Funky Function Transformation

Relevant wiki: Vieta's Formula Problem Solving - Intermediate

Given that $a$ , $b$ , $c$ , and $d$ are roots of $3x^4+mx-1=0$ , by Vieta's formula we have $a+b+c+d = 0$ and $abcd = - \dfrac 13$ . Then the roots of $f(x)$ are:

$\begin{cases} \dfrac {\color{#3D99F6}a+b+c}{d^2} = \dfrac {\color{#3D99F6}-d}{d^2} = - \dfrac 1d & \small \color{#3D99F6} \text{Since }a+b+c+d = 0 \\ \dfrac {a+b+d}{c^2} = - \dfrac 1c & \small \color{#3D99F6} \text{As before} \\ \dfrac {a+c+d}{b^2} = - \dfrac 1b & \small \color{#3D99F6} \text{and} \\ \dfrac {b+c+d}{a^2} = - \dfrac 1a \end{cases}$

Since $f(x)$ is monic, the coefficient of its first term is 1. The coefficient of its last term is the product of all roots or $\dfrac 1{abcd} = - 3$ , since $abcd = - \dfrac 13$ . Therefore, their sum is $1-3= \boxed{-2}$ .

Because the coefficient in front of the $x^3$ term in the original equation is $0$ , we can use one of Vieta's Formulas to find that $a + b + c + d = 0$ . Therefore, the roots can be rewritten as,

$-\frac{1}{d}, -\frac{1}{c}, -\frac{1}{b}, -\frac{1}{a}.$

If we treat the original equation as a function, we can start making progress by finding the function with roots that are the reciprocal of the original equations roots. To do this we reverse the order of the coefficients, so the function with reciprocal roots is,

$f(x) = -x^{4} + mx^{3} + 3.$

Now we just need the our function to have roots that are the opposite of what they currently are. To do this we multiply each of the coefficients in order by $(-1)^0, (-1)^1, (-1)^2, (-1)^3...$ , so our new function is,

$f(x) = -x^{4} - mx^{3} + 3.$

Finally since we want a monic function (where the leading coefficient is $0$ ), we multiply this function by $-1$ (which will not change the roots), so the sum of the first and last coefficients is $\boxed{-2}$ .

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Proof: To find polynomial $g(x)$ with roots that are the reciprocal of the roots of polynomial $f(x)$ you just reverse the order of the coefficients of $f(x)$ .

Label the roots of $f(x) = a_0x^n + a_1x^{n-1} + ... + a_{n-1}x + a_n$ as $r_1, r_2, r_3, ... r_i$ . Note that if $f(r_i) = 0, f(\frac{1}{1 / r_i}) = 0$ , so $g(x) = f(1/x)$ . Therefore,

$g(x) = f(1/x) = a_0(\frac{1}{x})^n + a_1(\frac{1}{x})^{n-1} + ... + a_{n-1} (\frac{1}{x})^1 + a_n.$

Finally since multiplying a polynomial function by any non-zero number doesn't change its roots, we can multiply this entire function by $x^n$ to get,

$g(x) = a_0 + a_1x + ... + a_{n-1} x^{n-1} + a_nx^n,$

which is the original function with reversed coefficients.

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Proof: To find polynomial $g(x)$ with roots that are $m$ times the roots of polynomial $f(x)$ you multiply each coefficient of $f(x)$ by $m^0, m^1, m^2, ..., m^n$ , where n is the degree of $f(x)$ .

Label the roots of $f(x) = a_0x^n + a_1x^{n-1} + ... + a_{n-1}x + a_n$ as $r_1, r_2, r_3, ... r_i$ . Note that if $f(r_i) = 0, f(\frac{m \times r_i}{m}) = 0$ , so $g(x) = f(x/m)$ . Therefore,

$g(x) = f(x/m) = a_0(\frac{x}{m})^n + a_1(\frac{x}{m})^{n-1} + ... + a_{n-1} (\frac{x}{m})^1 + a_n.$

Similarly to the above proof we can multiply the function by a non-zero number without changing its roots, so by multiplying the equation by $m^n$ simplifies to,

$g(x) = a_0x^n + a_1x^{n-1}(m^1) + ... + a_{n-1}x(m^{n-1}) + a_n(m^n),$

which is what we were trying to prove.