Convergence to pi^2/6 using Fourier Series and f(x) = x^2

• libelec
In summary: I appreciate your help but I don't understand what you want me to do about the borders. Do you want me to use the Dirichlet theorem or something like that?In summary, the conversation discusses using the Fourier trigonometrical series to prove the equation \sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} = \frac{{{\pi ^2}}}{6} for the function f(x) = x^2 on the interval [0, 2\pi]. The conversation mentions changing the limits of integration and using the Dirichlet theorem to find the Fourier coefficients, and also discusses using the periodic extensions of the functions and integrating over the period
libelec

Homework Statement

Using the Fourier trigonometrical series for $$f(x) = {x^2},{\rm{ }}0 \le x < 2\pi$$, prove that $$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} = \frac{{{\pi ^2}}}{6}$$

The Attempt at a Solution

This is more of a "what am I doing wrong question". First, because I'm not in the period defined by the trigonometrical coefficients, I have to change the limits using that:

$$0 \le x < 2\pi$$
$$- \pi \le x - \pi < \pi$$
$$\begin{array}{l} - \pi \le t < \pi , \\ t = x - \pi , \\ f(t) = f(x - \pi ) \\ \end{array}$$

Then I find the Fourier coefficients with this in hand:

$${a_0} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f(t).dt = } \frac{1}{\pi }\int\limits_0^{2\pi } {f(x).dx = } \frac{1}{\pi }\int\limits_0^{2\pi } {{x^2}.dx = } \frac{8}{3}{\pi ^2}$$

$${a_n} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f(t).\cos (nt)dt = } \frac{1}{\pi }\int\limits_0^{2\pi } {f(x).\cos (n(x - \pi ))dx = } \frac{1}{\pi }\int\limits_0^{2\pi } {{x^2}.\left[ {\cos (nx)\cos (n\pi ) - \sin (nx)\sin (n\pi )} \right].dx = }$$
$${a_n} = \frac{1}{\pi }\int\limits_0^{2\pi } {{x^2}.\cos (nx).{{( - 1)}^n}dx = } \frac{{{{( - 1)}^n}}}{\pi }\left( {\left. {\frac{{2x.\cos (nx)}}{{{n^2}}}} \right|_0^{2\pi }} \right) = \frac{{2{{( - 1)}^n}}}{{{n^2}}}$$

$${b_n} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f(t).\sin (nt).dx = } \frac{1}{\pi }\int\limits_0^{2\pi } {f(x).\sin (n(x - \pi )).dx = } \frac{1}{\pi }\int\limits_0^{2\pi } {{x^2}.\left[ {\sin (nx)\cos (n\pi ) + \sin (n\pi )\cos (nx)} \right].dx = }$$
$${b_n} = \frac{{{{\left( { - 1} \right)}^n}}}{\pi }\int\limits_0^{2\pi } {{x^2}.\sin (nx).dx = } \frac{{{{\left( { - 1} \right)}^n}}}{\pi }\left( {\left. {(\frac{2}{{{n^3}}} - \frac{{{x^2}}}{n})\cos (nx)} \right|_0^{2\pi }} \right) = \frac{{4{{\left( { - 1} \right)}^{n + 1}}}}{n}$$

Then, using Parseval's identity:

$${\left\| {f(x)} \right\|^2} = \frac{{{a_0}^2}}{2} + \sum\limits_{n = 1}^\infty {{a_n}^2 + {b_n}^2}$$
$$\frac{1}{\pi }\int\limits_0^{2\pi } {{{({x^2})}^2}.dx} = \frac{{{{\left( {\frac{8}{3}{\pi ^2}} \right)}^2}}}{2} + \sum\limits_{n = 1}^\infty {{{\left( {\frac{{2{{( - 1)}^n}}}{{{n^2}}}} \right)}^2} + {{\left( {\frac{{4{{( - 1)}^{n + 1}}}}{n}} \right)}^2}}$$
$$\frac{{32}}{5}{\pi ^4} = \frac{{32{\pi ^4}}}{9} + \sum\limits_{n = 1}^\infty {\frac{4}{{{n^4}}} + \frac{{16}}{{{n^2}}}}$$
$$\frac{{112}}{9}{\pi ^4} = \sum\limits_{n = 1}^\infty {\frac{{4 + 16{n^2}}}{{{n^4}}}}$$

And here I reach a deadpoint. What do I do?

