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Piecewise command...

November 20 2014 ctc 15

A function f is defined on R by

 

f(x):= (1+a|x|)^1/x      , x<0

         B                      ,  x=0

         ln(1+(a^2)|x|)/x , x>0

 

where α and β are constants. Investigate whether it is possible to choose α and β so
as to ensure that f is real-valued and continuous at x = 0. Compute any such values
for α and β correct to 10 significant figures. Make use of the piecewise command in
plotting a graph of any resulting continuous function(s) f over the range −20 ≤ x ≤ 20.

 

I used the help in Maple and manage to get 

f = piecewise(x < 0, (1+alpha*abs(x))^(1/x), x = 0, beta, x > 0, ln(1+alpha^2*abs(x))/x)

 

Not sure about how to compute a and B...

What does it mean by  f is real-valued and continuous at x = 0?

 

Hello,
Maple does not cancel out a variable.

Why is that?

Is there a way to solve this? 

(I pasted my code on the bottom of this message)

 

Thanks for your help/advice,

Stephan

restart:
M(x):=piecewise(x<=l,1/2*(q*x^2)/(EI)-3/8*(q*l*x)/(EI),l<x,1/2*(q*x^2)/(EI)-13/8*(q*l*x)/(EI)+5/4*(q*l^2)/(EI)):
M(x):=M(x)*(-EI);
# simplify() does not work.....?
M(x):=simplify(%) assuming EI>0;
# Wiht EI cancelled out by hand it schould look like:
M(x):=piecewise(x<=l,1/2*(q*x^2)-3/8*(q*l*x),l<x,1/2*(q*x^2)-13/8*(q*l*x)+5/4*(q*l^2));

 

Hi all

kx,ky is the wavenumber, how can I get the 4 cases of piecewise function according to kx=0,kx≠0 and ky=0,ky≠0. Thanks

J := `assuming`([4*(int(int(JJ*exp(-I*(kx*x+ky*y))*sin(2*l*pi*x/a)*sin(2*k*pi*y/b), x = 0 .. a, AllSolutions), y = 0 .. b, AllSolutions))/(a*b)], [k::posint, l::posint, a > 0, b > 0, JJ > 0])

Hi,

 

I'm trying to make a piecewise function that will have period 12. That is, it repeats every 12 units across the x-axis.

 

I managed to do one cycle successfully with

plot(piecewise(0<=t and t<2, 4,2<=t and t<12,0), t=0..13);

But I'm not sure how to do it periodically. I thought about using modular arithmetic in the conditions and setting up something like

H(t)=piecewise(0<=modp1(f(t),12) and modp1(f(t),12)<2, 4,2<=modp1(f(t),12) and modp1(f(t),12)<12,0);

but this clearly doesn't work.

 

Some direction or advice would be appreciated. Thanks!

 

Hi all

I have a mathematical problem and I asked it in various sites but the answers till yet are not correct.

Assume that we have:

T[m]:=t->t^m:
b[n,m]:=unapply(piecewise(t>=(n-1)*tj/N and t<n*tj/N, T[m](N*t-(n-1)*tj), 0), t):

where n,N,tj are known constants. furthermore assume that we want to comute the following integral:

for following approximations:

I have written the following code but it seems to be incorrect:

V1:=Vector([seq(seq(b[n,m](t),m=0..1),n=1..3)]);
V:=evalf(V1.Transpose(V1));

the original program is :

taaylor.mws

I will be so grateful if any one can help me to solve it by maple

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

hi all.

I have wrore the following program for optimization with bernstein and block pulse hybrid functions.

the program have some errors which i can't understand.

Bernestien1.mws

restart:

alias(C=binomial):
with(LinearAlgebra):
macro(LA= LinearAlgebra):


HybrFunc:=proc(N, M,  tj)               # N=Number of subintervals,  M=Number of functions in subintervals
 
local B, n, m;

global b;

for n from 1 to N do
for m from 0 to M-1 do

B := (i,m,t) -> C(m,i)*(1-t)^(m-i)*t^i:

b[n,m]:=unapply(piecewise(t>=(n-1)*tj/N and t<n*tj/N, B(m,2,N*t-(n-1)*tj), 0), t):
 od:od:


Array(1..N, 0..M-1, (n,m)->b[n,m](t)):

#convert(%,vector);
end proc:

HybrFunc(3, 3, 1);




                                       # End Of Definition
 
g2(t):=t;            #*exp(t-1):                      # Any other function can be replaced here
    

g1(t):=add(add(c[n,m]*b[n,m](t), m=0..2), n=1..3);
Optimization[Minimize](sqrt(int((g2(t)-g1(t))^2, t=0.. 1)));
assign(op(%[2]));
plot([g2(t),g1(t)], t=0..1, 0..5, color=[blue,red],thickness=[1,3],discont, scaling=constrained);

Array(1 .. 3, 0 .. 2, {(1, 0) = piecewise(0 <= t and t < 1/3, (1-3*t)^2, 0), (1, 1) = piecewise(0 <= t and t < 1/3, (6*(1-3*t))*t, 0), (1, 2) = piecewise(0 <= t and t < 1/3, 9*t^2, 0), (2, 0) = piecewise(1/3 <= t and t < 2/3, (2-3*t)^2, 0), (2, 1) = piecewise(1/3 <= t and t < 2/3, (2*(2-3*t))*(3*t-1), 0), (2, 2) = piecewise(1/3 <= t and t < 2/3, (3*t-1)^2, 0), (3, 0) = piecewise(2/3 <= t and t < 1, (3-3*t)^2, 0), (3, 1) = piecewise(2/3 <= t and t < 1, (2*(3-3*t))*(3*t-2), 0), (3, 2) = piecewise(2/3 <= t and t < 1, (3*t-2)^2, 0)}, datatype = anything, storage = rectangular, order = Fortran_order)

g2(t) := t

"g1(t):=c[1,0] ({[[(1-3 t)^2,0<=t and t<1/3],[0,otherwise]])+c[1,1] ({[[6 (1-3 t) t,0<=t and t<1/3],[0,otherwise]])+c[1,2] ({[[9 t^2,0<=t and t<1/3],[0,otherwise]])+c[2,0] ({[[(2-3 t)^2,1/3<=t and t<2/3],[0,otherwise]])+c[2,1] ({[[2 (2-3 t) (3 t-1),1/3<=t and t<2/3],[0,otherwise]])+c[2,2] ({[[(3 t-1)^2,1/3<=t and t<2/3],[0,otherwise]])+c[3,0] ({[[(3-3 t)^2,2/3<=t and t<1],[0,otherwise]])+c[3,1] ({[[2 (3-3 t) (3 t-2),2/3<=t and t<1],[0,otherwise]])+c[3,2] ({[[(3 t-2)^2,2/3<=t and t<1],[0,otherwise]])"

Error, (in Optimization:-NLPSolve) complex value encountered

Error, invalid left hand side in assignment

(1)



Download Bernestien1.mws

 I'll be so grateful if any one can help me.

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

Is it possible to solve piecewise differential equations directly instead of separating the pieces and solving them separately.

like for example if i have a two dimensional function f(t,x) whose dynamics is as follows:

dynamics:= piecewise((t,x) in D1, pde1, pde2); where D1 is some region in (t,x)-plane

now is it possible to solve this system with one pde call numerically?

pde(dynamics, boundary conditions, numeric); doesnot work

Hello Maple-Primers!

I am trying to evaluate a system at many different points.  I would like to include an interpolation function in this system, but have thusfar been unsuccessful.

Usually, I solve a system symbolically by using eliminate and unapply:

eq[1] := A = M^3;
eq[2] := C = A*2;
eq[3] := D = N+3;
eq[4] := B = piecewise(A = 0, 0,C);
eq[5] := E = B*D;
elimsol:=eliminate(convert(eq,list),[A,B,C,D,E])[1];

unappsol:=unapply(elimsol,[N,M]);

unappsol(1,2);
{A = 8, B = 16, C = 16, D = 4, E = 64} <--- great!

Now, I want to include an interpolation function in the system of equations.  They look like this (see worksheet for actual interpolation function):

B_interp := (W,T) -> CurveFitting:-ArrayInterpolation([FC_Map_W,FC_Map_T],FC_Map,Array(1 .. 1, 1 .. 1, 1 .. 2, [[[W, T]]]),method=linear);

eq[5] := E = B_interp(N,M);

Error, (in CurveFitting:-ArrayInterpolation) invalid input: coordinates of xvalues must be of type numeric <-- bad!

Anyone have any ideas?  I've tried to use polynomials, but I can't seem to get a fit close enough for my purposes.

Maple_2D_Interpolate_FC.mw

I have some data for a model in MapleSim that I would like to use a time look up table with.  I've found that the two options for interpolation are linear and 1st derivative, but the data was intended to be interpretted as piecewise constant.  Is there any way to acheive this option in MapleSim?

Hi Mapleprimers,

I'm using CodeGeneration to convert a procedure I obtained with unapply() into a Matlab function.  I'm having problems getting the outputting function to run correctly in Matlab.  I'm going to dynamically generating equations, so directly editing the Matlab code won't work here.  I'm having problems getting any output in Matlab.  Here is the code I'm working on:  Series_addGear_codegen.mw

Ideally I would like output in a matrix.  I've tried putting the unapplied procedure in another procedure, but the CodeGeneration doesn't work.

This is the maple output from the function:

unapp(1,2,3,4,5);
{BAT_A = -2.267032891, BAT_V = 271, EM2_A = .4615464218, EM2_P = 125.0790803, EM2_T = -1, EM2_V = 271, EM2_W = 2, GBa_T = 12, GBa_W = 5/3, GBb_T = -4, GBb_W = 5, GEN_A = 1.805486469, GEN_P = 489.2868330, GEN_T = -12, GEN_V = 271, GEN_W = 5/3, ICE_mdot_g = 20}

Since I'll know the order of variables, I want the Matlab function to output:

[-2.267032891,  271,  .4615464218,  125.0790803,  -1, ...]

This is the output after putting the Matlab function in Matlab:

>> unapp(1, 1, 1, 1, 1)
Undefined function or variable 'BAT_A'.

Error in unapp (line 39)
unappreturn = unique([BAT_A == -t35 / 0.271e3 - t73 / 0.271e3 BAT_V == 271 EM2_A == t35 /
0.271e3 EM2_P == t35 EM2_T == -FD_T EM2_V == 271 EM2_W == FD_W GBa_T == t44 GBa_W == t41 GBb_T
== -ICE_T GB

This is the Matlab code that is generated by Maple 18:

