IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

asish

if x is a real no. in [0,1],then the value of f(x)=_{[ m]}

_{[infinity ]} _{[ n]}

_[infinity _] {1+cos^2m (n!pi x)}is given by(even see rationality of x)
pls reply with explanation

asish

pls reply

anandghegde

I know only this
The Dirichlet function can be written analytically as

now I'm trying to solve your problem using this. Lemme see if I can do it.

anandghegde

can x be zero?

anandghegde

$\cos^{2m} (n!\pi x) =[\cos^2 (n!\pi x) ]^m$

we know that $0 \le \cos^2 (n!\pi x) \le 1$

case 1: $\cos^2 (n!\pi x)= 1$

implies $n!\pi x = k\pi$

or $x= \frac{k}{n!}$ = rational

therefore when x is rational, f(x) = 2

anandghegde

I'll try to continue wait......

anandghegde

case 2:

$\displatstyle \lim_{m\to\infty} \lim_{n\to\infty}\cos^{2m} (n!\pi x) =0$

here x is irrational.
Hence f(x) = 1 when x is irrational.

Now when cos^2 (n!pi x) =0

, this case is confusing me......ask experts like koni, elasti, bhatsir....

konichiwa2x

Case 1:

Limit

as

when

Case 2:

Limit

, where

is an integer.

which is a rational number.

For the other case, when

'x' is irrational.

Hence, Limit = 2 when 'x' is rational.
Limit = 1 when 'x' is irrational.

anandghegde

asish

thanx koni

konichiwa2x

You mean to say

? I think its quite safe to assume 'x' is not dependant on 'n' .

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