Given the function:

$\displaystyle f(x)=5e^{-x}-29x$

When, $\displaystyle x_{0}=\ln(6)$

$\displaystyle x_{1}=?$

I know...

$\displaystyle x_{n+1}=x_{n}-\frac{5e^{-x}-29x}{-5e^{-x}-29}$

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- Oct 15th 2012, 09:48 AMalane1994Newton's Method
Given the function:

$\displaystyle f(x)=5e^{-x}-29x$

When, $\displaystyle x_{0}=\ln(6)$

$\displaystyle x_{1}=?$

I know...

$\displaystyle x_{n+1}=x_{n}-\frac{5e^{-x}-29x}{-5e^{-x}-29}$ - Oct 15th 2012, 09:59 AMMarkFLRe: Newton's Method
You will have:

$\displaystyle x_1=\ln(6)-\frac{5e^{-\ln(6)}-29\ln(6)}{-5e^{-\ln(6)}-29}$

Now, use the two identities:

$\displaystyle -\ln(y)=\ln\left(\frac{1}{y} \right)$ and $\displaystyle e^{\ln(y)}=y$ to simplify. - Oct 15th 2012, 10:02 AMalane1994Re: Newton's Method
Thanks, I think it was the ln's that threw me... I hate those things...

- Oct 15th 2012, 10:09 AMMarkFLRe: Newton's Method
I advise you to learn to love the natural log function, as you will get to know it well during your study of the calculus. (Nod)(Cool)

- Oct 15th 2012, 10:12 AMalane1994Re: Newton's Method
Could you give me an answer. I have plugged it in two different times and gotten.

-80.169265 and 23.752784.

I have to go to 6 decimal places. - Oct 15th 2012, 10:28 AMMarkFLRe: Newton's Method
I get:

$\displaystyle x_1=\ln(6)-\frac{\frac{5}{6}-29\ln(6)}{-\frac{5}{6}-29}\approx0.077982108079$ - Oct 16th 2012, 04:49 AMalane1994Re: Newton's Method
When you put it into your calculator, do you put it in all at once? Or do you break it up into segments to put in your calculator. Because I plugged the same thing you have there and I got a different answer...

- Oct 16th 2012, 09:28 AMMarkFLRe: Newton's Method
I enter the entire expression. Exactly what are you entering?

- Oct 16th 2012, 09:42 AMalane1994Re: Newton's Method
I believe that it was just misplaced parentheses.