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The idea consists of the following: consider the unit sphere $\mathbb{S}^2 \subset \Bbb R^3$, and look at $$\Bbb C \equiv\Bbb R^2 \cong \{(x,y,0)\mid x,y \in \Bbb R \}\subset \Bbb R^3$$ Consider the north pole $N= (0,0,1)$. Consider the stereographic projection from $\Bbb S^2 \setminus N$ to $\Bbb C$. Then you take a object $\infty \not\in \Bbb C$ and ...


Whether or not things are "undefined" largely depeneds on what framework you are working in. If we are working in the naturals, we might say that $3-5$ is undefined. There are many systems where it makes sense to assign $\frac{n}{0}$ some value. In this particular example, it is defined to be complex infinity, which can be thought of as follows: suppose we ...


Sometimes it is useful in complex analysis to consider the complex numbers plus the "point at infinity". See this wiki article for details: Riemann Sphere


According to the help in Mathematica 10, you could do something like this: ListAnimate[Table[WolframAlpha[StringJoin["contour plot x^2+2xy+3y^2+x+y==", ToString[t]]], {t,0,5,1}]] This shows a plot for each of the values $t=0,1,2,3,4,5$.


A matrix is diagonalisable if its eigen values are distinct, $0$ and $k+2$ are the only eigen values..Now $0$ has multiplicity of $2$ and $k+2$ has $1$. If $k=-2$ then eigen value $0$ will have multiplicity of $3$ and now we can just check its eigen vector corresponding to $0$, if it produces $3$ linearly independent eigen vectors then the matrix is ...


Hint: Substitute $r = -k$ and $t = 0$ to get the first eigenvector Wolfram alpha is giving you, and $r = -1$ and $t = 1$ to get the second one. Wolfram alpha isn't giving you the entire two-dimensional eigenspace, but only an eigenbasis.


Wolfram Mathematica has a great help with many examples. Also it does have its own community on StackExchange http://mathematica.stackexchange.com/ . You'd better address questions regarding Mathematica there G1 = CompleteGraph[{7, 2}]; G2 = EdgeDelete[G1, {1 <-> 8, 2 <-> 8}]; GraphPlot[G1] GraphPlot[G2]


Replace $p$ and $p-1$ in this query by the appropriate values: Sort[Map[Mod[#^3,p]&, Range[p-1]]] This will give a multiset (couting how many elements a particular element is a cube of), but it will list all and work within the computation time limit. Breakdown: Sort[ Sort the result of Map[ ...

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