7

I have numerically determined that the sequence $\{f_x\} = \frac{\sin(x^2)}{x^2}$ approaches $1$ (as $x$ approaches $0$) faster than the sequence $\{g_x\} = \frac{\sin^2(x)}{x^2}$. However, I am stuck for determing the rate of convergence of each sequence.

Should I use the small angle approximation $\sin(x)\approx x$ and the fact that $|\sin(x)|\leq 1$? Thanks.

2 Answers2

5

Use Taylor expansion: $$ f_x=\frac{1}{x^2}\Bigl(x^2-\frac{x^6}{6}+O(x^{10})\Bigr)=1-\frac{x^4}{6}+O(x^8). $$

$$ g_x=\frac{1}{x^2}\Bigl(x-\frac{x^3}{6}+O(x^{5})\Bigr)^2=1-\frac{x^2}{3}+O(x^4). $$

Added detail on the expansion of $g_x$:

$$ \Bigl(x-\frac{x^3}{6}+O(x^{5})\Bigr)^2=x^2-2\cdot x\cdot\frac{x^3}{6}-2\cdot x\cdot O(x^5)+\frac{x^6}{36}+O(x^{10})=x^2-\frac{x^4}{3}+O(x^6) $$

Julián Aguirre
  • 76,354
  • Okay. But ${f_x}$ is supposed to converge to $1$ faster than ${g_x}$, and this expansion doesn't seem to show that? –  Sep 25 '13 at 14:25
  • 1
    @JuliánAguirre I expanded your expression for ${g_x}$ but I got $O(x^8)$. Could you please show more work for your second line? –  Sep 25 '13 at 14:38
  • So does this mean that ${f_x}$ converges faster because it is $O(x^8)$? –  Sep 25 '13 at 16:05
  • No. The important terms are $x^4$ in $f_x$ and $x^2$ in $g_x$:$$\frac{f_x-1}{g_x-1}=\frac{x^2}{2}+O(x^4).$$ – Julián Aguirre Sep 25 '13 at 16:16
  • SO can I say that ${f_x}$ converges to $1$ at a quartic rate faster than ${g_x}$? How would I word this properly? –  Sep 25 '13 at 17:56
  • You could say that $f_x-1$ converges to $0$ as $x^4$, which is faster than $x^2$, the rate of convergence of $g_x-1$. Or that $(f_x-1)/(g_x-1)$ converges to $0$ as $x^2$. – Julián Aguirre Sep 25 '13 at 18:45
4

Let's start with the definition of rate of convergence of functions (There is also definition of rate of convergence of sequences):

Definition. Let $f$ be a function defined on the interval $(a,b)$ that contains $x=0$, and suppose $\lim_{x\to 0}{f(x)=L}$. If there exists a function $g$ for which $\lim_{x\to 0}{g(x)=0}$ and a positive constant $K$ such that $$|f(x)-L| \le K|g(x)|$$ for all sufficiently small values of $x$, then $f(x)$ is said to converge to $L$ with rate of convergence $O(g(x))$.

Let $f(x)=\sin{x^2}/{x^2}$ and $g(x)=(\sin{x})^2/{x^2}$. From Taylor's Theorem, we know that

$$\sin{x}=x-\frac{1}{6}x^{3}\cos{\xi}$$

for some $\xi$ between $0$ and $x$. Hence,

$$f(x)=\frac{x^2-\frac{1}{6}x^{6}\cos{\xi}}{x^2}=1-\frac{1}{6}x^{4}\cos{\xi}$$

We know $\sin{x}=(1-\cos{2x})/{2}$ and

$$\cos{x}=1-\frac{1}{2}x^2+\frac{1}{24}x^4\cos{\xi}$$

for some $\xi$ between $0$ and $x$. Therefore,

$$g(x)=\frac{(1-1+2x^2-\frac{2}{3}x^4\cos{\xi})/2}{x^2}=\frac{x^2-\frac{1}{3}x^4\cos{\xi}}{x^2}=1-\frac{1}{3}x^2\cos{\xi}$$

Finally

$$|\frac{\sin{x^2}}{x^2}-1|=\frac{1}{6}|x^{4}\cos{\xi}|\le \frac{1}{6}|x^{4}|$$

and

$$|\frac{(\sin{x})^2}{x^2}-1|=\frac{1}{3}|x^2\cos{\xi}|\le \frac{1}{3}|x^2|$$

It follows that the rate of convergence of $f(x)$ is $O(x^4)$ and the rate of convergence of $g(x)$ is $O(x^2)$. It is common to express this in shorthand by writing $$\frac{\sin{x^2}}{x^2}=1+O(x^4) \qquad and \qquad \frac{(\sin{x})^2}{x^2}=1+O(x^2)$$ That means an error term with rate of convergence $O(x^4)$ approaches zero faster than an error term with rate of convergence $O(x^2)$, as the value of $x$ approaches zero. Therefore $f(x)$ approaches one faster than $g(x)$ as $x$ approaches zero.

Source: A Friendly Introduction to Numerical Analysis by Brian Bradie.

Dante
  • 1,908