# Proving that two representations of a Fourier series are the same

I have to show that

$$\sum_{n=0}^\infty A_n\cos\left({xn\frac{2\pi}{T}-\theta_n}\right) \equiv \sum_{n=-\infty}^\infty c_n \mathrm{e}^{\left({ixn\frac{2\pi}{T}}\right)}$$

I have tried two approaches: firstly I started with the right hand side, used $e^{i\theta} = \cos\theta + i\sin\theta$ to convert the complex exponential into a trigonometric form and then tried to express this in harmonic form to obtain the left hand side. I obtained two inconsistent equations, namely $c_nR\cos\alpha = 1$ and $c_nR\sin\alpha = i$ for some constants $R$ and $\alpha$.

My second approach was to start with the left hand side and use the fact that $\cos\theta = \frac{e^{i\theta}+e^{-i\theta}}{2}$ to obtain

$$\sum_{n=0}^\infty \frac{A_n}{2} \left({e^{i\left({xn\frac{2\pi}{T}+\theta_n}\right)}+e^{-i\left({xn\frac{2\pi}{T}+\theta_n}\right)}}\right)$$

which I cannot simplify to being equivalent to $\sum_{n=-\infty}^\infty c_n \mathrm{e}^{\left({ixn\frac{2\pi}{T}}\right)}$.

Are my approaches correct, or do I need to rethink the problem? Please can someone give me a hint as to how I can continue?

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Identity not holds because left-hand side depends on $\theta_n$ whereas right-hand side does not depend of them. – M. Strochyk Oct 15 '12 at 14:44
Hint: In your last displayed summation, $n$ varies from $0$ to $\infty$ but the summands contain both positive and negative values of $n$. So break up the sum into two different sums, where in one sum $n$ ranges from $1$ to $\infty$ and in the other $n$ ranges from $-1$ to $-\infty$, plus a separate term for the case $n=0$. Then define $c_n$ appropriately. Note that $c_n$ are complex numbers, and so, for example, it is perfectly acceptable to set $c_n = \frac{A_n}{2}e^{i\theta_n}$ for $n > 0$. – Dilip Sarwate Oct 15 '12 at 14:54
M. Strochyk: Is that correct? Can't $\theta_n$ just be incorporated into $c_n$ or something like that? Dilip Sawate: Thanks! I'll try that now. – Froskoy1 Oct 15 '12 at 14:55
Dilip Sawate: Thanks so much for your help! I've got the correct answer now. – Froskoy1 Oct 15 '12 at 15:00