## Line of best fit....

Hi Guys,

I will like to know how I can use Maple to plot a line of best fit given some data.

e.g for data like [x = 21.2, 24.7, 20.8, 20.8, 20.3] and [y = 16.6, 19.7,16.4,16.8,16.9].

Thanks.

## How to avoid the "no solution found" warning when ...

Morning all,

I repeatedly solve (command solve) a collection of systems of inequalies. Some of them can have no solution, but I am able to check if a solution has been found or not, and then take some decision about the system in question.

I have placed a few print commands at different critical locations within the loop where those systems are constructed and possibly solved.
Every time solve fails finding a solution it returns me a "no solution found" warning.
In order to keep my printings readable, could it be possible to avoid those warnings ?
Is there some "try & catch" like mechanism to manage warnings ?

## How to install the MAPLE package "SADE"?...

I can't understand how to install this package "SADE" in Maple 17. Secondly, I am new to Maple. So Please guide me the process to install this package and how to use it. Thanks!

## Difficult with Lambert W in implicitplot...

I'm trying plot  with implicitplot a expresion which involve Lambert W, but the result is very confuse, is it normal? Could i improve the result?, how?. Thank!

## solve with a variable or fsolve without variable...

I've encountered a very strange issue with Maple.

The result returns differently with solve and fsolve after/before a variable is given a certain value. See attachment.

The result comes from solve (with variable epsilon) returns value of the same variable with imaginary part while the fsolve returns the correct answer.

Now how can I achieve the same result as fsolve via solve?

Thanks!

Maple_Question_Solve_Fsolve.mw

Maple_Question_Solve_Fsolve.pdf  (exported PDF from Maple)

## Statistics Social - Cronbach - Pearson

Maple 2015

Application that allows us to measure the reliability of a group of data through a row and columns called cronbach alpha at the same time to measure the correlation of items through the pearson correlation of even and odd items. It can run on maple 18 to maple 2017. This will be useful when we are developing a thesis in the statistical part.

In Spanish

StatisticsSocialCronbachPearson.zip

Lenin Araujo Castillo

## How to get maple to simplify a trig expression...

Hi everyone,

I'm to simplify the expression cos(x)/sin(x) to cot(x). None of the "simplify" commands seem to work, I've tried assigning the expression to a name "a", then using the "simplify(a,trig)" command and it doesn't work either.

Anyone have any ideas on how I can tell maple to simplify this?

## Find equation of line which has minimum distance f...

I've got some points:

I have to find the (equation of) line which has minimum distance from these points but the distance formula that I have to use is:

I think we should settle with a for loop.

## Complicated integration from a simple function...

Hey guyz, I am in trouble with calculation attached integral. it is a simple function but a bit long. I can't solve it with maple, Do U have any idea?

intg.mw

## plotting problem...

hello, i just try to plot the relation between my outputs (u, and phat) with i from 0 to 10 , but i have aproblem any suggestions?

 (1)

 (2)

 (3)

 (4)

 (5)

 (6)

 (7)

 (8)

 (9)

 >
 > p:=Array(0..10):   p[0]:=0:   for i from 0 to 4 do       p[i+1]:=50*sin(3.14*(i+1)*(4)/0.6):       phat[i+1]:= p[i+1]+((7)*u[i])+((8)*u__dot[i])+((9)*u__doubledot[i]):       u[i+1]:= phat[i+1]/(5):       u__dot[i+1]:=(20*(u[i+1]-u[i]))-u__dot[i]:       u__doubledot[i+1]:= ((400*(u[i+1]-u[i]))-(40*u__dot[i])-(u__doubledot[i])): end do; for i from 5 to 9 do       p[i+1]:=0.0:       phat[i+1]:= p[i+1]+((7)*u[i])+((8)*u__dot[i])+((9)*u__doubledot[i]):       u[i+1]:= phat[i+1]/(5):       u__dot[i+1]:=(20*(u[i+1]-u[i]))-u__dot[i]:       u__doubledot[i+1]:= ((400*(u[i+1]-u[i]))-(40*u__dot[i])-(u__doubledot[i])):   end do;
 >
 >
 (10)

 > fd_table:=eval(seq[u(i),phat(i)],i=0..N);
 (11)

 >
 >
 >

## How do I solve a system with boundary conditions t...

Hello everyone,

I am having trouble trying to solve a system of differential equations. The modeling was made from the equilibrium equations of a pressure vessel. The set of equations is shown below:

As you see it is a set of two second-order partial differential equations. So, we need four boundary conditions. This one is the first. It means that the left end of the pressure vessel is fixed.

This one is the second boundary condition. It means that the right end of the pressure vessel is free.

This one is the third boundary condition. It means that the inner surface of the pressure vessel is subject to an external load:

At last, we have the fourth boundary condition. It means that the outer surface of the pressure vessel is free.

The first test I have been trying to do is the static case. In this case, the time terms (the right side of the two equations shown) is zero.