You are extending f(x) periodically with period $2\pi$. You get the same thing when you integrate a periodic function over any period. That is, if g(x) has period P, then

$$\int_a^{a+P} g(x)\, dx = \int_b^{b+P} g(x)\, dx$$

That means you don't have to use $-\pi$ to $\pi$ in your formulas for the coefficients; you can use 0 to $2\pi$, so you don't have to fiddle with changing the x2.

The other thing you are missing is using the Dirichlet theorem which tells you what the series will converge to at 0 and $2\pi$. Notice that your periodic extension of your function has a jump discontinuity at those points.

Also, one other question. Are you sure your function x2 wasn't given on $(-\pi,\pi)$ in the first place?

LCKurtz said:
You are extending f(x) periodically with period $2\pi$. You get the same thing when you integrate a periodic function over any period. That is, if g(x) has period P, then

$$\int_a^{a+P} g(x)\, dx = \int_b^{b+P} g(x)\, dx$$

That means you don't have to use $-\pi$ to $\pi$ in your formulas for the coefficients; you can use 0 to $2\pi$, so you don't have to fiddle with changing the x2.

The other thing you are missing is using the Dirichlet theorem which tells you what the series will converge to at 0 and $2\pi$. Notice that your periodic extension of your function has a jump discontinuity at those points.

Let me see if I get it: you're saying that x^2.cos(nx) and x^2.sin(nx) are 2Pi periodic, so that change of limits that produced the (-1)^n was wrong? I didn't have to do anything, just integrate them between 0 and 2Pi?

And what do you want me to do with the borders? I know that they will converge to the lateral limits average but I don't know what you mean.

And yes, x^2 was defined in (0, 2Pi).

libelec said:
Let me see if I get it: you're saying that x^2.cos(nx) and x^2.sin(nx) are 2Pi periodic, so that change of limits that produced the (-1)^n was wrong? I didn't have to do anything, just integrate them between 0 and 2Pi?

No, I am not saying that x2cos(nx) and x2sin(nx) are periodic. I am saying that the periodic extensions of them are periodic, and that is what you are expanding in a FS. I didn't go through your steps to see if you had any mistakes an your calculation of an and bn. I am just saying you can use the usual formulas on any period you wish, and in the present case, it is much easier to use $(0,2\pi)$.
And what do you want me to do with the borders? I know that they will converge to the lateral limits average but I don't know what you mean.

And yes, x^2 was defined in (0, 2Pi).

The reason I asked this is because, if I recall correctly, evaluating the even extension of x2 on $(-\pi,\pi)$ at 0 is the way I have seen the problem done.

LCKurtz said:
No, I am not saying that x2cos(nx) and x2sin(nx) are periodic. I am saying that the periodic extensions of them are periodic, and that is what you are expanding in a FS. I didn't go through your steps to see if you had any mistakes an your calculation of an and bn. I am just saying you can use the usual formulas on any period you wish, and in the present case, it is much easier to use $(0,2\pi)$.

But still, the expressions for an and bn, etc, are made for (-$$\pi$$, $$\pi$$), I still have to make some correction when I calculate them for (0, 2$$\pi$$), right?

No. You don't have to change anything but the limits. If f(x) has period $2\pi$ then

$$\frac 1 {\pi}\int_{-\pi}^{\pi} f(x)\cos{nx}\, dx = \frac 1 {\pi}\int_{0}^{2\pi} f(x)\cos{nx}\, dx$$

and ditto for the sine. In the current problem the latter is much preferred because it is the only one in which you can use $f(x) = x^2$.

And looking ahead, I'm guessing you might want to use the even periodic extension of x2, which would be $4\pi$ periodic. As I mentioned before, the usual series for this is the periodic extension of x2 on $(-\pi,\pi)$, but that is apparently not what you have been asked to do.

LCKurtz said:
No. You don't have to change anything but the limits. If f(x) has period $2\pi$ then

$$\frac 1 {\pi}\int_{-\pi}^{\pi} f(x)\cos{nx}\, dx = \frac 1 {\pi}\int_{0}^{2\pi} f(x)\cos{nx}\, dx$$

and ditto for the sine. In the current problem the latter is much preferred because it is the only one in which you can use $f(x) = x^2$.

