Calculus 1120, Class 42 Dan Barbasch April 30, 2012 Dan Barbasch - - PowerPoint PPT Presentation

Calculus 1120, Class 42

Dan Barbasch April 30, 2012

Dan Barbasch () Calculus 1120, Class 42 April 30, 2012 1 / 1

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. We know how to determine the radius and interval of convergence; we have to do it for each series separately.

Dan Barbasch () Calculus 1120, Class 42 April 30, 2012 2 / 1

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, Class 42 April 30, 2012 2 / 1

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 value ξ between a and x. One way to show that the series converges to the function is to show Rn(x) → 0 as n → ∞.

Dan Barbasch () Calculus 1120, Class 42 April 30, 2012 3 / 1

y = (1 + x)m at a = 0, the binomial series

a = 0 y (1 + x)m 1 y(1) m(1 + x)m−1 m y(2) m(m − 1)(1 + x)m−2 m(m − 1) . . . . . . . . . y(n) m . . . (m − n + 1)(1 + x)m−n m . . . (m − n + 1) y(n+1) m . . . (m − n)(1 + x)m−n−1 m . . . (m − n)(1 + ξ)m−n−1 The series is (1 + x)m ≈

∞

• n=0

m(m − 1) . . . (m − n + 1) n! xn For m > 0 an integer, the series stops, and becomes the binomial formula. We write m

k

• = m(m−1)...(m−k+1)

k!

. The radius of convergence is R = 1.

Dan Barbasch () Calculus 1120, Class 42 April 30, 2012 4 / 1

y = (1 + x)m at a = 0, the binomial series

Why does the series converge to (1 + x)m for |x| < 1? Answer: The function y = (1 + x)m satisfies y(0) = 1 and y′ = m(1 + x)m−1 = m

y (1+x) ⇔ (1 + x)y′ = my. This is a separable

differential equation, and it has a unique solution with the given initial

• condition. You can use term by term differentiation to verify that the

series satisfies (1 + x)y′ = my. PoP QuiZ:

1 Estimate √1 + x over the interval −0.5 < x < 0.5 using the Taylor

polynomial P2. Find a bound for the error.

2 Estimate

√ 2 with accuracy three decimals using the binomial series.

3 Check that the series for y = (1 + x)m satisfies the equation listed,

exercise 64.

4 We know from the notes that

Erf (x) = x

0 e−t2 dt = ∞ n=0 (−1)nx2n+1 (2n+1)n! . How many terms do we need

in order to be sure to approximate Erf (1) with accuracy < 10−3?

Dan Barbasch () Calculus 1120, Class 42 April 30, 2012 4 / 1

y = (1 + x)1/2

(1 + x)1/2 =

∞

• n=0

1/2 n

• xn = 1 + 1/2

1! x + 1/2(1/2 − 1) 2! x2+ +1/2(1/2 − 1)(1/2 − 2) 3! x3 + . . . for |x| < 1. We know (1 + x)1/2 = P2(x) + R2(x) = 1 + 1 2x − 1 8x2 + R2(x). R2(x) = 3 8(1 + ξ)−5/2 x3 6 = 1 16(1 + ξ)−5/2x3.

Dan Barbasch () Calculus 1120, Class 42 April 30, 2012 5 / 1

y = (1 + x)1/2

To compute R2 we use the formula R2(x) = f (3)(ξ)

3!

• x3. The formula on the

previous page is y(3)(ξ) = (1/2)(−1/2)(−3/2) (1 + ξ)3/2 = 3 8(1 + ξ)−3/2. |R2(x)| = 1 16 1 (1 + ξ)5/2 |x|3. We estimate, if 0.5 ≤ x ≤ 0.5 then so is ξ, because it is between 0 and x. Thus 1 − 0.5 ≤ 1 + ξ ≤ 1 + 0.5 ⇒ 1 1.55/2 ≤ 1 (1 + ξ)5/2 ≤ 1 (0.5)5/2 = √ 32 < 6. |R2(x)| ≤ 3 16|x|3. We also know |x|3 ≤ (0.5)3, but this also tells us that the closer x is to 0, the better the estimate.

