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mmcdara

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6 years, 289 days
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These are replies submitted by mmcdara

@vs140580 The reason is probably th...

mmcdara 4500
February 09 2023
0 0

@vs140580 

The reason is probably the word "+" in the file name, try this one and let me know.
PLS_PCR_Linnerund.mw

I advise you to start by searching for t...

mmcdara 4500
February 09 2023
0 0

I advise you to start by searching for the word homotopy on this site.
You will probably find some interesting things.

I think if you want someone to help you,...

mmcdara 4500
February 09 2023
0 0

I think if you want someone to help you, you should provide a Maple code instead of an excerpt from a PDF file that nobody here is willing to code themselves.

@Kitonum Both @tomleslie and I...

mmcdara 4500
February 07 2023
0 0

@Kitonum 

Both @tomleslie and I have already answered the same thing to a previous question
https://www.mapleprimes.com/questions/235674-Chevrons-Lemma.
It seems that @JAMET ​ does not read the answers it's been provided.

@JAMET

I also notice, as @tomleslie pointed out recently, that yours questions never provide any worksheet, a remark that obviously did not change your attitude..
I find this latter contemptuous and would personally make it a point to never respond to you again.

@vs140580  By the way, it happens ...

mmcdara 4500
February 06 2023
0 0

@vs140580 

By the way, it happens that one of the best book ever (IMO) on data mining, inference and prediction is freely avaliable:
https://hastie.su.domains/Papers/ESLII.pdf

In my paper version Chapter 13 is entirely devoted to Nearest-Neighbors methods.
Several algorithms (sometimes in pseudo code are provided too).

PCR is described in pp 79-80, PLS in pp 80-82 and, more generaly, regression methods in Chapter 3.

@Carl Love  Yeah, it was indeed cl...

mmcdara 4500
February 06 2023
0 0

@Carl Love 

Yeah, it was indeed clustering.
I don't remember exactly what you did but KNN is just a particular type of clustering.
Maybe this could help @vs140580

@Delali Accolley  I consider that ...

mmcdara 4500
February 05 2023
0 0

@Delali Accolley 

I consider that I have spent enough time trying to explain to you why your problem is not correctly set.
You have obviously paid no attention to my arguments and keep getting the same error again and again.
Please consider me finished with you.
I hope you find a more conciliatory ear than mine.

@tomleslie once again no uploaded w...

mmcdara 4500
February 05 2023
0 0

@tomleslie 

once again no uploaded worksheet.

There are always guys like that who think that there will be a kind soul to deal with all their mess

@Delali Accolley I would have hoped...

mmcdara 4500
February 05 2023
0 0

@Delali Accolley 

I would have hoped that you would have solved the question about "v" yourself.
Some points you have missed:

  • I see you have removed the option "numeric in pdsolve, but do you really think that your equations can be solved formally?
  • In your case you do use the option"numeric" (that is not that simple that what I showed in my previous reply).
    Which mean you have to transform your system in a differential (delayed) equation (see the file below).
  • You need to define l(t, v) (more likely l(t))

There are things only you can do, in particular:

  • What is l(t)?
  • As I guess your problem reducesto solving a couple of delayed differential equations of order two, you need to specify additional IC or BC ob a(t) and c(t).
     
> 

restart
> 

# Keep alpha unassigned to get a simpler set of equations
# alpha := .3306341:
# Keep delta unassigned to get a simpler set of equations
# delta := solve(1+0.4877489e-1 = exp(delta_a));
T := 1:
> 

l0 := 42.375*(33.16667/(52*(24-9.95)*7));
L0 := l0*T:

.2748108485

 (1)
> 

x0:=0:
> 

IOR := .2093723865:
> 

KOR := IOR/delta:
> 

r0 := alpha/KOR-delta:
# Keep alpha unassigned to get a simpler set of equations
# rho := r0:

# This is useless
# fsolve({K0 = Y0*KOR, w0 = (1-alpha)*Y0/L0, alpha*exp(x0)*K0^(alpha-1)*L0^(1-alpha) = r0+delta}); # # assign(%);
> 

