I have to say, I'm stumped. I've tried solving this problem but it's being a complete gimp and I'm completely fed up with it. Here it is in all its glory:

 

Use Taylor's theorem to prove that the function [math]f:\mathbb{R} \to \mathbb{R}[/math] given by [math]f(x) = e^{ax}\cos(bx+c)[/math] (for [math]a, b, c \in \mathbb{R}[/math]) can be written as the power series [math]f(x) = f(0) + f'(0) x + \cdots + \tfrac{1}{n!} f^{(n)}(0) x^n + \cdots[/math] for all [math]x\in\mathbb{R}[/math].

 

Any help is much appreciated as the question is worth 14 marks/25 in a past paper

(for anyone that's wondering, it's basically Maclaurin's theorem).

But I thought that was Taylor's theorem? You want to find the Taylor series about x=0; that's all there is to it, right?

No, Taylor's theorem is defined over a closed interval (i.e. [math]f:[a,b] \to \mathbb{R}[/math]) and over a finite number of terms and has a remainder at the end of it: i.e.

 

[math]\exists\, c \in [a,b] \mbox{ such that } f(b) = \sum_{i=0}^{n} \frac{f^{(i)}(a)}{i!} + \frac{(b-a)^{n+1}}{(n+1)!} f^{(n+1)}© [/math]

 

Or at least, that's the version I have.

Oh well, it's a bit late now, the exam's tomorrow

 

If anyone has any sudden thoughts, please let me know

 

(all I know is that you have to show the remainder function tends to zero as n->infinity and something about the radius of convergence of that series)