- #1
thegreenlaser
- 525
- 16
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.
[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.