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I want to solve this equation and looking at the plot there are at least 3 solutions. I want the greatest/smallest negative solution. Unfortunately using solve with assumptions produces no results and solve without assumptions only finds two solutions.

Can you please help me?

#select greatest negative value from solution


expr:= ax*cos(lambda)+ay*sin(lambda)-(a+b*lambda)



ax:=1:ay:=2:a:=0.5:b:=0.25: #examplanatory values




assume(-2*Pi<lambda,lambda<0): #does not work


sol_lambda:=[solve(expr=0,lambda, useassumptions)];# returns empty list even though without assumption one solution is found

Warning, solutions may have been lost




sol_lambda:=[solve(expr=0,lambda)]; #returns only two solutions even though looking at the plot 3 are there

Warning, solve may be ignoring assumptions on the input variables.


Warning, solutions may have been lost


[2.190357220, -.2688724573]


sol_l_v:=evalb~(sol_lambda<~0); #dirty workaraound

[false, true]


sol_l_add:=[ListTools:-SearchAll(true,sol_l_v)] ; #this seems overly complicated




lambda:=sol_lambda[sol_l_add[-1]];  #to select the last entry




expr; #test









I try to assign a value "y" function "g" but not working. Should be g(0.5,y)=800, but it is not appear in screen. Regards.






g(.5, y) := 200*y;



y := 4;



g(.5, y)

g(.5, 4)






I try to find the exact (symbolic) value of


I tried 'simplify' with different options and 'convert'. It would be pi=3.141... as numerical approximation suggests.

Many thanks.

I want Positive values of Arcsin but maple give to me Negative values of Arcsin?

my function is f=arcsin(1-x), 0<x<2 and f>0. and f(2)=3 pi/2 but in maple f(2)=-pi/2.

I want positive values of f Without adding anything to f.


f=2*x*(int(sqrt(1+y^2*(t*x-1)^2/(1-(t*x-1)^2)), t = 0 .. 1))    ,0<x<2,   0<y<1,

I want to Determine the value of f for the various values of x,y For a domain So that x , y Changing with steps 0.1. for example:

x y f
0.1 0.1  
0.2 0.1  
0.3 0.1  
0.4 0.1  
0.5 0.1  
0.6 0.1  
0.7 0.1  
0.8 0.1  
0.9 0.1  
0.2 0.2  
0.3 0.2  

I'm solving a problem described in this topic.

It's sufficient for me to obtain N variable via this manual procedure:

However when I want to quantify an expression (f) for any N (e.g. N=40), Maple shows a value with some I variable:


What does it mean?



f := x->x^2:

AB(f, 0, 5);

But my proc is not true for below example:

f := x -> x^3-2*x;
AB(f, -1, 4);
I think c has two value and it is not true.

How can I add a condition to my proc that c be between a and b (a<c<b)?



How to use The  value of lambda in other operation ?

solve({alpha = (1/2)*lambda*alpha*Pi, beta = (1/2)*lambda*beta*Pi}, lambda)


for exampel 



I use Maple 18. How can I produce a table of 50 values of the fanction f(x) in the interval [a,b]? size step is (b-a)/n


I want to compute some matrix multiplications and i need this expression to be 1 always, i.e,


for every calculation I do.

I have tried x^2+y^2+z^2+w^2:=1 and assign(x^2+y^2+z^2+w^2,1) but it doesn't work.

What I should type to make it work?


Thank you



I'm looking at Maple as a possible alternative to Mathcad (which I've been using for years, but is now very jaded compared to other options like Maple and Mathematica).  I'm a civil engineer and for what I do, one of the better features of Mathcad is the way it handles units.  For example, if I specify an angle in degrees (say phi=30 degrees) and then ask for sin(phi), I get 0.5.  At face value, I though Maple would do the same kind of thing.  However, this doesn't appear to be the case (see attached worksheet).  The only workaround that I can see is to specify the angle in degrees (but without assigning ) and then multiply the specified value by pi/180 (to convert to radians) before passing it to the sin function.  Which is all a bit messy and not at all an attractive solution.

Am I misunderstanding the way units work in Maple and is there a clean way of specifying angles in degrees (which is what engineers work with) and using these values directy in trig functions?

Thanks in anticipation,


hello, i went solve these equation ,with a & b take any value


thank you

If I were to evaluate a single numerical integral, I can use evalf( Int(,method = _d01amc)).

But when the expression say is created by a built in function, Student[VectorCalculus][Hessian], from a complicated expression involve integrals. The resulting expression does not have the option "method = _d01amc". It then takes a long time to evaluate.

See this "HE" variable for example. HE.txt

value(HE); # takes a long time

evalf(HE); # takes a long time


Is there a way to evaluate "HE", using ",method = _d01amc" wherever necessary?



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