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)

  • $\begingroup$ Hm, your series suffers from index sickness. $\endgroup$ – k.stm May 6 '15 at 7:12
  • 2
    $\begingroup$ 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. $\endgroup$ – robjohn May 6 '15 at 7:13
  • $\begingroup$ @k.stm: whoops i typoed. the top limit is infinity not n. $\endgroup$ – One Winged Pterodactyl May 6 '15 at 7:16
  • $\begingroup$ @OneWingedPterodactyl And the index in the sum is $n$, not $i$. $\endgroup$ – k.stm May 6 '15 at 7:18
  • $\begingroup$ @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? $\endgroup$ – One Winged Pterodactyl May 6 '15 at 7:20

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 $$ \left(\sum_{n=0}^\infty\binom{1/2}{n}X^n\right)^2=1+X\tag{3} $$


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