# Tag Info

7

The main question here is, there are too many approaches to perform the same operation. And normally, I didn't know which approach is the most optimal way in terms of efficiency. Mathematica's performance is hard to predict, even more so than that of other high-level languages. There is no simple guideline you can follow. There will always be surprises and ...

6

MichaelE2 already has answered the question in a comment: To use Method -> "ImplicitRungeKutta", differentiate the second equation and add a corresponding boundary condition for y. However, the OP expressed the concern that doing so might produce an inaccurate answer. Out of curiosity, I tried it. So, the following actually is an extended ...

5

FindMinimum seems to automatically choose the midpoint for a second starting point to specify the search direction (see tutorial). But the midpoint differs from the specified starting point by one bit. Therefore 55.00000000000001 == 55. is true since the default Internal`\$EqualTolerance (2.10721 = Log10[2.^7]) specifies that two machine reals are equal if ...

4

Simplify gives the correct result. Simplify[Sin[(4 π^2)/3] + Sin[2/3 π (-(3/2) + 2 π)]] (* 0 *)

4

The source of the NIntegrate error messages can be seen from a factor of the integrand, x^d/Sqrt[1-c x^d z^d], of toroot. For c > z^-3, the integrand is singular for some point in the domain, {x, 0, 1}. Moreover, if NIntegrate could integrate through the singularity (and, with help, it can), the result would be a complex number, which (presumably) is ...

4

The solution for any number of steps (for instance, 5) is With[{a = 1, τ = 3, m = 4}, tm = (m + 1) τ; s = NDSolveValue[{f'[t] == -a*Sum[f[t - n τ] UnitStep[t - n τ], {n, 0, m}], f[t /; t <= 0] == 1}, f, {t, 0, tm}]; Plot[s[t], {t, 0, tm}, ImageSize -> Large, AxesLabel -> {t, f}, LabelStyle -> {15, Bold, Black}]] This ...

3

Your sol1 is a list containing the desired function of t. Either extract the first part of that list, or use NDSolveValue with f[t] in the second argument to return said function of t directly. The "non-numerical value for a derivative at t == 3" message arises from HeavisideTheta not having a numerical value. Use UnitStep instead. Altogether: ...

3

Method 1: By using theorem 4 on page 219 of this book, we can easily obtain the convergence rates of these two iterative methods: φ1[x_] := Power[x + 1, (3)^-1] (D[φ1[x], x] /. x -> SuperStar[x]) // FullSimplify[#, (SuperStar[x])^3 - (SuperStar[x]) - 1 == 0] & φ2[x_] := (2 (x)^3 + 1)/(3 (x)^2 - 1) (D[φ2[x], x] /. x -> SuperStar[x]) // ...

3

This is an interpolation problem due to the course grid in h. For simplicity, label the solution s instead of testmodel001. Then, the grid in h is s["Coordinates"][] (* {100., 106.25, 112.5, 118.75, 125., 131.25, 137.5, 143.75, 150., 156.25, 162.5, 168.75, 175., 181.25, 187.5, 193.75, 200., 206.25, 212.5, 218.75, 225., 231.25, 237.5, ...

2

In the calculation of the Precision[] of 2 a - a, the error in 2 a and a are treated as independent. Therefore the error bounds, 2 Δa and Δa, are added. Thus the new error bound is estimated as 3 Δa, and the Precision[] is reduced by Log10 each iteration. Therefore we run out of Precision[] in Precision[a]/Log10 steps: a = 1.11111111111111111111; ...

1

Well It took a few minutes but I was able to run N[Log,6*10^6] on my laptop (Intel Core i7-8750H @ 2.2 GHz) in 330 seconds. Looking into the problem I found the OEIS sequence for the digits of Log(10) https://oeis.org/A002392 and in the references of this page the current world record of 1,200,000,000,100 digits: https://ehfd.github.io/world-record/...

1

Memoization is the way to go here. Your problem is that when Mathematica sees wp, it will rewrite it using your rule for wp[n_], and then it will try to evaluate wp, wp, and wp. But each of those evaluations depend on, e.g., wp, and so Mathematica will end up evaluating wp three times. This ends up exploding combinatorially: ...

1

I'm not sure what you are trying to model here, but there's not enough initial and boundary conditions for the system to be solved. I tried to just add a few, this works but I'm not sure exactly which initial and boundary conditions you need for your problem (*your eqs and conditions*) INC4 = Derivative[1, 0][T][0, τ]; INC5 = Derivative[2, 0][u][0, τ]; INC6 =...

1

using my myRootSearch and myRootSearchShow ClearAll[f]; f[x_, a_] := BesselJ[1, x]^2 + BesselK[1, x]^2 - Sin@Sin[a*x] Manipulate[ Column[{Quiet@myRootSearchShow[f[#, a] &, 0, 50, ImageSize -> 250], Quiet@myRootSearch[f[#, a] &, 0, 50]}], {a, 0, 4}]

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