# Section 1.3 of Burden and Faires: Big O Convergence

## Burden and Faires, 6c, p. 37

Let's consider the sequence $\left.\left(\sin\left(\frac{1}{n}\right)\right)^2\right.$

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

We need to remember those Taylor series polynomials, and think about what's happening as $n \rightarrow \infty$. 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 $\{\beta_n\}_{n=1}^\infty$ is a sequence which converges to zero, and $\{\alpha_n\}_{n=1}^\infty$ converges to a number α. If $\exists K > 0$ with $|\alpha_n - \alpha| \le K|\beta_n|$ for large n, then $\{\alpha_n\}_{n=1}^\infty$ converges to α with rate of convergence On). $\sin(x)=x-\frac{\cos(\xi(x))x^3}{3!}$

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

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

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

## Burden and Faires, 7d, p. 37

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

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

First we might demonstrate that the limit is, in fact, -1. Use L'Hopital's rule: $\lim_{h \rightarrow 0}F(h)=\lim_{h \rightarrow 0}\frac{1-e^h}{h}=\lim_{h \rightarrow 0}\frac{-e^h}{1}=-1$

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 $\left.T_1\right.$ for $\left.e^h\right.$: $F(h)=\frac{1-e^h}{h}=\frac{1-\left(1+h+e^{\xi(h)}\frac{h^2}{2}\right)}{h}$. $F(h)=\frac{1-\left(1+h+e^{\xi(h)}\frac{h^2}{2}\right)}{h}=\frac{-h-e^{\xi(h)}\frac{h^2}{2}}{h}=-1-e^{\xi(h)}\frac{h}{2}$

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

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

Hence, for sufficiently small h ( $\left.h \le 1\right.$), we can choose $\left.K=\frac{e^1}{2}\right.$. Then $|F(h)-(-1)|=|-e^{\xi(h)}\frac{h}{2}| \le Kh$, and $\left.F(h)= -1 + O(h)<\right.$ as $\left.h \rightarrow \infty \right.$.