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MaplePrimes Posts are for sharing your experiences, techniques and opinions about Maple, MapleSim and related products, as well as general interests in math and computing.

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  • So I have recently finished up a project that took different sounds found in nature, and through the Spectrogram command, plotted the frequency of each sound over time with some really cool results!

    https://www.maplesoft.com/applications/view.aspx?SID=154346 

    The contrast between sounds produced by the weather such as tornadoes, thunder, and hail versus something as innocuous as a buzzing bee, a chorus of crickets, or a purring cat really shows the variance in the different sounds we hear in our day to day life, while also creating some very interesting imagery.

    My personal favourite was the cricket chorus, producing a very ordered image with some really cool spikes through many different frequencies as the crickets chirped, as shown here:

    Using this plot, we can do some interesting things, like count the number of chirps in 8 seconds, which turns out to be 18-18.5. Why Is this important? Well, there’s a law known as Dolbear’s Law(shown here: https://en.wikipedia.org/wiki/Dolbear%27s_law)  which uses the number of chirps in a minute for Fahrenheit, or 8 seconds for Celsius to calculate the temperature. Celsius is very simple, and just requires adding 5 to the number of chirps in 8 seconds, which gets us a temperature of 23C.

    Tc= 5 + N8

    For Fahrenheit, it’s a bit more complicated, as we need the chirps in a minute. This is around 132 chirps in our case. Putting that into the formula:

    TF= 50 +((N60 – 40)/4)

    Which gets us 73F, or 22.7C, so you can see that it works pretty well! Pretty cool, huh?

     

    There was also some really cool images that were produced, like the thunder plot:

    Which I personally really like due to the contrasting black and yellow spike that occurs. Overall this was a very fun project to do, getting to tweak the different colours and scales of each spectrogram, creating a story out of a sound. Hope you all enjoy it!

     I accidentally stumbled on this problem in the list of tasks for mathematical olympiads. I quote its text in Russian-English translation:

    "The floor in the drawing room of Baron Munchausen is paved with the identical square stone plates.
     Baron claims that his new carpet (made of one piece of a material ) covers exactly 24 plates and
     at the same time each vertical and each horizontal row of plates in the living room contains 
    exactly 4 plates covered with carpet. Is not the Baron deceiving?"

    At first glance this seems impossible, but in fact the Baron is right. Several examples can be obtained simply by hand, for example

                                            or        

     

    The problem is to find all solutions. This post is dedicated to this problem.

    We put in correspondence to each such carpet a matrix of zeros and ones, such that in each row and in each column there are exactly 2 zeros and 4 ones. The problem to generate all such the matrices was already discussed here and Carl found a very effective solution. I propose another solution (based on the method of branches and boundaries), it is less effective, but more universal. I've used this method several times, for example here and here.
    There will be a lot of such matrices (total 67950), so we will impose natural limitations. We require that the carpet be a simply connected set that has as its boundary a simple polygon (non-self-intersecting).

    Below we give a complete solution to the problem.


    restart;
    R:=combinat:-permute([0,0,1,1,1,1]);
    # All lists of two zeros and four units

    # In the procedure OneStep, the matrices are presented as lists of lists. The procedure adds one row to each matrix so that in each column there are no more than 2 zeros and not more than 4 ones

    OneStep:=proc(L::listlist)
    local m, k, l, r, a, L1;
    m:=nops(L[1]); k:=0;
    for l in L do
    for r in R do
    a:=[op(l),r];
    if `and`(seq(add(a[..,j])<=4, j=1..6)) and `and`(seq(m-add(a[..,j])<=2, j=1..6)) then k:=k+1; L1[k]:=a fi;
    od; od;
    convert(L1, list);
    end proc:

    # M is a list of all matrices, each of which has exactly 2 zeros and 4 units in each row and column

    L:=map(t->[t], R):
    M:=(OneStep@@5)(L):
    nops(M);

                                                67950

    M1:=map(Matrix, M):

    # From the list of M1 we delete those matrices that contain <1,0;0,1> and <0,1;1,0> submatrices. This means that the boundaries of the corresponding carpets will be simple non-self-intersecting curves

    k:=0:
    for m in M1 do
    s:=1;
    for i from 2 to 6 do
    for j from 2 to 6 do
    if (m[i,j]=0 and m[i-1,j-1]=0 and m[i,j-1]=1 and m[i-1,j]=1) or (m[i,j]=1 and m[i-1,j-1]=1 and m[i,j-1]=0 and m[i-1,j]=0) then s:=0; break fi;
    od: if s=0 then break fi; od:
    if s=1 then k:=k+1; M2[k]:=m fi;
    od:
    M2:=convert(M2, list):
    nops(M2);

