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I have two equations that are valid under the substitution sin <-> cos, so a simple way to generate the second equation is to replace all occurrences of sin with cos. But Maple gets the wrong answer when I do this, because of its built-in simplification. Here is an example. Z1 and Y1 shows the problem; Z2 and Y2 shows that my attempt to overcome the problem doesn't work.  Z3 doesn't work either, proving that the problem is internally generated by Maple because Maple insists on ordering variables in its own way, no matter how I write them.

________simplified example from Maple 15

restart;
Z1:=sin(-a*x+b);
Z2:='sin(-a*x+b)';
Z3:=sin(b-a*x);



                         -sin(a x - b)
                         sin(-a x + b)
                         -sin(a x - b)
Y1:=subs(sin=cos,Z1);
Y2:=subs(sin=cos,Z2);


                         -cos(a x - b)
                         -cos(a x - b)
 correct answer should be cos(-a*x+b) but the calculated results are off by a minus sign.

Ans1:=evalf(subs(a=1,b=2,x=3,[Y1,Y2,cos(-a*x+b)]));
          [-0.5403023059, -0.5403023059, 0.5403023059]

Question: How do I override Maple's desire to stick the "-" sign outside the sin function?

Hi

 

I have some data:

Matrix(10, 2, {(1, 1) = 0, (1, 2) = 0, (2, 1) = .5, (2, 2) = 3.25, (3, 1) = 1.0, (3, 2) = 5.82, (4, 1) = 1.5, (4, 2) = 7.50, (5, 1) = 2.0, (5, 2) = 8.79, (6, 1) = 2.5, (6, 2) = 9.83, (7, 1) = 3.0, (7, 2) = 10.66, (8, 1) = 3.5, (8, 2) = 11.35, (9, 1) = 4.0, (9, 2) = 11.94, (10, 1) = 4.5, (10, 2) = 12.46})

 

I want Maple to make a trendline fitting a Logarithmic function. I can make it output some function with this:

LeastSquares(`<,>`(.5, 1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5), `<,>`(3.25, 5.82, 7.50, 8.79, 9.83, 10.66, 11.35, 11.94, 12.46), x, curve = a+b*ln(x))

It outputs:

5.96497783539274+4.25309474196387*ln(x)

 

But please notice, the dataset in the function does not have the first 0 and 0. If i do that:

LeastSquares(`<,>`(0, .5, 1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5), `<,>`(0, 3.25, 5.82, 7.50, 8.79, 9.83, 10.66, 11.35, 11.94, 12.46), x, curve = a+b*ln(x))

It outputs: 

Error, (in Matrix) numeric exception: division by zero

 

Besides that, i need the R-squard value for determinating how well it fits.

 

If i do the same thing i Excel the data set will give a formular: 5.5464ln(x)-0.2175 with a R-sward value of 0.9985.

 

How can i do this i maple?

 

Thanks in Advance!

 

----

Emil Kristensen

 

 

so we have to Write a maple function with -> that takes an integer N and a boolean function

F: {(i,j) l 0<= i,j<= N} -> {true,false} 


and returns a list containing all [i,j] such that F(i,j). A procedure that does this
would be


proc(N,F) local i, j, RV;
RV:=NULL;
for i from 1 to N do for j from 1 to N do
if F(i,j) then RV:=RV,[i,j] ; end if ;
end do ; end do ;
return RV ;
end proc ;


The problem is to do this inline, i.e. you have to write
(i,j)-> ...

 

please help...

Hi,

I have a 4-D function that I have found an analytic expression for in Maple, but I want to generate a numerical 4-D array, which I can export for use in Matlab. However, I don't think I'm using a very good solution, because it is extremely slow (several minutes for even 10x10x10x10 elements). My code is

NumRange:=10:
NumElements:=10:
dx:=2*NumRange/(NumElements-1):
A:=Array(1..NumElements,1..NumElements,1..NumElements,1..NumElements,[seq([seq([seq([seq(evalf(rho(ws1,ws2,ws1p,ws2p)),ws2p=-NumRange..NumRange,dx)],ws1p=-NumRange..NumRange,dx)],ws2=-NumRange..NumRange,dx)] , ws1=-NumRange..NumRange,dx)],datatype=float);

Rho is an exponential with 10 terms in the exponents.

I also tried using four for-loops but that was even slower!

Thanks for any input

Hello Maplers, i have encountered a little annoyance with Maple, that i would like to ask, whether it can be solved. 

