Series for $Ψ$

By Brad Rodgers on Wednesday, June 20, 2001 - 02:26 am:

What is the series expansion for

Ψ (x) [Ψ (x) = d / dx (\ln (Γ (x)))]

? Please explain how you obtain it.

$Γ (x) = \int_{0}^{\infty} t^{x - 1} e^{- t} dt$
Thanks,

Brad

[Editor: See Gamma and Beta functions for a discussion of the above function.]

By David Loeffler on Thursday, June 21, 2001 - 10:31 pm:

Ψ (x) = - 1 / x - γ + π^{2} x / 6 - ζ (3) x^{2} + ζ (4) x^{3} . . .

according to MAPLE.

(MAPLE also offers $Ψ (x + 1) = Ψ (x) + 1 / x$ , which can quite easily be proved from the recurrence for the gamma function. So for integer $n$ , $Ψ (n)$ is the sum of $n$ terms of the harmonic series minus $γ$ .

By Michael Doré on Friday, June 22, 2001 - 10:21 pm:

Interesting. Well we know that:

$\cot x = lim_{N \to \infty} \sum_{n = - N}^{N} 1 / (x + n π)$

(this is provable by taking logs with Euler's infinite sine product and then differentiating).

If you break up each term in the series using a geometric progression you arrive at:

$\cot x = 1 / x - (ζ (2) / π^{2}) x - (ζ (4) / π^{4}) x^{3} - . . .$

Now using David's formula we get:

$x (Ψ (- x) - Ψ (x)) / 2 = 1 - ζ (2) x^{2} - ζ (4) x^{4} - . . .$

Also:

$x \cot x = 1 - (ζ (2) / π^{2}) x^{2} - (ζ (4) / π^{4}) x^{4} - . . .$

So it looks to me like:

$π \cot (π x) = (Ψ (- x) - Ψ (x)) / 2$

Is this true?

By David Loeffler on Friday, June 22, 2001 - 11:10 pm:

Are you sure about that expansion of the cot function? There seems to be some problem with the constants (did you forget the negative n terms in the product?) The result is actually

Ψ (- x) - Ψ (x) = 1 / x + π \cot (π x)

(again with some help from MAPLE and its ever-useful series() function.)

Can anyone see how we would prove that $Ψ (1) = - γ$ ? If we could show that $lim_{n \to \infty} Ψ (n) - \ln (n) = 0$ it would follow (and this avoids using any complicated properties of gamma). Any ideas?

By David Loeffler on Friday, June 22, 2001 - 11:32 pm:

Err - sorry to answer my own question, but I have found an argument that sort of suggests why it might be true but an analyst would probably pick holes in it very rapidly.

We have

lim_{x \to \infty} Ψ (x) - \ln (x)

$= lim_{x \to \infty} d / dx (\ln Γ (x)) - \ln x$

$= lim_{x \to \infty} d / dx (\ln Γ (x) - x \ln x + x)$

$= lim_{x \to \infty} d / dx (\ln (Γ (x) e^{x} / x^{x}))$

By Stirling's formula $Γ (x) e^{x} / x^{x} = 1 / x \times (x! e^{x} / x^{x})$ is asymptotically $\sqrt{2 π / x}$ , so this is

$- 1 / 2 lim_{x \to \infty} d / dx (\ln x)$

$= - 1 / 2 lim_{x \to \infty} 1 / x$

$= 0$

so $lim_{x \to \infty} Ψ (x) - \ln (x) = 0$ as required.
Any ideas how we make this properly rigorous?

David

By Brad Rodgers on Saturday, June 23, 2001 - 05:14 am:

I doubt that this shows any promise, but its at least a bit interesting:

From

Ψ = γ' / γ

;

$Ψ (1) = \int_{0}^{\infty} e^{- t} \ln (t) dt$

which by integration by parts,

$= lim_{t \to 0} [e^{- t} \ln (t)] + \int_{0}^{\infty} e^{- t} / t dt$

As $e^{- t} / t = 1 / t - 1 + t / 2! - t^{2} / 3! + . . .$

And as

$\int_{0}^{\infty} (- 1)^{a + 1} t^{a} / (a + 1)! = lim_{t \to \infty} t^{a + 1} (- 1)^{a + 1} / [(a + 1)! (a + 1)]$

#(sparing the case of $1 / t$ )#

We know that

$Ψ (1) = lim_{t \to 0} [e^{- t} \ln (t)] + \int_{0}^{\infty} e^{- t} / t dt$

$= lim_{t \to \infty} (\ln (1 / t) + lim_{t \to \infty} (\ln (t)) + \lim_{t \to \infty} (\sum_{a = 0}^{\infty} t^{a + 1} (- 1)^{a + 1} / [(a + 1)! (a + 1)])$

$= lim_{t \to \infty} \sum_{a = 0}^{\infty} t^{a + 1} (- 1)^{a + 1} / [(a + 1)! (a + 1)]$

Which, (aside from being a peculiar result) if someone can relate to $γ$ , we can have a reasonably rigorous proof.

