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1. Newton's Method 
A root of a function f is a
point x_{0} in dom( f ) such that f(x_{0}) = 0. It's a
solution or root of the equation f(x) =
0, ie, a
point where the graph of f intersects the xaxis.
It's also called a zero of f. Let f
be a differentiable function and suppose
f has a
root. Newton's method is used to find a sequence of approximations a_{1}, a_{2}, a_{3}, ... to the
root that approaches the
root (ie, a_{n} is closer to the root than a_{n}_{–1} is). The idea
is as follows.
Refer to Fig. 1.1. Guess a_{1}. The tangent
line to f at a_{1} intersects the xaxis
at a_{2}. The tangent
line to f at a_{2} intersects the
xaxis
at a_{3}. And so on.
This process can be repeated over and over to get closer and closer to the
root. The equation of
Fig. 1.1 f(x_{0}) = 0, 
the tangent line at a_{1} is y = f(a_{1}) + f '(a_{1})(x – a_{1}). Since (a_{2}, 0) is on that line, we have 0 = f(a_{1}) + f '(a_{1})(a_{2} – a_{1}), so:
In general, since (a_{n}, 0) is on the tangent line at a_{n}_{–1}, we have:

Thus:

Recursion Formula
The formula a_{n} = a_{n}_{–1} – f (a_{n}_{–1})/f '(a_{n}_{–1}) is a
recursion formula. In mathematics, a recursion formula is one in which a
value in a sequence is calculated in terms of one or more previous values.
Let f(x) = x^{3} + x – 1.
a. Show that f has a root.
b. Use Newton's method to find an approximate value of that root accurate
to 6 decimal places. You may use a calculator.
Solution
a. f(0) = –1 < 0 and f(1) = 1 > 0. So by the intermediatevalue theorem f has a root x_{0}, which is in (0, 1).
b. f(x) = x^{3} + x – 1, f '(x) = 3x^{2} + 1. Let's take the initial guess a_{1} = 0.5. Then:
EOS
As f(0) = –1 < 0 and f(1) = 1 > 0, we have f(0) < 0 < f(1), so, as f
is continuous, by the intermediatevalue theorem
there exists c in (0, 1)
such that f(c) = 0, ie f has a root in (0, 1).
We take the midpoint of the
small interval in which the root is located as the initial guess.
The required accurate number of
decimal places of the approximation is 6. In each calculation of a_{n} we
should keep more
than 6 decimal places, for example 7 or 8, in order to get more accuracy in
subsequent calculations and to see whether
the last a_{n}
should be rounded up or down to be used as the answer. In this example we keep
8. In general if the required
accurate number of decimal places is k,
you should keep k or
k + 1 decimal places in the
calculations of the a_{n}'s
and
round the last one to get an approximate value of the root x_{0}.
The approximations stabilize at
6 decimal places after 5 iterations. So we're confident that they approach a
limit, which
is a root x_{0} of f. Thus we're confident that
the root x_{0} of f is approximately 0.682328
rounded up and accurate to 6
decimal places.
a_{5} and a_{6} have the same
value up to at least 6 decimal places. We say that the approximations stabilize
at 6 decimal
places at a_{5}. We then stop
and use the rounded value of a_{6} as the
approximate value of the root. In general the
approximations stabilize when 2 consecutive a_{n}'s
have the same value up to at least the required accurate number of
decimal places. We then stop and use the rounded value of the last a_{n} as
the approximate value of the root. Yes we use
the rounded value of the last a_{n},
say a_{m},
not that of the previous a_{m}_{–1}, as a_{m} and
a_{m}_{–1} may have
different digits after the
required accurate number of decimal places. Note that in our example a_{6} and a_{5} have the same value up to at least 8
decimal places.
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2. Solving The Equation f (x) = 0 
The equation x^{5} = 20 – 7x^{3} can be expressed as x^{5} + 7x^{3} – 20 = 0. The latter is of the
form f(x) = 0 where f(x) = x^{5} +
7x^{3} – 20. In
general an equation in an unknown x
can always be expressed in the form f(x) = 0 by moving everything to
the lefthand side. The solutions of for example the equation x^{5} = 20 – 7x^{3} are the same as those of the equation x^{5} +
7x^{3} – 20 = 0 and
thus are the same as the roots of the function f(x)
= x^{5} + 7x^{3} – 20. Mathematics is often used
to
determine a quantity by finding an equation that the quantity satisfies and
then solving the equation. In algebra we
learned to solve simple equations such as x^{2} – 2x – 3 = 0. However we didn't
learn to solve equations such as x^{5} + 7x^{3} –
20 = 0, which should be regarded as quite simple too. In the example below
we'll utilize Newton's method to solve the
equation x^{5} = 20 – 7x^{3}.
