Question

A useful substitution Replace x with x−1 in the series ln (1+x) = ∑ₖ₌₁∞ ((−1)ᵏ⁺¹ xᵏ)/k to obtain a power series for ln x centered at x = 1. What is the interval of convergence for the new power series?

Accepted Answer

Start with the given power series for $$ \ln(1+x) $$:

$$ \ln(1+x) = \sum_{k=1}^{\infty} (-1)^{k+1} \frac{x^k}{k} $$

which is centered at $$ x=0 $$ and converges for $$ -1 \u003c x \leq 1 . $$Make the substitution $$ x 	o x - 1  to $$express $$ \ln x  in $$terms of $$ x - 1 $$:

$$ \ln x = \ln(1 + (x-1)) = \sum_{k=1}^{\infty} (-1)^{k+1} \frac{(x-1)^k}{k} $$

This gives a power series for $$ \ln x $$ centered at $$ x = 1 . $$Determine the interval of convergence for the new series by considering the original interval $$ -1 \u003c x \leq 1 $$ for $$ \ln(1+x) .

$$Since the substitution is $$ x = t + 1 $$, the original variable $$ x  in $$the series corresponds to $$ t = x - 1 .

$$The original interval $$ -1 \u003c x \leq 1 $$ becomes $$ -1 \u003c t \leq 1 $$, so for $$ t = x - 1 $$, the interval is $$ -1 \u003c x - 1 \leq 1 . $$Solve the inequalities for $$ x $$:

$$ -1 \u003c x - 1 \leq 1 \implies 0 \u003c x \leq 2 $$

This means the power series for $$ \ln x $$ centered at $$ x=1 $$ converges for $$ x  in $$the interval $$ (0, 2] . $$Check the endpoints for convergence:

- At $$ x = 0 $$, the series does not converge because $$ \ln x  is $$not defined at zero.

- At $$ x = 2 $$, substitute into the series and verify convergence using the alternating series test or other convergence tests.

Thus, the interval of convergence for the new power series is $$ (0, 2] .$$

A useful substitution Replace x with x−1 in the series ln (1+x) = ∑ₖ₌₁∞ ((−1)ᵏ⁺¹ xᵏ)/k to obtain a power series for ln x centered at x = 1. What is the interval of convergence for the new power series?

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Key Concepts

Power Series and Centering

Interval of Convergence

Substitution in Power Series