Use geometric series to write power series representation

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The discussion focuses on proving that ln(1+x) equals its Maclaurin series for -1 < x ≤ 1, specifically using the geometric series to derive the power series representation for 1/(1+x) when |x| < 1. The initial step involves recognizing that 1/(1-x) can be expressed as a sum of an infinite series, which is valid for |x| < 1. The participant seeks clarification on how the left-hand expression equals the summation on the right, particularly in the context of the geometric series formula. The explanation emphasizes the importance of the convergence condition |x| < 1 and connects the series for 1/(1+x) to that of 1/(1-x) through substitution. Understanding this relationship clarifies the proof process.
NihalRi
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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
 
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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##?
 
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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).
 
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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
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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