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Ok so I just realized there is a very simple operator calculus reason for this (I apologize if my answers are annoying since they are very closely related but I am using this to document my ideas and also learning for myself how this works).

Consider the operator $H: f(s) \rightarrow xf(s+1)$. It then follows quite naturally that

$$ f(s) + xf(s+1) + x^2 f(s+2) + ... = \sum_{n=0}^{\infty} H^n[f(s)] = \frac{1}{1-H}$$

But recall the geometric series $\frac{1}{1-x}$ has two series of convergence $1+x+x^2 + x^3 ... $ and the laurent series $-\frac{1}{x} - \frac{1}{x^2} - \frac{1}{x^3} ... $

Therefore:

$$ \frac{1}{1-H} = -H^{-1} - H^{-2} - H^{-3} .... = -\frac{f(s-1)}{x}-\frac{f(s-2)}{x^2} - \frac{f(s-3)}{x^3} ... $$

Evaluating both series at $s=0$ then gives us that

$$ \sum_{n=0}^{\infty} f(n) = \left[\frac{1}{1-H}\right]_{@s=0} = -\sum_{n=1}^{\infty} f(-n)x^{-n} $$

And that is the connection that you found. I'm not sure what the ramifications of this formula entail but there is a (non optimistic) chance it could be deep.

For example the Jacobi theta function $\sum_{n=0}^{\infty} x^{n^2} = \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} \frac{e^{2i\pi n} - 1}{e^{2i\pi \sqrt{n}} - 1}x^n $ so we could try to bash it with this thing. What's curious is that once $n$ goes negative then $\sqrt{n}$ goes imaginary and $e^{2i\pi \sqrt{n}} $ starts to grow exponentially fast so this series somewhat shockingly actually will converge and produce something (up to a choice of branch cut). Is that the "natural" continuation of the Jacobi theta function? I'm not sure, but if it doesn't work its not obvious to me why.

Ok so I just realized there is a very simple operator calculus reason for this (I apologize if my answers are annoying since they are very closely related but I am using this to document my ideas and also learning for myself how this works).

Consider the operator $H: f(s) \rightarrow xf(s+1)$. It then follows quite naturally that

$$ f(s) + xf(s+1) + x^2 f(s+2) + ... = \sum_{n=0}^{\infty} H^n[f(s)] = \frac{1}{1-H}$$

But recall the geometric series $\frac{1}{1-x}$ has two series of convergence $1+x+x^2 + x^3 ... $ and the laurent series $-\frac{1}{x} - \frac{1}{x^2} - \frac{1}{x^3} ... $

Therefore:

$$ \frac{1}{1-H} = -H^{-1} - H^{-2} - H^{-3} .... = -\frac{f(s-1)}{x}-\frac{f(s-2)}{x^2} - \frac{f(s-3)}{x^3} ... $$

Evaluating both series at $s=0$ then gives us that

$$ \sum_{n=0}^{\infty} f(n) = \left[\frac{1}{1-H}\right]_{@s=0} = -\sum_{n=1}^{\infty} f(-n)x^{-n} $$

And that is the connection that you found. I'm not sure what the ramifications of this formula entail but there is a (non optimistic) chance it could be deep.

For example the Jacobi theta function $\sum_{n=0}^{\infty} x^{n^2} = \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \frac{e^{2i\pi n} - 1}{e^{2i\pi \sqrt{n}} - 1}x^{-n} $ so we could try to bash it with this thing. What's curious is that once $n$ goes negative then $\sqrt{n}$ goes imaginary and $e^{2i\pi \sqrt{n}} $ starts to grow exponentially fast so this series somewhat shockingly actually will converge and produce something (up to a choice of branch cut). Is that the "natural" continuation of the Jacobi theta function? Edit: i spoke too soon. I have the resultant series is just 0 everywhere and so isn’t a very satisfying continuation at all.

Ok so I just realized there is a very simple operator calculus reason for this (I apologize if my answers are annoying since they are very closely related but I am using this to document my ideas and also learning for myself how this works).

Consider the operator $H: f(s) \rightarrow xf(s+1)$. It then follows quite naturally that

$$ f(s) + xf(s+1) + x^2 f(s+2) + ... = \sum_{n=0}^{\infty} H^n[f(s)] = \frac{1}{1-H}$$

But recall the geometric series $\frac{1}{1-x}$ has two series of convergence $1+x+x^2 + x^3 ... $ and the laurent series $-\frac{1}{x} - \frac{1}{x^2} - \frac{1}{x^3} ... $

Therefore:

$$ \frac{1}{1-H} = -H^{-1} - H^{-2} - H^{-3} .... = -\frac{f(s-1)}{x}-\frac{f(s-2)}{x^2} - \frac{f(s-3)}{x^3} ... $$

Evaluating both series at $s=0$ then gives us that

$$ \sum_{n=0}^{\infty} f(n) = \left[\frac{1}{1-H}\right]_{@s=0} = -\sum_{n=1}^{\infty} f(-n)x^{-n} $$

And that is the connection that you found. I'm not sure what the ramifications of this formula entail but there is a (non optimistic) chance it could be deep.