C0 := -K0*delta+Y0:
c0 := C0/T:
> 

# this expression is too complex to avoid a concise representation of eq2
# sigma := w0*(1-l0)/c0:
> 

eq1 := diff(c(t), t) = (r(t)-rho)*c(t);
> 

eq3 := isolate( r(t)*a(t)+w(t)*l(t) = c(t)+diff(a(t), t), diff(a(t), t));

print():
> 

eq2 := sigma*c(t) = w(t)*(1-l(t));
> 

eq4 := Y(t) = exp(x(t))*K(t)^alpha*L(t)^(1-alpha);
> 

eq5 := r(t)-delta = alpha*Y(t)/K(t);
> 

eq6 := w(t) = (1-alpha)*Y(t)/L(t);
> 

eq7 := K(t) = int(a(v), v = t-T .. t);
> 

eq8 := L(t) = int(l(v), v = t-T .. t);
> 

eq9 := x(t) = piecewise(t = 0, 0, t > 0, 0.1e-1);

eq1 := diff(c(t), t) = (r(t)-rho)*c(t)

  

eq3 := diff(a(t), t) = r(t)*a(t)+w(t)*l(t)-c(t)

  

NULL

  

eq2 := sigma*c(t) = w(t)*(1-l(t))

  

eq4 := Y(t) = exp(x(t))*K(t)^alpha*L(t)^(1-alpha)

  

eq5 := r(t)-delta = alpha*Y(t)/K(t)

  

eq6 := w(t) = (1-alpha)*Y(t)/L(t)

  

eq7 := K(t) = int(a(v), v = t-1 .. t)

  

eq8 := L(t) = int(l(v), v = t-1 .. t)

  

x(t) = piecewise(t = 0, 0, 0 < t, 0.1e-1)

 (2)
> 

SV := {a(0) = 0, c(0) = c0};

{a(0) = 0, c(0) = -K0*delta+Y0}

 (3)
> 

pdsolve(eq, SV, numeric, 'time' = t, 'range' = 0 .. T)

Error, (in pdsolve/numeric/process_PDEs) number of dependent variables and number of PDE must be the same

  


Given that the initial condition is set at time t=0 and that the differential equation are to be
verified on (0, +oo), there is no need to use a piecewise fnction to define x(t)

> 

eq9 := x(t) =0.01

x(t) = 0.1e-1

 (4)

 

"Simplification" of differential equation eq1

 

> 

# Express r(t) (lhs of eq1) in terms of other quantities
#
# Note this expression contains l(t, v) which is never defined elsewhere.
# So you can expect to have problems.

expr_r := isolate(eval(eval(eval(eq5, eq4), [eq7, eq8]), eq9), r(t));

Eq1 := eval(eq1, expr_r);

r(t) = 1.010050167*alpha*(int(a(v), v = t-1 .. t))^alpha*(int(l(v), v = t-1 .. t))^(1-alpha)/(int(a(v), v = t-1 .. t))+delta

  

diff(c(t), t) = (1.010050167*alpha*(int(a(v), v = t-1 .. t))^alpha*(int(l(v), v = t-1 .. t))^(1-alpha)/(int(a(v), v = t-1 .. t))+delta-rho)*c(t)

 (5)

 

"Simplification" of differential equation eq3

 

> 

expr_w := isolate(eq2, w(t)):

Eq3 := eval(eq3, [expr_r, expr_w]):

# rewritten as (beta=alpha-1):

Eq3 := diff(a(t), t) = (1.010050167*alpha*(int(a(v), v = t-1 .. t))^beta*(int(l(v), v = t-1 .. t))^(1-alpha)+delta)*a(t)-sigma*c(t)*l(t)/(-1+l(t))-c(t):

 

Eq1 and Eq3 are a couple of integro-differential equations with unknowns "a" and "c".
(The question of what "l" is still open.)


Beyond the nature of "l": what are the integrals in the rhs of Eq1 and Eq3?

How to transform an integro-differential equation in to a delayed differential equation?