And looking ahead, I'm guessing you might want to use the even periodic extension of x2, which would be $4\pi$ periodic. As I mentioned before, the usual series for this is the periodic extension of x2 on $(-\pi,\pi)$, but that is apparently not what you have been asked to do.

So I use the even periodic extension for f(x): that is, f(x) = x2 for -2$$\pi$$<x<2$$\pi$$, and then I make the following calculations?:

$${a_0} = \frac{1}{{2\pi }}\int\limits_{ - 2\pi }^{2\pi } {{x^2}dx}$$

$${a_n} = \frac{1}{{2\pi }}\int\limits_{ - 2\pi }^{2\pi } {{x^2}\cos (\frac{1}{{2\pi }}nx)dx}$$

$${b_n} = \frac{1}{{2\pi }}\int\limits_{ - 2\pi }^{2\pi } {{x^2}\sin (\frac{1}{{2\pi }}nx)dx}$$

Is that what you're saying?

There shouldn't be any $\pi$ in your cosines and sines. $2p=4\pi$ so
$$\sin{\frac{n\pi x}{p}} = sin{\frac n 2 x}$$

But that's the way?

LCKurtz said:
There shouldn't be any $\pi$ in your cosines and sines. $2p=4\pi$ so
$$\sin{\frac{n\pi x}{p}} = sin{\frac n 2 x}$$

And, once more, I will add that I suspect you are doing the wrong problem. Maybe there is a typo somewhere. Because if you take the periodic extension of x2 on $(-\pi,\pi)$ and evaluate its series at $x = \pi$ you will get your desired answer.

OK, thanks. One question, though, when do I know when to use the periodic extensions (even an odd extensions)?

libelec said:
OK, thanks. One question, though, when do I know when to use the periodic extensions (even an odd extensions)?

That's a good question. It depends on what you intend to do with the series, but generally, the smoother the periodic function is, the faster the series will converge. In this case, if you take the even extension of x2 you get a continuous function whereas if you take the odd extension you get jump discontinuities at the ends. And the even extension has no sine terms so that's half the work. Another hint in this problem would be the series you are asked about has terms like 1/n2 in it which suggests you need a continuous periodic function.

Thank you.

1. How is the value of pi^2/6 derived using Fourier Series?

The value of pi^2/6 can be derived using the Fourier Series representation of the function f(x) = x^2. This involves representing the function as a sum of sine and cosine functions with specific coefficients, and then using the properties of Fourier Series to evaluate the integral of f(x) over a specific interval. This integral will equal pi^2/6, providing a way to derive the value using the Fourier Series.

2. What is the significance of the value pi^2/6 in relation to the function f(x) = x^2?

The value of pi^2/6 is significant because it represents the sum of the infinite series of squared reciprocals of natural numbers. This is known as the Basel problem, which was famously solved by Leonhard Euler using the Fourier Series representation of f(x) = x^2. The value of pi^2/6 also appears in other mathematical contexts, such as the Riemann zeta function.

3. Can the convergence to pi^2/6 using Fourier Series be proven rigorously?

Yes, the convergence to pi^2/6 using Fourier Series can be proven rigorously using mathematical techniques such as the Weierstrass M-test and the Dirichlet test. These tests can show that the Fourier Series representation of f(x) = x^2 converges uniformly, and therefore the value of pi^2/6 can be derived accurately.

4. Are there any other functions that can be used to derive the value of pi^2/6 using Fourier Series?

Yes, there are many other functions that can be used to derive the value of pi^2/6 using Fourier Series. Some examples include the functions f(x) = x, f(x) = x^3, and f(x) = sin(x). As long as the function satisfies certain conditions, such as being continuous and piecewise differentiable, it can be represented using a Fourier Series and used to derive the value of pi^2/6.

5. What practical applications does the convergence to pi^2/6 using Fourier Series have?

The convergence to pi^2/6 using Fourier Series has many practical applications in mathematics and engineering. It can be used to solve various differential equations, approximate functions, and analyze data. It also has applications in signal processing, image compression, and other fields that involve analyzing periodic functions. Understanding the convergence to pi^2/6 using Fourier Series can help in developing more accurate and efficient methods for solving mathematical problems.

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