?I am not able to produce it with the usual tricks for manipulating binomials.
This arises in the following problem: In dimension D=2n, this shows that Fierzing works, that is that the product of two chiral Weyl spinors can be reexpressed as a linear combination of even dimensional forms. Maple gives me confidence that the above expression is correct, but how can I convince myself (and worse: students) that it holds?
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That's easy...
Sorry, I don't know how to display math equations on this blog.
(a+b)^{2n}=\sum_{k=0}^{n}\begin{pmatrix}2n\\2k\end{pmatrix}a^{2n-2k}b^{2k}+\sum_{k=0}^{n-1}\begin{pmatrix}2n\\2k+1\end{pmatrix}a^{2n-2k-1}b^{2k+1}
So,
4^n=(1+1)^{2n}=\sum_{k=0}^{n}\begin{pmatrix}2n\\2k\end{pmatrix}+\sum_{k=0}^{n-1}\begin{pmatrix}2n\\2k+1\end{pmatrix}
and
0=(1-1)^{2n}=\sum_{k=0}^{n}\begin{pmatrix}2n\\2k\end{pmatrix}-\sum_{k=0}^{n-1}\begin{pmatrix}2n\\2k+1\end{pmatrix}
So,
\sum_{k=0}^{n}\begin{pmatrix}2n\\2k\end{pmatrix}=\sum_{k=0}^{n-1}\begin{pmatrix}2n\\2k+1\end{pmatrix}=4^n/2
You obviously have a lower standard of proof for yourself compared to your students... :)
You ask why the sum of 2n choose 2k from k = 0 to
k = n is 4^n/2.
Well, this is the number of subsets of even size of
a 2n-element set.
The number of subsets of a 2n-element set is 4^n,
and half these are even. That does it.
By the way, I'm John Baez. I'm only posting
"anonymously" because I couldn't manage to log in
on that blogger website. I gave them a username,
password, my name and an email address, and it all
seemed to work fine.
But when I try to "sign in" to post something, I keep
getting looped around back to the same page where I'm
supposed to type my username and password.
Any hints?
Hmm, this time I was able to log on!
Anyway, there's a tricky step in the proof I just gave.
The trick is to show that precisely half the subsets of
any finite set have an even number of elements.
While plausible (and true), your students may still have
some fun showing it.
(1/2)((1+x)^(2n)+(1-x)^(2n))
setting x=1 gives the proof.
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