For an arbitrary function $f(x,y)$ I am defining functions LogMT1
and LogMT2
as follows,
Nn = 5;
LogMT1 = Sum [ f[x^n, y^n]/(n*(1 - x^(2*n))), {n, 1 , Nn}];
LogMT2 =Sum[Log[(1 + x^n)/((1 - x^(n/2)*y^(n/2))*(1 - x^(n/2)*y^(-n/2)))], {n, 1 , Nn}];
Now I want to know the power series expansion of both these functions LogMT1 and LogMT2 as a power series in x - and I expect them to come as $\sqrt{x}$, $x$, $x\sqrt{x}$, $x^2$ and so on and each of them should be multiplied by a function of $y$ as a coefficient.
- I want to know how this can be done?
(..of course my eventual goal is to be able to determine $f(x,y)$ such that LogMT1 = LogMT2
for arbitrarily large values of $Nn$ and it would be great if someone can suggest a Mathematica way of being able to do that...)
Here is a function $f$ which seems to solve the above equation for arbitrarily large values of $Nn$ to arbitrarily large values of powers of $x$ as far as one can see this way,
(..the point is that I don't know how this function $f$ can be derived..)
$Assumptions = y > 0;
f[x_, y_] = Sqrt[x] (Sqrt[y] + 1/Sqrt[y]) + x (1 + y + 1/y) + x^(3/2) (y^(3/2) + 1/y^(3/2)) + x^2 (y^2 + 1/y^2) + ((x y)^(5/2) (1 - 1/y^2))/(
1 - Sqrt[x y]) + (x/y)^(5/2)/(1 - Sqrt[x/y]) (1 - y^2) // Simplify;
Nn = 30;(*you can increase this but it takes longer time*)
LogMT1 = Sum[f[x^n, y^n]/(n (1 - x^(2 n))), {n, 1, Nn}];
LogMT2 = Sum[ Log[(1 + x^n)/((1 - x^(n/2) y^(n/2)) (1 - x^(n/2) y^(-(n/2))))], {n, 1, Nn}];
Series[LogMT1 - LogMT2, {x, 0, Nn/2}] // Simplify
O(x^{31}) is the output showing that the equation is satisfied to that order.
eqn = Thread[Series[LogMT1 - LogMT2, {x, 0, 3}][[3]] == 0]
has as a 4th equation(-4*(1 + y^3))/(3*y^(3/2)) == 0
for instance? $\endgroup$