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monkey*
 one year ago
Find the first three iterates of f(x)=3x7 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*
 one year ago
Find the first three iterates of f(x)=3x7 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

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xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0What? i suppose this is newton's method appilication

mathmate
 one year ago
Best ResponseYou've already chosen the best response.3This is called a fixed point iteration, where you iterate \(x_{n+1}=f(x_n)\) Read about it here: https://en.wikipedia.org/wiki/Fixedpoint_iteration

mathmate
 one year ago
Best ResponseYou've already chosen the best response.3For 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*
 one year ago
Best ResponseYou've already chosen the best response.0Still a little confused about how you got everything.?

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0@mathmate that sequence is 2^n1 divergent right?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.3Yes, in this case. It can be used to find roots, for example Solve f(x)=x^24x+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
 one year ago
Best ResponseYou've already chosen the best response.3@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*
 one year ago
Best ResponseYou've already chosen the best response.0Oh, okay. Thanks for explaining it.
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