# Taylor Series/Euler's Method

by thegreenlaser
Tags: method, series or euler, taylor
 P: 474 I'm trying to learn about finite difference methods to solve differential equations. I'm using Advanced Engineering Mathematics 9th Ed., and in explaining Euler's method he claims the following Taylor series: $$y(x+h) = y(x) + hy'(x) + \dfrac{h^2}{2}y''(x) + \cdots$$ He then truncates that series, and because the equation to be solved is $$\dfrac{dy}{dx}=f(x,y)$$ he substitutes in f(x,y) for y'(x) in the Taylor series and goes on from there. My question is, isn't the y'(x) in the Taylor series actually $$\dfrac{dy}{dh}$$ and not $$\dfrac{dy}{dx}$$ as the substitution would imply? It seems to me that the variable in that Taylor series is h, with centre 0. I understand Euler's method geometrically, but if someone could explain this Taylor series issue, that would be greatly appreciated.
 P: 474 Taylor Series/Euler's Method Okay, I think I may have got it, but I'm not sure. So the maclaurin series of y(x+h) would be defined as: $$y(x+h) = \displaystyle \sum_{n=0}^{\infty} \frac{h^n}{n!}\cdot \left. \frac{d^n }{dh^n}y(x+h)\right|_{h=0} = y(x) + h \cdot \left. \frac{d }{dh}y(x+h)\right|_{h=0} + \frac{h^2}{2}\cdot \left. \frac{d^2 }{dh^2}y(x+h)\right|_{h=0} + \cdots$$ However, $$\displaystyle \frac{d}{dh}f(x+h) = \lim_{k \to 0} \frac{f(x+h+k)-f(x+h)}{k} = \frac{d}{dx}f(x+h)$$ So the taylor series can be written as: $$\therefore y(x+h) = \displaystyle \sum_{n=0}^{\infty} \frac{h^n}{n!}\cdot \left. \frac{d^n }{dx^n}y(x+h)\right|_{h=0} = y(x) + h \cdot \left. \frac{d }{dx}y(x+h)\right|_{h=0} + \frac{h^2}{2}\cdot \left. \frac{d^2 }{dx^2}y(x+h)\right|_{h=0} + \cdots$$ $$y(x+h) = \displaystyle y(x) + h \cdot \frac{d }{dx}y(x) + \frac{h^2}{2}\cdot\frac{d^2 }{dx^2}y(x) + \cdots$$ So, essentially, because it's a function of (x+h), the x derivative is the same as the h derivative. Is this correct, or is there another reason for the taylor series having the x derivative instead of the h derivative?