> 

diff(u(t), t) = int(u(v), v = t-1 .. t);
diff(%, t);

diff(u(t), t) = int(u(v), v = t-1 .. t)

  

diff(diff(u(t), t), t) = u(t)-u(t-1)

 (6)


Applying this strategy to (Eq1, Eq3) should potentially produce a couple of delayed
differential equations of order 2.


Until you solve the problem concerning l(t), the following remains theoretical.

Apply the principle above to Eq3 (for instance).

> 

dEq3 := diff(Eq3, t):
> 

isolate(Eq3, int(a(v), v = t-1 .. t)^beta):
> 

int_a := isolate(Eq3, int(a(v), v = t-1 .. t)):
VisuallySimpler := rhs(int_a) = psi;

exp((-0.9999999949e-2+ln((sigma*c(t)*l(t)-1.*delta*a(t)*l(t)+c(t)*l(t)+(diff(a(t), t))*l(t)+delta*a(t)-1.*c(t)-1.*(diff(a(t), t)))*(int(l(v), v = t-1 .. t))^(-1.+alpha)/((-1.+l(t))*a(t)*alpha)))/beta) = psi

 (7)
> 

eval( eval(dEq3, int_a), VisuallySimpler);

diff(diff(a(t), t), t) = (1.010050167*alpha*psi^beta*beta*(a(t)-a(t-1))*(int(l(v), v = t-1 .. t))^(1-alpha)/psi+1.010050167*alpha*psi^beta*(int(l(v), v = t-1 .. t))^(1-alpha)*(1-alpha)*(l(t)-l(t-1))/(int(l(v), v = t-1 .. t)))*a(t)+(1.010050167*alpha*psi^beta*(int(l(v), v = t-1 .. t))^(1-alpha)+delta)*(diff(a(t), t))-sigma*(diff(c(t), t))*l(t)/(-1+l(t))+sigma*c(t)*l(t)*(diff(l(t), t))/(-1+l(t))^2-sigma*c(t)*(diff(l(t), t))/(-1+l(t))-(diff(c(t), t))

 (8)


Here psi only contains a(t) and diff(a(t), t) meaning the equation above is a
delayed (highly non linear) differential equation.
Do the same for Eq1.

Once you have validated what I did ad fixed the "l(t) issue", come back again on this thread.

Download WhereYourErrorComesFrom_2.mw

Very interesting and well done.I would l...

mmcdara 4500
February 03 2023


Very interesting and well done.

I would like to recommend this excellent site, unfortunately (for you?) in French, which gathers a lot of information about 2D and 3D curves and surfaces, polygons and polyhedra: https://mathcurve.com/index.htm

The stellated dodecahedron is here https://mathcurve.com/polyedres/granddodecaedreetoile/granddodecaedreetoile.shtml

@Carl Love  I agree that "nol...

mmcdara 4500
February 02 2023
0 0

@Carl Love 

I agree that "nolist" not being the default option is a nuisance.

@Carl Love  Don't you mean&nbs...

mmcdara 4500
February 01 2023
0 0

@Carl Love 

Don't you mean 
plots:-display([entries(IP, nolist)])
or
plots:-display(entries(IP, nolist))
?

But maybe the 'nolist' option is the default choice in the newest Maple version

@FilippoYour second case is treated here...

mmcdara 4500
January 29 2023
0 0

@Filippo

Your first case is treated here in a more complete way (two resampling methods and a lot of additional plots including confidence interval on the prediction)

NonlinearFit_CorrelationMatrix.mw

Here is your second case.

NonlinearFit_CorrelationMatrix_Model_2.mw

BTW: I realized I didn't answer your two final questions :

  • Also, is there a command that returns the errors for the variaous fit parameters? (i.e. a,b,c,d)
    Answer is NO

  • Or at least, someone knows the methods and formulas to find them?
    Answer is YES, there exist methods to find them, but NO, there is no formula in the general case
    (unless you can find a transformation that linearizes the model w.r.t. the parameters)

@Q190504  Thanks @vv, I did n...

mmcdara 4500
January 28 2023
0 0

@Q190504 

Thanks @vv, I did nothing

@acer It(s likely I didn't read...

mmcdara 4500
January 28 2023
0 0

@acer 

It(s likely I didn't read the full question and focused on the many simplification attempts.

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