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# Fourier series expansion - calculus

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A function is defined as followed:

see attachment

Where f(t+2)=f(t) that is, f(t) has period 2.
i) Draw a plot of the function f(t). Comment fully on whether the function is even or odd or none of these.
ii) Find the first four non-zero coefficients for the Fourier series expansion of the function f(t)
iii) Using eg. excel, draw a plot of the partial sum of the Fourier series which uses just there first four non-zero terms and comment on it in relation to your plot (i)

https://brainmass.com/math/basic-calculus/fourier-series-expansion-calculus-567236

## SOLUTION This solution is FREE courtesy of BrainMass!

The function looks like:

This function is neither even function since , nor it is an odd function since
.
For example, while
However we can write the function as:
(1.1)
Where
(1.2)
And is an odd function and it looks like

The general expansion of a function over the symmetric interval is given by:
(1.3)
Where:
(1.4)
Since the cosine function is even and is odd, then the integrand is odd.
When we integrate an odd function over a symmetric interval, the result is identically 0.
So if we want to expand into a Fourier series, we need only calculate the coefficients of the sine series.
In this case :

(1.5)
This is a simple integral to evaluate:
(1.6)
Since we can write:
(1.7)

However, we see that for even n the coefficient is zero, hence we can write for the non-zero coefficients:
(1.8)
And
(1.9)
But since we are interested in the function , we can write:
(1.10)

The first four terms are:
(1.11)
It looks like:

However, if we use the first 30 terms it looks like:

Note the large oscillations next to the points of discontinuity in the function, known as the "Gibbs phenomenon".

By the way if we would have tried to calculate the cosine coefficients of directly we would have gotten:
(1.12)
But for we get:
(1.13)
And the leading coefficient in the expansion is , as expected.

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