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On November 11th, Canada and other Commonwealth member states will celebrate Remembrance Day, also known as Armistice Day. This holiday commemorates the armistice signed by Germany and the Entente Powers in Compiègne, France on November 11, 1918, to end the hostilities on the Western Front of World War I. The armistice came into effect at 11:00 am that morning – the “eleventh hour of the eleventh day of the eleventh month”. 

Similar to how November 11th – which can be written as 11/11 – is a palindromic date that reads the same forward and backward, last year there was “Twosday” – February 22, 2022, also written 22/2/22. 

Palindromic dates like November 11th that consist only of a day and a month happen every year, but how long will we have to wait until the next “Twosday”? We can use Maple Learn’s new Calendar Calculator to find out!


To use this document, simply input two dates and press ‘Calculate’ to find the amount of time between them, presented in a variety of units. For example, here are the results for the number of days left until Christmas from November 11th of this year:


If we return to our original question, which concerns how long we’ll have to wait until the next “Twosday”, we can use this document to find our answer:

You can use this document as a countdown to find out how much time is left until your favorite holiday, your next birthday, or the time between now and any past or future date; try out the countdown document here!

This is an successfull attempt to simulate space frames in MapleSim using the relatively new Rod component.

At t=3s, a lateral force component is applied to make the simulation more interesting.

The structure collapses/folds in an origami style fashion.

To build the model, MapleSim needs additional components.

For example, an equilateral triangle

requires the addition of rigid body frames at the connection of two rods.

Additionally, the rigid body frames must have initial position conditions (ICs) that match the intended structure.

Interestingly the ICs do not have to be set to Treat as Guess. It is only required to put an approximate coordinate. Leaving the ICs on Ignore was sufficient for the attached model.

Rod components can be replaced by Flexible Beam components which require considerably more simulation time and either Revolute joints at their ends or a rather complex connection with Rigid Body frames (of zero length) to adjacent Flexible Beam components.

 

Spaceframe_2.msim

 

 

Almost 300 years ago, a single letter exchanged between two brilliant minds gave rise to one of the most enduring mysteries in the world of number theory. 

In 1742, Christian Goldbach penned a letter to fellow mathematician Leonhard Euler proposing that every even integer greater than 2 can be written as a sum of two prime numbers. This statement is now known as Goldbach’s Conjecture (it is considered a conjecture, and not a theorem because it is unproven). While neither of these esteemed mathematicians could furnish a formal proof, they shared a conviction that this conjecture held the promise of being a "completely certain theorem." The following image demonstrates how prime numbers add to all even numbers up to 50:

From its inception, Goldbach's Conjecture has enticed generations of mathematicians to seek evidence of its legitimacy. Though weaker versions of the conjecture have been proved, the definitive proof of the original conjecture has remained elusive. There was even once a one-million dollar cash prize set to be awarded to anyone who could provide a valid proof, though the offer has now elapsed. While a heuristic argument, which relies on the probability distribution of prime numbers, offers insight into the conjecture's likelihood of validity, it falls short of providing an ironclad guarantee of its truth.

The advent of modern computing has emerged as a beacon of progress. With vast computational power at their disposal, contemporary mathematicians like Dr. Tomàs Oliveira e Silva have achieved a remarkable feat—verification of the conjecture for every even number up to an astonishing 4 quintillion, a number with 18 zeroes.

Lazar Paroski’s Goldbach Conjecture Document on Maple Learn offers an avenue for users of all skill levels to delve into one of the oldest open problems in the world of math. By simply opening this document and inputting an even number, a Maple algorithm will swiftly reveal Goldbach’s partition (the pair of primes that add to your number), or if you’re lucky it could reveal that you have found a number that disproves the conjecture once and for all.

# ----------------------------------THE DESIGN OF THE MAPLET SCREEN---------------------
with(Maplets[Elements]):
HCC:=Maplet(Window('title'="HEAT CONDUCTIVITY CONTROL",["WITH THIS APPLICATION THE CONDUCTIVITY COEFFICIENT OF A ONE-DIMENSIONAL OBJECT, APPROXIMATING THE TEMPRATURE OF THE OBJECT TO A TARGET TEMPRETURE AT A CERTAIN FINAL TIME, IS CONTROLLED. ",[["l",TextField[l](3)],["T",TextField[T](3)],["f(x,t)",TextField[f](15)],["phi(x)",TextField[ph](5)]],[["k(0)",TextField[k0](3)],["g0(t)",TextField[g0](10)],["k(l)",TextField[kl](3)],["g1(t)",TextField[g1](10)],["mu(x)",TextField[mu](10)]],[["alpha",TextField[alpha](3)],["kaplus(x)",TextField[kaplus](5)],["N",TextField[N](3)],["kstart(x)",TextField[kstart](3)],["beta",TextField[beta](3)],["eps",TextField[eps](3)]] ,[Button("Calculate the Control",Evaluate('kutu'=ms(N,l,alpha,T,ph,f,g0,g1,mu,kaplus,k0,kl,kstart,beta,eps))),[TextBox['kutu'](30..30)],Button("Draw the Control",Evaluate('Draw'='plot(kutu,x=0..l)')),Plotter['Draw'](),[[Button("Distance to
Target",Evaluate('kutu2'=ms8(N,l,alpha,T,ph,f,g0,g1,mu,kaplus,k0,kl,kstart,beta,eps))),TextField['kutu2'](12)],[Button("Approximation to kaplus",Evaluate('kutu3'='evalf(int((kutu-kaplus)^2,x=0..l))')),TextField['kutu3'](12)]]],Button("Shutdown",Shutdown())])):
# -------------------------PROCEDURE FOR CALCULATION OF THE CONTROL FUNCTION-----------
with(inttrans):
with(linalg):
ms:=proc(N,l,alpha,T,ph,f,g0,g1,mu,kaplus,k0,kl,kstart,beta,eps):
with(inttrans):
with(linalg):
w:=simplify(x^2/2*g1/(l*kl)+(x^2/2-x*l)*g0/(l*k0)):
phdal:=ph-subs(t=0,w):
fdal:=simplify(f-diff(w,t)+diff(kaplus*diff(w,x),x)):
# ---------------------------------Solution of the Heat Problem------------------------------------
dp:=proc(ka)
with(inttrans):
with(linalg):
phi:=Vector(1..N):
phi[1]:=1/sqrt(l):
for i from 2 to N do
phi[i]:=evalf(sqrt(2/l)*cos((i-1)*Pi*x/l)):
od:
K:=Array(1..N,1..N):
for j from 1 to N do
for k from 1 to N do
K[j,k]:=evalf(-int(ka*diff(phi[k],x$2)*phi[j],x=0..l)):
od:
od:
F:=Vector(1..N):
for n from 1 to N do
F[n]:=evalf(int(fdal*phi[n],x=0..l)):
od:
A:=Vector(1..N):
for m from 1 to N do
A[m]:=evalf(int(phdal*phi[m],x=0..l)):
od:
KL:=Matrix(1..N,1..N):
for j1 from 1 to N do
for k1 from 1 to N do
if (j1=k1) then KL[j1,k1]:=s+K[j1,k1] else KL[j1,k1]:=K[j1,k1] fi:
od:
od:
FL:=Vector(1..N):
for i1 from 1 to N do
FL[i1]:=evalf(laplace(F[i1],t,s));
od:
S:=Vector(1..N):
for i2 from 1 to N do
S[i2]:=(A[i2]+FL[i2]);
od:
C:=Vector(1..N):
C:=evalm(inverse(KL)&*S):
c:=Vector(1..N):
for i3 from 1 to N do
c[i3]:=evalf(invlaplace(C[i3],s,t)):
od:
v:=evalf(add(c[n1]*phi[n1],n1=1..N)):
uyak:=v+w;
end:
# ---------------------------------Solution of the Adjoint Problem------------------------------------
ap:=proc(ka)
with(inttrans):
with(linalg):
utau:=evalf(subs(t=T-tau,dp(ka))):
phe:=evalf(2*(subs(tau=0,utau)-mu));
phie:=Vector(1..N):
phie[1]:=1/sqrt(l):
for i4 from 2 to N do
phie[i4]:=evalf(sqrt(2/l)*cos((i4-1)*Pi*x/l)):
od:
Kc:=Array(1..N,1..N):
for j2 from 1 to N do
for k2 from 1 to N do
Kc[j2,k2]:=evalf(-int(ka*diff(phie[k2],x$2)*phie[j2],x=0..l)):
od:
od:
Fc:=Vector(1..N):
for m1 from 1 to N do
Fc[m1]:=0:
od:
Ac:=Vector(1..N):
for cm1 from 1 to N do
Ac[cm1]:=evalf(int(phe*phie[cm1],x=0..l)):
od:
KLC:=Matrix(1..N,1..N):
for cj1 from 1 to N do
for ck1 from 1 to N do
if (cj1=ck1) then KLC[cj1,ck1]:=s+Kc[cj1,ck1] else KLC[cj1,ck1]:=Kc[cj1,ck1] fi:
od:
od:
FLC:=Vector(1..N):
for ci1 from 1 to N do
FLC[ci1]:=evalf(laplace(Fc[ci1],tau,s));
od:
Sc:=Vector(1..N):
for ci2 from 1 to N do
Sc[ci2]:=(Ac[ci2]+FLC[ci2]);
od:
CC:=Vector(1..N):
CC:=evalm(inverse(KLC)&*Sc):
cc:=Vector(1..N):
for ci3 from 1 to N do
cc[ci3]:=evalf((invlaplace(CC[ci3],s,tau))):
od:
ve:=evalf(add(cc[cn]*phie[cn],cn=1..N)):
eta:=evalf(subs(tau=T-t,ve));
end:
# ---------------------------------Calculation of the Gradient----------------------------------
T�rev:=proc(alpha,ka)
T�re:=simplify(evalf(-int(diff(dp(ka),x)*diff(ap(ka),x),t=0..T)+2*alpha*(ka-kaplus)));
end:
# ----------------------------Calculation of the Cost Functional--------------------------------
Jka:=proc(ka)
IJ1:=evalf(int((subs(t=T,dp(ka))-mu)^2,x=0..l));
end:
Sta:=proc(ka)
IJ2:=simplify(evalf((int((ka-kaplus)^2,x=0..l))));
end:
II:=proc(ka)
IJ:=simplify(evalf(Jka(ka)+alpha*Sta(ka))):
end:# 
# -----------------------------------Minimizing Process--------------------------------------------
a[0]:=kstart:
ka[0]:=kstart:
say�:=0:
for im from 0 to 60 do
a[im+1]:=simplify(evalf(ka[im-say�]-beta*T�rev(alpha,ka[im-say�]))): 
fark:=evalf(II(ka[im-say�])-II(a[im+1])): 
if(fark>0 and fark<eps) then break elif (fark>0) then 
j:=im+1: ka[j-say�]:=a[im+1]:   elif(fark<=0) then  say�:=say�+1: beta:=beta/(1.2): ka[im-say�+2]:=ka[im-say�+1]:   else fi:
od:
optcont:=a[im+1]:
end:
# -------------------------END OF THE PROCEDURE FOR CALCULATION OF THE CONTROL FUNCTION-----------
# ------PROCEDURE FOR CALCULATION OF THE DISTANCE TO THE TARGET FUNCTION-----------
ms8:=proc(N,l,alpha,T,ph,f,g0,g1,mu,kaplus,k0,kl,kstart,beta,eps):
with(inttrans):
with(linalg):
w8:=simplify(x^2/2*g1/(l*kl)+(x^2/2-x*l)*g0/(l*k0)):
phdal8:=ph-subs(t=0,w8):
fdal8:=simplify(f-diff(w8,t)+diff(kaplus*diff(w8,x),x)):
phi8:=Vector(1..N):
phi8[1]:=1/sqrt(l):
for i8 from 2 to N do
phi8[i8]:=evalf(sqrt(2/l)*cos((i8-1)*Pi*x/l)):
od:
K8:=Array(1..N,1..N):
for j8 from 1 to N do
for k8 from 1 to N do
K8[j8,k8]:=evalf(-int(ms(N,l,alpha,T,ph,f,g0,g1,mu,kaplus,k0,kl,kstart,beta,eps)*diff(phi8[k8],x$2)*phi8[j8],x=0..l)):
od:
od:
F8:=Vector(1..N):
for m28 from 1 to N do
F8[m28]:=evalf(int(fdal8*phi8[m28],x=0..l)):
od:
A8:=Vector(1..N):
for m8 from 1 to N do
A8[m8]:=evalf(int(phdal8*phi8[m8],x=0..l)):
od:
KL8:=Matrix(1..N,1..N):
for j18 from 1 to N do
for k18 from 1 to N do
if (j18=k18) then KL8[j18,k18]:=s+K8[j18,k18] else KL8[j18,k18]:=K8[j18,k18] fi:
od:
od:
FL8:=Vector(1..N):
for i148 from 1 to N do
FL8[i148]:=evalf(laplace(F8[i148],t,s));
od:
S8:=Vector(1..N):
for i48 from 1 to N do
S8[i48]:=(A8[i48]+FL8[i48]);
od:
C8:=Vector(1..N):
C8:=evalm(inverse(KL8)&*S8):
c8:=Vector(1..N):
for i58 from 1 to N do
c8[i58]:=evalf(invlaplace(C8[i58],s,t)):
od:
v8:=evalf(add(c8[n8]*phi8[n8],n8=1..N)):
uyak8:=v8+w8;
IJ18:=evalf(int((subs(t=T,uyak8)-mu)^2,x=0..l));
end:
# ------END OF THE PROCEDURE FOR CALCULATION OF THE DISTANCE TO THE TARGET FUNCTION-----------
Maplets[Display](HCC):


