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256 Chapter 8 • Applications of Integrals §8.1 Example 8.7 y = ex2 y 1 2 3 x 0 1 2 3 x2 + y2 = 9 y = 2cosx2 Use Monte Carlo integration to approximate the area of the region in the first quadrant above the curves y = ex2 and y = 2 cos x2, and inside the circle x2 + y2 = 9. Solution: The region is the shaded area shown in the figure on the right, enclosed in a rectangle of width 2 and height 3. The area of the rectangle is 6, and a point (x, y) in the rectangle is inside the region if only if the following conditions are met: y > ex2 and y > 2 cos x2 and x2 + y2 < 9 Notice that these conditions are all in the form of inequalities. The Monte Carlo integration is then simple to perform in Octave, using 10 million random points: octave> N = 1e7; octave> x = 2*rand(1,N); octave> y = 3*rand(1,N); octave> 6*(sum(y > exp(x.^2) & y > 2*cos(x.^2) & x.^2 + y.^2 < 9))/N ans = 0.94612 The actual value—accurate to 5 decimal places—is 0.94606. The rand(1,N) command returns an array of N random numbers between 0 and 1. So the statement x = 2*rand(1,N) stores N random numbers between 0 and 2 in an array for the x-coordinates, and the statement y = 3*rand(1,N) stores N random numbers between 0 and 3 in an array for the y-coordinates. The statement y > exp(x.^2) returns a value of 1 if the condition y > ex2 is met, 0 otherwise. Similarly for the statements y > 2*cos(x.^2) and x.^2 + y.^2 < 9. Joining those three statements with & sym- bols returns a value of 1 if all three conditions are met, 0 otherwise. The sum command thus counts how many of the N points are inside the region. Dividing that count by N gives the ratio of points inside the region, then multiplying by 6 (the area of the rectangle) gives the approximate area of the region. Note that the size of the rectangle can affect the approximation—generally the larger the rectangle the more points must be used. Note also in this example that finding the area by using definite integrals would require numerical integration methods, since f (x) = ex2 and f (x) = 2 cos x2 cannot be integrated in a closed form. In fact, even finding the points of intersection of the three curves—in order to split the integrals—would require a numerical root-finding method (e.g. Newton’s method). Exercises A For Exercises 1-6, find the area of the region bounded by the given curves. 1. y = x2 and y = 2x+3 2. x = −y2 +2y and x = 0 3. y = x2 −1 and y = x3 −1 4. y = x4 and y = x 5. x = y2 and x = y+2 6. y = 4−4x2 and y = 1−x2 7. Find the area between y = 4x−x2 and y = x over [0,4]. 8. Find the area between y = cosh x and y = sinh x over [0,∞). 9. Find the area of the region defined by the inequalities 0 ≤x ≤y−x ≤1−y ≤1. | ElementaryCalculus_Page_266_Chunk3801 |
Area Between Curves • Section 8.1 257 10. Find the area between r = 1+cos θ and r = 2+2cos θ. 11. Find the area between r = 1+cos θ and r = 2+cos θ. B x2 −y2 = 1 y 2 x 1 2 x2 + y2 = 4 O P 12. Find the area of the region in the first quadrant between the unit hyperbola x2 −y2 = 1 and the circle x2 + y2 = 4 (i.e. the shaded region shown in the figure on the right) in two ways: (a) Integration using formula (8.1). (b) Draw a line segment from the origin O to the point of intersection P on the hyperbola, then use the areas of the resulting circular sector and hyperbolic sector, without resorting to integration. Does your answer agree with part (a)? Explain. 