Reversing the order of coefficients of a polynomial

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Flesh this out, make a 0 condition, write the polynomial as [ilmath]\sum_{i\eq 0}^n a_ix^i[/ilmath] and so forth Alec (talk) 13:45, 18 December 2017 (UTC)

Suppose we have:

• [ilmath]f(x):\eq a+bx+cx^2+dx^3+ex^4+\cdots+\alpha x^n[/ilmath]

Then

• [math]\frac{1}{x^n}f(x)\eq \frac{a}{x^n}+\frac{b}{x^{n-1} }+\frac{c}{x^{n-2} }+\frac{d}{x^{n-3} }+\frac{e}{x^{n-4} }+\cdots + \alpha [/math]

and lastly

• [math]\frac{1}{x^n}f\left(\frac{1}{x}\right)\eq ax^n + bx^{n-1}+cx^{n-2}+dx^{n-3}+ex^{n-4}+\cdots+\alpha[/math]

Say [ilmath]n\eq 5[/ilmath] then

• [math]\frac{1}{x^n}f\left(\frac{1}{x}\right)\eq ax^5+bx^4+cx^3+dx^2+ex+\alpha[/math] - we've flipped the coefficients!