Teaching and learning about math, Maple and MapleSim

How much did you weigh when you were born? How tall are you? What is your current blood pressure? It is well documented that in the general population, these variables – birth weight, height, and blood pressure – are normally or approximately normally distributed. This is the case for many variables in the natural and social sciences, which makes the normal distribution a key distribution used in research and experiments. 

The Maple Learn Examples Gallery now includes a series of documents about normal distributions and related topics in the Continuous Probability Distributions subcollection.

The Normal Distribution: Overview will introduce you to the probability density function, cumulative distribution function, and the parameters of the distribution. This document also provides an opportunity for you to alter the parameters of a normal distribution and observe the resulting graphs. Try out a few real life examples to see the graphs of their distributions! For example, according to Statology, diastolic blood pressure for men is normally distributed with a mean of 80 mmHg and a standard deviation of 20 mmHg.

Next, the Normal Distribution: Empirical Rule document introduces the empirical rule, also referred to as the 68-95-99.7 rule, which describes approximately what percentage of normally distributed data lies within one, two, and three standard deviations of the distribution’s mean.

The empirical rule is frequently used to assess whether a set of data might fit a normal distribution, so Maple Learn also provides a Model Checking Exploration to help you familiarize yourself with applications of this rule. 

In this exploration, you will work through a series of questions about various statistics from the data – the mean, standard deviation, and specific intervals – before you are asked to decide if the data could have come from a normal distribution. Throughout this investigation, you will use the intuition built from exploring the Normal Distribution: Overview and Normal Distribution: Empirical Rule documents as you analyze different data sets.

Once you are confident in using the empirical rule and working with normal distributions, you can conduct your own model checking investigations in real life. Perhaps a set of quiz grades or the weights of apples available at a grocery store might follow a normal distribution – it’s up to you to find out!

A new feature has been released on Maple Learn called “collapsible sections”! This feature allows for users to hide content within sections on the canvas. You can create a section by highlighting the desired text and clicking this icon in the top toolbar:

“Well, when can I actually use sections?” you may ask. Let me walk you through two quick scenarios so you can get an idea.

For our first scenario, let’s say you’re an instructor. You just finished a lesson on the derivatives of trigonometric functions and you’re now going through practice problems. The question itself is not long enough to hide the answers, so you’re wondering how you can cover the two solutions below so that the students can try out the problem themselves first.


Before, you might have considered hyperlinking a solution document or placing the solution lower down on the page. But now, collapsible sections have come to the rescue! Here’s how the document looks like now:  


You can see that the solutions are now hidden, although the section title still indicates which solution it belongs to. Now, you can 1) keep both solutions hidden, 2) show one solution at a time, or 3) show solutions side-by side and compare them!

Now for the second scenario, imagine you’re making a document which includes a detailed visualization such as in Johnson and Jackson’s proof of the Pythagorean theorem. You want the focus to be on the proof, not the visualizations commands that come along with the proof. What do you do?

It’s an easy solution now that collapsible sections are available!

Now, you can focus on the proof without being distracted by other information—although the visualization commands can still be accessed by expanding the section again.

So, take inspiration and use sections to your advantage! We will be doing so as well. you may gradually notice some changes in existing documents in the Maple Learn Gallery as we update them to use collapsible sections. 

Happy document-making!


A new collection has been released on Maple Learn! The new Pascal’s Triangle Collection allows students of all levels to explore this simple, yet widely applicable array.

Though the binomial coefficient triangle is often referred to as Pascal’s Triangle after the 17th-century mathematician Blaise Pascal, the first drawings of the triangle are much older. This makes assigning credit for the creation of the triangle to a single mathematician all but impossible.

Persian mathematicians like Al-Karaji were familiar with the triangular array as early as the 10th century. In the 11th century, Omar Khayyam studied the triangle and popularised its use throughout the Arab world, which is why it is known as “Khayyam’s Triangle” in the region. Meanwhile in China, mathematician Jia Xian drew the triangle to 9 rows, using rod numerals. Two centuries later, in the 13th century, Yang Hui introduced the triangle to greater Chinese society as “Yang Hui’s Triangle”. In Europe, various mathematicians published representations of the triangle between the 13th and 16th centuries, one of which being Niccolo Fontana Tartaglia, who propagated the triangle in Italy, where it is known as “Tartaglia’s Triangle”. 