 

Here's a few commands you can use within Maple to collect information about your computer.  This is on a Windows 7 machine but should work for most Win7+ systems.  Not sure how far back the WMIC commands can be used, and it won't work on Mac or Linux. 

kernelopts(version)

`Maple 2022.0, X86 64 WINDOWS, Mar 8 2022, Build ID 1599809`

(1)

interface(version)

`Standard Worksheet Interface, Maple 2022.0, Windows 7, March 8 2022 Build ID 1599809`

(2)

ssystem("WMIC CPU Get Name, NumberOfCores, NumberOfLogicalProcessors")[2]

"Name                                             NumberOfCores  NumberOfLogicalProcessors  

Intel(R) Core(TM)2 Duo CPU     P8700  @ 2.53GHz  2              2                          


"

(3)

ssystem("WMIC computersystem get totalphysicalmemory")[2]

"TotalPhysicalMemory  

8517836800           


"

(4)

ssystem("WMIC memorychip get devicelocator, capacity, speed")[2]

"Capacity    DeviceLocator  Speed  

4294967296  DIMM 0         800    

4294967296  DIMM 1         800    


"

(5)

````

Download Maple_-_computer-info.mw

This post is in response to a question regarding the speed of Maple 2023 on different computers. 

I'm asking users to suggest a few benchmark problems for Maple to calculate for testing.  Basic system information (Processor speed, RAM, video card etc..) perform the calculation, get the timing and then we can collect all the information into a chart where we can update it in the original post of this thread. 

All input is welcome, it doesn't have to be confined to Maple 2023.  We can expand to as many older versions as we wish. 

we have recieved lots of great sumissions, but we want your great submission and now you have more time.

 

The deadline for submissions to the Art Gallery and Showcase for the 2023 Maple Conference is rapidly approaching. We really want to see your art! It doesn't have to be incredibly impressive or sophisticated, we just want to see what our community can create! If you've been working on something or have a great idea, you still have a few days to get it together to submit.

A penrose tiling mosaic of that famous Windows 95 background

Submission can be made by email to gallery@maplesoft.com but be sure to visit the visit our Call for Creative Works for details on the format of the submission.

 

The Proceedings of the Maple Conference 2022 are up at mapletransactions.org and I hope that you will find the articles interesting.  There is a brief memorial to Eugenio Roanes-Lozano, whom some of you will remember from past meetings. 

The cover image was the "People's Choice" from the Art Gallery, by Paul DeMarco.

This provides a nice excuse to remind you to register at the conference page for the Maple Conference 2023 and in particular to remind you to submit your entries for the Art Gallery.  See you there!  The conference will take place October 26 and 27, and features plenary talks by our own Laurent Bernardin and by Tom Crawford (Oxford, but more widely known as "The Naked Mathematician" for his incredibly popular YouTube videos on mathematical topics). See Tom Rocks Maths for more (or less :)

The deadline for submission to the Proceedings (which will again be published in Maple Transactions) will be Nov 27, one month after the conference ends.  We have put new processes in place to ensure a more timely publication schedule, and we anticipate that the Proceedings will be published in early Spring 2024.

 

Advanced Engineering Mathematics with Maple (AEM) by Dr. Lopez is such an art.

Mathematics and Control Theory talks easily in Maple...

Thanks Prof. Lopez. You are the MAN !!

Dr. Ali GÜZEL

The concept of “Maple Learn art” debuted on the MaplePrimes blog in December 2021.  Since then, we’ve come a long way with new Maple Learn features and ever-growing creative minds.  Creating art using mathematical expressions and shapes is a great way to hone both your mathematical skills and your creativity, and is the perfect break from a bout of studying or the like.

I started my own Maple Learn art journey over one year ago.  Let’s see how one’s art can improve over time using new and advanced features!

Art with Shapes, March 2022

This pi-themed pie is simple and cute, but could use some additional features:

Adding Shaded() around Maple Learn shape commands colors them in!

Fun fact: I hand-picked all of the coordinates for that pi symbol.  It was an arduous but rewarding process.  Nowadays, I recommend a new method.  When you create a table in Maple Learn with two number columns, the values are plotted as points.  These points can be clicked and dragged across the plot window, and the table updates automatically to display the new coordinates.  How can you use this to make art?

  1. Create a table as described above.
  2. Move the points with your mouse to create an outline of the desired shape.
  3. Use the coordinates from your table in your geometry command.

Let’s apply these techniques in a newer piece: a full recreation of the spaghetti emoji!

Art with Shapes, August 2023

Would you look at that?!  Fully-shaded colors, a background, and lines of spaghetti noodles that weren’t painstakingly guesstimated combine to create a wonderfully improved piece of art.

Art with Animation, March 2022

Visit the document to see its animation.  Animation is an invaluable feature in Maple Learn, frequently utilized to observe how changing variables affect functions or model a concept.  We’ve harnessed its power for animated artwork!  This animation is cute, using parametric functions and more to change the image as the animation variable changes.  Like the previous piece, it’s missing a background, and the leaves overlap the stem awkwardly in some places.

Art with Animation, August 2023

 

This piece has a simple background made with a large black square, but it enhances the overall effect.

The animation here comes from piecewise functions, which display different values based on a given criterion.  In this case, the criterion is the current value of the animation variable.

There are 32 individual polygons in this image (including 8 really tiny ones along the edges!) and 8 rainbow colors.  Each color is associated with a different piecewise function, and displays four random squares in that color in each frame of the animation.

This image isn’t that much more advanced than the animated flower, but I think the execution has vastly improved.

Whether you’ve been following these blog posts since December 2021 or are new to Maple Learn, we hope you give Maple Learn art a try.

And don’t forget that Maple is also a goldmine of artistic potential.  Maple’s bountiful collection of packages such as Fractals, ColorTools, plottools and more are great places to start for math that is as aesthetically pleasing as it is informative.

This week, our staff participated in a series of art challenges using either Maple Learn or Maple itself, each featuring a suggested theme and suggested mathematical content.  Check out the challenges and some of our employees’ entries below, and try out a challenge for yourself!

 

Tuesday’s Art Theme: Pasta

Mathematical Content: Shapes

Example: Lazar Paroski’s spiraling take on spaghetti

 

Wednesday’s Art Theme: Nature

Mathematical Content: Fractals

Example: John May’s Penrose tiling landscape (in Maple!)

 

Thursday’s Art Theme: Disco

Mathematical Content: Animation

Example: Paulina Chin’s disco ball (in Maple!)

 

Friday’s Art Theme: Space

Mathematical Content: Color

Example: that’s today!  Who knows what our staff will create…?

 

We hope these prompts have inspired you! If you create some art you’re really proud of, consider submitting it to be featured in the 2023 Maple Conference’s Creative Works Showcase.

