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Fourier Series Formula

I get the reason why the Fourier Series sums up the sine and cosine function, but I do not understand what the purpose of the $a_0$ in the front? Is the $a_0$ there in case the original $f(x)$ function does not go through $(0, 0)$. I am no math genius so I am sorry if this a stupid question

T C
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user510
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  • $a_0$ represents the zero-frequency $a_0 \cos(0x) = a_0$. We could also try to add a term for $b_0 \sin(0x)$, but that would always be equal to zero so it would be pointless to include. – Erick Wong Jan 01 '17 at 08:38

3 Answers3

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For periodic sine and cosine functions, the average value of the function over a period is zero. i.e. $$\frac 1T\int_T \sin\left(\frac{2\pi}T x\right)dx=0$$ This means if the fourier series of a function is totally expressed in terms of pure sine and cosine functions, then its average value is zero. But sometimes we have a periodic function whose average value is not zero. For example:

wave

So here comes the mighty $a_0$ to get rid of that bias and express the rest of function in terms of sines and cosines.

polfosol
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You're quite right; you might find it an enlightening exercise to try and work out what would the Fourier series of $x \mapsto 1$ would be if you weren't allowed access to the $n=0$ term of the cosine expansion.

Patrick Stevens
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$b_{n} sin(nx) + a_{n} cos(nx)$ put $n = 0$.

Rajesh D
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  • I have never seen negative indices used for the sine coefficients of a Fourier series. From where does this convention originate? – Erick Wong Jan 01 '17 at 11:10
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    @ErickWong: Let's not be mean, it's obviously a typo: on standard US keyboards "_" (used for lower indices) is on the same key as "-", so here it seems that both symbols have been input by mistake. – Alex M. Jan 01 '17 at 11:18
  • edited. Thanks @ErickWong – Rajesh D Jan 01 '17 at 11:35
  • @AlexM. It wasn't obviously a typo to me, and I was genuinely curious. I could very easily imagine some author, perhaps in an applied field, saying "let's take two sets of coefficients, one indexed by $\mathbb Z^{\ge 0}$ and the other indexed by $\mathbb Z^{>0}$, and merge them into one coefficient set indexed by $\mathbb Z$ by negating one of them". There is an attractive parsimony in that arrangement, even though it makes for a non-uniform definition of coefficients. – Erick Wong Jan 01 '17 at 11:50
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    in Engineering! – Rajesh D Jan 01 '17 at 11:53