Question

Use the power series for 1 1−x to find a power series representation of f(x) = ln(1−x). What is the radius of convergence? (Note: you don’t need to use the ratio test here because we know the radius of convergence of the series P∞ n=0 x n .) (b) Use part (a) to find a power series for f(x) = x ln(1 − x). (c) By putting x = 1 2 in your result from part (a), express ln 2 as the sum of an infinite series

Answer 1

a. Recall that

$\displaystyle\int\frac{\mathrm dx}{1-x}=-\ln|1-x|+C$

For $|x|<1$, we have

$\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n$

By integrating both sides, we get

$\displaystyle-\ln(1-x)=C+\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}$

If $x=0$, then

$\displaystyle-\ln1=C+\sum_{n=0}^\infty\frac{0^{n+1}}{n+1}\implies 0=C+0\implies C=0$

so that

$\displaystyle\ln(1-x)=-\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}$

We can shift the index to simplify the sum slightly.

$\displaystyle\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}n$

b. The power series for $x\ln(1-x)$ can be obtained simply by multiplying both sides of the series above by $x$.

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

c. We have

$\ln2=-\dfrac\ln12=-\ln\left(1-\dfrac12\right)$

$\displaystyle\implies\ln2=\sum_{n=1}^\infty\frac1{n2^n}$

Answer 2

The power series of f(x) = ln(1 - x) is  $\rm -\sum^{\infty}_{n=1}\dfrac{x^{n} }{n}$, the power series of xln(1 - x) is  $\rm -\sum^{\infty}_{n=1}\dfrac{x^{n+1} }{n}$ and the value of ln(2) is $\rm \sum^{\infty}_{n=0}\dfrac{1}{n2^n}$.

Given :

f(x) = ln (1−x)

a) The integration of 1/(1 - x) is given by:

$\rm \int \dfrac{1}{1-x}dx=-ln|1-x| + C$

When |x| >1 :

$\dfrac{1}{1-x} = \sum^{\infty}_{n=0} x^n$

Now, integrate on both sides in the above equation.

$\rm -ln(1-x) = C+\sum^{\infty}_{n=0}\dfrac{x^{n+1} }{n+1}$   --- (1)

Now, at (x = 0) the above expression becomes:

$\rm -ln(1-0) = C+\sum^{\infty}_{n=0}\dfrac{0^{n+1} }{n+1}$

By simplifying the above expression in order to get the value of C.

C = 0

Now, substitute the value of C in expression (1).

$\rm ln(1-x) =-\sum^{\infty}_{n=0}\dfrac{x^{n+1} }{n+1}$

Now, by shifting the index the above expression becomes:

$\rm ln(1-x) =-\sum^{\infty}_{n=1}\dfrac{x^{n} }{n}$

b) Now, multiply by 'x' in the above expression in order to get the power series of (x ln(1 - x)).

$\rm xln(1-x) =-\sum^{\infty}_{n=1}\dfrac{x^{n+1} }{n}$

c) Now, substitute the value x = 1/2 in the expression (1).

$\rm ln2 = \sum^{\infty}_{n=0}\dfrac{1}{n2^n}$

For more information, refer to the link given below:

https://brainly.com/question/16407904