]
where the period separating the integer part of the number from the fractional part is in general called a radix point. For base 10, of course, the radix point is called a decimal point, and for base 2, a binary point.
Now, consider the Cantor set. In the first stage, all numbers lying between \(1 / 3\) and \(2 / 3\) are removed. The remaining numbers, then, must be of the form \(0.0 \ldots\) (base 3 ) or \(0.2 \ldots\) (base 3 ), since (almost) all the numbers having the form \(0.1 \ldots\) (base 3) have been removed. The single exception is the endpoint \(1 / 3=0.1\) (base 3 ), but this number can also be written as \(1 / 3=0.02\) (base 3 ), where the bar over a number or numbers signifies an infinite repetition. In the second stage, all numbers lying between \(1 / 9\) and \(2 / 9\) , say, are removed, and these numbers are of the form \(0.01 \ldots\) (base 3). We can see, then, that the Cantor set can be defined as the set of all numbers on the unit interval that contain no 1’s in their base 3 representation.
Using base 3 notation, it is easy to find a number that is not an endpoint, yet is in the Cantor set. For example, the number \(1 / 4\) is not an endpoint. We can convert \(1 / 4\) to base 3 by the following calculation:
\[\begin{aligned} \frac{1}{4} \times 3 &=0.75 \\ 0.75 \times 3 &=2.25 \\ 0.25 \times 3 &=0.75 \\ 0.75 \times 3 &=2.25 \end{aligned} \nonumber \]
and so on, from which we find that \(1 / 4=0 . \overline{02}\) (base 3 ). The number \(1 / 4\) , therefore, has no 1 ’s in its base 3 expansion and so is an element of the Cantor set.
We can now prove that the Cantor set is an uncountable set. The proof is similar to that used to prove that the real numbers are uncountable. If we suppose that the Cantor set is countable, then we can list all of its elements using a base 3 expansion; that is,
\[\begin{aligned} &x_{1}=0 . x_{11} x_{12} x_{13} \ldots(\text { base } 3), \\ &x_{2}=0 . x_{21} x_{22} x_{23} \ldots(\text { base } 3), \\ &x_{3}=0 . x_{31} x_{32} x_{33} \ldots(\text { base } 3), \end{aligned} \nonumber \]
and so on, where all the \(x_{i j}\) ’s must be either a 0 or a 2 (i.e., no 1 ’s are allowed). It is now simple to name a number \(y\) not on this list. We have
\[y=0 . y_{11} y_{22} y_{33} \cdots \nonumber \]
where \(y_{11} \neq x_{11}, y_{22} \neq x_{22}, y_{33} \neq x_{33}\) , etc.. That is, if \(x_{11}=0\) , then \(y_{11}=2 ;\) if \(x_{11}=2\) , then \(y_{11}=0\) , and so on. By constructing a number not on the list, we have obtained a reductio ad absurdum, and we can conclude that the Cantor set is uncountable.
Note that the endpoints of the Cantor set are just those numbers written in base 3 that end with either all 0 ’s or all 2 ’s, and these indeed form only a very small subset of the entire Cantor set.
From the base 3 representation, we can now see that the Cantor set has the following interesting characteristic. On the one hand, any small interval around a point in the Cantor set contains another point in the Cantor set. On the other hand, there is an interval between any two points in the Cantor set that is not in the Cantor set. For example, take the point \(1 / 4=0 . \overline{02}\) in the Cantor set, and the interval \(\left[1 / 4-3^{-4}, 1 / 4+3^{-4}\right] .\) In base 3 , the interval is given by \([0.0201 \overline{02}, 0.0210 \overline{02}]\) . There are, of course, an infinite number of points in the Cantor set in this interval, one of them being \(0.0202\) to the left of \(1 / 4\) , and \(0.02022\) to the right of \(1 / 4\) . If the interval was established using \(3^{-n}\) , for any \(n\) , we can still find other points in the interval that are in the Cantor set. Also, between any two points in the Cantor set, there is an interval that was removed during the construction of the Cantor set. So the Cantor set consists of neither line segments nor a set of discrete points.
The Cantor set is also self-similar, meaning that it contains smaller copies of itself at all scales.
The Cantor set is called a fractal, and is said to have a fractal dimension. Integer dimensions are more familiar to us: we say that a line has dimension 1 ; an area, dimension 2 ; and a volume, dimension \(3 .\)
For self-similar sets, we can place the definition of dimension on a more mathematical basis. A line segment, a square, and a cube can also be considered self-similar. Suppose that a selfsimilar set \(S\) is composed of \(m\) copies of itself scaled down by a factor of \(r\) . As examples, the line segment \([0,1]\) is composed of two copies of itself scaled down by a factor of two; namely, the segments \([0,1 / 2]\) and \([1 / 2,1]\) . A square is composed of four copies of itself scaled down by a factor of two. And a cube is composed of eight copies of itself scaled down by a factor of two. We write
\[m=r^{D}, \nonumber \]
where \(D\) is called the similarity dimension of the set. With \(r=2\) , the line segment has \(m=2\) , the square has \(m=4\) , and the cube has \(m=8\) , yielding the usual dimensions \(D=1,2\) and 3 , respectively.
Now the Cantor set is composed of two copies of itself scaled down by a factor of three. Therefore,
\[2=3^{D}, \nonumber \]
and the similarity dimension of the Cantor set is given by \(D=\log 2 / \log 3 \approx 0.6309\) , which has dimension smaller than that of a line, but larger than that of a point.
Sierpinski triangle
Other classical fractals can be constructed similarly to the Cantor set. The Sierpinski triangle starts with an equilateral triangle, \(S_{0}\) , with unit sides. We then draw lines connecting the three midpoints of the three sides. The just formed equilateral middle triangle of side length \(1 / 2\) is then removed from \(S_{0}\) to form the set \(S_{1}\) . This process is repeated on the remaining equilateral triangles, as illustrated in Fig. 12.6. The Sierpinski triangle \(S\) is defined as the limit of this repetition; that is, \(S=\lim _{n \rightarrow \infty} S_{n}\) .
