How to find the sum of two series using the Weierstrass M-test?

[Hummingbird25]

Hi

Given the function $$f(t) = t^2$$, were $$t \in ]- \pi, pi[$$, and is continious find the Fourier series for f(t). $$L = 2 \pi$$.

Then

$$A_0 = \frac{1}{2 \pi} \int \limit_{-\pi} ^{\pi} t^2 dt = \frac{\pi ^2}{3}$$

$$A_n = \int \limit_{-\pi} ^{\pi} t^2 \cdot cos(\matrm{n} \pi \mathrm{t}) dt$$

$$A_n = \int \limit_{-\pi} ^{\pi} u^2 \cdot cos(u) du$$, where u = n \pi t,

The new limit gives:

$$A_n = \int \limit_{0} ^{2 \pi^2 n} u^2 \cdot cos(u) du$$

$$A_n = [u^2 \cdot sin(u) - 2sin(u) + 2u sin(u)]_{0} ^{2 \pi ^ 2 n}$$

Then

$$B_n = \int \limit_{0} ^{2n \pi^2} u^2 \cdot sin(u) du$$

which gives

$$B_n = (2-4n^2 * \pi ^4 * cos(2n * \pi ^2) + 4n *sin (2n * \pi ^2) \pi ^2 -2$$

Therefore the Fourier series for f(t) is:

$$\frac{\pi ^ 2}{3} + \sum \limit_{n=1} ^{\infty} A_n cos(nt) + B_n sin(nt)$$

Could someone please me if my calculations are correct?

Sincerley

Yours
Hummingbird

 

[benorin]

you may check here that the Fourier coefficients ought to be

$$A_0=\frac{1}{\pi}\int_{-\pi}^{\pi}f(t) dt$$

$$A_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\cos (nt) dt$$

and

$$B_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\sin (nt) dt$$

so that

for $$f(t)=t^2$$ we have

$$A_0=\frac{1}{\pi}\int_{-\pi}^{\pi}t^2 dt = \frac{2}{\pi}\int_{0}^{\pi}t^2 dt = \frac{2\pi ^3}{3}$$

where the second move is from the integral of an even function over a symmetric interval (i.e. [-a,a]) is twice the integral over [0,a], also note that the same applies to A_n, hence

$$A_n=\frac{1}{\pi}\int_{-\pi}^{\pi}t^2 \cos (nt) dt = \frac{2}{\pi}\int_{0}^{\pi}t^2 \cos (nt) dt$$

integrate by parts twice to get EDIT: forgot to multiply by the $$\frac{1}{\pi}$$!

$$A_n=\frac{2}{\pi}\int_{0}^{\pi}t^2 \cos (nt) dt = \frac{2}{\pi}\left[ \frac{2t^2}{n}\sin (nt) + \frac{4t}{n^2}\cos (nt)-\frac{4}{n^3}\sin (nt)\right]_{t=0}^{\pi} = \frac{1}{\pi}\left( \frac{2\pi ^2}{n}\sin (n\pi ) + \frac{4\pi}{n^2}\cos (n\pi )-\frac{4}{n^3}\sin (n\pi )\right)$$
$$= \frac{2\pi }{n}\sin (n\pi ) + \frac{4}{n^2}\cos (n\pi )-\frac{4}{\pi n^3}\sin (n\pi )$$

know also that $$t^2\sin (nt)$$ is an odd function, and that the integral of an odd function over a symmetric interval (i.e. [-a,a]) is zero, hence

$$B_n=\frac{1}{\pi}\int_{-\pi}^{\pi}t^2\sin (nt) dt=0$$

 

[benorin]

Notably, you may perform definite integration on this http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=calculus&s2=integrate&s3=advanced

 

[Hummingbird25]

Hello ben,

I guess it mixed up the details from a textbook example. Sorry,

Then the Fourier series is expressed:

$$\frac{2\pi ^3}{3} + \sum \limit_{n=1} ^{\infty} \frac{2\pi ^2}{n}\sin (n\pi ) + \frac{4\pi}{n^2}\cos (n\pi )-\frac{4}{n^3}\sin (n\pi )$$ ?

Sincerely
Hummingbird25

you may check here that the Fourier coefficients ought to be

$$A_0=\frac{1}{\pi}\int_{-\pi}^{\pi}f(t) dt$$

$$A_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\cos (nt) dt$$

and

$$B_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\sin (nt) dt$$

so that

for $$f(t)=t^2$$ we have

$$A_0=\frac{1}{\pi}\int_{-\pi}^{\pi}t^2 dt = \frac{2}{\pi}\int_{0}^{\pi}t^2 dt = \frac{2\pi ^3}{3}$$

where the second move is from the integral of an even function over a symmetric interval (i.e. [-a,a]) is twice the integral over [0,a], also note that the same applies to A_n, hence

$$A_n=\frac{1}{\pi}\int_{-\pi}^{\pi}t^2 \cos (nt) dt = \frac{2}{\pi}\int_{0}^{\pi}t^2 \cos (nt) dt$$

integrate by parts twice to get

$$A_n=\frac{2}{\pi}\int_{0}^{\pi}t^2 \cos (nt) dt = \left[ \frac{2t^2}{n}\sin (nt) + \frac{4t}{n^2}\cos (nt)-\frac{4}{n^3}\sin (nt)\right]_{t=0}^{\pi} = \frac{2\pi ^2}{n}\sin (n\pi ) + \frac{4\pi}{n^2}\cos (n\pi )-\frac{4}{n^3}\sin (n\pi )$$

know also that $$t^2\sin (nt)$$ is an odd function, and that the integral of an odd function over a symmetric interval (i.e. [-a,a]) is zero, hence

$$B_n=\frac{1}{\pi}\int_{-\pi}^{\pi}t^2\sin (nt) dt=0$$

 

