Just looking at the following piece of math from Strogaz's dynamic / chaos book.
What I don't understand is the last part, where he claims that O($Ƞ^{2}$) is negligible if $f^{'}(x^{*})!=0$.
I guess I'm also not so sure about why the taylor expansion $f(x^{*}+Ƞ)$ = $f(x^{*}) + Ƞf^{'}(x^{*}) + O(Ƞ^{2}) $
It does make sense to think of it as : $f(x+Ƞ)$ is equal to $f(x^{*})$ + some change induced by Ƞ and some error term. Where the change is approximated by $Ƞ * df/dx$. But how come the second term of the expansion isn't equal to $(x-(x^{*}+Ƞ)) f^{'}(x^{*}+Ƞ)$ where $x^{*}+Ƞ = x$ so the whole thing should just be $0$?
Thanks a lot in advance!
Best regards,
Chen
Best Answer
The Taylor expansion is just the general formula $f(x + h) = f(x) + hf'(x) + \frac{h^2}{2!}f''(x) + ...$
This is not a rigorous proof of the claim, but "intuitively" you can discount higher order $\eta$ terms as $\eta$ is a small perturbation - the square of it (and cube, etc) can be taken to be negligibly small. Then $\dot{\eta} = f'(x^*) \eta$ can be used to see how perturbations grow with time (what is the solution to this ODE)?
This issue is when $f'(x^*) = 0$ - here stability cannot be determined without examining higher order terms which are themselves non-linear. For example, consider $f(x) = x^3 + c$. Linear stability anslysis can be used if $c = 1$, but not $c =0 $ (why?)