DIRECTED INVESTIGATION

COMPLEX NUMBERS, POLYNOMIALS & ITERATIONS

PART B Completed Outside of Class Time

Introduction

The aim of Part B is to explore the behavior of the number sequences produced by the iterative process known as Newton’s Method by applying it to different mathematical situations. Given a polynomial function f(x), the zeroes of the function occur solutions of the equations f(x) = 0. Mathematical formulas exist to find these solutions for polynomials of order 4 and below, such as the quadratic formula, cubic formula and quartic formula, however for polynomials of order 5 or greater, no such formula exists.

In this case mathematical methods are used to produce an approximation of a zero of the polynomial. The iterative process being explored in this Investigation is Newton’s Method, which involves guessing an initial value believed to be close to the zero, and applying an algorithm that results in a number sequence that converges to a zero of the specified function. The method achieves this by using the gradient of the tangent to f ( x ) to calculate an x-intercept that is closer to the actual zero. This process is then repeated starting with the new value, each iteration giving a closer approximation to the zero than the last. The number sequence produced by this method

The zero that the sequence targets varies with the initial value according to a fractal pattern of convergence regions but usually converges to the zero with value closest to the initial guess. Through Part B the convergence of number sequences created by Newton’s Method iterations from polynomial while comparing their behavior with convergence regions graphed by computer software.

In “Part A” of the investigation, an approximation to a zero of f ( x )=x3−8 was to be found by using

four iterations of Newton’s formula, starting with x0=1

.

a) The regions of convergence using Newton’s Method are displayed below. Clearly explain, in your own words, what the diagram is showing you.

The regions of convergence of Newton’s Method for the function f ( x )=x3−8 are represented in the Argand plane diagram above. The zeroes of the function occur whenf ( x )=0, the values of x that cause this are represented by the yellow dots. These yellow dots are situated at the 3 zeroes of the function which are x=1 ±√3 iandx=2and are located in regions shaded a particular colour, in this case cyan blue or red. These coloured regions illustrate which zero and starting point will iterate to using Newton’s Method. In this diagram the zero at x=2 is surrounded by a region shaded dark blue

which represents the field of all starting values which converge to x=2. This also happens for all starting points in cyan and red regions which converge to x=1+√3 i and x=1−√3 i respectively. These shaded regions have boundaries that form a fractal with increasingly smaller sections of the 3 attraction regions but still influence the convergence of the iterations.

Display the orbit plot for the iterations of Newton’s Method for f ( x )=x3−8 using x0=1−i

. Give sufficient results to clearly determine the behaviour.

Newton’s Method has the formula:

xn+1=x0−f (x0)f ' (x0)

Substituting the given function into the algorithm gives:

xn+1=x0−x0

3−8

3 x02

Using a starting point of x0=1−igives the iterative process

xn+1=(1−i)−((1−i)3−8

3 (1−i)2 )The results of iterations using this formula are recorded in the table below and can also be displayed as an orbit plot.

𝑛 Value afterxn

0 1.0+1.0i1 0.66666667+0.66666667i

2 0.44444444-2.55555556i3 -0.07676321-1.56989679i

4 -1.12544359-1.15190657i

5 -0.77418818-1.79586687i6 -0.9948382-1.70419899i

7 -0.99993352-1.7324538i8 -0.99999994-1.73205086i

9 -1.0-1.73205081i10 -1.0-1.73205081i11 -1.0-1.73205081i12 -1.0-1.73205081i13 -1.0-1.73205081i14 -1.0-1.73205081i560 -1.0-1.73205081i

The orbit plot has been constructed by plotting the points of The number sequence produced starting from x0=1−i converged to a value of x= -1.0-1.73205081i after 9 iterations and stayed constant through 560 iterations calculated by the plotting software. This providing an approximation of a zero at x= -1.0-1.73205081ifor the given function. It is also noted that this value is accurate to 10 significant figures to x=−1−√3 i, therefore it is likely x=−1−√3 i is a zero of the function f ( x )=x3−8.

