Tag Info

Hot answers tagged

34

In the first case PowerExpand comes to the rescue: PowerExpand@Power[p^(n m) q, 1/n] (* Out: p^m q^(1/n) *) Note however that "the transformations made by PowerExpand are correct only if $c$ is an integer or $a$ and $b$ are positive real numbers". Generally speaking, your assumptions can be listed in Reduce, Simplify, or FullSimplify using the ...


15

I can't help you with functions beyond Reduce, Simplify, ... but I can offer you a tip to help with reducing errors from manual translation. In order to validate that your manual transformation is correct, one can subtract the original expression from the manual transformation and then use Simplify on that expression. Sometimes it will return zero using ...


13

One thing is to make sure you have all the assumptions stated properly. For instance, the first two cases can be handled by passing all the assumptions to FullSimplify FullSimplify[Power[p^(n m) q, 1/n], Assumptions -> {q > 0, p > 0, n ∈ Integers, m ∈ Integers}] p^m q^(1/n) and FullSimplify[(p^n q)/(p^m r), Assumptions -> {q > 0, p ...


12

I use Mathematica in much the same way as you, although in the context of multi-stage physics derivations. Since no one has mentioned it, I'll describe an obvious approach to successive rewrites of an expression. I label each step a calculation with an indexed symbol that I can refer to later. This is preferable to In's and Out's whose numbers can change if ...


8

I often find that Simplify, etc work better when applied to parts of an expression. As I mentioned in this post, using something like Collect[expression, {selectedvariables}, Simplify[#, options -> values]& ] can work well in this respect. More generally, if you wish to simplify a particular Part, try expression[[index]] = ...


5

From solution provided by kirma to your previous question Sum[24 n + 12 n^2, {n, 1, Infinity}, Regularization -> "Dirichlet"] (* -2 *) From solution provided by xzczd to your previous question ramanujanSum[f_] := Block[{x, n}, FullSimplify[-Sum[ BernoulliB[n, 1]/n SeriesCoefficient[ f[x], {x, 0, n - 1}], {n, \[Infinity]}], n >= ...


4

At first glance it seems hopeless - you have all those hyperbolic trig and regular trig functions in your exponent. But then you notice that that is all superfluous since the only two variables you care about, X2 and P2 don't go into any of those functions, so that other stuff is just a distraction. {X1,P1,r, φ, θ, η} are all just constants! What you are ...


3

Using the Zeta function you can calculate arbitrary expression like in your example. g = 24 Zeta [-1 ] + 12 Zeta [-2] (* -2 *)



Only top voted, non community-wiki answers of a minimum length are eligible