David, what's Stirling's Theorem?

Thanks,

Brad

By Michael Doré on Saturday, June 23, 2001 - 12:34 pm:

Stirling's theorem is:

$n!$ is asymptotic to $n^{n} e^{- n} \sqrt{2 π n}$

(We say $f (n)$ is asymptotic to $g (n)$ if and only if $f (n) / g (n) \to 1$ as $n \to \infty$ )

Proof of Stirling's theorem is as follows.

For $x > 0$ we have:

$1 - x^{4} < 1 < 1 + x^{3}$

Factorise:

$(1 + x) (1 - x + x^{2} - x^{3}) < 1 < (1 + x) (1 - x + x^{2})$

Divide through by $1 + x$ which is positive:

$1 - x + x^{2} - x^{3} < 1 / (1 + x) < 1 - x + x^{2}$

Integrate this from 0 to $y$ to get:

$y - y^{2} / 2 + y^{3} / 3 - y^{4} / 4 < \ln (1 + y) < y - y^{2} / 2 + y^{3} / 3$ (*)

(Can you see why this step is valid? It may help to draw a diagram.)

Now let $a_{n} = \ln (n! e^{n} / n^{n + 1 / 2})$ .

Using (*) we get:

$1 / (12 n^{2}) - 1 / (12 n^{3}) < a_{n} - a_{n + 1} < 1 / (12 n^{2}) + 1 / (6 n^{3})$

It follows that for $n \geq 2$ we have:

$0 < a_{n} - a_{n + 1} < 1 / n^{2} < 1 / n (n - 1) = 1 / (n - 1) - 1 / n$

Therefore $a_{n}$ is decreasing and if you add the inequality to itself for successive values of $n$ you obtain:

$a_{2} - a_{n + 1} < 1 - 1 / (n + 1) < 1$

so $a_{n}$ is bounded below. But it is an axiom of real analysis that any sequence which is bounded below and decreasing is convergent. Hence $a_{n} \to L$ for some $L$ . We then have:

$n! e^{n} / (n^{n + 1 / 2}) \to e^{L} = M$

so $n!$ is asymptotic to $n^{n} e^{- n} M$ (**)

for some $M$ . It suffices to show that $M = \sqrt{2 π}$ . To do this let $I_{n} = \int_{0}^{π / 2} \sin^{r} t dt$ . Integration by parts gives $I_{r} = (r - 1) / r I_{r - 2}$ . So we have: $I_{2 n} = [(2 n)! / (2^{n} n!)^{2}] \times π / 2$ and $I_{2 n + 1} = (2^{n} n!)^{2} / (2 n + 1)!$ Since $I_{n}$ is decreasing we get:

$1 < I_{2 n} / I_{2 n + 1} < I_{2 n - 1} / I_{2 n + 1} < 1 + 1 / 2 n \to 1$

Hence $I_{2 n} / I_{2 n + 1} \to 1$ as $n \to \infty$ . Plug in (**) and you get $M = \sqrt{2 π}$ as required.

By David Loeffler on Saturday, June 23, 2001 - 11:24 pm:

Brad,

You have lost a

\ln (t)

somewhere, since your series seems to converge to

- \ln (t) - γ

. In fact it converges alarmingly fast, with error about

10^{- 6}

for

t = 10

(As for the presentation of the proof I think you have to let the upper limit of the integrals be $x$ , then let $x \to \infty$ at the end, otherwise you are adding a series of terms all of which are actually infinity. But that's a minor quibble.)

However, to show rigorously that your series is $- γ - \ln (t)$ looks like it will be very difficult as so little is known about $γ$ other than its definition. That is why I was forced to prove that $Ψ (1) = - γ$ above without actually mentioning $γ$ in the main body of the proof.

By David Loeffler on Saturday, June 23, 2001 - 11:54 pm:

Sorry, please ignore my comment about the presentation of the proof as that's what you've already done.

By Brad Rodgers on Sunday, June 24, 2001 - 01:56 am:

Are you sure I've forgot a ln somewhere? When you try the sum for t=10, it works very well with the ln in there, but for t=100, no such luck.(though it doesn't seem to work any better without the ln) I wouldn't expect the series to converge all that rapidly. I'll double check my work though.

Just wondering, what's unrigorous about your proof. Now that I understand Stirlings theorem, it seems perfectly fine to me (I'm certainly no analyst, though).

Thanks,

Brad

By Brad Rodgers on Sunday, June 24, 2001 - 03:31 am:

Sorry for the earlier post; I did forgot to put in a ln: where one evaluates integral from infinity to 0 of 1/t, two ln(infinity)'s are produced. So the result is

constant

Brad

By Brad Rodgers on Sunday, June 24, 2001 - 03:34 am:

It's interesting that the calculations are so close for t=10, but for t=100, they end up being so far off. It must be because we have to wait for the a!a to be larger than the t^a, which ends up giving too large a number for the t=100 for my computer to use.