Robert Israel

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18 years, 208 days
University of British Columbia
Associate Professor Emeritus
North York, Ontario, Canada

MaplePrimes Activity


These are answers submitted by Robert Israel

If gfun[guessgf] can find the generating function, convert/FPS may be able to find an explicit formula for the coefficients.  For example:

> L := [6, 24,60, 120, 210, 336]:
   g:= gfun[guessgf](L,t, [ogf]);
   convert(g[1],FPS,t);

Sum((6+11*k+6*k^2+k^3)*t^k,k = 0 .. infinity)

> factor(%);

Sum((k+3)*(k+2)*(k+1)*t^k,k = 0 .. infinity)

You're missing a multiplication sign * (space allowed if you're using 2D math input) in xy(x), and I have no idea what your i^() is supposed to be.

 

Hints:

1) you need some multiplication signs (spaces allowed if you're using 2D math input).

2) on the other hand, sin is a function.  You want sin(something), not sin * something.

3) diff does differentiation.

You could try, for example,

piecewise(expr, 1, 0);

or

charfcn[{true}](is(expr));

 

It looks like you're giving the numeric version of pdsolve a boundary condition at infinity.  That's not possible: pdsolve,numeric is only for problems on a bounded region (which can be approximated using a finite grid). 

On the other hand, the symbolic version of pdsolve does not return a general solution, but only some particular solutions, typically obtained using separation of variables. 

 

 

v := a[1]*b[2]+a[2]*b[1];

expand(q/v)*v;

Here is a function that takes a list of real constants and returns the position of the least positive member (if any), or 0 if there is none.   

leastpos:= proc(L::list) 
   local P, j;
   P:= select(type,L,positive);
   if P = [] then 0
   else member(min(P), L, j); j
   end if
end proc;

For example:

> leastpos([0, -20, 12]);

                  3

See the .. ?  It looks to me like c1 is a range.  I don't know what you mean by "constant of three terms".  But you can't multiply a range by something.  That's what is likely to be producing the "invalid terms in product".

add is for adding a specific number of terms.  You want a limit as n -> infinity, which requires n to be a symbolic variable.  So you'd need to use sum rather than add.  This will not work with `if`, but you could try piecewise.
Still, I don't expect Maple will be able to find the limit.

You can also use the angle-bracket shortcut

< ($1..4) >;

(the parentheses are required in 1D Maple input, but not 2D math input)

Hint for the "paper" proof: what do you know about the relation between the cosines of those angles (which I assume are the angles between the vector and the coordinate axes) and the components of the vector?

As for Maple: you might use VectorAngle in the LinearAlgebra package.  You'll want to assume that A1, A2, A3 are real.

 

Just for fun, here it is done with the elementwise operator ~. 

> K := [2, 4, 6, 8];
   [fsolve]~(K*~x^5-~11*x^4-~7*x^3+~12*x^2-~4*x,x);


Note that I'm putting [] around the fsolve so that solutions for different values of K will be in separate lists.

Sure, we can help.  What part are you stuck on?  What have you tried so far? 

Do you know how to:
   1) write a procedure?
   2) make an Array (which is probably what you should use rather than "array")?
   3) make a loop?
   4) compute a Fourier approximation?

Does the condition sqrt(X^2 + Y^2 + Z^2) = Radius uniquely define theta3 as a function of theta1 and theta2?  If it does, you might try something like

P1:= proc(theta1, theta2)
      global theta3;
      local t;
      theta3:= fsolve(f(theta1,theta2,t)^2 + g(theta1,theta2,t)^2 + h(theta1,theta2,t)^2 = Radius^2, t);
     P2(theta1,theta2):= g(theta1,theta2,theta3);
     P3(theta1,theta2):= h(theta1,theta2,theta3);
     f(theta1,theta2,theta3)
end proc;

plot3d([P1,P2,P3], 0 .. Pi, 0 .. Pi);

First of all, do you really mean an array? 

Your A is actually a list of lists.  If that's what you want, you can use select:

> select(t -> (t[3] < 3), A);

If  you really mean an Array, it's a bit more complicated.

> B:= Array(A):
 
   B[ select(j -> (B[j,3] < 3), [ $ ArrayDims(B)[1] ]), .. ];

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