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Questions on manipulating complicated expressions and making them look simpler using Simplify, FullSimplify and Reduce.

3
votes
You can use RootApproximant[N[val, 10000], 20] (14700 - 12000 Sqrt[2])/29929 So newval=(14700 - 12000 Sqrt[2])/29929; To test it: N[newval, 100000] - N[val, 100000] 0.*10^-100001 …
answered Mar 24 '16 by kiara
1
vote
You can simplify the expression as H12[k] is actually constant: H12[k_]={{-π, -π/4}, {(-π/4), -π}} So the derivatives should be zero.
answered Oct 20 '17 by kiara
-1
votes
The problem is that Normalize takes only vectors as input: (*Matrix definition*) QEG = {{-(Exp[-I k]/Sqrt[2]), -((I Exp[-I k])/ Sqrt[2])}, {-((I Exp[I k])/Sqrt[2]), -(Exp[I k]/Sqrt[2])}}; (*M …
answered Sep 28 '17 by kiara
2
votes
One way is to substitute it manually as exp = Sum[coeff[i] Sqrt[ab^2], {i, 1, 20000}]; AbsoluteTiming[exp2 = exp /. Sqrt[ab^2] -> ab;] 0.028188, Null}
answered Mar 9 '17 by kiara
2
votes
Following the comments you can write $x[t]$ as x[t_] = FullSimplify[Solve[(a x + b)*t^2/2 + x0 == x, x][[1, 1, 2]]]; With a=(-42 Na + (252 Ni)/25)/m and b=-g + (420000 Na + 100 Ch Nh …
answered Nov 22 '16 by kiara
2
votes
First tmp=(Conjugate[Sqrt[( 1 + 1/Sqrt[1 + Abs[\[Beta][qx, qy]]^2])/\[Beta][qx, qy]]] Sqrt[(1 + Abs[\[Beta][qx, qy]]^2 - Sqrt[ 1 + Abs[\[Beta][qx, qy]]^2]) Conjugate[\[Beta][qx, qy]]] …
answered Nov 7 '17 by kiara
3
votes
Just: qq /. FIinAlpha31''''[t] -> 0
answered May 17 '18 by kiara