                                                 394

    # We find the list T of all segments from which the boundary consists

    T:='T':
    n:=0:
    for m in M2 do
    k:=0: S:='S':
    for i from 1 to 6 do
    for j from 1 to 6 do
    if m[i,j]=1 then
    if j=1 or (j>1 and m[i,j-1]=0) then k:=k+1; S[k]:={[j-1/2,7-i-1/2],[j-1/2,7-i+1/2]} fi;
    if i=1 or (i>1 and m[i-1,j]=0) then k:=k+1; S[k]:={[j-1/2,7-i+1/2],[j+1/2,7-i+1/2]} fi;
    if j=6 or (j<6 and m[i,j+1]=0) then k:=k+1; S[k]:={[j+1/2,7-i+1/2],[j+1/2,7-i-1/2]} fi;
    if i=6 or (i<6 and m[i+1,j]=0) then k:=k+1; S[k]:={[j+1/2,7-i-1/2],[j-1/2,7-i-1/2]} fi; 
    fi;
    od: od:
    n:=n+1; T[n]:=[m,convert(S,set)];
    od:
    T:=convert(T, list):

    # Choose carpets with a connected border

    C:='C': k:=0:
    for t in T do
    a:=t[2]; v:=op~(a);
    G:=GraphTheory:-Graph([$1..nops(v)], subs([seq(v[i]=i,i=1..nops(v))],a));
    if GraphTheory:-IsConnected(G) then k:=k+1; C[k]:=t fi;
    od:
    C:=convert(C,list):
    nops(C);
                                                 
     208

    # Sort the list of border segments so that they go one by one and form a polygon

    k:=0: P:='P':
    for c in C do
    a:=c[2]: v:=op~(a);
    G1:=GraphTheory:-Graph([$1..nops(v)], subs([seq(v[i]=i,i=1..nops(v))],a));
    GraphTheory:-IsEulerian(G1,'U');
    U; s:=[op(U)];
    k:=k+1; P[k]:=[seq(v[i],i=s[1..-2])];
    od:
    P:=convert(P, list):

    # We apply AreIsometric procedure from here to remove solutions that coincide under a rotation or reflection

    P1:=[ListTools:-Categorize( AreIsometric, P)]:
    nops(P1);

                                                     28


    We get 28 unique solutions to this problem.

    Visualization of all these solutions:

    interface(rtablesize=100):
    E1:=seq(plottools:-line([1/2,i],[13/2,i], color=red),i=1/2..13/2,1):
    E2:=seq(plottools:-line([i,1/2],[i,13/2], color=red),i=1/2..13/2,1):
    F:=plottools:-polygon([[1/2,1/2],[1/2,13/2],[13/2,13/2],[13/2,1/2]], color=yellow):
    plots:-display(Matrix(4,7,[seq(plots:-display(plottools:-polygon(p,color=red),F, E1,E2), p=[seq(i[1],i=P1)])]), scaling=constrained, axes=none, size=[800,700]);

     

     

    Carpet1.mw

    The code was edited.

     

     

    The Railway Challenge is a competition designed by the Institute of Mechanical Engineers (IMechE), aimed at engaging young engineers with the rail industry.  The challenge, now in its seventh successive year, brings together teams of university students, as well as apprentices and graduates working in industry across the world to test their business knowledge, design ability and technical skills in a live test environment.

    The Railway Challenge at Sheffield (RCAS) is an extracurricular student-led activity within the Mechanical Engineering department at the university of Sheffield, that designs, codes and manufactures a 10 1/4 inch gauge miniature locomotive to compete in the IMechE’s  Railway Challenge.  The locomotive is assessed in accordance with a set of strict rules and a detailed technical specification, such as traction, ride comfort, and a business case. The locomotives are tested live at a competition, which takes place in June at the Stapleford Miniature Railway in Leicestershire, where several categories of winners and an overall Railway Challenge champion is crowned.

    The team consists of around twenty members, and students studying Mechanical Engineering and even cross discipline can get involved as soon as they come to the University, getting into to the design of components within the suspension or braking systems for example, before proceeding to manufacture and test; allowing the students to experience all the stages of an engineering product as well as skills gained by working in the team such as effective communication, time management and financial planning.

    Last year the team was granted a sponsorship from Maplesoft, and as a result, huge improvements were made within the team. Overall the team jumped from finishing in 7th place to in the summer winning the maintainability challenge and finishing in 4th place overall – mostly down to the electronics working for the first year ever!