2.Consider the function f ( x )=x4+16a) What are the exact values of the zeroes of f ( x )

The zeroes of f ( x )=x 4+16 occur when f ( x )=0i.e. when

x4+16=0

x4=−16 Converted to Polar Form:

x4=16 cis(π+k 2π )

x=[16 cis(π+k 2 π )]1/4

Using DeMoivres Theorem:

x=2 cis( π+k 2 π4 )

x=2 cisπ4

,2 cis3 π4

,2 cis5 π4

,2 cis7 π4

x=2(cosπ4

+isinπ4),2(cos

3 π4

+isin3 π4

),2(cos5 π4

+isin5 π4

),2(cos7 π4

+ isin7 π4

),

x=2( 1

√2+ 1

√2i),2(−1

√2+ 1

√2i),2(−1

√2+−1

√2i),2( 1

√2+−1

√2i),

x=2( 1

√2+ 1

√2i),2(−1

√2+ 1

√2i),2(−1

√2+−1

√2i),2( 1

√2+−1

√2i),

x=√2+√2i,−√2+√2i ,−√2−√2 i ,√2−√2 i

Exact value of zeroes of f ( x ) are:

x=±√2±√2i

b) Clearly explain what happens to iterations of Newton’s Method if you start with:

i) an initial value of x0=2+2i

.

Newton’s Method has the formula:

xn+1=x0−f (x0)f '(x0)

Substituting the given function into the algorithm gives:

xn+1=x0−x0

4+16

4 x03

Iterations of this process using initial value x0=2+2 iwere calculated using computer software, yielding the number sequence in the following table.

Diagram 2

The number sequence produced by Newton’s Method for the function f ( x )=x4+16 from a starting point of x0=2+2 i converges to a value of x=1.41421356+1.41421356i and remains constant for all calculated iterations. Given that this value is equal√2+√2i to nine significant figures, the conjecture can be made that the number sequence is

converging to the√2+√2i, an exact zero of f ( x )=x4+16 found in question 2a.

This is supported by the position of x0 in the attraction fields of the given function as graphed by the computer program and displayed in Diagram 1. The initial value is clearly situated in the attraction field of the zero√2+√2i, shaded dark blue in the diagram. This explains the converging behavior of the number sequence.

i) an initial value of x0=1+i

.In the previous question applying Newton’s Method for the function f ( x )=x 4+16 yielded the

iterative process:

xn+1=x0−x0

4+16

4 x03

Iterating from the initial value x0=1+i produces the following table of results:

x0: 2+2i

√2+√2i

√2+√2i

x0:1+i

Diagram 1

𝑛 Value afterxn

0 2.0+2.0i1 1.625+1.625i

2 1.45179506+1.45179506i3 1.41564781+1.41564781i

4 1.41421574+1.41421574i

5 1.41421356+1.41421356i6 1.41421356+1.41421356i

7 1.41421356+1.41421356i8 1.41421356+1.41421356i

9 1.41421356+1.41421356i10 1.41421356+1.41421356i560 1.41421356+1.41421356i

𝑛 Value afterxn

0 1+1i1 1.75+1.75i

2 1.49908892+1.49908892i3 1.42115354+1.42115354i

4 1.41426423+1.41426423i

5 1.41421357+1.41421357i6 1.41421356+1.41421356i

7 1.41421356+1.41421356i8 1.41421356+1.41421356i

9 1.41421356+1.41421356i10 1.41421356+1.41421356i560 1.41421356+1.41421356i

By the 6thiteration, the number sequence produced from a starting value of x0=1+i converged to a value accurate to nine significant figures to the exact function zero of x=√2+√2i. It is reasonable to assume the sequence was targeting this zero, which can again be explained from the functions attraction fields. As illustrated in Diagram 2, the starting value is situated in a large region of the attraction field of zerox=√2+√2i, shaded grey in this diagram and so will converge to the zero as found in the table of results.

c) Investigate (in detail using both computer software and algebra) what happens to the

iterations of Newton’s Method if you start with an initial value of x0=k+ki

for k being a positive number.

The behavior of Newton’s Method Iterations from a starting value of x0=k+ki

can be investigated

in a variety of ways, such as the number sequence produced in terms of variable k , the number

sequence produced for various values of k as well as the attraction fields of f ( x )=x 4+16.

