Polymath15 test problem - Revision history

Teorth: /* Large negative values of t */

2019-01-02T06:26:33Z

Large negative values of t

← Older revision		Revision as of 23:26, 1 January 2019
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	== Large negative values of <math>t</math> ==		== Large negative values of <math>t</math> ==

			See also [[Second attempt at computing H_t(x) for negative t]].

	We heuristically compute <math>H_t(x)</math> in the regime where <math>x</math> is large and <math>t</math> is large and negative with <math>\|t\|/x \asymp 1</math>. We shall only be interested in the zeroes and so we discard any multiplicative factor which is non-zero: we write <math>X \sim Y</math> if X is equal (or approximately equal) to Y times something that is explicit and non-zero.		We heuristically compute <math>H_t(x)</math> in the regime where <math>x</math> is large and <math>t</math> is large and negative with <math>\|t\|/x \asymp 1</math>. We shall only be interested in the zeroes and so we discard any multiplicative factor which is non-zero: we write <math>X \sim Y</math> if X is equal (or approximately equal) to Y times something that is explicit and non-zero.

Teorth: /* Large negative values of t */

2019-01-01T05:39:37Z

Large negative values of t

← Older revision		Revision as of 22:39, 31 December 2018
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	we thus have		we thus have
	:<math> H_t(x) \sim \sum_n \exp( - \frac{2 \pi u (n-N)^2}{8 \pi - iu} - \pi i n^2 + 2 \pi i (n-N)^2 )\quad (3.23)</math>		:<math> H_t(x) \sim \sum_n \exp( - \frac{2 \pi u (n-N)^2}{8 \pi - iu} - \pi i n^2 + 2 \pi i (n-N)^2 )\quad (3.23)</math>
	:<math> \sim \sum_n \exp( \frac{16 \pi^2 i (n-N)^2}{8 \pi - iu} ) e( \pi i n )\quad (3.24)</math>		:<math> \sim \sum_n \exp( \frac{16 \pi^2 i (n-N)^2}{8 \pi - iu} ) e^{\pi i n}\quad (3.24)</math>
	:<math> \sim \theta_{01}( \frac{16 \pi N}{8\pi - iu}, \frac{16 \pi}{8\pi - iu} )\quad (3.25)</math>		:<math> \sim \theta_{01}( \frac{16 \pi N}{8\pi - iu}, \frac{16 \pi}{8\pi - iu} )\quad (3.25)</math>
	where <math>\theta_{01}</math> is the theta function defined in [https://en.wikipedia.org/wiki/Theta_function#Auxiliary_functions this Wikipedia page]. Using the Jacobi identity we then have		where <math>\theta_{01}</math> is the theta function defined in [https://en.wikipedia.org/wiki/Theta_function#Auxiliary_functions this Wikipedia page]. Using the Jacobi identity we then have

Teorth: /* Large negative values of t */

2018-12-27T05:03:01Z

Large negative values of t

Teorth: /* Large negative values of t */

2018-12-25T00:11:14Z

Large negative values of t

← Older revision		Revision as of 17:11, 24 December 2018
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	:<math> \sim \theta_{01}( \frac{16 \pi N}{8\pi - iu}, \frac{16 \pi}{8\pi - iu} )</math>		:<math> \sim \theta_{01}( \frac{16 \pi N}{8\pi - iu}, \frac{16 \pi}{8\pi - iu} )</math>
	where <math>\theta_{01}</math> is the theta function defined in [https://en.wikipedia.org/wiki/Theta_function#Auxiliary_functions this Wikipedia page]. Using the Jacobi identity we then have		where <math>\theta_{01}</math> is the theta function defined in [https://en.wikipedia.org/wiki/Theta_function#Auxiliary_functions this Wikipedia page]. Using the Jacobi identity we then have
	:<math> H_t(x) \sim \theta_{01}(N, \frac{iu - 8\pi}{16 \pi} )</math>		:<math> H_t(x) \sim \theta_{10}(N, \frac{iu - 8\pi}{16 \pi} )</math>
	:<math> = \~~sum_n~~ (-1)^n \exp( - \pi i n^2 / 2 ) e^{2\pi i n N} e^{-u n^2/16}.</math>		:<math> \sim \theta( N + \frac{1}{2} \frac{iu - 8\pi}{16 \pi}, \frac{iu - 8\pi}{16 \pi})</math>
			:<math> \sim \sum_n \exp( - \pi i n(n+1) / 2 ) e^{2\pi i (n+1/2) N} e^{-u n(n+1)/16}.</math>
	As a sanity check, one can verify that the RHS is real-valued, just as <math>H_t(x)</math> is (by the functional equation).		As a sanity check, one can verify that the RHS is real-valued, just as <math>H_t(x)</math> is (by the functional equation).

