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The Fourier series $$f(t) = d + \sum_{n=-1}^\infty a_n \cos(nt) + b_n \sin(nt),$$

can be written in the form

$$\sum_{n=-\infty}^\infty c_n e^{int},$$

where

$c_n = \begin{cases} d \quad \text{for} \ n = 0 \\ (a - ib_n)/2, \quad \text{for} \ n \ \text{positive} \\ (a_{-n} + ib_{-n})/2, \quad \text{for} \ n \ \text{negative}. \\ \end{cases} $

Now if I expand this I get

\begin{align} f(t) \sum_{n=-\infty}^\infty c_n e^{int} & = \sum_{n=-\infty}^\infty c_n \bigg(\cos(nt) + i \sin(nt)\bigg) \\ & = c_0 + \sum_{n=1}^\infty c_n \bigg(\cos(nt) + i \sin(nt) + \cos(-nt) + i \sin(-nt)\bigg) \\ & = c_0 + \sum_{n=1}^\infty c_n \bigg(\cos(nt) + i \sin(nt) + \cos(nt) - i \sin(nt)\bigg) \\ & = c_0 + \sum_{n=1}^\infty c_n 2\cos(nt). \end{align}

But now the Fourier series has lost its dependence on sin functions! Am I incorrect or is this telling me that we only need $\cos$ functions to represent any periodic function using a Fourier series?

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    $\begingroup$ Is $c_n=c_{-n}$.? $\endgroup$
    – Nosrati
    Mar 8, 2017 at 7:55
  • $\begingroup$ @MyGlasses I have edited the post to specify the $c_n$'s. $\endgroup$ Mar 8, 2017 at 8:02

2 Answers 2

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The cosines are even functions and any (possibly infinite) linear combination of even functions is itself an even function.

This means that if your function is not even, then cosines will not suffice. Conversely, it can be shown that if a function is even, then its Fourier series will contain no sine terms.

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  • $\begingroup$ In the other answer to this question MyGlasses has derived another representation in which, yet again, $f(t)$ is only composed of combinations of $\cos$ functions. But that conflicts with your answer? $\endgroup$ Mar 11, 2017 at 11:53
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    $\begingroup$ The representation he derived is only true if $c_n=c_{-n}$. This happens only if $b_n=0$ for all $n>0$, which means the Fourier series contains no sine terms. $\endgroup$ Mar 11, 2017 at 18:26
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I rewrite your solution: \begin{align} f(t)& = \sum_{n=-\infty}^\infty c_n e^{int}\\ & = \sum_{n=-\infty}^\infty c_n \bigg(\cos(nt) + i \sin(nt)\bigg) \\ & = c_0 + \sum_{n=1}^\infty c_n \bigg(\cos(nt) + i \sin(nt) \bigg)+\sum_{n=1}^\infty c_{-n} \bigg(\cos(-nt) + i \sin(-nt)\bigg) \\ & = c_0 + \sum_{n=1}^\infty \color{blue}{c_n \bigg(\cos(nt) + i \sin(nt) + \cos(nt) - i \sin(nt)\bigg)} \\ & = c_0 + \sum_{n=1}^\infty c_n 2\cos(nt). \end{align} the blue line is true if $c_n=c_{-n}$ or by definition $(a - ib_n)/2=(a_{-n} + ib_{-n})/2$ or $b_n=0$ that means $c_n$'s are real, with other notation $$f(t)=\sum_{n=-\infty}^\infty c_n e^{int}=d+\sum_{n>0}^\infty c_n e^{int}+\sum_{n<0}^\infty c_n e^{int}=d+\sum_{n>0}^\infty( c_n e^{int}+ c_{-n} )e^{-int}=d+\sum_{n>0}^\infty (a_n e^{int}+ a_n e^{-int})=d+\sum_{n>0}^\infty a_n 2\cos nt$$

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  • $\begingroup$ By writing it in the other notation you have ended up with only cos functions also. So does that mean you have shown we only need cos functions to represent any periodic functions $f(t)$? $\endgroup$ Mar 11, 2017 at 11:51
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    $\begingroup$ This result obtain by $b_n=0$ and this will be only for even functions!. $\endgroup$
    – Nosrati
    Mar 11, 2017 at 11:56

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