1
$\begingroup$

I'm struggling about finding a way to find the upper bound of the error of Taylor polynomial approximation. I will explain better using a solved example I found...

$f: ]-3;+\infty[ \rightarrow \mathbb{R} $
$f(x)=ln(x+3) +1 $

Find the upper bound of the error approximating the function in $[1,3]$ using a second degree Taylor polynomial with $(x-2)$ powers.

$\textbf{Solution:}$

By Taylor's theorem $\mid{f(x)-P_2(x)}\mid=\mid{R_2(x)}\mid$ and $R_2(x)$ is the Lagrange's rest.

So $\mid{R_2(x)}\mid=\mid{\frac{f'''(c)(x-2)^3}{3!}}\mid$ with $c \in[2,x]$

$\mid{R_2(x)}\mid< \mid{\frac{2(x-2)^3}{6(c+3)^3}}\mid=\frac{1}{3^4}$

So I'm having a lot of trouble with this question:

1 - How can I know in which set is $c$? I know that $c \in[2,x]$ and $x \in [1,3]$ but how do I pick this 2 sets and find one for the values of $c$ (the numerical values)... I hope I'm being clear...

2 - Ok I understand that we need to choose the biggest value for $x$ because we need the biggest numerator possible and we need to choose the smallest value for $c$ because we need the smallest denominator. And I understand why choosing $3$ for $x$ (it's the obvious option)... But why choosing $0$ for $c$? Again I think my doubts about this kind of exercises is to know the values for my $c$...

Can anyone help me to be more clarified about this? Thanks!

$\endgroup$
1
$\begingroup$

You know that $\displaystyle R_2(x)=\frac{f^{\prime\prime\prime}(c)(x-2)^3}{3!}=\frac{2(x-2)^3}{6(c+3)^3}=\frac{(x-2)^3}{3(c+3)^3}$ with $1\le x\le 3$ and $c$ between 2 and $x$,

$\hspace{.6 in}$so $\displaystyle\big|R_2(x)\big|=\frac{|x-2|^3}{3(c+3)^3}\le\frac{1}{3(c+3)^3}<\frac{1}{3(4^3)}=\frac{1}{192}\;$

since $|x-2|\le1,\;$ and $\displaystyle1<c<3\implies 4<c+3<6\implies\frac{1}{4}>\frac{1}{c+3}\implies\frac{1}{(c+3)^3}<\frac{1}{4^3}$

$\endgroup$
  • $\begingroup$ Hi! I thought exactly like you did but as you see in my question the answer shouldn't be the answer $\endgroup$ – Granger Obliviate Apr 30 '16 at 22:06
  • $\begingroup$ @GrangerObliviate I don't think the answer given in your question is correct, since $1<c<3$. Where did you find this solution? $\endgroup$ – user84413 Apr 30 '16 at 22:18
  • $\begingroup$ I took this from solved exercises of my professor :/ $\endgroup$ – Granger Obliviate Apr 30 '16 at 23:11
  • $\begingroup$ @GrangerObliviate Thanks for your reply -- I think you might want to check with your professor about the solution to this exercise. $\endgroup$ – user84413 Apr 30 '16 at 23:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.