dazedANDconfused's profile - activity

There are several ways to do it and you'll have to decide on which suits you best. First method: Direct use of sympy. Th

With respect to sequences, look into the Lagrange (interpolation) polynomial for a set points which is here in the docum

With respect to sequences, look into the Lagrange (interpolation) polynomial for a set points which is here in the docum

With respect to sequences, look into the Lagrange (interpolation) polynomial for a set points which is here in the docum

Thanks, I didn't notice that. I use circular layout when I am not happy with positioning. I've edited my answer to show

I use 3 useful tweaks in making graphs: 1. Adjust the vertex size, 2. Adjust the distance between vertices, 3. arrange t

I'm not sure what you mean by overlapping vertices. The issue I see is that the labels are too big for the vertices. Wit

Good idea! I revised code to do that.

You haven't made it clear how big integers can be and still be small. The following code can easily be modified as neede

You haven't made it clear how big integers can be and still be small. The following code can easily be modified as neede

You haven't made it clear how big integers can be and still be small. The following code can easily be modified as neede

Nice, +1: I especially like the one using trig functions.

Yes, I was going to add another while loop which kept adding 2 for values <-1 if I needed to graph the negative porti

I want to plot a function that is abs(x) for -1<= x <= 1 and repeats to create a sawtooth shaped function. With the code

def h(x):
    while x>1:
        x=x-2
    return abs(x)
h(4.3)

The Sage sell server tells me the value is 0.300000000000000, which is what I want as h(4.3)=h(2.3)=h(.3)=abs(.3) However, when I tried to plot h(x) with

def h(x):
    while x>1:
        x=x-2
    return abs(x)
plot(h(x),x,0,5)

The plot looks like abs(x), so that h(4.3) now appears to be 4.3. What has gone wrong? How can I properly plot the h(x) I want?

EDIT: A points plot gives me the output I am expecting, which is different than plot(h(x),x,0,5).

def h(x):
    while x>1:
        x=x-2
    return abs(x)

 points([(x, h(x)) for x in srange(0,5,.01)], pointsize=20)

Good explanation! It clears up some misunderstandings I had.

Plotting a periodic function I want to plot a function that is abs(x) for -1<= x <= 1 and repeats to create a sawt

Plotting a periodic function I want to plot a function that is abs(x) for -1<= x <= 1 and repeats to create a sawt

I'd like to get the list of points that make up a sage plot so I can process it through LaTeX (eg pgfplots) and get a better looking plot. This post here is essentially what I'm looking. I didn't really understand the code but came up with this:

x,y=var('x,y')
f(x,y)=x^2-y^2
p = implicit_plot(f(x,y)==3,(x,-3,3),(y,-3,3),plot_points=300)
P = p.matplotlib()
R = P.get_children()[1]
S = R.collections[0]
r = S.get_paths()[0]
v = r.vertices
xvals = v[:,0]
yvals = v[:,1]

Change the function to f(x,y)=x^2+y^2 and the code works properly. But the example above only gives me the list of points for the left half of the hyperbola. How can I get the list of points when the graph is disconnected? Can that be extended to get the list of points if there are multiple graphs in the same plot?

LaTeX make results look beautiful, Sage handles the math, and sagetex provides a way of using Sage en route to producing a graph like that. The quality of the image you have given is not possible in a Sage Cell Server. You said, "I do not understand how to set the vertex attributes, the size of the nodes, and so on.". That is covered in the documentation here. This page will show you the sort of output you should expect with just Sage. At the top of that page it gives you a link for LaTeX drawing of graphs. That page gives you output of LaTeX code. The output can be realized through LaTeX using sagetex which I linked to in my earlier comment. A Sage Cell server won't do this.

LaTeX is used to make results look beautiful, Sage handles the math, and sagetex provides a way of using Sage en route to producing a graph like that. The quality of the image you have given is not possible in a Sage Cell Server. You said, "I do not understand how to set the vertex attributes, the size of the nodes, and so on.". That is covered in the documentation here. This page will show you the sort of output you should expect with just Sage. At the top of that page it gives you a link for LaTeX drawing of graphs. That page gives you output of LaTeX code. The output can be realized through LaTeX using sagetex which I linked to in my earlier comment. A Sage Cell server will not give you the LaTeX image resulting from the code.

