# [WSG21] Daily Study Group: Differential Equations (begins November 29)

Posted 6 months ago
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 A new study group devoted to Differential Equations begins next Monday! A list of daily topics can be found on our Daily Study Groups page. This group will be led by one of our outstanding Wolfram certified instructors, Luke Titus, and will meet daily, Monday to Friday, over the next three weeks. Luke will share the excellent lesson videos created by him for the upcoming Wolfram U course "Introduction to Differential Equations". Study group sessions include time for exercises, discussion and Q&A. This study group will help you achieve the "Course Completion" certificate for the "Introduction to Differential Equations" course after you complete the course quizzes.Sign up: Study group registration page
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Posted 6 months ago
 In the email giving a link to the first recording. a link to the Series Landing Page is given as Link to the Series Landing Pagehttps://wolfr.am/YxPcS9oQThis is the landing page for the previous session ("Introduction to Notebooks")What is the correct landing page ?
Posted 6 months ago
 Here is the link to the Differential Equations series: https://www.bigmarker.com/series/daily-study-group-intro-to-differential-equations/series_details
Posted 5 months ago
 Where are the mathematica notebooks for the sessions stored?
Posted 5 months ago
 A link to the download folder containing all the lecture notebooks (as well as any additional materials used during the sessions) is included in the reminder email you receive from BigMarker. It will also be shared in the live session everyday of the study group series.
Posted 5 months ago
 Thank you very much!
Posted 5 months ago
 When does today's session start? Thanks.
Posted 5 months ago
 Friday sessions for this series will start at 10:30 AM CT with the review session for the week running from 10:30-11:00 and the day's lesson beginning at 11:00 AM CT.
Posted 5 months ago
 Lesson 17, slide 1."When the coefficients of a linear second-order differential equation become polynomials, the equation appears in the form:P(x) y''(x)+Q(x)y'(x)+R(x)y(x)=0 (1)"This statement is correct but also slightly confusing. It may sound as if the form (1) relies on P, Q, and R being polynomial (while each of then can me (almost) any function of x). Im my view, it could be better to say "When the coefficients P, Q and R in the equation P(x) y''(x) + Q(x)y'(x) + R(x) y(x) =0 are the polynomial functions of x, the solution can be found by Taylor series expansion of y(x) ....". Well. it's just about wording, sorry for being a nag.
Posted 6 months ago
 I notice for the form y'(x)=f(x,y(x))we can use VectorPlot[{1,rhs}...] as direction field, how to derive this? thanks,
Posted 6 months ago
 ok, the first list is slope, got it,
Posted 5 months ago
 Thanks, Vincent. Could you please explain what's the meaning of "1" in the list. Is it the module (length) of the vector which is, in fact, variable (but this is apparently handled by color coding) or something else? The lecture said that the "1" is D[x,x] , but why are we taking this derivative? Thank you.
Posted 5 months ago
 Hi I am trying to resolve this trivial problem trying to verify a solution . In the first run I get the incorrect answer but in the second run if I substitute the solution directly into the 4th line I get the correct answer . Any suggestions ?
Posted 5 months ago
 Function does not evaluate its second argument automatically, so you could use Evaluate as shown below to achieve the desired result in the first case: In[1]:= sol = DSolveValue[p''[x] + 2 p'[x] - 3 p[x] == 0, p[x], x] Out[1]= E^(-3 x) C[1] + E^x C[2] In[2]:= solution = sol /. {C[1] -> 1, C[2] -> 0} Out[2]= E^(-3 x) In[3]:= verify = p''[x] + 2 p'[x] - 3 p[x] == 0 /. {p -> Function[{x}, Evaluate[solution]]} Out[3]= True 
Posted 5 months ago
 Thank you. I wonder which is a better way to avoid a naming conflict: localizing p[x] using , say, Module, or by using /. {p -> Function[{x}, Evaluate[solution]]} Could you please provide an alternative example? Thanks. M
Posted 5 months ago
 Thank you, Devendra. When you said "Function does not evaluate its second argument automatically" was it a general statement, or one related to a particular example? What would the "second argument" mean in general? M
Posted 5 months ago
 Yes, I also find it strange that the "Function does not evaluate its second argument automatically". When I first ran into the problem I tried a number of alternatives and the error will occur without using "Function" . I cannot fathom the reason for Wolfram doing this ....
Posted 5 months ago
 Hello Michael,In Function[x, body], the variable 'x' is the first 'argument' and the expression 'body' is the second 'argument'.The variable 'x' is effectively a 'local' (Module) variable inside Function and hence, as shown below, the function evaluation does not depend upon its name. IIn[1]:= f = Function[x, x^2]; In[2]:= g = Function[a, a^2]; In[3]:= f[5] Out[3]= 25 In[4]:= g[5] Out[4]= 25 In[5]:= f[x] Out[5]= x^2 In[6]:= g[x] Out[6]= x^2 Also, technically, Function has the attribute HoldAll: In[7]:= Attributes[Function] Out[7]= {HoldAll, Protected} In practice, this means that the second argument 'body' is not evaluated by Function, so Evaluate is required below (since 'body' is defined outside Function). In[8]:= body = x^2; In[9]:= Function[x, body][5] (*incorrect*) Out[9]= x^2 In[10]:= Function[x, Evaluate[body]][5] (*correct*) Out[10]= 25 Hope this helps.
Posted 5 months ago
 Could you please remind how to send questions about the quizzes? Thanks. M
Posted 5 months ago
Posted 5 months ago
 OK. Thanks. Btw, yesterday I sent there message about the MathOne , but did not hear back yet. Best. M
Posted 5 months ago
 Great! Thank you, Devendra, for opening the hood. Also, I did not realize that "body" is the second argument. Apparently, I had in mind a function of several variables. All the best. M
Posted 5 months ago
 If there is just one argument the { } can be omitted. For multiple arguments f = Function[{x, y}, x*y]; f[3, 4] (* 12 *) 
Posted 5 months ago
 Hello ! For some reason the GeneratedParameters function does not work for me Could you help me ? thank you
Posted 5 months ago
 Is it fair to say that DSolveValue [...] is simply equivalent to DSolve [...] [[1,1,2]]?
Posted 5 months ago
 Hello Michael,Yes, that is correct for single ODEs but for a system you could use [[1,All,2]], as shown below. In[1]:= DSolve[y'[x] == 1, y[x], x][[1, 1, 2]] Out[1]= x + C[1] In[2]:= DSolveValue[y'[x] == 1, y[x], x] Out[2]= x + C[1] In[3]:= DSolve[{y'[x] == 1, z'[x] == 2}, {y[x], z[x]}, x][[1, All, 2]] Out[3]= {x + C[1], 2 x + C[2]} In[4]:= DSolveValue[{y'[x] == 1, z'[x] == 2}, {y[x], z[x]}, x] Out[4]= {x + C[1], 2 x + C[2]} 
Posted 5 months ago
 Thank you, Devendra. Very nice: With "All", it displays the second element of all the solutions/assignments. MPS It seems to me that in the early versions there was a spot reserved for the arrow "->" , so that {a -> 3.}[[1,2]] would return "->". Am I mistaken?
Posted 5 months ago
 From Lesson # 3, Verify a Solution: Why do we need to use Functon & Evaluate:solution = 90 + C1 Exp[1/3 t];verify = (p'[t] == 1/3*p[t] - 30) /. p -> Function[t, Evaluate[solution]]Why do the following two statements FAIL?1) verify = (p'[t] == 1/3*p[t] - 30) /. p -> Function[t, solution]OR 2) verify = (p'[t] == 1/3*p[t] - 30) /. p -> solutionOK, the first one fails, perhaps, because of the HoldAll attribute for Function. But why does the second one fail when there is no HoldAll attribute there?
Posted 5 months ago
 Hello Zbigniew,The second one fails because Derivative requires a Function input as shown in In[3] below. In[1]:= solution = 90 + C1 Exp[1/3 t]; In[2]:= (p'[t] == 1/3*p[t] - 30) /. p -> solution Out[2]= Derivative[1][(90 + C1 E^(t/3))][t] == -30 + 1/3 (90 + C1 E^(t/3))[t] In[3]:= f'[t] // InputForm Out[3]//InputForm= Derivative[1][f][t] 
Posted 5 months ago
 Thank you, Devendra, I didn't realize that all the different ways of expressing derivatives in Mathematica are based on the Derivative[n, m, ...][f][x, y, ...] function. All this makes perfect sense now.Best, Zbigniew
Posted 5 months ago
 In the Lesson 7, the first equation 5 e^y(t)+5 t e^y(t) dy/dt =0 is equivalent to dy/dt = - 1/t; which is trivially solved in one line: y(t) = - ln(t) + c What is the purpose of multiplying both sides by "5 e^y(t)" and making the simple solution more complicate? Is it done for purely *didactic * reasons?
Posted 5 months ago
 Hi Michael. That term is there mainly to demonstrate the concept of an exact equation and the methods to solve them. You could definitely cancel that term and solve the equation much easier, but it wouldn't demonstrate how exact equations can be solved. Since you can cancel that term to get a simpler equation, it is a nice way to check the result obtained when using the methods of exact equations.
Posted 5 months ago
 OK. I suspected that this was a didactic tool. Thanks for confirmation.
Posted 5 months ago
 I am getting an error message when I active DSolveValue for the verifying the solution in Exercise 4 of the second order ODE Exercise 9 set: "DSolveValue::deqn: Equation or list of equations expected in the first argument." The equation is clearly displayed. My example notebook is attached. Please advise how to fix this. Thanks. Attachments:
Posted 5 months ago
 The problem is that you have assignments (=) instead of equatity (==) for y[0] and y'[0]. This should work: DSolveValue[{4*y''[t] + 6*y'[t] + 2*y[t] == 0, y[0] == 5, y'[0] == 1/2}, y[t], t] // Expand Note, however, that after you have evaluated the wrong DSolveValue expression, then Mathematica has saved the two assignments (y[0] = 5 and y'[0] = 1/2) and will interfere when evaluating the corrected expression. One way to fix this is to restart the kernel after fixing the expression (e.g. using the Quit[] command).
Posted 5 months ago
 My oversight, thank you.
Posted 5 months ago
 Exercise 10(4) y1[x_] := x^(1/2 (1 - Sqrt[5])); y2[x_] := x^(1/2 (1 + Sqrt[5])); What are the advantages of using [29]-[30] instead of [38]-[39]? n[29]:= (x^2 y''[x]-y[x]==0)/.**y->Function[x,y1[x]]**//Simplify Out[29]=True In[30]:= (x^2 y''[x]-y[x]==0)/.**y->Function[x,y2[x]]**//Simplify Out[30]=True  In[38]:= (x^2 y''[x]-y[x]==0)/.**y->y1**//Simplify Out[38]=True In[39]:= (x^2 y''[x]-y[x]==0)/.**y->y2**//Simplify Out[39]=True  And similarly with the exercise 9(3) Thanks
Posted 5 months ago
 Both ways will get the job done. However, using Function is a little more versatile as it can handled pretty much any form of the function. For example, if you had defined y1[x] instead as y1 = x^(1/2 (1 - Sqrt[5])), then y -> y1 wouldn't work and you would need to use y -> Function[x, Evaluate[y1]].
Posted 5 months ago
 From what you said, I take that using the "Function" is a more secure, bullet-proof approach. It is good to know! Thank you.
Posted 5 months ago
 In reviewing Lesson 7 on Exact Equation, Example 1, Slide 9, I'm confused about a particular step.After integrating "M" & "N", expressions, we are told that the result of combining them to create the auxiliary function is: [Phi](x,g(x)) = 2 g(x) ln(x) + 3x^2 - 2g(x).However, when you integrate "M", you get: 3 x^2 + g Log[x]; when you integrate "N", you get: - 2 g + g Log[x].In the lesson, Luke indicates that, when combining "M" & "N", you consider the term that is common to them, g Log[x], only one time. This would give you the expression above for PHI.Including it would give an expression with a "2" in front of the g(x) Ln(x) term: [Phi](x,g(x)) = 2 g(x) ln(x) + 3x^2- 2g(x). Using that expression, I got a slightly different answer for g[x] that had a "2" in front of the Log[x] term in the denominator.My question: What allows us to ignore the second instance of g Log[x] when combining "M" & "N" to create PHI?
Posted 5 months ago
 Sorry, the original expression in the lesson is: [Phi](x,g(x)) = g(x)ln(x)+3x^2-2g(x).
Posted 5 months ago
 Hi William. Thank you for your question. It is an important point. What we are doing in that step is identifying the terms terms that compose Phi(x,g). Notice that Phi(x,g) will consist of three terms. The first is a function of x and g, the second is a function of only x, and the third is a function of only g. To calculate M, we take the partial derivative of Phi with respect to x. Therefore the part of Phi that is a function of only g will drop out because that partial derivative will be zero. Therefore, when we integrate M with respect to x we only get back the function dependent on g and x and the function dependent on x. Similarly for N, that is calculated by taking the partial derivative of Phi with respect to g. Therefore, the part of Phi that depends only on g drops out when taking that partial derivative, so after integrating we only get back the function dependent on g and x and the function dependent on g. We used integration to identify the parts of the function Phi. These are: a function dependent on g and x, a function dependent on only x, and a function dependent on only g. To obtain Phi, we do not add the results of the integrals, we just pick out these functions from the integrals.
