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0

A graph having edges with real weights has an adjacency matrix $W$ with real entries. The example graph given in the Wolfram page has the adjacency matrix shown below. \begin{equation*} \begin{bmatrix} 0 & 1.6 & 1.4 & 0\\ 1.6 & 0 & 1.2 & 0\\ 1.4 & 1.2 & 0 & 2.5\\ 0 & 0 & 2.5 & 0 \end{bmatrix} \end{equation*} ...


2

This is called fractional part: $\text{frac}[x]$. See here for an example.


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You can use the fractional part function frac(). NB there are actually three different functions sometimes called the fractional part. All three agree at nonnegative real and negative integral arguments, but in general they differ for negative nonintegral arguments. Their implementations in W.A. are (along with suggestive examples): frac(x): frac(-1/3) ...


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For positive numbers, you can always do $x-\text{floor}(x)$.


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The given constraints on $y_1, y_2, y_3$ are such that there are only three cases: $$(y_1, y_2, y_3) \in \{(1,0,0), (0,1,0), (0,0,1)\}.$$ In each case, two of the three variables $$x_1, x_2, x_3$$ are constrained to $0$ and the resulting function $z$ becomes univariate on the remaining nontrivial variable. Note that WolframAlpha does not accept/parse the ...


0

You specified in your query that y1,y2,y3 are all positive definite. y1+y2+y3=1 and y3 is and integer. Or y1,y2 =0 and y3 =1, this forces x1,x2 = 0, 2 <= x3 <= 3 z reaches its minimum under this condition at x3 =2, z=1 WA is confused because of the length of the query. Appending any additional argument to the end of minimize ...


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Actually I got it working by using the proper notation: A && ~B && (~B && C || ~C && A && (D || B && ~A || D && A && B || C && ~D))


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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 ...


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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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