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

[cheater1]

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.

Homework Statement

Homework Equations

The Attempt at a Solution

 

[Dick]

That's actually a good start. So (1+x)*(sum x^n) should be (sum x^n)+x*(sum x^n), right? Start writing terms out starting with n=0. It should become pretty obvious soon.

 

[cheater1]

ohh i see it now, thanks you. My class just started this section and I'm kinda new to this. Is there a shortcut of doing this or I have to write out the terms to find power series?

 

[Dick]

ohh i see it now, thanks you. My class just started this section and I'm kinda new to this. Is there a shortcut of doing this or I have to write out the terms to find power series?

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.

 

[Sourabh N]

Dick is right. Also, when you get used to writing down the terms, you'll start seeing the patterns and won't have to write down all the terms.

 

[cheater1]

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

 

[Mark44]

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

 

[seeker247]

I'm very thankful for this tread, as I ran into the same problem as cheater1.
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.

 

[qspeechc]

1-x | 1+x

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

 

[Mark44]

I'm very thankful for this tread, as I ran into the same problem as cheater1.
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.

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.

 

[seeker247]

There must be something I'm missing, since the result I get from computing the division "(1+x)/(1-x)" is "-1+2/(1-x)"; so that the power series representation is "-1 + 2(sum x^n)" for n from 0 to infinity...
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.

 

[Mark44]

You are doing the division incorrection. When you divide 1 + x by 1 - x, the first term you get is 1, not -1.

 

[seeker247]

You are doing the division incorrection. When you divide 1 + x by 1 - x, the first term you get is 1, not -1.

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)"

 

[Mark44]

In the series expansion on the Wolframalpha page, it has 1 + 2x + 2x² + ...

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.

 

[seeker247]

It all makes sense now: the quotient is the series.
Thanks Mark44, and all who contributed. I'm glad I joined this forum.

 

[seeker247]

If you can entertain one more follow up query, I'd like to know if the method of division you just explained is for special cases (as I've been taught to arrange the dividend and divisor in increasing powers of x when dividing polynomials)? ..

 

[Mark44]

You mean that you've been taught to arrange them by decreasing powers of x. No, I don't think this is for special cases. The algorithm works exactly the same as what you've been taught.

If you perform the division with (x + 1) / (-x + 1), you get -1 + 2/(-x + 1) = -1 + 2/(1 - x) = -1 + 2(1 + x + x² + ... + xⁿ + ...)
= -1 + 2 + 2x + 2x² + ... + 2xⁿ + ...
= 1 + 2x + 2x² + ... + 2xⁿ + ...

 

[seeker247]

Thanks, once again, for the clarification.

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.