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1

TraditionalForm@FullSimplify@Exp[FullSimplify@PowerExpand[ Log[((Ft Bcr)^(2 p + 4)*Ft^(23 p - 4))/(Bcr^(1/(2.5 p - 4))* Ft^((24 p + 4)/(3 p - 1)))^(3 p - 6)]]] $\text{Ft}^{p+\frac{60}{3 p-1}+36} \text{Bcr}^{-\frac{1.2}{4.\, -2.5 p}+2. p+2.8}$


2

Reduce has its own syntax for specifying constrains and domains. Please try this: Reduce[2*a*b*n*c*Q - 3*a*b*n*p*c*Q + a*b*n*p^2*c*Q - 4*a^2*b*n*p*Q^2 - 4*a*b^2*n*p*Q^2 + 6*a^2*b*n*p^2*Q^2 + 6*a*b^2*n*p^2*Q^2 - 2*a^2*b*n*p^3*Q^2 - 2*a*b^2*n*p^3*Q^2 + a^2*b*c*Q^2 + a*b^2*c*Q^2 - 2*a^2*b*p*c*Q^2 - 2*a*b^2*p*c*Q^2 == 0 && 0 < a ...


0

Try this: << VectorAnalysis` SetCoordinates[Cartesian[x, y, z]]; R = {x, y, z}; r0 = {x0, y0, z0}; r = Sqrt[(R - r0).(R - r0)]; Y = 1/r*Exp[-(r/L)]; Now Simplify[Grad[Y, {x, y, z}] /. x -> x0 + Sqrt[ r1^2 - y^2 + 2 y y0 - y0^2 - z^2 + 2 z z0 - z0^2], {x0 > 0, y0 > 0, z0 > 0}] yielding (* {-((E^(-(Sqrt[r1^2]/L)) (L + ...


4

The Package HypExp does exactly that. Here is the link to paper for what I believe was the last extension. After digging around a bit, the package files should be available here ( Edit freely available link) Several years ago, there has been some work on the simplification of polylogarithms into a Hopt Algebras, which simplifies the reduction of the ...


3

Please check the result by doing the following eqn = FullSimplify[ 2*a*b*n*c*Q - 3*a*b*n*p*c*Q + a*b*n*p^2*c*Q - 4*a^2*b*n*p*Q^2 - 4*a*b^2*n*p*Q^2 + 6*a^2*b*n*p^2*Q^2 + 6*a*b^2*n*p^2*Q^2 - 2*a^2*b*n*p^3*Q^2 - 2*a*b^2*n*p^3*Q^2 + a^2*b*c*Q^2 + a*b^2*c*Q^2 - 2*a^2*b*p*c*Q^2 - 2*a*b^2*p*c*Q^2 == 0] // TraditionalForm This returns ...


2

Others are certainly correct in pointing out that (x^3)^(1/3) is only equal to x under certain circumstances. But if you know that, and just want to simplify while assuming that x is positive, then just use PowerExpand PowerExpand[(x^3)^(1/3)] (* x *) PowerExpand[(x^3 y^3 z)^(1/3)] (* x y z^(1/3) *)


2

Well, it's not generally true. For example: ((-1)^3)^(1/3) N[%] (* Out: (-1)^(1/3) 0.5 + 0.866025 I *) I suppose you could do FullSimplify[(x^3 y^3 z)^(1/3), Assumptions -> {x > 0, y > 0, z > 0}] (* Out: x y z^(1/3) *)


1

It's really simple. Just use Together. f1 = 1/x; f2 = x/(x^2 - 1); f1 + f2 1/x + x/(-1 + x^2) Together[f1 + f2] (-1 + 2 x^2)/(x (-1 + x^2))


0

Try this: f1[x_, y_] = (x + y)^5; f2[x_, y_] := (x - 3 y)^5 Simplify[Expand /@ (f1[x, y] + f2[x, y])] (* 2 (x^5 - 5 x^4 y + 50 x^3 y^2 - 130 x^2 y^3 + 205 x y^4 - 121 y^5) *)


1

I'm afraid there's no simple way or shortcut to properly implement tensor symmetries. Manual rules that move indices might work in simple situations, but inevitably fall short when dealing with more complicated symmetries, more indices, or bigger contractions. That being said, we can take @Szabolcs' cue and rewrite everything in terms of the (as of ...


12

This is not a bug; it is an implicit representation of the domain. Note that condition $1-x^3 \neq 0$ cannot be simplified further over the complexes or, at least, Reduce doesn't simplify it further: Reduce[1 - x^3 != 0, x, Reals] (* Out: x < 1 || x > 1 *) Reduce[1 - x^3 != 0, x, Complexes] (* Out: -1 + x^3 != 0 *) A similar thing can happen ...


0

My way is: (no guarantee to be the best way) a = Cos[_]; rules = a -> Hold[a] Cos[t] + 2 Cos[u] + 3 /. rules It returns 3 + 3 Hold[a]. If you don't put a with Hold, Mathematica will evalute it, which we do not expect. You can also write a function to generate a list of rules when you need several such replaces. The benefit of writing rules = a -> ...


2

One way to do that is as follow: Unprotect[Cos, Sin]; Cos[t_] = a; Sin[t_] = b; c = 2*Cos[t] + Sin[t]^2 (*2 a + b^2*) But be careful if you are going to use Cos or Sin to do calculations. they will return a and b.


