Maple Questions and Posts

How to Plot a Rational Algebraic Function...

Asked by: paulmcquad 0

2 hours ago

0 1

Hi,

I'm trying to plot:

y = x(x^4 - 10*x^2 + 39)/30

I keep getting a what looks like a straight line. it should be a curved line.

I got it to work on :

y = x(x^4 - 10*x^2 + 39)/30

https://www.mathway.com/Algebra

help me to plot time series? in x=0 and all the t...

Asked by: permanoon123 25

3 hours ago

0 0

mapleprime.mw

Nonlinear least squares...

Asked by: Jak 5

4 hours ago

1 2

Hi,

I'm wanting to use nonlinear least squares fittting to minimise the error between two curves.

One curve, f₁, I know the parameters for and the other, f2 , I do not, which is what I am wanting to find through this methood.

f₁=(2*x)/(3+5*x)

f₂=a₁*(2-exp(-a₂*x))

Any help would be muchly appricated, thank you in advance!

StepProperties returning endefined, StepResponsePl...

Asked by: gkokovidis 2113

6 hours ago

0 2

I am trying to recreate an example, See attached worksheet below. I can't figure out what I am doing wrong here. I have tried using the command StepProperties on other discrete transfer functions, without any problems. Having a function that plots the Step Response of a system, Continuous or Discrete, should be built in.

Trying to recreate step response of discrete time system as in this example below:

https://www.mathworks.com/help/control/examples/creating-discrete-time-models.html

>	restart:with(DynamicSystems):interface(version);

(1)

>

>	sysz:=TransferFunction((z-1)/(z^2-1.85*z+0.9),discrete,sampletime=0.1);

"module() ... end module"

(2)

>	propz:=StepProperties(sysz);

(3)

>

Download DiscreteStepResponseError.mw

Regards,

Georgios

PracticeSheet command...

Asked by: jalal 45

6 hours ago

0 0

I explore the "PracticeSheet" command of the "Basic" package, and I want to build two sheets on the Discriminant of polynomial ( send degree) and tangent to curves ( Derivative application). Do you have any ideas?

Many thanks!

DrillerConceptMaple.mwDrillerConceptMaple.mw

Defects in PolynomialIdeals package....

Asked by: Seonhwa Kim 15

13 hours ago

3 0

Hello,

I am Seonhwa Kim, a mathematical researcher in Korea. Recently, I have extensively used Maple to compute character varieties of 3-manifolds. Several months ago, I obtained some strange results in Maple which implies a contradiction in theory. I have been struggling with these issues since it is usually about enormous polynomial systems. Eventually, I could figure out that the issues are caused by a defect in Maple and were able to construct a minimal working example to produce wrong computations in Maple. I am writing this post to report them.

This is mainly about the PolynomialIdeal package. Along with the documentation in Maple, If an ideal J is radical, PrimeDecomposition and PrimaryDecomposition should have the same result. But, as we see the following, the result of PrimeDecomposition and PrimaryDecomposition are different although J is a radical ideal.

The problem seems to be that the PrimaryDecomposition command in Maple sometimes produces incorrect results.

We can compute the primary decomposition of J by hand, It should be <x> and <y, x-1>.

I double-checked this by the other software;Macauley2, Singular, and Magma, for example, you can see it as follows.

Secondly, not only for PrimaryDecomposition but also PrimeDecomposition may produce an incorrect result.

Here is a minimal working example.

Maple tells us a compatible result of prime and primary decomposition of a radical ideal J.

But the first component of J, < b-1, c-a+1 >, contains the third component < a, b-1, c+1 >.

It contradicts with the definition of Primary decomposition. So the correct answer should be < b - 1, c - a +1 >.

I also checked that Macaluey2, Singular and Magma. They all say that my hand computation is correct. as follows.

I have used Maple 2017 by the license of my institute (Korea Institute for Advanced Study).

When I noticed these defects, I thought it would be fixed in the newest Maple version.

So, I have tried my examples by Maple2019 free trial, but It also has the same problem.

I guess this problem is not reported or recognized yet.

I hope this problem will be fixed as soon as possible.

