If a function is extended to make it periodic then must the integration limits also be extended?
up vote
3
down vote
favorite
The following extract is taken from "Riley, Hobson and Bence - Mathematical methods for physics and engineering", Section 12.5 - "Non-periodic functions", page 422 and 423:
Find the Fourier series of $f(x)=x^2,$ for $,,0lt x le 2$
We must first make the function periodic. We do this by extending the range of interest to $−2 lt x le 2$ in such a way that $f(x)=f(-x)$ and then letting $f(x+4k) = f(x)$, where $k$ is any integer. This is shown in figure 12.5.
Now we have an even function of period 4. The Fourier series will faithfully represent $f(x)$ in the range, $−2 lt x le 2$, although not outside it. Firstly we note that since we have made the specified function even in $x$ by extending the range, all the coefficients $b_n$ will be zero.
Where $$b_n=frac{2}{L}int_{x_0}^{x_0+L}f(x)sinleft(frac{2pi nx}{L}right)dxtag{12.5}$$
similarly $$a_n=frac{2}{L}int_{x_0}^{x_0+L}f(x)cosleft(frac{2pi nx}{L}right)dxtag{12.6}$$ where $x_0$ is arbitrary but is often taken as $0$ or $-frac{L}{2}$ and $L$ is the period of $f(x)$.
Now we apply $(12.5)$ and $(12.6)$ with $L = 4$ to determine the remaining coefficients: $$a_n=frac{2}{4}int_{-2}^{2}x^2cosleft(frac{2pi nx}{4}right)dx=frac{4}{4}int_{0}^{2}x^2cosleft(frac{pi nx}{2}right)dx$$
The textbook then goes on the evaluate the Fourier series for $x^2$ on $]0,2]$ (which I won't go through here as that is not what this question is about).
The following related question I asked on "Chegg Study" website is here:
Show that the Fourier series of $f(x)=cosleft(frac{x}{2}right)$ for $-pilt x lt pi$ is given by
$$f(x)=frac{2}{pi}+frac{4}{pi}sum_{n=1}^{infty}frac{(-1)^{n-1}cos(nx)}{4n^2 - 1}$$
So my attempt at the finding the Fourier series for $cosleft(frac{x}{2}right)$ on the interval $-pi lt x lt pi$ is by using the same method for the previous case $f(x) = x^2$ above.
Since $cosleft(frac{x}{2}right)$ is not periodic on $-pi lt x lt pi$,$,$ I will extend the interval to $-2pi lt x lt 2pi$
A plot of the extended function is shown below:
Then by $(12.6)$,
$$begin{align}a_n &=frac{2}{L}int_{x_0}^{x_0+L}f(x)cosleft(frac{2pi nx}{L}right)dx \&=frac{2}{4pi}int_{-2pi}^{2pi}cosleft(frac{x}{2}right)cosleft(frac{2pi nx}{4pi}right)dx\&=frac{1}{2pi}int_{-2pi}^{2pi}cosleft(frac{x}{2}right)cosleft(frac{nx}{2}right)dx\&=0end{align}$$
and this has been verified here by Wolfram Alpha.
But according to this answer given by "Chegg Study" website which gives the correct answer but the author does not extend the range of integration or change the argument of the cosine. So basically the author does not make the function periodic. Now it was my understanding that the Dirichlet conditions for convergence demand that a function must be periodic. Are the Dirichlet conditions wrong then?
Finally, my question is; why did the limits of integration need to be changed in the first example I gave, $f(x)=x^2$, but not for the second example of $f(x)=cosleft(frac{x}{2}right)$?
I have researched this topic extensively in order to find an answer, here is a short extract from "Mathematical Methods in the Physical Sciences", third edition by Mary L. Boas, Section 8, page 360,
The method used in the $x^2$ and $cos(x/2)$ case was general and did not specify any restrictions on the function being extended.
So, put in another way, is there some kind of rule that says extending a trigonometric function to a full period is always forbidden since integrating over the period will be zero?
integration trigonometry fourier-analysis fourier-series trigonometric-integrals
|
show 1 more comment
up vote
3
down vote
favorite
The following extract is taken from "Riley, Hobson and Bence - Mathematical methods for physics and engineering", Section 12.5 - "Non-periodic functions", page 422 and 423:
Find the Fourier series of $f(x)=x^2,$ for $,,0lt x le 2$
We must first make the function periodic. We do this by extending the range of interest to $−2 lt x le 2$ in such a way that $f(x)=f(-x)$ and then letting $f(x+4k) = f(x)$, where $k$ is any integer. This is shown in figure 12.5.
