1
$\begingroup$

I'm trying to find the power series representation for $ \ln(1-10x) $

Attempt at solution:

$$ \ln(1-10x) = \int {-10\over1-10x} \ dx = -10 \int \sum_{n=0}^\infty (10x)^n dx $$ $$ = -10 \sum_{n=0}^\infty {10^n x^{n+1}\over n+1} + C $$

$$ C = \ln(1) = 0, \; \text {letting x = 0 to find value of C} $$

The answer is not being accepted, I think I might be making some mistakes with coefficients, but not sure how. I'm leaving $10^n$ since the integration is over $ x$.

$\endgroup$
0

1 Answer 1

0
$\begingroup$

Note that $$ \dfrac{1}{1 - 10x} = \sum_{n=0}^{+\infty}(10x)^n \quad \text{if} \quad |10x|< 1 $$ integrating both sides we get, $$ \int \dfrac{1}{1 - 10x}dx = \int \sum_{n=0}^{+\infty}(10x)^n = \sum_{n=0}^{+\infty} 10^n\int x^ndx = \sum_{n=0}^{+\infty} 10^n\dfrac{x^{n+1}}{n+1} + C $$ But, $$ \int \dfrac{1}{1 - 10x}dx = -\dfrac{1}{10}\int \dfrac{(-10)dx}{1 - 10x} = \ln(1 - 10x) $$ So that $$ \ln(1 - 10x) = -\sum_{n=0}^{+\infty} 10^{n+1}\dfrac{x^{n+1}}{n+1} + C $$ For $x = 0$, it follows that $C = 0$. Thus, $$ \ln(1 - 10x) = -\sum_{n=0}^{+\infty} 10^{n+1}\dfrac{x^{n+1}}{n+1} $$ Note that x = 0 is contained in the interval of convergence. Therefore, its replacement is valid.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .