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Thus, in uniform motions, and only in uniformmotions, is the acceleration equal to zero.But if we wish for a body moving under inertia in empty space to move uniformlyin a straight line when observed in two different coordinate systems, it is necessaryfor the transition formulas from one inertial system to the other to be linear. Thatis the reason why, in Example 15, the linear formulas (5.30) were chosen for thecoordinate transformations.Example 29 (The second derivative of a simple implicit function) Let y = y(t) andx = x(t) be twice-differentiable functions. Assume that the function x = x(t) has adifferentiable inverse function t = t (x).
Then the quantity y(t) can be regarded asan implicit function of x, since y = y(t) = y(t (x)). Let us find the second derivative assuming that x (t) = 0.yxxBy the rule for differentiating such a function, studied in Sect. 5.2.5, we haveyx =yt,xt2105Differential Calculusso thatyyxx( xt )t (y )= yx x = x t = t =xtxtytt xt −yt xtt(xt )2xt=xt ytt − xtt yt.(xt )3We remark that the explicit expressions for all the functions that occur here, , depend on t, but they make it possible to obtain the value of y atincluding yxxxxthe particular point x after substituting for t the value t = t (x) corresponding to thevalue x.For example, if y = et and x = ln t, thenyx =ytet= tet ,=xt1/tyxx=(yx )t et + tet= t (t + 1)et .=xt1/tWe have deliberately chosen this simple example, in which one can explicitlyexpress t in terms of x as t = ex and, by substituting t = ex into y(t) = et , findxthe explicit dependence of y = ee on x.
Differentiating this last function, one canjustify the results obtained above.It is clear that in this way one can find the derivatives of any order by successivelyapplying the formula(n)yx n =(yx(n−1)n−1 )txt.5.2.7 Problems and Exercises1. Let α0 , α1 , . . . , αn be given real numbers. Exhibit a polynomial Pn (x) of degree(k)n having the derivatives Pn (x0 ) = αk , k = 0, 1, . . . , n, at a given point x0 ∈ R.2. Compute f (x) if!exp(− x12 ) for x = 0,a) f (x) =0for x = 0;!12x sin x for x = 0,b) f (x) =0for x = 0.c) Verify that the function in part a) is infinitely differentiable on R, and thatf (n) (0) = 0.d) Show that the derivative in part b) is defined on R but is not a continuousfunction on R.e) Show that the function!11for − 1 < x < 1,exp(− (1+x)2 − (1−x)2 )f (x) =0for 1 ≤ |x|is infinitely differentiable on R.5.3 The Basic Theorems of Differential Calculus2113.
Let f ∈ C (∞) (R). Show that for x = 0 n 11(n) 1n dn−1x f= (−1).fnn+1xdxxx4. Let f be a differentiable function on R. Show thata) if f is an even function, then f is an odd function;b) if f is an odd function, then f is an even function;c) (f is odd) ⇔ (f is even).5.
Show thata) the function f (x) is differentiable at the point x0 if and only if f (x) −f (x0 ) = ϕ(x)(x − x0 ), where ϕ(x) is a function that is continuous at x0 (and inthat case ϕ(x0 ) = f (x0 ));b) if f (x) − f (x0 ) = ϕ(x)(x − x0 ) and ϕ ∈ C (n−1) (U (x0 )), where U (x0 ) is aneighborhood of x0 , then f (x) has a derivative (f (n) (x0 )) of order n at x0 .6. Give an example showing that the assumption that f −1 be continuous at the pointy0 cannot be omitted from Theorem 3.7.
a) Two bodies with masses m1 and m2 respectively are moving in space underthe action of their mutual gravitation alone. Using Newton’s laws (formulas (5.1)and (5.2) of Sect. 5.1), verify that the quantity m1 m21122m1 v1 + m2 v2 + −G=: K + U,E=22rwhere v1 and v2 are the velocities of the bodies and r the distance between them,does not vary during this motion.b) Give a physical interpretation of the quantity E = K + U and its components.c) Extend this result to the case of the motion of n bodies.5.3 The Basic Theorems of Differential Calculus5.3.1 Fermat’s Lemma and Rolle’s TheoremDefinition 1 A point x0 ∈ E ⊂ R is called a local maximum (resp.
local minimum)and the value of a function f : E → R at that point a local maximum value (resp.local minimum value) if there exists a neighborhood UE (x0 ) of x0 in E such that atany point x ∈ UE (x0 ) we have f (x) ≤ f (x0 ) (resp. f (x) ≥ f (x0 )).Definition 2 If the strict inequality f (x) < f (x0 ) (resp. f (x) > f (x0 )) holds atevery point x ∈ UE (x0 )\x0 = ŮE (x0 ), the point x0 is called strict local maximum(resp.
strict local minimum) and the value of the function f : E → R a strict localmaximum value (resp. strict local minimum value).2125Differential CalculusFig. 5.8Definition 3 The local maxima and minima are called local extrema and the valuesof the function at these points local extreme values of the function.Example 1 Let!f (x) =x 2 , if − 1 ≤ x < 2,4, if 2 ≤ x(see Fig. 5.8). For this functionx = −1 is a strict local maximum;x = 0 is a strict local minimum;x = 2 is a local maximum;the points x > 2 are all local extrema, being simultaneously maxima and minima,since the function is locally constant at these points.Example 2 Let f (x) = sin x1 on the set E = R\0.The points x = ( π2 + 2kπ)−1 , k ∈ Z, are strict local maxima, and the pointsx = (− π2 + 2kπ)−1 , k ∈ Z, are strict local minima for f (x) (see Fig.
