Section 1.3 of Burden and Faires: Big O Convergence

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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..

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