Thank you for attention.

Sincerely,

Seonhwa

How add legend to a dataplot? ...

Asked by: emendes 145

Yesterday at 9:53 AM

0 3

Hello

I need to add a legend to a figure using dataplot (I am not even sure that is the right option). In what follows I show what I did.

dataplot([28,28,28],[.6481496576, .648149657615473, .6512873548],style='point',colorscheme=["Blue","Orange","Red"],	
labels = ["k", "y(k)"], legend = ["10-digit precision", "15-digit precision", "Floating-point iteration"] ,legendstyle = [font = ["HELVETICA", 9], location = right]);

The outcome is

As can be seen some parts of the legend are missing.

How can I get this right? Do I have other ways to do the same thing?

Many thanks

Ed

Adjusting the range of the axes in 3D plot...

Asked by: student_md 210

Yesterday at 7:04 AM

0 4

Suppose that we have a 3D plot in Maple like above. How can we adjust the ranges of the axes as follows?

The ranges of axes x and axes t must be decimal like (0.1, 0.2 ,...,1)
The range of axes z must be scientific as figure above.

Maple sometimes automatically does it but sometimes doesn't it. I don't understand the logic behind it.

Change output `<=` to `>=`...

Asked by: Fzen 50

Yesterday at 2:44 AM

0 6

Hi all, why maple output alway change from `> =` to `<=` ? example

How to config to `>=`. Thank you very much

Why Minus an Integras Sign Places parentheses Arou...

Asked by: mapleatha 105

Yesterday at 8:43 PM

0 3

I am using an integal sign using Int. If there is a minus sign in front of it, then

I get the expression -(Int ... dx). Why is the parenthesis there? Can we avoind it?

Thank you!

mapleatha

How to get the real and imaginary parts of complex...

Asked by: Alger 489

Yesterday at 7:38 PM

0 3

Hi all,

How to get the real and imaginary parts of this complex expression.

Thank you in advance

real_imag_parts.mw

A problem for function or expression evaluation in...

Asked by: Alireza1982 0

Yesterday at 7:37 PM

0 1

Hello every one
I have a question!
My code starts with assuming the time variable (t) to be specified in a an interval for each time that the for loop executes as follows:

discontinuity := [0.403e-8, 0.45e-8, 0.478e-8, 0.55e-8];
for j from 1 to 3 do
assume(discontinuity[j]<=t,t<discontinuity[j+1]);

After that the code runs and calculate every thing. For the second time, I mean for J=2, it does not work properly.
T[1] is the result of its first running (for j=1):
T[1]:=3.000023586*10^(-6)*exp(-4977.085344*t~)-1.325122648*10^(-6)*exp(-4.015800624*10^9*t~)
When it wants to evaluate this expression at specific time such as discontinuity[2] it cannot evaluate it. I tried to use unapply command to consider it as a function of t but it did not work. Here is the results:
AA:=simplify(eval(T[1], t = discontinuity[1]));
This is its result: AA:=3.000023586*10^(-6)*exp(-4977.085344*t~)-1.325122648*10^(-6)*exp(-4.015800624*10^9*t~)
Without unapply command: AA:=T[1]
Please answer my question.
Thank you so much

Maybe a bug in MAPLE 2015's Explore ?...

Asked by: mmcdara 892

February 20 2020

1 6

Hi,

Here is the minimal situation that reveals which that could be a (little) bug in the MAPLE 2015 version of Explore.

In the attached file y is a list of numbers and val(r) a procedure that just print the value the rth element of y.
Changing the value of r is done with Explore (of course of no interest at all).
If I define the parameter r as a list ( r=[1..numelems(y)] ), only the value of y[1] is displayed: changing the value of r generates an error.
This doesn(t happen if r is defined as a slider ( r=1..numelems(r) ).

I discovered than the initial instance of Explore defines r as an integer while all the others (due to a change of r) define r as a string.
In the last command of the attached file you will see hjow I have circumvent this problem.

Is it a bug in Explore or does it exist some way to force the values of r to the implicit type they have in r=[$1..numelems(y)] ?

