Exercise 1.6 from Stein and Shakarchi’s Fourier Analysis

The problem is as follows:

We have

    \begin{align*}g'(t) &= f'(t) \cos ct - c f(t) \sin ct - c^{-1} f''(t) \sin ct - f'(t) \cos ct \\&= - c f(t) \sin ct - c^{-1} f''(t) \sin ct \\&= - \frac{1}{c} \left( \sin ct) (f''(t) + c^2 f(t) \right) \\&= 0\end{align*}

and

    \begin{align*}h'(t) &= f'(t) \sin ct + c f(t) \cos ct + c^{-1} f''(t) \cos ct - f'(t) \sin ct \\&=  c f(t) \cos ct + c^{-1} f''(t) \cos ct \\&= \frac{1}{c} \left( \cos ct) (f''(t) + c^2 f(t) \right) \\&= 0.\end{align*}

Each equality with 0 follows from the assumption f''(t) + c^2 f(t) = 0. Therefore g'(t) = h'(t) = 0 and g(t), h(t) are constant.

The problem asks for constants a and b. Could it be that g(t) and h(t) are these constants? That would be a natural guess. Let’s just try computing the sum

    \begin{align*}g(t) \cos ct + h(t) \sin ct &= f(t) \cos^2 ct + f(t) \sin^2 ct \\&= f(t).\end{align*}

.

So indeed a = g(t) and b = h(t).

May 10th, 2023 | Posted in Math

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