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It consists of the fact that the full length L of thepath traversed cannot exceed the maximal absolute value of the velocity multipliedby the time interval of the displacement. What has just been said can be written inthe following more precise form:r(b) − r(a) ≤ sup ṙ(t)|b − a|.(5.47)t∈]a,b[As will be shown later, this natural inequality does indeed always hold. It is alsocalled Lagrange’s finite-increment theorem, while relation (5.46), which is validonly for numerical-valued functions, is often called the Lagrange mean-value theorem (the role of the mean in this case is played by both the value f (ξ ) of thevelocity and by the point ξ between a and b).40 .
Lagrange’s theorem is important in that it connects the increment of a function over a finite interval with the derivative of the function on that interval. Up tonow we have not had such a theorem on finite increments and have characterizedonly the local (infinitesimal) increment of a function in terms of the derivative ordifferential at a given point.Corollaries of Lagrange’s TheoremCorollary 1 (Criterion for monotonicity of a function) If the derivative of a functionis nonnegative (resp. positive) at every point of an open interval, then the function isnondecreasing (resp. increasing) on that interval.Proof Indeed, if x1 and x2 are two points of the interval and x1 < x2 , that is, x2 −x1 > 0, then by formula (5.46)f (x2 ) − f (x1 ) = f (ξ )(x2 − x1 ),where x1 < ξ < x2 ,and therefore, the sign of the difference on the left-hand side of this equality is thesame as the sign of f (ξ ).Naturally an analogous assertion can be made about the nonincreasing (resp.decreasing) nature of a function with a nonpositive (resp.
negative) derivative.Remark By the inverse function theorem and Corollary 1 we can conclude, in particular, that if a numerical-valued function f (x) on some interval I has a derivative2165Differential Calculusthat is always positive or always negative, then the function is continuous and monotonic on I and has an inverse function f −1 that is defined on the interval I = f (I )and is differentiable on it.Corollary 2 (Criterion for a function to be constant) A function that is continuouson a closed interval [a, b] is constant on it if and only if its derivative equals zero atevery point of the interval [a, b] (or only the open interval ]a, b[).Proof Only the fact that f (x) ≡ 0 on ]a, b[ implies that f (x1 ) = f (x2 ) for allx1 , x2 , ∈ [a, b] is of interest.
But this follows from Lagrange’s formula, accordingto whichf (x2 ) − f (x1 ) = f (ξ )(x2 − x1 ) = 0,since ξ lies between x1 and x2 , that is, ξ ∈ ]a, b[, and so f (ξ ) = 0.Remark From this we can draw the following conclusion (which as we shall see,is very important for integral calculus): If the derivatives F1 (x) and F2 (x) of twofunctions F1 (x) and F2 (x) are equal on some interval, that is, F1 (x) = F2 (x) onthe interval, then the difference F1 (x) − F2 (x) is constant.The following proposition is a useful generalization of Lagrange’s theorem, andis also based on Rolle’s theorem.Proposition 2 (Cauchy’s finite-increment theorem) Let x = x(t) and y = y(t) befunctions that are continuous on a closed interval [α, β] and differentiable on theopen interval ]α, β[.Then there exists a point τ ∈ [α, β] such thatx (τ ) y(β) − y(α) = y (τ ) x(β) − x(α) .If in addition x (t) = 0 for each t ∈ ]α, β[, then x(α) = x(β) and we have the equalityy(β) − y(α) y (τ )=.x(β) − x(α) x (τ )(5.48)Proof The function F (t) = x(t)(y(β) − y(α)) − y(t)(x(β) − x(α)) satisfies thehypotheses of Rolle’s theorem on the closed interval [α, β].
Therefore there existsa point τ ∈ ]α, β[ at which F (τ ) = 0, which is equivalent to the equality to beproved. To obtain relation (5.48) from it, it remains only to observe that if x (t) = 0on ]α, β[, then x(α) = x(β), again by Rolle’s theorem.Remarks on Cauchy’s Theorem 10 . If we regard the pair x(t), y(t) as the lawof motion of a particle, then (x (t), y (t)) is its velocity vector at time t, and(x(β) − x(α), y(β) − y(α)) is its displacement vector over the time interval [α, β].The theorem then asserts that at some instant of time τ ∈ [α, β] these two vectors5.3 The Basic Theorems of Differential Calculus217are collinear. However, this fact, which applies to motion in a plane, is the same kindof pleasant exception as the mean-velocity theorem in the case of motion along aline.
Indeed, imagine a particle moving at uniform speed along a helix. Its velocitymakes a constant nonzero angle with the vertical, while the displacement vector canbe purely vertical (after one complete turn).20 . Lagrange’s formula can be obtained from Cauchy’s by setting x = x(t) = t,y(t) = y(x) = f (x), α = a, β = b.5.3.3 Taylor’s FormulaFrom the amount of differential calculus that has been explained up to this pointone may obtain the correct impression that the more derivatives of two functionscoincide (including the derivative of zeroth order) at a point, the better these functions approximate each other in a neighborhood of that point.
