Hi,

I get the error message "Error, invalid input: diff expects 2 or more arguments, but received 1" from the following program. Could you please help me? Thank you i,n advance for your help!

som:=0:

for b1 from 10 to 10 by 1 do
for b2 from 1 to 2 by 1 do
for alpha from 0.5 to 0.5 by 0.1 do
for beta from 0.33 to 0.5 by 0.1 do
for c from 1 to 1 by 1 do
for f from 1 to 10 by 1 do
for g from 1 to 10 by 0.1 do
for lambdaj from 0.2 to 0.4 by 0.1 do
for gammaj from 0.2 to 0.4 by 0.1 do

p:='p';

aiSQ:=(alpha*b1)/(alpha*b2+beta*b2+c);
ajSQ:=(beta*b1)/(alpha*b2+beta*b2+c);
UiSQ:=(1/2)*alpha*b1^2*(alpha^2*b2+2*alpha*beta*b2+c*alpha+beta^2*b2+2*beta*c)/(alpha*b2+beta*b2+c)^2;
UjSQ:=(1/2)*beta*b1^2*(alpha^2*b2+2*alpha*beta*b2+2*c*alpha+beta^2*b2+beta*c)/(alpha*b2+beta*b2+c)^2;
USQ:=(1/2)*b1^2*(alpha+beta)*(alpha*b2+beta*b2+2*c)/(alpha*b2+beta*b2+c)^2;
UTSQ:=UiSQ+UjSQ+USQ;

ai:=(-alpha*b2*f*p+alpha*b1*c-b2*f*p+b1*c)/(c*(alpha*b2+b2*beta+b2+c));
aj:=(alpha*b2*f*p+b1*beta*c+b2*f*p+c*f*p)/(c*(alpha*b2+b2*beta+b2+c));

aineg:=-(alpha*b2*f*lambdaj*p+b2*f*lambdaj*p+alpha*b1*c+b1*c)/(c*(alpha*b2*lambdaj-2*alpha*b2-b2*beta+b2*lambdaj-2*b2-c));
ajneg:=(alpha*b2*f*lambdaj*p+alpha*b1*c*lambdaj+b2*f*lambdaj*p+c*f*lambdaj*p-alpha*b1*c-b1*beta*c+b1*c*lambdaj-b1*c)/(c*(alpha*b2*lambdaj-2*alpha*b2-b2*beta+b2*lambdaj-2*b2-c));
uj:=beta*(b1*(aineg+ajneg)-(b2/2)*(aineg+ajneg)^2)-(c/2)*ajneg^2-p*f*(ajneg-aj);
uL:=(alpha+1)*(b1*(aineg+ajneg)-(b2/2)*(aineg+ajneg)^2)-(c/2)*aineg^2+p*f*(ajneg-aj);
eqtj:=gammaj*(uL-USQ)-((1-gammaj)/(1-lambdaj))*(uj-UjSQ)-tj;
tj:=solve(eqtj,tj);

dai:=diff(ai,p);
daj:=diff(aj,p);
daineg:=diff(aineg,p);
dajneg:=diff(ajneg,p);
dtj:=diff(tj,p);
Ujp:=beta*(b1*(ai+aj)-(b2/2)*(ai+aj)^2)-(c/2)*aj^2-p*f*(ajneg-aj)+(1-lambdaj)*tj;
ULp:=(alpha+1)*(b1*(ai+aj)-(b2/2)*(ai+aj)^2)-(c/2)*ai^2+p*f*(ajneg-aj)-tj-g*((p^2)/2);
eqp:=diff(ULp,p);
eqpp:=diff(eqp,p);
p:=solve(eqp,p);

CSQ:=b1-b2*(aiSQ+ajSQ);
Cabat:=b1-b2*(ai+aj);
Cneg:=b1-b2*(aineg+ajneg);

if (ai>aineg) then f*(aineg-ai)=0
else if (aj>ajneg) then f*(ajneg-aj)=0
else if (CSQ>0 and Cabat>0  and Cneg>0 and eqpp<0 and p>0 and p<1 and beta<alpha and aiSQ>0 and ajSQ>0 and ai>0 and aj>0 and aineg>0 and ajneg>0 and tj>0)
then
#print(b1,b2,alpha,beta,c,f,g,lambdaj,gammaj,p);
som:=som+1;
fi;fi;fi;
od;od;od;od;od;od;od;od;od;
som;
 

How to make maple sheet background transparent?

and change word color ?

