## c – Evaluation of π using Monte Carlo methods – Serial vs OMP

I wrote this simple code to evaluate the π using the Monte Carlo method. This is the serial version:

``````long double compute_pi_serial (long interval const) {

srand (time (NULL));

double x, y;
int i, circle = 0;

for (i = 0; i <interval; i ++) {
x = (double) (rand () / (double) RAND_MAX);
y = (double) (rand () / (double) RAND_MAX);

if (pow (x, 2) + pow (y, 2) <= 1.0) circle ++;
}

return (double long) circle / interval * 4.0;
}
``````

Afterwards, I wrote a parallel version using `OpenMP` And this is the result:

``````long double compute_pi_omp (const long interval, const int threads) {

double x, y;
int i, circle = 0;

{

srand (SEED);

#pragma omp for reduction (+: circle)
for (i = 0; i <interval; i ++) {
x = (double) (rand () / (double) RAND_MAX);
y = (double) (rand () / (double) RAND_MAX);

if (pow (x, 2) + pow (y, 2) <= 1) circle ++;
}
}

return (double long) circle / interval * 4.0;
}
``````

This is an efficient method or there is a more efficient version using always `OpenMP`?

## real analysis – Is \$ π: mathcal {C} ^ ∞ (M, N) → mathcal {C} ^ ∞ (S, N) \$, \$ π (f) = f | _S \$ a quotient map in the \$ mathcal {C} ^ 1 \$ topology?

This question was previously published in MSE.

Leave $$M, N$$ be connected smooth manifolds (without limit), where $$M$$ It is a compact variety, so we can put a topology in space. $$mathcal C ^ infty (M, N)$$ using $$mathcal {C} ^ 1$$ Topology of Whitney.

Now consider $$S subset M$$ a compact submanifold of $$M$$ with a limit such that $$text {dim} S = text {dim} M$$, using the same process we can put a topology in $$mathcal C ^ infty (S, N)$$ using the $$mathcal {C} ^ 1$$ Topology of Whitney. There is a natural continuous projection of $$mathcal C ^ infty (M, N)$$ in $$mathcal C ^ infty (S, N)$$Defined by

begin {align *} pi: mathcal C ^ infty (M, N) and to mathcal C ^ infty (S, N) \ f & mapsto left.f right | _ {S}. end {align *}

My question: Is $$pi$$ An open map or at least a quotient map?

$$mathcal {C} ^ 1$$-Whitney Topology is also called $$mathcal {C} ^ 1$$Strong topology

As noted by user Adam Chalumeau, the following exercise is found in the book "Differential Topology of Morris W. Hirsh"

[Exercise 16, page 41]: Leave $$M, N$$ be $$mathcal {C} ^ r$$ collectors. Leave $$V⊂M$$ be an open set then

• The restriction map $$δ: mathcal {C} ^ r (M, N) → mathcal {C} ^ r (V, N)$$ $$δ (f) = f | V$$ It is continuous for the weak topology, but not always for the strong one.

• $$δ$$ It is open for strong topologies, but not always for the weak. "

From our $$M$$ It's weak compact topology = strong topology. However, I do not know how to solve this exercise, much less adapt that test to the case I want.

## trigonometry: show that the arcosine (X) + arccos (y) = π / 2, Si and only if x = y

arcosine (X) + arccos (X) = π / 2;
That a thought came to my mind that in general
arcsin (X) + arccos (Y) = π / 2, implied that X = Y.
I have a hunch that it's true and I've done a kind of self-test but it's illegal to use a test method, then I also tried to use graphics but I'm still stuck. I would be grateful if someone could help me please.

## \$ Π (x) ge log log x \$ holds \$ 2 le x e e {{e ^ 3} <5.3 times 10 ^ 8 \$?

The book Theory of Numbers by G H Hardy, et al. test $$π (x) ge log log x$$ for $$x> e ^ {e ^ 3}$$. There is a way to try out that also goes for $$2 le x e e ^ {e ^ 3}$$ otherwise (in the worst case) some valuable source to verify this numerically?

## calculation: How do I find x when θ = 1/4 π of dx / dθ = (x + 2) without ^ 2 2θ if x is 0 when θ is 0?

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