Now we have an even function of period 4. The Fourier series will faithfully represent $f(x)$ in the range, $−2 lt x le 2$, although not outside it. Firstly we note that since we have made the specified function even in $x$ by extending the range, all the coefficients $b_n$ will be zero.
Where $$b_n=frac{2}{L}int_{x_0}^{x_0+L}f(x)sinleft(frac{2pi nx}{L}right)dxtag{12.5}$$
similarly $$a_n=frac{2}{L}int_{x_0}^{x_0+L}f(x)cosleft(frac{2pi nx}{L}right)dxtag{12.6}$$ where $x_0$ is arbitrary but is often taken as $0$ or $-frac{L}{2}$ and $L$ is the period of $f(x)$.
Now we apply $(12.5)$ and $(12.6)$ with $L = 4$ to determine the remaining coefficients: $$a_n=frac{2}{4}int_{-2}^{2}x^2cosleft(frac{2pi nx}{4}right)dx=frac{4}{4}int_{0}^{2}x^2cosleft(frac{pi nx}{2}right)dx$$
The textbook then goes on the evaluate the Fourier series for $x^2$ on $]0,2]$ (which I won't go through here as that is not what this question is about).
The following related question I asked on "Chegg Study" website is here:
Show that the Fourier series of $f(x)=cosleft(frac{x}{2}right)$ for $-pilt x lt pi$ is given by
$$f(x)=frac{2}{pi}+frac{4}{pi}sum_{n=1}^{infty}frac{(-1)^{n-1}cos(nx)}{4n^2 - 1}$$
So my attempt at the finding the Fourier series for $cosleft(frac{x}{2}right)$ on the interval $-pi lt x lt pi$ is by using the same method for the previous case $f(x) = x^2$ above.
Since $cosleft(frac{x}{2}right)$ is not periodic on $-pi lt x lt pi$,$,$ I will extend the interval to $-2pi lt x lt 2pi$
A plot of the extended function is shown below:
Then by $(12.6)$,
$$begin{align}a_n &=frac{2}{L}int_{x_0}^{x_0+L}f(x)cosleft(frac{2pi nx}{L}right)dx \&=frac{2}{4pi}int_{-2pi}^{2pi}cosleft(frac{x}{2}right)cosleft(frac{2pi nx}{4pi}right)dx\&=frac{1}{2pi}int_{-2pi}^{2pi}cosleft(frac{x}{2}right)cosleft(frac{nx}{2}right)dx\&=0end{align}$$
and this has been verified here by Wolfram Alpha.
But according to this answer given by "Chegg Study" website which gives the correct answer but the author does not extend the range of integration or change the argument of the cosine. So basically the author does not make the function periodic. Now it was my understanding that the Dirichlet conditions for convergence demand that a function must be periodic. Are the Dirichlet conditions wrong then?
Finally, my question is; why did the limits of integration need to be changed in the first example I gave, $f(x)=x^2$, but not for the second example of $f(x)=cosleft(frac{x}{2}right)$?
I have researched this topic extensively in order to find an answer, here is a short extract from "Mathematical Methods in the Physical Sciences", third edition by Mary L. Boas, Section 8, page 360,
The method used in the $x^2$ and $cos(x/2)$ case was general and did not specify any restrictions on the function being extended.
So, put in another way, is there some kind of rule that says extending a trigonometric function to a full period is always forbidden since integrating over the period will be zero?
integration trigonometry fourier-analysis fourier-series trigonometric-integrals
2
The root of your problem lies in this statement: "Since $cos(x/2)$ is not periodic on $-pi < x < pi$..." That statement is non-sensical in this context. You can extend any function on $[-pi,pi)$ to a $2pi$ periodic function on $mathbb{R}$ by making copies of the graph on $[-pi,pi)$.
– DisintegratingByParts
Nov 13 at 13:32
@DisintegratingByParts Thanks, can you please elaborate on this by making an answer explaining why it is nonsensical for $cos(x/2)$, but makes sense to extend $x^2$ from $0lt x le 2$ to $-2 lt x le 2$? It is understanding the distinction between these two cases that is causing me all the confusion here.
