Power Series and Convergence for ln(x+1)

In summary: I'm sorry, I don't understand what you are asking for. Could you please explain in more detail what you need help with specifically? From the conversation, it seems like you are trying to relate a power series to the function ln(x+1) and also find the sum of the series. Is that correct? In summary, the conversation revolves around finding the power series for ln(x+1) and how to find the sum of an infinite power series. The power series given in the conversation is not equivalent to ln(x+1) and the correct expression is given in a link. The connection between the function and power series is not clear and the poster is seeking clarification. The poster also mentions not being able to use a calculator for this
  • #1
Eric Song
12
0

Homework Statement


What is the power series for the function ln (x+1)? How do you find the sum of an infinite power series?

Homework Equations


sigma from n=1 to infinity (-1)^n+1 (1/n2^n)
That is the power series, how is that equivalent to ln (x+1)?
How do you find the sum, or what does it converge to?

The Attempt at a Solution


I know that the series converges via the ratio test: (-1)^n+2 (1/n2^n+1)/ (-1)^n+1 (1/n2^n) < 1.
So it converges, but how do I go from the power series to the function of ln(x+1)? I tried looking at known series, but still not sure. Also, how do I find the sum?
 
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  • #2
Help please.
 
  • #3
What connection is there between the function and the power series, given that x does not occur in the power series? (Can it be that the expression for the power series is meant for the value of x = 2?)
Assuming that x > 1, try writing x+1 = x(1+1/x) and taking the log of that.
 
  • #4
Hello Eric, :welcome:

Here at PF we always encourage a bit of initiative from the posters. Nowadays folks google like crazy and you may have done that too. Did you find anything like e.g. this ?

Eric Song said:
$$\sum_{n=1}^\infty {(-1)^{n+1}\over (n2)^n}$$
That is the power series, how is that equivalent to ln (x+1)?
It's not. I don't know n2 and there is no ##x## in your expression. See the link for the right expression

Eric Song said:
How do you find the sum, or what does it converge to?
Well, you use your calculator to calculate ##\ln(1+x)## or you use a big computer to do the summation term by term with very high precision ... it of course converges to ##\ln(1+x)##
(in other words: what do you mean with this question ? :rolleyes:)

Note that by posting #2 you bumped yourself off the 'https://www.physicsforums.com/unanswered/threads' list that structurally gets quite a bit of attention from helpers :frown: !
 
  • #5
No I did not google it, it is directly from one of my assignments. Also, I'm not allowed to use a calculator for this question.
 
  • #6
mjc123 said:
What connection is there between the function and the power series, given that x does not occur in the power series? (Can it be that the expression for the power series is meant for the value of x = 2?)
Assuming that x > 1, try writing x+1 = x(1+1/x) and taking the log of that.
Yeah, that's what I'm confused about too. I don't know how to relate the expression to the function because there is no x in the expression.
 
  • #7
BvU said:
Hello Eric, :welcome:

Here at PF we always encourage a bit of initiative from the posters. Nowadays folks google like crazy and you may have done that too. Did you find anything like e.g. this ?

It's not. I don't know n2 and there is no ##x## in your expression. See the link for the right expression

Well, you use your calculator to calculate ##\ln(1+x)## or you use a big computer to do the summation term by term with very high precision ... it of course converges to ##\ln(1+x)##
(in other words: what do you mean with this question ? :rolleyes:)

Note that by posting #2 you bumped yourself off the 'https://www.physicsforums.com/unanswered/threads' list that structurally gets quite a bit of attention from helpers :frown: !
also, thanks for letting me know for the future.
I didn't google it.
it's supposed to say n(2^n)
 
  • #8
Eric Song said:
it's supposed to say n(2^n)
Why didn't you type that, then ?

Anyway, it's not the power series for ln(1+x). So where does that expression come from ?

Do you know how to generate a Taylor series ? How to differentiate ln(1+x) ? Many (namely n) times over ?
 
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  • #9
BvU said:
Why didn't you type that, then ?

Anyway, it's not the power series for ;n(1+x). So where does that expression come from ?

Do you know how to generate a Taylor series ? How to differentiate ln(1+x) ? Many (namely n) times over ?
No, could you please help.
I was told that it was for the function ln(x+1)
 
  • #10
BvU said:
Do you know how to generate a Taylor series ? How to differentiate ln(1+x) ?

Eric Song said:
No, could you please help.
If you are studying power series, presumably you have a textbook that covers how the terms in a Taylor series or a Maclaurin series is generated. What does your textbook say about how to do this?
 
  • #11
I think that you should carefully read the problem statement and post it correctly. The "Relevant Equations" should be equations that will help to solve the stated problem. But the ones that you posted seem to be more problem questions about things that may be answered after the stated problem is solved.
 
  • #12
Mark44 said:
If you are studying power series, presumably you have a textbook that covers how the terms in a Taylor series or a Maclaurin series is generated. What does your textbook say about how to do this?
It says that a maclaurin series is a special type of power series.
 
  • #13
Eric Song said:
It says that a maclaurin series is a special type of power series.
And that's all it says?
Does your book mention what particular type of power series it is?
Does your book mention how to find the terms in a Taylor series or Maclaurin series?
 
  • #14
Mark44 said:
And that's all it says?
Does your book mention what particular type of power series it is?
Does your book mention how to find the terms in a Taylor series or Maclaurin series?
Jesus, you sound peeved off, NO ONE is forcing you to help me. It's completely voluntary.
 
  • #15
Mark44 said:
And that's all it says?
Does your book mention what particular type of power series it is?
Does your book mention how to find the terms in a Taylor series or Maclaurin series?
If I knew everything, why would I even ask?
 
