When I write

evalf(Pi);

I get 3.14159...

But, when I want to visualize the value "e", after writing

evalf(e);

I get e .

Why I cannot see the digits of "e"?

Did I instal not properly my Matlab14.0?

Hi,

Just a simple question:

How do I unassume (remove the assume) all the variables in my code in the easiest way?

Thanks a lot for your help.

In a loop (variables set to some calculation) when maple produces an error how can I set it to a value of zero?

I want to find the first positive and first negative root of a bessel function

j2 := (x, a) -> k(x)^2*a^2*(diff((diff(1000000000*sin(1/1000000000*k(x)*a)/(k(x)*a), x))/(k(x)*a), x))/(k(x)*a)

I couldn't use the built in bessel function with two variables, so I had to enter it by hand.

I want to find the values of x where j2 is zero as a function of a:

f := (a) -> RootOf(j2(x, a), x, 0 .. 2)

plot(f(a), a = 2 .. 2.25, y = -3 .. 3)

Hey folks. I have a massively complicated equation I'm trying to solve and the internal memory can't take it.

Here is the problem.

Let f(x) := 1 - lambda*x^2

alpha := 1/f(1)

By solving the following equation,

alpha*f(f(x/alpha)) = f(x) + O(x^4)

We can determine a value of lambda and hence alpha (for which the real value is roughly -2.5 and the calculated value here is roughly -2.7).

hello everybody

please heeeelp

In the name of God

Dear friends

Hi

I have a question about releasing a part of memory occupied by some lists, sequences and matrices, for example matrix "M" or list "L".

If we use the following commands to allocate amount of memory for a matrix or list:

M:=Matrix(2000,2000,float[8]);L:=[seq(equ[i],i=1..10000)]:

where equ[i...

hello everybody can you help me please

how can i canvert this list [1,2,3,4,5,6] to a number 123456

hi ,friends

now ,i get a matrix of symbolic expression,it's difficulty to find the simplest forms using the common command as like simplify or combine

p := Matrix(4, 4, {(1, 1) = (1/2)*exp((1/2)*J[1]-(1/2)*J[2]+(1/2)*J[3])+(1/2)*exp(-(1/2)*J[1]+(1/2)*J[2]+(1/2)*J[3]), (1, 2) = 0, (1, 3) = 0, (1, 4) = (1/2)*exp((1/2)*J[1]-(1/2)*J[2]+(1/2)*J[3])-(1/2)*exp(-(1/2)*J[1]+(1/2)*J[2]+(1/2)*J[3]), (2, 1) = 0, (2, 2) = (1/2)*exp(-(1/2)*J[1...

Hello friends,

When I solve the optimization problem with restriction (below):

NLPSolve( ln(x1) +ln(x2+5), {x1 +x2 <=4}, assume = nonnegative); [-Float(infinity), [x1 = HFloat(0.0), x2 = HFloat(0.8333333333333334)]]

I get the expression "-Float(infinity)"..

What do Float(infinity) and HFloat(0.833333) mean?

Thanks for the attention,

Hello everybody!

I'm working on a statistical problem. I need to plot confidence regions for parameters in the model, but I don't know what I should do. I tried to get some programs from the internet, but either they don't work or I don't understand details of them to change for my work. Can someone introduce a good program to me?, or does anyone have another useful idea for me?

Thanks a lot!

I meet a problem in the use physcis package.

for example,

restart; with(Physics); Setup(mathematicalnotation = true):

...

I am analyzing a system that was at steady state and then subjected to a step change. I have the steady state modeled with a system of ODEs.

I am going to model the step change with a system of PDE's. The system will be discretized in space and time (z,t). I know how to setup one side of the boundary conditions (0,t) for all variables, but the other (i.e A(z,0)) I would like to use the results of the steady analysis. Does anybody have some ideas on...

I would like to do the following

A[1,1],A[2,2],A[3,3]:=seq(i+2,i=1..3)

using seq on the left-hand-side. I know I have done it in the past but don't get it this time. End up with error messages. Not sure what am I missing.

thanks a lot

Dimitrios

I typed x:=a*cos(t)^3

For a while it was handling this correctly.

But then it started replacing the multiplication sign by ~ every time it used this or anything similar.

Something has changed the status of the a .

If I use b instead of a, it works fine.

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