Posted 5 months ago
 Luke,Thanks!That made it a lot clearer!By the way, great course! Well-developed, well-executed!William
Posted 5 months ago
 Thanks William! I really appreciate it.
Posted 5 months ago
 I second that , The best WolframU course so far ......
Posted 5 months ago
 Exercise 12: Repeated characteristic roots.Slide 5 ("exercise 4"), after the equationy (x) = c1 exp (r x) + c2 x exp (rx),it is stated: "Thus, we can conclude that the solution behaves like C exp (r x) at x-> Infinity". I guess this is a typo: it would be wrong in general to neglect the (dominant) contribution ~ x exp (r x) in comparison to ~ exp(r) in the limit of large x.
Posted 5 months ago
 Hi Michael. Yes, while the exponential term dominates the long term behavior, the factor of x should still be kept. I will fix that in the notebook for that exercise.
Posted 5 months ago
 Thank you, Luke. I agree that exp (rx) is the decisive factor, but in our case we are not comparing exp (r x) with x, but rather x with 1.
Posted 5 months ago
 Exercise 13(1). Just for practice, I tried to do it without "manual labor" while still reproducing all the steps of defining the "undetermined coefficients" . However, the implementation is sloppy. For instance , using "DeleteCases" to remove a bunch of 0s produced on the previous step, seems redundant. Apparently, I did not find the right version of the "CoefficientList[]" Please, advise. eq =y''[x] + 2 y'[x] + y[x] - x^2 - 3 Sin[x] == 0; parSol[x_]:= a Sin[x] + b Cos[x] + c x^2 + d x + e; collConst = Collect[eq[[1]] /. y -> Function[x, parSol[x]], {x, Sin[x], Cos[x]}] Out[39]= 2 c+2 d+e+(4 c+d) x+(-1+c) x^2+2 a Cos[x]+(-3-2 b) Sin[x] In[40]:= Solve[DeleteCases[Flatten[CoefficientList[collConst, {x, Sin[x], Cos[x]}]], 0] == Table[0, {5}], {a, b, c, d, e}] Out[40]= {{ a->0, b->-(3/2), c->1, d->-4, e->6 }} 
Posted 5 months ago
 I think the way you are doing it is probably the best way simply. Deleting the zeros will be necessary since CoefficientList will consider all combinations of powers of the terms. The combinations of powers that do not appear will be given the value of zero, and those will need to be deleted.
Posted 5 months ago
 Thank you. Still, it does not look pretty. It would be good to tune up the action of CoefficientList, so that it considers only non-zero elements. Am I asking too much ? :) PS: It can be done by using If[], but it would not look neat either.
Posted 5 months ago
 I joined the study group today. How to download the previous lecture series ?
Posted 5 months ago
 Welcome to the study group. The reminder emails for the study group sessions will contain all the links you need for this study group--including the link to download all the notebooks and the links to the recordings.Also you can access the previous recordings by clicking on the "Start this section" links for each day at the link posted by Michael above: https://www.bigmarker.com/series/daily-study-group-intro-to-differential-equations/series_details
Posted 5 months ago
Posted 5 months ago
 Sorry for asking, but where should we look for the Quizzes? Is it at the usual "amoeba site" or anywhere else? Thank you.
Posted 5 months ago
 Unfortunately we have not been able to release the quizzes yet. We hope to release the quizzes on Friday along with the beta version of the entire course (including all the videos and the final exam). We'll clarify further at tomorrow's session.
Posted 5 months ago
 I'm looking at the overview of Lesson 15. There is a nice working Manipulate implementation of the oscillating spring and a block attached to it. The code is hidden very well--I can't find it. In fact, the cell is marked as though there is no code associated with it. Where is the code for this Manipulate function? Please advise. Attachments:
Posted 5 months ago
 Just above the output cell bracket you will see a tiny little cell bracket. Select that, then (on Windows) go to the menu Cell -> Cell Properties then select "Open"
Posted 5 months ago
 Wow! I would have never guessed it. So far I have been closing the code cells by double-clicking their output cell bracket. I've been a Mathematica user for a long time, but I didn't know that there was another option to do the same by selecting the code cell and then going to Cell >> Cell Properties and clicking or unclicking the box next to "Open" (in Linux). Again, I find it very interesting and, I bet, so would many other users. If I may, I suggest you point out this option during the Friday review. Many thanks for this pointer!
Posted 5 months ago
 I'll comment on that at the beginning of the session today.
Posted 5 months ago
 In lesson 11, slide 12, the solution of the problem in Example 1 is defined by means of the ASSIGNMENT, "=", whereas in slide 14, the analogous solution of the problem in Example 2 is defined by means of the DELAYED ASSIGNMENT, ": =". Traditionally, functions in Mathematica are defined via the delayed assignment. Is there a reason for the cited difference, or it simply doesn't matter? Please advise! Attachments:
Posted 5 months ago
 In my notebook "Lesson 11", slide 12, the first sol you mentioned is also a delayed assignment: Now you have almost found the same solution as that found by DSolveValue:sol[x_] := c1 y1[x] + c2 y2[x] Perhaps you have accidentally changed that := to = and the saved the notebook?
Posted 5 months ago
 No, I didn't do it. But, an excellent hypothesis, Hakan! It prompted me to investigate this matter further. I originally downloaded the materials before the meetings started. Then additional materials were added, so after one week into the course, I downloaded all the materials again (and I keep both zip files). It is there where this change took place. I downloaded the materials again a few minutes ago, and the "error" is there--you can confirm it yourself.My question stands: is it an error or a correction? Or, perhaps, it doesn't matter, as both versions seem to work equally well.
Posted 5 months ago
 Ah, this is interesting. I downloaded the files early as well, before they was zipped into a file. And I can confirm your findings: The new/current Lesson 11 (dates 2021-12-02 in the zip file) has indeed sol2 = .... I also noted that the notebook has been converted to presentation notebooks, not the "continuous notebook" as I had earlier. (I tend to prefer the non continuous versions since they are easier to browse.)Thanks for pointing this out. Now I've updated the notebooks.
Posted 5 months ago
 Thank you, Hakan for confirming my observation.
Posted 5 months ago
 Hi Zbigniew. Using the Set vs SetDelayed (sol2[x]= vs sol3[x]:=) is just for presentation reasons. Using just = will display the output while using := will not display the output. They are both accomplishing the same task though.
