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## New answers tagged eulers-constant

1

The natural definition of $e$ is not a definition of that single number, but rather of a special function $x\mapsto \exp(x)$, which has the property that $\exp(0)=1$ and that it is its own derivative: $\exp'(x)=\exp(x)$. It turns out, that $\exp$ is uniquely determined by this. It follows that for any constant $k$, $\exp'(kx)=k\exp(kx)$ and ...

3

The most intuitive explanation I know involves a combination of three facts: If a particle has position $p(t)$ proportional to its velocity $p'(t)$, say $p'(t) = kp(t)$, then $p(t) = A e^{kt}$ for some constant $A$. We can take this as a definition of the exponential. In the complex plane, multiplication by $i$ is the same as a counter-clockwise rotation ...

8

Just think about it this way: $\pi$ is related to the circle, whose equation is $x^2+y^2=r^2$. Euler's constant e is related to the hyperbola, whose equation is $x^2-y^2=r^2$. In order to turn $y^2$ into $-y^2$ we need a substitution of the form $y\mapsto iy$.

1

For a given $y>0$, the equation $x\ln(y) = y\ln(x)$ is equivalent to $\ln(x)/x=\ln(y)/y$. Here is the graph of the function $f(x)=\displaystyle\frac{\ln(x)}{x}$. The function has a global maximum at $x=e$. If $y\leq 1$, then $f(y)\leq0$ and the equation has only one solution: $x=y$. If $y>1$, then $f(y)>0$ and $y\neq e$ and the equation has ...

1

Here is a relevant paper. They find a parametric equation for the non-$“x=y”$ branch of the graph (Desmos link), and then prove that the branches intersect at $(e,e)$. The rest of this answer is my own input. The fact that the two branches of the graph of $x^y=y^x$ intersect at $(e,e)$ is related to the fact that the graphs of $y=x^e$ and $y=e^x$ are ...

0

This is a derivation of Claude's answer. The definition of the Lambert-$W$ function is the solution to: $$x = y e^y$$ Our equation is: $$x \ln y = y \ln x$$ $$\implies - \frac {\ln x}{x} = -\frac {\ln y}y = - \ln y e^{- \ln y}$$ $$\implies W\left({-\frac {\ln x}{x}}\right) = - \ln y = - y \frac{\ln x}{x}$$ $$\implies y = - \frac{x}{\ln ... 2 For most points on y=x, you have dy/dx=1. There is one point where dy/dx could be 1 or -1. Call that point (x,y)=(a,a).$$x\ln y=y\ln x\\ \ln y+\frac xy\frac{dy}{dx}=\frac{dy}{dx}\ln x+\frac yx\\ \ln a+\frac{dy}{dx}=\frac{dy}{dx}\ln a+1$$This has a unique solution unless \ln a=1 2 Any equation which can be written$$A+Bx+C\log(D+Ex)=0$$has solutions expressed in terms of Lambert function. In the case of x^y=y^x, this writes$$y = -\frac{x}{\log x}\,W\left(-\frac{\log x}{x}\right)$$and what Lambert and Euler showed is that, in the real domain, the function W(z) exists if z\geq -\frac 1e. So, for the argument -\frac{\log ... 8 Here is an approach. We may observe that we have$$ \int_0^{+\infty}\frac 1{(x+t)(\log^2 x +\pi^2)}dx=\frac 1{\log t}+\frac 1{1-t},\quad 0<t<1.\tag1 $$Proof. Let t be any real number such that 0<t<1. Set$$ f(z):=\frac 1{(z+t)(\log z +i\pi)} $$where \log is the determination of the logarithm defined on the complex plane cut along ... 6 From the standard integral representation of the Euler-Mascheroni constant we have:$$ \gamma = \int_{0}^{1}\left(\frac{1}{\log(1-x)}+\frac{1}{x}\right)\,dx \tag{1}$$hence if we define the Gregory coefficients C_n through the Taylor series of \frac{z}{\log(1-z)} in a neighbourhood of zero:$$ \frac{z}{\log(1-z)}=\sum_{n\geq 0}C_n\,x^n \tag{2} we have ...

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