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 $\alpha$. If $\exists K>0$ with $|\alpha _{n}-\alpha |\leq K|\beta _{n}|$ for large $n$, then $\{\alpha _{n}\}_{n=1}^{\infty }$ converges to $\alpha$ with rate of convergence $O(\beta _{n})$.

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

Therefore

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

and

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

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 $O({\frac {1}{n^{2}}})$.

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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 $|F(h)-L|\leq K|G(h)|$ 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

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

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

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