Blaise Pascal had no association with the triangle until years after his 1662 death, when his book, Treatise on Arithmetical Triangle, which compiled various results about the triangle, was published. In fact, the triangle was not named after Pascal until several decades later, when it was dubbed so by Pierre Remond de Montmort in 1703.

The Maple Learn collection provides opportunities for students to discover the construction, properties, and applications of Pascal’s Triangle. Furthermore, students can use the triangle to detect patterns and deduce identities like Pascal’s Rule and The Binomial Symmetry Rule. For example, did you know that colour-coding the even and odd numbers in Pascal’s Triangle reveals an approximation of Sierpinski’s Fractal Triangle?

See Pascal’s Triangle and Fractals

Or that taking the sum of the diagonals in Pascal's Triangle produces the Fibonacci Sequence?

See Pascal’s Triangle and the Fibonacci Sequence

Learn more about these properties and discover others with the Pascal’s Triangle Collection on Maple Learn. Once you are confident in your knowledge of Pascal’s Triangle, test your skills with the interactive Pascal’s Triangle Activity



Many everyday decisions are made using the results of coin flips and die rolls, or of similar probabilistic events. Though we would like to assume that a fair coin is being used to decide who takes the trash out or if our favorite soccer team takes possession of the ball first, it is impossible to know if the coin is weighted from a single trial.


Instead, we can perform an experiment like the one outlined in Hypothesis Testing: Doctored Coin. This is a walkthrough document for testing if a coin is fair, or if it has been doctored to favor a certain outcome. 


This hypothesis testing document comes from Maple Learn’s new Estimating collection, which contains several documents, authored by Michael Barnett, that help build an understanding of how to estimate the probability of an event occurring, even when the true probability is unknown.

One of the activities in this collection is the Likelihood Functions - Experiment document, which builds an intuitive understanding of likelihood functions. This document provides sets of observed data from binomial distributions and asks that you guess the probability of success associated with the random variable, giving feedback based on your answer. 



Once you’ve developed an understanding of likelihood functions, the next step in determining if a coin is biased is the Maximum Likelihood Estimate Example – Coin Flip activity. In this document, you can run as many randomized trials of coin flips as you like and see how the maximum likelihood estimate, or MLE, changes, bearing in mind that if a coin is fair, the probability of either heads or tails should be 0.5. 



Finally, in order to determine in earnest if a coin has been doctored to favor one side over the other, a hypothesis test must be performed. This is a process in which you test any data that you have against the null hypothesis that the coin is fair and determine the p-value of your data, which will help you form your conclusion.

This Hypothesis Testing: Doctored Coin document is a walkthrough of a hypothesis test for a potentially biased coin. You can run a number of trials on this coin, determine the null and alternative hypotheses of your test, and find the test statistic for your data, all using your understanding of the concepts of likelihood functions and MLEs. The document will then guide you through the process of determining your p-value and what this means for your conclusion.

So if you’re having suspicions that a coin is biased or that a die is weighted, check out Maple Learn’s Estimating collection and its activities to help with your investigation!

The Maple Conference starts tomorrow Oct. 26 at 9am EDT! It's not too late to register: Even if you can't attend all the presentations, registration will allow you to view the recorded videos after the conference. 

Check out the detailed conference program here:

With Halloween right around the corner, we at Maplesoft wanted to celebrate the occasion with an activity where you can carve your own pumpkin… using math! 


Halloween is said to have originated a few hundred years back in ancient Celtic festivals, specifically one called Samhain. This was celebrated from October 31st to November 1st to mark the end of harvesting season and the beginning of winter, or the "darker quarter" of the year. Since then, Halloween has evolved into a fun celebration of candy and costumes in many countries!


With that said, here’s my take on the pumpkin carving activity: 



The great thing is, if you mess up, you can always go back; unlike carving pumpkins in real life. My design is pretty simple (although cute), so let’s see what you all can impress us with!


