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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.

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  • $\begingroup$ 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? $\endgroup$
    – chris
    Nov 2, 2012 at 17:15
  • $\begingroup$ so what's the question now? $\endgroup$
    – chris
    Nov 3, 2012 at 9:18
  • $\begingroup$ @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 $x$. $\endgroup$
    – user6818
    Nov 5, 2012 at 3:06

2 Answers 2

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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 *)
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  • $\begingroup$ 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! $\endgroup$
    – user6818
    Nov 2, 2012 at 18:54
  • $\begingroup$ @user6818 are you sure you did not mistype your set of equations above? $\endgroup$
    – chris
    Nov 2, 2012 at 18:56
  • $\begingroup$ You can take my "answer" above and substitute in the equations and check to arbitrary powers of $x$ that it solves. $\endgroup$
    – user6818
    Nov 2, 2012 at 19:02
  • $\begingroup$ 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 $\endgroup$
    – chris
    Nov 2, 2012 at 19:06
  • $\begingroup$ 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 $Nn$. $\endgroup$
    – user6818
    Nov 2, 2012 at 22:09
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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)) *)
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  • $\begingroup$ 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...) $\endgroup$
    – user6818
    Nov 2, 2012 at 18:59
  • $\begingroup$ 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]. $\endgroup$
    – b.gates.you.know.what
    Nov 2, 2012 at 19:11
  • $\begingroup$ 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. $\endgroup$
    – user6818
    Nov 2, 2012 at 22:11
  • $\begingroup$ 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}] $\endgroup$
    – user6818
    Nov 2, 2012 at 22:58

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