TIA

Download explore.mw

how to write this ODE in Maple?...

Asked by: nm 2781

February 20 2020

0 6

I'd like to solve this ODE

But not sure how to write it in Maple. In Mathematica, this is what I do

ClearAll[F, x, y, t];
ode = D[F[x[t], y[t]], x[t]]*D[x[t], t] + D[F[x[t], y[t]], y[t]]*D[y[t], t] == 0;
DSolve[ode, F[x[t], y[t]], {x[t], y[t]}]

and it gives

In Maple, I tried

ode:=diff(F(x(t),y(t)),x(t))*diff(x(t),t)+diff(F(x(t),y(t)),y(t))*diff(y(t),t)=0

Error, invalid input: diff received x(t), which is not valid for its 2nd argument

ode:=D[1](F)(x(t),y(t))*diff(x(t),t)+D[2](F)(x(t),y(t))*diff(y(t),t)=0;
dsolve(ode,F(x(t),y(t)))

Error, (in dsolve) expected the indeterminate function as, say, F(x) where x is of type "name" - and also cannot be a procedure name. Received: [F(x(t), y(t))]

What is the correct way to write this in Maple?

How do I sort a formula in the following form...

Asked by: ing123 0

February 20 2020

0 0

Hello,

I want to sort the formulae to Psi and Beta, but I don't know how it works. I have tried it with sort, simplify, isolate, but that isn't what I'm searching.

It should looks like the simplier formula in the picture.

ab := (diff(Psii(t), t, t))*J-l[f]*(F[s, f, l]+F[s, f, r])+l[r]*(F[s, r, l]+F[s, r, r])-(1/2)*b[r]*(-F[s, r, l]*delta[l]+F[s, r, r]*delta[r]) = 0;
/ d / d \\
|--- |--- Psii(t)|| J - l[f] (F[s, f, l] + F[s, f, r])
\ dt \ dt //

+ l[r] (F[s, r, l] + F[s, r, r])

1
- - b[r] (-F[s, r, l] delta[l] + F[s, r, r] delta[r]) = 0
2
bc := (diff(betaa(t), t, t))*m*v*betaa(t)+F[s, r, l]*delta[l]+F[s, r, r]*delta[r]-(diff(Psii(t), t)) = 0;
/ d / d \\
|--- |--- betaa(t)|| m v betaa(t) + F[s, r, l] delta[l]
\ dt \ dt //

/ d \
+ F[s, r, r] delta[r] - |--- Psii(t)| = 0
\ dt /
cd := (diff(betaa(t), t))*m*v+F[s, r, l]+F[s, r, r]+F[s, f, l]+F[s, f, r]-(diff(Psii(t), t)) = 0;
/ d \
|--- betaa(t)| m v + F[s, r, l] + F[s, r, r] + F[s, f, l]
\ dt /

/ d \
+ F[s, f, r] - |--- Psii(t)| = 0
\ dt /
F[s, f, l] := c[fl]*alpha[fl];
c[fl] alpha[fl]
F[s, f, r] := c[fr]*alpha[fr];
c[fr] alpha[fr]
F[s, r, l] := c[rl]*alpha[rl];
c[rl] alpha[rl]
F[s, r, r] := c[rr]*alpha[rr];
c[rr] alpha[rr]
alpha[fl] := (-v*betaa-l[f]*(diff(Psii(t), t)))/(-v+(1/2)*b[f]*(diff(Psii(t), t)));
/ d \
-v betaa - l[f] |--- Psii(t)|
\ dt /
-----------------------------
1 / d \
-v + - b[f] |--- Psii(t)|
2 \ dt /
alpha[fr] := (-v*betaa-l[f]*(diff(Psii(t), t)))/(v-(1/2)*b[f]*(diff(Psii(t), t)));
/ d \
-v betaa - l[f] |--- Psii(t)|
\ dt /
-----------------------------
1 / d \
v - - b[f] |--- Psii(t)|
2 \ dt /
alpha[rl] := delta[l]+(-v*betaa+l[r]*(diff(Psii(t), t)))/(-v+(1/2)*b[r]*(diff(Psii(t), t)));
/ d \
-v betaa + l[r] |--- Psii(t)|
\ dt /
delta[l] + -----------------------------
1 / d \
-v + - b[r] |--- Psii(t)|
2 \ dt /
alpha[rr] := delta[r]+(-v*betaa+l[r]*(diff(Psii(t), t)))/(-v-(1/2)*b[r]*(diff(Psii(t), t)));
/ d \
-v betaa + l[r] |--- Psii(t)|
\ dt /
delta[r] + -----------------------------
1 / d \
-v - - b[r] |--- Psii(t)|
2 \ dt /

