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Prove that sums of trig functions can be expressed as sums of complex exponentials

by RoRi

Prove that sums of the form

    \[ S_n (x) = \frac{1}{2}a_0 + \sum_{k=1}^n (a_k \cos (kx) + b_k \sin (kx)) \]

are equal to sums of the form

    \[ S_n (x) = \sum_{k=-n}^n c_k e^{ikx}. \]

Proof. We compute this directly, substituting in the formulas for sine and cosine in terms of the complex exponential,

    \[ \sin (kx) = \frac{e^{ikx} - e^{-ikx}}{2i}, \quad \text{and} \quad \cos (kx) = \frac{e^{ikx} + e^{-ikx}}{2}, \]

that we derived in this exercise. So, we have

    \begin{align*} S_n (x) &= \frac{1}{2}a_0 + \sum_{k=1}^n (a_k \cos (kx) + b_k \sin (kx)) \\[9pt] &= \frac{1}{2}a_0 + \sum_{k=1}^n \left( a_k \frac{e^{ikx} + e^{-ikx}}{2} + b_k \frac{e^{ikx} - e^{-ikx}}{2i} \right) \\[9pt] &= \frac{1}{2}a_0 + \frac{1}{2} \sum_{k=1}^n \left( a_k e^{ikx} + a_k e^{-ikx} - ib_k e^{ikx} + i b_k e^{-ikx} \right) \\[9pt] &= \frac{1}{2} a_0 + \frac{1}{2} \sum_{k=1}^n \left( a_k - ib_k\right)e^{ikx} + \frac{1}{2} \sum_{k=1}^n \left( a_k + ib_k \right)e^{-ikx} \\[9pt] &= \frac{1}{2}a_0 + \frac{1}{2} \sum_{k=1}^n (a_k - i b_k )e^{ikx} + \frac{1}{2} \sum_{k=-n}^{-1} (a_{-k} + ib_{-k})e^{ikx} \\[9pt] &= \frac{1}{2} a_0 + \frac{1}{2} \sum_{k=1}^n c_k e^{ikx} + \frac{1}{2} \sum_{k=-n}^{-1} c_{-k}e^{ikx} \\[9pt] &= \sum_{k=-n}^n c_k e^{ikx} \end{align*}

where

    \[c_k = \frac{1}{2} (a_k - ib_k), \quad c_{-k} = \frac{1}{2}(a_{-k} + ib_{-k}), \quad c_0 = 0. \qquad \blacksquare \]

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