0.1 The Slope of Nonlinear Function

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1 WEEK Reading [SB],.3-.7, pp Te Slope of Nonlinear Function If we want approimate a nonlinear function y = f() by a linear one around some point 0, te best approimation is te line tangent to te grap of te function y = f() at te point ( 0, f( 0 )). Te slope of tis tangent line is te derivative of y = f() at 0 and is denoted as f ( 0 ) or df d ( 0). More precisely: Te tangent line of te function y = f() at a point 0 is te limit of secant wic passes troug two points ( 0, f( 0 )) and (, f()), wen 0. Wat is te slope of tis secant? Tis is f() f( 0 ) 0. Tus te slope of te tangent line is f() f( 0 ) lim. 0 0 Definition Te derivative of a function y = f() at 0 is f ( 0 ) = lim 0 f() f( 0 ) 0, equivalently f f( 0 + ) f( 0 ) ( 0 ) = lim. 0

2 Eample. Let us calculate using te definition te derivative of quadratic function f() = at a point 0 : f ( 0 ) = lim 0 f( 0 +) f( 0 ) = lim 0 ( 0 +) 0 = lim = lim = lim 0 ( 0 + ) = 0. Eample. In previous proof we ave used te formula (a + b) = a + ab + b. Tis is a particular case of general Newton Binom formula (a + b) k = Cka 0 k + Cka k b + Cka k b Ck k ab k + Ck k b k were C i k = k! i!(k i)! are binomial coefficients given by C i k = k! i!(k i)!,

3 tat is C 0 k =, C k = k, C k = In particular (k ) k,..., Ck k = k, Ck k =. C 0 =, C =, tus (a + b) = a + b (wow!) Furtermore C 0 =, C =, C =, tus (a + b) = a + ab + b. And furtermore C 0 3 =, C 3 = 3, C 3 = 3, C 3 3 =, tus (a + b) 3 = a 3 + 3a b + 3ab + b 3. Te binomial coefficients C j k form Pascal s triangle were eac number is te sum of te two directly above it. We use tis formula to find te derivative of te function f() = k : f f( ( 0 ) = lim 0 +) f( 0 ) ( 0 = lim 0 +) k k 0 0 = lim k 0 +kk 0 +Ck k k 0 k + k k 0 0 = k lim k 0 +Ck k k 0 k + k 0 = lim 0 (k k 0 + Ck k k 0 k + k ) = k k Rules for Computing Derivatives (a) (f ± g) ( 0 ) = f ( 0 ) ± g ( 0 ), (b) (kf) ( 0 ) = kf ( 0 ), (c) (f g) ( 0 ) = f ( 0 ) g( 0 ) + f( 0 ) g ( 0 ), (d) ( f g ) ( 0 ) = f ( 0 ) g( 0 ) f( 0 ) g ( 0 ) g( 0 ), (e) ( k ) = k k, (f) ((f() n ) = n(f()) n f (), Eercises.0-. 3

4 0. Continuous Functions A function is continuous if its grap as no brakes. Precise definition: a function y = f() is continuous at if for any sequence,,..., n,...} wic converges to te sequence converges to f(), tat is Eample. Te function f( ), f( ),..., f( n ),...} lim n = lim n n f( n ) = f(). f() = is discontinuous at = 0: for a sequence, 0 +, > 0 n = n } =,, 3,..., n,...}, wic converges to = 0 from te left, te sequence f( n ) = n } =,, 3,..., n,...} converges to 0 = f(0), but for te sequence n = n } =,, 3,..., n,...} wic converges to = 0 from te rigt, te sequence f( n ) = n + } = +, +, 3 +,..., +,...} n converges to f(0). We write in tis case lim f() = 0 = f(0) = lim f()

5 > plots[multiple](plot, [, =..0], [ +, = 0..]); Eample. Te function f() =, 0 0, = 0 is discontinuous at = 0: lim =, f(0) = 0, 0 lim 0 + =. Te function is continuous at eac point of its domain (, 0) (0, ) but not at = Differentiability A function y = f() is called differentiable if it as te derivative at every point of its domain. Te grap of suc function as tangent everywere, tat is its grap is a smoot curve. A function y = f() is called continually differentiable function (a C function in sort) if (a) f() is continuous, (b) f() is differentiable, (c) f () is continuous. Eample. Te function y = as no tangent at = 0, so it as no derivative at tis point, it is not differentiable, it is not smoot. Eample. Te function f() =, 0, > 0 is continuous at = 0: te left limit at = 0 is lim f() = lim ) = ( 5

6 as well as te rigt limit lim f() = lim = But it is not differentiable at = 0: te left derivative is and te rigt derivative is lim 0 f () = ( ) =0 = =0 = 0 lim 0 +f () = () =0 = =0 =. > plots[multiple](plot, [, =..0], [, = 0..]); 0 3 f() 3 0 f () 6

7 Eample. Te function f() =, 0 3, > 0 is differentiable at = 0: te left derivative is and te rigt derivative is lim 0 f () = ( ) =0 = =0 = 0 lim 0 +f () = ( 3 ) =0 = 3 =0 = f() f () 7

