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2

My solution is in the experimental mathematics style: Series[int - intpaper, {τ, \[Infinity], 12}] // Normal // FullSimplify (* 0 *) You have to believe that if two series expansions are equal up to 12th order this is an identity. Advantage of the approach is its unbeatable simplicity. Of course, sceptics with faster computers can verify even higher ...


3

This solution gets a trivial step away from the answer. (Strikeout after addressing the comments.) You can consider the difference diff = int - intpaper, and check that it vanishes. Rather than having Mathematica take the imaginary part, do it "by hand": intpaperz = 2/Sqrt[-1 - (Sqrt[-1 + Sqrt[3]] l - 2 I \[Tau])^2/(3 + Sqrt[3])]; intpaperzc = 2/Sqrt[-1 ...


0

NIntegrate[expr,{\[Theta],0,2*Pi}]// Timing {0.008, 0.0151049} When you integrating it within a limit, it will always give you a number, not a function. Now, say you are interested in finding the integration as a function of the upper limit. In that case you can get an InterpolatingFunction which you can use in any further calculation. data = Table[{...


5

You can do something like this: Simplify[Sqrt[x^2]] (* Sqrt[x^2] *) $Assumptions = _ ∈ Reals (* _ ∈ Reals *) Simplify[Sqrt[x^2]] (* Abs[x] *) This tells those functions that have an Assumptions option that any expression is considered real. Caveat: This refers to any expression, not just any variable! So you get this now: Simplify[Sqrt[x] ∈ Reals] (* ...


2

To some extent (and with some care) this can be done with FeynCalc. At least I used it several times when I needed to compute gradients and divergences of Cartesian vectors. The trick is to work with D-dimensional 4-vectors and take the limit $D \to 3$ at the end. Since FeynCalc doesn't distinguish between upper and lower indices, the results are the same as ...


6

Everything you want in the question can be done by defining the derivative of the Norm: Derivative[1][Norm][z_] := z/Norm[z] D[Norm[x - y], {x}] (* ==> (x - y)/Norm[x - y] *) Simplify[D[Norm[x - y], {y}]] (* ==> (-x + y)/Norm[x - y] *) Here, the syntax I used for the derivatives is such that it would remain valid if x or y were replaced by vectors ...



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