smalltalk: newton's method, chaos, and fractalssmalltalk: newton’s method, chaos, and...
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Smalltalk: Newton’s Method, Chaos, andFractals
Brian Heinold
Mount St. Mary’s University
February 2, 2017
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A Question
Solve x2 + 7x + 1 = 0.
Answer? Use the quadratic formula:
−b±√b2 − 4ac
2a=−7±
√72 − 4
2
Simplifies to 12(−7 +
√45) and 1
2(−7−√
45)
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A Question
Solve x2 + 7x + 1 = 0.
Answer? Use the quadratic formula:
−b±√b2 − 4ac
2a=
−7±√
72 − 4
2
Simplifies to 12(−7 +
√45) and 1
2(−7−√
45)
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![Page 4: Smalltalk: Newton's Method, Chaos, and FractalsSmalltalk: Newton’s Method, Chaos, and Fractals Brian Heinold Mount St. Mary’s University February 2, 2017 1/82 A Question Solve](https://reader034.vdocuments.mx/reader034/viewer/2022042806/5f7003eed528942d625aa4e0/html5/thumbnails/4.jpg)
A Question
Solve x2 + 7x + 1 = 0.
Answer? Use the quadratic formula:
−b±√b2 − 4ac
2a=−7±
√72 − 4
2
Simplifies to 12(−7 +
√45) and 1
2(−7−√
45)
4 / 82
![Page 5: Smalltalk: Newton's Method, Chaos, and FractalsSmalltalk: Newton’s Method, Chaos, and Fractals Brian Heinold Mount St. Mary’s University February 2, 2017 1/82 A Question Solve](https://reader034.vdocuments.mx/reader034/viewer/2022042806/5f7003eed528942d625aa4e0/html5/thumbnails/5.jpg)
A Question
Solve x2 + 7x + 1 = 0.
Answer? Use the quadratic formula:
−b±√b2 − 4ac
2a=−7±
√72 − 4
2
Simplifies to 12(−7 +
√45) and 1
2(−7−√
45)
5 / 82
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Let’s Check Our Work. . .
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A Harder Question
Solve x5 + 7x + 1 = 0.
Answer? Hmmm...
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A Harder Question
Solve x5 + 7x + 1 = 0.
Answer? Hmmm...
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Let’s Try Wolfram. . .
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Look Carefully
Why = in the first and ≈ in the second?
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Look Carefully
Why = in the first and ≈ in the second?
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Solving Equations
Quadratic equations — solved via quadratic formula oralgebra
Fifth-order equations — there’s no formula and no algebrathat works
Another equation that can’t be solved analytically:cosx = x
High school math’s dirty little secret: Many (if not most)equations from real applications can’t be solved by thetechniques you learn
And this includes integrals and differential equations
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Solving Equations
Quadratic equations — solved via quadratic formula oralgebra
Fifth-order equations — there’s no formula and no algebrathat works
Another equation that can’t be solved analytically:cosx = x
High school math’s dirty little secret: Many (if not most)equations from real applications can’t be solved by thetechniques you learn
And this includes integrals and differential equations
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![Page 14: Smalltalk: Newton's Method, Chaos, and FractalsSmalltalk: Newton’s Method, Chaos, and Fractals Brian Heinold Mount St. Mary’s University February 2, 2017 1/82 A Question Solve](https://reader034.vdocuments.mx/reader034/viewer/2022042806/5f7003eed528942d625aa4e0/html5/thumbnails/14.jpg)
Solving Equations
Quadratic equations — solved via quadratic formula oralgebra
Fifth-order equations — there’s no formula and no algebrathat works
Another equation that can’t be solved analytically:cosx = x
High school math’s dirty little secret: Many (if not most)equations from real applications can’t be solved by thetechniques you learn
And this includes integrals and differential equations
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![Page 15: Smalltalk: Newton's Method, Chaos, and FractalsSmalltalk: Newton’s Method, Chaos, and Fractals Brian Heinold Mount St. Mary’s University February 2, 2017 1/82 A Question Solve](https://reader034.vdocuments.mx/reader034/viewer/2022042806/5f7003eed528942d625aa4e0/html5/thumbnails/15.jpg)
Solving Equations
Quadratic equations — solved via quadratic formula oralgebra
Fifth-order equations — there’s no formula and no algebrathat works
Another equation that can’t be solved analytically:cosx = x
High school math’s dirty little secret: Many (if not most)equations from real applications can’t be solved by thetechniques you learn
And this includes integrals and differential equations
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![Page 16: Smalltalk: Newton's Method, Chaos, and FractalsSmalltalk: Newton’s Method, Chaos, and Fractals Brian Heinold Mount St. Mary’s University February 2, 2017 1/82 A Question Solve](https://reader034.vdocuments.mx/reader034/viewer/2022042806/5f7003eed528942d625aa4e0/html5/thumbnails/16.jpg)
Solving Equations
Quadratic equations — solved via quadratic formula oralgebra
Fifth-order equations — there’s no formula and no algebrathat works
Another equation that can’t be solved analytically:cosx = x
High school math’s dirty little secret: Many (if not most)equations from real applications can’t be solved by thetechniques you learn
And this includes integrals and differential equations
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![Page 17: Smalltalk: Newton's Method, Chaos, and FractalsSmalltalk: Newton’s Method, Chaos, and Fractals Brian Heinold Mount St. Mary’s University February 2, 2017 1/82 A Question Solve](https://reader034.vdocuments.mx/reader034/viewer/2022042806/5f7003eed528942d625aa4e0/html5/thumbnails/17.jpg)
Solving Equations Numerically
So what do we do?