Posted 5 months ago
 Thank you, Luke. I knew there must be a reason. I'm glad to know it now. Thanks again!
Posted 5 months ago
 Is there a Level 1 Certification available for this study group?
Posted 5 months ago
 A Level 1 certification will be available by taking the Final Exam available in the course. We hope to share the pre-release version of the course today with all our study-group attendees.
Posted 5 months ago
 Congratulations and many thanks for the pre-release version of the course. Excelent!Greetings from Barranquilla Colombia. Jorge O
Posted 5 months ago
 [WSG21] Daily Study Group: Differential Equations (begins November 29)
Posted 5 months ago
 How to copy a Quiz in the local NoteBook?
Posted 5 months ago
Posted 5 months ago
 OK, that's why I really need to get an account ! (?) Now I will follow your and Arben's advice and write Wolfram-U a letter :)PS Just did it.
Posted 5 months ago
Posted 5 months ago
 Hi Jürgen, We will address this more details at today's session to clarify all questions about accessing the lesson videos and the quizzes within the online course.
Posted 5 months ago
 I tried to post it before, but for some reason it did not workLesson 17, slide 1."When the coefficients of a linear second-order differential equation become polynomials, the equation appears in the form:P(x) y''(x)+Q(x)y'(x)+R(x)y(x)=0 (1)"This statement is correct but also slightly confusing. It may sound as if the form (1) relies on P, Q, and R being polynomial (while each of then can me (almost) any function of x). In my view, it would be better to say"When the coefficients P, Q and R in the equation  P(x) y''(x) + Q(x)y'(x) + R(x) y(x) =0 are the polynomial functions of x, the solution can be found by Taylor series expansion of y(x) ....". Well. it's just about wording, sorry for being a nag.
Posted 5 months ago
 That was just to simplify the discussion. The method works when they are general analytic functions, but the term analytic wasn't defined at that point.
Posted 5 months ago
 Thank you, Luke. Just double-checking: If the functions are polynomial, then, in order to find the coefficients in the series expansion of y(x), you can use them (the functions) "as is". If they are analytic but not polynomial, you need fist to expand them and then to find the (coefficients in the expansion of ) y(x). Is it correct? Thank you. M
Posted 5 months ago
 That's right. The polynomial assumption at the beginning was just to guarantee that we are working with analytic functions so we wouldn't have to worry about any of that.
Posted 5 months ago
 That's right. The polynomial assumption at the beginning was just to guarantee that we are working with analytic functions so we wouldn't have to worry about any of that.
Posted 5 months ago
 Section 5, Exercise 2 The radius of convergence can be calculated using SumConvergence:In[2]:= SumConvergence[n (2 x + 1)^n, n, Assumptions -> x [Element] Reals]Out[2]= -1 < x < 0Therefore, the series converges whenever -1<= x <= 0, so the radius of convergence is 1/2.In the previous statement the original inequality -1 < x < 0 is substituted by -1<= x <= 0*. However, they are not equivalent, because the Sum [n,{n,0,Infinity }] does not converge. To avoid the confusion, the "=" sign should be dropped. The situation is slightly different in Exercise 1. Addressing this difference would be helpful for revealing a nuance in the definition of the radius of convergence.
Posted 5 months ago
 Thanks for pointing that out. I'll make the change in the notebook.
Posted 5 months ago
 Even though I'm registered for this study group, I stopped receiving notifications about the upcoming meetings as well as the follow-ups about their recordings. I would appreciate receiving those announcements.
Posted 5 months ago
 Hi, Zbigniew. The mist important was about opening the course at the Wolfram-U site, with all the lectures, Quizzes and Exam posted there. With this you do not need to get the weekly invites (well, they will still send them as reminders). Did you get that one?If you don't, please contact the team. I was cited a few times for sending the links through this forum.
Posted 5 months ago
 Hi Zbigniew, Could you please email wolfram-u@wolfram.com from the email you used to register for the study group? We can check to make sure you receive the reminders.
Posted 5 months ago
 Hi Michael, Oh, I have missed this one. Thanks for the pointer. Best, Zbigniew
Posted 5 months ago
 [WSG21] Daily Study Group: Differential Equations (begins November 29)
Posted 5 months ago
 When clicking on the Feedback button in the online version of the course, it gives an "HTTP Error Code 404 - Oops, the page you're looking for can't be found.". Before that is fixed, should I report comments / questions on the online version via the usual Wolfram support channels (i.e. email)?
Posted 5 months ago
 Hi Hakan, Please email wolfram-u@wolfram.com directly about any feedback related to the course.
Posted 5 months ago
 In lesson 20, Special Functions, on slide 12, you whetted our appetites with the following statement:  It can be shown that given a mass initially displaced at an angle theta_o on a pendulum, its period of oscillation will follow the hypergeometric function: 2F1[1/2, 1/2, 1, sin(theta_o)^2] Well, it can be shown? SHOW US, please! Below, I comment on this statement. Attachments:
Posted 5 months ago
 Again, I would like to know how equation (1), in lesson 20, Special Functions, on slide 12 (see also my previous note), was derived.Below is my attempt at it, in which I ended up with an expression involving the EllipticF function. Wow! Surprisingly, it turns out that this expression is actually equivalent to the one involving the hypergeometric function. By the way, I'm amazed by the symbolic power of Mathematica in finding the integral and even more in finding that the two expressions are equal to each other. If I were to do it by paper and pencil, it would be ...But the question still remains: how did you come up with (1)? Attachments:
Posted 5 months ago
 Hi Zbigniew. I did not derive that equation, but you can find a reference here: https://aip.scitation.org/doi/10.1063/1.4922268
Posted 5 months ago
 Thanks, Luke, I really appreciate the pointer. Cheers, Zbigniew
Posted 5 months ago
 Our College math professor called HGF "The mother of all bombs". And he really meant it, being in his youth a part of the "Russian Manhattan Project". I should also mention Dr. Eric Weinstein and his MathWorld (a part of WR) - the outstanding Encyclopedia of Math, which (in conjunction with Wolfrfam Demo projects) allows to test many concepts using the Mathematica. PS: Morse and Feshbach, in my view, is still a great resource to be kept in the vicinity of your desk :)
Posted 5 months ago
 Thank you Zbigniew for this insightful post.