     

    Using Maplesoft’s donation the team switched form a central CRIO control system to a distributed network using I2C protocols and Arduino hardware. This did away with some of the electrical teething problems the team has suffered in previous years. It also introduced our Mechanical Engineers to coding that they would otherwise not do in their course.

    This year Maplesoft have again sponsored RCAS. The team is hoping to use the licenses to perform their structures calculations in an easy way to keep track of them for use in the design report. They are also hoping to use MapleSim for dynamics modelling, to assist with suspension design, and designing any electronics or control elements, such as filter design and motor control.

    We’re so excited to bring you guys #MapleOfficeHours! This is a program we’ve designed for students (but open to everyone) to connect via social media for help with Maple. Maple can play a really important part in your courses and can sometimes be intimidating for new users. We get it, there’s a lot of ground to cover. With #MapleOfficeHours, we will use social media as a live Q&A platform to help you figure out how to use Maple for your homework, assignments and more. Having trouble with a command or function? #MapleOfficeHours. Need help finding a specific resource or app? #MapleOfficeHours. You get the idea…

    Just like the office hours your professors hold, #MapleOfficeHours is going to be available on a regular basis for support. More events will be scheduled soon, but look out for Twitter chats, mini-webinars, Facebook live events and more. Once we get going, we’d love to hear your feedback on what other types of events we can offer and what topics you’d like to see covered.

    Our first #MapleOfficeHours event will be a live Twitter chat. On October 16th at 2PM EDT, I will be joined with Maple Product Manager, @DanielSkoog and Tech Support Team Lead, Dr. Matt Calder to answer as many questions about Maple that we can possibly fit into an hour’s time.

    To join the Twitter chat, use the hashtag #MapleOfficeHours when posting your questions and/or mention us with @maplesoft.

    Looking forward to seeing everyone at our first #MapleOfficeHours event on October 16th, 2PM EDT!

    Using the syntax in Maple we develop the energy with conservation equations here we are applying the commands int, factor, solve among others. We also integrate vector functions through the scalar product and finally we calculate conservative fields applying the rotational to a field of force. Exclusive for engineering students. In spanish.

    Work_of_a_Force.mw

    Lenin Araujo castillo

    Ambassafor of Maple


     

    Maple 2017.1 and 2017.2 introduced several improvements in the solution of PDE & Boundary conditions problems (exact solutions).  Maple 2017.3 includes more improvements in this same area.

     

    The following is a set of 25 examples of different PDE & Boundary Conditions problems that are solvable in Maple 2017.3 but not in Maple 2017.2 or previous releases. In the examples that follow, in some cases the PDE is different, in other cases the boundary conditions are of a different kind or solving the problem involves different computational strategies.

    As usual, at the end there is a link pointing to the corresponding worksheet.

     

    pde[1] := diff(u(x, t), t) = k*(diff(u(x, t), x, x)); bc[1] := u(x, 0) = 6+4*cos(3*Pi*x/L), (D[1](u))(0, t) = 0, (D[1](u))(L, t) = 0

    diff(u(x, t), t) = k*(diff(diff(u(x, t), x), x))

     

    u(x, 0) = 6+4*cos(3*Pi*x/L), (D[1](u))(0, t) = 0, (D[1](u))(L, t) = 0

    (1)

    `assuming`([pdsolve([pde[1], bc[1]], u(x, t))], [L > 0, k > 0])

    u(x, t) = 6+4*cos(3*Pi*x/L)*exp(-9*k*Pi^2*t/L^2)

    (2)

    pde[2] := diff(g(t, x), t) = diff(g(t, x), x, x)+a*g(t, x); bc[2] := g(0, x) = 1

    diff(g(t, x), t) = diff(diff(g(t, x), x), x)+a*g(t, x)

     

    g(0, x) = 1

    (3)

    pdsolve([pde[2], bc[2]])

    g(t, x) = exp(a*t)

    (4)

    pde[3] := diff(u(x, t), t) = diff(u(x, t), x, x)

    bc[3] := u(x, 0) = f(x), u(-1, t) = 0, u(1, t) = 0

    diff(u(x, t), t) = diff(diff(u(x, t), x), x)

     

    u(x, 0) = f(x), u(-1, t) = 0, u(1, t) = 0

    (5)

    pdsolve([pde[3], bc[3]], u(x, t))

    u(x, t) = Sum((Int(f(x)*sin(n*Pi*x), x = -1 .. 1))*sin(n*Pi*x)*exp(-Pi^2*n^2*t), n = 1 .. infinity)

