Don't answer 13

Algebra Level 4

f ( 2013 ) = 1 f ( 2014 ) = 1 f ( 2015 ) = 2 f ( 2016 ) = 3 f ( 2017 ) = 5 f ( 2018 ) = 8 \large{\begin{aligned} f(2013)&=1\\ f(2014)&=1\\ f(2015)&=2\\ f(2016)&=3\\ f(2017)&=5\\ f(2018)&=8 \end{aligned}}

Let f ( x ) f(x) be a 5 th 5^\text{th} degree polynomial satisfying the equations above. Find f ( 2019 ) f(2019) .


The answer is 8.

Hamza A by Hamza A , 19, USA

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1 solution

We have 6 values provided for a 5 t h 5^{th} degree polynomial so we can use the method of finite differences.

n f ( n ) D 1 ( n ) D 2 ( n ) D 3 ( n ) D 4 ( n ) D 5 ( n ) 2013 1 0 1 − 1 2 − 3 2014 1 1 0 1 − 1 − 3 2015 2 1 1 0 − 4 2016 3 2 1 − 4 2017 5 3 − 3 2018 8 0 2019 8 \begin{array} {lrrrr} n & f(n) & D_1(n) & D_2(n) & D_3 (n) & D_4(n) & D_5(n) \\ 2013 & 1 & 0 & 1 & -1 & 2 & -3 \\ 2014 & 1 & 1 & 0 & 1 & -1 & -3 \\ 2015 & 2 & 1 & 1 & 0 & -4 \\ 2016 & 3 & 2 & 1 & -4 \\ 2017 & 5 & 3 & -3 \\ 2018 & 8 & 0 \\ 2019 & 8 \\ \end{array}

This is because we know that the 5 t h 5^{th} difference is constant and comes out to be -3 .

Hence f ( 2019 ) = 8 \boxed{f(2019)=8}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Wow, and I used Lagrange interpolation. SMH. Nice solution.

James Wilson - 5 months ago

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