Calculus 1120, Fall 2012 Dan Barbasch November 15, 2012 Dan - - PowerPoint PPT Presentation

Calculus 1120, Fall 2012

Dan Barbasch November 15, 2012

Dan Barbasch () Calculus 1120, Fall 2012 November 15, 2012 1 / 10

Power Series

∞

• n=0

an(x − a)n. a is (sometimes) called the center. an are called the coefficients. Issues/Summary of what we will learn For what values of x does the power series converge? Answer: There is an open interval (centered at a) (a − R, a + R) so that the series converges absolutely for x in this interval. R is called the radius of convergence. At the endpoints the series may or may not converge absolutely or conditionally. The totality of values for which the series converges is called the interval of convergence. What are its uses? (Short) Answer: Within the radius of convergence a − R < x < a + R you can add/substract/divide/multiply, differentiate and integrate the series term by term as if it were a polynomial.

Dan Barbasch () Calculus 1120, Fall 2012 November 15, 2012 2 / 10

Power Series

Examples:

1 For what values of x does ∞

n=0 xn converge?

2 Same for ∞

n=1 nxn−1 and ∞ n=1 xn n .

3 Same for ∞

n=1(−1)n(x − 3)n.

4 For what values of x does ∞

n=1(−1)n (x−1)n n

converge? What is the value?

5 For what values of x does ∞

n=0 xn n! converge? What is the sum?

6 For what values of x does ∞

n=0 n!xn converge? What is the sum?

7 Find a power series expansion for

1 1−x about a = 2.

8 Find a power series expansion for

1 (1−x)2 about a = 0.

Dan Barbasch () Calculus 1120, Fall 2012 November 15, 2012 2 / 10

General Principles

The radius of convergence is (usually) computed by using the root/ratio test. For the interval of convergence, you have to consider the endpoints

• separately. Usually/often the series are alternating or have positive
• terms. You use the n−term test, alternating series test. Sometimes

the series is absolutely convergent, sometimes conditionally, sometimes divergent. Within the radius of convergence the series is absolutely convergent. You can perform all of the usual operations, adding/multiplying/dividing, differentiating and integrating term by term. But only within the radius of convergence. Often you can find an explicit expression for the series within the radius of convergence. But this does not guarantee anything about what happens at the endpoints.

Dan Barbasch () Calculus 1120, Fall 2012 November 15, 2012 3 / 10

Radius of Convergence

Theorem: (corrollary to theorem 18, page 596). The convergence of a power series ∞

n=0 an(x − a)n is described by one of the following three

cases:

1 There is a number R > 0 so that the series converges absolutely for

|x − a| < R, and diverges for |x − a| > R.

2 The series converges for x = a only. 3 The series converges for all x.

In case (2) we write R = 0. In case (3) we write R = ∞. This number R is called the radius of convergence. (you can see from the formulas that R is lim

n→∞

|an| |an+1| or 1

n

• |an|

). This says nothing about x = a − R or x = a + R. In practice, to find the radius of convergence, you apply the ratio/root test. The condition lim

n→∞

|an+1||x − a|n+1 |an||x − a|n = |x − a| lim

n→∞

|an+1| |an| < 1

Dan Barbasch () Calculus 1120, Fall 2012 November 15, 2012 4 / 10

Warm Up For Next Time

Problem 1: Find the power series expansion of 1 (1 − x)3 about a = 0. Use it to express

∞

• n=0

(n + 2)(n + 1)xn as a fraction (rational function). Problem 2: Find the radius of convergence and interval of convergence of f (x) =

∞

• n=0

(−1)n(x − 2)2n (2n)! . What are the values f (2), f (1)(2), f (2)(2) and f (3)(2)? Problem 3: Let f (x) = ∞

n=0 (−1)nx2n+1 (2n+1)! . Compute f (2)(x). Can you guess

what function f (x) might be? Hints: Problem 1 is related to example 8. Problems 3 is related to example 5; but this is not an ordinary differential equation, something involving higher derivatives. Problem 2 is a preview of new material; look ahead in the sections, see what is to come.

