To find area between two functions you have to find integral of module of differences of them.
But in current situation it's a bit easier:
sin(x) > -sin(x) on [0, \pi]
v = \int_o^\pi sin(x)-(-sin(x)) dx = 2 \int_0^\pi sin(x) dx
My English is also not really good :wink:
Hi Faiza,
If you have only few point of your curve, than you may use two ways:
1. Find approximation polynomial and than evaluate integral;
2. Just use following formula (by integral definition):
\sum_{i=1}^{n-1} f_i(x_{i+1}-x_i) , where f_i=f(x_i)
There are some other methods you may...
robert Ihnot:
Thanks for correcting me.
It's the first method I've remembered. The hardest thing is to factorize P-1, and It's not a problem for a not large P and if you know factorization of P-1 you may check g to be primitive root for polynomial time.
And I did not say that it's EASY...
Dodo:
You started in interesting direction, but f(n) = k! where k=\sum(k_i)
Let's start iterative theorem prove:
k=1
then t(n)=1 and 1<2\sqrt{n} for any n>1.
Prove for k+1
t(n)=(k+1)!=k!(k+1)
Now we have to prove that k!(k+1) < 2\sqrt{n}
Let's suppose that it's not true. Divide both...
Hi, there is one method but it required knowing expansion of the number P-1 on the simple factors.
So, if P-1 = p_{1}^{a_1} \times \cdots \times p_{n}^{a_n} where each p_i is prime and for each i=1...n : \frac{g^{P-1}}{p_i} \equiv 1 (mod P), than g - is primitive root (generator) by modulo...