Week #6 - Taylor Series, Derivatives and Graphs Section 10.1

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1 Week #6 - Taylor Series, Derivatives and Graphs Section 10.1 From Calculus, Single Variable by Hughes-Hallett, Gleason, McCallum et. al. Copyright 005 by John Wiley & Sons, Inc. This material is used by permission of John Wiley & Sons, Inc. QUIZ PREPARATION PROBLEMS For Exercises 1-10, find the Taylor polynomials of degree n approximating the given functions for x near 0. (Assume p is a constant.) x, n = 3,5,7 f(x) = (1 x) 1 f(0) = 1 f (x) = (1 x) ( 1) = (1 x) f (0) = 1 f (x) = (1 x) 3 ( 1) = (1 x) 3 f (0) = f (x) = 3(1 x) 4 f (0) = 6. f (n) (x) = n!(1 x) n 1 f (n) (0) = n! Using this information, P 3 (x) = 1 + (x 0) + (x 0) + 6 (x 0)3 3! = 1 + x + x + x 3 Similarly, P 5 (x) = 1 + x + x + 6 3! x3 + x4 + 5! 5! x5 = 1 + x + x + x 3 + x 4 + x 5 Following the same pattern, x, n =,3,4 P 7 (x) = 1 + x + x + x 3 + x 4 + x 5 + x 6 + x 7 1

2 f(x) = (1 + x) 1/ f(0) = 1 f (x) = 1 (1 + x) 1/ f (0) = 1 ( )( ) 1 1 f (x) = (1 + x) 3/ f (0) = 1 4 ( )( )( ) f (x) = (1 + x) 5/ f (0) = 3 8 ( )( )( )( ) f (x) = (1 + x) 7/ f (0) = Using this information, Similarly, P (x) = f(0) + f (0)(x 0) + f (0) (x 0) = x x P 3 (x) = x x = x 1 8 x x3 1 3! x3 P 4 (x) = x 1 4 x ! x = x 1 4 x x x4 x4 8. tan x, n = 3,4 Recall the useful identities that sin(0) = 0 and cos(0) = 1. f(x) = tan(x) f(0) = 0 f (x) = (cos x) f (0) = 1 f (x) = (cos x) 3 ( sin x) = (cos x) 3 (sin x) f (0) = 0 f (3) (x) = 6(cos x) 4 ( sin x)(sin x) + (cos x) 3 cos(x) = 6(cos x) 4 (sin x) + (cos x) f (3) (0) = f (4) (x) = 4(cos x) 5 ( sin x)(sin x) + 1(cos x) 4 (sin x)(cos x) 4(cos x) 3 ( sin x) = 4(cos x) 5 (sin x) (cos x) 3 (sin x) f (4) (0) = 0

3 Because the fourth derivative is zero, both P 3 (x) and P 4 (x) will be equal: P 4 (x) = x x + 1 3! x x4 = x x3 P 3 (x) = x x3 For Exercises 11-16, find the Taylor polynomial of degree n for x near the given point a. 11. e x,a = 1, n = 4 f(x) = e x f (x) = e x f (x) = e x f (x) = e x f (x) = e x f(1) = e f (1) = e f (1) = e f (1) = e f (1) = e Recall, e is just a number, with a value close to.718. We build the Taylor polynomial using the usual formula. P 4 (x) = e + e(x 1) + e (x 1) + e 3! (x 1)3 + e (x 1)4 = e + e(x 1) + e (x 1) + e 6 (x 1)3 + e (x 1) x, a = 1, n = 3 f(x) = (1 + x) 1/ f(1) = ( ) 1 f (x) = (1 + x) 1/ f (1) = 1 ( )( ) 1 1 f (x) = (1 + x) 3/ f (1) = 1 1 = 4 3/ 8 ( )( ) ( ) f (x) = (1 + x) 5/ f 3 (1) = 8 5/ = 3 3 P 3 (x) = + 1 (x 1) 1 8! (x 1) (x 1)3 3! 3

4 13. sin x, a = π/, n = 4 Recall from the graphs of sin and cos that sin(π/) = 1 and cos(π/) = 0. f(x) = sin x f(π/) = 1 f (x) = cos x f (π/) = 0 f (x) = sin x f (π/) = 1 f (x) = cos x f (π/) = 0 f (x) = sin x f (π/) = 1 P 4 (x) = 1 + 0(x π/) 1! (x π/) + 0 3! (x π/)3 + 1 (x π/)3 = 1 1! (x π/) + 1 (x π/)4 14. cos x, a = π/4, n = 3 π/4 radians is 45 o. From the standard triangles, this gives sin(π/4) = cos(π/4) = 1. f(x) = cos x f(π/4) = 1 f (x) = sin x f (π/4) = 1 f (x) = cos x f (π/4) = 1 f (x) = sin x f (π/4) = 1 P 3 (x) = 1 1 (x π/4) 1! (x π/4) + 1 3! (x π/4) ln(x ), a = 1, n = 4 f(x) = ln(x ) f(1) = 0 f (x) = 1 x x = x 1 f (1) = f (x) = x f (1) = f (x) = 4x 3 f (1) = 4 f (x) = 1x 4 f (1) = 1 4

