Assume that a[1],a[2],..a[k] are positive integer numbers. Let n be a positive integer number. 

Suppose that igcd(a[1],a[2],..a[k])=1. 

My question: Is there a command in Maple such that the output of the this command be true provided that there are "non-negative integer numbers" x[1],x[2],..x[k] which satisfy the following condition:

a1*x1+a2*x2+...+ak*xk=n

Thanks in advance

Recently I examined a piece of code of mine in an attempt to possibly convert it to another language as it is a numeric code and as such slower in Maple than I'd like it to run. In doing this I ran across the following strangeness, here reproduced in a minimum working example (file attached).

Consider this trivial integral:

x1:=Int(3.52*10^8, ti = 0 .. 1);
(4)

and also this one:

x2:=sin(2*Pi*x1);
(5)

I can then evaluate (4) and take sine(2Pi * the evaluation of (4)):

value(x1);
(6)

sin(2*Pi*(6));
(7)

Hmm... let's evaluate x2, which should be the same, right

value(x2);
0.00000556012229902952                                                              (8)

Oddly enough, it is not. Now the reason they are not 0 is due to round-off error (running the same sheet with Digits := 40 confirms that); but at the same time, (6) is in fact exact. More oddly, if I wrap the input leading to (7) in evalf() then it outputs 0., i.e. exact and correct. I suspect the problem must lie in the different treatments of Pi in the three cases.

I am not ready to call this behaviour a bug since I can see that different ways of evaluating what is essentially the same expression leads to a diffferent round-off. What strikes me is the relatively large errors in this case. The sheet was run with Digits being 15 (my default set in my .mapleinint), I initially expected somewhat more accuracy in the sine function than a mere 6 digits or so. On second thought, however, what is going on seems to be that the evaluation of the integral must be numerical and the large no. of cycles limits the accuracy; if one replaces 3.52E8 (a float) with 352E6 (an exact number) then (7) becomes 0 (exact) while (8) remains at the above value. Why

evalf(sin(2*Pi*(6)))

yields an exact value I do not quite understand.

So, caveat computor once again. This example, while it may look contrived, actually arose from a real-world case I was dealing with (the 352E6 is a frequency in Hz, in my actual application it can vary in time therefore the integration to get the no. of cycles in a given time interval). One annoyance here is that the "right" way to do this is not obvious, at least not to me.

M.D.

integration_test.mw

Is there any facility to apply Finite Volume Method to Partial idifferential equation on MAPLE?
Any comand?

Any Code?

How can we remove 0=0 from the above by single comand if it lie in any position from the set?

How I can solve algebraic differential equation of index 2 in Mae 15?

How to make maple sheet background transparent?

and change word color ?

>>> maple = pywinauto.application.Application().start(r'C:\Program Files\Maple 2015\bin.win\maplew.exe')
C:\Users\mas\AppData\Local\Programs\Python\Python36-32\lib\site-packages\pywinauto\application.py:1044: RuntimeWarning: Application is not loaded correctly (WaitForInputIdle failed)
  warnings.warn('Application is not loaded correctly (WaitForInputIdle failed)', RuntimeWarning)
>>> maple.Maple.PrintControlIdentifiers()
__main__:1: DeprecationWarning: Method .PrintControlIdentifiers() is deprecated, use .print_control_identifiers() instead.
Traceback (most recent call last):
  File "C:\Users\mas\AppData\Local\Programs\Python\Python36-32\lib\site-packages\pywinauto\application.py", line 246, in __resolve_control
    criteria)
  File "C:\Users\mas\AppData\Local\Programs\Python\Python36-32\lib\site-packages\pywinauto\timings.py", line 453, in wait_until_passes
    raise err
pywinauto.timings.TimeoutError

During handling of the above exception, another exception occurred:

