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Taylor Series/Euler's Method

by thegreenlaser
Tags: method, series or euler, taylor
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thegreenlaser
#1
Jul8-11, 04:23 PM
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:

[tex]y(x+h) = y(x) + hy'(x) + \dfrac{h^2}{2}y''(x) + \cdots[/tex]

He then truncates that series, and because the equation to be solved is [tex]\dfrac{dy}{dx}=f(x,y)[/tex] 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 [tex]\dfrac{dy}{dh}[/tex] and not [tex]\dfrac{dy}{dx}[/tex] 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.
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gb7nash
#2
Jul8-11, 05:10 PM
HW Helper
P: 805
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
#3
Jul8-11, 05:26 PM
P: 474
Quote Quote by gb7nash View Post
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
#4
Jul8-11, 06:09 PM
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:
[tex]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[/tex]

However,
[tex]\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)[/tex]

So the taylor series can be written as:
[tex]\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[/tex]
[tex]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[/tex]

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?


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