Newton's Method - Calculus 1 LecturesMethod12.pdf · NEWTON’S METHOD REAL NEWTON METHOD EXAMPLES COMPLEX NEWTON’S METHOD How do we nd roots, i.e., solve f(x) = 0? f(x) = ax2 +

NEWTON’S METHOD

Newton’s Method - Calculus 1 Lecture

Dr. Rich StankewitzBall State University

Dr. Rich Stankewitz Ball State University Newton’s Method - Calculus 1 Lecture

NEWTON’S METHOD

Outline Talk

I. Newton’s Method for approximating roots

i) Real Newton’s methoda. f (x) = x − cos xb. f (x) = (x − 1)(x − 2)c. f (x) = x(x − 1)(x + 1)

ii) Complex Newton’s methoda. f (z) = (z − α)(z − β)b. f (z) = z3 − 1c. f (z) = z(z − 1)(z + 1)


NEWTON’S METHODREAL NEWTON METHOD EXAMPLESCOMPLEX NEWTON’S METHOD

How do we find roots, i.e., solve f (x) = 0?

f (x) = ax2 + bx + c

x = −b±√b2−4ac2a




f (x) = ax2 + bx + c

x = −b±√b2−4ac2a




f (x) = rx4 + dx3 + ax2 + bx + c

A formula (procedure) exists to get answers in terms of radicals.




f (x) = rx4 + dx3 + ax2 + bx + c

A formula (procedure) exists to get answers in terms of radicals.




f (x) = sx5 + rx4 + dx3 + ax2 + bx + c

NO formula (procedure) exists to get answers in terms of radicals.Proved by Galois Theory.




f (x) = sx5 + rx4 + dx3 + ax2 + bx + c

NO formula (procedure) exists to get answers in terms of radicals.Proved by Galois Theory.




f (x) = cos x − x

NO formula. This is transcendental .




f (x) = cos x − x

NO formula. This is transcendental .



When a formula/procedue does not exist to EXACTLY find roots,we can approximate roots to any degree of accuracy (up tocomputer limitations) using Newton’s Method, BUT ONLY IF WECAN START WITH A REASONABLE GUESS.

1.0 1.5 2.0 2.5 3.0

-5

0

5

X

α

f̃ (x)

x1 x0

f (x) L

Figure: An illustration of the first step in Newton’s method.



1.0 1.5 2.0 2.5 3.0

-5

0

5

X

α

f̃ (x)

x1 x0

f (x) L

y − f (x0) = f ′(x0)(x − x0)

0− f (x0) = f ′(x0)(x1 − x0)

x1 = x0 −f (x0)

f ′(x0)What if f ′(x0) = 0?



1.0 1.5 2.0 2.5 3.0

-5

0

5

X

α

f̃ (x)

x1 x0

f (x) L

y − f (x0) = f ′(x0)(x − x0)

0− f (x0) = f ′(x0)(x1 − x0)

x1 = x0 −f (x0)

f ′(x0)What if f ′(x0) = 0?



1.0 1.5 2.0 2.5 3.0

-5

0

5

X

α

f̃ (x)

x1 x0

f (x) L

y − f (x0) = f ′(x0)(x − x0)

0− f (x0) = f ′(x0)(x1 − x0)

x1 = x0 −f (x0)

f ′(x0)What if f ′(x0) = 0?



1.0 1.5 2.0 2.5 3.0

-5

0

5

X

α

f̃ (x)

x1 x0

f (x) L

y − f (x0) = f ′(x0)(x − x0)

0− f (x0) = f ′(x0)(x1 − x0)

x1 = x0 −f (x0)

f ′(x0)What if f ′(x0) = 0?



If one seeks to find a root α of a differentiable real valued functionf (x) defined for a real variable x , then one can apply Newton’smethod as follows. We start with an initial guess x0 close to αand define

x1 = x0 −f (x0)

f ′(x0)x2 = x1 −

f (x1)

f ′(x1)

x3 = x2 −f (x2)

f ′(x2)xn+1 = xn −

f (xn)

f ′(xn).

Definition

The function F (x) = x − f (x)f ′(x) is called the Newton Map for f .

Thus, the sequence of iterates F (x0),F (F (x0)),F (F (F (x0))), ... isthe same as the sequence xn generated above, and it will be proventhat xn converges to the sought after root α whenever our originalguess x0 is “close enough” to α.



If one seeks to find a root α of a differentiable real valued functionf (x) defined for a real variable x , then one can apply Newton’smethod as follows. We start with an initial guess x0 close to αand define

x1 = x0 −f (x0)

f ′(x0)x2 = x1 −

f (x1)

f ′(x1)

x3 = x2 −f (x2)

f ′(x2)xn+1 = xn −

f (xn)

f ′(xn).

Definition

The function F (x) = x − f (x)f ′(x) is called the Newton Map for f .

Thus, the sequence of iterates F (x0),F (F (x0)),F (F (F (x0))), ... isthe same as the sequence xn generated above, and it will be proventhat xn converges to the sought after root α whenever our originalguess x0 is “close enough” to α.



Example: Let f (x) = x2 − 3x + 2 = (x − 1)(x − 2) and so

F (x) = x − x2 − 3x + 2

2x − 3=

x2 − 2

2x − 3.

