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Is $$ \sum_{n=0}^\infty {x^n \over \sqrt{n!}} > 0 $$ for all real $x$? (I think it is.) If so, how would one prove this? (To confirm: This is the power series for $e^x$, except with the denominator replaced by $\sqrt{n!}$.)

Thank you and Happy New Year!

J. Russell

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I don't know why people are voting to close this. I would be interested in learning about methods that could be used to prove that an entire function, given by convergent power series, had no real zeroes, so I would like this question to stay open. –  David Speyer Jan 5 '12 at 15:07
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I completely agree with @David. –  Igor Rivin Jan 5 '12 at 15:33
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$1/\sqrt{n!}$ for the absolute value of coefficients is an interesting choice for random polynomials (called Weyl polynomials), their roots are roughly uniformly distributed in a large disc. –  Roland Bacher Jan 5 '12 at 17:09
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This is very far from a solution, but might be of interest. Let $F(x) = \sum_{n=0}^\infty x^n/\sqrt{n!}$ and let $H(x) = F(x)F(-x)$. Then $H(x) = \sum_{n=0}^\infty c_nx^n/\sqrt{n!}$ where $c_n = \sum_{r=0}^n (-1)^r \sqrt{\binom{n}{r}}$. It is clear that $c_n = 0$ if $n$ is odd, and it would be sufficient to prove that $c_n > 0$ if $n$ is even, since then $F(-x) = H(x)/F(x)$ is a positive function. I have checked this is true for $n \le 100$. –  Mark Wildon Jan 5 '12 at 17:34
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@Mark Wildon You should really post that observation as a separate question, because it is absolutely bizarre that it works. The largest term in the sum defining $c_{1000}$ is about $1.4 \times 10^{300}$. The fact that you get such nice cancellation to get an answer near $0.02$ seems like a miracle to me. Of course, $\sum (-1)^r \binom{n}{r}$ has even larger terms cancelling, but that is for a very good reason; I can't see any reason for your sum to be so small. –  David Speyer Jan 5 '12 at 22:45
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7 Answers 7

up vote 67 down vote accepted

Looks like the computers really spoiled us :)

GH gave a perfectly valid answer already but the cheapest way to prove positivity is to write $\int_0^1(1-t^n)\log(\frac 1t)^{-3/2}\,\frac{dt}t=c\sqrt n$ with some positive $c$ (just note that the integral converges and the integrand is positive, and make the change of variable $t^n\to t$). Hence $\int_0^1 (f(x)-f(xt))\log(\frac 1t)^{-3/2}\,\frac{dt}t=cxf(x)$. If $x$ is the largest zero of $f$ (which must be negative), then plugging it in, we get $0$ on the right and a negative number on the left, which is a clear contradiction. Thus, crossing the $x$-axis is impossible. Of course, there is nothing sacred about $1/2$. Any power between $0$ and $1$ works just as well.

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Very slick! I had not seen that trick before. –  David Speyer Jan 6 '12 at 13:26
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Fedja, thank you. Because $d e^x / dx = e^x$, you can show that $e^x$ is positive by asking what would happen at a zero crossing. I've tried (unsuccessfully) to analyze $f'(x)$ with the hope of doing something similar. Your proof strikes me as having the same flavor as that sort of manipulation. Could you offer any motivation for how you came up with your integral "operator"? –  J Russell Jan 6 '12 at 14:35
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@ J Russell. Yes, I started with the proof for the exponent you mentioned. The main difference is that you need half-derivative here, not the full one. The half-derivative (or, rather, quarter-Laplace) operator is well-known: it is just (regularized) convolution with $|t|^{-3/2}$. We needed the one-sided version here (the other side is not integrable), which works especially well on the negative exponents: $\int_0^\infty \frac{e^{-nx}-e^{n(x+t)}}{t^{1/2}}\frac{dt}{t}=n^{1/2}e^{-nx}$, so I just made the logarithmic change of variable to make it act on powers as needed. –  fedja Jan 6 '12 at 16:53
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Nice!! For what its worth, $c$ is $\int_0^\infty (1-e^{-x}) \phantom. dx^{3/2} = 2 \Gamma(1/2) = 2\sqrt{\pi}$ by integration by parts. [My previous comment, now deleted, reported an incorrect answer because I left the factor $1/t$ out of the integrand in the numerical calculation.] –  Noam D. Elkies Jan 7 '12 at 23:02
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Do you really think so? I don't. –  fedja Jan 8 '12 at 15:59
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The affirmative answer follows from my response to this related question.

