# Homework Help: Proof of a function

1. Mar 29, 2005

### stunner5000pt

Show for every real x, that $$f(x) = \lim_{n \rightarrow \infty} ( \lim_{m \rightarrow \infty} (cos n! \pi x)^{2m})$$ exists and compute it. (show as a familiar rational number) Provein deatil taht your calculations are correct.What is a more familiar name for the function f?

The argument of the Cosine is a real number, and since the cosine fuctin maps from Reals to Reals then the cosine part exists and is real. But how owuldi deal with the limit parts? L'Hopital's rule doesnt apply here...

Doesnt the cosine function wave up and down?? So hgow would this limit exist then?

ANy help would be greatly appreciated! Thanks!

2. Mar 29, 2005

### HallsofIvy

Is that $$cos(n!\pi x)$$? I assume it is although it could mean $$cos(n!\pi)x$$ or even $$cos(n!)\pi x$$

Hint: cos(x) always lies between -1 and 1 so for every fixed n $$cos(n!\pi x)$$ is a number between -1 and 1. What happens to a number between -1 and 1 when raised to higher and higher powers? (Pay special attention to those values of n and x that make $$cos(n!\pi x)$$= 1 or -1.)

3. Mar 29, 2005

### bogdan

Sorry to disturb...but nothing happens to a number between -1 and 1 raised to higher and higher powers...think of e=(1-1/n)^n...(sorry i can't help anymore...)

4. Mar 29, 2005

### dextercioby

N-are treaba "e"-ul.That limit should be 0,but unfortunately,there are values for which the cosine is "+" or "-" 1,so then the limit would not exist.

So i'd say that we should let the OP to explain to us what the initial function really is...

Daniel.

5. Mar 29, 2005

### bogdan

Here's an idea...

cos(n!*pi*x)^(2*m)=[cos(n!*pi*x)^2]^m=[1-sin(n!*pi*x)^2]^m...
if only one could prove that sin(n!*pi*x)^2 -> 0 as n -> infinity...it would be quite simple from here...it's "classical" (1-1/u)^u where u->infinity...
? did help ?

6. Mar 29, 2005

### dextercioby

No matter the argument (incidentally,it tends to infinity for every nonzero "x"),the sine still oscilates between "-1" & "1"...Sine squared only between 0 & 1...

Daniel.

7. Mar 29, 2005

### bogdan

What tends to infinity ?
I know sin*sin is between 0 and 1...but for the argument n!*pi*x i hope it (sin*sin) tends to 0...
something like n!*pi*x tends to an even multiple of pi for n great enough...so that sin*sin -> 0...that's maybe why it's 2*m...not only m...

8. Mar 29, 2005

### dextercioby

Then why did they put a limit for "n--->infty"...?You're basically saying that it doesn't matter,whether that limit exists or not.Sides,even if it didn't matter (that u're taking a limit over "n!"),that "x" is still real and that argument of "sine" would not be an integer multiple of "pi"...

Daniel.

9. Mar 29, 2005

### bogdan

That argument will never be an integer multiple of pi...maybe it will come close enough to it...so that sin->0...anyway...I don't think the limits should be taken separately...not first lim m and then lim n...but at the same time...
I don't know if this helps anyone so I'll stop here...:)
Sorry again if I mislead some of you...or said very stupid things...

10. Mar 29, 2005

### dextercioby

U didn't & needn't worry about that.It's a HW forum & everyone might get to learn something

Daniel.

11. Mar 29, 2005

### saltydog

Well, I'm interested but can't prove it. I wish to propose the following though:

Let:

$$f(x)=\lim_{n\rightarrow\infty}(\lim_{m\rightarrow\infty}[Cos(n!\pi x)]^{2m})=k$$

Where:

$$k=\left\{\begin{array}{cc}1,&\mbox{ if }x\in\mathbb{N}}\\0,&{ if }x\notin \mathbb{N}\end{array}\right$$

12. Mar 29, 2005

### dextercioby

How do you know it's zero,for $x\notin N$...?