    (6)

    pde[4] := diff(u(x, t), t) = k*(diff(u(x, t), x, x)); bc[4] := u(0, t) = 0, u(L, t) = 0, u(x, 0) = piecewise(0 < x and x <= (1/2)*L, 1, (1/2)*L < x and x < L, 2)

    pde[4] := diff(u(x, t), t) = k*(diff(u(x, t), x, x))

     

    u(0, t) = 0, u(L, t) = 0, u(x, 0) = piecewise(0 < x and x <= (1/2)*L, 1, (1/2)*L < x and x < L, 2)

    (7)

    `assuming`([pdsolve([pde[4], bc[4]], u(x, t))], [L > 0])

    u(x, t) = Sum((2*cos((1/2)*Pi*n)+2+4*(-1)^(1+n))*sin(n*Pi*x/L)*exp(-k*Pi^2*n^2*t/L^2)/(Pi*n), n = 1 .. infinity)

    (8)

    pde[5] := diff(u(x, t), t) = k*(diff(u(x, t), x, x))

    bc[5] := (D[1](u))(0, t) = 0, (D[1](u))(L, t) = 0, u(x, 0) = -3*cos(8*Pi*x/L)

    diff(u(x, t), t) = k*(diff(diff(u(x, t), x), x))

     

    (D[1](u))(0, t) = 0, (D[1](u))(L, t) = 0, u(x, 0) = -3*cos(8*Pi*x/L)

    (9)

    `assuming`([pdsolve([pde[5], bc[5]], u(x, t))], [0 < L, 0 < k])

    u(x, t) = -3*cos(8*Pi*x/L)*exp(-64*k*Pi^2*t/L^2)

    (10)

    pde[6] := diff(u(x, t), t) = k*(diff(u(x, t), x, x))+f(x, t); bc[6] := u(0, t) = 0, u(l, t) = 0, u(x, 0) = g(x)

    diff(u(x, t), t) = k*(diff(diff(u(x, t), x), x))+f(x, t)

     

    u(0, t) = 0, u(l, t) = 0, u(x, 0) = g(x)

    (11)

    pdsolve([pde[6], bc[6]], u(x, t))

    u(x, t) = Sum(2*(Int(g(tau1)*sin(Pi*n1*tau1/l), tau1 = 0 .. l))*sin(Pi*n1*x/l)*exp(-k*Pi^2*n1^2*t/l^2)/l, n1 = 1 .. infinity)+Int(Sum(2*(Int(f(x, tau1)*sin(Pi*n*x/l), x = 0 .. l))*sin(Pi*n*x/l)*exp(-k*Pi^2*n^2*(t-tau1)/l^2)/l, n = 1 .. infinity), tau1 = 0 .. t)

    (12)

    pde[7] := diff(u(x, t), t) = diff(u(x, t), x, x); bc[7] := u(x, 0) = f(x), u(-1, t) = 0, u(1, t) = 0

    diff(u(x, t), t) = diff(diff(u(x, t), x), x)

     

    u(x, 0) = f(x), u(-1, t) = 0, u(1, t) = 0

    (13)

    pdsolve([pde[7], bc[7]], u(x, t))

    u(x, t) = Sum((Int(f(x)*sin(n*Pi*x), x = -1 .. 1))*sin(n*Pi*x)*exp(-Pi^2*n^2*t), n = 1 .. infinity)

    (14)

    pde[8] := diff(u(x, t), t) = k*(diff(u(x, t), x, x))-h*u(x, t); bc[8] := u(x, 0) = sin(x), u(-Pi, t) = u(Pi, t), (D[1](u))(-Pi, t) = (D[1](u))(Pi, t)

    diff(u(x, t), t) = k*(diff(diff(u(x, t), x), x))-h*u(x, t)

     

    u(x, 0) = sin(x), u(-Pi, t) = u(Pi, t), (D[1](u))(-Pi, t) = (D[1](u))(Pi, t)

    (15)

    pdsolve([pde[8], bc[8]], u(x, t))

    u(x, t) = sin(x)*exp(-t*(k+h))

    (16)

    pde[9] := diff(u(x, t), t) = diff(u(x, t), x, x)

    bc[9] := u(0, t) = 20, u(1, t) = 50, u(x, 0) = 0

    diff(u(x, t), t) = diff(diff(u(x, t), x), x)

     

    u(0, t) = 20, u(1, t) = 50, u(x, 0) = 0

    (17)

    pdsolve([pde[9], bc[9]], u(x, t))

    u(x, t) = 20+Sum((-40+100*(-1)^n)*sin(n*Pi*x)*exp(-Pi^2*n^2*t)/(Pi*n), n = 1 .. infinity)+30*x