– BLAZE
Nov 13 at 20:58
2
A function can be periodic on $mathbb{R}$ with period $2pi$. But that's different than saying it is periodic on $[-pi,pi]$, which doesn't make sense in this context.
– DisintegratingByParts
Nov 13 at 22:17
1
I can see how that could lead to confusion. You are extending $cos(x/2)$ on $[-pi,pi]$ to all of $mathbb{R}$ so that it has period $2pi$. He should be cautioning you not to confuse this periodic extension with the natural graph of $cos(x/2)$, which has period $4pi$.
– DisintegratingByParts
Nov 14 at 4:28
1
Another thing to keep in mind is this: If a function $f$ is defined on $mathbb{R}$ and is periodic with period $L$, then $int_{a}^{a+L}f(x)dx$ has the same value, regardless of the value of $a$.
– DisintegratingByParts
Nov 14 at 4:31
|
show 1 more comment
up vote
3
down vote
favorite
up vote
3
down vote
favorite
The following extract is taken from "Riley, Hobson and Bence - Mathematical methods for physics and engineering", Section 12.5 - "Non-periodic functions", page 422 and 423:
Find the Fourier series of $f(x)=x^2,$ for $,,0lt x le 2$
We must first make the function periodic. We do this by extending the range of interest to $−2 lt x le 2$ in such a way that $f(x)=f(-x)$ and then letting $f(x+4k) = f(x)$, where $k$ is any integer. This is shown in figure 12.5.
Now we have an even function of period 4. The Fourier series will faithfully represent $f(x)$ in the range, $−2 lt x le 2$, although not outside it. Firstly we note that since we have made the specified function even in $x$ by extending the range, all the coefficients $b_n$ will be zero.
Where $$b_n=frac{2}{L}int_{x_0}^{x_0+L}f(x)sinleft(frac{2pi nx}{L}right)dxtag{12.5}$$
similarly $$a_n=frac{2}{L}int_{x_0}^{x_0+L}f(x)cosleft(frac{2pi nx}{L}right)dxtag{12.6}$$ where $x_0$ is arbitrary but is often taken as $0$ or $-frac{L}{2}$ and $L$ is the period of $f(x)$.
Now we apply $(12.5)$ and $(12.6)$ with $L = 4$ to determine the remaining coefficients: $$a_n=frac{2}{4}int_{-2}^{2}x^2cosleft(frac{2pi nx}{4}right)dx=frac{4}{4}int_{0}^{2}x^2cosleft(frac{pi nx}{2}right)dx$$
The textbook then goes on the evaluate the Fourier series for $x^2$ on $]0,2]$ (which I won't go through here as that is not what this question is about).
The following related question I asked on "Chegg Study" website is here:
Show that the Fourier series of $f(x)=cosleft(frac{x}{2}right)$ for $-pilt x lt pi$ is given by
$$f(x)=frac{2}{pi}+frac{4}{pi}sum_{n=1}^{infty}frac{(-1)^{n-1}cos(nx)}{4n^2 - 1}$$
So my attempt at the finding the Fourier series for $cosleft(frac{x}{2}right)$ on the interval $-pi lt x lt pi$ is by using the same method for the previous case $f(x) = x^2$ above.
Since $cosleft(frac{x}{2}right)$ is not periodic on $-pi lt x lt pi$,$,$ I will extend the interval to $-2pi lt x lt 2pi$
A plot of the extended function is shown below:
Then by $(12.6)$,
$$begin{align}a_n &=frac{2}{L}int_{x_0}^{x_0+L}f(x)cosleft(frac{2pi nx}{L}right)dx \&=frac{2}{4pi}int_{-2pi}^{2pi}cosleft(frac{x}{2}right)cosleft(frac{2pi nx}{4pi}right)dx\&=frac{1}{2pi}int_{-2pi}^{2pi}cosleft(frac{x}{2}right)cosleft(frac{nx}{2}right)dx\&=0end{align}$$
and this has been verified here by Wolfram Alpha.
