Section 1.3 of Burden and Faires: Big O Convergence

Burden and Faires, 6c, p. 37

Let's consider the sequence

${\displaystyle \left.\left(\sin \left({\frac {1}{n}}\right)\right)^{2}\right.}$

and try to determine its rate of convergence to zero as ${\displaystyle n\rightarrow \infty }$.

We need to remember those Taylor series polynomials, and think about what's happening as ${\displaystyle n\rightarrow \infty }$. The argument to sine is getting really small, so sine is approaching 0.

We want to know the rate at which it is approaching zero.

Definition 1.18: Suppose ${\displaystyle \{\beta _{n}\}_{n=1}^{\infty }}$ is a sequence which converges to zero, and ${\displaystyle \{\alpha _{n}\}_{n=1}^{\infty }}$ converges to a number ${\displaystyle \alpha }$. If ${\displaystyle \exists K>0}$ with ${\displaystyle |\alpha _{n}-\alpha |\leq K|\beta _{n}|}$ for large ${\displaystyle n}$, then ${\displaystyle \{\alpha _{n}\}_{n=1}^{\infty }}$ converges to ${\displaystyle \alpha }$ with rate of convergence ${\displaystyle O(\beta _{n})}$.

${\displaystyle \sin(x)=x-{\frac {\cos(\xi (x))x^{3}}{3!}}}$

Therefore

${\displaystyle \sin({\frac {1}{n}})={\frac {1}{n}}-{\frac {\cos(\xi ({\frac {1}{n}}))({\frac {1}{n}})^{3}}{3!}}}$

and

${\displaystyle |\left(\sin({\frac {1}{n}})\right)^{2}|=|\left({\frac {1}{n}}-{\frac {\cos(\xi ({\frac {1}{n}}))({\frac {1}{n}})^{3}}{3!}}\right)^{2}|=|\left({\frac {1}{n}}\right)^{2}-2{\frac {\cos(\xi ({\frac {1}{n}}))({\frac {1}{n}})^{4}}{3!}}+\left({\frac {\cos(\xi ({\frac {1}{n}}))({\frac {1}{n}})^{3}}{3!}}\right)^{2}|\leq \left({\frac {1}{n}}\right)^{2}}$

Hence we have found a ${\displaystyle \left.\kappa \right.}$ and a ${\displaystyle \beta _{n}={\frac {1}{n^{2}}}}$ that work, and ${\displaystyle \sin({\frac {1}{n}})}$ converges to ${\displaystyle 0}$ with rate of convergence ${\displaystyle O({\frac {1}{n^{2}}})}$.

Burden and Faires, 7d, p. 37

Consider the function ${\displaystyle F(h)={\frac {1-e^{h}}{h}}}$. We want to find the rate of convergence to -1 as ${\displaystyle \left.h\rightarrow \infty \right.}$.

Definition 1.19: Suppose that ${\displaystyle \lim _{h\to 0}G(h)=0}$ and ${\displaystyle \lim _{h\to 0}F(h)=L}$. If ${\displaystyle \exists K>0}$ such that ${\displaystyle |F(h)-L|\leq K|G(h)|}$ for sufficiently small ${\displaystyle h}$, then ${\displaystyle \left.F(h)=L+O(G(h))\right.}$.

First we might demonstrate that the limit is, in fact, -1. Use L'Hopital's rule:

${\displaystyle \lim _{h\rightarrow 0}F(h)=\lim _{h\rightarrow 0}{\frac {1-e^{h}}{h}}=\lim _{h\rightarrow 0}{\frac {-e^{h}}{1}}=-1}$

Again we use the Taylor series (Maclaurin series, really) to help us out: only in this case we're going to need to go to ${\displaystyle \left.T_{1}\right.}$ for ${\displaystyle \left.e^{h}\right.}$:

${\displaystyle F(h)={\frac {1-e^{h}}{h}}={\frac {1-\left(1+h+e^{\xi (h)}{\frac {h^{2}}{2}}\right)}{h}}}$.

${\displaystyle F(h)={\frac {1-\left(1+h+e^{\xi (h)}{\frac {h^{2}}{2}}\right)}{h}}={\frac {-h-e^{\xi (h)}{\frac {h^{2}}{2}}}{h}}=-1-e^{\xi (h)}{\frac {h}{2}}}$

So

${\displaystyle F(h)-(-1)=-e^{\xi (h)}{\frac {h}{2}}}$

Therefore

${\displaystyle |F(h)-(-1)|=|-e^{\xi (h)}{\frac {h}{2}}|\leq {\frac {e^{h}}{2}}h}$

Hence, for sufficiently small h (${\displaystyle \left.h\leq 1\right.}$), we can choose ${\displaystyle \left.K={\frac {e^{1}}{2}}\right.}$. Then

${\displaystyle |F(h)-(-1)|=|-e^{\xi (h)}{\frac {h}{2}}|\leq Kh}$, and ${\displaystyle \left.F(h)=-1+O(h)<\right.}$ as ${\displaystyle \left.h\rightarrow \infty \right.}$.