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Sincesin x = x −(−1)n 2n+11 3 1 5+ 0 · x 2n+2 + r2n+2 (0; x),x + x − ··· +x3!5!(2n + 1)!where by Lagrange’s formular2n+2 (0; x) =sin(ξ + π2 (2n + 3)) 2n+3,x(2n + 3)!we have, for |x| ≤ 1,|r2n+2 (0; x)| ≤1,(2n + 3)!1but (2n+3)!< 10−3 for n ≥ 2. Thus the approximation sin x ≈ x −required precision on the closed interval |x| ≤ 1.13!+1 55! xhas theExample 12 We shall show that tan x = x + 13 x 3 + o(x 3 ) as x → 0. We havetan x = cos−2 x,tan x = 2 cos−3 x sin x,tan x = 6 cos−4 x sin2 x + 2 cos−2 x.Thus, tan 0 = 0, tan 0 = 1, tan 0 = 0, tan 0 = 2, and the relation now followsfrom the local Taylor formula.1Example 13 Let α > 0. Let us study the convergence of the series ∞n=1 ln cos nα .1For α > 0 we have nα → 0 as n → ∞.
Let us estimate the order of a term of theseries:1 111111= − · 2α + o 2α .ln cos α = ln 1 − · 2α + o 2αn2! n2 nnnThus we have a series of terms of constant sign whose terms are equivalent to1−1those of the series ∞n=1 2n2α . Since this last series converges only for α > 2 , whenα > 0 the original series converges only for α > 12 (see Problem 15b) below).1 41 6Example 14 Let us show that ln cos x = − 12 x 2 − 12x − 45x + O(x 8 ) as x → 0.This time, instead of computing six successive derivatives, we shall use thealready-known expansions of cos x as x → 0 and ln(1 + u) as u → 0:2285Differential Calculus 111ln cos x = ln 1 − x 2 + x 4 − x 6 + O x 8 = ln(1 + u) =2!4!6! 11= u − u2 + u3 + O u4 =23 81 2 1 4 1 6= − x + x − x +O x−2!4!6! 8 81 41 611 61x−+=x−2·+Oxx+Ox−2 (2!)22!4!3(2!)3 111= − x2 − x4 − x6 + O x8 .21245Example 15 Let us find the values of the first six derivatives of the function ln cos xat x = 0.sin xWe have (ln cos) x = −cosx , and it is therefore clear that the function has derivatives of all orders at 0, since cos 0 = 0.
We shall not try to find functional expressionsfor these derivatives, but rather we shall make use of the uniqueness of the Taylorpolynomial and the result of the preceding example.If f (x) = c0 + c1 x + · · · + cn x n + o x n as x → 0,thenck =f (k) (0)k!and f (k) (0) = k!ck .Thus, in the present case we obtain(ln cos)(0) = 0,(ln cos)(3) (0) = 0,(ln cos)(5) (0) = 0,(ln cos) (0) = 0,1(ln cos) (0) = − · 2!,21· 4!,121(ln cos)(6) (0) = − · 6!.45(ln cos)(4) (0) = −Example 16 Let f (x) be an infinitely differentiable function at the point x0 , andsuppose we know the expansionf (x) = c0 + c1 x + · · · + cn x n + O x n+1of its derivative in a neighborhood of zero.
