# Taylor polynomail approximation

• Dec 13th 2009, 09:28 AM
stones44
Taylor polynomail approximation
What is/how can I find the taylor polynomial for a semi circle, such as sqrt(25-x^2) ?

when i calculate the derivatives at 0, i just get 0's so i don't know how I can do it.

Thanks
• Dec 13th 2009, 09:41 AM
TheEmptySet
Quote:

Originally Posted by stones44
What is/how can I find the taylor polynomial for a semi circle, such as sqrt(25-x^2) ?

when i calculate the derivatives at 0, i just get 0's so i don't know how I can do it.

Thanks

$f(x)=\sqrt{25-x^2} \implies f(0)=5$

$f'(x)=\frac{-x}{\sqrt{25-x^2}} \implies f'(0)=0$

I agree that the first derivative is zero,but what about the 2nd?

You will find that all of the even derivaitves are equal to zero at zero, but not the odd.

i.e

$f''(x)=\frac{-x^2}{(\sqrt{25-x^2})^3}-\frac{1}{\sqrt{25-x^2}} \implies f''(0)=-\frac{1}{5}$
• Dec 13th 2009, 09:43 AM
iPod
$\sqrt{25-x^2}=0$

$25-x^2=0$

$x^2=25$

$x=5$

$x=-5$
• Dec 13th 2009, 09:50 AM
stones44
wait...i have that the second derivative = -x^2 * (25-x^2)^-1.5
• Dec 13th 2009, 09:58 AM
TheEmptySet
$f(x)=(25-x^2)^{1/2}$

$f'(x)=-x(25-x^2)^{-1/2}$

Now by the product rule

$f''(x)=-1(25-x^2)^{-1/2}-x^2(25-x^2)^{-3/2}$