For example the Jacobi theta function $\sum_{n=0}^{\infty} x^{n^2} = \sum_{n=0}^{\infty} \frac{\sin(2\pi n)}{2\sqrt{n} \sin(2\pi \sqrt{n})}x^n $ so we could try to bash it with this thing. What's curious is that once $n$ goes negative then $\sqrt{n}$ goes imaginary and $\sin(2\pi \sqrt{n})$ starts to grow exponentially fast so this series somewhat shockingly actually will converge and produce something (up to a choice of branch cut). Is that the "natural" continuation of the Jacobi theta function? I'm not sure, but if it doesn't work its not obvious to me why.

Ok so I just realized there is a very simple operator calculus reason for this (I apologize if my answers are annoying since they are very closely related but I am using this to document my ideas and also learning for myself how this works).

Consider the operator $H: f(s) \rightarrow xf(s+1)$. It then follows quite naturally that

$$ f(s) + xf(s+1) + x^2 f(s+2) + ... = \sum_{n=0}^{\infty} H^n[f(s)] = \frac{1}{1-H}$$

But recall the geometric series $\frac{1}{1-x}$ has two series of convergence $1+x+x^2 + x^3 ... $ and the laurent series $-\frac{1}{x} - \frac{1}{x^2} - \frac{1}{x^3} ... $

Therefore:

$$ \frac{1}{1-H} = -H^{-1} - H^{-2} - H^{-3} .... = -\frac{f(s-1)}{x}-\frac{f(s-2)}{x^2} - \frac{f(s-3)}{x^3} ... $$

Evaluating both series at $s=0$ then gives us that

$$ \sum_{n=0}^{\infty} f(n) = \left[\frac{1}{1-H}\right]_{@s=0} = -\sum_{n=1}^{\infty} f(-n)x^{-n} $$

And that is the connection that you found. I'm not sure what the ramifications of this formula entail but there is a (non optimistic) chance it could be deep.

For example the Jacobi theta function $\sum_{n=0}^{\infty} x^{n^2} = \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} \frac{e^{2i\pi n} - 1}{e^{2i\pi \sqrt{n}} - 1}x^n $ so we could try to bash it with this thing. What's curious is that once $n$ goes negative then $\sqrt{n}$ goes imaginary and $e^{2i\pi \sqrt{n}} $ starts to grow exponentially fast so this series somewhat shockingly actually will converge and produce something (up to a choice of branch cut). Is that the "natural" continuation of the Jacobi theta function? I'm not sure, but if it doesn't work its not obvious to me why.

Ok so I just realized there is a very simple operator calculus reason for this (I apologize if my answers are annoying since they are very closely related but I am using this to document my ideas and also learning for myself how this works).

Consider the operator $H: f(s) \rightarrow xf(s+1)$. It then follows quite naturally that

$$ f(s) + xf(s+1) + x^2 f(s+2) + ... = \sum_{n=0}^{\infty} H^n[f(s)] = \frac{1}{1-H}$$

But recall the geometric series $\frac{1}{1-x}$ has two series of convergence $1+x+x^2 + x^3 ... $ and the laurent series $-\frac{1}{x} - \frac{1}{x^2} - \frac{1}{x^3} ... $

Therefore:

$$ \frac{1}{1-H} = -H^{-1} - H^{-2} - H^{-3} .... = -\frac{f(s-1)}{x}-\frac{f(s-2)}{x^2} - \frac{f(s-3)}{x^3} ... $$

Evaluating both series at $s=0$ then gives us that

$$ \sum_{n=0}^{\infty} f(n) = \left[\frac{1}{1-H}\right]_{@s=0} = -\sum_{n=1}^{\infty} f(-n)x^{-n} $$

And that is the connection that you found. I'm not sure what the ramifications of this formula entail but there is a (non optimistic) chance it could be deep.

Ok so I just realized there is a very simple operator calculus reason for this (I apologize if my answers are annoying since they are very closely related but I am using this to document my ideas and also learning for myself how this works).

Consider the operator $H: f(s) \rightarrow xf(s+1)$. It then follows quite naturally that

$$ f(s) + xf(s+1) + x^2 f(s+2) + ... = \sum_{n=0}^{\infty} H^n[f(s)] = \frac{1}{1-H}$$

But recall the geometric series $\frac{1}{1-x}$ has two series of convergence $1+x+x^2 + x^3 ... $ and the laurent series $-\frac{1}{x} - \frac{1}{x^2} - \frac{1}{x^3} ... $

Therefore:

$$ \frac{1}{1-H} = -H^{-1} - H^{-2} - H^{-3} .... = -\frac{f(s-1)}{x}-\frac{f(s-2)}{x^2} - \frac{f(s-3)}{x^3} ... $$

Evaluating both series at $s=0$ then gives us that

$$ \sum_{n=0}^{\infty} f(n) = \left[\frac{1}{1-H}\right]_{@s=0} = -\sum_{n=1}^{\infty} f(-n)x^{-n} $$

And that is the connection that you found. I'm not sure what the ramifications of this formula entail but there is a (non optimistic) chance it could be deep.

For example the Jacobi theta function $\sum_{n=0}^{\infty} x^{n^2} = \sum_{n=0}^{\infty} \frac{\sin(2\pi n)}{2\sqrt{n} \sin(2\pi \sqrt{n})}x^n $ so we could try to bash it with this thing. What's curious is that once $n$ goes negative then $\sqrt{n}$ goes imaginary and $\sin(2\pi \sqrt{n})$ starts to grow exponentially fast so this series somewhat shockingly actually will converge and produce something (up to a choice of branch cut). Is that the "natural" continuation of the Jacobi theta function? I'm not sure, but if it doesn't work its not obvious to me why.