Space. The final frontier. A frontier we wouldn’t stand a chance of exploring if it weren’t for the work of one Albert Einstein and his theories of special relativity. After all, how are we supposed to determine at what speed an alien spaceship is traveling towards Earth if we can’t understand how Newtonian physics break down at high velocities? That is precisely the question that this Maple MathApp asks. Using the interactive tool, you can see how the relative velocities change depending on your reference point. Just what you need for the next time you see a UFO rocketing through the sky!

But what if you don’t have the MathApp on hand when the aliens visit? (So rare to travel anywhere without a copy of Maple on you, I know, but it could happen.) You’ll have to just learn more about special relativity so that you can make those calculations on the fly. And luckily, we have just what you need to do that: our new Maple Learn collection on modern physics, created by Lazar Paroski. Still not quite sure how to wrap your head around the whole thing? Check out this document on the postulates of special relativity, which explains and demonstrates some of the fundamentals of special relativity with lively animations.

 

Screenshot of a Maple Learn document. The right side shows a paused animation of an observer, a moving car, and a moving bird. The left side shows calculations for the relative speeds.

Once you’ve gotten familiar with the basics, it’s time to get funky. This document on time dilation shows how two observers looking at the same event from different frames of reference can measure different times for that event. And of course once you start messing with time, everything gets weird. For an example, check out this document on length contraction, which explains how observers in different frames of reference can measure different lengths for the same moving object. Pretty wild stuff.

 

Screenshot of a Maple Learn document, showing a paused animation of two observers, one inside a moving car, and one outside. Light inside the car is moving up and down.Screenshot of a Maple Learn document, showing a paused animation of two observers, one inside a moving bus and one outside. There is light moving back and forth inside the bus.

So now, armed with this collection of documents, we’ll all be ready for the next time the aliens come down to Earth—ready to calculate the relative speed of their UFOs from the perspectives of various observers. That’ll show ‘em!

The pendulum and the cantilever share simple-looking ordinary differential equations (ODEs), but they are challenging to solve:

This post derives solutions from Lawden and Bisshopp by Maple commands, which (to the best of my knowledge) have not been published providing not only results but also the accompanying computer algebra techniques. A tabulated format has been chosen to better highlight similarities and differences.

Both solutions have in common that in a first step, the unknown function is integrated and then in a second step the inverse of the unknown function (i.e., the independent variable) is integrated. Only in combination with a well-chosen set of initial/boundary conditions solutions are possible. This makes these two cases difficult to handle by generic integration methods.

Originally, I was not looking for this insight. I was more interested in an exact solution for nonlinear deformations to benchmark numerical simulation results.  Relatively quickly, I was able to achieve this with the help of this forum, but after that I was left with some nagging questions:

Why does Maple not provide a solution for the pendulum although one exists?

Why isn’t there an explicit solution for the cantilever when there is one for the pendulum?

Why is it so difficult to proof that elliptic expressions are equal?

Repeatedly, whenever there was time, I came back to these questions and got more and more a better understanding of the two problems and the overall context. It also required me to learn more of Maple, and I had to revisit fundamentals of functions, differential equations, and integration, which was entirely possible within Maples help system. Today, I am satisfied to the point that I think it is too much to expect Maple to provide a high-level general integration method for such problems.

I am also satisfied that I was able to combine all my findings scattered across many documents and Mapleprime questions into a single executable textbook-style document with hidden Maple code that:

Exclusively uses and manipulates equation expressions (no assignment operators := were required),

Avoids differentials that are often used in textbooks but (for good reasons) are not supported by Maple,

Exclusively applies high-level commands (i.e. no extraction of subexpression, manipulation
           and subsequent re-assembly of expressions was needed).

The solutions for the pendulum and the cantilever are substantially different although the ODEs and essential derivation steps are similar. I think that an explicit solution for the cantilever, as it exists for the pendulum, is impossible (using elliptic integrals and functions). I leave it open to comments: whether this is correct and whether it is attributable to the set of initial and boundary conditions, the different symmetry of the sine and cosine functions in the ODEs, or both. I hope that the tabular presentation will provide an easy overview, allowing to form an own opinion.

If you are patient enough to work through the table, you will find a link between the cantilever and the pendulum that you are probably not expecting. 

Finally, I have to give credit to Bisshopp, who was probably the first to provide a solution for the cantilever. The clarity and compactness on only 3 pages and the way how the inverse of functions was determined before the age of computers makes this paper worth studying. Also, Lawden has to be mentioned, who did the same on 3 pages for the pendulum in a marvelous book on elliptic functions and applications. It happens that he is overlooked in more recent publications and it’s unclear to me if he was the first who published an explicit solution. His book might be one of the last of its kind in this age of computers, and for that reason alone, it is worth enjoying as he enjoyed writing it.

 

The Pendulum and the Cantilever Side by Side

C_R, Summer 2023

• 

To better compare the pendulum and the cantilever, the symbol `&varphi;` was chosen for the angle of the pendulum for the simple reason that this comparison started with bending theory, where `&varphi;` is often used to denote a deflection angle.

• 

Leibniz and Newton notation was not used to make functional dependencies of variables visible. Instead functional notation  `&varphi;` = `&varphi;`(t) and `&varphi;` = `&varphi;`(s)is used.

• 

To create an executable document that maintains a clear representation, it is necessary to use functional notation for differential equations and remove functional notation for integration. To avoid using the same symbol for both the integration variable and the upper limit of integration, this document uses two ways to express when the upper limit of integration varies (i.e., depends on the dependent variable of the functions being searched, namely `&varphi;` = `&varphi;`(t) and `&varphi;` = `&varphi;`(s)). Both ways have their pros and cons.

 

Typesetting:-EnableTypesetRule({"EllipticE", "EllipticE2", "EllipticF", "EllipticK", "InverseJacobiAM", "InverseJacobiSN", "JacobiSN"})

 

©_®

Pendulum

Cantilever

Independent variable

Time t

Arclength s

Dependent variable

Angle of the pendulum with respect to direction of gravity `&varphi;`(t)

varphi(t)

(1)

The slope of the cantilever with respect to the unbend state `&varphi;`(s)

varphi(s)

(2)

Parameters

• 

Length l

• 

Gravitational constant g 

• 

Length L

• 

A force F at the free end

• 

A bending moment M at the free end

