MaplePrimes Questions

Hi,

I am a newbie and therefore have a rather simple question for the community:

By reading in my measurement data (topographic data) I have generated column vectors:
for i from 1 to k do
    B[i] := Vector(A[1 .. k, i]);
end do:
where k is a predefined rank of the original square matrix. The arithmetic mean was then calculated for each individual column vector:
for i from 1 to k do
    Am[i] := add(B[i][n], n = 1 .. k)/k;
end do:

Now I try to create new column vectors by subtracting the corresponding mean value from each entry of the original vectors, something like that:
for i from 1 to k do
C[i]:=B[i][n]-Am[i],n=1..k

But I just can't get it right. Can anyone help me? Thanks in advance!

Cheers
Christian

 

For example, the column dimension is 6 and I ran the following command,

A := Mod(2,Matrix(3,6,(i,j)->if i<2 and j < 3 then 1  elif i = 2 and j > 2 and j <5 then 1 else 0 end if),float[8])

the result is the following matrix with the last row containing all 0s. 

A := Matrix(3, 6, [[1., 1., 0., 0., 0., 0.], [0., 0., 1., 1., 0., 0.], [0., 0., 0., 0., 0., 0.]])
Actually, I want the last row containing all 1s but I do not know how.
Any help?

 

i want to write a expression without any quotes . for example in this case i want to wrate E*1    E*2 

I can remove this error.?

I if a run the CompleteSquare command 

 

CompleteSquare(x^2 + y^2 - 2*x - y - 2 = 10, x, y)

the output is (y - 1/2)^2 + (x - 1)^2 - 13/4 = 10

Why does it places the y's before the x's? 

How do I compute the gcd of polynomials with scalars e.g. how to compute gcd (x^9-a, bx^6-c )  as polynomials in x.   https://www.maplesoft.com/support/help/Maple/view.aspx?path=Task/GCDofPolynomials   here I don't see how to specify the polynomials are in x.

Is the linux version of Maple architecture dependent? 

Hi everyone, I have a problem in the code solving coupled partial differential equations. I could not find out the solution. Please help me out with this. Find the code in the attachment.

I didn't understand what the page on implicit differentiation meant by 2nd derivatives are just like in using diff    Like this  implicitdiff(y^2 = x^3+a*x+b, y, x, [x$n]) ??  idk 

I have the following ODE which I would like to solve with Maple rather than solving by hand (having solved this type of equation by hand many times now):

diff(f(x,y,z),z$2) = A - B*e^(-A*z)

where A and B are constants and I have indicated the second derivative of a function of x,y and z with respect to z.  

Dear maple users,
Greetings.
I am solving an ode problem with an analytical solution.
programming running properly, but my plot not exact with the already existing article plot. 
how to get the exact plot.

Thanking you.

Code:JVB.mw
 

restart

N := 3;

3

 

1

(1)

dsolve(diff(f(x), `$`(x, 3)));

f(x) = (1/2)*_C1*x^2+_C2*x+_C3

(2)

Rf := 2*(diff(f[m-1](x), x, x, x))-(2*mh*mh)*(diff(f[m-1](x), x))+sum(f[m-1-n](x)*(diff(f[n](x), x)), n = 0 .. m-1)-bet*(sum(sum(2*f[m-1-n](x)*(diff(f[n-t](x), x))*(diff(f[t](x), x, x))+f[m-1-n](x)*f[n-t](x)*(diff(f[t](x), x, x, x))+x*(diff(f[m-1-n](x), x))*(diff(f[n-t](x), x))*(diff(f[t](x), x, x)), t = 0 .. n), n = 0 .. m-1));

2*(diff(diff(diff(f[m-1](x), x), x), x))-2*(diff(f[m-1](x), x))+sum(f[m-1-n](x)*(diff(f[n](x), x)), n = 0 .. m-1)-.2*(sum(sum(2*f[m-1-n](x)*(diff(f[n-t](x), x))*(diff(diff(f[t](x), x), x))+f[m-1-n](x)*f[n-t](x)*(diff(diff(diff(f[t](x), x), x), x))+x*(diff(f[m-1-n](x), x))*(diff(f[n-t](x), x))*(diff(diff(f[t](x), x), x)), t = 0 .. n), n = 0 .. m-1))