Download int-too-many-levels-of-recursion.mw

how can the graphic as .EPS

f(x,y)=sin(x)*cos(x) , -1<x,y<1

Hello again, I posted a thread here earlier and received some great responses. I've made some progress in Maple since then but once again I've ended up on a question where I am completely stuck. I am only meant to solve this one in maple as the answer is written below the question. The question is:

I realise this is alot to ask for but I'm studying from afar and I don't have as many options for help as other students at the moment. I have managed to solve a), and I know how to solve b) the traditional way (pen and paper), but I have no idea how to do it in maple. My solution for a) is attached below. c) and d) strictly rely on the answer from b) so I'd greatly appreciate if I could have some help with it so I atleast could attempt the others on my own. This is the final question I have so once I'm done with this I'll pretty much be a master Maple... or not.. :P

I have commented the maple document so it is easier to understand what I've done and what I want help with. Also, I very much apologise if something that I write don't make sense, English is not my native language.

Linear-Algebra-Transformations.mw
 

Download Linear-Algebra-Transformations.mw

Hello, I just started working in Maple and I am struggling with a question that I found in my book. The question goes as follows:

Consider the set of vectors T = {v₁, v₂, v_{3​​​​,} v₄} in ℝ⁴ with

v₁ = (2,5,-2,0), v₂ = (-2,-4,b,4), v₃ = (-1,2,-2,-5), v₄ = (b,2,5,3)

Find all values of b, such that T is a linearly independent set

I am supposed to first solve it by hand and then in maple. I can solve it by hand but I have no clue how to do it in maple.The only thing I've done so far is create a matrix that contains the vectors in the set S as column entries.

I know that the columns of A are linearly independent if  det A ≠ 0. This is where my near non-existent knowledge in Maple stops me...

The answer is for all b {2}

I appreciate any tips...

Hi maple users

I'm about to do an assignment on the Joukowski transformation, where I have modelled and airfoil resembling the NACA-64015 airfoil. The parameters of the circle in z-plane are as follows: r=1.1241, x=-0.1241, y=0.0, where r is the radius of the circle, and x and y is the parameters used to offset the center of the circle. The problem is that I can't seem to plot the airfoil with contuours showing the fluid flow. I would therefore like to know if any of you guys has solved this before? Any help is much appreciated.

I have a function as below,

Download Quesions.mw

How I can normalize it in range [0,1]

Hello

I have problem to calculate a numerical integration. I guess, I can't use integration cammands correctly or singularity is the problem.
The equation has a singularity at y=0 (exp(1E-9*I*x/y)). The limits are semi_infinite.
integration:

f1 := (-(1.671130827*10^(-21)*I)*((0.6132469529e-1-.5253537767*I)*x^3+(.2894208344-2.479398027*I)*y*x^2+(0.1107017247e-1+(-.2377238011-0.2774956673e-1*I)*y^2)*x+(-.1167302837+.9999999999*I)*y^3+(0.2912028092e-1-.2494663781*I)*y)*exp(-1.732070784*10^(-15)*x*y)*((-0.6132469529e-1-.5253537767*I)*x^3+(.3115611798+2.669068985*I)*y*x^2+(-0.1107017247e-1+(-.3699321954+0.4318229002e-1*I)*y^2)*x+(.1455017102+1.246477826*I)*y^3+(0.2512356422e-1+.2152274756*I)*y)*exp(1.732070784*10^(-15)*x*y)*exp(-2.178669712*10^(-15)*y^2-3.534838336*10^(-16)*x^2+(1.000000000*10^(-9)*I)*x/y)/x)*y:

evalf(Int(f1,y=0..infinity));



Thank you
Test4_MaplePrimes.mw

 I am using applyrule since algsubs cannot do for more than one substitution.Also, I would be glad if any other alternative is provided.
thank you
 

Can any one do the addition of values (in string format) x in textarea0, y in textarea1 and get the mathcontainer result in the text box2.

