Download Imaginary numbers, unsolveable equations, and Newton`s method

Imaginary numbers, unsolveable equations, and
Newton’s method
How a mathematician copes
Jeffrey Diller
Department of Mathematics
University of Notre Dame
March 24, 2012
Mathematicians study
Mathematicians study
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numbers.
Mathematicians study
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numbers.
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equations.
Mathematicians study
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numbers.
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equations.
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geometry.
Equations we can solve
Equations we can solve
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3x = 2.
Renée Descartes
Equations we can solve
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3x = 2.
Renée Descartes
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More generally, any linear equation Ax = B.
Equations we can’t solve
Equations we can’t solve
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5x = 50x.
Equations we can’t solve
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5x = 50x.
cos x = x.
Equations we can’t solve
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5x = 50x.
cos x = x.
x 5 − 10x 3 + 3 = 0.
Equations we can’t solve
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5x = 50x.
cos x = x.
x 5 − 10x 3 + 3 = 0.
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x 2 = 2.
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Equations we can’t solve
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5x = 50x.
cos x = x.
x 5 − 10x 3 + 3 = 0.
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x 2 = 2.
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Essentially any other, non-linear equation.
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Equations we can’t solve
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5x = 50x.
cos x = x.
x 5 − 10x 3 + 3 = 0.
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x 2 = 2.
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Essentially any other, non-linear equation.
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DRAT!
Fortunately, close is often good enough when it comes to solving
equations.
x = 1, 1.4, 1.41, 1.414, . . .
is an increasingly good list of approximations for a solution of
x 2 = 2.
Fortunately, close is often good enough when it comes to solving
equations.
x = 1, 1.4, 1.41, 1.414, . . .
is an increasingly good list of approximations for a solution of
x 2 = 2.
In fact, 1.4142 = 1.999396, so x = 1.414 is for most purposes
indistinguishable from an actual solution of x 2 = 2.
Fortunately, close is often good enough when it comes to solving
equations.
x = 1, 1.4, 1.41, 1.414, . . .
is an increasingly good list of approximations for a solution of
x 2 = 2.
In fact, 1.4142 = 1.999396, so x = 1.414 is for most purposes
indistinguishable from an actual solution of x 2 = 2.
There are many methods to approximate solutions to equations.
Newton’s Method
Newton’s idea, better expressed by Raphson: pretend equation is
linear in order to get closer to solving it.
Newton’s Method
Newton’s idea, better expressed by Raphson: pretend equation is
linear in order to get closer to solving it.
Given a guess xn at a solution of f (x) = 0, Newton’s method
generates a new guess
xn+1 = N(xn ) := xn −
f (xn )
.
f 0 (xn )
A competitor
Interval bisection is another way to approximate solutions of
equations.
A competitor
Interval bisection is another way to approximate solutions of
equations. Take for example f (x) = x 2 − 2.
A competitor
Interval bisection is another way to approximate solutions of
equations. Take for example f (x) = x 2 − 2.
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Since f (1) = −1 < 0 and f (2) = 2 > 0, the intermediate
value theorem promises a root of f (x) in the interval
[a0 , b0 ] := [1, 2].
A competitor
Interval bisection is another way to approximate solutions of
equations. Take for example f (x) = x 2 − 2.
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Since f (1) = −1 < 0 and f (2) = 2 > 0, the intermediate
value theorem promises a root of f (x) in the interval
[a0 , b0 ] := [1, 2].
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Consider the value f (1.5) = .25 at the midpoint 1.5 of [1, 2].
A competitor
Interval bisection is another way to approximate solutions of
equations. Take for example f (x) = x 2 − 2.
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Since f (1) = −1 < 0 and f (2) = 2 > 0, the intermediate
value theorem promises a root of f (x) in the interval
[a0 , b0 ] := [1, 2].
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Consider the value f (1.5) = .25 at the midpoint 1.5 of [1, 2].
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Since f (1.5) has the same sign as f (b0 ), we replace b0 with
b1 = 1.5 and leave the other endpoint a1 = a0 alone.
A competitor
Interval bisection is another way to approximate solutions of
equations. Take for example f (x) = x 2 − 2.
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Since f (1) = −1 < 0 and f (2) = 2 > 0, the intermediate
value theorem promises a root of f (x) in the interval
[a0 , b0 ] := [1, 2].
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Consider the value f (1.5) = .25 at the midpoint 1.5 of [1, 2].
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Since f (1.5) has the same sign as f (b0 ), we replace b0 with
b1 = 1.5 and leave the other endpoint a1 = a0 alone.
