Suppose we wish to compute the value of the continuous function f(x) for a given value of x. This is the number y = f(x). (7.113) For example, if f(x) = /x, then y is the square root of x. If f(x) = 1/x, then y is the reciprocal of x. In general, the practical solution to evaluating f(x) is to construct an appropriate iteration procedure for successively calculating a sequence of approximate values yk that rapidly approach the exact value given by equation (7.113); i.e., lim Yk = Y. k→∞ (7.114) APPLICATIONS 229 To proceed with the construction of an iteration procedure, first rewrite equa- tion (7.113) in implicit form: F(x, y) = 0. (7.115) for representing Note that, in general, there are an unlimited number of equation (7.113) in implicit form. Each of these forms leads to a different iteration scheme for calculating yk. For example, let ways y = Vx, x > 0. (7.116) Then all of the following relations are valid implicit forms: F1 (x, y) = y – Vx = 0, F2(x, y) = y? – x = 0, (7.117) F3(x, y) -1= 0. y? Let Yk be the value of kth approximation to y. Application of the mean value theorem gives F(x, Yk) = F(x, Yk) – F(x, y) = (Yk – y) ƏF(x, Tk) ду (7.118) Now solve equation (7.118) for y; doing where j is a value between this gives the relation Yk and Y. F(x, Yk) F, (x, Jk)' y = Yk (7.119) where the partial derivative is indicated by the notation ƏF(x, §k) ду F,(x, Tk) = (7.120) Now the value of y is unknown. Therefore, our iteration scheme will consist of replacing yk by yk and replacing the exact value y by the corrected value Yk+1. Carrying out these replacements in equation (7.119) gives the following iteration scheme for computing approximations to the values of y for equation (7.113): F(x, Yk) Fy(x, Yk) Yk+1 = Yk = D(x, Yk). (7.121)

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7.2.5
Newton's Method
Suppose we wish to compute the value of the continuous function f(x) for a
given value of x. This is the number
y = f(x).
(7.113)
For example, if f(x) = Vx, then y is the square root of x. If f (x) = 1/x, then
y is the reciprocal of x. In general, the practical solution to evaluating f(x)
is to construct an appropriate iteration procedure for successively calculating
a sequence of approximate values yk that rapidly approach the exact value
given by equation (7.113); i.e.,
lim Yk = Y.
(7.114)
APPLICATIONS
229
To proceed with the construction of an iteration procedure, first rewrite equa-
tion (7.113) in implicit form:
F(x, y) = 0.
(7.115)
Note that, in general, there are an unlimited number of
equation (7.113) in implicit form. Each of these forms leads to a different
iteration scheme for calculating yk. For example, let
ways
for representing
y = Vx, x > 0.
(7.116)
Then all of the following relations are valid implicit forms:
F1 (x, y) = y – Vx = 0,
F2 (x, y) = y² – x = 0,
(7.117)
F3(x, y) = 7
- 1= 0.
Let yk be the value of kth approximation to y. Application of the mean value
theorem gives
ƏF(x, Tk)
F(x, Yix) = F(x, Yk) – F(x, y) = (yk – y):
(7.118)
dy
where jk is a value between yk and y. Now solve equation (7.118) for y; doing
this gives the relation
F(x, Yk)
y = Yk
(7.119)
F,(x, ÿk)'
where the partial derivative is indicated by the notation
ƏF(x, Tk)
ду
Fy(x, Tk) =
(7.120)
Now the value of g is unknown. Therefore, our iteration scheme will consist
of replacing Jk by yk and replacing the exact value y by the corrected value
Yk+1. Carrying out these replacements in equation (7.119) gives the following
iteration scheme for computing approximations to the values of y for equation
(7.113):
F(x, Yk)
F,(r, Yx)
Yk+1 = Yk
= ¤(x, Yk).
(7.121)
The iteration procedure given by equation (7.121) is called Newton's method.
Historically, it was first used for calculating approximations to the roots of
Transcribed Image Text:7.2.5 Newton's Method Suppose we wish to compute the value of the continuous function f(x) for a given value of x. This is the number y = f(x). (7.113) For example, if f(x) = Vx, then y is the square root of x. If f (x) = 1/x, then y is the reciprocal of x. In general, the practical solution to evaluating f(x) is to construct an appropriate iteration procedure for successively calculating a sequence of approximate values yk that rapidly approach the exact value given by equation (7.113); i.e., lim Yk = Y. (7.114) APPLICATIONS 229 To proceed with the construction of an iteration procedure, first rewrite equa- tion (7.113) in implicit form: F(x, y) = 0. (7.115) Note that, in general, there are an unlimited number of equation (7.113) in implicit form. Each of these forms leads to a different iteration scheme for calculating yk. For example, let ways for representing y = Vx, x > 0. (7.116) Then all of the following relations are valid implicit forms: F1 (x, y) = y – Vx = 0, F2 (x, y) = y² – x = 0, (7.117) F3(x, y) = 7 - 1= 0. Let yk be the value of kth approximation to y. Application of the mean value theorem gives ƏF(x, Tk) F(x, Yix) = F(x, Yk) – F(x, y) = (yk – y): (7.118) dy where jk is a value between yk and y. Now solve equation (7.118) for y; doing this gives the relation F(x, Yk) y = Yk (7.119) F,(x, ÿk)' where the partial derivative is indicated by the notation ƏF(x, Tk) ду Fy(x, Tk) = (7.120) Now the value of g is unknown. Therefore, our iteration scheme will consist of replacing Jk by yk and replacing the exact value y by the corrected value Yk+1. Carrying out these replacements in equation (7.119) gives the following iteration scheme for computing approximations to the values of y for equation (7.113): F(x, Yk) F,(r, Yx) Yk+1 = Yk = ¤(x, Yk). (7.121) The iteration procedure given by equation (7.121) is called Newton's method. Historically, it was first used for calculating approximations to the roots of
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