Taylor Polynomials. James K. Peterson. Department of Biological Sciences and Department of Mathematical Sciences Clemson University

James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University September 24, 2013

Outline 1 First Order Approximation s Second Order Approximations 2 Approximation Over Long Times

Abstract This lecture is about function approximation and what are called.

Let s consider a function which is defined locally at the point p. This means there is at least an interval (a, b) containing p where f is defined. Of course, this interval could be the whole x axis! Let s also assume f exists locally at p in this same interval. Now pick any x is the interval [p, b) (we can also pick a point in the left hand interval (a, p] but we will leave that discussion to you!). From Calculus I, recall Rolle s Theorem and the Mean Value Theorem. These are usually discussed in Calculus I, but we really prove them carefully in course called Mathematical Analysis (but that is another story).

Rolle s Theorem Theorem Rolle s Theorem Let f : [a, b] R be a function defined on the interval [a, b] which is continuous on the closed interval [a, b] and is at least differentiable on the open interval (a, b). If f (a) = f (b), then there is at least one point c, between a and b, so that f (c) = 0. Proof. Take a piece of rope in your hand and make sure you hold the right and left side at the same height. When the rope is stretched tight, it represents the graph of a constant function: at all points c, f (c) = 0. If we pull up on the rope in between our hands, the graph goes up to a maximum and at that point c, again we have f (c) = 0. A similar argument holds if we pull down on the rope to get a minimum.

The Mean Value Theorem Theorem Let f : [a, b] R be continuous on [a, b] and is at least differentiable on (a, b). Then there is at least one point c, between a and b, so that f (c) = [f (b) f (a)]/[b a]. Proof. Define M = (f (b) f (a))/(b a) and the function g(t) = f (t) f (a) M(t a). Note g(a) = 0 and g(b) = f (b) f (a) [(f (b) f (a))/(b a)](b a) = 0. Apply Rolle s Theorem to g. So there is a < c < b so that g (c) = 0. But g (t) = f (t) M, so g (c) = 0 tells us f (c) = M. We conclude there is a point c, between a and b so that f (c) = [f (b) f (a)]/(b a)].

The Mean Value Theorem Theorem Let f : [a, b] R be continuous on [a, b] and is at least differentiable on (a, b). Then for each x, there is at least one point c, between a and x, so that f (c) = [f (x) f (a)]/[x a]. Proof. Define M = (f (x) f (a))/(x a) and the function g(t) = f (t) f (a) M(t a). Note g(a) = 0 and g(x) = f (x) f (a) [(f (x) f (a))/(x a)](x a) = 0. Apply Rolle s Theorem to g. So there is a < c < x so that g (c) = 0. But g (t) = f (t) M, so g (c) = 0 tells us f (c) = M. We conclude there is a point c, between a and x so that f (c) = [f (x) f (a)]/(x a)].

Zeroth Order Taylor Polynomial Theorem Let f : [a, b] R be continuous on [a, b] and be at least differentiable on (a, b). Then for each x, there is at least one point c, between a and x, so that f (x) = f (a) + f (c)(x a). The constant f (a) is called the zeroth order Taylor Polynomial for f at a. and we denote it by p 0 (x; a). The point a is called the base point. Note we are approximating f (x) by the constant f (a) and the error we make is f (c)(x a). If f (t) = t 3, by the theorem above, we know on the interval [1, 3] that f (t) = f (1) + f (c)(t 1) where c is some point between 1 and 3. Thus, t 3 = 1 + (3c 2 )(t 1) for some 1 < c < 3. So here the zeroth order Taylor Polynomial is p 0 (t, 1) = 1 and the error is (3c 2 )(t 1).

