Sum of infinite Fourier series

[bobred]

Homework Statement

Show that

\sum_{r=0}^\infty\frac{1}{(2r+1)^2}=\frac{\pi^2}{8}

Homework Equations

The equation of the function is

F(t)&=&\dfrac{\pi}{4}-\dfrac{2}{\pi}\left(\cos t+\dfrac{\cos3t}{3^{2}}+\dfrac{\cos5t}{5^{2}}+\cdots\right)-\left(\sin t-\dfrac{\sin2t}{2}+\dfrac{\sin3t}{3}-\cdots\right)

The Attempt at a Solution

We are given the condition that t=0, so the cos terms are all 1 giving

\left(\cos t+\dfrac{\cos3t}{3^{2}}+\dfrac{\cos5t}{5^{2}}+\cdots\right)=\dfrac{1}{1^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{5^{2}}+\dfrac{1}{7^{2}}\cdots=\frac{\pi^2}{8}

 

[xcvxcvvc]

Homework Statement

Show that

\sum_{i=0}^\infty\frac{1}{(2r+1)^2}=\frac{1}{8}

Homework Equations

The Attempt at a Solution

Did you write the question incorrectly? The expression that you're summing doesn't have the indexing variable in it anywhere, so it diverges toward infinity:
\sum_{i=0}^\infty\frac{1}{(2r+1)^2}=\frac{1}{(2r+1)^2}\sum_{i=0}^\infty1=\frac{1}{(2r+1)^2}\infty = \infty

 

[hunt_mat]

I think it's just a typo.

 

[bobred]

Hi

The i should have been r.

I know from elsewhere that the sum is \frac{\pi^{2}}{8}, but I haven't really shown that, any hints?

Thanks

James

 

[hunt_mat]

You have to know the original function which the series you have written down represents. Evaluate this function at t=0 and do a little algebra.

Mat

 

[bobred]

The original function is piecewise

f(t)=\begin{cases}<br /> -t &amp; \left(-\pi&lt;t\leq0\right)\\<br /> 0 &amp; \left(0&lt;t\leq\pi\right)\end{cases}

With a period of 2\pi

James

 

[hunt_mat]

You have the value of it's Fourier series at t=0, now calculate what f(0) is and equate the two values.

 

[bobred]

Hi

Of course, duh.

Thanks for your help

James

 

[bobred]

Hi

I am next asked to choose an appropriate value for t and find the value of convergence of the following,

<br /> \sum_{r=0}^\infty\frac{(-1)^r}{2r+1}=\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}<br />

I know this converges to \frac{\pi}{4}, but I am not sure where to start, the function is still the same as at the beginning of the thread

Any ideas where to start?

Thanks, James

 

[Gib Z]

Integrate both the original function and its Fourier Series term by term. Then try to substitute some value of x into get that result.

 

[hunt_mat]

Don't integrate but differentiate.

 

[Gib Z]

Don't integrate but differentiate.

Whoops, my mistake.

 

[bobred]

I differentiate the approximation and set t=-\frac{\pi}{2} giving

\frac{\pi}{2}=\frac{2}{\pi}\sum_{r=0}^\infty\frac{1}{2r+1}\sin((2r+1)t) as sin alternates sign we get

\frac{\pi}{2}=\frac{2}{\pi}\sum_{r=0}^\infty\frac{(-1)^r}{2r+1}

To which I find

\sum_{r=0}^\infty\frac{(-1)^r}{2r+1}=\frac{\pi^2}{4} and not \frac{\pi}{4}

Any ideas where I'm going wrong?

Thanks

 

[Dickfore]

<br /> \sum_{r = 0}^{\infty}{\frac{1}{(2 r + 1)^{2}}} = \sum_{r = 0}^{\infty}{\frac{1}{r^{2}}} - \sum_{r = 1}^{\infty}{\frac{1}{(2 r)^{2}}} = \left(1 - \frac{1}{4}\right) \, \sum_{r = 1}^{\infty}{\frac{1}{r^{2}}}<br />

The sum:

<br /> \zeta(2) \equiv \sum_{r = 1}^{\infty}{\frac{1}{r^{2}}}<br />

can be found by expanding the function f(x) = x^{2} in Fourier series in the region [-\pi, \pi] and taking its value for x = \pi.

 

[bobred]

Hi

Sorry for my ignorrance, but I can't see how this helps with the question (see post#9)

Jamees

 

[Mute]

I differentiate the approximation and set t=-\frac{\pi}{2} giving

\frac{\pi}{2}=\frac{2}{\pi}\sum_{r=0}^\infty\frac{1}{2r+1}\sin((2r+1)t) as sin alternates sign we get

\frac{\pi}{2}=\frac{2}{\pi}\sum_{r=0}^\infty\frac{(-1)^r}{2r+1}

To which I find

\sum_{r=0}^\infty\frac{(-1)^r}{2r+1}=\frac{\pi^2}{4} and not \frac{\pi}{4}

Any ideas where I'm going wrong?

Thanks

The problem is that you differentiated f(t), so your series converges to f'(t), which is either equal to -1 or 0, but you set to it f'(t=-pi/2) = pi/2 and equated the pi/2 to the series instead of the actual value of f(t=-pi/2) = -1.

 

[vela]

Why not just set t=pi/2 in the original function?

 

[bobred]

Hi

If I put t=\pi/2 the f(\pi/2)=0 and we get

\frac{\pi}{4}=\sum_{r=0}^\infty\frac{(-1)^r}{2r+1}

Thanks everyone for your help.

James