You can also make your own original art and publish it to your channel so that anyone can see your own artistic creations. You can also attend the Maple Conference next week on October 26 and 27, an event filled with two days of presentations from members of the Maplesoft Community. Participants will also be able to see all the artwork submitted for the Art Gallery and Creative Showcase, where you can draw inspiration for your own submissions to next year’s showcase! The conference is virtual and free of charge, and you can register here.


Looking forward to seeing you there!

The Maple Conference will be starting in two weeks! The detailed agenda, which includes abstracts of invited and contributed talks, is available here:

Please join us on October 26 and 27 for two days of presentations from our staff members and the larger Maple community, a look at our Art Gallery and Creative Showcase, opportunities for networking with other Maple enthusiasts, and more! The conference is virtual and free of charge, and you can register at

We look forward to seeing you at the conference!


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.

A salesperson wishes to visit every city on a map and return to a starting point. They want to find a route that will let them do this with the shortest travel distance possible. How can they efficiently find such a route given any random map?

Well, if you can answer this, the Clay Mathematics Institute will give you a million dollars. It’s not as easy of a task as it sounds.

The problem summarized above is called the Traveling Salesman Problem, one of a category of mathematical problems called NP-complete. No known efficient algorithm to solve NP-complete problems exists. Finding a polynomial-time algorithm, or proving that one could not possibly exist, is a famous unsolved mathematical problem.

Over years of research, many advances have been made in algorithms that can solve the problem, not in perfectly-efficiently time, but quickly enough for many smaller examples that you can hardly notice. One of the most significant Traveling Salesman Problem solutions is the Concorde TSP Solver. This program can find optimal routes for maps with thousands of cities.

Traveling Salesman Problems can also be used outside of the context of visiting cities on a map. They have been used to generate gene mappings, microchip layouts, and more.

The power of the legendary Concorde TSP Solver is available in Maple. The TravelingSalesman command in the GraphTheory package can find the optimal solution for a given graph. The procedure offers a choice of the recently added Concorde solver or the original pure-Maple solver.

To provide a full introduction to the Traveling Salesman Problem, we have created an exploratory document in Maple Learn! Try your hand at solving small Traveling Salesman examples and comparing different paths. Can you solve the problems as well as the algorithm can?


The Proceedings of the Maple Conference 2022 are up at 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.

What are planes? Are they aircraft that soar through the sky, or flat surfaces you'd come across in your geometry textbook? By definition yes, but they can be so much more. In the world of math, observing a system of equations with three variables allows us to plot them as planes in ℝ3. As we plot planes, these geometric entities can start intersecting, creating captivating visualizations. However, the intersection of planes is not just an abstract mathematical concept present only in the classroom. Throughout our daily lives, we come into contact with intersecting planes everywhere. Have you ever noticed the point where two walls and the floor in your room converge? That’s an intersection in its simplest form! And the line where the pages of a book are bound together? Another everyday intersection!

Room image:, Book image: 

However, just seeing plane intersections is but a tiny piece of the puzzle. After all, how can we delve into the intriguing properties of these intersections without quantifying them? Enter the focus of Maple Learn's newest collection: Intersection of Planes. Not sure about how you can identify the different scenarios that three planes can form in ℝ3? Check out the eight documents that provide complete walk-throughs for solving each individual case that three planes can form! With cases ranging from three parallel and distinct planes to three planes forming a triangular prism to three planes intersecting in a line, you’ll gain a mastery of understanding the intersection of planes by the time you’re finished with the examples.


Once you’ve gained an understanding of how to identify and solve the cases that three planes can form, it’s time to test your knowledge! This quiz-like document takes you through the steps of solving for the intersection of three planes with guiding questions and comprehensive feedback. Once you successfully find the intersection or identify the case, you’ll be provided with an interactive 3D plot that allows you to see what the math you’ve been doing looks like. This opportunity to solve any of the 16 different possible systems of equations allows you to prove that you’re on another level!

