Math Help Forum - View Single Post

Laurent · #10 November 11th, 2008, 04:57 AM

Quote:

Originally Posted by tasos

Thank you very much Laurent for your answer. It is very thorough.

However, I have a little question.

How do you go from the left part to the right one in the following expression:
Thanks

This is because of this result:

Quote:

Originally Posted by Laurent

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.

In fact, as soon as a function $f$ is differentiable at $0$ , we have $f(x)=f(0)+O(x)$ as $x$ tends to 0 (because $\frac{f(x)-f(0)}{x}\to f'(0)$ so that $\frac{f(x)-f(0)}{x}=O(1)$ as $x\to0$ ).

And if $g(u)=O(u)$ as $u\to0$ , then $g(u)\to_{u\to 0} 0$ , so that we can compose: $f(g(u))=f(0)+O(g(u))=f(0)+O(u)$ . This can be written $f(O(u))=f(0)+O(u)$ as $u$ tends to 0.

Now, in this expansion, you can replace $u$ by any sequence which converges to 0. For instance, $f(O\left(\frac{1}{n^2}\right))=f(0)+O\left(\frac{1}{n^2}\right)$ . If $f=\exp$ , you get what I wrote and used.

(I'm thinking of something that may have been confusing: when I wrote $e\left( 1+O\left(\frac{1}{n^2}\right)\right)$ , it meant $e\times \left( 1+O\left(\frac{1}{n^2}\right)\right)$ , not exponential of the parenthesis)