# [SOLVED]differentiating a fourier series

#### dwsmith

##### Well-known member
What are rules for differentiating a Fourier series?

For example, given
$$f = \frac{4}{\pi}\sum_{n=1}^{\infty}\frac{\sin(2n-1)\theta}{2n-1} = \begin{cases} 1, & 0 < \theta < \pi\\ 0, & \theta = 0, \pm\pi\\ -1, & -\pi < \theta < 0 \end{cases}$$

Can this be differentiating term wise? If so, what conditions must be met?

#### dwsmith

##### Well-known member
In order to differentiate a Fourier sine series,

$f(\theta)$ is piecewise smooth on $[0,\pi]$

$f(\theta)$ is piecewise continuous on $[0,\pi]$.

$f(0) = f(\pi)$

All three are met.

$$\frac{4}{\pi}\sum_{n = 1}^{\infty}\cos(2n-1)\theta$$

Is this right?

#### Sudharaka

##### Well-known member
MHB Math Helper
What are rules for differentiating a Fourier series?

For example, given
$$f = \frac{4}{\pi}\sum_{n=1}^{\infty}\frac{\sin(2n-1)\theta}{2n-1} = \begin{cases} 1, & 0 < \theta < \pi\\ 0, & \theta = 0, \pm\pi\\ -1, & -\pi < \theta < 0 \end{cases}$$

Can this be differentiating term wise? If so, what conditions must be met?
Hi dwsmith, Refer the following links to find the criteria that must be met in order for the derivative/integral of a Fourier series to be equal to the derivative/integral of the associated function.

http://www.aerostudents.com/files/partialDifferentialEquations/fourierSeries.pdf

Pauls Online Notes : Differential Equations - Convergence of Fourier Series

Kind Regards,
Sudharaka.

#### dwsmith

##### Well-known member
Hi dwsmith, Refer the following links to find the criteria that must be met in order for the derivative/integral of a Fourier series to be equal to the derivative/integral of the associated function.

http://www.aerostudents.com/files/partialDifferentialEquations/fourierSeries.pdf

Pauls Online Notes : Differential Equations - Convergence of Fourier Series

Kind Regards,
Sudharaka.
I did. The derivative should be 0 but when theta is 0 the series is 4/Pi. What is wrong?

And I am lost on this part too.

Does the Fourier series in part (b) converge for any value of $\theta$ other than $\theta = \pm\frac{\pi}{2}$?
$$f'(\theta) = \frac{4}{\pi}\sum_{n = 1}^{\infty}\cos(2n - 1)\theta = \frac{2}{\pi}\text{Re}\left(\exp\left[i(2n - 1)\theta\right]\right)$$

#### Sudharaka

##### Well-known member
MHB Math Helper
I did. The derivative should be 0 but when theta is 0 the series is 4/Pi. What is wrong?

And I am lost on this part too.

Does the Fourier series in part (b) converge for any value of $\theta$ other than $\theta = \pm\frac{\pi}{2}$?
$$f'(\theta) = \frac{4}{\pi}\sum_{n = 1}^{\infty}\cos(2n - 1)\theta = \frac{2}{\pi}\text{Re}\left(\exp\left[i(2n - 1)\theta\right]\right)$$
The function $$f$$ is not continuous at $$x=0$$. Therefore the derivative series converges to $$f'$$ only in the intervals, $$(-\pi,0)\mbox{ and }(0,\pi)$$.

What do you mean by "part (b)" ?

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#### dwsmith

##### Well-known member
The function $$f$$ is not continuous at $$x=0$$. Therefore the derivative series is only valid in the intervals, $$(-\pi,0)\mbox{ and }(0,\pi)$$.

What do you mean by "part (b)" ?
Part b is the derivative series.

#### Sudharaka

##### Well-known member
MHB Math Helper
In order to differentiate a Fourier sine series,

$f(\theta)$ is piecewise smooth on $[0,\pi]$

$f(\theta)$ is piecewise continuous on $[0,\pi]$.

$f(0) = f(\pi)$

All three are met.

$$\frac{4}{\pi}\sum_{n = 1}^{\infty}\cos(2n-1)\theta$$

Is this right?
Sorry for the confusion, but I think I had overlooked something here. Note that the second condition should read, "$f(\theta)$ is continuous on $[0,\pi]$". Refer >>this<<. This is not satisfied by $$f$$ and hence it is not differentiable term by term.

Kind Regards,
Sudharaka.

#### dwsmith

##### Well-known member
Sorry for the confusion, but I think I had overlooked something here. Note that the second condition should read, "$f(\theta)$ is continuous on $[0,\pi]$". Refer >>this<<. This is not satisfied by $$f$$ and hence it is not differentiable term by term.

Kind Regards,
Sudharaka.
Since there is no Fourier series, it doesn't converge for theta = 0?

#### Sudharaka

##### Well-known member
MHB Math Helper
Since there is no Fourier series, it doesn't converge for theta = 0?
The Fourier series of $$f$$ is,

$f(\theta) = \frac{4}{\pi}\sum_{n=1}^{\infty}\frac{\sin(2n-1)\theta}{2n-1} = \begin{cases} 1, & 0 < \theta < \pi\\ 0, & \theta = 0, \pm\pi\\ -1, & -\pi < \theta < 0 \end{cases}$

You cannot differentiate this series term by term to obtain the Fourier series of $$f'$$ since the function $$f$$ is not right/left continuous at the endpoints of $$[-\pi,\pi]$$. Does this clarify things for you?

#### dwsmith

##### Well-known member
The Fourier series of $$f$$ is,

$f(\theta) = \frac{4}{\pi}\sum_{n=1}^{\infty}\frac{\sin(2n-1)\theta}{2n-1} = \begin{cases} 1, & 0 < \theta < \pi\\ 0, & \theta = 0, \pm\pi\\ -1, & -\pi < \theta < 0 \end{cases}$

You cannot differentiate this series term by term to obtain the Fourier series of $$f'$$ since the function $$f$$ is not right/left continuous at the endpoints of $$[-\pi,\pi]$$. Does this clarify things for you?
No. The question ask if the derivative series converges when theta is 0. Since there is no derivative series, the series doesn't converge when theta is zero, correct?

#### Sudharaka

##### Well-known member
MHB Math Helper
No. The question ask if the derivative series converges when theta is 0. Since there is no derivative series, the series doesn't converge when theta is zero, correct?
It is possible that in the question the term "derivative series" means the series,

$\sum_{n=1}^{\infty}\cos(2n-1)\theta$

This series diverges at $$\theta=0$$.