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Assume that $\epsilon$ is a very small positive number (for example $\epsilon$ is of order $10^{-200})$. Also $n$ numbers are given in form $\delta_i$ $(1\leq i\leq n)$ such that for every $i$, $|\delta_i|<\epsilon$. Now assume this expression is given: $$(1+\delta_1)(1+\delta_2)\dots(1+\delta_n)$$ The question is, is it possible to find some $|a|<\epsilon$, such that $(1+a)^n$ is equal to the above expression. I mean: $$(1+\delta_1)(1+\delta_2)\dots(1+\delta_n)=(1+a)^n$$ If yes, how can we prove its existence?

I guess the answer is yes but It's just a guess. I tried using $\ln(1+x)$ and its taylor series, but I couldn't get to something useful. I also tried to use induction, but this approach also didn't give me anything! Maybe we can use Bernoulli's inequality or binomial coeffiients, but I couldn't prove this using it. Maybe you can!

Any help is so much appreciated!

Mason Rashford
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    When you posted this before, I suggested using continuity. With $a=-\epsilon$ the right hand is less than the left, if $a=\epsilon$ it's greater. – lulu Oct 28 '23 at 22:57
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    @lulu Oh what a genius answer! Thank you so much. I'm sorry for deleting the post, I thought it's better to post a more precise question. I didn't get your answer at that time. Now it's solved. Thanks again. – Mason Rashford Oct 28 '23 at 23:02

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The answer is yes, but the proof is non-constructive. There is a theorem called the intermediate value theorem: for any continuous function $f: \mathbb{R} \rightarrow \mathbb{R}$, if $a < b$, then for any value $c$ between $f(a)$ and $f(b)$, there exists some value $d$ such that $f(d) = c$. Or, to put it another way, $f$ maps intervals to intervals.

Using this, the answer becomes obvious: (1 + x)^n is a continuous function, and, if we indeed assume $\epsilon$ to be sufficiently small (i.e. smaller than $1$), we can write $(1 - \epsilon)^n < \prod_{i = 1}^{n}(1 + \delta_i) < (1 + \epsilon)^n$, which completes the proof.

kabel abel
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