Posted 5 months ago
 Thanks, Michael for your pointers and kind words. I appreciate them. Also, I'm fascinated by your lab. I can't wait till the next Wolfram conference, hopefully in person, and am looking forward to meeting you. Cheers, Zbigniew
Posted 5 months ago
 Thank you, Zbigniew. It would be a pleasure to meet in person. However, I have an additional suggestion. As we are "here and now", and already found a common denominator, why don't we try doing together something useful for our learning of Differential Equations and Mathematica, and may be even beyond. I was thinking of writing a "computational essay" on the topic dealing with the non-linear oscillations (and, naturally, with the Differential Equations). And your last question sounded to me like a natural part of this project. The problem that I have in mind deals with a simple "(almost) High school level" system, which may display, (almost) in spirit of NKS, very complex behavior. A while ago, we have done some ground work with my son, and I can send a short note on the subject (for anyone interested in this endeavor).PS This is not just a personal message. Anyone on the forum, who is interested, is welcome to join this discussion (some background in Math and physics would be very helpful). I am very grateful to the Wolfram-U team and personally to Luke for creating such an inspiring and productive atmosphere.
Posted 5 months ago
 Michael,Awesome idea, and I'm game. Definitely! I'll be mostly free after I give and grade the final exam in my undergraduate Fluid Mechanics, i.e., probably in two days. I would love to get involved in such a project. I'm not sure how long this discussion board will be available. So for this kind of collaboration, I propose that we deal with each other directly via email and/or skype or zoom. My email address is below.Best wishes, ZbigniewGiles and West House Faculty in Residence919-667-5150 (c)Zbigniew J. Kabala, PhD, PE, Associate ProfessorDuke UniversityDepartment of Civil & Environmental EngineeringE-mail: kabala@duke.eduhttp://www.cee.duke.edu/faculty/zbigniew-j-kabala
Posted 5 months ago
 Good morning,I tried to submit Quiz 1 of the on-line course but I get an error during the grading (btw, it looks like no answer is correct!). I attach the image of the reported errors. Please, let me know if you have any ideas on what's going wrong.Regards - - - Fabio
Posted 5 months ago
 Same here
Posted 5 months ago
 Hello Wolfram U team,I'm not able to submit Quiz answers without errors. And all answers are incorrect, no matter which I choose.For Quiz 1, the web site reports error messages, see attached image, similar to those reported above by Fabio and Luis.John Burgers. Attachments:
Posted 5 months ago
 Guys, No need to post ALL your error messages ... We got it. The quizzes are messed up, and we have to wait until they are fixed.
Posted 5 months ago
 Awsome, Zbigniew. Welcome to the team! I will send you some background information and a sort of initial proposal. Later, I am sure, it will expand from both sides :)PS I do not want to overload this forum, and will send directly to your email.
Posted 5 months ago
 Thank you everyone for reporting the errors. We are looking into this and will hopefully be able to share a resolution at tomorrow's session.-Wolfram U Team
Posted 5 months ago
 Just wanted to post an update: We have received quiz submissions from 39 users for at least one or more of the 8 quizzes and 36 successful submissions (with score >- 60%) for at least one or more quizzes.We are still looking into what might be causing the errors posted in the screenshots above.
Posted 5 months ago
 Hi Abrita, I tried the quiz again and it doesn't work from my Wolfram U Course page, but if I access it directly from wolframcloud it seems to work fine, I attached the screen capture, hope it is of some help to solve the issue. By the way thanks to you, Luke and others in Wolfram U team for this course.
Posted 5 months ago
 UPDATE: Everything seem to be working now, thanks!
Posted 5 months ago
 Please let us know where on Wolfram U is this course with its quizzes, lectures, and recordings ...I tried to find it but to no avail: when I googled for "wolfram u differential equations course" I got the following link (somehow the link can't be copied, so please repeat the search to see the issue):  [Differential Equations, Interactive Online Video Course][1] A comprehensive introduction to fundamental concepts and solution methods for differential equations, including video lessons and interactive notebooks. Follow ... and below is the screenshot of what I get when I click the link, "Comming Soon." Again, where is the course?
Posted 5 months ago
 I have a question about downloading Wolfram University notebooks that does not relate to this course directly but more to Wolfram University's course Visual Explorations in Data Science. I have had a lot of trouble downloading the notebooks from the course page with the link at the bottom. Is there another place I can download the notebooks for Visual Explorations in Data Science?
Posted 5 months ago
 Is there a way to convert a series into summation notation? For example, is there a way to convert the following? Series[Exp[x],{x,0,10}] into x^n/n! in summation notation?
Posted 5 months ago
 Use the Normal function to do that. Sometimes you need also the Activate function, but not in the example below:  In[4]:= Series[Exp[x], {x, 0, 6}] // Normal Out[4]= 1 + x + x^2/2 + x^3/6 + x^4/24 + x^5/120 + x^6/720 In[5]:= % /. x -> 1.5 Out[5]= 4.47754 
Posted 5 months ago
 For some reason DSolve has an issue with the equation for Problem 5 Quiz 5. Wolfram Alpha found an answer but DSolve returned an error message.
Posted 5 months ago
 Don't know what is in the quiz but the code in your question is invalid, should be DSolve[{-3 y[x] - x y'[x] + y''[x] == 0, y[0] == 0, y'[0] == 1}, y[x], x] (* {{y[x] -> ( E^(x^2/2) Sqrt[\[Pi]] + E^(x^2/2) Sqrt[\[Pi]] x^2 - 8 HermiteH[-3, x/Sqrt[2]])/(2 Sqrt[2])}} *) 
Posted 5 months ago
 The equation in DSolve is not the same as the one in WolframAlpha.It should instead be: DSolve[{ y''[x] + x y'[x] - 3 y[x] == 0, y[0] == 0, y'[0] == 1}, y[x], x]
Posted 5 months ago
 Will the participants in this course have online access to the course lessons and exercises after 12/27?
Posted 5 months ago
 Yes, this course will be available for everyone on the Wolfram U site once it has been released. We have shared a pre-release version of the course with the participants of this particular study group.
Posted 5 months ago
Posted 5 months ago
 Very interesting. Thank you, Hakan, for sharing. M
Posted 5 months ago
 [Comment removed by the author since he was confused.]