The Sierpinski triangle is composed of three copies of itself scaled down by a factor of two, so that
\[3=2^{D}, \nonumber \]
and the similarity dimension is \(D=\log 3 / \log 2 \approx 1.5850\) , which has dimension between a line and an area.
Similar to the Cantor set, the Sierpinksi triangle, though existing in a two-dimensional space, has zero area. If we let \(A_{i}\) be the area of the set \(S_{i}\) , then since area is reduced by a factor of \(3 / 4\) with each iteration, we have
\[A_{n}=\left(\frac{3}{4}\right)^{n} A_{0} \nonumber \]
so that \(A=\lim _{n \rightarrow \infty} A_{n}=0\) .
Sierpinski carpet
The Sierpinski carpet starts with a square, \(S_{0, \text { with unit sides. Here, we remove a center }}\) square of side length \(1 / 3\) ; the remaining set is called \(S_{1}\) . This process is then repeated as illustrated in Fig. 12.7.
Here, the Sierpinski carpet is composed of eight copies of itself scaled down by a factor of three, so that
\[8=3^{D} \text {, } \nonumber \]
and the similarity dimension is \(D=\log 8 / \log 3 \approx 1.8928\) , somewhat larger than the Sierpinski triangle. One can say, then, that the Sierpinski carpet is more space filling than the Sierpinski triangle, though it still has zero area.
Koch curve
Like the Cantor set, we start with the unit interval, but now we replace the middle one-third by two line segments of length \(1 / 3\) , as illustrated in \(12.8\) , to form the set \(S_{1}\) . This process is then repeated on the four line segments of length \(1 / 3\) to form \(S_{2}\) , and so on, as illustrated in Fig. \(12.8\) .
Here, the Koch curve is composed of four copies of itself scaled down by a factor of three, so that
\[4=3^{D} \nonumber \]
and the similarity dimension is \(D=\log 4 / \log 3 \approx 1.2619 .\) The Koch curve therefore has a dimension lying between a line and an area. Indeed, with each iteration, the length of the Koch curve increases by a factor of \(4 / 3\) so that its length is infinite.
Koch snow flake
Here, we start with an equilateral triangle. Similar to the Koch curve, we replace the middle one-third of each side by two line segments, as illustrated in 12.9. The boundary of the Koch snow flake has the same fractal dimension as the Koch curve, and is of infinite length. The area bounded by the boundary is obviously finite, however, and can be shown to be \(8 / 5\) of the area of the original triangle. Interestingly, here an infinite perimeter encloses a finite area.
Correlation Dimension
Modern studies of dynamical systems have discovered sets named strange attractors. These sets share some of the characteristics of classical fractals: they are not space filling yet have structure at arbitrarily small scale. However, they are not perfectly self-similar in the sense of the classical fractals, having arisen from the chaotic evolution of some dynamical system. Here, we attempt to generalize the definition of dimension so that it is applicable to strange attractors and relatively easy to compute. The definition and numerical algorithm described here was first proposed in a paper by Grassberger and Procaccia (1993)
Definitions
Consider a set \(S\) of \(N\) points. Denote the points in this set as \(\mathbf{x}_{i}\) , with \(i=1,2, \ldots, N\) and denote the distance between two points \(\mathbf{x}_{i}\) and \(\mathbf{x}_{j}\) as \(r_{i j} .\) Note that there are \(N(N-1) / 2\) distinct distances. We define the correlation integral \(C(r)\) to be
\[C(r)=\lim _{N \rightarrow \infty} \frac{2}{N(N-1)} \times\left\{\text { number of distinct distances } r_{i j} \text { less than } r\right\} . \nonumber \]
If \(C(r) \propto r^{D}\) over a wide range of \(r\) when \(N\) is sufficiently large, then \(D\) is to be called the correlation dimension of the set \(S\) . Note that with this normalization, \(C(r)=1\) for \(r\) larger than the largest possible distance between two points on the set \(S\) .
The correlation dimension agrees with the similarity dimension for an exactly self-similar set. As a specific example, consider the Cantor set. Suppose \(N\) points that lie on the Cantor set are chosen at random to be in \(S\) . Since every point on the set is within a distance \(r=1\) of every other point, one has \(C(1)=1\) . Now, approximately one-half of the points in \(S\) lie between 0 and \(1 / 3\) , and the other half lie between \(2 / 3\) and 1. Therefore, only 1/2 of the possible distinct distances \(r_{i j}\) will be less than \(1 / 3\) (as \(N \rightarrow \infty\) ), so that \(C(1 / 3)=1 / 2\) . Continuing in this fashion, we find in general that \(C\left(1 / 3^{s}\right)=1 / 2^{s}\) . With \(r=1 / 3^{s}\) , we have
\[s=-\frac{\log r}{\log 3} \nonumber \]
so that
\[\begin{aligned} C(r) &=2^{\log r / \log 3} \\ &=\exp (\log 2 \log r / \log 3) \\ &=r^{\log 2 / \log 3}, \end{aligned} \nonumber \]
valid for small values of \(r .\) We have thus found a correlation dimension, \(D=\log 2 / \log 3\) , in agreement with the previously determined similarity dimension.
Numerical computation
Given a finite set \(S\) of \(N\) points, we want to formulate a fast numerical algorithm to compute \(C(r)\) over a sufficiently wide range of \(r\) to accurately determine the correlation dimension \(D\) from a log-log plot of \(C(r)\) versus \(r .\) A point \(\mathbf{x}_{i}\) in the set \(S\) may be a real number, or may be a coordinate pair. If the points come from a Poincaré section, then the fractal dimension of the Poincaré section will be one less than the actual fractal dimension of the attractor in the full phase space.
