7

MinimalBy[z /. Simplify@First@Solve[Reduce[z==num/den^2, #], z] & /@ {x, y}, LeafCount] $\frac{\left(6-5 \sqrt{2}\right) x-12 \left(\sqrt{2}-2\right) y}{\left(23 \sqrt{2}-55\right) x-12 y}$ Another way Factor[num/den^2, Extension -> Automatic] With[{expr = (a x + b y)/(c x + d y)}, expr /. RootReduce@SolveAlways[% == expr, {x, y}]] // FullSimplify


6

Here is an indirect way to prove this. Applying DifferentialRootReduce[] to your res gives a tenth-order linear ODE: Short[ode = First[Head[DifferentialRootReduce[res, x]]][y, x][[1]]] (-8741760 - 1918960 x^3 + 143800 x^6 - 2560 x^9) y[x] + (-7160448 x + 2599248 x^4 - 337056 x^7 + 6144 x^10) y'[x] + (11531328 x^2 - 1639768 x^5 + 265156 x^8 - 4864 x^...


5

Let y be your expression. Here is the way to establish the desired fact: a = Assuming[x > 0, D[y, x] // FullSimplify] (* 1 *) b = Limit[y, x -> 1] (* 1 *) Y = DSolveValue[{z'[x] == a, z[1] == b}, z, x]; Y[x] (* x *) Of course, it is an indirect albeit mathematically rigorous way. Explanation Assume we would like to simplify a function $y(x)$.The ...


5

Change the complexity function so that it discourages log powers and log sums, which in turn converts $\log(a)+\log(b)$ into $\log(a b)$ and $\log(a)^2-\log(b)^2$ into $(\log(a)+\log(b))(\log(a)-\log(b))$: FullSimplify[E^((-Log[x]^2 + Log[1 + x]^2)/Log[1 + 1/x]) - x^2, Assumptions -> x > 0, ComplexityFunction -> Function[LeafCount[#] + 100 ...


4

As stated in the documentation for Simplify: Simplify tries expanding, factoring, and doing many other transformations on expressions, keeping track of the simplest form obtained. For some simplifications, such as yours, the required transformations actually increase complexity (as measured by Mathematica) before the cancellations can occur. This can be ...


3

$Version (* "12.2.0 for Mac OS X x86 (64-bit) (December 12, 2020)" *) Clear["Global`*"] expr1 = Cos[θ] ((Sqrt[-((g H)/(-1 + Sin[θ]))] Sqrt[-g H (-1 + Sin[θ])])/g - (H Sin[θ])/(-1 + Sin[θ])); expr2 = Cos[θ] (H - (H Sin[θ])/(-1 + Sin[θ])); The two expressions are not generally equal. To find instances where they are not ...


3

I would try FullSimplify[TrigToExp[yourfunction]] Now that you have been sufficiently flogged for not typing out your input... FullSimplify[ TrigToExp[ Sin[ArcCos[(Sqrt[2/(1 + Tan[\[Theta]/2]^4)] + Sqrt[2/( 1 + Tan[\[Theta]/2]^-4)])/2]]]] gives Sqrt[1/2 - Sqrt[Sin[\[Theta]]^4/(3 + Cos[2 \[Theta]])^2]] and FullSimplify[ Sin[ArcCos[(Sqrt[2/(1 + Tan[...


3

expr = Exp[(Log[1 + x]^2 - Log[x]^2)/Log[1 + 1/x]] - x^2; FullSimplify[expr, x > 0, ComplexityFunction -> StringLength@*ToString] FullSimplify[expr, TransformationFunctions -> {Automatic, PowerExpand[#, Assumptions -> x > 0] &}] x x


3

Probably the simplest approach is: Simplify[-(1/2) Df[x - x4] J[x] - 1/2 Df[-x + x4] J[x], Df[x - x4] == Df[-(x - x4)]] -Df[-x + x4] J[x]


2

Clear["Global`*"] delFunc[funcform_, lstofparams_, lstoferrs_] := √Total[ Table[(D[funcform, lstofparams[[i]]]*lstoferrs[[i]])^2, {i, Length[lstofparams]}]] funcanderr[funcform_, lstofparams_, lstofvals_] := {funcform, delFunc[funcform, lstofparams, #[[2]] & /@ lstofvals]} /. Table[lstofparams[[i]] -> lstofvals[[i, 1]], {i, ...