	[[Category:Polymath15]]		[[Category:Polymath15]]

Teorth: /* Large negative values of t */

2018-12-24T23:33:54Z

Large negative values of t

← Older revision		Revision as of 16:33, 24 December 2018
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	:<math> \sim \theta_{01}( \frac{16 \pi N}{8\pi - iu}, \frac{16 \pi}{8\pi - iu} )</math>		:<math> \sim \theta_{01}( \frac{16 \pi N}{8\pi - iu}, \frac{16 \pi}{8\pi - iu} )</math>
	where <math>\theta_{01}</math> is the theta function defined in [https://en.wikipedia.org/wiki/Theta_function#Auxiliary_functions this Wikipedia page]. Using the Jacobi identity we then have		where <math>\theta_{01}</math> is the theta function defined in [https://en.wikipedia.org/wiki/Theta_function#Auxiliary_functions this Wikipedia page]. Using the Jacobi identity we then have
	:<math> H_t(x) \sim \theta_{01}(-N, \frac{iu - 8\pi}{16 \pi} )</math>		:<math> H_t(x) \sim \theta_{01}(N, \frac{iu - 8\pi}{16 \pi} )</math>
	:<math> = \sum_n (-1)^n \exp( - \pi i n^2 / 2 ) e^{-2\pi i n N} e^{-u n^2/16}.</math>		:<math> = \sum_n (-1)^n \exp( - \pi i n^2 / 2 ) e^{2\pi i n N} e^{-u n^2/16}.</math>
	As a sanity check, one can verify that the RHS is real-valued, just as <math>H_t(x)</math> is (by the functional equation).		As a sanity check, one can verify that the RHS is real-valued, just as <math>H_t(x)</math> is (by the functional equation).

	[[Category:Polymath15]]		[[Category:Polymath15]]

Teorth: /* Large negative values of t */

2018-12-24T23:29:36Z

Large negative values of t

← Older revision		Revision as of 16:29, 24 December 2018
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	:<math> H_t(x) \sim \theta_{01}(-N, \frac{iu - 8\pi}{16 \pi} )</math>		:<math> H_t(x) \sim \theta_{01}(-N, \frac{iu - 8\pi}{16 \pi} )</math>
	:<math> = \sum_n (-1)^n \exp( - \pi i n^2 / 2 ) e^{-2\pi i n N} e^{-u n^2/16}.</math>		:<math> = \sum_n (-1)^n \exp( - \pi i n^2 / 2 ) e^{-2\pi i n N} e^{-u n^2/16}.</math>
			As a sanity check, one can verify that the RHS is real-valued, just as <math>H_t(x)</math> is (by the functional equation).

	[[Category:Polymath15]]		[[Category:Polymath15]]

Teorth: /* Large negative values of t */

2018-12-24T23:28:10Z

Large negative values of t

← Older revision		Revision as of 16:28, 24 December 2018
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	:<math> H_t(x) \sim \int_{\bf R} \exp( \frac{i \|t\|^{1/2} v}{2} \log \frac{\tilde x}{4\pi} - \pi \|t\|^{1/2} v/4 + i \|t\| v^2 / 2\tilde x) \zeta(\frac{1+i\tilde x}{2} + i \|t\|^{1/2} v) e^{-v^2 + \pi \|t\|^{1/2} v / 4}\ dv.</math>		:<math> H_t(x) \sim \int_{\bf R} \exp( \frac{i \|t\|^{1/2} v}{2} \log \frac{\tilde x}{4\pi} - \pi \|t\|^{1/2} v/4 + i \|t\| v^2 / 2\tilde x) \zeta(\frac{1+i\tilde x}{2} + i \|t\|^{1/2} v) e^{-v^2 + \pi \|t\|^{1/2} v / 4}\ dv.</math>
	The two factors of <math>\exp( \pi \|t\|^{1/2} v/4 ) </math> cancel. If we now write		The two factors of <math>\exp( \pi \|t\|^{1/2} v/4 ) </math> cancel. If we now write
	:<math>N := \sqrt{\tilde x}{4\pi}</math>		:<math>N := \sqrt{\frac{\tilde x}{4\pi}}</math>
	and		and
	:<math>u := \|t\|/N^2 = 4\pi \|t\|/\tilde x,</math>		:<math>u := \|t\|/N^2 = 4\pi \|t\|/\tilde x,</math>

