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. , hk ) := Δk−1 gk (x0 ; h1 , . . . , hk−1 ),where gk (x) = Δ1 f (x; hk ) = f (x + hk ) − f (x).a) Let f ∈ C (n−1) [a, b] and suppose that f (n) (x) exists at least in the open interval ]a, b[. Show that if all the points x0 , x0 + h1 , x0 + h2 , x0 + h1 + h2 , . . . , x0 +h1 + · · · + hn lie in [a, b], then inside the smallest closed interval containing all ofthem there is a point ξ such thatΔn f (x0 ; h1 , . . . , hn ) = f (n) (ξ )h1 · · · hn .b) (Continuation.) If f (n) (x0 ) exists, then the following estimate holds: nΔ f (x0 ; h1 , . . . , hn ) − f (n) (x0 )h1 · · · hn ≤≤ sup f (n) (x) − f (n) (x0 ) · |h1 | · · · |hn |.x∈]a,b[c) (Continuation.) Set Δn f (x0 ; h, . .

. , h) =: Δn f (x0 ; hn ). Show that if f (n) (x0 )exists, thenΔn f (x0 ; hn ).h→0hnf (n) (x0 ) = limd) Show by example that the preceding limit may exist even when f (n) (x0 ) doesnot exist.Hint: Consider, for example, Δ2 f (0; h2 ) for the function!x 3 sin x1 , x = 0,f (x) =0,x = 0,and show thatΔ2 f (0; h2 )= 0.h→0h2lim2345Differential Calculus15. a) Applying Lagrange’s theorem to the function x1α , where α > 0, show thatthe inequality1111<−α (n − 1)α nαn1+αholds for n ∈ N and α > 0.b) Use the result of a) to show that the series ∞n=11nσconverges for σ > 1.5.4 The Study of Functions Using the Methods of DifferentialCalculus5.4.1 Conditions for a Function to be MonotonicProposition 1 The following relations hold between the monotonicity properties ofa function f : E → R that is differentiable on an open interval ]a, b[ = E and thesign (positivity) of its derivative f on that interval:f (x) > 0 ⇒f is increasing⇒ f (x) ≥ 0,f (x) ≥ 0 ⇒ f is nondecreasing ⇒ f (x) ≥ 0,f (x) ≡ 0 ⇒f ≡ const.⇒ f (x) ≡ 0,f (x) ≤ 0 ⇒f is nonincreasing⇒ f (x) ≤ 0,f (x) < 0 ⇒f is decreasing⇒ f (x) ≤ 0.Proof The left-hand column of implications is already known to us from Lagrange’stheorem, by virtue of which f (x2 ) − f (x1 ) = f (ξ )(x2 − x1 ), where x1 , x2 ∈ ]a, b[and ξ is a point between x1 and x2 .

It can be seen from this formula that for x1 < x2the difference f (x2 ) − f (x1 ) is positive if and only if f (ξ ) is positive.The right-hand column of implications can be obtained immediately from thedefinition of the derivative. Let us show, for example, that if a function f that isdifferentiable on ]a, b[ is increasing, then f (x) ≥ 0 on ]a, b[. Indeed,f (x + h) − f (x).h→0hf (x) = limIf h > 0, then f (x + h) − f (x) > 0; and if h < 0, then f (x + h) − f (x) < 0.Therefore the fraction after the limit sign is positive.Consequently, its limit f (x) is nonnegative, as asserted.Remark 1 It is clear from the example of the function f (x) = x 3 that a strictlyincreasing function has a nonnegative derivative, not necessarily one that is alwayspositive.

In this example, f (0) = 3x 2 |x=0 = 0.5.4 Differential Calculus Used to Study Functions235Remark 2 In the expression A ⇒ B, as we noted at the appropriate point, A is asufficient condition for B and B a necessary condition for A. Hence, one can makethe following inferences from Proposition 1.A function is constant on an open interval if and only if its derivative is identicallyzero on that interval.A sufficient condition for a function that is differentiable on an open interval tobe decreasing on that interval is that its derivative be negative at every point of theinterval.A necessary condition for a function that is differentiable on an open interval tobe nonincreasing on that interval is that its derivative be nonpositive on the interval.Example 1 Let f (x) = x 3 − 3x + 2 on R. Then f (x) = 3x 2 − 3 = 3(x 2 − 1), andsince f (x) < 0 for |x| < 1 and f (x) > 0 for |x| > 1, we can say that the functionis increasing on the open interval ]−∞, −1[, decreasing on ]−1, 1[, and increasingagain on ]1, +∞[.5.4.2 Conditions for an Interior Extremum of a FunctionTaking account of Fermat’s lemma (Lemma 1 of Sect.

5.3), we can state the following proposition.Proposition 2 (Necessary conditions for an interior extremum) In order for a pointx0 to be an extremum of a function f : U (x0 ) → R defined on a neighborhood U (x0 )of that point, a necessary condition is that one of the following two conditions hold:either the function is not differentiable at x0 or f (x0 ) = 0.Simple examples show that these necessary conditions are not sufficient.Example 2 Let f (x) = x 3 on R.

Then f (0) = 0, but there is no extremum atx0 = 0.Example 3 Let!f (x) =x2xfor x > 0,for x < 0.This function has a bend at 0 and obviously has neither a derivative nor an extremum at 0.Example 4 Let us find the maximum of f (x) = x 2 on the closed interval [−2, 1].It is obvious in this case that the maximum will be attained at the endpoint −2, buthere is a systematic procedure for finding the maximum.

