Inverse polynomials

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

Inverse of Linear polynomial function

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

Proof:

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

Inverse of Quadratic polynomial function

Let there be a quadratic polynomial f(x)=ax2+bx+cf(x)=ax^2+bx+c; it's inverse function g(x)=b+b2+4a(xc)2ag(x)=\frac{-b{^{+}_{-}}\sqrt{b^2+4a(x-c)}}{2a}

Proof:

Let inverse of f(x)f(x) be g(x)g(x) f(g(x))=x\Rightarrow f(g(x))=x ag2(x)+bg(x)+c=x\Rightarrow ag^2(x)+bg(x)+c=x g2(x)+bag(x)+ca=xa\Rightarrow g^2(x)+\frac{b}{a}g(x)+\frac{c}{a}=\frac{x}{a} g2(x)+bag(x)=xca\Rightarrow g^2(x)+\frac{b}{a}g(x)=\frac{x-c}{a} (g(x)+b2a)2b24a2=xca\Rightarrow (g(x)+\frac{b}{2a})^2-\frac{b^2}{4a^2}=\frac{x-c}{a} (g(x)+b2a)2=xca+b24a2=4a(xc)+b24a2\Rightarrow (g(x)+\frac{b}{2a})^2=\frac{x-c}{a}+\frac{b^2}{4a^2}=\frac{4a(x-c)+b^2}{4a^2} g(x)+b2a=+b2+4a(xc)2a\Rightarrow g(x)+\frac{b}{2a}=\frac{{^{+}_{-}}\sqrt{b^2+4a(x-c)}}{2a} g(x)=b+b2+4a(xc)2a\Rightarrow g(x)=\frac{-b{^{+}_{-}}\sqrt{b^2+4a(x-c)}}{2a}

Inverse of any nn degree polynomial

Let there be a nn degree polynomial f(x)=k=0nakxkf(x)=\sum_{k=0}^{n}a_kx^k and let there be a function z(x0,x1...,xn)z(x_0,x_1...,x_n) which takes the the coefficients of any nn 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)f(x) is g(x)=z(a0x,a1,a2...,an)g(x)=z(a_0-x,a_1,a_2...,a_n)

Proof:

Let inverse of f(x)f(x) be g(x)g(x) f(g(x))=x\Rightarrow f(g(x))=x angn(x)+an1gn1(x)...a1g(x)+a0=x\Rightarrow a_ng^n(x)+a_{n-1}g^{n-1}(x)...a_1g(x)+a_0=x angn(x)+an1gn1(x)...a1g(x)+a0x=0\Rightarrow a_ng^n(x)+a_{n-1}g^{n-1}(x)...a_1g(x)+a_0-x=0 g(x)=z(a0x,a1,a2...,an)\Rightarrow g(x)=z(a_0-x,a_1,a_2...,a_n)


Example of last statement:

For degree 2: z(x0,x1,x2)=x1+x124x2x02x2z(x_0,x_1,x_2)=\frac{-x_1{^{+}_{-}}\sqrt{x_1^2-4x_2x_0}}{2x_2} \therefore For a function f(x)=ax2+bx+cf(x)=ax^2+bx+c with inverse function as g(x)g(x) g(x)=z(cx,b,a)=b+b24a(cx)2a=b+b2+4a(xc)2ag(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
3 months, 2 weeks ago

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Really good! @Zakir Husain.

I have another idea for a note of yours:

A proof of Euler's identity!

A Former Brilliant Member - 3 months, 2 weeks ago

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