Fourier series question

[Krupski69]

I have completed part (a). For part (b) I have tried letting x = 0 and 2pi to cancel the sin terms. however I cannot get the (pi^2)/6 term.

 

[LCKurtz]

You need to be more clear about exactly what you are expanding. Apparently you are taking [MATH]f(x) = x^2,~0 < x < 2\pi[/MATH] and giving the [MATH]2\pi[/MATH] periodic extension of that. Try evaluating that at [MATH]x=0[/MATH] or [MATH]x = 2\pi[/MATH]. Particularly note whether the FS represents a continuous function at either point. Note this is not the same problem as if you took [MATH]f(x) = x^2, ~-\pi < x < \pi[/MATH] and did the periodic extension of that.

 

[Romsek]

The problem is clear about what's being expanded. Take \(\displaystyle f(x) =\begin{cases} x^2 &0\leq x < 2\pi\\0 &\text{else}\end{cases}\) and repeat it every \(\displaystyle \pm 2\pi\)

It's the periodic extension of \(\displaystyle x^2\) with period \(\displaystyle 2\pi\)

The Fourier series coefficients as given in the problem do indeed converge to this.

Now here's the trick. The Fourier series converges to the average of the values at discontinuities.
We have discontinuities at every \(\displaystyle 2\pi k, ~k \in \mathbb{Z}\) and at \(\displaystyle x=0\) the series will converge to
\(\displaystyle \dfrac{4\pi^2 + 0}{2} = 2\pi^2\)

So we have

\(\displaystyle 2\pi^2 = \dfrac{4\pi^2}{3} + 4 \cdot \text{(the desired sum)}\)

\(\displaystyle \dfrac{\pi^2}{2} - \dfrac{\pi^2}{3} = \text{(the desired sum)}\)

\(\displaystyle \dfrac{3\pi^2}{6} - \dfrac{2\pi^2}{6} = \dfrac{\pi^2}{6}\)

 

[LCKurtz]

Begin {Nitpicking}

The problem is clear about what's being expanded. Take \(\displaystyle f(x) =\begin{cases} x^2 &0\leq x < 2\pi\\0 &\text{else}\end{cases}\) and repeat it every \(\displaystyle \pm 2\pi\)

That description is worse than the original (never mind that the tex doesn't render correctly, not your fault). You have defined a function for all [MATH]x[/MATH] and it doesn't make any sense to talk about extending such a function periodically.

It's the periodic extension of \(\displaystyle x^2\) with period \(\displaystyle 2\pi\)

Again, that isn't well defined unless you provide a particular interval of length [MATH]2\pi[/MATH] for which the function is defined. Then you can talk about extending it periodically. That is why I made the comment about the original post not being clear. Any interval of length [MATH]2\pi[/MATH] will contain a period of a [MATH]2\pi[/MATH] periodic function. The original statement of the problem identifies a period but not the domain. That's why the problem is poorly stated.

The Fourier series coefficients as given in the problem do indeed converge to this.

Now here's the trick. The Fourier series converges to the average of the values at discontinuities.
We have discontinuities at every \(\displaystyle 2\pi k, ~k \in \mathbb{Z}\) and at \(\displaystyle x=0\) the series will converge to
\(\displaystyle \dfrac{4\pi^2 + 0}{2} = 2\pi^2\)

So we have

\(\displaystyle 2\pi^2 = \dfrac{4\pi^2}{3} + 4 \cdot \text{(the desired sum)}\)

\(\displaystyle \dfrac{\pi^2}{2} - \dfrac{\pi^2}{3} = \text{(the desired sum)}\)

\(\displaystyle \dfrac{3\pi^2}{6} - \dfrac{2\pi^2}{6} = \dfrac{\pi^2}{6}\)

End{Nitpicking}

And I thought the idea here was not to give complete solutions...

 

[Romsek]

\(\displaystyle f(t)= t^2,~0\leq t < 2\pi\)


\(\displaystyle
T=2\pi\\
f_T(t) = \sum \limits_{k=-\infty}^\infty~f(t-k T)\left(u(t - k T) - u(t - (k+1)T)\right)
\)

\(\displaystyle f_T(t) \text{ is the periodic extension of the domain restricted $f(t)$ with period $2\pi$}\)

any better?

 

[Deleted member 4993]

And I thought the idea here was not to give complete solutions...

You are correct - especially when the OP did not show any work - just claims.

However, the total work was very interesting to me. Made me dust up my old Kaplan's book....

 

[LCKurtz]

\(\displaystyle f(t)= t^2,~0\leq t < 2\pi\)


\(\displaystyle
T=2\pi\\
f_T(t) = \sum \limits_{k=-\infty}^\infty~f(t-k T)\left(u(t - k T) - u(t - (k+1)T)\right)
\)

\(\displaystyle f_T(t) \text{ is the periodic extension of the domain restricted $f(t)$ with period $2\pi$}\)

any better?

Yes. I never doubted you understood it, but whether or not the OP does, who knows. We may never know if he/she doesn't return to the thread. It wasn't clear to me the OP knew the FS sum wasn't continuous, let alone what it converged to at the jumps.