integration – Integral of $intlimits_0^{2pi } {{e^{acos (theta – b) + ccos (d – theta )}}dtheta } $?

I know that the integral of

$intlimits_0^{2pi } {{e^{acos (theta – b))}}dtheta } = 2pi {I_0}(a)$

Where, ${I_0}(a)$ is the Modified bessel function of the first kind.

I am trying to find the integral of $intlimits_0^{2pi } {{e^{acos (theta – b) + ccos (d – theta )}}dtheta } $. Can I transform this integral into Bessel function or some known function?

binance – Sent old ERC-20 THETA tokens to mainnet wallets

I recently sent my old Theta token to Binance as well as the Theta Wallet online (wallet.thetatoken.org), and have yet to see a balance in either wallet despite confirmations. I realize now that I sent the old tokens that hadn’t been swapped to mainnet wallets. Is there any hope of recovering my old tokens and swapping them for mainnet THETA and TFUEL?

Theta functions – Is $ sum_ {n in mathbb {Z}} e ^ {- (n- mu) ^ 2/2 sigma ^ 2} le sum_ {n in mathbb {Z} } e ^ {- n ^ 2/2 sigma ^ 2} $ for all $ mu $ and all $ sigma $?

I looked at discrete Gaussian distributions and came to the following assumption. I would be very happy to have proof (or proof).

Guess. To let $ mu in (0.1) $ and $ sigma ^ 2> 0 $. Then $$ sum_ {n in mathbb {Z}} e ^ {- (n- mu) ^ 2/2 sigma ^ 2} le sum_ {n in mathbb {Z}} e ^ { -n ^ 2/2 sigma ^ 2}. $$

Numerical evidence supports this assumption.

In the continuous analogue (i.e. replace $ sum _ { mathbb {Z}} $ With $ int _ { mathbb {R}} $), that's an equality. Indeed as $ sigma ^ 2 to 0 $the discrete sum becomes a closer approximation to the continuous integral and the inequality almost becomes an equality. (Even for $ sigma ^ 2 = 1 $is the difference only $ 2.7 times 10 ^ {- 8} $.)

On the other hand if we take $ sigma ^ 2 to infty $then the infinite sum is dominated by a single term. The inequality becomes approximate $$ forall mu ~~~~~ e ^ {- mu ^ 2/2 sigma ^ 2} le e ^ 0, $$ that is trivially true.

So we see it in both extremes ($ sigma ^ 2 to 0 $ and $ sigma ^ 2 to infty $) the presumption applies. This is further proof that it applies to all values ​​of $ sigma ^ 2 $.

These sums can be expressed in Jacobi theta functions. However, I don't see how helpful this is.

Algorithms – What is the big theta of $ ( log n) ^ 2 + 2n + 4n + log n + 50 $?

$ f (n) = ( log n) ^ 2 + 2n + 4n + log n + $ 50

I am trying to prove mathematically that $ f (n) $ falls under the temporal complexity of $ theta (( log n) ^ 2) $.

I have to come to the end $ f (n) leq C ( log n) ^ 2 $for a positive constant $ C $ and $ x geq k $.

What I tried is:

$ ( log n) ^ 2 + log n leq 2 ( log n) ^ 2 $

i want to add $ 6n + $ 50 to both sides, but cannot find the constant $ c $ by algebra in $ c ( log n) ^ 2 $.

I was just trying to set random values ​​from $ c $ close $ c ( log n) ^ 2 geq 6n $ true and $ c = $ 100 works for $ n $ larger than a value, but there is a mathematical way to find that value from $ c $ and therefore find the big theta of this function?

partial derivative – get $ z = f (x, y) $ with $ x = rcos theta $ and $ y = rsen theta $ prove it

To get $ z = f (x, y) $ With $ x = rcos theta $ and $ y = rsen theta $
Prove that
$$ frac { partially ^ 2 z} { partially r ^ 2} = frac { partially ^ 2 z} { partially x ^ 2} cos ^ 2 theta + 2 frac { partially ^ 2 z } { partial x partial y} sen theta cos theta + frac { partial ^ 2 z} { partial y ^ 2} sen ^ 2 theta $$


Find the solution, but I don't know if it's right

Find $$ frac { partial z} { partial r} = frac { partial z} { partial x} cos theta + frac { partial z} { partial y} sen theta
$$

If this is the only thing it does, ^ 2 to get the result. Is this procedure okay?

