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Hi,

I am trying to simplify the expression s as given below. (I am not sure why it comes up with all the vector caclulus notation in it but it should display okay when you enter it)

Because of the presence of the exponential imaginary fucntions I thought evalc might be useful but when I use it I get a huge expression with csgn appearing in it. To my knowledge csgn appears when assumptions are not correctly specified - is this so? I can't see any assumption...

Maple evaluates

> sum(1/z^2, z = 1 .. infinity)

to (1/6)*Pi^2. What do I do when sumation is over all even positive integers? Is there any  closed-form symbolic formula for this case?

Above is the minimal example. What I need is to compute 

> sum(2*binomial(m-1, k-1)*binomial(n-1, k-1)/binomial(m+n, m), k = 1 .. infinity)

where k belongs to a set of only even positive integers, not any posint. The second expression evaluates to 2*m*n/((m+n-1...

Test.mw

If I use a binary {0,1} linear program the cardinal constraint ie control the
number of 1's (long positions in portfolio) works beautiful. See attached worksheet.

The problem starts when I convert the problem into a constrained integer {-1,0,1}
linear problem where -1= short position, 0=no position and 1=long position.

In the first example when we add the constraint  add(w[i], i = 1 .. NC) = 4

I can solve a [0,1] linear programming problem by using:

LPSolve(Ob, con, assume = binary);


I managed to solve [-1,1] linear programming problem by using the transformation
2*w[i]-1 ie   2*0-1=-1 and 2*1-1=1 as follows:

LPSolve(add(PP[i]*(2*w[i]-1), i = 1 .. NC)-add(Draw[i]*(2*w[i]-1), i = 1 .. NC), {add(w[i], i = 1 .. NC) = PS}, maximize = true, assume = binary);


Now my question is how can I solve a [-1,0,1] problem?
I tried to use:

 x := proc (i::integer)

      return i^2;

end proc:

sum(x(m),m=1..3);

Error, invalid input: x expects its 1st argument, i, to be of type integer, but received i.

 

If I want to reserve the ::integer declaration, How to solve this problem. I know if there is no

I have the simplified version of what I want to do:

restart: with(LinearAlgebra):
d:=4:

product(MatrixExponential(I*Matrix(d,{(l,l)=1})*A[l,l]),l=1..d):

 

But I get the following error:

Error, (in Matrix) integer indices required for Matrix

Help appreciated.

 

 

 

Hi e-friends,

I want to minimize a function subject to a set of S restrictions.

The restrictions are related to matrices V, W, X and Y:

 

V = [v1, .., vS]  order L x S

W = [w1, .., wS] order L x S

 

X = [x1, .., xS]  order LxS

Y = [y11,.. yS] order L x S

 

How may I write in MAPLE in compact form  the following S inequalities (for any arbitrary integers L and S)?. ...

Hey peeps

 

I'm quite new to maple, and to this site, so I hope I do this properly.

I'm trying to make a procedure, that througout the procedure divides 2 numbers. But I need it to round down to the nearest integers, at all time, as I need an integer as output. I have searched quite a lot, and haven't found anything that helps me to do this.

The procedure is the following:

walla := proc (x) local y, z, w;

y := x;

i have an image and would like to perform a canny edge detection. one of the first steps is the convolution of the image with a smoothed derivative filter, i.e. a gaussian. the problem is, that i don`t know, how to convolve my collection of discrete pixels (the image) with a 2D gaussian. how do i get an integer-valued convolution kernel that approximates a Gaussian with a variable sigma?

I use the interactive plot builder (at least Maple14) and want to define a parameter to be an integer.

How do I carry this into execution?

 

If a function is differentiable at some point c of its domain, then it is also continuous at c. However here we extend the notion of differentiability to be valid for individual points on the real number line, specifically positive integers.

 f(n)=(-1)^n* n^(1/n)

THEOREM MRBK 8.0

f=f' / (I*Pi+(1-ln(n))/n^2)| n ∈ {1,2,3,...}

By THEOREM MRBK 4.0, When n is in the set of (positive) integers the derivative of f is exactly I*Pi*f+(1-ln(n))*f/n^2.

So f' = I*Pi*f+(1-ln(n))*f/n^2| n ∈ {1,2,3,...}

Solving for f, we have the following:

f' = I*Pi*f+(1-ln(n))*f/n^2

f' = f*(I*Pi+(1-ln(n))/n^2)

f=f' / (I*Pi+(1-ln(n))/n^2)

 

For more on this click here (W/A).

find integer...

June 02 2010 yasser 10

how can i found the integer number after any divident using maple commands

like

(13/4)

 

Am I missing some justification for this last one?

> zip(`/`,Array([3]),Array([9]));
                                     [1/3]
 
> zip(`/`,Array([3],datatype=integer[4]),Array([9]));
                                     [1/3]
 
> zip(`/`,Array([3]),Array([9],datatype=integer[4]));
                                     [1/3]
 
> zip(`/`,Array([3],datatype=integer[4]),Array([9],datatype=integer[4]));
                            [0.333333333333333315]

In the list of how irrational, rational, and from small to large integers to sort.Can not calculate irrational. Can you help? Thanks

Since the MRB constant is an alternating sum of positive integers to their own roots, f(n)=(-1)^n* n^(1/n); a thorough understanding of the changes in f, as n changes, is important.
In this blog we will begin to explore the derivative of f at integer values of n, and as n-> infinity. I am not sure weather this will help us in computing more digits of the MRB constant since we already know so many,

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