Taylor polymomial/series of -ln(1-x)

careless25 · 2012-06-11 13:53:36

I have to find the series for g(x) = -ln(1-x) using the function f(x) = 1/(1-x).
Seems simple as f(x) is just the derivative of g(x).

So I start by finding the series for 1/(1-x) =

k=0 to inf.

Now I am dont know where to go from here, I can integrate the whole series but that doesnt get me the right answer.

I also tried to work the series from ln(x) to -ln(1-x) but also got stuck.

anonimnystefy · 2012-06-11 14:08:46

What do you get when you integrate the series for 1/(1-x) ?

Bob · 2012-06-11 19:09:23

hi careless25,

Your integration method does work and your series for 1/(1-x) is correct.

So I have the same question as anonimnystefy, what did you get and why did you think it was wrong?

Bob

bobbym · 2012-06-11 20:03:16

Hi careless25;

careless25 · 2012-06-12 05:03:10

Yea I get your answer bobbym but wolfram gives another representation.
http://www.wolframalpha.com/input/?i=-l … esentation

Bob · 2012-06-12 05:47:04

If you mean this (see picture below), then it is the same. There are 2k minus signs there and that will work out to be +

That's the problem with getting a computer to work it out ... you may not get the simplest form.

It happened to me recently with an integration involving trig. Took 3 lines of working to change Wolfram's version into mine.

Bob

careless25 · 2012-06-12 05:50:32

Thanks! That explains a lot.

anonimnystefy · 2012-06-12 06:49:17

As Bob said,you will not always get the simplest form from a computer, but, speaking in bobbym's name, more often you will. So don't discard computers right away.

bobbym · 2012-06-12 08:13:47

Hi;

Yea I get your answer bobbym but wolfram gives another representation.

Alpha's representation are often a little different. It does not reason the same way as we do. Different methods can sometimes produce the same answer in different forms.

But the series for 1 / ( 1 - x ) is a well know one in combinatorics so you will not need a computer. You can just integrate the general term.

anonimnystefy · 2012-06-12 08:41:55

Ah,the most beautiful generating function.

bobbym · 2012-06-12 08:51:39

Hi;

One last beef:

Although I applaud Wolfram for doing something for everyone to benefit by I have repeatedly emailed them about getting Alpha to understand Mathematica's syntax!

anonimnystefy · 2012-06-12 08:56:20

You mean,so that you can use either?

bobbym · 2012-06-12 08:58:12

Yes, Wolfram is a bit like a guy I knew who would use an English to English book to watch Benny Hill!

Math Is Fun Forum

#1 2012-06-11 13:53:36

Taylor polymomial/series of -ln(1-x)

#2 2012-06-11 14:08:46

Re: Taylor polymomial/series of -ln(1-x)

#3 2012-06-11 19:09:23

Re: Taylor polymomial/series of -ln(1-x)

#4 2012-06-11 20:03:16

Re: Taylor polymomial/series of -ln(1-x)

#5 2012-06-12 05:03:10

Re: Taylor polymomial/series of -ln(1-x)

#6 2012-06-12 05:47:04

Re: Taylor polymomial/series of -ln(1-x)

#7 2012-06-12 05:50:32

Re: Taylor polymomial/series of -ln(1-x)

#8 2012-06-12 06:49:17

Re: Taylor polymomial/series of -ln(1-x)

#9 2012-06-12 08:13:47

Re: Taylor polymomial/series of -ln(1-x)

#10 2012-06-12 08:41:55

Re: Taylor polymomial/series of -ln(1-x)

#11 2012-06-12 08:51:39

Re: Taylor polymomial/series of -ln(1-x)

#12 2012-06-12 08:56:20

Re: Taylor polymomial/series of -ln(1-x)

#13 2012-06-12 08:58:12

Re: Taylor polymomial/series of -ln(1-x)

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