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i want to prove this identity:

$(1 + \sum\limits_{n=1}^\infty {1/2 \choose n} X^n)^2 = 1+X$

in the formal power series ring Q[[X]]. (so i can't just quote the binomial expansion for the square root)

the only thing i can think of is to calculate the coefficients of each power of x on the left hand side and work them out directly, but it gets very messy and painful and life's too short for it.

anyone care to bash through the calculations or provide insight? (not homework, by the way)

One Winged Pterodactyl
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  • Hm, your series suffers from index sickness. – k.stm May 06 '15 at 07:12
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    Why can't you just quote the binomial expansion? In any case, we still have the Chu-Vandermonde Identity $$\sum_{k=0}^n\binom{1/2}{k}\binom{1/2}{n-k}=\binom{1}{n}$$ which shows the identity above. – robjohn May 06 '15 at 07:13
  • @k.stm: whoops i typoed. the top limit is infinity not n. – One Winged Pterodactyl May 06 '15 at 07:16
  • @OneWingedPterodactyl And the index in the sum is $n$, not $i$. – k.stm May 06 '15 at 07:18
  • @robjohn: i can't quote the binomial theorem because the derivation for fractional powers i'm aware of relies on calculus. thanks for the link, i wasn't aware of vandermonde's identity generalising to non integer binomial coefficients. can you give a reference for the proof? – One Winged Pterodactyl May 06 '15 at 07:20
  • @OneWingedPterodactyl: it is simply computing the coefficients of $(1+x)^{1/2}(1+x)^{1/2}=1+x$. Since the series for $(1+x)^{1/2}$ converges absolutely for $|x|\lt1$, we get the Chu-Vandermonde Identity quoted above. However, it will require that all the eigenvalues of $X$ have absolute value less than $1$ to ensure convergence. – robjohn May 06 '15 at 07:27
  • @robjohn: hmm, but you're in a formal power series ring, where X is an indeterminate (not a matrix or anything), and convergence isn't a valid concept. how do you expand $(1+x)^{1/2}$ in this case? if i was working in the real numbers i could use the binomial theorem which relies on calculus, which i can't do here. in fact, in a power series ring, $(1+x)^{1/2}$ doesn't make sense unless you define it to be the power series that when you square it gives you 1+x – One Winged Pterodactyl May 06 '15 at 07:34
  • @OneWingedPterodactyl: since you are working in a formal power series ring, we don't need to worry about convergence at all. We only need to compute the coefficients using the product formula from the ring structure, and that follows from the Chu-Vandermonde Identity. – robjohn May 06 '15 at 07:42
  • @robjohn: yeah i see what you mean. but then, what all this comes down to is looking for a proof of chu-vandermonde that doesn't make the assumption that you're in R or C, but can work in a formal power series ring, where things like $(1+x)^{1/2}$ don't really make sense. – One Winged Pterodactyl May 06 '15 at 07:52
  • @OneWingedPterodactyl: the Chu-Vandermonde Identity is an identity on real/complex numbers, and the coefficients of the formal power series are real/complex numbers – robjohn May 06 '15 at 07:55
  • @robjohn: OHHHHHHH ok i see what you mean now. i was too busy looking at the proof given in section 1 here: https://proofwiki.org/wiki/Chu-Vandermonde_Identity and i never realised that the x in that proof doesn't have to be the same as the X in the identity i'm trying to prove. ok i get it now thanks – One Winged Pterodactyl May 06 '15 at 07:59

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The Chu-Vandermonde Identity gives that $$ \sum_{k=0}^n\binom{1/2}{k}\binom{1/2}{n-k}=\binom{1}{n}\tag{1} $$ Computing the square of the series using $(1)$ and the product formula for the ring structure of the formal power series ring $$ (a_n)_{n\in\mathbb{N}}\times(b_n)_{n\in\mathbb{N}}=\left(\sum_{k=0}^na_kb_{n-k}\right)_{n\in\mathbb{N}}\tag{2} $$ gives us that $$ \left(\sum_{n=0}^\infty\binom{1/2}{n}X^n\right)^2=1+X\tag{3} $$

orangeskid
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robjohn
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