Serge Lang: Increasing/Decreasing Function

1. Jun 7, 2015

Cosmophile

1. The problem statement, all variables and given/known data
Consider only values of $x \geq 0$, and let $f(x) = -x + \frac {x^3}{3!} + \sin x$. Show whether $f$ is increasing or decreasing.

2. Relevant equations
$$f(x) = -x + \frac {x^3}{3!} + \sin x$$
$$f'(x) = -1 + \frac {x^2}{2} + \cos x$$
3. The attempt at a solution
I know that $f$ is increasing whe $f' > 0$, and that $f$ is decreasing when $f' < 0$. In the first case, I have $$f'(x) = -1 + \frac {x^2}{2} + \cos x > 0$$
$$x^2 + 2 \cos x > 2$$
$$|x| > 2|(1- \cos x)|$$

Unfortunately, I'm having a hard time making any sense of this solution. For some reason, trigonometric functions are the only functions I've dealt with that give me any issue with these types of problems. It's demoralizing, really. Anyway, any help is greatly appreciated, as always!

2. Jun 7, 2015

Staff: Mentor

I don't see that approach as bearing fruit. Have you studied Maclaurin and Taylor series, yet? This problem seems to be aimed in that direction.
Here's the Maclaurin series for sin(x):
$$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} -+ \dots + (-1)^n\frac{x^{2n + 1}}{(2n + 1)!} + \dots$$

3. Jun 8, 2015

Cosmophile

I've seen them, but I haven't studied them. I think Lang may be hinting to them for future use, though. Here is the full problem (problem number 11):

4. Jun 8, 2015

Staff: Mentor

Can you show that f1 in prob. 11a is increasing? It looks like Lang's intent is for you to work through the various subparts of this problem before tackling the inequality in your first post.

BTW, this is definitely leading up to the Maclaurin series for sin(x) and cos(x).

5. Jun 8, 2015

Ray Vickson

You should first try to determine the behavior of $f(x)$ for $0 \leq x \leq 2 \pi$. For $x > 2 \pi$ the polynomial part continues to increase, while the $\sin$ part just repeats its behavior on the initial segment.

6. Jun 8, 2015

Cosmophile

$$f(x) = x - \sin x$$
$$f'(x) = 1 - \cos x$$

$f$ is increasing when $f'>0$, so we have $1- \cos x > 0 \to 1 \geq \cos x$, which is true $\forall x \in \mathbb R$ because the range of $\cos x$ is $[-1,1]$.

This is the same method I used on the problem I posted, but the outcome was not nearly as friendly (and also is almost definitely wrong).

7. Jun 8, 2015

Cosmophile

I'm working on establishing that level of intuition; my schooling has always been algorithmic. What tips do you have for doing this? Thanks!

8. Jun 8, 2015

Staff: Mentor

I would suggest looking at the graphs of simple polynomials, such as y = x, y = x2, y = x3, and so on, to get a visual feel (as opposed to only algebraic, which is how I interpret your use of "algorithmic") for how these functions behave, both near the origin as well as farther away.

For polynomials with more terms, satisfy yourself that, for large x, the dominant term of y = x3 - 2x2 + 3 (for example) is the x3 term. The larger x is, the less influence the other terms have on the behavior.

9. Jun 9, 2015

Ray Vickson

Well, if $p(x) = x^3/3! - x = x^3/6 - x$, then $p'(x) = x^2 / 2 - 1 > 0$ if $x > \sqrt{2}$. So, since $\pi/2 > \sqrt{2}$, the function $p(x)$ is strictly increasing for $x > \pi/2$, which is even stronger than saying for $x > 2 \pi$. As for the $\sin$ part repeating, that just follows from the fact that the trig functions are periodic, with period $2 \pi$.

I agree with Mark44, that drawing graphs can be helpful. Looking at a graph may not constitute an actual proof, but it helps you to gain insight and to (sometimes, at least) see what type of "analytical" reasoning might be best.