3158 Reputation

18 Badges

15 years, 257 days

MaplePrimes Activity

These are answers submitted by MrMarc

thank you but that did not work for me. I also tried

assign(seq(f(i, i) = 1, i = 1 .. 5)):
Matrix(5, 5, f);

but it does not work very well. I get a matrix with unknown(i,j) but the diagonal is correct

all right thanx Doug :-)  I was not aware that you could turn of the warning messages interface( warnlevel=0 ):
That solves my problem because everytime I write a procedure I just execute it and then wait for Maple to
output the local variables warning messages and then I just define those variables as local.
By turning the warning messages of I dont have to bother with the last step, which is great :-)

One interesting follow up question might be:  Does the above methods work for recursive checking ie 


add(`if`(i+1 =4 and i= 2, 1, 0), i=L);

Does there exist any simple method for that or are you forced to write a loop ?!

ok, thanx Robert I will try that :-)  
I also think that it should be something that Maple should be able to do by default.

Yes I want to know that as well :-)

I cant really find a good way ! I can display lists in MathContainer but I cant print expressions and then get a new row ie  \n

It should look something like:


Could anyone explain this for me ie how this can be done in Maple:

"Assuming you are working with two time series, say f(t) and g(t), then I
would use the Fourier Transform to convert to the frequency domain, resulting
in F(w) and G(w). Then convolution is simply multiplication in the frequency
domain. Don't forget to back-transform (apply the FFT again) the convolved
sequence to get back to the time domain and see what the convolved signal
actually looks like."

Is it possible to analyze a system of two variables that are cross correlated? So for example if
you have an inpulse shock (as I understand is the dirac unction) you can look at the respons for the
other variable and then use such a respons to confirm cross correlation properties between the
two variables ?

Does this impulse respons plot have anything to do with stability? Like differential equations
that can be stable or unstable over time ? or maybe we are not looking over time ..or ?! 

Lets take my initial example again:

sys1 := TransferFunction(s*(s-1)/((s+2)*(s+4)*(s+6))):
ImpulseResponsePlot(sys1, 2);
Why does this ImpulseResponsPlot have this shape ? Is it possible to explain the shape?
Is this ImpulseResponsPlot always bounded between -1 and 1 ?
Is there an easy way to use an slider to illustrate the shape of this plot given certain input values ?
How many and what type of transfer functions exists ? 
As I understand it  TransferFunction(1) and  TransferFunction(1/s) must represent the most extreme
cases. Are all the rest of the transfer functions in between these two extremes ?!
Do you know a simple reference where I can learn more ? Preferably something for dummies, he he
Does there exist a relationship to impulse response and vector autoregressive (VAR) models ?!

all right thanx ! This is the annoying part with Maple. Maple is supposed to be good at algebra.
Which is far from the truth. Most of the time it is easier to do the algebraic manipulation on paper .

I got :   U := (L^p+K^p)^(1/p);
and    A = simplify(-(diff(U, K))/(diff(U, L)), symbolic);

I want to show that
A = (K/L)^(p-1);

I sympathize with you :-)  I have been in your shoes a million times. Unfortunately I dont have an answers.
Sometime a pen and paper is the best solution

As I said I know the slope of these thick blacklines but I dont know the intercept...humm

Below is an example where vertical tickmarks on the x-axis would be prefered. The string is quite lenghty
which means that the x-axis become very crowded quite quickly if you have standard horizontal string tickmarks.
How can I plot vertical tickmarks ?

listplot([seq(i, i = 1 .. 50)], tickmarks = [[seq(i = "QQWWWW", i = 1 .. 50)], default], thickness = 3, color = green);

If you have a Maplet you can simply type :  Set(('DD')(itemlist) = [x1, x2, x3, x4, x5, etc etc] );

where DD is the DropDownBox. However this does not seem to work with embeeded components such as a ComboBox ?!


"Just a quick try: isnt that a differential equation for g : = f @ e ?"

I dont know what you mean ?!


The basic idea is to maximize profits given a profit function  F(e(w)*L)-w*L  where  e(w) is an effort function ie the

higher wage the more effort etc


profit := F(e(w)*L)-w*L:

foc1 := convert(diff(profit, L), diff) = 0:

foc2 := convert(diff(profit, w), diff) = 0:

A := solve(foc1, eval(diff(F(t1), t1), {t1 = e(w)*L})):

B := solve(foc2, eval(diff(F(t1), t1), {t1 = e(w)*L})):

C := A = B:

lhs(C)*(diff(e(w), w))*e(w) = rhs(C)*(diff(e(w), w))*e(w);

                       w*(diff(e(w), w)) = e(w)

The maximum will appear when w*diff(e(w),w))=e(w) 


So for example if we assume that the effort function is  e(w) = ((w-b)/b)^lambda  then we get:


e := ((w-b)/b)^lambda:

ee := w*(diff(e, w)):

Ap := ee = e:

wage = solve(Ap, w);

effort = subs(w = rhs(%), e);

b := 2:   lambda := .5:

wage = solve(Ap, w);

effort = subs(w = rhs(%), e);

plot([e, ee], w = 0 .. 10, view = [0 .. 7, 0 .. 2]);

                             wage = - b / (lambda -1)
                             effort =  .........
                             wage = 4.
                             effort = 1.


The profit maximum is when w=4 and effort=1


Now the problem is that the profit function is not specified so we need to find a profit function which has

a maximum where isolate(diff(prof, w) = 0, w) = -b/(lambda-1) . I mean I know some arbitrary function like 

profit := (32*e/lambda)^lambda-w*L:   which has a maximum at w=4  but when you change lambda such a

relationship disappear so it is not a valid solution. So the question still stands:  How do I find a concave

profit function that has these properties ( ie maximum at w=4  and parameter stability with respect to lambda ) ???



e := ((w-b)/b)^lambda:

b := 2:   lambda := .5:   L := 1:   alpha := .6:

profit := (32*e/lambda)^lambda-w*L:

isolate(diff(profit, w) = 0, w);

plot(profit, w = 0 .. 12);

                               w = 4.000000000

5 6 7 8 9 10 11 Last Page 7 of 24