Purcell EM Problem 1.2: Theory Behind Numerical Solution?

yucheng
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Homework Statement
(Purcell Electricity and Magnetism Problem 1.2)
Relevant Equations
N/A
The author start of with $\frac{1}{(y+\sqrt{3})^2} - 2 \cdot \frac{1}{1 + y^2} \left( \frac{y}{\sqrt{1+y^2}} \right) = 0$ and arrives at the equation $y = \frac{(1+y^2)^{3/2}}{2(y+\sqrt{3})^2}$ The solution is merely by iterating (use an initial guess value of y, calculate the RHS, then use this answer for the next calculation) Then the answer forms a converging sequence, with the limit ##y \approx 0.1463##. Which method is this supposed to be? Newton's Method?
 
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yucheng said:
Homework Statement:: N/A
Relevant Equations:: N/A

The author start of with $$\frac{1}{(y+\sqrt{3})^2} - 2 \cdot \frac{1}{1 + y^2} \left( \frac{y}{\sqrt{1+y^2}} \right) = 0$$ and arrives at the equation $$y = \frac{(1+y^2)^{3/2}}{2(y+\sqrt{3})^2}$$ The solution is merely by iterating (use an initial guess value of y, calculate the RHS, then use this answer for the next calculation) Then the answer forms a converging sequence, with the limit ##y \approx 0.1463##. Which method is this supposed to be? Newton's Method?
No, not really. It's the fixed-point-iteration because we are looking for a fixed point ##y=f(y)## where ##f(y)## is the function on the right.

Newton is looking for zeroes. Originally by approximating with ##y_0+p## and using only the constant and linear term for ##p##, and ##y_0## the starting guess: ##0=y-f(y)=y_0+p-f(y_0+p)=a_1+b_1p+ r(p)## then solve for ##0=a_1+b_1p_1## and continue with ##y_1 :=p_1+p_2## and solve ##0=a_2+b_2p_3## etc.
 
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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