Then, from the uniqueness of the Taylorexpansion we have (k)f(0) = k!ck ,5.3 The Basic Theorems of Differential Calculus229and so f (k+1) (0) = k!ck . Thus for the function f (x) itself we have the expansionf (x) = f (0) +c01!c1 2n!cnx+x + ··· +x n+1 + O x n+2 ,1!2!(n + 1)!or, after simplification,f (x) = f (0) +c0ccx + 1 x 2 + · · · + n x n+1 + O x n+2 .12n+1Example 17 Let us find the Taylor expansion of the function f (x) = arctan x at 0.12 −1 = 1 − x 2 + x 4 − · · · + (−1)n x 2n + O(x 2n+2 ),Since f (x) = 1+x2 = (1 + x )by the considerations explained in the preceding example,11(−1)n 2n+11x+ O x 2n+3 ,f (x) = f (0) + x − x 3 + x 5 − · · · +1352n + 1that is,1(−1)n 2n+11x+ O x 2n+3 .arctan x = x − x 3 + x 5 − · · · +352n + 1Example 18 Similarly, by expanding the function arcsin x = (1 − x 2 )−1/2 by Taylor’s formula in a neighborhood of zero, we find successively,(1 + u)−1/2 = 1 +− 1 (− 1 − 1) 2− 12u+ 2 2u + ··· +1!2!− 12 (− 12 − 1) · · · (− 12 − n + 1) nu + O un+1 ,n!11·3x4 + · · · +(1 − x)−1/2 = 1 + x 2 + 222 · 2!1 · 3 · · · (2n − 1) 2nx + O x 2n+2 ,+n2 · n!1·31 3x + 2x5 + · · · +arcsin x = x +2·32 · 2! · 5(2n − 1)!!+x 2n+1 + O x 2n+3 ,(2n)!!(2n + 1)+or, after elementary transformations,arcsin x = x +1 3 [3!!]2 5[(2n − 1)!!]2 2n+1x +x + ··· +x+ O x 2n+3 .3!5!(2n + 1)!Here (2n − 1)!! := 1 · 3 · · · (2n − 1) and (2n)!! := 2 · 4 · · · (2n).2305Differential CalculusExample 19 We use the results of Examples 5, 12, 17, and 18 and find[x − 13 x 3 + O(x 5 )] − [x −arctan x − sin x= limx→0 tan x − arcsin xx→0 [x + 1 x 3 + O(x 5 )] − [x +3lim= lim− 16 x 3 + O(x 5 )1 36xx→0+ O(x 5 )1 33! x1 33! x+ O(x 5 )]+ O(x 5 )]== −1.5.3.4 Problems and Exercises21.
Choose numbers a and b so that the function f (x) = cos x − 1+axis an in1+bx 2finitesimal of highest possible order as x → 0.x x) ].2. Find limx→∞ x[ 1e − ( x+13. Write a Taylor polynomial of ex at zero that makes it possible to compute thevalues of ex on the closed interval −1 ≤ x ≤ 2 within 10−3 .4. Let f be a function that is infinitely differentiable at 0. Show thata) if f is even, then its Taylor series at 0 contains only even powers of x;b) if f is odd, then its Taylor series at 0 contains only odd powers of x.5.
Show that if f ∈ C (∞) [−1, 1] and f (n) (0) = 0 for n = 0, 1, 2, . . . , and thereexists a number C such that sup−1≤x≤1 |f (n) (x)| ≤ n!C for n ∈ N, then f ≡ 0 on[−1, 1].6. Let f ∈ C (n) (]−1, 1[) and sup−1<x<1 |f (x)| ≤ 1. Let mk (I ) = infx∈I |f (k) (x)|,where I is an interval contained in ]−1, 1[. Show thata) if I is partitioned into three successive intervals I1 , I2 , and I3 and μ is thelength of I2 , thenmk (I ) ≤1mk−1 (I1 ) + mk−1 (I3 ) ;μb) if I has length λ, thenmk (I ) ≤2k(k+1)/2 k k;λkc) there exists a number αn depending only on n such that if |f (0)| ≥ αn , thenthe equation f (n) (x) = 0 has at least n − 1 distinct roots in ]−1, 1[.Hint: In part b) use part a) and mathematical induction; in c) use a) and prove byinduction that there exists a sequence xk1 < xk2 < · · · < xkk of points of the openinterval ]−1, 1[ such that f (k) (xki ) · f (k) (xki+1 ) < 0 for 1 ≤ i ≤ k − 1.7.
Show that if a function f is defined and differentiable on an open interval I and[a, b] ⊂ I , then5.3 The Basic Theorems of Differential Calculus231a) the function f (x) (even if it is not continuous!) assumes on [a, b] all thevalues between f (a) and f (b) (the theorem of Darboux);13b) if f (x) also exists in ]a, b[, then there is a point ξ ∈ ]a, b[ such that f (b) −f (a) = f (ξ )(b − a).8. A function f (x) may be differentiable on the entire real line, without having acontinuous derivative f (x) (see Example 7 in Sect. 5.1.5).a) Show that f (x) can have only discontinuities of second kind.b) Find the flaw in the following “proof” that f (x) is continuous.Proof Let x0 be an arbitrary point on R and f (x0 ) the derivative of f at the pointx0 .