We find f (x) = 2x, thenwe find all points of the open interval ]−2, 1[ at which f (x) = 0. In this case, theonly such point is x = 0. The maximum of f (x) must be either among the points2365Differential Calculuswhere f (x) = 0, or at one of the endpoints, about which Proposition 2 is silent.Thus we need to compare f (−2) = 4, f (0) = 0, and f (1) = 1, from which weconclude that the maximal value of f (x) = x 2 on the closed interval [−2, 1] equals4 and is assumed at −2, which is an endpoint of the interval.Using the connection established in Sect.

5.4.1 between the sign of the derivative and the nature of the monotonicity of the function, we arrive at the followingsufficient conditions for the presence or absence of a local extremum at a point.Proposition 3 (Sufficient conditions for an extremum in terms of the first derivative)Let f : U (x0 ) → R be a function defined on a neighborhood U (x0 ) of the point x0 ,which is continuous at the point itself and differentiable in a deleted neighborhoodŮ (x0 ). Let Ů − (x0 ) = {x ∈ U (x0 ) | x < x0 } and Ů + (x0 ) = {x ∈ U (x0 ) | x > x0 }.Then the following conclusions are valid:a) (∀x ∈ Ů − (x0 ) (f (x) < 0)) ∧ (∀x ∈ Ů + (x0 ) (f (x) < 0)) ⇒ (f has noextremum at x0 );b) (∀x ∈ Ů − (x0 ) (f (x) < 0)) ∧ (∀x ∈ Ů + (x0 ) (f (x) > 0)) ⇒ (x0 is a strictlocal minimum of f );c) (∀x ∈ Ů − (x0 ) (f (x) > 0)) ∧ (∀x ∈ Ů + (x0 ) (f (x) < 0)) ⇒ (x0 is a strictlocal maximum of f );d) (∀x ∈ Ů − (x0 ) (f (x) > 0)) ∧ (∀x ∈ Ů + (x0 ) (f (x) > 0)) ⇒ (f has noextremum at x0 ).Briefly, but less precisely, one can say that if the derivative changes sign in passing through the point, then the point is an extremum, while if the derivative does notchange sign, the point is not an extremum.We remark immediately, however, that these sufficient conditions are not necessary for an extremum, as one can verify using the following example.Example 5 Let!f (x) =2x 2 + x 2 sin x1for x = 0,0for x = 0.Since x 2 ≤ f (x) ≤ 3x 2 , it is clear that the function has a strict local minimumat x0 = 0, but the derivative f (x) = 4x + 2x sin x1 − cos x1 is not of constant signin any deleted one-sided neighborhood of this point.

This same example shows themisunderstandings that can arise in connection with the abbreviated statement ofProposition 3 just given.We now turn to the proof of Proposition 3.Proof a) It follows from Proposition 2 that f is strictly decreasing on Ů − (x0 ). Sinceit is continuous at x0 , we have limŮ − (x0 )x→x0 f (x) = f (x0 ), and consequentlyf (x) > f (x0 ) for x ∈ Ů − (x0 ). By the same considerations we have f (x0 ) > f (x)5.4 Differential Calculus Used to Study Functions237for x ∈ Ů + (x0 ). Thus the function is strictly decreasing in the whole neighborhoodU (x0 ) and x0 is not an extremum.b) We conclude to begin with, as in a), that since f (x) is decreasing on Ů − (x0 )and continuous at x0 , we have f (x) > f (x0 ) for x ∈ Ů − (x0 ). We conclude from theincreasing nature of f on Ů + (x0 ) that f (x0 ) < f (x) for x ∈ Ů + (x0 ).

Thus f has astrict local minimum at x0 .Statements c) and d) are proved similarly.Proposition 4 (Sufficient conditions for an extremum in terms of higher-orderderivatives) Suppose a function f : U (x0 ) → R defined on a neighborhood U (x0 )of x0 has derivatives of order up to n inclusive at x0 (n ≥ 1).If f (x0 ) = · · · = f (n−1) (x0 ) = 0 and f (n) (x0 ) = 0, then there is no extremum atx0 if n is odd. If n is even, the point x0 is a local extremum, in fact a strict localminimum if f (n) (x0 ) > 0 and a strict local maximum if f (n) (x0 ) < 0.Proof Using the local Taylor formulaf (x) − f (x0 ) =1 (n)f (x0 )(x − x0 )n + α(x)(x − x0 )n ,n!(5.82)where α(x) → 0 as x → x0 , we shall reason as in the proof of Fermat’s lemma.

Werewrite Eq. (5.82) as1 (n)f (x0 ) + α(x) (x − x0 )n .f (x) − f (x0 ) =(5.83)n!Since f (n) (x0 ) = 0 and α(x) → 0 as x → x0 , the sum f (n) (x0 ) + α(x) has thesign of f (n) (x0 ) when x is sufficiently close to x0 . If n is odd, the factor (x − x0 )nchanges sign when x passes through x0 , and then the sign of the right-hand side ofEq. (5.83) also changes sign. Consequently, the left-hand side changes sign as well,and so for n = 2k + 1 there is no extremum.If n is even, then (x − x0 )n > 0 for x = x0 and hence in some small neighborhoodof x0 the sign of the difference f (x) − f (x0 ) is the same as the sign of f (n) (x0 ), asis clear from Eq.

(5.83).Let us now consider some examples.Example 6 (The law of refraction in geometric optics (Snell’s law)15 ) According toFermat’s principle, the actual trajectory of a light ray between two points is suchthat the ray requires minimum time to pass from one point to the other comparedwith all paths joining the two points.It follows from Fermat’s principle and the fact that the shortest path between twopoints is a straight line segment having the points as endpoints that in a homogeneous and isotropic medium (having identical structure at each point and in eachdirection) light propagates in straight lines.15 W.Snell (1580–1626) – Dutch astronomer and mathematician.2385Differential CalculusFig.

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