But according to this answer given by "Chegg Study" website which gives the correct answer but the author does not extend the range of integration or change the argument of the cosine. So basically the author does not make the function periodic. Now it was my understanding that the Dirichlet conditions for convergence demand that a function must be periodic. Are the Dirichlet conditions wrong then?
Finally, my question is; why did the limits of integration need to be changed in the first example I gave, $f(x)=x^2$, but not for the second example of $f(x)=cosleft(frac{x}{2}right)$?
I have researched this topic extensively in order to find an answer, here is a short extract from "Mathematical Methods in the Physical Sciences", third edition by Mary L. Boas, Section 8, page 360,
The method used in the $x^2$ and $cos(x/2)$ case was general and did not specify any restrictions on the function being extended.
So, put in another way, is there some kind of rule that says extending a trigonometric function to a full period is always forbidden since integrating over the period will be zero?
integration trigonometry fourier-analysis fourier-series trigonometric-integrals
The following extract is taken from "Riley, Hobson and Bence - Mathematical methods for physics and engineering", Section 12.5 - "Non-periodic functions", page 422 and 423:
Find the Fourier series of $f(x)=x^2,$ for $,,0lt x le 2$
We must first make the function periodic. We do this by extending the range of interest to $−2 lt x le 2$ in such a way that $f(x)=f(-x)$ and then letting $f(x+4k) = f(x)$, where $k$ is any integer. This is shown in figure 12.5.
Now we have an even function of period 4. The Fourier series will faithfully represent $f(x)$ in the range, $−2 lt x le 2$, although not outside it. Firstly we note that since we have made the specified function even in $x$ by extending the range, all the coefficients $b_n$ will be zero.
Where $$b_n=frac{2}{L}int_{x_0}^{x_0+L}f(x)sinleft(frac{2pi nx}{L}right)dxtag{12.5}$$
similarly $$a_n=frac{2}{L}int_{x_0}^{x_0+L}f(x)cosleft(frac{2pi nx}{L}right)dxtag{12.6}$$ where $x_0$ is arbitrary but is often taken as $0$ or $-frac{L}{2}$ and $L$ is the period of $f(x)$.
Now we apply $(12.5)$ and $(12.6)$ with $L = 4$ to determine the remaining coefficients: $$a_n=frac{2}{4}int_{-2}^{2}x^2cosleft(frac{2pi nx}{4}right)dx=frac{4}{4}int_{0}^{2}x^2cosleft(frac{pi nx}{2}right)dx$$
The textbook then goes on the evaluate the Fourier series for $x^2$ on $]0,2]$ (which I won't go through here as that is not what this question is about).
The following related question I asked on "Chegg Study" website is here:
Show that the Fourier series of $f(x)=cosleft(frac{x}{2}right)$ for $-pilt x lt pi$ is given by
$$f(x)=frac{2}{pi}+frac{4}{pi}sum_{n=1}^{infty}frac{(-1)^{n-1}cos(nx)}{4n^2 - 1}$$
So my attempt at the finding the Fourier series for $cosleft(frac{x}{2}right)$ on the interval $-pi lt x lt pi$ is by using the same method for the previous case $f(x) = x^2$ above.
Since $cosleft(frac{x}{2}right)$ is not periodic on $-pi lt x lt pi$,$,$ I will extend the interval to $-2pi lt x lt 2pi$
A plot of the extended function is shown below:
Then by $(12.6)$,
$$begin{align}a_n &=frac{2}{L}int_{x_0}^{x_0+L}f(x)cosleft(frac{2pi nx}{L}right)dx \&=frac{2}{4pi}int_{-2pi}^{2pi}cosleft(frac{x}{2}right)cosleft(frac{2pi nx}{4pi}right)dx\&=frac{1}{2pi}int_{-2pi}^{2pi}cosleft(frac{x}{2}right)cosleft(frac{nx}{2}right)dx\&=0end{align}$$
and this has been verified here by Wolfram Alpha.
But according to this answer given by "Chegg Study" website which gives the correct answer but the author does not extend the range of integration or change the argument of the cosine. So basically the author does not make the function periodic. Now it was my understanding that the Dirichlet conditions for convergence demand that a function must be periodic. Are the Dirichlet conditions wrong then?
Finally, my question is; why did the limits of integration need to be changed in the first example I gave, $f(x)=x^2$, but not for the second example of $f(x)=cosleft(frac{x}{2}right)$?
