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I'm trying to prove

$$x-\frac{x^3}{6} < \sin(x) < x - \frac{x^3}{6} + \frac{x^5}{120} ~~ \forall x \in \Bbb R^{+}$$

I wrote $\sin(x)=x-x^3/3!+x^5/5!-x^7/7!+\dots$ and then the expression for $\sin(x)-(x-\frac{x^3}{6})$ i.e. $x^5/5!-x^7/7!+x^9/9!+\dots$.

I don't see why $x^5/5!-x^7/7!+x^9/9!+\dots$ should be positive for all positive real $x$. Any idea?

Simply Beautiful Art
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    That certainly has to be invalid for $x \le 0$. – Daniel Schepler Aug 15 '17 at 21:55
  • @DanielSchepler I think my book is wrong then. Anyhow, how do you prove it for $x>0$ ? –  Aug 15 '17 at 21:55
  • @DanielSchepler I've modified the question after your comment –  Aug 15 '17 at 21:56
  • you can try to prove a more general result: suppose that $(a_n)\downarrow 0$ and $s:=\sum_{k=0}^\infty (-1)^k a_k<\infty$, then $$|s_n-s|\le a_{n+1},\quad\forall n\in\Bbb N_{>0}$$ where $s_n:=\sum_{k=0}^n(-1)^k a_k$. – Masacroso Aug 15 '17 at 21:58
  • @Masacroso That argument will only work for small $x$. – Daniel Schepler Aug 15 '17 at 22:03
  • If $0 < x < \frac{\pi}{2}$ then another argument would be to use the Taylor series theorem with remainder term: $\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} \cos c$ for some $c$ between 0 and $x$. And similarly, $\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \frac{x^7}{7!} \cos c_2$ for some $c_2$ between 0 and $x$. – Daniel Schepler Aug 15 '17 at 22:06
  • Just a note that for $x<0$ inequalities are reversed. This is easy to see since all the functions are odd. – Ennar Aug 15 '17 at 22:08
  • @DanielSchepler For large values of $x$, the argument is trivial, since the lower bound goes to $-\infty$ while the upper bound goes to $+\infty$. So indeed, Masacroso's approach is quite valid. – Simply Beautiful Art Aug 15 '17 at 22:45
  • @SimplyBeautifulArt Yes, it's certainly possible another argument such as yours could supplement it for large $x$. All I was saying was that as it stands, it requires $\frac{x^7}{7!} < \frac{x^5}{5!}$, or $x < \sqrt{6 \cdot 7}$. – Daniel Schepler Aug 15 '17 at 22:59
  • Here's a similar way: https://math.stackexchange.com/questions/2316231/prove-that-x-frac16x3-sinxx-frac16x3-frac1120x5-for-so/2316465#2316465 – marty cohen Aug 16 '17 at 04:39
  • Related : https://math.stackexchange.com/questions/390899/proving-that-x-fracx33-sin-x-x-for-all-x0 – Arnaud D. Jun 04 '20 at 09:06

5 Answers5

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For any $x>0$ we have $\sin(x)<x$, hence by applying $\int_{0}^{t}(\ldots)\,dx$ to both sides we get $1-\cos t < \frac{t^2}{2}$. By applying $\int_{0}^{x}(\ldots)\,dt$ to both sides we get $x-\sin x<\frac{x^3}{6}$, which can be rearranged as $\sin(x)>x-\frac{x^3}{6}$. By performing the same trick again we also get $\sin(x)<x-\frac{x^3}{6}+\frac{x^5}{120}$ and the wanted inequality is proved for any $x>0$.

An equivalent approach is noticing that $$ \iint_{0\leq a \leq b \leq x}(a-\sin a)\,da\,db,\qquad \iiint_{0\leq a\leq b\leq c\leq x}(a-\sin a)\,da\,db\,dc $$

are clearly positive.
We are dealing with odd functions, hence the reversed inequality holds over $\mathbb{R}^-$.

Jack D'Aurizio
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  • Thanks. This is a nice trick. Any idea how to directly prove $x^5/5!-x^7/7!+x^9/9!-....>0$ (for positive $x$) though? –  Aug 15 '17 at 22:07
  • @Blue: $$ \frac{x^5}{5!}-\frac{x^7}{7!}+\ldots = \int_{0}^{x}\int_{0}^{a}(u-\sin u),du,da $$ is the integral of a positive function. Just the same argument in disguise. – Jack D'Aurizio Aug 15 '17 at 22:10
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hint

Prove that

$$1-\frac {x^2}{2}\le \cos (x)\le 1-\frac {x^2}{2}+\frac {x^4}{24} $$

and integrate.

to prove this, it is easier to show that

$$-x\le -\sin (x)\le -x+\frac {x^3}{6} $$

or the easiest

$$0\le 1-\cos (x)\le \frac {x^2}{2} $$

then integrate twice between $0$ and $x $.

Taylor expansion give only local information.

hamam_Abdallah
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Define $$f(x) = \sin x - x + \frac{1}{6}x^3$$ and $$g(x) = x - \frac{1}{6}x^3 + \frac{1}{120}x^5 - \sin x$$ We want to show $f(x) >0$ and $g(x) > 0$ for $x > 0$.

Notice that $f$ and its first four derivatives vanish at $x=0$. If $f$ has a positive zero, say $f(a_0) = 0$, then by Rolle's Theorem $f'$ has a zero $a_1$ with $0 < a_1 < a_0$. Repeating this argument, we have $0 < a_4 < a_3 < a_2 < a_1 < a_0$ with $f^{(n)}(a_n) = 0$. Since $f^{(4)}(x) = \sin x$, we must have $\pi \le a_4$, so $\pi < a_0$. This shows $f(x)$ does not change sign for $0<x<\pi$. Since $-x + (1/6) x^3 > 3/2$ and is increasing for $x > 3$ while $\sin x \ge -1$, we see that $f(x) > 0$ for all $x > 0$.

To see that $g(x) > 0$ for $x>0$, observe that $g(0) = g'(0) = 0$, $g''(x) = f(x)$, and $f(x) > 0$ for $x>0$, as we just showed.

awkward
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Hint: The Maclaurin expansion of $\sin$ is alternating, and for "small enough" values of $x$, the absolute value of the terms are monotonically decreasing. What can you say in general about bounding an alternating series who's terms monotonically decrease?

For "large enough" values of $x$, it is good enough to notice the lower and upper bounds are not in $[-1,1]$ and to then observe that the lower bound is monotonically decreasing while the upper bound is monotonically increasing.

Simply Beautiful Art
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All of them fit in this infinite sequence of inequalities:

$$\frac{x^2}{2!} - \frac{x^4}{4!} + \frac{x^6}{6!} - \cdots \ge 0 \ \forall x \in \mathbb{R} \\ \frac{x^3}{3!} - \frac{x^5}{5!} + \frac{x^7}{7!} -\cdots > 0 \ \forall x>0 \\ \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \cdots > 0 \ \forall x \in \mathbb{R^*}\\ \cdots \cdots \\ $$

It's enough to show the first sum is $\ge 0$ and $>0$ for $0<|x|< \epsilon$, and notice that the derivative of any of those sums is the previous sum. The first inequality follows from the identity of power series $$\frac{x^2}{2!} - \frac{x^4}{4!} + \frac{x^6}{6!} - \cdots= 2 \left( \sum_{k \ge 0}\,(-1)^k\frac{x^{2k+1}}{2^{2k+1}(2k+1)!}\right)^2$$ which one can check directly.

orangeskid
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