# Let $f(x)$ be a polynomial satisfying $\lim_{x\to \infty} \frac {x^4 f(x)}{x^8+1} =3$, $f(2)=5$, $f(3)=10$,$f(-1)=2$,$f(-6)=37$. Find $f(0)$

It’s clear that $$f(x)$$ is a 4th degree polynomial.

If $$f(x)=ax^4+bx^3+cx^2+dx+e$$, then $$a=3$$

From the rest of the given data, I can form four linear equations, which should give me the value of $$a,b,c,d$$, but that’s far too tedious and time consuming, and I don’t think the question is meant to be solved that way. Is there any alternative method?

It is clear that $$f(x)$$ is of deegree 4 and \begin{align} f(2) &= 1 + 2^2 \\ f(-1) &= 1 + (-1)^2 \\ f(3) &= 1+ 3 ^ 2 \\ f(-6) &= 1+ (-6)^2 \end{align} so $$f(x) - (1+x^2)$$ has roots $$2,3,-1,-6$$ $$\Rightarrow f(x) - (1+ x^2) = A(x+1)(x-2)(x+6)(x-3)$$

now you can continue

• Shouldn’t $A$ be known for finding $f(0)$? – Aditya Oct 11 '20 at 9:24
• @Aditya There you have to use the limiting condition. That should give you your $A$. – Yuzuriha Inori Oct 11 '20 at 9:45
• $f(0)=A\cdot(-1)\cdot2\cdot(-6)\cdot3+1$, and $A=3$ – am301 Oct 11 '20 at 9:56
• $\lim_{x\to \infty} \frac {x^4 f(x)}{x^8+1} =3=A$ @am301 – Aryadeva Oct 11 '20 at 10:17

Assume:

$$f(x) = \sum_{i=0}^k a_i x^k$$

From the three linear system in the question, we can write:

$$\begin{bmatrix} 5 \\ 10 \\ 2 \\ 37 \end{bmatrix} = \begin{bmatrix} \alpha_0 & \alpha_1 2^1 & \alpha_2 2^2 & \alpha_3 2^3 &\alpha_4 2^4 \\ \alpha_0 & \alpha_1 3^1 & \alpha_2 3^2 & \alpha_3 3^3 &\alpha_4 3^4 \\ \alpha_0 & \alpha_1 (-1)^1 & \alpha_2 (-1)^2 & \alpha_3 (-1)^3 &\alpha_4 (-1)^4 \\ \alpha_0 & \alpha_1 (-6)^1 & \alpha_2 (-6)^2 & \alpha_3 (-6)^3 &\alpha_4 (-6)^4 \\ \end{bmatrix}$$

With our linear system factorisation:

$$\begin{bmatrix} 5 \\ 10 \\ 2 \\ 37 \end{bmatrix} = \begin{bmatrix} 1 & 2^1 & 2^2 & 2^3 & 2^4 \\ 1 & 3^1 & 3^2 & 3^3 &3^4 \\ 1 & (-1)^1 &(-1)^2 & (-1)^3 & (-1)^4 \\ 1 & (-6)^1 & (-6)^2 & (-6)^3 & (-6)^4 \\ \end{bmatrix} \begin{bmatrix} \alpha_o\\ \alpha_1 \\ \alpha_2 \\ \alpha_3 \\\alpha_4 \end{bmatrix}$$

From the limit condition, $$\alpha_4 =3$$

$$\begin{bmatrix} 5 \\ 10 \\ 2 \\ 37 \end{bmatrix} = \begin{bmatrix} 1 & 2^1 & 2^2 & 2^3 & 2^4 \\ 1 & 3^1 & 3^2 & 3^3 &3^4 \\ 1 & (-1)^1 &(-1)^2 & (-1)^3 & (-1)^4 \\ 1 & (-6)^1 & (-6)^2 & (-6)^3 & (-6)^4 \\ \end{bmatrix} \begin{bmatrix} \alpha_o\\ \alpha_1 \\ \alpha_2 \\ \alpha_3 \\\ 3 \end{bmatrix}$$

We could rewrite this system as (by doing some algebra under the hood):

$$\begin{bmatrix} 1 & 2^1 & 2^2 & 2^3 \\ 1 & 3^1 & 3^2 & 3^3 \\ 1 & (-1)^1 &(-1)^2 & (-1)^3 \\ 1 & (-6)^1 & (-6)^2 & (-6)^3 \\ \end{bmatrix} \begin{bmatrix} \alpha_o\\ \alpha_1 \\ \alpha_2 \\ \alpha_3 \end{bmatrix} = \begin{bmatrix} 5- 3 \cdot 2^4 \\ 10 - 3 \cdot 3^5 \\ 2 -3 (-1)^4 \\ 37-3(-6)^4 \end{bmatrix}$$

