# Points of Convergence for Lagrange Remainder Theorem

• bonfire09
In summary, the Lagrange Remainder Theorem can be used to verify the expansion of ln(1+x) for all x values in the interval (-1,1]. The series will converge in the entire interval and can even converge for a more specific interval such as [-1/1.0001, 1]. The book chose to use the interval [-1/2, 1] as it is a convenient and commonly used interval.
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.

Last edited:
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".

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.

bonfire09 said:
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.

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}##

bonfire09 said:
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.

bonfire09 said:
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.

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

bonfire09 said:
So the series will converge in the entire interval ##(-1,1]##?

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

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.

bonfire09 said:
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##.

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

bonfire09 said:
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.

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

## 1. What is the Lagrange Remainder Theorem?

The Lagrange Remainder Theorem, also known as the Taylor Remainder Theorem, is a mathematical theorem that states the error between a function and its Taylor polynomial approximation can be bounded by the value of the next term in the Taylor series. It is a powerful tool in calculus and is used to estimate the accuracy of polynomial approximations.

## 2. Who was Lagrange and why is this theorem named after him?

Joseph-Louis Lagrange, also known as Count of Turin, was an Italian mathematician and astronomer who made significant contributions to the fields of mathematics and physics. He is best known for his work in calculus, number theory, and mechanics. The Lagrange Remainder Theorem is named after him in recognition of his contributions to the field of calculus.

## 3. How is the Lagrange Remainder Theorem used in real-world applications?

The Lagrange Remainder Theorem has applications in fields such as engineering, physics, and economics. It is used to approximate solutions to equations that cannot be solved exactly, as well as in numerical analysis to estimate the error in algorithms and models. It is also used in the study of differential equations and in optimization problems.

## 4. Can the Lagrange Remainder Theorem be applied to all functions?

No, the Lagrange Remainder Theorem can only be applied to functions that are infinitely differentiable, meaning that they have derivatives of all orders. This includes most commonly used functions such as polynomials, trigonometric functions, and exponential functions. However, it cannot be applied to functions with singularities or discontinuities.

## 5. Are there any limitations to the Lagrange Remainder Theorem?

While the Lagrange Remainder Theorem is a powerful tool in calculus, it has some limitations. It only provides an upper bound for the error in a Taylor polynomial approximation and does not give an exact value. Additionally, it assumes that the higher-order derivatives of the function are continuous, which may not always be the case in real-world applications. It is important to use the theorem with caution and to consider its limitations when applying it to a problem.

• Calculus and Beyond Homework Help
Replies
14
Views
2K
• Calculus and Beyond Homework Help
Replies
3
Views
490
• Calculus and Beyond Homework Help
Replies
1
Views
335
• Calculus and Beyond Homework Help
Replies
3
Views
409
• Calculus and Beyond Homework Help
Replies
1
Views
597
• Calculus and Beyond Homework Help
Replies
2
Views
707
• Calculus and Beyond Homework Help
Replies
3
Views
259
• Calculus and Beyond Homework Help
Replies
2
Views
178
• Calculus and Beyond Homework Help
Replies
4
Views
299
• Calculus and Beyond Homework Help
Replies
1
Views
928