Define the distance \(r_{i j}\) between two points \(\mathbf{x}_{i}\) and \(\mathbf{x}_{j}\) in \(S\) to be the standard Euclidean distance. To obtain an accurate value of \(D\) , one needs to throw away an initial transient before collecting points for the set \(S\) . Since we are interested in a graph of \(\log C\) versus \(\log r\) , we compute \(C(r)\) at points \(r\) that are evenly spaced in \(\log r .\) For example, we can compute \(C(r)\) at the points \(r_{s}=2^{s}\) , where \(s\) takes on integer values.
First, one counts the number of distinct distance \(r_{i j}\) that lie in the interval \(r_{s-1} \leq r_{i j}<r_{s}\) . Denote this count by \(M(s)\) . An approximation to the correlation integral \(C\left(r_{s}\right)\) using the \(N\) data points is then obtained from
\[C\left(r_{s}\right)=\frac{2}{N(N-1)} \sum_{s^{\prime}=-\infty}^{s} M\left(s^{\prime}\right) \nonumber \]
One can make use of a built-in MATLAB function to speed up the computation. The function call
returns the floating point number \(\mathrm{F}\) and the integer \(\mathrm{E}\) such that \(\mathrm{x}=\mathrm{F} * 2^{\wedge} \mathrm{E}\) , where \(0.5 \leq|\mathrm{F}|<1\) and \(\mathrm{E}\) is an integer. To count each \(r_{i j}\) in the appropriate bin \(M(s)\) , one can compute all the values of \(r_{i j}\) and place them into the Matlab array \(r\) . Then one uses the vector form of \(10 \mathrm{~g} 2 . \mathrm{m}\) to compute the corresponding values of \(s\) :
One can then increment the count in a Matlab array \(M\) that corresponds to the particular integer values of s. The Matlab function cumsum.m can then be used to compute the Matlab array C.
Finally, a least-squares analysis of the data is required to compute the fractal dimension \(D\) . By directly viewing the log-log plot, one can choose which adjacent points to fit a straight line through, and then use the method of least-squares to compute the slope. The MATLAB function polyfit.m can determine the best fit line and the corresponding slope. Ideally, one would also compute a statistical error associated with this slope, and such an error should go to zero as the number of points computed approaches infinity.


21.2: Virus Infections and Hosts
Skills to Develop
- List the steps of replication and explain what occurs at each step
- Describe the lytic and lysogenic cycles of virus replication
- Explain the transmission and diseases of animal and plant viruses
- Discuss the economic impact of animal and plant viruses
Viruses can be seen as obligate, intracellular parasites. A virus must attach to a living cell, be taken inside, manufacture its proteins and copy its genome, and find a way to escape the cell so that the virus can infect other cells. Viruses can infect only certain species of hosts and only certain cells within that host. Cells that a virus may use to replicate are called permissive . For most viruses, the molecular basis for this specificity is that a particular surface molecule known as the viral receptor must be found on the host cell surface for the virus to attach. Also, metabolic and host cell immune response differences seen in different cell types based on differential gene expression are a likely factor in which cells a virus may target for replication. The permissive cell must make the substances that the virus needs or the virus will not be able to replicate there.
Steps of Virus Infections
A virus must use cell processes to replicate. The viral replication cycle can produce dramatic biochemical and structural changes in the host cell, which may cause cell damage. These changes, called cytopathic (causing cell damage) effects, can change cell functions or even destroy the cell. Some infected cells, such as those infected by the common cold virus known as rhinovirus, die through lysis (bursting) or apoptosis (programmed cell death or “cell suicide”), releasing all progeny virions at once. The symptoms of viral diseases result from the immune response to the virus, which attempts to control and eliminate the virus from the body, and from cell damage caused by the virus. Many animal viruses, such as HIV (human immunodeficiency virus), leave the infected cells of the immune system by a process known as budding , where virions leave the cell individually. During the budding process, the cell does not undergo lysis and is not immediately killed. However, the damage to the cells that the virus infects may make it impossible for the cells to function normally, even though the cells remain alive for a period of time. Most productive viral infections follow similar steps in the virus replication cycle: attachment, penetration, uncoating, replication, assembly, and release (Figure \(\PageIndex{1}\)).
Attachment
A virus attaches to a specific receptor site on the host cell membrane through attachment proteins in the capsid or via glycoproteins embedded in the viral envelope. The specificity of this interaction determines the host—and the cells within the host—that can be infected by a particular virus. This can be illustrated by thinking of several keys and several locks, where each key will fit only one specific lock.
Entry
The nucleic acid of bacteriophages enters the host cell naked, leaving the capsid outside the cell. Plant and animal viruses can enter through endocytosis, in which the cell membrane surrounds and engulfs the entire virus. Some enveloped viruses enter the cell when the viral envelope fuses directly with the cell membrane. Once inside the cell, the viral capsid is degraded, and the viral nucleic acid is released, which then becomes available for replication and transcription.
Replication and Assembly
The replication mechanism depends on the viral genome. DNA viruses usually use host cell proteins and enzymes to make additional DNA that is transcribed to messenger RNA (mRNA), which is then used to direct protein synthesis. RNA viruses usually use the RNA core as a template for synthesis of viral genomic RNA and mRNA. The viral mRNA directs the host cell to synthesize viral enzymes and capsid proteins, and assemble new virions. Of course, there are exceptions to this pattern. If a host cell does not provide the enzymes necessary for viral replication, viral genes supply the information to direct synthesis of the missing proteins. Retroviruses, such as HIV, have an RNA genome that must be reverse transcribed into DNA, which then is incorporated into the host cell genome. They are within group VI of the Baltimore classification scheme. To convert RNA into DNA, retroviruses must contain genes that encode the virus-specific enzyme reverse transcriptase that transcribes an RNA template to DNA. Reverse transcription never occurs in uninfected host cells—the needed enzyme reverse transcriptase is only derived from the expression of viral genes within the infected host cells. The fact that HIV produces some of its own enzymes not found in the host has allowed researchers to develop drugs that inhibit these enzymes. These drugs, including the reverse transcriptase inhibitor AZT , inhibit HIV replication by reducing the activity of the enzyme without affecting the host’s metabolism. This approach has led to the development of a variety of drugs used to treat HIV and has been effective at reducing the number of infectious virions (copies of viral RNA) in the blood to non-detectable levels in many HIV-infected individuals.