Daniel.

13. Mar 29, 2005

### saltydog

Well, in an unorthodox manner: I plotted the function:

$$Cos^{2n}[n! \pi x]$$

in Mathematica and noticed that it tends to zero for any non-integer values of x. I know, that's crummy but it's something to work with.

Last edited: Mar 29, 2005
14. Mar 29, 2005

### dextercioby

Daniel.

EDIT:If Mathematica is so smart,can't it compute that limit,so that a graph would be useless...? :uhh:

15. Mar 29, 2005

### saltydog

Yea, and it cost more than my machine too and it's got other problems as well but I digress. It can't though.

Here's a plot for x=3.1. All the plots for non-integer numbers that I've tested are similar.

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Last edited: Mar 29, 2005
16. Mar 29, 2005

### dextercioby

Okay,what about the limits over "n" & "m"...?

Daniel.

17. Mar 29, 2005

### HallsofIvy

I believe they have medication for that!

I said "a number". 1- 1/n is not "a number".

18. Mar 29, 2005

### dextercioby

Ouch,that escaped me.Good thing Halls decided to quote.

$$e=:\lim_{n\rightarrow +\infty} \left(1+\frac{1}{n}\right)^{n}$$ ,okay...?

U've missed the limit (that's wrong) and the "-" (together with the limit) would yield $e^{-1}$

Daniel.

19. Mar 29, 2005

### bogdan

Ok...you are both right...
but i believe you are still wrong about the problem...
And what kind of medication is there for my condition ?
HallsofIvy...I did not attack your person...but a statement...so...please...at least the idea I was trying to convey was not wrong...
Of course...this being an hw forum...you just had to point that out...about e...
I am very curious to see the solution...

20. Mar 29, 2005

### saltydog

Alright, I think I have it:

For:

$$f(x)=\lim_{n\rightarrow\infty}(\lim_{m\rightarrow\infty}[Cos(n!\pi x)]^{2m})=k$$

Where:

$$k=\left\{\begin{array}{cc}1,&\mbox{ if }x\in\mathbb{Q}}\\0,&{ if }x\notin \mathbb{Q}\end{array}\right$$

Case 1:

x is rational:

Eventually, (n!)x becomes an integer and then remains so with increasing n. Thus (n!)x$\pi$ becomes k$\pi$ with k an integer. The cosine of this is then 1, or -1 which raised to any even power is 1.

Case 2:

x is irrational:

In this case, (n!)x never becomes an integer which means that the cosine of the number is always less than one and this number raised to a large power tends to zero.

What do you guys think?

Last edited: Mar 29, 2005
21. Mar 29, 2005

### dextercioby

So you claim that is Dirichlet's function...

I dunno whether the order of the limits matters,i think it doesn't.

I think your line of reasoning is correct.

Daniel.

22. Mar 29, 2005

### bogdan

The only problem is that you don't know how "less than one" the cosine is...it's not a "fixed" number...
And...why should be there 2m ? (the cosine can't be -1 for n large enough...n!x -> even)
The best thing is that you reduced it to irrational numbers...
That's what I think of your solution...:)

23. Mar 29, 2005

### saltydog

Hum, I don't think it matters how less than one it is. You know, for any number less than one, then use epsilon and delta and Cauchy, to finish the proof.

That's confusing. The n!x quantity, for rationals, when multiplied by pi will be npi and the cosine will then be either -1 or 1.

24. Mar 29, 2005

### saltydog

My goodness. I checked Dirichlet's function on the web and PlanetMath gives the limit expression above for it. Guess you knew that already.

25. Mar 29, 2005

### bogdan

Let's take x=p/q; p,q - integers; q<>0;
For n=2q we have n!x=(2q-1)!*2q*p/q=(2q-1)!*2p which is an even number...so...for n large enough the product is an even number of pi's...
I didn't say that was a number less then one...it's a sequence (or whatever it's called)...it's not fixed...n still varies...
If I didn't make myself clear this time I'm gonna shoot myself...:)...with a bazooka...