Prove the formula of squareroot 2

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The forum discussion focuses on proving the formula for the square root of 2 using a series expansion approach. The user successfully applies the theorem (1+x)a and the series expansion for (1-x)-1/2 to derive the relationship between square root values. The final result demonstrates that 5 times the square root of 2 divided by 7 equals the infinite series expansion. This method effectively utilizes convergence conditions and factorial terms to validate the formula.

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[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
 
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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].
 
LCKurtz said:
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].

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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