# Taylor polynomial problem

• August 4th 2012, 07:51 AM
leonardthecow
Taylor polynomial problem
Hi -

I'm given a problem where I must find the "minimum order of the Taylor polynomial required to approximate the quantity with an absolute error no greater than 10^-3", where the given quantity is e^-0.5

I've calculated the remainder to be R(-0.5)=[0.5^(n+1)]/(n+1)!, yet I do not know how to solve the inequality 10^-3>[0.5^(n+1)]/(n+1)!

I know the answer is supposed to be n=4, but I have no clue how to solve for n. Any help???

Thanks!!
• August 4th 2012, 09:13 AM
FernandoRevilla
Re: Taylor polynomial problem
Quote:

Originally Posted by leonardthecow
I've calculated the remainder to be R(-0.5)=[0.5^(n+1)]/(n+1)!, yet I do not know how to solve the inequality 10^-3>[0.5^(n+1)]/(n+1)!

We have

$\left |{R_n(x)}\right |=\left |\dfrac{e^{\xi}}{(n+1)!}x^{n+1}\right |\quad (\xi\in (-1/2,0))$

$\left |{R_n(x)}\right |\leq \dfrac{1}{(n+1)!}\;\dfrac{1}{2^{n+1}}$

$\dfrac{1}{(n+1)!\;2^{n+1}}\leq 10^{-3} \Leftrightarrow 1000 \leq (n+1)!\;2^{n+1}$

Now, verify that $n=4$ is the smallest natural number satisfying the last inequality.
• August 4th 2012, 01:49 PM
HallsofIvy
Re: Taylor polynomial problem
Quote:

Originally Posted by leonardthecow
Hi -

I'm given a problem where I must find the "minimum order of the Taylor polynomial required to approximate the quantity with an absolute error no greater than 10^-3", where the given quantity is e^-0.5

I've calculated the remainder to be R(-0.5)=[0.5^(n+1)]/(n+1)!, yet I do not know how to solve the inequality 10^-3>[0.5^(n+1)]/(n+1)!

I know the answer is supposed to be n=4, but I have no clue how to solve for n. Any help???

Thanks!!

Did you try direct computation? If n= 1, $.5^{1+1}/(1+1)!= (1/4)(1/2)= 1/8= 0.125> 0.001$. If n= 2, $.5^{2+1}/(2+1)!= (1/8)(1/6)= 1/48= 0.0208333...> 0.001$. If n= 3, $.5^{3+1}/(3+1)!= (1/16)(1/24)= 1/384= 0.002604...> 0.001$. If n= 4, $.5^(4+1)/(4+1)!= (1/32)(1/120)= 1/3840= 0.000260...< 0.001$.