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monkey*

  • one year ago

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

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  1. monkey*
    • one year ago
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    Help please.

  2. monkey*
    • one year ago
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    Anyone?

  3. xapproachesinfinity
    • one year ago
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    What? i suppose this is newton's method appilication

  4. mathmate
    • one year ago
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    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

  5. xapproachesinfinity
    • one year ago
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    oh i see

  6. mathmate
    • one year ago
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    For example, if f(x)=2x+1, and \(x_0\)=0, then \(x_1 = f(x_0) = 2(0)+1=1\) \(x_2 = f(x_1) = 2(1)+1=3\) \(x_3 = f(x_2) = 2(3)+1=7\) ....

  7. monkey*
    • one year ago
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    Still a little confused about how you got everything.?

  8. xapproachesinfinity
    • one year ago
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    @mathmate that sequence is 2^n-1 divergent right?

  9. mathmate
    • one year ago
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    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.

  10. mathmate
    • one year ago
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    @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.

  11. monkey*
    • one year ago
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    Oh, okay. Thanks for explaining it.

  12. mathmate
    • one year ago
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    You're welcome! :)

  13. jdoe0001
    • one year ago
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    testing something

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