# Newton's Method

• October 15th 2012, 09:48 AM
alane1994
Newton's Method
Given the function:

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

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

$x_{1}=?$

I know...

$x_{n+1}=x_{n}-\frac{5e^{-x}-29x}{-5e^{-x}-29}$
• October 15th 2012, 09:59 AM
MarkFL
Re: Newton's Method
You will have:

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

Now, use the two identities:

$-\ln(y)=\ln\left(\frac{1}{y} \right)$ and $e^{\ln(y)}=y$ to simplify.
• October 15th 2012, 10:02 AM
alane1994
Re: Newton's Method
Thanks, I think it was the ln's that threw me... I hate those things...
• October 15th 2012, 10:09 AM
MarkFL
Re: 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)
• October 15th 2012, 10:12 AM
alane1994
Re: 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.
• October 15th 2012, 10:28 AM
MarkFL
Re: Newton's Method
I get:

$x_1=\ln(6)-\frac{\frac{5}{6}-29\ln(6)}{-\frac{5}{6}-29}\approx0.077982108079$
• October 16th 2012, 04:49 AM
alane1994
Re: 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...
• October 16th 2012, 09:28 AM
MarkFL
Re: Newton's Method
I enter the entire expression. Exactly what are you entering?
• October 16th 2012, 09:42 AM
alane1994
Re: Newton's Method
I believe that it was just misplaced parentheses.