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[SOLVED] Fourier coefficient theta^2

dwsmith

Well-known member
Feb 1, 2012
1,673
We have $f(\theta) = \theta^2$ for $-\pi < \theta\leq \pi$.

$$
a_0 = \frac{1}{\pi}\int_{-\pi}^{\pi}\theta^2 d\theta = \frac{2\pi^2}{3}
$$

$$
a_n = \frac{1}{\pi}\left[\left.\frac{\theta^2}{n}\sin n\theta\right|_{-\pi}^{\pi}-\frac{1}{n}\left[\left.\frac{-\theta}{n}\cos n\theta\right|_{-\pi}^{\pi}+\int_{-\pi}^{\pi}\cos n\theta d\theta\right]\right] = \frac{2}{n\pi}\left[\pi^2\sin n\pi - \frac{1}{n}\sin n\pi\right]
$$

However, since I have only sine, all the coefficients would be zero. Is that correct?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
We have $f(\theta) = \theta^2$ for $-\pi < \theta\leq \pi$.

$$
a_0 = \frac{1}{\pi}\int_{-\pi}^{\pi}\theta^2 d\theta = \frac{2\pi^2}{3}
$$

$$
a_n = \frac{1}{\pi}\left[\left.\frac{\theta^2}{n}\sin n\theta\right|_{-\pi}^{\pi}-\frac{1}{n}\left[\left.\frac{-\theta}{n}\cos n\theta\right|_{-\pi}^{\pi}+\int_{-\pi}^{\pi}\cos n\theta d\theta\right]\right] = \frac{2}{n\pi}\left[\pi^2\sin n\pi - \frac{1}{n}\sin n\pi\right]
$$

However, since I have only sine, all the coefficients would be zero. Is that correct?
I should have this instead
$$\frac{1}{\pi}\left[\frac{2\pi^2}{n}\sin n\pi - \frac{1}{n}\left[\frac{-2\pi}{n}\cos n\pi - \frac{2}{n}\sin n\pi\right]\right]= \frac{2}{n^2}(-1)^n =
\begin{cases}
\frac{2}{n^2} & \text{if n even}\\
\frac{-2}{n^2} & \text{if n odd}
\end{cases}
$$

Correct?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
We have $f(\theta) = \theta^2$ for $-\pi < \theta\leq \pi$.

$$
a_0 = \frac{1}{\pi}\int_{-\pi}^{\pi}\theta^2 d\theta = \frac{2\pi^2}{3}
$$

$$
a_n = \frac{1}{\pi}\left[\left.\frac{\theta^2}{n}\sin n\theta\right|_{-\pi}^{\pi}-\frac{1}{n}\left[\left.\frac{-\theta}{n}\color{red}{\cos n\theta}\right|_{-\pi}^{\pi}+\int_{-\pi}^{\pi}\cos n\theta d\theta\right]\right] = \frac{2}{n\pi}\left[\pi^2\sin n\pi - \frac{1}{n}\sin n\pi\right]
$$

However, since I have only sine, all the coefficients would be zero. Is that correct?
You don't "have only sine"!
 

dwsmith

Well-known member
Feb 1, 2012
1,673
You don't "have only sine"!
I made the correction is post 2.
Is this the solution?
$$
\frac{\pi^2}{3} + 2\sum_{n = 1}^{\infty}\left[\frac{(-1)^n}{n}\cos n\theta\right].
$$
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
I made the correction is post 2. (I didn't see that when making my previous post.)
Is this the solution?
$$
\frac{\pi^2}{3} + 2\sum_{n = 1}^{\infty}\left[\frac{(-1)^n}{n}\cos n\theta\right].
$$
Correct except that I think that the 2 should be a 4. It looks as though you left out the 2 in the derivative of $\theta^2$, when doing the integration by parts.