How to find the upper bound of an error by Taylor polynomial approximation

[Granger]

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!

 

[jvicens]

Hello there,

I can help you understand the process of finding the upper bound of the error in Taylor polynomial approximation. Let's break down the steps to make it easier to understand.

1. The value of $c$ in the given interval $[2,x]$ can be any number between 2 and $x$. This is because the Lagrange's remainder theorem states that there exists a value $c$ in the given interval such that the remainder term $R_n(x)$ is equal to the $n+1$th derivative of the function at that point. In this case, we are using a second degree Taylor polynomial, so we need to consider the third derivative of the function $f(x)$. This means that $c$ can be any value between 2 and $x$.

2. The reason we choose the biggest value for $x$ and the smallest value for $c$ is to ensure that the remainder term $R_2(x)$ is as small as possible. This is because the remainder term is in the denominator of the inequality, so a smaller value for $R_2(x)$ will result in a larger upper bound for the error.

3. The reason we choose $x=3$ is because it is the maximum value in the given interval $[1,3]$. This means that it will give us the largest possible numerator for $R_2(x)$. As for choosing $c=0$, it is not a specific value that we are choosing, it is just a placeholder to represent any value between 2 and $x$. So we can say that $c$ is a value between 2 and 3, which is essentially the same as saying $c=0$.

I hope this helps clarify your doubts. If you have any further questions, please feel free to ask.