• 

The bending stiffness EI 

ODE

diff(varphi(t), t, t)+C*sin(varphi(t)) = 0

diff(diff(varphi(t), t), t)+C*sin(varphi(t)) = 0

(3)

(for derivation see for example Wikipedia [1] or Lawden [2])

diff(varphi(s), s, s)+C*cos(varphi(s)) = 0

diff(diff(varphi(s), s), s)+C*cos(varphi(s)) = 0

(4)

(for derivation see for example Bisshopp [3] or Beléndez [4])

Definitions

`&varphi;`(0) = 0, `&varphi;`((1/4)*T) = `&varphi;__0`

varphi(0) = 0, varphi((1/4)*T) = varphi__0

(5)

`&varphi;`(t)is periodic with the oscillation period T (i.e., the movement is bounded):

0 < abs(`&varphi;__0`) and abs(`&varphi;__0`) < Pi

0 < abs(varphi__0) and abs(varphi__0) < Pi

(6)

`&varphi;`(L) = `&varphi;__0`, Eval(diff(varphi(s), s), s = L) = 1/rho

varphi(L) = varphi__0, Eval(diff(varphi(s), s), s = L) = 1/rho

(7)

For a downward force:

0 < `&varphi;__0` and `&varphi;__0` < (1/2)*Pi

0 < varphi__0 and varphi__0 < (1/2)*Pi

(8)

Parameter C

"C=omega^(2)", where omegais the angular frequency of the pendulum for small anglular excursions

C = g/l

C = g/l

(9)

"Specific" Load

C = F/EI

C = F/EI

(10)

Initial/

boundary
conditions

Eval(varphi(t), t = -(1/4)*T) = -`&varphi;__0`, Eval(diff(varphi(t), t), t = -(1/4)*T) = 0

Eval(varphi(t), t = -(1/4)*T) = -varphi__0, Eval(diff(varphi(t), t), t = -(1/4)*T) = 0

(11)

Eval(varphi(s), s = 0) = 0, varphi(L) = varphi__0, Eval(diff(varphi(s), s), s = L) = 1/rho

Eval(varphi(s), s = 0) = 0, varphi(L) = varphi__0, Eval(diff(varphi(s), s), s = L) = 1/rho

(12)

Only the second condition is essential.
Additional essential condition: The length L of the cantilever beam is constant (not a boundary condition in its common sense but essential for the solution).

#1 integration step
with the second condition

Method: Integration with an integration factor (and converting to D notation, not shown)

DEtools:-intfactor(diff(diff(varphi(t), t), t)+C*sin(varphi(t)) = 0, `&varphi;`(t)); DETools:-firint((diff(diff(varphi(t), t), t)+C*sin(varphi(t)) = 0)*%, `&varphi;`(t))

-2*C*cos(varphi(t))+(diff(varphi(t), t))^2+c__1 = 0

(13)

Substituting initial conditions (11)

eval(convert(-2*C*cos(varphi(t))+(diff(varphi(t), t))^2+c__1 = 0, D), t = -(1/4)*T); convert(value({Eval(diff(varphi(t), t), t = -(1/4)*T) = 0, Eval(varphi(t), t = -(1/4)*T) = -varphi__0}), D); isolate(eval(`%%`, %), c__1)

c__1 = 2*C*cos(varphi__0)

(14)

and isolating diff(`&varphi;`(t), t)in (13) yields

convert(isolate(subs(c__1 = 2*C*cos(varphi__0), -2*C*cos(varphi(t))+(diff(varphi(t), t))^2+c__1 = 0), diff(`&varphi;`(t), t)), radical)

diff(varphi(t), t) = (2*C*cos(varphi(t))-2*C*cos(varphi__0))^(1/2)

(15)

Alternative method [5]: Integration to an implicit representation

dsolve({(Eval(varphi(s), s = 0) = 0, varphi(L) = varphi__0, Eval(diff(varphi(s), s), s = L) = 1/rho)[3], diff(diff(varphi(s), s), s)+C*cos(varphi(s)) = 0}, `&varphi;`(s), implicit)[1]

Int(1/(-2*C*sin(_a)+(2*C*sin(varphi(L))*rho^2+1)/rho^2)^(1/2), _a = 0 .. varphi(s))-s-c__2 = 0

(16)

and differentiation w.r.t. to s 

diff(Int(1/(-2*C*sin(_a)+(2*C*sin(varphi(L))*rho^2+1)/rho^2)^(1/2), _a = 0 .. varphi(s))-s-c__2 = 0, s); isolate(%, diff(`&varphi;`(s), s)); expand(subs(varphi(L) = varphi__0, Eval(diff(varphi(s), s), s = L) = 1/rho, %))

diff(varphi(s), s) = (-2*C*sin(varphi(s))+2*C*sin(varphi__0)+1/rho^2)^(1/2)

(17)

(This method works only if rho <> infinity; i.e., only with curvature/bending moment at the free end.)

Continuing now without bending momenteval(diff(varphi(s), s) = (-2*C*sin(varphi(s))+2*C*sin(varphi__0)+1/rho^2)^(1/2), rho = infinity)

diff(varphi(s), s) = (-2*C*sin(varphi(s))+2*C*sin(varphi__0))^(1/2)

(18)

#2 integrating the inverse function

With the chain rule

(diff(`&varphi;`(t), t))*(diff(t(`&varphi;`), `&varphi;`)) = 1

(diff(varphi(t), t))*(diff(t(varphi), varphi)) = 1

(19)

isolate((diff(varphi(t), t))*(diff(t(varphi), varphi)) = 1, diff(`&varphi;`(t), t))

diff(varphi(t), t) = 1/(diff(t(varphi), varphi))

(20)

(15) can be written as

isolate((diff(varphi(t), t) = 1/(diff(t(varphi), varphi)))*(1/(diff(varphi(t), t) = (2*C*cos(varphi(t))-2*C*cos(varphi__0))^(1/2))), diff(t(`&varphi;`), `&varphi;`)); subs(`&varphi;`(t) = `&varphi;`, %)

diff(t(varphi), varphi) = 1/(2*C*cos(varphi)-2*C*cos(varphi__0))^(1/2)

(21)

where the functional notation `&varphi;`(t)was removed for integration

map(Int, diff(t(varphi), varphi) = 1/(2*C*cos(varphi)-2*C*cos(varphi__0))^(1/2), `&varphi;` = 0 .. `&varphi;`(t), continuous); (`@`(value, lhs) = rhs)(%)

-t(0)+t(varphi(t)) = Int(1/(2*C*cos(varphi)-2*C*cos(varphi__0))^(1/2), varphi = 0 .. varphi(t), continuous)

(22)

This relation allows to determine the time t = t(`&varphi;`(t)) the pendulum takes to swing to a certain angle `&varphi;` = `&varphi;`*t__.

Similarly, with

(diff(`&varphi;`(s), s))*(diff(s(`&varphi;`), `&varphi;`)) = 1

(diff(varphi(s), s))*(diff(s(varphi), varphi)) = 1

(23)

isolate((diff(varphi(s), s))*(diff(s(varphi), varphi)) = 1, diff(`&varphi;`(s), s))

diff(varphi(s), s) = 1/(diff(s(varphi), varphi))

(24)

(16) can be written as

isolate((diff(varphi(s), s) = 1/(diff(s(varphi), varphi)))*(1/(diff(varphi(s), s) = (-2*C*sin(varphi(s))+2*C*sin(varphi__0))^(1/2))), diff(s(`&varphi;`), `&varphi;`))

diff(s(varphi), varphi) = 1/(-2*C*sin(varphi(s))+2*C*sin(varphi__0))^(1/2)

(25)

and integrated over s

map(int, diff(s(varphi), varphi) = 1/(-2*C*sin(varphi(s))+2*C*sin(varphi__0))^(1/2), `&varphi;` = 0 .. `&varphi;__s`, continuous)

-s(0)+s(varphi__s) = int(1/(-2*C*sin(varphi(s))+2*C*sin(varphi__0))^(1/2), varphi = 0 .. varphi__s, continuous)

(26)

This expression relates the arclength s to the slope `&varphi;__s` at the location s (i.e., `&varphi;`(s) = `&varphi;__s`). It is the inverse of what we intend to solve (i.e., `&varphi;` = `&varphi;`(s)) but it is essential to derive a solution. Unlike for the pendulum, an indexed symbol `#msub(mi("&varphi;",fontstyle = "normal"),mi("s"))`has been chosen to avoid formally correct but uncommon expressions like "t(`&varphi;`(t))."

Special cases

Oscillation period T

When `&varphi;`*t__ = `&varphi;__0` the pendulum has swung a quater of the period. With
t(0) = 0, `&varphi;`(t) = `&varphi;__0`, t(`&varphi;__0`) = (1/4)*T

t(0) = 0, varphi(t) = varphi__0, t(varphi__0) = (1/4)*T

(27)

(22) becomes

subs(t(0) = 0, varphi(t) = varphi__0, t(varphi__0) = (1/4)*T, -t(0)+t(varphi(t)) = Int(1/(2*C*cos(varphi)-2*C*cos(varphi__0))^(1/2), varphi = 0 .. varphi(t), continuous))

(1/4)*T = Int(1/(2*C*cos(varphi)-2*C*cos(varphi__0))^(1/2), varphi = 0 .. varphi__0, continuous)

(28)

where `&varphi;`(t)was replaced in the integrant by `&varphi;` to create input that can be processes by the int command. After evaluation

isolate(`assuming`([simplify(value((1/4)*T = Int(1/(2*C*cos(varphi)-2*C*cos(varphi__0))^(1/2), varphi = 0 .. varphi__0, continuous)))], [C > 0, 0 < `&varphi;__0` and `&varphi;__0` < Pi]), T)

T = 4*2^(1/2)*InverseJacobiAM((1/2)*varphi__0, 2^(1/2)/(1-cos(varphi__0))^(1/2))/(C^(1/2)*(1-cos(varphi__0))^(1/2))

(29)

where 1/am = InverseJacobiAM denotes the inverse Jacobian amplitude function.

Expression for `&varphi;__0`
Calculation of the length L in order to get an expression to determine the unknown slope `&varphi;__0` at the free end of the cantilever as a function of the load parameter C. With
s(0) = 0, s(`&varphi;__s`) = L, `&varphi;__s` = `&varphi;__0`, `&varphi;`(s) = `&varphi;`

s(0) = 0, s(varphi__s) = L, varphi__s = varphi__0, varphi(s) = varphi

(30)

(26) becomessubs(s(0) = 0, s(varphi__s) = L, varphi__s = varphi__0, varphi(s) = varphi, -s(0)+s(varphi__s) = int(1/(-2*C*sin(varphi(s))+2*C*sin(varphi__0))^(1/2), varphi = 0 .. varphi__s, continuous))

L = int(1/(-2*C*sin(varphi)+2*C*sin(varphi__0))^(1/2), varphi = 0 .. varphi__0, continuous)

(31)

where `&varphi;`(s)is replaced by `&varphi;` to create input that can be processes by the int command. After evaluation