EDIT. Noam Elkies gave a nicer and more general argument here.

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GH, thanks to you and Noam (and to de Bruijn) for this nice piece of analysis. –  J Russell Jan 6 '12 at 14:34
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Here is another non-answer. In "Asymptotic Methods in Analysis", chapter 6, de Bruijn proves that $$S(s,n)=\frac{2}{\pi}\Gamma(s)(2ns\log 2n)^{-s}\left(\sin(\pi s)+O\left((\log n)^{-1}\right)\right)$$ where $$S(s,n)= \sum_{k=0}^{2n} (-1)^k \binom{2n}{k}^s$$ for all $0\le s\le\frac{3}{2}$. So at least this explains things asymptotically.

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Actually he proves this for $0<s<\frac{3}{2}$. –  GH from MO Jan 6 '12 at 4:22
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Inspired by your response I answered the question in the affirmative. See my response. –  GH from MO Jan 6 '12 at 5:57
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Additional data for Liviu's plots. I used Pari/GP with 1200 digits dec prec, documenting also the required number of terms after which the absolute values of the summands of the series decrease below 1e-100. There seems to be no local minimum...

$\small \begin{array}{rl|r} & & \text{# of terms}\\ x & f(x) & \text{ required} \\ \hline \\ -1 & 0.438599896749 & 201 \\ -2 & 0.247539616819 & 201 \\ -3 & 0.162554775870 & 211 \\ -4 & 0.117399404501 & 257 \\ -5 & 0.0903120618145 & 304 \\ -6 & 0.0726061182760 & 354 \\ -7 & 0.0602796213492 & 407 \\ -8 & 0.0512783927864 & 464 \\ -9 & 0.0444561508357 & 525 \\ -10 & 0.0391295513879 & 589 \\ -11 & 0.0348689168813 & 658 \\ -12 & 0.0313919770798 & 730 \\ -13 & 0.0285063993737 & 808 \\ -14 & 0.0260770215882 & 889 \\ -15 & 0.0240063146159 & 976 \\ -16 & 0.0222222780410 & 1067 \\ -17 & 0.0206706877888 & 1162 \\ -18 & 0.0193099849974 & 1263 \\ -19 & 0.0181078191003 & 1369 \\ -20 & 0.0170386561852 & 1479 \\ -21 & 0.0160820905671 & 1595 \\ -22 & 0.0152216309789 & 1715 \\ -23 & 0.0144438135509 & 1841 \\ -24 & 0.0137375438980 & 1972 \\ -25 & 0.0130936024884 & 2108 \\ -26 & 0.0125042681404 & 2250 \\ -27 & 0.0119630281606 & 2396 \\ -28 & 0.0114643528377 & 2548 \\ -29 & 0.0110035182996 & 2705 \\ -30 & 0.0105764661081 & 2867 \\ -31 & 0.0101796910429 & 3035 \\ -32 & 0.00981015071575 & 3208 \\ -33 & 0.00946519223932 & 3386 \\ -34 & 0.00914249232841 & 3569 \\ -35 & 0.00884000806032 & 3758 \\ -36 & 0.00855593615550 & 3953 \\ -37 & 0.00828867911422 & 4152 \\ -38 & 0.00803681690505 & 4357 \\ -39 & 0.00779908317617 & 4567 \\ -40 & 0.00757434517200 & 4783 \\ -41 & 0.00736158670179 & 5004 \\ -42 & 0.00715989363457 & 5231 \\ -43 & 0.00696844149585 & 5462 \\ -44 & 0.00678648482039 & 5700 \\ -45 & 0.00661334797911 & 5942 \\ -46 & 0.00644841724806 & 6190 \\ -47 & 0.00629113392871 & 6444 \\ -48 & 0.00614098836080 & 6703 \\ -49 & 0.00599751469633 & 6967 \\ -50 & 0.00586028632445 & 7236 \\ -51 & 0.00572891185489 & 7511 \\ -52 & 0.00560303158255 & 7792 \\ -53 & 0.00548231436720 & 8078 \\ -54 & 0.00536645487311 & 8369 \\ -55 & 0.00525517112099 & 8666 \\ -56 & 0.00514820231209 & 8968 \\ -57 & 0.00504530688991 & 9275 \\ -58 & 0.00494626080983 & 9588 \\ -59 & 0.00485085599129 & 9907 \\ -60 & 0.00475889893049 & 10230 \\ -61 & 0.00467020945455 & 10560 \\ -62 & 0.00458461960073 & 10894 \\ -63 & 0.00450197260623 & 11234 \\ -64 & 0.00442212199624 & 11580 \\ -65 & 0.00434493075923 & 11931 \\ -66 & 0.00427027059992 & 12287 \\ -67 & 0.00419802126157 & 12649 \\ -68 & 0.00412806991028 & 13016 \\ -69 & 0.00406031057475 & 13388 \\ -70 & 0.00399464363573 & 13766 \end{array} $

