# Thread:

1. ## 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

2. 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

$f''(x)=\frac{-x^2}{(\sqrt{25-x^2})^3}-\frac{1}{\sqrt{25-x^2}} \implies f''(0)=-\frac{1}{5}$

3. $\sqrt{25-x^2}=0$

$25-x^2=0$

$x^2=25$

$x=5$

$x=-5$

4. wait...i have that the second derivative = -x^2 * (25-x^2)^-1.5

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

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

Now by the product rule

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