4.1).Definition 4 An extremum x0 ∈ E of the function f : E → R is called an interior extremum if x0 is a limit point of both sets E− = {x ∈ E | x < x0 } andE+ = {x ∈ E | x > x0 }.In Example 2, all the extrema are interior extrema, while in Example 1 the pointx = −1 is not an interior extremum.Lemma 1 (Fermat) If a function f : E → R is differentiable at an interior extremum, x0 ∈ E, then its derivative at x0 is 0 : f (x0 ) = 0.Proof By definition of differentiability at x0 we havef (x0 + h) − f (x0 ) = f (x0 )h + α(x0 ; h)h,where α(x0 ; h) → 0 as h → x, x0 + h ∈ E.Let us rewrite this relation as follows:)*f (x0 + h) − f (x0 ) = f (x0 ) + α(x0 ; h) h.(5.45)5.3 The Basic Theorems of Differential Calculus213Since x0 is an extremum, the left-hand side of Eq.
(5.45) is either non-negativeor nonpositive for all values of h sufficiently close to 0 and for which x0 + h ∈ E.If f (x0 ) = 0, then for h sufficiently close to 0 the quantity f (x0 ) + α(x0 ; h)would have the same sign as f (x0 ), since α(x0 ; h) → 0 as h → 0, x0 + h ∈ E.But the value of h can be both positive or negative, given that x0 is an interiorextremum.Thus, assuming that f (x0 ) = 0, we find that the right-hand side of (5.45)changes sign when h does (for h sufficiently close to 0), while the left-hand sidecannot change sign when h is sufficiently close to 0. This contradiction completesthe proof.Remarks on Fermat’s Lemma 10 .
Fermat’s lemma thus gives a necessary conditionfor an interior extremum of a differentiable function. For noninterior extrema (suchas the point x = −1 in Example 1) it is generally not true that f (x0 ) = 0.20 . Geometrically this lemma is obvious, since it asserts that at an extremum of adifferentiable function the tangent to its graph is horizontal. (After all, f (x0 ) is thetangent of the angle the tangent line makes with the x-axis.)30 . Physically this lemma means that in motion along a line the velocity must bezero at the instant when the direction reverses (which is an extremum!).This lemma and the theorem on the maximum (or minimum) of a continuousfunction on a closed interval together imply the following proposition.Proposition 1 (Rolle’s10 theorem) If a function f : [a, b] → R is continuous on aclosed interval [a, b] and differentiable on the open interval ]a, b[ and f (a) = f (b),then there exists a point ξ ∈ ]a, b[ such that f (ξ ) = 0.Proof Since the function f is continuous on [a, b], there exist points xm , xM ∈[a, b] at which it assumes its minimal and maximal values respectively.
If f (xm ) =f (xM ), then the function is constant on [a, b]; and since in that case f (x) ≡ 0, theassertion is obviously true. If f (xm ) < f (xM ), then, since f (a) = f (b), one of thepoints xm and xM must lie in the open interval ]a, b[. We denote it by ξ .
Fermat’slemma now implies that f (ξ ) = 0.5.3.2 The Theorems of Lagrange and Cauchy on FiniteIncrementsThe following proposition is one of the most frequently used and important methodsof studying numerical-valued functions.Theorem 1 (Lagrange’s finite-increment theorem) If a function f : [a, b] → R iscontinuous on a closed interval [a, b] and differentiable on the open interval, ]a, b[,10 M.Rolle (1652–1719) – French mathematician.2145Differential CalculusFig. 5.9there exists a point ξ ∈ ]a, b[ such thatf (b) − f (a) = f (ξ )(b − a).(5.46)Proof Consider the auxiliary functionF (x) = f (x) −f (b) − f (a)(x − a),b−awhich is obviously continuous on the closed interval [a, b] and differentiable on theopen interval ]a, b[ and has equal values at the endpoints: F (a) = F (b) = f (a).Applying Rolle’s theorem to F (x), we find a point ξ ∈ ]a, b[ at whichF (ξ ) = f (ξ ) −f (b) − f (a)= 0.b−aRemarks on Lagrange’s Theorem 10 . In geometric language Lagrange’s theoremmeans (see Fig.
5.9) that at some point (ξ, f (ξ )), where ξ ∈ ]a, b[, the tangent tothe graph of the function is parallel to the chord joining the points (a, f (a)) and(a).(b, f (b)), since the slope of the chord equals f (b)−fb−a20 . If x is interpreted as time and f (b) − f (a) as the amount of displacementover the time b − a of a particle moving along a line, Lagrange’s theorem says thatthe velocity f (x) of the particle at some time ξ ∈ ]a, b[ is such that if the particlehad moved with the constant velocity f (ξ ) over the whole time interval, it wouldhave been displaced by the same amount f (b) − f (a). It is natural to call f (ξ ) theaverage velocity over the time interval [a, b].30 . We note nevertheless that for motion that is not along a straight line theremay be no average speed in the sense of Remark 20 .
Indeed, suppose the particle ismoving around a circle of unit radius at constant angular velocity ω = 1. Its law ofmotion, as we know, can be written asr(t) = (cos t, sin t).Thenṙ(t) = v(t) = (− sin t, cos t)#and |v| = sin2 t + cos2 t = 1.5.3 The Basic Theorems of Differential Calculus215The particle is at the same point r(0) = r(2π) = (1, 0) at times t = 0 and t = 2π ,and the equalityr(2π) − r(0) = v(ξ )(2π − 0)would mean that v(ξ ) = 0. But this is impossible.Even so, we shall learn that there is still a relation between the displacement overa time interval and the velocity.
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