Teorth: /* Large negative values of t */

2018-12-24T23:23:37Z

Large negative values of t

← Older revision		Revision as of 16:23, 24 December 2018
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	where <math>\theta_{01}</math> is the theta function defined in [https://en.wikipedia.org/wiki/Theta_function#Auxiliary_functions this Wikipedia page]. Using the Jacobi identity we then have		where <math>\theta_{01}</math> is the theta function defined in [https://en.wikipedia.org/wiki/Theta_function#Auxiliary_functions this Wikipedia page]. Using the Jacobi identity we then have
	:<math> H_t(x) \sim \theta_{01}(-N, \frac{iu - 8\pi}{16 \pi} )</math>		:<math> H_t(x) \sim \theta_{01}(-N, \frac{iu - 8\pi}{16 \pi} )</math>
	:<math> = \sum_n (-1)^n \exp( - \pi i n^2 / 2 ) e^{2\pi i n N} e^{-u n^2/16}.</math>		:<math> = \sum_n (-1)^n \exp( - \pi i n^2 / 2 ) e^{-2\pi i n N} e^{-u n^2/16}.</math>

	[[Category:Polymath15]]		[[Category:Polymath15]]

Teorth: /* Small values of x */

2018-12-24T23:20:04Z

Small values of x

← Older revision		Revision as of 16:20, 24 December 2018
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	Here is a [https://pastebin.com/TiFk6CfF script] for verifying the absence of zeroes of <math>H_t</math> in a rectangle. It can eliminate zeros in the rectangle <math>\{0 \leq x \leq 1000, 0.4 \leq y \leq 0.45\}</math> when t = 0.4.		Here is a [https://pastebin.com/TiFk6CfF script] for verifying the absence of zeroes of <math>H_t</math> in a rectangle. It can eliminate zeros in the rectangle <math>\{0 \leq x \leq 1000, 0.4 \leq y \leq 0.45\}</math> when t = 0.4.


			== Large negative values of <math>t</math> ==

			We heuristically compute <math>H_t(x)</math> in the regime where <math>x</math> is large and <math>t</math> is large and negative with <math>\|t\|/x \asymp 1</math>. We shall only be interested in the zeroes and so we discard any multiplicative factor which is non-zero: we write <math>X \sim Y</math> if X is equal (or approximately equal) to Y times something that is explicit and non-zero.

			From equation (35) of the writeup we have

			:<math>H_t(x) = \int_{\bf R} \frac{1}{8} \xi(\frac{1+ix}{2} + i \|t\|^{1/2} v) \frac{1}{\pi} e^{-v^2}\ dv</math>
			:<math> \sim \int_{\bf R} \xi(\frac{1+ix}{2} + i \|t\|^{1/2} v) e^{-v^2}\ dv.</math>