The maple commands that I am using are the following:

```restart; E := 200*10^9; nu := .33; G := E/(2*(1+nu)); RI := 0.254e-1; RO := 2*RI; p := proc (x) options operator, arrow; 50000000 end proc; sys := [E*(nu*(diff(v(x, r), x))/r+nu*(diff(diff(v(x, r), x), r))+(1-nu)*(diff(diff(u(x, r), x), x)))/(-2*nu^2-nu+1)+G*(diff(diff(u(x, r), r), r)+diff(diff(v(x, r), x), r)+(diff(u(x, r), r))/r+(diff(v(x, r), x))/r) = 0, E*((1-nu)*(diff(diff(v(x, r), r), r))+nu*(diff(diff(u(x, r), x), r))+(1-nu)*(diff(v(x, r), r))/r-(1-nu)*v(x, r)/r^2)/(-2*nu^2-nu+1)+G*(diff(diff(u(x, r), r), x)+diff(diff(v(x, r), x), x)) = 0]; BCs := {E*(nu*v(L, r)/r+nu*(D[2](v))(L, r)+(1-nu)*(D[1](u))(L, r))/(-2*nu^2-nu+1) = 0, E*(nu*v(x, RI)/RI+(1-nu)*(D[2](v))(x, RI)+nu*(D[1](u))(x, RI))/(-2*nu^2-nu+1) = -p(x), E*(nu*v(x, RO)/RO+(1-nu)*(D[2](v))(x, RO)+nu*(D[1](u))(x, RO))/(-2*nu^2-nu+1) = 0, u(0, r) = 0}

sol := pdsolve(sys, BCs, numeric)```

I am getting the following error:

Error, (in pdsolve/numeric/process_IBCs) initial/boundary conditions must depend upon exactly one of the independent variables: 0.1459531181e12*v(L, r)/r+0.1459531181e12*(D[2](v))(L, r)+0.2963290579e12*(D[1](u))(L, r) = 0

In this case, my boundary conditions do depend on more than one independent variable. How do I proceed?

Pedro Guaraldi

## Maple 2017: What to expect?...

Is there anyone who has seen maple 2017 provide some details about what new features are being introduced. Is there a platform where we can suggest what features we would like to be added or enhanced?

## Bad help and needless command in SignalProcessing

Maple

Let us consider the help to RectangleWindow SignalProcessing-RectangleWindow.pdf SignalProcessing-RectangleWindow.mw.
Let us execute the example, taking N:=4 (in order to display the outputs).

```with(SignalProcessing):
N := 4;
a := GenerateUniform(N, -1, 1);
Matrix(1, 4, [[.396167882718146, -.826878267806025, -0.908376742154361e-2, .324899681378156]])
RectangleWindow(a);
Vector[row](4, [.396167882718146, -.826878267806025, -0.908376742154361e-2, .324899681378156])
c := Array(1 .. N, 'datatype' = 'float'[8], 'order' = 'C_order'):
RectangleWindow(Array(1 .. N, 'fill' = 1, 'datatype' = 'float'[8], 'order' = 'C_order'), 'container' = c);
Vector[row](4, [1., 1., 1., 1.])
u := `~`[log](FFT(c)):
plots:-display(Array([plots:-listplot(Re(u)), plots:-listplot(Im(u))]));
```

We see an uncommented code which (intentionally or unintentionally) produces two empty plots.
The questions arise:

• What is the aim of the RectangleWindow command which does nothing
but the conversion of a Matrix(1,N,...) /Array(1..N,...) to a Vector[row](N,...)?
• Could such help be called friendly to Maple users?

There are many questions to Maplesoft and there are no answers from them: strategic silence.
RectangleWindow.mw

## Problem aplying the chain rule with dchange...

I have a problem using dchange when my variable depend on two (or more variables) and I would like to apply the chain rule.

For example, when I use the command

I would expect something like

But I get an error saying that the number of new variables and transformation equations must be the same.

Any idea how I could solve it?

Thanls a lot for your help.

## Float Infinity instead of integer...

Hi I have this code

``` psi:=proc(n,x);
(1/sqrt(sqrt(pi)*2^n*factorial(n))*exp(-x^2/2)*HermiteH(n,x))
end proc;
psi := proc(n, x) exp(-1/2*x^2)*HermiteH(n, x)/sqrt(sqrt(pi)*2^n*n!) end proc

psi2=proc(a,x);
psi(a,x):=(1/sqrt(sqrt(pi)*2^a*factorial(a))*exp(-x^2/2)*HermiteH(a,x))
end proc;
psi2 = (proc(a, x)
psi(a, x) := exp(-1/2*x^2)*HermiteH(a, x)/sqrt(sqrt(pi)*2^a*a!)

end proc)
for n from 0 to 2 do;
for a from 0 to 2 do;
result=proc(n,a);
result(n,a)=psi*psi2
end proc;
print(evalf(int(result(n,a),x=0..infinity)));
od;
od;```

it returns

Float(infinity) signum(result(0, 0))

Float(infinity) signum(result(0, 1))

Float(infinity) signum(result(0, 2))

Float(infinity) signum(result(1, 0))

Float(infinity) signum(result(1, 1))

Float(infinity) signum(result(1, 2))

Float(infinity) signum(result(2, 0))

Float(infinity) signum(result(2, 1))

Float(infinity) signum(result(2, 2))

I know the results for (0,0), (1,1) and (2,2) should be 1 and the rest should be 0.

Can anybody help fix this please

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