I have researched this topic extensively in order to find an answer, here is a short extract from "Mathematical Methods in the Physical Sciences", third edition by Mary L. Boas, Section 8, page 360,
The method used in the $x^2$ and $cos(x/2)$ case was general and did not specify any restrictions on the function being extended.
So, put in another way, is there some kind of rule that says extending a trigonometric function to a full period is always forbidden since integrating over the period will be zero?
integration trigonometry fourier-analysis fourier-series trigonometric-integrals
integration trigonometry fourier-analysis fourier-series trigonometric-integrals
edited Nov 19 at 8:49
asked Nov 11 at 0:06
BLAZE
6,030112754
6,030112754
2
The root of your problem lies in this statement: "Since $cos(x/2)$ is not periodic on $-pi < x < pi$..." That statement is non-sensical in this context. You can extend any function on $[-pi,pi)$ to a $2pi$ periodic function on $mathbb{R}$ by making copies of the graph on $[-pi,pi)$.
– DisintegratingByParts
Nov 13 at 13:32
@DisintegratingByParts Thanks, can you please elaborate on this by making an answer explaining why it is nonsensical for $cos(x/2)$, but makes sense to extend $x^2$ from $0lt x le 2$ to $-2 lt x le 2$? It is understanding the distinction between these two cases that is causing me all the confusion here.
– BLAZE
Nov 13 at 20:58
2
A function can be periodic on $mathbb{R}$ with period $2pi$. But that's different than saying it is periodic on $[-pi,pi]$, which doesn't make sense in this context.
– DisintegratingByParts
Nov 13 at 22:17
1
I can see how that could lead to confusion. You are extending $cos(x/2)$ on $[-pi,pi]$ to all of $mathbb{R}$ so that it has period $2pi$. He should be cautioning you not to confuse this periodic extension with the natural graph of $cos(x/2)$, which has period $4pi$.
– DisintegratingByParts
Nov 14 at 4:28
1
Another thing to keep in mind is this: If a function $f$ is defined on $mathbb{R}$ and is periodic with period $L$, then $int_{a}^{a+L}f(x)dx$ has the same value, regardless of the value of $a$.
– DisintegratingByParts
Nov 14 at 4:31
|
show 1 more comment
2
The root of your problem lies in this statement: "Since $cos(x/2)$ is not periodic on $-pi < x < pi$..." That statement is non-sensical in this context. You can extend any function on $[-pi,pi)$ to a $2pi$ periodic function on $mathbb{R}$ by making copies of the graph on $[-pi,pi)$.
– DisintegratingByParts
Nov 13 at 13:32
@DisintegratingByParts Thanks, can you please elaborate on this by making an answer explaining why it is nonsensical for $cos(x/2)$, but makes sense to extend $x^2$ from $0lt x le 2$ to $-2 lt x le 2$? It is understanding the distinction between these two cases that is causing me all the confusion here.
– BLAZE
Nov 13 at 20:58
2
A function can be periodic on $mathbb{R}$ with period $2pi$. But that's different than saying it is periodic on $[-pi,pi]$, which doesn't make sense in this context.
– DisintegratingByParts
Nov 13 at 22:17
1
I can see how that could lead to confusion. You are extending $cos(x/2)$ on $[-pi,pi]$ to all of $mathbb{R}$ so that it has period $2pi$. He should be cautioning you not to confuse this periodic extension with the natural graph of $cos(x/2)$, which has period $4pi$.
– DisintegratingByParts
Nov 14 at 4:28
1
Another thing to keep in mind is this: If a function $f$ is defined on $mathbb{R}$ and is periodic with period $L$, then $int_{a}^{a+L}f(x)dx$ has the same value, regardless of the value of $a$.
– DisintegratingByParts
Nov 14 at 4:31
2
2
The root of your problem lies in this statement: "Since $cos(x/2)$ is not periodic on $-pi < x < pi$..." That statement is non-sensical in this context. You can extend any function on $[-pi,pi)$ to a $2pi$ periodic function on $mathbb{R}$ by making copies of the graph on $[-pi,pi)$.
