1. ## power series representation!

I'm studying for my final exam in calculus and I'm not exactly sure how to turn this particular natural log function into a power series representation..

f(x) = (x+1)ln(1−x)

I know what the answer is because it's just a practice question but I have no idea how to get it! help! thank you!

2. Originally Posted by khood
I'm studying for my final exam in calculus and I'm not exactly sure how to turn this particular natural log function into a power series representation..

f(x) = (x+1)ln(1−x)

I know what the answer is because it's just a practice question but I have no idea how to get it! help! thank you!
Assume that the logarithm can be expressed as a polynomial. I.e.

$\ln{(1 - x)} = a + bx + cx^2 + dx^3 + \dots$

If $x = 0$ then

$\ln{1} = a \implies a = 0$

Take the derivative, you should find

$-(1 - x)^{-1} = b + 2cx + 3dx^2 + 4ex^3 + \dots$

If $x = 0$ then

$-1 = b$.

Take the derivative again, you should find

$-(1 - x)^{-2} = 2c + 6dx + 12ex^2 + 20fx^3 + \dots$

If $x = 0$ then

$-1 = 2c \implies c = -\frac{1}{2}$.

Take the derivative again, you should find

$-2(1 - x)^{-3} = 6d + 24ex + 60fx^2 + \dots$

If $x = 0$ then

$-2 = 6d \implies d = -\frac{1}{3}$

Follow this procedure and you should find

$\ln{(1 - x)} = -x - \frac{1}{2}x^2 - \frac{1}{3}x^3 - \frac{1}{4}x^4 + \dots$.

So $(x + 1)\ln{(1 - x)} = (x + 1)\left(-x - \frac{1}{2}x^2 - \frac{1}{3}x^3 - \frac{1}{4}x^4 + \dots\right)$.

I'll let you do the expanding.

3. Find the power series for ln(1-x) Multiply the result by (x+1) collect like terms

4. i did that but I don't understand how to put the (x+1) back into the series representation for ln(1-x).

5. The series expansion of ln (1-x) is well known...

$\ln (1-x) = - x - \frac {x^{2}}{2} - \frac{x^{3}}{3} - \dots = - \sum_{n=1}^{\infty} \frac {x^{n}}{n}$

Using simple multiplication rule you obtain...

$(1+x)\cdot \ln (1-x)= - x - (1+\frac{1}{2})\cdot x^{2} - (\frac{1}{2} + \frac{1}{3})\cdot x^{3} - \dots = -x - \sum_{n=2}^{\infty} \frac{2n-1}{n\cdot (n-1)}\cdot x^{n}$

Kind regards

$\chi$ $\sigma$

6. Like what chisigma did --- if you are interested I took a slighlty different approach based on what I used to use to find power series representations to solutions of differential equation, playing with the indices

See attachment