       Re: pair sums applied to trignometry sums

• To: mathgroup at smc.vnet.net
• Subject: [mg52537] Re: pair sums applied to trignometry sums
• From: Roger Bagula <tftn at earthlink.net>
• Date: Wed, 1 Dec 2004 05:57:59 -0500 (EST)
• References: <200411290622.BAA27977@smc.vnet.net> <cohigl\$1i5\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```
Daniel Lichtblau wrote:

>
>It occurs to me that these functions might be simplified, as they are
>each themselves sums of pairs of functions with terms satisfying simple
>recurrences. For example, fs can be written as the sum of n-even + n-odd
>terms, and these are just the sums of terms 1/(2*k+1)*x^(4*k+1)/(4*k+1)!
>and (-1)*(2*k+1)/(2*k+2)*x^(4*k+3)/(4*k+3)! respectively.
>
>In more detail we get the function below.
>
>InputForm[fsin2[x_] = Together[-Sum[(2*k+1)/(2*k+2)*x^(4*k+3)/(4*k+3)!,
>{k,0,Infinity}] +
>   Sum[1/(2*k+1)*x^(4*k+1)/(4*k+1)!, {k,0,Infinity}]]]
>
>Out//InputForm= (-4 + 4*Cosh[x] + x*Sin[x] - x*Sinh[x])/(2*x)
>
>(Isn't it great to have a symbolic math engine at ones fingertips?)
>
>As a quick check:
>
>In:= InputForm[Max[Abs[Table[fsin2[x]-fsin[x], {x,-Pi,Pi,.1}]]]]
>Out//InputForm= 3.372302437298913*^-15
>
>(Isn't it great to have a numeric math engine at ones fingertips?)
>
>The advantage to using the closed form is twofold. One is that numeric
>computations are better, and the other is that they are significantly
>faster. To see the latter:
>
>In:= Timing[Plot[fsin[x],{x,-Pi,Pi}]]
>Out= {0.3 Second, -Graphics-}
>
>In:= Timing[Plot[fsin2[x],{x,-Pi,Pi}]]
>Out= {0.01 Second, -Graphics-}
>
>For the former, just notice what happens when we get outside the range
>-Pi<x<Pi, for example the interval {15*Pi,16*Pi}.
>
>
>Daniel Lichtblau
>Wolfram Research
>
>
>
>
Dear Daniel Lichtblau ,
There is a reason for using the specific pair {1/(n+1),n/(1+n)}.
It stems from group theory and functional inversion.
The pair function {1/(1+x),x/(1+x)} is connected to
the Farey tree functions by functional inversion:
x/(1-x) if the  functional inverse to x/(1+x)
Solve[z-x/(1-x)==0,x]
(1-x)/x if the  functional inverse to 1/(1+x)
Solve[z-(1-x)/x==0,x]

There is a larger group that contains these Farey tree transforms too
called the anharmonic group:
{x, 1/x, 1/(x-1),x-1,x/(1-x),(1-x)/x}
and it "implies"
an functional inversion group of:
{x,1/x,(1+x)/x,x+1,x/(1+x),1/(1+x)}

I have done a lot of work in this area in the last few years.

I'm glad someone realized that these functions are important besides me.
There should be a number of ways to do them as is usual
when dealing with fundamentals in Mathematics.
I chose the ones I did because they are a torus under
the inverse substitution:
{1/(1+x),x/(1+x)}/. x->1/x

I thank you for your work and I'll see if they work on my version of
Mathematica.

The anharmonic group connection makes me believe that there are further
"subharmonic" functions yet to be discovered.
Respectfully, Roger L. Bagula