MATH 6102 Spring 2009 A Bestiary of Calculus Special Functions
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1 MATH 6102 Spring 2009 A Bestiary of Calculus Special Functions
2 Transcendental Functions Last time we discussed eponential, logarithmic, and trigonometric functions. Theorem 1: If f : R R is a continuous function that satisfies f ( + y)=f () f (y), then f is an eponential function. If f (1) =, then > 0 and 1. Theorem 2: If f : R + R is a continuous function that satisfies g(y)=g() ( ) ( ) + g(y), ( ) then g is an logarithmic function. If g(b) = 1, then b > 0 and b Feb-2009 MATH
3 Transcendental Functions Theorem 3: There eists eactly one pair of real functions s and c such that for all real numbers and y, the following equations hold: 2 2 [ s ( )] [ c ( )] 1 s ( y ) scy ( ) ( ) csy ( ) ( ) c ( y) ccy ( ) ( ) ssy ( ) ( ) 16-Feb-2009 MATH
4 Transcendental Functions Theorem 4: There eists eactly one pair of real functions s and c such that for all real numbers and y, the following equations hold: 2 2 [ s ( )] [ c ( )] 1 s ( y ) scy ( ) ( ) csy ( ) ( ) c ( y) ccy ( ) ( ) ssy ( ) ( ) 16-Feb-2009 MATH
5 Elliptic Computations 2 2 y a b The area is πab which (0,b) can be computed by integrals. (a,0) What is the length of the curve??? Why did we care? (Planetary orbits) 16-Feb-2009 MATH
6 a Elliptic Computations y b Parameterization: (0,b) Let = a cos t and y = b sin t This works well!! (a,0) Can we parameterize the parabola or the hyperbola similarly? Not unless we use comple numbers! 16-Feb-2009 MATH
7 Hyperbolic Functions Studied hyperbolic functions Used them to obtain solutions of cubics Found standard addition formulae for hyperbolic functions, derivatives and relation to the eponential function. Vincenzo Riccati i Feb-2009 MATH
8 Riccati s Work Developed the hyperbolic functions and proved their consistency using only geometry of the unit hyperbola 2 y 2 = 1 or 2y = 1. Followed his father s interests in differential i equations arising from geometrical problems. Led to study of the rectification of the conics in Cartesian coordinates and interest in the areas under the unit hyperbola. Developed properties of the hyperbolic functions from purely geometrical considerations 16-Feb-2009 MATH
9 Unit Hyperbolas P (-1,0) (1,0) Q y 1 2y Feb-2009 MATH
10 Circular Relation Arclength relation to area in a circle 2 2A r and Ar 2 r Does this hold in hyperbolas? Not eactly! 16-Feb-2009 MATH
11 Hyperbolic Areas B The graph is y = k. Area of ΔOAA Area y k 2 2 O A C B' A' C' Note then that Area(ΔOAA ) = Area(ΔOBB ) = Area(ΔOCC ) 16-Feb-2009 MATH
12 Area considerations for y=1 1 O P Q P' 1. Area(ΔOAA )= Area(ΔOAA ) 2.Area(ΔOPQ) = Area( AA PP ) (Subtract Area(ΔOQP ) from above) 3. Area(APP A ) = Area(OAP) (Add Area(QAP) to above) 4. a d D Area( AOP) ln( a) u 1 A A' 16-Feb-2009 MATH
13 Hyperbolic Trigonometric u/2 P Let u/2 = area bounded d by -ais, i y =, and curve. Define the coordinates of the point P by = ch(u) y = sh(u) (0,0) (1,0) 16-Feb-2009 MATH
