Hi

Dear friends, I am a relatively new user and I have a problem in entering and calculating the 2F1 hypergeometric function. My question is how to enter this function in Maple in equations so that Maple recognizes it? Because hypergeom ([1], [2], [3]) Maple itself is a 3-element function, while 2F1 hypergeometric ([1], [2], [3], [4]) is a four-element!

Almost i did 10 method for this ode equation all of them are succes but this one is giving me some confusing and i am looking for  get my answer, the mothod say if we have the auxilary equation if substitute the solution of this auxilary equation in our series solution then substitute in ode equation must be satisfy but it is not satisfy so when he did assumption for the auxilary equation he say it satisfy if we sabstitute this assumption in our series solution!

My question is this how we get thus assumption ? and why finding exact  solution of auxilary equation not satisfy?

and i hope mapleprimes don't delete this question becuase of this pictures also it help for undrestanding

there is other picture for different auxilary equation just  add one multiply term for G(xi)^4 in case anyone needed i will upload

Download odetest.mw

how fixed that?

Download we.mw

In many papers, I've noticed that the solution to an ODE (ordinary differential equation) often emerges directly when there's only a single function involved. My question is: is there a way to generate solutions to an ODE by producing specific parameters?

Download get_all_solution_of_ode_by_generation.mw

i can do one by one for all case but i am intrested for this idea how we can do that for each equation automatically calculate all case without use one by one case ?
there is any other shorter way for get solution of this mw

all_case.mw

Download find_all_case_solution_of_ode_.mw

've written in Maple to set up and solve the differential equation for a simple harmonic oscillator and plot the solution. To incorporate the Hamiltonian into my worksheet, I needed to define the kinetic energy and potential energy of the system, and then combine them to express the total Hamiltonian. After that, I can calculate and plot the kinetic, potential, and total energies over time. Then I plotted  the kinetic energy, potential energy, and total energy over time

Download Hamiltonian_Energy.mw
Now, in the second worksheet, I tried to create a procedure to calculate the Hamiltonian Energy for a general example. I have set delta = 0: This removes the damping effect, leaving only the spring constant k and mass m as factors. Now the equation represents a pure simple harmonic oscillator:

The Hamiltonian (total energy) should be conserved, meaning it will remain constant over time without damping. The plot of H(t) should show a straight line, indicating that the kinetic and potential energy sum does not change. But as you can see the energy increases over time. I can not figured out where is my mistake in this code

Any guidance or examples of implementation would be greatly appreciated!

Thank you!

Download Hamiltoninan_procedure_.mw

I have this problem:

(x'+x''+epsilon/x)(x'-(x')^3/3!)(x''-(x'')^3/3!)(epsilon/x -(epsilon/x)^3/3!)=1

x(0)=epsilon , x'(epsilon)=epsilon+epsilon^{-1}

0<epsilon<<1

Now I want to try the ansatz: x(epsilon,t)=x_{-1}(t)/epsilon+x_0(t)+x_1(t)epsilon.

Can you suggest me a way to implement this ansatz (i.e finding x_{-1}(t),x_0(t),x_1(t)).

Appreciate your time!

i did all the time like that and don't have any issue but i don't know why not take derivative by x and t have a problem when i remove t is ok but when i take duble derivative by x and t not run and did't give me error too what is problem with this?

Download non_sense.mw

every time i have a small problem 2  why not cancel number 2 in denominator , i don't want see a number with fraction like 1/3  3/4 how i fix this

Download cancelation.mw

Download Heat_Equation_1D_Robin_Conditions.mw

What is the correct way to find all terms of the form  _Cn^m where n is non negative integer and m is exponent which can be 1?

The problem is finding _Cn with no exponent. Maple's   _Cn^anything does not match _Cn but only when there is an actual exponent present.

It is easy to find _Cn^m with m present and easy to find _Cn with no exponent, but how to combine the two is the problem. Here is an example.

Given e:=_C1^2+_C1+_C2^3+a+b; I want to obtain the set {_C1^2,_C1,_C2^3}  but I get instead {_C1, _C2, _C1^2, _C2^3} using ther following code

e:=_C1^2+_C1+_C2^3+a+b;
indets(e,   Or(  And(symbol, suffixed(_C, nonnegint)),
                 And(symbol, suffixed(_C, nonnegint))^anything  ));

If instead I do this

e:=_C1^2+_C1+_C2^3+a+b;
indets(e, And(symbol, suffixed(_C, nonnegint))^anything);

This gives {_C1^2, _C2^3} , so it did not pick the _C1 term since exponent is missing. If I do this instead

e:=_C1^2+_C1+_C2^3+a+b;
indets(e, And(symbol, suffixed(_C, nonnegint)));

This it gives {_C1, _C2} and if I do 

e:=_C1^2+_C1+_C2^3+a+b;
indets(e,   And(symbol, suffixed(_C, nonnegint))^Or(anything,identical(1)))

it gives {_C1^2, _C2^3}

What to do to obtain {_C1^2,_C1,_C2^3} 

I am trying to find all _Cn^m in an expression where m can be anything including not being there (or 1 basically) but in Maple, exponent 1 is not counted as exponent.

Update

Another input example is e:=_C1^2+_C1+ _C1*_C2^3+a+b;  This should be produce the set {_C1^2,_C1,_C2^3} 

Any trick to use for this?

Maple 2024.1

What is the relationship between font size and screen pixels on plots? My screen are 2560 x1440 and 1920 x1200. Is for exanple size 12 font 12 pixels high?

Also when a diplay plot is say 800 x 600 pixels in size does the plot  does it use the max horizontal or vertical number of pixels? I know it may not fill the plot area in both directions. Or are some pixels trimmed of around the sides. Is it that the bounding box is the 800 x 600? I know when a caption is added the plot area is reduced in size vertically. 

Download Heat_Eq_2D.mw

Any ideas, guys?

While Maple provides built-in functions for handling continued fractions, such as NumberTheory[ContinuedFraction], I employ custom implementations to better suit educational purposes. The custom code allows for step-by-step demonstrations, making it easier for students to understand how each convergent is calculated. Additionally, it offers more flexibility in modifying
the process of exploring generalized or specific continued fractions, such as those used in solving
quadratic equations or demonstrating the convergence properties of irrational numbers.

If the logarithmic error plot is approaching zero, the errors are decreasing as the number of iterations increases. This behavior shows that the continued fraction is indeed converging to the actual value.

However, a logarithmic scale that approaches zero might be a bit misleading. Given that the current code produces a logarithmic error plot that may not appear to go sufficiently negative, it might indicate that the approximations are not yet close enough.

The issue I face with the error plot stopping after 12 terms might be due to the logarithmic transformation attempting to take the logarithm of very small or zero errors, which can cause issues when plotting. However, I don't know how to fix this problem

Download continued-fraction.mw