5 P 3 (x) = 0 + (x 1)! (x 1) + 4 3! (x 1)3 1 (x 1)4 3! = (x 1) (x 1) + 3 (x 1)3 1 (x 1)4 16. sin(x), a = π/4, n = 4 Recall from the graphs of sin and cos that sin(π/) = 1 and cos(π/) = 0. f(x) = sin(x) f(π/4) = sin(π/) = 1 f (x) = cos(x) f (π/4) = 0 f (x) = 4sin(x) f (π/4) = 4 f (x) = 8cos(x) f (π/4) = 0 f (x) = 16sin(x) f (π/4) = 16 P 4 (x) = 1 + 0(x π/4) 4! (x π/4) + 0 3! (x π/4) (x π/4)4 = 1 (x π/4) + (x π/4)4 3 For Problems 17-0, suppose P (x) = a+bx+cx is the second degree Taylor polynomial for the function f about x = 0. What can you say about the signs of a, b, c if f has the graph given below? 18. Since P (x) is the second degree Taylor polynomial for f(x) about x = 0, we know from the definition that P(0) = a = f(0) y-intercept P (0) = b = f (0) slope at x = 0 P (0) = c = f (0) concavity at x = 0 As with the previous problem, a > 0, b < 0 and c < As with the original problem, a < 0, b < 0 and c > 0. 5

6 1. The function f(x) is approximated near x = 0 by the second degree Taylor polynomial P (x) = 5 7x + 8x Give the value of (a) f(0) (b) f (0) (c) f (0) Using the fact that f(x) P (x) = f(0) + f (0)x + f (0) x and identifying coefficients with those given for P (x), we obtain the following: (a) f(0) = constant term = 5, (b) f (0) = coefficient of x = -7, (c) f (0) = coefficient of x = 8, so f (0) = Find the second-degree Taylor polynomial for f(x) = 4x 7x + about x = 0. What do you notice? f(x) = 4x 7x + f(0) = f (x) = 8x 7 f (0) = 7 f (x) = 8 f (0) = 8, so P (x) = + ( 7)x + 8 x = 4x 7x +. We notice that f(x) = P (x) in this case. 6. Find the third-degree Taylor polynomial for f(x) = x 3 + 7x 5x + 1 about x = 0. What do you notice? f (x) = 3x + 14x 5, f (x) = 6x + 14, f (x) = 6. Thus, about a = 0, P 3 (x) = ! x + 14! x + 6 3! x3 = 1 5x + 7x + x 3 = f(x). Once again, we notice that the Taylor polynomial of a function which already is a polynomial gives us the same polynomial back. 7. (a) Based on your observations in Problems 5-6, make a conjecture about Taylor approximations in the case when f is itself a polynomial. (b) Show that your conjecture is true. 6

7 f (0) = (a 1 + a x + + na n x n 1 ) x=0 = a 1 ; f (0) = (a + 3 a 3 x + + n(n 1)a n x n ) x=0 =! a. (a) We ll make the following conjecture: If f(x) is a polynomial of degree n, i.e. f(x) = a 0 + a 1 x + a x + + a n 1 x n 1 + a n x n ; then P n (x), the n th degree Taylor polynomial for f(x) about x = 0, is f(x) itself. (b) All we need to do is to calculate P n (x), the n th degree Taylor polynomial for f about x = 0 and see if it is the same as f(x). f(0) = a 0 ; If we continue doing this, we ll see in general Therefore, f (k) (0) = k! a k, k = 1,,3,...,n. P n (x) = f(0) + f (0) x + f (0) x + + f(n) (0) 1!! n! = a 0 + a 1 x + a x + + a n x n = f(x). 33. (a) Find the Taylor polynomial approximation of degree 4 about x = 0 for the function f(x) = e x. (b) Compare this result to the Taylor polynomial approximation of degree for the function f(x) = e x about x = 0. What do you notice? (c) Use your observation in part (b) to write out the Taylor polynomial approximation of degree 0 for the function in part (a). (d) What is the Taylor polynomial approximation of degree 5 for the function f(x) = e x? (a) f(x) = e x. f (x) = xe x,f (x) = (1 + x )e x,f (x) = 4(3x + x 3 )e x, f (4) = 4(3 + 6x )e x + 4(3x + x 3 )xe x. The Taylor polynomial about x = 0 is x n P 4 (x) = ! x +! x + 0 3! x3 + 1 x4 = 1 + x + 1 x4. 7

8 (b) f(x) = e z. The Taylor polynomial of degree is Q (x) = 1 + x 1! + x! = 1 + x + 1 x. If we substitute x for x in the Taylor polynomial for e x of degree, we will get P 4 (x), the Taylor polynomial for e x of degree 4: Q (x ) = 1 + x + 1 (x ) = 1 + x + 1 x4 = P 4 (x). (c) Let Q 10 (x) = 1 + x 1! + x! + + x10 10! be the Taylor polynomial of degree 10 for e x about x = 0. Then P 0 (x) = Q 10 (x ) (d) Let e x Q 5 (x) = 1 + x 1! + + x5 5!. Then = 1 + x 1! + (x ) + + (x ) 10! 10! = 1 + x 1! + x4! + + x0 10!. e x Q 5 ( x) = 1 + x 1! + ( x)! + ( x)3 3! + ( x)4 = 1 x + x 4 3 x3 + 3 x x5. + ( x)5 5! 8