Points of Convergence for Lagrange Remainder Theorem

[bonfire09]

Homework Statement

At what points $x$ in the interval $(-1,1]$ can one use the Lagrange Remainder Theorem to verify the expansion

$ln(1+x)=\sum_{k=1}^{\infty} (-1)^{k+1}{\frac{x^k}{k!}}$

Homework Equations

The Attempt at a Solution

Now I know that $ln(1+x)=\sum_{k=1}^{\infty} (-1)^{k+1}{\frac{x^k}{k}}$ when $x\in(-1,1]$. If we let $P_n$ denote the nth Taylor polynomial for $ln(1+x)$ then $f(x)-P_n(x)=\frac{(-1)^{n+1}}{(n+1)(1+c)^{n+1}}x^{n+1}$ centered at $0$.
The only difference between these two forms is the $k!$ which I'm not sure how to deal with. Also if you could provide me with some hint on how to go about this problem that be great thanks.

 

[STEMucator]

The function is equal to its Taylor expansion if and only if the remainder term $|R_n(x)| → 0$ as $n → ∞$. Otherwise I'm not quite sure what you mean in particular by the "Lagrange Remainder theorem".

 

[bonfire09]

The book wants be to use the Lagrange remainder theorem for this problem. But I'm not sure how to find for which x values satisfy $ln(1+x)=\sum_{k=1}^{\infty}(-1)^{k+1}\frac{x^k}{k!}$. Basically I need to find the some interval that this equation is satisfied.

 

[STEMucator]

The book wants be to use the Lagrange remainder theorem for this problem. But I'm not sure how to find for which x values satisfy $ln(1+x)=\sum_{k=1}^{\infty}(-1)^{k+1}\frac{x^k}{k!}$. Basically I need to find the some interval that this equation is satisfied.

Could you perhaps write the theorem out.

Otherwise you can use any $x$ in the interval $(-1, 1]$ and the series will be valid.

 

[bonfire09]

Lagrange Remainder Theorem- Let $I$ be an open interval containing the point $x_0$ and let $n$ be a nonegative integer. Suppose that the function $f:I-->\mathbb{R}$ has $n+1$ derivatives. Then for each point $x$ in $I$ there is a point $c$ strictly between $x$ and $x_0$ such that $f(x)=\sum_{k=0}^{n} \frac{f^k}{k!} (x-x_0)^k+\frac{f^{n+1}}{(n+1)!}(x-x_0)^k$.

Theorem 8.8- For each natural number $n$ and each number $x>-1$ there is a number $c$ strictly between $0$ and $x$ such that $ln(1+x)=x-\frac{x^2}{2}+...+\frac{(-1)^{n+1}}{n} x^n+ \frac{(-1)^n}{(n+1)(1+c)^{n+1}}x^{n+1}$

 

[STEMucator]

Lagrange Remainder Theorem- Let $I$ be an open interval containing the point $x_0$ and let $n$ be a nonegative integer. Suppose that the function $f:I-->\mathbb{R}$ has $n+1$ derivatives. Then for each point $x$ in $I$ there is a point $c$ strictly between $x$ and $x_0$ such that $f(x)=\sum_{k=0}^{n} \frac{f^k}{k!} (x-x_0)^k+\frac{f^{n+1}}{(n+1)!}(x-x_0)^k$.

I see. And $f(x)$ is equal to its expansion if and only if the remainder term goes to zero.

Theorem 8.8- For each natural number $n$ and each number $x>-1$ there is a number $c$ strictly between $0$ and $x$ such that $ln(1+x)=x-\frac{x^2}{2}+...+\frac{(-1)^{n+1}}{n} x^n+ \frac{(-1)^n}{(n+1)(1+c)^{n+1}}x^{n+1}$

Something to notice here... you don't actually know whether or not $x < c < 0$ or $0 < x < c$. All you know is that $x \in I$. So you know the series will converge in the interval $(-1, 1]$. If you try it by plugging in the endpoint 1, you will get a convergent series for example.

 

[bonfire09]

So the series will converge in the entire interval $(-1,1]$?

 

[STEMucator]

So the series will converge in the entire interval $(-1,1]$?

Yes. Although it would be good to show that it actually does.

 

[bonfire09]

Oh ok. I did try doing that at first but when I compared it against the answer in the book I was off. They stated that the interval was supposed to be $\frac{-1}{2}\leq x\leq 1$ which I have no idea how they got that.

 

[STEMucator]

Oh ok. I did try doing that at first but when I compared it against the answer in the book I was off. They stated that the interval was supposed to be $\frac{-1}{2}\leq x\leq 1$ which I have no idea how they got that.

Well, if they want a particular interval, then what's the closest point to zero from -1? That's one of the endpoints. The other endpoint is of course given. They just happened to use $-1/2$.

 

[bonfire09]

So they just chose some random point close to $0$?

 

[STEMucator]

So they just chose some random point close to $0$?

What I meant to say was the convergence would be valid for an interval as specific as $[-\frac{1}{1.0001}, 1]$ if you wanted.

$-1/2$ is a nice number though.

 

[bonfire09]

Oh Ok i see thanks. I just was confused why they chose that -1/2.