8 Remark. Here is an eample of differentiable but not C function: f() = sin(/), 0 0, = 0. Eercises. Ceck te continuity and te differentiability of f() = 3 <. Solution. Ceck te continuity at = : lim f() = lim 3 = 3 =, lim f() = lim =, + + so te function is continuous. Te derivative of our function is f () = 3 <., tus lim f () = lim 3 = 3, and lim + f () = lim + f =, so f () does not eist at =, te function is not C.. Ceck te continuity and te differentiability of f() = 3 < 3. Solution. Ceck te continuity at = : lim f() = lim 3 = 3 =, lim f() = lim + +(3 ) = 3 =, so te function is continuous. Te derivative of our function is f () = 3 < 3., tus lim f () = lim 3 = 3, and lim + f () = lim + 3 = 3, so f () is continuous, te function is C. 3. Ceck te continuity and te differentiability of f() = 3. Eercises

9 0. Higer order derivatives Te second derivative of a function y = f() is te derivative of te derivative f (). Notation f () or d d ( df d ()) = d f d (). For eample ( 3 ) = (( 3 ) ) = ( ) =. A C function is a twice continuously differentiable function. Te k-t derivative of f is denoted by f [k] = dk f d k (). If tis k-t derivative is continuous, ten we say f is C k. If f as continuous f [k] -s for all k, ten we say f is C. All polynomials are C. Eercises. Ceck te continuity and te differentiability of f() = 0 > 0. Solution. Te derivative of our function is f () = 0 > 0. =, tus te function is continuous, differentiable, but te second derivative does not eists at = 0. So tis function is C but not C. f() 9

10 3 f () f (). We ave already cecked tat te function 3 < f() = 3. is C. But is it C? Te second derivative of our function is f () = 6 < 0., so at = te left second derivative is lim f () = 6 = = 6 = 6 0

11 and te rigt second derivative is lim +f () = 0 = = 0 tus f () does not eists, i.e. tis function is not C f() f ()

12 f () 3. We ave already cecked tat te function f() =, 0 3, > 0 is C, but is it C?. Construct a function wic is C but not C 3. Eercises Approimation by Differential By definition of te derivative tus f ( 0 ) f( 0 + ) f( 0 ), f( 0 + ) f ( 0 ) + f( 0 ). Equivalently, taking = 0 + we obtain f() f( 0 ) + f ( 0 ) ( 0 ). Tis allows to approimate f() by te linear function f( 0 )+f ( 0 ) ( 0 ) around a point 0, for wic f( 0 ) and f ( 0 ) are easy to calculate. Denote f() f( 0 ) = f and 0 =, ten te above can be rewritten as f f ( 0 ).

13 Write df instead of f and d instead of. Ten df is called differential of f. df = f ( 0 ) d, Eample. Estimate 90. Solution. Consider te function f() =. Te point nearest to 90 for wic we can calculate f() (and f ()) is = 900: f(900) = 900 = 30, furtermore, te derivative of f() = is f () = =, tus f (900) = 60. So f(90) can be approimated as f(90) f(900) + f (900) 0 = 90 = = = > f() := sqrt(); df() := diff(f(), ); > 0 := 900.; k := eval(df(), = 0); f(0) := eval(f(), = 0); > g() := k ( 0) + f(0); > eval(f(), = 90.); eval(g(), = 90); > plot(f(), g(), = ); Eercises Taylor Formula Te linear approimation f() f( 0 ) + f ( 0 ) ( 0 ). 3

14 is a particular case of more general approimation of a function wit Taylor polynomials P n () f() P n () = f( 0 ) + f ( 0 ) ( 0 ) + f ( 0 )! ( 0 ) f [n] ( 0 ) n! ( 0 ) n were n! is te factorial n! = 3... n. Te Taylor series of f is infinite Taylor polynomial P () = f( 0 ) + f ( 0 ) ( 0 ) + f ( 0 )! ( 0 ) f [n] ( 0 ) n! ( 0 ) n Equivalent form P ( 0 + ) = f( 0 ) + f ( 0 ) + f ( 0 )! f [n] ( 0 ) n! n +... Te particular case of tis series wen 0 = 0 P () = f(0) + f (0) + f (0) f [n] (0) n +...! n! is called MacLaurin series. Eample. Estimate 90 now using te second order Taylor polynomial. Solution. f() f(900) + f (900) ( 900) + f (900) ( 900).! Tus f () =, f (900) = = f () =, f (900) = = f(90) = = = Compare tis by obtained by linear approimation and te value 90 = given by calculator. By MAPLE > f := sqrt(); > T := taylor(f, = 900, 3); T := 30 + ( 900) ( ) + O(( 900) 3 ) > P := convert(t, polynom); P := 5 + ( 900) > t := eval(p, = 90); > evalf(t);

15 Eercises. Find te MacLaurin polynomial P () for te functions f() = + and f() =.. Estimate e using te MacLaurin polynomials P (), P (), P 3 (). 3. Estimate ln using Taylor polynomials P (), P (), P 3 () at =.. Find te MacLaurin polynomials P (), P (), P 3 (), P ()) for te a polynomial f() = a 3 + b + c + d. Homework. Problem.8(f) from [SB].. Problem.(b) from [SB]. 3. Give an eample of a function wic is C 3 but not C.. Problem.3 from [SB]. 5. Estimate ln by Taylor polynomials P (), P () at =. 5