  • #16
Eric Song said:
Jesus, you sound peeved off, NO ONE is forcing you to help me. It's completely voluntary.
Eric Song said:
If I knew everything, why would I even ask?
Sorry. Please forgive us. We may sound agrivated, but we are not. For homework-type questions, we are only allowed to ask leading questions and give hints. So people are fishing for what you are told in your book/class so they can lead you from there to the solution. This is not a simple problem and, hopefully, the book/class has already given you a lot to work with.

PS. Actually, in a very real way, the process of these questions are more important to learn than the answer to this specific problem. So you are probably learning good things just by answering the questions and following the comments.
 
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  • #17
Eric Song said:
If I knew everything, why would I even ask?
We don't expect you to know everything, but we do expect you to put in some preliminary effort, which in this case entails reading your book.

Here are the questions you asked in post #1.
Eric Song said:
What is the power series for the function ln (x+1)? How do you find the sum of an infinite power series?
I'm reasonably sure that your textbook explains a process for finding the terms of a power series, devoting a number of pages on how to do this, plus several examples. Since this is a lengthy explanation, it's really beyond the scope of what we do here to reproduce what's in virtually every calculus textbook.

Eric Song said:
How do you find the sum of an infinite power series?
This is a question that takes many pages to explain in a textbook, so is too broad to be answered in a web forum.

Eric Song said:
Jesus, you sound peeved off, NO ONE is forcing you to help me. It's completely voluntary.
You're right, it's completely voluntary -- none of the homework helpers or mentors gets paid a cent. That aside, as was already mentioned, forum rules require that question posters show some effort at solving their problems. For the question you asked, "What is the power series for the function ln (x+1)?", the most obvious resource for the answer to this question is your textbook, but you seem reluctant to see what it says.
 
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  • #18
My 2 cents: you find the coefficients of the Taylor series of f(x) by repeatedly differentiation f(x) and plugging in the number you're expanding about, x=0. So do that, look for a pattern and derive the general form of the coefficients and you're done.
 
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  • #19
Mark44 said:
We don't expect you to know everything, but we do expect you to put in some preliminary effort, which in this case entails reading your book.

Here are the questions you asked in post #1.
I'm reasonably sure that your textbook explains a process for finding the terms of a power series, devoting a number of pages on how to do this, plus several examples. Since this is a lengthy explanation, it's really beyond the scope of what we do here to reproduce what's in virtually every calculus textbook.

This is a question that takes many pages to explain in a textbook, so is too broad to be answered in a web forum.

You're right, it's completely voluntary -- none of the homework helpers or mentors gets paid a cent. That aside, as was already mentioned, forum rules require that question posters show some effort at solving their problems. For the question you asked, "What is the power series for the function ln (x+1)?", the most obvious resource for the answer to this question is your textbook, but you seem reluctant to see what it says.
Okay, I'm sorry, I'll just figure it out myself. Thanks for a GREAT first experience on physics forums.
 
  • #20
Eric Song said:
Okay, I'm sorry, I'll just figure it out myself. Thanks for a GREAT first experience on physics forums.
Maybe you didn't actually read the forum rules (even though you agreed to them when you signed up). Here's a link to the forum rules: https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/

From the section titled "Homework Guidelines"
Attempts are mandatory
You MUST show that you have attempted to answer your question in order to receive help. You MUST make use of the homework template, which automatically appears when a new topic is created in the homework help forums. Once your question or problem has been responded to, do not go back and delete (or edit) your original post.

And you don't have to "figure it out for yourself." You just need to make a good-faith start, and we can help you with the rest.
 
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  • #21
Mark44 said:
Maybe you didn't actually read the forum rules (even though you agreed to them when you signed up). Here's a link to the forum rules: https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/

From the section titled "Homework Guidelines"And you don't have to "figure it out for yourself." You just need to make a good-faith start, and we can help you with the rest.
Okay, I read the rules when I signed up. I did give it a try. I'll get back when I have tried again.
 
  • #22
From post #1:
Eric Song said:
What is the power series for the function ln (x+1)?
See https://en.wikipedia.org/wiki/Taylor_series, in the section titled "Definition". I'm almost certain that your textbook has the same information.
 
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  • #23
Mark44 said:
From post #1:
See https://en.wikipedia.org/wiki/Taylor_series, in the section titled "Definition". I'm almost certain that your textbook has the same information.
I just took my calc bc exam today.
 

Related to Power Series and Convergence for ln(x+1)

1. What is a power series for ln(x+1)?

A power series for ln(x+1) is an infinite sum of terms where each term is a constant multiplied by a power of (x+1). It is written as ∑n=0^∞ an(x+1)^n, where an is the coefficient of the nth term.

2. How is a power series for ln(x+1) derived?

A power series for ln(x+1) is derived by using the Taylor series expansion for ln(x+1). This involves taking derivatives of ln(x+1) and evaluating them at a specific value of x, usually 0. The coefficients of the resulting terms are then used to construct the power series.

3. What is the radius of convergence for a power series for ln(x+1)?

The radius of convergence for a power series for ln(x+1) is 1. This means that the series will converge for all values of x within a distance of 1 from the center, which is x=0.

4. How accurate is a power series for ln(x+1)?

The accuracy of a power series for ln(x+1) depends on the value of x and the number of terms included in the series. Generally, the more terms included, the more accurate the approximation will be. However, for larger values of x, a larger number of terms may be needed to achieve a desired level of accuracy.

5. In what applications are power series for ln(x+1) used?

Power series for ln(x+1) are commonly used in mathematical and scientific calculations that involve logarithms. They can also be used to approximate ln(x+1) in cases where the actual value of ln(x+1) may be difficult to calculate directly.

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