A Question on Polynomials, part 4

Since the $x$ -values are consecutive, we can just consider the triangle of differences: $\begin{array}{lcccccccc} \text{Row 0} & & 3 & & 3^2 & & 3^3 & & 3^4 \\ \text{Row 1} & & & 6 & & 18 & & 54 & \\ \text{Row 2} & & & & 12 & & 36 & & \\ \text{Row 3} & & & & & 24 & & & \end{array}$

And since the interpolating polynomial has degree $3$ , all the rows below Row 3 are zero. This gives the next value by working from the bottom and moving up the triangle $\begin{array}{lcccccccccc} \text{Row 0} & & 3 & & 3^2 & & 3^3 & & 3^4 & & \color{#3D99F6}{195}\\ \text{Row 1} & & & 6 & & 18 & & 54 & & \color{#3D99F6}{114} & \\ \text{Row 2} & & & & 12 & & 36 & & \color{#3D99F6}{60} & & \\ \text{Row 3} & & & & & 24 & & \color{#3D99F6}{24} & & & \\ \color{#3D99F6}{\text{Row 4}} & & & & & & \color{#3D99F6}{0} & & & & \end{array}$

which gives $f(5) = \boxed{195}$ .

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Clarification:

• To move down the triangle, at each step we use $\begin{array}{ccc} a & & b \\ & \color{#3D99F6}{b-a} & \end{array}$
• To move up the triangle, at each step we use $\begin{array}{ccc} a & & \color{#3D99F6}{a+b} \\ & b & \end{array}$
• We can also read off the polynomial from the left-most side of the triangle as $f(x) = \frac{3}{0!} + \frac{6}{1!}(x-1) + \frac{12}{2!}(x-1)(x-2) + \frac{24}{3!}(x-1)(x-2)(x-3)$ and we could do similarly even if the $x$ -values aren't consecutive, but then we'd have to use Newton's divided differences instead.

For variety, I am using matrix to solve the problem.

Let $f(x) = ax^3+bx^2+cx+d$ . Then we have:

$\begin{bmatrix} 1 & 1 & 1 & 1 \\ 8 & 4 & 2 & 1 \\ 27 & 9 & 3 & 1 \\ 64 & 16 & 4 & 1 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} = \begin{bmatrix} 3 \\ 9 \\ 27 \\ 81 \end{bmatrix}$

$\implies \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 8 & 4 & 2 & 1 \\ 27 & 9 & 3 & 1 \\ 64 & 16 & 4 & 1 \end{bmatrix}^{-1} \begin{bmatrix} 3 \\ 9 \\ 27 \\ 81 \end{bmatrix} = \begin{bmatrix} -\frac 16 & \frac 12 & -\frac 12 & \frac 16 \\ \frac 32 & -4 & \frac 72 & -1 \\ -\frac {13}3 & \frac {19}2 & -7 & \frac {11}6 \\ 4 & -6 & 4 & -1 \end{bmatrix} \begin{bmatrix} 3 \\ 9 \\ 27 \\ 81 \end{bmatrix} = \begin{bmatrix} 4 \\ -18 \\ 32 \\ -15 \end{bmatrix}$

Therefore, $f(x) = 4x^3 -18x^2 + 32x -15$ , $\implies f(5) = 4(125) -18(25) + 32(5) -15 = \boxed{195}$ .

Let $f(x)= 3\left({x-1 \choose 0} +2{x-1 \choose 1}+2^2{x-1 \choose 2}+2^3{x-1 \choose 3}\right) \\ =3\left(1+2\left(x-1\right)+2\left(x-1\right)\left(x-2\right) +\frac{4\left(x-1\right)\left(x-2\right)\left(x-3\right)}{3}\right)$

We see that this the degree of $f$ is 3 and it fulfills and the condition $f(k)=3^k$ for $k=1, 2, 3,4$ .

Now $f(5)= 3\left({4 \choose 0} +2{4 \choose 1}+2^2{4 \choose 2}+2^3{4 \choose 3}\right) =3\left(3^{4}-2^4 \right) = 195$ .