Posted 5 months ago
 Hi Hakan, I second Michael's comment: very interesting!But if something looks to good to be true, it usually is NOT true!I think the problem is that your apparent Laplace Transform solution is actually NOT a solution. I still don't understand where it came from, and why is there y[0] in this spurious solution, Exp[-(2a+c) t /b] y[0] that is clearly different than the one by DSolve. The term y[0] shouldn't be there after you already applied the initial conditions! I still have to think about it.If I may, I think that in your enthusiasm, as we all often do, you violated the forgotten Roman proverb: "Hurry slowly!" and you wrapped your solutions into elegant Module-based functions, so that you could later test other ODEs subject to other initial conditions. But it seems that you did not evaluate the Module-based functions carefully enough, perhaps by peppering them initially with the Print command after each statement.So, let's take a step back, and hurry slowly in analyzing the Laplace Transform solution.First, that solution is NOT a solution of the ODE. Indeed, the following statement confirms this assertion, as it does NOT yield TRUE:  eqn /. y -> Function[t, Evaluate[Exp[-(2 a + c) t/b] ]] // FullSimplify Second, going step-by-step in obtaining the solution leads to the following:  In[15] := leqn = LaplaceTransform[eqn, t, s] /. {y[0] -> 0, y'[0] -> 1} Out[15] := c LaplaceTransform[y[t], t, s] + b s LaplaceTransform[y[t], t, s] + a LaplaceTransform[t^2 y''[t], t, s] == 0 In[16] := SolveValues[leqn, LaplaceTransform[y[t], t, s]] Out[16] := {-((a LaplaceTransform[t^2 y''[t], t, s])/(c + b s))} Note that in both Out[15] and Out[16] the term LaplaceTransform[t^2 y''[t], t, s] is UNRESOLVED. Consequently, the step SolveValues cannot really solve the equation leqn, even though it deceptively gives an impression that it does.Again, thank you for a very interesting case that highlights the problem with Mathematica, which, without warning, produces here a clearly spurious solution.
Posted 5 months ago
 Thanks for a very interesting and illuminating analysis of this, Zbigniew!I willingly admit that I have missed something here (as you put it: in my enthusiasm :-)) , especially the part that Luke showed many times: to check that the "solution" actually solves the stated problem, Mea culpa! See below for the result when testing the solution "á la "Luke".One comment: the y[0] term is only in the last test of LaplaceTransform[], i.e. the one without any initial values, but the one with initial values has no such term.In your code, it seems that you tested the version without initial values. See below for the test with outputs on all steps on the problem with initial values, which shows that the solution is not a proper solution. Also, when I'm trying to repeat your experiment (given that eqn is the same as my eq), I got the same result as in my code in the above (the las test): there is no unevaluated term LaplaceTransform[y[t],t,s]. I'm not sure what the difference between out codes are. Hmm. I tested this also in Mathematica v12.3.1 (instead of v 13.0), and then I got different result where the solution from SolveValues includes a lot of LaplaceTransform[y[t],t,s] terms. Which Mathematica version do you use?Regarding wrapping the test into a Module: I first tested this line by line and got the following which is not the same as your output, but you use another approach which might explain this difference, or perhaps I missed something. In the last step we can see that the obtained "solution" is not a solution to the problem.
Posted 5 months ago
 Hakan,Let me second Michael again: I love the forum and the exchanges it stimulates!Now, I looked again at the code I posted at 4:30 am my time (EST), and I did mess up copying parts of it (In[15] and Out[15]), NOT applying the initial condtions. I have now corrected these two lines in that post. Here is my code again, this time carefully run and copied:  In[1] := eqn = a t^2 y''[t] + b y'[t] + c y[t] == 0; In[2] := leqn = LaplaceTransform[eqn, t, s] /. {y[0] -> 0, y'[0] -> 1} Out[2] := c LaplaceTransform[y[t], t, s] +b s LaplaceTransform[y[t], t, s] +a LaplaceTransform[t^2 y'' [t], t, s] == 0 In[3] := sol = SolveValues[leqn, LaplaceTransform[y[t], t, s]] Out[3] := {-a LaplaceTransform[t^2 y'' [t], t, s])/(c + b s)} Note that the term LaplaceTransform[t^2 y'' [t], t, s] is unresolved. Mathematica doesn't know what to do with it. You missed this part again by not printing out the output between your lines starting with lt2 and sol. I'm using Mathematica 12.3 right now. But I doubt that version 13 can resolve Laplace transform of a product t^2 y'' [t]. Here is what I get:  In[4] := LaplaceTransform[t^2 y''[t], t, s] Out[4] := LaplaceTransform[t^2 y''[t], t, s] Which means: "I can't do it!" Please let us know what you get!Again, thanks, Hakan, for finding this fascinating problem. I can't wait for Luke to weigh in ...Best, Zbigniew
Posted 5 months ago
 Hi again, Zbigniew.Your equation, or rather the initial values is not the same as mine. I have {y'[0] -> 0, y[0] -> 1} while you have {y'[0] -> 1, y[0] -> 0}. When I use your initial values SolveValues[] returns 0, as does InverseTransformTransform[], which would indicate that there is no solution?In version 12.3.1 both variants behaves as your example. So there seems to be a version issue.But now Luke starts to talk...Best,Hakan
Posted 5 months ago
 Since I don't know how to get the nice copyable outputs as Zbigniew's, here's a notebook showing the different experiments using Mathematica version 13.0 (in Mathematica Online).The notebook is available for further experiments here: euler equation laplace transform.nb
Posted 5 months ago
 That is absolutely awesome, Hakan! I will definitely download and install Mathematica 13.
Posted 5 months ago
 Hello Hakan et al.,Long time (6 days) no see. I finally got Mathematica 13 and was able to return to the old problem of solving Euler's equation with the Laplace Transform. The crucial point that we left hanging is the ability of SolveValues to solve the equation generated by Hakan and denoted leqn in his post. To make life simple I imbed my notebook here and remind us of the dilemma (I also attach it--see below).By the way, Hakan, it would be nice to have your contact info and keep in touch. Please post it or send it to my email at Duke: kabala@duke.edu.Cheers,Zbigniew Attachments:
Posted 5 months ago
 Hi again, Zbigniew.Thanks for your new analysis of this problem. I can file an issue about this to Wolfram support or perhaps to Wolfram Community. But - as you point out - Luke or someone else from Wolfram might comment about it in this discussion.My email is hakank@gmail.com (and my home page is http://hakank.org/ ). I'm just a math hobbyist, but very fascinated in various aspects of mathematics.Best, Hakan
Posted 5 months ago
 Hi Hakan,I'm not sure how long this forum would stay open beyond the quiz and final exam submission deadline of Dec 27. So thanks for your contact info. I'm also a math enthusiast, although I have a math-related job at Duke University (teaching fluid mechanics and other courses). We tend to trust systems like Mathematica, but they are all products made by humans, so they are prone to mistakes. This was an interesting exercise and a reminder about the need to be vigilant and skeptical. Best,Zbigniew
Posted 5 months ago
 Hi again, Zbigniew.Unless they change the policy for some reason, these Study Group discussions are not closed. The study group discussions from the last year are still open.Best, Hakan
Posted 5 months ago
 Good to know, Hakan!I wish they made it clear from the beginning and, especially at the end of the course. I find the discussions stimulating and can see myself returning to these groups over and over again.Best wishes, Zbigniew
Posted 5 months ago
 I have a question that does not have any possible choices on the final exam. Is it okay for me to post a question with no answers on the final exam, or should I not post it?
