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I want to prove that:

$$\wp(2z) = \frac14 \left(\frac{\wp''(z)}{\wp'(z)}\right)^2 - 2\wp(z)$$

and I am asked to do it using the poles of both sides. I think that $\wp(2z)$ has exactly 4 poles, at the half lattice points. and the right side kind of looks like it has 4 poles in the same places too, because those are the zeroes of $\wp'$, but $\wp$ has a pole at $0$, so the singularity at zero could be removable, not sure how to prove it isn't.

But even then, just because they have the same poles doesn't tell me much (I think?).

Jean Marie
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pizzaroll
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1 Answers1

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I would express the functions involved in power series.

$$\wp(2z) \!=\! \frac{1}{4z^2}\!+\!O(z)^2\!,\, 2\wp(z) \!=\! \frac{2}{z^2}\!+\!O(z)^2\!,\, \wp''(z) \!=\! \frac{6}{z^4}\!+\!O(z)^0,\, \wp'(z) \!=\! -\frac{2}{z^3}\!+\!O(z)^1\!, \tag{1} $$ and now $$\wp(2z) + 2\wp(z) =\frac{9}{4z^2}+O(z)^2,\quad \frac14 \left(\frac{\wp''(z)}{\wp'(z)}\right)^2 = \frac{9}{4z^2}+O(z)^2, \tag{2}$$ so your equation is true up to $\;O(z)^2\;$ for the pole at the origin in the period parallelogram. Similar considerations apply at the other three poles at half lattice points. Now the two sides have the same poles and residues which implies that the difference of the two sides has no poles, and since $\wp$ has a period parallelogram, by Liouville's theorem the two sides must differ by a constant. Using equation $(2)$ that constant is zero.

Somos
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  • So if the functions on both sides are the same near the poles, that means they're the same everywhere because of the Identity Theorem, I think? Do you need to do the other poles, or is one enough then? – pizzaroll Apr 29 '18 at 10:38
  • @pizzaroll Read the last part of my answer about Louvilles's theorem – Somos Apr 29 '18 at 19:14
  • Louivile's theorem says the difference between the two sides is constant. Evaluate in a special point to see that this constant equals zero. – Hrodelbert Jan 01 '20 at 11:15