    (18)

    pde[10] := diff(u(x, y), x, x)+diff(u(x, y), y, y) = 0; bc[10] := u(x, 0) = 0, u(x, 1) = 0, u(0, y) = y^2-y, u(1, y) = 0

    diff(diff(u(x, y), x), x)+diff(diff(u(x, y), y), y) = 0

     

    u(x, 0) = 0, u(x, 1) = 0, u(0, y) = y^2-y, u(1, y) = 0

    (19)

    pdsolve([pde[10], bc[10]], u(x, y))

    u(x, y) = Sum(-4*((-1)^n-1)*sin(Pi*y*n)*(exp(Pi*n*(2*x-1))-exp(Pi*n))*exp(-Pi*n*(x-1))/((exp(2*Pi*n)-1)*Pi^3*n^3), n = 1 .. infinity)

    (20)

    pde[11] := diff(u(x, y), x, x)+diff(u(x, y), y, y) = 0

    bc[11] := (D[1](u))(0, y) = 0, (D[1](u))(L, y) = 0, u(x, H) = f(x), u(x, 0) = 0

    diff(diff(u(x, y), x), x)+diff(diff(u(x, y), y), y) = 0

     

    (D[1](u))(0, y) = 0, (D[1](u))(L, y) = 0, u(x, H) = f(x), u(x, 0) = 0

    (21)

    `assuming`([pdsolve([pde[11], bc[11]], u(x, y))], [0 < L, 0 < H])

    u(x, y) = Sum(2*(Int(cos(Pi*x*n/L)*f(x), x = 0 .. L))*cos(Pi*x*n/L)*exp(Pi*n*(H-y)/L)*(exp(2*Pi*y*n/L)-1)/(L*(exp(2*Pi*H*n/L)-1)), n = 1 .. infinity)

    (22)

    pde[12] := diff(u(x, y), x, x)+diff(u(x, y), y, y) = 0

    bc[12] := (D[1](u))(L, y) = 0, u(x, H) = 0, u(x, 0) = 0, (D[1](u))(0, y) = g(y)

    diff(diff(u(x, y), x), x)+diff(diff(u(x, y), y), y) = 0

     

    (D[1](u))(L, y) = 0, u(x, H) = 0, u(x, 0) = 0, (D[1](u))(0, y) = g(y)

    (23)

    `assuming`([pdsolve([pde[12], bc[12]], u(x, y))], [0 < x, x <= L, 0 < y, y <= H])

    u(x, y) = Sum(-2*(Int(sin(Pi*y*n/H)*g(y), y = 0 .. H))*sin(Pi*y*n/H)*(exp(-Pi*n*(L-2*x)/H)+exp(Pi*L*n/H))*exp(Pi*n*(L-x)/H)/(Pi*n*(exp(2*Pi*L*n/H)-1)), n = 1 .. infinity)

    (24)

    pde[13] := diff(u(x, y), x, x)+diff(u(x, y), y, y) = 0

    bc[13] := (D[1](u))(0, y) = 0, u(x, 0) = 0, u(x, H) = 0, u(L, y) = g(y)

    diff(diff(u(x, y), x), x)+diff(diff(u(x, y), y), y) = 0

     

    (D[1](u))(0, y) = 0, u(x, 0) = 0, u(x, H) = 0, u(L, y) = g(y)

    (25)

    `assuming`([pdsolve([pde[13], bc[13]], u(x, y))], [0 < L, 0 < H])

    u(x, y) = Sum(2*(Int(sin(Pi*y*n/H)*g(y), y = 0 .. H))*sin(Pi*y*n/H)*exp(Pi*n*(L-x)/H)*(exp(2*Pi*x*n/H)+1)/(H*(exp(2*Pi*L*n/H)+1)), n = 1 .. infinity)

    (26)

    pde[14] := diff(u(x, y), x, x)+diff(u(x, y), y, y) = 0

    bc[14] := u(0, y) = g(y), u(L, y) = 0, (D[2](u))(x, 0) = 0, u(x, H) = 0

    diff(diff(u(x, y), x), x)+diff(diff(u(x, y), y), y) = 0

     

    u(0, y) = g(y), u(L, y) = 0, (D[2](u))(x, 0) = 0, u(x, H) = 0

    (27)