 

It's when i try to define a function, like f(x)=2x and try to define it with a command f(x):=2x, a pop-up box comes up, asking me whether i'd like to use a 'function definition', or 'remember table assignment', and i would like to make Maple remember my choice that i want a function. 

 

I know i can write it like f:=x->2x, but i hate to look at that, to be frankly..

 

So, is there any way to solve this?

 

so we have to Write a function that diagonalises a complex (2x2) matrix if possible,  

we need the argument to be a (2x2) matrix say A.   and we need the return value to be a list [a1 ;a2 ;b1;b2] of two complex numbers followed by two 2-vectors such that {b1,b2} is a basis for C^2 and so that  

Ab1 =a1b1 , Ab2=a2b2  if these exist. if not then the function should return an empty list []

also, the thing is that we're not allowed to load any maple packages, we have to do it by hand :'(

thanks <3

 

Dear Maple users,

 

i have a set of 2 Lines: L1 (determined by the intersection of plane x + y -1=0 and plane x - z - 1=0), 

L2 ( intersection of plane x + y-7=0 and plane x-y+1 = 0 ).

which functions or commands of maple should I use "visualize" those 2 lines L1 and L2?

 

thanks for your help,

 

JJ

I am trying to create a procedure that can solve integrals using the Composite Simpson's 3/8 rule. However when I test my procedure against maple's ApproximateInt I am getting the wrong results.

Here is my attempt:

restart;


f:= x -> exp(x)*sin(4*x); # function I am using

simp := proc(a, b, n)
  local h, sum, i, single:
  h := (b-a)/n:
  sum := 0:
  single := (3*h/8) * (f(a) + f(b)): # this is the end points
    for i from a+h by h to b-h do
       sum := sum + (3*h/8) * (3*f(i)):
    end do:
print(evalf(sum + single));
end proc:


simp(0,1,12);
                                                                                0.6224486445
evalf(Student:-Calculus1:-ApproximateInt(f(x), 0..1, method = simpson[3/8], partition=12));

                                                                                0.5323516717

 

As you can see my answer is not very close to the answer given by Maple. I am not sure why my procedure simp is wrong.

Hi there,

I'm quite new to Maple so please forgive me! I have a system of partial differential equations I'm trying to solve in Maple as such below 

 

df/dt = f(1-f) - f * h

dg/dt = g(1-g) * Gradient(1-f * gradient(g))

dh/dt = (g - h) + Laplacian(h),

where f,g,h are functions of space and time (i.e. f(x,y,z,t)). I guess my first question is - is this possible in Maple to evaluate? (I'm currently unsure on ICs as I'm figuring it out from the model - it's a model for cancer growth I'm trying to evaluate but have a rough idea of what I'd use).

If it is possible, can you please share how I'd write this? Everytime I've tried I seem to be failing to define anything properly, so your expertise would be greatly appreciated!

and  x have range=(0 to 2),the y have range = (2,3),follow when the x=(3 to 4),so the y=(3 to -3) how to implement the title's function 

still the x and y have variability value and variability number

 

is it Morphism?...

February 13 2014 rit 235

i assign 

seta := [x+1, x^2]

setb := [x^3, 2*x+5]

 

does morphism mean that

i use card_prod

to get

(x+1, x^3)

(x+1, 2*x+5)

(x^2, x^3)

(x^2, 2*x+5)

such that i composite each of 4 sets still satisfy F(f o g) = f o g

example

subs(x=x^3, x+1)

How to find inverse function of a multivariable function

for example

f := x^2 + y + z^3

f := x^2 + y^3

How to test  associativity?

How to determine which of below has associativity?

 

The definition x*(y*z) = (x*y)*z.

asso := -(1/2)*(x+y+sqrt(x^2+2*x*y-3*y^2))/y;
asso := -(1/4)*(2*x+y+sqrt(4*x^2+4*x*y-7*y^2))/y;
asso := -(2*x+y)/(y+z);
asso := (1/2)*(-y-z+sqrt(y^2-2*z*y+z^2-8*z*x))/z;
asso := (1/2)*(-z+sqrt(z^2-4*z*x-4*z*y))/z;

what is the difference between endomorphism and identity map?

http://mathfaculty.fullerton.edu/mathews//n2003/hornermod.html

i find a program above, 

if no general method, how about specific to convert it

if have general method, how is it?

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