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Newton’s method

mark

(10 pts)

In this problem, you’re going to apply Newton’s method to find the largest root of a cubic polynomial. Specifically - given your cubic polynomial, your task will be to:

Find the corresponding Newton’s method iteration function,
Use a graph to find an integer that just a little larger than the largest root to use as a starting point, and
Compute the first ten Newton steps as decimal numbers, and
Include your graph with the root labeled.

To get your personal cubic, just choose your name from the following list:

I strongly encourage you to use a computer on this problem! Feel free to use the code you find on our Newton’s method demo page.

mearing

My function is f(x)=x^3+2x^2-3x-1

N(x)=x-\frac{x^3+2x^2-3x-1}{3x^2+6x-3}

Starting point x=2

First ten Newton steps
1.47058823529412
1.24713267877313
1.20069873242387
1.19869492645648
1.19869124352843
1.19869124351600
1.19869124351600
1.19869124351600

cmoore14

My personal function is f(x)=x^3 + 4x^2 + 4x - 1

N(x)=x-\frac{x^3+4x^2+4x-1}{3x^2+8x+4}
Starting point will be x=1
First ten steps:
0.466666666666667
0.247341459106165
0.206900350351841
0.205570845172493
0.205569430402191
0.205569430400590
0.205569430400590
0.205569430400590
0.205569430400590
0.205569430400590

chowell1

my function is x^3 + 5x^2 + 6x - 1 and my starting value is x = 1.
x - \frac{x^3 + 5x^2 + 6x - 1}{3x^2 + 10x + 6}
first ten steps:
0.973684210526316
0.408294412001828
0.186113568685012
0.148911122096412
0.147899773982271
0.147899035705181
0.147899035704787
0.147899035704787

rserra

My function is f(x)=x^3-3x^2+2x-1

1.N(x)=x-\frac{x^3-3x^2+2x-1}{3x^2-6x+2}

2.Starting point is x=3

3.First ten Newton steps:

2.5454545454545
2.3596149159152
2.3258013450058
2.3247190494171
2.3247179572459
2.3247179572447
2.3247179572447
2.3247179572447
2.3247179572447
2.3247179572447

bserra

My function is f(x)=x^3+2x^2-3x-1

N(x)=x-\frac{x^3+2x^2-3x-1}{3x^2+4x-3}
Starting point will be at x=2
First ten Newton steps:
1.4705882352941
1.2471326787731
1.2006987324239
1.1986949264565
1.1986912435284
1.198691243516
1.198691243516
1.198691243516
1.198691243516
1.198691243516

jbarry1

My function is f(x)=x^3-4x-1 and my starting point is x=3.

\frac{\left(x^3-4x-1\right)}{3x^2-4}

First ten Newton steps:
2.39130434782609
2.15496354423737
2.11594549472371
2.11490826644907
2.11490754147711
2.11490754147676
2.11490754147676
2.11490754147676
2.11490754147676
2.11490754147676

abjorkma

My function is: f(x)=x^3+4x^2+3x-1

N(x)= x-\frac{x^3+4x^2+3x-1}{3x^2+8x+3}
Starting point is at x=1
The first ten Newton steps are:
0.500000000000000
0.290322580645161
0.248606244506255
0.246982029751572
0.246979603722876
0.246979603717467
0.246979603717467
0.246979603717467
0.246979603717467
0.246979603717467
Screen Shot 2021-10-30 at 9.48.14 PM1920×1097 74.8 KB

mbradle7

My cubic function is: f(x) = x^3 - 2x^2 - 3x - 1.

N(x) = \frac{x^3 - 2x^2 - 3x - 1}{3x^2 - 4x - 3}
The starting point is: x = 4.
The first ten Newton steps are:
3.34482758620690
3.11140087996411
3.08013890533442
3.07959578610095
3.07959562349145
3.07959562349144
3.07959562349144
3.07959562349144
3.07959562349144
3.07959562349144

lliberty

My function is: f(x)=x^3+2x^2+x−1

N(x)=x-\frac{x^3+2x^2+x-1}{3x^2+4x+1}
Starting point: x=0.5
First 10 Newton Steps:
0.466666666666667
0.465572390572391
0.465571231878066
0.465571231876768
0.465571231876768
0.465571231876768
0.465571231876768
0.465571231876768
0.465571231876768
0.465571231876768

acole6

My cubic function is:

f(x)=x^3-x-1

starting point:

x=2

N(x)=x-\frac{\left(x^3-x-1\right)}{3x^2-1}

First ten Newton steps:
1.54545454545455
1.35961491591518
1.32580134500585
1.32471904941713
1.32471795724586
1.32471795724475
1.32471795724475
1.32471795724475
1.32471795724475
1.32471795724475

dkey

My function is f(x)=x^3+3x^2+2x−1.

