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

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- Dec 13th 2009, 09:28 AMstones44Taylor 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 AMTheEmptySet
$\displaystyle f(x)=\sqrt{25-x^2} \implies f(0)=5$

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

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

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

$\displaystyle 25-x^2=0$

$\displaystyle x^2=25$

$\displaystyle x=5$

$\displaystyle x=-5$ - Dec 13th 2009, 09:50 AMstones44
wait...i have that the second derivative = -x^2 * (25-x^2)^-1.5

- Dec 13th 2009, 09:58 AMTheEmptySet
$\displaystyle f(x)=(25-x^2)^{1/2}$

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

Now by the product rule

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