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Verifying that a symbolic expression in two variables is 0.

asked 11 years ago

caffeinemachine gravatar image

I want to verify using SAGE that the expression:

(  n +1k) 2n 1( nk) 2n+( n k ) 2n 1+( n k1) 2n  1

is identically zero.

The SAGE script for this is:

binomial(n+1,k)2^(-n-1) - binomial(n,k)2^(-n) + binomial(n,k)2^(-n-1) + binomial(n,k-1)2^(-n-1)

I have tried using full_simplify() but it doesn't boil the expression down to zero.

Please help.

Thanks.

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answered 11 years ago

tmonteil gravatar image

updated 11 years ago

The reason is that your expression is not equal to zero !!!

If you multiply your expression by 2n+1, and simplify the two bionmials in the middle, you get (n+1k)(nk)+(nk1)

This can not be zero since (n+1k)+( nk1) is usually bigger than (nk) (unless k>n+1).

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Ah yes! My bad. The expression I was supposed to wrestle with was: binomial(n+1,k)2^(-n-1) - binomial(n,k)2^(-n) + binomial(n,k)2^(-n-1) - binomial(n,k-1)2^(-n-1) and yes sage is able to bring it down to zero. Thanks so much. I made a sign mistake. Really sorry. And thanks for your kind reply.

caffeinemachine gravatar imagecaffeinemachine ( 11 years ago )

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Asked: 11 years ago

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Last updated: May 24 '13