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And repeat: [a0 , b0 ] ⊃ [a1 , b1 ] ⊃ [a2 , b2 ] ⊃ . . . are intervals
that shrink to the root x.
Square root of 2 via interval bisection
Square root of 2 via interval bisection (cont)
Square root of 2 via interval bisection (cont)
Square root of 2 via Newton’s Method
Now we approximate
√
2 using Newton’s Method.
Square root of 2 via Newton’s Method
Now we approximate
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√
2 using Newton’s Method.
Let x0 = 2 be our first guess at a root of f (x) = x 2 − 2.
Square root of 2 via Newton’s Method
Now we approximate
√
2 using Newton’s Method.
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Let x0 = 2 be our first guess at a root of f (x) = x 2 − 2.
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Get x1 , x2 , . . . from xn+1 = N(xn ) =
1
2
(xn + 2/xn ) .
Square root of 2 via Newton’s Method
Now we approximate
√
2 using Newton’s Method.
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Let x0 = 2 be our first guess at a root of f (x) = x 2 − 2.
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Get x1 , x2 , . . . from xn+1 = N(xn ) =
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The results:
1
2
(xn + 2/xn ) .
Square root of 2 via Newton’s Method
Now we approximate
√
2 using Newton’s Method.
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Let x0 = 2 be our first guess at a root of f (x) = x 2 − 2.
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Get x1 , x2 , . . . from xn+1 = N(xn ) =
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The results:
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Conclusion: there’s a reason interval bisection wasn’t named
after anyone famous.
1
2
(xn + 2/xn ) .
A closer look
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In interval bisection, each interval is half as long as its
predecessor.
A closer look
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In interval bisection, each interval is half as long as its
predecessor.
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I.e. approximations of a root are guaranteed to improve by
exactly 1 (binary) digit at each step.
A closer look
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In interval bisection, each interval is half as long as its
predecessor.
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I.e. approximations of a root are guaranteed to improve by
exactly 1 (binary) digit at each step.
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In Newton’s method, if f (x0 ) = 0 and x = x0 + δ is a good
guess at x0 , then we can use Taylor approximation
A closer look
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In interval bisection, each interval is half as long as its
predecessor.
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I.e. approximations of a root are guaranteed to improve by
exactly 1 (binary) digit at each step.
I
In Newton’s method, if f (x0 ) = 0 and x = x0 + δ is a good
guess at x0 , then we can use Taylor approximation
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f (x) ≈ f 0 (x0 )δ + 12 f 00 (x0 )δ 2 to obtain
A closer look
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In interval bisection, each interval is half as long as its
predecessor.
I
I.e. approximations of a root are guaranteed to improve by
exactly 1 (binary) digit at each step.
I
In Newton’s method, if f (x0 ) = 0 and x = x0 + δ is a good
guess at x0 , then we can use Taylor approximation
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f (x) ≈ f 0 (x0 )δ + 12 f 00 (x0 )δ 2 to obtain
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N(x) ≈ x0 + C δ 2 , where C is a constant involving f 0 (x0 ) and
f 00 (x0 ) but not δ.
A closer look
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In interval bisection, each interval is half as long as its
predecessor.
I
I.e. approximations of a root are guaranteed to improve by
exactly 1 (binary) digit at each step.
I
In Newton’s method, if f (x0 ) = 0 and x = x0 + δ is a good
guess at x0 , then we can use Taylor approximation
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f (x) ≈ f 0 (x0 )δ + 12 f 00 (x0 )δ 2 to obtain
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N(x) ≈ x0 + C δ 2 , where C is a constant involving f 0 (x0 ) and
f 00 (x0 ) but not δ.
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So if x is accurate to k digits (i.e. δ ≈ 2−k ), then N(x) is
accurate to 2k digits (i.e. δ 2 = 2−2k ).
A closer look
I
In interval bisection, each interval is half as long as its
predecessor.
I
I.e. approximations of a root are guaranteed to improve by
exactly 1 (binary) digit at each step.
I
In Newton’s method, if f (x0 ) = 0 and x = x0 + δ is a good
guess at x0 , then we can use Taylor approximation
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f (x) ≈ f 0 (x0 )δ + 12 f 00 (x0 )δ 2 to obtain
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N(x) ≈ x0 + C δ 2 , where C is a constant involving f 0 (x0 ) and
f 00 (x0 ) but not δ.
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So if x is accurate to k digits (i.e. δ ≈ 2−k ), then N(x) is
accurate to 2k digits (i.e. δ 2 = 2−2k ).
