Question

Given `f(x) = x^4+ax^3+bx^2+cx+d` with `f(1) = 1`, `f(2) = 5`, `f(3) = 19` and `f(6) = 665`, what is the value of `f(4)+f(5)` ?

Answer 1

Solution

Explanation:

`f(x) = x^4+ax^3+bx^2+cx+d`
Given
`f(1) = 1 =>`
`a+b+c+d+1= 1`
`a+b+c+d=0`

`f(2) = 5 =>`
`8a+4b+2c+d+16 = 5`
`8a+4b+2c+d = -11`

`f(3) = 19 =>`
`27a+9b+3c+d = -62`

`f(6) = 665 =>`
`216a+36b+6c+d = -631`
solving gives the solution as `a = -2.9333, b=-2.4, c = 16.7333, d = -11.4`

`f(4)+f(5) = 256+64a+16b+4c+d+625+125a+25b+5c+d`
`= 881+189a+41b+9c+2d`
Substitution will give you
`f(4)+f(5) = 356`

Answer 2

`f(4)+f(5) = 356`

Explanation:

Since we're dealing with `f(1), f(2),.., f(6)` this is like matching a finite sequence with a polynomial.

Define:

`a_n = n^4 - f(n) = -(an^3+bn^2+cn+d)`

Since this is a cubic, then starting with `a_1, a_2,..,a_6`, if we construct the sequence of differences, then of differences of differences, etc. we will after `4` steps get to a sequence whose terms should both be `0`.

`{ (a_1 = 1^4 - f(1) = 1-1 = 0), (a_2 = 2^4 - f(2) = 16 - 5 = 11), (a_3 = 3^4 - f(3) = 81 - 19 = 62), (a_4 = 4^4 - f(4) = 256 - f(4)), (a_5 = 5^4 - f(5) = 625 - f(5)), (a_6 = 6^4 - f(6) = 1296 - 665 = 631) :}`

So our initial sequence is:

`0, 11, 62, a_4, a_5, 631`

The sequence of differences between successive pairs of terms is:

`11, 51, (a_4 - 62), (a_5 - a_4), (631 - a_5)`

The sequence of differences of those differences is:

`40, (a_4 - 113), (a_5 - 2a_4 + 62), (a_4 - 2a_5 + 631)`

The sequence of differences of those differences is:

`(a_4 - 153), (a_5 - 3a_4 + 175), (3a_4 - 3a_5 + 569)`

The sequence of differences of those differences is:

`(a_5 - 4a_4 + 328), (6a_4-4a_5+394)`

So:

`{ (a_5 - 4a_4 + 328 = 0), (6a_4 - 4a_5 + 394 = 0) :}`

Dividing the second of these equations by `2` we find:

`3a_4-2a_5+197 = 0`

Adding this to the first equation, we get:

`-a_4-a_5+525=0`

Hence:

`f(4)+f(5) = (256-a_4)+(625-a_5) = -a_4-a_5+525+356 = 356`