13. Find the area common to the four circles of radius 5 shown in Figure 8.1.4. (Hint: Use symmetry.) 5 5 5 5 Figure 8.1.4 Exercise 13 y = mx+ b y = ax2 A (a) Area A B C D T y = mx+ b y = ax2 (b) Area T Figure 8.1.5 Exercise 14 14. Let A be the area of the region bounded by the parabola y = ax2 and the line y = mx + b, where a, m, and b are positive constants (see Figure 8.1.5(a)). Let T be the area of the triangle △BCD, where B and C are where the line intersects the parabola, and the point D on the parabola has the same x-coordinate as the midpoint of the line segment BC (see Figure 8.1.5(b)). Show that A = 4 3T. 15. In Example 8.7 the region was enclosed in the rectangle R = { (x, y) : 0 ≤x ≤2 , 0 ≤y ≤3 }. Use Monte Carlo integration to approximate the area of the region again, using a different enclosing rectangle R: (a) R = { (x, y) : 0 ≤x ≤2 , 1 ≤y ≤3 } (b) R = { (x, y) : 0 ≤x ≤3 , 0 ≤y ≤3 } Are the results significantly different than before? 16. Approximate the area of the region bounded by y = x2 and y = cos x in two different ways: (a) Use Monte Carlo integration with 10 million points. (b) Use a numerical root-finding method to find the points of intersection of the curves, then use those points in a numerical integration method to find the area. C 17. A dog is chained to a fixed point at the circular base of a cylindrical silo. The silo’s radius is 50 ft and the chain can wrap exactly halfway around the silo. How much total area can the dog roam, not counting the area inside the silo? | ElementaryCalculus_Page_267_Chunk3802 |
258 Chapter 8 • Applications of Integrals §8.2 8.2 Average Value of a Function planet d ✸ Sun Figure 8.2.1 According to Kepler’s laws of planetary motion, a planet revolv- ing around the Sun follows an elliptical orbit, with the Sun at one focus of the ellipse, as in Figure 8.2.1. The distance d between the planet and the Sun varies over the ellipse, reaching a min- imum distance and a maximum distance (by the Extreme Value Theorem). How would you find the average distance between the planet and the Sun over one complete orbit? The idea is to gener- alize the usual notion of an average of numbers. Recall that for n numbers x1, x2, ... , xn the average, denoted by ¯x, is simply the sum of the numbers divided by how many numbers there are, namely ¯x = x1 + x2 + ··· + xn n . In statistics ¯x is called the mean of x1, x2, ... , xn. This definition makes sense for a finite set of numbers, but in the case of a planet revolving around the Sun, there are an uncountably infinite number of distances between the planet and the Sun, making the above definition impossible to use. A way of taking a sum over an infinite continuum of values is needed instead. Such a method has already been encountered: the definite integral, which is merely a sum of a continuum of infinitesimal quantities. f (x0) f (x1) f (x2) f (xn−1) f (xn) y x a = x0 x1 x2 ... xn−1 xn = b ... Figure 8.2.2 ∆xi = xi −xi−1 = (b−a)/n To motivate the definition of the average value of a function f over a closed interval [a,b], denoted by 〈f 〉, consider a partition P = {a = x0 < x1 < x2 < ··· < xn−1 < xn = b} that divides [a,b] into n subintervals [xi−1, xi] of equal length ∆xi = xi −xi−1 = (b −a)/n, as in Fig- ure 8.2.2. The n function values f (x1), f (x2), ... , f (xn) constitute only a finite subset of all the func- tion values f (x) over [a,b], so their average would be an approximation of the true function average 〈f 〉, namely: 〈f 〉≈f (x1) + f (x2) + ··· + f (xn) n = nX i=1 f (xi) n By properties of summations, divide the entire sum by the constant b −a and multiply each term in the sum by b −a to get: | ElementaryCalculus_Page_268_Chunk3803 |
