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Given the equation $$ \frac{1}{x^4}+e^{x-100}=10^8 $$ that has one positive root > 1, formulate Newton's method for finding the root. Make one iteration with starting value = 1. Try to make another iteration with starting value 0. Explain the different results.

"The two starting values 0 and 1 are too bad to receive good approximation."

Suggest a better starting guess between 0 and 1. Rationalize how you make the assumption. Approximate the error of the starting guess. Clue: Use method-independent error estimation.

My solution

$$ f(x)=\frac{1}{x^4}+e^{x-100}-10^8 $$ $$ f'(x)=-\frac{4}{x^5}+e^{x-100} $$ $$ x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} $$ Use starting value 1 $$ x_1=1-\frac{1+e^{-99}-10^8}{-4+e^{-99}}=-2.5*10^8 $$ Use starting value 0

$x_{1} =$ division by zero!

Starting guess can be rationalized by dropping the small exp term and using 10^-2 as starting guess but that didn't converge for me either. Why not? How do I estimate the error of the starting guess by "method-independent error estimation"?

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There is an error in the derivative: $(x^{-4})'=-4\,x^{-5}$. –  Julián Aguirre Sep 7 '12 at 9:23
    
"dropping the small exp term"? No way. You are looking for the root that is greater than 1. So, $1/x^4$ contributes less than $1$ to the sum (which must be $10^8$). Clearly, $e^{x-100}$ is the dominating term here, and the root is actually greater than $100$... There is another positive root near $0$, where $1/x^4$ dominates. But this is not the one you were told to find. –  user31373 Sep 9 '12 at 2:37

1 Answer 1

up vote 2 down vote accepted

There is a test for the initial guess, but you must know a interval where the root is.

Supose you know this interval, lets call it $[a,b]$. Then you must verify 3 conditions.

1) $f'(x)$ cant change signal for all $x\in [a,b]$;

2) $f''(x)\neq 0$ for all $x\in [a,b]$;

3) $f(x)\cdot f''(x) > 0$ for all $x > 0$.

If this 3 coditions are true, you can choose $a$ or $b$ to apply Newton's Method, convergence will be guaranteed.

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