Egress
The last stage of viral replication is the release of the new virions produced in the host organism, where they are able to infect adjacent cells and repeat the replication cycle. As you’ve learned, some viruses are released when the host cell dies, and other viruses can leave infected cells by budding through the membrane without directly killing the cell.
Exercise \(\PageIndex{1}\)
Influenza virus is packaged in a viral envelope that fuses with the plasma membrane. This way, the virus can exit the host cell without killing it. What advantage does the virus gain by keeping the host cell alive?
- Answer
-
The host cell can continue to make new virus particles.
Link to Learning
Watch a video on viruses, identifying structures, modes of transmission, replication, and more.
Different Hosts and Their Viruses
As you’ve learned, viruses are often very specific as to which hosts and which cells within the host they will infect. This feature of a virus makes it specific to one or a few species of life on Earth. On the other hand, so many different types of viruses exist on Earth that nearly every living organism has its own set of viruses that tries to infect its cells. Even the smallest and simplest of cells, prokaryotic bacteria, may be attacked by specific types of viruses.
Bacteriophages
Bacteriophages are viruses that infect bacteria (Figure \(\PageIndex{2}\)). When infection of a cell by a bacteriophage results in the production of new virions, the infection is said to be productive . If the virions are released by bursting the cell, the virus replicates by means of a lytic cycle (Figure \(\PageIndex{3}\)).
An example of a lytic bacteriophage is T4, which infects Escherichia coli found in the human intestinal tract. Sometimes, however, a virus can remain within the cell without being released. For example, when a temperate bacteriophage infects a bacterial cell, it replicates by means of a lysogenic cycle (Figure \(\PageIndex{3}\)), and the viral genome is incorporated into the genome of the host cell. When the phage DNA is incorporated into the host cell genome, it is called a prophage . An example of a lysogenic bacteriophage is the λ (lambda) virus, which also infects the E . coli bacterium. Viruses that infect plant or animal cells may also undergo infections where they are not producing virions for long periods. An example is the animal herpesviruses, including herpes simplex viruses, the cause of oral and genital herpes in humans. In a process called latency , these viruses can exist in nervous tissue for long periods of time without producing new virions, only to leave latency periodically and cause lesions in the skin where the virus replicates. Even though there are similarities between lysogeny and latency, the term lysogenic cycle is usually reserved to describe bacteriophages. Latency will be described in more detail below.
Exercise \(\PageIndex{2}\)
Which of the following statements is false?
- In the lytic cycle, new phage are produced and released into the environment.
- In the lysogenic cycle, phage DNA is incorporated into the host genome.
- An environmental stressor can cause the phage to initiate the lysogenic cycle.
- Cell lysis only occurs in the lytic cycle.
- Answer
-
C
Animal Viruses
Animal viruses, unlike the viruses of plants and bacteria, do not have to penetrate a cell wall to gain access to the host cell. Non-enveloped or “naked” animal viruses may enter cells in two different ways. As a protein in the viral capsid binds to its receptor on the host cell, the virus may be taken inside the cell via a vesicle during the normal cell process of receptor-mediated endocytosis. An alternative method of cell penetration used by non-enveloped viruses is for capsid proteins to undergo shape changes after binding to the receptor, creating channels in the host cell membrane. The viral genome is then “injected” into the host cell through these channels in a manner analogous to that used by many bacteriophages. Enveloped viruses also have two ways of entering cells after binding to their receptors: receptor-mediated endocytosis, or fusion . Many enveloped viruses enter the cell by receptor-mediated endocytosis in a fashion similar to some non-enveloped viruses. On the other hand, fusion only occurs with enveloped virions. These viruses, which include HIV among others, use special fusion proteins in their envelopes to cause the envelope to fuse with the plasma membrane of the cell, thus releasing the genome and capsid of the virus into the cell cytoplasm.
After making their proteins and copying their genomes, animal viruses complete the assembly of new virions and exit the cell. As we have already discussed using the example of HIV, enveloped animal viruses may bud from the cell membrane as they assemble themselves, taking a piece of the cell’s plasma membrane in the process. On the other hand, non-enveloped viral progeny, such as rhinoviruses, accumulate in infected cells until there is a signal for lysis or apoptosis, and all virions are released together.
As you will learn in the next module, animal viruses are associated with a variety of human diseases. Some of them follow the classic pattern of acute disease , where symptoms get increasingly worse for a short period followed by the elimination of the virus from the body by the immune system and eventual recovery from the infection. Examples of acute viral diseases are the common cold and influenza. Other viruses cause long-term chronic infections , such as the virus causing hepatitis C, whereas others, like herpes simplex virus, only cause intermittent symptoms. Still other viruses, such as human herpesviruses 6 and 7, which in some cases can cause the minor childhood disease roseola, often successfully cause productive infections without causing any symptoms at all in the host, and thus we say these patients have an asymptomatic infection .
In hepatitis C infections, the virus grows and reproduces in liver cells, causing low levels of liver damage. The damage is so low that infected individuals are often unaware that they are infected, and many infections are detected only by routine blood work on patients with risk factors such as intravenous drug use. On the other hand, since many of the symptoms of viral diseases are caused by immune responses, a lack of symptoms is an indication of a weak immune response to the virus. This allows for the virus to escape elimination by the immune system and persist in individuals for years, all the while producing low levels of progeny virions in what is known as a chronic viral disease. Chronic infection of the liver by this virus leads to a much greater chance of developing liver cancer, sometimes as much as 30 years after the initial infection.