Use a numerical method to get an approximate solution.
What?
It’s a series of steps to follow. Each step hopefully gets us alittle closer to the exact answer.
Newton’s Method is one such numerical method.
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Solving Equations Numerically
So what do we do?
Use a numerical method to get an approximate solution.
What?
It’s a series of steps to follow. Each step hopefully gets us alittle closer to the exact answer.
Newton’s Method is one such numerical method.
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Solving Equations Numerically
So what do we do?
Use a numerical method to get an approximate solution.
What?
It’s a series of steps to follow. Each step hopefully gets us alittle closer to the exact answer.
Newton’s Method is one such numerical method.
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Solving Equations Numerically
So what do we do?
Use a numerical method to get an approximate solution.
What?
It’s a series of steps to follow. Each step hopefully gets us alittle closer to the exact answer.
Newton’s Method is one such numerical method.
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Solving Equations Numerically
So what do we do?
Use a numerical method to get an approximate solution.
What?
It’s a series of steps to follow. Each step hopefully gets us alittle closer to the exact answer.
Newton’s Method is one such numerical method.
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Solving Equations
Solving an equation like x5 + 7x+ 1 = 0 is looking for where thegraph crosses the x-axis.
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Newton’s Method
We start by making a guess at what the solution is.
The closer our guess is to a solution, the better.
Then follow the following sequence of steps.
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Newton’s Method
We start by making a guess at what the solution is.
The closer our guess is to a solution, the better.
Then follow the following sequence of steps.
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Newton’s Method
We start by making a guess at what the solution is.
The closer our guess is to a solution, the better.
Then follow the following sequence of steps.
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Step 1
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Step 2
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Step 3
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Step 4
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Step 5
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Steps 6, 7. . .
Repeat these steps a few more times.
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The Overall Process
1 Make an initial guess
2 Keep repeating the following until we get tired:1 Go to the function2 Follow the tangent line to the axis
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The Overall Process
1 Make an initial guess
2 Keep repeating the following until we get tired:1 Go to the function2 Follow the tangent line to the axis
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Algebraic Version
Instead of graphically, we do things symbolically.
Following the tangent line means using the derivative. A littlealgebra tells us where the tangent line meets the axis:
improved guess = old guess− f(old guess)
f ′(old guess)
Or, in more common notation:
xn+1 = xn −f(xn)
f ′(xn)x0 = initial guess
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Algebraic Version
Instead of graphically, we do things symbolically.
Following the tangent line means using the derivative. A littlealgebra tells us where the tangent line meets the axis:
improved guess = old guess− f(old guess)
f ′(old guess)
Or, in more common notation:
xn+1 = xn −f(xn)
f ′(xn)x0 = initial guess
35 / 82
![Page 36: Smalltalk: Newton's Method, Chaos, and FractalsSmalltalk: Newton’s Method, Chaos, and Fractals Brian Heinold Mount St. Mary’s University February 2, 2017 1/82 A Question Solve](https://reader034.vdocuments.mx/reader034/viewer/2022042806/5f7003eed528942d625aa4e0/html5/thumbnails/36.jpg)
Algebraic Version
Instead of graphically, we do things symbolically.
