# 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) ?

Apr 20, 2016

Solution

#### Explanation:

$f \left(x\right) = {x}^{4} + a {x}^{3} + b {x}^{2} + c x + d$
Given
$f \left(1\right) = 1 \implies$
$a + b + c + d + 1 = 1$
$a + b + c + d = 0$

$f \left(2\right) = 5 \implies$
$8 a + 4 b + 2 c + d + 16 = 5$
$8 a + 4 b + 2 c + d = - 11$

$f \left(3\right) = 19 \implies$
$27 a + 9 b + 3 c + d = - 62$

$f \left(6\right) = 665 \implies$
$216 a + 36 b + 6 c + d = - 631$
solving gives the solution as $a = - 2.9333 , b = - 2.4 , c = 16.7333 , d = - 11.4$

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

May 9, 2016

$f \left(4\right) + f \left(5\right) = 356$

#### Explanation:

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

Define:

${a}_{n} = {n}^{4} - f \left(n\right) = - \left(a {n}^{3} + b {n}^{2} + c n + d\right)$

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 $0$.

$\left\{\begin{matrix}{a}_{1} = {1}^{4} - f \left(1\right) = 1 - 1 = 0 \\ {a}_{2} = {2}^{4} - f \left(2\right) = 16 - 5 = 11 \\ {a}_{3} = {3}^{4} - f \left(3\right) = 81 - 19 = 62 \\ {a}_{4} = {4}^{4} - f \left(4\right) = 256 - f \left(4\right) \\ {a}_{5} = {5}^{4} - f \left(5\right) = 625 - f \left(5\right) \\ {a}_{6} = {6}^{4} - f \left(6\right) = 1296 - 665 = 631\end{matrix}\right.$

So our initial sequence is:

$0 , 11 , 62 , {a}_{4} , {a}_{5} , 631$

The sequence of differences between successive pairs of terms is:

$11 , 51 , \left({a}_{4} - 62\right) , \left({a}_{5} - {a}_{4}\right) , \left(631 - {a}_{5}\right)$

The sequence of differences of those differences is:

$40 , \left({a}_{4} - 113\right) , \left({a}_{5} - 2 {a}_{4} + 62\right) , \left({a}_{4} - 2 {a}_{5} + 631\right)$

The sequence of differences of those differences is:

$\left({a}_{4} - 153\right) , \left({a}_{5} - 3 {a}_{4} + 175\right) , \left(3 {a}_{4} - 3 {a}_{5} + 569\right)$

The sequence of differences of those differences is:

$\left({a}_{5} - 4 {a}_{4} + 328\right) , \left(6 {a}_{4} - 4 {a}_{5} + 394\right)$

So:

$\left\{\begin{matrix}{a}_{5} - 4 {a}_{4} + 328 = 0 \\ 6 {a}_{4} - 4 {a}_{5} + 394 = 0\end{matrix}\right.$

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

$3 {a}_{4} - 2 {a}_{5} + 197 = 0$

Adding this to the first equation, we get:

$- {a}_{4} - {a}_{5} + 525 = 0$

Hence:

$f \left(4\right) + f \left(5\right) = \left(256 - {a}_{4}\right) + \left(625 - {a}_{5}\right) = - {a}_{4} - {a}_{5} + 525 + 356 = 356$