14 Properties of the functions ch(u) and sh(u) 1. ch(u) 2 sh(u) 2 = 1 This is immediately obvious. 2. ch(u + v) ) = ch(u)ch(v) ) + sh(u)sh(v) ) 3. sh(u + v) = sh(u)ch(v) + ch(u)sh(v) We will prove 2 & 3 shortly. 16-Feb-2009 MATH
15 Second Fundamental Property A'' A' O P C' P' C'' B'' B' O Area(PC A )=Area(ΔOC A )-u/2 Rotate counterclockwise through π/4. We do not change area!! Rotation carries 2 y 2 = 1 to 2y y = 1 16-Feb-2009 MATH
16 Second Fundamental Property B =(ch(u), sh(u)) = (c 1,s 1 ) B = (,1/2), P = (/ (1/ 2, 1/ 2) A'' B C = B C = s 1 and OC =OC = c 1 Area bounded db by OP, y = 1/2 and OB : C'' P' B'' O d K d ln ln / / 2 2 u 1 ln u u e 1 e 20 0 and y0 2 2e 16-Feb-2009 MATH u 0
17 Second Fundamental Property Now we need to find C. B C OC so it has slope -1 and equation 1 y ( 0) 2 0 Since C lies on the diagonal, = y and 0 1 y O A'' P' C'' B'' 16-Feb-2009 MATH
18 Second Fundamental Property Thus, the distance B C is given by the distance formula BC /2 This last term is positive if 0 > 1/2 16-Feb-2009 MATH
19 Second Fundamental Property Recall that 0 = e u /2 and B C = sh(u). Therefore e 2 e e sh( u ) 2 u 2 2 4e 2 u u u 16-Feb-2009 MATH
20 Second Fundamental Property OC /2 e 2 e e ch( u) 2 u e 2 u u u 16-Feb-2009 MATH
21 Hyperbolic Trigonometric Traditionally, we have: Functions ch(u) = cosh(u) sh(u) = sinh(u) Define the remaining 4 hyperbolic trig functions as epected: tanh(u), coth(u), sech(u), csch(u) 16-Feb-2009 MATH
22 Hyperbolic Trig Functions The hyperbolic trigonometric functions satisfy if the fll following properties 2 2 cosh( ) sinh( ) 1 sinh( y) sinh( )cosh( y) cosh( )sinh( y) cosh( y) cosh( )cosh( y) sinh( )sinh( y) cosh( ) cosh( ) sinh( ) sinh( ) 16-Feb-2009 MATH
23 Hyperbolic Trig Identities 2 2 cosh sinh tanh sech sinh( y ) sinh cosh ycosh sinh y cosh( y) cosh cosh ysinh sinh y tanh tanh y tanh( y) 1 tanhtanhy 2 cosh 1 2 cosh 1 sinh and cosh Feb-2009 MATH
24 Hyperbolic Trig Functions Since the eponential function has a power series epansion e n n n0! The hyperbolic trig functions have power series epansions 2n 2n1 cosh( ) and sinh( ) (2 n)! (2 n 1)! n 0 n 0 16-Feb-2009 MATH
25 Hyperbolic Trig Functions Recall that the Maclaurin series for the sine and cosine are: ( 1) ( 1) cos( ) and sin( ) (2 n)! (2n 1)! n 2n n 2n1 n0 n0 VERY SIMILAR!!!! 16-Feb-2009 MATH
26 Hyperbolic Trig Functions Replace by i, where i 2 = -1.: n 2n n 2n 2n 2n ( 1) ( i) ( 1) i cos( i) cosh (2 n )! (2 n )! (2 n )! n 0 n 0 n 0 n 2n1 n 2n1 2n1 2n1 ( 1) ( i) ( 1) i i sin( i) i sinh (2 n1)! (2n1)! (2n1)! n 0 n 0 n 0 cosh cos( i) cos i sinh isin( i) isin i 16-Feb-2009 MATH
27 The Graphs y y y = sinh() y = cosh() The catenary 16-Feb-2009 MATH
28 The Graphs y 16-Feb-2009 MATH
29 The Derivatives d d sinh d cosh cosh sinh d 16-Feb-2009 MATH
30 The Derivatives d d tanh d d 2 2 sech coth csch d sech sech tanh d d csch csch coth d 16-Feb-2009 MATH
31 The Inverse Functions cosh 1 ln sinh ln tanh ln 1 coth ln 1 sech ln csch ln 16-Feb-2009 MATH
32 Graphs of the Inverses 16-Feb-2009 MATH