    `assuming`([pdsolve([pde[14], bc[14]], u(x, y))], [0 < x, x <= L, 0 < y, y <= H])

    u(x, y) = Sum(-2*(exp(-(L-2*x)*(1/2+n)*Pi/H)-exp((1/2)*Pi*(1+2*n)*L/H))*cos((1/2)*Pi*(1+2*n)*y/H)*(Int(cos((1/2)*Pi*(1+2*n)*y/H)*g(y), y = 0 .. H))*exp((1/2)*Pi*(1+2*n)*(L-x)/H)/(H*(exp(Pi*(1+2*n)*L/H)-1)), n = 0 .. infinity)

    (28)

    pde[15] := diff(u(x, y), x, x)+diff(u(x, y), y, y) = 0

    bc[15] := u(0, y) = 0, u(L, y) = 0, u(x, 0) = (D[2](u))(x, 0), u(x, H) = f(x)

    diff(diff(u(x, y), x), x)+diff(diff(u(x, y), y), y) = 0

     

    u(0, y) = 0, u(L, y) = 0, u(x, 0) = (D[2](u))(x, 0), u(x, H) = f(x)

    (29)

    `assuming`([pdsolve([pde[15], bc[15]], u(x, y))], [0 < x, x <= L, 0 < y, y <= H])

    u(x, y) = Sum(2*exp(Pi*n*(H-y)/L)*sin(Pi*x*n/L)*((Pi*n+L)*exp(2*Pi*y*n/L)+Pi*n-L)*(Int(sin(Pi*x*n/L)*f(x), x = 0 .. L))/(L*((Pi*n+L)*exp(2*Pi*H*n/L)+Pi*n-L)), n = 1 .. infinity)

    (30)

    pde[16] := diff(u(x, t), t, t) = c^2*(diff(u(x, t), x, x))

    bc[16] := u(0, t) = 0, (D[1](u))(L, t) = 0, (D[2](u))(x, 0) = 0, u(x, 0) = f(x)

    diff(diff(u(x, t), t), t) = c^2*(diff(diff(u(x, t), x), x))

     

    u(0, t) = 0, (D[1](u))(L, t) = 0, (D[2](u))(x, 0) = 0, u(x, 0) = f(x)

    (31)

    `assuming`([pdsolve([pde[16], bc[16]], u(x, t))], [0 < x, x <= L])

    u(x, t) = Sum(2*cos((1/2)*c*Pi*(1+2*n)*t/L)*(Int(sin((1/2)*Pi*(1+2*n)*x/L)*f(x), x = 0 .. L))*sin((1/2)*Pi*(1+2*n)*x/L)/L, n = 0 .. infinity)

    (32)

    pde[17] := diff(w(x1, x2, x3, t), t) = diff(w(x1, x2, x3, t), x1, x1)+diff(w(x1, x2, x3, t), x2, x2)+diff(w(x1, x2, x3, t), x3, x3)

    bc[17] := w(x1, x2, x3, 0) = exp(x1)+x2*x3^5

    diff(w(x1, x2, x3, t), t) = diff(diff(w(x1, x2, x3, t), x1), x1)+diff(diff(w(x1, x2, x3, t), x2), x2)+diff(diff(w(x1, x2, x3, t), x3), x3)

     

    w(x1, x2, x3, 0) = exp(x1)+x2*x3^5

    (33)

    pdsolve([pde[17], bc[17]])

    w(x1, x2, x3, t) = Sum(t^n*((proc (U) options operator, arrow; diff(diff(U, x1), x1)+diff(diff(U, x2), x2)+diff(diff(U, x3), x3) end proc)@@n)(exp(x1)+x2*x3^5)/factorial(n), n = 0 .. infinity)

    (34)

    pde[18] := diff(u(x, t), t)+u(x, t)*(diff(u(x, t), x)) = -x

    bc[18] := u(x, 0) = x

    diff(u(x, t), t)+u(x, t)*(diff(u(x, t), x)) = -x

     

    u(x, 0) = x

    (35)

    pdsolve([pde[18], bc[18]], u(x, t))

    u(x, t) = x/tan(t+(1/4)*Pi)

    (36)

    pde[19] := diff(u(x, t), t)-u(x, t)^2*(diff(u(x, t), x)) = 3*u(x, t)

    bc[19] := u(x, 0) = x

    diff(u(x, t), t)-u(x, t)^2*(diff(u(x, t), x)) = 3*u(x, t)

     

    u(x, 0) = x

    (37)

    pdsolve([pde[19], bc[19]], u(x, t))

    u(x, t) = -exp(3*t)*((-6*exp(6*t)*x+6*x+9)^(1/2)-3)/(exp(6*t)-1)

