Download Math 237. Calculus II Solutions to the HW on Newton`s Method (3.7

Math 237. Calculus II
Solutions to the HW on Newton's Method (3.7)
Assigned: 5, 9, 11, 14, and find the zeros of y = ex – 3x.
Selected for Grading: 9, 14
Solutions.
5. Find the largest root of x3 + 6x2 + 9x + 1 = 0.
Let f (x) = x3 + 6x2 + 9x + 1 = 0. Then Newton's formula for this setting is
A quick peek at its graph shows that this function has its largest zero somewhere between –1 and zero.
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Starting off with x0 = –0.5, I got the following approximations.
x0 = –0.5
x1 = 0.0666666667
x2 = –0.0991243961
x3 = –0.1202800956
x4 = –0.1206146753
x5 = –0.1206147584
At this point I started getting the same value over and over again.
So the largest root of x3 + 6x2 + 9x + 1 = 0 is approximately –0.1206147584.
9. For this one I will use Newton's method and the function f (x) = cos(x) – 2x. Here is this function's graph.
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There is a zero between x = 0 and x = 1. I'll start with x1 = 0.5.
First, though, I need to build the formula to be used.
f '(x) = –sin(x) – 2 and
So the formula to use is
Here are the successive approximations that I got using this formula with my calculator.
x1 = 0.5
x2 = 0.4506266931
x3 = 0.4501836476
x4 = 0.4501836113
x5 = 0.4501836113
The solution to the equation cos x = 2x is approximately x = 0.4501836113.
11. To find all real zeros of f (x) = x4 – 8x3 + 22x2 – 24x + 8, I'll use Newton's method as many times as
necessary, once for each zero. I know that since this function is a polynomial function of degree four it has
at most four zeros. A peek at its graph . . .
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shows that there are only three. It looks like the middle one is exactly 2. A quick check . . .
f (2) = 16 – 64 + 88 – 48 + 8 = 112 – 112 = 0
. . . shows that this is true. So I need to use Newton's Method only twice.
For each use, I will use the formula:
First, with x1 = 0.5:
x2 = 0.575
x3 = 0.5855859612
x4 = 0.5857863666
x5 = 0.5857864376, and all the subsequent approximations were equal to this fifth one.
And using x1 = 3.5:
x2 = 3.425
x3 = 3.414414039
x4 = 3.414213633
x5 = 3.414212562, and all the rest were this same value.
Summary: The roots of x4 – 8x3 + 22x2 – 24x + 8 = 0 are x ≈ 0.5857864376, x = 2, and x ≈ 3.414212562.
14. We seek the smallest positive root of 2 cot(x) = x. Let f (x) = 2 cot(x) – x. In order to begin the process, I
need a first guess. A look at the graph of f . . .
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shows that it lies very close to x = 1. So I'll start off with the guess x1 = 1. I'll use the formula
Here are the estimates that I got.
x2 = 1.074305226
x3 = 1.076871416
x4 = 1.076873986, and all successive approximations had this same value.
The smallest positive root of 2 cot(x) = x is approximately 1.076873986.
The Additional Question
I'll use the function f (x) = ex – 3x. Here is its graph.
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As you can see, it has zeros between 0 and 1 and between 1 and 2.
To find the zero between 0 and 1 I started off with x0 = 0.5. And I'll use the recursive formula
Here are the successive values that I got.
x0 = 0.5
x1 = 0.610059655
x2 = 0.6189967797
x3 = 0.6190612834
x4 = 0.6190612867
At this point I started getting the same value over and over again.
So the first zero for f (x) = ex – 3x is approximately 0.6190612867.
To find the zero between 1 and 2 I'll use the same recursive formula as above, but I'll start off with x0 = 1.5.
Here are the succissive values that I got.
x0 = 1.5
x1 = 1.512358146
x2 = 1.512134625
x3 = 1.512134552
At this point I started getting the same value over and over again.
So the first zero for f (x) = ex – 3x is approximately 1.512134552.