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Unformatted text preview: Onedimensional Taylor’s Series Approximation (Taylor1.doc) Let f be a function which has continuous derivatives to order (n+1) in an interval [51, b].
The Taylor’s series polynomial expansion, 1;, off at some point x0 in [a, b] is given by _ 2 — n
Tl(x;xo)=f(xo)+f’(Iu)(I—Io)+f”(xo)(x 2x0) +.+f(ﬂ)(xo)(x n—TO) where f’, f ”, and f m are ﬁrst, second, and nEh derivatives of the function f with respect to
x. The error between the true function f at x and the approximation I], at x is given by n+1
R x:x = (n+1) f Em.
n( o) f ( ) (n + I)! where 5c is a point between x and x0. In general we don’t know the exact value of the
point i but this remainder or error term frequently allows for an upper bound on the error to be computed.
Example: Consider approximating the function 2‘ about the point IO = 0. That is, we want a polynomial approximation for e‘. (The special case of Taylor’s series using x0 = 0 as the expansion point is called a lVlaclautin2 expansion.) The approximation is developed as (ex) :33, (ex) = 1"”,‘1 (en)_ez'
(dX)
yielding
__ 2 L n _ __ n+1
e‘=e°+e°(x—0)+e°(x 0) ++e°(x 0) i—ei‘——(I 0) ,
n! (n+1)!
which reduces to the well known series
1 x2 x3 Id In 2 xn+l
e 21+x+—+—+—+~+—+e .
2 3! 4! n! (n+1)! The error of the approximation, that is the error if we omit the remainder term, for
x e [1, l] is bounded by ‘ 1841 1 e
e = (n+1)! '(n+1)1‘ ZWBECanuse this ideaito develop inte‘grlation*~forrnul‘a§ for. difﬁculti'funcloris; {Consider H integrating e'x over [0, ":3. Then substituting x for x and integrating the expansion term
by term we obtain Note: If we apply Taylor’s expansion using only the constant term with the error term
being the linear term, we have f(X)= f(xo)+f’(i)(x—xo)i with it being some point between x and 1:0. This yields the standard derivative form of the
well known mean value theorem (Buck, page 43), where i is a point between x and x0, f(x)f(xo)
(it—10) I f '(i) = Homework: 1. Use the methods of Taylor’s series expansion, substitution and term by term integration
to integrate Answer: te"’1dx=tt—+——u+.
° 3 10 42 2. Expand a’ in a Taylor’s series about the point 0.
3. Expand sin(x) (x in degrees) about the point 0. Footnotes:
1 Brook Taylor (16851731), English mathematican.
2 Colin Maclaurin (1698—1746), Scotch mathematican. References:
Kreyszig, Erwin. 1962. Advanced Engineering Mathematics. John Wiley and Sons, Inc.,
New York. For short biosketches of famous contributors to mathematics. Stationary Points: maximum, minimum, and saddle points off
(StanionaryPointsldoc) The deﬁnition of a derivative of a function f at a point x° is I _  f(xo+h)—f(xo)
f (x°)—}‘1_gg———h——— when this limit exists. Note that the direction of approaching the limit point is immaterial
(here it approaches zero from above or below) and the limit must be the same from any
direction for the derivative to exist. (In multiple dimensions the same property must hold,
but there are an inﬁnite number of directions from which to approach the limit point.) We
use the limit concept to show that if a function with continuous ﬁrst derivative, has a local minimum at a point, say x°, then f ’(x0) = 0. (The following proof follows that of Protter
and Morrey, 1964.) A point x° is a local minimum of a function f with continuous ﬁrst derivative, if for all x in
a neighborhood of x0, denoted N,3 (x0) = {xlix — x° < E} , the following is true f(x) 2 for“), for allx e N£(x°).
Assuming that f ’ exists and is continuous on N€(x°) , we have that f(x° +h) 2 ﬁx”), for allx = x° Ht 6 N,(x°)
and, thus, o .. 0
W20, h>0andforallx=xo+h€Ne(xo) The limit must also be bounded and thus, f ’(xo) 2 0. Now considering I: < 0, we have
due to the division by a negative number causes the inequality sign to be reversed , _ 5
W50, h<0andforallx=xo+hENe(xo) and this implies that f ’(xo) 5 0. Hence, since the limits must be the same, the only
possibility is that f ’(x0) = 0. Therefore, if x° is a minimum of a function f with a
continuous ﬁrst derivative on the interior of any interval for x, then f ’(x0) = 0. The derivative equal to zero property is a necessary but not sufﬁcient condition for a point
to be a minimum. It turns out that the same condition, f ’(x0) = 0, applies to local
maximum points as well. This is also true for inﬂection points. Hence, we have a class of
points, S f, called stationary points which satisfy s, = {arm = 0}. Figure I is a graph of a function showing the tangent planes to the curve with zero slope.