(3)

dsolve(diff(f[m](x), x, x, x)-CHI[m]*(diff(f[m-1](x), x, x, x)) = h*H*Rf, f[m](x));

f[m](x) = Int(Int(Int(CHI[m]*(diff(diff(diff(f[m-1](x), x), x), x))+2*h*(diff(diff(diff(f[m-1](x), x), x), x))-2*h*(diff(f[m-1](x), x))+h*(sum(f[m-1-n](x)*(diff(f[n](x), x)), n = 0 .. m-1))-(1/5)*h*(sum(sum(2*f[m-1-n](x)*(diff(f[n-t](x), x))*(diff(diff(f[t](x), x), x))+f[m-1-n](x)*f[n-t](x)*(diff(diff(diff(f[t](x), x), x), x))+x*(diff(f[m-1-n](x), x))*(diff(f[n-t](x), x))*(diff(diff(f[t](x), x), x)), t = 0 .. n), n = 0 .. m-1)), x), x)+_C1*x, x)+_C2*x+_C3

(4)

f[0](x) := 1-exp(x);

1-exp(x)

(5)

for m to N do CHI[m] := `if`(m > 1, 1, 0); f[m](x) := int(int(int(2*CHI[m]*(diff(f[m-1](x), x, x, x))-(2*h*H*mh*mh)*(diff(f[m-1](x), x))+h*H*(sum(f[m-1-n](x)*(diff(f[n](x), x)), n = 0 .. m-1)), x)-h*H*(sum(sum(2*f[m-1-n](x)*(diff(f[n-t](x), x))*(diff(f[t](x), x, x))+f[m-1-n](x)*f[n-t](x)*(diff(f[t](x), x, x, x))+x*(diff(f[m-1-n](x), x))*(diff(f[n-t](x), x))*(diff(f[t](x), x, x)), t = 0 .. n), n = 0 .. m-1))*bet, x)+_C1*x, x)+_C2*x+_C3; s1 := evalf(subs(x = 0, f[m](x))) = 0; s2 := evalf(subs(x = 0, diff(f[m](x), x))) = 0; s3 := evalf(subs(x = 1, f[m](x))) = 0; s := {s1, s2, s3}; f[m](x) := simplify(subs(solve(s, {_C1, _C2, _C3}), f[m](x))) end do:

f(x) := sum(f[l](x), l = 0 .. N);