    (38)

    pde[20] := diff(u(x, t), t)-x*u(x, t)*(diff(u(x, t), x)) = 0

    bc[20] := u(x, 0) = x

    diff(u(x, t), t)-x*u(x, t)*(diff(u(x, t), x)) = 0

     

    u(x, 0) = x

    (39)

    pdsolve([pde[20], bc[20]], u(x, t))

    u(x, t) = -LambertW(-t*x)/t

    (40)

    pde[21] := diff(u(x, t), t)+u(x, t)*(diff(u(x, t), x)) = 0

    bc[21] := u(x, 0) = x

    diff(u(x, t), t)+u(x, t)*(diff(u(x, t), x)) = 0

     

    u(x, 0) = x

    (41)

    pdsolve([pde[21], bc[21]], u(x, t))

    u(x, t) = x/(t+1)

    (42)

    pde[22] := (t+u(x, t))*(diff(u(x, t), x))+t*(diff(u(x, t), t)) = 0; bc[22] := u(x, 1) = x

    (t+u(x, t))*(diff(u(x, t), x))+t*(diff(u(x, t), t)) = 0

     

    u(x, 1) = x

    (43)

    pdsolve([pde[22], bc[22]], u(x, t))

    u(x, t) = (t-x-1)/(ln(1/t)-1)

    (44)

    pde[23] := (diff(r*(diff(u(r, theta), r)), r))/r+(diff(u(r, theta), theta, theta))/r^2 = 0; bc[23] := u(a, theta) = f(theta), u(r, -Pi) = u(r, Pi), (D[2](u))(r, -Pi) = (D[2](u))(r, Pi)

    (diff(u(r, theta), r)+r*(diff(diff(u(r, theta), r), r)))/r+(diff(diff(u(r, theta), theta), theta))/r^2 = 0

     

    u(a, theta) = f(theta), u(r, -Pi) = u(r, Pi), (D[2](u))(r, -Pi) = (D[2](u))(r, Pi)

    (45)

    `assuming`([pdsolve([pde[23], bc[23]], u(r, theta), HINT = boundedseries)], [a > 0])

    u(r, theta) = (1/2)*(2*(Sum(r^n*(sin(n*theta)*(Int(f(theta)*sin(n*theta), theta = -Pi .. Pi))+cos(n*theta)*(Int(f(theta)*cos(n*theta), theta = -Pi .. Pi)))*a^(-n)/Pi, n = 1 .. infinity))*Pi+Int(f(theta), theta = -Pi .. Pi))/Pi

    (46)

    pde[24] := diff(g(t, x), t) = diff(g(t, x), x, x)+a*g(t, x); bc[24] := g(0, x) = f(x)

    diff(g(t, x), t) = diff(diff(g(t, x), x), x)+a*g(t, x)

     

    g(0, x) = f(x)

    (47)

    pdsolve([pde[24], bc[24]])

    g(t, x) = exp(a*t)*invfourier(fourier(f(x), x, s1)*exp(-t*s1^2), s1, x)

    (48)

    pde[25] := diff(u(x, y), x, x)+diff(u(x, y), y, y) = 0; bc[25] := u(0, y) = y*(-y+1), u(1, y) = 0, (D[2](u))(x, 0) = 0, (D[2](u))(x, 1) = 0

    diff(diff(u(x, y), x), x)+diff(diff(u(x, y), y), y) = 0

     

    u(0, y) = y*(-y+1), u(1, y) = 0, (D[2](u))(x, 0) = 0, (D[2](u))(x, 1) = 0

    (49)

    pdsolve([pde[25], bc[25]], u(x, y))

    u(x, y) = Sum(2*((-1)^n+1)*cos(Pi*y*n)*(exp(Pi*n*(2*x-1))-exp(Pi*n))*exp(-Pi*n*(x-1))/((exp(2*Pi*n)-1)*Pi^2*n^2), n = 1 .. infinity)

    (50)

    ``


     

    Download ImprovementsInPdsolve.mw

    Edgardo S. Cheb-Terrab
    Physics, Differential Equations and Mathematical Functions, Maplesoft

    And the Nobel prize in physics 2017 went for work in General Relativity ! Actually, experimental work involving sophisticated detectors and Numerical Relativity, one of the branches of GR. The prize was awarded to Rainer Weiss (85 years old, 1/2 of the prize), Barry Barish (81 years old, 1/4 of the prize) and Kip Thorne (77 years old, 1/4 of the prize) who have "shaken the world again" with their work on Ligo experiment, which was able to detect ripples in the fabric of spacetime.

    General Relativity continues to be at the center of work in theoretical and experimental physics. I take this opportunity to note that, in Maple 2017, among the several improvements in the Physics package regarding General Relativity, there is a new package, Physics:-ThreePlusOne, all dedicated to the symbolic manipulations necessary to formulate problems in Numerical Relativity.

