Give the quadratic approx. to sqrt(25 + h)...

• Nov 30th 2012, 01:58 PM
sfspitfire23
Give the quadratic approx. to sqrt(25 + h)...
Hi guys, am wondering if my reasoning is correct -

Give the quadratic approx. to sqrt(25 + h) and discuss how this approx compares to the first order approximation-

I start the problem by Taylor expanding f(x) = sqrt(x) around the initial starting point x_0 = 25. I then get

First order: f(x_0 + (x - x_0)) = f(x_0) + f ' (x_0) * (x - x_0)

This is the same as f(x_0 + h) = f(x_0) + f ' (x_0) * h

Second order: f(x_0 + h) = f(x_0) + f ' (x_0) * h + (f '' (x_0) / 2 ) * h^2

The second order approximation is much closer to the actual value x as the error is o(h^2) which implies that the error goes to 0 faster than o(h) as h goes to 0

Thanks
• Nov 30th 2012, 08:00 PM
hollywood
Re: Give the quadratic approx. to sqrt(25 + h)...
I think if you look at Taylor's Theorem, you'll find that the error in the linear approximation is \$\displaystyle O(h^2)\$ and the error in the quadratic approximation is \$\displaystyle O(h^3)\$. Otherwise it sounds like you have the right idea.

You'll have to give the actual Taylor expansion (5 + (number)h + (number)h^2) - that's what "Give the quadratic approx." means. You probably know that, but it wasn't clear from your post.

- Hollywood