1-0.7644444444e-1*exp(5.*x)*h^2*x-0.1333333333e-1*x^2*exp(5.*x)*h^2-2.675700596*exp(2.*x)*h^2*x-0.5876096022e-1*exp(6.*x)*h^3*x-0.9282030175e-2*x^2*exp(6.*x)*h^3+.9962792493*exp(3.*x)*h^3*x+.1647896790*exp(5.*x)*h^3*x+0.2066962962e-1*x^2*exp(5.*x)*h^3+3.357118680*exp(2.*x)*h^3*x-.3264340965*exp(4.*x)*h^3*x+0.3999999998e-1*exp(2.*x)*ln(exp(x))*h^2+58.61348006*h^3+1.023148148*h^2*x^3+0.1364197531e-1*ln(exp(x))*h^3*x^3-0.8954734530e-1*exp(2.*x)*h^3*x^4-.1353159884*x^3*exp(4.*x)*h^3+.7542645986*exp(3.*x)*h^3*x^2-0.2830138323e-1*x^3*h^3*exp(3.*x)-0.6455420536e-1*exp(x)*h^3*ln(exp(x))*x+0.4775858416e-1*exp(x)*h^3*ln(exp(x))*x^2+0.8888888887e-3*exp(x)*h^3*ln(exp(x))^2+8.400000000*h*exp(x)-exp(x)-0.6666666666e-1*h*ln(exp(x))+.1416666666*exp(4.*x)*h^2*x-.4790123458*exp(3.*x)*h^2*x+.1333333333*exp(3.*x)*h*x+.3791666665*exp(4.*x)*h^2-1.340020575*exp(3.*x)*h^2+.3111111109*exp(3.*x)*h+5.570191338*h^2*exp(2.*x)-.4500000000*h*exp(2.*x)-0.9874869443e-1*exp(6.*x)*h^3+.4125877323*exp(3.*x)*h^3-4.984787877*h^3*exp(2.*x)-.8010958741*exp(4.*x)*h^3+.3215641638*exp(5.*x)*h^3-5.930474628*h^2*x+36.04284024*exp(x)*h^3*x+8.324321524*x^2*h^2-.5362260993*h^3*x^3-6.207072379*exp(x)*x^2*h^3+1.664189246*exp(x)*h^3*x^3-8.237962963*h+.1200000000*exp(x)*h^2*ln(exp(x))+0.2222222222e-1*exp(3*x)*h*x+24.00299428*h^3*x-2.098561083*x^2*h^3-53.48457977*h^3*exp(x)+0.9949705035e-2*ln(exp(x))*h^3*x^4-0.7308641971e-2*ln(exp(x))*exp(4.*x)*h^3+0.8984910834e-2*ln(exp(x))*exp(3.*x)*h^3-0.3741666666e-1*ln(exp(x))*h^3*exp(2.*x)-.1188740741*exp(5.*x)*h^2-12.53662834*x^2*h+25.90916526*h^2*exp(x)-30.39962862*h^2-0.7499999999e-1*h*exp(2*x)+0.5185185185e-1*exp(3*x)*h+5.372840718*exp(x)*x^2*h^2-25.09181716*exp(x)*h^2*x+0.8976305409e-1*h^3*x^5+0.2158026099e-1*exp(7.*x)*h^3+0.8606919260e-1*h^3*x^4+0.5079365079e-3*x^3*exp(7.*x)*h^3-.3215468487*x^2*exp(4.*x)*h^3+0.1762236380e-1*exp(7.*x)*h^3*x+0.5048727639e-2*exp(7.*x)*x^2*h^3-3.116709690*exp(2.*x)*x^2*h^3+.1066289908*exp(2.*x)*h^3*x^3-8.527777777*h*x-0.2814814814e-2*ln(exp(x))*exp(4.*x)*h^3*x-0.1053497943e-2*ln(exp(x))*exp(3.*x)*h^3*x+0.4848332783e-1*h^3*x^6+.7462278773*h^2*x^4+.5519508187*exp(x)*h^3*x^4+0.9367631194e-1*exp(x)*h^3*ln(exp(x))+3.581893812*exp(2.*x)*x^2*h^2

 

 

NULL


 

Download JVB.mw

 

Analytical solution approach:

 

 

 

 

Dear all

I hope to find the supremum of the sequence of the function using maple 18, but when I run the code there is no results

maximize.mw

Many thanks

 

 

Dear, 

Please, I would like to ask your help with the following situation: 

If we have to solve in Maple a linear system like A.x =b, we employ the command Linsolve. However, how can I solve a system like x.A=b? 

The only method I know is to compute x=b.A-1. Is this an efficient method or there is a better one you recommend? 

Many thanks for your help.  

phi:=(j,x)->piecewise(j=0,exp(x),1/(j-1)!*Integrate(exp((1-theta)*x)*theta^(j-1),theta=0..1))

I want 15 digits after the dot, so I set Digits:=15

Setting j=3, x=0.01 I received

phi(j,x)=0.167084168060000

 

Doing the same in python I receive

0.16708416805754214

Hi, 

Here are two sequences of commands that should give the same kind of plots. But, while the first one returns the expected display, the second doesn't (look to the labels on the histogram plot).

There is no hidden character that could explain this second display. Just that I proceeded this way:

  1. I executed the second sequence once.
  2. Then I told myself that displaying the histogram was superfluous, so I replaced its final semicolon with a colon.
  3. And I finally thought that, no, the histogram had a real interest and that I should display it; and I restored the semcolon (this is what you can see on the second sequence).
    And this add to the histogram the labels inherited from the second plot ...

Nothing dramatic here, but was the development team aware of this curiosity?
 

 

restart:

N := 10:
S := Statistics:-Sample(Normal(0, 1), 10)^+:
Statistics:-Histogram(S);
plots:-logplot(<  < [$1..N]> | S >, labels=["A", "B"]);
 

 

 

restart:

N := 10:
S := Statistics:-Sample(Normal(0, 1), 10)^+:
Statistics:-Histogram(S);
plots:-logplot(<  < [$1..N]> | S >, labels=["A", "B"]);

 

 

 


 

Download Strange_display.mw

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