As already discussed, herpes simplex virus can remain in a state of latency in nervous tissue for months, even years. As the virus “hides” in the tissue and makes few if any viral proteins, there is nothing for the immune response to act against, and immunity to the virus slowly declines. Under certain conditions, including various types of physical and psychological stress, the latent herpes simplex virus may be reactivated and undergo a lytic replication cycle in the skin, causing the lesions associated with the disease. Once virions are produced in the skin and viral proteins are synthesized, the immune response is again stimulated and resolves the skin lesions in a few days by destroying viruses in the skin. As a result of this type of replicative cycle, appearances of cold sores and genital herpes outbreaks only occur intermittently, even though the viruses remain in the nervous tissue for life. Latent infections are common with other herpesviruses as well, including the varicella-zoster virus that causes chickenpox. After having a chickenpox infection in childhood, the varicella-zoster virus can remain latent for many years and reactivate in adults to cause the painful condition known as “shingles” (Figure \(\PageIndex{4}\)).
Some animal-infecting viruses, including the hepatitis C virus discussed above, are known as oncogenic viruses : They have the ability to cause cancer. These viruses interfere with the normal regulation of the host cell cycle either by either introducing genes that stimulate unregulated cell growth (oncogenes) or by interfering with the expression of genes that inhibit cell growth. Oncogenic viruses can be either DNA or RNA viruses. Cancers known to be associated with viral infections include cervical cancer caused by human papillomavirus (HPV) (Figure \(\PageIndex{5}\)), liver cancer caused by hepatitis B virus, T-cell leukemia, and several types of lymphoma.
Link to Learning
Visit the interactive animations showing the various stages of the replicative cycles of animal viruses and click on the flash animation links.
Plant Viruses
Plant viruses, like other viruses, contain a core of either DNA or RNA. You have already learned about one of these, the tobacco mosaic virus. As plant viruses have a cell wall to protect their cells, these viruses do not use receptor-mediated endocytosis to enter host cells as is seen with animal viruses. For many plant viruses to be transferred from plant to plant, damage to some of the plants’ cells must occur to allow the virus to enter a new host. This damage is often caused by weather, insects, animals, fire, or human activities like farming or landscaping. Additionally, plant offspring may inherit viral diseases from parent plants. Plant viruses can be transmitted by a variety of vectors, through contact with an infected plant’s sap, by living organisms such as insects and nematodes, and through pollen. When plants viruses are transferred between different plants, this is known as horizontal transmission , and when they are inherited from a parent, this is called vertical transmission .
Symptoms of viral diseases vary according to the virus and its host (Table \(\PageIndex{1}\)). One common symptom is hyperplasia , the abnormal proliferation of cells that causes the appearance of plant tumors known as galls . Other viruses induce hypoplasia , or decreased cell growth, in the leaves of plants, causing thin, yellow areas to appear. Still other viruses affect the plant by directly killing plant cells, a process known as cell necrosis . Other symptoms of plant viruses include malformed leaves, black streaks on the stems of the plants, altered growth of stems, leaves, or fruits, and ring spots, which are circular or linear areas of discoloration found in a leaf.
| Symptom | Appears as |
|---|---|
| Hyperplasia | Galls (tumors) |
| Hypoplasia | Thinned, yellow splotches on leaves |
| Cell necrosis | Dead, blackened stems, leaves, or fruit |
| Abnormal growth patterns | Malformed stems, leaves, or fruit |
| Discoloration | Yellow, red, or black lines, or rings in stems, leaves, or fruit |
Plant viruses can seriously disrupt crop growth and development, significantly affecting our food supply. They are responsible for poor crop quality and quantity globally, and can bring about huge economic losses annually. Others viruses may damage plants used in landscaping. Some viruses that infect agricultural food plants include the name of the plant they infect, such as tomato spotted wilt virus, bean common mosaic virus, and cucumber mosaic virus. In plants used for landscaping, two of the most common viruses are peony ring spot and rose mosaic virus. There are far too many plant viruses to discuss each in detail, but symptoms of bean common mosaic virus result in lowered bean production and stunted, unproductive plants. In the ornamental rose, the rose mosaic disease causes wavy yellow lines and colored splotches on the leaves of the plant.
Summary
Viral replication within a living cell always produces changes in the cell, sometimes resulting in cell death and sometimes slowly killing the infected cells. There are six basic stages in the virus replication cycle: attachment, penetration, uncoating, replication, assembly, and release. A viral infection may be productive, resulting in new virions, or nonproductive, which means that the virus remains inside the cell without producing new virions. Bacteriophages are viruses that infect bacteria. They have two different modes of replication: the lytic cycle, where the virus replicates and bursts out of the bacteria, and the lysogenic cycle, which involves the incorporation of the viral genome into the bacterial host genome. Animal viruses cause a variety of infections, with some causing chronic symptoms (hepatitis C), some intermittent symptoms (latent viruses such a herpes simplex virus 1), and others that cause very few symptoms, if any (human herpesviruses 6 and 7). Oncogenic viruses in animals have the ability to cause cancer by interfering with the regulation of the host cell cycle. Viruses of plants are responsible for significant economic damage in both agriculture and plants used for ornamentation.
