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We wish show

$$ \sin x = x - \frac{x^3}{6} + O(x^5) \text{ when } x \to 0 $$

I am rather new to big $O$ notation. I understand this means proving that

$$ \sin x - \left(x - \frac{x^3}{6}\right) = O(x^5) \text{ when x } \to 0 $$

This, in turn, by definition entails proving that there is $C > 0$ and a neighbourhood $\mathcal{N} = (0 - r, 0 + r)$, $r > 0$, s.t.

$$ \forall x \in \mathcal{N} : |\sin x - \left(x - \frac{x^3}{6}\right) | \leq C |x^5| $$

My guess is that I need to use the Taylor approximation of $\sin x$ in some way, but I can't figure out how nor exactly what purpose it would serve.

Gonçalo
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lafinur
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  • I don't know if you ask https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lh%C3%B4pital-rule-or-series-expansion, actually Newton figured out the expansion of trigonometric without using any later tools. Those kind expansions of basic functions can be done by finding its terms of every order from limit like $\lim_{x\to 0}\frac{\sin x - x + x^3/6}{x^5}$, which is a example of your problem. – Nanayajitzuki Nov 24 '24 at 21:23
  • Do you know what property the $n$th degree Taylor polynomial has? What does Taylor’s Theorem actually say? – Ted Shifrin Nov 24 '24 at 21:26
  • Well, my knowledge extends to the following fact: $\sin x - (x - x^3/6)$ is the error term of the Taylor approximation of $\sin x$ with $n=4$ terms, and it simplifies to $x^5(1/5! - x^2/7! + x^4/9! - \ldots)$. I know the series within the parenthesis converges, but I don't know to what value, so that doesn't seem to help in finding the value of $C$. – lafinur Nov 24 '24 at 21:31
  • Generally speaking, the Taylor’s Theorem tells us, you have $\sin x = x - x^3/6 + C(x)x^5$ where $C(x)$ is NOT a constant, which equals to $(\sin x)^{(5)}(\xi(x))/5!$ where $\xi(x)$ is still a function of $x$, $C(x)$ will reach the series coefficient at its limit to $0$. – Nanayajitzuki Nov 24 '24 at 21:44
  • You don't have to find the value of $C$; you just need to show that $1/5! - x^2/7! + \ldots$ converges and is bounded near $0$. – Rob Arthan Nov 24 '24 at 21:45
  • Use Taylor's formula with Lagrange's remainder. – Gribouillis Nov 24 '24 at 22:16

3 Answers3

2

As you are considering the limit as $x$ goes to zero, you are permitted to demand $|x| < 1.$ With that condition, the Taylor series for $\sin x$ is strictly alternating decreasing, while the value $\sin x$ lies between any two consecutive partial sums.

Recall $$ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} ... $$

With our $|x| < 1,$ when also $x > 0$ we have $$ x - \frac{x^3}{3!} < \sin x < x - \frac{x^3}{3!} + \frac{x^5}{5!} $$ so that

$$ 0 < \sin x - x + \frac{x^3}{3!} < \frac{x^5}{5!} $$

Very similar for $x < 0$

This topic seems now to be called the Alternating Sign Test, or AST. At the same time, I didn't find a discussion that applies it to power series as I do here.

Let me define a series, strictly alternating decreasing, and prove basic facts. Let $t_1 > t_2 > t_3 > t_4... > 0$ and the series be $$ t_1 - t_2 + t_3 - t_4 + t_5 - t_6 +-+- $$ It is positive as it is $$ (t_1 - t_2) + (t_3 - t_4) + (t_5 - t_6) +-+- $$ It is smaller than $t_1$ because it is $$ t_1 - (t_2 - t_3) - (t_4 - t_5) - (t_6 - t_7) +-+- $$ It is bigger than $t_1 - t_2$ since it is $$ (t_1 - t_2) + (t_3 - t_4) + (t_5 - t_6) +-+- $$ It is smaller than $t_1 - t_2 + t_3 $ because it is $$ t_1 - (t_2 - t_3) - (t_4 - t_5) - (t_6 - t_7) +-+- $$ AND SO ON

Will Jagy
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2

Here's another way to prove what you want: \begin{align} \left|\sin x - x + \frac{x^3}{6}\right| &= |x|^5\left| \frac1{5!} - \frac{x^2}{7!} + \frac{x^4}{9!} - \cdots \right| \\ &\le |x|^5\left( \frac1{5!} + \frac{|x|^2}{7!} + \frac{|x|^4}{9!} + \cdots \right) \\ &\le |x|^5\left( \frac1{5!} + \frac1{7!} + \frac1{9!} + \cdots \right) \\ &\le e |x|^5, \end{align} where, in the third line, I used $|x|\le 1$, and in the last line, I used $e = \sum_{n=0}^\infty \frac1{n!}$.

Pranay
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1

Here's a very elementary solution:

Consider the following limit and apply L'Hôpital's rule several times:

\begin{align} \lim_{x\to 0}\frac{\sin x-x+x^3/6}{x^5} &=\lim_{x\to 0}\frac{\cos x-1+x^2/2}{5\ x^4}\\ &=\lim_{x\to 0}\frac{-\sin x+x}{20\ x^3}\\ &=\lim_{x\to 0}\frac{-\cos x+1}{60\ x^2}\\ &=\lim_{x\to 0}\frac{\sin x}{120\ x}=\lim_{x\to 0}\frac{\cos x}{120}\\ &=\frac{1}{120} \end{align}

Now just apply the definition of limit to $\varepsilon=1$ (or any other specific value):

There exists $\delta>0$ such that if $|x|<\delta$ then $\displaystyle\left|\frac{\sin x-x+x^3/6}{x^5}-\frac{1}{120}\right|<1$

It follows that $$\left|\frac{\sin x-x+x^3/6}{x^5}\right|<1+\frac{1}{120}=\frac{121}{120}$$ $$\left|\sin x-x+x^3/6\right|<\frac{121}{120}|x|^5$$

jjagmath
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