Q1P Prove (11.4)聽for a function of ... [FREE SOLUTION]

Chapter 7: Q1P (page 377)

Prove (11.4)for a function of period 2Lexpanded in a sine-cosine series.

Short Answer

Expert verified

The required equation that is to be proven is $<f^{2} (x)> = {(\frac{a_{0}}{2})}^{2} + \frac{1}{2} {鈭�}_{n = 1}^{鈭�} (a_{n}^{2} + b_{n}^{2})$

Step by step solution

Definition of Fourier series

A Fourier series is a sum that represents a periodic function as a sum of sine and cosine swells. The frequence of each surge in the sum, or harmonious, is an integer multiple of the periodic function's abecedarian frequence. Each harmonious phase and breadth can be determined using harmonious analysis

Given Parameters

Given the period of a function as 2L

It is to be proven that equation (11.4) , that is $<f^{2} (x)> = {(\frac{a_{0}}{2})}^{2} + \frac{1}{2} {鈭�}_{n = 1}^{鈭�} (a_{n}^{2} + b_{n}^{2})$ stands true for given period of function

Use formula for standard form of Fourier series of function to find the average value of square of function

It is known that Standard form for Fourier series is

$f (x) = \frac{a_{0}}{2} + {鈭�}_{n = 1}^{鈭�} [a_{n} \cos \frac{n 蟺 x}{L} + b_{n} \sin \frac{n 蟺 x}{L}]$

Calculate the average value of square of the function for the given period as

$<f^{2} (x)> = \frac{1}{2 L} {鈭�}_{- L}^{L} {[\frac{a_{0}}{2} + {鈭�}_{n = 1}^{鈭�} [a_{n} \cos \frac{n 蟺 x}{L} + b_{n} \sin \frac{n 蟺 x}{L}]]}^{2} d x = \frac{1}{2 L} {鈭�}_{- L}^{L} {(\frac{a_{0}}{2})}^{2} d x + \frac{1}{2 L} {鈭�}_{- L}^{L} a_{0} {鈭�}_{n = 1}^{鈭�} [a_{n} \cos \frac{n 蟺 x}{L} + b_{n} \sin \frac{n 蟺 x}{L}] d x + \frac{1}{2 L} {鈭�}_{- L}^{L} {[{鈭�}_{n = 1}^{鈭�} [a_{n} \cos \frac{n 蟺 x}{L} + b_{n} \sin \frac{n 蟺 x}{L}]]}^{2} d x$

Here, on further solving observe that the second term of integral will become zero and in the last integral, the sum of square of sine and cosine will integrate to 1/2

Further solve to get average value of square of function as

$<f^{2} (x)> = {(\frac{a_{0}}{2})}^{2} + \frac{1}{2 L} {鈭�}_{- L}^{L} {鈭�}_{n = 1}^{鈭�} (\frac{1}{2} a_{n}^{2} + b_{n}^{2}) d x <f^{2} (x)> = {(\frac{a_{0}}{2})}^{2} + \frac{1}{2} {鈭�}_{n = 1}^{鈭�} (a_{n}^{2} + b_{n}^{2})$

Therefore, the equation (11.4) , that is $<f^{2} (x)> = {(\frac{a_{0}}{2})}^{2} + \frac{1}{2} {鈭�}_{n = 1}^{鈭�} (a_{n}^{2} + b_{n}^{2})$ stands true for given period of function 2l expanded in a sine-cosine series

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91影视

Prove (11.4)for a function of period 2Lexpanded in a sine-cosine series.

Short Answer

Step by step solution

Definition of Fourier series

Given Parameters

Use formula for standard form of Fourier series of function to find the average value of square of function

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