Euler's Method?

Question

y' = F(x,y)
dy/dx = F(x,y)

T1(x) = 1st degree taylor polynomial
--> y0= yn-1 + ln F(xn+1,yn-1)

Modify euler's method to use 2nd degree taylor polynomials

Answer 1

I can't figure out what you are doing, especially wher you get the "ln" and the F(xn+1,yn-1)

Check out:
http://en.wikipedia.org/wiki/Euler_integration

The first degree Taylor (actually McLauren) polynomial has the form:

T1(x) = f(0) + xf'(0)

In the Euler method, this becomes:

y[n+1] = y[n] + hF(x[n], y[n])
x[n+1] = x[n] + h = x[0] + nh

For the second degree Taylor polynomial we would have:

T2(x) = f(0) + xf'(0) + (1/2)(x^2)f''(0)

If you can differentiate F with respect to x to get F'(x,y), then the translation to solving the differential equation becomes:

y[n+1] = y[n] + hF(x[n], y[n]) + (1/2)(h^2)F'(x[n], y[n])

If you can't differentiate F then you have to approximate F'.

But Euler's method is bad enough for solving ordinary differential equations. Extending it in this way would be making things worse rather than better. Choose another method instead.