Taylor polynomail approximation

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December 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
December 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

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

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

$25-x^2=0$

$x^2=25$

$x=5$

$x=-5$
December 13th 2009, 09:50 AM
stones44

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

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