Timeline for Alternating sum of square roots of binomial coefficients
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9 events
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| Jan 7, 2012 at 5:33 | history | edited | Noam D. Elkies | CC BY-SA 3.0 |
Update for fedja's direct proof in mo.84958; remove conflicting indices $k$; justify Taylor convergence per D.Speyer's comment
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| Jan 6, 2012 at 18:37 | comment | added | GH from MO | In the previous comment $a_m$ should be doubled. | |
| Jan 6, 2012 at 18:19 | comment | added | GH from MO | @Noam: I meant that you are really arguing by induction on $n$, and it would be cleaner to say it that way. Here is something more fun: your argument also gives $\sum_{r=0}^n {n\choose r}^\alpha a^r b^{n-r}\geq 0$ for any real $a$ and $b$, even $n$, and $\alpha\leq 1$. Indeed, this can be reduced to $\sum_{r=0}^n (-1)^r{n\choose r}^\alpha f(n,r;x,y)\geq 0$ for $x,y>0$, where $f(n,r;x,y):=(x/y)^{r-\frac{n}{2}}+(y/x)^{r-\frac{n}{2}}=\sum_{m=0}^\infty a_m\left(r-\frac{n}{2}\right)^{2m}$ with $a_m:=(\log x-\log y)^{2m}/(2m)!\geq 0$. The statement follows as in your original post. | |
| Jan 6, 2012 at 17:23 | comment | added | Noam D. Elkies | @GH & @David Speyer & @Mark Wildon: Thanks! \\ @David: Yes, I'll add the remark about the circle of convergence when I edit this later today to fix or improve some other details. \\ @GH: I don't see that either approach is obviously clearer (once we correct $g(x) - g(x+1)$ to $g(x+1) - g(x)$). My initial $g$ doesn't have $g'\geq 0$ everywhere. Basically it's an intermediate-value result, or a formula for the $n$-th finite difference as a convolution of the $n$-th derivative with the convolution power $\chi_{[0,1]}^{*n}$ where $\chi=$ characteristic function. But that may mystify some too... | |
| Jan 6, 2012 at 14:55 | vote | accept | Mark Wildon | ||
| Jan 6, 2012 at 13:24 | comment | added | David E Speyer | Just to fill in a gap that bothered me, you need to know that the Taylor series converges in the relevant range. That's true because the first pole of the $\Gamma$ function is at $0$ and it has no zeroes, so $\log \Gamma(r+1)+ \log \Gamma(n-r+1)$ is analytic on a disc centered at $n/2$ of radius $n/2+1$, which encloses the region we care about. Other than that, gorgeous argument! | |
| Jan 6, 2012 at 8:11 | comment | added | GH from MO | Personally I find the last sentence a bit confusing. I would say: if a function $g$ has positive derivative, then its first finite difference $g(x)-g(x+1)$ is positive; repeating this argument $n$ times, we find that if the $n$-th derivative is positive then the $n$-th finite difference is also positive. | |
| Jan 6, 2012 at 7:51 | comment | added | GH from MO | Beautiful argument. | |
| Jan 6, 2012 at 6:53 | history | answered | Noam D. Elkies | CC BY-SA 3.0 |