Glossary
- acute disease
- disease where the symptoms rise and fall within a short period of time
- asymptomatic disease
- disease where there are no symptoms and the individual is unaware of being infected unless lab tests are performed
- AZT
- anti-HIV drug that inhibits the viral enzyme reverse transcriptase
- bacteriophage
- virus that infects bacteria
- budding
- method of exit from the cell used in certain animal viruses, where virions leave the cell individually by capturing a piece of the host plasma membrane
- cell necrosis
- cell death
- chronic infection
- describes when the virus persists in the body for a long period of time
- cytopathic
- causing cell damage
- fusion
- method of entry by some enveloped viruses, where the viral envelope fuses with the plasma membrane of the host cell
- gall
- appearance of a plant tumor
- horizontal transmission
- transmission of a disease from parent to offspring
- hyperplasia
- abnormally high cell growth and division
- hypoplasia
- abnormally low cell growth and division
- intermittent symptom
- symptom that occurs periodically
- latency
- virus that remains in the body for a long period of time but only causes intermittent symptoms
- lysis
- bursting of a cell
- lytic cycle
- type of virus replication in which virions are released through lysis, or bursting, of the cell
- lysogenic cycle
- type of virus replication in which the viral genome is incorporated into the genome of the host cell
- oncogenic virus
- virus that has the ability to cause cancer
- permissive
- cell type that is able to support productive replication of a virus
- productive
- viral infection that leads to the production of new virions
- prophage
- phage DNA that is incorporated into the host cell genome
- vertical transmission
- transmission of disease between unrelated individuals


10.2: Chi-Square Goodness of Fit
In probability, you calculated probabilities using both experimental and theoretical methods. There are times when it is important to determine how well the experimental values match the theoretical values. An example of this is if you wish to verify if a die is fair. To determine if observed values fit the expected values, you want to see if the difference between observed values and expected values is large enough to say that the test statistic is unlikely to happen if you assume that the observed values fit the expected values. The test statistic in this case is also the chi-square. The process is the same as for the chi-square test for independence.
Hypothesis Test for Goodness of Fit Test
-
State the null and alternative hypotheses and the level of significance
\(H_{o}\): The data are consistent with a specific distribution
\(H_{A}\): The data are not consistent with a specific distribution
Also, state your \(\alpha\) level here. -
State and check the assumptions for the hypothesis test
- A random sample is taken.
- Expected frequencies for each cell are greater than or equal to 5 (The expected frequencies, E , will be calculated later, and this assumption means \(E \geq 5\)).
-
Find the test statistic and p-value
Finding the test statistic involves several steps. First the data is collected and counted, and then it is organized into a table (in a table each entry is called a cell). These values are known as the observed frequencies, which the symbol for an observed frequency is O . The table is made up of k entries. The total number of observed frequencies is n . The expected frequencies are calculated by multiplying the probability of each entry, p , times n .
\(\text{Expected frequency( entry }i )=E=n^{*} p\)
Test Statistic :
\(\chi^{2}=\sum \dfrac{(O-E)^{2}}{E}\)
where O is the observed frequency and E is the expected frequency.
Again, the test statistic involves squaring the differences, so the test statistics are all positive. Thus a chi-squared test for goodness of fit is always right tailed.
p-value:
Using the TI-83/84: \(\chi \text { cdf }(\text { lower limit, } 1 \mathrm{E} 99, d f)\)
Using R: \(1-\text { pchisq }\left(\chi^{2}, d f\right)\)
Where the degrees of freedom is df = k - 1
4. Conclusion
This is where you write reject \(H_{o}\) or fail to reject \(H_{o}\). The rule is: if the p-value < \(\alpha\), then reject \(H_{o}\). If the p-value \(\geq \alpha\), then fail to reject \(H_{o}\),
5. Interpretation
This is where you interpret in real world terms the conclusion to the test. The conclusion for a hypothesis test is that you either have enough evidence to show \(H_{A}\) is true, or you do not have enough evidence to show \(H_{A}\) is true.
Example \(\PageIndex{1}\) goodness of fit test using the formula
Suppose you have a die that you are curious if it is fair or not. If it is fair then the proportion for each value should be the same. You need to find the observed frequencies and to accomplish this you roll the die 500 times and count how often each side comes up. The data is in Example \(\PageIndex{1}\). Do the data show that the die is fair? Test at the 5% level.
| Die values | 1 | 2 | 3 | 4 | 5 | 6 | Total |
|---|---|---|---|---|---|---|---|
| Observed Frequency | 78 | 87 | 87 | 76 | 85 | 87 | 100 |
Solution
1. State the null and alternative hypotheses and the level of significance
\(H_{o}\): The observed frequencies are consistent with the distribution for fair die (the die is fair)
\(H_{A}\): The observed frequencies are not consistent with the distribution for fair die (the die is not fair)
\(\alpha\) = 0.05
2. State and check the assumptions for the hypothesis test
- A random sample is taken since each throw of a die is a random event.
- Expected frequencies for each cell are greater than or equal to 5. See step 3.
3. Find the test statistic and p-value
First you need to find the probability of rolling each side of the die. The sample space for rolling a die is {1, 2, 3, 4, 5, 6}.
Now you can find the expected frequency for each side of the die. Since all the probabilities are the same, then each expected frequency is the same.
\(\text{Expected Frequency} =E=n^{*} p=500 * \dfrac{1}{6} \approx 83.33\)
Test Statistic:
It is easier to calculate the test statistic using a table.
| O | E | O-E | \((O-E)^{2}\) | \(\dfrac{(O-E)^{2}}{E}\) |
|---|---|---|---|---|
| 78 | 83.33 | -5.22 | 28.4089 | 0.340920437 |
| 87 | 83.33 | 3.67 | 13.4689 | 0.161633265 |
| 87 | 83.33 | 3.67 | 13.4689 | 0.161633265 |
| 76 | 83.33 | -7.33 | 53.7289 | 0.644772591 |
| 85 | 83.33 | 1.67 | 2.7889 | 0.033468139 |
| 87 | 83.33 | 3.67 | 13.4689 | 0.161633265 |
| Total | 0.02 | \(\chi^{2} \approx 1.504060962\) |
The test statistic is \(\chi^{2} \approx 1.504060962\)
The degrees of freedom are df = k - 1 = 6 - 1 = 5
Using TI-83/84: \(p-\text {value}=\chi^{2} \operatorname{cdf}(1.50406096,1 E 99,5) \approx 0.913\)
Using R: \(p-\text {value}=1-\text { pchisq }(1.50406096,5) \approx 0.9126007\)
4. Conclusion
Fail to reject \(H_{o}\) since the p-value is greater than 0.05.