			To cancel off an exponential decay factor in the <math>\xi</math> function, it is convenient to shift the v variable by <math>\pi \|t\|^{1/2}/8</math>, thus
			:<math> H_t(x) \sim \int_{\bf R} \xi(\frac{1+ix}{2} + i \|t\|^{1/2} v - \pi i \|t\|/8) e^{-(v - \pi \|t\|^{1/2}/8)^2}\ dv</math>
			:<math> \sim \int_{\bf R} \xi(\frac{1+i\tilde x}{2} + i \|t\|^{1/2} v) e^{-v^2 + \pi \|t\|^{1/2} v / 4}\ dv</math>
			where
			:<math> \tilde x := x - \pi \|t\|/4 = x + \frac{\pi t}{4}.</math>
			Now from the definition of <math>\xi</math> and the Stirling approximation we have
			:<math> \xi(s) \sim M_0(s) \zeta(s)</math>
			where <math>M_0</math> is defined in (6) of the writeup. Thus
			:<math> H_t(x) \sim \int_{\bf R} M_0(\frac{1+i\tilde x}{2} + i \|t\|^{1/2} v) \zeta(\frac{1+i\tilde x}{2} + i \|t\|^{1/2} v) e^{-v^2 + \pi \|t\|^{1/2} v / 4}\ dv.</math>
			By Taylor expansion we have
			:<math> M_0(\frac{1+i\tilde x}{2} + i \|t\|^{1/2} v) \sim M_0(\frac{1+i\tilde x}{2}) \exp( \alpha( \frac{1+i\tilde x}{2} ) i \|t\|^{1/2} v + \alpha'(\frac{1+i \tilde x}{2}) \frac{-\|t\| v^2}{2} )</math>
			:<math> \sim \exp( \alpha( \frac{1+i\tilde x}{2} ) i \|t\|^{1/2} v + \alpha'(\frac{1+i \tilde x}{2}) \frac{-\|t\| v^2}{2} )</math>
			where <math>\alpha</math> is defined in equation (8) of the writeup. We have the approximations
			:<math> \alpha(\frac{1+i\tilde x}{2} ) \approx \frac{1}{2} \log \frac{\tilde x}{4\pi} + \frac{i\pi}{4}</math>
			and
			:<math> \alpha'(\frac{1+i\tilde x}{2} ) \approx \frac{-i}{\tilde x}</math>
			and hence
			:<math> H_t(x) \sim \int_{\bf R} \exp( \frac{i \|t\|^{1/2} v}{2} \log \frac{\tilde x}{4\pi} - \pi \|t\|^{1/2} v/4 + i \|t\| v^2 / 2\tilde x) \zeta(\frac{1+i\tilde x}{2} + i \|t\|^{1/2} v) e^{-v^2 + \pi \|t\|^{1/2} v / 4}\ dv.</math>
			The two factors of <math>\exp( \pi \|t\|^{1/2} v/4 ) </math> cancel. If we now write
			:<math>N := \sqrt{\tilde x}{4\pi}</math>
			and
			:<math>u := \|t\|/N^2 = 4\pi \|t\|/\tilde x,</math>
			we conclude that
			:<math> H_t(x) \sim \int_{\bf R} \exp( i \|t\|^{1/2} v \log N + i u v^2 / 8 \pi) \zeta(\frac{1+i\tilde x}{2} + i \|t\|^{1/2} v) e^{-v^2}\ dv.</math>
			If we formally write <math>\zeta(s) = \sum_n \frac{1}{n^s}</math> (ignoring convergence issues) we obtain
			:<math> H_t(x) \sim \sum_n \int_{\bf R} \exp( i \|t\|^{1/2} v \log N + i u v^2 / 8 \pi) n^{-\frac{1+i\tilde x}{2} - i \|t\|^{1/2} v} e^{-v^2}\ dv</math>
			:<math> \sim \sum_n \int_{\bf R} \exp( - i \|t\|^{1/2} v \log \frac{n}{N} + i u v^2 / 8 \pi -\frac{1+i\tilde x}{2} \log \frac{n}{N} ) e^{-v^2}\ dv</math>
			We can compute the <math>v</math> integral to obtain
			:<math> H_t(x) \sim \sum_n \exp( - \frac{\|t\| \log^2 \frac{n}{N}}{4 (1 - iu / 8 \pi)} -\frac{1+i\tilde x}{2} \log \frac{n}{N}).</math>
			Using the Taylor approximation
			:<math> \log \frac{n}{N} \approx \frac{n-N}{N} - \frac{(n-N)^2}{2N^2} </math>
			and dropping some small terms, we obtain
			:<math> H_t(x) \sim \sum_n \exp( - \frac{\|t\| (n-N)^2}{4 N^2 (1 - iu/8\pi)} -\frac{i\tilde x}{2} \frac{n-N}{N} + \frac{i \tilde x (n-N)^2}{4N^2} ).</math>
			Writing <math>\tilde x = 4\pi N^2</math> and <math>\|t\| = u N^2</math>, this becomes
			:<math> H_t(x) \sim \sum_n \exp( - \frac{2\pi u (n-N)^2}{8\pi - iu} -2 \pi i N(n-N) + \pi i (n-N)^2 ).</math>
			Writing
			:<math> N(n-N) = \frac{1}{2} n^2 - \frac{1}{2} (n-N)^2 - \frac{1}{2} N^2 </math>
			we thus have
			:<math> H_t(x) \sim \sum_n \exp( - \frac{2 \pi u (n-N)^2}{8 \pi - iu} - \pi i n^2 + 2 \pi i (n-N)^2 )</math>
			:<math> \sim \sum_n \exp( \frac{16 \pi^2 i (n-N)^2}{8 \pi - iu} ) e( \pi i n )</math>
			:<math> \sim \theta_{01}( \frac{16 \pi N}{8\pi - iu}, \frac{16 \pi}{8\pi - iu} )</math>
			where <math>\theta_{01}</math> is the theta function defined in [https://en.wikipedia.org/wiki/Theta_function#Auxiliary_functions this Wikipedia page]. Using the Jacobi identity we then have
			:<math> H_t(x) \sim \theta_{01}(-N, \frac{iu - 8\pi}{16 \pi} )</math>
			:<math> = \sum_n (-1)^n \exp( - \pi i n^2 / 2 ) e^{2\pi i n N} e^{-u n^2/16}.</math>

