Maple 2020 Questions and Posts

These are Posts and Questions associated with the product, Maple 2020

I want to solve the equation x^(1/x) =1.2
solve(x^(1/x)=1.2)

The output is 1.257734541376526421
How do i get the other solution, numerically.

Is it possible to restrict or stipulate the domain to look for roots over a specific interval, some rootfinding command over the real numbers.
I also tried solve(x^(1/x)=1.2,AllSolutions,explicit)
Wolfram immediately gives me the second solution. https://snipboard.io/Nqt74K.jpg https://snipboard.io/Nqt74K.jpg

Hi

I am trying to use Explore command to model the CSA  of a trough.

So I want to  be able to change the variables: length of the walls (l:, slider 0..20) and the width (w:slider 0..20) and the angle (theta: slider 0..Pi/2)

Want to be able to see the diagram change and also display the area using the formula [output (15)] on the graph

Here is my attempt.

trough.mw

How to change the distance (spacing) between 2D Input cells and 2D Output cells so that this procedure can be inserted into a style sheet and thus applied to multiple documents?

Regards

Oliveira

Run these codes

with(inttrans)

y := t -> invlaplace(exp(-s)/s, s, t)

The following gives correct answer

y(t - 1)

The following gives junk answer

y(t - 1.1)

and I cannot plot added shifted signals in time domain if their shifts are not integer numbers. Maple goes into infinite calculation loop.

In the worksheet below is an equation for non-Euclidean lines. How can Maple plot these?

NonEuclideanLines.mw

I want to know how to remove an error message from the last line of using Maple.  I must have made some kind of error, but I want to move on but I can't get past the error message.  How can I remove the error message.  I have tried using "restart" but nothing happened.  I'm stuck until I clear this error message from the last line of my worksheet.  Can you help please?

Thanks

Dave

PS: as you can guess I'm pretty new to Maple.

Hello,

I have a problem regarding the physics package, which concerns, more specifically, the Define command.

It seems to be the case that it provides the wrong answer, and blatantly so, and I can't see what I am doing wrong for the life of me.

I want to build a fast way of doing coordinate transformations in general relativity, for which the basis vectors are called E[mu,~nu]. XX[~alpha] is the old coordinates expressed in terms of the new (in this case, the spherical coordinates are expressed in terms of cylindrical ones, i.e. [t,r, theta, phi]->[t, rho, phi,z].) As can be seen, only the first component in XX depends on time, but E contains weird elements (look at, for example, E[1,~2].) But when I do the calculation manually, it returns the expected results.

What am I doing wrong? I really appreciate any help you can provide.

Benzema

This worksheet creates geodesics in the Poincare disk by transformation of a series of circles of diminishing radii in the complex plane.

The intersections of the geodesics are meant to create the first few pentagonal uniform tilings in the Poincare disk.

I do not know the mathematically correct way to create such a display, so the radii of the circles are only a trial and error approximation.

What Maple code will provide the radii of the complex circles which produce an accurate uniform pentagonal tiling?

Is there a better overall strategy for producing uniform tilings of the Poincare disk? 

HyperbolicTiling.mw

I've been trying to implement numeric integration by using different methods of Int, for example _d01ajc, but when I use it, the following error comes up:

W := -16*(x - 1/2)*x*(t + 2*w)*(ln((m^2 + w*x*(-1 + x))/m^2)*ln(-1/(w*(t + w)*x^2 - w*(t + w)*x + m^2*t)) + ln(-x*w^2*(-1 + x))*ln((x^2 - x)*w + m^2) + ln(1/m^2)*ln((t + w)*(-1 + x)) + ln(-1/(x*w))*ln(m^2) + dilog(x*w^2*(-1 + x)/(w*(t + w)*x^2 - w*(t + w)*x + m^2*t)) - dilog(x*w*(-1 + x)*(t + w)/(w*(t + w)*x^2 - w*(t + w)*x + m^2*t)))/t^3 + 8*(x - 1/2)*((t + w)*x - t/2)*(-t*(w*x^2 + m^2 - w*x)*(t + w)*ln((x^2 - x)*w + m^2) + w*(x*w*(-1 + x)*(t + w)*ln((t + w)/w) + m^2*t*ln(m^2)))*(2*w*x + t)/(w*(t + w)*(w*(t + w)*x^2 - w*(t + w)*x + m^2*t)*t^3*x) + (16*x^2 - 8*x)/(t*w*(-1 + x)) + (2*t*x + 2*w*x - t - 2*w)*(2*w*x + t - 2*w)*(1 - 2*x)*ln((t*x + (-1 + x)*w)/((-1 + x)*w))/(((-1 + x)^2*w^2 + w*t*x*(-1 + x) + m^2*t)*t^2) + (-1 + 2*x)*(2*m^2 + w*x - w)*(2*w*x^2 + 2*m^2 - 3*w*x + w)*ln(m^2/(m^2 + w*x*(-1 + x)))/(w^2*(-1 + x)^2*((-1 + x)^2*w^2 + w*t*x*(-1 + x) + m^2*t)) + (-1 + 2*x)*(2*m^2 + w*x - w)*(2*w*x^2 + 2*m^2 - 3*w*x + w)*ln(m^2/(m^2 + w*x*(-1 + x)))/(w^2*(-1 + x)^2*(w*(t + w)*x^2 - w*(t + w)*x + m^2*t)) + (2*w*x + t)*((t + w)*x - t/2)*(x - 1/2)*ln(w^4/(t + w)^4)/(t^2*(w*(t + w)*x^2 - w*(t + w)*x + m^2*t)) - 8*(x - 1/2)*((-2 + 2*x)*w + t)*((-1 + x)*w + (x - 1/2)*t)*((t*x + (-1 + x)*w)*w^2*(-1 + x)^2*ln((t*x + (-1 + x)*w)/((-1 + x)*w)) + (-(t*x + (-1 + x)*w)*((x^2 - x)*w + m^2)*ln((x^2 - x)*w + m^2) + ln(m^2)*m^2*w*(-1 + x))*t)/((t*x + (-1 + x)*w)*w*(-1 + x)*t^3*((-1 + x)^2*w^2 + w*t*x*(-1 + x) + m^2*t)) + 16*(x - 1/2)*(ln(1/m^2)*ln((t*x + (-1 + x)*w)*(-1 + x)*w/((-1 + x)^2*w^2 + w*t*x*(-1 + x) + m^2*t)) + ln((x^2 - x)*w + m^2)*ln(1/((-1 + x)^2*w^2 + w*t*x*(-1 + x) + m^2*t)) + ln((-1 + x)^2*w^2)*ln((x^2 - x)*w + m^2) - dilog((t*x + (-1 + x)*w)*(-1 + x)*w/((-1 + x)^2*w^2 + w*t*x*(-1 + x) + m^2*t)) + dilog((-1 + x)^2*w^2/((-1 + x)^2*w^2 + w*t*x*(-1 + x) + m^2*t)))*(t*x + 2*(-1 + x)*w)/t^3

f3 := (mm, ss, tt, ww) -> Int(eval(W, [s = ss, t = tt, w = ww, m = mm]), x = 0 .. 1, _d01ajc)

A3 := evalf(f3(1, 3, -2, -1))

Error, (in evalf/int) number expected for float[8] parameter, got 0.+0.*I

What's wrong?

restart; with(LinearAlgebra); with(plots); with(SolveTools); eq1 := m1*m2*x^4+(c1*m2+c2*m1+c2*m2)*x^3+(c1*c2+k1*m2+k2*m1+k2*m2)*x^2+(c1*k2+c2*k1)*x+k1*k2

m1*m2*x^4+(c1*m2+c2*m1+c2*m2)*x^3+(c1*c2+k1*m2+k2*m1+k2*m2)*x^2+(c1*k2+c2*k1)*x+k1*k2

(1)

m1 := 2;
m2 := 4;
c1 := 1;
c2 := 0;
k1 := 3;
k2 := 2;

2

 

4

 

1

 

0

 

3

 

2

(2)

eq1

8*x^4+4*x^3+24*x^2+2*x+6

(3)

``

sol1 := evalf(solve(8*x^4+4*x^3+24*x^2+2*x+6 = 0))

-0.2293604281e-1+.5263956354*I, -.2270639572+1.627879522*I, -.2270639572-1.627879522*I, -0.2293604281e-1-.5263956354*I

(4)

assign(sol1)

Error, invalid left hand side in assignment

 

NULL

``

Download complex_numbers.mw

Can I specify a unit for a symbol using the Units package, so that when I substitute a value for that symbol into an expression, the units for the value are already known?

Hello,

Problem description

1) Define a function in terms of variables and parameters,

2) FInd the optimal solution

3) Plot the optimal solution in terms of a single parameter

 

Solution

What I have done is very basic since I haven't use Maple for many years. Here is an example

1) f(x;a,b,c):= ax^2+bx+c; (I don't knpw how to tell Maple that {a,b,c} is a set of parameters)

2) solve(f(x)=0,x); (I could use something like solve(f(x)=0,x,'parametric','real') but I am not interested in so detailed solution)

3) Here is my main problem. I want to save the first solution of x in optx(a;b,c) and the second in opttx(a;b,c) but I don't know how to do it (again, here a is my variable and b,c my parameters)

4) I also don't know hot to plot optx(a) as a increases, whereas the values of {b,c}={e.g., 2,3}

 

I would appreciate any resources/guidance 

Hello

I purchased the license of Maple 2019 and I am using the same version up to now. I am wondering if I download the latest version, can I use the former license? When I click on "check for updates" there is nothing for updating but when I want to run a Maple file written in the latest version, I face an error. 

Hello colleagues

I use Maple 2020 and I try to create Histogram of Poisson with a mean 200 [ Histogram(Sample(Poisson(200),10000))]. but I get an incorrect result in the form of a histogram as uniform distribution. But In my Maple 2015 I get a correct result.what is the problem?