`assuming`([value(L = int(1/(-2*C*sin(varphi)+2*C*sin(varphi__0))^(1/2), varphi = 0 .. varphi__0, continuous))], [C > 0, 0 < `&varphi;__0` and `&varphi;__0` < (1/2)*Pi])

L = EllipticK((1/2)*(2*sin(varphi__0)+2)^(1/2))/C^(1/2)-EllipticF(1/(sin(varphi__0)+1)^(1/2), (1/2)*(2*sin(varphi__0)+2)^(1/2))/C^(1/2)

(32)

where K = EllipticK and F = EllipticF denote the complete and incomplete elliptic integrals of the first kind.

#3 Solutions of particular interest  

Explicit solution for `&varphi;`(t) in bounded motion.
Rearranging (22)

`assuming`([simplify(2*C*(-t(0)+t(varphi(t)) = Int(1/(2*C*cos(varphi)-2*C*cos(varphi__0))^(1/2), varphi = 0 .. varphi(t), continuous)))], [C > 0])/(sqrt(C)*sqrt(2))

-C^(1/2)*2^(1/2)*(t(0)-t(varphi(t))) = Int(1/(cos(varphi)-cos(varphi__0))^(1/2), varphi = 0 .. varphi(t), continuous)

(33)

and substituting this essential identity (expression 5.1.3 from Lawden [2])

-cos(`&varphi;__0`)+cos(`&varphi;`) = 2*sin((1/2)*`&varphi;__0`)^2-2*sin((1/2)*`&varphi;`)^2

cos(varphi)-cos(varphi__0) = 2*sin((1/2)*varphi__0)^2-2*sin((1/2)*varphi)^2

(34)

yields

subs(cos(varphi)-cos(varphi__0) = 2*sin((1/2)*varphi__0)^2-2*sin((1/2)*varphi)^2, -C^(1/2)*2^(1/2)*(t(0)-t(varphi(t))) = Int(1/(cos(varphi)-cos(varphi__0))^(1/2), varphi = 0 .. varphi(t), continuous))

-C^(1/2)*2^(1/2)*(t(0)-t(varphi(t))) = Int(1/(2*sin((1/2)*varphi__0)^2-2*sin((1/2)*varphi)^2)^(1/2), varphi = 0 .. varphi(t), continuous)

(35)

`assuming`([value(-C^(1/2)*2^(1/2)*(t(0)-t(varphi(t))) = Int(1/(2*sin((1/2)*varphi__0)^2-2*sin((1/2)*varphi)^2)^(1/2), varphi = 0 .. varphi(t), continuous))], [0 < `&varphi;__0` and `&varphi;__0` < Pi])

-C^(1/2)*2^(1/2)*(t(0)-t(varphi(t))) = 2^(1/2)*InverseJacobiAM((1/2)*varphi(t), csc((1/2)*varphi__0))/sin((1/2)*varphi__0)

(36)

which simplifies further with

t(0) = 0, t(`&varphi;`(t)) = t

t(0) = 0, t(varphi(t)) = t

(37)

to

subs(t(0) = 0, t(varphi(t)) = t, ((-C^(1/2)*2^(1/2)*(t(0)-t(varphi(t))) = 2^(1/2)*InverseJacobiAM((1/2)*varphi(t), csc((1/2)*varphi__0))/sin((1/2)*varphi__0))*(1/sqrt(2)))*sin((1/2)*`&varphi;__0`))

sin((1/2)*varphi__0)*C^(1/2)*t = InverseJacobiAM((1/2)*varphi(t), csc((1/2)*varphi__0))

(38)

Mapping now sn = JacobiSN with the modulus csc((1/2)*`&varphi;__0`) to the expression above (Maple converts InverseJacobiAM to InverseJacobiSN and simplifies automatically)

map(JacobiSN, sin((1/2)*varphi__0)*C^(1/2)*t = InverseJacobiAM((1/2)*varphi(t), csc((1/2)*varphi__0)), csc((1/2)*`&varphi;__0`))

JacobiSN(sin((1/2)*varphi__0)*C^(1/2)*t, csc((1/2)*varphi__0)) = sin((1/2)*varphi(t))

(39)

the angle `&varphi;`as a function of time is obtained explicitly

convert(isolate(JacobiSN(sin((1/2)*varphi__0)*C^(1/2)*t, csc((1/2)*varphi__0)) = sin((1/2)*varphi(t)), `&varphi;`(t)), sincos)

varphi(t) = 2*arcsin(JacobiSN(sin((1/2)*varphi__0)*C^(1/2)*t, 1/sin((1/2)*varphi__0)))

(40)

Bending curve of the cantilever for a given load (i.e., for a given `&varphi;__0`, obtainable from (32)).

To obtain a parametric form x(p), y(p)of the bending curve, the following two ODEs have to be integrated

diff(x(s), s) = cos(`&varphi;`(s)), diff(y(s), s) = sin(`&varphi;`(s))

diff(x(s), s) = cos(varphi(s)), diff(y(s), s) = sin(varphi(s))

(41)

Applying

diff(x(s), s) = (diff(x(`&varphi;`), `&varphi;`))*(diff(`&varphi;`(s), s)), diff(y(s), s) = (diff(y(`&varphi;`), `&varphi;`))*(diff(`&varphi;`(s), s))

diff(x(s), s) = (diff(x(varphi), varphi))*(diff(varphi(s), s)), diff(y(s), s) = (diff(y(varphi), varphi))*(diff(varphi(s), s))

(42)

in the following way to (41)

subs({diff(x(s), s) = (diff(x(varphi), varphi))*(diff(varphi(s), s)), diff(y(s), s) = (diff(y(varphi), varphi))*(diff(varphi(s), s))}, diff(varphi(s), s) = 1/(diff(s(varphi), varphi)), {diff(x(s), s) = cos(varphi(s)), diff(y(s), s) = sin(varphi(s))})[]

(diff(x(varphi), varphi))/(diff(s(varphi), varphi)) = cos(varphi(s)), (diff(y(varphi), varphi))/(diff(s(varphi), varphi)) = sin(varphi(s))

(43)

isolate(((diff(x(varphi), varphi))/(diff(s(varphi), varphi)) = cos(varphi(s)), (diff(y(varphi), varphi))/(diff(s(varphi), varphi)) = sin(varphi(s)))[1], diff(x(`&varphi;`), `&varphi;`)), isolate(((diff(x(varphi), varphi))/(diff(s(varphi), varphi)) = cos(varphi(s)), (diff(y(varphi), varphi))/(diff(s(varphi), varphi)) = sin(varphi(s)))[2], diff(y(`&varphi;`), `&varphi;`))

diff(x(varphi), varphi) = cos(varphi(s))*(diff(s(varphi), varphi)), diff(y(varphi), varphi) = sin(varphi(s))*(diff(s(varphi), varphi))

(44)

yields two ODEs where x and y depend on the variable `&varphi;`(s). Substituting (25) and `&varphi;`(s) = `&varphi;` 

subs(diff(s(varphi), varphi) = 1/(-2*C*sin(varphi(s))+2*C*sin(varphi__0))^(1/2), `&varphi;`(s) = `&varphi;`, [diff(x(varphi), varphi) = cos(varphi(s))*(diff(s(varphi), varphi)), diff(y(varphi), varphi) = sin(varphi(s))*(diff(s(varphi), varphi))])[]

diff(x(varphi), varphi) = cos(varphi)/(-2*C*sin(varphi)+2*C*sin(varphi__0))^(1/2), diff(y(varphi), varphi) = sin(varphi)/(-2*C*sin(varphi)+2*C*sin(varphi__0))^(1/2)

(45)

map(Int, (diff(x(varphi), varphi) = cos(varphi)/(-2*C*sin(varphi)+2*C*sin(varphi__0))^(1/2), diff(y(varphi), varphi) = sin(varphi)/(-2*C*sin(varphi)+2*C*sin(varphi__0))^(1/2))[1], `&varphi;` = 0 .. `&varphi;__s`, continuous), map(Int, (diff(x(varphi), varphi) = cos(varphi)/(-2*C*sin(varphi)+2*C*sin(varphi__0))^(1/2), diff(y(varphi), varphi) = sin(varphi)/(-2*C*sin(varphi)+2*C*sin(varphi__0))^(1/2))[2], `&varphi;` = 0 .. `&varphi;__s`, continuous)

Int(diff(x(varphi), varphi), varphi = 0 .. varphi__s, continuous) = Int(cos(varphi)/(-2*C*sin(varphi)+2*C*sin(varphi__0))^(1/2), varphi = 0 .. varphi__s, continuous), Int(diff(y(varphi), varphi), varphi = 0 .. varphi__s, continuous) = Int(sin(varphi)/(-2*C*sin(varphi)+2*C*sin(varphi__0))^(1/2), varphi = 0 .. varphi__s, continuous)

(46)

results in a parametric solution with parameter `#mrow(mi("p"),mo("&equals;"),mi("\`&varphi;__s\`"))` where "0<= `&varphi;__s`<=`&varphi;__0`." 
For the x coordinate

subs(`assuming`([x(0) = 0, simplify(value((Int(diff(x(varphi), varphi), varphi = 0 .. varphi__s, continuous) = Int(cos(varphi)/(-2*C*sin(varphi)+2*C*sin(varphi__0))^(1/2), varphi = 0 .. varphi__s, continuous), Int(diff(y(varphi), varphi), varphi = 0 .. varphi__s, continuous) = Int(sin(varphi)/(-2*C*sin(varphi)+2*C*sin(varphi__0))^(1/2), varphi = 0 .. varphi__s, continuous))[1]))], [0 < `&varphi;` and `&varphi;` < (1/2)*Pi, C > 0]))

x(varphi__s) = (2^(1/2)*sin(varphi__0)^(1/2)-(-2*sin(varphi__s)+2*sin(varphi__0))^(1/2))/C^(1/2)

(47)

For the y coordinate a long expression with the following structure

subs(`assuming`([y(0) = 0, simplify(value((Int(diff(x(varphi), varphi), varphi = 0 .. varphi__s, continuous) = Int(cos(varphi)/(-2*C*sin(varphi)+2*C*sin(varphi__0))^(1/2), varphi = 0 .. varphi__s, continuous), Int(diff(y(varphi), varphi), varphi = 0 .. varphi__s, continuous) = Int(sin(varphi)/(-2*C*sin(varphi)+2*C*sin(varphi__0))^(1/2), varphi = 0 .. varphi__s, continuous))[2]), radical)], [0 < `&varphi;` and `&varphi;` < (1/2)*Pi, 0 < `&varphi;__0` and `&varphi;__0` < (1/2)*Pi, 0 < `&varphi;__s` and `&varphi;__s` < (1/2)*Pi, C > 0])); y(`&varphi;__s`) = (A__1*EllipticE(z__1, k)+A__2*EllipticF(z__1, k)+A__3*EllipticE(z__2, k)+A__4*EllipticF(z__2, k))/sqrt(C)

y(varphi__s) = (A__1*EllipticE(z__1, k)+A__2*EllipticF(z__1, k)+A__3*EllipticE(z__2, k)+A__4*EllipticF(z__2, k))/C^(1/2)

(48)

is obtained where A__i = A__i(`&varphi;__s`, `&varphi;__0`), z__i = z__i(`&varphi;__s`, `&varphi;__0`) and k = k(`&varphi;__0`).

NULL

Not required in the above: To derive an explicit solution, Lawden performed a change of variable of this kind

sin((1/2)*`&varphi;`) = sin((1/2)*`&varphi;__0`)*sin(u)

sin((1/2)*varphi) = sin((1/2)*varphi__0)*sin(u)

(49)

 

which is not needed with Maple commands.

 

