f:=sin((x-1)*(y+3))/(exp((x-1)^2+(y+3)^2)-1);
limit(f,{x=1,y=-3});

On running the above code, I don't get a result. I don't understand why, and what are some of the underlying math principles that I would have to use to evaluate the limit of f at (1,-3) if I were to solve this question on paper?

I need to find where the limit of the function:

{1-cos(x*y^3)}/(x^2+y^6)^(1+a) exists/does not exists for different values of a, given 0<=a<=1 and as (x,y)->(0,0).

I've tried writing a procedure that changes a in particular increments but that is clearly not the most efficient way. Are there any rules of limits that I should be using?

I've got a set

E:={(x,y,z): x^2+y^2=-2*z-x, z^2+y^2=1} and need to find points of E which have minimal or maximal distance from (0,0,0). I've set up the Lagrangian as F:=sqrt(x^2+y^2+z^2) + L1(x^2+y^2+2z+x)+L2(z^2+y^2-1)

and consequently obtained the equations:

x/sqrt(x^2+y^2+z^2) + 2*x*L1+L1=0

y/sqrt(x^2+y^2+z^2) + 2*y*L1+2*y*L2=0

z/sqrt(x^2+y^2+z^2)+2*L1+2*L2*z=0

for which I've set up
eqn1,eqn2,eqn3 as the three equations and vars:=x,y,z

and used solve() but I'm not getting the right answer( I need to first express x,y,z in terms of L1, L2 and then get values for L1 and L2 by substituting in the constraints and eventually get values of x,y,z.)

How should I implement that?

For example, given a 3d point p(x,y,z), with (x,y,z) as its coordinates. Then it is transformed by rotation and translation, as 

p'=R(p)*p+t(p), where R(p) is a 3x3 rotation matrix that is a matrix of functions of p, and t(p) is a 3x1 vector function of p. 

My question is how to derive dp'/dp (as a 3x3 matrix) using maple? 

To make it clear,I want to do it in a way that dp'/dp = ∂p'/∂p + ∂p'/∂R*∂R/∂p +  ∂p'/∂t*∂t/∂p

And I'd like to know each intermediate quantity, such as ∂p'/∂R, ∂R/∂p.

Anyone can help?

Thanks a lot. 

What Maple15 commands will display a Rouleaux tetrahedron or a Meissner tetrahedron?

Hi all,

I'm having a challenge solving an integration that has some error functions associated with it.

Description:

I am trying to find the Riemann midpoint of the integral. All the parameters are known except the u and du terms which came into the expression following a laplace transform.

the output I get from the above input is shown below, and as it can be seen, the u and du terms coupled with a complimentary error function still exist. I have tried the "simplify" function and it returns similar output expression.

Any help will really be appreciated.

Thanks :)

Download Maple_Worksheet.mw

Hi,

I'm using Maple 18 and none of my worksheets that I've developed on earlier versions work on it. I tried starting a new worksheet with a very simple command (attached) and even that's giving me problems. As you can see, I have no trouble plotting constant functions, but as soon as I put in a variable it breaks down. It won't solve equations either.

FYI I'm using an HP G72 laptop with Windows 7 and no installation issues came up.

Can anyone tell me what's going on here?

Thanks,

Tom

Hi all,

I'm trying to use logplot but it's very time consuming. Any suggestion please!

Cheers!

restart:with(plots):

S1:=x+x^2*Sum(t^(k*alpha+1)/GAMMA(k*alpha+2),k=0..infinity);

S2:=x+x^2*Sum(t^(k*alpha+1)/GAMMA(k*alpha+2),k=0..n);

alpha:=0.75:x:=0.5:n:=10:

logplot(abs(S1-S2),t=0..1);

So here is the issue: I have a 50 by 50 tridiagonal matrix. The entries in the first row, first column are -i*x and the last row last column is -i*x; these are along the main diagonal, where i is complex and x is a variable. Everything in between these two entries is 0. Above and below the main diagonal the entries are -1. My issue is that I have to find a conditon on x that makes the eigenvalues real. I am completely new to maple and have no programming experience.. Can someone show me how to this?

Using Maple 18, I solved for minimum and maximum price. Instead of using fsolve I wanna use procedure programming structure in order to get the same results. How can I do it?

min_sol := fsolve([bc_cond, slope_cond, x[G, 1] = w[aggr, 1]], {p = 0 .. 1, x[G, 1] = 0 .. w[aggr, 1], x[G, 2] = 0 .. w[aggr, 2]}); p_min := subs(min_sol, p); max_sol := fsolve([bc_cond, slope_cond, x[G, 2] = w[aggr, 2]], {p = 0 .. 1, x[G, 1] = 0 .. w[aggr, 1], x[G, 2] = 0 .. w[aggr, 2]}); p_max := subs(max_sol, p);
{p = 0.3857139820, x[G, 1] = 127.8000000, x[G, 2] = 38.99045418}
0.3857139820
{p = 0.8841007104, x[G, 1] = 44.30160890, x[G, 2] = 164.2000000}
0.8841007104

Hi,

I am trying to realize the following calculation in Maple.

$
  \left[\sum_{i=0}^n y_i(x) \partial_x^i , \sum_{j=0}^m z_j(x) \partial_x^j \right]  \\
=   \sum_{i=0}^n \sum_{j=0}^m \sum_{l=0}^i  \binom il y_i(x) \left( \partial_x^{i-l} z_j(x)\right) \partial_x^{l+j} \\
- \sum_{j=0}^m \sum_{i=0}^n \sum_{l=0}^j  \binom jl z_j(x) \left( \partial_x^{j-l} z_i(x)\right) \partial_x^{l+i} \ .

$

Is there a way to make maple understand d/dx as a differential opperator and calculate with it? When i for example try to calculate diff(d/dx, x) it should give me d^2/dx^2 as a result. Unfortunately i don't know how to realize this.

Basic problem is i don't know how to realize operator expressions in maple like for example:

f(x) d/dx      ( f(x) is a smooth function of x here )

where when applied to a function h(x) it should result in f(x) d/dx h(x) .

Is that possible?

Thank you very much in advance.

I am trying to expand out the terms  of equation 13.  The expand command causes the lhs to be zero?

Download Question_R12.mw

I want to know with what x,y, z,  function f is minimum, whereas function g is constant.

regards

Hi

I want to know with what x/y, z,  function f is minimum, whereas function g is constant.

regards

Hello every one,

Is any one knows how to solve the following inequality with assumptions that all parameters are real positive and k<1 and delta > c*alpha

(1/2)*((alpha*k^2-3*alpha*k-2*beta)*sqrt(delta^2*(k-1)*(k-2)*(c*alpha-delta)^2)-k*delta*(alpha*k^2-3*alpha*k+2*alpha-2*beta)*(c*alpha-delta))/(delta^2*(alpha*k^2-3*alpha*k-2*beta))<0

I tried the following code but it  dosn't make sense:

u:=(1/2)*((alpha*k^2-3*alpha*k-2*beta)*sqrt(delta^2*(k-1)*(k-2)*(c*alpha-delta)^2)-k*delta*(alpha*k^2-3*alpha*k+2*alpha-2*beta)*(c*alpha-delta))/(delta^2*(alpha*k^2-3*alpha*k-2*beta))

solve({u < 0,alpha > 0, beta > 0, c > 0, delta > 0, delta > c*alpha, k > 0, k < 1, })

In fact I want to know under which circumastances the above inequality is negative.

THX