How is called the class of functions whose inverse function is a polynomial? Is there any study of such functions?

Sonia Ayers
2022-07-13
Answered

How is called the class of functions whose inverse function is a polynomial? Is there any study of such functions?

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Zackery Harvey

Answered 2022-07-14
Author has **21** answers

The functions you're interested in are a particular subset of the algebraic functions. To be specific, an algebraic function $f(x)$ is a function that satisfies

${a}_{n}(x)f(x{)}^{n}+{a}_{n-1}(x)f(x{)}^{n-1}+\cdots +{a}_{0}(x)=0$

where the ${a}_{i}(x)$ are polynomials. By contrast, let

$P(x)={b}_{n}{x}^{n}+{b}_{n-1}{x}^{n-1}+\cdots +{b}_{0}$

be an arbitrary polynomial. Its inverse, ${P}^{-1}(x)$, satisfies

$P({P}^{-1}(x))={b}_{n}{P}^{-1}(x{)}^{n}+{b}_{n-1}{P}^{-1}(x{)}^{n-1}+\cdots +{b}_{0}=x.$

So inverses of polynomials are those algebraic functions for which the defining coefficient functions ${a}_{i}$ are constants, except for the last, which is a constant minus $x$.

${a}_{n}(x)f(x{)}^{n}+{a}_{n-1}(x)f(x{)}^{n-1}+\cdots +{a}_{0}(x)=0$

where the ${a}_{i}(x)$ are polynomials. By contrast, let

$P(x)={b}_{n}{x}^{n}+{b}_{n-1}{x}^{n-1}+\cdots +{b}_{0}$

be an arbitrary polynomial. Its inverse, ${P}^{-1}(x)$, satisfies

$P({P}^{-1}(x))={b}_{n}{P}^{-1}(x{)}^{n}+{b}_{n-1}{P}^{-1}(x{)}^{n-1}+\cdots +{b}_{0}=x.$

So inverses of polynomials are those algebraic functions for which the defining coefficient functions ${a}_{i}$ are constants, except for the last, which is a constant minus $x$.

asked 2022-06-04

Is it true that for $F:{\mathbb{R}}^{n}\to {\mathbb{R}}^{m}$ which has an inverse function ${F}^{-1}:{\mathbb{R}}^{m}\to {\mathbb{R}}^{n}$, says $F$ is diﬀerentiable at $a\in {\mathbb{R}}^{n}$ and ${F}^{-1}$ is diﬀerentiable at $b=F(a)\in {\mathbb{R}}^{m}$，then $m$ must be equal to $n$?

asked 2022-04-07

the notation $(f+g{)}^{-1}$ for the inverse function

asked 2022-05-31

Let $f(x)={x}^{3}+x$. If h is the inverse function of $f$, then ${h}^{\prime}(2)=$=?

The derivative of $f$ is: $3{x}^{2}+1$. But, how do I find the ${f}^{-1}x$?

The derivative of $f$ is: $3{x}^{2}+1$. But, how do I find the ${f}^{-1}x$?

asked 2022-07-02

Prove that if $X$ is a subset of ${\mathbb{R}}^{n}$ and $Y$ is a subset of ${\mathbb{R}}^{m}$, and $X$ and $Y$ are closed and bounded, then if $f:X\to Y$ is continuous and has a inverse function, than the inverse function is also continuous.

asked 2022-07-12

Is there any branch of constructing inverse function exist that is inverse of $n$ functions?

asked 2022-04-07

Im trying to find $2$ functions $f,g$ such that $f$ and $g$ both not linear, and $f,g:\mathbb{R}\to \mathbb{R}$ and onto, such that $f\left(g\left(x\right)\right)=x,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{mediummathspace}{0ex}}g\left(f\left(x\right)\right)=x$

($f$ is the inverse function of $g$).

There exists such functions?

($f$ is the inverse function of $g$).

There exists such functions?

asked 2022-04-30

Let $\mathsf{f}\mathsf{(}\mathsf{x}\mathsf{)}$ be a differential and an invertible function, such that ${\mathsf{f}}^{\u2033}\mathsf{(}\mathsf{x}\mathsf{)}\mathsf{>}\mathsf{0}$ and ${\mathsf{f}}^{\prime}\mathsf{(}\mathsf{x}\mathsf{)}\mathsf{>}\mathsf{0}$.

Prove

${\mathsf{f}}^{\mathsf{-}\mathsf{1}}\left(\frac{{\mathsf{x}}_{\mathsf{1}}\mathsf{+}{\mathsf{x}}_{\mathsf{2}}\mathsf{+}{\mathsf{x}}_{\mathsf{3}}}{\mathsf{3}}\right)\mathsf{>}\frac{{\mathsf{f}}^{\mathsf{-}\mathsf{1}}\mathsf{(}{\mathsf{x}}_{\mathsf{1}}\mathsf{)}\mathsf{+}{\mathsf{f}}^{\mathsf{-}\mathsf{1}}\mathsf{(}{\mathsf{x}}_{\mathsf{2}}\mathsf{)}\mathsf{+}{\mathsf{f}}^{\mathsf{-}\mathsf{1}}\mathsf{(}{\mathsf{x}}_{\mathsf{3}}\mathsf{)}}{\mathsf{3}}$

Prove

${\mathsf{f}}^{\mathsf{-}\mathsf{1}}\left(\frac{{\mathsf{x}}_{\mathsf{1}}\mathsf{+}{\mathsf{x}}_{\mathsf{2}}\mathsf{+}{\mathsf{x}}_{\mathsf{3}}}{\mathsf{3}}\right)\mathsf{>}\frac{{\mathsf{f}}^{\mathsf{-}\mathsf{1}}\mathsf{(}{\mathsf{x}}_{\mathsf{1}}\mathsf{)}\mathsf{+}{\mathsf{f}}^{\mathsf{-}\mathsf{1}}\mathsf{(}{\mathsf{x}}_{\mathsf{2}}\mathsf{)}\mathsf{+}{\mathsf{f}}^{\mathsf{-}\mathsf{1}}\mathsf{(}{\mathsf{x}}_{\mathsf{3}}\mathsf{)}}{\mathsf{3}}$