5. Interpretation
There is not enough evidence to show that the die is not consistent with the distribution for a fair die. There is not enough evidence to show that the die is not fair.
Example \(\PageIndex{2}\) goodness of fit test using technology
Suppose you have a die that you are curious if it is fair or not. If it is fair then the proportion for each value should be the same. You need to find the observed frequencies and to accomplish this you roll the die 500 times and count how often each side comes up. The data is in Example \(\PageIndex{1}\). Do the data show that the die is fair? Test at the 5% level.
Solution
1. State the null and alternative hypotheses and the level of significance
\(H_{o}\): The observed frequencies are consistent with the distribution for fair die (the die is fair)
\(H_{A}\): The observed frequencies are not consistent with the distribution for fair die (the die is not fair)
\(\alpha\) = 0.05
2. State and check the assumptions for the hypothesis test
- A random sample is taken since each throw of a die is a random event.
- Expected frequencies for each cell are greater than or equal to 5. See step 3.
3. Find the test statistic and p-value
Using the TI-83/84 calculator:
Using the TI-83:
To use the TI-83 calculator to compute the test statistic, you must first put the data into the calculator. Type the observed frequencies into L1 and the expected frequencies into L2. Then you will need to go to L3, arrow up onto the name, and type in \((L 1-L 2)^{\wedge} 2 / L 2\). Now you use 1-Var Stats L3 to find the total. See Figure \(\PageIndex{1}\) for the initial setup, Figure 11.2 .2 for the results of that calculation, and Figure \(\PageIndex{3}\) for the result of the 1-Var Stats L3.
Figure \(\PageIndex{1}\): Input into TI-83 Figure \(\PageIndex{2}\): Result for L3 on TI-83 Figure \(\PageIndex{3}\): 1-Var Stats L3 Result on TI-83The total is the chi-square value, \(\chi^{2}=\sum x \approx 1.50406\).
The p-value is found using \(p-\text {value}=\chi^{2} \operatorname{cdf}(1.50406096,1 E 99,5) \approx 0.913\), where the degrees of freedom is df = k - 1 = 6 - 1 = 5.
Using the TI-84:
To run the test on the TI-84, type the observed frequencies into L1 and the expected frequencies into L2, then go into STAT, move over to TEST and choose \(\chi^{2}\) GOF-Test from the list. The setup for the test is in Figure \(\PageIndex{4}\) .
Figure \(\PageIndex{4}\): Setup for Chi-Square Goodness of Fit Test on TI-84Once you press ENTER on Calculate you will see the results in Figure \(\PageIndex{5}\) .
Figure \(\PageIndex{5}\): Results for Chi-Square Test on TI-83/84The test statistic is \(\chi^{2} \approx 1.504060962\)
The p-value \(\approx 0.913\)
The CNTRB represent the \(\dfrac{(O-E)^{2}}{E}\) for each die value. You can see the values by pressing the right arrow.
Using R:
Type in the observed frequencies. Call it something like observed.
observed<- c(type in data with commas in between)
Type in the probabilities that you are comparing to the observed frequencies.
Call it something like null.probs.
null.probs <- c(type in probabilities with commas in between)
chisq.test(observed, p=null.probs) – the command for the hypothesis test
For this example (Note since you are looking to see if the die is fair, then the probability of each side of a fair die coming up is 1/6.)
observed<-c(78, 87, 87, 76, 85, 87)
null.probs<-c(1/6, 1/6, 1/6, 1/6, 1/6, 1/6)
chisq.test(observed, p=null.probs)
Output:
Chi-squared test for given probabilities
data: observed
X-squared = 1.504, df = 5, p-value = 0.9126
The test statistic is \(\chi^{2}=1.504\) and the p-value = 0.9126.
4. Conclusion
Fail to reject \(H_{o}\) since the p-value is greater than 0.05.
5. Interpretation
There is not enough evidence to show that the die is not consistent with the distribution for a fair die. There is not enough evidence to show that the die is not fair.
Homework
Exercise \(\PageIndex{1}\)
In each problem show all steps of the hypothesis test. If some of the assumptions are not met, note that the results of the test may not be correct and then continue the process of the hypothesis test.
-
According to the M&M candy company, the expected proportion can be found in Example \(\PageIndex{3}\). In addition, the table contains the number of M&M’s of each color that were found in a case of candy (Madison, 2013). At the 5% level, do the observed frequencies support the claim of M&M?
Blue Brown Green Orange Red Yellow Total Observed Frequencies 481 371 483 544<PHONE_NUMBER> Expected Proportion 0.24 0.13 0.16 0.20 0.13 0.14 Table \(\PageIndex{3}\): M&M Observed and Proportions -
Eyeglassomatic manufactures eyeglasses for different retailers. They test to see how many defective lenses they made the time period of January 1 to March 31. Example \(\PageIndex{4}\) gives the defect and the number of defects. Do the data support the notion that each defect type occurs in the same proportion? Test at the 10% level.
Defect type Number of defects Scratch 5865 Right shaped - small 4613 Flaked 1992 Wrong axis 1838 Chamfer wrong 1596 Crazing, cracks 1546 Wrong shape 1485 Wrong PD 1398 Spots and bubbles 1371 Wrong height 1130 Right shape - big 1105 Lost in lab 976 Spots/bubble - intern 976 Table \(\PageIndex{4}\): Number of Defective Lenses -
On occasion, medical studies need to model the proportion of the population that has a disease and compare that to observed frequencies of the disease actually occurring. Suppose the end-stage renal failure in south-west Wales was collected for different age groups. Do the data in Example \(\PageIndex{5}\) show that the observed frequencies are in agreement with proportion of people in each age group (Boyle, Flowerdew & Williams, 1997)? Test at the 1% level.
