In my previous post I wrote that I was not sure if the expression:

a*sin(b*x)-c*tan(pi/2*x)+d*(cos(e*x)-1)+f*(1/cos(pi/2*x)-1) (1)

still could be used in some form as an ordinary generating function for the Möbius function..

I have now proven using Wolfram|Alpha that it is wrong, it can’t be used as an o.g.f, and here is the proof starting by entering this line into Wolfram|Alpha:

expand a*sin(b*x)-c*tan(pi/2*x)+d*(cos(e*x)-1)+f*(1/cos(pi/2*x)-1)

Then from the block: Series expansion at x=0:

form the six equations: (a*b-(pi*c)/2)=1, (pi^2*f-4*d*e^2)/8=-1, 1/24*(-4*a*b^3-pi^3*c)=-1, 1/384*(16*d*e^4+5*pi^4*f)=0, 1/240*(2*a*b^5-pi^5*c)=-1, (61*pi^6*f-64*d*e^6))/46080=1

I.e the first six values of the Möbius function with six unknowns. Separate them into two groups to be entered separately into Wolfram|Alpha:

(a*b-(pi*c)/2)=1, 1/24*(-4*a*b^3-pi^3*c)=-1, 1/240*(2*a*b^5-pi^5*c)=-1

and

(pi^2*f-4*d*e^2)/8=-1, 1/384*(16*d*e^4+5*pi^4*f)=0, (61*pi^6*f-64*d*e^6))/46080=1

In the first case more than one solution is possible. But in the second case no solutions exist and thereby expression (1) can’t be expanded into the Möbius function, which was to be proven.

Mats Granvik, mats.granvik(AT)abo.fi

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