Why is $(1+\frac{x}{n})^n = e^x(1-\frac{x^2}{2n}) + O(\frac{1}{n^2})$?

Question

I'm reading a paper by Edelman and Kostlan (https://arxiv.org/abs/math/9501224, page 8 at the bottom) about real random polynomials for my bachelors thesis. He uses the identity mentioned in the title.

Although I think it is probably very elementary to show this, I couldnt figure out, why it holds. My idea was to use the logarithm and the taylor expansion of $ln(1+x)$. I would get \begin{align} \ln \left( \left(1+\frac{x}{n} \right)^n \right)&=n \cdot \ln \left(1+\frac{x}{n} \right)\\ &= n \cdot \left(\frac{x}{n} - \frac{x^2}{2n^2}+\frac{x^3}{3n^3}+ \dots \right)\\ &= x-\frac{x^2}{2n} + O \left(\frac{1}{n^2} \right). \end{align}

But now I get \begin{align} (1+\frac{x}{n})^n &= e^{\ln \left( \left(1+\frac{x}{n} \right)^n \right)}\\ &= e^{x-\frac{x^2}{2n} + O \left(\frac{1}{n^2} \right)}\\ &= e^{x \cdot \left(1-\frac{x}{2n}+O \left(\frac{1}{n^2} \right) \right)}, \end{align} which is not at all what I wanted. I also tried using Taylors theorem but could not find a function that would work. I also tried using the binomial theorem, but it ended in a mess (maybe there is a certain way of index shifting that I missed). I would greatly appreciate help or a hint.

Answer 1

What you did is right, now you continue by expanding $e^y$ around $y=x$. You have $e^{y+\Delta y}=e^y + e^y \Delta y + O(\Delta y^2)$. Replace $y$ with $x$ and $\Delta y$ with $-x^2/(2n)+O(1/n^2)$.

Answer 2

$\newcommand{\bra}[1]{\left(#1\right)}$ Here's a proof of what you need, missing very few details that you will require if you need to reproduce it for your thesis. I found some duplicate questions, but frankly they're poorly written and don't have the best answers either.

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I assume $x\in \mathbb R,n \in \mathbb N$ are such that $\frac{x}{n}>-1$ : this holds because in the paper, we fix $x$ and look at arbitrarily large values of $n$. Start with $$ \bra{1+\frac xn}^n = \exp\bra{n\log\bra{1+\frac xn}} $$ Now, because $x$ is fixed and $n$ arbitrarily large, $\frac xn$ is close to $0$. By assuming $n$ large enough, we see that $\frac xn$ is as close to $0$ as necessary. Hence,$$ \log\bra{1+\frac xn} = \frac xn - \frac {x^2}{2n^2} + O\bra{\frac{1}{n^3}} $$ where the constant involved in the definition of the $O$ notation depends upon the fixed value of $x$, but is certainly the same for all large enough values of $n$. (If you wish, I can rewrite the above without the $O$). As a consequence, $$ n\log\bra{1+\frac xn} = x - \frac {x^2}{2n} + O\bra{\frac{1}{n^2}} $$ and hence $$ \exp\bra{n\log\bra{1+\frac xn}} = \exp \bra{x - \frac {x^2}{2n} + O\bra{\frac{1}{n^2}}} = e^x\exp\bra{-\frac{x^2}{2n} + O\bra{\frac 1{n^2}}} \tag{eq1} \label{eq1} $$ At this point, we know what we need : the Taylor series of the exponential $$ \exp\bra{-\frac{x^2}{2n} + O\bra{\frac 1{n^2}}} = 1 +\bra{-\frac{x^2}{2n} + O\bra{\frac 1{n^2}}} + O\bra{\bra{-\frac{x^2}{2n} + O\bra{\frac 1{n^2}}}^2} $$

However, observe that the third term of the RHS contains powers of $n$ only greater than $2$ (and terms involving the fixed value of $x$). Therefore, combining this with the $O\bra{\frac 1{n^2}}$ from the second term, $$ \exp\bra{-\frac{x^2}{2n} + O\bra{\frac 1{n^2}}} = 1 -\frac{x^2}{2n} + O\bra{\frac 1{n^2}} \tag{eq2}\label{eq2} $$ Combining $\ref{eq1}$ and $\ref{eq2}$, \begin{align*}\bra{1+\frac xn}^n &= e^x\bra{1 -\frac{x^2}{2n}} + e^xO\bra{\frac 1{n^2}}\\ &= e^x\bra{1 -\frac{x^2}{2n}} + O\bra{\frac 1{n^2}} \end{align*}

because $e^x$ is a constant as $x$ is fixed, and is absorbed inside the $O$. Hence, the result is obtained.