Calculus 1120, Class 44 Dan Barbasch May 4, 2012 Dan Barbasch - - PowerPoint PPT Presentation

Calculus 1120, Class 44

Dan Barbasch May 4, 2012

Dan Barbasch Calculus 1120, Class 44 May 4, 2012 1 / 1

Φ(x)

Φ(x) =

∞

• n=0

(−1)nx2n+1 (2n + 1)n! . For x > 0 this is an alternating series. By the Error Estimate for Alternating Series, |S − sn| ≤ an+1,

• Φ(x) −

n

• i=0

(−1)ix2i+1 (2i + 1)i!

• ≤

x2n+3 (2n + 3)(n + 1)! For example Φ(1) = 1 + −1 3 + 1 5 · 2! + −1 7 · 3! + 1 9 · 4! + . . . . So Φ(1) ≈ 1 − 1

3 + 1 10 with error less than 1 42, and Φ(1) ≈ 1 − 1 3 + 1 10 − 1 42

with error less than

1 216.

Dan Barbasch Calculus 1120, Class 44 May 4, 2012 2 / 1

Error Estimates

1 Trapezoidal Rule, Simpson’s Rule, Left/Right Endpoint Rule 2 Alternating Series 3 Taylor’s formula 4 Improper Integrals/Comparison Test 5 Integral Test Dan Barbasch Calculus 1120, Class 44 May 4, 2012 3 / 1

Erf (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 Pn(A) approximates Φ(A) by less than 1/200. Together the error will be less than 1/100. |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 44 May 4, 2012 4 / 1

Erf (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 conditions for the alternating series test. I found that you need n ≥ 96. 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 decreasing, and go to 0. May depend on the value for x. The terms may increase for a while before decreasing.

Dan Barbasch Calculus 1120, Class 44 May 4, 2012 4 / 1

Erf (x)

The actual value you are estimating is

√π 2 .

Review of Volumes as Integrals

1 Slices: V =

b

a A(x) dx.

2 Washers: V =

b

a π[r2(x)2 − r1(x)2] dx

3 Shells: V =

b

a 2πr(x)h(x) dx.

Problem: Compute the volume obtained by rotating the region below y = e−x2, to the right of x = 0, and above the x−axis, about the y−axis. Answer: V = 2π ∞ xe−x2 dx. A xe−x2 dx = −e−x2 2

• A

=

• 1

2 − e−A2 2

• −

→ 1 2 as A → ∞. So V = 2π/2 = π.

Dan Barbasch Calculus 1120, Class 44 May 4, 2012 4 / 1

Erf (x)

Problem: Compute the same volume using slices perpendicular to the x−axis. Answer: The axis perpendicular to the xy−plane is z: The region is below the surface y = e−(x2+z2).

Dan Barbasch Calculus 1120, Class 44 May 4, 2012 4 / 1

Erf (x)

So the slice at x is the region 0 ≤ y ≤ e−(x2+z2). The volume is V = ∞

−∞

∞

−∞

e−(x2+z2) dz

• dx =

∞

−∞

e−x2 ∞

−∞

e−z2 dz

• dx =

= ∞

−∞

e−t2 dt 2 . Putting the two together, ∞

−∞

e−t2 dt = √π. We conclude ∞ e−t2 dt = √π 2 .

Dan Barbasch Calculus 1120, Class 44 May 4, 2012 4 / 1

Erf (x)

The Error function Erf (x) is defined to be Erf (x) = 2 √π x e−t2 dt. Problem: Graph this function. Fundamental Theorem of Calculus: If F(x) = x

a f (t) dt, then

F ′(x) = f (x). So Erf (x)′ =

2 √πe−x2, and Erf (x)(2) = − 4 √πxe−x2. The function is

increasing, concave up for x ≤ 0 and concave down for x ≥ 0. Furthermore limx→∞ Erf (x) = 1, and Erf (−x) = −Erf (x), so also limx→−∞ Erf (x) = −1.

Dan Barbasch Calculus 1120, Class 44 May 4, 2012 4 / 1

Erf (x)

Normal Distribution: N(µ, σ) = 1 √ 2πσ e− 1

2( x−µ σ ) 2

. The normal distribution is considered the most prominent distribution in

• statistics. The

√ 2π makes it so that ∞

−∞

N(µ, σ)(x) dx = 1 ⇔ ∞

−∞

e− x2

2 dx =

√ 2π.

Dan Barbasch Calculus 1120, Class 44 May 4, 2012 4 / 1

Extra Credit

1

∞

−∞

dx x2(1 + ex).

2

lim

x→0+ x

1

x

cos t t2 dt.

3

∞

• n=2

ln n ln ln n.

4

lim

n→∞

Arctan n √n .

5

• 3dt

100 + 2t .

Dan Barbasch Calculus 1120, Class 44 May 4, 2012 5 / 1