Maple Questions and Posts

Calibrated my equations before attempting to solve for them (search for "Calib_1" in my script)
Split the original 6x6 system into two 3x3 sub-systems (since 3 out of 6 variables only appear in 3 out of 6 equations) and solved for one sub-system

What else can you think of? Should I instead use the parallel solver on the whole 6x6 system rather than just the unsolved 3x3 sub-system?

How to find, if exist, singular solutions? That is, some valuation of some parameters that will yield a solution that cannot be obtained by applying the same valuation to a general solution. Carl Love (who I cannot tag) once mentioned: "The parameter valuations that lead to singular solutions can often be guessed by using valuations that would produce zeros in denominators in the general solution. A singular solution can't be expressed as any instantiation of a generic symbolic solution. By instantiation I mean an assigment of numeric values to some parameters. Here's an example:"

#2x2 matrix and 2x1 vector. 5 parameters (a, b, d, x, y). The 2 decision variables are
#unseen and unnamed in this pure matrix-vector form. Their values are the two entries 
#in the solution vectors S0 and S1.

A:= <a, b; 0, d>;  B:= <x, y>;

#Get a generic solution:
S0:= LinearAlgebra:-LinearSolve(A, B);

#Instantiate 3 parameters (a, d, y) to 0 and solve again:
S1:= LinearAlgebra:-LinearSolve(eval([A, B], [a, d, y]=~ 0)[]);

#Note that no possible instantiation of S0 can produce S1.

Thank you!

How can we find a shortest closed walk passing thr...

Asked by: lcz 994

May 12 2023

1 3

A Hamiltonian walk on a connected graph is a closed walk of minimal length which visits every vertex of a graph (and may visit vertices and edges multiple times). (See https://mathworld.wolfram.com/HamiltonianWalk.html)

Please note that there is a distinction between a Hamiltonian walk and a Hamiltonian cycle. Any graph can have a Hamiltonian walk, but not necessarily a Hamiltonian cycle. The Hamiltonian number h(n) of a connected G is the length of a Hamiltonian walk.

For example, let's consider the graph FriendshipGraph(2).

g:=GraphTheory[SpecialGraphs]:-FriendshipGraph(2);
DrawGraph(g)

Then its Hamiltonian number is 6. A Hamiltonian walk is as shown in the following picture.

PS: After being reminded by others, I learned that a method in the following post seems to work, but I don't understand why it works for now.

https://cs.stackexchange.com/questions/43731/travelling-salesman-which-can-repeat-cities

it's possible to formulate this as a Travelling Salesman Problem by augmenting the sparse graph into a complete graph by adding edges. These new edges represent shortest paths between any two vertices in the original graph and have weight equal to the path length.

It seems that the correctness of this transformation requires rigorous proof.

How to solve this one?...

Asked by: Harry Xu 20

May 12 2023

0 2

not working dsolve comand for ODEs...

Asked by: Tamour_Zubair 80

May 11 2023

0 1

Hello Everyone;

Hope you are fine. I I have make a code but there is error and dsolve comand is working. I also need to plot. Kindly help me. I am waiting kind response. Code is attached.

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$restart; Cm := 1.0; ENa := 50.0; EK := -77.0; ELeak := -54.387; gNa := 120.0; gK := 36.0; gLeak := .3; II := piecewise(`and`(t >= 10, t <= 40), 10.0, 0.); alpha_m := proc (V) .1*(V+40.0)/(1.0-exp((-1)*(V+40.0)/10.0)) end proc; beta_m := proc (V) 4.0*exp((-1)*(V+65.0)/18.0) end proc; alpha_h := proc (V) 0.7e-1*exp((-1)*(V+65.0)/20.0) end proc; beta_h := proc (V) 1.0/(1.0+exp((-1)*(V+35.0)/10.0)) end proc; alpha_n := proc (V) 0.1e-1*(V+55.0)/(1.0-exp((-1)*(V+55.0)/10.0)) end proc; beta_n := proc (V) .125*exp((-1)*(V+65.0)/80.0) end proc; eq1 := diff(V(t), t) = (II-gNa*m(t)^3*h(t)*(V(t)-ENa)-gK*n(t)^4*(V(t)-EK)-gLeak*(V(t)-ELeak))/Cm; eq2 := diff(m(t), t) = alpha_m(V(t))*(1-m(t))-beta_m(V(t))*m(t); eq3 := diff(h(t), t) = alpha_h(V(t))*(1-h(t))-beta_h(V(t))*h(t); eq4 := diff(n(t), t) = alpha_n(V(t))*(1-n(t))-beta_n(V(t))*n(t); ics := {V(0) = -65, h(0) = .6, m(0) = 0.5e-1, n(0) = .32}$

diff(V(t), t) = 3.683900000-120.0000000*m(t)^3*h(t)*(V(t)-50.0)-36.00000000*n(t)^4*(V(t)+77.0)-.3000000000*V(t)

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sol := dsolve({eq1, eq2, eq3, eq4, ics})

Error, (in dsolve) invalid input: solve expects its 1st argument, eqs, to be of type {`and`, `not`, `or`, algebraic, relation(algebraic), ({list, set})({`and`, `not`, `or`, algebraic, relation(algebraic)})}, but received {{V(0) = -65, h(0) = 3/5, m(0) = 1/20, n(0) = 8/25}}

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Download Question1.mw

Thanks

How can I collect(,diff) in the Physics package?...

