# Show $f(x) = x^3 - \sin^2{x} \tan{x} < 0$ on $(0, \frac{\pi}{2})$

This is the last of a homework problem set from Principles of Mathematical Analysis (Ch. 8 #18(a)) that I've been working/stuck on for a few days:

Define $f(x) = x^3 - \sin^2{x}\tan{x}.$ Find out whether it is positive or negative for all $x \in (0, \frac{\pi}{2})$, or whether it changes sign. Prove your answer.

I've thought of a couple possible ways to solve it but have gotten stuck each time.

1. Power series: This would be super easy, since the polynomial is gone and all of the other terms are negative. The problem: I'd have to calculate the $\tan$ power series and show various properties of the Bernoulli numbers. Big hassle.
2. Straight up approximation: I've spent most of the time on this method, but I always overestimate, causing me to lose the strict inequality. This was using various trig identities and just basic stuff like $\sin{x} < x < \tan{x}$ on $(0, \frac{\pi}{2})$.
3. Integrating and more approximating. Running into the same difficulty as above; overestimation.

I'm kind of just running around in circles at this point, probably missing something simple. Can I get a hint?

• Clearly the function is $0$ at $0$ and $\to\infty$ as $x\to\pi/2$, and is continuous, so it's enough to show it has no zeros in the open interval. I wonder if some numerical methods of finding zeros could be shown to diverge to the endpoints? Apr 1, 2012 at 22:04
• I was just considering that, in fact. I've gained a lot of respect for numerical/asymptotic methods recently so I'd love to use them! Apr 1, 2012 at 22:06

You can take the derivative to show that it is always negative except at $0$ where it's zero. Though that's not obvious, so you take the derivative again to show the same thing. Until you get something obvious, then work your way back up.

Ok, I think you got it. So I'll add the answer now.

The first three derivatives are:

$$f'(x)=3 x^2 - 2\sin^2(x) - \tan^2(x)$$ $$f''(x)=6 x - 4\cos(x)\sin(x) - 2\sec^2(x)\tan(x)$$ $$f'''(x)=-2(5+2\cos(2x))\tan^4(x)$$

Just plug and play to see that $f(0)=0$, $f'(0)=0$, and $f''(0)=0$.

You can see that $5+2\cos(2 x)\ge 3$ and $\tan^4(x)\ge 0$. Therefore $f'''(x)\le 0$.

As long as $f'''(x)$ is continuous, which is true up to $x<{\pi\over 2}$, then we have $f''(x)<0$ for $0<x<{\pi\over 2}$.

Rinse and repeat. $f'(x)$ is zero at zero and then decreasing, since $f''(x)$ is negative after zero. So $f'(x)<0$ for $0<x<{\pi\over 2}$. Now $f(x)$ is zero at zero and decreasing, since $f'(x)$ is negative after zero.

Therefore $f(x)<0$ for $0<x<{\pi\over 2}$.

Update:

@Aryabhata asked how I got that expression for the third derivative. It was by factoring a polynomial. Without any messing around, you get for the third derivative:

$$6 - 4\cos^2(x) + 4\sin^2(x) - 2\sec^4(x) - 4\sec^2(x)\tan^2(x)$$

You then convert everything to sines, where here we replace $\sin(x)$ with $s$:

$$6 - 4(1-s^2) + 4 s^2 - {2\over (1-s^2)^2} - {4\over (1-s^2)}{{s^2}\over(1-s^2)}$$

Now you pull out a common numeric factor and denominator, and then expand to get:

$$-{2\over(1-s^2)^2}(7 s^4 - 4 s^6)$$

and factoring:

$$-{2 s^4\over(1-s^2)^2}(7 - 4 s^2)$$

There are some obvious trig identities we can apply to simplify that, but for the purposes of this problem, we'll stop right there. We can already see by inspection that $f'''(x)\le 0$, which is all that is needed for the answer above.

(Note: when I did this problem originally, I replaced everything with cosines instead of sines. That made factoring the polynomial much harder. When I saw the answer, I realized I should have used sines instead. So when I wrote up this answer, I used sines to make myself look smarter than I really am. We all do that when we write up proofs, right? We look for the shortest path, probably taking $17$ different dumb detours before writing the final elegant-looking solution. Or maybe that's just me.)