Furthermore: Formally, it would have been nice to have the pendulum start its movement at t=0 at an angle -`&varphi;__0`. However, this leads to an output in (36) with two inverse elliptic functions where `&varphi;`(t) is difficult if not impossible to isolate.

 

Solution for the free end of the cantilever (i.e., `&varphi;__s` = `&varphi;__0`and s = L)

subs(x(0) = 0, `&varphi;__s` = `&varphi;__0`, x(`&varphi;__0`) = x(L), x(varphi__s) = (2^(1/2)*sin(varphi__0)^(1/2)-(-2*sin(varphi__s)+2*sin(varphi__0))^(1/2))/C^(1/2))

x(L) = 2^(1/2)*sin(varphi__0)^(1/2)/C^(1/2)

(50)

subs(`assuming`([y(0) = 0, y(`&varphi;__0`) = y(L), simplify(value(subs(`&varphi;__s` = `&varphi;__0`, (Int(diff(x(varphi), varphi), varphi = 0 .. varphi__s, continuous) = Int(cos(varphi)/(-2*C*sin(varphi)+2*C*sin(varphi__0))^(1/2), varphi = 0 .. varphi__s, continuous), Int(diff(y(varphi), varphi), varphi = 0 .. varphi__s, continuous) = Int(sin(varphi)/(-2*C*sin(varphi)+2*C*sin(varphi__0))^(1/2), varphi = 0 .. varphi__s, continuous))[2])))], [0 < `&varphi;` and `&varphi;` < (1/2)*Pi, 0 < `&varphi;__0` and `&varphi;__0` < (1/2)*Pi, 0 < `&varphi;__s` and `&varphi;__s` < (1/2)*Pi, C > 0]))

y(L) = (EllipticK((1/2)*(2*sin(varphi__0)+2)^(1/2))-EllipticF(1/(sin(varphi__0)+1)^(1/2), (1/2)*2^(1/2)*(sin(varphi__0)+1)^(1/2))-2*EllipticE((1/2)*2^(1/2)*(sin(varphi__0)+1)^(1/2))+2*EllipticE(1/(sin(varphi__0)+1)^(1/2), (1/2)*2^(1/2)*(sin(varphi__0)+1)^(1/2)))/C^(1/2)

(51)

``

Some remarks

NULL

NULL

 

Comparison to the solution from Lawden

Expression (39) compared to sin((1/2)*theta(t)) = sin((1/2)*alpha)*sn(t+K, sin((1/2)*alpha))

sin((1/2)*theta(t)) = sin((1/2)*alpha)*sn(t+K, sin((1/2)*alpha))

(52)

(pasted from DLMF 22.19.2 for the case C = 1). The same with adapted variables  subs(alpha = `&varphi;__0`, sn = JacobiSN, K = 0, theta(t) = `&varphi;__t`, sin((1/2)*theta(t)) = sin((1/2)*alpha)*sn(t+K, sin((1/2)*alpha)))

sin((1/2)*varphi__t) = sin((1/2)*varphi__0)*JacobiSN(t, sin((1/2)*varphi__0))

(53)

Now equating the left hand side of (39) to JacobiSN(z, k)and identifying the parameters z and k

subs(C = 1, lhs(JacobiSN(sin((1/2)*varphi__0)*C^(1/2)*t, csc((1/2)*varphi__0)) = sin((1/2)*varphi(t)))) = JacobiSN(z, k)

JacobiSN(sin((1/2)*varphi__0)*t, csc((1/2)*varphi__0)) = JacobiSN(z, k)

(54)

map(op, JacobiSN(sin((1/2)*varphi__0)*t, csc((1/2)*varphi__0)) = JacobiSN(z, k)); solve([(rhs-lhs)(%)], {k, z})[]

k = csc((1/2)*varphi__0), z = sin((1/2)*varphi__0)*t

(55)

Using the following identity from Maple's FunctionAdvisor and the correspondence in (55)

FunctionAdvisor(identities, JacobiSN(z, 1/k))[5]

JacobiSN(z, k) = JacobiSN(z*k, 1/k)/k

(56)

yields

convert(subs(k = csc((1/2)*varphi__0), z = sin((1/2)*varphi__0)*t, JacobiSN(z, k) = JacobiSN(z*k, 1/k)/k), sincos)

JacobiSN(sin((1/2)*varphi__0)*t, 1/sin((1/2)*varphi__0)) = sin((1/2)*varphi__0)*JacobiSN(t, sin((1/2)*varphi__0))

(57)

Comparing this with (53) confirms that (40) is correct.

Equivalent expressions to determine `&varphi;__0` 

As returned by value:

normal(L = EllipticK((1/2)*(2*sin(varphi__0)+2)^(1/2))/C^(1/2)-EllipticF(1/(sin(varphi__0)+1)^(1/2), (1/2)*(2*sin(varphi__0)+2)^(1/2))/C^(1/2))

L = (EllipticK((1/2)*(2*sin(varphi__0)+2)^(1/2))-EllipticF(1/(sin(varphi__0)+1)^(1/2), (1/2)*(2*sin(varphi__0)+2)^(1/2)))/C^(1/2)

(58)

simplify instead of value:

convert(`assuming`([simplify(L = int(1/(-2*C*sin(varphi)+2*C*sin(varphi__0))^(1/2), varphi = 0 .. varphi__0, continuous))], [0 < `&varphi;__0` and `&varphi;__0` < (1/2)*Pi]), sincos)

L = -I*2^(1/2)*EllipticF(I*sin(varphi__0)^(1/2)/(1-sin(varphi__0))^(1/2), I*(1-sin(varphi__0))^(1/2)/(sin(varphi__0)+1)^(1/2))/(C^(1/2)*(sin(varphi__0)+1)^(1/2))

(59)

With integration tools and change of variables using x = sin(`&varphi;`):

Int(1/sqrt(-2*C*sin(`&varphi;`)+2*C*sin(`&varphi;__0`)), `&varphi;` = 0 .. `&varphi;__0`); L = IntegrationTools:-Change(%, x = sin(`&varphi;`), x); simplify(subs(isolate(x__0 = sin(`&varphi;__0`), `&varphi;__0`), %)); subs(x__0 = sin(`&varphi;__0`), `assuming`([value(%)], [0 < x and x < 1, 0 < x__0 and x__0 < 1]))

L = 2^(1/2)*EllipticF(sin(varphi__0)^(1/2)/(sin(varphi__0)+1)^(1/2), I*(-sin(varphi__0)^2+1)^(1/2)/(-1+sin(varphi__0)))/(C^(1/2)*(1-sin(varphi__0))^(1/2))

(60)

Without having a Maple way: Christian Wolinski has provided 3 additional expressions where one is of particular simplicity [6]

  L = EllipticF(sqrt(1-1/(sin(`&varphi;__0`)+1))*sqrt(2), (1/2)*sqrt(2)*sqrt(sin(`&varphi;__0`)+1))/sqrt(C)

L = EllipticF((1-1/(sin(varphi__0)+1))^(1/2)*2^(1/2), (1/2)*2^(1/2)*(sin(varphi__0)+1)^(1/2))/C^(1/2)

(61)

 

 

Useful identities

FunctionAdvisor(identities, InverseJacobiSN)[3]

InverseJacobiSN(z, k) = InverseJacobiAM(arcsin(z), k)

(62)

InverseJacobiSN(z, k) = InverseJacobiAM(arcsin(z), k)

InverseJacobiSN(z, k) = EllipticF(z, k)

(63)

FunctionAdvisor(identities, InverseJacobiSN)[1]

JacobiSN(InverseJacobiSN(z, k), k) = z

(64)

FunctionAdvisor(identities, JacobiSN)[5]

JacobiSN(z, 1/k) = k*JacobiSN(z/k, k)

(65)

 

Explicit solution for `&varphi;__0`

Since Maple's solve does not solve (32) for `&varphi;__0`, one could try to isolate `&varphi;__0` in expression (32) by combining "somehow" the two elliptic expression into a single expression and to apply an inverse operation to it.

Maple's simplify or combine do not seem to be able to help in this respect. There might be addition theorems that could be applied but identifying them in Maple or in DLMF requires expertise in this field of special functions.

Easier is to try to evaluate (31) in different ways (see above) and hope for success. This yielded equivalent expressions with only one elliptic integral EllipticF.

Using identities such elliptic integrals can be converted to inverse elliptic functions where elliptic functions can be applied to. Trying this exemplarily for (61)  

 

L/sqrt(C) = InverseJacobiSN(sqrt(1-1/(sin(`&varphi;__0`)+1))*sqrt(2), (1/2)*sqrt(2)*sqrt(sin(`&varphi;__0`)+1))

L/C^(1/2) = InverseJacobiSN((1-1/(sin(varphi__0)+1))^(1/2)*2^(1/2), (1/2)*2^(1/2)*(sin(varphi__0)+1)^(1/2))

(66)

map(JacobiSN, L/C^(1/2) = InverseJacobiSN((1-1/(sin(varphi__0)+1))^(1/2)*2^(1/2), (1/2)*2^(1/2)*(sin(varphi__0)+1)^(1/2)), (1/2)*sqrt(2)*sqrt(sin(`&varphi;__0`)+1))

JacobiSN(L/C^(1/2), (1/2)*2^(1/2)*(sin(varphi__0)+1)^(1/2)) = (1-1/(sin(varphi__0)+1))^(1/2)*2^(1/2)

(67)

does not isolate `&varphi;__0`in the same way `&varphi;`(t)could be separated for the pendulum. The reason why such attempts are deemed to fail is simple. Differently to the pendulum, `#msub(mi("&varphi;",fontstyle = "normal"),mi("0"))` is not a constant "input" to the system but, causally speaking, an "output". While C in the case of the pendulum is constant and independent of `&varphi;__0`, C and `&varphi;__0`functionally depend on each other for the cantilever; i.e., `&varphi;__0` = `&varphi;__0`(C). This fundamentally makes the two cases different although many derivation steps are similar.

Student:-ODEs

Student:-ODEs:-ODESteps(diff(diff(varphi(t), t), t)+C*sin(varphi(t)) = 0)

Error, (in Student:-ODEs:-OdeSolveOrder2) ODE is not supported

 

Student:-ODEs:-ODESteps(diff(diff(varphi(s), s), s)+C*cos(varphi(s)) = 0)gives a solution for `&varphi;` = `&varphi;`(s)with two integration constants C1 and C2, but determining the integration constants with the first two boundary conditions of (12) is probably impossible. odetest confirms that the solution is correct but one of the arguments of an elliptic function is not unitfree (which raises more questions): JacobiSN((1/2)*sqrt(2*C-2*C1)*(-s+C2), sqrt(-(C1+C)/(-C1+C))).

Links

https://www.mapleprimes.com/questions/232863-Testing-Maples-Solution-Of-The-Nonlinear

https://www.mapleprimes.com/questions/131520-Animation-Of-A-Simple-Pendulum

 

Applying a horizontal load instead of a vertical bends the cantilever in an arc-like fashion. For this load case the corresponding ODE is that of the pendulum (see [2], chapter 5, exercise 8). The parametric equation (bending curve) of the arc becomes simpler but still no explicit solution `&varphi;__0` = `&varphi;__0`(C)seems possible.