Average Value of a Function • Section 8.2 259 〈f 〉≈ 1 b −a nX i=1 f (xi) · b −a n = 1 b −a n X i=1 f (xi)∆xi Note that the last summation on the right is just a Riemann sum for the definite integral Rb a f (x)dx, with the points x∗ i chosen to be the right endpoints of the intervals [xi−1, xi] for i = 1 to n. Thus, taking the limit of that sum as n →∞(which means including more and more function values in the average) yields the following definition: The average value 〈f 〉of a function f over a closed interval [a,b] is:1 〈f 〉= 1 b −a Zb a f (x) dx (8.3) Example 8.8 Find the average value of f (x) = x2 over [0,1]. Solution: By definition, with a = 0 and b = 1, 〈f 〉= 1 1−0 Z1 0 f (x) dx = Z1 0 x2 dx = x3 3 ¯¯¯¯ 1 0 = 13 3 −03 3 = 1 3 Note that this says that if you took all the numbers between 0 and 1 and squared them, then the average of those squares would be 1/3. Example 8.9 Find the average value of f (x) = x2 over [−1,1]. Solution: By definition, with a = −1 and b = 1, 〈f 〉= 1 1−(−1) Z1 −1 f (x) dx = 1 2 Z1 −1 x2 dx = x3 6 ¯¯¯¯ 1 −1 = 13 6 −(−1)3 6 = 1 3 Note that this is the same as the average over [0,1], as shown in the previous example. This should make sense, since the function f (x) = x2 is symmetric about the y-axis, so the values of f (x) from [−1,0] are the same as those from [0,1]. The values from [−1,1] just duplicate the values from [0,1] and hence do not change the average. Example 8.10 Find the average value of f (x) = sin x over [0,π]. Solution: By definition, with a = 0 and b = π, 〈f 〉= 1 π−0 Zπ 0 f (x) dx = 1 π Zπ 0 sin x dx = −1 π cos x ¯¯¯¯ π 0 = −1 π (cos π −cos 0) = −1 π (−1−1) = 2 π 1In some statistics or mathematics texts you might see the notation ¯f for the average value. | ElementaryCalculus_Page_269_Chunk3804 |
260 Chapter 8 • Applications of Integrals §8.2 Example 8.11 x y 0 5 −5 3 −3 (x, y) (4,0) d Figure 8.2.3 x2 25 + y2 9 = 1 Find the average distance from the ellipse x2 25+ y2 9 = 1 to the point (4,0). Solution: Let d represent the distance from any point (x, y) on the ellipse to the point (4,0), as in Figure 8.2.3. If (x, y) is on the ellipse x2 25 + y2 9 = 1 then y2 = 9(1−x2 25) = 9 25(25−x2). So by the distance formula, d is given by d2 = (x−4)2 + (y−0)2 = (x−4)2 + y2 = (x−4)2 + 9 25(25−x2) = 25(x−4)2 + 9(25−x)2 25 = 25x2 −200x + 400 + 225 −9x2 25 = 16x2 −200x + 625 25 d2 = (4x−25)2 25 , and so taking square roots gives d = ± 4x−25 5 = −4x−25 5 = 25−4x 5 for −5 ≤x ≤5, since d = (4x −25)/5 < 0 on [−5,5] and the distance cannot be negative. Note that by symmetry of the ellipse about the x-axis, only the upper half of the ellipse is needed for the average distance, since the lower half just duplicates the distances. Hence, the average distance is 〈d〉= 1 5−(−5) Z5 −5 25−4x 5 dx = 1 50 (25x−2x2) ¯¯¯¯ 5 −5 = 1 50 (125−50 −(−125−50)) = 5 . Notice that the point (4,0) is a focus of the ellipse x2 25 + y2 9 = 1 (why?), which as it turns out makes the calculation of the average distance fairly simple. What if you wanted the average value of a function f that is not easily integrable? One alternative to numerical integration techniques is the Monte Carlo method. The idea behind it is simple: go back to the usual definition of an average, by taking a large number N of random numbers x1, x2, ... , xN in [a,b] and then using the approximation 〈f 〉≈f (x1) + f (x2) + ··· + f (xN) N . This might seem like taking a step backward from calculus, and it is, but it is surprisingly useful, as well as simple to implement with a computer. In addition, it can be shown that as the number of random points in [a,b] increases, the approximations converge to the actual average. | ElementaryCalculus_Page_270_Chunk3805 |