4

One way to do something similar is to use rules instead of using equals. rules = {Cos[t] -> a, Sin[t] -> b} 2 Cos[t]^2 + 3 Sin[t] //. rules 2 a^2 + 3 b


7

This is an example related to limitations of standard symbolic capabilities of Mathematica, there are many similar issues ( usually they come from arbitrary choices of branches of complex functions), see e.g. analogous problems Why does Integrate declare a convergent integral divergent? or Bug in mathematica analytic integration?. Functions like Log and ...


3

Use Assuming[a > 0, a^2 + b^2 + 2 a b // FullSimplify] or a^2 + b^2 + 2 a b // FullSimplify[#, a > 0] &


2

How about this? ArcTan[Simplify[TrigExpand[Tan[5 ArcTan[a] + ArcTan[b]]]]] Could save yourself some work! Explanation: The trick here is to understand the identity you mentioned is about Tan, not ArcTan: Tan[x+y] = (Tan[x]+Tan[y])/(1 -Tan[x]Tan[y]). TrigExpand does the above type of expansion for trig functions. Replace x and y with ArcTan[a] and ...


1

How about: 5 ArcTan[a] + ArcTan[b] /. {(m_Integer: 1) ArcTan[a_] + (n_Integer: 1) ArcTan[b_] /; Positive[m] && Positive[n] :> With[{min = Min[m, n]}, (m - min) ArcTan[a] + (n - min) ArcTan[b] + min ArcTan[(a + b)/(1 - ab)]]} (* 4 ArcTan[a] + ArcTan[(a + b)/(1 - ab)] *) Works on positive integral coefficients, which is how I read ...


2

Kxx1 = -((\[Pi]^2 (x - x0) Cosh[\[Pi] (x - x0)])/(2 (-Cos[\[Pi] (y - y0)] + Cosh[\[Pi] (x - x0)]))) + (\[Pi]^2 (x - x0) Cosh[\[Pi] (x - x0)])/(2 (-Cos[\[Pi] (-1 + y + y0)] + Cosh[\[Pi] (x - x0)])) + (\[Pi]^2 (x - x0) Sinh[\[Pi] (x - x0)]^2)/(2 (-Cos[\[Pi] (y - y0)] + Cosh[\[Pi] (x - x0)])^2) - ...


1

Maybe this type of visualization will help understand the evaluation order better. Things are printed in InputForm here, but please "think" FullForm when you look at expressions. In[2]:= On[] Plus[3^3,6*9] Off[] During evaluation of In[2]:= On::trace: On[] --> Null. >> During evaluation of In[2]:= Power::trace: 3^3 --> 27. >> During ...


1

This particular problem can dealt with algebraically by completing squares, equating coefficients and subtracting constants. (I am making the assumptions this is a left hand side whose right hand side is 0). pol = -0.433284 - 0.758719 x + 0.00289158 x^2 - 0.443672 y + 0.00149027 y^2; cr = CoefficientRules[pol, {x, y}] {p, q, r, s, t} = cr[[All, 2]]; {h, ...


4

Although kguler posted an answer using a nice internal function that does this (almost) directly I find this kind of expression manipulation interesting in itself so I'm going to see what can be done without it. expr = a[b[c, d[e][f], g], h]; edges = Reap[Replace[expr, h_[___, c_[___] | c_?AtomQ, ___] /; Sow[h -> c] :> 1, {0, -1}]][[2, 1]]; ...


7

An alternative method to WReach's method is to use SparseArray`ExpressionToTree which produces the same output without string wrappers: expr = a[b[c, d[e][f], g], h]; ett = SparseArray`ExpressionToTree[expr] (* {{a,0,a[b[c,d[e][f],g],h]}->{b,1,b[c,d[e][f],g]}, {b,1,b[c,d[e][f],g]}->{c,2,c}, {b,1,b[c,d[e][f],g]}->{d[e],3,d[e][f]}, ...


0

Is this what you are seeking? expr = a[b[c, d[e][f], g], h] boxes = ToBoxes@TreeForm[expr] positions = Cases[boxes, LineBox[{x__}] -> x, Infinity] nodes = Cases[ boxes, StyleBox[x_, __] :> ToExpression@x, Infinity] /. {t_Times :> First@t, Verbatim[HoldForm][x_] -> x} Rule @@@ Extract[nodes, List /@ positions] {a -> b, a -> ...


3

There are two problems. One is that you want SolveAlways to find values for the parameters for which the expressions are equivalent. The second is that, as stated, it's not possible because the system is overdetermined. You'll want to allow for a constant term in your second expression. expr1 = -0.433284 - 0.758719 x + 0.00289158 x^2 - 0.443672 y + ...


1

e[k_, t_] := Cos[Pi (k - 1) t]; cosIntRaw[k_, l_, m_] := Integrate[e[k, t] e[l, t] e[m, t], {t, 0, 1}]; cosIntRaw[k, l, m] // Simplify -((Sin[(k - l - m)*Pi]/ (1 + k - l - m) + Sin[(k + l - m)*Pi]/ (-1 + k + l - m) + Sin[(k - l + m)*Pi]/ (-1 + k - l + m) + Sin[(k + l + ...


6

The problem is that b^(1/3) has three roots. You are choosing the real root, -(3/2^(2/3)), while Mathematica is choosing, (3 (-1)^(1/3))/2^(2/3) which is complex. To get the result you want, rewrite your expressions for Y and Z like this: Y = 12 Q/ Sqrt[A (2 - (3 2^(1/3) a)/Surd[Sqrt[b^2 - 4 c^3] + b, 3] - Surd[Sqrt[b^2 - 4 ...



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