Carl Love

Carl Love

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12 years, 361 days
Himself
Wayland, Massachusetts, United States
My name was formerly Carl Devore.

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These are replies submitted by Carl Love

@J4James 

If I change the x and y ranges as you have done, I also get an empty plot. I don't know why that is. Perhaps the initial point needs to be within the ranges? If you simply want to crop the plot, use your original ranges and include the option view= [5.2..7.2, 0.2..1.6].

Do you have any Maple code to accompany your Question?

@Jahani_21 

Is this simple enough?

Sin:= proc(x,n)
local k;
     Sum((-1)^k*x^(2*k+1)/(2*k+1)!, k= 0..n-1)
end proc:

@acer 

If you or the OP don't like the summation symbol appearing in gray when using Sum, then you could simply set

`print/Sum`:= ()-> 'sum'(args);

After that, all summation signs should appear the same.

The eigenvalues are purely imaginary and the equilibrium point is a center. This is clearly shown by your phase plot.

@Chia I'm not sure that I understand your question. Are you saying that you want to restrict the solutions to being multiples of 10? Then just use e = 10.

The technique for the second problem is essentially the same as that for the first. The plotting is a little more sophisticated because there are two parameters.

 

restart:

g:= b*x^2+2*a*x*y+b*y^2+4*x*z-2*a^2*y*z+2*b*z^2:
V:= [x,y,z]:

G:= VectorCalculus:-Gradient(g, V):

solve(convert(G,list), V);

[[x = 0, y = 0, z = 0]]

(1)

So (0,0,0) is a critical point regardless of a or b.

H:= VectorCalculus:-Hessian(g, V);

H := Matrix(3, 3, {(1, 1) = 2*b, (1, 2) = 2*a, (1, 3) = 4, (2, 1) = 2*a, (2, 2) = 2*b, (2, 3) = -2*a^2, (3, 1) = 4, (3, 2) = -2*a^2, (3, 3) = 4*b})

(2)

The critical point will be a relative minimum if all three eigenvalues of the Hessian are positive.

E:= LinearAlgebra:-Eigenvalues(H):

Any region that remains white in the following plot is a region of the ab-plane where all three eigenvalues are positive.

plots:-implicitplot(
     convert(E <=~ 0, list), a= -10..10, b= -10..10,
     gridrefine=3, filled
);

 

I didn't show the plot for it, but the symmetrically corresponding region below the horizontal axis is the region for which (0,0,0) is a relative maximum.

Download 2nd_derivative_test_2.mw

No. It's a big problem that's been discussed here several times. No workaround is known.

The complexity of the answer depends heavily on n. Try the following for several small integer values of n, including zero and some negatives (n = -1 is most interesting).

rsolve({f(c)=(2*n-c)*f(c-1)-(c-1)*n*f(c-2), f(0)=1, f(1)=2*n-1}, f(c));

@sazroberson 

The GRID package, when used on a single computer, is a part of regular Maple, at no extra cost. The GRID package can also be for distributed processing, across multiple computers. To use that aspect requires an additional purchase.

@maple2015 

To implement Markiyan's suggestion in the way that you expect it to work, you need to reassign the value of U:

U:= map(rhs, U);

After this, U[2,1] will return the matrix entries without the u(t) =, as you expect.


I see in a comment in your code a reference to to the command evalm. Do not ever use this command under any circumstances. Matrix multiplication is simply phi.U;

 

@acer 

If you explicitly specify gridlines= false in the plot command, then the gridlines won't appear when the worksheet is uploaded to MaplePrimes.

Great work on the baseball problem!

@Axel Vogt 

Every rank 1 matrix can be generated as x.y^%T for nonzero vectors x and y. Since there are 15 nonzero 4-vectors over Z[2], there are at most 15^2 = 225 rank 1 matrices. That still doesn't answer for me why Kitonum's code works, but his 225 seems plausible.

@GPY 

Yes, and make alpha a parameter to dsolve. And I hope that you'd only be doing that for fixed values of v0.

@Axel Vogt What's the point? The answer is unique, so interpolation gives the same thing.

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