It is hard to say whether Maple is right or wrong when its answer is FAIL. In this case (the n::posint case), we know that there is enough information to give the definitive answer false. We may say that Maple should know that the answer is false. But it hasn't been programmed to analyze the situation deep enough. My guess is that it boils down to this: Maple knows that the inequality is false for real n between real 2 and real 4, but it doesn't know that there is an integer between those bounds.

There is no data attached to your question.

It may be possible to improve the output from NonlinearFit by restricting the signs of the model parameters. I may be able to tell when I have the data. In the meantime, I suggest that you read the Wikipedia arcticle "Gompertz function." It is an S-shaped curve, so it is an increasing function, but neither exponential growth nor decay.

@brian bovril I think a procedure for nested radicals would be rather difficult. First, you'd need a standardized input format for the radicals. Next, all the solutions that I've seen are based on "tricks"; there are no algorithms. For continued fractions there are some algorithms (but they wouldn't apply to the "continued fraction" at the start of this thread because of the way it "splits" at multiple places).

@brian bovril I think a procedure for nested radicals would be rather difficult. First, you'd need a standardized input format for the radicals. Next, all the solutions that I've seen are based on "tricks"; there are no algorithms. For continued fractions there are some algorithms (but they wouldn't apply to the "continued fraction" at the start of this thread because of the way it "splits" at multiple places).

Did you remember to convert the 11.5 degrees to radians?

@Nuraz First, note that your procedure M can be simplified to the following:

M := proc (message) local k; [seq(numb[k], k = message)] end proc;

It produces a list as output. In order to apply the binary conversion to the elements of the list, you need to ?map the convert command:

map(convert, PL, binary);
              [1, 10, 11, 101101, 101110, 101111]

This process is so common in Maple that there is a symbol abbreviation for it:

convert ~ (PL, binary);
              [1, 10, 11, 101101, 101110, 101111]

So what do you intend to do with them in binary? If you are going to encrypt, it would probably be easier keeping them as regular integers.

@Nuraz First, note that your procedure M can be simplified to the following:

M := proc (message) local k; [seq(numb[k], k = message)] end proc;

It produces a list as output. In order to apply the binary conversion to the elements of the list, you need to ?map the convert command:

map(convert, PL, binary);
              [1, 10, 11, 101101, 101110, 101111]

This process is so common in Maple that there is a symbol abbreviation for it:

convert ~ (PL, binary);
              [1, 10, 11, 101101, 101110, 101111]

So what do you intend to do with them in binary? If you are going to encrypt, it would probably be easier keeping them as regular integers.

@acdah Try executing your code again from the restart. I get your expected answer and not 0 when I run it.

The body of your question is blank: No text, no attahced file.

You have 7 polynomials, but only 4 variables. Do you expect some of the polynomials to be redundant?

@Nuraz Look closely at the characters:

Bits :- Split ~ ([10, 4, 90, 6]);

@Nuraz Look closely at the characters:

Bits :- Split ~ ([10, 4, 90, 6]);

You wrote:

You can do this at the beginning of your worksheet.

with(Bits):

then your code

> nums := [10, 3, 90, 6];
> Bits := Split(nums);

will work.

It won't work, and it'd be a foolish thing to do. You'd be overwriting the module name. It won't work because the command needs to be mapped over the list argument.

You wrote:

You can do this at the beginning of your worksheet.

with(Bits):

then your code

> nums := [10, 3, 90, 6];
> Bits := Split(nums);

will work.

It won't work, and it'd be a foolish thing to do. You'd be overwriting the module name. It won't work because the command needs to be mapped over the list argument.

@brian bovril This one is easy because the inside of the expression contains a copy of the whole thing.

eq:= (x^2-22)^3-22 = x:
fsolve(%, x);
                     -4.95716044054056, 5.
eval(eq, x=5);
                             5 = 5

Obviously the solution is positive, so we can ignore the negative result from fsolve.

Another way to get the equation, perhaps more intuitive, is

x = (22+(22+x)^(1/3))^(1/2);