Posted 5 months ago
 Wow guys! It develops faster than the speed of thought! I watch it as a thrilling (and thought-provoking) show, holding my breath. Thanks to all the participants! M
Posted 5 months ago
 I have a question that has a duplicate answer I am taking the Calculus Final Exam for Wolfram U and this is a question
Posted 5 months ago
 Hi, Peter. Is this a typo? It's probably worth reporting to Wolfram_U team. Best. M
Posted 5 months ago
 Hi Luke, One suggestion for Lesson 17 Slide 11. As a reminder for some students would it be worth to state the relation Gamma[n+1] = Factorial[n] for integer n > 0 ?
Posted 5 months ago
 I know the landing page website was given but, I'd like to know how to navigate there from my Wolfram Cloud account landing page. Could someone demonstrate navigating from a user's Wolfram Cloud.com landing page to the course we are taking? That may help answer how to arrive at other courses.I enjoyed the course immensely, it was 40 years ago that I was last instructed on this topic. It's wonderful how the technology has advanced! Hats off to Luke, superb work. I was able to complete the quizzes and exam. Can't tell you how happy that makes me. Your probably wondering why bother? I've got some engineering problems to work on.Thank-you again, John
Posted 5 months ago
 Hi John,Right now there is no way to navigate from your Wolfram Cloud home directory to the Wolfram U courses. This is something we will consider for the future.The easiest way to get to all Wolfram U courses right now is from the Wolfram U site https://www.wolfram.com/wolfram-u/all-classes-courses/Of course once you have taken some of the quizzes and exercises from a course, copies of these quiz and exercise notebooks will start appearing in your "Copied Files" folder within your Wolfram Cloud home directory.
Posted 5 months ago
 Small suggestion for improvement to the videos (and examples) - pause for a few seconds at the final result so the student has a chance to read and digest the result. Thanks.
Posted 5 months ago
 Thanks very much for the introduction to the Airy Functions. I picture multi dose radiotherapy for prostate cancer as a simple geometric progression in descent. Problem with this model, a cure is often obtained well in advance of the moment the tumor cell population is reduced to one cell in a million (or less). My intuition has always been a prostate cancer cure by means of radiotherapy is a perfect example of a 'catastrophe'. Now, with your explanations / demonstrations of the Airy Ai(z) and Bi(z) equations I have a perfect model for what happens. At some fortunate point in time the tumor cell death rate reaches up towards infinity. While, at the same moment in time the tumor cell reorganization and reproduction rates descend towards zero. Thanks to all of you, Eric Weisstein and George Biddell Airy.
Posted 5 months ago
 Thank you, Alan, for bringing up this interesting and important topic. At first glance it seems to me that the situation you describe (and surely the catastrophes in general) require nonlinearity,* while the Airy equation (AE) is boringly linear. How do you relate your story to the systems that AE can describe? Which particular feature of AE you consider catastrophic? In my little experience with AE **, its solutions behave quite modestly (in my case they usually lead to asymptotically decaying oscillations) and smoothly reacted to variation of boundary conditions. By definition, catastrophes are related to strong changes in the systems' behavior in reaction to miniscule variation of the parameters (such as the boundary/initial conditions).** It was related to the (decaying) elastic field introduced by cylindrical inclusions in lipid bilayers.
Posted 5 months ago
 Sorry, "boring" was a joke. The function is amazing, and even more amazing is that such a simple equation could make such a big contribution in almost everything.
Posted 5 months ago
 O.K. Right now I think Professor Airy's equations 'describe' tumor death. Right now, I have no clear idea as to the actual mechanism. I am not able to fully explain tumor death with Airy's equations. Tumors shrink during a course of repeated episodes of radiotherapy. It is possible; therefore the 'turning point' event is a physical collapse like with the world trade center buildings. I wish I could prove the energy field brings about caustics and that these areas of concentrated energy overwhelm the DNA molecule. Bringing about double strand breaks in regions thought to be immune to double strand breaks. You are correct. Airy's equations are an early perhaps the earliest possible descriptor of a catastrophe. It is very possible you might be able to arrive at a better analytical solution.
Posted 5 months ago
 Thanks Alan. After reading for a while, I now seem to understand what you mean (I just didn't find it in the lecture). I see the spikes on the curves of AFs e.g. in this article: https://en.wikipedia.org/wiki/Airy_function. The reading also reminded me that my first exposure to AF came in a basic QM course - solving the Schrödinger equation for a triangular barrier. Thanks for the push.Best, M
Posted 5 months ago
 A remark on a final exam question: The question involving the oscillating spring needs a specification of the positive direction of the coordinate system, is it up or down?
Posted 5 months ago
 I encountered only one problem with "spring". t is the time. Unfortunately, it never comes back (unless you discuss it with Dr. Smolin). So, the direction is positive..
Posted 5 months ago
Posted 5 months ago
 Before I got deeper into this after the class, does it work only for "named" functions? Thank you. M
Posted 5 months ago
 DifferentialRootReduce is not restricted to named functions, it can reduce quite complex functions. I'm not sure how complex function it can manage, though. The help page ( https://reference.wolfram.com/language/ref/DifferentialRootReduce.html ) just states: DifferentialRootReduce will attempt to represent any expression as a DifferentialRoot object.
Posted 5 months ago
 Hi, Hakan. Thanks for your insightful contributions. Trying to digest them, I had issues with creating local copies (possibly, because I do not have Math. One account). Could you please attach the NB file to a couple of most recent, dealing with Root reduce. Thank you and hope to meet you in new classes. Best. M
Posted 5 months ago
 Hi Michael. Below the post there is a button "Make Your own Copy" which should show a Wolfram Cloud version of the note book. Here's the explicit link: https://www.wolframcloud.com/env/c0022b5b-98c8-4085-a1a8-c3c83cd0818b . Does that work for you? If not, I've also attached my original notebook, "laplacetodiffeq.nb". I've registered for the two upcoming Wolfram U Study Groups: "Introduction to Cryptography" and "Quantum Computation Framework", and the Wolfram U Webinar Series "New in Wolfram Language 13". Perhaps we'll meet again in the Study Groups? It will be a busy January and February...All the best and thanks for your great contributions here.Hakan Attachments:
Posted 5 months ago
 Great thanks, Hakan. Thank you for being our Forty Niner! Will probably see you in one of the classes u've mentioned. I am choosing between QC and Crypto. What do you think? Which on would be more conceptual rather than (mostly) package-learning? Is their content available? Best. Michael
Posted 5 months ago
 Hi Michael.(I first thought that 'Forty-Niner' is a baseball/football term, and had to google it. :-) Thanks.)What I understand, the Quantum Study Group is a study group, but not a course (https://www.bigmarker.com/series/daily-study-group-quantum-computation-framework/series_details). The Framework is presented here: https://community.wolfram.com/groups/-/m/t/2416125 . It's is over my head for the moment, and that is why I would like to learn more.OTOH, the Cryptography study group (https://www.bigmarker.com/series/daily-study-group-intro-to-cryptography/series_details?utm_bmcr_source=wolfram-u ) is both a Study Group and a course and seems to be more conceptual - it's over 3 weeks - and with much Wolfram Language coding :-). Here is a Wolfram Technology Conference talk on the Cryptograhy course: https://www.youtube.com/watch?v=MATcvnsfSS8 .
