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I assume you mean assigning a new value to some variable? In that case, $:=$ is often used, especially in computer science. The symbol $\leftarrow$ is also often used, and is perhaps more intuitively understandable, as the value on the right-hand side is "put into" the variable on the left-hand side. Closely related, := is also sometimes used as the symbol ...


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Online and free (but using neither Mathematica nor R), the code $$\texttt{streamplot{{-2x*ln(x/y),-y+5sqrt(xy)-3yx^(2/3)},{x,0,5},{y,0,5}}}$$ produces the diagram: The fixed point in the middle is a stable node (sink), it might attract every trajectory starting from some $V(0)>0$, $K(0)>0$, and it is located at ...


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(*p=2;n=5; set according to need*) Select[ FromDigits[#, x] & /@ IntegerDigits[ Table[counter, {counter, p^(n + 1)}], p, n + 1], IrreduciblePolynomialQ[#, Modulus -> p] &] // Expand Explanation: create a table with all the numbers from $0$ to $p^{n+1}-1$ written in base $p$, convert each such number to the list of its digits in ...


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Use the sieve of Eratosthenes. Create a map $m$ from monic polynomial to bools, and then starting with the polynomial of smallest degree and moving up, if you find a polynomial $p$ such that $m_p = true$, set $m_{fp}$ equal to $false$ for all $f$ where $\deg f > 0$.


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I built the game in Mathematica using the rules I think you're trying to accomplish. Here is the code if you have Mathematica: $\hspace{3cm}$ bound = {{2, -1}, {2, 5}, {1, 0}, {3, 0}, {0, 1}, {4, 1}, {0, 2}, {4, 2}, {-1, 3}, {5, 3}, {-1, 4}, {5, 4}, {-1, 5}, {5, 5}, {0, 6}, {1, 6}, {3, 6}, {4, 6}} DynamicModule[{pos1 = {x1, y1} = {2, 2}, pos2 = ...


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Try filling in $13/2$ instead of $6.5$. When using points, Mathematica will automatically treat everything numerically, and hence also gives rounded answers.


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There are packages available for MATLAB do solve partial differential equations (e.g., Partial Differential Equation Toolbox) but I believe they are add-ons that you have to pay for.


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The complex values arise because you get phase information from the signal as well as frequency composition. If your interested in a power spectrum all you have to do is use the absolute value (some square it - depending on application) of the fourier transformed signal.


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Note that in the Kronecker product of graphs (also called the Tensor product), the sequence of valencies (also called degrees) of the vertices result from all of the different products of the valencies of the two graph factors. Thus, if your graph was factorisable into a 5-vertex and 2-vertex part then it must have an even number of vertices of each valency ...



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