Hello, 

I use Syrup to solve for the transfer function of a circuit, and in particular, the voltage at a given node.   If I use the "trans" option, the solution generates the expected result, in the expected format.  Syrup solves the differential equations and the resulting time domain equation is as expected.

If I then use the 'AC'  option, I get the result in the s-domain.  I then solve for the step response, and take the inverse Laplace transform.  The solution is given in terms of sinh and cosh.   

If I plot the voltage in either solution as a function of time, the plots are identical.  So, the sinh/cosh solution must be somehow simplifiable so that it "looks" like the time domain solution I get from the transient analysis.
 

restart

with(inttrans)

with(Syrup)NULL

with(DynamicSystems)

interface('displayprecision' = 8)

with(plots)

NULL

assum := R1::real, C1::real, R2::real, C2::real, R1 > 0, R2 > 0, C1 > 0, C2 > 0, t::realvin::real, vin > 0

vin::real, 0 < vin

(1)

ckt := [vin(5), Rm(1.0*10^7), Cm(1.8*10^(-11), ic = 0), `&//`(R1(9.*10^6), C1(1.5*10^(-11), ic = 0)), `&//`(R2(1.*10^6), C2(1.5*10^(-11), ic = 0))]

[vin(5), Rm(10000000.0), Cm(0.1800000000e-10, ic = 0), `&//`(R1(9000000.), C1(0.1500000000e-10, ic = 0)), `&//`(R2(1000000.), C2(0.1500000000e-10, ic = 0))]

(2)

NULL

NULL

ckt1 := [vin(0), Rm(0.10e8), Cm(0.18e-10, ic = 2.5), `&//`(R1(0.9e7), C1(0.15e-10, ic = 2.25)), `&//`(R2(0.1e7), C2(0.15e-10, ic = .25))]

[vin(0), Rm(0.10e8), Cm(0.18e-10, ic = 2.5), `&//`(R1(0.9e7), C1(0.15e-10, ic = 2.25)), `&//`(R2(0.1e7), C2(0.15e-10, ic = .25))]

(3)

Draw(ckt)

Solution doing Transient Analysis

deqs, rest := Solve(ckt, 'tran', 'returnall')

sol1 := dsolve(deqs):

Reassign rest to include the solution.

rest := eval(rest,sol1):

Compute the voltage across the capacitor Cm.

tmp1:=simplify(eval(v[Cm](t), sol1));

(1/32164)*(-415*8041^(1/2)-40205)*exp((100000/459)*(-119+8041^(1/2))*t)+5/2+(1/32164)*(415*8041^(1/2)-40205)*exp(-(100000/459)*(119+8041^(1/2))*t)

(4)

evalf[5]((1/32164)*(-415*8041^(1/2)-40205)*exp((100000/459)*(-119+8041^(1/2))*t)+5/2+(1/32164)*(415*8041^(1/2)-40205)*exp(-(100000/459)*(119+8041^(1/2))*t))

-2.4070*exp(-6389.4*t)+2.5000-0.92993e-1*exp(-45461.*t)

(5)

solve(tmp1=2.5*(1-exp(-1)),t)

p10 := plot(tmp1, t = 0 .. 0.00050000, color = 'blue', linestyle = 'dash')

p10 := plot(tmp1, t = 0 .. 0.5e-3, color = 'blue', linestyle = 'dash')

 

Solution from AC analysis:

sol, rest := Solve(ckt, 'returnall')

vnode2 := eval(v[2], sol)

1000000*(27*s+1000000)/(1377*s^2+71400000*s+400000000000)

(7)

1000000*(27*s+1000000)/(1377*s^2+71400000*s+400000000000)

(8)

``

NULLNULL

Impulse Response

temp1 := invlaplace(vnode2, s, t)

(1000000/24123)*exp(-(700000/27)*t)*(3*8041^(1/2)*sinh((100000/459)*t*8041^(1/2))+473*cosh((100000/459)*t*8041^(1/2)))

(9)

"(->)"

41.454*exp(-25926.*t)*(269.02*sinh(19536.*t)+473.*cosh(19536.*t))

(10)

Step Response

temp2 := invlaplace(vnode2/s, s, t)

5/2-(5/16082)*exp(-(700000/27)*t)*(83*8041^(1/2)*sinh((100000/459)*t*8041^(1/2))+8041*cosh((100000/459)*t*8041^(1/2)))

(11)

p1 := plot({temp2, (1-exp(-1))*2.5}, t = 0 .. 0.5e-3)

temp3 := solve(temp2 = (1-exp(-1))*2.5, t)

0.1505875131e-3

(12)

p2 := plot([temp3, temp3], [0, 2.5])

display({p1, p10, p2})

 
   

``

What is going on?  Where are the sinh and cosh terms in the solution coming from and why wont it simplify so that it looks like the other solution if they are identical?  Is there something special I need to do when using the invlaplace function?

Thanks!

Download help_maple.mw

 

 

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