	[[Category:Polymath15]]		[[Category:Polymath15]]

Teorth at 15:08, 30 March 2018

2018-03-30T15:08:59Z

← Older revision		Revision as of 08:08, 30 March 2018
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	Here is a [https://pastebin.com/TiFk6CfF script] for verifying the absence of zeroes of <math>H_t</math> in a rectangle. It can eliminate zeros in the rectangle <math>\{0 \leq x \leq 1000, 0.4 \leq y \leq 0.45\}</math> when t = 0.4.		Here is a [https://pastebin.com/TiFk6CfF script] for verifying the absence of zeroes of <math>H_t</math> in a rectangle. It can eliminate zeros in the rectangle <math>\{0 \leq x \leq 1000, 0.4 \leq y \leq 0.45\}</math> when t = 0.4.

			[[Category:Polymath15]]

← Older revision		Revision as of 22:03, 26 December 2018
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	From equation (35) of the writeup we have		From equation (35) of the writeup we have

	:<math>H_t(x) = \int_{\bf R} \frac{1}{8} \xi(\frac{1+ix}{2} + i \|t\|^{1/2} v) \frac{1}{\pi} e^{-v^2}\ dv</math>		:<math>H_t(x) = \int_{\bf R} \frac{1}{8} \xi(\frac{1+ix}{2} + i \|t\|^{1/2} v) \frac{1}{\pi} e^{-v^2}\ dv \quad (3.1)</math>
	:<math> \sim \int_{\bf R} \xi(\frac{1+ix}{2} + i \|t\|^{1/2} v) e^{-v^2}\ dv.</math>		:<math> \sim \int_{\bf R} \xi(\frac{1+ix}{2} + i \|t\|^{1/2} v) e^{-v^2}\ dv. \quad (3.2)</math>