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "C:\Users\mas\AppData\Local\Programs\Python\Python36-32\lib\site-packages\pywinauto\__init__.py", line 50, in wrap
    return method(*args, **kwargs)
  File "C:\Users\mas\AppData\Local\Programs\Python\Python36-32\lib\site-packages\pywinauto\application.py", line 585, in print_control_identifiers
    this_ctrl = self.__resolve_control(self.criteria)[-1]
  File "C:\Users\mas\AppData\Local\Programs\Python\Python36-32\lib\site-packages\pywinauto\application.py", line 249, in __resolve_control
    raise e.original_exception
  File "C:\Users\mas\AppData\Local\Programs\Python\Python36-32\lib\site-packages\pywinauto\timings.py", line 431, in wait_until_passes
    func_val = func(*args, **kwargs)
  File "C:\Users\mas\AppData\Local\Programs\Python\Python36-32\lib\site-packages\pywinauto\application.py", line 191, in __get_ctrl
    dialog = self.backend.generic_wrapper_class(findwindows.find_element(**criteria[0]))
  File "C:\Users\mas\AppData\Local\Programs\Python\Python36-32\lib\site-packages\pywinauto\findwindows.py", line 84, in find_element
    elements = find_elements(**kwargs)
  File "C:\Users\mas\AppData\Local\Programs\Python\Python36-32\lib\site-packages\pywinauto\findwindows.py", line 303, in find_elements
    elements = findbestmatch.find_best_control_matches(best_match, wrapped_elems)
  File "C:\Users\mas\AppData\Local\Programs\Python\Python36-32\lib\site-packages\pywinauto\findbestmatch.py", line 533, in find_best_control_matches
    raise MatchError(items = name_control_map.keys(), tofind = search_text)
pywinauto.findbestmatch.MatchError: Could not find 'Maple' in 'dict_keys([])'
>>> maple.Maple.print_control_identifiers()
Traceback (most recent call last):
  File "C:\Users\mas\AppData\Local\Programs\Python\Python36-32\lib\site-packages\pywinauto\application.py", line 246, in __resolve_control
    criteria)
  File "C:\Users\mas\AppData\Local\Programs\Python\Python36-32\lib\site-packages\pywinauto\timings.py", line 453, in wait_until_passes
    raise err
pywinauto.timings.TimeoutError

During handling of the above exception, another exception occurred:

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "C:\Users\mas\AppData\Local\Programs\Python\Python36-32\lib\site-packages\pywinauto\application.py", line 585, in print_control_identifiers
    this_ctrl = self.__resolve_control(self.criteria)[-1]
  File "C:\Users\mas\AppData\Local\Programs\Python\Python36-32\lib\site-packages\pywinauto\application.py", line 249, in __resolve_control
    raise e.original_exception
  File "C:\Users\mas\AppData\Local\Programs\Python\Python36-32\lib\site-packages\pywinauto\timings.py", line 431, in wait_until_passes
    func_val = func(*args, **kwargs)
  File "C:\Users\mas\AppData\Local\Programs\Python\Python36-32\lib\site-packages\pywinauto\application.py", line 191, in __get_ctrl
    dialog = self.backend.generic_wrapper_class(findwindows.find_element(**criteria[0]))
  File "C:\Users\mas\AppData\Local\Programs\Python\Python36-32\lib\site-packages\pywinauto\findwindows.py", line 84, in find_element
    elements = find_elements(**kwargs)
  File "C:\Users\mas\AppData\Local\Programs\Python\Python36-32\lib\site-packages\pywinauto\findwindows.py", line 303, in find_elements
    elements = findbestmatch.find_best_control_matches(best_match, wrapped_elems)
  File "C:\Users\mas\AppData\Local\Programs\Python\Python36-32\lib\site-packages\pywinauto\findbestmatch.py", line 533, in find_best_control_matches
    raise MatchError(items = name_control_map.keys(), tofind = search_text)
pywinauto.findbestmatch.MatchError: Could not find 'Maple' in 'dict_keys([])'
>>>

alpha+{6*RK[1]*alpha+2+(96/5)*R^2*K^2*alpha^2-(1/6)*R*alpha+64*R^3*K^3*alpha^3}*beta+{(24/5)*RK+44*R^2*K^2*alpha^3}*beta^2=0

Hello,

I have the follownig set of inequality:

{0 < p[1, 2], p[1, 1] < 2*p[2, 2]+(3/2)*p[1, 2], p[1, 2]^2/p[2, 2] < p[1, 1], (2/3)*p[1, 2] < p[2, 2]}

Now I need to find value of p11,p12,p22 that satisfy the above inequality. Is there any easy way to find

parameters p11, p12, p22 in maple?

