MATH 1225 Lecture 5: Sec 2.4 (2)

A useful remark: Since the derivative function is defined through limits, it is easy to show the following result: If f(s) and g(x) are equal on an open interval (a, b), and f?(c) exists for some c E (a, b), then g?(c) also exists and g(c) = f (c). this allows one to calculate the derivative of functions Defined piecewise, except at the x values where the definition changes from one formula to another. For example: if then h (x) = 4x on (- infinity , 5) (because it agrees with (2x^2 + 4)? there) and h?(x) = 2e^2x on (5, infinity) (because it agrees with (e^2x) there). Whether or not h?(5) exists requires more work and we will see how to answer this in the following two questions. (1) Consider the function (a) Show that f(s) is continuous at x = 0 by calculating the left and right hand limits of f(s) as it approaches 0 and showing they are both equal to f(0). (b) Show that f(s) is not differentiable at 0 by calculating and showing the two sided limit does not exist (note: calculate these limits explicitly, do not take shortcuts using derivative formulas). These one - sided limits are called the left and right - sided derivatives of f(x) at 0. By definition, f?(0) exists if and only if the two one - sided derivatives both exists and are equal (note: there is nothing special about 0; one sided derivatives are defined similarity for any s value). (c) Using the Power Rule Theorem from class and the remark preceding this question, calculate f(x) (note that by (b), 0 should not be in the domain of .f?(x)). Sketch a graph of f(s) and f(x) on the same XY - axis (you do not need to be precise, just sketch time shape).

42 Metric Spaces Proof. The proof is somewhat lengthy but straightforward. We subdivide it into four steps (a) to (d). We construct (a) X=(X, d) (b) an isometry T of X onto W, where W Then we prove: (c) completeness of X. (d) uniqueness of X, except for isometries. Roughly speaking, our task will be the assignment of suitable limits to Cauchy sequences in X that do not converge. However, we should not introduce "too many" limits, but take into account that certain se- quences "may want to converge with the same limit" since the terms o those sequences "ultimately come arbitrarily close to each other intuitive ide " This a can be expressed mathematically in terms of a suitable equivalence relation [see (1), below). This is not artificial but is suggested by the process of completion of the rational line mentioned at the beginning of the section. The details of the proof are as follows. X-(X, d). (%) and (x) be (a) Construction of Let Cauchy sequences in Х. Define (4) to be equivalente, to (x), written Let X be the set of all equivalence classes,of Cauchy sequences thus obtained wewrite.(%)eito mean that (x.) is a member of i (a representative of the class , we now set where (t)e i and (y)e y. We show that this limit exists. We have hence we obtain and a similar inequality with m and n interchanged. Together, 1e review of the concept of equivalence, see Al 4 in Appendix 1

1.6 Completion of Metric Spaces 43 Since (%) and (%) are Cauchy, we can make the right side as small as we please. This implies that the limit in (2) exists because Ris complete. We must also show that the limit in (2) is independent of the particular choice of representatives. In fact, if ( )-(x) and (%)-(y"), then by (1), as n-a, which implies the assertion We prove that d in (2) is a metric on X. Obviously, d satisfies (MI) in Sec. 1.1 as well as d(i,i)s0 and (M3). Furthermore, gives (M2), and (M4) for d follows from TU ) by letting n-00, with (b) Construction of an isometry T: X-Wc each bex we associate the class bex which contains the constant Cauchy sequence (b, b,). This defines a mapping T X_W onto the subspace W= TX) ,. The mapping T is given by b-→ b= T, where (b, b,.Jeb. We see that T is an isometry since (2) becomes simply メ here é is the class of (ya) where y- c for all n. Any isometry is injective, and T: X-→ W is surjective since T(X)=w. Hence W and X are isometric; cf. Def. 1.6-1(b). We show that W is dense in X. We consider any iex. Let ()e i. For every e>0 there is an N such that in> N)

