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I tried several different ways with things like this: Sum(Binomial(n-1,n-i)*Sum(Binomial(k+i,i)*Binomial(n-1,n-k),{k,0,n}),{i,0,n}) Unfortunately, none of those succeeded. I finally used brute force calculation for n=0,n=1,...n=8 to get ...
I am thinking that the combination of `Floor` and `Log` is making this difficult. Is there any way that you could eliminate those and write the problem in terms of `2*n+1
Does this do what you want? ReplaceAll[{0.710411+3.67653s+s^2, 1.42082+3.67653s+s^2, 2.13123+3.67653s+s^2, 2.84164+3.67653s+s^2, 3.55206+3.67653s+s^2, 4.26247+3.67653s+s^2, 4.97288+3.67653s+s^2, ...
This F[p_]:=(Print[{p,Precision[p]}];If[p==0,1,FindRoot[Erf[Sqrt[p*r]]==((2/Sqrt[\[Pi]])*Sqrt[p]),{r,2},WorkingPrecision->100,PrecisionGoal->16][[1]][[2]]]); Plot[F[p],{p,0,\[Pi]/4},PlotRange->{0,30},PlotPoints->5000] shows that `Plot`...
In your notebook that you attached, thank you for that, would be impossible to help otherwise, 1: You have M[r] and P[r] and R[r] but you are not solving for those and have not defined those. 2: In eq3 you use := and = which should perhaps be =...
If you try Simplify[p,q] Mathematica will try to simplify p given the information supplied in q. For your example Simplify[x*x==4,x==2] returns True See if that will do the kinds of things you are looking for.
Perhaps look at [docs for DiscreteLimit][1] [1]: https://reference.wolfram.com/language/ref/DiscreteLimit.html
Try Integrate[x^2/(r-x),{x,0,R},Assumptions->0
If you Simplify[(A12 P312 + A13 P213 + A23 P123)/P123123 + A12 A13 A23 == 0] you should see it is a very large numerator/denominator==0. If you replace that with numerator==0 then the problem is about half as big and may be easier to...
Does this do exactly what you want for this specific example? poly=x2 y0(-b^2-a^2 λ)+x0 y2(-b^2-a^2 λ)+(x2 y1-x1 y2)(b^2-a^2 λ)+x1 y0(b^2+a^2 λ)+x0 y1(b^2+a^2 λ); term=(x2 y1-x1 y2)(b^2-a^2 λ); term==term-poly which returns ...