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## New answers tagged constants

1

Given that $C$ is a constant (meaning it does not change with any variable change), we can indeed substitute $C=\ln(k)$ for some other constant $k$. Though this constant is related to $C$, it is still a constant - because $C$ is in no way related to a variable, $C$ will not change, and neither will $k$. Your simplified form is thus correct.

1

We have, $$\ln(y) = -{x^2\over 2y^2} + C$$ if $\ln(k) = C$ then $$\ln(y) = -{x^2\over 2y^2} + \ln (k)$$ $$\implies \frac{x^2}{2y^2}=-\ln (y)+\ln(k)$$ $$\implies \frac{x^2}{2y^2}=\ln\left(\frac{k}{y}\right)$$ $$\implies \frac{k}{y}=e^{\frac{x^2}{2y^2}}$$ $$\implies \frac{y}{k}=e^{-\frac{x^2}{2y^2}}$$ $$\implies y=k\cdot e^{-\frac{x^2}{2y^2}}$$ Your ...

3

Yes, this is correct. To me, it is simpler to argue that $$y = \exp \left( -\frac{x^2}{2y^2}+C \right) = e^C \exp \left( -x^2/(2y^2) \right) = K e^{-x^2/(2y^2)},$$ where $K$ is some constant since $C$ is a constant.

0

$$\frac { d }{ dx } \left( f\left( x \right) \left( x+c \right) ^{ -1 } \right) =0\\ \frac { df\left( x \right) }{ dx } \left( x+c \right) ^{ -1 }+f\left( x \right) \frac { d\left( \left( x+c \right) ^{ -1 } \right) }{ dx } =0\\ \frac { df\left( x \right) }{ dx } \frac { 1 }{ \left( x+c \right) } -f\left( x \right) \frac { 1 }{ { \left( x+c \right) }^{ ... 8 Use the chain rule. Define u = x + c then use the fact that$$\frac{d\cdot}{dx} = \frac{du}{dx} \frac{d\cdot}{du}$$where the \cdot represents any function, so$$\frac{df}{dx} = \frac{du}{dx} \frac{df}{du}$$It also follows that$$ \begin{array}{rcll} \frac{d^2f}{dx^2} &=& \frac{d}{dx} (\frac{df}{dx}) &\quad\mbox{definition of 2nd ...

1

Introducing $y=x+c$ one gets $$0 = \frac{d(f(x))}{d(x+c)} = \frac{d(f(y-c))}{dy} = f'(y-c) \cdot 1 = f'(x) = \frac{d(f(x))}{dx}$$ For the second derivative we get: $$0 = \frac{d^2(f(x))}{d(x+c)^2} = \frac{d}{d(x+c)} \frac{d(f(x))}{d(x+c)} = \frac{d(f'(x))}{d(x+c)} = (f')'(x) = f''(x) = \frac{d^2(f(x))}{dx^2}$$

1

Let's start off with the case of the Fibonacci sequence. We have $$\sum_{n = 1}^k \left( \frac{F_{n+2}}{F_{n+1}} -\frac{F_{n+3}}{F_{n+2}} \right) = \left(\frac{F_3}{F_2} - \frac{F_4}{F_3}\right) + \left(\frac{F_4}{F_3} - \frac{F_5}{F_4}\right) + \cdots + \left(\frac{F_{k+2}}{F_{k+1}}- \frac{F_{k+3}}{F_{k+2}}\right)\\ = \frac{F_3}{F_2} - ... 10 A probability distribution of the continued fraction expansion terms follows the Gauss-Kuzmin distribution for almost all irrational numbers:$$p(k)=-\log_2\left(1-\frac{1}{(k+1)^2}\right)=\log_2\frac{(k+1)^2}{k(k+2)}$$All generalized Khinchin's constants (including K=K_0, the geometric mean), are derived from this distribution. In this case, you seek ... 2 All real zeroes of the Riemann Zeta function (also called the trivial zeroes) are of the form:$$x=-2n Where $n\in \Bbb N_0$ All the other zeroes are supposed to lie on a critical strip in the Complex Plane.

1

The function $\Omega(n)$ is, by definition, the number of prime factors of $n$, counted with multiplicity (that is, not necessarily distinct). For example, $\Omega(720) = \Omega(2^43^25) = 4+2+1 = 7$. The summatory function for $\Omega$ is indeed $F(x) = \sum_{n\le x} \Omega(n)$. For example, \$F(8) = \Omega(1) + \Omega(2) + \cdots + \Omega(8) = 0 + 1 + 1 + ...

1

This constant appears in the conjectured asymptotic estimate on the number of twin primes below a given size, see here. It's also noteworthy that this constant wasn't just taken out of thin air, this can be actually derieved heuristically via Cramer's random model.

1

It appears to be called the "twin prime constant" because it comes up a lot in theorems and conjectures about twin primes. For example, from the relevant MathWorld page:

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