Alternative symbols for 4

θ

phi, theta

References

[1] https://en.wikipedia.org/wiki/Pendulum_(mechanics)
[2] Lawden, Derek F. “Elliptic Functions and Applications.” Acta Applicandae Mathematica 24 (1989): 201-202.
[3] Bisshopp, K.E. and Drucker, D.C. (1945) Large Deflection of Cantilever Beams. Quarterly of Applied Mathematics, 3, 272-275.
[4] BELÉNDEZ, Tarsicio; NEIPP, Cristian; BELÉNDEZ, Augusto. "Large and small deflections of a cantilever beam". European Journal of Physics. Vol. 23, No. 3 (May 2002). ISSN 0143-0807, pp. 371-379
[5] Rouben Rostamian, https://www.mapleprimes.com/questions/236511-How-To-Integrate-This-Ode-And-How-To#answer295192
[6] Christian Wolinski, https://www.mapleprimes.com/questions/233304-How-To-Find-The-Inverse-Function-Of#comment283638

 

 

Download Cantilever_and_pendulum_side_by_side_-_input_hidden.mw

aRandStep2D := proc(X0, Y0, dx, dy)
  local X, Y, P, R;
  P := Array(1 .. 2);
  R := rand(1 .. 8)();
  if R = 1 then X := X0 - dx; Y := Y0 + dy; end if;
  if R = 2 then X := X0; Y := Y0 + dy; end if;
  if R = 3 then X := X0 + dx; Y := Y0 + dy; end if;
  if R = 4 then X := X0 - dx; Y := Y0; end if;
  if R = 5 then X := X0 + dx; Y := Y0; end if;
  if R = 6 then X := X0 - dx; Y := Y0 - dy; end if;
  if R = 7 then X := X0; Y := Y0 - dy; end if;
  if R = 8 then X := X0 + dx; Y := Y0 - dy; end if;
  P[1] := X; P[2] := Y;
  return P;
end proc 

SetStart := proc(b)
  local alpha, R, P;
  P := Array(1 .. 2);
  alpha := rand(1 .. b)();
  P[1] := alpha*b;
  P[2] := alpha*b;
  return P;
end proc 

RandomFactTpq := proc(N, pb, dx, dy)
  local alpha, X, Y, f, P, counter, B, n, T;
  P := Array(1 .. 2);
  counter := 0; f := 1;
  B := floor(evalf(sqrt(N))); #Set maximal searching steps
  T := floor(evalf(sqrt(N))); #For SetStart's use
  P := SetStart(T);
  X := P[1]; Y := P[2];
  while f = 1 and counter < B do  #loop
    n := pb - X - Y;
    f := gcd(N, n);
    if f > 1then break; end if;
    P := aRandStep2D(X, Y, dx, dy); #A random move
    X := P[1]; Y := P[2];
    if X < 1 or Y < 1 or N - pb - 1 < X or X <= Y then
      P := SetStart(T);       # Restart when out of borders
      X := P[1]; Y := P[2];
    end if;
    counter := counter + 1;    #Counting the searched steps
  end do;
  if  f>1  then print(Find at point (X, Y), found divisor = f, searching steps = counter);
  else print(This*time*finds*no*result, test*again!); end if;
end proc


wxbRandWalkTpqNew4.pdf

Hi
It's been years since expressions like A %* B %+ C involving inert arithmetic operators used in infix form are correctly understood (parsed) when written on a 1D-Math input line. The idea is simple: have the operators %. %*, %+, %-, %^, %/ work on input as infix operators the same way their active forms: ., *, +, -, ^, / do. This useful functionality, however, remained elusive when using 2D-Math input notation, so one would have to resort to using the functional form of the operators. E.g., input the above as `%+`(`%*`(A, B) ,C), which for me is really ugly. Besides being a bit demoralizing: we do all this fuzz about how great computer algebra and 2D-Math input notation is, and then input things in that way …

So this is to mention that this elusive functionality of inert arithmetic operators used in infix form within a 2D-Math input line now works. The novelty is present in the latest Maplesoft Physics Updates for Maple 2023, which is version 1490. As usual, to install the Updates open a Maple worksheet and input Physics:-Version(latest).

Here is an image (worksheet at the end) showing the new thing


The implementation is pretty new; reports of anything related to these inert operators not displayed/working as you'd expect are much appreciated. 


Download Inert_arithmetic_operators_in_2D_Math.mw

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

Hi everyone,

I'd like to draw your attention to a package we recently uploaded to the Maple Cloud, here. You can download the package from the linked Cloud page, or directly from here as a workbook file: NaturalLanguage.maple. I'll include a lightly edited version of the "cover sheet" that introduces the package below. I have left the first four sections folded closed - you can see those in the linked Maple Cloud page or the workbook - because I think the last section is the most interesting.
 

Using natural language models in Maple


This package explores using large language models such as ChatGPT for processing natural language in Maple. Let's load the package from source, save it in this workbook, and load it.

restart

read "this:///lib/NaturalLanguage.mpl"

savelib('NaturalLanguage', "this://")

with(NaturalLanguage)

Warning, the NaturalLanguage package is an experimental package designed as an interface to publicly available large language models such as ChatGPT. Use of this package is entirely at the user's risk. Results may be inconsistent, misleading, or flat out incorrect. 

[Explain, GetCommand, GetMath, Query, RawQuery]

(1)

We note there is a warning: we will see output of large language models, which will often include inaccurate statements. Please keep this in mind - this is not (yet?) technology that you want to use to build a bridge!

We also note that there are five publically exposed commands. Let's look at them in turn.

RawQuery

   

Query

   

GetMath

   

GetCommand

   

Explain

 
• 

The Explain command asks the given model to explain the item in the query. After an explanation, you can ask for more detail by issuing the command Explain(more), Explain("go on"), Explain(elaborate), Explain(further), Explain("continue"), or minor variations. The item you ask about can be a mathematical expression, equation, list of equations, or a string.

• 

The default model is GPT-4; you can select the other model by using the model = ChatGPT option, or forcing use of the GPT-4 model by using the model = GPT4 option. Note that OpenAI may deprecate and disable models, so the set of models supported may change in the future.

display(Explain(x^2+y^2=1));display(Explain(x^2+y^2 = 1))

This equation represents a circle with a radius of 1 centered at the origin (0,0) on a coordinate plane.

 

display(Explain(more));display(Explain(more))

The equation x^2 + y^2 = 1 represents a circle in a two-dimensional plane (specifically, a Cartesian plane).

The center of the circle is at the origin of the coordinates (0,0) and its radius is 1. This is because any point (x, y) on the circle is a distance r (the radius) away from the center, and by the Pythagorean theorem, this distance is given by the square root of (x^2 + y^2). The fact that x^2 + y^2 = 1 implies that the radius r = sqrt(1) = 1.

This fundamental equation is also the unit circle in trigonometry, where angles are measured in radians. The coordinates (x, y) represent the cosine and sine of the angle respectively. The unit circle is a crucial concept in trigonometry, complex number theory, and calculus. It simplifies many mathematical concepts and provides a geometric interpretation of a variety of phenomena in physical sciences and engineering.

 

display(Explain(Re(exp(x*I + y*I)) = Re(exp(x*I)) * Re(exp(y*I)) - Im(exp(x*I)) * Im(exp(y*I))));display(Explain(Re(exp(I*x+I*y)) = Re(exp(I*x))*Re(exp(I*y))-Im(exp(I*x))*Im(exp(I*y))))

The equation seems to be using Euler's formula, which states that e^(ix) = cos(x) + i*sin(x). The 'Re' refers to the real part of a complex number, and 'Im' refers to the imaginary part. So, if you put Euler's formula in there your equation is equivalent to cos(x + y) = cos(x)cos(y) - sin(x)sin(y), which is basically the well-known trigonometric formula for the cosine of a sum. In a nutshell, it's dealing with the trigonometric identities and properties of exponential functions with complex numbers.

 