33 The Lambert W Function This is also known as the omega function () or the ProductLog function (in Mathematica). In Maple it is known as the Lambert W function. Arose from Lambert s work to solve: = q + m Euler etended it to - = ( - )v + 16-Feb-2009 MATH
34 The Lambert W Function From = q + m Substitute - for, set m = / and q = ( - )v m q ( ) v ( ) v 1 1 ( )v ( ) v 16-Feb-2009 MATH
35 The Lambert W Function Divide by - and let go to. lim lim v d v 2 d 2 log v log v 16-Feb-2009 MATH
36 The Lambert W Function If we can solve this for = 1, then we can solve it for any. log v log v log v Set z and uv log z uz Then apply Euler s solution to get: log v v v v v 2! 3! 4! 5! 16-Feb-2009 MATH
37 The Lambert W Function From the work of Wright et al after 1925, more is known. Consider the product function y = e Local min: (1, 1/e) Decreasing: (,1) Increasing: (1,) Has an inverse on those two intervals. 16-Feb-2009 MATH
38 The Lambert W Function For > 1, we define the Lambert W function W() by. y = W() if and only if = ye y [RECALL: y = log if and only if = e y ] y = W 0 () y = W 1 () 16-Feb-2009 MATH
39 The Lambert W Function W() solves a broader class of equations: 1. Solve y = e - for. The graph is the graph of y = e flipped across both the - and y-aes. Thus, we should have that = W(y). Check: = W(y) if and only if = W(y) if and only if y = e if and only if y = e 16-Feb-2009 MATH
40 The Lambert W Function W() solves a broader class of equations: 2. Solve a = b for. Rewrite this to the form z = ue u whose solution is u = W(z) a b a e lnb ln b a ln b lnbe Set z aln b and u lnb u W ( z ) ln b W( aln b) W( aln b)/lnb 16-Feb-2009 MATH
41 The Lambert W Function 3. Solve b = a for. Rewrite this to the form z = ue u whose solution is u = W(z) e b b a a aln b e e e ln (ln ) e a ( aln ) e e a ln b ( a ln ) e aln aln aln b ( aln ) e aln Set z a ln b and u a ln u W( z) a ln W ( a ln b ) e W( aln b)/ a 16-Feb-2009 MATH
42 The Lambert W Function 4. Solve + b = a for. b a. Put u b a a ua a a b b ub b u ua u ua u b v a va, where vu By 2 b v W ( a ln a )/ln a bw( a b ln a)/lna 16-Feb-2009 MATH
43 The Lambert W Function 5. Consider h() = - an infinite eponential tower. When does it converge? Definitely at = 1 and for 0 < < 1. Anywhere else? Euler showed that it converged for e -e < < e 1/e. Wright showed that when it converges it converges to W( ln ) ln 16-Feb-2009 MATH
44 The Lambert W Function Its Derivative i d d W( ) e W ( ) W( ) e W( ) W( ) e W( ) W( ) W ( ) W ( ) 1 W ( ) e W ( e ) 1 1 W( ) W ( ) W ( ) W ( ) e W ( ) e e 1 W ( ) W( ) 1 1 W( ) W( ) W ( ) ;, 1/ 1 W( ) o e 16-Feb-2009 MATH
45 The Lambert W Function Its Maclaurin Series We find that the power series epansion is W( ) n1 ( n) n! n1 n Not from first principles, rather from series reversion of ze z z n n1 n0! 16-Feb-2009 MATH
46 The Lambert W Function Its Antiderivative i i Using Integration by Parts it can be shown that W( ) d W 2 ( ) W ( ) 1 W( ) C 16-Feb-2009 MATH
47 The Lambert W Function W( e) 1 W (0) 0 Some of its Values 1 W 1 e ln2 W ln2 2 W (1) e or ln i W Feb-2009 MATH
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