(10 pts)

For this first Discourse assignment, you’re going to explore Newton’s iteration to find the largest root of your own personal cubic polynomial. Find your personal problem by choosing your name from the menu below.

An archived instance of a Chaos and Fractals forum

mark

(10 pts)

For this first Discourse assignment, you’re going to explore Newton’s iteration to find the largest root of your own personal cubic polynomial. Find your personal problem by choosing your name from the menu below.

adrienne

My polynomial is f(x) = x^3 - 4x^2 + 2x +4 so my Newton’s method iteration function is

x - \frac{f(x)}{f'(x)} = x - \frac{x^3-4x^2+2x+4}{3x^2 - 8x+2}.

I graphed f using Desmos and came up with the following picture:

Using the graph, I picked x_0=3 (shown in green in the graph) and after three iterates found x_3 \approx 2.732.

scowart

My polynomial is f(x)=x^3-4x^2+x. Plugging my function into Newton’s Method iteration function yields

N(x)=x - \frac{x^3-4x^2+x}{3x^2 - 8x+1}.

Using desmos I graphed my function whiched showed:

The x_0 I chose was x_0=3. Iterating through Newton’s method yielded sucessively x_1 = 4.5, x_2 \approx 3.93204, x_3 \approx 3.75113, and finally x_4 \approx 3.73225.

ofeldman

My polynomial is f(x) = x^3 - 8x^2 + 15x + 4, so my Newton’s method iteration function is

N(x) = x - \frac{f(x)}{f'(x)} = x - \frac{x^3 - 8x^2 + 15x +4}{3x^2 - 16x + 15}

Graphing f using Desmos shows:

Starting with x_0=4.5, I iterated Newton’s method to result in x_1 \approx 4.33333333333, x_2 \approx 4.25925925926, x_3 \approx 4.23807248108

pikenber

My polynomial is f(x) = x^3 - 7x^2 + 12x - 6, so my Newton’s Method iteration function is

x - \frac{f(x)}{f'(x)} = x - \frac{x^3-7x^2+12x-6}{3x^2 - 14x+12}.

I graphed f using Desmos which showed:

I chose 5 as my x_0 (the green point above) and after 2 iterates, I had a close approximation, but after 3 it was even closer (the purple point above). x_1 \approx 4.76470588235, x_2 \approx 4.73262860528, x_3 \approx 4.7320509933.

badler

My polynomial is f(x)=x^3-2x^2-4x so my Newton’s method iteration function is:

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

I graphed my polynomial using Desmos and came up with the following picture:

Using the graph my initial estimate was x_0=3.

After 3 iterations:

x_1\approx 3.2727

x_2\approx 3.2367

x_3\approx 3.2361

mreyeslo

My polynomial is

f(x)=x^3-10x^2+29x-26

so my Newton’s method iteration function is

x - \frac{f(x)}{f'(x)} = x - \frac{x^3-10x^2+29x-26}{3x^2 - 20x+29}.%

This is my function on desmos

The x_0 I chose was x_0=6.The three iterates are as shown below

x_1\approx 5.7647, x_2\approx5.7326, x_3\approx 5.73205

jwilso13

My polynomial is f(x) = x^{3}-12x^{2}+40x-40. The corresponding Newton’s iteration function is,

N(x)=x- \frac{x^{3}-12x^{2}+40x-40}{3x^{2}-24x+40}.

I graphed f using Desmos and had the following result:

My initial estimate of x_0 = 7. After three iterations I found that x_1 \approx 7.263, x_2 \approx 7.23636, and x_3 \approx 7.2360. Also, I don’t know why LaTeX hates me with that last prt.

pstedina

My polynomial is f(x)=x^3-13x^2+52x-66 , so my Newton’s method iteration function is

N(x)=x-\frac{f(x)}{f'(x)}=x-\frac{x^3-13x^2+52x-66}{3x^2-26x+52}.

Using desmos, I graphed f:

I chose x_0=7 as my estimate and iterated three times

x_1\approx6.765

x_2\approx6.733

x_3\approx6.732

lee7

Given the polynomial f(x) = x^3 - 6x^2 + 4x , Newton’s method iteration function would take the form

N(x)=x-\frac{f(x)}{f'(x)}=x-\frac{x^3-6x^2+4x}{3x^2-12x+4}.

Graphing f(x) using Desmos gives us the following:

From this graph, we can estimate the root of the function to be approximately 5.

Using x_0=5 , I iterated the function 3 times to get the estimates:

x_1\approx5.2632

x_2\approx5.2364

x_3\approx5.2361

mark