Carl Love

Carl Love

28055 Reputation

25 Badges

12 years, 361 days
Himself
Wayland, Massachusetts, United States
My name was formerly Carl Devore.

MaplePrimes Activity


These are replies submitted by Carl Love

This brute force technique is far from optimal, especially considering that the sum is converted to a continuous increasing function of N.

It is important for you to note that Markiyan changed your Sum to sum (with lowercase s). It is easy for a reader to miss that detail. The fsolve will not work without that. With the evaluated sum, it converts the summation to a continuous increasing function of N.

Also, he changed your pi to Pi. The latter is the well-known constant, while the former is just another variable.

It is important for you to note that Markiyan changed your Sum to sum (with lowercase s). It is easy for a reader to miss that detail. The fsolve will not work without that. With the evaluated sum, it converts the summation to a continuous increasing function of N.

Also, he changed your pi to Pi. The latter is the well-known constant, while the former is just another variable.

@Przemek I can't give any more advice until you post the equations.

You did not enter the procedure exactly as I typed it. It's Statistics :- CDF, not Statistics - CDF. Cut-and-paste the boldface code below. The non-boldface part is Maple's responses.

normalcdf:= proc(x1,x2,mu,sigma)
local t,dist:= Statistics:-CDF(Normal(mu, sigma), t);
     eval(dist, t= x2) - eval(dist, t= x1)
end proc:
normalcdf(2100,2200,2150,50);
                            /1  (1/2)\
                         erf|- 2     |
                            \2       /
evalf(%);
                       0.682689492137088


You did not enter the procedure exactly as I typed it. It's Statistics :- CDF, not Statistics - CDF. Cut-and-paste the boldface code below. The non-boldface part is Maple's responses.

normalcdf:= proc(x1,x2,mu,sigma)
local t,dist:= Statistics:-CDF(Normal(mu, sigma), t);
     eval(dist, t= x2) - eval(dist, t= x1)
end proc:
normalcdf(2100,2200,2150,50);
                            /1  (1/2)\
                         erf|- 2     |
                            \2       /
evalf(%);
                       0.682689492137088


I edited the tags on your Question.

@rkm You can use labels if you want. The percentage sign means, literally, "the result of the previous command". There's also %% and %%% for "the result of the second-most-recent command" and "the result of the third-most-recent command." But that's as far back as you can go without using labels.

@rkm You can use labels if you want. The percentage sign means, literally, "the result of the previous command". There's also %% and %%% for "the result of the second-most-recent command" and "the result of the third-most-recent command." But that's as far back as you can go without using labels.

I picked a subset of the four shortest polynomials, {f0, f1, f5, f6}, and tried to solve just those. After running for about an hour (with apparently insignificant memory usage--38 M), I had a loss of kernel.

Are you looking for an exact solution, or would a decimal approximation be okay? If the latter, try fsolve.

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.

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).

First 603 604 605 606 607 608 609 Last Page 605 of 709