[benorin]

Hello ben,

I guess it mixed up the details from a textbook example. Sorry,

Then the Fourier series is expressed:

$$\frac{2\pi ^3}{3} + \sum \limit_{n=1} ^{\infty} \frac{2\pi ^2}{n}\sin (n\pi ) + \frac{4\pi}{n^2}\cos (n\pi )-\frac{4}{n^3}\sin (n\pi )$$ ?

Sincerely
Hummingbird25

Rather it is expressed:

$$\frac{1}{2}A_0+\sum_{n=1}^{\infty} A_n\cos (nt)+ \sum_{n=1}^{\infty} B_n\sin (nt)=\frac{\pi ^3}{3} + \sum \limit_{n=1} ^{\infty} \left( \frac{2\pi ^2}{n}\sin (n\pi ) + \frac{4\pi}{n^2}\cos (n\pi )-\frac{4}{n^3}\sin (n\pi )\right) \cos (nt)$$

 

[Hummingbird25]

Okay I get that,

By the way,

f(t) has continious deriatives, and is periodic $$2 \pi$$

Then the Fourier series of f(t) converge to f(t) Uniformly on $$]-\pi, \pi[$$ ??

Or am I missing a condition?

Sincerely

Hummingbird

Rather it is expressed:

$$\frac{1}{2}A_0+\sum_{n=1}^{\infty} A_n\cos (nt)+ \sum_{n=1}^{\infty} B_n\sin (nt)=\frac{\pi ^3}{3} + \sum \limit_{n=1} ^{\infty} \left( \frac{2\pi ^2}{n}\sin (n\pi ) + \frac{4\pi}{n^2}\cos (n\pi )-\frac{4}{n^3}\sin (n\pi )\right) \cos (nt)$$

 

[benorin]

I forgot to distribute the $$\frac{1}{\pi}$$ in the calculation of A_n in my first post, I fixed it: look for the EDIT,

very important simplifications: $$\sin (n\pi ) = 0\mbox{ for }n=1,2,3,...$$

and

$$\cos (n\pi ) = (-1)^{n}\mbox{ for }n=1,2,3,...$$

and hence $$A_n = \frac{2\pi }{n}\sin (n\pi ) + \frac{4}{n^2}\cos (n\pi )-\frac{4}{\pi n^3}\sin (n\pi ) = \frac{4}{n^2}(-1)^{n}$$

so the series becomes

$$\frac{\pi ^3}{3} + \sum \limit_{n=1} ^{\infty} \left( \frac{2\pi }{n}\sin (n\pi ) + \frac{4}{n^2}\cos (n\pi )-\frac{4}{\pi n^3}\sin (n\pi )\right) \cos (nt ) = \frac{\pi ^3}{3} + \sum \limit_{n=1} ^{\infty} (-1)^{n}\frac{4}{n^2}\cos (nt )$$

that is

$$\boxed{t^2 \sim \frac{\pi ^3}{3} + 4\sum \limit_{n=1} ^{\infty} (-1)^{n}\frac{\cos (nt)}{n^2} = \frac{\pi ^3}{3} -4\left( \frac{\cos (t)}{1^2}-\frac{\cos (2t)}{2^2}+\frac{\cos (3t)}{3^2}-\mdots \right)}$$​

 

[benorin]

Uniform convergence can be proved by the Weierstrass M-test, just note that

$$\left| (-1)^{n}\frac{\cos (nt)}{n^2} \right| \leq \frac{1}{n^2}=M_n$$ for all n

and $$\sum \limit_{n=1} ^{\infty}M_n=\sum \limit_{n=1} ^{\infty}\frac{1}{n^2}$$ converges, so the given series converges uniformly on $$\left[ -\pi ,\pi \right]$$ by the Weierstrass M-test.

 

[Hummingbird25]

Uniform convergence can be proved by the Weierstrass M-test, just note that

$$\left| (-1)^{n}\frac{\cos (nt)}{n^2} \right| \leq \frac{1}{n^2}=M_n$$ for all n

and $$\sum \limit_{n=1} ^{\infty}M_n=\sum \limit_{n=1} ^{\infty}\frac{1}{n^2}$$ converges, so the given series converges uniformly on $$\left[ -\pi ,\pi \right]$$ by the Weierstrass M-test.

Okay thank You then I only have one final question.

Given the two series

$$\sum_{k=1} ^{\infty} \frac{1}{k^4}$$

and

$$\sum_{k=1} ^{\infty} (-1)^{k-1} \frac{1}{k^2}$$

I need to find the sum of these two.

In series number 1:

I can see that by the test of comparison, that it converges

$$\frac{1}{k^{2+t}} < \frac{1}{k^{2}}$$

But what is the next step in finding the sum here?

In series two:

What do I here? Do I test for convergens ?

Sincerely
Hummingbird