- #1

- 34

- 0

## Homework Statement

Represent (1+x)/(1-x) as a power series.

## Homework Equations

## The Attempt at a Solution

I started with 1/ (1-x) = sum (x)^n n= 0 - infinity

(1 + x) sum x^n

and this is where I am stuck.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter cheater1
- Start date

- #1

- 34

- 0

Represent (1+x)/(1-x) as a power series.

I started with 1/ (1-x) = sum (x)^n n= 0 - infinity

(1 + x) sum x^n

and this is where I am stuck.

- #2

Dick

Science Advisor

Homework Helper

- 26,263

- 619

- #3

- 34

- 0

- #4

Dick

Science Advisor

Homework Helper

- 26,263

- 619

You got that the power series is 1+2x+2x^2+2x^3+..., I hope? Writing out terms never hurts. And if you've just started on the subject, I'd say you should always do it until it becomes clearer. No, no shortcut really except for adjusting indices between series with different starting points.

- #5

- 631

- 0

- #6

- 34

- 0

alright, thanks a lot guys. I'm going to practice some more.

- #7

Mark44

Mentor

- 35,439

- 7,308

Another approach is to divide 1 + x by 1 - x using polynomial long division.

- #8

- 6

- 0

I would like to point out, however, that while with Dick's method the 1st term of the series is 1, with Mark44's the first term appears to be -1. Any help in understanding why this is so would be greatly appreciated.

- #9

- 840

- 14

The first term is 1 as far as I can see. Don't do it -x+1 | x+1, if the notation makes sense.

- #10

Mark44

Mentor

- 35,439

- 7,308

No, if you use the technique I suggested, you get a first term of +1, not -1. Dick's approach and mine yield the same results.

I would like to point out, however, that while with Dick's method the 1st term of the series is 1, with Mark44's the first term appears to be -1. Any help in understanding why this is so would be greatly appreciated.

- #11

- 6

- 0

As a side note, I recognize that the 1st term of the correct representation is 1, but simply don't understand why the method involving division seems to yields -1 as the 1st term. Thanks.

- #12

Mark44

Mentor

- 35,439

- 7,308

- #13

- 6

- 0

If you could tell me how I may be doing this wrongly, I'd really appreciate it ..

Also, I did try to evaluate the problem using Wolfram Alpha, and my answer showed up as the first of several "alternate forms" listed ..I've included the link.

http://www.wolframalpha.com/input/?i=(1+x)/(1-x)"

Last edited by a moderator:

- #14

Mark44

Mentor

- 35,439

- 7,308

It's hard to lay out division problems using just text, so I'll try to explain the first few steps.

1 - x ) 1 + x

Divide 1 by 1, getting a partial answer of 1.

1 - x ) 1 + x ( 1

Multiply the divisor 1 - x by the partial answer 1, getting 1 - x. Put it under the dividend (the thing being divided).

1 - x ) 1 + x ( 1

______1 - x

Subtract 1 - x from 1 + x, getting 2x

1 - x ) 1 + x ( 1

______1 - x

_________2x

Divide 2x by 1, getting 2x

1 - x ) 1 + x ( 1 + 2x

______1 - x

_________2x

Keep repeating this process.

- #15

- 6

- 0

Thanks Mark44, and all who contributed. I'm glad I joined this forum.

- #16

- 6

- 0

- #17

Mark44

Mentor

- 35,439

- 7,308

If you perform the division with (x + 1) / (-x + 1), you get -1 + 2/(-x + 1) = -1 + 2/(1 - x) = -1 + 2(1 + x + x

= -1 + 2 + 2x + 2x

= 1 + 2x + 2x

- #18

- 6

- 0

The whole problem, as I have discovered, was that I thought I could simply represent the result of the division[(x + 1) / (-x + 1)], which is "-1 + 2/(-x + 1)", by the power series "-1 + 2(sum x^n)" for n from 0 to infinity.

Again, thanks.

Share: