is it possible to generalize a function to a combinatorial level for approximate axioms

for example, first 100 or 1000 data points satisfy axioms

or 100% satisfy a axioms which means satisfy to infinity

because i find data always not exactly satisfy the axioms,
i guess it only satisfy to some limit, this may explain why data has decimal number

or conversely is it possible to generalize some axioms which approximate the original exact axioms
then data can exactly satisfy the approximate axioms

can generalize a nested forloop to achieve this goal?

how can it be done in algebra point of view?

For example:

x*y = for loop -> for loop -> i*j

it can change for loop expression into algebra

for i from 1 to 10 do
for j from 1 to 10 do
print i*j
od:
od:

Hi

 Mt := 1-(sum(4*l^2*exp(-beta[n]^2*tau)/(beta[n]^2*(l^2+beta[n]^2)), n = 1 .. infinity))

where the beta[n]s   are  roots of :

beta[n]*BesselJ(1, beta[n])-l*BesselJ(0, beta[n]) = 0

for  l=1,10,20,40,50,100

I want to plot Mt vs. tou for these l 's  in one diagram

Hi,

I have a problem solving two equations.  They are as follows:

s := 1/(273.16+50); s1 := 1/(273.16+145); s3 := 1/(273.16+250); s2 := 1/(273.16+197.5); gamma0 := 0.1e-3; gamma1 := .5; gamma2 := 0.15e-2; beta := -3800:

c := 300; n := 200; tau1 := 99; tau2 := 120;

Delta := solve(1-exp(-(gam0*tau1+(1/2)*gam1*tau1^2)*exp(beta*s1)) = 1-exp(-(gam0*a+(1/2)*gam1*a^2)*exp(beta*s2)), a);
a := Delta[1];

Theta := solve(1-exp(-(gam0*(a+tau2-tau1)+(1/2)*gam1*(a+tau2-tau1)^2)*exp(beta*s2)) = 1-exp(-(gam0*b+(1/2)*gam1*b^2)*exp(beta*s3)), b);
b := Theta[1];

n1 := int((gam1*t+gam0)*exp(beta*s1)*exp(-(gam0*t+(1/2)*gam1*t^2)*exp(beta*s1)), t = 0 .. tau1);
n22 := (n-n1)*(int((gam1*t+gam0)*exp(beta*s2)*exp(-(gam0*t+(1/2)*gam1*t^2)*exp(beta*s2)), t = a1 .. a1+tau2-tau1));
n2 := eval(n22, a1 = a);
n33 := (n-n1-n2)*(Int((gam1*t+gam0)*exp(beta*s3)*exp(-(gam0*t+(1/2)*gam1*t^2)*exp(beta*s3)), t = b1 .. c));
n3 := eval(n33, a1 = a);
n4 := n-n1-n2-n3;

g1 := -n1*(Int((1/(gam1*t+gam0)-t*exp(beta*s1))*(gamma2*t^2+gamma1*t+gamma0)*exp(beta*s1)*exp(-(gamma0*t+(1/2)*gamma1*t^2+(1/3)*gamma2*t^3)*exp(beta*s1)), t = 0 .. tau1))-n2*(Int((1/(gam0+gam1*(a+t-tau1))-(a+t-tau1)*exp(beta*s2))*(gamma0+gamma1*(a+t-tau1)+gamma2*(a+t-tau1)^2)*exp(beta*s2)*exp(-(gamma0*(a+t-tau1)+(1/2)*gamma1*(a+t-tau1)^2+(1/3)*gamma2*(a+t-tau1)^3)*exp(beta*s2)), t = tau1 .. tau2))-n3*(Int((1/(gam0+gam1*(b+t-tau2))-(b+t-tau2)*exp(s3))*(gamma0+gamma1*(b+t-tau2)+gamma2*(b+t-tau2)^2)*exp(beta*s3)*exp(-(gamma0*(b+t-tau2)+(1/2)*gamma1*(b+t-tau2)^2+(1/3)*gamma2*(b+t-tau2)^3)*exp(beta*s3)), t = tau2 .. c))+(n-n1-n2-n3)*(1/(gam0+gam1*(b+c-tau2))-(b+c-tau2)*exp(s3))*(gamma0+gamma1*(b+c-tau2)+gamma2*(b+c-tau2)^2)*exp(beta*s3)*exp(-(gamma0*(b+c-tau2)+(1/2)*gamma1*(b+c-tau2)^2+(1/3)*gamma2*(b+c-tau2)^3)*exp(beta*s3));

g2 := -n1*(Int((t/(gam1*t+gam0)-(1/2)*t^2*exp(beta*s1))*(gamma2*t^2+gamma1*t+gamma0)*exp(beta*s1)*exp(-(gamma0*t+(1/2)*gamma1*t^2+(1/3)*gamma2*t^3)*exp(beta*s1)), t = 0 .. tau1))-n2*(Int(((a+t-tau1)/(gam0+gam1*(a+t-tau1))-(1/2)*(a+t-tau1)^2*exp(beta*s2))*(gamma0+gamma1*(a+t-tau1)+gamma2*(a+t-tau1)^2)*exp(beta*s2)*exp(-(gamma0*(a+t-tau1)+(1/2)*gamma1*(a+t-tau1)^2+(1/3)*gamma2*(a+t-tau1)^3)*exp(beta*s2)), t = tau1 .. tau2))-n3*(Int(((b+t-tau2)/(gam0+gam1*(b+t-tau2))-(1/2)*(b+t-tau2)^2*exp(s3))*(gamma0+gamma1*(b+t-tau2)+gamma2*(b+t-tau2)^2)*exp(beta*s3)*exp(-(gamma0*(b+t-tau2)+(1/2)*gamma1*(b+t-tau2)^2+(1/3)*gamma2*(b+t-tau2)^3)*exp(beta*s3)), t = tau2 .. c))+(n-n1-n2-n3)*((b+c-tau2)/(gam0+gam1*(b+c-tau2))-(1/2)*(b+c-tau2)^2*exp(s3))*(gamma0+gamma1*(b+c-tau2)+gamma2*(b+c-tau2)^2)*exp(beta*s3)*exp(-(gamma0*(b+c-tau2)+(1/2)*gamma1*(b+c-tau2)^2+(1/3)*gamma2*(b+c-tau2)^3)*exp(beta*s3));

solve({g1 = 0, g2 = 0}, {gam0, gam1});

Warning, solutions may have been lost.

What do I do wrong?

Thanks for advice in advance.

Hi there,

I am trying to maximize a function given a set of values to a parameter in the function. The function is an differential equation belonging to a system of two differential equations.

I have a for loop to state different values to the parameter.

Maple yields the error:

Error, (in Optimization:-NLPSolve) cannot evaluate the solution further right of 0.17757507e-4, probably a singularity

When trying to maximize the function.

Supposed that I was doing something wrong in the loop, if I reproduce the contents of the loop outside, and set a value for the parameter. If I plot the solution of the ordinary differential equation, I can see where the maximum lies.

Having plot it, the Optimizamtion:-Maximize works as expected.

However, omitting the plot has a weird effect: I only get the same result depending on the bounds I set for the Maximization:

de1 := diff(A(t), t) = r*m*(1-g)*A(t)-piecewise(t < 8, r*A(t), t >= 8, (r+k)*A(t));
de2 := diff(G(t), t) = r*m*g*A(t)-l*G(t);

ics := A(0) = 25.0, G(0) = 0.;
num := dsolve({de1, de2, ics}, {A(t), G(t)}, type = numeric, output = listprocedure, parameters = [g]);

num(parameters = [g = .15]);
val := eval(G(t), num);

# odeplot(val, [t, G(t)], t = 0 .. 100);

Maximize(val);
Error, (in Optimization:-NLPSolve) cannot evaluate the solution further right of 0.17757507e-4, probably a singularity

val2 := Maximize(val);

Error, (in Optimization:-NLPSolve) cannot evaluate the solution further right of 0.17757507e-4, probably a singularity

val3 := Maximize(val(t), t = 0 .. 60);

  [10267.824035766165, [t = 8.25727747134303]]

val4 := Maximize(val(t), t = 0 .. 100);

[6.863211343195069e-9, [t = 59.84184367042171]]

The right answer is [10267.824035766165, [t = 8.25727747134303]]: Why do I get two different answers even if in that range there is only one relative maximum?

I ignore whether the way I am specifying the arguments for the Maximize function is correct. val is a procedure.

What am I missing?

Attached is the worksheet: MaplePrimes_malaria_param_variation_2.mw

Thanks,

jon

Hi all!

F is a delta function:

F:=delta(x-x[0])*delta(y-y[0])

I want it be expaned through trigonometric series:

F:=sum(sum(Q[k*l]*sin(l*Pi*x/a)*sin(k*Pi*y/b), k = 1 .. infinity), l = 1 .. infinity)

So I want to get every Q:

Q[k, l] := `assuming`([4*(int(int(f[z1]*sin(l*Pi*x/a)*sin(k*Pi*y/b), x = 0 .. b), y = 0 .. a))/(a*b)], [k::posint, l::posint, a > 0, b > 0])

But it result in (when x[0]:=a/2, y[0]:=b/2):

4*(int(int(F[0]*exp(I*omega*t)*delta(x-x[0])*delta(y-y[0])*sin(l*Pi*x/a)*sin(k*Pi*y/b), x = 0 .. b), y = 0 .. a))/(a*b)

I wonder HOW CAN I GET THE EXACT RESULT:Q[k, l] := 4*sin(l*Pi/a)*sin(k*Pi/b)/(a*b)

THANKS!

http://en.wikipedia.org/wiki/Hypersurface

http://people.cs.uchicago.edu/~niyogi/papersps/surfacesampling.pdf

hypersurface is a homogenous polynomial

f(x,y) = 0

i do not understand how sampling hypersurface can generate this kind of polynomial

Dear All,

i am solving a system of pde with boundar conditons then i got this error...

Error, (in pdsolve/numeric/plot) unable to compute solution for tau>HFloat(0.0):

Thank.

jeffrey_fluid.mw

Download jeffrey_fluid.mw

Good day everyone, could you please help use Gauss Elimination method for these system of equations. See the worksheet here F1.mw

Thanks.

restart; with(linalg); with(stats); with(plots); with(Statistics); with(LinearAlgebra); 

s := 1/(273.16+50); s1 := 1/(273.16+145); s3 := 1/(273.16+250); s2 := 1/(273.16+197.5); gamma0 := 0.1e-3; gamma1 := .5; gamma2 := 0.15e-2; beta := -3800;
c := 300; n := 200; tau1 := 99; tau2 := 120;

Delta := solve(1-exp(-(gam0*tau1+(1/2)*gam1*tau1^2)*exp(beta*s1)) = 1-exp(-(gam0*a+(1/2)*gam1*a^2)*exp(beta*s2)), a);
a := Delta[1];

Theta := solve(1-exp(-(gam0*(a+tau2-tau1)+(1/2)*gam1*(a+tau2-tau1)^2)*exp(beta*s2)) = 1-exp(-(gam0*b+(1/2)*gam1*b^2)*exp(beta*s3)), b);
b := Theta[1];

n1 := n*(int((gam1*t+gam0)*exp(beta*s1)*exp(-(gam0*t+(1/2)*gam1*t^2)*exp(beta*s1)), t = 0 .. tau1));
200. - 200. exp(-0.01119474511 gam0 - 0.5541398828 gam1)
n2 := (n-n1)*(int((gam1*t+gam0)*exp(beta*s2)*exp(-(gam0*t+(1/2)*gam1*t^2)*exp(beta*s2)), t = a .. a+tau2-tau1));

g1 := -n1(gam0, gam1)*(int((1/(gam1*t+gam0)-t*exp(beta*s1))*(gamma2*t^2+gamma1*t+gamma0)*exp(beta*s1)*exp(-(gamma0*t+(1/2)*gamma1*t^2+(1/3)*gamma2*t^3)*exp(beta*s1)), t = 0 .. tau1))-evalf(n2(gam0, gam1)*(int((1/(gam0+gam1*(a+t-tau1))-(a+t-tau1)*exp(beta*s2))*(gamma0+gamma1*(a+t-tau1)+gamma2*(a+t-tau1)^2)*exp(beta*s2)*exp(-(gamma0*(a+t-tau1)+(1/2)*gamma1*(a+t-tau1)^2+(1/3)*gamma2*(a+t-tau1)^3)*exp(beta*s2)), t = tau1 .. tau2)))

g2 := -n1*(int((t/(gam1*t+gam0)-(1/2)*t^2*exp(beta*s1))*(gamma2*t^2+gamma1*t+gamma0)*exp(beta*s1)*exp(-(gamma0*t+(1/2)*gamma1*t^2+(1/3)*gamma2*t^3)*exp(beta*s1)), t = 0 .. tau1))-evalf(n2*(int(((a+t-tau1)/(gam0+gam1*(a+t-tau1))-(1/2)*(a+t-tau1)^2*exp(beta*s2))*(gamma0+gamma1*(a+t-tau1)+gamma2*(a+t-tau1)^2)*exp(beta*s2)*exp(-(gamma0*(a+t-tau1)+(1/2)*gamma1*(a+t-tau1)^2+(1/3)*gamma2*(a+t-tau1)^3)*exp(beta*s2)), t = tau1 .. tau2)))

solve({g1 = 0, g2 = 0}, {gam0, gam1})

I want to find the answer of gam0 and gam1. It takes me 20 hours until now...and still evaluating...

Please Help ..

I can't directly copy Maple code into messages but have to insert into Word first, apply some changes that for better readability and copy it from there via Word-paste. Is there a direct way?

I am trying separation of variables in Maple. I get an equation that has the form   

And now I want to tell Maple to assign the terms with 1/R in them to one variable, say eq21, and the term with 1/Z to second variable, say eq22 The idea is that I can later more easily work with each separate ode. 

I do not know to separate those apart. I could offcourse copy and paste by hand, but I want to automate this.

I tried match() and patmatch() but I am not seeing the way. Here is the code:

restart;
T:=  (r,z)-> Z(z)*R(r);
eq1:= diff(T(r,z),r$2)+1/r*diff(T(r,z),r)+diff(T(r,z),z$2);
eq2:=expand(eq1/(Z(z)*R(r)));

#need now way to break the above into 2 different variables.

ps. I know I can do this:

restart;
T:=  (r,z)-> Z(z)*R(r);
eq1:= diff(T(r,z),r$2)+1/r*diff(T(r,z),r)+diff(T(r,z),z$2);
eq2:=expand(eq1/(Z(z)*R(r)));
eq3:=collect(eq2,1/R(r));
eq21:=op(1,eq3);
eq22:=op(2,eq3);

But this for me is not the right way to do it. I think there should be more algebraic way.

there is a solution of equation,so the equation can be divided by the solution,but because the equation is complex,it can't be simplify by the soution,can anyone give me some help?thanks a lot.

Buenas

Alguien tiene el planteamento de la ecuacion 1D HEAT EQUATION en maple?

Me pueden enviar?

Gracias

Saludos

En anexo esta el enunciado

a curve has residual p if it is linked, in a complete intersection, to a curve with residual p-1

0 residual if is a complete intersection of two surfaces

do complete intersection means two surfaces totally overlapped?

why they are not the same one if complete intersection?

test.mw

In this file, I tried my best to solve the pde. But the answer is still rather non-informative. I need some help to simplify it.

I did notice that my Maple might need reinstallation, due to a "bug" in the 18.02 update.

My ultimate aim is try to use some similar techniques to solve this, test2.mw, which has a similar type pde.

The standard pdsolve(pde) would just not work.

UPDATE:

I used the same file in Maple 17 on a differnt Machine, which can be solved by pdsolve. So I guess it's just that  the 18.02 update package is broken itself. I have tried to uninstall and reinstall twice.

Thanks,

casper