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Expand a limit form containing sum to series?

Hi, my problem is as follows?

Define $F(q,\lambda)$ as

$$F(q,\lambda):=\lim\limits_{z\rightarrow 0}\left (\sum_{i=1/2}^\infty \ln [1+z^{-1}q^{i}\exp({-i^{2} \lambda})][1+zq^{i}\exp({i^{2} \lambda})]-\ln z\right ) , $$
where the summation index $i$ are half integers and $|q|<1$. Expand $F(q,\lambda)$ first as series of $\lambda$,

$$ F(q,\lambda)=F1(q)+F2(q)\lambda^2+...$$ Then expand $F_2(q)$ in $q$

$$ F2(q)=a0+a1q+a2q^2+...$$ What I want is to compute the coefficients $a_0,a_1,a_2$ using Mathematica, and higher term's coefficients if posible. I tried the following code(rather straightforward) and it failed(got stuck in calculation)

Series[SeriesCoefficient[
Limit[Sum[Log[(1 + z q^i Exp[\[Lambda] i^2]) (1 + z^(-1) q^i Exp[-\[Lambda] i^2])], {i, 1/2, Infinity}] - 
Log[z], z -> 0], 
{\[Lambda], 0, 2}], {q, 0, 2}]

It seems mathematica is struggling to first do the infinite sum in the limit which possibly doesn't have a simpler expression at all.

Any help or comment would be appreciated.
*********** Background(hopefully useful to solve the problem):$F(q,\lambda) $ is the free energy of a type of fermion system, $F(q,\lambda)=\ln Z(q,\lambda) $ where $Z$ is the partition function.

POSTED BY: Wei Yu
2 Replies

That seems to be difficult.

Look at (you forgot the Log in your code above)

sum[n_] := 
 Sum[Log[(1 + z q^i Exp[\[Lambda] i^2]) (1 + 
       z^(-1) q^i Exp[-\[Lambda] i^2])], {i, 1/2, n}] // Expand

and (you may want to chose different values for q, lambda and z )

Plot[Evaluate[
  sum[m] - Log[z] /. q -> .5 /. \[Lambda] -> 5 /. z -> 2], {m, 1, 15}]

Now look at

ser = Series[sum[3], {\[Lambda], 0, 3}] // FullSimplify

Perhaps you get some ideas when considering

Coefficient [ser, \[Lambda]^1]
Coefficient [ser, \[Lambda]^2]
POSTED BY: Hans Dolhaine

I plot sum[n] using different q, lambda and z, and I find that when n goes to infinity, it simply diverges. So F(z,q,lambda) is not well defined. But my problem now is solved (in completely different way). Thank you for your advice.

POSTED BY: Wei Yu
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