N(x)=x-\frac{x^3+3x^2+2x−1}{3x^2+6x+2}

Starting point is x=1

First ten Newton steps
1.11538461538462
0.604247758432780
0.377421440964612
0.327133118530718
0.324723375139264
0.324717957272100
0.324717957244746
0.324717957244746
0.324717957244746
0.324717957244746

lfountai

My function is:

f(x)=x^3+3x^2+2x-1

Newton’s method of iteration function:

N(x)=x-\frac{x^3+3x^2+2x-1}{3x^2+6x+2}

Starting point it:

x=1

First ten Newton steps:
0.545454545454545
0.359614915915184
0.325801345005845
0.324719049417125
0.324717957245858
0.324717957244746
0.324717957244746
0.324717957244746
0.324717957244746
0.324717957244746

cbrowni1

My function is f(x) = x^3 - 9x - 1
N(x) = x - (x^3 - 9x - 1)/(3x^2 - 9)
Starting point x = 1
First ten steps are…
-0.500000000000000
-0.0909090909090909
-0.111250627825213
-0.111264157583397
-0.111264157590214
-0.111264157590214
-0.111264157590214
-0.111264157590214

mabbott

My personal function is f(x)=x^3-x^2-2x-1

N(x)=x-\frac{x^3-x^2-2x-1}{3x^2-2x-2}
Starting point is x=3
First ten Newton steps:
2.42105263157895
2.18950354224912
2.14909446114016
2.14790006548345
2.14789903570555
2.14789903570479
2.14789903570479
2.14789903570479
2.14789903570479
2.14789903570479

mwilli21

My cubic function is: f(x)=x^3-x-1

N(x)=x-\frac{x^3-x-1}{3x^2-1}

Starting point is at x=2

The first ten Newton Steps are:
1.54545454545455
1.35961491591518
1.32580134500585
1.32471904941713
1.32471795724586
1.32471795724475
1.32471795724475
1.32471795724475
1.32471795724475
1.32471795724475

My graph with the labeled root:

sbindas

My function is f(x) = x^3 - 4x^2 +3x -1
N(x) = x - \frac{x^3 - 4x^2 +3x -1}{3x^2 - 8x + 3}
Starting point x=4
First 10 Newton steps
3.42105263157895
3.18950354224912
3.14909446114016
3.14790006548345
3.14789903570555
3.14789903570479
3.14789903570479
3.14789903570479
3.14789903570479
3.14789903570479

cmarsha4

My function is

f(x) = x^3 - 6x^2 + 9x -1

N(x) = x - (x^3 - 6x^2 + 9x - 1)/(3x^2 - 12x + 9)

Starting point will be x = 4.

first ten steps:
3.666666667
3.548611110
3.532390160
3.532088980
3.532088880
3.532088880
3.532088880
3.532088880
3.532088880
3.532088880

jlytle2

My function is f (x)= x^3 - x^2 - 6x -1.
N (x) = x - (x^3+2x^2-3x-1)/(3x^2 - 2x -6)
Starting point x = 0
First ten newton steps
-0.166666666667
-0.172470978441
-0.17248009311
-0.172480093133
-0.172480093133
-0.172480093133
-0.172480093133
-0.172480093133
-0.172480093133
-0.172480093133

vspargo

my cubic function is:

f(x)=x^3-4x-1

N(x)=x-\frac{\left(x^3-4x-1\right)}{3x^2-3}

starting point is

x=3

first ten Newton steps:
2.39130434782609
2.15496354423737
2.11594549472371
2.11490826644907
2.11490754147711
2.11490754147676
2.11490754147676
2.11490754147676
2.11490754147676
2.11490754147676