f(θ) = (8)/(π)∑n=0^∞(sen[(2n+1)θ])/((2n+1)^3)

Name: f(θ) = (8)/(π)∑n=0^∞(sen[(2n+1)θ])/((2n+1)^3)
Uploaded: 2023-07-25T13:47:13-08:00
Duration: 27 s
Channel: Numerade
Description: f(θ) = (8)/(π)∑n=0^∞(sen[(2n+1)θ])/((2n+1)^3)

Question

Chai Santi · Answer

So for this question first we have that f x is an odd function so f minus x is equal to minus fx

Sri K · Answer

Let&#x27;s do the solution of the given question. So we are having fx is equals to pi by 2 for 0 less

Adi S · Answer

as fx is equals to zero when x is greater than zero and less than no and fx will be 18 if x is l

Adi S · Answer

We have f of x equals 2 when x is greater than minus 2 and less than 0, x when x is greater than

Adi S · Answer

In this question we have to find the Fourier series of the function f of x equals to mod of x wh

Question

$f(\theta) = \frac{8}{\pi} \sum_{n=0}^{\infty} \frac{\text{sen}[(2n+1)\theta]}{(2n+1)^3}$

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