Math 114 Notes. 1 Reference. David Stein. February 29, Trigonometric Identities. 1.2 Logarithmic Identities. sin(u) = 1.

Transcription

1 Math 4 Notes Davi Stein February 9, 06 eference. Trigonometric Ientities sin(u) = csc(u) cos(u) = sec(u) tan(u) = csc(u) csc(u) = sin(u) sec(u) = cos(u) cot(u) = tan(u) tan(u) = sin(u) cos(u) sin(u) = sin(u) cos(u) sin( u) = cos(u) cos( u) =sin(u) tan( u) = cot(u) csc( u) =sec(u) sec( u) =csc(u) cot( u) = tan(u) sin( u) =sin(u) cos(u) cos( u) = cos (u) sin (u) = cos (u) = sin (u) tan( u) = tan(u) tan (u) sin () = +cos() cos () = cos(). Logarithmic Ientities b = e ln(b) log b () = ln() ln(b)

2 ln( )=ln() ln(y)=ln()+ln(y) ln( y )=ln() ln(y) ln( y )=y ln(). Basic Derivatives uv = uv0 + vu 0 u v = vu0 uv 0 v Chain ule: y = y u u.4 Trigonometric Derivatives sin() = cos() cos() = sin() tan() =sec () cot() = csc () sec() =sec() tan() csc() = csc() cot() sin () = p sin ( a )= p a tan () = + a tan ( a )= a + sec ( p a ) = p p a a sec ( a )= Eponential an Logarithmic Derivatives: b = b ln(b)(b>0) ln() = (>0) ln( ) = ( 6= 0) log b() = ln(b) (>0) log b = ln(b) ( 6= 0) ln(f()) = f 0 () f().5 Basic Integrals bf() = b f() (a + b) n = (a+b)n+ a(n+) + C

3 .6 Trigonometric Integrals sin() = cos()+c sin(a) = cos() a + C cos() =sin()+c cos(a) = sin() a + C tan() = ln cos() + C =ln sec() + C cot() =ln sin() + C sec() =ln sec() + tan() + C csc() =ln csc() cot() + C p =sin a ( a )+C a + = a tan ( a )+C p = a a sec ( a )+C.7 Eponential an Logarithmic Integrals e = e + C e a = e a + C a = a ln(a) + C ln = ln + C January, 0 Lecture. Section 6. Eample : ( p +) 4 p = p ( +) 4 p Defineu = p + ; u = u 4 u = = u 5 = = 5 + C 5 u5 + C 5 (p + ) 5 + C p Eample : sin 0 () Define u =sin(); u = cos() = u 0 u

4 = u + C = sin () + C Eample : p (( + ) + ) Define u = + ; u = = (u p (u ) + ) u... oesn t work Define u = + ; u = = ( (u+) p u) u January, 0 ecitation p. 64, n. 9: ( p ) 4 Define u = 4 ; u = = ( p u ) u = ( p u ( )) u = 6 ( )+C ( +)u( +) = u + C = p u + C = p 4 + C (Hints for choosing u: () chose the most i cult thing in the problem; () always better to choose something in the enominator; () u =,which compares well with the in the numerator.) p. 64, n. 9: p ( +) 4 p ) Define u = p += + ; u = u = p = (u 4 + )u = 5 u5 + C = 5 (sqrt + )5 + C Net eample: ( 7 ) Define u = 7; u = = ( u ) u... this is a problem but we know that u = 7 = ( u+7 u )u = ( u u + 7 u )u 4

5 = u +7ln( u )+C = 7+7ln( 7 )+C Net eample: ( p ) 5 Define u = 5 ; u = 50; = = ( pu 50 ) = ( (50 pu) ) = ( ( (u+) pu) Define u = sqrt5 )... an we re stuck. u = sqrt u = sqrt5 = sqrt5 u = ( 5 ( u) = ( (5 p 5 5 u) ) uu = ( (5 ) ) u )u 50 u 4 January 8, 0 Lecture 4. Section Integral represents area, such as area uner a curve (for f() must take where f() < 0) 0; otherwise, More generally, if f() apple g() for a apple apple b, then a : b (g()f()) is the area boune by y = f(),y = g(), = a, = b. (The above statement is a special case of this restatement where f() = 0.) Note that f() can be < 0in this restatement. P n i= ((g( i ) f( i )) i ) Alternatively: If f(y) apple g(y) for c apple y apple, then b a(g(y f(y)))y = area boune by = f(y), = g(y),y = c, y =. (i.e., reflecteover = y; i.e., integrating up the y-ais instea of along the -ais.) This give us two options for integrating any relate functions. (e.g., integrating along one ais may require separately integrating two segments, but he other ais may only feature one segment.) Eample: = y ; = y + creates a sieways parabola with a iagonal line running through it - probably easier to integrate vertically - first, fin intersections (y = y +; y y = 0) (use quaratic formula) - points of intersection: = y +=± p 5 Chapter 6, problem 8: Fin the area between the intersection of these curves: 5

6 y = sec () 4 y =4 cos () Both graphs are asymmetric over the Y ais, so we on t have to integrate across the whole range; can just fin the range from 0 to the positive intersection point, an multiply by. So where is the intersection point? - i.e., where is the following true: Answer: sec () 4 =4 cos () 6 cos 4 () = cos 4 () = 6 cos() = = 0 (4 cos () ( sec 4 )) = P 0 (4 ( + sin() (tan )) ) =(4 ( + p 4 0) (p 0) =4 + p Chapter 6, problem 4: Fin area boune by, y, y =,y =ln() - on t want to integrate over, because y =ln() hits the X-ais at = ; woul have to integrate in two parts - instea, integrate over y-ais - left curve: = 0; right curve: = e y Answer: 0 e y 0y = 0e y = P 0e y = e e 0 = e 4. 6.: Volume - To Transcribe 5 February 04, 0 Lecture 5. Section 7. - Integration by Parts Integration by parts is an integration technique - really just the prouct rule for integrals. (f()g()) = f 0 ()g()+f()g 0 () By taking the antierivative (inefinite integral) of both sies, we get: 6

7 f() g() = f 0 () g() + f() g 0 () f() g 0 () = f() g() g() f 0 () uv = uv vu Eample: e u = ; v = e u = ; v = e = e uv = uv vu= e e = e e + C Check: (e e )=( e )+( e ) e Sometimes you may have to apply integration by parts more than once to get the answer - eample: e Define u = ; v = e ; u =; v = e = uv = uv vu = e e Define u = ; v = e ; u = ; v = e = e = ( e e ) = e 9 e + 9 e = e 9 e + 7 e + C (However, note that you have to be consistent: use the same u an v each time through.) Another eample: ln() Define u = v; v =ln(); u = ; v =?... we on t know the integral of ln() Define u =ln(); v = ; u = ; v = = ln() = ln() = ln() + C Another eample: ln Define u =ln; v = = Define u = ; v = = ln = ln + C Another eample: p ln Define u =ln; v = p Define u = ; v = 7

8 = uv vu = ln = + C = = = Another eample: sin () Define u =sin ; v = Define u = p ; v = sin p p Define u = ; u = = pu u = u + C = p u + C = p + C Another eample: e cos() Define u = e ; v = cos() (oesn t matter which is which; coul o it the other way) Define u = e ; v = sin = e cos() = e sin() e sin() Define u = e ; v =sin() Define u =e ; v = cos() = e sin() ( e cos()+ e cos ) This looks ba, but it can be simplifie: = e sin()+ 9 e 4 cos() 9 e cos 9 e cos() = e sin()+ 9 e cos()+c e cos() = e sin()+ e cos()+c Sometimes, i erent types of integration can lea to i erent results that are actually equivalent. Consier: sin() cos() There are at least three i erent ways to o this integral. Metho : Define u =sin(); u = cos() = uu = u + C = sin + C 8

9 Metho : Define u = cos(); u = sin() = uu = u + C = cos ()+C Metho use an ientity: sin() cos() = sin() = ( cos )+C = 4 cos()+c These look i erent, but they re all the same - consier: sin ()+ cos () = cos () = sin () Integration by parts can be useful as a reuction formula: n e a, a is a constant: Define u = n ; v = e a Define u = n n ; v = a ea = a n e a = n a ( n e a ) n e a = a n e a n a e a e.g. : e = e e = e e = e ( e 0 e ) = e e = e = e e e + C Another reuction formula: (ln ) n Define u =(ln) n ; v = Define u = n(ln ) n ; v = ln() n = (ln ) n n (ln ) n Eample: Use the reuction formula to evaluate (ln ) 4 : n =4: (ln ) 4 = (ln ) 4 4 (ln ) = (ln ) 4 4((ln ) (ln ) ) = (ln ) 4 =4(ln ) + (ln ) =...+ ((ln ) (ln ) ) =...+ (ln n) 4( ln )+C Eample: (ln ) Define u = (ln ) ; v = Define u = (ln ) ; v = 9

10 = (ln ) (ln ) = (ln ) ( ln ) ln Define u = ln, v = Define u =, v = = (ln ) = (ln ) = ln = (ln ) ln = (ln ) 9 ln C 7.. Trigonometric Integration sin m cos n where m, n are positive integers case (i): At least one of m, n is o: sin 7 cos = sin 6 cos sin = (sin ) cos sin = ( cos ) cos sin Define u = cos(),u= sin() = ( u ) u u = ( u +u 4 u 6 )u u = u eu 4 +u 6 u 8 u = u 8 u 6 +u 4 u u = 9 u9 7 u7 + 5 u5 u + C = 9 cos9 7 cos7 + 5 cos5 cos + C Say n is o: sin m cos n sin m cos n cos (say n =k) =sin m cos k cos = sin m ( sin ) k cos Now substitute u =sin() : u m (u ) k u 6 February 06, 0 Lecture 6. Section 7. - Integral Trigonometry Functions eview: sin() = cos()+c cos() =sin()+c tan() =ln sec() + C 0

11 cot() =ln sin() + C sec() =ln sec() + tan() + C csc() =ln csc() cot() + C New functions: sin m () cos n () where m, n 0 an are integers Options: If m is o, to split one power of m to get, an write sin() interms of cos () usingsin () = cos (). If n is o, to split one power of n to get, an write sin() in terms of cos () usingsin () = cos (). Eample: sin 5 ()(= sin 5 cos 0 ) = sin 4 ()sin() = (sin () )sin() Now use sin () = cos () : = ( cos ()) sin() Now use integration by parts: Define u = cos(); u = sin() = ( u ) u = ( u + u 4 )u = (u u + 5 u5 )+C u + u 5 u5 + C = cos()+ cos () 5 cos5 ()+C Another eample: sin () cos () = sin () cos () cos() Now use cos () = sin () : sin ()( sin ()) cos() Define u =sin(),u= cos() = u ( u ) u = (u u 4 ) u = u 5 u5 + C = sin () 5 sin5 ()+C Another eample: sin () cos 5 () First way (split ): = sin () cos 5 ()sin() Use sin () = cos () ( cos () cos 5 ()sin() Define u = cos(),u= sin()

12 = ( u )u 5 u = u 5 u 7 u = u 7 u 5 u = 8 u8 6 u6 + C = 8 cos8 () 6 cos6 ()+C Secon way (split 5): = sin () cos 4 () cos() = sin ()(cos ()) cos() = sin ()( sin ()) cos() Use cos () = sin () : Define u =sin(),u= cos() = u ( u ) u = u ( u + u 4 )u = u u 5 + u 7 u = 4 u4 u6 + 8 u8 + C = 4 sin4 () sin6 ()+ 8 sin8 ()+C What happens when both m, n are even (an 0? Then use the following ientities: sin () = ( cos()) cos () = ( + cos()) eview: cos(a + b) = cos(a) cos(b) sin(a)sin(b) What if a = b? cos() = cos () sin () Substitute sin ()forcos () : cos() = sin () sin () cos() = sin () ) sin () = cos() ) sin () = ( cos()) An for the other one: Substitute cos 6 ()forsin () in () : cos() = cos () ( cos ()) = cos () +cos () = cos () ) + cos() = cos () cos () = ( + cos()) Eample: cos () = + cos() = + sin()+c An eample where you have to o it twice because both m, n are even:

13 sin 4 () cos () = (sin ()) cos (),. = ( ( cos()) ( + cos()) = 4 ( cos()) ( + cos()) = 8 ( cos()) ( + cos()) ecognize (a b)(a + b)pattern here, an use it to save some work ( cos())( cos())(+cos())ecognize cos () = = 8 sin () = 8 sin () sin () cos() sin () = ( cos(4)) = ( sin(4) 4 ) sin () cos() Define u =sin(),u= cos() u u = = u = 6 u = 6 sin () = 8 ( ( sin(4) 4 6 (sin ())) + C = 6 ( 4 sin(4) sin ()) + C Eample: 4 0 sin4 () cos () = 6 ( = 4 sin(4) sin ()) ( 4 0 ) = 6 ( 4 ) Net topic: tan m () sec n () - also a stanar way to o it: eview: Trig ientities of tan an sec : tan ()+=sec () (Proof: Take cos ()+sin () = an ivie both sies by cos () ) tan() =sec () ) sec () = tan()+c cot() = csc () ) csc () = cot()+c sec() =sec() tan() ) sec() tan() =sec()+c csc() = csc() cot() ) csc() cot() = csc()+c Using these ientities for tan m ()sec n () where m, n are nonnegative integers: Case : nis even an : Split powers of m to, then rewrite sec () in terms of tan () (using sec () = +tan ()) Eample: tan ()sec 4 () = tan ()sec ()sec ()

14 = tan ()( + tan ()) sec () Define u = tan(),u=sec () = u ( + u ) u = u + u 4 u = u + 5 u5 + C = tan ()+ 5 tan5 ()+C If n is even, this will always work. Eample: sec 6 () (note: m =0,n= 6) = sec 4 ()sec () = (sec ()) sec () = ( + tan ()) sec () Define u = tan(),u=sec () = ( + u ) u = ( + u + y 4 ) u = u + u + 5 u5 + C = tan()+ tan ()+ 5 tan5 ()+C Net case: m is o (an n ): Split one power of m an one power of n to, the rewrite the rest in terms of sec(), then substitute u =sec(). Eample: tan ()sec () = tan ()sec () tan()sec() ecognize tan () =sec () =(sec () ) sec () tan()sec() Define u =sec() = (u )u u = u 4 u u = 5 u5 u + C = 5 sec5 () sec ()+C euction formula for sec n ()(n ) : sec n () = sec n sec () Define u =sec n (),v =sec () Define u =(n ) sec n sec() tan() ; v = tan() =(n ) sec n () tan() =sec n () tan() (n ) (sec n () tan () ) =sec n () tan() (n ) sec n ()(sec () ) =sec n () tan() (n )( sec n () sec ( n )() ) =sec ( n )() tan() (n )( sec n () )+(n )( sec n () ) ) (n ) (sec n () ) =sec n () tan()+(n ) sec n ) sec n () = n secn () tan()+ n n sec n () Last case: m is even, n is o: In this case, rewrite tan m () in terms of sec () using tan () =sec (), an use the reuction formula for sec m (). Eam- 4

15 ple: tan 4 ()sec () = (sec ) sec () = sec 4 () sec () + ) sec () = sec 7 () sec 5 ()+sec () euce sec 7 to sec 5,etc.: sec 7 () sec 5 () + sec () Use reuction formula: = 6 sec5 () tan()+ 5 6 sec 5 () sec 5 () + sec () Now combine the sec 5 ()es, like this: = 6 sec5 7 () tan() 6 sec 5 () + sec () Use reuction formula again: = 6 sec5 7 () tan() 6 ( 4 sec () tan()+ 4 sec () )+ sec () = 6 sec5 7 () tan() 4 sec 7 () tan() 8 sec () )+ sec () = 6 sec5 7 () tan() 4 sec () tan() 8 sec () ) = 6 sec5 7 () tan() 4 sec () tan() 8 ( sec() tan()+ sec() = 6 sec5 7 () tan() 4 sec () tan() 6 sec() tan()+ 6 ln sec()+ tan() + C Finally, what happens if n = 0? tan m () If m is o, let s say, m =: tan () = tan () sec() tan()sec() = sec () sec() tan() sec() Define u =sec(),u= tan()sec() = u u u = u u u = u ln uu + C =sec () ln sec() + C 7 February 06, 0 ecitation Eample isc/washer/shell methos: Fin area of curve y = p (4 ) rotate aroun the Y-ais - first question: o we want to use or y? - answer: let s use, since the equation is alreay in that format - net question: o we want washers or shells? - answer: we re rotating aroun Y to create cyliners, so let s use shells 4 (top - bottom) 0 = 4 0 (p 4 = 0) 5

16 = 4 0 p 4 This is a problem because we on t know how to integrate the latter term - let s try switching to Y an using washers instea: 0 (right left )y = 0 (4 y )y = 0 6 8y + y4y 8 = (6y y + y5 0) Net topic: Arc length - b a p +(F 0 ()) Problem 6.5 5: y = [0, 60] y 0 = = (y 0 ) = 4 q u =+ 4 ; u = 4;4u = 6 p 0 u 4u = 4u u =4 u 6 = 8 (u ) 6 = 8 (64 ) = 68 Problem 6.5 9: y = [, ] y 0 = 4 (y 0 ) =( 4 ) = = q = q (In cases like this, look for someth + 6 6,whichis( 4 ) ) - can factor it that way.) Integration by parts: e Define u =, v = e, u =, v = e = e e = e e + C L I P E T: Log functions, Inverse trigonometry, Polynomials, Eponentials, Trigonometry - this is the orer of preference of choosing U - i.e., when face with two factors, choose the one for U that comes earlier in this list over the one that comes later - e.g.: ln(): is a polynomial; ln() is a log function - choose ln() for u: Define u =ln(); v = ; u = ; v = ln() ln() 4 =ln 0+ =ln 4 e sin() Define y = e,v =sin(), u = e, v = cos() 6

17 e sin() = e cos() cos()e e sin() = e cos()+ cos()e Define u = e,v = cos(), v = e, v =sin() e sin() = e cos()+e sin() e sin() e sin() = e sin() e cos() e sin() = e sin() e cos() + C 8 February, 0 Lecture Final Eam: May 8, 4:0pm-7:5pm Fin a reuction formula for tan m () (finishing last week s lecture): Solution: tan m () = tan m () tan () = tan m ()(sec () ) = tan m ()sec () tan m For the first integral, efine u = tan(),u=sec () = tan m ()sec () = u m u = um m + C = m tanm ()+C For the secon integral: tan m () = m tanm () tan m () Quick review: tan() =ln sec() + C tan () = sec () = tan() + C 8. 7.: Trigonometric substitution p 9 Let =sin( ) Then = cos( ) ) p 9 = q 9 9 sin ( ) cos( ) Now 9 9sin ( ) = 9( sin ( )) = 9 cos ( ) ) p 9 9sin = p 9 cos ( ) =9 cos ( ) = cos( ) cos( ) =9 cos ( ) 7

18 =9 ( + cos( ) = 9 sin ( + )+C However, this nees to be written in terms of : =sin( )! =sin ( ) sin( ) = sin( )cos( ) =sin( ) cos(theta) = = 9 cos ( ) =9 cos ( ) =9 sin( ) )+C 9) p (9 ( + cos( ) = 9 ( + Eample: Fin the area boune by ( =0, =,y =0, + y =9) y = p 9 p 0 9 = 9 sin ( )+ p 9 0 = 9 sin ( )+ p5 = p 5+ 9 sin ( ) You can integrate this, but if you sketch it, this is the area of a right triangle + the area of a sector - easier to just break it up an compute the areas of each shape. Another eample: q 4 Define =sin( ); = cos( ) p 4 4sin ( ) sin( ) cos( ) = p 4cos ( ) sin( ) cos( ) = cos( )cos( ) sin(theta) = cos ( ) sin( ) Notice cos ( ) sin( ) = sin ( ) sin( ) = sin( ) sin( ) =csc( ) sin( ) = (csc( ) sin( )) = (ln csc( ) cot( ) + cos( )) + C = (ln Another eample: p 9 use u-substitution. q 4 + p 4 )+C =ln p 4 + p (4 + C... on t bother with trig substitutions; just Another eample: p 4 9 Define yu =, u = = p 4 u u u 8

19 = p 4 u u u u =sin( )... -O- Let =sin( ) i.e.,, = sin( ) = cos( ) ) p 4 9 = p 4 () = p 4 9 = p 4 9 = cos( ) sin( ) = cos ( ) cos( ) q 4 4sin ( ) = cos( ) Another eample: p 9+ Let = tan( ) =sec ( ) p 9+ = p tan ( ) = p 9( + tan ( )) = p 9sec ( ) =sec( ) p 9+ = sec( ) sec ( ) =9sec ( ) Can also use a reuction formula: sec n () = n (secn () tan()+ n p n 9+ = 9( sec() tan()+ sec n sec( ) ) () = 9 (sec( ) tan( )+ln sec( ) + tan( ) )+C Now use trig properties to figure out sec( ), etc.: sec( ) = q 9+ 9 ( p 9+ +ln p 9+ + )+C = p ln +p 9+ + C Note: We can rop the from the ln term, because ln( y )=ln() we re just aing a ln() term that gets absorbe into C. Eample: +4 - just use u-substitution with u = +4. Eample: +4 4 = = Define u = +4,u= u u = = ln u + C = ln( + 4) + C ln(y),so 9

20 Eample: p ( 9) Define p =sec( ), sec( ) tan( ) 9= p 9sec ( ) 9= p 9(sec ( ) ) = p 9 tan ( ) = tan( ) sec n ( ) = n secn tan( )+ n n sec n sec ( ) sec( ) = sec( ) tan(theta)+ (sec( ) ) (sec( ) = sec( ) tan( ) sec( ) = sec( ) tan( ) ln sec( ) + tan( ) 9 ( p 9 ln + p 9 )+C = p 9 9 ln + p 9 + C What o you o if you have? Just o the same thing, but treat it as ( p ). Eample 7. #40: ( apple 4) ( 6) Define =4sec( ),=4sec( ) tan( ) 6 = 6 sec ( ) 6 = 6 tan ( ) =4 tan ( ) therefore( 6) =4 tan ( ) = r sec ( ) 4 tan ( ) 4sec( ) tan( ) ( 6) =4 sec 4 ( ) tan ( ) =4 sec ( ) tan ( ) sec ( ) Define u = tan( ),u=sec ( ) =4 +u u u =4 u +u = 4( u + u)+c = 4( u + u)+c = 4( cot( ) + tan( )) + C 4 = 4( p + p )+C 6 = p + p 6 + C 6 Summary: ule #: If a is involve, try substituting = a sin( ). ule #: If a + is involve, try substituting = a tan( ). ule #: If a is involve, try substituting = a sec( ). 0

21 9 February 8, 0 Lecture Last time: Integrating rational polynomials - toay s topic: what to o with irreucible polynomials - eample: a + b + c is irreucible if b 4ac < 0 (factoring results in comple numbers) - eample: +4 + On the other han, +4 is reucible - can either use the quaratic equation (solve for roots, then factor into ( r ) ( r )): b± p b 4ac a = 4±p 40 4 = ±p 0 ) r = +p 0,r = p 0 Note that given the term, we nee to multiply the factore pair by : ( r )( r ), or carry the into either term Alternatively, completing the square: +4 = ( +) = ( + + ) = ( + ) 5 = (( + ) 5 q ) = (( + ) + ) q q 5 = ( + ) ( ++ 5 ) p q =( + 0 ( ++ 5 ) Any polynomial can be factore into the prouct of linear factors an irreucible quaratic factors - from p() q() where p() is linear an q() is an irreucible quaratic Eample: (note: +4 +isirreucible) Completing the square: +4 + = ( +)+ = ( + + ) + = ( + ) + ) integral = + (+) + Define u = +,u= = (u )+ u + u = u u + u = u u + u + u u Looking at the first part: u + Define v =u +,v =4uu = u + 4uu v v = = ln v

22 = ln(u + ) Looking at the secon part: u + -usetrigientity: a tan ( a ) = = u + p tan ( p u) u = p tan ( p u) u Final answer (substituting + back in for u): = ln( ( + p ) + ) tan ( p ( + )) + C +a = What happens with something like this: (u +) u? - use the tan substitution: p p u = tan( ) u =sec ( ) ( p u) + = tan ( )+=sec ( ) = p = p = (sec ( )) (sec ( )) sec ( ) p cos ( ) = p ( + cos( )) sin( ) ( + = p = p ) ( +sin( ) cos( )) = p (tan ( p u)+ p p u p ) u + u + = p (tan ( p u)+ p u u + ) Final answer: ( +4+) p (tan ( p (+))+ A partial fraction is a rational fraction of the form where a + b + c) isirreucible: A, B, a, b, c are constants; m p (+) +4+ )+C A A+B (a+b) or m (a +b+c), m Eample:, +,, (+), +, +4+, +4+), + ( +4+) 4 is an integer Every rational function p() q(),whereegp<eg q, can be written as a sum of partial fractions, such that if (a + b) m is a factor of q, then the sum contains A a+b + A (a+b) Am (a+b) ; an if (a + b + c) m is a factor of q, then m the sum contains A i,b i s. A +B a +b+c + A+B (a +b+c)...+ Am+Bm (a +b+c) m Eample: Fin the partial fraction ecomposition of + + ( +4+)(+) = A + + B+C +4+. Solution: +=A( +4 + ) + (B + C)( + ) Substitute = : =A( 4 + ) + 0 ) A = Substitute =0:=A +C ) C =4 for some constants ( +4+) (+) such that

23 You can also collect the coe cients of :0=A + B ) B = + ) ( +4+)(+) = = (Can check by just taking the common enominator: ( + 4) ( + ) ( +4 + ) = +). Another eample: Fin the partial fraction ecomposition of f() = + Solution: Fin A, B, C, D, E ) + ( +4+) (+) = A + + B+C ( +4+) (+) D+E ( +4+) +=A( +4 + ) +(B + C)( +4 + )( + ) + (D + E)( + ) Substitute some value: = : =A( 4 + ) ) A = Substitute 0 : = A 9+C +E ) = 9+C + E ) C + E = 0 Compare coe cients at highest power: 0 4 = 4 (4A +B) ) B = Compare coe cients at net highest power: 0 = ( 6 A +6 B +C) ) C =) E =4 Substitute any other value of D: = =( ) ( ) + ( + ) ( ) () + (D + 4) () (0) = D +8) D = Now just put everything back: f() = ( +4+) Eample: (+) - solution: fin A + B + C + + E (+) Solution: =A ( + ) + B ( + ) + C ( + ) + D Substitute = :B = Substitute = :D = 5 Compare highest coe cient: 0 = A + C Compare net highest coe cient: = A +B -substituteb = := A 6 ) A =8 8 Final answer: (+) = (+) + (+) February 8, 0 Lecture 0. ational fractions - see section 7.4, problems 7-78 Eample: sec() Note: We alreay know the integral of this - = ln sec()+ tan() +C - but let s erive it accoring to the techniques we ve evelope Define u = tan = cos() = u +u u +u = u u = u,u u = ( u)(+u) = A u + B +u =A( + u)+b( u)

24 A =,B = ) u = u + +u = ln u +ln +u + C =ln +u u + C =ln +tan tan + C Now multiply numerator an enominator by cosine: =ln cos +sin + C =ln cos sin (cos +sin ) (cos sin )(cos +sin ) + C =ln cos + cos sin +sin cos sin + C =ln +sin cos =ln cos + sin cos + C =ln sec + tan + C + C 0. Section 7. #55-59 Integrals of the form sin m sin n ; sin m cos n ; cos m cos n (where m, n are constants) - these are important for Fourier transforms - can be performe using trig ientities: cos(a + B) = cos A cos B sin A sin B cos(a B) = cos A cos B +sinasin B sin(a + B) =sina cos B + cos B sin A sin(a B) =sina cos B cos B sin A If we take ientity # an subtract ientity #, we get: (sin A sin B) + (cos A cos B cos A cos B) = cos(a B) cos(a + B) (sin A sin B) = cos(a B) cos(a + B) Similarly: (sin A cos B)+(sinAcos B sin A cos B) =sin(a + B) sin(a B) (cos A cos B) = cos(a + B) cos(a B) We etrapolate those eamples to the cases above: sin cos = sin( +)+sin( = sin 5 +sin( ) = sin 5 sin = ( cos cos )+C An: sin cos ) = cos 4 sin 4 = cos C = 8 cos 4 + C Thus, sin A =sinacos A 4

25 0. Section 7.7: Improper integrals Consier: f(), wheref is continuous on [a, ) a or: f() Definition: a f() =lim b b! f() a or a f() =lim a b! f() b In either case, the limit eists an is finite, then the integral converges - otherwise, the integral iverges. Eample of a convergent improper integral: b = + + b = b = b + b =lim b! =lim b! b + ) = Eample of a ivergent improper integral: = b =ln b =lnb ln =lnb lim b!;lnb = ) this integral iverges to. Generalizing: Proof: Eample: 0 = P p = if p apple P b = P + P p+ b = b p+ b 0 + if p> p+! 0+ p as b! + = tan b 0 = tan b tan 0 = tan b! as b! ) 0 Eample: + = 5

26 Fin 0 e a where a>0 b 0 e a = e a a b 0 = eab a = a! 0 a = a as b! ) e a = 0 a a Eample: Fin the volume of the object obtaine by revolving the region boune by e a,= 0,y = 0 about the X-ais Using the isc metho: Volume = y 0 = e (a) 0 = e a 0 = Eample of integral of unboune functions: on 0 <apple,p>0, on 0 <apple P Suppose f() is efine on a<appleb an is unboune as! a: b a f() =lim c!a+ f() (with c approaching a from the right) b Eample: 0 p p is unboune near =0 c Eample: p = + += + =c = p = =c = p = =c = p c ) lim c!0+ c 0 =lim c!0+ c =lim c!0+ ln c =lim c!0+ ln ln c =lim c!0+ ln c = Therefore this integral iverges. Eample: 0 =lim c!0+ c p =lim c!0+ p c = 6

27 =lim c!0+ c =lim c!0+ c =lim c!0+ + c = To sum up - fact: 0 0 P =?(0 <p<) P = (p ) February 0, 0 ecitation Three situations where trigonometric substitution is helpful: a! = a sin a +! = a tan a! = a sec Eample: First, check to see if we can o this without trig substitution If not, choose one of the above cases - e.g., we re in case here: =sin ; = cos cos( ) sin ( ) cos ( ) cos( ) = sec( ) =ln sec( ) + tan( ) + C =ln p + p + C Partial fractions: ( )(+) = A + B + =A( + ) B( ) =:A = = :B = ) ( )(+) = =ln + + C + + = ln + ln + +C Special variations in case the fractions are repeate or irreucible: ( )( +) = A + B+C + =A( + ) + (B + C)( ) = =A( + ) + (B + C)( ) =A; A = 7

28 Eample: Eample: = ++B B + B C 0=+B =C B = C B = C = = ( ) = A + B + C = ( )( ) +( +)+( ++) = A + B + C ( ) + D+E + + F+G H+I ( ++) + J+K ( ++)...ugly, but entirely solvable February 7, 0 Lecture ecurrence efinition of a sequence Special sequence r n + n = where r is a real number: r =,,,...(a n = for all n) r>:lim n! r n = r =0, 0, 0, <r<:lim n! r n =0 <r<0: thenr = a for some 0 <a< So lim n! r n =lim n! r n =lim n! ( ) n a n = 0 because lim n! a n =0 r = :,,,,,... lim n! r n =lim n! ( ) n oes not eist r< : thenr = a for some a> So r n =( a) n =( ) n a n lim n! r n =lim n! ( ) n a n oes not eist because lim n! a n = 8

29 ules of limits:. lim n! (a n + b n )=lim n! a n ± lim n! b n. lim n! a n b n =(lim n! a n ) (lim n! b n ). lim n! a n bn = limn! an lim n! b n 4. lim n! c a n = c lim n! a n for any constant Eample: lim n! ( n n+ )=lim n! n lim n! n+ =0 0=0 Squeeze Theorem: If a n apple b apple c n for all (but possibly finitely) n, an if lim n! a n =lim a! c n = L, then lim n! b n = L More properties:. lim n! a n = lim n! a n if lim n! a n eists.. if f() is a continuous function, an if lim n! a n = L is finite, then lim n! f(a n )=f(l). if a n 0 (an if limit eists) then lim n! a n 0 if a n apple 0 (an if limit eists) then lim n! a n apple 0 4. if a n apple b n for all (but possibly finitely many) n, then lim n! a n apple lim n! b n Eample: Fin the limit of sin(n) n apple sin(n) apple ) sin(n) n apple n apple n lim n! n =lim n! n =0 sin(n) So lim n! n = 0 by Squeeze Theorem (even though lim n! sin(n) oes not eist) Eample: Fin lim n! n sin( n ) Solution: n sin n = sin n n sin lim!0 = sin ) lim n n! = n More properties:. if lim!a f() =L an lim n! a n then lim n! f(a n )=L. if lim n! a n = L an k is a fie positive integer, then lim n! a n+k = L 9

30 . if a n = b n for all, but possibly finitely ns, then lim n! a n =lim n! b n (or both limits o not eist) Convergence Theorem: If a n is monotonic increasing an is boune above (a n apple M for some number M), then lim n! a n eists (an is finite) Eample: Prove that the sequence efine recursively by a = p p,a n+ = +an for n has a limit. Fin the limit. First, show that a n is monotonically increasing We can prove that a n apple : a k+ = p +a k apple p +=5apple Because a n is monotonically increasing an is finite, we know that it has a limit - but how to fin the limit? If lim n! a n = L an limn!a n+ = L, then from a n+ = p +a n we get lim n! a n+ = p +lim n! a n L = p +L ) L =+L ) L L =0 ) (L + )(L ) = 0 ) L = or L = But we also know that L 0 (because a n 0) ) L =. Growth rate: We say that a n << b n if a n is much smaller than n n if n is very large, or more precisely: lim n(!) ( an b b n ) = 0, or equivalently, lim n n! an = Last time: ln n<<n for any fie (a >0) ) ln Proof :lim! =lim n!! =lim! n =0 Can we generalize? Is (ln n) k << n a? Yes, for any positive integer of k. Proof: ln(n) << n aoverk ) (ln n) k << (n a k ) k = n a Eample: For any positive integer a>0 an b>, n a << b n. Proof: lim! b (apply L Hopital s ule) a lim a! b ln b =0ifaapple a (a ) lim a! b (ln b) if <aapple 0

31 February 7, 0 ecitation (+) =lim a! a (+) Use U-substitution: lim a! a+ ( 8 )= 8 - converges to 8 u u =lim a! u a+ =lim a! (a+) Another eample: 0 4 = lim b b! 0 4 = lim b 4 b! 4 u u = lim b! 4 ln u b4 =lim b! 4 ln b4 4 ln =lim b! 4 ln b4 = Another eample: a =lim a!0 +lim 5 b!0 + =lim a!0 =lim a!0 4 a 4 +lim b!0 + 4a lim b!0 + = + = Does Not Eist Eample: e =lim a! lim b! = b a e b 4 4 b 4 + 4b 4 =lim a! lim b! a b e u u =lim a! lim b! e u b a Known limits: lim! e =0 lim!0 + ln = lim! ln = 5 4 March 04, 0 Lecture Eam: Covers up to en of chapter 7 - only formula that will be given is reuction formula Toay: Eam review Eam #, problem #8: cos() cos() : cos(a + B) = cos(a) cos(b) sin(a)sin(b) cos(a B) = cos(a) cos(b) + sin(a) sin(b) ) cos(a + B) + cos(a B) = cos(a) cos(b) ) cos(a) cos(b) = (cos(a + B) + cos(a B)) = cos( +) + cos( ) = cos(4) + cos( ) cos(4) + cos() =

32 = sin 4 sin ( 4 + ) = 8 (sin 4 +sin)+c Eam #, problem #9: = Let u = +4,u=, = 5 4 u u = (tan 0) = tan = ln u 5 4 u = Eam #, problem #: ( + 4) = tan( ); =sec ( ) + 4 = 4 tan ( ) + 4 = 4(tan ( ) + ) = 4 sec ( ) ( + 4) =(4sec ( )) =8sec ( ) ) 8sec ( ) sec ( ) = 6 sec 5 ( ) Eam #, problem #4: Completing the square: + +9= =( + ) +8 = + (+) +8 Let u = +,u= = (u )+ u +8 u = u+ u +8 u = u u +8 u + u +8 u First part: Let v = u +8,v =uu v v =ln v Secon part: p tan p u (8) 8 =ln(( + ) + 8) + p 8 tan + p 8 + C =ln( + + 9) +... Eam #, problem #5: 0 cos ( ) = ( + cos( )) 0 = sin( ) ( + ) 0 = ( 0) = 4 5 March 8, 0 Lecture Miterm Wenesay

33 Last time: Sequences - toay: series :finiteseries n this is an infinite series Fact: + r + r r n = rn r Eample: n =( ) n = n+ = n+ 0 = n Another proof: n = ( = ( n ) = n n ) Series: A series is a + a a n +...= P n= a n =lim n! P n k= a k For eample, say you have this series: S = a S = a + a S = a + a + a... lim n! S n, if eists, is calle the sum of the series P n= a n. A series converges if its sum eists an is finite. For eample: + ( ) + + ( ) this series iverges, as it oes not converge on either 0 or. Geometric series: a + ar + ar ar n +... = P n= arn if <r<, otherwise series iverges. = a r Eamples: n +... = = =

34 =.5 + = ( ) = + = 4 = 4 =0.75 (a =,r = ) +...+( ) n +... n Eample: Write.4... as a fraction n m. Answer: Define a =0.00,r =0.0 =.4 + a + r + ar +... = = =.4 + = = = = Eample problem 8. #: P k= k 4 k+ = P k= k 4 4 k = P k k= 4 a = first term = 6 ; r = 4 = 6 4 = 6 = 4 Eample problem 8. #: k 5 6 k P k=0 4 = P k=0 k k k 5 6 = P k=0 0 a =5 6 ; r = 0 ; a r = 56 0 = Another eample: not a geometric series. How about this series ( telescoping series ): (n+)(n+) S = = S = + 4 = note that terms cancel: = S = = = 5 S n = n+ 4 4

35 ) lim n! S n =lim n! Eample problem 8. #50: P k=0 (k+)(k+4) n+ = Using partial fractions: A = ; B = ) (k+)(k+4) = k+ k+4 = ( k+ S n = P n k=0 k =0: P = ( k =: P = ( 4...= ( n+4 ) = k+4 ) (k+)(k+4) = 4 ) 7 Eample P problem 8. #55: k= ( p k+ S = p p4 p k+ ) S = p p4 + p p5 P n k=0 ( k+ ) (terms cancel) S = p p4 + p p5 + p 4 p6 k+4 ) S 4 = p p4 + p p5 + p 4 p6 + p 5 p7...s n = p + p p p n+ n+ lim n! = S n = p + p Surprisingly ivergent series: n March 5, 0 Lecture Basic properties of series: Aseries P n= a n is calle convergent if lim n! S n = S eists an is finite. S is also calle the sum of the series. If lim n! S = ±, then we say that the series iverges to + or. If lim n! oes not eist, we simply say that the series iverges. If P n= a n = A an P n= b n = B, then P n= a n ± b n = A ± B. If P n= a n = A, an C is a constant, then P n= C a n = C A. If lim n! a n oes not eist, or eists but is not zero, then the series P n= a n iverges. The contrapositive of the above: If P n= a n converges, then lim n! a n = 0. 5