If we let x0 = 0.5, then

x1 = F (x0) = 0.875 x2 = F (F (x0)) = F (x1) = 0.9875

x3 = F (F (F (x0))) = F (x2) = 0.999847561 xn → 1.

If we let x0 = 3, then

x1 = 2.333333 x2 = 2.066666

x3 = 2.0039215 xn → 2.

Some starting points (choices for x0) will find the root 1 whileothers find the root 2. How do we know which root we’ll find?




F (x) = x − x2 − 3x + 2

2x − 3=

x2 − 2

2x − 3.


x1 = F (x0) = 0.875 x2 = F (F (x0)) = F (x1) = 0.9875

x3 = F (F (F (x0))) = F (x2) = 0.999847561 xn → 1.


x1 = 2.333333 x2 = 2.066666

x3 = 2.0039215 xn → 2.





F (x) = x − x2 − 3x + 2

2x − 3=

x2 − 2

2x − 3.


x1 = F (x0) = 0.875 x2 = F (F (x0)) = F (x1) = 0.9875

x3 = F (F (F (x0))) = F (x2) = 0.999847561 xn → 1.


x1 = 2.333333 x2 = 2.066666

x3 = 2.0039215 xn → 2.





F (x) = x − x2 − 3x + 2

2x − 3=

x2 − 2

2x − 3.


x1 = F (x0) = 0.875 x2 = F (F (x0)) = F (x1) = 0.9875

x3 = F (F (F (x0))) = F (x2) = 0.999847561 xn → 1.


x1 = 2.333333 x2 = 2.066666

x3 = 2.0039215 xn → 2.




Go to Real Newton Method Applet.


http://www.jimrolf.com/java/realNewtTool/realNewtTool.html


Investigate f (x) = x(x − 1)(x + 1) with applet.



“The shortest route between two truths in the real domain passesthrough the complex domain.” - Jacques Salomon Hadamard

Complex numbers:z = a + bi ∈ C where i =

√−1.

Let α, β ∈ C.Let f (z) = (z − α)(z − β) and so

F (z) =z2 − αβ

2z − (α + β).

Go to Complex Newton Method Applet.


http://www.jimrolf.com/java/complexNewtTool/complexNewtTool.html


f (z) = z3− 1 = [z − 1][z − (−0.5 + i0.5√

3)][z − (−0.5− i0.5√

3)].

e−2πi/3

1

e2πi/3

Figure: A reasonable (but false) guess for the picture description of theglobal dynamics of F (z), the Newton Map for f (z) = z3 − 1.



f (z) = z3− 1 = [z − 1][z − (−0.5 + i0.5√

3)][z − (−0.5− i0.5√

3)].

e−2πi/3

1

e2πi/3

Figure: A reasonable (but false) guess for the picture description of theglobal dynamics of F (z), the Newton Map for f (z) = z3 − 1.



Theorem

(Common Boundary Condition) Let f (z) be an analytic function

such that its Newton Map F (z) = z − f (z)f ′(z) is a rational map. If

w1 and w2 are roots of f (z), then ∂AF (w1) = ∂AF (w2).

Above, AF (w) = {z ∈ C : F n(z)→ w} is the attracting basin ofw , i.e., the set of all starting values such that Newton’s methodconverges to w .

A moment’s thought tells us that this Common BoundaryCondition forces the picture to necessarily be very complicatedwhen f has 3 or more roots.



∆(1, r) ⊂ AF (1)

∆(e2πi/3, r) ⊂ AF (e2πi/3)

∆(e−2πi/3, r) ⊂ AF (e−2πi/3)

·0

Figure: Local Attracting Property of Newton’s Method for f (z) = z3 − 1.

How can one color in the rest of the red, blue, and turquoise pointsso that each color shares the same boundary??



Attracting basins of Newton’s method for f (z) = z3 − 1.



Show applet for quartic Newton Method.



A menagerie of FRACTALS to see and investigate, includingRabbits, Dragons, and Elephants, Star Clusters, and Galaxies.



(Sumi) Let h1(z) = z2/4 and h2(z) = z2 − 1 and then using thesecond iterates of these maps we set f (z) = g2

1 and g(z) = g22 .

Probability P(z) of tending to ∞under random iteration.Devil’s Colosseum(Dante’s Inferno)

Graph of 1− P(z).Inverted Devil’s Colosseum(Dante’s Purgatory)(Devil’s sand castle)(Fractal wedding cake)



(Sumi) Let h1(z) = z2/4 and h2(z) = z2 − 1 and then using thesecond iterates of these maps we set f (z) = g2

1 and g(z) = g22 .

Probability P(z) of tending to ∞under random iteration.Devil’s Colosseum(Dante’s Inferno)

Graph of 1− P(z).Inverted Devil’s Colosseum(Dante’s Purgatory)(Devil’s sand castle)(Fractal wedding cake)






Documents

Newton's Method - Calculus 1 LecturesMethod12.pdf · NEWTON’S METHOD REAL NEWTON METHOD EXAMPLES COMPLEX NEWTON’S METHOD How do we nd roots, i.e., solve f(x) = 0? f(x) = ax2 +