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5

Yes, in some rare cases it is indeed convenient to use the individual simplification functions, even though they are used inside FullSimplify. I think the general idea is that FullSimplify may not use the desired simplification at the exact step during the simplification where you want it to be used. This (probably) happens when the expression has so many ...


2

I think the closest you can get to what you looking for is Assuming[{a, b, c} ∈ Reals, With[{x = {a, b, c}}, x.x == Norm[x]^2 // Simplify]] True


1

The difference between both results is caused by a subtle difference in the inputs. The first replacement has a simple list of replacement rules: expr /. { f[v3, w3] -> a1 c1 f[v1, w1] + a1 c2 f[v1, w2] + a2 c1 f[v2, w1] + a2 c2 f[v2, w2], f[v4, w4] -> b1 d1 f[v1, w1] + b1 d2 f[v1, w2] + b2 d1 f[v2, w1] + b2 d2 f[v2, w2]} The second replacement ...


0

k[q_] := (EllipticTheta[2, 0, q]/EllipticTheta[3, 0, q])^2 kc[q_] := (EllipticTheta[4, 0, q]/EllipticTheta[3, 0, q])^2 G[n_] := (2 k[E^(-Pi Sqrt[n])] kc[E^(-Pi Sqrt[n])])^(-1/12) TrueQ[G[25] == N@GoldenRatio] (*True*) If you want a more exact solution, you might want to check this out.


1

func[z_] := ((-2 + 4 I) Cos[0.0628319 z] + (2 - I) Sin[ 0.0628319 z])/((1 + 2 I) Cos[0.0628319 z] + (4 + 2 I) Sin[ 0.0628319 z]) ComplexExpand@func[0] Show[ Plot[ Im@func[z], {z, 0, 100}, PlotStyle -> Blue ], Plot[ Re@func[z], {z, 0, 100}, PlotStyle -> Red ], PlotRange -> All ] (*1.2 + 1.6 I*)


1

You're not tryng to get the imaginary part of a number. You are trying to get the imaginary part of an expression. Let's say we define your expression to be equivalent to eq: eq==(2 + 4 I) Cos[0.0628319 z] + (2 - I) Sin[0.0628319 z])/ ((1 + 2 I) Cos[0.0628319 z] + (4 + 2 I) Sin[0.0628319 z]) Solving the above for z gives 4 solutions. Lets say we take one ...


2

Observing strictly that the domain of x as the upper limit of the summation index is the integers, the limit exists, it can be calculated easily with Mathematica and it is different from zero. We need to consider this sum \[Sigma]WH[x_] := Sqrt[\[Pi]^2/12 + Simplify[Sum[(x!*x!)/(k!*(2 x - k)!)*(-1)^(x - k)/(x - k), {k, 0, x - 1}], x \[Element] ...


0

This function lst[inp_] := Module[{apt, expr}, expr = inp; apt[x_] := {x[[1]], x[[2, 2, 1]]}; SetAttributes[apt, Listable]; expr[[0]] = List; SortBy[apt[expr], Last] ] lst[a*Exp[k1*t] + b*Exp[k2*t] + c*Exp[k3*t]] returns a list sorted by k1,k2,...: {{a, k1}, {b, k2}, {c, k3}}. This may only work when a,b,c,k1,k2,k3 are numbers due to ...


0

You can simply include the domain restriction in your system of equations instead: Eq4 = -1 == (Cos[δ] Sin[t])/(Cos[φ ] Sin[δ] - Cos[t] Cos[δ] Sin[φ]) Reduce[{Eq4, 0 < t < Pi/2} /. {δ -> -0.401426, φ -> 0.841248699}, t] (* Out: t == 0.869424 *) Reduce complains that it "was unable to solve the system with inexact coefficients. The answer ...


0

Try this: Coefficient[a, E^(-0.03056065153782187` t)] Coefficient[b, E^(-0.03056065153782187` t)] (* {0.678839} {-0.660328} *) Have fun!


