Carl Love

## 27209 Reputation

11 years, 343 days
Himself
Wayland, Massachusetts, United States
My name was formerly Carl Devore.

## value...

There is no hope for solving this equation in Maple. I would guess that it far beyond the capability of any CAS.

It is not a differential equation (so dsolve doesn't apply), because it is evaluated at a particular point, z=0. It is not a regular equation (so solve doesn't apply), because you don't want to solve for a particular x

However, there are a few functions for which Maple can take an arbitrary-order derivative, although this facility is quite buggy. You can use this to test out a few simple cases for g(z).

`T:= Eval(Diff(g(z)^x, z\$(x-1)), z=0) = n*x*x!:#I moved x! to other side of equation.#Test g(z) = exp(z) (so g(z)^x = exp(z*x))value(subs(g(z)^x= exp(z*x), T));                   (x - 1)                                     x        = n x x!`

I am not sure yet (need to work with the worksheet), but I think that you need to use unapply on sigvm and then compose (@) the resulting function with a coordinate transformation function.

I'm having some trouble wrapping the 3D processor of my mind around your coordinate transformations: I think of the normal cylindrical coordinates as x = r*cos(theta), y= r*sin(theta), z=z. You seem to have some permutation of the x,y,z. (I am not saying that your transformations are wrong, just that they are difficult for me the visualize.)

## trace...

It's `dsolve/numeric/DAE/sm_intersect`, not `dsolve/dae/numeric/sm_intesect`. The word order and capitalization matter. You can view the procedure via showstat(`dsolve/numeric/DAE/sm_intersect`). It's fairly short, but I don't know what it does. You may gain a little insight into the situation by issuing the command

trace(`dsolve/numeric/DAE/sm_intersect`);

## diffeo-functional equation?...

Maple does not have a general facility for regular functional equations (AFAIK), let alone a diffeo-functional equation (which may be the coinage of a new term) like this. Although I wouldn't be surprised if the answer in this particular case could be found after a few guess trials with different g(z).

As to the variable with respect to (wrt) which the derivative is taken: It is probably wrt z for two reasons:

1. The text you cite says that x is a natural number.
2. If it were wrt x, then there would be no need for z at all; it could just as well be G^x for some constant G.

The proper way to enter such an expression into Maple is with Eval (inert form) or eval (active form) rather than subs:

1/x!*Eval(Diff(g(z)^x, z\$(x-1)), z=0) = n*x;

## x_n > 0?...

Are the x_n restricted to positive reals?

## power series independent of n also...

Combining Axel's idea with my power series idea given in another Answer, we see that the power series expressed with lower limit n is also independent of n, and is the same as if the lower limit were 0. However, oddly enough, though the two expressions are equal, they are not identical as expressions.

S1:= sum(x^k/(6*k+1)!, k= 0..infinity);

S2:= sum(x^k/(6*k+1)!, k= n..infinity);

simplify(S1-S2);
0

## power series independent of n also...

Combining Axel's idea with my power series idea given in another Answer, we see that the power series expressed with lower limit n is also independent of n, and is the same as if the lower limit were 0. However, oddly enough, though the two expressions are equal, they are not identical as expressions.

S1:= sum(x^k/(6*k+1)!, k= 0..infinity);

S2:= sum(x^k/(6*k+1)!, k= n..infinity);

simplify(S1-S2);
0

## ...but suppose we view it as a power ser...

But it seems that the series starting at k = 0 is viewed as a power series evaluated at 1:

sum(x^k/(6*k+1)!, k= 0..infinity):
eval(%, x=1);

## ...but suppose we view it as a power ser...

But it seems that the series starting at k = 0 is viewed as a power series evaluated at 1:

sum(x^k/(6*k+1)!, k= 0..infinity):
eval(%, x=1);

## Which regions? Which colours?...

It is not clear what you want.

• Let A be the region enclosed by r = 1.
• Let B be the region enclosed by r = 2*cos(theta)

Then, proceeding left to right, there are three disjoint regions:

• Let R1 = A minus B
• Let R2 = A intersect B
• Let R3 = B minus A

What color do you want R1? What color do you want R2? What color do you want R3? What color do you want the boundary of A? What color do you want the boundary of B? Do you want to see the gridlines underneath the regions?

## a_k(t) not uniquely determined...

@litun Consider a simpler example: z(t) = t, G = 1 + t + t^2. Then G = z^0 + z^1 + z^2, so a_0(t) = a_1(t) = a_2(t) = 1. Or G = (1+t)*z^0 + t*z^1, so a_0(t) = 1 + t, a_1(t) = t, a_2(t) = 0. And there are numerous other possibilities. So the a_k(t) are not uniquely determined.

## Stirling numbers?...

@tuGUTS Like I said, I don't know of any case where rsolve solves an equation with more than one independent variable; although I'm not absolutely sure that it will never solve them. Hopefully someone who knows more will comment on this.

This is just a seat-of-the-pants guess, but the answer to your equation may involve Stirling numbers.

## fill= NULL...

...it is automatically filled with zeros on the empty places.

Zero is the default, but the fill value can be anything, including NULL.

b:= Matrix([a||(1..4)], fill= NULL);

The advantage of NULL is that it is automatically removed when converted to a list.

convert(b^%T, listlist);

[[1, 5, 4, 2], [2, 3, 8, 2], [4, 4, 5], [5, 5, 4], [3, 7], [2, 8], [3], [4], [5]]

## fill= NULL...

...it is automatically filled with zeros on the empty places.

Zero is the default, but the fill value can be anything, including NULL.

b:= Matrix([a||(1..4)], fill= NULL);

The advantage of NULL is that it is automatically removed when converted to a list.

convert(b^%T, listlist);

[[1, 5, 4, 2], [2, 3, 8, 2], [4, 4, 5], [5, 5, 4], [3, 7], [2, 8], [3], [4], [5]]

## Commas and restrictive and nonrestrictiv...

@Christopher2222 wrote:

On interpretations of the same sentence.
He didn't marry her because she was rich - (he didn't marry her)
He didn't marry her because she was rich - (he did marry her)

When the subordinate clause ("because she was rich") does not change the meaning of the main clause ("he didn't marry her"), it needs to be set off with a comma.

He didn't marry her, because she was rich. (He didn't marry her.)
He didn't marry her because she was rich. (He did marry her.)

 First 642 643 644 645 646 647 648 Last Page 644 of 698
﻿