Question

Hi, I'm trying to solve a taylor series problem (actually, a maclaurin series in this case) for..

f(x)=ln(1-2x)
I need to find the interval of convergence and the radius of convergence. I don't know what to do past the step where I write down the series itself.

...ln(1-2x)= 0 + -2(x-0)/1! + -4(x-0)^2/2! + -16(x-0)^3/3! + ...

Answer 1

how did you get that series?

Answer 2

For the function f(x)=ln(1-2x), I found out the first few derivatives of f(x), then the values when x=0. Then, I wrote these values in the ''pattern'' for the maclaurin series.
And that's where I'm stuck...

Answer 3

These where my first derivatives:
f(x)=ln(1-2x)
f'(x)=-2/(1-2x)
f''(x)=-4/(1-2x)^2
f'''(x)=-16/(1-2x)^2
and it keeps going...

f(0)=0
f'(0)=-2
f''(0)=-4
f'''(0)=-16...

These are the values

Answer 4

looks good to me. it would be prettier if you wrote
\[x\] instead of
\[x-0\] for each term

Answer 5

Yes, I know. I just keep the 0's to remind me of where ''a'' goes in a taylor series form.

Answer 6

1/x-1 = Sum of x^n from 0 to infinity



1/(2x-1) = \[\sum _{n=0}^{\infty } (2x)^n\]

integrate that

Answer 7

1+ 2x + 4x^2 +8 x^3

integrate into

x + 2/2x^2+ 4/3 x^3 + 8/4 x^4 =

Answer 8

which is \[\sum _{n=1}^{\infty } \frac{2^{n-1}}{n} x^n\]=\[-\frac{1}{2} \text{Log}[1-2 x]\]

Answer 9

\[-2\sum _{n=1}^{\infty } \frac{2^{n-1}}{n} x^n\]= ln(1-2x)

Answer 10

Ok, thanks for the help ! I'll write these down and try to understand them.

Answer 11

Integral of 1/(2x-1) is
−1/2Log[1−2x]