Difference between revisions of "Bounding the derivative of H t - second approach"

From Polymath1Wiki
Jump to: navigation, search
(Created page with "We have :<math>H_t(z) = \frac{1}{2} \int_{\bf R} e^{tu^2} \Phi(u) e^{izu}\ du</math> and hence on differentiation under the integral sign :<math>H'_t(z) = \frac{i}{2} \int_...")
(No difference)

Revision as of 12:42, 12 March 2018

We have

[math]H_t(z) = \frac{1}{2} \int_{\bf R} e^{tu^2} \Phi(u) e^{izu}\ du[/math]

and hence on differentiation under the integral sign

[math]H'_t(z) = \frac{i}{2} \int_{\bf R} e^{tu^2} u \Phi(u) e^{izu}\ du.[/math]

For any [math]0 \leq \theta \lt \pi/8[/math], we may shift the contour to [math]i\theta + {\bf R}[/math] and then use the even nature of [math]\Phi[/math] to reflect the left half of that contour around the origin to obtain the identities

[math]H_t(z) = \frac{1}{2} \int_{i\theta}^{i\theta+\infty} e^{tu^2} \Phi(u) e^{izu}\ du + \frac{1}{2} \overline{\int_{i\theta}^{i\theta+\infty} e^{tu^2} \Phi(u) e^{i\overline{z}u}\ du}[/math]

and similarly

[math]H'_t(z) = \frac{i}{2} \int_{i\theta}^{i\theta+\infty} e^{tu^2} \Phi(u) u e^{izu}\ du - \frac{i}{2} \overline{\int_{i\theta}^{i\theta+\infty} e^{tu^2} \Phi(u) u e^{i\overline{z}u}\ du}.[/math]

Since

[math]\Phi(u) = \sum_{n=1}^\infty (2\pi^2 n^4 e^{9u} - 3 \pi n^2 e^{5u}) \exp(-\pi n^2 e^{4u})[/math]

we thus have

[math]H_t(z) = \frac{1}{2} \sum_{n=1}^\infty 2\pi^2 n^4 I_{t,\theta}(z-9i, \pi n^2) - 3\pi n^2 I_{t,\theta}(z-5i, \pi n^2) + 2\pi^2 n^4 \overline{I_{t,\theta}(\overline{z}-9i, \pi n^2)} - 3\pi n^2 \overline{I_{t,\theta}(\overline{z}-5i, \pi n^2)} \quad (0)[/math]

and

[math]H'_t(z) = \frac{1}{2} \sum_{n=1}^\infty 2\pi^2 n^4 J_{t,\theta}(z-9i, \pi n^2) - 3\pi n^2 J_{t,\theta}(z-5i, \pi n^2) - 2\pi^2 n^4 \overline{J_{t,\theta}(\overline{z}-9i, \pi n^2)} + 3\pi n^2 \overline{J_{t,\theta}(\overline{z}-5i, \pi n^2)} \quad (1)[/math]

where

[math]I_{t,\theta}(b,\beta) := \int_{i\theta}^{i\theta+\infty} \exp( tu^2 - \beta e^{4u} + ibu)\ du[/math]

and

[math]J_{t,\theta}(b,\beta) := \int_{i\theta}^{i\theta+\infty} \exp( tu^2 - \beta e^{4u} + ibu) u\ du.[/math]

It is thus of interest to find good estimates for [math]I_{t,\theta}(b,\beta), J_{t,\theta}(b,\beta)[/math].

If [math]z = x+yi[/math] with [math]y\gt0[/math], we expect the dominant term here to be [math]2\pi^2 n^4 \overline{J_{t,\theta}(\overline{z}-9i, \pi n^2)}[/math]. For [math]b = \overline{z}-9i = x - (y+9)i[/math] and [math]\beta = \pi n^2[/math], the phase [math]-\beta e^{4u} + ibu[/math] has a stationary point at

[math]u = \frac{1}{4} \log \frac{ix+y+9}{4\pi n^2} = \frac{1}{4} \log \frac{|ix+y+9|}{4\pi n^2} + i (\frac{\pi}{8} - \arctan \frac{y+9}{x}).[/math]

It is thus natural to select [math]\theta[/math] to equal

[math] \theta := \frac{\pi}{8} - \arctan \frac{y+9}{x} \quad (1)[/math]

as this should remove most of the oscillation. But one can select other values too. (For the purposes of numerically bounding [math]H'_t(z)[/math] for [math]x[/math] in some interval, one should probably pick a single [math]\theta[/math] for this interval that is close to, but not necessarily equal to, the value given in (1). Note though that one has to use the same value of [math]\theta[/math] to evaluate all the terms in the above sum (e.g. one needs the same [math]\theta[/math] to handle [math]b = z-9i, \overline{z}-9i, z-5i, \overline{z}-5i[/math]).

