vv

13922 Reputation

20 Badges

10 years, 9 days

MaplePrimes Activity


These are replies submitted by vv

@Ugurgozutok 

I don't quite understand the style of your questions. It is neither mathematics (no context, no proper notations etc) nor Maple (not a single line of code). For example, when computing gcd (or need irreducibility as in your other thread) the exact ring for your polynomials is essential.

@Ugurgozutok 

As I have mentioned, the problem is delicate in general and not fully stated (you gave 0 details). The linearity in d in my example is not important, but d in Z is essential. If your polynomial p is in R[x,y] and the degree is just 20, the chances for a solution are about 0% (even if p is linear in d).

@Carl Love 

It will be difficult to shrink it further. Vote up.

@nm 

Yes, I know that Mathematica knows better than the user what the user wants. But you must admit that a rule is necessary in order to choose correctly. I hope that the rule is other than "accept what mma does".

It don't see a simple and general solution here. Your line cannot be parsed in Maple due to embedded double quotes (without "\").
There is an inherent ambiguity, without "rules". For example, the line:

1,2,3,"x+y, algorithm=[",123,"]","OK",5

can be interpreted as having 6 or 8 fields.

For a solution (assuming the rules are made clear)  a dedicated parser must be programmed. Or, regular expressions.

If the user has an international keyboard, the foreign characters are obtained without problem (they use the UTF-8 encoding)  for symbols and strings.

In Windows, an easy way to obtain the missing characters is using Windows'  charmap.exe

But at least for Maple<2020, StringTools does not support non-Ascii.

 

The complex roots are the roots of the polynomyal in lambda

P:=(lambda,t) -> 1+1/3*lambda+1/18*lambda^2-1/324*lambda^3+1/1944*lambda^4 - exp(I*t)

for t in [0,2*Pi)
A sequence of 40 random roots are obtained with:

r:=rand(0..2.*Pi):
seq(fsolve(P(lambda,r()),complex), 1..10);

 

@Carl Love

Carl, it's not only about 1/4 versus 1/2 line. But I think that a beginner will be intimidated by the "1/2" - one.

I definitely like your coding style, but when the question comes from a beginner, an "elementary" version should be added (a fortiori if it's 1/4).

@Carl Love 

Isn't this overkill? Why not simply

`&||` := (x,y) -> 1/(1/x+1/y);

if we use it only as operator (not as function `&||`(a,b,...)).

@Carl Love 

M:= f(output((J:= Iterator:-Combination(6,4))), 0, [2,2]):
(in J do M+~1 od);

f(Vector[row](4, [0, 1, 2, 3]), 0, [2, 2]) + 1, f(Vector[row](4, [0, 1, 2, 3]), 0, [2, 2]) + 1, f(Vector[row](4, [0, 1, 2, 3]), 0, [2, 2]) + 1, f(Vector[row](4, [0, 1, 2, 3]), 0, [2, 2]) + 1, f(Vector[row](4, [0, 1, 2, 3]), 0, [2, 2]) + 1, f(Vector[row](4, [0, 1, 2, 3]), 0, [2, 2]) + 1, f(Vector[row](4, [0, 1, 2, 3]), 0, [2, 2]) + 1, f(Vector[row](4, [0, 1, 2, 3]), 0, [2, 2]) + 1, f(Vector[row](4, [0, 1, 2, 3]), 0, [2, 2]) + 1, f(Vector[row](4, [0, 1, 2, 3]), 0, [2, 2]) + 1, f(Vector[row](4, [0, 1, 2, 3]), 0, [2, 2]) + 1, f(Vector[row](4, [0, 1, 2, 3]), 0, [2, 2]) + 1, f(Vector[row](4, [0, 1, 2, 3]), 0, [2, 2]) + 1, f(Vector[row](4, [0, 1, 2, 3]), 0, [2, 2]) + 1, f(Vector[row](4, [0, 1, 2, 3]), 0, [2, 2]) + 1

@Carl Love 

Do you mean that in your code Alias is invoked only once? It creates a 2x2 Matrix.

(Actually I am a little afraid of what Alias does and try to avoid it).

@Carl Love 

That's interesting, but I find the resulting code unintuitive.
Probably most users would prefer:

J:=Iterator:-Combination(6,4):
seq(ArrayTools:-Alias(x[]+~1, [2,2]), x in J);

 

@emendes 

The denominator of R[n] is  4*2^(2^n).

Actually, using rsolve ==>  R[n] = -1/2*cos(arccos(3/4)*2^n)+1/2

Using this, R can of course be formed even for n >> 1000.

 

 

 

@JanBSDenmark 

If you want an explanation. In mathematics sqrt and arcsinh are multivalued functions (defined on C).
All CASes have problems in choosing the wanted branches for such functions.
E.g. with your notation, sol[2] is a solution for another branch:  - x*sqrt(4*x^2 + 1) + arcsinh(2*x) - 5 = 0.

@David Sycamore 

P calls PP (from my first answer), so, PP must be present and executed in order to use P.

Just execute (use the (!!!) icon) the attached:

PP.mw

 

First 49 50 51 52 53 54 55 Last Page 51 of 176