If f (t) = e 1.2t, by the theorem above, we know on the interval [0, 5] that f (t) = f (0) + f (c)(t 0) where c is some point between 0 and 5. Thus, e 1.2t = 1 + ( 1.2)e 1.2c (t 0) for some 0 < c < 5 or e 1.2t = 1 1.2e 1.2c t So here the zeroth order Taylor Polynomial is p 0 (t, 1) = 1 and the error is 1.2e 1.2c t. If f (t) = e.00231t, by the theorem above, we know on the interval [0, 8] that f (t) = f (0) + f (c)(t 0) where c is some point between 0 and 8. Thus, e.00231t = 1 + (.00231)e.00231c (t 0) for some 0 < c < 8 or e.00231t = 1.00231e.00231c t. So here the zeroth order Taylor Polynomial is p 0 (t, 1) = 1 and the error is.00231e.00231c t.

The Order One Taylor Polynomial Theorem Let f : [a, b] R be continuous on [a, b] and is at least two times differentiable on (a, b). Then for each x, there is at least one point c, between a and x, so that f (x) = f (a) + f (a)(x a) + (1/2)f (c)(x a) 2. Proof. Define M = (f (x) f (a) f (a)(x a))/[(x a) 2 ] and the function g(t) = f (t) f (a) f (a)(t a) M(t a) 2. So g(a) = 0 and

Proof. g(x) = f (x) f (a) f (a)(x a) ( f (x) f (a) f ) (a)(x a) (x a) 2 (x a) 2 = f (x) f (a) f (a)(x a) (f (x) f (a) f (a)(x a)) = 0. Apply Rolle s Theorem to g which tells us there is a point d so that g (d) = 0 with d between a and x. Next, we find g (t) = f (t) f (a) 2M(t a). Then g (a) = 0 and we know g (d) = 0. Apply Rolle s Theorem to g on [a, d]. So there is a < c < d with g (c) = 0. But g (t) = f (t) 2M, so f (c) = 2M. We conclude there is a point c, between a and b so that (1/2)f (c) = [f (x) f (a) f (a)(x a)]/(x a) 2 ].

Proof. We conclude there is a point c between a and x so that f (x) = f (a) + f (a)(x a) + 1 2 f (c)(x a) 2. The function f (a) + f (a)(x a) is called the first order Taylor Polynomial for f at base point a and is also called p 1 (t, a). It is also the traditional tangent line to f at a from first semester calculus. The error is (1/2)f (c)(x a) 2.

First Order Approximation s For f (t) = e 1.2t on the interval [0, 5] find the tangent line approximation, the error and maximum the error can be on the interval. f (t) = f (0) + f (0)(t 0) + (1/2)f (c)(t 0) 2 = 1 + ( 1.2)(t 0) + (1/2)( 1.2) 2 e 1.2c (t 0) 2 = 1 1.2t + (1/2)(1.2) 2 e 1.2c t 2. where c is some point between 0 and 5. The first order Taylor Polynomial is p 1 (t, 0) = 1 1.2t which is also the tangent line to e 1.2t at 0. The error is (1/2)( 1.2) 2 e 1.2c t 2. Now let AE denote absolute value of the actual error and ME be maximum absolute error on a given interval.

First Order Approximation s continued: The largest the error can be on [0, 5] is when f (c) is the biggest it can be on the interval. Here, AE = (1/2)(1.2) 2 e 1.2c t 2 (1/2)(1.2) 2 1 (5) 2 = (1/2)1.44 25 = ME. Do this same problem on the interval [0, 10]. The approximations are the same but now 0 < c < 10 and AE = (1/2)(1.2) 2 e 1.2c t 2 (1/2)(1.2) 2 1 (10) 2 = (1/2)1.44 100 = ME.

First Order Approximation s Do this same problem on the interval [0, 100]. The approximations are the same, 0 < c < 100 and AE = (1/2)(1.2) 2 e 1.2c t 2 (1/2)(1.2) 2 (1)(100) 2 = (1/2)1.44 10 4 = ME. Do this same problem on the interval [0, T ]. The approximations are the same, 0 < c < T and AE = (1/2)(1.2) 2 e 1.2c t 2 (1/2)(1.2) 2 (1)(T ) 2 = (1/2)1.44 T 2 = ME.