The integral [math]I_{t,\theta}(b,\beta)[/math] is improper and so cannot be directly evaluated numerically. However, we can always cut it off as

[math]I_{t,\theta}(b,\beta) = I_{t,\theta,\leq X}(b,\beta) + I_{t,\theta, \geq X}(b,\beta)[/math]

for any cutoff [math]X \geq 0[/math], where

[math]I_{t,\theta,\geq X}(b,\beta) := \int_{i\theta}^{i\theta+X} \exp( tu^2 - \beta e^{4u} + ibu}\ du[/math]

and

math>I_{t,\theta,\geq X}(b,\beta) := \int_{i\theta+X}^{i\theta+\infty} \exp( tu^2 - \beta e^{4u} + ibu}\ du.</math>

Writing [math]u = i\theta + X + v[/math] and setting [math]b = x - i a[/math] (where in practice [math]a = 9+y, 9-y, 7+y, 7-y[/math]), the tail can be written as

[math]I_{t,\theta,\geq X}(b,\beta) = \int_0^\infty \exp( t (-\theta^2 + X^2 + v^2 + 2i\theta X + 2i\theta v) - \beta e^{4X} e^{4v} \cos \theta - i \beta e^{4X} e^{4v} \sin \theta - \theta x + iX x + iv x+ aX + av + ia\theta)\ dv[/math]

and hence by the triangle inequality

[math]|I_{t,\theta,\geq X}(b,\beta)| \leq \exp( -t \theta^2 + tX^2 - \beta e^{4X} \cos \theta - \theta x + aX ) \int_0^\infty \exp( tv^2 - \beta e^{4X} (e^{4v}-1) \cos \theta + av )\ dv.[/math]

The exponent [math]tv^2 - \beta e^{4X} (e^{4v}-1) \cos \theta + av[/math] equals [math]0[/math] when [math]v=0[/math] and has derivative

[math]2tv - 4 \beta e^{4X} \cos \theta e^{4v} + a.[/math]

Note that [math]e^{4v} \geq 1+4v[/math]. Thus if [math]X[/math] is so large that

[math] \beta e^{4X} \cos \theta \gt \max( \frac{t}{2}, \frac{a}{4}) [/math]

or equivalently

[math]X \gt \frac{1}{4} \log \frac{\max(\frac{t}{2}, \frac{a}{4})}{\beta \cos \theta} \quad (2)[/math]

then this derivative will be bounded from below by the negative quantity [math]-(4 \beta e^{4X} \cos \theta - a)[/math]. This implies that

[math]\exp( tv^2 - \beta e^{4X} (e^{4v}-1) \cos \theta + av ) \leq \exp( - v (4 \beta e^{4X} \cos \theta - a) )[/math]

thus giving the tail bound

[math]|I_{t,\theta,\geq X}(b,\beta)| \leq \frac{\exp( -t \theta^2 + tX^2 - \beta e^{4X} \cos \theta - \theta x + aX )}{4 \beta e^{4X} \cos \theta - a}\quad (3) [/math]

whenever (2) holds. This will start decaying rapidly when [math]X[/math] is large enough (basically one needs [math]e^{4X} \cos \theta[/math] to become large).

For large values of [math]n[/math] (and hence of [math]\beta = \pi n^2[/math]), one can set [math]X=0[/math] and estimate the entire integral. Indeed, we conclude that if

[math]\pi n^2 \cos \theta \gt \max(\frac{t}{2}, \frac{a}{4} )[/math]

then

[math]|I_{t,\theta}(b,\pi n^2)| \leq \frac{\exp( -t \theta^2 - \pi n^2 \cos \theta - \theta x )}{4 \pi n^2 \cos \theta - a}.[/math]

For instance, if

[math]\pi n_0^2 \cos \theta \gt \max(\frac{t}{2}, \frac{a}{4} )[/math]

then

[math] |\sum_{n \geq n_0} 2\pi^2 n^4 I_{t,\theta}(x-ia, \pi n^2)| \leq \sum_{n \geq n_0} 2\pi^2 n^4 \frac{\exp( -t \theta^2 - \pi n^2 \cos \theta - \theta x )}{4 \pi n^2 \cos \theta - a} [/math]
[math] \leq \frac{2\pi^2 \exp(-t \theta^2 - \theta x)}{4\pi n_0^2 \cos \theta - a} \sum_{n \geq n_0} n^4 \exp( - \pi n n_0 \cos \theta) [/math]
[math] = \frac{2\pi^2 \exp(-t \theta^2 - \theta x - \pi n_0^2 \cos \theta)}{4\pi n_0^2 \cos \theta - a} \sum_{m=0}^\infty (n_0^4 + 4 n_0^3 m + 6 n_0^2 m^2 + 4 n_0 m^3 + m^4) \alpha^m [/math]
[math]= \frac{2\pi^2 \exp(-t \theta^2 - \theta x - \pi n_0^2 \cos \theta)}{4\pi n_0^2 \cos \theta - a} ( \frac{n_0^4}{1 - \alpha} + \frac{4 n_0^3 \alpha}{(1 - \alpha)^2} + \frac{6 n_0^2 (\alpha^2+\alpha)}{(1-\alpha)^3} + \frac{4n_0 (\alpha^3+4\alpha^2+\alpha)}{(1-\alpha)^4} + \frac{\alpha^4 + 11 \alpha^3 + 11 \alpha^2 + \alpha}{(1-\alpha)^5 ) \quad (4)[/math]