• Well after 9 (8? 10?) derivatives, I think I have solved the entire problem. I guess I just wasn't taking enough of them before! Apr 2, 2012 at 4:10
• It only takes three derivatives. Apr 2, 2012 at 4:57
• Mark: +1. You might wish to use \cos, \sin, and so on.
– Did
Apr 2, 2012 at 6:13
• I think the basic question here is: what led me to take this approach? The answer is quite simple. I plotted the function and it's first three derivatives. I could see the behavior that they all were decreasing and negative between $0$ and $\pi/2$. I went out to three derivatives for exactly the reason that @Didier mentioned, which was to get rid of the $x$ term. Once I saw the behavior, I looked for a way to manipulate the third derivative to make it obvious that it's negative. That didn't have to work out, but it did. Apr 12, 2012 at 1:33
• I would not expect this approach to work in general, since most functions that are merely negative over some range won't have the observed behavior of decreasing as well as the derivatives decreasing. Apr 12, 2012 at 1:33

Power series approach works, though it requires some numerical computation. Since you seemed to be interested in asymptotics and approximations, you might find this interesting.

It is enough to show that

$$\sin ^3 x \gt x^3 \cos x,\quad 0 \lt x \lt \frac{\pi}{2}$$

Now the power series for $\displaystyle \sin^3 x$ is

$$\sin^3 x = x^3 - \frac{x^5}{2} + \frac{13x^7}{120} - \frac{41x^9}{3024} + \frac{671x^{11}}{604800}- \frac{73x^{13}}{1140480} + \dots$$

Note that $\displaystyle \sin^3 x = -\frac{1}{4} (\sin 3x - 3 \sin x)$ and so the $\displaystyle (2k+1)^{th}$ derivative at $\displaystyle 0$ is

$$c_{2k+1} = \frac{(-1)^{k+1}}{4}(3^{2k+1} - 3)$$

Now for $\displaystyle k \ge 4$, it we can show that

$$\left|\frac{c_{2k+1}}{(2k+1)!}\right| \gt \left|\frac{\pi^2}{4} \frac{c_{2k+3}}{(2k+3)!}\right|$$

(It is enough to verify for $\displaystyle k=4$, as $\displaystyle 4|c_{2k+1}| (2k+2)(2k+3) - \pi^2|c_{2k+3}|$ in an increasing function of $\displaystyle k$).

Notice that $\displaystyle \frac{c_{2k+1}}{(2k+1)!}$ is nothing but the coefficient in the Taylor series.

Thus for $\displaystyle 0 \lt x \lt \frac{\pi}{2}$, we have that (alternating series with decreasing absolute value of the terms)

$$\sin ^3 x \gt x^3 - \frac{x^5}{2} + \frac{13x^7}{120} - \frac{41x^9}{3024}$$

Similarly for $\displaystyle 0 \lt x \lt \frac{\pi}{2}$ we have that

$$x^3 \cos x \lt x^3 - \frac{x^5}{2} + \frac{x^7}{24}$$

Now it is easy to see that for $\displaystyle 0 \lt x \lt \frac{\pi}{2}$ we have that

$$x^3 - \frac{x^5}{2} + \frac{13x^7}{120} - \frac{41x^9}{3024} \gt x^3 - \frac{x^5}{2} + \frac{x^7}{24}$$

• Actually, computing a Taylor series requires an infinite number of derivatives. And you had to compute two Taylor series. :) Apr 2, 2012 at 5:12
• @MarkAdler: No, the derivatives are easy to calculate (the $c_k$ for $\sin^3 x$ and $\pm 1$ for $\cos x$). Also, as I showed, you don't need an infinite number as you are actually interested in truncating the series, and getting the inequalities. The point of my first statement was to emphasise that it is a 'simple' enough proof (i.e. less error prone) which can be verified easily by a human. Apr 2, 2012 at 5:44
• (which is what OP seemed to be trying to get at) Apr 2, 2012 at 5:50
• You missed the smiley face. Apr 2, 2012 at 5:55
• Anyway, you said your approach does not require computing derivatives of multiple functions, where obviously it does. Now you're backtracking and saying it's easy to calculate the derivatives of the functions. :) (Note the smiley face.) Apr 2, 2012 at 6:01