36 (Note: The converse oes not hol: lim n! a n may eist an be 0, but P n= a n may still iverge. For eample, the harmonic series converges on 0, but the sum iverges to infinity.) Integral Test: Suppose f() is a continuous, ecreasing, an nonnegative function efines for >0. Suppose a n = f(n). Then P n= a n converges if an only if f() converges. Eample P #: n= n f() =, = ) n= n =, an P n= n iverges. Eample P #: n= n = f() =, = = =0 ( ) = < ) P n= n converges, accoring to the Integral Test. ecall that converges if P>an iverges if P apple. ) P P n= n P converges if P>an iverges if P apple, accoring to the Integral Test. If n f() =lim n! f() is finite, then lim n!(s n a )= (lim n! S n ) a apple f() < ; ) lim n! S n <. 7 March 7, 0 Lecture a atio test: Suppose a n 0 for all n. Consier: =lim n+ n! a n (provie it eists or equals ):. Series P n= a n converges if <;. Series P n= a n iverges if >;. Series P n= a n may converge or iverge if = - must use other tests. Eample: P n=0 n n! where is any fie real number. First: Verify that a n 0 for all n -true. Net, consier: a n+ a n = n+ (n+)! n n! = n+ ) lim n! n+ = 0. This proves that a n << n!, i.e., lim n! a n n! = 0, where a is a constant. Eample: P n= n! n n 6

37 a n+ a n = = (n+!) n n n (n+) n (n+)! (n+) n+ n! = n+)! n n (n+) n! n+ n n n n What o we o now? - remember that lim n! ( + n )n = e =( n n+ )n =( + n n ) lim n n! (n+) = n e ) n! << n n,i.e.,lim n! n! n n =0 oot Test: Suppose a n or is ): ) n = + n) n <! the series converges.. Series P n= a n converges if <;. Series P n= a n iverges if >; 0 for all n. Let =lim n! n p a n (provie it eists. Series P n= a n may converge or iverge if = - must use other tests. Easy P eample: n n n= n+ - oes this converge? - use root test: q np an = n n n n+ = n n+ lim p n n n! a n =lim n! n+ = <! series converges. Eample: P n= n a n+ = n+,a n = n a n+ a n = n+ n = n n+ n ) lim n! n+ = - therefore, we on t know if the series converges or iverges; nee to use other tests. Problem #5: P l= k k 99 a k+ k+ (k+),a 99 k = k a k+ a k = k+ k9 9 k+ 99 lim k! k k+ k k+ k = k 9 =>! series iverges. Basic Comparison Test: Suppose a n 0,b n 0 for all n:. If a n apple k b n for some constant k, an P n= b n converges, then P converges.. If a n k b n for some constant k, an P n= b n iverges, then P iverges. Problem #7: P k= k +4 n= a n n= a n 7

38 k +4 < k P converges! k k +4 Problem #8: P k= (k ln(k)) (k ln(k)) apple P k= k ) lim k= (k ln(k)) also converges. k if ln k converges also converges. a Limit Comparison Test: Suppose a n 0,b n 0 for all n, an lim n n! bn (where 0 <c< ): P n= a n converges i P n= b n converges. Problem #8: P k= k +k k 4 +4k b k = k l = 4 k a lim k k k! =lim +k b k k! k 4 +4k =lim l! Problem #4: P k= k k k Can t use the basic comparison test, e.g.,, because that s not true k k k Limit comparison test - choose b k by just ropping the slower-growing term: Let b k = a k b k = k k k = k = k k k ( )k a ) lim k k! b k = ) P k= =, 0 < < converges (geometric with r = k < ) ) P k= a k converges by the limit comparison test. Problem #6: P k= k pk+ apple k, which we know converges! P k= Problem #9: P k= k k +4 b k = k k = k, which iverges a lim k k! b k =! a k iverges by limit comparison test. Problem #4: P k= k + k b P k = k, which iverges k= k + k ) P k= k + k also iverges. Problem #48: P k= k ln(k) k k pk+ converges. = c 8

39 k ln(k) apple P k k when k converges also converges. ) P k ln(k) Eample: P k= ln(k) Fact: ln(k) << k when a>0) Fact: ln(k) apple p (k) for large k Fact: ln(k) apple k for large k for large k, an therefore iverges. ) (ln k) k (Also true for any ln(k) n.) 8 April 0, 0 Lecture Toay s topic: Alternating Series Alternating series are generally of the form P n= ( )n a n - eample of an alternating series: Alternating Series Test: Suppose lim n! a n = 0 an the sequence {a n } n= is ecreasing:. P n= ( )n a n or P n= ( )n+ a n converges.. S an apple S apple S m+ for all positive integers n, m.. S is between S n an S n+ for any n: i.e., S S n apple S n+! S n = a n+ Eample: ( )n+ n +... converges to ln(). Consier: P n= ( )n+ a n = a a + a a S = a S = a a S = a a + a S 4 = a a + a a 4 S 5 = a a + a a 4 + a 5 Eample: Consier P ( ) k ln(k) k= k. a k = ln k k. lim k! ln(k) k 0 for k = : yes = 0: yes. S ln(k+) k+ apple ln(k) k :? 9

40 f() = ln() for f 0 () = ln() = ln() ln() > >e = e =.6 ) f() is ecreasing for >e ) ln(k) is ecreasing for k So P! P k= ( ) k k= k ( ) k ln k k Alternating Series Test:. P k= ( )k k p < 0if>e converges by A.S.T. converges. converges if p>.. P k= ( )k k! converges. Eample: Let S be the sum of ( ) n+ n +... Estimate the error: S ( + Answer: The error is < 4 = 6 Some series have positive an negative terms, but aren t alternating - e.g.: : has positive an negative terms, but isn t alternating P n= sin n n Absolute convergence - efinition: A series P n= a n is calle absolutely convergent if P n= a n converges. Facts about absolutely convergent series:. If P n= is absolutely convergent, then P the converse is false: P n= n= a n is convergent. (Note that iverges. ( ) n n converges, but P Eample: Does P n= sin n converge? - yes: consier P P n= n converges; ) P iverges. n= sin(n) n n= ( )n n n= sin(n) n = sin(n) n apple n Definition: P n= a n is calle conitionally convergent if it converges but P n= a n iverges. Problem #46. Determine whether P ( ) k+ e k k= (k+)! converges absolutely or conitionally. Answer: P = P k= k=0 e k (k+)! a k+ a k ( ) l+ e k (k+)! atio test: = ek+ k+! (k+)! e k = e k+ a lim k+ k! a k =0<! series converges, ) absolute convergence. Net topic: 9. - Power Series y = f( 0 )+f 0 ( 0 )( 0 ) y f( 0 )=f 0 ( 0 )( 0 ) 40

41 f() f( 0 )+f 0 ( 0 )( 0 ) for near 0 Note that y = f() an y = f( 0 )+f 0 ( 0 )( 0 ) have the same erivative f 0 () at 0. This is linear approimation - pretty goo estimate - however, we can o better: quaratic estimation - consier approimating f() by a quaratic function a + b( 0 )+c( 0 ) : we want P ( 0 )=f( 0 ); P( 0 0 )=f 0 ( 0 ); P 00 ( = 0) = f 00 ( 0 ). P ( 0 )=a = f( 0 ), so a must be equal to f( 0 ).. Since P () =b + c ( 0 ), ) P 0 ( 0 )=b +0=b.. b = f 0 ( 0 ). 4. Now P 00 () =c =c 5. c = f 00 ( 0 ), i.e., c = f 00 ( 0) ) P () =f( 0 )=f 0 ( 0 )( 0 )+ f 00 ( 0) ( 0 ) 9 April 0, 0 Lecture Eam eview Taylor Polynomials: Choosing functions to approimate other functions - eample: P (): Polygon of egree = a + b( a )+c( a ) Use P n () =a 0 + a ( 0 )+...+ a n ( 0 ) n to approimate f() Want P n ( a )=f( a )! a 0 = f( 0 ) Pn( 0 a )=f 0 ( a )! a = f 0 ( 0 ) Pn 00 ( a )=f 00 ( a )! a = f 00 ( 0) Pn 000 ( a )=f 000 ( a )! a = f 000 ( 0)!... P n ()( a )=f (n) ( a )! a n = f (n) ( 0) n! The polynomial P n () with a i = f (n )( n) i!,i = 0,...,n is calle the Taylor polynomial of egree n centere at = 0. Eample: Fin Taylor polynomial of egree at = 0 for sin() Answer: f() =sin() f( 0 )=f(0) = 0 4

