Find the first three iterates of f(x)=-3x-7 if x(o at the bottom)=-8

X1=17, x2=-58, x3=167
X1=21, x2=-61, x3=170
X1=-17, x2=58, x3=-167
X1=-21, x2=61, x3=-170

Question

Find the first three iterates of f(x)=-3x-7 if x(o at the bottom)=-8

X1=17, x2=-58, x3=167
X1=21, x2=-61, x3=170
X1=-17, x2=58, x3=-167
X1=-21, x2=61, x3=-170

monkey* · Answer

Help please.

monkey* · Answer

Anyone?

xapproachesinfinity · Answer

What? i suppose this is newton's method appilication

mathmate · Answer

This is called a fixed point iteration, where you iterate $x_{n+1}=f(x_n)$ Read about it here: https://en.wikipedia.org/wiki/Fixed-point_iteration

xapproachesinfinity · Answer

oh i see

mathmate · Answer

For example, if f(x)=2x+1, and $x_0$=0, then
$x_1 = f(x_0) = 2(0)+1=1$
$x_2 = f(x_1) = 2(1)+1=3$
$x_3 = f(x_2) = 2(3)+1=7$
....

monkey* · Answer

Still a little confused about how you got everything.?

xapproachesinfinity · Answer

@mathmate  that sequence is 2^n-1 divergent right?

mathmate · Answer

Yes, in this case.
It can be used to find roots, for example
Solve f(x)=x^2-4x+3=0
We rearrange to have
g(x)=x=(x^2+3)/4
with $x_0$=1.5
then
g($x_0$)=1.3125
g(1.3125=1.18066
g(1.1866)=1.09849
...
eventually it will converge (slowly) to the root x=1.

mathmate · Answer

@monkey* 
you can see another example I just posted previously.  This second example converges slowly to x=1.  The first example (and your question) will not converge, but that's normal.

monkey* · Answer

Oh, okay. Thanks for explaining it.

mathmate · Answer

You're welcome!   :)

jdoe0001 · Answer

testing something