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lst = Table[{a, NIntegrate[(-(a^2/2) + x (-x + Sqrt[a^2 + x^2]))* Tanh[\[Pi]*x], {x, 0, \[Infinity]}]}, {a, 0.2, 1, 0.05}]; ListPlot[lst]


3

This is at least how I might start such a problem: First define a function that calculates the numerical integral (using your definition) from 0 to some number: res[a_?NumericQ, xmax_?NumericQ] := res[a, xmax] = NIntegrate[ x Tanh[Pi x] Sqrt[x^2 + a^2] - (a^2/2 + x^2) Tanh[\[Pi] x], {x, 0, xmax}, WorkingPrecision -> 50] You might notice ...


1

Assuming that you have decided to call the coordinates x, y, and z throughout, you could do this, building on the definition you already have (I shortened your momentum definition, and in addition, you could also use $\hbar$ instead of h by entering ESChbESC): r = {x, y, z}; p= -I h D[#,{{x,y,z}}]& L = Function[{f}, r\[Cross]p[f]]; L[f[x, y, z]] ...


2

Alternate answer, this is an exact analytic approach to the nearest point problem: (not i think precisely what @martin was after, but its an interesting problem and others may find it useful) lb = -1;ub = 1; pts0 = Select[Flatten[ Table[ {i, j}, {i, 2 lb, 2 ub , .2}, {j, 2 lb , 2 ub , .2}], 1] ,Norm[#] < 1 &]; intv[ p_, pn_] := If[(pn[[1]] != ...


4

Here you are with the bands -- note also an (I think) improvement over the brute force fine discretization of the line: (I'm Not sure if that improved performance, but it didn't hurt and it looks cleaner) caveat I think my little trick thinning down the lndat list is not guaranteed to find all of the strictly nearest points. It seems to work for the ...


10

So you guys know - quasicrystals are cool structures that can consist of finite number of parts which can be arranged in never repeating - aperiodic - pattern. Thing here is called projection method from a regular lattice. http://www.nature.com/nmat/journal/v3/n11/fig_tab/nmat1244_F3.html Interestingly if you know Fibonacci rabbits problem - that is also ...


2

You can try to improve it "by hand", as follows. Assume that the model is like this: model = (a*(Exp[c*t] - 1))/(1 + b*Exp[c*t]) and assume that you have in mind that your data correspond to {{1, intensity1}, {2, intensity2},...} (see my comment above) and assume that norm is the name of your data. Try the following: Clear[model, a, b, c]; model = ...


4

ff = FindFit[normalized, {fitEquation2}, {y0, a, k}, t] Show[ListLinePlot@normalized, Plot[fitEquation2 /. ff, {t, 0, 20}, Evaluated -> True]]


6

You can do this without a NDSolve by calculating the distance from the follower cranks joint to the end of the driving cranks end. Then use this distance with law of cosines to calculate the deviation angle. This is also pretty easy to implement on ANY hardware capable of doing a ArcCos and Atan2 operation (note that in c atan2 parameters are swapped). ...



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