Use geometric series to write power series representation

Click For Summary

Homework Help Overview

The discussion revolves around proving that ln(1+x) equals its Maclaurin series for the interval -1 < x ≤ 1. Participants are specifically focused on using the geometric series to derive the power series representation for 1/(1+x) within the constraint |x| < 1.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between the geometric series and the power series representation, particularly questioning the derivation of the series from the expression 1/(1-x). There is an emphasis on understanding the convergence criteria and the implications of the series representation.

Discussion Status

Some participants have provided insights into the geometric series formula and its application, while others express confusion about specific steps in the derivation. There appears to be a productive exchange regarding the definitions and properties of geometric series, with some participants gaining clarity on the topic.

Contextual Notes

Participants note the importance of the condition |x| < 1 for the convergence of the series, and there is a reference to the definitions and limits associated with infinite series.

NihalRi
Messages
134
Reaction score
12

Homework Statement


Complete the proof that ln (1+x) equals its Maclaurin series for -1< x ≤ 1 in the following steps.

Use the geometric series to write down the powe series representation for 1/ (1+x) , |x| < 1

This is the part (b) of the question where in part (a)I proved that ln (1+x) equals its Maclaurin series for 0< x ≤ 1by showing the limit of the errom is zero (hense converges).

Homework Equations

The Attempt at a Solution


The solution is actually given, I just couldn't understand it. Shown below

1. 1/(1-x) =∑ x^k for |x|<1
2. 1/(1+x) =∑ (-x)^k
3. = ∑(-1)^k x^k
4. = 1 - x + x^2 - x^3 ...for |-x| < 1 which is for |x|<1

To be clear, I only don't understand step one, the rest is just rearranging. Specifically I don't see how the expression on the left equals the summation on the right.

Note - all the summations are to infinity starting with k= 0
 
Physics news on Phys.org
NihalRi said:
Specifically I don't see how the expression on the left equals the summation on the right.
Do you know the formula for a convergent, infinite geometric series starting from unity as the first term and with the ratio ##x##?
 
Last edited:
  • Like
Likes   Reactions: NihalRi
NihalRi said:

Homework Statement


Complete the proof that ln (1+x) equals its Maclaurin series for -1< x ≤ 1 in the following steps.

Use the geometric series to write down the powe series representation for 1/ (1+x) , |x| < 1

This is the part (b) of the question where in part (a)I proved that ln (1+x) equals its Maclaurin series for 0< x ≤ 1by showing the limit of the errom is zero (hense converges).

Homework Equations

The Attempt at a Solution


The solution is actually given, I just couldn't understand it. Shown below

1. 1/(1-x) =∑ x^k for |x|<1
2. 1/(1+x) =∑ (-x)^k
3. = ∑(-1)^k x^k
4. = 1 - x + x^2 - x^3 ...for |-x| < 1 which is for |x|<1

To be clear, I only don't understand step one, the rest is just rearranging. Specifically I don't see how the expression on the left equals the summation on the right.

Note - all the summations are to infinity starting with k= 0

By DEFINITION,
\sum_{k=0}^{\infty} x^k = \lim_{n \to \infty} \sum_{k=0}^n x^k
if that limit exists.

You can apply high-school formulas for the finite sum, and so examine in detail what happens when ##n \to \infty##. You will see immediately why having ##|x| < 1## is important. Of course, the summation ##\sum (-1)^k x^k## is obtained from that of ##\sum y^k## just by putting ##y = -x## (in case that was what was bothering you).
 
  • Like
Likes   Reactions: NihalRi
blue_leaf77 said:
Do you know the formula for a convergent, infinite geometric series starting from unity as the first term and with the ratio ##x##?

Ahh now that you mention it... totally get it now. Thank you
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
7
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
8
Views
2K
  • · Replies 22 ·
Replies
22
Views
4K