Average Value of a Function • Section 8.2 261 Example 8.12 The Monte Carlo method is easy to implement in Octave/MATLAB. Typically only a “one-liner” is needed, owing to Octave’s vectorization—i.e. the ability to perform mathematical operations on entire arrays of objects all at once. For example, recall from Example 8.8 that the average value of f (x) = x2 over [0,1] is 1/3 = 0.33333.... Approximate the average value using an array of 100 million (108) random numbers in [0,1]: octave> mean(rand(1,1e8).^2) ans = 0.3333292094741531 The dot in the rand(1,1e8).^2 command applies the squaring operation (^2) to each of the 108 random numbers in the array returned by the rand(1,1e8) command. The aggregate mean function then calcu- lates the array’s mean. Trigonometric, exponential and other functions can be applied to arrays, with the function evaluating each array element individually. In general the command (b−a).*rand(1,N)+a will return an array of N random numbers in the interval (a,b). For example, the function f (x) = sin(x2) cannot be integrated in a closed form, but its average value over [π,2π] can be approximated easily in Octave (actual average = -0.04154374531416104): octave> mean(sin((pi.*rand(1,1e8)+pi).^2)) ans = -0.04153426177596753 Exercises A For Exercises 1-9, find the average value of the function f (x) over the given interval. 1. f (x) = 1, over [0,3] 2. f (x) = x, over [0,1] 3. f (x) = x2, over [0,2] 4. f (x) = x3, over [0,2] 5. f (x) = sin 2x, over [0,π/2] 6. f (x) = ex, over [−1,4] 7. f (x) = x3, over [−1,1] 8. f (x) = sin x, over [−π/2,π/2] 9. f (x) = 1 x, over [1,3] 10. Electrical signals are commonly represented by a periodic waveform x(t), which is a function of time t and has period T (i.e. T is the smallest positive number such that x(t+T) = x(t) for all t). The average power of the waveform is defined as the average value of its square over a single period: x2(t) ® = 1 T ZT 0 x2(t) dt . (a) Find the average power of the waveform x(t) = Acos(ωt+φ), where A > 0 and ω > 0 and φ are all constants. (b) The root mean square of a waveform, abbreviated as rms, is the square root of the average power. Calculate the rms of the waveform from part (a). Write your answer in decimal form as a percentage of the amplitude A. | ElementaryCalculus_Page_271_Chunk3806 |
262 Chapter 8 • Applications of Integrals §8.2 E R s C 11. An electric circuit with a supplied voltage (electromotive force) E, a capacitor with capacitance C, and a resistor with resistance R, is shown in the picture on the right. When a switch s in the circuit is opened at time t = 0 the current I through the circuit begins to decrease exponentially as a function of time t (mea- sured in seconds after the switch is opened), given by I = E R e−t/RC for t ≥0. (a) Sketch a rough graph of I as a function of t. (b) Note that at time t = 0 the current is I = E R (measured in amperes), which is the familiar formula from Ohm’s Law. That is the peak value of I. What is the current I at time t = 5RC? Write your answer in decimal form as a percentage of the peak current E R (e.g. 0.42 E R , which would be 42% of the peak current). (c) Find the average current in the circuit over the time interval [0,5RC]. Write your answer in decimal form as a percentage of the peak current. 