42 f 0 ( 0 )=f 0 (0) = cos() =0 = f 00 ( 0 )=f 00 (0) = sin() =0 =0 f () ( 0 )=f 000 (0) = cos() =0 = P ( )=f( 0 )+f 0 ( 0 )( 0 )+ f 00 ( 0) ( 0 ) + f 000 ( 0)! ( 0 ) =0+ ( 0) + 0 ( 0) + 6 ( 0) = 6 Eample: Fin Taylor polynomial of egree at = 0 for e Answer: f() =e,f(0) = e 0 = f 0 () =e,f 0 (0) = e 0 = f 00 () =e,f 0 (0) = e 0 = f 000 () =e,f 0 (0) = e 0 = P () =f(0) + f 0 (0) + f 00 (0) + f 000 (0) 6 = Taylor s emainer Theorem: Estimate f() P n () : f() P n () = f (n+) (c) (n+)! ( 0 ) n+ for some c = c between an 0. (c epens on.) Note: For n =0,P n () =f( 0 ).f() P n () =f() f( 0 ) f (n+) (c) (n+)! ( 0 ) n+ = f 0 (c)! ( 0 ) = f 0 (c)( 0 ) i.e., f() f( 0 )=f 0 (c)( 0 ) Eample: Suppose P () is use to approimate e in the interval apple apple. Estimate the maimum error e P () on the interval. Answer: f() =e e P () = f (4) (c) 4! ( 0) where c is between 0 an = ec 4 4 = ec 4 4 apple e 4 apple 4 = 8 Power Series ( centere at 0 ): P n=0 a n( 0 ) n = a 0 + a ( 0 )+a ( 0 ) a n ( 0 ) n +... Eample: n +...( 0 =0, n = for all n) This is just a geometric series with r = n - only converges to for << Question for power series: For what oes the series converge? i.e., what is the interval of convergence? Try using atio Test: n th term of series is a n ( 0 ) n : (n+) term lim n! n th term a =lim n+ n! a n 0 n+ 0 n 0 4

43 =lim n! a n+ a n 0 = 0 lim n! a n+ a n Eample: For what oes this power series converge: Answer: P ] n= nn n (n ) th -term = n th -term n+ (n+) n n lim n! n (n+) = = = n+ n (n+) = n n (n+) So by the atio Test, the series converges if <, i.e., if << P At = an =, atio Test fails. However: n= n n = P n P n= n = n= n converges. (p-series, where p =) Series converges absolutely; therefore, series converges at = an = Conclusion: Series converges on apple apple (this is the interval of convergence); iverges for all other. Eample: Fin the raius an interval of convergence for the power series: P n=0 n n! =+! +! + n! n! +... Answer: n+ (n+)! n! = n n+ ) lim n! n+ = 0 for all ) Series converges absolutely for all. Eample: Fin the raius an interval of convergence for the power series: P n= nn n Answer: n+ (n+) n+ lim n! n n n = (n+)n+ n n = (n+)n n. (n + ) =(+ ( n )n (n + ) 0if =0 = = if 6= 0 ) interval of convergence = = 0 Or use oot Test: n n n n = n ( = n 0if =0 lim n! n = = if 6= 0 4

44 ) interval of convergence = = 0 Eample: Fin the raius an interval of convergence for the power series: Answer: P n= ( ) n n n+ n n+ = n n n+! as n! ) series converges for < i.e., the raius of convergence is. At enpoints,, : =:Series = P ( ) n n= =:Series = P ( ) n ( ) n n n = P n=! series converges. n= n = P n= n! series iverges. ) interval of convergence = apple < or apple <. Suppose f() = P n=0 a n( 0 ) n (e.g., = , <<) Then f 0 () = P n= a n n( 0 ) n = P P k=0 a k+ (k + )( 0 ) k f() = a n n=0 n+ ( 0) n+ + C - i.e., power series can be integrate Note: aius of convergence oesn t change, though the enpoints of the interval of convergence may change. 0 April 8, 0 Lecture Toay: Continue power series For the power series n +...= Substituting for : (( ) n )( n )+...= Integrating both sies from 0 to t: t 0 = t t = 0 t 0 = t, t 0 = t 0 = t,..., t 0 n = tn+ n+ ) t 0 + =ln(+) t 0 =ln(+t) ln() = ln( + t) 0=ln(+t) t ) t + t t ( )n+ tn n +...=ln(+t), <tapple When you integrate a power series, it has the same raius of convergence, but may have a i erent interval of convergence. Di erentiating both sies of n +...= : n n +... =( )( ) ( ) = ( ), << = : ! iverges to. =: ! iverges (n th term test) ) raius of convergence = (same as original series); interval of convergence is <<. eplacing by in ( ) n n +...= +, we get: ( ) n +...= + 44

45 t Integrate both sies from 0 to t: t + t5 t = t 0 + = tan t 0 = tan t tan 0 = tan t eplacing t with : ( )n n+ n+ +...= tan () Same raius of convergence ( <<), but still nee to check the enpoints as the interval may converge: =: + 5 7! alternating series convergence (actually converges on tan () = 4 ) = : ! also alternating series convergence ) interval of convergence = apple apple Eample: Fin a power series for ln( Answer: ). ln( + ) = +..., < apple Substituting for : ln( ) =..., < apple = ( ), > Eample: Fin a power series for +. Answer: We know that + = ( ) n n +... Consier + = ( ) n n +... Consier substituting for = + ( ) ( ) +... =( ) n ( ) n +... = + ( 4 ) ( 6 )+...+( ) n n n +... n+ Interval of integration: apple <! p ( ) <<p ( ) Net topic: Taylor series (section 9.) If f() is a function, the Taylor series for f() centere at = 0 is, by efinition, the power series P f (n) ( 0) n=0 n! ( 0 ) n = f( 0 )+ f 0 ( 0)! ( 0 ) + f 00 ( 0)! ( 0 ) f ( n) 0) n! ( 0 ) n If 0 = 0, you get: f(0) + f 0 (0)! + f 00 (0)! f (n )(0) n! n +... Some functions cannot be approimate by a Taylor series - e.g., cos() - must use a i erent technique Eample: Fin Maclaurin series for cos() Answer: f() = cos(),f(0) = cos(0) = f 0 () = sin(),f 0 (0) = 0 45

46 f 00 () = cos(),f 00 (0) = f () () =sin(),f () (0) = 0 (epeats P from here) n=0 =+0 +! ! =! ! 6! +... This series converges for <<. Note: This is why cosine is even - in general, functions for which the Maclaurin series only contains even powers are even functions; functions for which the Maclaurin series only contains o powers are o functions. Eample: Fin Maclaurin series for sin() Answer: f() =sin(),f(0) = sin(0) = 0 f 0 () = cos(),f 0 (0) = f 00 () = sin(),f 00 (0) = 0 f () () = cos(),f () (0) = (epeats from here) P n=0 =! ! 7! +... This series converges for <<. Eample: Fin Maclaurin series for e Answer: f() =e,f(0) = e 0 = f 0 () =e,f 0 (0) = e 0 =...f (n) (0) = for all n + +! +! + 4 4! +... << Eample: Fin Maclaurin series for e Answer: Substitute for above: f() =e,f(0) = e 0 = f 0 () =e,f 0 (0) = e 0 =...f (n) (0) = for all n + 4! << 6! + 8 4! 0 5! +... Eample: Fin Maclaurin series for cos() Answer: Substitute for in cos() series above: 4! ! << 6 6 6! +...+( ) n n n ()! +... Eample: Prove that the Maclaurin series for cos() converges to cos(). Answer: Take the partial sum of the series: S n =! + 4 4! 6 6! +...+( )n n (n)!