MCode:=CodeGeneration[Matlab](unapp, declare = [FD_T::float, FD_W::float, GB_R::float, ICE_T::float, ICE_W::float],defaulttype=float,optimize,defaulttype = numeric);
Warning, could not preprocess. Found `abs` or similar in the 'While'/'For' conditions.
Warning, procedure/module options ignored
function unappreturn = unapp(FD_T, FD_W, GB_R, ICE_T, ICE_W)
t2 = 0.1e1 * FD_T * FD_W;
t5 = abs(FD_W);
t7 = abs(FD_T);
t9 = t5 ^ 2;
t13 = t7 ^ 2;
t15 = t9 * t5;
t21 = t13 * t7;
t23 = t9 ^ 2;
t31 = t13 ^ 2;
t33 = 0.1483000000e3 - 0.4267000000e1 * t5 - 0.1277000000e2 * t7 + 0.3640000000e-1 * t9 - 0.1160000000e1 * t5 * t7 + 0.2580000000e0 * t13 - 0.1181000000e-3 * t15 + 0.5994000000e-3 * t9 * t7 - 0.1171000000e-3 * t5 * t13 - 0.1739000000e-2 * t21 + 0.1245000000e-6 * t23 - 0.1200000000e-5 * t15 * t7 + 0.1584000000e-5 * t9 * t13 - 0.4383000000e-6 * t5 * t21 + 0.2947000000e-5 * t31;
if (-t2 == 0.0e0)
t35 = 0.0e0;
elseif (-t2 < 0.0e0)
t35 = t33;
else
t35 = -t33;
end
t36 = ICE_T * ICE_W;
t37 = 0.1e1 * t36;
t41 = ICE_W / GB_R;
t42 = abs(t41);
t44 = ICE_T * GB_R;
t45 = abs(t44);
t47 = t42 ^ 2;
t51 = t45 ^ 2;
t53 = t47 * t42;
t59 = t51 * t45;
t61 = t47 ^ 2;
t69 = t51 ^ 2;
t71 = 0.5280000000e-11 - 0.3849000000e-13 * t42 + 0.7190000000e2 * t45 + 0.1168000000e-15 * t47 - 0.1296000000e1 * t42 * t45 - 0.2489000000e1 * t51 - 0.1451000000e-18 * t53 - 0.1326000000e-3 * t47 * t45 + 0.8141000000e-2 * t42 * t51 + 0.4539000000e-2 * t59 + 0.6325000000e-22 * t61 + 0.2091000000e-6 * t53 * t45 - 0.3455000000e-5 * t47 * t51 - 0.2499000000e-4 * t42 * t59 + 0.5321000000e-4 * t69;
if (-t37 == 0.0e0)
t73 = 0.0e0;
elseif (-t37 < 0.0e0)
t73 = t71;
else
t73 = -t71;
end
unappreturn = unique([BAT_A == -t35 / 0.271e3 - t73 / 0.271e3 BAT_V == 271 EM2_A == t35 / 0.271e3 EM2_P == t35 EM2_T == -FD_T EM2_V == 271 EM2_W == FD_W GBa_T == t44 GBa_W == t41 GBb_T == -ICE_T GBb_W == ICE_W GEN_A == t73 / 0.271e3 GEN_P == t73 GEN_T == -t44 GEN_V == 271 GEN_W == t41 ICE_mdot_g == t36]);

 

 

 

i am using maple to solve a system of ordinary differential equations , 3 unknows (x,y, x ), and 3 equations (dx/dt,dy/dt,dz/dt)

there is one known variable denpendent on x and z

# code begins here

if x(t) <= z(t) then Q(t) := 8 end if;

if x(t) > z(t) then Q(t) := 10 end if;

 

eq1 := diff(x(t), t) = 3*x(t)-1;

eq2 := diff(y(t), t) = y(t)+Q(t);

eq3 := diff(z(t), t) = z(t);

eqs := {eq1, eq2, eq3};

 

# code ends here

 

above i put the system of ODEs, the code maybe illegal in maple, but i wrote in this way to make it clear.

Q is dependent on x and z.

 

in the past, when i was trying to solve ODEs, normally, eqs contains with only x,y,z as unknowns. but in this eqs, clearly, Q is included as an unknown. 

 

i've tried to use piecewise function to express Q(t), but failed.

 

how could i solve a system like this? thanks 

 

 

Hi all,

As we know that the indefinite integral in Maple is defined up to a piecewise constant.

For example,

 

Due to the indeterminate range of variable n, int returns the piecewise function.

But in the other similar case, we get a generic solution rather than a piecewise function,

 

Why not to returns the result like this,

 

Thanks for any help.

f:=piecewise(x>0,1+x,x=0,2,x<0,1+x^2)

in my work i must use some parts of this piecewise function.

But i don't know how can i call the part that i will use.

How can i do that ?

Thanks for help

 

Hi all;

Hope all of you  be in good health

I want to construct a special function b_{nm}(t) like:

with piecewise command i did it but the result is incorrect.

any one can help me to do it?

Best wishes

 

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

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);




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