Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Taylor Series/Euler's Method

  1. Jul 8, 2011 #1
    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.
     
  2. jcsd
  3. Jul 8, 2011 #2

    gb7nash

    User Avatar
    Homework Helper

    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).
     
  4. Jul 8, 2011 #3
    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.
     
  5. Jul 8, 2011 #4
    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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?