Wikipedia:Reference desk/Archives/Mathematics/2023 January 24

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January 24

Proving the limit of xⁿ by definition‏‏

This question bugs me for quite a while now, and I would like to hear an honest opinion:
Is this a valid proof, or is it somewhat "circular" by assuming the nth root function is defined and continuous?

${\begin{aligned}|x^{n}-a^{n}|&={\Big |}(x-a+a)^{n}-a^{n}{\Big |}\\&=\left|\sum _{k=0}^{n-1}{\tbinom {n}{k}}a^{k}(x-a)^{n-k}\right|\\&\leq \sum _{k=0}^{n-1}{\tbinom {n}{k}}|a|^{k}|x-a|^{n-k}\\&<\sum _{k=0}^{n-1}{\tbinom {n}{k}}|a|^{k}\delta ^{n-k}=\varepsilon \\\delta &={\sqrt[{n}]{\varepsilon +|a|^{n}}}-|a|\end{aligned}}$

יהודה שמחה ולדמן (talk) 11:57, 24 January 2023 (UTC)

Of which statement is this meant to be a proof? $\lim _{x\to a}x^{n}=a^{n}$? If so, some conditions on $a$ and $n$ may be needed. The appeal to the binomial theorem already implies that $n$ is a non-negative integer. If $a$ and $n$ can both be equal to $0$, you run into $0^{0}$ being (depending on who you talk to) iffy. Apart from that, the statement follows from the continuity of multiplication. If a proof is required using the $\forall \varepsilon \,\exists \,\delta$ definition of limit, the last steps need more care and finesse. Even assuming definedness and continuity of the $n$-th root function, I don't see where the expression for $\delta$ comes from and how it guarantees the validity of the preceding equality. Using that $\delta ^{n{-}k}\leq \delta$ for $0<\delta \leq 1$, the last inequality can be extended to

$\sum _{k=0}^{n-1}{\tbinom {n}{k}}|a|^{k}\delta ^{n-k}\leq \left(\sum _{k=0}^{n-1}{\tbinom {n}{k}}|a|^{k}\right)\delta .$

Assuming $a\neq 0$ and $n\geq 1,$ this expression allows you to find a solution for $\delta$ that pushes this upper bound down to $\varepsilon$.  --Lambiam 12:43, 24 January 2023 (UTC)

It seems I am not helping anyone by skipping steps.

${\begin{aligned}&\sum _{k=0}^{n-1}{\tbinom {n}{k}}|a|^{k}\delta ^{n-k}+|a|^{n}=\varepsilon +|a|^{n}\\&(\delta +|a|)^{n}=\varepsilon +|a|^{n}\\&\delta ={\sqrt[{n}]{\varepsilon +|a|^{n}}}-|a|\end{aligned}}$

So again: is this a valid proof or not?

יהודה שמחה ולדמן (talk) 15:45, 24 January 2023 (UTC)

The last step, in this form, obviously requires that the $n$th root of a positive number is defined. If this is not given and may not be assumed, the proof is lacking something. The proof of the lacking bit will normally be based on the continuity of not the $n$th root, but the $n$th power. Since this appears to be what the theorem is about, appealing to the conclusion of the unfinished proof for its lacking bit will indeed introduce a circularity. Strictly speaking we can do with the formally weaker existence of some $y$ such that for $x>0$ and $\varepsilon >0$ we have $x^{n}<y^{n}\leq x^{n}+\varepsilon .$ This is not true in the domain of integers, so it requires something like continuity. An awkward proof can perhaps be built on the fact that the real numbers can be approximated arbitrarily closely by rational numbers.  --Lambiam 18:43, 24 January 2023 (UTC)

I saw this in a video. Could this be filling the gap we were missing?

${\begin{aligned}{\bigl |}{\sqrt[{n}]{x}}-{\sqrt[{n}]{a}}{\bigr |}&={\frac {|x-a|}{\displaystyle \left|\sum _{k=0}^{n-1}\left({\sqrt[{n}]{x}}\right)^{n\,-\,1\,-\,k}\left({\sqrt[{n}]{a}}\right)^{k}\right|}}\\&\leq {\frac {|x-a|}{\left({\sqrt[{n}]{a}}\right)^{n\,-\,1}}}<{\frac {\delta }{\left({\sqrt[{n}]{a}}\right)^{n\,-\,1}}}=\varepsilon \end{aligned}}$

יהודה שמחה ולדמן (talk) 20:54, 26 January 2023 (UTC)

I don't personally feel like I'm missing a gap. This too obviously require the $n$th root of a positive number to be defined. If that is not given, the gap is this very lack of the $n$th root being defined, so this derivation won't help to fill the gap. The question is ultimately what one may and may not assume at this stage, which depends on an unknown context.  --Lambiam 22:46, 26 January 2023 (UTC)

I agree with Lambian. It is a complicated way of going around in a loop and not a proof of anything. The big trick one does with epsilon delta to get it to work easily is not to find an accurate bound but a simple bound which can be much larger. In this case if |a-x| < δ and 0 < δ < 1 then as Lambian pointed out |a-x|ⁿ < δ so you simply needs δ < ε/(|a|+1)ⁿ in your argument, or a bit less than 1 if that is greater than 1. NadVolum (talk) 00:06, 27 January 2023 (UTC)