The points of tangency are the stationary points of the function. That is, points where the
ﬁrst derivative is zero. Figure 1. Graph of a function with tangent planes
with zero slepes identiﬁed The stationary point condition is a method for isolating maximum, minimum and inﬂection
points of a function. But we need to be able to distinguish between the stationary points
according to type (max, min, inﬂection). A multipledimensional inﬂection point is called
a saddle point and is the term most commonly applied to this category of stationary points. This discrimination process is based on the Taylor’s series expansion previously
developed. Recall the Taylor’s series expansion of the function f with a second order error term is (x x) f(X)= f(x°’°—)+f(X)(x X°)+f”(x)—— 2 wherej? is a point between x and x". When x" is a stationary point, we have, since flcrﬂ) = 0’ (x— X)2 f(X)= f(x°)+f”(X)———— 2 and if x0 is a strict—local minimum point then f (x) > f (x0). This implies that (x I")2
2 f ”(x) > 0. This condition holds if f "(2) > 0. Note here that we do not know the exact value of 5:;
only that we can make it as near to x° as needed since the local minimum condition holds for all x in an E neighborhood about x°, that is, for all x E (x° €,x° + 5). However,
since f ” is assumed to be a continuous function, then the sign of f " at x0 and at i will
be the same. This continuity condition allows us to test the sign of f ” at the known
stationary point x0 rather than the unknown point 1?. This leads us to the discriminate test
for stationary points, where x° is a point such that f ’(x°)= 0 and if _ f ”(x°) > 0, then x° is a local minimum,
f ”(x°) < 0, then x0 is a local maximum, f ”(x°) = 0, need to test higher order approximations. When the second derivative at the stationary point is zero, then we must go to the third
derivative term. If the third derivative term is zero, f ”’(x°) = 0, then we continue with the
fourth derivative term, etc., until we obtain a nonzero term at the stationary point. Ifthe
ﬁrst nonzero tennis an odd power, f w (x0) at Owith nodd, then the point is an inﬂection
point (or saddle point). If the power it is even, then the cases are similar to that of
f”(x°). That is, if n is even and f“’(x°) > 0 then x0 is a local minimum and if n is even and f‘")(x°) < 0 then x° is a local maximum. Example 1: 4 3 2
Find the stationary point(s) of the function f(x) = x? — 3’3“" — 925 + 6x + 5 and determine their type(s). This function is graphed in Figure 2. The derivative function f’, graphed in
Figure 3, is f’(x)=x3—2x25x+6. This function can be factored as
f'(x) = x3 —2x2 5x+ 6 ._. (x—l)(x+2)(x—3) and the stationary points are x, = 1, x2 = —2, and x3 = 3. To determine what type they are,
we evaluate f ”(x,) for each stationary point x, and classify the points according to the
sign. The function f”(x) = 3x2  4x + 5 and thus, (1') at the point x =1, f”(l) = 3— 4— 5 = —6 < 0, and is a local maximum point,
(ii) at the point x = 2, f”(—2) = 12+ 8 5 =15 > O, and is a local minimum point, and
(iii) at the point x = 3, f”(3) z 27*12— 5 =10 > 0, and is a local minimum point. Figure 2. Plot of the function f over [5, 5]. f'tx) Figure 3. Plot of f’ over [5, 5]. Notice that the zero’s of the f ' function coincide with maximum and minimum points of
the function. Thus, our solution method is to find the roots of the function f’ which,
hopefully, can be obtained with less work than directly searching for the maximum and  minimum points. Example 2: Find the stationary point(s) of the function f (x) = (x — 1)5 and determine their type(s).
The derivative function f’ is f’(x) = 5(x — 1)4 = 0, at the point x =1.
The derivatives yield f ”(x) = 20(x — D3, which is zero at the point x = 1, f ”'(x) = 600: 1)2, which is zero at the point x = 1,
f (4) (x) = 120(x  1), which is zero at the point x =1, and f ‘5’ (x) = 120, which is non  zero at the point x = 1. Thus, since the 5‘h derivative is nonzero, then the stationary pointx = 1 is an inflection
point. Homework: Find the stationary points, if they exist, and determine their type.