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Thanks for the table. After I wrote that there is a minimum I did some additional computations and I realized I was wrong. Your table suggests that the question is far form trivial. –  Liviu Nicolaescu Jan 6 '12 at 0:13
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Note again the amazing cancellation. To compute that −70 term, the largest term in the sum is $(70)^{70^2}/\sqrt{(70)^2!} \approx e^{70^2/2}$. –  David Speyer Jan 6 '12 at 0:23
    
This function is indeed miraculous. –  Liviu Nicolaescu Jan 6 '12 at 0:57
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Gottfried, thank you for the numerical evaluation. This is consistent with how I believe the series behaves and with the numerical calculations I have done, although I didn't have the tools to push it out anywhere near as far as you did. A comment and an open-ended question: Presumably one could use interval arithmetic to produce a computer-aided proof of positivity for some negative values of $x$. Suppose you could prove positivity down to some large negative value. How large in magnitude would that $x$ have to be for you to really believe, absent a proof, that the series is positive? –  J Russell Jan 6 '12 at 1:50
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This comment serves to record a partial attempt, which didn't get very far but might be useful to others. Following a suggestion of Mark Wildon and Arthur B, define $$f_n(\alpha) := \sum (-1)^r \binom{n}{r}^{\alpha}.$$ This is zero for $n$ odd, so we will assume $n$ is even from now on.

Mark Wildon shows that it would be enough to show that $f_n(1/2) \geq 0$ for all $n$. It is easy to see that $f_n(0) = 1$ and $f_n(1)=0$. Arthur B notes that, experimentally, $f_n(\alpha)$ appears to be decreasing on the interval $[0,1]$. If we could prove that $f_n$ was decreasing, that would of course show that $f_n(1/2) > f_n(1) =0$.

I had the idea to break this problem into two parts, each of which appears supported by numerical data:
1. Show that $f_n$ is convex on $[0,1]$.
2. Show that $f'_n(1) < 0$.
If we establish both of these, then clearly $f_n$ is decreasing.

I have made no progress on part 1, but here is most of a proof for part 2. We have $$f'_n(1) = \sum (-1)^r \binom{n}{r} \log \binom{n}{r} = \sum (-1)^r \binom{n}{r} \left( \log(n!) - \log r!- \log (n-r)! \right)$$ $$=-2 \sum (-1)^r \binom{n}{r} \left( \log(1) + \log(2) + \cdots + \log (r) \right)$$ $$=-2 \sum (-1)^r \binom{n-1}{r} \log r.$$ At the first line break, we combined the $r!$ and the $(n-r)!$ terms (using that $n$ is even); at the second, we took partial differences once.