Newton’s method: the caveat
WARNING: the reasoning for Newton’s method is local; it
requires that x be close to x0 .
This is especially applicable in the event that
Newton’s method: the caveat
WARNING: the reasoning for Newton’s method is local; it
requires that x be close to x0 .
This is especially applicable in the event that
Life is cruel, part 1: Your equation doesn’t have a solution. E.g.
x 2 = −1.
Newton’s method: the caveat
WARNING: the reasoning for Newton’s method is local; it
requires that x be close to x0 .
This is especially applicable in the event that
Life is cruel, part 1: Your equation doesn’t have a solution. E.g.
x 2 = −1.
Solution: create some more numbers.
Complex numbers
There is no real number x that solves x 2 = −1. We define i to be
a new ‘imaginary’ number with the property that i 2 = −1. A
complex
is a number with real and imaginary parts: e.g.
√ number
1
2 + i, 2 − 2 i.
Complex numbers
There is no real number x that solves x 2 = −1. We define i to be
a new ‘imaginary’ number with the property that i 2 = −1. A
complex
is a number with real and imaginary parts: e.g.
√ number
1
2 + i, 2 − 2 i.
How is this OK?
Complex numbers
There is no real number x that solves x 2 = −1. We define i to be
a new ‘imaginary’ number with the property that i 2 = −1. A
complex
is a number with real and imaginary parts: e.g.
√ number
1
2 + i, 2 − 2 i.
How is this OK?
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Logically consistent.
Complex numbers
There is no real number x that solves x 2 = −1. We define i to be
a new ‘imaginary’ number with the property that i 2 = −1. A
complex
is a number with real and imaginary parts: e.g.
√ number
1
2 + i, 2 − 2 i.
How is this OK?
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Logically consistent.
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Convenient for describing physical phenomena. E.g.
electromagnetic waves.
Complex numbers
There is no real number x that solves x 2 = −1. We define i to be
a new ‘imaginary’ number with the property that i 2 = −1. A
complex
is a number with real and imaginary parts: e.g.
√ number
1
2 + i, 2 − 2 i.
How is this OK?
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Logically consistent.
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Convenient for describing physical phenomena. E.g.
electromagnetic waves.
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Many equations without real solutions have complex solutions.
E.g x 100 = −34 has 100 complex solutions but no real ones!
More generally,
Complex numbers
There is no real number x that solves x 2 = −1. We define i to be
a new ‘imaginary’ number with the property that i 2 = −1. A
complex
is a number with real and imaginary parts: e.g.
√ number
1
2 + i, 2 − 2 i.
How is this OK?
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Logically consistent.
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Convenient for describing physical phenomena. E.g.
electromagnetic waves.
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Many equations without real solutions have complex solutions.
E.g x 100 = −34 has 100 complex solutions but no real ones!
More generally,
Theorem (Fundamental Theorem of Algebra)
Every polynomial of degree d has exactly d complex roots
Complex numbers
There is no real number x that solves x 2 = −1. We define i to be
a new ‘imaginary’ number with the property that i 2 = −1. A
complex
is a number with real and imaginary parts: e.g.
√ number
1
2 + i, 2 − 2 i.
How is this OK?
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Logically consistent.
I
Convenient for describing physical phenomena. E.g.
electromagnetic waves.
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Many equations without real solutions have complex solutions.
E.g x 100 = −34 has 100 complex solutions but no real ones!
More generally,
Theorem (Fundamental Theorem of Algebra)
Every polynomial of degree d has exactly d complex roots
counted with multiplicity).
(Yes, yes:
Picturing complex numbers
The complex plane C: the real and imaginary parts of a complex
number may be regarded as horizontal and vertical coordinates of a
point in the plane.
Newton meets Cayley
In the late 1800’s, Arthur Cayley looked at using Newton’s method
with an arbitrary complex initial guess to find solutions to
polynomial equations.
Newton meets Cayley
In the late 1800’s, Arthur Cayley looked at using Newton’s method
with an arbitrary complex initial guess to find solutions to
polynomial equations.
He tackled the case of degree two polynomials immediately.
Newton meets Cayley (cont)
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Let p(z) = az 2 + bz + c be a quadratic polynomial with
complex coefficients a, b, c and distinct roots r1 , r2 .
Newton meets Cayley (cont)
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Let p(z) = az 2 + bz + c be a quadratic polynomial with
complex coefficients a, b, c and distinct roots r1 , r2 .
