How is the remainder for Taylor polynomials calculated?

[mathmari]

Hey!

I want to calculate the Taylor polynomial of order $n$ for the funktion $ f(x) = \frac{1}{ 1−x}$ for $x_0=0$ and $0 < x < 1$ and the remainder $R_n$.

We have that \begin{equation*}f^{(k)}(x)=\frac{k!}{(1-x)^{k+1}}\end{equation*}
I have calculated that \begin{equation*}P_{0,n}(x)=\sum_{k=0}^n\frac{f^{(k)}(0)\cdot x^k}{k!}=\sum_{k=0}^n x^k \end{equation*}

Is the remainder \begin{equation*}R_n=\frac{\frac{(n+1)!}{(1-\xi)^{n+2}}}{(n+1)!}\cdot (x-0)^{n+1}=\frac{x^{n+1}}{(1-\xi)^{n+2}}\end{equation*} ? Or do we have to use an other formula? (Wondering) In some notes I read that in general the remainder doesn't converge to $0$ for all $\xi\in (0,x)$. Can you give me an example for this? (Wondering)

 

[I like Serena]

Is the remainder \begin{equation*}R_n=\frac{\frac{(n+1)!}{(1-\xi)^{n+2}}}{(n+1)!}\cdot (x-0)^{n+1}=\frac{x^{n+1}}{(1-\xi)^{n+2}}\end{equation*} ? Or do we have to use an other formula?

Hey mathmari! (Smile)

Yep. We can use that remainder.
There are also other formulas for the remainder, but this one should do fine.

In some notes I read that in general the remainder doesn't converge to $0$ for all $\xi\in (0,x)$. Can you give me an example for this?

To converge, the remainder term has to be well-defined, doesn't it?
For which $\xi$ is it ill-defined, and therefore not convergent? (Wondering)

Furthermore, for the remainder term to converge to zero, we also need a condition on $x$ don't we?
And with that condition it becomes impossible to pick a $\xi$ so that the remainder doesn't converge. (Thinking)

 

[mathmari]

To converge, the remainder term has to be well-defined, doesn't it?
For which $\xi$ is it ill-defined, and therefore not convergent? (Wondering)

It is ill-defined if $\xi=1$, right? (Wondering)

Furthermore, for the remainder term to converge to zero, we also need a condition on $x$ don't we?

So that the remainder converges to $0$ it must hold that $x<1$, right? (Wondering)

 

[I like Serena]

It is ill-defined if $\xi=1$, right?

So that the remainder converges to $0$ it must hold that $x<1$, right?

Yep. (Nod)
And it must also hold that $x>-1$.

Anyway, since it is given that $0<x<1$, the remainder converges for all $\xi \in (0,x)$. (Nerd)

 

[mathmari]

Yep. (Nod)
And it must also hold that $x>-1$.

Anyway, since it is given that $0<x<1$, the remainder converges for all $\xi \in (0,x)$. (Nerd)

So, that means that for all $\xi \in (0,x)$ the remainder converges to $0$, right?

 

[I like Serena]

So, that means that for all $\xi \in (0,x)$ the remainder converges to $0$, right?

Yes. (Nod)

 

[mathmari]

At the next subquestion it says:

Using the geometric series we can calculate $\xi$ as a function of $n$ and $x$. Do that and show that it really holds that $R_n\rightarrow 0$ for $n\rightarrow \infty$.

Do we have to set equal $R_n=f(x)-P_n$ and calculate the value of $\xi$ ? Or what do I have to do here? (Wondering)

 

[I like Serena]

At the next subquestion it says:

Using the geometric series we can calculate $\xi$ as a function of $n$ and $x$. Do that and show that it really holds that $R_n\rightarrow 0$ for $n\rightarrow \infty$.

Do we have to set equal $R_n=f(x)-P_n$ and calculate the value of $\xi$ ? Or what do I have to do here?

Yes. I believe that is what we need to do.

 

[mathmari]

Yes. I believe that is what we need to do.

We have that $$R_n=\frac{1}{1-x}-\sum_{k=0}^n x^k =\frac{1}{1-x}-\frac{x^{n+1}-1}{x-1}=\frac{1}{1-x}+\frac{x^{n+1}-1}{1-x}=\frac{x^{n+1}}{1-x} \\ \Rightarrow \frac{x^{n+1}}{(1-\xi)^{n+2}}=\frac{x^{n+1}}{1-x} \Rightarrow \frac{1}{(1-\xi)^{n+2}}=\frac{1}{1-x} \Rightarrow (1-\xi)^{n+2}=1-x \\ \Rightarrow 1-\xi=\sqrt[n+2]{1-x}\Rightarrow \xi=1-\sqrt[n+2]{1-x}$$

If we substitute this in the remainder we get $$R_n=\frac{x^{n+1}}{1-x}\rightarrow 0$$ or not? (Wondering)

 

[I like Serena]

Yep. All correct. (Nod)

 

[mathmari]

Yep. All correct. (Nod)

Great! Thank you so much! :-)