Utilize Newton's method to solve
the equation x^{5} = 20 – 7x^{3} with the solution or solutions accurate
to 8 decimal places.
You many need a calculator.
Solution
EOS
To solve an equation we must
find all of its solutions if they exist. If there's a unique solution then we
must show this
uniqueness
As f(1) = –12 < 0 and f(2) = 68 > 0, we have f(1) < 0 < f(2), so, as f
is continuous, by the intermediatevalue theorem
there exists c in (1, 2)
such that f(c) = 0, ie f has a root in (1, 2).
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3. Approximations Of Roots Of Numbers 
discussed in Section
8.3. Newton's method can also be used to approximate such roots.
Solution
EOS
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4. Newton's Method Doesn't Always Work 
Newton's method doesn't always
work. For example, as seen in Fig. 4.1, the choice of a_{1} as shown
causes the sequence
{a_{n}}
to move farther away from x_{0} rather than closer to it. In
this case the sequence {a_{n}}
diverges to infinity. For another
example, as seen in Fig. 4.2, the choice of a_{1} as shown
causes the sequence {a_{n}} to
oscillate between a_{1} and a_{2} without
converging to x_{0}.
Fig. 4.1 This choice of a_{1} causes 
Fig. 4.2 This choice of a_{1} causes 
Example 4.1
Let f(x)
= x^{4} + 4x^{3} + 4x^{2} – x
– 1.
a. Show that the equation f(x) = 0 has a solution in (–1,
0).
b. Attempt to use Newton's method to find a solution of f(x) = 0 starting with a_{1} = 0. What happens?
Solution
a. f(–1) = 1
> 0, f(0) =–1
< 0, so by the intermediatevalue theorem there exists x_{0} in (–1, 0) such that f(x_{0}) = 0, ie the
equation f(x)
= 0 has a solution x_{0} in (–1, 0).
b. f(x) = x^{4} + 4x^{3} + 4x^{2} – x – 1, f '(x) = 4x^{3} + 12x^{2} + 8x – 1, a_{1} = 0,
Thus the a_{n}'s
oscillate between –1 and 0 and consequently don't converge to x_{0}.
EOS
Note that the x_{0} in this example can't be either
–1 or 0. However even if it were either –1 or 0, we would still be correct
to say that the a_{n}'s
oscillate between –1 and 0 and don't converge to x_{0}, because they don't converge to either –1 or 0.
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5. Various Approximations 
Section
8.3 discusses the approximation of values of functions using tangentline
approximation. Section
8.4 deals
with the approximation of errors in measurement. This section, Section 5.3.3,
handles the approximation of roots of
functions utilizing Newton's method.
Problems & Solutions 
1. Let f(x) = x – e^{–}^{x}.
a. Show that the function f
has a root.
b. Use Newton's method to find an approximation of that root accurate to
6 decimal places. You may use a calculator.
Solution
a. f(0)
= –1 < 0, f(1) = 1 – e^{–1} > 0, and f
is continuous. So by the intermediatevalue theorem f has root x_{0}, which is in
(0, 1).
b. We have f '(x) = 1 + e^{–}^{x}. Let the initial guess be a_{1} = 0.5. Then:
2. Utilize Newton's
method to solve the equation 3t^{5} = 10t^{3} – 15t + 15 accurate to 10 decimal places. You may use
a
calculator.
Solution
4. Let f(x) = 2x^{3} – 3x^{2} – x + 1.
a. Show that the equation f(x) = 0 has a solution in (0,
1).
b. Attempt to use Newton's method to find a solution of f(x) = 0 starting with a_{1} = 1. What happens?
a. f(0) = 1
> 0, f(1) = –1
< 0, so by the intermediatevalue theorem there exists x_{0} in (0, 1) such that f(x_{0}) = 0, ie the
equation f(x)
= 0 has a solution x_{0} in (0, 1).
b. f(x) = 2x^{3} – 3x^{2} – x + 1, f '(x) = 6x^{2} – 6x – 1, a_{1} = 1,
Thus the a_{n}'s oscillate
between 0 and 1 and consequently don't converge to x_{0}.
Solution
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