By definition of the derivative and Lagrange’s theoremf (x0 ) = limx→x0f (x) − f (x0 )= lim f (ξ ) = lim f (ξ ),x→x0ξ →x0x − x0where ξ is a point between x0 and x and therefore tends to x0 as x → x0 .9. Let f be twice differentiable on an interval I . Let M0 = supx∈I |f (x)|, M1 =supx∈I |f (x)| and M2 = supx∈I |f (x)|. Show thata) if I = [−a, a], then M0 x 2 + a 2f (x) ≤+M2 ;a2a!√√M1 ≤ 2 M0 M2 , if the length of I is not less than 2 M0 /M2 ,√b)M1 ≤ 2M0 M2 , if I = R;√c) the numbers 2 and 2 in part b) cannot be replaced by smaller numbers;d) if f is differentiable p times on R and the quantities M0 and Mp =supx∈R |f (p) (x)| are finite, then the quantities Mk = supx∈R |f (k) (x)|, 1 ≤ k < p,are also finite and1−k/pMk ≤ 2k(p−k)/2 M0Mρk/p .Hint: Use Exercises 6b) and 9b) and mathematical induction.10.
Show that if a function f has derivatives up to order n + 1 inclusive at a point x0and f (n+1) (x0 ) = 0, then in the Lagrange form of the remainder in Taylor’s formularn (x0 ; x) =1 (n) fx0 + θ (x − x0 ) (x − x0 )n ,n!1where 0 < θ < 1 and the quantity θ = θ (x) tends to n+1as x → x0 .11. Let f be a function that is differentiable n times on an interval I . Prove thefollowing statements.13 G.Darboux (1842–1917) – French mathematician.2325Differential Calculusa) If f vanishes at (n + 1) points of I , there exists a point ξ ∈ I such thatf (n) (ξ ) = 0.b) If x1 , x2 , .
. . , xp are points of the interval I , there exists a unique polynomialL(x) (the Lagrange interpolation polynomial) of degree at most (n − 1) such thatf (xi ) = L(xi ), i = 1, . . . , n. In addition, for x ∈ I there exists a point ξ ∈ I suchthat(x − x1 ) · · · (x − xn ) (n)f (x) − L(x) =f (ξ ).n!c) If x1 < x2 < · · · < xp are points of I and ni , 1 ≤ i ≤ p, are natural numberssuch that n1 + n2 + · · · + np = n and f (k) (xi ) = 0 for 0 ≤ k ≤ ni − 1, then thereexists a point ξ in the closed interval [x1 , xp ] at which f (n−1) (ξ ) = 0.d) There exists a unique polynomial H (x) (the Hermite interpolating polynomial)14 of degree (n − 1) such that f (k) (xi ) = H (k) (xi ) for 0 ≤ k ≤ ni − 1. Moreover, inside the smallest interval containing the points x and xi , i = 1, .
. . , p, thereis a point ξ such thatf (x) = H (x) +(x − x1 )n1 · · · (x − xp )np (n)f (ξ ).n!This formula is called the Hermite interpolation formula. The points xi , i =1, . . . , p, are called interpolation nodes of multiplicity ni respectively. Special casesof the Hermite interpolation formula are the Lagrange interpolation formula, whichis part b) of this exercise, and Taylor’s formula with the Lagrange form of the remainder, which results when p = 1, that is, for interpolation with a single node ofmultiplicity n.12. Show thata) between two real roots of a polynomial P (x) with real coefficients there is aroot of its derivative P (x);b) if the polynomial P (x) has a multiple root, the polynomial P (x) has thesame root, but its multiplicity as a root of P (x) is one less than its multiplicity as aroot of P (x);c) if Q(x) is the greatest common divisor of the polynomials P (x) and P (x),P (x)has the roots ofwhere P (x) is the derivative of P (x), then the polynomial Q(x)P (x) as its roots, all of them being roots of multiplicity 1.13.
Show thata) any polynomial P (x) admits a representation in the form c0 + c1 (x − x0 ) +· · · + cn (x − x0 )n ;b) there exists a unique polynomial of degree n for which f (x) − P (x) = o((x −x0 )n ) as E x → x0 . Here f is a function defined on a set E and x0 is a limit pointof E.14 Ch. Hermite (1822–1901) – French mathematician who studied problems of analysis; in particular, he proved that e is transcendental.5.3 The Basic Theorems of Differential Calculus23314. Using induction on k, 1 ≤ k, we define the finite differences of order k of thefunction f at x0 :Δ1 f (x0 ; h1 ) := Δf (x0 ; h1 ) = f (x0 + h1 ) − f (x0 ),Δ2 f (x0 ; h1 , h2 ) := ΔΔf (x0 ; h1 , h2 ) == f (x0 + h1 + h2 ) − f (x0 + h2 ) −− f (x0 + h1 ) − f (x0 ) == f (x0 + h1 + h2 ) − f (x0 + h1 ) − f (x0 + h2 ) + f (x0 ),...Δk f (x0 ; h1 , . .
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