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Laurent · #8 November 10th, 2008, 01:36 PM

Quote:

Originally Posted by tasos

Hello to everyone,

Could somebody please tell me how do we solve exercises like the one I describe below:

$exp(1+(O(1/n))^2)$ = e + O(1/n)

Thanks a lot,
Tasos

Are you sure you put the parentheses at the right place? Because this is correct, but not optimal, as you shall see:

I'll apply usual properties of the "big O" notation. If you don't know them, just ask, the proofs are very short.
First you have $\left(O\left(\frac{1}{n}\right)\right)^2=O\left(\frac{1}{n^2}\right)$ , and it is more usual to write it this way.
Then $e^{1+O\left(\frac{1}{n^2}\right)}=e e^{O\left(\frac{1}{n^2}\right)}= e\left( 1+O\left(\frac{1}{n^2}\right)\right)$ $=e + e\, O\left(\frac{1}{n^2}\right)=e+O\left(\frac{1}{n^2}\right)$ .
I used the following expansion of the exponential at 0: $e^{O(u)}=1+O(u)$ when $u$ tends to 0, composed with the sequence $\left(\frac{1}{n^2}\right)_n$ which tends to 0.

Now, $0\leq\frac{1}{n^2}\leq \frac{1}{n}$ , hence the previous big O can be replaced by $O\left(\frac{1}{n}\right)$ , and this is what you need.

Let us now suppose that the question was $e^{\left(1+O\left(\frac{1}{n}\right)\right)^2}=1+O\left(\frac{1}{n}\right)$ , which seems more plausible to me.
Then we would have: $\left(1+O\left(\frac{1}{n}\right)\right)^2=1+O\left(\frac{1}{n}\right)+O\left(\frac{1}{n^2}\right)=1+O\left(\frac{1}{n}\right)$ (as before, the second big O can be included in the first one), and by the same computation as above we conclude $e^{\left(1+O\left(\frac{1}{n}\right)\right)^2}=e+O\left(\frac{1}{n}\right)$ .

ps: about the way to write symbols, there's a section in the forum related to "LaTeX", and this is where you should look for help about this.