Maple 2019 Questions and Posts

These are Posts and Questions associated with the product, Maple 2019

Hello! I would like to ask a short question about particularsol function. Could someone please explain to me, why the function interprets the cosinus term as particular solution, even though it is homogeneous solution? Is it a bug in Maple or I unterstand something wrong?

As the title states, I want to have an equation f(x), and f(x) = 0 if x < 0, f(x) = x if x >= 0. How could I accomplish this?

I'm actually trying to generating a differential equation something like y'(x) + k*h(y) = sin(x) where h(y) is what I described above. Is there any convenient way to do this?


I'm new to Maple and was wondering if anyone could help me with how to put a discrete distribution such as Pr(X=1)=0.25, Pr(X=2)=0.65, Pr(X=3)=0.1, into Maple.


Dear community, please help me to verify that the obtained solution by using the fsolve command of maple is correct or wrong? and one more question how to generate interval in which our solution should be contained. dear admin if my question is duplicate please do not delete. please have a look on my maple file

To solve this, I got this far but am not sure where to go next?


f2 := x^2 - 3;
f2d := diff(f2, x);
                             x  - 3
                              2 x

set value for x0, number precision

x0 := 3;
eps := 0.1*10^(-5);


in the ThermophysicalData[Chemicals] package that compute the coefficients for different species how I can find that coefficients for seven coefficients not nine of them

in other words, I am seeking to find Databases for the NASA Seven-Coefficient Polynomial Fits for Calculating Thermodynamic Properties of Individual Species.



My equations are not getting solved.

Please tell me, is there any other command to solve these type of equations.



I have an issue where I believe the values given by evalf (I have independent verification), but the plot of the same function is incorrect. Specifically, g(a) plotted for a <=0 is correct, and incorrect for a > 0.

What am I doing wrong?

plot(g(a), a=-2..5) #I don't believe anything for a > 0

evalf(g(-1)) #I believe all of these values













The Fitting Procedure

The data are T,CP,H and S. I need to determine a polynomial with seven Coefficients.

The essential input to the fitting procedure is a table of specific heat, enthalpy, and entropy as a function of temperature. 

The constrained linear least-squares fitting procedure is a three-stage process. The first step is to determine simultaneously a1, through a5 for temperature ranges by fitting the specific heats to Eq. (1).

 The second step is to determine a6 for temperature ranges by fitting the H data. In this state, a1-a5 are held fixed, and as in the cp fitting, equality constraints on the fit and its first derivative are imposed at the common temperature. Finally, the a7 coefficients are determined by fitting the entropy (S) data.



Maple Worksheet - Error

Failed to load the worksheet /maplenet/convert/ .


Hello experts,I need your help to obtain the solution of the mentioned equation in the attached picture by using the newton method by supposing the random value of involved constant like D  etc. I need an algorithm for the newton method for the mentioned equation. note h=1+x^2/2

Hello everyone, I want to the expression for Q after putting T=0, pls help me to solve it.pls have the attachment



(1/12)*6^(1/2)*((n-1)*(-15*k^2*n+15*k^2-30*k*n+30*k-15*n+15+(1920*n*q*sigma^6-1920*q*sigma^6+225*k^4*n^2+5760*Q*n*sigma^3-450*k^4*n+900*k^3*n^2-5760*Q*sigma^3+225*k^4-1800*k^3*n+1350*k^2*n^2+900*k^3-2700*k^2*n+900*k*n^2+1350*k^2-1800*k*n+225*n^2+900*k-450*n+225)^(1/2)))^(1/2)/(sigma^2*(n-1)), -(1/12)*6^(1/2)*((n-1)*(-15*k^2*n+15*k^2-30*k*n+30*k-15*n+15+(1920*n*q*sigma^6-1920*q*sigma^6+225*k^4*n^2+5760*Q*n*sigma^3-450*k^4*n+900*k^3*n^2-5760*Q*sigma^3+225*k^4-1800*k^3*n+1350*k^2*n^2+900*k^3-2700*k^2*n+900*k*n^2+1350*k^2-1800*k*n+225*n^2+900*k-450*n+225)^(1/2)))^(1/2)/(sigma^2*(n-1)), (1/12)*(-6*(n-1)*(15*k^2*n-15*k^2+30*k*n+(1920*n*q*sigma^6-1920*q*sigma^6+225*k^4*n^2+5760*Q*n*sigma^3-450*k^4*n+900*k^3*n^2-5760*Q*sigma^3+225*k^4-1800*k^3*n+1350*k^2*n^2+900*k^3-2700*k^2*n+900*k*n^2+1350*k^2-1800*k*n+225*n^2+900*k-450*n+225)^(1/2)-30*k+15*n-15))^(1/2)/(sigma^2*(n-1)), -(1/12)*(-6*(n-1)*(15*k^2*n-15*k^2+30*k*n+(1920*n*q*sigma^6-1920*q*sigma^6+225*k^4*n^2+5760*Q*n*sigma^3-450*k^4*n+900*k^3*n^2-5760*Q*sigma^3+225*k^4-1800*k^3*n+1350*k^2*n^2+900*k^3-2700*k^2*n+900*k*n^2+1350*k^2-1800*k*n+225*n^2+900*k-450*n+225)^(1/2)-30*k+15*n-15))^(1/2)/(sigma^2*(n-1))





After running the program without any error, no plot obtained.

What is the reason?



I try using the DeepLearning package.
I use the function Classify and, even in the simplest test case presented in the its help page (please look at it), I regularly get connection errors to the mpython server as soon as I execute classifier := Classify(...) or classifier(...) more than once.
Errors are one of these twos

Error, (in Train) unable to communicate with mpython server
Error (in Python:-EvalFunction) unable to communicate with mpython server

I work with Windows 7 Enterprise, on an 8 proc PC and 64 GB of memory. The worst situation happened when Maple didn't even return these errors and that I saw inflating the consumed memory in 2 minutes, forcing me to manually shut down my PC because the task manager wasn't no longer  operational.

Is it a known problem?
Could it be an installation problem?

Even if it's not the point here, I would like to say that trying to use the DeepLearning package is really challenging considering the poverty of the help pages.

Hello all

I wanna solve an optimal control problem and I have searched the Internet but I could not find any tutorial or video course on how to solve it with the Pontryagin maximum principle method. It is my first time that I want to use MAPLE for solving an optimal control problem and I would be thankful if someone can help me.

$$\max \int_{0}^{1} x_{2} [u(t)-u(t)^2] dt        $$

$$  \dot{x}_{0} = -(1-u(t)) x_{0}(t)+2 x_{1}(t) $$

$$  \dot{x}_{1}(t) = (1-u_{t}) x_{0}(t) +2 x_{2}(t) -[3-u(t)]x_{1}(t)  $$

$$   \dot{x}_{2}(t) = (1-u(t))x_{1}(t) -2 x_{2}(t)   $$

$$  0 \le u_{t}  \le \frac{1+t^2}{1+t}  $$



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