# Homework Help: Tayor series with truncation error

1. Aug 27, 2011

### roldy

1. The problem statement, all variables and given/known data
Consider dy/dx=x+y, a function of both x and y subject to initial condition, y(x0)=y0.
Use Taylor series to determine y(x0+$\Delta$x) to 4th order accuracy.

Initial condition: x0=0, y(x0)=1
step size: $\Delta$x=0.1

Show 5 significant digits in the answer.

2. Relevant equations
$\epsilon$=O($\Delta$x5)
Do the calculations for only one step.

3. The attempt at a solution
dy/dx=f(x,y)

Taylor series:
y(x0+$\Delta$x)=y(x0)+$\Delta$xf(x0,y(x0))+$\epsilon$

My solution:

f(x0,y(0))=f(0,1)=0+1=1

y(0+0.1)=1+0.1(1)+.00001=1.10001

Does this seem correct. It feels like I missed something. On the other hand it makes sense. Did I miss something or mess up a step?

2. Aug 27, 2011

### Ray Vickson

To 4th order in h (= Delta x), y(x0 + h) = y(x0) + sum{D^k y(x0) * h^k/k!: k=1..4}, where D^k y(x0) = (d/dx)^k y(x)|_{x = x0} = kth derivative of y(x) at x0. Can you figure out (d/dx)^2 y, (d/dx)^3 y, etc?

RGV

3. Aug 27, 2011

### roldy

I thought that doing calculations for one step meant that I didn't need to take any derivatives. For example, a calculation using 3 steps would mean the Taylor series would have the original function, 1st and 2nd derivatives. Wouldn't I just get zeros beyond the first derivative of x+y anyways?

4. Aug 28, 2011

### roldy

Can anyone provide some insight on this. This is review and I don't ever remember doing a Taylor series on a differential equation.