Exponential Functions and Taylor Series

Transcription

1 MATH 4530: Aalysis Oe Expoetial Fuctios ad Taylor Series James K. Peterso Departmet of Biological Scieces ad Departmet of Mathematical Scieces Clemso Uiversity March 29, 2017

2 MATH 4530: Aalysis Oe Outlie 1 Revistig the Expoetial Fuctio 2 Taylor Series

3 MATH 4530: Aalysis Oe Revistig the Expoetial Fuctio Theorem lim k (x k /) = 0 for all x. There is a k 0 with x /k 0 < 1. Thus, x k0+1 (k 0 + 1)! x k0+2 (k 0 + 2)!. x k0+j (k 0 + j)! = = x k x k x k0 k 0! x k ( ) j x x k0 k 0! k 0 x k 0 x k0 k 0! x k0 k 0! ( ) 2 x x k0 < k 0! k 0 Sice x k 0 < 1, this shows lim k (x k /) 0 as k.

4 MATH 4530: Aalysis Oe Revistig the Expoetial Fuctio Although we do ot have the best tools to aalyze the fuctio ( S(x) = lim 1 + x ) let s see how far we ca get. We start with this Theorem: Theorem lim k=0 x coverges for all x. Let f (x) = lim k=0 x k ad f (x) = these limits exist ad are fiite. k=0 x k. We eed to show We will follow the argumet we used i the earlier Theorem s proof with some chages.

5 MATH 4530: Aalysis Oe Revistig the Expoetial Fuctio N =k+1 x k = x k (k + 1)! + x (k + 2)! x N k (k + (N k) )! { x (k + 1) +... x N k (k + 1)... (k + (N k) ) = x k x k { x (k + 1) +... x N k (k + 1) N k ) } x k } N k k=0 r k where r = x (k+1). We ca the use the same expasio of powers of r as before to fid ( ) N k+1 1 x /(k + 1) N =k+1 x k x k 1 x /(k + 1)

6 MATH 4530: Aalysis Oe Revistig the Expoetial Fuctio Sice there is a K 1 so x /(k + 1) < 1/2 for k > K 1, for all k > K 1, 1/(1 x /(k + 1)) < 2 ad so lim N { N =k+1 } x k x k lim 2(1 N (1/2)N k+1 ) = 2 x k Now pick ay k > K 1, the we see f (x), the limit of a icreasig sequece of terms, is bouded above which implies it coverges to its supremum. f ( x ) = 1 + x + x /2! x k / + lim N N =k x + x /2! x k / + 2 x k. x k

7 MATH 4530: Aalysis Oe Revistig the Expoetial Fuctio So we kow f ( x ) has a limit for all x. To see f (x) coverges also, pick a arbitrary ɛ > 0 ad ote sice x k 0 as k, K k > K implies x k < ɛ/2. So for k > K, we have lim N N =k+1 x k = 2 x k < ɛ The, for k > K, k f (x) x k =0 lim N N =k+1 x k lim N N =k+1 x k < ɛ. This says lim k k =0 x k = f (x).

8 MATH 4530: Aalysis Oe Revistig the Expoetial Fuctio Next, let s look at S (x) = (1 + x/) ad S(x) = lim S (x). S (x) = 1 + = 1 + = 1 + ( ) x k k k = 1 +! ( k)! x k k (( 1)( 2)... ( (k 1)) (1 1/)... (1 (k 1)/) x k Cosider (1 1/)... (1 (k 1)/) x k x k = ((1 1/)... (1 (k 1)/) 1) x k < ((1 + 1/) 1) x k x k k

9 MATH 4530: Aalysis Oe Revistig the Expoetial Fuctio The we have (1 1/)... (1 (k 1)/) x k x k x k < (1/) x k = f ( x ) 0 as. as sice f ( x ) is fiite, the ratio f ( x ) goes to zero as. So we kow Theorem S(x) = lim S (x) = 1 + lim x k = f (x). We just did this argumet.

10 MATH 4530: Aalysis Oe Revistig the Expoetial Fuctio We usually abuse otatio for these sorts of limits ad we will start doig that ow. The otatio k=0 x k lim k=0 x k. Also, sice we are always lettig go to, we will start usig lim to deote lim for coveiece. Theorem S(x) is cotiuous for all x Fix x 0 ad cosider ay x B 1 (x 0 ). Note S (x) = (1 + x/) 1 (1/) = (1 + x/) 1. We see S (x) = S (x) (1 + x/) 1. For each, apply the MVT to fid c x betwee x 0 ad x so that S(x) S(x0) x x 0 = S (c x ).

