Taylor Polynomial approximation

zjhok2004
Messages
8
Reaction score
0

Homework Statement


obtain the number r = √15 -3 as an approximation to the nonzero root of the equation x^2 = sinx by using the cubic Taylor polynomial approximation to sinx

Homework Equations


cubic taylor polynomial of sinx = x- x^3/3!

The Attempt at a Solution


Sinx = x-x^3/3! + E(x)
x^2 = x-x^3/6+ E(x)How do I able to obtain the r?
 
Last edited:
Physics news on Phys.org
In order to get exact equality you need
$$\sin(x) = x - x^3/6 + E(x)$$
where ##E(x)## is the error due to using only two terms. As an approximation, we assume this error to be zero, so we pretend that
$$\sin(x) = x - x^3/6$$
and solve the equation
$$x^2 = x - x^3/6$$
 
jbunniii said:
In order to get exact equality you need
$$\sin(x) = x - x^3/6 + E(x)$$
where ##E(x)## is the error due to using only two terms. As an approximation, we assume this error to be zero, so we pretend that
$$\sin(x) = x - x^3/6$$
and solve the equation
$$x^2 = x - x^3/6$$
But how do I able to obtain the number r?
 
zjhok2004 said:
But how do I able to obtain the number r?
As the problem statement says, ##r## is a nonzero root of the equation ##x^2 = \sin(x)##. You will find an approximation to this by solving the equation ##x^2 = x - x^3/6##.
 
jbunniii said:
As the problem statement says, ##r## is a nonzero root of the equation ##x^2 = \sin(x)##. You will find an approximation to this by solving the equation ##x^2 = x - x^3/6##.
Solved, thanks!
 
Last edited:
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...
Back
Top