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How do I use the difference root in Wolfram|Alpha?

Posted 1 year ago

Hello!

I have a difference root here: solve the DifferenceRoot (through RSolve) in the following sum: https://www.wolframalpha.com/input?i=sum+%5B+polygamma%280%2C+n+%2B+3%29%5D%2F%28n%2B1%2Fa%29+from+n+%3D+2+to+m

How am I supposed to use the difference root once I solve it (using RSolve for example)? Do I just plug in the portion in the square brackets: "[m+1]"? What exactly do I do with this once I have obtained a formula for the difference root?

2 Replies
Posted 1 year ago

It looks like the size of the intermediate results won't fit in the buffer space in Wolfram|Alpha, or I haven't been able to find a trick to coax Wolfram|Alpha into doing this.

But Mathematica has different product, slightly different notation.

FullSimplify[FunctionExpand[Sum[PolyGamma[0,n+3]/(n+1/a),{n,2,m}]]]

returns

((a*(-14+25*a-14*a^2-9*a^3+6*(-1+a)^2*(2+a)*EulerGamma))/(-1+a)- 
  6*(1+a)*(1+a*(-2+EulerGamma)-EulerGamma)*HarmonicNumber[a^(-1)+m]+
  6*(-1+a^2)*HarmonicNumber[3+m]*HarmonicNumber[a^(-1)+m]-
  6*(1+a)*(EulerGamma+PolyGamma[0,a^(-1)])*(-1+2*a+(-1+a)*
  PolyGamma[0,4+m])+6*(-1+a^2)*Sum[(PolyGamma[0,a^(-1)]-
  PolyGamma[0,1+a^(-1)+K[1]])/(3+K[1]),{K[1],-1,m}])/(6*(-1+a^2))

but that still contains a sum that it is not able to completely reduce, perhaps because of the symbolic upper bound in your original sum.

Perhaps this might give a little more insight into what the first few terms of your sum look like

Table[FullSimplify[PolyGamma[0,n+3]/(n+1/a)],{n,2,12}]

returns

{(a*(25/12 - EulerGamma))/(1 + 2*a),
 (a*(137/60 - EulerGamma))/(1 + 3*a), 
 (a*(49/20 - EulerGamma))/(1 + 4*a),
 (a*(363/140 - EulerGamma))/(1 + 5*a), 
 (a*(761/280 - EulerGamma))/(1 + 6*a),
 (a*(7129/2520 - EulerGamma))/(1 + 7*a), 
 (a*(7381/2520 - EulerGamma))/(1 + 8*a),
 (a*(83711/27720 - EulerGamma))/(1 + 9*a), 
 (a*(86021/27720 - EulerGamma))/(1 + 10*a),
 (a*(1145993/360360 - EulerGamma))/(1 + 11*a),
 (a*(1171733/360360 - EulerGamma))/(1 + 12*a)}

but I don't find a nice compact form for that sum.

POSTED BY: Bill Nelson
Posted 1 year ago

Hello! This is actually quite helpful, you're saying that the difference root is just solved (as a recurrence relation for example) and then you plug in "m+1"?

If that's true, this is actually extremely helpful.

Is the sum that can't be reduced using a recurrence relation?

POSTED BY: Updating Name
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