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 ) ∞ ( − 1) n x 2 n +1 � Φ( x ) = (2 n + 1) n ! . n =0 For x > 0 this is an alternating series. By the Error Estimate for Alternating Series, | S − s n | ≤ a n +1 , � n � ( − 1) i x 2 i +1 x 2 n +3 � � � � Φ( x ) − � ≤ � � (2 i + 1) i ! (2 n + 3)( n + 1)! � � i =0 For example Φ(1) = 1 + − 1 5 · 2! + − 1 1 1 3 + 7 · 3! + 9 · 4! + . . . . So Φ(1) ≈ 1 − 1 3 + 1 42 , and Φ(1) ≈ 1 − 1 1 3 + 1 10 − 1 10 with error less than 42 1 with error less than 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( ∞ ) = lim x →∞ 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 P n ( A ) approximates Φ( A ) by less than 1 / 200 . Together the error will be less than 1 / 100 . � ∞ � ∞ e − t 2 dt ≤ e − t dt = e − A . | Φ( ∞ ) − Φ( A ) | = A 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: � n � ( − 1) i 6 2 i +1 6 2 n +3 1 � � � � Φ(6) − � ≤ (2 n + 3)( n + 1)! ≤ � � � (2 i + 1) i ! � 200 i =0 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 x 2 n +1 test hold. Precisely show that for a fixed x > 0 , the terms (2 n + 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 ) √ π The actual value you are estimating is 2 . Review of Volumes as Integrals � b 1 Slices: V = a A ( x ) dx . � b a π [ r 2 ( x ) 2 − r 1 ( x ) 2 ] dx 2 Washers: V = � b 3 Shells: V = a 2 π r ( x ) h ( x ) dx . Problem: Compute the volume obtained by rotating the region below y = e − x 2 , to the right of x = 0, and above the x − axis, about the y − axis. � ∞ xe − x 2 dx . Answer: V = 2 π 0 A � A � xe − x 2 dx = − e − x 2 � 2 − e − A 2 � 1 → 1 � = − 2 as A → ∞ . � 2 2 � 0 � 0 So V = 2 π/ 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 − ( x 2 + z 2 ) . Dan Barbasch Calculus 1120, Class 44 May 4, 2012 4 / 1

Erf ( x ) So the slice at x is the region 0 ≤ y ≤ e − ( x 2 + z 2 ) . The volume is � ∞ �� ∞ � ∞ e − x 2 �� ∞ � � e − z 2 dz e − ( x 2 + z 2 ) dz V = dx = dx = −∞ −∞ −∞ −∞ �� ∞ � 2 e − t 2 dt = . −∞ Putting the two together, � ∞ e − t 2 dt = √ π. −∞ √ π � ∞ e − t 2 dt = We conclude . 2 0 Dan Barbasch Calculus 1120, Class 44 May 4, 2012 4 / 1

Erf ( x ) � x 2 e − t 2 dt . √ π The Error function Erf ( x ) is defined to be Erf ( x ) = 0 Problem: Graph this function. � x Fundamental Theorem of Calculus: If F ( x ) = a f ( t ) dt , then F ′ ( x ) = f ( x ). So Erf ( x ) ′ = √ π e − x 2 , and Erf ( x ) (2) = − 4 √ π xe − x 2 . The function is 2 increasing, concave up for x ≤ 0 and concave down for x ≥ 0 . Furthermore lim x →∞ Erf ( x ) = 1, and Erf ( − x ) = − Erf ( x ), so also lim x →−∞ Erf ( x ) = − 1 . Dan Barbasch Calculus 1120, Class 44 May 4, 2012 4 / 1

Erf ( x ) 1 2 e − 1 2 ( x − µ σ ) Normal Distribution: N ( µ, σ ) = √ . 2 πσ The normal distribution is considered the most prominent distribution in √ statistics. The 2 π makes it so that � ∞ � ∞ √ e − x 2 2 dx = N ( µ, σ )( x ) dx = 1 ⇔ 2 π. −∞ −∞ Dan Barbasch Calculus 1120, Class 44 May 4, 2012 4 / 1

Extra Credit � ∞ dx x 2 (1 + e x ) . 1 −∞ � 1 cos t lim dt . x → 0 + x 2 t 2 x ∞ ln n � ln ln n . 3 n =2 Arctan n lim √ n . 4 n →∞ � 3 dt 100 + 2 t . 5 Dan Barbasch Calculus 1120, Class 44 May 4, 2012 5 / 1