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 Answer 1

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 $\displaystyle1<c<3\implies 4<c+3<6\implies\frac{1}{4}>\frac{1}{c+3}\implies\frac{1}{(c+3)^3}<\frac{1}{4^3}$

$\endgroup$
4
  • $\begingroup$ Hi! I thought exactly like you did but as you see in my question the answer shouldn't be the answer $\endgroup$ Commented Apr 30, 2016 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
    Commented Apr 30, 2016 at 22:18
  • $\begingroup$ I took this from solved exercises of my professor :/ $\endgroup$ Commented Apr 30, 2016 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
    Commented Apr 30, 2016 at 23:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .