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Hello,

I have to simplify a piecewise function and Maple gets a more complicated solution than needed.




I don't know how to handle this kind of problems with Maple?
I don't understand why Maple doesn't see this?
Am I doing something wrong?

Thanks in advance for your help / advice.


# the code of my example
restart:
Mf(x):=piecewise(x<=L/2,1/2*x*F,x>1/2*L,1/2*x*F-F*(x-1/2*L));
# Make a dimensionless function:
# -    Mf(x):= Mf(xi)*F*L
# -    variable ξ  ( xi:=x/L )
eq[1]:=Mf(xi)*F*L=Mf(x);
Mf(xi):=solve(eq[1],Mf(xi));
Mf(xi):=subs(x=xi*L,Mf(xi));
# F is the Force and L is the Length of the beam:
Mf(xi):=simplify(Mf(xi)) assuming F>0,L>0;
print("When I simplify this function by hand it will be");
Mf(xi):=piecewise(xi<=1/2,1/2*xi,xi>1/2,-1/2*xi+1/2);




Hi all,

I will use the following dummy example.

with function,f

f:=(xid,yid)->sum(x[i],i=1..xid)*sum(y[i],i=1..yid);

 

and a complicated term, myterm


myterm:=(f(3,4)+f(2,2))*f(1,1):
myterm:=expand(myterm);

 

'if' i have some previous knowledge, or know a bit of the term, i can find the structure by doing this


repar:=[f(1,1),f(2,2),f(3,4)];  # Or with more f(xid,yid) terms

tmp:=seq(repar[i]=ff[i],i=1..3);
simplify(myterm,{tmp});          # This is fine, gives me what i want

 

But, can we go further, and more 'obvious'

 

Given the fucntion f, same as before, and the same 'myterm'

can I have this
restart:
iwant:=(f(3,4)+f(2,2))*f(1,1);  # as a result, straightforward

so I dont have to go back to 'repar' and find that the terms exactly are.

 

Thanks,

 

Hi say I have the vector V1.

V1:=Vector([a,b,c,d,e,f,g]):

and function myfun.

 

how do i use it as the input to the function my fun, by taking away each element in turn?
myfun(V1[2..]);              # 1st element removed
myfun(V1[[1,3..]]);         # 2nd element removed
myfun(V1[[1,2,4..]]);      # 3rd element removed

and so go

is there a more efficient way?

 

Many thanks,

This has been buggin me all day, but if i define a variable with an expresion say "a:=3 x+5" and i then want to assign this expression to a function like so "f:=x->a". if i then call f(7), maple return with "3 x+5" so just returning the value of a without substituting x with 7.

Is there any way in which i can define a function this way?

I am new user of Maple. I have an expression like f(x,y)/g(x,y) and I want to have Maple name f(x,y) as something simple like "h" so that later equations write h/g(x,y) instead of the complete f(x,y)/g(x,y), and also to be able to simplify other equations in terms of h.

For example, a=(x+1)(y-5)^2/(x+y-3). Can I get Maple to let h=(x+1)(y-5)^2, so that I get h/(x+y-3) when I ask for a?

Sorry if this question is basic (and my math terminology incorrect), but I have been trying for a while.

Hi all

In matlab software we have a command namely fmincon which minimizes any linear/nonlinear algebric equations subject to linear/nonlinear constraints.

Now my question is that: what is the same command in maple?or how can we minimize linear/nonlinear function subject to linear/nonlinear constraints in maple?

thanks a lot

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

Hi all,
I need to find mina(w) then a=-1,0,1 and

w := piecewise((x-a)/t <= -10, 202*a-200*x-(202001/101)*t, (x-a)/t <= -910/101, 3*a-x-(1011/101)*t, (x-a)/t <= 910/101, 2*a-t, (x-a)/t <= 10, a+x-(1011/101)*t, 10 < (x-a)/t, -198*a+200*x-(202001/101)*t)

Also need to plot it, i think it should look like this 

Thanks

 

 

 

Hello fellow maple fans :)

 

I have tried to plot this continuous Piecewise function

plot(piecewise(0 <= t and t < 2, 2-t, 2 <= t and t <= 3, 2*t-4))

On the interval [0,6]

When i try it plot it it won't show the entre internal, is there a way to tell maple that the funktion should continuous until [0,6]

Please see this http://snag.gy/NPrjG.jpg if you are not sure what i mean in the end it should look like this http://snag.gy/ScgtF.jpg

Greetings David Thanks in advance :)

 

I've been poking around with convolutions on Maple, and some weird behavior came up---if I let it compute the convolution of a piecewise function, then take the convolution of that, it comes out differently than if I enter a function from scratch as the middle step---file attached (PiecewiseProblem.mw).  I'm not really a Maple pro, so am I'm doing something crazy here?

Thanks!

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

 

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