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Prove the formula of squareroot 2

  1. Sep 13, 2010 #1
    [SOLVED] prove the formula of squareroot 2

    Verify the following formula:
    Squareroot (2)=7/5*(1+ 1/100 + 1*3/(100*200) + 1*3*5/(100*200*300) + etc..)

    Use the Fact that 50=5^2*2=7^2+1

    Theorem: (1+x)^a=1 +a*x/1 + (a(a-1)*x^2)/(1*2)+ (a(a-1)(a-2)*x^3)/(1*2*3)+ etc...

    Hint: use the theorem with (1-x)^(-1/2)

    I used the series expansion for (1-x)^(-1/2) and got

    (1-x)^(-1/2)=1+ (1*x)/(1*2)+(1*3*X^2)/(1*2*4)+ (1*3*5*x^3)/(1*2*4*6)+etc ....

    I do understand that the squareroot (50)=5*squareroot(2) But I am stuck

    Please help, thanks
     
    Last edited: Sep 13, 2010
  2. jcsd
  3. Sep 13, 2010 #2

    LCKurtz

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    Remember you need |x| < 1 for convergence. See if this will get you started:

    [tex]\frac 1 7 = \frac 1 {\sqrt{49}} = (50-1)^{-\frac 1 2}=50^{-\frac 1 2}(1-\frac 1 {50})^{-\frac 1 2}[/tex]

    [tex]\frac 1 7 = \frac 1 {5\sqrt {2}}(1-\frac 2 {100})^{-\frac 1 2}[/tex]

    Now solve for [itex]\sqrt{2}[/itex].
     
  4. Sep 13, 2010 #3
    Thanks for helping!

    I solved it by using the fact that (1-1/50)^(-1/2)=squareroot(50/49)=1+1/(2*50) + 1*3/(2*4*5*50^2)+1*3*5/(1*2*3*50^3)

    Then 5*squareroot(2)/7=(1+1/100+1*3/(100*200)+1*3*5/(100*200*300))
     
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