1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Serge Lang: Increasing/Decreasing Function

  1. Jun 7, 2015 #1
    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
    [tex] f(x) = -x + \frac {x^3}{3!} + \sin x [/tex]
    [tex] f'(x) = -1 + \frac {x^2}{2} + \cos x [/tex]
    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 [tex] f'(x) = -1 + \frac {x^2}{2} + \cos x > 0 [/tex]
    [tex] x^2 + 2 \cos x > 2 [/tex]
    [tex] |x| > 2|(1- \cos x)| [/tex]

    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. jcsd
  3. Jun 7, 2015 #2

    Mark44

    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):
    $$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} -+ \dots + (-1)^n\frac{x^{2n + 1}}{(2n + 1)!} + \dots$$
     
  4. Jun 8, 2015 #3
    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):

    ImageUploadedByPhysics Forums1433767329.063119.jpg
     
  5. Jun 8, 2015 #4

    Mark44

    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).
     
  6. Jun 8, 2015 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  7. Jun 8, 2015 #6
    [tex] f(x) = x - \sin x [/tex]
    [tex] f'(x) = 1 - \cos x [/tex]

    ##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).
     
  8. Jun 8, 2015 #7
    I'm working on establishing that level of intuition; my schooling has always been algorithmic. What tips do you have for doing this? Thanks!
     
  9. Jun 8, 2015 #8

    Mark44

    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.
     
  10. Jun 9, 2015 #9

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Serge Lang: Increasing/Decreasing Function
Loading...