Expoetial Fuctios ad Taylor Series James K. Peterso Departmet of Biological Scieces ad Departmet of Mathematical Scieces Clemso Uiversity March 29, 207 Outlie Revistig the Expoetial Fuctio Taylor Series
limk x k /) = 0 for all x. There is a k0 with x /k0. Thus, x k0+ k0 + )! x k0+2 k0 + 2)!. x k0+j k0 + j)! = = x k0 + x k0 x x k0 k0! k0 k0! x x k0 + 2 k0 + ) j x x k0 k0 k0! x k0 k0! ) 2 x x k0 k0 k0! Sice x, this shows limk x k /) 0 as k. k0 Although we do ot have the best tools to aalyze the fuctio Sx) = lim + x ) let s see how far we ca get. We start with this : lim k=0 x coverges for all x. Let f x) = lim k=0 x k ad fx) = these limits exist ad are fiite. k=0 x k. We eed to show We will follow the argumet we used i the earlier s proof with some chages.
N x k x k = k + )! + x k + 2)! +... + x N k k + N k) )! { = x k x k + ) +... x N k } k + )... k + N k) ) { x k x k + ) +... x N k } k + ) N k x k N k r k ) k=0 where r = x k+). We ca the use the same expasio of powers of r as before to fid ) N k+ N x k x /k + ) x k x /k + ) Sice there is a K so x /k + ) /2 for k > K, for all k > K, / x /k + )) 2 ad so lim N { N } x k x k lim 2 N /2)N k+ ) = 2 x k Now pick ay k > K, the we see f x), the limit of a icreasig sequece of terms, is bouded above which implies it coverges to its supremum. N f x ) = + x + x /2! +... + x k / + lim N + x + x /2! +... + x k / + 2 x k. x k
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 x k = 2 x k ɛ The, for k > K, k f x) x k =0 N lim x k N lim N N x k ɛ. k This says limk =0 x k = f x). Next, let s look at Sx) = + x/) ad Sx) = lim Sx). Sx) = + = + ) x k k k = +! x k k)! k ) 2)... k )) = + /)... k )/) x k x k k Cosider /)... k )/) x k x k = /)... k )/) ) x k + /) ) x k
The we have /)... k )/) x k x k x k /) x k = f x ) 0 as. as sice f x ) is fiite, the ratio f x ) goes to zero as. So we kow Sx) = lim Sx) = x k + lim = f x). We just did this argumet. 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. Sx) is cotiuous for all x Fix x0 ad cosider ay x Bx0). Note S x) = + x/) /) = + x/). We see S x) = Sx) + x/). For each, apply the MVT to fid c x betwee x0 ad x so that Sx) Sx0) x x0 = S c x ).
Now lim + c x /) =, so for sufficietly large, + c x / 2. Hece, if sufficietly large + c x /) S c x ) 2 + c x /) Also, if you look at Sx) for x0 x x0 +, + x0 + x + x0 + implyig Sx0 ) Sx) Sx0 + ). Sice these umbers could be egative, we see Scx Dx0) = max{ Sx0 ), Sx0 ) } which we ca choose to be positive. Thus, sice x0 cx x0 +, S c x ) 2Dx0) for sufficietly large. This tells us lim Sx) Sx0) x x0 2Dx0). Lettig, we fid Sx) Sx0) x x0 2Dx0). This shows Sx) is cotiuous at x0 as if ɛ > 0 is chose, if x x0 ɛ/dx0) we have Sx) Sx0) ɛ. Now pick ay x that is irratioal. The let xp) be ay sequece of ratioal umbers which coverges to x. The sice Sx) = limp Sxp), we have Sx) = limp + xp/) = limp e xp. Hece, we ca defie e x = limp e xp ad we have exteded the defiitio of e x off of Q to IR. We coclude ) e x = lim + x/) = k=0 x k 2) e = lim + /) = k=0 3) e x is a cotiuous fuctio of x.
