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Homework Help: Root Finding of complex trig function

  1. Jan 21, 2008 #1
    1. The problem statement, all variables and given/known data

    a) Show that for small values of x, xtanx is approximately equal to x^2 and 2 - coshx is approximately equal to 1 - 0.5(x^2). Draw a conclusion from this regarding the probable number and approximate locations of roots on the interval [-1,1].

    b) Use Newton's method to identify one of these roots to three decimal places.

    2. Relevant equations

    xtanx = 2 - coshx

    3. The attempt at a solution

    My main problem is with the first part.

    How does xtanx approximate to x^2 for small values of x? Better yet, how would I show that? Testing out 10^-10, xtanx = 1.745 x 10^-22 and x^2 = 10^-20, which is still 2 orders of magnitude off. Based on the second part of the question, this approximation should be fairly accurate for x's in [-1,1], but I wouldn't really say it is.

    Even if this difference in order of magnitude is acceptable and still considered approximately equal, how exactly would I show that? (besides subbing in numbers and doing a comparison chart or something).

    I'm assuming once I accurately show that these approximations are true, I can estimate the number and approximate locations of the roots of the equation by finding how many times x^2 intersects 1 - 0.5(x^2) then use the substituted equation in Newton's method to approximate the actual roots.

    - Thanks!
  2. jcsd
  3. Jan 21, 2008 #2
    Looking at the 2 - coshx terms now, I see that for values of x in [-1,1] it does approximate well to 2 - 0.5(x^2) so it's really just the xtanx term that's bother me.
  4. Jan 21, 2008 #3

    Gib Z

    User Avatar
    Homework Helper

    I've uploaded a scan of page 48 from Courant, Volume 1.


    Now, in the second inequality on the page, we multiply every term by cos x;

    [tex] \cos x < \frac{x}{\tan x} < 1[/tex].

    Now, as x approaches zero, the left hand side becomes 1, and the right hand side is 1.

    So the term in the middle must be approximated very well by 1, when x is near zero.

    Another way of saying that is that tan x is approximately x when x is near zero. Hence we can see, x tan x, near zero, can be approximated by x*x = x^2.
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