# Inverse polynomials

If $g(x)$ is inverse of a polynomial $f(x)$ then $f(g(x))=x$

Inverse of Linear polynomial function

Let there be a linear polynomial $f(x)=ax+b$; it's inverse function $g(x)=\frac{x-b}{a}$

Proof:

Let the inverse of $f(x)$ be $g(x)$ $\Rightarrow f(g(x))=x$ $\Rightarrow ag(x)+b=x$ $\Rightarrow g(x)=\frac{x-b}{a}$

Let there be a quadratic polynomial $f(x)=ax^2+bx+c$; it's inverse function $g(x)=\frac{-b{^{+}_{-}}\sqrt{b^2+4a(x-c)}}{2a}$

Proof:

Let inverse of $f(x)$ be $g(x)$ $\Rightarrow f(g(x))=x$ $\Rightarrow ag^2(x)+bg(x)+c=x$ $\Rightarrow g^2(x)+\frac{b}{a}g(x)+\frac{c}{a}=\frac{x}{a}$ $\Rightarrow g^2(x)+\frac{b}{a}g(x)=\frac{x-c}{a}$ $\Rightarrow (g(x)+\frac{b}{2a})^2-\frac{b^2}{4a^2}=\frac{x-c}{a}$ $\Rightarrow (g(x)+\frac{b}{2a})^2=\frac{x-c}{a}+\frac{b^2}{4a^2}=\frac{4a(x-c)+b^2}{4a^2}$ $\Rightarrow g(x)+\frac{b}{2a}=\frac{{^{+}_{-}}\sqrt{b^2+4a(x-c)}}{2a}$ $\Rightarrow g(x)=\frac{-b{^{+}_{-}}\sqrt{b^2+4a(x-c)}}{2a}$

Inverse of any $n$ degree polynomial

Let there be a $n$ degree polynomial $f(x)=\sum_{k=0}^{n}a_kx^k$ and let there be a function $z(x_0,x_1...,x_n)$ which takes the the coefficients of any $n$ degree polynomial and gives it's zeroes as output (the first input is the constant term, next is the linear and so on), then the inverse function of $f(x)$ is $g(x)=z(a_0-x,a_1,a_2...,a_n)$

Proof:

Let inverse of $f(x)$ be $g(x)$ $\Rightarrow f(g(x))=x$ $\Rightarrow a_ng^n(x)+a_{n-1}g^{n-1}(x)...a_1g(x)+a_0=x$ $\Rightarrow a_ng^n(x)+a_{n-1}g^{n-1}(x)...a_1g(x)+a_0-x=0$ $\Rightarrow g(x)=z(a_0-x,a_1,a_2...,a_n)$

Example of last statement:

For degree 2: $z(x_0,x_1,x_2)=\frac{-x_1{^{+}_{-}}\sqrt{x_1^2-4x_2x_0}}{2x_2}$ $\therefore$ For a function $f(x)=ax^2+bx+c$ with inverse function as $g(x)$ $g(x)=z(c-x,b,a)=\frac{-b{^{+}_{-}}\sqrt{b^2-4a(c-x)}}{2a}=\frac{-b{^{+}_{-}}\sqrt{b^2+4a(x-c)}}{2a}$ Note by Zakir Husain
1 year, 1 month ago

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Really good! @Zakir Husain.

I have another idea for a note of yours:

A proof of Euler's identity!

- 1 year, 1 month ago