# Taylor's Theorem

• Nov 10th 2011, 01:52 PM
veronicak5678
Taylor's Theorem
Consider the forward difference approx. to f'

f'(0)= f(h)-f(0) / h

Let the error be E = f'(0)- (f(h)-f(0) / h)

Derive a formula for E by Taylor expanding f about x0=0 and using the Remainder thm.

So far I have f(x)=f(x0) + f'(x0)h+f''(x0)h^2 /2 + f'''(c)h^3 /6

So the error will be f'''(c)h^3 /6 for some c in the interval

What do I do now?
• Nov 10th 2011, 03:03 PM
Deveno
Re: Taylor's Theorem
shouldn't it be f(h) = f(0) + f'(0)h + f"(c)h^2/2, for some c in (0,h)?
• Nov 10th 2011, 03:21 PM
veronicak5678
Re: Taylor's Theorem
You're right. But my error term is the same, f'''(0)h^3 /6

My error is supposed to be of the form K1h+k2h^2 and I'm not sure how to get there...
• Nov 10th 2011, 04:13 PM
Deveno
Re: Taylor's Theorem
if f(h) = f(0) + f'(0)h + f"(0)h^2/2 + f'''(c)h^3/6, then

then

f(h) - f(0) = f'(0)h + f"(0)h^2/2 + f'''(c)h^3/6

[f(h) - f(0)]/h = f'(0) + f"(0)h/2 + f'''(c)h^2/6

f'(0) - [f(h) - f(0)]/h = -f"(0)h/2 - f'''(c)h^2/6
• Nov 10th 2011, 06:45 PM
veronicak5678
Re: Taylor's Theorem
I was making a very stupid mistake. Thanks for helping!