The first iteration in terms of k can be found algebraically:

Newton’s Method has the formula:

xn+1=x0−f (x0)f '(x0)

Substituting f ( x )=x 4+16 into the algorithm gives:

xn+1=x0−x0

4+16

4 x03

From the starting value of x0=k+kithe first iteration is given by the equation:

x1=(k+ki)−((k+ki)4+16

4 (k+ki)3 )x1=(k+ki)−(−4 k4+16

8k3i−8 k3 )x1=¿¿¿

x1=8 k4 i−8 k 4+8 k 4i2−8 k4 i+4 k 4−16

8 k3 i−8 k3

x1=8 k4 i−8 k 4−8 k4−8 k 4i+4 k 4−16

8 k3i−8 k3

x1=−16 k4 +4 k 4−16

8 k3i−8 k3

x1=−3 k4−42 k3i−2k 3

Diagram 3 Diagram 4

Converting to Polar form:

x1=(3k¿¿ 4+4)cisπ

(√8k 3 ) cis3 π4

¿

x1=( 3k 4+4

√8 k3 )cisπ4

This shows the first iteration for all values of koccurs on the line of π4

and has a modulus given by

the value of( 3k4+4

√8k3 ). As k is any positive number there are an infinite number of possible moduli so

all possible locations for this first iteration is given by a line in the direction π4

from the origin and so

have the form k+ki also. The first iteration showed that a value of this form will iterate another on the

same line of π4

, as a result all iterations will have the form kcisπ4

.This is shown below on an

Argand Diagram along with the attraction fields off ( x )=x 4+16.

A[100 x 100] Argand Diagram (3) suggest that all possible values of the first iteration of x0=k+ki for all positive k occur in the region of attraction for the zero of x=√2+√2i as shaded grey. This appears to be the case for small values of k as shown by another attraction field graph (Diagram 4)

For this to be true, all initial values in form x0=k+ki must converge to the zero x=√2+√2i which can be tested by substituting various positive values of k and finding the resultant number

sequences. It is noted that the zero x=√2+√2i lies on the line ofπ4

also.To investigate if this holds

true for inaccurate guesses of the functions zero, extremely small and large values for k will be explored, as well as when k is an irrational number.The resulting number sequences are listed in the tables below.

k=1× 10−6=1 E−6𝑛 Value afterxn

0 1E-6+1E-6i1 1E21+1 E21i

2 7.5 E20+7.5 E20i3 5.625 E20+5.625 E20i

4 4.21875 E20+4.21875 E20i

5 3.164062 E20+3.164062 E20i6 2.373047 E20+2.373047 E20i

7 1.779785 E20+1.779785 E20i8 1.334839 E20+1.334839 E20i

171 1.41421397+1.41421397i172 1.41421356+1.41421356i560 1.41421356+1.41421356i

k=1× 106=1 E 6𝑛 Value afterxn

0 1E6+1E6i1 7.5 E5+7.5 E5i

2 5.625 E5+5.625 E53 4.21875 E5+4.21875 E5i

4 3.164062 E5+3.164062 E5i

5 2.373047 E5+2.373047 E5i6 1.779785 E5+1.779785 E5i

7 1.334839 E5+1.334839 E5i8 1.001129 E5+1.001129 E5i

50 1.41421378+1.41421378i51 1.41421356+1.41421356i560 1.41421356+1.41421356i

The resultant number sequences of the positive values for k tested all converged to the predicted zero ofx=√2+√2i. Using an initial value of

x0=(1 ×10−6 )+(1×10−6)i , 172 iterations of Newtons Method were required before the results converged to a value accurate to 9 significant figures of x=√2+√2i . This also occurred for

x0=(1 ×106 )+(1×106) i after 51 iterations and it was found

x0=√8+√8 i converged toward x=√2+√2i after 6 iterations. All continued to converge for a large number of iterations. From these results it can be deduced that all for all positive values of k , initial values in form x0=k+ki when iterated using Newton’s Method for the

function f ( x )=x4+16 , will converge to the zero of x=√2+√2i .

3.Consider the function f ( x )=x4−16

k=√8𝑛 Value afterxn

0 2.82842712+2.82842712i1 2.16551452+2.16551452

2 1.72260878+1.72260878i3 1.48758865+1.48758865i

4 1.41946607+1.41946607i

5 1.41424264+1.41424264i6 1.41421356+1.41421356i

7 1.41421356+1.41421356i8 1.41421356+1.41421356i

9 1.41421356+1.41421356i10 1.41421356+1.41421356i560 1.41421356+1.41421356i

a) What are the exact values of the zeros of f ( x )