1

Cases[a, coeff_.*e^(-0.0305607 t) :> coeff, 2] {0.678839} Cases[b, coeff_.*e^(-0.0305607 t) :> coeff, 2] {-0.660328} The pattern coeff_. would work even if the coefficient were 1, e.g. Cases[{e^(-0.0305607 t)}, coeff_.*e^(-0.0305607 t) :> coeff, 2] {1} because the dot, signifying default value, defaults to 1 in the case of a ...


3

Eliminate[{ k == (a b c t)/(Sqrt[(f + t)^2]), p == t/(t + f) }, {t, f}] a^2 b^2 c^2 p^2 == k^2


2

k = (a b c t)/(Sqrt[(f + t)^2]); Assuming that {t > 0, p > 0} Simplify[k /. t + f -> t/p, {t > 0, p > 0}] (* a b c p *) Or, Simplify[k /. f -> t (1 - p)/p, {t > 0, p > 0}] (* a b c p *) Or, assuming {0 <= p < 1, f > 0} Simplify[k /. t -> f p/(1 - p), {0 <= p < 1, f > 0}] (* a b c p *)


2

Try this: (-1 + x) Sqrt[(1 - x) (-1 + y^2)] + Sqrt[(1 - x)^3 (-1 + y^2)] /. Sqrt[a_^3*b_] -> a*Sqrt[a*b] // Simplify (* 0 *) Have fun!


3

If you translate the variables to be positive, you can use PowerExpand: ClearSystemCache[] Simplify[(-1 + x) Sqrt[(1 - x) (-1 + y^2)] + Sqrt[(1 - x)^3 (-1 + y^2)], -1 <= x <= 1 && -1 <= y <= 1, TransformationFunctions -> {Automatic, Simplify[ PowerExpand[# /. {x -> -1 + a, y -> -1 + b}] /. {a -> x + 1, b ...


1

Interesting is that a graph Plot3D[(-1 + x) Sqrt[(1 - x) (-1 + y^2)] + Sqrt[(1 - x)^3 (-1 + y^2)], {x, -1, 1}, {y, -1, 1}] is a non-homogenous plain


1

Altough I cannot tell you what goes wrong with Simplify I would suggest implementing the simplification semi-manually. For example, with te = (-1 + x) Sqrt[(1 - x) (-1 + y^2)] + Sqrt[(1 - x)^3 (-1 + y^2)] you could use te /. a_.*Sqrt[b_] :>Simplify[Sign[a], -1 < x < 1 && -1 < y < 1]*Sqrt[Expand[a^2 b]] This brings the summands in ...


3

For problems like these, I like to take small steps and check each one, so please bear with me. First, we write the two exponential functions, which I am calling f1 and f2. Then write the next two more complicated functions, called gSing for singlet and gTrip for triplet. Then write an expression for the denominator and an expression for the volume ...


3

I think you will have to ask a mathematician if the limit is really 0 (or even real) since in Mathematica you can get this s[x_] := Sqrt[ Pi^2/12 + Sum[(x!*x!)/(k!*(2 x - k)!)*(-1)^(x - k)/(x - k), {k, 0, x - 1}]] s[x] // FullSimplify Limit[s[x], x -> Infinity] // FullSimplify So the limit seems to be complex, but I am not sure, I am not a ...


3

Is this what you are seeking? FullSimplify[f, u > 1] (* 1/2 u (-3 + 4 u) Sqrt[2 + 1/u - u^2] *) Response to edit in question The LeafCount of this result is LeafCount[%] (* 24 *) On the other hand, the LeafCount of the outcome desired by the OP is g = 1/2 (-3 + 4 u) Sqrt[-u (u + 1) (u^2 - u - 1)] LeafCount[%] (* 27 *) Note also that ...


3

Read the documentation about Rationalize Now starting from the complex form you can build you own function nicef = Exp[Rationalize[Im[Chop[Log[#]]], 1/16] I] & nicef[-0.735145 + 0.67791 I] E^((12 I)/5) or if the input is $a$ then just Rationalize[a,1/16]



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