1^()circ Develop power series for f(x)=e^(x cot a)cos x
$1^\circ$ Develop power series for $f(x)=e^{x\cot a}\cos x$, $\varphi(x)=e^{x\cot a}\sin x$. $2^\circ$ Sum $\sum_{k=1}^\infty\frac{\sin ka}{k!}$ and $\sum_{k=1}^\infty\frac{\cos kx}{k!}$. $3^\circ$ Develop Fourier series for $e^{\cos x}\cos(\sin x)$ and $e^{\cos x}\sin(\sin x)$.
Explanation of Method
All three parts flow from a single complex identity: $e^{e^{ix}}=\sum_{n\ge0}e^{inx}/n!$, whose real and imaginary parts are $e^{\cos x}\cos(\sin x)$ and $e^{\cos x}\sin(\sin x)$. Part $1^\circ$ uses the same trick with $x(\cot a+i)=x\cdot e^{ia}/\sin a$ in place of $ix$; Part $2^\circ$ specializes the Part $3^\circ$ identity at particular points.
- Expand $e^{x(\cot a+i)}$ as a power series and separate real/imaginary parts.
- Specialize the complex exponential series $e^{e^{ix}}=\sum e^{inx}/n!$ to sum the two numerical series.
- Extract real and imaginary parts of $e^{e^{ix}}$ directly as the requested Fourier series.
Part 1° · Power series for $f,\varphi$
Since $\cot a+i=\dfrac{\cos a+i\sin a}{\sin a}=\dfrac{e^{ia}}{\sin a}$: $$ e^{x(\cot a+i)} = e^{x\cot a}(\cos x+i\sin x) = \sum_{n=0}^\infty\frac{x^n(\cot a+i)^n}{n!} = \sum_{n=0}^\infty\frac{x^ne^{ina}}{n!\sin^na}. $$ Separating real and imaginary parts: $$ \boxed{f(x)=e^{x\cot a}\cos x = \sum_{n=0}^\infty\frac{x^n\cos(na)}{n!\sin^na}, \qquad \varphi(x)=e^{x\cot a}\sin x = \sum_{n=0}^\infty\frac{x^n\sin(na)}{n!\sin^na}} $$
Part 2° · Summing the two series
From $\sum_{k=0}^\infty\dfrac{e^{ika}}{k!}=e^{e^{ia}}=e^{\cos a}[\cos(\sin a)+i\sin(\sin a)]$, taking imaginary parts (the $k=0$ term contributes $0$): $$ \boxed{\sum_{k=1}^\infty\frac{\sin ka}{k!} = e^{\cos a}\sin(\sin a)} $$ From $\sum_{k=0}^\infty\dfrac{e^{ikx}}{k!}=e^{\cos x}[\cos(\sin x)+i\sin(\sin x)]$, taking real parts (the $k=0$ term contributes $1$): $$ \boxed{\sum_{k=1}^\infty\frac{\cos kx}{k!} = e^{\cos x}\cos(\sin x)-1} $$ (Verified numerically at $a=0.9,x=1.2$: both identities match to 12+ decimal places.)
Part 3° · Fourier series
Directly from $e^{e^{ix}}=\sum_{n=0}^\infty\dfrac{\cos(nx)+i\sin(nx)}{n!}$: $$ \boxed{e^{\cos x}\cos(\sin x) = 1+\sum_{n=1}^\infty\frac{\cos(nx)}{n!}}, \qquad \boxed{e^{\cos x}\sin(\sin x) = \sum_{n=1}^\infty\frac{\sin(nx)}{n!}} $$ (Verified numerically at $x=1.2$: both series match their closed forms to 12+ decimal places.)
Summary Table
| Quantity | Value |
|---|---|
| $f(x)=e^{x\cot a}\cos x$ | $\sum\dfrac{x^n\cos(na)}{n!\sin^na}$ |
| $\varphi(x)=e^{x\cot a}\sin x$ | $\sum\dfrac{x^n\sin(na)}{n!\sin^na}$ |
| $\sum\sin(ka)/k!$ | $e^{\cos a}\sin(\sin a)$ |
| $\sum\cos(kx)/k!$ | $e^{\cos x}\cos(\sin x)-1$ |
| Fourier series (cos) | $e^{\cos x}\cos(\sin x)=1+\sum\cos(nx)/n!$ |
| Fourier series (sin) | $e^{\cos x}\sin(\sin x)=\sum\sin(nx)/n!$ |
Additional Notes
- The single generating identity $e^{e^{ix}}=\sum e^{inx}/n!$ unifies all three parts of this problem — Part $1^\circ$ is the same identity with $ix$ replaced by the more general complex number $x(\cot a+i)$, and Part $2^\circ$ is Part $3^\circ$ evaluated at specific points.
- Because the coefficients $1/n!$ decay faster than any power of $n$, these are among the most rapidly convergent series in the entire collection — a direct consequence of $e^{\cos x}\cos(\sin x)$ and its sine companion being entire (analytic on all of $\mathbb C$), the strongest possible smoothness condition.
- This factorial-coefficient Fourier series is the natural $2\pi$-periodic analogue of the Bessel-function generating identity $e^{z\cos\theta}=\sum J_n(z)$-type expansions, though here the exponent's linearity in $e^{ix}$ (rather than $\cos\theta$) keeps the coefficients elementary rather than requiring Bessel functions.