Does the limit of these sums converge/diverge?

Question

I am looking at these two sums:

$$s_N=\sum_{n=0}^N\frac{\sin n}{n!}x^n, \qquad \text{and} \qquad c_N=\sum_{n=0}^N\frac{\cos n}{n!}x^n, \quad \text{for} \quad x\in \mathbb{R}. $$

I am interested in the the expression $Q(x,N)=\sqrt{s_N^2+c_N^2}$ and the question of whether it converges towards a function or not. In particular, I have the suspicion that:

$$\lim_{N \to \infty}Q(x,N) = e^{qx}, \qquad \text{where} \qquad 0.5403 < q < 0.5404.$$

I only suspect this because I plugged $Q$ into a graphics calculator and nudged $q$ around for a bit. But is my assumption true? How can I prove that $Q$ diverges or converges?

When trying to calculate $s_N^2+c_N^2$, I tried the formula

$$\left( \sum_{i=0}^N a_i \right)^2= \sum_{i=0}^N a_i^2+2\sum_{i<j}^N a_ia_j$$

which, by using $\sin^2n+\cos^2n =1,$ gives me

$$s_N^2+c_N^2 = \sum_{n=0}^N\left( \frac{x^{2n}}{n!^2} \right) + 2\sum_{n<m}^N \frac{\sin(n)\sin(m)+\cos(m)\cos(n)}{n!m!}x^{n+m},$$

but I don't know how to continue from here.

Answer 1

We have $$g(x)+if(x)=\sum_{n=0}^\infty {e^{ni}x^n\over n!}=e^{e^ix}=e^{x\cos 1+ix\sin 1}\\ =e^{x\cos 1}e^{ix\sin 1}$$ Then $$h(x)=|g(x)+if(x)|=e^{x\cos 1}$$

Answer 2

The power series $\sum_{n=0}^\infty \frac{\sin n}{n!} x^n$, $\sum_{n=0}^\infty \frac{\cos n}{n!} x^n$ both have radius of convergence $r=\infty$. Hence for each $x \in \mathbb{R}$ we have $$ s_N(x) \to \sum_{n=0}^\infty \frac{\sin n}{n!} x^n =:f(x), \quad c_N(x) \to \sum_{n=0}^\infty \frac{\cos n}{n!} x^n =:g(x) \quad (N \to \infty). $$ Hence $Q_x(x,N) \to \sqrt{f(x)^2+g(x)^2} =:h(x)$. To evaluate $h$ consider: $$ f'(x)=\sum_{n=1}^\infty \frac{\sin n}{(n-1)!} x^{n-1} = \sum_{n=0}^\infty \frac{\sin (n+1)}{n!} x^n = \cos(1) f(x) + \sin(1) g(x), \quad f(0)= 0, $$ $$ g'(x)=\sum_{n=1}^\infty \frac{\cos n}{(n-1)!} x^{n-1} = \sum_{n=0}^\infty \frac{\cos (n+1)}{n!} x^n = - \sin(1) f(x)+\cos(1) g(x), \quad g(0)=1. $$ Thus $$ (f^2+g^2)'=2ff'+2gg'= 2f(\cos(1) f + \sin(1) g) + 2g(- \sin(1) f+\cos(1) g) $$ $$ =2\cos(1)(f^2 + g^2), \quad f^2(0)+g^2(0)=1. $$ This is a linear IVP for $f^2+g^2$ with solution $$ f^2(x)+g^2(x)=e^{2\cos(1)x}. $$ Thus $h(x)= e^{\cos(1)x}$.

Answer 3

You could even continue and show that $$\Big(Q(x,N)\Big)^2=e^{2 x \cos (1)}-$$ $$\sqrt{\frac{2}{\pi }}\,\frac{ x^{N+1}\,e^N}{N^{\frac{2N+3}{2}}}\, \cos \big(N+1-x \sin (1)\big)\, e^{x \cos (1)}+\cdots$$