Dan Barbasch () Calculus 1120, Class 42 April 30, 2012 5 / 1

y = (1 + x)1/2

So we can use the approximation √1 + x ≈ 1 + x/2 − x2/8 for this

• interval. For example

√ 1.4 is approximated by 1.18 The error is less than

(0.4)3 16 . Note that here 0 ≤ ξ ≤ 1, the estimate is better than over the

interval −0.5 ≤ x ≤ 0.5. Question: Is this an over or underestimate? Answer: It depends whether x ≥ 0 or x ≤ 0. Problem: Estimate √ 2. The same calculations apply, the answer is 1 + 1

2 − 1 8 = 1.375. The error is

less than 1/16 = 0.0625. This is better than the previous estimate because x = 1 > 0. Instead of 3 we get 1. From other considerations we know that √ 2 = 1.4142.

Dan Barbasch () Calculus 1120, Class 42 April 30, 2012 5 / 1

Φ(x)

Φ(x) =

∞

• n=0

(−1)nx2n+1 (2n + 1)n! . For x > 0 this is an alternating series. So the error is

• Φ(x) −

n

• i=0

(−1)ix2i+1 (2i + 1)i!

• ≤

x2n+1 (2n + 1)(n + 1)! For example Φ(1) is approximated by 1 − 1

3 = 2 3 with error less than 1 10.

Also 1 − 1

3 + 1 10 with error less than 1 42.

Dan Barbasch () Calculus 1120, Class 42 April 30, 2012 6 / 1

Φ(x)

Question: What about Φ(∞) = limx→∞ Φ(x)? How large do we need to take x to have an error less than say 10−2? Answer: We do it in two steps. First we find an A > 0 so that the error betwen Φ(∞) and Φ(A) is less than 1/200, then we find n so that the Taylor polynomial approximates Φ(A) by less than 1/200. Together the error will be less than 1/200. |Φ(∞) − Φ(A)| = ∞

A

e−t2 dt ≤ ∞

A

e−t dt = e−A. So we can make this error less than 1/200 by taking A > ln 200 say A ≥ 6.

Dan Barbasch () Calculus 1120, Class 42 April 30, 2012 6 / 1

Φ(x)

We now check how many terms we need from the series:

• Φ(6) −

n

• i=0

(−1)i62i+1 (2i + 1)i!

• ≤

62n+3 (2n + 3)(n + 1)! ≤ 1 200 Plug in values of n until you get below 1/200. For this to be justified, you need to verify the alternating series test. Extra Credit (harder): Verify that the conditions of the alternating series test hold. Precisely show that for a fixed x > 0, the terms x2n+1 (2n + 1)n! are decresing, and go to 0. The actual value you are estimating is

√π 2 .

Dan Barbasch () Calculus 1120, Class 42 April 30, 2012 6 / 1

Review

Fundamental Theorem of Calculus Techniques of Integration

◮ Substitution ◮ Integration by parts ◮ Trigonometric Integrals

Applications

◮ Areas of regions between curves ◮ Volumes by slicing, disks/washers, shells ◮ Arclength ◮ Surface area ◮ Distance traveled, displacement ◮ Mass and Moments (not covered in any depth) Dan Barbasch () Calculus 1120, Class 42 April 30, 2012 7 / 1

Review

Integration, chapter 8

1

Techniques of Integration 8.1-8.5

2

Numerical Integration 8.6 (see notes for left and right endpoint approximation)

3

Improper Integrals 8.7

review problems are on page 507 Differential equations, 7.2 parts of 9.1-9.4

1

Graphic methods, Slope fields 9.1, parts of 9.4 draw slope fields and some solutions for problems on page 540

2

Separable and Linear Equations 7.2, 9.3 page 433 1-8, 9-22, page 526 1-21

3

Applications, parts of 9.1-9.4. page 434 23-46, page 540 13-18

Sequences and Series, 10.1 -10.5. page 623 1-18, 19-40 Power Series, Taylor and MacLaurin Series, 10.7-10.8 page 600, 2-5, 8-12, 30-32, 42-44, 50-54. page 606 6-10, 18-20, 30-32, 37-38. Convergence of Taylor series and Applications, section 10.9 and 10.10. page 613 6-10, 16-20, 36-38, 45/ page 620 4-6, 15-16, 23, 25-27, 33-35, 45-47, 53-55.

Dan Barbasch () Calculus 1120, Class 42 April 30, 2012 7 / 1