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Show that ∑_(n=1)^∞ cos^n (2^n x) converges for a.e. x, but diverges on a dense set of x’s .

Jack

New member
Nov 8, 2012
9
Show that ∑_(n=1)^∞ cos^n (2^n x) converges for a.e. x, but diverges on a dense set of x’s .

Show that \(\displaystyle \sum_{n=1}^\infty \cos^n (2^n x)\) converges for a.e. x, but diverges on a dense set of x’s .
 
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Jameson

Administrator
Staff member
Jan 26, 2012
4,043
Re: Show that ∑_(n=1)^∞ cos^n (2^n x) converges for a.e. x, but diverges on a dense set of x’s .

Hi Jack,

Did you know that you can use Latex on MHB? The way you write is pretty close already to the correct Latex syntax so if you just learn a few common pieces of code you'll be able to use it immediately.

I rewrote the sum in your OP as:

\sum_{n=1}^\infty \cos^n (2^n x)

Jameson
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,706
Re: Show that ∑_(n=1)^∞ cos^n (2^n x) converges for a.e. x, but diverges on a dense set of x’s .

Show that \(\displaystyle \sum_{n=1}^\infty \cos^n (2^n x)\) converges for a.e. x, but diverges on a dense set of x’s .
If $x$ is of the form $\dfrac{a\pi}{2^b}$ (where $a$ and $b$ are integers) then $\cos^n (2^n x)$ will take the value 1 infinitely often. That deals with showing that the series diverges on a dense set.

Convergence a.e. looks harder. I will pass on that for now.