12. A spring with spring constant k and damping constant ν connects two point particles with mass m in a gravitational wave detector. A gravitational wave passes through the detector at time t = 0 and induces oscillation in the spring, with a period of 2π/Ωand energy E at time t ≥0 given by E(t) = 1 4 mR2 ¡ Ω2 sin2 (Ωt+φ) + ω2 0 cos2(Ωt+φ) ¢ , where ω2 0 = 2k/m, φ = tan−1 (2νΩ/(m(ω2 0 −Ω2)), and R is a constant. (a) Show that the average energy 〈E〉over one period [0,2π/Ω] of oscillation is 〈E〉= 1 8 mR2 (ω2 0 + Ω2) . (b) Suppose a large number of identical detectors of this type are uniformly distributed in a planar array at a density of σ detectors per unit area. The energy Eσ(t) imparted to each detector at time t ≥0 by a gravitational wave is Eσ(t) = νΩ2R2 sin2 (Ωt+φ) . Show that the average energy 〈Eσ〉over one period [0,2π/Ω] of oscillation is 〈Eσ〉= 1 2νΩ2R2 . B 13. For the ellipse x2 a2 + y2 b2 = 1, with a > b > 0, the foci are the points (c,0) and (−c,0), where c = p a2 −b2. Find the average distance from the ellipse to either of its foci in terms of the constants a, b, and c. 14. Write a computer program to use the Monte Carlo method with 1 million random points to approx- imate the average distance from the ellipse x2 25 + y2 9 = 1 to the point (0,0). Use symmetry to choose the smallest interval for the points. Could you have used formula (8.3) instead? Explain. | ElementaryCalculus_Page_272_Chunk3807 |
Arc Length and Curvature • Section 8.3 263 8.3 Arc Length and Curvature Just like the area of a plane region can be found using calculus, so too can the length of a plane curve. Along the way the mystery mentioned in a footnote in Chapter 1 will finally be solved: what is the length of the hypotenuse of a right triangle with infinitesimal sides? For a function y = f (x) denote by s the length of the piece of that curve over an interval [a,b], as in Figure 8.3.1 (a). Call s the curve’s arc length over [a,b]. y x a b s y = f (x) (a) Arc length s dx dy ds (b) Infinitesimal triangle 1 dy dx ds dx (c) Noninfinitesimal triangle Figure 8.3.1 Arc length By the Microstraightness Property, for a ≤x < b the curve is a straight line of length ds over the infinitesimal interval [x, x + dx], as in Figure 8.3.1 (b), where ds > 0 is the infinitesimal change in s over that interval. Notice that you cannot simply apply the Pythagorean Theorem here, since that would make ds = p (dx)2 +(dy)2 = p 0+0 = 0, which is false. The trick is to divide all sides of this infinitesimal right triangle by dx, which yields the similar—and noninfinitesimal—right triangle shown in Figure 8.3.1(c). The Pythagorean Theorem can then be applied to that triangle: ds dx = s 12 + µ dy dx ¶2 ⇒ ds = s 1+ µ dy dx ¶2 dx Summing up those infinitesimal lengths ds then yields the arc length s: The arc length s of a curve y = f (x) over [a,b] is: s = Zb a ds = Zb a s 1+ µdy dx ¶2 dx (8.4) That such a formula exists is of course good news, but as you have probably guessed, the in- tegral cannot be evaluated in a closed form except for a few functions.2 In most cases numerical integration methods will be required. 2This will be the last “good news/bad news” scenario in this book. That’s the good news. | ElementaryCalculus_Page_273_Chunk3808 |
