Chapter 16: Return To Roots
- Chapter 16: The Mad God. Study Guide for Seattle Christian School (9th Grade) Learn with flashcards, games, and more — for free.
- Therefore, the roots of the given quadratic equation are real, irrational and unequal. Example 2: Without solving, examine the nature of roots of the equation 4x 2 – 4x + 1 = 0? Solution: The discriminant D of the given equation is D = b 2 – 4ac = (-4) 2 – (4 x 4 x 1) = 16-16=0 Clearly, the discriminant of the given quadratic equation is.
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Calculus Of One Real Variable – By Pheng Kim Ving |
8.5 |
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1.Newton's Method |
Bible Deuteronomy Chapter 16 Verse 7. Root it in the place. In the morning you are to return to your tents.
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 x-axis. 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 x-axis at a_{2}. The tangent line to f at a_{2} intersects the
x-axis 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 tangentline 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.
Example 1.1
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 intermediate-value 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 intermediate-value 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 left-hand 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}.
Example 2.1
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 intermediate-value 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.
Example 3.1
Solution
EOS
Remarks 3.1
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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 intermediate-value 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 tangent-line 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 intermediate-value 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:
Chapter 16: Return To Roots Worksheet
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
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?
Solution
a.f(0) = 1 > 0, f(1) = –1 < 0, so by the intermediate-value 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
Return To My Roots
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