The zeroes of f ( x )=x 4−16 occur when f ( x )=0 i.e. when

x4−16=0

x4=16 Converted to Polar Form:

x4=16 cis(0+k 2 π )

x=[16 cis(k 2π )]1/4

Using DeMoivres Theorem:

x=2 cis( kπ2 )

x=2 cis 0,2 cisπ2

,2 cisπ ,2 cis3 π2

x=2,2(cosπ2+ isin

π2),2(cosπ+isinπ ),2(cos

3 π2

+isin3 π2

),

x=2,2(0+ i),2(−1),2(0−i),

x=2,2 i,−2 ,−2 i

Exact value of zeroes of f ( x ) are: x=± 2 ,± 2 i

b) Clearly explain what happens toiterations of Newton’s Method if you start with:

ii) an initial value of x0=2+2i

.

Newton’s Method has the formula:

xn+1=x0−f (x0)f ' (x0)

Substituting the given function into the algorithm gives:

xn+1=x0−x0

4−16

4 x03

Iterations of this process using initial value x0=2+2 i were calculated using computer software, yielding the number sequence in the following table.

𝑛 Value afterxn

0 2.0+2.0i1 1.375+1.375i

2 0.64657682+0.64657682i3 -3.21453818-3.21453818i

4 -2.38079825-2.38079825i

5 -1.71149629-1.71149629i6 -1.08415474-1.08415474i

7 -0.02837534-0.02837534i8 4.377002 E4+4.377002 E4i

45 0.15605935+0.15605935i

46 -262.98888016-262.98888017i

121 0.00000001-2.00000001i122 0.0-2.0i560 0.0-2.0i

Diagram 5

Diagram 6

Using computer software it was calculated that the number sequence produced only converged after a very significant number of iterations. For the first 45 iterations, the real and imaginary parts of the values are equal, accurate to at least 8 sig.figs according to the computer software and

therefore lie on the line of π4

It is noted that the initial value, with polar form √8ciscisπ4

is positioned

on a boundary line of the basins of attractive which appears to run in the direction of cisπ4

as shown

in Diagram 5.After 45 iterations the real and imaginary components of the iterations diverge, giving values diverging from the boundary line, and toward zerox=−2 i. This divergence from the line of

cisπ4

may be the result of rounding by computer software. Consequently the number sequence

produced by Newton’s Method for the functionf ( x )=x 4−16 from a starting point of x0=2+2 i ‘converges’ to a value of x=2 iafter 121 otherwise chaotic iterations. This value remains constant for all subsequent iterations.

iii) an initial value of x0=1+i

.In the previous question applying Newton’s Method for the function f ( x )=x 4−16 yielded the

iterative process:

xn+1=x0−x0

4−16

4 x03

Iterating from the initial value x0=1+i produces the following table of results:

𝑛 Value afterxn

0 1.0+1.0i1 -0.25-0.25i

2 63.8125+63.8125i3 -47.85937115+47.85937115i

4 35.89451924+35.89451924i

5 26.92086781+26.92086781i6 20.1905996+20.1905996i

7 15.14282821+15.14282821i8 11.35683316+11.35683316i

42 1.59017331+1.59017331i

43 0.94393524+0.94393523

98 2.0000289-0.00000473i99 2.0-0.0i560 2.0-0.0i

The number sequence produced by Newton’s Method for the function f ( x )=x 4−16 from a starting

point of x0=1+i only converges to a value of x=2after the high number of 99 otherwise chaotic iterations. This value remains constant for all subsequent iterations. As in the previous question, the

starting value also occurs on the fractal boundary line of cisπ4

which the first 45 iterations lie on

because the real and imaginary parts of the values are equal accurate to at least 8 significant figures. After this the real and imaginary components of the iterations slowly diverge, giving values diverging from the boundary line, and toward zerox=2. It is possible this occurs due to rounding errors in the computer software.

c) Investigate (in detail using both computer software and algebra) what happens to iterations

of Newton’s Method if you start with an initial value of x0=k+ki

for k being a positive number.