With your newfound mastery of solving systems of equations, check out similar documents in the recently added linear algebra collection! Try calculating the volume of a parallelepiped or deriving the formula for the distance between a point and a plane

What are you waiting for? Gear up and join us on Maple Learn today! Whether you're diving deep into the world of linear algebra or merely dabbling, there’s a world of discovery waiting for you.

Jill is walking on some trails after a long and stressful day at work. Suddenly, her stress seems to be lifted off her shoulders as her attention gets drawn to nature's abundant beauty. From the way the flowers blossom to the way the leaves grow on their stems, it is stunning.

When many think of mathematics, what comes to mind is often numbers, equations, and calculations. While these aspects are essential to math, they only scratch the surface of a profoundly creative discipline. Mathematics is much more than mere numerical manipulation. It is a rich and intricate realm that influences everything from art and science to philosophy and technology.

Just as Jill was stunned by the beauty of nature, you too can be amazed by the beauty of math! The golden ratio is one math concept that garners a reputation for being particularly beautiful, perhaps due to its presence in different parts of nature. You can explore it through some Maple Learn documents.

Check out the Fibonacci sequence and golden ratio document to better understand the golden ratio and its relationships. Perhaps, once you have a good grasp on the basics, you would like to check out the golden spiral document. Notice how the spiral that results resembles the outline of a nautilus shell and the arms of a spiral galaxy!

The spiral generated in the maple learn document on the golden spiral. A nautilus shell whose shape resembles the golden spiral generated in the maple learn document.A spiral galaxy whose arms resemble the spiral generated in the Maple Learn document on the golden spiral.

Nautilus shell image: -- Spiral galaxy image:

Next, you may want to understand why the golden ratio is considered the most irrational number. You can do that by checking out the most irrational number document. Or you could explore this golden angle document to see how the irrationality of the number can be used to reproduce structures found in nature, such as the arrangement of seeds in a sunflower's centre!

An image generated in the golden angle Maple Learn document where points are packed around the center of a circle using the golden angle. The points are tightly packed around the center.The previous image is superimposed on top of an image of the center of a sunflower. The superimposed image matches the seeds' packing in the sunflower's center.

Sunflower image:

That's all for this post! No worries, though. Maple Learn has hundreds of documents that can aid you in exploring the abundant beauty of math. Enjoy!


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 !!


Registration for Maple Conference 2023 is now open! This year’s conference will again be a free virtual event. Please visit our site to see more information about the event and to register.

Our call for presentations has now concluded, but it is not too late to submit to our Maple Conference Art Gallery and Creative Works Showcase.

The Agenda section, where you’ll find information about the conference format and an overview schedule, has been added. This will be updated as the details are finalized.  

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 `ϕ` was chosen for the angle of the pendulum for the simple reason that this comparison started with bending theory, where `ϕ` 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  `ϕ` = `ϕ`(t) and `ϕ` = `ϕ`(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 `ϕ` = `ϕ`(t) and `ϕ` = `ϕ`(s)). Both ways have their pros and cons.


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





Independent variable

Time t

Arclength s

Dependent variable

Angle of the pendulum with respect to direction of gravity `ϕ`(t)



The slope of the cantilever with respect to the unbend state `ϕ`(s)





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 


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

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


(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


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


`ϕ`(0) = 0, `ϕ`((1/4)*T) = `ϕ__0`

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


`ϕ`(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


`&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


For a downward force:

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

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


Parameter C

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

C = g/l

C = g/l


"Specific" Load

C = F/EI

C = F/EI




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


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


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


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)


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)


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


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)


(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)


#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


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

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


(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)


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)


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


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

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


(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)


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)


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


(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)


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))


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


(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)


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)


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)


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



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)


`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)


which simplifies further with

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

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



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))


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))


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)))


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))



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))


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))


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))


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)


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)


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)


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)


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


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)



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)


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)



Some remarks




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))


(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))


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)


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


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



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))


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)


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))


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))


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)




Useful identities

FunctionAdvisor(identities, InverseJacobiSN)[3]

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


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

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


FunctionAdvisor(identities, InverseJacobiSN)[1]

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


FunctionAdvisor(identities, JacobiSN)[5]

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



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))


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)


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:-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))).



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


phi, theta


[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,
[6] Christian Wolinski,




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