11 MATH 4530: Aalysis Oe Revistig the Expoetial Fuctio Now lim (1 + c x /) = 1, so for sufficietly large, 1 < 1 + c x / 1 < 2. Hece, if sufficietly large (1 + c x /) < S (c x ) < 2 (1 + c x /) Also, if you look at S (x) for x 0 1 < x < x 0 + 1, 1 + x 0 1 < 1 + x < 1 + x implyig S (x 0 1) < S (x) < S (x 0 + 1). Sice these umbers could be egative, we see S (cx D(x 0 ) = max{ S (x 0 1), S (x 0 1) } which we ca choose to be positive. Thus, sice x 0 1 < cx < x 0 + 1, S (c x ) < 2D(x 0 ) for sufficietly large. This tells us lim S (x) S (x 0) x x 0 2D(x 0 ). Lettig, we fid S(x) S(x 0 ) x x 0 2D(x 0 ).

12 MATH 4530: Aalysis Oe Revistig the Expoetial Fuctio This shows S(x) is cotiuous at x 0 as if ɛ > 0 is chose, if x x 0 < ɛ/d(x 0 ) we have S(x) S(x 0 ) < ɛ. Now pick ay x that is irratioal. The let (x p ) be ay sequece of ratioal umbers which coverges to x. The sice S(x) = lim p S(x p ), we have S(x) = lim p (1 + x p /) = lim p e xp. Hece, we ca defie e x = lim p e xp defiitio of e x off of Q to IR. We coclude (1) e x = lim (1 + x/) = k=0 x k (2) e = lim (1 + 1/) = k=0 1 (3) e x is a cotiuous fuctio of x. ad we have exteded the

13 MATH 4530: Aalysis Oe Revistig the Expoetial Fuctio Theorem (e x ) = e x Let (x p ) be ay sequece covergig to x 0. The, we have where c,p x S (x p ) S (x 0 ) = S x p x (c x,p ) = S (cx,p ) (1 + cx,p /) 1. 0 is betwee x 0 ad x p implyig lim p c,p x = x 0. Give ɛ > 0 arbitrary, P p > P x 0 ɛ < cx,p have 1 + x 0 ɛ ( 1 + x ) 0 ɛ < < 1 + c,p x < 1 + x 0 + ɛ (1 + c,p x = ) ( < 1 + x 0 + ɛ < x 0 + ɛ. We the )

14 MATH 4530: Aalysis Oe Revistig the Expoetial Fuctio This also tells us x0+ɛ < c,p x < x0 ɛ ad so ( 1 + x ) 0 ɛ x0+ɛ < < (1 + c,p x ( 1 + x 0 + ɛ ) c,p x ) x 0 ɛ Now let p ad ad fid e x0 ɛ 1 < lim p lim (1 + c,p x ) 1 < e x0+ɛ c,p x

15 MATH 4530: Aalysis Oe Revistig the Expoetial Fuctio Now, lettig ɛ 0 +, we fid e x0 = lim ɛ 0 + ex0 ɛ lim p lim ) (1 + c,p x c,p x = lim p lim S (x p ) S (x 0 ) x p x 0 as e x is a cotiuous fuctio. We coclude { } lim lim S (cx,p S(xp ) S(x 0 ) ) = lim p p x p x 0 lim ɛ 0 + ex0+ɛ = e x0 = e x0. Sice we ca do this for ay such sequece (x ) we have show e x is differetiable with (e x ) = e x.

16 MATH 4530: Aalysis Oe Revistig the Expoetial Fuctio Theorem (1) e x+y = e x e y (2) e x = 1/e x (3) e x e y = e x y = e x /e y (1): e x e y = lim ( (1 + x/) ) lim ((1 + y/) ) = lim ((1 + x/) (1 + y/)) = lim ( 1 + (x + y)/ + xy/ 2 ) ) = lim (1 + (x + y)/ + (xy/)(1/)). Pick a ɛ > 0. The N 1 > N 1 xy / < ɛ ad N 2 > N 2 1 (x + y)/ > 0.