From the starting value of x0=k+ki the first iteration is given by the equation:

x1=(k+ki)−((k+ki)4−16

4(k+ki)3 )x1=(k+ki)−(−4 k 4−16

8k3i−8 k3 )x1=¿¿¿

x1=8 k4 i−8 k 4+8 k 4i2−8k4 i+4k 4+16

8k3 i−8 k3

x1=8 k4 i−8 k 4−8 k4−8 k 4i+4k 4+16

8 k 3i−8 k3

x1=−16 k4 +4 k 4+16

8 k 3i−8 k 3

x1=−3 k4+4

2 k3i−2 k 3

Converting to Polar form:

x1=(3k¿¿ 4−4)cisπ

(√8 k3 ) cis3 π4

¿

x1=( 3k 4−4

√8 k3 )cisπ4

This shows that for any initial guess in form x0=k+ki

for k being a positive number, the

first iteration will have its exact value on the line of cisπ4

. As k is any positive number there

are an infinite number of possible moduli so all possible locations for this first iteration is

given bykcisπ4

. Because of this it can be shown the 2nd iteration also lies on cisπ4

.

Using an initial value ofx0=kcisπ4

:

x2=(kcisπ4)−((kcis

π4)

4

−16

4 (kcisπ4)

3 )Using DeMoivres theorem:

x2=((4 kcis

3 π4 )(kcis

π4 ))−kcis

π4+16 cis 0

4 kcis3 π4

x2=4 k2 kcisπ−kcis

π4+16 cis 0

4 kcis3 π4

x2=1k (cis

π4 )− 1

4 k2 (cisπ4 )+ 16

4 k3(cis

5 π4

)

x2=4k 2

4√2 k3 +4 k2

4√2 k3 i− k4 √2 k3 −

k4 √2k3 i− 16

4√2k 3−16

4√2k 3 i

x2=4 k2−k−16

4 √2k3 + 4k2−k−164√2 k3 i

This shows the second iteration also has the form x0=k+ki

which was previously shown to iterate to a value of the same form also. For this reason it is proven that all iterations produced from

a starting value of x0=k+ki

from Newton’s Method of the function f ( x )=x 4−16 will lie on the line

cisπ4

.The zeroes of this function were earlier shown to bex=± 2 ,± 2 i, all of which do not occur on

the line kcisπ4

therefore the iterations of Newton’s Method will not converge to them.Newtons

Method also possesses the property of converging initial guesses at a greater rate the closer they are to the real root of a given function.

‘Newton’s method may converge slowly at first. However, as the iterates comecloser to the root, the speed of convergence increases.’

University of Pittsburgh. 2009. Rootfinding for Nonlinear Equations

This property can be used to prove that the iterations provided by the Jim Rolf software used is incorrectly converging to the zero and contradicting the algebra shown above. The two initial guesses of x0=1+i and x0=150+150 i were iterated to produce the two tables below:

𝑛 Value afterxn

0 1.0+1.0i1 -0.25-0.25i

2 63.8125+63.8125i3 -47.85937115+47.85937115i

4 35.89451924+35.89451924i

5 26.92086781+26.92086781i6 20.1905996+20.1905996i

7 15.14282821+15.14282821i8 11.35683316+11.35683316i

98 2.0000289-0.00000473i99 2.0-0.0i

The above tables show that an initial guess of x0=150+150 i which is significantly further from a root of the function, converges to the root of x=2 after 96 iterations compared to a greater number of 98 for the initial guess of x0=1+i. This is clear violation of the properties of Newton’s Method by the Jim Rolf software, highlighting its inaccuracy and limitations. It also backs up the notion that the incorrect convergence is due to the software rounding off iterations.

1. Consider the function f ( x )=x5−1Investigate (in detail using both computer software and algebra) what happens to iterations

of Newton’s Method if you start with an initial value of x0=kcis( π

5) for k being a positive

number.

The behavior of iterations of Newton’s Method in the form

x0=kcis( π5) for the function f ( x )=x5−1 can be

investigated by finding the first iteration In the general terms

of k .

Newton’s Method has the formula:

xn+1=x0−f (x0)f '(x0)

Substituting f ( x )=x5−1 into the algorithm gives:

xn+1=x0−x0

5−1

5 x04

From the starting value of x0=k+ki the first iteration is given by the equation:

𝑛 Value afterxn

0 150+150i1 112.4999997+112.4999997i

2 84.37499908+84.37499908i3 63.28124764+63.28124764i

4 47.46093179+47.46093179i

5 35.59568949+35.59568949i6 26.69674494+26.69674494i

7 20.02250615+20.02250618 15.01675503+15.01675503

95 1.99999979+0.00000068i96 2.0-0.0i

x1=(kcisπ5)−((kcis

π5)

5

−1

5 (kcisπ5

)4 )