This last sum is evaluated asymptotically in this math.SE thread. The leading term is $\log \log n$, so the sum is positive for $n$ large, and $f'_n$ is negative, as desired. The sole gap in this argument is that the math.SE thread doesn't give explicit bounds, so this proof might only be right for large enough $n$.

This answer becomes much more interesting if someone can crack that convexity claim.

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Here is a plot of

$$\frac{1}{100}\left(\sum_{k=0}^{16}\frac{x^k}{\sqrt{k!}}\right)$$

on the interval $[-4,0]$. (Above I added the terms up to degree $16$.)

alt text

Next, is a plot of

$$\frac{1}{100}\left(\sum_{k=0}^{15}\frac{x^k}{\sqrt{k!}}\right)$$

on the interval $[-3,0]$. (Above I added the terms up to degree $15$)

alt text

This is one strange series.

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Degree $n=16$ is of little use for $x=-10$, for example. The terms increase in size up to about $n=50$ and only then start to decrease. –  Gerald Edgar Jan 5 '12 at 20:10
    
@Gerard I know that is why I chose shorter intervals. The function seems to have a local minimum at $-2.44$ and the value there is very small $\approx 0.202$. It's a curious function. –  Liviu Nicolaescu Jan 5 '12 at 20:30
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Another "not-yet-answer"...

I've tried another idea. Assume the function f(x) is expressed by the following composition: $$\small x' = \exp(x)-1 $$ $$\small f(x) = g(x') = g(exp(x)-1) $$ The idea is, that the unavoidable big "hump" in the partial sums, after which the sequence of partial sums begins to decrease, may be absorbed by the function $\small g(x)$ - because $\small \exp(x) $ is really small for large negative x and x' is then very little above -1. I did not yet arrive at a conclusive result; but the power series for $\small g(x) $ begins with the smooth looking form (and gives the partial sums for $\small x'=\exp(-100)-1 $):
$\qquad \small \begin{array} {r|r} \text{powerseries} & \text{partial sums for x' } \\ \hline \\ 1.00000000000 & 1.00000000000 \\ +1.00000000000x & 3.72007597602E-44 \\ +0.207106781187x^{2} & 0.207106781187 \\ +0.0344748426106x^{3} & 0.172631938576 \\ -0.0100670743762x^{4} & 0.162564864200 \\ +0.00821765977664x^{5} & 0.154347204423 \\ -0.00654357122833x^{6} & 0.147803633195 \\ +0.00537330847179x^{7} & 0.142430324723 \\ -0.00451702185603x^{8} & 0.137913302867 \\ +0.00386915976824x^{9} & 0.134044143099 \\ -0.00336528035075x^{10} & 0.130678862748 \\ +0.00296428202807x^{11} & 0.127714580720 \\ -0.00263893325448x^{12} & 0.125075647465 \\ +0.00237058888853x^{13} & 0.122705058577 \\ -0.00214611388717x^{14} & 0.120558944690 \\ +0.00195602261228x^{15} & 0.118602922077 \\ -0.00179331457091x^{16} & 0.116809607506 \\ +0.00165272361723x^{17} & 0.115156883889 \\ -0.00153022060566x^{18} & 0.113626663284 \\ +0.00142267593977x^{19} & 0.112203987344 \\ -0.00132762563657x^{20} & 0.110876361707 \\ +0.00124310598493x^{21} & 0.109633255722 \\ -0.00116753462507x^{22} & 0.108465721097 \\ +0.00109962364925x^{23} & 0.107366097448 \\ \end{array} $

The the question is, for some large negative x, say $\small x=-100 \qquad x'=exp(-100)-1 = -1+ \epsilon $ the series $\ g(x') $ converges to zero. Unfortunately - although we've translated the original problem to one with nice small numbers I don't see, how to really come nearer a solution, because the convergence of $\small g(-1+\epsilon) $ is really slow - if it converges at all to a positive value... So this is not yet a solution, but perhaps a suggestion for a path to try...

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