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| 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 ... |
| I am sorry. I had no idea that you were trying to do this using WolframAlpha. In WolframAlpha If I simplify your integral to ... |
| Try Simplify[Subscript[H,SS]*Subscript[H,KK],Subscript[H,x_]*Subscript[H,y_]==0] which returns 0 because x_ matches something named x while y_ matches something different named y |
| I am VERY impressed! I thought it might likely be the case that you would never be able to track down the real reason for this. You taught me and you and perhaps others one more thing to watch out for. Congratulations and thank you. |
| Thank you for catching my mistake. When I correct that and I remove my 10^300 scaling factor then the plot matches and the integral is almost exactly 1.0 |
| That did seem oddly formatted. Perhaps this will be close enough Function[K57068,ConditionalExpression[Root[256.Re[K57068]^2-576.#1+432.#1^2- 108.#1^3+9.#1^4 &,1],0 x,Re[_]->0} which returns 0. - 576.*x + 432.*x^2 - 108.*x^3 +... |
| l[r_]:=Tr[MatrixPower[N[p[r].p[r]],1/2]]-1; Plot[l[r],{r,0,0.2}] t=Table[{r,l[r]},{r,1/100,80/100,1/100}] ListLinePlot[t] ![enter image description here][1] [1]:... |
| I understood that q and q1 had the same value. I was only looking for any and every tiny difference in the way the two sets of equations were written, hoping any tiny difference might uncover the problem. Now to hopefully the more important part. ... |