Function can be tricky (2)
⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ f ( x ) f ( 2 0 1 6 ) f ( 1 ) f ( 0 ) = k 1 x 2 0 1 6 + k 2 x 2 0 1 5 + ⋯ + k 2 0 1 5 x 2 + k 2 0 1 6 x + k 2 0 1 7 = 2 0 1 6 = 2 0 1 6 = 2 0 1 6
If f ( x ) is defined that f ( a + b ) = f ( a b ) , where a and b for any real numbers .
find ( k 1 k 2 k 3 … k 2 0 1 6 k 2 0 1 7 ) + ( k 1 + k 2 + k 3 + ⋯ + k 2 0 1 6 + k 2 0 1 7 ) .
The answer is 2016.
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1 solution
Uhh, I'm not quite sure whether we can safely state that f ( x ) = 2 0 1 6 . Infinitely many polynomial functions satisfy the given conditions, not only f ( x ) = 2 0 1 6 .
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Thanks for your comment , I added a statement : f ( x ) is defined that f ( a + b ) = f ( a b ) , where a and b for any real numbers . This problem would be better . Then f ( x ) = f ( x + 0 ) = f ( x × 0 ) = f ( 0 ) for any x . f ( x ) = 2 0 1 6 satisfies the statement only .
By some theorem (I forgot), the general polynomial P(x)=an^x+a(n-1)^(x-1)...+a0 where an, a(n-1), are "a sub n, a sun n-1" can have MORE than n roots if 0=an=a(n-1)= a(n-2)...=a0.
In this case, we see from Tommy's analysis that f(x) is equal for all x's. With the fact that f(x) has more than 2017 roots, we know that that a potential candidate for f(x) is f(x) = 2016. Solve for 0 and find that f(x)-2016 =0. f(x) -2016 is degree 2017 and notice that for any x, it it is equal to zero. Using the aforementioned theorem above, f(x) -2016 is FOR SURE equal to 0. So f(x) =2016. And then the rest may proceed...
f ( x ) = f ( x + 0 ) = f ( x × 0 ) = f ( 0 ) for any x
⇒ f ( x ) = 2 0 1 6
⇒ k 1 = k 2 = k 3 = ⋯ = k 2 0 1 5 = k 2 0 1 6 = 0 , k 2 0 1 7 = 2 0 1 6
( k 1 k 2 k 3 … k 2 0 1 6 k 2 0 1 7 ) + ( k 1 + k 2 + k 3 + ⋯ + k 2 0 1 6 + k 2 0 1 7 )
= 0 + 2 0 1 6
= 2 0 1 6