Following the tangent line means using the derivative. A littlealgebra tells us where the tangent line meets the axis:
improved guess = old guess− f(old guess)
f ′(old guess)
Or, in more common notation:
xn+1 = xn −f(xn)
f ′(xn)x0 = initial guess
36 / 82
![Page 37: Smalltalk: Newton's Method, Chaos, and FractalsSmalltalk: Newton’s Method, Chaos, and Fractals Brian Heinold Mount St. Mary’s University February 2, 2017 1/82 A Question Solve](https://reader034.vdocuments.mx/reader034/viewer/2022042806/5f7003eed528942d625aa4e0/html5/thumbnails/37.jpg)
Algebraic Version
Instead of graphically, we do things symbolically.
Following the tangent line means using the derivative. A littlealgebra tells us where the tangent line meets the axis:
improved guess = old guess− f(old guess)
f ′(old guess)
Or, in more common notation:
xn+1 = xn −f(xn)
f ′(xn)x0 = initial guess
37 / 82
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Example
f(x) = x5 + 7x + 1
Let’s use x = 1 as our initial guess.
Iterating = x− x5 + 7x + 1
5x4 + 7:
x0 = 1
x1 = 1−15 + 7 · 1 + 1
5 · 14 + 7= .25
x2 = .25−.255 + 7 · .25 + 1
5 · .254 + 7= −0.1419 . . .
x3 = −.1419 · · · −(−0.1419 . . . )5 + 7 · (−0.1419 . . . ) + 1
5 · (−0.1419 . . . )4 + 7= −0.1428 . . .
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Example
f(x) = x5 + 7x + 1
Let’s use x = 1 as our initial guess.
Iterating = x− x5 + 7x + 1
5x4 + 7:
x0 = 1
x1 = 1−15 + 7 · 1 + 1
5 · 14 + 7= .25
x2 = .25−.255 + 7 · .25 + 1
5 · .254 + 7= −0.1419 . . .
x3 = −.1419 · · · −(−0.1419 . . . )5 + 7 · (−0.1419 . . . ) + 1
5 · (−0.1419 . . . )4 + 7= −0.1428 . . .
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Example
f(x) = x5 + 7x + 1
Let’s use x = 1 as our initial guess.
Iterating = x− x5 + 7x + 1
5x4 + 7:
x0 = 1
x1 = 1−15 + 7 · 1 + 1
5 · 14 + 7= .25
x2 = .25−.255 + 7 · .25 + 1
5 · .254 + 7= −0.1419 . . .
x3 = −.1419 · · · −(−0.1419 . . . )5 + 7 · (−0.1419 . . . ) + 1
5 · (−0.1419 . . . )4 + 7= −0.1428 . . .
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Example
f(x) = x5 + 7x + 1
Let’s use x = 1 as our initial guess.
Iterating = x− x5 + 7x + 1
5x4 + 7:
x0 = 1
x1 = 1−15 + 7 · 1 + 1
5 · 14 + 7= .25
x2 = .25−.255 + 7 · .25 + 1
5 · .254 + 7= −0.1419 . . .
x3 = −.1419 · · · −(−0.1419 . . . )5 + 7 · (−0.1419 . . . ) + 1
5 · (−0.1419 . . . )4 + 7= −0.1428 . . .
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Example
f(x) = x5 + 7x + 1
Let’s use x = 1 as our initial guess.
Iterating = x− x5 + 7x + 1
5x4 + 7:
x0 = 1
x1 = 1−15 + 7 · 1 + 1
5 · 14 + 7= .25
x2 = .25−.255 + 7 · .25 + 1
5 · .254 + 7= −0.1419 . . .
x3 = −.1419 · · · −(−0.1419 . . . )5 + 7 · (−0.1419 . . . ) + 1
5 · (−0.1419 . . . )4 + 7= −0.1428 . . .
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Example
f(x) = x5 + 7x + 1
Let’s use x = 1 as our initial guess.
Iterating = x− x5 + 7x + 1
5x4 + 7:
x0 = 1
x1 = 1−15 + 7 · 1 + 1
5 · 14 + 7= .25
x2 = .25−.255 + 7 · .25 + 1
5 · .254 + 7= −0.1419 . . .
x3 = −.1419 · · · −(−0.1419 . . . )5 + 7 · (−0.1419 . . . ) + 1
5 · (−0.1419 . . . )4 + 7= −0.1428 . . .
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Example
f(x) = x5 + 7x + 1
Let’s use x = 1 as our initial guess.
