Huge but small polynomial!
x 2 0 1 6 − 1 = 0 If the roots of the above equation are 1 , α 1 , α 2 , α 3 , … , α 2 0 1 5 , find the value of ( 1 − α 1 ) ( 1 − α 2 ) ( 1 − α 3 ) ⋯ ( 1 − α 2 0 1 5 ) .
The answer is 2016.
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3 solutions
Assuming that ( 1 − α 1 ) is just a factor once, then we can simplify the polynomial in question as
x − 1 x 2 0 1 6 − 1 .
which we cannot substitute x = 1 yet because it gives us an indeterminate form.
By L'Hopital's rule, we get the derivative of both numerator and denominator and get
1 2 0 1 6 x 2 0 1 5
So we can now substitute x and get 2016.
(x-1)(x-a1)(x-a2)..(x-a2015)=x^2016-1 Differentiate both sides and put x=1 That will give you (1-a1)(1-a2)...(1-a2015)=2016
x 2 0 1 6 − 1 = ( x − 1 ) ( x − α 1 ) ( x − α 2 ) . . . . . . ( x − α 2 0 1 5 ) = ( x − 1 ) ( x 2 0 1 5 + x 2 0 1 4 + x 2 0 1 3 + . . . + x + 1 ) ⇒ ( x − α 1 ) ( x − α 2 ) . . . . . . ( x − α 2 0 1 5 ) = ( x 2 0 1 5 + x 2 0 1 4 + x 2 0 1 3 + . . . + x + 1 ) ⇒ ( 1 − α 1 ) ( 1 − α 2 ) . . . . . . ( 1 − α 2 0 1 5 ) = ( 1 2 0 1 5 + 1 2 0 1 4 + 1 2 0 1 3 + . . . + 1 1 + 1 ) = 2 0 1 6