2

I'm not sure that your approach will give you much insight into how Refine works internally. Perhaps a better starting point is the internal implementation notes which state: When assumptions specify that variables are real, polynomial constraints are handled by cylindrical algebraic decomposition, while linear constraints are handled by the simplex ...


2

b /. First@ Solve[{f == x^2 + y^2, a == x + y, b == x y}, b, {x, y}] // InputForm (* (a^2 - f)/2 *)


2

By default FullSimplify and Simplify avoid large numbers through the default complexity function. Use LeafCount instead if you don’t mind the large numbers: FullSimplify[num/den^2, ComplexityFunction->LeafCount] (* (6432772386615749160*x + 4548621517377519337*Sqrt[2]*x - 42218157353358559728*y - 29852856498186887412*Sqrt[2]*y) / (...


1

You can get a step by step solution using WolframAlpha. For this you type == at the beginning of a cell. Then you type your expression and hit Return. In the result, there is a button labeled "Step-by-step solution". It looks like:


1

Try this: expr = Cos[\[Theta]] ((Sqrt[-((g H)/(-1 + Sin[\[Theta]]))] Sqrt[-g H (-1 + Sin[\[Theta]])])/ g - (H Sin[\[Theta]])/(-1 + Sin[\[Theta]])); Simplify[expr /. Sqrt[a_]*Sqrt[b_] :> Sqrt[a*b], {g > 0, H > 0}] (* (H Cos[\[Theta]])/(1 - Sin[\[Theta]]) *) Have fun!


1

Trying to implement your MyRefine through pattern matching alone is unlikely to work, as it would require explicitly considering a huge number of corner cases. Indeed, I very much doubt that Refine is based on this approach. It must have much more advanced internal "intelligence" similar to Simplify. I have no idea how that is implemented, but I am ...


1

First of all, both of your cases did not work as intended in my version of Mathematica (12.2). I'm not sure if it's what you're looking for, but here are two methods of defining an even function which is to be used in FullSimplify Using upvalues based on this answer: Df[x_] + Df[y_] /; x == -y ^:= 2 Df[x]; FullSimplify[-(1/2) Df[x - x4] J[x] - 1/2 Df[-x + x4]...


1

Here are a couple of heuristic approaches. One is to use the Wolfram Resource Function RadicalDenest. ResourceFunction["RadicalDenest"][1/(2^(1/2) - 2^(1/3))] (* Out[1443]= 1 + 1/2^(1/3) + 1/2^(1/6) + 2^(1/6) + 2^(1/3) + Sqrt[2] *) More roundabout is to use a variant of one of the methods internal to RadicalDenest. The idea is to find the ...


1

It looks like you're looking for the solution of an underdetermined equation system. eqn = {X == A1 S + N1, Y == A2 S + N1, Z == 2 A1 A2 S^2 + A1 N1 S + A2 N1 S + N1^2} sol=Solve[eqn, {A1, A2, S, N1} ]; N1/.sol (*{1/2 (X + Y - Sqrt[X^2 - 6 X Y + Y^2 + 4 Z]), 1/2 (X + Y + Sqrt[X^2 - 6 X Y + Y^2 + 4 Z])}*)


1

You could take a brute-force approach and provide your own transformation function to handle that case: Assuming[Element[n, Integers], Simplify[ Ceiling[n/2] + Floor[n/2] == n, TransformationFunctions -> {Automatic, ReplaceAll[Ceiling[a_] + Floor[a_] :> 2 a]} ] ] (* Out: True *)


1

Posting as an answer at the request of the OP There is probably no built-in way to achieve what you want because the result you are after is in general not correct: in order to get it you need to rewrite and reorder the summations, which is only possible under special convergence conditions. That being said, if you are sure beforehand that all your ...


1

This way it simlifies even more (or less) num/den^2 // Apart // FullSimplify // Factor // FullSimplify $$ \frac{7 \left(124944256776+87864579823 \sqrt{2}\right) x-12 \left(476402576068+337753093375 \sqrt{2}\right) y}{\left(691401+441484 \sqrt{2}\right) \left(\left(13669487+10004398 \sqrt{2}\right) x+12 \left(616179+439573 \sqrt{2}\right) y\right)} $$


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