Using DeMoivres theorem:

x1=(kcisπ5)−( k5 cisπ−1cis 0

5k 4 cis4π5 )

x1=((5 k4 cis

4 π5 )(kcis

π5 ))−k5 cisπ+1 cis 0

5k4 cis4 π5

x1=5 k5 cisπ−k5 cisπ−1cisπ

5 k 4 cis4 π5

x1=(4 k5−1)cisπ

5 k4 cis4 π5

x1=( 4 k5−15k 4 )cis

π5

The above calculation shows that the iterations of Newtons Method from a starting value of kcisπ5

will continue to have values on the line ofcisπ5

. This suggests that the iterations of kcisπ5

will

converge to the zero of function f ( x )=x5−1 that lies on this line. The exact zeroes of this function can be found using the roots of unity.

Using the functionf ( x )=x5−1 , the zeroes are given when f ( x )=0

x5−1=0x5=1

x ¿[cis (k 2 π )]15

Using DeMoivres Theorem:

x=cis ( k 2 π5 )

x=1 , cis( 2 π5 ) ,cis ( 4 π

5 ) , cis( 6 π5 ) , cis( 8 π

5 )

Diagram 7

None of the exact zeroes of the function have a value on the line cisπ5

however it was previously

shown that all iterations from a starting value of kcisπ5

also have the formkcisπ5

. When the value

for k becomes negative the iterations cross the origin and enter into the 3rd quadrant on the line

kcis6 π5

as shown in Diagram 7. For this reason iterations from a starting point of kcisπ5

will

converge to the zero ofcis( 6 π5 ) for the function f ( x )=x5−1

x0=.5 cisπ5𝑛 Value afterxn

0 0.4045085+0.29389263i1 -2.26524758-1.64579871i

2 -1.81483049-1.31855153i3 -1.458254-1.05948354i

4 -1.1819312-0.85872328i

5 -0.98106303-0.71278402i6 -0.85967264-0.62458874i

7 -0.81464518-0.59187437i8 -0.80909423-0.58784136i

9 -0.80901701-0.58778526i10 -0.80901699-0.58778525i560 -0.80901699-0.58778525i

For this to be true, all initial values in form x0=kcis( π

5) must converge to the zero

x=cis6 π5 which

can be tested by substituting various positive values of k and finding the resultant number sequences. To investigate if this holds true for even inaccurate guesses of the functions zero, extremely small and large values for k will be explored, as well as when k is an irrational number. The resulting number sequences are listed in the tables below. This is supported by the basins of

attraction which are green along the line of kcis6 π5

, however there are points where boundary lines

converge that are shown as black dots. It was found that an initial value in this zone still converged

to cis6 π5

.

x0=√5 cisπ5𝑛 Value afterxn

0 1.80901699+1.31432778i1 1.44074146+1.04675994i

2 1.13650621+0.82572009i3 0.8676588+0.63039102i

4 0.5718282+0.41545751i

5 -0.19080745-0.13862973i6 -52.44484487-38.10341018i

7 -41.95587591-30.48272815i8 -33.56470075-24.38618253i

28 -0.80901703-0.58778528i29 -0.80901699-0.58778525i560 -0.80901699-0.58778525i

x0=50 cisπ5𝑛 Value afterxn

0 40.45084972+29.38926261i1 32.36067975+23.51141007i

2 25.88854374+18.809128013 20.71083483+15.0473023i

4 16.56866749+12.03784156i

5 13.25493307+9.63027258i6 10.60394421+7.70421644i

7 8.48314989+6.16336917i8 6.78650653+4.93068561i

30 -0.80901702-0.58778527i31 -0.80901699-0.58778525i560 -0.80901699-0.58778525i

These tables of values show that for the tested initial values in formxo=kcisπ5

, the number

sequences produced using Newton’s Method converged to a value of xo=¿-0.80901699-0.58778525i

after a low number of iterations. It is noted that this value is accurate to 8 significant figures of the

zero x=cis6 π5

which supports the notion that all initial values in form xo=kcisπ5

will converge to

this zero for the given function. For the most inaccurate initial guess ofx0=50 cisπ5

the number

sequence converged after 29 iterations, suggesting this convergence is due to Newton’s Method and not rounding error like previous questions.

