Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|improve this question
It seems it can't work because 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? – chris Nov 2 '12 at 17:15
so what's the question now? – chris Nov 3 '12 at 9:18
@chris The question still remains as to if and how one can solve the equation LogMT1 = logMT2 ... at least perturbatively for large values of $Nn$ and order by order in powers of $x$. – user6818 Nov 5 '12 at 3:06

In its current form (which I suspect is wrong) you can get a series solution near the origin as follows. Let's first Taylor expand the difference between the two series and find a set of equation corresponding to requiring the series is identically null.

eqn = Thread[Series[LogMT1 - LogMT2, {x, 0, 5}][[3]] == 0]

This set of equations can only be satisfied for specific values of y, say y=-1, which seems dodgy to me. Moving on, let's assume f[x,y]=g[x] and find what constraints we have on g[x]

eqn2 = Select[eqn/. y->-1 //Release,(Length[#] > 1) &]/. f->Function[{x, y}, g[x]] // Simplify;

Let's solve for these equations:

sol=  Solve[eqn2, Table[D[g[x], {x, n}], {n, 0, 5}] /. x -> 0]

It follows that the Taylor expansion of g[x] near the origin is:

Normal[Series[g[x], {x, 0, 5}]] /. sol

(* -5 x^4+2 x^3+5 x^2-2 x *)
share|improve this answer
As far as I can check in powers of $x$ on Mathematica, the following solves the equation for arbitrarily large values of $Nn$, $f(x,y) = \sqrt{x}(y + 1/y) + x(1+y^2 + 1/y) + x^{3/2}(y^3+1/y^3) + x^2(y^4+1/y^4) + \sum_{n=5}^\infty x^{n/2}(y^n + 1/y^n - y^{n-4} - 1/y^{n-4})$. I don't know how to prove this and I would like to know if there is a way to derive this! – user6818 Nov 2 '12 at 18:54
@user6818 are you sure you did not mistype your set of equations above? – chris Nov 2 '12 at 18:56
You can take my "answer" above and substitute in the equations and check to arbitrary powers of $x$ that it solves. – user6818 Nov 2 '12 at 19:02
I just did and it doesn't! rF = f -> Function[{x, y}, Sqrt[x] (y + 1/y) + x (1 + y^2 + 1/y) + x^(3/2) (y^3 + 1/y^3) + x^2 (y^4 + 1/y^4) + Sum[x^(n/2) (y^n + 1/y^n - y^(n - 4) - 1/y^(n - 4)), {n, 5, Infinity}]] Series[LogMT1 - LogMT2 /. rF, {x, 0, 3}] // Simplify – chris Nov 2 '12 at 19:06
Apologies there seems to have been a typo! (..actually the typo was in the original paper from where this issue has arisen..) I have now typed into the original question a solution for $f(x,y)$ and how it can be perturbatively checked to be right - but I have no clue how that can be derived. Clearly there seems to exist a solution for the equation I wrote for arbitrarily high values of $Nn$. – user6818 Nov 2 '12 at 22:09

I think you can use Series on your expressions :

Series[LogMT1, {x, 0, 1}] 

Series[LogMT2, {x, 0, 2}] //Normal

(* (Sqrt[x] (1 + y))/Sqrt[y] + x (1 + 3/(2 y) + (3 y)/2) + 
    x^2 (1/2 + 7/(4 y^2) + (7 y^2)/4) + (4 x^(3/2) (1 + y^3))/(3 y^(3/2)) *)
share|improve this answer
I have tried doing what you are saying but the problem is that this "Series" command when used on LogMT1 doesn't generate the fractional powers of $x$ as it would when applied on LogMT2. Thats why a term by term comparison between the two functions turns out to be difficult. (...But I know otherwise that an expansion of LogMT1 about x =0 must have these fractional powers of x...) – user6818 Nov 2 '12 at 18:59
Then you can try to define LogMT12 = Sum[f[Sqrt[z]^(2 n), y^n]/(n*(1 - Sqrt[z]^(2 2*n))), {n, 1, Nn}], expand in powers of z and then substitute z -> Sqrt[x]. – b.gatessucks Nov 2 '12 at 19:11
Do you want to substitute \Sqrt[x] for z or just x? Also what kind of a series expansion is Mathematica doing here? This is neither a Laurent series nor a Taylor series. – user6818 Nov 2 '12 at 22:11
doing this didn't help, (..this too gave back a Taylor series with integer powers of x..) Nn = 5; LogMT1 = Sum [ f[Sqrt[x]^(2*n), y^n]/(n*(1 - (Sqrt[x])^(4*n))), {n, 1 , Nn}]; Series [LogMT1, {x, 0, Nn}] – user6818 Nov 2 '12 at 22:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.