We have mostly beeninterested in approximations of a function in the neighborhood of a point by a polynomial Pn (x) = Pn (x0 ; x) = c0 + c1 (x − x0 ) + · · · + cn (x − x0 )n , and that willcontinue to be our main interest. We know (see Example 25 in Sect. 5.2.6) that analgebraic polynomial can be represented asPn (x0 )Pn (x0 )(x − x0 ) + · · · +(x − x0 )n ,1!n!(n)Pn (x) = Pn (x0 ) +(k)that is, ck = Pn k!(x0 ) (k = 0, 1, . . .
, n). This can easily be verified directly.Thus, if we are given a function f (x) having derivatives up to order n inclusiveat x0 , we can immediately write the polynomialPn (x0 ; x) = Pn (x) = f (x0 ) +f (x0 )f (n) (x0 )(x − x0 ) + · · · +(x − x0 )n , (5.49)1!n!whose derivatives up to order n inclusive at the point x0 are the same as the corresponding derivatives of f (x) at that point.Definition 5 The algebraic polynomial given by (5.49) is the Taylor11 polynomialof order n of f (x) at x0 .We shall be interested in the value off (x) − Pn (x0 ; x) = rn (x0 ; x)(5.50)of the discrepancy between the polynomial Pn (x) and the function f (x), which isoften called the remainder, more precisely, the nth remainder or the nth remainder11 B.Taylor (1685–1731) – British mathematician.2185Differential Calculusterm in Taylor’s formula:f (x) = f (x0 ) +f (x0 )f (n) (x0 )(x − x0 ) + · · · +(x − x0 )n + rn (x0 ; x).
(5.51)1!n!The equality (5.51) itself is of course of no interest if we know nothing moreabout the function rn (x0 ; x) than its definition (5.50).We shall now use a highly artificial device to obtain information on the remainderterm. A more natural route to this information will come from the integral calculus.Theorem 2 If the function f is continuous on the closed interval with end-pointsx0 and x along with its first n derivatives, and it has a derivative of order n + 1at the interior points of this interval, then for any function ϕ that is continuous onthis closed interval and has a nonzero derivative at its interior points, there exists apoint ξ between x0 and x such thatrn (x0 ; x) =ϕ(x) − ϕ(x0 ) (n+1)f(ξ )(x − ξ )n .ϕ (ξ )n!(5.52)Proof On the closed interval I with endpoints x0 and x we consider the auxiliaryfunctionF (t) = f (x) − Pn (t; x)(5.53)of the argument t. We now write out the definition of the function F (t) in moredetail:%$f (n) (t)f (t)n(5.54)(x − t) + · · · +(x − t) .F (t) = f (x) − f (t) +1!n!We see from the definition of the function F (t) and the hypotheses of the theoremthat F is continuous on the closed interval I and differentiable at its interior points,and that$f (t)f (t) f (t)F (t) = − f (t) −+(x − t) −(x − t) +1!1!1!%f (n+1) (t)f (t)f (n+1) (t)+(x − t)2 − · · · +(x − t)n = −(x − t)n .2!n!n!Applying Cauchy’s theorem to the pair of functions F (t), ϕ(t) on the closedinterval I (see relation (5.48)), we find a point ξ between x0 and x at whichF (x) − F (x0 ) F (ξ )= .ϕ(x) − ϕ(x0 )ϕ (ξ )Substituting the expression for F (ξ ) here and observing from comparison offormulas (5.50), (5.53) and (5.54) that F (x) − F (x0 ) = 0 − F (x0 ) = −rn (x0 ; x),we obtain formula (5.52).5.3 The Basic Theorems of Differential Calculus219Setting ϕ(t) = x − t in (5.52), we obtain the following corollary.Corollary 1 (Cauchy’s formula for the remainder term)rn (x0 ; x) =1 (n+1)f(ξ )(x − ξ )n (x − x0 ).n!(5.55)A particularly elegant formula results if we set ϕ(t) = (x − t)n+1 in (5.52):Corollary 2 (The Lagrange form of the remainder)rn (x0 ; x) =1f (n+1) (ξ )(x − x0 )n+1 .(n + 1)!(5.56)We remark that when x0 = 0 Taylor’s formula (5.51) is often called MacLaurin’sformula.12Let us consider some examples.Example 3 For the function f (x) = ex with x0 = 0 Taylor’s formula has the formex = 1 +111x + x 2 + · · · + x n + rn (0; x),1!2!n!(5.57)and by (5.56) we can assume thatrn (0; x) =1eξ · x n+1 ,(n + 1)!where |ξ | < |x|.Thusrn (0; x) =1|x|n+1 |x|eξ · |x|n+1 <e .(n + 1)!(n + 1)!(5.58)|x|But for each fixed x ∈ R, if n → ∞, the quantity (n+1)!, as we know (see Example 12 of Sect.