Iterating = x− x5 + 7x + 1
5x4 + 7:
x0 = 1
x1 = 1−15 + 7 · 1 + 1
5 · 14 + 7= .25
x2 = .25−.255 + 7 · .25 + 1
5 · .254 + 7= −0.1419 . . .
x3 = −.1419 · · · −(−0.1419 . . . )5 + 7 · (−0.1419 . . . ) + 1
5 · (−0.1419 . . . )4 + 7= −0.1428 . . .
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Look familiar?
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Chaos Appears!
Try Newton’s method on 3
√3−4xx = 0.
The iterating formula simplifies to 4x(1− x).
The first 36 values, using a starting value of .4:.960, .154, .520, .998, .006, .025, .099, .358, .919, .298, .837, .547,
.991, .035, .135, .466, .995, .018, .071, .263, .774, .699, .842, .532,
.996, .016, .064, .241, .732, .785, .676, .876, .434, .983, .068, .252,
.754, .742, .765, .719, .808, .620, .942, .219, .683, .866, .464, .995
Looks pretty random. . .
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Chaos Appears!
Try Newton’s method on 3
√3−4xx = 0.
The iterating formula simplifies to 4x(1− x).
The first 36 values, using a starting value of .4:.960, .154, .520, .998, .006, .025, .099, .358, .919, .298, .837, .547,
.991, .035, .135, .466, .995, .018, .071, .263, .774, .699, .842, .532,
.996, .016, .064, .241, .732, .785, .676, .876, .434, .983, .068, .252,
.754, .742, .765, .719, .808, .620, .942, .219, .683, .866, .464, .995
Looks pretty random. . .
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Chaos Appears!
Try Newton’s method on 3
√3−4xx = 0.
The iterating formula simplifies to 4x(1− x).
The first 36 values, using a starting value of .4:.960, .154, .520, .998, .006, .025, .099, .358, .919, .298, .837, .547,
.991, .035, .135, .466, .995, .018, .071, .263, .774, .699, .842, .532,
.996, .016, .064, .241, .732, .785, .676, .876, .434, .983, .068, .252,
.754, .742, .765, .719, .808, .620, .942, .219, .683, .866, .464, .995
Looks pretty random. . .
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Chaos Appears!
Try Newton’s method on 3
√3−4xx = 0.
The iterating formula simplifies to 4x(1− x).
The first 36 values, using a starting value of .4:.960, .154, .520, .998, .006, .025, .099, .358, .919, .298, .837, .547,
.991, .035, .135, .466, .995, .018, .071, .263, .774, .699, .842, .532,
.996, .016, .064, .241, .732, .785, .676, .876, .434, .983, .068, .252,
.754, .742, .765, .719, .808, .620, .942, .219, .683, .866, .464, .995
Looks pretty random. . .
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Even Stranger. . .
Iterations for two very close starting values:
.400 .401
.960 .960
.154 .151
.520 .512
.998 .999
.006 .002
.025 .009
.099 .036
.358 .137
.919 .474
.400000 .400001
.960 .960
.154 .154
.520 .520
.998 .998
.006 .006
.025 .025
.099 .099
.358 .357
.919 .919
.298 .299
.837 .838
.547 .543
.991 .993
.035 .029
.135 .113
.466 .400
.995 .960
.018 .153
.071 .519
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Even Stranger. . .
Iterations for two very close starting values:
.400 .401
.960 .960
.154 .151
.520 .512
.998 .999
.006 .002
.025 .009
.099 .036
.358 .137
.919 .474
.400000 .400001
.960 .960
.154 .154
.520 .520
.998 .998
.006 .006
.025 .025
.099 .099
.358 .357
.919 .919
.298 .299
.837 .838
.547 .543
.991 .993
.035 .029
.135 .113
.466 .400
.995 .960
.018 .153
.071 .519
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Chaos
This is an example of chaos.
Irregular, but not totally random
Sensitive dependence on initial conditions
Even if our starting values were vanishingly close, say only10−20 apart, it would only take several dozen iterations forthem to start to diverge.
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Chaos
This is an example of chaos.
Irregular, but not totally random
Sensitive dependence on initial conditions
Even if our starting values were vanishingly close, say only10−20 apart, it would only take several dozen iterations forthem to start to diverge.
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Chaos
This is an example of chaos.
