Skip to main content

equation solving - Finding and plotting roots with parameter


I have the following function:



u(x,t)=k=1v0Lλkcsin(λk)sin(λkxL)sin(λkctL)


and this equation whose solutions are the λk needed for the function above:


ελ=cot(λ)


How can I feed the solutions of this (not analytically solvable) equation into my function? If possible, I'd like to create a function u(x,t,ε) to create 3D plots. What I have found so far:


Clear[findRoots]
Options[findRoots] = Options[Reduce];
findRoots[gl_Equal, {x_, von_, bis_},
prec : (_Integer?Positive | MachinePrecision | Infinity) :
MachinePrecision, wrap_: Identity, opts : OptionsPattern[]] :=
Module[{work, glp, vonp,

bisp}, {glp, vonp, bisp} = {gl, von, bis} /.
r_Real :> SetPrecision[r, prec];
work = wrap@Reduce[{glp, vonp <= x <= bisp}, opts];
work = {ToRules[work]};
If[prec === Infinity, work, N[work, prec]]];

uk = (v0 L)/(lambdak c Sin[lambdak]) Sin[lambdak x/L] Sin[lambdak c t/L];

This function finds the roots in an interval manually defined since FindRoots or NSolve only finds one root. Removed the ε for a test:


lambdalist = x /. findroots[x == Cot[x], {x, 1, 100}];


This gives me a list like this to use:


{0.860334, 3.42562...}

Which works with my function


u[x_,t_,L_,c_,v0_] = Sum[uk, {lambdak, lambdalist}];

My question is:


Is it possible to add ε as a parameter or do I have to define it beforehand to find the values?




Comments

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]