Let (쯔스-k h. Then This shows that every s-neighborhood of the arbitrary i e X contains an element of W. Hence W is dense in X. (c) Completeness of X. Let () be any Cauchy sequence in X. Since W is dense in X, for every there is a &ne W such that Pl Hence by the triangle inequality and this is less than any given ε>0 for sufficiently large m and n because (辶) is Cauchy. Hence (2-) is Cauchy. Since T: X-w is isometric and i W, the sequence (z), where zm-Tim, is Cauchy in X. Let te X be the class to which (.) belongs. We show that i is the limit of () By (4) rt Since (m)e i (see right before) and i e W, so that (2 the inequality (5) becomes ei nm to and the right side is smaller than any given e>0 for sufficiently large n. Hence the arbitrary Cauchy sequence (L) in X has the limit ieX, and X is complete.

1.6 Completion of Metric Spaces 45 っn+ワ Fig. 11. Notations in part (d) of the proof of Theorem 1.6-2 (d) Uniqueness of X except for isometries. If (X, d) is another complete metric space with a subspace W dense in X and isometric with X, then for any i, fe X we have sequences ( ), (SJ in W such thatn and -y; hence follows from [the inequality being similar to (3)1. Since W is isometric with We X and W-X, the distances on X and X must be the same. Hence X and X are isometric. We shall see in the next two chapters (in particular in 2.3-2, 3.1- and 3.2-3) that this theorem has basic applications to individual incomplete spaces as well as to whole classes of such spaces.

Page 3 of 5 which gives us y as a function of Theoretically. the integration on the right of t= 1/ - g + alpha f(V)dv can be performed and the result could provide velocity. v,. as a function of t; Then we could perform the integration in y=y0 + V/- g + alpha f(V)dv and substitute v(t) to obtain y as a function of t. Theoretically, that is. Let's look at in example An Example Suppose the air resistance model for a particular object in free fall has where units for v area feet/second Note how f is defined in terms of the sign of v. As usual. downward has been selected as the negative direction so velocities v of an object in free fall will have negative values Thus -v will be positive and the square root makes. sense Similarly if the object is moving upward, v> 0. the value of f(v)) is negative. The object is dropped from a height: of 500 feet with no initial velocity Using g = 32, the integral models from above becomes and where we have used the definition of f(v) for negative values of v since the object is in free fall With a little hard work or the use of a CAS, the first equation can be written as Page 4 of 5 and the second as Now if we could solve the first expression for v in terms of r and substitute the result for v in the expression for y, the result would be very unwieldy to work with. Actually it is impossible to solve the first expression to get a nice closed form for v. Thanks to technology. however, these expressions can still be used to study the object's motion Choose any value of v. say v=-1. Substituting -1 for v in the equations for t and y tells us that t = 0.032 and y = 499.984 so the point (0.032, 499.984) would lie on a graph of distance versus time Similarly, setting v = -50 provides the point (1 839, 452 462) By using a wide range of values for v. we can generate the complete graph of the distance function as a function of time Fortunately, most CAS provides the capability of plotting the points (t(v), y(v)) foe a range of values of v without having to compute each point separately Such a graph appears below. One final note. It is also possible that the integrals representing t and y cannot be integrated into a closed form No problem In this case, simply use the expressions t= 1/ - g + alpha f(V)dV y=y0+ V/ - g + alpha f(V)dV Page 5 of 5 and the numerical integrators provided by a CAS to determine values of t und y Exercises In all of the exercises you will need a CAS or related tool In the example above the graph indicates the object strikes the ground some time between 6 and 7 seconds from release Determine this time to three decimal places In the case where alpha = 1/2 ,f(v) = -v, v0 = 0, y0=100 determine closed form. expressions for v and y as function. Of t. An object is thrown upward at 40 ft/sec from a height of 100 feet The air resistance can be modeled by alpha =0.001 and f(v) = - |v| v Using the trajectory model determine the speed of the object when it strike the ground, determine the time the object strikes the ground after takeoff, construct the graph of the object's height versus time.