Hey,

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

restart:

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

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

plot(expr)

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.

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

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

Download select_solution.mw

Thanks!

Honigmelone

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.

value_function.mw

Download value_function.mw

I try to find the exact (symbolic) value of

(-2*sqrt(7)-4)*EllipticK((1/8)*sqrt(2)*(-3+sqrt(7)))^2+4*EllipticE(-(1/8)*sqrt(2)*(-3+sqrt(7)))*sqrt(7)*EllipticK((1/8)*sqrt(2)*(-3+sqrt(7)))

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.

Thanks.

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:

I'm solving a problem described in this topic.

It's sufficient for me to obtain N variable via this manual procedure: file4.mw

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?

Hi

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

lambda^6;

Hello,

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,

x^2+y^2+z^2+w^2=1

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,

Ian

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

b*x*ln(x)-x*ln(a)+a=0

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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