Wikipedia:Reference desk/Archives/Mathematics/2011 May 1

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May 1

Continuity

Is $f(x)={\begin{cases}{\frac {1}{x}},&x<0\\x,&x\geq 0\end{cases}}$ continuous at 0? Because $\lim _{x\to 0+}f(x)=f(0)$ and $\lim _{x\to 0-}f(x)$ does not exist. Widener (talk) 07:04, 1 May 2011 (UTC)

It's right-continuous. But it's not continuous because for that you need both directional limits to exist and be equal to one another and the function value at the point. -- Meni Rosenfeld (talk) 08:19, 1 May 2011 (UTC)

Probability

how to prove that sum (i=1 to n) P(Ai) > n-1 implies that intersection (i=1 to n) P(Ai) >0? 131.111.222.12 (talk) 10:34, 1 May 2011 (UTC)

Hint: Decompose the A_i's into disjoint pieces. If $P\left(\bigcap _{i=1}^{n}A_{i}\right)=0$, then every piece with positive probability is contained in at most n-1 A_i's. -- Meni Rosenfeld (talk) 11:00, 1 May 2011 (UTC)

Thanks, Meni!

Identifying a quotient

Let A be the subgroup of $Z^{4}$ with all $(n_{1},n_{2},n_{3},n_{4})$ such that $n_{1}+n_{2}=0$, and B be the subgroup of all $(-m_{1}+m_{2}-m_{3}+m_{4},m_{1}-m_{2}+m_{3}-m_{4},m_{1}+m_{2},m_{3}+m_{4})$. What's the quotient A/B? Money is tight (talk) 15:42, 1 May 2011 (UTC)

Take the mapping from A to $Z^{3}$ that forgets about the $n_{2}$ coordinate. Under this mapping, B goes over to the lattice generated by $(2,0,0),(0,2,0),(0,0,2),(1,1,0),(1,0,1)$. Let C be the lattice generated by $(2,0,0),(0,2,0),(0,0,2)$. Then $A/C=Z_{2}^{3}$, and thus $A/B$ is the a quotient of $Z_{2}^{3}$ by the subgroup generated by $(1,1,0),(1,0,1)$, which is isomorphic to $Z_{2}^{2}$. Hence $A/B=Z_{2}$. (Hopefully right, caveat emptor and all that.) Sławomir Biały (talk) 16:14, 1 May 2011 (UTC)

Another way to see the same result is to consider for given integers $y_{1},y_{2},y_{3}$ the integer solutions of

${\begin{aligned}y_{1}&=-x_{1}+x_{2}-x_{3}+x_{4}\\y_{2}&=x_{1}+x_{2}\\y_{3}&=x_{3}+x_{4}\end{aligned}}$

This system has a solution if and only if $y_{2}+y_{3}\equiv y_{1}{\pmod {2}}$. So, let $\phi (n_{1},n_{2},n_{3},n_{4})=n_{1}+n_{3}+n_{4}{\pmod {2}}$. We see that ${\text{ker}}\phi |_{A}=B$, so $A/B=\phi (A)=\mathbb {Z} _{2}$. Sławomir Biały (talk) 20:51, 1 May 2011 (UTC)

Thanks for your help. I see you work tirelessly to help with other people's problem, I'll never have that kind of generosity :D. Money is tight (talk) 10:25, 2 May 2011 (UTC)

Tensors

I'm having trouble understanding where the transformation formula for second order tensors comes from. We have that $\displaystyle \ V_{i}'=\sum _{j=1}^{3}a_{ij}V_{j}$ for vectors, where the $a_{ij}$'s are the cosines of the appropriate angles. My textbook then says that it is "straightforward to define [higher order] tensors", and gives the formula $\displaystyle \ T'_{kl}=\sum _{i=1}^{3}\sum _{j=1}^{3}a_{ki}a_{jl}T_{ij}$ for second order tensors. Could someone give me a clearer understanding behind the motivation for giving tensors this transformation rule? Thanks. 74.15.138.241 (talk) 20:41, 1 May 2011 (UTC)

It should be

$T'_{kl}=\sum _{i=1}^{3}\sum _{j=1}^{3}a_{ki}a_{lj}T_{ij}$

You can see this by considering the special case of tensors that are tensor products of vectors. Then, you can use the fact that any tensor can be expanded as a linear combination of such tensor products. Count Iblis (talk) 21:34, 1 May 2011 (UTC)

Alright, but why can we say that the tensor product between vectors is a tensor? That is, how can it be shown that the tensor product is covariant?74.15.138.241 (talk) 22:52, 1 May 2011 (UTC)

That follows directly from the transformation rule for vectors. So, if x and y are two vectors and xy denotes the tensor product, then in the new frame we have the transformed vectors x' and y' and in that frame we have the tensor product x'y', which is thus already defined. Count Iblis (talk) 02:58, 2 May 2011 (UTC)

Sorry, one last question: is the tensor product the same as the direct product? The index in my textbook says that they are (Tensor, product. see Direct product) but the wikipedia articles look very different. 74.15.138.241 (talk) 04:09, 2 May 2011 (UTC)

It is explained in detail here: Tensor product, but me being a physicist, I don't bother with this :) . You can just as well define tesnors as any object that transforms in a certain way and motivate that definition heuristically by looking at examples from physics where a tensor arises. Such examples can then involve e.g. vectors whose components are also vectors... Count Iblis (talk) 14:47, 2 May 2011 (UTC)

The real motivation is that there are physically important quantities that transform according to that rule, such as the strain tensor. In fact this is not the only possibility for a second order tensor -- tensors that follow that rule are called contravariant, but there are also covariant and mixed second order tensors that transform according to a different rule. Looie496 (talk) 16:08, 2 May 2011 (UTC)

Could you link me to a proof that the strain tensor transforms like above? Thanks 74.15.138.241 (talk) 20:35, 2 May 2011 (UTC)

That's trivial, just apply the chain rule for differentiation (

dY/dx_j = dY/dx'_r dx'_r/dxj, with summation over repeated indices). In case of the stress tensor, you can use that the stress in a particular direction is given by the contraction of the stress tensor and the unit vector pointing in that direction. That the stress tensor is indeed a tensor then follows from the quotient theorem. Count Iblis (talk) 20:53, 2 May 2011 (UTC)

Thanks. 74.15.138.241 (talk) 23:18, 2 May 2011 (UTC)