This is the Fourier Series representation for a periodic function with period 2p, given in my lecture note.

$\dfrac{a_0}{2} + \sum_{n=1}^{\infty}(a_n cos(\dfrac{n\pi t}{p})+b_nsin(\dfrac{n\pi t}{p}))$

with the coefficients

$a_0 = \dfrac{1}{p} \int_c^{c+2p} f(t) dt$

$a_n = \dfrac{1}{p} \int_c^{c+2p} f(t) cos(\dfrac{n\pi t}{p})dt$

$b_n = \dfrac{1}{p} \int_c^{c+2p} f(t) sin(\dfrac{n\pi t}{p})dt$

From my understding, if the given function is even, there will be only $cosine$ term in the series, thus $b_n = 0$. But if the function is odd, there will only be $sine$ term in the series which means $a_0$ and $a_n$ are $0$.

By applying the above method, I can only find just the necessary coefficients and avoid finding all unless the function is neither odd nor even.

So, basically, what I understood was, if the function is odd, then it can be represented by the Fourier Sine Series and Cosine Series if it is even (correct me if i am wrong).

Everything was clear until I came across this following definition for Odd function:

If $f(t)$ is an odd function defined on the inverval $-p\leq t\leq p$ with a period of 2p then

$f(t) = \sum_{n=1}^{\infty}b_n sin(\dfrac{n\pi t}{p})$

with coefficient of

$b_n = \dfrac{2}{p}\int_0^p f(t) sin(\dfrac{n\pi t}{p})dt$

Looks like it is only integrating from 0 to p (half of period) and multiply by 2 which result in having $\dfrac{2}{p}$ instead of $\dfrac{1}{p}$ like the previous definition.

But I don't think we can simply multiply by 2 after integrating from 0 to p as the area of positive region times 2 is not equal to the area of positive region plus area of negative region.

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Now I am confused between whether to use $\dfrac{1}{p}$ or $\dfrac{2}{p}$ while finding coefficients.If both are correct then when do i use the each of them?


1 Answer 1


Note that the function $f(t)$ is odd, but the integrand $f(t)\sin(n\pi t/p)$ is even ('odd' times 'odd' is 'even'). This is why you can integrate over half the interval and just multiply by 2.


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