	To cancel off an exponential decay factor in the <math>\xi</math> function, it is convenient to shift the v variable by <math>\pi \|t\|^{1/2}/8</math>, thus		To cancel off an exponential decay factor in the <math>\xi</math> function, it is convenient to shift the v variable by <math>\pi \|t\|^{1/2}/8</math>, thus
	:<math> H_t(x) \sim \int_{\bf R} \xi(\frac{1+ix}{2} + i \|t\|^{1/2} v - \pi i \|t\|/8) e^{-(v - \pi \|t\|^{1/2}/8)^2}\ dv</math>		:<math> H_t(x) \sim \int_{\bf R} \xi(\frac{1+ix}{2} + i \|t\|^{1/2} v - \pi i \|t\|/8) e^{-(v - \pi \|t\|^{1/2}/8)^2}\ dv \quad (3.3)</math>
	:<math> \sim \int_{\bf R} \xi(\frac{1+i\tilde x}{2} + i \|t\|^{1/2} v) e^{-v^2 + \pi \|t\|^{1/2} v / 4}\ dv</math>		:<math> \sim \int_{\bf R} \xi(\frac{1+i\tilde x}{2} + i \|t\|^{1/2} v) e^{-v^2 + \pi \|t\|^{1/2} v / 4}\ dv \quad (3.4)</math>
	where		where
	:<math> \tilde x := x - \pi \|t\|/4 = x + \frac{\pi t}{4}.</math>		:<math> \tilde x := x - \pi \|t\|/4 = x + \frac{\pi t}{4}. \quad (3.5)</math>
	Now from the definition of <math>\xi</math> and the Stirling approximation we have		Now from the definition of <math>\xi</math> and the Stirling approximation we have
	:<math> \xi(s) \sim M_0(s) \zeta(s)</math>		:<math> \xi(s) \sim M_0(s) \zeta(s)\quad (3.6)</math>
	where <math>M_0</math> is defined in (6) of the writeup. Thus		where <math>M_0</math> is defined in (6) of the writeup. Thus
	:<math> H_t(x) \sim \int_{\bf R} M_0(\frac{1+i\tilde x}{2} + i \|t\|^{1/2} v) \zeta(\frac{1+i\tilde x}{2} + i \|t\|^{1/2} v) e^{-v^2 + \pi \|t\|^{1/2} v / 4}\ dv.</math>		:<math> H_t(x) \sim \int_{\bf R} M_0(\frac{1+i\tilde x}{2} + i \|t\|^{1/2} v) \zeta(\frac{1+i\tilde x}{2} + i \|t\|^{1/2} v) e^{-v^2 + \pi \|t\|^{1/2} v / 4}\ dv.\quad (3.7)</math>
	By Taylor expansion we have		By Taylor expansion we have
	:<math> M_0(\frac{1+i\tilde x}{2} + i \|t\|^{1/2} v) \sim M_0(\frac{1+i\tilde x}{2}) \exp( \alpha( \frac{1+i\tilde x}{2} ) i \|t\|^{1/2} v + \alpha'(\frac{1+i \tilde x}{2}) \frac{-\|t\| v^2}{2} )</math>		:<math> M_0(\frac{1+i\tilde x}{2} + i \|t\|^{1/2} v) \sim M_0(\frac{1+i\tilde x}{2}) \exp( \alpha( \frac{1+i\tilde x}{2} ) i \|t\|^{1/2} v + \alpha'(\frac{1+i \tilde x}{2}) \frac{-\|t\| v^2}{2} )\quad (3.8)</math>
	:<math> \sim \exp( \alpha( \frac{1+i\tilde x}{2} ) i \|t\|^{1/2} v + \alpha'(\frac{1+i \tilde x}{2}) \frac{-\|t\| v^2}{2} )</math>		:<math> \sim \exp( \alpha( \frac{1+i\tilde x}{2} ) i \|t\|^{1/2} v + \alpha'(\frac{1+i \tilde x}{2}) \frac{-\|t\| v^2}{2} )\quad (3.9)</math>
	where <math>\alpha</math> is defined in equation (8) of the writeup. We have the approximations		where <math>\alpha</math> is defined in equation (8) of the writeup. We have the approximations
	:<math> \alpha(\frac{1+i\tilde x}{2} ) \approx \frac{1}{2} \log \frac{\tilde x}{4\pi} + \frac{i\pi}{4}</math>		:<math> \alpha(\frac{1+i\tilde x}{2} ) \approx \frac{1}{2} \log \frac{\tilde x}{4\pi} + \frac{i\pi}{4} \quad (3.10)</math>
