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This might be a stupid question: I recently did a question concerning some polynomials and I was supposed to calculate $(X^n-1) \div (X-1)$. It was supposed to be done by induction on $n$ but I kind of overlooked that and did it by doing just normal polynomial division and writing some "..." in between my steps.

I recognize that this is not a formally nice way to do it. But is it actually wrong? Does it not calculate the solution correctly (in my case it did)?

I thought maybe the problem could be that since I did not use induction, I know (through my "proof") the solution for one fixed (?) $n$ but not for all $n \in \mathbb{N}$, meaning that I don't know if it holds for example for $n+1$. So maybe similarly to showing the solution just for $n=3$, I have just shown it for one number (even if I just called it $n$)?

  • In my mind $...$ is really short hand for induction. If you think about it is the same concept... "we see if it holds for one it will hold for the next and so it holds for all values of n". But... it's informal. And an informal proof can be declared not really acceptable even if it isn't wrong. – fleablood May 13 '24 at 17:24
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    In contexts like this, writing ". . ." is often a way of suggesting an induction without writing it out but with confidence that the induction could be written out in detail on request. So if a mathematician wrote a computation using ". . ." I wouldn't object, but if students who are just learning about induction wrote the same computation, I'd suspect that they didn't really understand induction and how it relates to ". . .". – Andreas Blass May 13 '24 at 17:24
  • The thing is the insight that $x^n-1 = (x-1)(x^{n-1}+ ..... +1)$ is not what is being tested or challenged. That's presumably well known to the student. What is being tested is the students ability to apply induction formally. – fleablood May 13 '24 at 17:31
  • This type of induction can be viewed as a special case of telescopy - see here in the linked dupe. So you can replace the nonrigorous ellipses "..." by an invocation of telescopy. – Bill Dubuque May 20 '24 at 17:25

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Showing that

$~(x-1) \times (x^{n-1} + x^{n-2} + \cdots + x^1 + 1) = x^n - 1~$

is generally done in one of two ways.

One way is to avoid induction, and simply multiply everything out. This approach analogizes to doing polynomial long division of $~(x^n - 1) \div (x-1).~$

Edit
See also the comments immediately following this answer.
Bill Dubuque makes the point that the above approach merely hides the use of (for example) induction (or recursion). Personally, lacking any formal training here, I am not in a position to dispute his viewpoint.

The alternative (induction) approach is to prove that it applies when $~n = 1,~$ assume that it applies when $~n=N,~$ and then prove that this implies that the formula must then hold for $~n = N+1.~$ This implication is normally demonstrated by focusing on $~x^{N+1} - x^N = x^N \times (x - 1).$

user2661923
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  • "simply multiply everything out" doesn't avoid induction - it just hides (encapsulates) it in some Lemma or Theorem that is (implicitly) employed when the argument is made fully rigorous, e.g. in the recursive definition of $,\sum_{k=1}^n x^k.\ $ – Bill Dubuque May 20 '24 at 17:31
  • @BillDubuque Are you sure? My intent was that $$\left( ~x^{n-1} + x^{n-2} + \cdots + x + 1 ~\right) \times (x - 1)$$ $$= \left( ~x^n + x^{n-1} + \cdots + x^2 + x^1 ~\right) - \left( ~x^{n-1} + x^{n-2} + \cdots + x + 1 ~\right)$$ $$ = x^n - 1.$$ I see no use of induction or recursion here. Instead, I see a formula for the standard multiplication of two polynomials, which holds for any fixed $~n \in \Bbb{Z^+}.~$ What am I missing? – user2661923 May 20 '24 at 18:56
  • Yes, I am sure. You have not rigorously defined those expressions containing ellipses and symbolic exponents. See also Blass's comment on the quesion. – Bill Dubuque May 20 '24 at 19:02
  • @BillDubuque I have edited my answer, to accommodate your comments. – user2661923 May 20 '24 at 19:05