Dan Barbasch () Calculus 1120, Fall 2012 November 15, 2012 5 / 10

Taylor and MacLaurin Series

Fact: For a power series ∞

n=0 an(x − a)n with positive radius of

convergence R > 0, f (n)(a) = n!an. We now go the other direction: Definition: The Taylor series at a of y = f (x) is

∞

• n=0

f (n)(a) n! (x − a)n. For a = 0 it is called MacLaurin series. Warning: This series may or may not converge, and it is not clear that it converges to the value of the function. In principle we know how to determine the radius and interval of convergence of a series; root/ratio test.

Dan Barbasch () Calculus 1120, Fall 2012 November 15, 2012 6 / 10

Taylor and MacLaurin Series

To see whether the series converges to the function: Taylor polynomial: Pn(x) =

n

• i=0

f (i)(a) n! (x − a)i. You must show that f (x) − Pn(x) → 0 as n → ∞. We will learn some theorems that let us conclude this. For example: P0(x) = f (a) P1(x) = f (a) + f ′(a)(x − a) P2(x) = f (a) + f ′(a)(x − a) + f (2)(a)

2

(x − a)2. The function is being approximated by polynomials of various orders. Pn(x) is the n−th partial sum of the Taylor series.

Dan Barbasch () Calculus 1120, Fall 2012 November 15, 2012 6 / 10

Examples

Find the Taylor series for the following functions:

1 f (x) = ln x at a = 2. 2 f (x) = cos x at a = π/3. 3 f (x) = ex at a = 1. 4 f (x) =

x e−t2 dt at a = 0.

5

1 1 + x2 at a = 0. Note: The problems on the material in section 10.8 will be due after we have covered the material.

Dan Barbasch () Calculus 1120, Fall 2012 November 15, 2012 7 / 10

y = ex at a = 1.

a = 1 y ex e . . . . . . . . . y(n) ex e The series is ex ≈ e + e 1!(x − 1) + . . . e n!(x − 1)n + . . ..

Dan Barbasch () Calculus 1120, Fall 2012 November 15, 2012 8 / 10

y = cos x at a = π/3

a = π/3 y cos x 1/2 y(1) − sin x − √ 3/2 y(2) − cos x −1/2 y(3) sin x √ 3/2 y(4) cos x 1/2 The series is cos x ≈1/2 + − √ 3/2 1! (x − π/3) + −1/2 2! (x − π/3)2+ + √ 3/2 3! (x − π/3)3 + 1/2 4! (x − π/3)4 + . . .

Dan Barbasch () Calculus 1120, Fall 2012 November 15, 2012 9 / 10

y = cos x at a = π/3

Inside the radius of convergence we can rearrange the terms: cos x ≈

∞

• n=0
• 1/2

(4n)!(x − π/3)4n + − √ 3/2 (4n + 1)!(x − π/3)4n+1+ + −1/2 (4n + 2)!(x − π/3)4n+2 + √ 3/2 (4n + 3)!(x − π/3)4n+3

• Dan Barbasch ()

Calculus 1120, Fall 2012 November 15, 2012 9 / 10

Taylor’s formula

Theorem: f (x) = Pn(x) + Rn(x). where Rn(x) = f (n+1)(ξ) (n + 1)! (x − a)n+1 for a ξ between a and x. One way to show that the series converges to the function is to show Rn(x) → 0 as n → ∞. Example: y = cos x at a = π/3. f (n+1)(x) = ± sin x

• r

± cos x Rn(x) = ± cos ξ, ± sin ξ (n + 1)! (x − π/3)n+1. Need to estimate Rn(x): |Rn(x)| ≤ |x − π/3|n+1 (n + 1)! .

Dan Barbasch () Calculus 1120, Fall 2012 November 15, 2012 10 / 10