product of a series

Hi,

how should I write this production in matlab?

p(k)=(m1−m2)*(m2−m3)*,...*,(mN−2−mN−1)*(m99−m100), where k from 1 to 100.

thx

Answer 1

[Star Strider]

Numerically:

m = rand(1, 100);
m = rand(1, 100);                               % Row Vector
rm = reshape(m, 2, []);
p = prod(diff(rm,[],1))                         % Desired Result

EDIT — (21 Dec 2019 at 15:46)

Alternatively:

p = prod(-diff(rm,[],1))                         % Desired Result

in the event that the first row was supposed to be subtracted from the second, instead of the second being subtracted from the first.

[Marco Riani]

I think the solution Star provided (given a vector or 2n elements) computes

(x(2)-x(1))*((x(4)-x(3))*...*(x(2n)-x(2n-1))

For example suppose x is

x=[1 5 6 11 12 4] % Row Vector

rm = reshape(x, 2, []); gives

rm =

1 6 12

5 11 4

and

diff(rm,[],1)

ans =

4 5 -8

and the solution given by Star is the product of (x(2)-x(1))*((x(4)-x(3))*...*(x(2n)-x(2n-1))

Furthermore reshape(x, 2, []) assumes that x has a number of elements which is even.

In order to obtain

(x(1)-x(2))*((x(2)-x(3))*...*(x(n-1)-x(n))

assuming x is a row vector the correct solution (if I am not mistaken) is

prod(diff(fliplr(x)))

In the example above, if x=[1 5 6 11 12 4]

fliplr(x) is

4 12 11 6 5 1

diff(fliplr(x)) is

8 -1 -5 -1 -4

(x(n-1)-x(n))* .... ((x(2)-x(3))*(x(1)-x(2))*

Of course if x is a column vector, it is enough to replace fliplr with flipud.

The code is given below

x=[1 5 6 11 12 4]; % Row Vector
rm = reshape(x, 2, []);
% Solution given by Star
prod(diff(rm,[],1))   

% New solution in presence of a row vector 
prod(diff(fliplr(x)))

% New solution in presece of a column vector
x=[1 5 6 11 12 4]'; % Column Vector
prod(diff(flipud(x)))