1 f (x) = x3 + x 2. f(x) = x4 +Jt2 3. f(x) = 615 —4x+lO References: Protter, Murray H. and Charles B. Morrey, Jr. 1964. College Calculus with Analytic
Geomegg. AddisonWesley Publishing Company, Inc., Reading, Mass. Generalization to n Dimensions: The necessary conditions for a stationary point for a function of n variables is that all the
ﬁrst—order partial derivative are zero. This means that Wzo, foralli='1,2i"'=n‘ (2) L To see why this result is true, consider the ndimensional function f(x1,x2,   ,x_) which
has a local minimum at the point (xﬂxg, ,x,?) if 0 0 O 0 0 0
f(x1.x2,~,x,.)2f(x,,x2,,x,), forall(x1,x2,'~,x,)eNE(x1Jewax.) Again, a point is in ane neighborhood of another point if the distance between the two
points is less than E, or 2(xi—xf’r <6. {:1 Censider a variation in the 1"“ coordinate about the point (xf,x§ , ,x? ,   ,xf). This new point is of the form (xlogé', ~,x? +h, ,xS) where his small; technically «I? < G.
Since this point is within thee neighborhood of the test point, then 0 D 0 G
f(x,°,x2,,x?+h,~,x3)2f(x1.x2., f,‘,x,). This implies for h > 0, that. 0 U 0 0 0 0 0 0
f(x12xzs”'sx; +h:"'axn)—f(x1:xzan'axl's'uvxﬂ) >0
Wk ._ . Taking the limit as h approaches zero as a positive number this becomes 0 0 D 0
Qf(xl,x2,',xi,,x,,) 8x. I 20. Taking the limit as h approaches zero as a negative number yields the following, note that
dividing by a negative number changes the direction of the inequality, 0 U 0 0
J(x1,12,...,xi,.n’xn) 3x I 50. These two conditions holding simultaneously imply that the ﬁrst partial of f with respect
to xi must be zero. Thus, we have the general multiple dimensional local minimum point condition stated above. By the same argument, with inequalities ﬂipped, we ﬁnd the same
condition must hold at a maximum point. So the general multipledimension stationary
point condition is that of equation (2). OneDimensional NewtonRaphson Method
Our problem is to ﬁnd an 1? such that go?) = 0. We approach this problem by making a guess for the root, call it x1, and then use Taylor’s Series to deve10p a linear
approximation to 3 about this point: 30:) = g(x‘) + g’(x1)(x  x1).
The idea is to ﬁnd g(x) = 0, so we set the linear appr0ximation to zero, the righthand side
of the above equation, and then solve for the value of x, call it x2, which satisﬁes this
condition
3(x‘)+ g’(X‘)(xx1) = 0, which yields the new estimate of the root, x2, as x2 =x1_ 30511)
g’(x ) If we continue this process, we have the general iteration formula xk+l =xk _ 3(1")
3’06") e The sequence of points, {xk};1 frequently converges to a point, call it )2, and this point
has the property that go?) = 0. That is, the limit point it is a root of the function 3. Note that convergence is not guaranteed. When this sequence converges, it does so very fast, so that we can truncate the sequence when we get within an acceptable tolerance of the
root. This process is illustrated below. Figure 2. The N ewtonRaphson rootfinding process. The function gl is the linear
approximation to g at the point x1 = 3.5. The function gll is the linear approximation to g
at the second iteration point x2 = 4.75 . Example. Find a root of the function g(x) = x3 —11x2 + 39x: 45. This function has two distinct roots: 3 and 5. The following table displays two sets of iterations of the process
using the starting points 2 and 6. From the initial point x1 = 2, the process converges to
the root SE = 3 . From the initial point x1 = 6, the process converges to the root )2 = 5.
The initial and undated points as well as the function value are displayed for each of 10
iterations. k+1 ) 30:
6.000 5.400 2.3040 5.400 5.100 0.4410
5.100 5.009 0.0351
5.009 5.000 0.0003
5.000 5.000 0.0000
5.000 5.000 0.0000
5.000 5.000 0.0000
5.000 5.000 0.0000
5.000 5.000 0.0000
5.000 5.000 0.0000 \DOOHJONUlhUJNHR‘ l—'
0 Problems 1. Find a root of the function x2  2x—8 using the NewtonRaphson method with starting
 1 pomt x = 3. 2. For Problem 1, can you find a starting point which will cause the Newton—Raphson
method to fail? ...
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