Question

Use a power series and the fact that \(i^{2}=-1\) to verify the formula. $$ g(x)=\frac{1}{2}\left(e^{i x}+e^{-i x}\right)=\cos x $$

Short answer

\(\frac{1}{2} \left(e^{i x}+e^{-i x}\right) = \cos x\). The given formula is verified through the use of the power series of the exponential function and the property of \(i^{2} = -1\).

Step-by-step solution

Write the power series representation of \(e^{ix}\) and \(e^{-ix}\)

Power series representation of an exponential function \(e^{u}\) is given by \(\sum_{n=0}^{\infty}\frac{u^{n}}{n!}\). So we can write \(e^{ix}\) and \(e^{-ix}\) as power series: \(e^{ix}=\sum_{n=0}^{\infty}\frac{(ix)^{n}}{n!}\) and \(e^{-ix}=\sum_{n=0}^{\infty}\frac{(-ix)^{n}}{n!}\).

Break down the power series

Now break down these series by separating the even and odd powers terms. For \(e^{ix}\), the power series becomes \(\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!} + i \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}\). Similarly for \(e^{-ix}\), the power series becomes \(\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!} - i \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}\). Note here that \(\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!}\) is the power series for \(\cos x\) and \(\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}\) is the power series for \(\sin x\).

Add the two series

Now we add the two series \(e^{ix}\) and \(e^{-ix}\). All the imaginary parts cancel out and we get \(\frac{1}{2} \left(e^{i x}+e^{-i x}\right) = \frac{1}{2} \left(2 \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!}\right) = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!}\). This series is the power series representation of \(\cos x\).