3.1.3), tends to zero. Hence it follows from the estimate (5.58) andthe definition of the sum of a series thatn+1ex = 1 +111x + x2 + · · · + xn + · · ·1!2!n!for all x ∈ R.12 C.MacLaurin (1698–1746) – British mathematician.(5.59)2205Differential CalculusExample 4 We obtain the expansion of the function a x for any a, 0 < a, a = 1,similarly:ax = 1 +ln2 a 2ln alnn a nx+x + ··· +x + ··· .1!2!n!Example 5 Let f (x) = sin x. We know (see Example 18 of Sect.
5.2.6) thatf (n) (x) = sin(x + π2 n), n ∈ N, and so by Lagrange’s formula (5.56) with x0 = 0and any x ∈ R we findrn (0; x) =π1sin ξ + (n + 1) x n+1 ,(n + 1)!2(5.60)from which it follows that rn (0; x) tends to zero for any x ∈ R as n → ∞. Thus wehave the expansionsin x = x −(−1)n 2n+11 3 1 5x + x − ··· +x+ ···3!5!(2n + 1)!(5.61)for every x ∈ R.Example 6 Similarly, for the function f (x) = cos x, we obtainπ1cos ξ + (n + 1) x n+1rn (0; x) =(n + 1)!2(5.62)andcos x = 1 −(−1)n 2n1 2 1 4x + x − ··· +x + ···2!4!(2n)!(5.63)for x ∈ R.Example 7 Since sinh x = cosh x and cosh x = sinh x, formula (5.56) yields thefollowing expression for the remainder in the Taylor series of f (x) = sinh x:rn (0; x) =1f (n+1) (ξ )x n+1 ,(n + 1)!where f (n+1) (ξ ) = sinh ξ if n is even and f (n+1) (ξ ) = cosh ξ if n is odd.
In any case|f (n+1) (ξ )| ≤ max{| sinh x|, | cosh x|}, since |ξ | < |x|. Hence for any given valuex ∈ R we have rn (0, x) → 0 as n → ∞, and we obtain the expansionsinh x = x +valid for all x ∈ R.1 3 1 51x + x + ··· +x 2n+1 + · · · ,3!5!(2n + 1)!(5.64)5.3 The Basic Theorems of Differential Calculus221Example 8 Similarly we obtain the expansioncosh x = 1 +1 2 1 41 2nx + x + ··· +x + ··· ,2!4!(2n)!(5.65)valid for any x ∈ R.Example 9 For the function f (x) = ln(1 + x) we have f (n) (x) =that the Taylor series of this function at x0 = 0 is(−1)n−1 (n−1)!,(1+x)n1(−1)n−1 n1x + rn (0; x).ln(1 + x) = x − x 2 + x 3 − · · · +23nso(5.66)This time we represent rn (0; x) using Cauchy’s formula (5.55):rn (0; x) =1 (−1)n n!(x − ξ )n x,n! (1 + ξ )norx −ξrn (0; x) = (−1) x1+ξnn,where ξ lies between 0 and x.If |x| < 1, it follows from the condition that ξ lies between 0 and x that x − ξ |x| − |ξ | |x| − |ξ |1 − |x|1 − |x| 1 + ξ = |1 + ξ | ≤ 1 − |ξ | = 1 − 1 − |ξ | ≤ 1 − 1 − |0| = |x|.Thus for |x| < 1rn (0; x) ≤ |x|n+1 ,(5.67)(5.68)(5.69)and consequently the following expansion is valid for |x| < 1:11(−1)n−1 nx + ··· .ln(1 + x) = x − x 2 + x 3 − · · · +23n(5.70)We remark that outside the closed interval |x| ≤ 1 the series on the right-handside of (5.70) diverges at every point, since its general term does not tend to zero if|x| > 1.Example 10 For the function (1+x)a , where α ∈ R, we have f (n) (x) = α(α −1) · · ·(α − n + 1)(1 + x)α−n , so that Taylor’s formula at x0 = 0 for this function has theformα(α − 1) 2αx+x + ··· +1!2!α(α − 1) · · · (α − n + 1) nx + rn (0; x).+n!(1 + x)α = 1 +(5.71)2225Differential CalculusUsing Cauchy’s formula (5.55), we findα(α − 1) · · · (α − n)(1 + ξ )α−n−1 (x − ξ )n x,n!where ξ lies between 0 and x.If |x| < 1, then, using the estimate (5.68), we have rn (0; x) ≤ α 1 − α · · · 1 − α (1 + ξ )α−1 |x|n+1 .1n rn (0; x) =(5.72)(5.73)When n is increased by 1, the right-hand side of Eq.
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