Conclusion

It has been shown in Question 1 that iterations of Newton’s Method from an initial value located in a basin of attraction will converge to the zero of the basin it occurs in. This was shown in Question 1 where the Newton’s Method iterations of f ( x )=x3−8 behaved according to the basins of attraction and converged to a function zero of x=−1−√3 i .

In Question 2 the iterations produced by Newtons Method from the function f ( x )=x 4+16 were

investigated from a starting value in form x0=k+ki for k being any positive number. Using algebraic methods it was determined that all iterations from this starting value were also in this form with a

value ofkcisπ4

. This allowed the number sequence to converge from xo=√8 cisπ4

and xo=cisπ4

to

the exact function zero x=2 cisπ4

that lie on the line ofπ4

. This convergence also obeyed the basins

of attraction with both starting values occurring in the region for zero x=2 cisπ4

.

Question 3 dealt with iterations from initial values in form x0=k+ki for the even order function

f ( x )=x 4−16 which now lie on the boundary line of attraction regions. Once again it was determined

algebraically that all iterations from this value had an argument ofπ4

and never leave the line of π4

. It

was also found that the exact zeroes of the functionf ( x )=x 4−16 werex=± 2 ,± 2 i, none of which

have an arg ofπ4

. As a result of this the number sequence produced by Newton’s Method was

unable to converge to a zero, instead the values iterated chaotically in the form x=kcisπ4

including

negative k values. When k became negative the origin is crossed and the iterations entered the

third quadrant on the line of5 π4

.

The Jim Rolf computer software converged these starting values to zeroes of f ( x )=x 4−16 after a great number of iterations, however this was shown to be a result of the program rounding values.

For each iteration the software rounded the exact value, giving a result off the boundary line of π4

,

eventually landing in an attraction basin and converging to a zero.

Question 6 required investigating the behavior of Newtons Method iterations from an initial value of

xo=kcisπ5

for a function of odd orderf ( x )=x5−1. It was shown algebraically that the iterations

starting from this value had the form kcisπ5

including negative k where the iterations entered the

third quadrant and had values of kcis6 π5

. Using unity roots the exact zeroes of the function were

found including cis6 π5

which the iterations of Newton’s Method converged to. This is because this

value was the only zero that occurred with the form kcisπ5

as cis6 π5

can also be written as −cisπ5

.

It is noticed that the Newtons Method all iterations for functions of order 4 from an initial value

kcisπ4

also had a form of kcisπ4

. For the order 5 function of f ( x )=x5−1 it was shown that all

iterations remained in the same form if a starting value of kcisπ5

was used. It can also be shown

that this occurs for the function f ( x )=x3−8 from a starting value of kcisπ3

. From this it is possible to

make the conjecture that a function with the order of n when iterated using Newtons Method from a

starting value of kcisπn

will only have iterations occurring in form kcisπn

or in the third quadrant on

the opposite argument with a value in form kcisπ (n+1)

n.

From the results of Question 2 and 3 the conjecture can be expanded to address the behavior of iterations of Newton’s Method according to their initial value.

For an even order function with form xn+c (where c is the constant), iterations of Newtons Method

from a starting value of kcisπn

will converge to the closest function zero that is a multiple of cisπn

.

This is supported by the behavior of the iterations in Question 2.

Using an even order function with form xn−c the iterations of Newton’s Method from a starting

value ofx0=kcisπn

will not converge to a zero as the function has no zeroes on the line of cisπn

,

therefore the iterations which only occur on this line can never converge to a zero. Instead the

number sequence will have chaotic values in the form ± kcisπn

occurring in the first and third

quadrants. This is supported by the behavior of iterations in Question 2.

Odd order functions in form xn−c will have iterations from initial value x0=kcisπ5

that cross the

origin and converge to a zero on the line of π (n+1)

n. This holds true for Question 4 where it was

shown that for the function f ( x )=x5−1 , iterations of Newton’s method from

x0=.5 cisπ5

, x0=50 cisπ5

and x0=√5 cisπ5

converged to the function zero of cis6 π5

.

Iterations from starting value of kcisπn

for odd order functions in form xn+c will converge to the

function zero with on the line of πn

. If this conjecture holds true the function f ( x )=x5+1 when

iterated using Newton’s Method from any initial value in form x0=kcisπ5

will converge to the zero cis

π5

. This is shown occur using computer software in diagram 8.

These conjectures conclude the summary of findings for Part B of the Complex numbers, Polynomials and Iterations Directed Investigation.

Diagram 8