Understanding the Taylor Series in Euler's Method

[thegreenlaser]

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.

 

[gb7nash]

Is your problem with the taylor series itself or what you're plugging into y'(x) for the differential equation?

h is the step value here. You pick h, depending on how small of a step you want to make. So in this case, you can treat h as a constant, but x is the variable. Since y is a function of the variable x, y'(x) is the derivative of y with respect to x (y'(x) is also another way of writing dy/dx).

 

[thegreenlaser]

Is your problem with the taylor series itself or what you're plugging into y'(x) for the differential equation?

h is the step value here. You pick h, depending on how small of a step you want to make. So in this case, you can treat h as a constant, but x is the variable. Since y is a function of the variable x, y'(x) is the derivative of y with respect to x (y'(x) is also another way of writing dy/dx).

I guess my confusion is with the Taylor series itself. I get that h is a constant, but it seems to me that in the taylor series it's being treated as the variable.

Edit: I should note that I've done very little calculus with multivariable functions, so that could be where my confusion lies.

 

[thegreenlaser]

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?