Asked by: Hullzie16 182

May 11 2023

1 2

I would like to collect in derivatives of a function(s) in a worksheet where I have the Physics package loaded but It does not want to do it. When I bring the expression to a blank worksheet it does everything just fine. Am I missing something small? Or is there additional commands/options I need to input?

Thoughts?

CollectDiff.mw

How to evaluate a one-dimensional improper integra...

Asked by: sursumCorda 922

May 11 2023

1 2

The highly oscillatory integrand is:

(sin(x + sqrt(x)) + x*BesselJ(0, x^2))/(1 + x): # int(%, x = 0 .. infinity, numeric); # Note that the upper limit of `x` is NOT finite.

Unfortunately, Maple still cannot return a result in a few minutes. For instance,

>	restart; infolevel[`evalf/int`] := 1:

>

timelimit(0.1e3, evalf(Int((sin(x+sqrt(x))+x*BesselJ(0, x^2))/(1+x), x = 0 .. infinity)))

Control: Entering NAGInt

trying d01amc (nag_1d_quad_inf)
Control: d01amc failed
evalf/int/control: NAG failed result = result
evalf/int/improper: integrating on interval 0 .. infinity
evalf/int/improper: applying transformation x = 1/x
evalf/int/improper: interval is 0 .. 1 for the integrand:

                        /     (1/2)\            /    2\
                     sin\x + x     / + x BesselJ\0, x /
                     ----------------------------------
                                   1 + x

and interval is 0 .. 1 for the integrand:

                                                /   1 \
                                         BesselJ|0, --|
                        /       (1/2)\          |    2|
                        |1   /1\     |          \   x /
                     sin|- + |-|     | + --------------
                        \x   \x/     /         x
                     ----------------------------------
                                 /    1\ 2
                                 |1 + -| x
                                 \    x/

Control: Entering NAGInt
Control: trying d01ajc (nag_1d_quad_gen)
Control: d01ajc failed
evalf/int/control: NAG failed result = result

evalf/int/control: singularity at left end-point
evalf/int/transform: series contains {x^(1/2), x^(3/2), x^(5/2), x^(7/2), x^(9/2), x^(11/2), x^(13/2), x^(15/2), x^(17/2), x^(19/2), x^(21/2), x^(23/2), x^(25/2)}

evalf/int/singleft: applying transformation x = x^2
evalf/int/singleft: interval is 0 .. 1. for the integrand:

                  /   / 2            \    2        /    4\\
                2 \sin\x + csgn(x) x/ + x BesselJ\0, x // x
                ---------------------------------------------
                                        2
                                   1 + x

evalf/int/control: Applying simplify/ln, integrand is 2*(sin(x^2+x)+x^2*BesselJ(0,x^4))/(1+x^2)*x
evalf/int/CreateProc: Trying easyproc
evalf/int/CreateProc: Trying makeproc
evalf/int/ccquad: n = 2 integral estimate = .7758263476953
n = 6 integral estimate = .8097413335347
evalf/int/ccquad: n = 18 integral estimate = .8097470386638
error = .8097470386638e-11
From ccquad, result = .8097470386638 integrand evals = 19 error = .8097470386638e-11
Control: Entering NAGInt
Control: trying d01ajc (nag_1d_quad_gen)
Control: d01ajc failed
evalf/int/control: NAG failed result = result

evalf/int/control: singularity at left end-point

evalf/int/transform: series contains {1/x^(3/2), x^(1/2), cos(1/x^2+1/4*Pi), sin((x+x^(1/2))/x^(3/2)), sin(1/4*(4+Pi*x^2)/x^2)}
evalf/int/singleft: applying transformation x = x^2
evalf/int/singleft: interval is 0 .. 1. for the integrand:

                   /   /        /1\ \                    \
                   |   |1 + csgn|-| x|                    |
                   |   |        \x/ | 2          /   1 \|
                 2 |sin|-------------| x + BesselJ|0, --||
                   |   |      2      |             |    4||
                   \   \     x       /             \   x //
                 ------------------------------------------
                                 3 /     2\
                                x \1 + x /

evalf/int/control: Applying simplify/ln, integrand is 2*(sin((1+x)/x^2)*x^2+BesselJ(0,1/x^4))/x^3/(1+x^2)
evalf/int/control: Applying simplify/trig, integrand is (2*sin((1+x)/x^2)*x^2+2*BesselJ(0,1/x^4))/(x^3+x^5)
evalf/int/control: singularity at left end-point

evalf/int/transform: series contains {cos(1/x^4+1/4*Pi), sin((1+x)/x^2), sin(1/4*(4+Pi*x^4)/x^4)}
evalf/int/transform: no transform found
evalf/int/series: integrating on 0 .. .2493468135214 the series:

       2 (%2)     2 (%2) x   /2 (%2)      \ 3   / 2 (%2)      \ 5
      --------- - -------- + |------- - %3| x + |- ------- + %3| x
        (1/2)       (1/2)    | (1/2)     |      |    (1/2)     |
      Pi      x   Pi         \Pi          /      \ Pi          /

           /       2 (%2)      \ 7   /     2 (%2)      \ 9
         + |- %3 + ------- - %4| x + |%3 - ------- + %4| x
           |         (1/2)     |      |       (1/2)     |
           \       Pi          /      \     Pi          /


           /2 (%2)           \ 11   /          2 (%2) \ 13
         + |------- - %4 - %5| x   + |%4 + %5 - -------| x
           | (1/2)          |       |            (1/2)|
           \Pi               /       \          Pi     /

           /       2 (%2)      \ 15   /     2 (%2)      \ 17    / 19\
         + |- %5 + ------- + %6| x   + |%5 - ------- - %6| x   + O\x /
           |         (1/2)     |       |       (1/2)     |
           \       Pi          /       \     Pi          /


                      /        4\
             (1/2)    |4 + Pi x |
      %1 := 2      sin|---------|
                      |     4   |
                      \ 4 x    /
                    /1 + x\   (1/2)
      %2 := %1 + sin|-----| Pi
                    | 2 |
                    \ x   /
             (1/2)    /1    1   \
            2      cos|-- + - Pi|
                      | 4   4   |
                      \x        /
      %3 := ---------------------
                      (1/2)
                  4 Pi
                        /        4\
               (1/2)    |4 + Pi x |
            9 2      sin|---------|
                        |     4   |
                        \ 4 x    /
      %4 := -----------------------
                       (1/2)
                  64 Pi
                (1/2)    /1    1   \
            53 2      cos|-- + - Pi|
                         | 4   4   |
                         \x        /
      %5 := ------------------------
                        (1/2)
                  512 Pi
                           /        4\
                  (1/2)    |4 + Pi x |
            1371 2      sin|---------|
                           |     4   |
                           \ 4 x    /
      %6 := --------------------------
                          (1/2)
                  16384 Pi

evalf/int/CreateProc: Trying easyproc
evalf/int/CreateProc: Trying makeproc

evalf/int/CreateProc: Trying procmake
evalf/int/control: applying double-exponential method
evalhf mode unsuccessful -- retry in software floats
evalf/int/quadexp: applying double-exponential method

Error, (in tools/sign) time expired

>

=

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=

Download oscillatoryInt.mws

I tried to specify another method (as described in evalf/int), but it seems that this doesn't work. Any ideas?

How to replace variable names partially?...

Asked by: hoverstar 5

May 11 2023

1 7

For example, if I want to replace all "x" in t:={x1,x2,x3...x100} to get t:={y1,y2,y3...y100}, how can maple do this, please? Thanks a lot!

The book didnt even gave the right answer...

Asked by: The function 110

May 10 2023

0 4

The answers from the book dont make any sense after using their solution. Its not producing the expected results.

This is the translation by the way: "A battery that has a voltage U, is in series with a resistor R, and a condenser with capacity C. The current strength on a given time "t" in this circuit is given by i=i(t)=U/R*e^-(t/R*C). Determine R and C based on the data given in table 5.6, the current is U=100V."

Greetings,

The Function

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(2)

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(3)

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"v(t):=U/(R)*(e)^(-(t/(R*C)))"

proc (t) options operator, arrow, function_assign; U*exp(-t/(R*C))/R end proc

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(5)

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Matrix(%id = 18446746712187741238)

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Vector(5, {(1) = .1, (2) = .2, (3) = .3, (4) = .4, (5) = .5})

Vector[column](%id = 18446746712187761598)

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(8)

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(10)

Download Mapleprimes_Question_Book_2_Paragraph_5.12_Question_5.mw

How can I calculate the maximum number of rectangl...

Asked by: Teep 195

May 10 2023

1 4

Good day!

I have been wrestling with what appears to be a simple problem, but with no success and so, I thought I would reach out for support.