	and		and
	:<math> \alpha'(\frac{1+i\tilde x}{2} ) \approx \frac{-i}{\tilde x}</math>		:<math> \alpha'(\frac{1+i\tilde x}{2} ) \approx \frac{-i}{\tilde x} \quad (3.11)</math>
	and hence		and hence
	:<math> H_t(x) \sim \int_{\bf R} \exp( \frac{i \|t\|^{1/2} v}{2} \log \frac{\tilde x}{4\pi} - \pi \|t\|^{1/2} v/4 + i \|t\| v^2 / 2\tilde x) \zeta(\frac{1+i\tilde x}{2} + i \|t\|^{1/2} v) e^{-v^2 + \pi \|t\|^{1/2} v / 4}\ dv.</math>		:<math> H_t(x) \sim \int_{\bf R} \exp( \frac{i \|t\|^{1/2} v}{2} \log \frac{\tilde x}{4\pi} - \pi \|t\|^{1/2} v/4 + i \|t\| v^2 / 2\tilde x) \zeta(\frac{1+i\tilde x}{2} + i \|t\|^{1/2} v) e^{-v^2 + \pi \|t\|^{1/2} v / 4}\ dv.\quad (3.12)</math>
	The two factors of <math>\exp( \pi \|t\|^{1/2} v/4 ) </math> cancel. If we now write		The two factors of <math>\exp( \pi \|t\|^{1/2} v/4 ) </math> cancel. If we now write
	:<math>N := \sqrt{\frac{\tilde x}{4\pi}}</math>		:<math>N := \sqrt{\frac{\tilde x}{4\pi}}\quad (3.13)</math>
	and		and
	:<math>u := \|t\|/N^2 = 4\pi \|t\|/\tilde x,</math>		:<math>u := \|t\|/N^2 = 4\pi \|t\|/\tilde x,\quad (3.14)</math>
	we conclude that		we conclude that
	:<math> H_t(x) \sim \int_{\bf R} \exp( i \|t\|^{1/2} v \log N + i u v^2 / 8 \pi) \zeta(\frac{1+i\tilde x}{2} + i \|t\|^{1/2} v) e^{-v^2}\ dv.</math>		:<math> H_t(x) \sim \int_{\bf R} \exp( i \|t\|^{1/2} v \log N + i u v^2 / 8 \pi) \zeta(\frac{1+i\tilde x}{2} + i \|t\|^{1/2} v) e^{-v^2}\ dv.\quad (3.15)</math>
	If we formally write <math>\zeta(s) = \sum_n \frac{1}{n^s}</math> (ignoring convergence issues) we obtain		If we formally write <math>\zeta(s) = \sum_n \frac{1}{n^s}</math> (ignoring convergence issues) we obtain
	:<math> H_t(x) \sim \sum_n \int_{\bf R} \exp( i \|t\|^{1/2} v \log N + i u v^2 / 8 \pi) n^{-\frac{1+i\tilde x}{2} - i \|t\|^{1/2} v} e^{-v^2}\ dv</math>		:<math> H_t(x) \sim \sum_n \int_{\bf R} \exp( i \|t\|^{1/2} v \log N + i u v^2 / 8 \pi) n^{-\frac{1+i\tilde x}{2} - i \|t\|^{1/2} v} e^{-v^2}\ dv\quad (3.16)</math>
	:<math> \sim \sum_n \int_{\bf R} \exp( - i \|t\|^{1/2} v \log \frac{n}{N} + i u v^2 / 8 \pi -\frac{1+i\tilde x}{2} \log \frac{n}{N} ) e^{-v^2}\ dv</math>		:<math> \sim \sum_n \int_{\bf R} \exp( - i \|t\|^{1/2} v \log \frac{n}{N} + i u v^2 / 8 \pi -\frac{1+i\tilde x}{2} \log \frac{n}{N} ) e^{-v^2}\ dv\quad (3.17)</math>
	We can compute the <math>v</math> integral to obtain		We can compute the <math>v</math> integral to obtain
	:<math> H_t(x) \sim \sum_n \exp( - \frac{\|t\| \log^2 \frac{n}{N}}{4 (1 - iu / 8 \pi)} -\frac{1+i\tilde x}{2} \log \frac{n}{N}).</math>		:<math> H_t(x) \sim \sum_n \exp( - \frac{\|t\| \log^2 \frac{n}{N}}{4 (1 - iu / 8 \pi)} -\frac{1+i\tilde x}{2} \log \frac{n}{N}).\quad (3.18)</math>
	Using the Taylor approximation		Using the Taylor approximation
	:<math> \log \frac{n}{N} \approx \frac{n-N}{N} - \frac{(n-N)^2}{2N^2} </math>		:<math> \log \frac{n}{N} \approx \frac{n-N}{N} - \frac{(n-N)^2}{2N^2} \quad (3.19)</math>