47 Prove that this series converges to cos(), an then use Taylor remainer theorem: cos() S n = f n+ (c) (n+)! ( 0)n+ for some c such that 0 <c< Since f n+ () ± sin() or ± cos() ) f n+ (c) apple ) cos() S n apple (n+)! n+! 0 as n!for any ) lim n! S n = cos(). What can we o with Taylor series? Eample: Write the integral 0 e as a series (since its integral has no formula) Answer: Use a series: e = + 4! 6! + 8 4! 0 5! +... ) 0 e = !! +... = over+! 5! ( )n n! (n+) +... This alternating series converges, an converges to. 0 e Error of series is less than the value of the net term. If you use S = +! 5! 7 as an approimation for I, the error I S < 4! 9. Eample: Fin the Maclaurin series for sin (). Answer: Use sin () = ( cos()) = = = =! ( 4! ! + 4! 6 6 6! +...) 8 4 4! !... 0 cos() 8 4 4! !...+( ) (n+) n (n)! +... (Bottom line: Try not to take the erivative; try using substitutions or conversions first.) Binomial theorem: Consier ( + ) n = n 0 In this theorem, + n + n n n n k = n! n (n )...(n k+) (n k)! k!(n k)! = k! (n k)! = n (n )...(n k+) k! 5 = 5!!! = 5 4! = 0( + ) = p ( + ) = e.g.: n n = = = ( )! = ( ) = 8! = ( ) ( ) = ( ) ( )! = 8 6 =

48 April 8, 0 Lecture Binomial series - ( + ) P,whereP is any fie number other than 0 - using P P p (p )... (p k+) k notation: k = k!,k =,,... Eample: = ( ) ( ) ( ) 4 4! = ( ) ( ) ( 5 ) 5 4 = 4 4 Taylor series for binomial selection: ( + ) P = P P + P + P P n n converges for <<, but convergence of enpoints epens on P,sotestitwhen using Fin Maclaurin series for sin () - oing this by fining the erivative at each level is painful - instea: Answer: sin () = 0 p t t =( ) We can therefore start by fining the Taylor series for ( + ) ), an later substitute for : ( + ) = n n +... =+( ) + ( )( )! + ( =( 5! ) ( )+( ! ) ) ( + ) =+ P n= ) ( ) ( )=+P =+ P n= ( )n =+ P n= 5... (n ) n n! n )( )( )! +... n= ( )n 5... (n ) ( ) n n n!... (n ) 5... (n ) n n! n ) sin () = + P n= = + P n= n n! ( ) n n... (n ) n+ n n! n+... (n ) (n+) n n! n+ = C Section 9.4: What can we o with Taylor series? Use Taylor series to fin limits + e Eample: Problem #8. fin lim!0 4 + =lim!0 4 We know that e =+ +! +! +... an e = +!! e ) lim + +!0 4 =lim +!0 4 =lim! +!...!0 4 =lim !0 4 ) limit oes not eist.! +... Or use L Hopital s ule: lim!0 + e 4 (limit approaches 0 0 ) 48

49 =lim!0 e ( ) +e 8 =lim!0 8! - top goes to ; bottom goes to 0; limit oes not eist. Eample: Problem #: tan lim +!0 5 Note that + = , an that + = tan () =lim !0 =lim!0 =lim! = 5 Eample: Fin lim!0 ( sin ) Use a series: sin() =! + 5! 5... ln( sin() ) = ln( sin() )= ln( ) emember that ln( + ) =( +...)...an ln( ) = ( ) ) ln( sin() )= ( 6 + ) lim!0 = 6 ) lim!0 ( sin() ) = e (to powers higher than ) ) Sometimes there may be a function that can t be integrate, an a Taylor series can be use to approimate it Eample: Fin 0. sin( ) 0 Can t be integrate, but can be approimate with Taylor series, with error < P 0 4 () = (We ll start with just three terms; that may be enough for this error - if not, we can a more) ) sin( )= ) sin( )= ) 0. sin( ) = (0.) (0.) (0.) 0... (Alternating Series) As long as the thir term is < 0 4, then we re one (0.) ; (0.) = (an we re alreay much less than the error term) ) 0. sin( ) with error < Sometimes you can use a series to figure out the function represente by a given series Eample: Fin P k= = k k Notice that this resembles This also resembles ln( ) = ( ) Therefore, substitute = ) Answer = ln( )= ln( into ln( ) )= ( ln()) = ln() 49

50 Problem #57. Ientify the function represente by the series P = P k (k ) k k= = P k= k k (k ) k k Now let s consier = P k= kk... an k=0 k = k= kk = P k= ) ( )= ( ) P )= 9 k= k) k = o ver9 ( ) k(k )k k(k ) k k= k = 9 ( ) = 6 ( ) What is this vali for? - ( ) is vali for < ; reversing the substitution: sum is vali for <, an iverges for other Chapter 0: Parametric equations Eample: A unit circle can be represente by = cos(t); y =sin(t), 0 apple t apple - this equation makes the same unit circle as + y =, but the parametric equation has a irection: as t goes from 0 to Another eample: y =! = t, y = t, <t< - again, has a irection More complicate eamples: = t t, y =t + t +, 0 apple t apple 4 - graphing is i cult; can o it by han but it takes a while, or can o it by a computer How about fining a tangent line? - y = y t t = 9t + t Fin the slope of the line tangent to the curve efine by = t t, y =t +t+ 9() at the point where t =, i.e., at (, y) =(0, 5) - answer: + () = 0 How about secon erivative? - y = first erivative, an then ivie by t t y t t t - i.e., take the erivative of the Secon erivative of = t t, y =t + t + : (8t 9t 9t ) (t ) = (9t 9t ) (t ) t ( 9t + t t ) t = (t ) 8t (9t +) (t ) t = April 0, 0 ecitation eview of Taylor polynomials: An N th -orer polynomial of f() = P N f ( n)(a) n=0 n! ( a) n 50

51 Error rates: What orer polynomial o we nee to get within an error factor? n apple f n+ (c) (n+)! ( a)n+ An eample: Approimating cos() to within cos( 4 ): () Pick a center for approimation a that results in an f() close to the value that we want, an that results in a value that we know - in this case, we can use 4, an cos( 4 )= p First term: Secon term: Thir term: p() = p( 4 )= p () f( 4 ) 0! ( 4 )0 = p f 0 ( 4 )! ( 4 ) = p ( 4 ) f 00 ( 4 )! ( 4 ) = p ( p p p 4 ) () () ( 4 ) () ( 4 ) p () ( 4 p 4 ) () ( 4 4 ) emainer = sin(c)! ( 4 4 ) The largest that sin(c) can be is - thus, remainer apple ( 4 eview of power series: Page 60: Big list of power series that we shoul know e = P k=0 k! k =+ + +! ) 6 Taylor series provie alternative representations of functions - for eample, e = P k=0 k! k = P k=0 k! k+ Another eample: = P k=0 k - so if we want a representation of can use P k=0 ()k = P k]0 k k,we Another eample: 5 + = 5 + = 5 ( ) - now we can substitute for in our power series: 5 P k=0 ( )k k = P k=0 ( )k k+5 Another eample: ( ) - this is just the erivative of, so just take the erivative of each element in the series eview of raius of convergence: P P k k k=0 k=0 k! Fining the raius of convergence: lim k! a k+ a k (k+) (k+) (k+)! =lim k! =lim k! k+ k k k k! Another eample: P =lim k! k+ =0 k= k+ ( ) k+ k+ k ( ) k k k ( ) k k 5