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0

Indeed, WA missed some simplification. $$\log(1-e^{i2\pi x})-\log(1-e^{-i2\pi x})=\log\left(\frac{1-\cos(2\pi x)+i\sin(2\pi x)}{1-\cos(2\pi x)-i\sin(2\pi x)}\right)\\ =\log\left(\frac{2\sin^2(\pi x)-i2\cos(\pi x)\sin(\pi x)}{2\sin^2(\pi x)+i2\cos(\pi x)\sin(\pi x)}\right)=\log\left(-\frac{ie^{i\pi x}}{ie^{-i\pi x}}\right)=i2\pi x+i(2k+1)\pi.$$ So with the ...

2

The inequality as it currently stands has infinitely many solutions. For look for solutions with $p=q\ge 1$. Then each term on the right is $\le 3p^3$. There are $8$ of them, so the right-hand side is $\le 24p^3$. The left-hand side is $p^4$, which is $\ge 24p^3$ if $p\ge 24$.

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The following is taken from this answer. In common usage, $0^0$ is often encountered in set theory as the number of maps from the empty set to the empty set, or as $x^0$ in combinatorics and polynomials. In all of these cases, $0^0=1$ is the proper definition, since there is $1$ map from the empty set to the empty set, and because $$\lim_{x\to0}x^0=1$$ ...

1

This has been a subject of debate, and there is really no one resolution. One reason for this is that there is no one definition of exponentiation itself. There are several definitions corresponding to different domains of the exponent. We can prove (or at least, render) these definitions equivalent where the domains overlap for strictly positive bases, ...

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To me $0^0$ is very nicely defined as the cardinality of the set of maps from the empty set to the empty set, hence equals $1$. However, in the context of limits one must be aware that the binary operation of exponentiation is not continuous at $(0,0)$ (and not even defined in an open neighbourhood of $(0,0)$), which implies that $a_n\to 0$, $b_n\to 0$ gives ...

1

To find out all of the solutions: It's a kinda magical fact that you can encode max and min into mathematical equations if you use absolute value. $$\max(a,b)=\frac{(a-b)+|a-b|}{2}+b$$ With this, you can write an (albeit complicated) equation that wolfram alpha will recognize. To find out if $(x,y,z)$ is a solution for a particular set of values: If you ...

2

For the concavity part, the second derivative of $x$ with respect to $t$ is $$\frac{d(-xe^{-x^2})}{dx}\frac{dx}{dt}$$ This gives you $$-x(2x^2-1)e^{-x^2}$$ So the curve is concave up at $x<-\frac{1}{\sqrt{2}}$ and $0<x<\frac{1}{\sqrt{2}}$, and concave down elsewhere. This is consistent with the graph Wolfram Alpha gives.

2

Your sentence "It seems that it's increasing for $x < 0$ and decreasing $x > 0$" says it all. Because of this the phase portrait has an equilibrium at $0$ and the arrows on $(0,+\infty)$ and $(-\infty,0)$ are respectively $\leftarrow$ and $\rightarrow$. Nothing else is needed for the phase portrait.

3

Here is a parametrization using Mathematica. As @Patrick Stevens pointed out, you cannot plot this in Wolfram Alpha (yet).

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Expanding on my comment: The equation you have is, for example, $x^2+(y-2)^2=1$. The graph you want is the set of all points $(x,y,z)$ which satisfy this equality. Note that $z$ is not in the equation, which means that $z$ can be anything as long as the $x$ and $y$ work. If we looked at the same equation in two dimensions, we would say the graph is the set ...

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The only way I've got for making WolframAlpha recognise this as a request for a 3D plot in which one variable happens to be absent, is: plot [x^2]+[(y-2)^2] + 0.000000001z = 1 This will not quite produce the right plot, but it's very close; I've had to add a term which is nearly zero, to make WA understand the $z$ term. I have reported as a bug the ...

2

The two answers are equivalent. Remember that $sin$ and $tan$, from a trigonometric point of view, are just ratios of sides of a right angled triangle. The $\tfrac{1}{\sqrt{x^2-1}}$ in the wolfram result looks suspiciously like an application of Pythagoras's theorem, doesn't it? You can do the calculation yourself.

0

Euler's Formula is as follows; $$e^{i\theta} = \cos\theta + i \sin \theta.$$ You can think of the number $1$ as $e^{\ln1}$ and hence of $1^i$ as $e^{i\ln1}$. But $\ln 1 = 0,$ so really you have $e^{0i} = e^0 = 1$, as I'm sure you're aware. Even if we look at this by feeding $0$ into Euler's Formula, we have $$e^{0i} = \cos(0) + i\sin(0) = 1.$$

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