Age Group 16-29 30-44 45-59 60-75 75+ Total Observed Frequency 32 66 132 218 91 539 Expected Proportion 0.23 0.25 0.22 0.21 0.09 Table \(\PageIndex{5}\): Renal Failure Frequencies -
In Africa in 2011, the number of deaths of a female from cardiovascular disease for different age groups are in Example \(\PageIndex{6}\) ("Global health observatory," 2013). In addition, the proportion of deaths of females from all causes for the same age groups are also in Example \(\PageIndex{6}\). Do the data show that the death from cardiovascular disease are in the same proportion as all deaths for the different age groups? Test at the 5% level.
Age 5-14 15-29 30-49 50-69 Total Cardiovascular Frequency 9 16 56 433 513 All Cause Proportion 0.10 0.12 0.26 0.52 Table \(\PageIndex{6}\): Deaths of Females for Different Age Groups -
In Australia in 1995, there was a question of whether indigenous people are more likely to die in prison than non-indigenous people. To figure out, the data in Example \(\PageIndex{7}\) was collected. ("Aboriginal deaths in," 2013). Do the data show that indigenous people die in the same proportion as non-indigenous people? Test at the 1% level.
Prisoner Dies Prisoner Did Not Die Total Indigenous Prisoner Frequency 17 2890 2907 Frequency of Non-Indigenous Prisoner 42 14459 14501 Table \(\PageIndex{7}\): Death of Prisoners -
A project conducted by the Australian Federal Office of Road Safety asked people many questions about their cars. One question was the reason that a person chooses a given car, and that data is in Example \(\PageIndex{8}\) ("Car preferences," 2013).
Safety Reliability Cost Performance Comfort Looks 84 62 46 34 47 27 Table \(\PageIndex{8}\): Reason for Choosing a Car
- Answer
-
For all hypothesis tests, just the conclusion is given. See solutions for the entire answer.
1. Reject Ho
3. Reject Ho
5. Reject Ho


8.7.E: Problems on Integration of Complex and Vector-Valued Functions
Prove Corollary \(1(\text { iii })-\) (vii) in §4 componentwise for integrable maps \(f: S \rightarrow E^{n}\left(C^{n}\right) .\)
Prove Theorems 2 and 3 componentwise for \(E=E^{n}\left(C^{n}\right)\).
Do it for Corollary 3 in §6.
Prove Theorem 1 with
\[
\int_{A}|f|<\infty
\]
replaced by
\[
\int_{A}\left|f_{k}\right|<\infty, \quad k=1, \ldots, n .
\]
Prove that if \(f: S \rightarrow E^{n}\left(C^{n}\right)\) is integrable on \(A,\) so is \(|f| .\) Disprove the converse.
Disprove Lemma 1 for \(m A=\infty\).
Complete the proof of Lemma 3.
Complete the proof of Theorem 3.
Do Problem 1 and \(2^{\prime}\) for \(f: S \rightarrow E\).
Prove formula (1) from definitions of Part II of this section.
\(\Rightarrow 10\). Show that
\[
\left|\int_{A} f\right| \leq \int_{A}|f|
\]
for integrable maps \(f: S \rightarrow E .\) See also Problem 14.
[Hint: If \(m A<\infty,\) use Corollary \(1(\text { ii ) of } §4 \text { and Lemma } 1 . \text { If } m A=\infty,\) us imitate" the proof of Lemma \(3 .\) ]
Do Problem 11 in §6 for \(f_{n}: S \rightarrow E .\) Do it componentwise for \(E=\) \(E^{n}\left(C^{n}\right) .\)
Show that if \(f, g: S \rightarrow E^{1}(C)\) are integrable on \(A,\) then
\[
\left|\int_{A} f g\right|^{2} \leq \int_{A}|f|^{2} \cdot \int_{A}|g|^{2} .
\]
In what case does equality hold? Deduce Theorem \(4\left(\mathrm{c}^{\prime}\right)\) in Chapter \(3,\) §§1-3, from this result.
[Hint: Argue as in that theorem. Consider the case
\[
\left.\left(\exists t \in E^{1}\right) \quad \int_{A}|f-t g|=0 .\right]
\]
Show that if \(f: S \rightarrow E^{1}(C)\) is integrable on \(A\) and
\[
\left|\int_{A} f\right|=\int_{A}|f| ,
\]
then
\[
(\exists c \in C) \quad c f=|f| \quad \text { a.e. on } A.
\]
[Hint: Let \(a=\int_{A} f .\) The case \(a=0\) is trivial. If \(a \neq 0,\) let
\[
c=\frac{|a|}{a} ;|c|=1 ; c a=|a| .
\]
Let \(r=(c f)_{\mathrm{re}} .\) Show that \(r \leq|c f|=|f|\),
\[
\begin{aligned}\left|\int_{A} f\right| &=\int_{A} c f=\int_{A} r \leq \int_{A}|f|=\left|\int_{A} f\right| , \\ & \int_{A}|f|=\int_{A} r=\int_{A} (c f)_{\mathrm{re,}} \end{aligned}
\]
\(\left.(c f)_{\mathrm{re}}=|c f|(\mathrm{a.e.}), \text { and } c f=|c f|=|f| \text { a.e. on } A .\right]\)
Do Problem 10 for \(E=C\) using the method of Problem \(13 .\)
Show that if \(f: S \rightarrow E\) is integrable on \(A,\) it is integrable on each \(\mathcal{M}\)-set \(B \subseteq A .\) If, in addition,
\[
\int_{B} f=0
\]
for all such \(B,\) show that \(f=0\) a.e. on \(A .\) Prove it for \(E=E^{n}\) first.
[Hint for \(\left.E=E^{*}: A=A(f>0) \cup A(f \leq 0) . \text { Use Theorems } 1(\mathrm{h}) \text { and } 2 \text { from } §5 .\right]\)
In Problem \(15,\) show that
\[
s=\int f
\]
is a \(\sigma\)-additive set function on
\[
\mathcal{M}_{A}=\{X \in \mathcal{M} | X \subseteq A\} .
\]
(Note \(4 \text { in } §5) ; s\) is called the indefinite integral of \(f\) in \(A .\)