    The GR functionality implemented in Physics, Physics:-Tetrads and Physics:-ThreePlusOne is unique in computer algebra systems and reflects the Maplesoft intention, for several years now, to provide the very best possible computer algebra environment for Physics, regarding current research activity as well as related education in advanced mathematical-physics methods.

    For what is going on in theoretical physics nowadays and its connection with General Relativity, in very short: the unification of gravity with the other forces, check for instance this map from Aug/2015 (by the way a very nice summary for whoever is interested):

    It is also interesting the article behind this map of topics as well as this brilliant and accessible presentation by Nima Arkani-Hamed (Princeton):  Quantum Mechanics and Spacetime in the 21st Century, given in the Perimeter Institute for Theoretical Physics (Waterloo), November 2014. 

    Edgardo S. Cheb-Terrab
    Physics, Differential Equations and Mathematical Functions, Maplesoft

    We have released a small maintenance update to Maple. Maple 2017.3 provides enhancements in several areas, including mathematical typesetting, pdsolve, and the Physics package. It also provides improvements to the MapleCloud, including a fix for a problem that prevented some Mac users from logging on with their Google credentials.

    This update is available through Tools>Check for Updates in Maple, and is also available from our website on the Maple 2017.3 download page.

     

    Group of exercises solved using Maple scientific software, with the necessary considerations of some basic commands: evalf and convert that will show the solutions with the user-defined digits and the angular measurement in sexagesimal degrees. Important use of the law of the triangle through of vector position applied to vectors in vector spaces, vector force and vector moment for engineering students. In spanish.

    Exercises_of_vectors_forces_and_moment_with_Maple.mw

    Videotutorial:

    https://www.youtube.com/watch?v=DxpO0gc5GCA

    Lenin Araujo Castillo

    Ambassador of Maple

    # Riemann hypothesis is false! (simple proof)
     

    restart;
    assume( s>0, s<1/2, t>0 );
    coulditbe(abs(Zeta(s+I*t))=0);

                                  true

    # Q.E.D.

    Unfortunately coulditbe(Zeta(s+I*t)=0) returns FAIL, but our assertion is already demonstrated!

    The moral: the assume facility deserves a much more careful implementation.

    We’ve just released a major new version of MapleSim. The MapleSim 2017 family of products provides new and improved model development and analysis tools, expands modeling scope, introduces new deployment options, and strengthens toolchain connectivity.  Here are some highlights:

    • The new Initialization Diagnostics App further simplifies the initialization task by helping you determine how your initial values are computed and what you need to do to adjust them.
       
    • The new Modal Analysis App helps you explore and understand the natural vibration modes of your mechanism, so you can determine how to reduce the vibration in the final product. 
       
    • Over 100 new components include expansions to the Electrical and Magnetic libraries.
       
    • A new Modelica® code editor makes it easier to create Modelica-based custom components.
       
    • The MapleSim Heat Transfer Library from CYBERNET, a new add-on component library, provides a comprehensive view into heat transfer effects in your model, enabling you to refine your  design to improve performance and avoid overheating.
       
    • The new MapleSim Explorer product provides a cost-effective deployment solution that allows you to make MapleSim models available to more people in your organization.

     

    There’s more, of course.  See What’s New in MapleSim for lots more details.

    eithne

    Try More/GeneratePDF  in  the menu under  a post/question. See screen_26.09.17.docx as an example of a result. Also Adobe Acrobat Reader fails with it. That was submitted to MaplePrimes staff through the Contact  button at the bottom of this page. I obtained no feedback from them.

    This might be of interest to some of us here - a comparison of differential equation solvers between many different packages/tools/libraries:

    http://www.stochasticlifestyle.com/comparison-differential-equation-solver-suites-matlab-r-julia-python-c-fortran/

    The "analysis" of maple's capabilities are presented as somewhat limited in comparison to mathematica's - I wonder if this is a simple bias/misinformation of the author, or if his conclusions are correct. 

    My interface has frozen, but above is a screen shot of what is by far the most unusual response from the CAS in the i guess 8 or so years ive been using it in total.

     

    *updated situation its allowing me to interrupt evaluation

    The development of the calculation of moments using force vectors is clearly observed by taking a point and also a line. Different exercises are solved with the help of Maple syntax. We can also visualize the vector behavior in the different configurations of the position vector. Applications designed exclusively for engineering students. In Spanish.

    Moment_of_a_force_using_vectors_updated.mw

    Lenin Araujo Castillo

    Ambassador of Maple

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