## nt.number theory – Is this number theoretic quantity upper bounded?

It seems my thinking gets organized after posting the question.

The natural logarithm of the quantity $$pi(n)!$$ is near $$pi(n)log(pi(n)/e) + (log(taupi(n)))/2$$ (a number-theoretic use for $$tau$$, the circumference of a unit radius circle). Using an approximation to $$pi(n)$$ we get that this is less than $$An$$ for some $$A lt 2$$. But $$An/(n-h)$$ is bounded above by $$2A$$, and gets very close to $$A$$. So with some work the original quantity should be shown to be less than $$e^A$$.

Verification is still appreciated.

Gerhard “And Still Worth An Acknowledgement” Paseman, 2020.05.30.

## Apps related to the icons on the upper left corner on an Android Samsung Partial circle with inner circle arrow
and the rectangle with the clock_

## Ubuntu (Gnome and 18.04) always hides the upper LH corner close box under a menu – can I change this?

#### Stack Exchange Network

Stack Exchange network consists of 177 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Visit Stack Exchange

## inequality – Upper bound on x where \$2^x leq (ax)^4\$

We have some constant $$a > 1$$ and we know the following inequality:
$$2^x leq (ax)^4$$

And need to find an upper bound on $$x$$.

I thought of trying to calculate where $$2^x$$ intersects $$(ax)^4$$ and then the larger intersection would be an upper bound for $$x$$.
So this is what I did:I called the value where they intersects $$t$$ and solved:

$$2^t = (at)^4\ tln2 = 4ln(at)\ atln2=4aln(at)\ frac{ln2}{4a}=(at)^{-1}ln(at)\ -frac{ln2}{4a}=(at)^{-1}ln((at)^{-1})\ -frac{ln2}{4a}=e^{ln((at)^{-1})}ln((at)^{-1})\ Wleft(-frac{ln2}{4a}right)=ln((at)^{-1})\ t=frac{e^{-Wleft(-frac{ln2}{4a}right)}}{a}$$

And therefore:

$$xleq max left{frac{e^{-W_0left(-frac{ln2}{4a}right)}}{a},frac{e^{-W_{-1}left(-frac{ln2}{4a}right)}}{a}right}$$

But I don’t know how to continue from here. How can I bound this expression with $$W$$?

## java – Do not distinguish between upper and lower case

I have this exercise that asks for a word on the screen and outputs the first non-repeating character it finds, but I would like it not to be case sensitive since if it finds an a and an A it thinks they are two different characters.
Code:

``````System.out.println("Palabra:");
CaracterNoRepetido();
}
public static void CaracterNoRepetido(){
Scanner lector = new Scanner(System.in);
char caracter = 0;
for(int i = 0; i < cadena.length(); i++) {

boolean repetido = false;
for(int j = 0; !repetido && j < cadena.length(); j++) {
if(j != i)
}

if(!repetido) {
break;
}
}

System.out.println(caracter);
``````

}
}

## iPad 2 screen is unresponsive in the upper half

I have an iPad 2 that I got for free that I am going to fix (digitizer) and give to my nephew. Before fixing it, I was figuring out what I would have to do for him, and I noticed that while in portrait mode you couldn't click or touch half of the screen. Would you need another part or just the digitizer? I plan to fix this no matter what, so don't complain about how useless the iPad 2 is. Thanks in advance!

## algorithms: upper limit for the recurrence equation \$ T (n) = 4T left ( frac {n} {2} right) + n ^ 2 \$

It's been a while for me since my undergraduate class in algorithms. Could you help me find an upper limit for this recurrence equation?

$$T (n) = begin {cases} 14 & quad text {if} n <1 \ 4T left ( frac {n} {2} right) + n ^ 2 & quad text {if} n gt1 \ end {cases}$$

## integration – Upper limit of a complex integral

Working with a class of polynomials, I have found this integral

$$A_n (x) = frac {n!} {2 pi , i} int_C frac {e ^ {x , (e ^ t-1)}} {t ^ {n + 1}} dt$$
where $$C$$ h is a closed circuit described in the positive sense surrounding the
origin.

I wonder if it is possible to find an upper limit of this integral (depending on $$n$$)

## Complex analysis: exponential decay of the Fourier series of a periodic holomorphic function in the upper half plane

Leave $$f$$ be a periodic holomorphic function in the upper half plane $$mathbb H$$. Here newspaper means $$f (z + 1) = f (z)$$ for all $$z in mathbb H$$.

So $$f$$ equals your Fourier series
$$f (z) = sum_ {n in mathbb Z} c_n q ^ n, quad q = e ^ {2 pi i z}.$$

Suppose that the Fourier series of $$f$$ it has the form

$$sum_ {n geq n_0} c_n q ^ n$$

for some $$n_0 in mathbb Z$$.

It is true that

$$sum_ {n geq 1} c_n q ^ n ll e ^ {- epsilon y}$$
for some $$and$$, where $$z = x + iy$$?

This question originates from the following statement: & # 39; A weakly holomorphic modular form is a harmonic form of mass & # 39 ;. For the original question, see https://mathoverflow.net/questions/357309/a-weakly-holomorphic-modular-form-is-a-harmonic-maass-form

## Ordinary differential equations: example of exhaust flow from the lower half-plane to the upper half-plane

There is a full vector field $$v = (v_ {1}, v_ {2}): D rightarrow mathbb {R} ^ 2$$, where $$D$$ is an open subset of $$mathbb {R} ^ 2$$, $$v in C ^ 1$$, with the following properties:

1) say $$l: = {x_ {2} = 0 }$$, we have $$D cap l neq emptyset$$ and $$v_ {2} (x) leq 0$$ for all $$x in l cap D$$ and there is a point $$x in l$$ S t. $$v_ {2} (x) <0$$;

2) The whole $${x in l cap D s.t. v_ {2} (x) = 0 }$$ has an accumulation point $$A en l cap D$$

3) There is a point $$x in D$$ with $$x_ {2} <0$$ and a time $$t in mathbb {R}$$ S t. $$phi ^ {t} (x) 2> 0$$ (i.e., the integral curve from $$x$$ sometime it reaches half of the upper plane).

The motivation for the question lies in the following statement "if the vector field $$v$$ never points to the middle of the top plane along $$l$$, then each solution from the lower half plane remains in the closed lower half plane, "which I cannot say whether it is true or not. Actually, I know that if the vector field is tangent throughout $$l$$ or if the set of points where it is tangent has no accumulation point in $$l$$, then the claim is true. This motivates 1) and 2).

Finally let me say that I would take as an answer an example where $$v$$ it is not complete.