Here is an exercise similar to the old one:
How do I show 2.?
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Here is an exercise similar to the old one:
How do I show 2.?
This is a code to find inequality $ abc> 2 ( text {area of the triangle}) $
Where $ a $, $ b $ and $ c $ are the sides of a triangle.
The following and the corresponding code work as intended. However, they take a long time. Performance can be improved without reducing max_tries
?
import random
import math
hn=100000000
num=0
max_tries=10000000
tries = 0
while tries != max_tries:
a= random.randint(1,hn)
b= random.randint(1,hn)
c= random.randint(1,hn)
if a+b>c and c+b>a and a+c>b :
s=(a+b+c)/2
tries= tries +1
k =math.sqrt(s*(s-a)*(s-b)*(s-c))
if a*b*c> 2*k:
num= num+1
print(num/max_tries)
To let $ (a_1, a_2), (w, z), (u, v) in mathbb {R} ^ 2 $ so that $ vert (w, z) vert leq 1 $ and $ vert (u, v) vert geq R $, Where $ R $ is a constant greater than $ 1 $.
It's true that
begin {equation}
vert left langle (a_1, a_2), (w, z) right rangle vert leq vert left langle (a_1, a_2), (u, v) right rangle vert?
end {equation}
About me the answer is "yes". I argued like this:
begin {equation}
vert left langle (a_1, a_2), (w, z) right langle vert leq vert (a_1, a_2) vert leq vert (a_1, a_2) cdot R vert leq vert left langle (a_1, a_2), (u, v) right rangle vert
end {equation}
Could someone please help me understand if it's true or not?
This is discussed in Courant's introduction to Calculus and Analysis I. $ f (x) = x ^ 2 $ is continuous over a fixed finite interval (a, b).
It is written as:
$ | f (x) – f (x_ {0}) | = | x ^ 2 – x_ {0} ^ 2 | = | x – x_ {0} || x + x_ {0} | leqslant 2 | x – x_ {0} | (| b | + | a |) <2 sigma (| b | + | a |) <2 epsilon $
I have no idea what inequality sentences the author used:
$ | x – x_ {0} || x + x_ {0} | leqslant 2 | x – x_ {0} | (| b | + | a |) $
$ 2 | x – x_ {0} | (| b | + | a |) <2 sigma (| b | + | a |) $
Given a number of random variables, so for each $ i $ $ in $ $ mathbb N $::
$ Pr (X_i = -9) = 1/10 $
$ Pr (X_i = 1) = 9/10 $
And I define for everyone $ n $ $ in $ $ mathbb N $:: $ S_n = sum ^ n_ {i = 1} X_i $.
How can I use chernoff inequalitie to find an upper limit for? $ Pr (| S_n |> 10 sqrt {n}) $ ?
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I read about Dirichlet polynomials in the book Analytic Number Theory by the authors mentioned. Can anyone justify the following inequality? Assume that $ a (n), b (m) $ are sequences of non-negative numbers, so $ a (n) = 0 $ big enough for everyone $ n $ and $ b (m) = 0 $ big enough for everyone $ m $. How do you prove that? $$ sum _ { substack {n, m, n_1, m_1 in mathbb N \ nm = n_1 m_1}} a (n) b (m) a (n_1) b (m_1) leq sum_n sum_m a (n) ^ 2 b (m) ^ 2 tau (nm)? $$ Here $ dau $ is the function that counts the number of positive integer divisors. This is according to equation (9.7) on page 231.
Given that
| a | = 1
| b-2 | = 3
| c-5 | = 6
Find the maximum value of | 2a-3b-4c |
where a, b, c are complex numbers.
Given non-negative random variables $ X_1, X_2, … $ how to show that
$$ mathbb {E} exp (t max borders_ {1 leq i leq n} X_i) leq sum borders_ {1 leq i leq n} mathbb {E} exp (tX_i ). $$
I think we should start with that
$$ max limit_ {1 leq i leq n} X_i leq sum limit_ {1 leq i leq n} X_i $$
and that applies to Jensen's inequality, but I need help clarifying the details
To let $ T in B (E) $ be a limited linear operator on the Banach grid $ E $. Consider his addition $ T: E ^ * rightarrow E ^ * $ defined in the usual way as $ T ^ * (y (x)): = y (Tx) $ With $ y $ a limited function in $ E ^ * $.
Now leave $ mu in E ^ * $ be an eigenvector of $ T $ whose intrinsic value is $ alpha neq 0 $and define $ nu: = T ^ * R (r_1, T ^ *) | mu | $ With $ r_1> rho (T) $ ($ R $ denotes the resolving operator and $ rho $ the spectral radius of $ T $).
How to show that $ T ^ * nu le r_1 nu $?