Results 1 to 2 of 2

Math Help - Functions?

  1. #1
    Super Member fardeen_gen's Avatar
    Joined
    Jun 2008
    Posts
    539

    Functions?

    If f(x) = {k^x/(k^x + √k)} (k >0) then prove that:
    2n -1
    2f(r/2n) = (2n - 1)
    r = 1
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Apr 2009
    From
    Atlanta, GA
    Posts
    409

    Achtung Baby!

    Define f(x)=\frac{k^x}{k^x+k^{1/2}} for some k>0

    Define a_r=f(\frac r{2n})

    Show that 2\sum_{r=1}^{2n-1} a_r = 2n-1 for all n>1, k>0

    Rearrange a_1+a_2+a_3+...+a_{2n-2}+a_{2n-1} like so: a_{n}+(a_{n-1}+a_{n+1})+(a_{n-2}+a_{n+2})+...+(a_1+a_{2n-1})

    So 2\sum_{r=1}^{2n-1} a_r= 2a_n+2\sum_{p=1}^{n-1}(a_{n-p}+a_{n+p})

    Now 2a_n=2f(\frac n{2n})=2f(\frac12)=2\frac{k^{1/2}}{k^{1/2}+k^{1/2}}=1

    And a_{n-p}+a_{n+p}= \frac{k^{(n-p)/2n}}{k^{(n-p)/2n}+k^{1/2}}+\frac{k^{(n+p)/2n}}{k^{(n+p)/2n}+k^{1/2}}

    a_{n-p}+a_{n+p}=\frac{(k^{(n-p)/2n})(k^{(n+p)/2n}+k^{1/2})+(k^{(n+p)/2n})(k^{(n-p)/2n}+k^{1/2})}{(k^{(n-p)/2n}+k^{1/2})(k^{(n+p)/2n}+k^{1/2})}=1

    (Multiplying out, the numerator and denominator are equal for all n,p,k.)

    Therefore, 2a_n+2\sum_{p=1}^{n-1}(a_{n-p}+a_{n+p})=1+2(n-1)=2n-1

    QED

    **Where do you get these crazy problems???
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: April 15th 2010, 06:50 PM
  2. Replies: 3
    Last Post: February 23rd 2010, 05:54 PM
  3. Replies: 11
    Last Post: November 15th 2009, 12:22 PM
  4. Replies: 7
    Last Post: August 12th 2009, 05:41 PM
  5. Replies: 1
    Last Post: April 15th 2008, 10:00 AM

Search Tags


/mathhelpforum @mathhelpforum