Thanks

Ramakrishnan V

y := GetProperty(TextArea1,value);
Do(%TextArea2 = Do(%MathContainer0,x=%TextArea0,y=%TextArea1));

howTogetNumericAnswerinTextArea.mw

Download howTogetNumericAnswerinTextArea.mw

Has there been a option included in the plots:-setoptions3d that is equivalent to the size option found in plots:-setoptions? I can't seem to find one.  I would really like to be able to set all my 3d plots to the same size. Or am I missing something?

>>> 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([])'
>>>

Yesterday, I accidentally discovered a nasty bug in a fairly simple example:

restart;
Expr:=a*sin(x)+b*cos(x);
maximize(Expr, x=0..2*Pi);
minimize(Expr, x=0..2*Pi);
                                    

I am sure the correct answers are  sqrt(a^2+b^2)  and  -sqrt(a^2+b^2)  for any real values  a  and  b .  It is easy to prove in many ways. The simplest method does not require any calculations and can be done in the mind. We will consider  Expr  as the scalar product (or the dot product) of two vectors  <a, b>  and  <sin(x), cos(x)>, one of which is a unit vector. Then it is obvious that the maximum of this scalar product is reached if the vectors are codirectional and equals to the length of the first vector, that is, sqrt(a^2+b^2).

Bugs in these commands were noted by users and earlier (see search by keywords bug, maximize, minimize) but unfortunately are still not fixed. 

Hey there,

I'm trying to build a procedure that can function as an adapted form of Prim's algorithm. The idea is that on a graph with just vertices, the procedure has a starting point, and from there will find out which vertex is the cheapest to connect to (currently expressed purely by the lowest distance). Once this is found, the connected vertex is removed from a list that has vertices that aren't connected yet, and added to a list of vertices that are in the minimal spanning tree.

My problem is that I get an error returned that says "invalid Boolean expression", and I'm not sure how to solve it. Can anybody here point me in the right direction?

The procedure is defined as follows:

Primmetje := proc (aantal, posities, begin)
local knopenover, knopeninmst, huidig, V, kaart, e, a;
knopenover := [seq(i, i = 1 .. aantal)];
knopeninmst := {};
huidig := [0, 0];
if begin <> {} then
  V := [begin];
knopeninmst := knopeninmst union {V}
end if;
remove(V, knopenover);
kaart := Graph(aantal);
SetVertexPositions(kaart, posities);
while nops(knopeninmst) < aantal
do for e in knopeninmst
   do for a in knopenover
     do if huidig = [0, 0] or Distance(posities[e], posities[a]) < Distance(posities[huidig[1]], posities[huidig[2]]) then
   huidig := [e, a];
knopeninmst := knopeninmst union {a};
remove(a, knopenover);
AddEdge(kaart, huidig)
end if
end do
end do
end do
end proc

When I try to execute it with some parameters the return is this:

vp := [2.5, 21], [6, 13.5], [8, 10], [11, 24.5], [14.3, 19.4], [16.8, 26], [22, 21.5], [22, 17], [22.2, 12.5], [26.8, 23], [28, 20.5], [30, 25.5], [32, 21], [29.5, 16];
Primmetje(14, vp, 1);
Error, (in Primmetje) invalid boolean expression: [[6, 13.5]]

I think it has something to do with the double brackes, but I'm not sure how to solve it.