– DisintegratingByParts
Nov 13 at 13:32
The root of your problem lies in this statement: "Since $cos(x/2)$ is not periodic on $-pi < x < pi$..." That statement is non-sensical in this context. You can extend any function on $[-pi,pi)$ to a $2pi$ periodic function on $mathbb{R}$ by making copies of the graph on $[-pi,pi)$.
– DisintegratingByParts
Nov 13 at 13:32
@DisintegratingByParts Thanks, can you please elaborate on this by making an answer explaining why it is nonsensical for $cos(x/2)$, but makes sense to extend $x^2$ from $0lt x le 2$ to $-2 lt x le 2$? It is understanding the distinction between these two cases that is causing me all the confusion here.
– BLAZE
Nov 13 at 20:58
@DisintegratingByParts Thanks, can you please elaborate on this by making an answer explaining why it is nonsensical for $cos(x/2)$, but makes sense to extend $x^2$ from $0lt x le 2$ to $-2 lt x le 2$? It is understanding the distinction between these two cases that is causing me all the confusion here.
– BLAZE
Nov 13 at 20:58
2
2
A function can be periodic on $mathbb{R}$ with period $2pi$. But that's different than saying it is periodic on $[-pi,pi]$, which doesn't make sense in this context.
– DisintegratingByParts
Nov 13 at 22:17
A function can be periodic on $mathbb{R}$ with period $2pi$. But that's different than saying it is periodic on $[-pi,pi]$, which doesn't make sense in this context.
– DisintegratingByParts
Nov 13 at 22:17
1
1
I can see how that could lead to confusion. You are extending $cos(x/2)$ on $[-pi,pi]$ to all of $mathbb{R}$ so that it has period $2pi$. He should be cautioning you not to confuse this periodic extension with the natural graph of $cos(x/2)$, which has period $4pi$.
– DisintegratingByParts
Nov 14 at 4:28
I can see how that could lead to confusion. You are extending $cos(x/2)$ on $[-pi,pi]$ to all of $mathbb{R}$ so that it has period $2pi$. He should be cautioning you not to confuse this periodic extension with the natural graph of $cos(x/2)$, which has period $4pi$.
– DisintegratingByParts
Nov 14 at 4:28
1
1
Another thing to keep in mind is this: If a function $f$ is defined on $mathbb{R}$ and is periodic with period $L$, then $int_{a}^{a+L}f(x)dx$ has the same value, regardless of the value of $a$.
– DisintegratingByParts
Nov 14 at 4:31
Another thing to keep in mind is this: If a function $f$ is defined on $mathbb{R}$ and is periodic with period $L$, then $int_{a}^{a+L}f(x)dx$ has the same value, regardless of the value of $a$.
– DisintegratingByParts
Nov 14 at 4:31
|
show 1 more comment
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2993286%2fif-a-function-is-extended-to-make-it-periodic-then-must-the-integration-limits-a%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
The root of your problem lies in this statement: "Since $cos(x/2)$ is not periodic on $-pi < x < pi$..." That statement is non-sensical in this context. You can extend any function on $[-pi,pi)$ to a $2pi$ periodic function on $mathbb{R}$ by making copies of the graph on $[-pi,pi)$.
– DisintegratingByParts
Nov 13 at 13:32
@DisintegratingByParts Thanks, can you please elaborate on this by making an answer explaining why it is nonsensical for $cos(x/2)$, but makes sense to extend $x^2$ from $0lt x le 2$ to $-2 lt x le 2$? It is understanding the distinction between these two cases that is causing me all the confusion here.
– BLAZE
Nov 13 at 20:58
2
A function can be periodic on $mathbb{R}$ with period $2pi$. But that's different than saying it is periodic on $[-pi,pi]$, which doesn't make sense in this context.
– DisintegratingByParts
Nov 13 at 22:17
1
I can see how that could lead to confusion. You are extending $cos(x/2)$ on $[-pi,pi]$ to all of $mathbb{R}$ so that it has period $2pi$. He should be cautioning you not to confuse this periodic extension with the natural graph of $cos(x/2)$, which has period $4pi$.
– DisintegratingByParts
Nov 14 at 4:28
1
Another thing to keep in mind is this: If a function $f$ is defined on $mathbb{R}$ and is periodic with period $L$, then $int_{a}^{a+L}f(x)dx$ has the same value, regardless of the value of $a$.
– DisintegratingByParts
Nov 14 at 4:31