Asymptotics – Prove that for all functions g: N -> R> = 0 and all numbers a in R> = 0, if g in omega (1), then a + g in theta (g)

Here is a more readable version of the question:

Prove that for all functions $ g: mathbb {N} to mathbb {R} ^ { geq 0} $and all the numbers $ a in mathbb {R} ^ { geq 0} $, if $ g in Omega (1) $ then $ a + g in Theta (g) $

What I've done so far:

In order to $ a + g in Theta (g) $. $ a + g in mathcal {O} (g) Keil a + g in Omega (g) $,

If we expand our assumption, we get:

$ exists c_1, n_1 in mathbb {R} ^ {+}, forall n in mathbb {N}, n geq n_1 implies g (n) geq c_1 $

To prove $ a + g in Omega (g) $::

If we expand the definition, we get:

$ exists c_2, n_2 in mathbb {R} ^ {+}, forall n in mathbb {N}, n geq n_2 implies a + g (n) geq c_2 * g (n) $

To let $ c_1 = 1 $ and $ n_2 = n_1 $, To let $ n in mathbb {N} $, Accept $ n geq n_2 $, To prove $ a + g (n) geq c_2 * g (n) $,

$ g (n) = g (n) \ implies g (n) geq g (n) \ Leftrightarrow c_1 * g (n) geq c_1 * g (n) \ Leftrightarrow g (n) geq g (n) text {(since $ c_1 = 1 $)} \ Left right arrow a + g (n) geq g (n) text {(enlargement of the left side since $ a in mathbb {R} ^ { geq 0} $)} $

To prove $ a + g in mathcal {O} (g) $::

If we expand the definition, we get:

$ exists c_3, n_3 in mathbb {R} ^ +, forall n in mathbb {N}, n geq n_3 implies a + g (n) leq c_3 * g (n) $

I'm fighting here because I'm not really sure what its value is $ c_3 $ I should use or how to derive one. (I tried to use my assumption of $ g (n) geq c_1 $ but I don't really know where to go from there). Any help is greatly appreciated and I apologize for any formatting errors in advance. Thank you very much.

Number theory – perfect square problem $ a ^ 2 + b ^ 2 – 2ab $ cos ($ theta $)

I'm trying to find out if $ a ^ 2 + b ^ 2 – 2ab $cos ($ theta $) can always be a perfect square if $ a, b $ are clearly positive integers> $ 1 $With $ a $ < $ b $and cos ($ theta $) $ epsilon $ )$ 0, a / 2b $(. All values ​​of cos ($ theta $) in this interval must be irrational or rational numbers $ p / q $ and if they're rational, it seems to me $ q $ could be a factor of $ from $. $ a $, or $ b $what the expression would do $ a ^ 2 + b ^ 2 – 2ab $cos ($ theta $) an integer. I've been kicking this around for some time and I can't see a way into the problem. Suggestions would be very welcome.

Graphic – Why does Schlick's approximation contain a $ (1- cos theta) ^ 5 $ term?

The approximation writes the reflection coefficient as$$ R (θ) = R_0 + (1-R_0) (1-cosθ) ^ 5, R_0 = left ( frac {n_1-n_2} {n_1 + n_2} right) ^ 2. $$Why is the exponent 5? Schlick 1994 leads this exponent in Eq. (24) with the claim that it is the correct Fresnel approximation but without explanation.