ab;
/
|
/ d / d \\ |
|--- |--- Psii(t)|| J - l[f] |
\ dt \ dt // |
|
\

/ / d \\
c[fl] |-v betaa - l[f] |--- Psii(t)||
\ \ dt //
-------------------------------------
1 / d \
-v + - b[f] |--- Psii(t)|
2 \ dt /

/ / d \\\ / /
c[fr] |-v betaa - l[f] |--- Psii(t)||| | |
\ \ dt //| | |
+ -------------------------------------| + l[r] |c[rl] |delta[
1 / d \ | | |
v - - b[f] |--- Psii(t)| | | |
2 \ dt / / \ \

/ d \\
-v betaa + l[r] |--- Psii(t)||
\ dt /|
l] + -----------------------------|
1 / d \ |
-v + - b[r] |--- Psii(t)| |
2 \ dt / /

/ / d \\\ /
| -v betaa + l[r] |--- Psii(t)||| |
| \ dt /|| 1 |
+ c[rr] |delta[r] + -----------------------------|| - - b[r] |
| 1 / d \ || 2 |
| -v - - b[r] |--- Psii(t)| || |
\ 2 \ dt / // \
/ / d \\
| -v betaa + l[r] |--- Psii(t)||
| \ dt /|
-c[rl] |delta[l] + -----------------------------| delta[l]
| 1 / d \ |
| -v + - b[r] |--- Psii(t)| |
\ 2 \ dt / /

/ / d \\ \
| -v betaa + l[r] |--- Psii(t)|| |
| \ dt /| |
+ c[rr] |delta[r] + -----------------------------| delta[r]| =
| 1 / d \ | |
| -v - - b[r] |--- Psii(t)| | |
\ 2 \ dt / / /

0
bc;
/ d / d \\
|--- |--- betaa(t)|| m v betaa(t)
\ dt \ dt //

/ / d \\
| -v betaa + l[r] |--- Psii(t)||
| \ dt /|
+ c[rl] |delta[l] + -----------------------------| delta[l]
| 1 / d \ |
| -v + - b[r] |--- Psii(t)| |
\ 2 \ dt / /

/ / d \\
| -v betaa + l[r] |--- Psii(t)||
| \ dt /|
+ c[rr] |delta[r] + -----------------------------| delta[r]
| 1 / d \ |
| -v - - b[r] |--- Psii(t)| |
\ 2 \ dt / /

/ d \
- |--- Psii(t)| = 0
\ dt /
cd;
/ d \
|--- betaa(t)| m v
\ dt /

/ / d \\
| -v betaa + l[r] |--- Psii(t)||
| \ dt /|
+ c[rl] |delta[l] + -----------------------------|
| 1 / d \ |
| -v + - b[r] |--- Psii(t)| |
\ 2 \ dt / /

/ / d \\
| -v betaa + l[r] |--- Psii(t)||
| \ dt /|
+ c[rr] |delta[r] + -----------------------------|
| 1 / d \ |
| -v - - b[r] |--- Psii(t)| |
\ 2 \ dt / /

/ / d \\
c[fl] |-v betaa - l[f] |--- Psii(t)||
\ \ dt //
+ -------------------------------------
1 / d \
-v + - b[f] |--- Psii(t)|
2 \ dt /

/ / d \\
c[fr] |-v betaa - l[f] |--- Psii(t)||
\ \ dt // / d \
+ ------------------------------------- - |--- Psii(t)| = 0
1 / d \ \ dt /
v - - b[f] |--- Psii(t)|
2 \ dt /

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