Posted 5 months ago
 Thanks again, Hakan. - very valuable information. Frankly, I did not even recognize difference between the study group and the course. Hope to TTYL in the classes to come.
Posted 5 months ago
 Suggestion. I think, next time you should include a derivation of the equations used by Ernest Rutherford and his colleagues to describe the processes of radioactive birth and decay. Why? A graduate of your course, unaware of Rutherford's work is likely to explain / model a birth and decay process by the method of integration by parts or alternatively by some expression including the Heaviside step function. See: Rutherford E et al (1951) Radiations from Radioactive Substances. (Chapter I pages: 1 - 37) https://archive.org/details/radiationsfromra0000ruth_a2v4 I learned a lot. I enjoyed this course very much. Thanks. Attachments:
Posted 5 months ago
 Thanks for a really great and interesting Study Group and Course! I've learned a lot on this fascinating topic and ordered the Boyce & DiPrima book ("Elementary Differential Equations", the 2017 version) that Devendra talked about, so I can continue to learn more.Again, thanks!
Posted 5 months ago
 I hope that everyone will stay hooked for a while. There are still some interesting questions to discuss. In addition, it will be painful to part with a group of such wonderful and productive individuals. I am not well familiar with the social media, but there should be a way to stay in touch (?) Thanks to Wolfram Research and the Wolfarm-U team for making this possible.
Posted 5 months ago
 Hi, A request to the course organizers: Could you also please upload the video lessons of Wednesday, December 8 and Thursday, December 9 in the daily study group page https://www.bigmarker.com/series/daily-study-group-intro-to-differential-equations/seriesdetails?utmbmcr_source=moocpageThank you in advance.
Posted 5 months ago
 Hi Artin, I confirmed that both day's recordings are available on the BigMarker study group page. If you still continue to face issues watching these, please email us at wolfram-u@wolfram.com.
Posted 5 months ago
 Thank you Abrita. Maybe it was my mistake that I had not seen the videos.
Posted 5 months ago
 I am checking out the next study group "Quantum Computation Framework" I have just installed Mathematica 13 so I can run this paclet.I have attempted both methods to run the paclet below but get an error PacletInstall::notavail: No paclet named Wolfram__QuantumFramework is available for download from any currently enabled paclet sites. 
Posted 5 months ago
 There appear to be a number of problems with Quiz 8. For instance, it marks me incorrect when I say the matrix {{0,4},{0,0}} is not diagonalizable but DiagonalizableMatrixQ[{{0, 4}, {0, 0}}] yields false.
Posted 5 months ago
 Unwanted DashesHi! I hope you are still around :) I have a strange issue trying to plot solutions of equation. To demonstrate it, I use a plot of Sin[]: Plot[Evaluate[Table[Sin[v s], {v, 0.1, 0.7, 0.15}]], {s, 0, 15}, PlotStyle -> Table[{Hue[1 - 0.15 i], Thickness[0.01]}, {i, 0, 4}], AspectRatio -> 0.75, PlotRange -> {{0, 15}, {-1.2, 1.2}}, Frame -> True, GridLines -> Automatic]; How/why do the dashes appear and how to get rid of them? Thanks!
Posted 5 months ago
 Hi Michael.When I run this program (in v13.0 and v12.3.1) I don't see any dashes. Would you mind showing it in an image?/Hakan
Posted 5 months ago
 Hi, Hakan! Great to hear from you! I almost sure that it did not happen before. In the list plots it can be addressed by Joined->True., but here it does not work. Thank you! M Attachments:
Posted 5 months ago
 Ah, now I see what you mean. Have you tried to reset the session, e.g. with Quit[]? What version of Mathematica do you use?/Hakan
Posted 5 months ago
 Thank you, Hakan. Yes, I checked all the "usual suspects". And the version is 12.3.1. My actual question is how to control the Plot (line) style to set up the output exactly as desirable. There is definitely a way to do so. With mutual effort it will be discovered!Thank you again. Mo
Posted 5 months ago
 Hi Michael,If you are trying to reproduce the image in the plot_dashes.nb notebook you need to add a dash specification to the PlotStyle. Something like dashStyles = {Dashing[Medium], Dashing[{0, Small}], Dashing[{0, Small, Small, Small}], Dashing[{0.03, 0.01}], Dashing[{0.02, 0.01}]} Plot[Evaluate[Table[Sin[v s], {v, 0.1, 0.7, 0.15}]], {s, 0, 15}, PlotStyle -> Table[{dashStyles[[i + 1]], Hue[1 - 0.15 i], Thickness[0.01]}, {i, 0, 4}], AspectRatio -> 0.75, PlotRange -> {{0, 15}, {-1.2, 1.2}}, Frame -> True, GridLines -> Automatic] You will have to experiment with the Dashing to exactly reproduce the plot.
Posted 5 months ago
 Thank you, Rohit.You wrote: "If you are trying to reproduce the image in the plot_dashes.nb notebook ..." No, I do not. Sorry for ambiguity. I am solving a Diff. Eq. and making a plot of its solutions. In that plot, I define thickness and color of the lines, say PlotStyle-> {{Red, Thick},{Blue,Thick}}. In my plot, however, In addition to the required features, lines contain dots and dashes which are unwanted. To demonstrate, I sent a plot of Sin[] functions, which are also supposed to be just colored, but also contain unwanted dashes. I do not know how to to get rid of dashes.Thank you. M
Posted 4 months ago
 Perhaps we can help more if you show the complete code, including the differential equation./Hakan
Posted 4 months ago
 That was my original plan. However, after discovering that it has nothing to do with the code and shows up with a set of simplest function, I posted the simple version. Hope things are going well with you, and we will meet again soon.Best. Mo