Ask Experts Questions for FREE Help !
Ask
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
    Ultra Member
     
    #1

    Dec 27, 2006, 04:42 AM
    sinx/x is decreasing
    Here's one for my fellow 'mathies' if they want to tackle it.

    It'de be fun to see some different approaches.

    Prove is strictly decreasing on the interval
    asterisk_man's Avatar
    asterisk_man Posts: 476, Reputation: 32
    Full Member
     
    #2

    Dec 27, 2006, 07:15 AM
    Let me work through it.

    is strictly decreasing in the range if the first derivative is strictly negative in that range



    now I have to show that it is less than 0 in the range
    I think I'll try showing that it is less than 0 @ x=0 and the original equation has no points of inflection in the range.
    (using L'Hôpital's rule since we have 1/x situation)


    hmm... the first derivative is zero @ x=0. This point is in the range and it is not negative. I don't think this fits the definition of "strictly decreasing" since at this point it is neither increasing nor decreasing. Lets clear this one up and then I can try to continue from there. Maybe the range you mean is

    Anyway, this was a fun use of the new math markup!
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
    Ultra Member
     
    #3

    Dec 27, 2006, 08:50 AM
    Thanks for your input. BTW, I like the Koch snowflake avatar.

    Anyway, I thought maybe try the 2nd derivative to find the inflection pts.

    Finding the 2nd derivative, setting it to 0 and solving for x, we find the point

    2.0815789772 to the right of Pi/2. We are concave down from 0 to 2.08. Therefore, have

    negative slope from 0 to Pi/2. This may not be rigorous enough for some.


    Another thought I had was to use the Taylor series for sinx/x









    Therefore, the sequence is decreasing.

    This may need more tweaking. Whatcha think?
    Attached Images
     
    asterisk_man's Avatar
    asterisk_man Posts: 476, Reputation: 32
    Full Member
     
    #4

    Dec 27, 2006, 09:17 PM
    Yeah. I'm on board with the 2nd derivative but that still doesn't help with the fact that is not strictly decreasing in the range given since the 1st derivative is not negative at x=0. If it were negative @ x=0 then I think you've just finished it.

    As far as the Taylor series goes, I think that you're just showing that the terms of the series are strictly decreasing in absolute value.
    You need to prove something like

    And I'm pretty sure that you can not remove the summation and compare the terms so I'm not sure how you would do that.

    This is a good discussion so be sure to add your input.

    As an aside, are you using any application to help with some of your algebra? I have been looking for good free applications but I haven't been overly thrilled by what I've found yet and was hoping to hear your experiences.

    As another aside, the first time that I viewed your newest post I didn't see all of the final set of equations and none of the attached image. Keep it in mind if it looks like I may have lost my mind at some point!
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
    Ultra Member
     
    #5

    Dec 28, 2006, 05:28 AM
    Hey Asterisk Man:

    Thanks for the input. The Taylor series may be OK at a specific point, but it doesn't show

    anything over the domain 0 to Pi/2. I like the 2nd derivative thing better. BTW, I had a typo. It should, indeed, be

    . The slope is 0 at x=0.


    Anyway, I do the algebra myself. If I ever need any help, I use my TI-92

    or Maple. I am sorry, I do not know of any freeware algebra programs. I will scout

    around, though. As far as the images not displaying, I don't know.
    Elisha Grey's Avatar
    Elisha Grey Posts: 31, Reputation: 0
    Junior Member
     
    #6

    Jan 17, 2007, 10:52 AM
    d/dx(sinx/x) = (sin x - x cos x)/x^2 < 0 iff sin x - x cox s < 0 iff sin x < x cos x iff tan x < x, which is not true in this range. In fact the opposite is true: tanx >= x in this range, so sin x/x is strictly increasing in this range.

    Also at 0 lim d/dx(sin x/x) = lim((sinx-xcos x)/x^2) =
    (by L'Hospital's rule) lim((cos x - cos x + x sin x)/2x) = lim(sin x/2x) = 1/2.

    Whoa, baby! Mistake above! For the limit above everything is OK until the next to last form, which should be
    lim(sin x/2), and NOT lim(sin x/2x). So this limit is 0.

Not your question? Ask your question View similar questions

 

Question Tools Search this Question
Search this Question:

Advanced Search

Add your answer here.


Check out some similar questions!

Decreasing water pressure [ 7 Answers ]

For some reason the water pressure in my kitchen sink has recently been reduced to a slow trikle. The rest of the house is fine it's only the kitchen sink. I took the Moen handle off (just the top) and all of the washers etc appear to be good. I also took off and cleaned the spray attachment and...


View more questions Search