e x ) = e x Let xp) be ay sequece covergig to x0. The, we have Sxp) Sx0) xp x0 = S c x,p ) = Scx,p ) + cx,p /). where cx,p is betwee x0 ad xp implyig limp cx,p = x0. Give ɛ > 0 arbitrary, P p > P x0 ɛ cx,p x0 + ɛ. We the have + x0 ɛ + c,p x + x0 + ɛ = + x0 ɛ ) ) + c,p x + x0 + ɛ ) This also tells us + x0+ɛ + c,p x + x0 ɛ ad so + x0 ɛ ) + x0+ɛ + c,p x + x0 + ɛ ) + c,p x ) + x0 ɛ Now let p ad ad fid e x0 ɛ lim lim p + c,p x ) e x0+ɛ + c,p x
Now, lettig ɛ 0 +, we fid e x0 = lim ɛ 0 ex0 ɛ lim lim + p ) + c,p x + c,p x = lim p lim Sxp) Sx0) xp x0 as e x is a cotiuous fuctio. We coclude lim p lim Sc,p x ) = lim p { Sxp) Sx0) xp x0 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. ) e x+y = e x e y 2) e x = /e x 3) e x e y = e x y = e x /e y ): e x e y = lim + x/) ) lim + y/) ) = lim + x/) + y/)) = lim + x + y)/ + xy/ 2 ) ) = lim + x + y)/ + xy/)/)). Pick a ɛ > 0. The N > N xy / ɛ ad N2 > N2 x + y)/ > 0.
So if > max{n, N2}, we have + x + y)/ ɛ + x + y)/ + xy/ 2 + x + y)/ + ɛ/ ) + x + y)/ ɛ/) + x + y)/ + xy/ 2 ) + x + y)/ + ɛ/ because all terms are positive oce is large eough. Now let to fid Fially, let ɛ 0 + to fid which tells us ) is true. 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 2): e x e x = lim x/) ) lim + x/) ) = lim + x/) x/)) = lim x 2 / 2) = lim x 2 /)/) ) Now argue just like i ). Give ɛ > 0, N > N x 2 / ɛ. Thus, ɛ/ x 2 / 2 + ɛ/ ) x 2 / 2 ɛ/) + ɛ/) ) ) lim ɛ/) lim x 2 / 2 lim + 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.
Let s look at Taylor polyomials for some familiar fuctios. ): cosx) has the followig Taylor polyomials ad error terms at the base poit 0: cosx) = + cosx)) ) c0)x, c0 betwee 0 ad x cosx) = + 0x + cosx)) 2) c)x 2 /2, c betwee 0 ad x cosx) = + 0x x 2 /2 + cosx)) 3) c2)x 3 /3!, c2 betwee 0 ad x cosx) = + 0x x 2 /2 + 0x 3 /3! + cosx)) 4) c3)x 4 /4!, c3 betwee 0 ad x cosx) = + 0x x 2 /2 + 0x 3 /3! + x 4 /4! + cosx)) 5) c4)x 5 /5!,... cosx) = c3 betwee 0 ad x ) k x 2k /2k)! + cosx)) 2+) c2+)x 2+ /2 + )!, k=0 c2+ betwee 0 ad x Sice the 2) th Taylor Polyomial of cosx) at 0 is p2x, 0) = k=0 )k x 2k /2k)!, this tells us that cosx) p2x, 0) = cosx)) 2+) c2+)x 2+ /2 + )!. The biggest the derivatives of cosx) ca be here i absolute value is. Thus cosx) p2x, 0) ) x 2+ /2 + )! ad we kow lim cosx) p2x, 0) = lim x 2+ /2 + )! = 0 Thus, the Taylor polyomials of cosx) coverge to cosx) at each x. We would say cosx) = k=0 )k x 2k /2k)! ad call this the Taylor Series of cosx) at 0 We ca do this for fuctios with easy derivatives.
So we ca show at x = 0 ) six) = k=0 )k x 2k+ /2k + )! 2) sihx) = k=0 x 2k+ /2k + )! 3) coshx) = 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. Homework 28 28. 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. 28.2 Fid the th order Taylor polyomial of six) at x = 0 ad show the error term goes to zero as. 28.3 Fid the th order Taylor polyomial of coshx) at x = 0 ad show the error term goes to zero as.