Joe Riel

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These are answers submitted by Joe Riel

The frontend command can be wielded with nice effect here,
frontend(LinearAlgebra:-Eigenvectors, [A(p)], [{Matrix},{}]);

                               [a2(p)]  [0    1]
                               [     ], [      ]
                               [a1(p)]  [1    0]
If the jpeg is not a color image, then the resulting image can be converted to a Matrix (a color image cannot be converted to a Matrix because it requires at least 3 layers). One way is to use the Create command in ImageTools:
img := Read("your_picture.jpg"):
img := Create(img,format=Matrix,copy=true);
It is necessary to specify the format and copy options. Were you planning on directly multiplying the image and the filter matrix? There is a Convolution procedure that convolves an image with an rtable (Matrix or Array). You can use the View command (in ImageTools) to view the result. Here is a simple example:
img := Checkerboard():
View([img,Convolution(img, <<1|0|0|1>>)]);
That displays a side-by-side view of the original image and the convolved image.
Maple's invlaplace function does not know how to compute the inverse laplace of exp(-td*s). We can, however, use the addtable function to add an entry to the transform table. Alas, this doesn't work as well as it could. That is, we cannot replace Maple's partial knowledge about the inverse laplace of exp. Instead, use a new function name, say, Exp to represent the exponential. Then we can do
addtable(invlaplace, Exp(-k*s), Dirac(t-k), s, t, [k]):
H1 := s -> 1/s:
D1 := s -> (Exp(-td*s) - Exp(-s))/s:
T1 := s- > D1(s)*(H1(s)):
    (-Heaviside(-t + td) + Heaviside(td)) (t - td) - Heaviside(t - 1) (t - 1)
What you want to do is to combine the left and right sides of the equations into a single polynomial in terms of u and v, extract all the coefficients, equate them to zero, and solve for the parameters. This can be done in one line:
solve({coeffs((lhs-rhs)(eq1), {u,v})});
             {a3 = 0, a6 = 0, a4 = -3, a1 = 5, a2 = -5, a5 = 7}
However, more instructive is to do this one step at a time:
p := (lhs-rhs)(eq1);
        2                             2                                     2      
    a1 u  + (a4 + 3 + 2 a6) u v + a3 v  + (a1 + a2 + a6) u + a5 v + a6 - 5 u  - 7 v

p := collect(p, {u,v});
               2                                              2                  
    (-5 + a1) u  + ((a4 + 3 + 2 a6) v + a1 + a2 + a6) u + a3 v  + (a5 - 7) v + a6

eqs := {coeffs(p, {u,v})};
           {a1 + a2 + a6, a4 + 3 + 2 a6, a3, a6, -5 + a1, a5 - 7}


             {a3 = 0, a6 = 0, a4 = -3, a1 = 5, a2 = -5, a5 = 7}
Note the use of collect, which was omitted in the one-liner. While this isn't necessary for this example, in general it is required for coeffs to correctly compute the coefficients of the polynomial. A one argument call to solve was used because all the indeterminates of the set of expressions passed to solve were the unknowns to solve for.
simplify(Ans,symbolic), though possibly a bit heavy-handed (see ?simplify,details), works here to cancel the desired terms (as well as combine the 1/x and x terms outside the Sum).
The problem is that Maple is attempting to solve a polynomial with 365 terms. A better approach would be to solve this symbolically:
eq := (1 + j/n)^m = val:
sol := solve(eq, j);
                      sol := exp(ln(val)/m)*n-n
eval(sol, [n = 365, m = 365, val=1.1]);
True, but that only works if the scaling factor is a nonzero numeric.
The easy way to handle this is to input the expression as a piecewise expression then use convert/Heaviside to convert it to an expression with unit steps (Heaviside functions). Finally use inttrans[laplace] to find its Laplace transform:
f1 := piecewise(t <= 1, 0, t<=4, cos(Pi*t), 0):
f2 := convert(f1,Heaviside);
              f2 := cos(Pi t) Heaviside(-1 + t) - cos(Pi t) Heaviside(-4 + t)
inttrans[laplace](f2, t, s);
To input a less-than sign (<), use the html construct &lt; (the semicolon is necessary).
There shouldn't be any problem pasting multiple lines, the real issue, I suspect, is that Maple won't accept the 2D output as input. A simple way to workaround that is to use lprint (or set interface(prettyprint=1)). The output of lprint is in 1D notation, which can be pasted into an input line. If you do not want any line breaks in the output, call interface(screenwidth=infinity) before issuing lprint. You don't normally want screenwidth = infinity because it breaks Maple's normal output (which trys to center on the screenwidth). That isn't an issue with lprint, since it prints flush left. You can write a simple procedure to temporarily reassign screenwidth and print the expression in 1D format, flush left: Lprint := proc() local width; width := interface('screenwidth' = infinity); lprint(args); interface('screenwidth' = width); NULL end proc:
This is a classic case of premature evaluation. In the plot3d statement, the expression Shape(x,y) is evaluated as is, that is, with the symbolic constants x and y as arguments to Shape. Because their values are themselves, Maple cannot evaluate the conditional. There are a few ways to handle this. The simplest might be to just pass the procedure Shape to plot3d: plot3d(Shape, -100..100, -100); Note that the x and y ranges are passed as ranges, not as equations. Another possibility is to modify Shape to return an unevaluated function call to Shape when x and y are not numeric: Shape := proc(x,y) if x::numeric and y::numeric then `if`(x^2 + y^2 < (Diameter/2)^2, 1, 0); else 'Shape'(args); end if; end proc: A somewhat more robust approach would also check whether Diameter is numeric.
I'm unclear on what you are asking. Do you want the unevaluated function call myproc(x,y) to display as `x whatever y'? You can more or less accomplish that by creating a `print/myproc` procedure and using a neutral operator (see ?print and ?neutral for details).
`print/myproc` := proc(a,b) a &whatever b end proc:
                  a &whatever b
Try using combine/power: y := 12/b/(b^s)*s/sin(Pi*b)^2*Pi^2; combine(y, power); 12/b^(s+1)*s/sin(Pi*b)^2*Pi^2

In html input mode, use &lt; and &gt; to input < and >, respectively. For an ampersand, use &

Just add the independent variables as arguments to a function call: y1(u1,u2), y2(u1,u2). A slightly nicer way to do this is to use the alias command (see ?alias for details): alias(y1=y1(u1,u2), y2=y2(u1,u2) ): J := (y1-y10)^2+(y2-Y20)^2+k*(u1)^2+k*(u2)^2: diff(J,u1); 2*(y1-y10)*(diff(y1, u1))+2*(y2-Y20)*(diff(y2, u1))+2*k*u1
Use the frontend command; it freezes various subexpressions.
z := a*f(x)+b*g(y)+c*x+d*y:
frontend(diff, [z,f(x)]);
frontend(diff, [z, x]);
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