First Order Approximation s If f (t) = e βt, for β = 1.2 10 5, find the tangent line approximation, the error and the maximum error on [0, 5]. On the interval [0, 5], then f (t) = f (0) + f (0)(t 0) + 1 2 f (c)(t 0) 2 = 1 + ( β)(t 0) + 1 2 ( β)2 e βc (t 0) 2 = 1 βt + 1 2 β2 e βc t 2. where c is some point between 0 and 5. The first order Taylor Polynomial is p 1 (t, 0) = 1 βt which is also the tangent line to e βt at 0. The error is 1 2 β2 e βc t 2.

First Order Approximation s continued: The largest the error can be on [0, 5] is when f (c) is the biggest it can be on the interval. Here, AE = (1/2)(1.2 10 5 ) 2 e 1.2 10 5c t 2 (1/2)(1.2 10 5 ) 2 (1)(5) 2 = (1/2)1.44 10 10 (25) = ME Do this same problem on the interval [0, 10] The approximation is the same, c is between 0 and 10 and AE = (1/2)(1.2 10 5 ) 2 e 1.2 10 5c t 2 (1/2)(1.2 10 5 ) 2 (1)(10) 2 = (1/2)1.44 10 10 100 = ME.

First Order Approximation s Do this same problem on the interval [0, 100] The approximation is the same, c is between 0 and 100 and AE = (1/2)(1.2 10 5 ) 2 e 1.2 10 5c t 2 (1/2)(1.2 10 5 ) 2 (1)(100) 2 = (1/2)1.44 10 10 10 4 = ME. Do this same problem on the interval [0, T ] The approximation is the same, c is between 0 and T and AE = (1/2)(1.2 10 5 ) 2 e 1.2 10 5c t 2 (1/2)(1.2 10 5 ) 2 (1)(T ) 2 = (1/2)1.44 10 10 T 2 = ME.

First Order Approximation s If f (t) = cos(3t) + 2 sin(4t), then on the interval [0, 5] find the tangent line approximation, the error and the maximum error. We have f (0) = 1 and f (t) = 3 sin(3t) + 8 cos(4t) so that f (0) = 8. Further, f (t) = 9 cos(3t) 32 sin(4t) so f (c) = 9 cos(3c) 32 sin(4c). Thus, for some 0 < c < 5 f (t) = f (0) + f (0)(t 0) + 1 2 f (c)(t 0) 2 = 1 + (8)(t 0) + 1 2 ( 9 cos(3c) 32 sin(4c))(t 0)2, The first order Taylor Polynomial is p 1 (t, 0) = 1 + 8t which is also the tangent line to cos(3t) + 2 sin(4t) at 0. The error is (1/2)( 9 cos(3c) 32 sin(4c))(t 0) 2.

First Order Approximation s continued: the error is largest on [0, 5] when f (c) is the biggest it can be on [0, 5]. Here, AE = (1/2)( 9 cos(3c) 32 sin(4c)) t 2 (1/2) 9 cos(3c) 32 sin(4c) 5 2 (1/2) 9 + 32 5 2 = (1/2)(41)(25) = ME Do this same problem on the interval [0, 10]. The approximations are the same, 0 < c < 10 and AE = (1/2)( 9 cos(3c) 32 sin(4c)) t 2 (1/2) 9 cos(3c) 32 sin(4c) (10) 2 = (1/2)(41)(100) = ME.

First Order Approximation s Do this same problem on the interval [0, 100]. The approximations are the same, 0 < c < 100 and AE = (1/2)( 9 cos(3c) 32 sin(4c)) t 2 (1/2) 9 cos(3c) 32 sin(4c) (100) 2 = (1/2)(41)(10 4 ) = ME. Do this same problem on the interval [0, T ]. The approximations are the same, 0 < c < T and AE = (1/2)( 9 cos(3c) 32 sin(4c)) t 2 (1/2) 9 cos(3c) 32 sin(4c) T 2 = (1/2)(41)T 2 = ME.