display(Explain("go on", model = ChatGPT));display(Explain("go on", model = ChatGPT))

The given equation appears to be dealing with the real part of a complex exponential expression. Let's break it down step by step to understand its meaning:

1. The equation involves the expression e^(i*x+i*y), where e represents Euler's number (~2.718), i is the imaginary unit (√(-1)), and x and y are real numbers.
2. The expression e^(i*x+i*y) can be rewritten using Euler's formula: e^(i*x+i*y) = cos(x+y) + i*sin(x+y). This is a complex number in the form a + bi, where 'a' represents the real part (Re) and 'b' represents the imaginary part (Im).
3. The real part of e^(i*x+i*y) is, therefore, Re(e^(i*x+i*y)) = cos(x+y).
4. The right side of the equation consists of the product of two complex exponential expressions: Re(e^(i*x))*Re(e^(i*y))-Im(e^(i*x))*Im(e^(i*y)).
5. The expression Re(e^(i*x)) represents the real part of e^(i*x), which is cos(x), and Im(e^(i*x)) represents the imaginary part of e^(i*x), which is sin(x).
6. Similarly, Re(e^(i*y)) represents the real part of e^(i*y), which is cos(y), and Im(e^(i*y)) represents the imaginary part of e^(i*y), which is sin(y).
7. Therefore, the right side becomes: (cos(x)*cos(y)) - (sin(x)*sin(y)).

So, the equation Re(e^(i*x+i*y)) = Re(e^(i*x))*Re(e^(i*y))-Im(e^(i*x))*Im(e^(i*y)) essentially states that the real part of e^(i*x+i*y) (i.e., cos(x+y)) is equal to the real part of e^(i*x) (i.e., cos(x)) multiplied by the real part of e^(i*y) (i.e., cos(y)), minus the imaginary part of e^(i*x) (i.e., sin(x)) multiplied by the imaginary part of e^(i*y) (i.e., sin(y)).

In simpler terms, the equation is connecting the real part of the sum of two complex exponential expressions with the product of the real parts and the difference of the imaginary parts of those individual expressions.

 

display(Explain(Re(sigma) < 0));display(Explain(Re(sigma) < 0))

The inequality "Re(sigma) < 0" is a statement in complex number theory. It is saying that the real part (Re) of a complex number sigma is less than zero. This means that the complex number is to the left of the imaginary axis in the complex number plane. This kind of inequality is often used in the study of complex functions, in areas such as control systems or signal processing in electrical engineering, and also in number theory and the study of dynamical systems in mathematics.

 

display(Explain(continue));display(Explain(continue))

The inequality "Re(sigma) < 0" is typically related to the field of complex numbers. Here, "Re" stands for the real part of a complex number and "sigma" is usually used to represent a complex number. So, "Re(sigma) < 0" is saying that the real part of the complex number sigma is negative.

Complex numbers are usually written in the form a + bi, where a and b are real numbers and i is the imaginary unit. The "Re" in "Re(sigma)" refers to the a part of this form, i.e. the real part of the complex number.

This inequality is often found in fields like control systems engineering or signal processing. For instance, in control systems, all the poles of a stable system must have a negative real part. Another context you might come across this inequality is in complex analysis, one might examine the region where the real part of a complex function is negative. So, the inequality "Re(sigma) < 0" expresses that a complex number is in the left half of the complex plane, which is often crucial for the stability of many systems in engineering.

 

display(Explain("The Maple command simplify"));display(Explain("The Maple command simplify"))

The Maple command "simplify" is used to simplify mathematical expressions. In Maple, this command applies various rules to reduce an equation or expression to its simplest or most compact form.

For example, if you have an equation like `(x^3 - x^2)/x^2`, using the simplify command would reduce it to `x - 1`. It simplifies algebraic expressions, performs factorization, combines like terms, cancel out common factors, etc.

The general syntax is `simplify(expr)`, where `expr` is the expression that you want to simplify. The simplified expression is returned as output.

Please note that the simplifications made by this command are based on general rules and may not always give the expected results for some specific expressions due to the complexity of mathematical simplifications.

 

display(Explain(continue, model=ChatGPT));display(Explain(continue, model = ChatGPT))

The Maple command "simplify" is used to simplify mathematical expressions. It takes an expression as input and attempts to find a simplified form by applying various algebraic rules and simplification techniques.

When you use the "simplify" command, Maple applies a series of simplification steps, which can include:

1. Expanding and factoring polynomials: This step expands products of terms and factors out common terms.
2. Simplifying fractions: The command simplifies fractions by canceling common factors in the numerator and denominator.
3. Simplifying trigonometric expressions: Maple applies trigonometric identities to simplify expressions involving trigonometric functions.
4. Simplifying logarithmic and exponential expressions: The command applies rules of logarithms and exponents to simplify expressions with these functions.
5. Simplifying square roots and radicals: It simplifies square roots by expressing them in simplest radical form.
6. Simplifying expressions involving absolute values: Maple simplifies absolute value expressions by considering different cases based on the sign of the argument.
7. Applying mathematical identities and rules: The command uses various mathematical identities and rules to simplify expressions further.

The "simplify" command in Maple is very powerful and can handle a wide range of mathematical expressions. However, it may not always produce the most simplified form, as simplification can sometimes be a subjective task. In such cases, you can try using other commands or options available in Maple to refine the simplification process.

To use the "simplify" command in Maple, you can simply type "simplify(expression)" where "expression" represents the mathematical expression you want to simplify.

 

display(Explain("Weight of a W10X54 beam?"))

The weight of a W10X54 beam refers to the weight specification of a beam in the W Beam series, commonly used in construction and engineering projects. The nomenclature "W10X54" is decoded as follows:

- W: it stands for 'wide flange', a type of structural steel beam where the flanges are parallel to each other.
- 10: it indicates the nominal depth of the wide flange, which is about 10 inches.
- 54: it indicates the weight of the beam per linear foot, which is about 54 pounds.

So, a W10X54 beam weighs 54 pounds per foot.

 

display(Explain(further))

"Weight of a W10X54 beam" refers to the weight of a specific kind of structural steel beam used in construction and architecture. The "W" stands for "wide flange", which is a particular type of I-beam.

The "10" means the nominal depth of the beam in inches (also known as its height when standing upright).

The "54" represents the beam's weight per foot in pounds. In other words, for every foot of length the beam has, it weighs 54 pounds.

 So if you have a W10X54 beam that is 10 feet long, its total weight would be 540 pounds (10 feet times 54 pounds per foot).

Please note that these are nominal dimensions and may not exactly match the true dimensions of the beam due to manufacturing variables.

 

NULL

 

What do you think? Will this be an important part of mathematical software in the near future, or are we still far away from that point? We'd love to hear your opinions.

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