Results 1 to 6 of 6

Math Help - Sine Integral Function - Inflection Point

  1. #1
    Member
    Joined
    Jan 2011
    Posts
    156

    Sine Integral Function - Inflection Point

    Find the coordinates of the first inflection point to the right of the origin:

    Si(x) = \int^x_0 \frac{sin t}{t}dt

    I think that it's obvious that I have to find the second derivative of Si(x) and then set it equal to zero. Here is what I did:

    Let g(x) = Si(x) = \int^x_0 \frac{sin t}{t}dt

    g'(x) = \frac{dg}{dx} = \frac{d}{dx}[\int^x_0 \frac{sin t}{t}dt = \frac{sin(x)}{x} (according to FTC1)

    g''(x) = (\frac{sin(x)}{x})' = (\sin(x)(x^{-1}))' = \cos(x)(x^{-1}) + \sin(x)(-x^{-2}) = \frac{x \cos(x) - \sin(x)}{x^2}

    Assuming all is well to here (big assumption, I know), I get stuck on solving for zero:

    0 = x \cos(x) - \sin(x)

    When I graph this on my calculator, there isn't an obvious solution.

    For whatever it's worth, my calculator tells me that the answer to this is approximately \frac{2}{3}\pi

    Anybody see where I am going wrong or have a suggestion to get me unstuck?

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Sambit's Avatar
    Joined
    Oct 2010
    Posts
    355
    I think the only way is to use method of iteration.

    EDIT: Or the equation can be rewritten as x=\tan x. Draw the curves y=x and y=\tan x on same graph paper; the first point of intersection on the right of the origin will be your answer.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2011
    Posts
    156
    Thanks. Do you agree with my approach and my calculation of the 2nd derivative? I think that I can use Newton's method to approximate the root, but I need to make sure that I have the calculation correct up to that point.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member Sambit's Avatar
    Joined
    Oct 2010
    Posts
    355
    Yes your calculation is alright.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jan 2011
    Posts
    156
    OK, I've worked this problem out and have determined that the x-coordinate of the inflection point is 4.4934 (using Newton's method). Now I need to evaluate Si(4.4934).

    I'm applying FTC2 to find this answer:

    Let f(x) = \frac{sin(x)}{x} = F'(x)

    Then, if I can determine F(x), I should be able to find Si(4.4934) = F(4.4394) - F(0). But I don't know how to do this (other than with a calculator).

    What do I do? Thanks.
    Last edited by joatmon; April 3rd 2011 at 06:07 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member Sambit's Avatar
    Joined
    Oct 2010
    Posts
    355
    It does not have any closed form. You may use the following expansion: {\rm Si}(x)=\frac{\pi}{2} <br />
                 - \frac{\cos x}{x}\left(1-\frac{2!}{x^2}+\cdots\right)<br />
                 - \frac{\sin x}{x}\left(\frac{1}{x}-\frac{3!}{x^3}+\cdots\right)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Find the inflection point of this function
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: July 12th 2011, 05:36 AM
  2. Find inflection point(s) of the function
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 12th 2009, 07:36 PM
  3. Replies: 0
    Last Post: November 3rd 2009, 11:18 AM
  4. Inflection Point Fun
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 12th 2009, 09:31 PM
  5. Replies: 2
    Last Post: July 23rd 2007, 10:38 PM

Search Tags


/mathhelpforum @mathhelpforum