Irregular, but not totally random
Sensitive dependence on initial conditions
Even if our starting values were vanishingly close, say only10−20 apart, it would only take several dozen iterations forthem to start to diverge.
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The Double Pendulum: Another Chaotic System
A double pendulum is a pendulum attached to anotherpendulum.
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The Double Pendulum: Another Chaotic System
A double pendulum is a pendulum attached to anotherpendulum.
If we change the starting angle by something as small as justthe width of an atom (like a .0000000001 difference), after 30-60seconds the pendulum will be doing something completelydifferent.
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Another Chaotic System: The Magnetic Pendulum
No matter how hard you try to start it in the same location,each time you release it, after a few seconds, it will be doingsomething different.
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Another Chaotic System: Plinko
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Another Chaotic System: The Weather
The Butterfly Effect — A butterfly flapping its wings inJapan can mean the difference between a tornado and asunny day six months later in Texas
Weather is a chaotic system; this is why we can’t predictthe weather more than a few days out.
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Another Chaotic System: The Weather
The Butterfly Effect — A butterfly flapping its wings inJapan can mean the difference between a tornado and asunny day six months later in Texas
Weather is a chaotic system; this is why we can’t predictthe weather more than a few days out.
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Another Chaotic System: Life
Before leaving for school, I stop to look at myself in themirror for a few seconds. When I get to the intersection atthe road, there is now a car coming that I wouldn’t havemet. While waiting, I see a quarter next to my bike andpick it up. If I hadn’t picked it up, a few days later someonewalking by would have thrown out their back while trying topick it up. Later that day at the doctor’s office that personmet someone else. She was moving to Houston the next day,but she and the guy that threw out his back have a goodconversation and decide to stay in touch. Later they startdating, get married. That marriage would not havehappened if I hadn’t stopped to look in the mirror.
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Some Good Books on Chaos
Chaos: Making a New Science, James Gleick, 1987.
Does God Play Dice? Ian Stewart, 1989.
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Imaginary Numbers
Solutions of x3 − 1 = 0?
x = 1 is one, but there are two others: x = −12 ±
32 i
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Imaginary Numbers
Solutions of x3 − 1 = 0?
x = 1 is one, but there are two others: x = −12 ±
32 i
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Imaginary Numbers
Solutions of x3 − 1 = 0?
x = 1 is one, but there are two others: x = −12 ±
32 i
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Newton’s Method Works with Imaginary Numbers
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Trying different starting values
Let’s see how different starting values behave.
Some will go to one root, some to other.
The closer they are to a root, the more quickly they willget there.
Try each starting value in the range from say -2 to 2 in thereal and imaginary directions.
Color each one based on how long it takes until it is within.00001 of a root.
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Trying different starting values
Let’s see how different starting values behave.
Some will go to one root, some to other.
The closer they are to a root, the more quickly they willget there.
Try each starting value in the range from say -2 to 2 in thereal and imaginary directions.
Color each one based on how long it takes until it is within.00001 of a root.
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Trying different starting values
Let’s see how different starting values behave.
Some will go to one root, some to other.
The closer they are to a root, the more quickly they willget there.
Try each starting value in the range from say -2 to 2 in thereal and imaginary directions.
Color each one based on how long it takes until it is within.00001 of a root.
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Trying different starting values
Let’s see how different starting values behave.
Some will go to one root, some to other.
The closer they are to a root, the more quickly they willget there.
Try each starting value in the range from say -2 to 2 in thereal and imaginary directions.
Color each one based on how long it takes until it is within.00001 of a root.
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Trying different starting values
Let’s see how different starting values behave.
Some will go to one root, some to other.
The closer they are to a root, the more quickly they willget there.
Try each starting value in the range from say -2 to 2 in thereal and imaginary directions.
Color each one based on how long it takes until it is within.00001 of a root.
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This Is What We Get
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A Zoomed-In Version
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Example Orbits of Iterations
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Example Orbits of Iterations
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Example Orbits of Iterations
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Example Orbits of Iterations
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Newton’s Method on z9 − 1 = 0 (Roughly)
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Newton’s Method on zc − 1 = 0 (c is imaginary)
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Newton’s Method on zc − 1 = 0 (c is imaginary)
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Netwon’s Method on z−4.7+.3i + z − 1
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Good Book on Fractals
Chaos and Fractals: New Frontiers of Science, 2nd ed.,Heinz-Otto Peitgen, Hartmut Jurgens, & Dietmar Saupe, 2004
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Thanks for your attention!
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