Root Finding of complex trig function

Batmaniac
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Homework Statement



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.


Homework Equations



xtanx = 2 - coshx



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

[img=http://img244.imageshack.us/img244/5373/courantpage48extractdd8.th.jpg]

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

\cos x < \frac{x}{\tan x} < 1.

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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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