Sum of near infinite series?

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John Mitchell
John Mitchell el 22 de Oct. de 2019
Comentada: David Hill el 22 de Oct. de 2019

Having some trouble figuring out where to start with this problem:
Find the product of the first n (n=9999) terms of the following expression where n is an integer variable greater than one:
eq = 2/1 * 2/3 * 4/3 * 4/5 * 6/5 * 6/7 * 8/7 * 8/9 .... *n-1/n
Any help on figuring this out would be appreciated.

Respuestas (2)

David Hill
David Hill el 22 de Oct. de 2019
p=[];
for i=1:2:n+1
p=[p,(n+1)/n,(n+1)/(n+2)];
end
P=prod(p(1:n));
  2 comentarios
John Mitchell
John Mitchell el 22 de Oct. de 2019
Thank you for responding, but I forgot to mention that we can't use for loops in the problem, is there any way to do them without loops?
David Hill
David Hill el 22 de Oct. de 2019
a=2*[1:floor((n+1)/2);1:floor((n+1)/2)];
nu=a(:)';
d=[1,nu-1];
p=prod(nu(1:n)/d(1:n));

Iniciar sesión para comentar.


Stephan
Stephan el 22 de Oct. de 2019
Think about your row. You can rewrite it - always two elements are the same as n^2/(n^2-1).
With this knowledge read about cumprod. This should be enough to bring you to work the rest by yourself.

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