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