Is there any easy way to simplify this?
$$(x - 1)^n + (x + 1)^n \mod x^2$$
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Bill Dubuque
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You can expand by binomial theorem. Mod x^2 means you can ignore terms where powers of $x$ is 2 or higher. – P Vanchinathan Oct 27 '16 at 02:57
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Most terms of the binomial theorem expansions of $(x-1)^n$ and $(x+1)^n$ are divisible by $x^2$. The leftover terms are $$((-1)^n+(-1)^{n-1}nx)+(1+nx).$$
If $n$ is even, this equals $2$. If $n$ is odd, this is $2nx$.
Bill Dubuque
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pi66
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2Or $ $ Taylor expanding: $\bmod x^2!:\ f(x)\equiv \color{#0a0}{f(0)}+\color{#c00}{f'(0)},x\equiv \color{#0a0}{(-1)^n!+!1}+\color{#c00}{((-1)^{n-1}!+!1)n},x\ \ $ – Bill Dubuque Jul 08 '24 at 17:07
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1Or, equivalently, $,x^2\mid g(x)= f(x)-f(0)-f'(0)x,$ by $,g(0)=0=g'(0),,$ i.e. the Double Root Test. $\ \ $ – Bill Dubuque Jul 08 '24 at 17:43