# Taylor series homework check

1. Aug 30, 2011

### roldy

1. The problem statement, all variables and given/known data
I posted this already but decided to revive this thread since I re-worked the problem.

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.
Do the calculations for only one step.

2. Relevant equations
$\epsilon$=O($\Delta$x5)

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

Taylor series:
y(x0+$\Delta$x)=y(x0)+$\Delta$xf'(x0,y(x0))+
\Delta[/itex]x21/2!f''(x0,y(x0))+$\Delta$x31/3!f'''(x0,y(x0))+$\Delta$x41/4!f''''(x0,y(x0))+$\epsilon$+[itex]

Is this correct?

My solution:

The derivatives:
f'(x,y)=dy/dx=y+x=1+0=1
f''(x,y)=d2y/dx2=dy/dx+1=1+1=2
f'''(x,y)=d3y/dx3=dy2/dx2=2
f''''(x,y)=d4y/dx4=d3y/dx3=2

y(0+.1)=1+(.1)(1)+1/2!(.1)2(2)+1/3!(.1)3(2)+1/4!(.1)4(2)+(.1)5

y(.1)=1+.1+.01+.001/3+.0001/12+.00001=1.11035

Did I solve this correctly? I want to be able to have something decent when I meet with the professor tomorrow.

2. Aug 30, 2011

### PAllen

Everything looks right. I am not willing to double check all your arithmetic.

3. Aug 30, 2011

### roldy

That's fine. I just wanted to know if I was on the right track. Thanks for taking a look.