where [math]\alpha := e^{-\pi n_0 \cos \theta}[/math] (one can do better here by expressing [math]\frac{n^4}{4\pi n^2 \cos \theta - a}[/math] as an improper fraction). A similar argument gives

[math] |\sum_{n \geq n_0} 3\pi n^2 I_{t,\theta}(x-ia, \pi n^2)| \leq \frac{3\pi \exp(-t \theta^2 - \theta x - \pi n_0^2 \cos \theta)}{4\pi n_0^2 \cos \theta - a} ( \frac{n_0^2}{1-\alpha} + \frac{2n_0 \alpha}{(1-\alpha)^2} + \frac{\alpha^2+\alpha}{(1-\alpha)^3} ).[/math]

This lets us control the tail of the series in (0).

Similar arguments let us handle the derivative. Indeed by arguing as before we have

[math]|J_{t,\theta,\geq X}(b,\beta)| \leq \exp( -t \theta^2 + tX^2 - \beta e^{4X} \cos \theta - \theta x + aX ) \int_0^\infty \exp( tv^2 - \beta e^{4X} (e^{4v}-1) \cos \theta + av ) (|X+i\theta| + v)\ dv.[/math]

If (2) holds then again we have

[math]\exp( tv^2 - \beta e^{4X} (e^{4v}-1) \cos \theta + av ) \leq \exp( - v (4 \beta e^{4X} \cos \theta - a) )[/math]

and hence

[math]|J_{t,\theta,\geq X}(b,\beta)| \leq \exp( -t \theta^2 + tX^2 - \beta e^{4X} \cos \theta - \theta x + aX ) ({|X+i\theta|}{4 \beta e^{4X} \cos \theta - a} + \frac{1}{4 \beta e^{4X} \cos \theta - a)^2}) \quad (5) [/math]

which lets us handle the tail for any [math]J_{t,\theta}(b,\beta)[/math]. In particular, if [math]n \geq n_0[/math] and

[math]\pi n_0^2 \cos \theta \gt \max(\frac{t}{2}, \frac{a}{4} )[/math]

then

[math]|J_{t,\theta}(b,\pi n^2)| \leq \exp( -t \theta^2 - \pi n^2 \cos \theta - \theta x ) ( \frac{\theta}{4 \pi n^2 \cos \theta - a} + \frac{1}{(4 \pi n^2 \cos \theta - a)^2 )[/math]

and by arguing as before

[math] |\sum_{n \geq n_0} 2\pi^2 n^4 J_{t,\theta}(x-ia, \pi n^2)| \leq 2\pi^2 \exp( -t \theta^2 - \pi n^2 \cos \theta - \theta x ) ( \frac{\theta}{4 \pi n_0^2 \cos \theta - a} + \frac{1}{(4 \pi n_0^2 \cos \theta - a)^2 )[/math]
[math] \times ( \frac{n_0^4}{1 - \alpha} + \frac{4 n_0^3 \alpha}{(1 - \alpha)^2} + \frac{6 n_0^2 (\alpha^2+\alpha)}{(1-\alpha)^3} + \frac{4n_0 (\alpha^3+4\alpha^2+\alpha)}{(1-\alpha)^4} + \frac{\alpha^4 + 11 \alpha^3 + 11 \alpha^2 + \alpha}{(1-\alpha)^5 ) [/math]

and similarly

[math] |\sum_{n \geq n_0} 3\pi n^2 J_{t,\theta}(x-ia, \pi n^2)| \leq 3\pi \exp( -t \theta^2 - \pi n^2 \cos \theta - \theta x ) ( \frac{\theta}{4 \pi n_0^2 \cos \theta - a} + \frac{1}{(4 \pi n_0^2 \cos \theta - a)^2 )[/math]
[math] \times ( \frac{n_0^2}{1-\alpha} + \frac{2n_0 \alpha}{(1-\alpha)^2} + \frac{\alpha^2+\alpha}{(1-\alpha)^3} ).[/math]