Results 1 to 6 of 6

Math Help - [SOLVED] factorial manipulation

  1. #1
    Newbie
    Joined
    Dec 2008
    Posts
    4

    [SOLVED] factorial manipulation

    hi,
    I'm trying to do a proof by induction showing that (2x+1)!/(2^x (x)!) = ((2x+1)(2x-1)!)/(2^(x-1) (x-1)!) , but i don't quite remember factorial manipulation. I can recall that (x+1)! = (x+1)x! , but i can't remember if there's anything like this for (x-1)! or in this case, (2x-1)! as well.

    any help is appreciated!
    Last edited by spidey64; December 17th 2008 at 05:21 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    They aren't equal. Take x = 5.

    \begin{aligned} \text{LHS} & = \frac{(2x+1)!}{2^x \cdot x!} \\ & = \frac{(2(5)+1)!}{2^5 \cdot 5!} \\ & = \frac{11!}{32 \cdot 5!} \\ & = 10395 \end{aligned}.......... \begin{aligned} \text{RHS} & = \frac{(2x+1)!(2x-1)!}{2^{x-1} \cdot (x-1)!} \\ & = \frac{(2(5)+1)!(2(5)-1)!}{2^{5-1} \cdot (5-1)!} \\ & = \frac{11! \cdot 9!}{16 \cdot 4!} \\ & = 3.7721375 \times 10^{10} \end{aligned}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2008
    Posts
    4
    crap....

    the proposition is:

    1*3*5*...*(2x-1) = (2x-1)!/(2^(x-1) (x-1)!)

    so i took the inductive step to (where x->x+1) :

    1*3*5*...*(2x-1)*(2x+1) = (2x-1)!/(2^(x-1) (x-1)!) * (2x+1)


    so i'm trying to get (2x-1)!/(2^(x-1) (x-1)!) * (2x+1) to equal

    (2(x+1)-1)!/(2^((x+1)-1) ((x+1)-1)!) which equals (2x+1)!/(2^x (x)!), right?

    or am i way off here?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Dec 2008
    Posts
    4
    Quote Originally Posted by o_O View Post
    They aren't equal. Take x = 5.

    \begin{aligned} \text{LHS} & = \frac{(2x+1)!}{2^x \cdot x!} \\ & = \frac{(2(5)+1)!}{2^5 \cdot 5!} \\ & = \frac{11!}{32 \cdot 5!} \\ & = 10395 \end{aligned}.......... \begin{aligned} \text{RHS} & = \frac{(2x+1)!(2x-1)!}{2^{x-1} \cdot (x-1)!} \\ & = \frac{(2(5)+1)!(2(5)-1)!}{2^{5-1} \cdot (5-1)!} \\ & = \frac{11! \cdot 9!}{16 \cdot 4!} \\ & = 3.7721375 \times 10^{10} \end{aligned}


    AH, my mistake, in the original post, i placed an extra factorial in the numerator, my second post reflects the REAL problem. sorry.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    Quote Originally Posted by spidey64 View Post
    the proposition is:

    1 \cdot 3 \cdot 5 \cdots (2x-1) = \frac{(2x-1)!}{2^{x-1}(x-1)!}
    1*3*5*...*(2x-1) = (2x-1)!/(2^(x-1) (x-1)!)
    Assume it is true for n = k, i.e. 1 \cdot 3 \cdot 5 \cdots (2k-1) = \frac{(2k-1)!}{2^{k-1}(k-1)!}

    It remains to show that it is also true for n=k+1, i.e. 1 \cdot 3 \cdot 5 \cdots (2k-1)(2k+1) = \frac{(2k+1)!}{2^{k} \cdot k!}

    Looking at the LHS:
    \begin{aligned} \underbrace{1 \cdot 3 \cdot 5 \cdots (2k-1)}_{\text{Our assumption}}(2k+1) & = \frac{(2k-1)!}{2^{k-1}(k-1)!} \cdot (2k+1) \\ & = \frac{(2k+1)(2k-1)!}{2^{k-1}(k-1)!} \cdot {\color{red} \frac{2k}{2k}} \\ & = \frac{(2k+1){\color{red}2k}(2k-1)!}{{\color{red}2}\cdot 2^{k-1} \cdot {\color{red}k}(k-1)!}\end{aligned}

    Can you change the numerator into a single factorial? Simplify the denominator as well. Then you should be good.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Dec 2008
    Posts
    4
    Quote Originally Posted by o_O View Post
    Assume it is true for n = k, i.e. 1 \cdot 3 \cdot 5 \cdots (2k-1) = \frac{(2k-1)!}{2^{k-1}(k-1)!}

    It remains to show that it is also true for n=k+1, i.e. 1 \cdot 3 \cdot 5 \cdots (2k-1)(2k+1) = \frac{(2k+1)!}{2^{k} \cdot k!}

    Looking at the LHS:
    \begin{aligned} \underbrace{1 \cdot 3 \cdot 5 \cdots (2k-1)}_{\text{Our assumption}}(2k+1) & = \frac{(2k-1)!}{2^{k-1}(k-1)!} \cdot (2k+1) \\ & = \frac{(2k+1)(2k-1)!}{2^{k-1}(k-1)!} \cdot {\color{red} \frac{2k}{2k}} \\ & = \frac{(2k+1){\color{red}2k}(2k-1)!}{{\color{red}2}\cdot 2^{k-1} \cdot {\color{red}k}(k-1)!}\end{aligned}

    Can you change the numerator into a single factorial? Simplify the denominator as well. Then you should be good.

    Aha! yes! then the numerator merges into one (2k+1)! and the denom multiplies a 2^1 * 2^k-1 making it 2^k and merges the factorial to k! !!!
    thanks a ton!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] SI manipulation
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 15th 2009, 11:23 PM
  2. [SOLVED] Manipulation of power series
    Posted in the Calculus Forum
    Replies: 7
    Last Post: April 9th 2009, 11:15 PM
  3. [SOLVED] A limit with a factorial
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 29th 2008, 11:05 PM
  4. Replies: 5
    Last Post: October 5th 2008, 05:45 PM
  5. [SOLVED] matrix manipulation
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: August 8th 2007, 10:27 AM

Search Tags


/mathhelpforum @mathhelpforum