First Order Approximation s Homework 34 For these problems, find the Taylor polynomial of order one, i.e. the Tangent line approximation. state the error in terms of the second derivative. state the maximum absolute error on the given interval. 34.1 f (t) = e 2.3 10 4 t at base point 0, intervals [0, 10], [0, 100] and [0, T ]. 34.2 f (t) = e 6.8 10 6 t at base point 0, intervals [0, 20], [0, 200] and [0, T ]. 34.3 f (t) = 2 cos(5t) + 5 sin(3t) at base point 0, intervals [0, 30], [0, 300] and [0, T ]. 34.4 f (t) = 8 cos(2t) + 3 sin(7t) at base point 0, intervals [0, 40], [0, 400] and [0, T ].

Second Order Approximations We can do more: The Order Two Taylor Polynomial Theorem Let f : [a, b] R be continuous on [a, b] and is at least three times differentiable on (a, b). Then for each x, there is at least one point c, between a and x, so that f (x) = f (a) + f (a)(x a) + (1/2)f (a)(x a) 2 +(1/6)f (c)(x a) 3. The function f (a) + f (a)(x a) + (1/2)f (a)(x a) 2 is the second order Taylor polynomial to f at a. It is also called the second order approximation to f at a. The error term is now (1/6)f (a)(x a) 3.

Second Order Approximations If f (t) = e βt, for β = 1.2 10 5, find the second order approximation, the error and the maximum error on [0, 5]. On the interval [0, 5], then there is some 0 < c < 5 so that f (t) = f (0) + f (0)(t 0) + (1/2)f (0)(t 0) 2 +(1/6)f (c)(t 0) 3 = 1 + ( β)(t 0) + 1 2 ( β)2 (t 0) 2 + 1 6 ( β)3 e βc (t 0) 3 = 1 βt + (1/2)β 2 (1/6)β 3 e βc t 3. The second order Taylor Polynomial is p 2 (t, 0) = 1 βt + (1/2)β 2 t 2 which is also called the quadratic approximation to e βt at 0. The error is 1 6 β3 e βc t 3.

Second Order Approximations continued: the error is largest on [0, 5] when f (c) is the biggest it can be on the interval. Here, AE = (1/6)(1.2 10 5 ) 3 e 1.2 10 5c t 3 (1/6)(1.2 10 5 ) 3 e 1.2 10 5c t 3 (1/6)(1.2 10 5 ) 3 (1) (5) 3 = (1/6) 1.728 10 15 (125) = ME Do this same problem on the interval [0, T ]. The approximations are the same, 0 < c < T and AE = (1/6)(1.2 10 5 ) 3 e 1.2 10 5c t 3.

Second Order Approximations continued: hence, AE (1/6)(1.2 10 5 ) 3 e 1.2 10 5c t 3 (1/6)(1.2 10 5 ) 3 T 3 = (1/6) 1.728 10 15 (T 3 ) = ME. If f (t) = 7 cos(3t) + 3 sin(5t), then on the interval [0, 15] find the second order approximation, the error and the maximum error. We have f (0) = 7 and f (t) = 21 sin(3t) + 15 cos(5t) so that f (0) = 15. Further, f (t) = 63 cos(3t) 75 sin(5t) so f (0) = 63 and finally f (t) = 189 sin(3t) 375 cos(5t) and f (c) = 189 sin(3c) 375 cos(5c).

