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The Jacobi theta function satisfies $θ (t) = t^{- 1 / 2} θ (1 / t)$

Statement

Assume countable choice. For every $t > 0$ ,

$θ (t) = t^{- 1 / 2} θ (1 / t) .$

Facts & Assumptions

Given: Countable choice and a real number $t > 0$ .

[L1]

The Jacobi theta function is $θ (t) = \sum_{n \in Z} e^{- π n^{2} t}$ (The Jacobi theta function $θ (t) = \sum_{n \in Z} e^{- π n^{2} t}$ for $t > 0$ ).

[L2]

Assuming countable choice, for $g_{t} (x) = e^{- π t x^{2}}$ the Fourier transform with $e^{- 2 π i x ξ}$ normalization is ${\hat{g}}_{t} (ξ) = t^{- 1 / 2} e^{- π ξ^{2} / t}$ (Euclidean Gaussian transform with the 2π normalization).

[L3]

Assuming countable choice, Poisson summation gives $\sum_{n \in Z} f (n) = \sum_{m \in Z} \hat{f} (m)$ for every Schwartz function $f$ on $R$ (Poisson summation for Schwartz functions).

Proof

technique · direct

1.1givenL3algebra

Repeated differentiation of $g_{t} (x) = e^{- π t x^{2}}$ gives a polynomial times the same Gaussian. For every polynomial $P$ , $P (x) e^{- π t x^{2}} \to 0$ faster than any reciprocal power as $∣ x ∣ \to \infty$ , since $t > 0$ . Thus $g_{t}$ is Schwartz and [L3] applies.

2.1L1L2L3step 1.1algebra∎

By [L1], $θ (t) = \sum_{n \in Z} g_{t} (n)$ . Poisson summation from [L3], step 1.1 and the exact Gaussian transform [L2] therefore give $θ (t) = \sum_{m \in Z} {\hat{g}}_{t} (m) = t^{- 1 / 2} \sum_{m \in Z} e^{- π m^{2} / t} = t^{- 1 / 2} θ (1 / t) .$

Depends on

Used by

Dependency tree · two levels

32 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Elias M. Stein and Rami Shakarchi, Complex Analysis, Ch. 6 §2.1 (standard reference, not scraped)
K. Chandrasekharan, Lectures on the Riemann Zeta-Function, Lecture 12, The Theta Relation (standard reference, not scraped)

The Jacobi theta function satisfies θ(t)=t−1/2θ(1/t)θ(t)=t−1/2θ(1/t)