Results 1 to 4 of 4
Like Tree3Thanks
  • 2 Post By Prove It
  • 1 Post By MarkFL

Math Help - help in series

  1. #1
    Junior Member
    Joined
    Aug 2011
    Posts
    64

    help in series

    If x = 1 + a + a^2 + a^3 + ...

    y = 1 + b + b^2 + b^3 + ...

    then show that 1 + ab + a^2b^2 + a^3b^3 + ... = xy/(x + y - 1).

    How to go about this. Guidance required.

    Thanks.

    Aranga
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,829
    Thanks
    1602

    Re: help in series

    Quote Originally Posted by arangu1508 View Post
    If x = 1 + a + a^2 + a^3 + ...

    y = 1 + b + b^2 + b^3 + ...

    then show that 1 + ab + a^2b^2 + a^3b^3 + ... = xy/(x + y - 1).

    How to go about this. Guidance required.

    Thanks.

    Aranga
    I'm going to assume that \displaystyle \begin{align*} |a| < 1 \end{align*} and \displaystyle \begin{align*} |b| < 1 \end{align*}. Then

    \displaystyle \begin{align*} x &= 1 + a + a^2 + a^3 + \dots \\ &= \frac{1}{1 - a} \\ \\ y &= 1 + b + b^2 + b^3 + \dots \\ &= \frac{1}{1 - b} \\ \\ \frac{x \, y}{x + y - 1} &= \frac{\left( \frac{1}{1 - a} \right) \left( \frac{1}{1 - b} \right) }{ \frac{1}{1 - a} + \frac{1}{1 - b} - 1 } \\ &= \frac{\frac{1}{(1 - a)(1 - b)}}{\frac{1 - b + 1 - a - (1 - b)(1 - a) }{(1 - a)(1 - b)}} \\ &= \frac{1}{1 - b + 1 - a - (1 - b)(1 - a)} \\ &= \frac{1}{2 - a - b - (1 - a - b + ab)} \\ &= \frac{1}{1 - ab} \\ &= 1 + ab + (ab)^2 + (ab)^3 + \dots \\ &= 1 + ab + a^2b^2 + a^3b^3 + \dots \end{align*}
    Thanks from MarkFL and arangu1508
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: help in series

    We are given:

    x=1+a+a^2+a^3+\cdots

    Multiply through by a:

    ax=a+a^3+a^4+\cdots=x-1

    a=\frac{x-1}{x}

    Likewise, we find:

    b=\frac{y-1}{y}

    Now, use a similar method, where:

    f(x,y)=1+ab+(ab)^2+(ab)^3+\cdots

    Then, after multiplying through by ab solve for f(x,y), then use your values of a and b to get the desired result.
    Thanks from arangu1508
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,904
    Thanks
    765

    Re: help in series

    Hello, arangu1508!

    Another approach . . .


    \text{If }\,\begin{Bmatrix}x \:=\: 1 + a + a^2 + a^3 + \hdots & [1] \\ y \:=\: 1 + b + b^2 + b^3 + \hdots & [2] \end{Bmatrix}

    \text{then show that: }\:S \;=\;1 + ab + a^2b^2 + a^3b^3 + \cdots \:=\: \frac{xy}{x + y - 1}

    From [1]: . x \:=\:\frac{1}{1-a} \quad\Rightarrow\quad a \:=\:\frac{x-1}{x}\;\;[3]

    From [2]: . y \:=\:\frac{1}{1-b} \quad\Rightarrow\quad b \:=\:\frac{y-1}{y}\;\;[4]

    We have: . S \;=\;1 + (ab) + (ab)^2 + (ab)^3 + \cdots \;=\;\frac{1}{1-ab}\;\;[5]


    \text{Substitute [3] and [4] into [5]: }\:S \;=\;\frac{1}{1-\left(\frac{x-1}{x}\right)\left(\frac{y-1}{y}\right)} \;=\;\frac{1}{1 - \frac{(x-1)(y-1)}{xy}}

    \text{Multiply by }\frac{xy}{xy}\!:\;S \;=\;\frac{xy}{xy - (x-1)(y-1)} \;=\; \frac{xy}{xy - xy + x + y - 1}


    \text{Therefore: }\:S \;=\;\frac{xy}{x+y-1}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: May 22nd 2012, 06:57 AM
  2. Replies: 5
    Last Post: October 3rd 2011, 02:12 AM
  3. Replies: 3
    Last Post: September 29th 2010, 07:11 AM
  4. Replies: 0
    Last Post: January 26th 2010, 09:06 AM
  5. Replies: 1
    Last Post: May 5th 2008, 10:44 PM

Search Tags


/mathhelpforum @mathhelpforum