17 MATH 4530: Aalysis Oe Revistig the Expoetial Fuctio which tells us (1) is true. So if > max{n 1, N 2 }, we have 1 + (x + y)/ ɛ < 1 + (x + y)/ + xy/ 2 < 1 + (x + y)/ + ɛ/ ( ) 1 + (x + y)/ ɛ/) < (1 + (x + y)/ + xy/ 2 < ( ) 1 + (x + y)/ + ɛ/ because all terms are positive oce is large eough. Now let to fid Fially, let ɛ 0 + to fid e x+y ɛ e x e y e x+y+ɛ e x+y e x e y e x+y e x e y = e x+y

18 MATH 4530: Aalysis Oe Revistig the Expoetial Fuctio (2): e x e x = lim ( (1 x/) ) lim ((1 + x/) ) = lim ((1 + x/)(1 x/)) = lim ( 1 x 2 / 2) = lim ( 1 (x 2 /)(1/) ) Now argue just like i (1). Give ɛ > 0, N > N x 2 / < ɛ. Thus, 1 ɛ/ < 1 x 2 / 2 < 1 + ɛ/ ( ) 1 x 2 / 2 < ( 1 ɛ/) < ( 1 + ɛ/) ) ( ) lim (1 ɛ/) lim (1 x 2 / 2 lim 1 + ɛ/ This says e ɛ e x e x e ɛ ad as ɛ 0, we obtai (2). (3): e x e y = e x+( y) = e x y ad e x e y = e x /e y ad so (3) is true.

19 MATH 4530: Aalysis Oe Taylor Series Let s look at Taylor polyomials for some familiar fuctios. (1): cos(x) has the followig Taylor polyomials ad error terms at the base poit 0: cos(x) = 1 + (cos(x)) (1) (c 0 )x, c 0 betwee 0 ad x cos(x) = 1 + 0x + (cos(x)) (2) (c 1 )x 2 /2, c 1 betwee 0 ad x cos(x) = 1 + 0x x 2 /2 + (cos(x)) (3) (c 2 )x 3 /3!, c 2 betwee 0 ad x cos(x) = 1 + 0x x 2 /2 + 0x 3 /3! + (cos(x)) (4) (c 3 )x 4 /4!, c 3 betwee 0 ad x cos(x) = 1 + 0x x 2 /2 + 0x 3 /3! + x 4 /4! + (cos(x)) (5) (c 4 )x 5 /5!,... cos(x) = c 3 betwee 0 ad x ( 1) k x 2k /(2k)! + (cos(x)) (2+1) (c 2+1 )x 2+1 /(2 + 1)!, k=0 c 2+1 betwee 0 ad x

20 MATH 4530: Aalysis Oe Taylor Series Sice the (2) th Taylor Polyomial of cos(x) at 0 is p 2 (x, 0) = k=0 ( 1)k x 2k /(2k)!, this tells us that cos(x) p 2 (x, 0) = (cos(x)) (2+1) (c 2+1 )x 2+1 /(2 + 1)!. The biggest the derivatives of cos(x) ca be here i absolute value is 1. Thus cos(x) p 2 (x, 0) (1) x 2+1 /(2 + 1)! ad we kow lim cos(x) p 2 (x, 0) = lim x 2+1 /(2 + 1)! = 0 Thus, the Taylor polyomials of cos(x) coverge to cos(x) at each x. We would say cos(x) = k=0 ( 1)k x 2k /(2k)! ad call this the Taylor Series of cos(x) at 0 We ca do this for fuctios with easy derivatives.

21 MATH 4530: Aalysis Oe Taylor Series So we ca show at x = 0 (1) si(x) = k=0 ( 1)k x 2k+1 /(2k + 1)! (2) sih(x) = k=0 x 2k+1 /(2k + 1)! (3) cosh(x) = k=0 x 2k /(2k)! (4) e x = f (x) = k=0 x k / It is very hard to do this for fuctios whose higher order derivatives require product ad quotiet rules. So what we kow that we ca do i theory is ot ecessarily what we ca do i practice.

22 MATH 4530: Aalysis Oe Taylor Series Homework Prove the th order Taylor polyomial of e x at x = 0 is f (x) = k=0 x k / ad show the error term goes to zero as Fid the th order Taylor polyomial of si(x) at x = 0 ad show the error term goes to zero as Fid the th order Taylor polyomial of cosh(x) at x = 0 ad show the error term goes to zero as.