Hi Mapleprimes,

I have tried to search solutions online without luck. My problem is this. Suppose I have three equations:

f(x,y,z)

g(x,y,z)

h(x,y,z)

I would solve using solve({f,g,h},{x,y,z}) which would give me solutions x*, y* and z*. I need to assign the solutions to use them in subsequent computations. I would like to impose non-negativity constraints, however. Hence, the solution of x is max(x*,0). Obviously if a non-negativity constraint binds, this affects the solutions of y* and z*. I would like to assign the solutions taking this into account. How would you propose I do this? Bear in mind that the solution of x* might be a-b where the relative magnitudes of paramenters a and b are not given. Further down in my code, I would like to assign values to parameters and then let Maple give me the solutions with the non-negativity constraints. I hope I am making sense.

Thanks very much in advance.

Best,

Christian

I want to use Burr distribution and hence I am creating custom probability distribution. The pdf of Burr is not piecewise. It gives error.

burrpdf := alpha*k*((x-gamma)/beta)^(alpha-1)/beta(1+((x-gamma)/beta)^alpha)^(k+1)

burr_distribution := Statistics[Distribution](PDF = (alpha > 0, beta > 0, gamma > 0, k > 0), burrpdf)

Error, (in Statistics:-Distribution) invalid input: too many and/or wrong type of arguments passed to NewDistribution;

I've been using Maple 2018 only a few days now, was mostly using Maple 2016 and never had any issues with returning an output (at least one that wasnt my fault), but with 2018 I've seen a few times where a simple task would return an output of "__SELECTION" and then my input. No idea what this is or why its happening and all I really need to know is how to prevent it from happening. If its something with my preferences or settings that needs to be addressed, thats fine, but otherwise this is getting to be a real pain. See attached 

I have a function involving sinh(x)/cosh(x) to evaluate (see attached SKS77.mw and S77.pdf). I got different values depending on whether or not the function is algebraically simplified. What's the right way to evaluate the exponential function in this case? Thanks

Let

f(x)= 2x^2 -2, x >= 0

Find d/dx f^-1(x)|(subscript x=0.)

Note that f(1)=0.

Use d/dx f^-1(x)= 1/(f'[f^-1(x)])

Hello

I was trying to introduce vector r_vec that has 3 components in x,y,z

I've attached my file, I've 2 questions here

first, why isn't the vector shown in as r_vec = () ei + () ej + () ek instead appears as a column vector

second, why doesn't it accept differentating

thank you
 

Download tst1.mwtst1.mw

I have the following pertubation problem I want to use maple to expand for me.

We have epsilon := eps;

x(t,eps):= x_{-1}(t)/eps+x_0(t)+x_1(t)*eps

z(t,eps):=z_{-1}(t)/eps+z_0(t)+z_1(t)*eps

I want to expand a Taylor series of the following function upto some arbitray order of eps, i.e O(eps^3) or higher (depending on my mood :-)), around t=0, f(x(t,eps),z(t,eps),cos(t/eps),sin(t/eps)).

Anyone has any suggestion how to use maple 2017.3 to do this?

Thanks!

Hello Guys, I hope you are all fine. I have been struggling with creating an animation of the points (x,y) in maple. I have tried this example 
L := [[1, 1], [3, 2], [3.4, 6], [5, 3, 7], [3, 7, 9, 1], [2, 6, 8, 4, 5]];
animate(PointPlot, [L[trunc(t)]], t = 1 .. 6, frames = 150)
but in my case it shows two points at different location means it takes x and y seperate value and showed it on 1 and 2 on x axis but i want to animate it as the location of point. Please help me. 
Thank you in anticipation.

I have several functional equations in equally many unknown functions of at least two variables, plus parameters.  ("collect" works just for single equations, right?)

I know that for certain parameter ranges, all the functions involved will be quadratic, and I know some coefficients are zero.  That gives me some  coefficients to determine.  I want to

1. specify the functional equations [done in a very primitive low-tech way in the attachment, using atomic variables rather than indices ... have I done correctly?!?] 
2. get Maple to collect coefficients (the K's and the L's in the attachment; the variables are (y,z))
3. get Maple to state an equation system these coefficients have to satisfy (these will unfortunately be coupled quadratics)
4. get Maple to solve that equation system if possible, and if not: to tell me when (= for what parameter values, parameters being the "remaining letters" in the attachment) I have specified enough coefficients
5. in case of a solution, get Maple to tell me which coefficients are real and positive (for those that are solution of quadratic eq's: whether a positive solution exists)

