IIT JEE , AIEEE Forum - Mathematics , Algebra

konichiwa2x

Prove that

.

LAMPARD

,In maths, the Viète formula, named after Francois Viete, is the following infinite product type representation of the mathematical constant

1)divide the eqn by 2 and then take its reciprocal...
This resultant exp. matches wid exp. of Viete's formula..

The expression on the right hand side has to be understood as a limit expression

$\lim_{n \rightarrow \infty} \prod_{i=1}^n {a_i \over 2}=\frac2\pi$

where $a_n=\sqrt{2+a_{n-1}}$ with initial condition $a_1=\sqrt{2}.$

$\frac2\pi = \frac{\sqrt2}2 \cdot \frac{\sqrt{2+\sqrt2}}2 \cdot \frac{\sqrt{2+\sqrt{2+\sqrt2}}}2 \cdot \cdots\!$
and so on till ...

anchitsaini

i had no idea abt this but found this on wikipedia, when lampard gave the formula that --

$\frac2\pi = \frac{\sqrt2}2 \cdot \frac{\sqrt{2+\sqrt2}}2 \cdot \frac{\sqrt{2+\sqrt{2+\sqrt2}}}2 \cdot \cdots\!$
(http://en.wikipedia.org/wiki/Pi)

thus
2/pi = 1/root 2 * root[1/2 + 1/2 root(1/2)] *......
hence
pi =2 / [1/root 2 * root[1/2 + 1/2 root(1/2)] *......

konichiwa2x

lampard and anchit, anyone can state results...Can you give me an orginal method?

ananth_patri

i dont think these standard of ques are ever asked in JEE.....

konichiwa2x

this is well within JEE syllabi and not too difficult either.. I do agree it might not be asked in the upcoming years, but such questions were asked earlier...

feynmann

easy !!!!!

see the Dr is nothing but cos pi/4 , cos pi/8 , cos pi/16 .............

as

we know that

cos @/2 = sqrt ( 1/2 + 1/2 cos @ )

and cos pi/4 = 1/sqrt ( 2 )

the rest is cake walk !!

Let us denote the product (Dr ) upto m+1 term by Sm and pi/4 = @

so Sm = 2 / cos @ cos @/2 cos @/4 cos @/8 .... cos @/2^m

now multiplynig the Dr & Nr by sin @/2^m we get

Sm = 4 *@ (2^m/@ )sin @/2^m/ sin 2@

Now letting mtending to infinity and evaluating the limit we get

S( reqd ) = 4 * pi/4 *1 = pi

konichiwa2x

nice, feynmann!

here is my solution:

Since

.

Then

.

or

.

For

then

Since

.

Also,

.

We obtain

Thereore,

Ask & Discuss Questions with Community & Experts