By Crammaz rule:

$$\alpha_o = \frac{\begin{vmatrix}5- 3 \cdot 2^4 & 2^1 & 2^2 & 2^3 \\ 10 - 3 \cdot 3^4 & 3^1 & 3^2 & 3^3 \\ 2 -3 (-1)^4 & (-1)^1 &(-1)^2 & (-1)^3 \\ 37-3(-6)^4 & (-6)^1 & (-6)^2 & (-6)^3 \\ \end{vmatrix} }{\begin{vmatrix} 1 & 2^1 & 2^2 & 2^3 \\ 1 & 3^1 & 3^2 & 3^3\\ 1 & (-1)^1 &(-1)^2 & (-1)^3 \\ 1 & (-6)^1 & (-6)^2 & (-6)^3 \\ \end{vmatrix} }$$

Let's call,

$$J= \begin{vmatrix} 1 & 2^1 & 2^2 & 2^3 \\ 1 & 3^1 & 3^2 & 3^3\\ 1 & (-1)^1 &(-1)^2 & (-1)^3 \\ 1 & (-6)^1 & (-6)^2 & (-6)^3 \\ \end{vmatrix}$$

Now consider the numerator, from an almost unbelievable simplification by Zhang (see this post), we can write it as:

$$\begin{vmatrix}5- 3 \cdot 2^4 & 2^1 & 2^2 & 2^3 \\ 10 - 3 \cdot 3^4 & 3^1 & 3^2 & 3^3 \\ 2 -3 (-1)^4 & (-1)^1 &(-1)^2 & (-1)^3 \\ 37-3(-6)^4 & (-6)^1 & (-6)^2 & (-6)^3 \\ \end{vmatrix} = \begin{vmatrix} 2^4 & 2^1 & 2^2 & 2^3 \\ 3^4 & 3^1 & 3^2 & 3^3 \\ (-1)^4 & (-1)^1 &(-1)^2 & (-1)^3 \\ (-6)^4 & (-6)^1 & (-6)^2 & (-6)^3 \\ \end{vmatrix}\cdot (-3)+\begin{vmatrix}2^2 & 2^1 & 2^2 & 2^3 \\ 3^2 & 3^1 & 3^2 & 3^3 \\ (-1)^2 & (-1)^1 &(-1)^2 & (-1)^3 \\ (-6)^2 & (-6)^1 & (-6)^2 & (-6)^3 \\ \end{vmatrix} +\begin{vmatrix}1 & 2^1 & 2^2 & 2^3 \\ 1 & 3^1 & 3^2 & 3^3 \\ 1 & (-1)^1 &(-1)^2 & (-1)^3 \\ 1 & (-6)^1 & (-6)^2 & (-6)^3 \\ \end{vmatrix}$$

The second determinant is zero, and simplify the first one.

$$\begin{vmatrix}5- 3 \cdot 2^4 & 2^1 & 2^2 & 2^3 \\ 10 - 3 \cdot 3^4 & 3^1 & 3^2 & 3^3 \\ 2 -3 (-1)^4 & (-1)^1 &(-1)^2 & (-1)^3 \\ 37-3(-6)^4 & (-6)^1 & (-6)^2 & (-6)^3 \\ \end{vmatrix} = -3 \cdot 2 \cdot -1 \cdot -6 \begin{vmatrix} 2^3 & 1 & 2^1 & 2^2 \\ 3^3 & 1& 3^1 & 3^2 \\ (-1)^3 & 1 &(-1)^1 & (-1)^2 \\ (-6)^3 & 1 & (-6)^1 & (-6)^2 \\ \end{vmatrix}+\begin{vmatrix}1 & 2^1 & 2^2 & 2^3 \\ 1 & 3^1 & 3^2 & 3^3 \\ 1 & (-1)^1 &(-1)^2 & (-1)^3 \\ 1 & (-6)^1 & (-6)^2 & (-6)^3 \\ \end{vmatrix}$$

Rearranging the columns of first expression in sum,

$$\begin{vmatrix}5- 3 \cdot 2^4 & 2^1 & 2^2 & 2^3 \\ 10 - 3 \cdot 3^4 & 3^1 & 3^2 & 3^3 \\ 2 -3 (-1)^4 & (-1)^1 &(-1)^2 & (-1)^3 \\ 37-3(-6)^4 & (-6)^1 & (-6)^2 & (-6)^3 \\ \end{vmatrix} = 3 \cdot 2 \cdot 1 \cdot 6 J+J= ( 3 \cdot 2 \cdot 1 \cdot 6 +1) J$$

Hence,

$$\alpha_o = 3 \cdot 2 \cdot 1 \cdot 6 + 1 =37$$