Consider a two-dimensional strip of any given width, x, and depth, y.

I wish to maximize the quantity of strips that can fit inside a larger two-dimensional surface of any given width, X, and depth, Y. Overhanging is not permitted and all smaller strips must lie within the interior of the larger surface.

I'm looking for a solution method that can calculate the best configuration necessary to fit the maximum number of smaller strips in the interior of the larger rectangle. Can someone suggest a way (if possible) to solve for this?

Thanks for reading.

How do (can) I solve this ODE?...

Asked by: wjg 20

May 10 2023

2 5

diff(f(x), x) = a-a/((1+a*x)*(1/(1+a*x))^a)-a*f(x) where a is a constant and an initial condition f(0)= b is known Thank you.

Isolate variable...

Asked by: Andre Luis Alves Neves 15

May 10 2023

1 3

Hello,

How do I isolate the variable ng in the following expression:

rho := `ρg`*ng + `ρp`*np + `ρw`*nw;
B := np/Kp + nw/Kw + ng/Kg;

C := (B/rho)^(1/2);

As you can see, I am running the isolate command, but it is not including the variable C, or it returns the wrong result.

isolate(((np/Kp + nw/Kw + ng/Kg)/(`ρg`*ng + `ρp`*np + `ρw`*nw))^(1/2), ng);
( Kp nw + Kw np) Kg
ng = - -----------------

Kw Kp

isolate(C=((np/Kp + nw/Kw + ng/Kg)/(`ρg`*ng + `ρp`*np + `ρw`*nw))^(1/2), ng);
0 = 0

Thanks.

How to re-write expressions in terms of each other...

Asked by: MaPal93 90

May 10 2023

0 10

I think some form of simplify() would do but I am not sure how.

See the following script for more details:

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#Define the assumptions ex-ante (variances as real and positive, correlations in between -1 and +1 and so on...) - or Maple wouldn't know

>

t__1 := (`σ__v`[2]^2*(`ρ__v`[1, 2]^2-1)-`σ__ε2`^2)*`σ__v`[1]^2/((`σ__v`[2]^2*(`ρ__v`[1, 2]^2-1)-`σ__ε2`^2)*`σ__v`[1]^2-`σ__ε1`^2*(`σ__ε2`^2+`σ__v`[2]^2))

>

t__2 := -`σ__v`[1]*`ρ__v`[1, 2]*`σ__v`[2]*`σ__ε1`^2/((`σ__v`[2]^2*(`ρ__v`[1, 2]^2-1)-`σ__ε2`^2)*`σ__v`[1]^2-`σ__ε1`^2*(`σ__ε2`^2+`σ__v`[2]^2))

>

t__3 := (`σ__v`[1]*`ρ__v`[1, 2]*`σ__v`[2]*`σ__ε1`^2*`ν__0`[2]-`ν__0`[1]*`σ__ε1`^2*(`σ__ε2`^2+`σ__v`[2]^2))/((`σ__v`[2]^2*(`ρ__v`[1, 2]^2-1)-`σ__ε2`^2)*`σ__v`[1]^2-`σ__ε1`^2*(`σ__ε2`^2+`σ__v`[2]^2))

See for example that they share the same denominator...
# I want Maple to automatically/smartly choose the most compact way...e.g., (i) t2=t1+... or (ii) t2=t3*... (i) is equivalent to (ii) but I want Maple to pick (ii) if more compact than (i)
# I have 9 expressions like t1,t2,t3 and I want Maple to re-write them as function of each other smartly

>

Download rearrangingterms.mw

It must be something similar to this https://www.mapleprimes.com/maplesoftblog/201455-Rearranging-The-expression-Of-Equations, but my case is slighty different.

How to solve some polynomial equation over the dom...

Asked by: sursumCorda 922

May 09 2023

0 7

In certain tasks, I need to find all accurate positive (not just nonnegative) roots that exist of some multivariate polynomials like:

nsd := 16*a*b*c*(9 + a^2 + b^2 + c^2)*(b*c + a*(b + c) + 3*(a + b + c)) - (3 + a + b + c)^2*(a*b + 3*c)*(3*b + a*c)*(3*a + b*c): # assume((a, b, c) >~ 0);

According to fsolve/details, for one general equation, the fsolve command only computes "a single real root", so it is inadequate to tackle this question. But if I use the solve command with floating-point values, the computation cannot finish in ten minutes instead! (Maybe a longer time will suffice, yet this is rather unacceptable.)

>

restart;

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>	#assume((a, b, c) >~ 0); nsd := 16abc(a^2 + b^2 + c^2 + 9)(a(b + c) + 3(a + b + c) + bc) - (a + b + c + 3)^2(ab + 3c)(ac + 3b)(bc + 3*a):

>