Second Order Approximations continued: Thus, for some 0 < c < 15 f (t) = f (0) + f (0)(t 0) + (1/2)f (0)(t 0) 2 +(1/6)f (0)(t 0) 3 = 7 + (15)(t 0) + (1/2)( 63)(t 0) 2 +(1/6)(189 sin(3c) 375 cos(5c))(t 0) 3 = 7 + 15t (1/2)63t 2 +(1/6)(189 sin(3c) 375 cos(5c))(t 0) 3 The second order Taylor Polynomial is p 2 (t, 0) = 7 + 15t (63/2)t 2 which is also the quadratic approximation to 7 cos(3t) + 3 sin(5t) at 0. The error is (1/6)(189 sin(3c) 375 cos(5c))(t 0) 3.

Second Order Approximations continued: The largest the error can be on the interval [0, 15] is then AE = (1/6)(189 sin(3c) 375 cos(5c))(t 0) 3 (1/6) (189 sin(3c) 375 cos(5c))t 3 (1/6)(189 + 375) t 3 (1/6)(564)(15) 3 = ME. Do this same problem on the interval [0, T ]. The approximations are the same, 0 < c < T and AE = (1/6)(189 sin(3c) 375 cos(5c))(t 0) 3 (1/6)(189 + 375) t 3 (1/6)(564)T 3 = ME.

Second Order Approximations Homework 35 For these problems, find the Taylor polynomial of order two, i.e. the second order or quadratic approximation. state the error in terms of the third derivative. state the maximum error on the given interval. 35.1 f (t) = e 2.8 10 3 t at base point 0, intervals [0, 10], [0, 100] and [0, T ]. 35.2 f (t) = e 1.8 10 4 t at base point 0, intervals [0, 20], [0, 200] and [0, T ]. 35.3 f (t) = 6 cos(2t) 5 sin(4t) at base point 0, intervals [0, 30], [0, 300] and [0, T ]. 35.4 f (t) = 6 cos(5t) + 3 sin(2t) at base point 0, intervals [0, 40], [0, 400] and [0, T ].

Approximation Over Long Times We often want to approximate a functions like e βt over a long time scale line human life time.

Approximation Over Long Times We often want to approximate a functions like e βt over a long time scale line human life time. Let s get started on this. We ll graph the tangent line and the function and see what AE and ME look like

Approximation Over Long Times We often want to approximate a functions like e βt over a long time scale line human life time. Let s get started on this. We ll graph the tangent line and the function and see what AE and ME look like We want to know when is the tangent line approximation a reasonable one?

Approximation Over Long Times Homework 36 For these problems given the function f (t) and the base point t 0 : Tangent Line Approximations Write the approximation equation f (t) = P 1 (t, t 0 ) + (1/2)f (c t )(t t 0 ) 2. State the Tangent line function, TL(t) = P 1 (t, t 0 ). State where the point c t is located for the tangent line approximation. For a given time, t, state the absolute error. Draw on the same graph f (t) and the tangent line labeling important things. For the given final time, T F, determine the overestimate of the absolute error f (t) TL(t)

Approximation Over Long Times Homework 36 Continued Quadratic Approximations Write the approximation equation f (t) = P 2 (t, t 0 ) + (1/6)f (c t )(1/6)(t t 0 ) 3. State the Quadratic approximation Q(t) = P 2 (t, t 0 ). State where the point c t is located for the quadratic approximation. For a given time, t, state the absolute error. Draw on the same graph f (t) and the quadratic labeling important things. For the given final time, T F, determine the overestimate of the absolute error f (t) Q(t)

Approximation Over Long Times Homework 36 Continued 36.1 Measure t in days and use β = 0.0015, f (t) = e βt, t 0 = 0 and T F = 60 years. 36.2 Measure t in days and use β = 0.00025, f (t) = e βt, t 0 = 0 and T F = 70 years. 36.3 Measure t in days and use β = 0.0000145, f (t) = e βt, t 0 = 0 and T F = 80 years. 36.4 Measure t in days and use β = 0.0000015, f (t) = e βt, t 0 = 0 and T F = 90 years.