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8

Is there a way to get coordinate of just a particular point? You can convert Line into the corresponding set of Points each of which will be a Button which Prints the coordinates of that Point when you click on it (try this!): plot = BodePlot[1000/((1 + s/10^3)*(1 + s/10^6))]; plot /. Line[pts_] :> Map[Button[Point[#], Print[#]] &, pts, {-2}] ...


6

Below is an extended comment showing the effects that can happen when one ignores the correlations among parameter estimators (either on purpose or because journal authors or editors don't know enough about statistics to include necessary information). First I create some data with the same curve form as the function CDDHybrid and a set of predictor values ...


5

Introduction The first thing that I noticed is that the functional form for all of your CDD and LHS functions are identical. I propose that rather than defining six functions for each, we define one function and change the input arguments to cover the various cases. cdd[s_, a_, b_, c_] := 10^(a + b*c - b*s)*Exp[-10^(s - c)] lhs[s_?NumericQ, a_?NumericQ, ...


5

This does what I think you're after, fiddle with options as desired: With[{a = Interval[1.01 + .18 {-1, 1}], b = Interval[.92 + .11 {-1, 1}], c = Interval[2.2 + .2 {-1, 1}]}, Plot[{Min[a + b*x + c*x^2], 1.01 + .92 x + 2.2 x^2, Max[a + b*x + c*x^2]}, {x, -5, 5}, Filling -> {1 -> {3}}, FillingStyle -> Darker, PlotStyle -> {None, Red, ...


4

Could use Filling -> {4 -> {{1}, LightRed}, 1 -> {{3}, LightBlue}, 3 -> {{6}, LightRed}} and omit FillingStyle. Or Filling -> {1 -> {3}, 4 -> {6}}, FillingStyle -> {{LightBlue, Opacity[0.5]}, {LightRed, Opacity[0.5]}} which respectively produce the plots below.


4

bp[s_] := 1000/((1 + s/10^3)*(1 + s/10^6)) mybp = BodePlot[bp[s]] Have a look at Short Cases Short[Cases[Normal@mybp, Line[s_] :> s, Infinity], 25] ...


3

According to this MathGroup post the function SpaceForm was documented only via Information (i.e. the SpaceForm::usage Message) even in Mathematica 3.0. With current version 10.4.1 the situation is still the same: ? SpaceForm SpaceForm[n] prints as n spaces. So you shouldn't worry: this function is in the current situation right from the start, ...


3

Using Cases[] plot = BodePlot[1000/((1 + s/10^3)*(1 + s/10^6))] Cases[plot, Line[x___] :> x, Infinity] (*two group of data*)


2

To the best of my knowledge, the answer to your question is simply: no. Meta comments: I don't think your question ought to be closed as "out of scope", since you are not actually asking for anyone to write the function for you. My suggestion would be to write the function yourself and edit the question to include your attempt, making the focus of the ...


2

f = OpenWrite["test.txt"]; nsp[n_] := OutputForm[StringJoin[ConstantArray[" ", n]]] Write[f, 1, nsp[3], 2, nsp[1], 3]; Close[f] FilePrint["test.txt"] 1 2 3


2

Just for reference, the sum depicted in the OP has a representation in terms of the $q$-hypergeometric function. In particular, $$\sum_{n=0}^{\infty}\frac{(-1)^nq^{6n^2}}{(q^3;q^3)_n(-q;q)_{3n}}={}_1\phi_4\left({{0}\atop{-q,-q^2,-q^3,0}}\mid q^3,-q^6\right)$$ Compare: Series[QHypergeometricPFQ[{0}, {-q, -q^2, -q^3, 0}, q^3, -q^6], {q, 0, 15}] 1 - q^6 + ...


2

I learned a lot from WReach's and Leonid's answers and I'd like to make a small contribution: It seems worth emphasizing that the primary intention of the list-valued second argument of Flatten is merely to flatten certain levels of lists (as WReach mentions in his List Flattening section). Using Flatten as a ragged Transpose seems like a side-effect of ...


1

Activate@Replace[Inactive[f][3],3->1, Infinity] 1 or Replace[Hold@f[3], 3 -> 1, Infinity] // ReleaseHold 1



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