	and dropping some small terms, we obtain		and dropping some small terms, we obtain
	:<math> H_t(x) \sim \sum_n \exp( - \frac{\|t\| (n-N)^2}{4 N^2 (1 - iu/8\pi)} -\frac{i\tilde x}{2} \frac{n-N}{N} + \frac{i \tilde x (n-N)^2}{4N^2} ).</math>		:<math> H_t(x) \sim \sum_n \exp( - \frac{\|t\| (n-N)^2}{4 N^2 (1 - iu/8\pi)} -\frac{i\tilde x}{2} \frac{n-N}{N} + \frac{i \tilde x (n-N)^2}{4N^2} ).\quad (3.20)</math>
	Writing <math>\tilde x = 4\pi N^2</math> and <math>\|t\| = u N^2</math>, this becomes		Writing <math>\tilde x = 4\pi N^2</math> and <math>\|t\| = u N^2</math>, this becomes
	:<math> H_t(x) \sim \sum_n \exp( - \frac{2\pi u (n-N)^2}{8\pi - iu} -2 \pi i N(n-N) + \pi i (n-N)^2 ).</math>		:<math> H_t(x) \sim \sum_n \exp( - \frac{2\pi u (n-N)^2}{8\pi - iu} -2 \pi i N(n-N) + \pi i (n-N)^2 ).\quad (3.21)</math>
	Writing		Writing
	:<math> N(n-N) = \frac{1}{2} n^2 - \frac{1}{2} (n-N)^2 - \frac{1}{2} N^2 </math>		:<math> N(n-N) = \frac{1}{2} n^2 - \frac{1}{2} (n-N)^2 - \frac{1}{2} N^2 \quad (3.22)</math>
	we thus have		we thus have
	:<math> H_t(x) \sim \sum_n \exp( - \frac{2 \pi u (n-N)^2}{8 \pi - iu} - \pi i n^2 + 2 \pi i (n-N)^2 )</math>		:<math> H_t(x) \sim \sum_n \exp( - \frac{2 \pi u (n-N)^2}{8 \pi - iu} - \pi i n^2 + 2 \pi i (n-N)^2 )\quad (3.23)</math>
	:<math> \sim \sum_n \exp( \frac{16 \pi^2 i (n-N)^2}{8 \pi - iu} ) e( \pi i n )</math>		:<math> \sim \sum_n \exp( \frac{16 \pi^2 i (n-N)^2}{8 \pi - iu} ) e( \pi i n )\quad (3.24)</math>
	:<math> \sim \theta_{01}( \frac{16 \pi N}{8\pi - iu}, \frac{16 \pi}{8\pi - iu} )</math>		:<math> \sim \theta_{01}( \frac{16 \pi N}{8\pi - iu}, \frac{16 \pi}{8\pi - iu} )\quad (3.25)</math>
	where <math>\theta_{01}</math> is the theta function defined in [https://en.wikipedia.org/wiki/Theta_function#Auxiliary_functions this Wikipedia page]. Using the Jacobi identity we then have		where <math>\theta_{01}</math> is the theta function defined in [https://en.wikipedia.org/wiki/Theta_function#Auxiliary_functions this Wikipedia page]. Using the Jacobi identity we then have
	:<math> H_t(x) \sim \theta_{10}(N, \frac{iu - 8\pi}{16 \pi} )</math>		:<math> H_t(x) \sim \theta_{10}(N, \frac{iu - 8\pi}{16 \pi} )\quad (3.26)</math>
	:<math> \sim \theta( N + \frac{1}{2} \frac{iu - 8\pi}{16 \pi}, \frac{iu - 8\pi}{16 \pi})</math>		:<math> \sim \theta( N + \frac{1}{2} \frac{iu - 8\pi}{16 \pi}, \frac{iu - 8\pi}{16 \pi})\quad (3.27)</math>
	:<math> \sim \sum_n \exp( - \pi i n(n+1) / 2 ) e^{2\pi i (n+1/2) N} e^{-u n(n+1)/16}.</math>		:<math> \sim \sum_n \exp( - \pi i n(n+1) / 2 ) e^{2\pi i (n+1/2) N} e^{-u n(n+1)/16}.\quad (3.28)</math>
	As a sanity check, one can verify that the RHS is real-valued, just as <math>H_t(x)</math> is (by the functional equation).		As a sanity check, one can verify that the RHS is real-valued, just as <math>H_t(x)</math> is (by the functional equation).

	[[Category:Polymath15]]		[[Category:Polymath15]]