# 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!}}##

## 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.

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Homework Helper
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.

Homework Helper
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}##

Homework Helper
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]##?

Homework Helper
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.

Homework Helper
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##?

Homework Helper
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.