Newton's Method Problem: f(x) = xcosx + tan2x
- Thread starter kinerry
- Start date
pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,976
You do know what Newton’s method is?
We need a good guess for a seed number.
So graph the function to see an approximate a root.
In this case it appears to be about 3.7
\(\displaystyle \L
\begin{array}{rcl}
f(x) & = & x\cos (x) + \tan (2x) \\
f'(x) & = & \cos (x) - x\sin (x) + 2\sec ^2 (2x) \\
g_1 & = & 3.7 \\
g_2 & = & g_1 - \frac{{f(g_1 )}}{{f'(g_1 )}} \\
\end{array}\)
We need a good guess for a seed number.
So graph the function to see an approximate a root.
In this case it appears to be about 3.7
\(\displaystyle \L
\begin{array}{rcl}
f(x) & = & x\cos (x) + \tan (2x) \\
f'(x) & = & \cos (x) - x\sin (x) + 2\sec ^2 (2x) \\
g_1 & = & 3.7 \\
g_2 & = & g_1 - \frac{{f(g_1 )}}{{f'(g_1 )}} \\
\end{array}\)
Yes, you have to make a good initial guess or you'll end up with a zero out of your interval.
I chose 3.6 fo this. Besides, I wanted to play around with the animation.
Newton's Method is just interations. Find your derivative of \(\displaystyle xcos(x)+2tan(2x)\), then use the formula over and overn until the answers become very close together.
I chose 3.6 fo this. Besides, I wanted to play around with the animation.
Newton's Method is just interations. Find your derivative of \(\displaystyle xcos(x)+2tan(2x)\), then use the formula over and overn until the answers become very close together.