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Let z0 ∈ C be an initial guess and z1 , z2 , . . . be successive
guesses obtained by applying Newton’s method repeatedly to
z0 .
Newton meets Cayley (cont)
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Let p(z) = az 2 + bz + c be a quadratic polynomial with
complex coefficients a, b, c and distinct roots r1 , r2 .
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Let z0 ∈ C be an initial guess and z1 , z2 , . . . be successive
guesses obtained by applying Newton’s method repeatedly to
z0 .
Theorem (Cayley)
If ` ⊂ C is the line segment joining r1 to r2 , then lim zn = r1 if and
only if z0 and r1 lie on the same side of the perpendicular bisector
of `.
Newton meets Cayley (cont)
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Let p(z) = az 2 + bz + c be a quadratic polynomial with
complex coefficients a, b, c and distinct roots r1 , r2 .
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Let z0 ∈ C be an initial guess and z1 , z2 , . . . be successive
guesses obtained by applying Newton’s method repeatedly to
z0 .
Theorem (Cayley)
If ` ⊂ C is the line segment joining r1 to r2 , then lim zn = r1 if and
only if z0 and r1 lie on the same side of the perpendicular bisector
of `.
How to prove this?
Newton’s method: the caveat (cont)
Life is cruel, Part 2: What if you’re just clueless? That is, your
equation has solutions, but you have no idea at all where they are.
E.g. there are seven complex numbers that solve
x 7 − 21x 6 + 15x 3 + x − 5 = 0,
but where are they?
Newton meets Cayley (cont)
Alas!
Newton meets Cayley (cont)
Alas! Degree three (i.e. cubic) polynomial equations stumped
Cayley.
Newton meets Cayley (cont)
Alas! Degree three (i.e. cubic) polynomial equations stumped
Cayley.
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1879: “the case of a cubic polynomial appears to present
considerable difficulty.”
Newton meets Cayley (cont)
Alas! Degree three (i.e. cubic) polynomial equations stumped
Cayley.
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1879: “the case of a cubic polynomial appears to present
considerable difficulty.”
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1890: (translated from French) “I hope to apply this theory to
the case of a cubic equation, but the calculations are much
more difficult.”
Newton meets Cayley (cont)
Alas! Degree three (i.e. cubic) polynomial equations stumped
Cayley.
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1879: “the case of a cubic polynomial appears to present
considerable difficulty.”
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1890: (translated from French) “I hope to apply this theory to
the case of a cubic equation, but the calculations are much
more difficult.”
. . . and then nothing more from Cayley on this subject.
100 years later
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Computer pictures show why Cayley got stuck: tracking which
initial guesses lead to which solutions of a polynomial
equation results in complicated ‘fractal’ pictures.
100 years later
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Computer pictures show why Cayley got stuck: tracking which
initial guesses lead to which solutions of a polynomial
equation results in complicated ‘fractal’ pictures.
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Mathematicians studying complex dynamics begin to
understand these pictures in the 80’s and 90’s.
100 years later
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Computer pictures show why Cayley got stuck: tracking which
initial guesses lead to which solutions of a polynomial
equation results in complicated ‘fractal’ pictures.
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Mathematicians studying complex dynamics begin to
understand these pictures in the 80’s and 90’s.
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In 2001, Hubbard, Schleicher, and Sutherland show how to
reliably choose initial guesses that will find all solutions of a
polynomial equation.
Life’s not so bad after all
Theorem (Hubbard, Schleicher, Sutherland)
Applying Newton’s method to d 2 evenly spaced complex initial
guesses on a large circle will suffice to ‘find’ all solutions of a
polynomial equation of degree d.
Some further questions
Some further questions
1. How big can the set of bad initial guesses be for a polynomial
of given degree? (Apparently not very).
Some further questions
1. How big can the set of bad initial guesses be for a polynomial
of given degree? (Apparently not very).
2. Is there an analogue for the HSS Theorem for other
approximation schemes related to Newton’s Method: e.g. the
secant method?
Some further questions
1. How big can the set of bad initial guesses be for a polynomial
of given degree? (Apparently not very).
2. Is there an analogue for the HSS Theorem for other
approximation schemes related to Newton’s Method: e.g. the
secant method?
3. Same question for Newton’s method applied to systems of
polynomial equations?
Some further questions
1. How big can the set of bad initial guesses be for a polynomial
of given degree? (Apparently not very).
2. Is there an analogue for the HSS Theorem for other
approximation schemes related to Newton’s Method: e.g. the
secant method?
3. Same question for Newton’s method applied to systems of
polynomial equations?
Thank you!