Best

Is there any way to integrate this in maple?

lambda^2*t*(diff(theta(t), t, t)) = lambda^2*(diff(theta(t), t))-Pr*s*lambda*(diff(theta(t), t))+Pr*(diff(theta(t), t))-Pr*t*(diff(theta(t), t))

Dear Users!

Hope you would be fine with everything. I want to evaluate the following expression for k = 3, j = 0, r = 1.

I am waiting for your positive reply. Thanks in advance

Hello,
I want to collect a function into terms without using ?expand() since this expands everything which I dont want.

f:=GAMMA(L+2*q-3-k)/(GAMMA(L-k)*k)*((GAMMA(-2*q+L)*GAMMA(L+2*q-3-k)-GAMMA(L+2*q-3)*GAMMA(L-2*q-1-k)*(L+2*k-1-(4*k+2)*q))/((2*(-1+2*q))*(4*q-3)*GAMMA(L+2*q-3)*GAMMA(L+2*q-3-k)));
collect(f,[k,GAMMA])

then has 1 term which still contains a denominator, but I want them seperate so I can use ?op() for all additive terms.

Is there an option without expanding the entire thing to enforce termwise selection?

Of course I could do it in a second step, but I want to avoid it and think it should be simpler.

Thanks everyone for helping me over the years. I've just handed in my PhD- and I really considered Maple Primes like a supervisor.

Currently I am relearning Financial maths - as depending on grants I may leave academia :(

Today I am learning utility functions and risk aversion and thought to make a graphso i could visualise them

 Here is a graph of the log of the utility of x - with two utility functions - constant absolute risk aversion (lower surface) - and constant relative risk aversion (disjoint surface above); for both functions  g (and in the attached worksheet R) is a parameter of these functions; annoyingly for these versions of these functions to be plotted on the same axis - they are so different in scale that it is hard to see anything interesting.

However one of the key features of utility functions is that we consider them to be unaffected by scalling- i.e. that if U_2(x)=c*U_1(x) for all x then U_2(x) and U_1(x) are considered to be the same function.

This means that scalling can be done in a much more useful way than what I have done. Instead of plotting f(x;R)=x^(1-R)/(1-R) on the interval I (x=1..100), i'd like to plot g(x;R)=f(x;R)/max(f(x;R),I)  on the interval I.

I worked out that on a 2d graph this can be done using maximise. But I'd like to plot g(x;R) in 3d as both x and R vary and i cant think of how to do that! 

Cara_functions.mw

restart;

with(VectorCalculus);

with(LinearAlgebra);

r1 := Vector([0, 0, 1]);

r2 := Vector([sin(theta1), 0, cos(theta1)]);

r3 := Vector([VectorCalculus:-`*`(sin(theta2), cos(phi2)), VectorCalculus:-`*`(sin(theta2), sin(phi2)), cos(theta2)]);

M := Matrix([r1, r2, r3]); ex := `assuming`([simplify(VectorCalculus:-`*`(Determinant(M), 1/VectorCalculus:-`+`(VectorCalculus:-`+`(VectorCalculus:-`+`(1, DotProduct(r1, r2)), DotProduct(r1, r3)), DotProduct(r2, r3))))], [theta1 > 0, theta2 > 0, phi2 > 0]);

dex := eval(simplify(diff(arctan(ex), phi2)), phi2 = t);

VectorCalculus:-`*`(2, Int(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(2, Int(dex, t = 0 .. phi2)), 1/VectorCalculus:-`*`(4, Pi)), VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(2, Pi), sin(theta1)), sin(theta2)), 1/VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(4, Pi), 4), Pi))), [phi2 = 0 .. Pi, theta2 = 0 .. Pi, theta1 = 0 .. Pi], method = _CubaCuhre, epsilon = 0.5e-2));

evalf(%)

Ok I deleted my other question, since there was a mistake. I actually want to integrate the following expression. The arctan is not every positive in my integral there, so I needed to go this way to make it continuous. The problem here is the nested integral inside Int(...,t=0..phi2) which leads to maple not being able to evaluate.

How can you get maple to evaluate i^i?

when i type in
I^I

i just get

I^I

and similarly when i raise numbers to complex powers i get results like 2^(2I+6)