If you're interested in producing professional looking graphs then LaTeX is the way to go using the packages tkz-graph and tkz-berge. These can work with Sage via the sagetex package. You can see some examples here or here or, for digraphs here. In these cases, the edges vertex sizes, labels, etc are set through LaTeX. The cooperation between Sage and LaTex is not "incorporated completely" but does a decent job.

There's always Cocalc as well if you have an internet connection.

Those two roots are both approximations of 0. If possible, you should always seek to solve theoretically and numerically. If the answers agree then you feel confident everything was done correctly. If the answers disagree then you have to look deeper. The roots of the function are when it is equal to 0, so you need to solve the equation (x^2)*cos(2*x)=0. That means either x=0 or cos(2*x)=0. Now cos(x)=0 at +/- (2k+1)pi/2 so cos(2x)=0 at +/- (2k+1)pi/4. Since we're looking for roots over [-10,10], this means our roots are 0, +/- pi/4, 3pi/4,....11pi/4. Your answer +/- 0.7853981633974483, 2.356194490192345, 3.9269908169872405, 5.497787143782152, 7.0685834705770345, 8.639379797371932 are the roots +/- pi/4, 3pi/4,....11pi/4. What's missing is 0. In your search for roots over [-.1,0] and [0,.1] you've found -7.755114791616843e-09 and 7.755114791616843e-09.

You mention inflection points but usually we calculate inflection points by taking the second derivative but from your answers I see you are setting the first derivative equal to 0. Where the derivative is 0 (or undefined) are usually called the critical values. If you rewrite the second answer as x=cot(2x) and think about the graphs of cot(2x) and x and where the intersect you'll get an idea of the number of answers. Try plotting

A = plot(cot(2*x),-10,10,ymax=10, ymin=-10, detect_poles=True)
B = plot(x,-10,10,ymax=10, ymin=-10,color='red')
(A+B).show()

Which looks like:

There is no way to solve that equation analytically. However, since you know where, approximately, the roots are you can use find_root , documentation here to find them. Your first positive solution would be found by, for example, find_root(cot(2*x)-x,.01,.8) to get 0.5384369931559019. What should become apparent is that learning mathematical theory can help you to use SAGE like an expert.

Following the answers, both plot(sgn(x)*abs(x)^(1/3),(x,-10,10)), which uses sgn once and the more natural plot(lambda x: RR(x).nth_root(3), (-10, 10)) work. The RR is telling Sage that real roots are expected. See documentation here.

This question has been covered before, see here. Remember your order of operation: exponentiation doesn't apply to the negative. You've taken a positive cube root and then made it negative. If you try (-10)^(1/3).n() then you get 1.07721734501594 + 1.86579517236206*I.

Most equations have no clean solution and must be evaluated with numerical approximations. This is one of them. Even polynomials can cause problems.

I fiddled around and got it to work by changing the legend_label:

var("x_1","x_2")
eq1=solve(3*x_1+4*x_2==14,x_2)
eq2=solve(5*x_1+6*x_2==-8,x_2)
eq3=solve(4*x_1-7*x_2==18,x_2)
eq4=solve(2*x_1-3*x_2==-8,x_2)
b=10
p=plot([eq1[0].rhs(),eq2[0].rhs(),eq3[0].rhs(),eq4[0].rhs()], (x_1, 0,b),color=["#c72b91","#2b37c7", "#cb7b2b","#2bcb4b"],axes_labels=["$x_1$","$x_2$"],legend_label=[r"$%s$"%latex(eq1[0].rhs()),r"$%s$"%latex(eq2[0].rhs()),r"$%s$"%latex(eq3[0].rhs()),r"$%s$"%latex(eq4[0].rhs())])
show(p)

This is the result:

Since there are several equations in the legend, I use a list. Looking at that part of the code, r"$%s$"%latex(eq1[0].rhs()) accomplishes the task by first getting the latex code for eq1[0].rhs() with latex(eq1[0].rhs()) followed by inserting that in for the string r"$%s$. The r is for a raw string and since its a formula, LaTeX needs it between dollar signs.

It seems a bit awkward, maybe others have a more efficient way.

I've emailed Cocalc support for help. Should I delete this question?