Phew. I am still a complete newbie. Edit: Attachment link: STcoeff2match.mw where the equations themselves are EQ0, EQ1 and EQ2 at the bottom. Copying and pasting them, they look like this (download STcoeff2pastedEQs.mw)

Hi guys, I am trying to solve a system of differential equations, I have done the hand written calculations and I know the answer however I need to put it in a maple code for a generic system which I will work on over time. Here is what I have so far, 

restart;

eqn[1]:=-1/8*D[4](a)(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t))=0;

eqn[2]:=-1/8*D[5](a)(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t))=0;

eqn[3]:=-1/8*D[6](a)(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t))=0;

eqn[4]:=-1/8*D[7](a)(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t))=0;

eqn[5]:=-1/8*D[8](a)(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t))=0;

eqn[6]:=-1/8*D[9](a)(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t))=0;

eqn[7]:=-1/8*D[10](a)(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t))=0;

eqn[8]:=-1/8*D[11](a)(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t))-1/2=0;

eqn[9]:=-1/8*D[12](a)(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t))=0;

dsolve({seq(eqn[i],i=1..9)},a(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t)));

Then I get an error return which says:

Error, (in dsolve) too many arguments; some or all of the following are wrong: [{u(x, y, t)}, a(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(diff(u(x, y, t), x), x), diff(diff(u(x, y, t), x), y), diff(diff(u(x, y, t), t), x), diff(diff(u(x, y, t), t), y), diff(diff(u(x, y, t), t), t))].

I know that if I replace u(x,y,t) with a dummy variable U, and its derivative with Ux,Uy,... and so on then it will work, but I need the function u(x,y,t) to be part of the solution.

I know the solution should give me:

a(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(diff(u(x, y, t), x), x), diff(diff(u(x, y, t), x), y), diff(diff(u(x, y, t), t), x), diff(diff(u(x, y, t), t), y), diff(diff(u(x, y, t), t), t)) = -4*diff(u(x,y,t),x,x) + F(x,y,t),

where F(x,y,t) is the constant function.

Please any help would be great!!
 

The summation takes too long time. Please help me
 

Hello people in mapleprimes,

This time question is a sequel of the previous one:
https://www.mapleprimes.com/questions/224346-Batch-File-And-Directory
, where I could obtain the file with the output.
Maple has been terminated after exhausting the output file.

But, I want  to do the next calculation while using the result in the output file.
For that, I think it might be a good way to have maple calculate a mpl file again, after having added 
new expressions to the original mpl file. But, copying and pasting expressions seen in the output file to the original mpl file with my hand, seems a little messy.

So, what I want to know is whether there isn't a good way for continuing calculations one after another,
while having maple termination after each calculation.

I will be very glad if you will give me an answer to this question.

 Thanks in advance.

I wish to calculate connection, curvature, Ricci curvature etc. for a

Riemannian metric given as follows: there is an orthogonal frame of vector

fields with stipulated Lie bracket relations between them. The frame is

orthogonal but not orthonormal, and the lengths of its vector fields are functions

of a single function on the manifold. Given these metric values on the frame and the

Lie bracket relations, the covariant derivatives are in principle computable from the

Koszul formula, hence connection and curvature are all determined.

When I try to define the metric using a dual coframe in ATLAS's Metric

routine, it allows me to define it but claims there is not actual curvature.

From the help it seems the coframes used in this routine are always given

as differentials of coordinates. Is there a way to get the metric via the data

given above without putting in by hand all the different Koszul formulas etc.?

I am considering a Fourier series

$cos (\alpha x) = \frac{1}{2}a_0 + \sum_{k=1}^{\infty}a_k cos(kx)$ for x between -pi and pi.

I have also shown using a different Fourier series that cos (\alpha x) has an alternative representation:

\frac{cos(\alpha x)}{\sin \alpha \pi} = \frac{1}{\pi \alpha} (1 + \frac{(\alpha \ pi)^2}{6} - \frac{\alpha x^2}{2 \pi} + \frac{2*\alpha^3}{\pi}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^2(k^2 - \alpha^2}*cos(kx)$.

To show that the second representation is a better approximation, I need to find the number of terms for this series and the original Fourier series needed for there to be a difference of 10^{-3} from the exact value of cos(\alpha \pi), assuming that \alpha = 0.75.  Could someone advise how I might do this?

Hi!

Is it possible to get step by step solution for:

- Laplace

- taylor

- fourie

Thanks for answers!