264 Chapter 8 • Applications of Integrals §8.3 Example 8.13 y x 0 1 y = coshx s 1 Find the arc length of the curve y = cosh x over [0,1]. Solution: Since dy dx = sinh x, then the arc length s is: s = Z1 0 s 1+ µ dy dx ¶2 dx = Z1 0 p 1+sinh2 x dx = Z1 0 cosh x dx = sinh x ¯¯¯¯ 1 0 = sinh 1 −sinh 0 = sinh 1 ≈1.1752 Example 8.14 y x 0 −L/2 L/2 y = acosh x a s h a A catenary—a hanging uniform cable whose ends are fastened at the same height h a distance L apart—has its lowest point—the apex—a distance a > 0 above the ground. It can be shown3 that with the apex at (0,a), the equation of the catenary is y = a cosh x a. Find the arc length of the catenary. Solution: The figure on the right shows the catenary. By symmetry the total arc length s is twice the arc length over [0,L/2]: s = 2 ZL/2 0 s 1+ µ dy dx ¶2 dx = 2 ZL/2 0 q 1+sinh2 x a dx = 2 ZL/2 0 cosh x a dx = 2a sinh x a ¯¯¯¯ L/2 0 = 2a sinh L 2a Elliptic Integrals x y x2 a2 + y2 b2 = 1 −a a b −b Suppose you tried to find the circumference s of the ellipse x2 a2 + y2 b2 = 1, with a > b > 0, which has eccentricity e = c a, where c = p a2 −b2. By symmetry, s is quadruple the arc length of the upper hemisphere y = b q 1−x2 a2 over [0,a]: s = 4 Za 0 s 1+ µ dy dx ¶2 dx = 4 Za 0 s 1+ µ − bx a p a2 −x2 ¶2 dx = 4 Za 0 s a2(a2 −x2)+ b2x2 a2(a2 −x2) dx = 4 a Za 0 s a4 −(a2 −b2)x2 a2 −x2 dx Now try the trigonometric substitution x = asin θ, dx = acos θ dθ: 3See pp.162-163 in SMITH, C.E., Applied Mechanics: Statics, New York: John Wiley & Sons, Inc., 1976. | ElementaryCalculus_Page_274_Chunk3809 |
Arc Length and Curvature • Section 8.3 265 s = 4 a Zπ/2 0 s a4 −(a2 −b2)a2sin2 θ a2 −a2sin2 θ a cos θ dθ = 4 Zπ/2 0 s a2 −(a2 −b2) sin2 θ ✘✘✘✘✘ 1−sin2 θ ✘✘✘ cos θ dθ = 4a Zπ/2 0 s 1−a2 −b2 a2 sin2 θ dθ s = 4a Zπ/2 0 p 1−e2 sin2 θ dθ (since e2 = c2 a2 = a2 −b2 a2 ) The last integral is a special case of the elliptic integral of the second kind E(k,φ) = Zφ 0 p 1−k2 sin2 θ dθ , with k = e and φ = π 2 . This special case is denoted by E(k) = E(k, π 2 ). Thus, the circumference s of the ellipse is: s = 4aE(e) = 4aE(e, π 2) The integral E(k) for 0 < k < 1 cannot be evaluated in a closed form. There are tables4 for certain values of k between 0 and 1, but a number of scientific computing applications have built-in functions to evaluate elliptic integrals. For example, suppose you want to find the circumference s of the ellipse x2 25 + y2 9 = 1. Then a = 5, b = 3, c = p a2 −b2 = 4, and e = c a = 0.8, so s = 4aE(e) = 20E(0.8). In the Python-based open-source mathematical software system Sage5, the elliptic integral E(k,φ) is provided by the function elliptic_e(φ,k2). Use k = e = 0.8 and φ = π 2: In [1]: 20*elliptic_e(pi/2,0.8^2) Out[1]: 25.5269988633981 The circumference is thus approximately 25.5269988633981. In Octave/MATLAB the function ellipke(e2) evaluates the elliptic integral E(e), with one extra step: MATLAB>> [K,E] = ellipke(0.8^2); MATLAB>> 20*E ans = 25.526998863398131 4See p.609 in ABRAMOWITZ, M. AND I.A. STEGUN, Handbook of Mathematical Functions, New York: Dover Publications, Inc., 1965. 5Available at https://www.sagemath.org | ElementaryCalculus_Page_275_Chunk3810 |
266 Chapter 8 • Applications of Integrals §8.3 The parametric formula for arc length can be derived by dividing all sides of the infinitesimal right triangle in Figure 8.3.1(b) by dt, then applying the Pythagorean Theorem to the resulting noninfinitesimal right triangle: ds dt = sµdx dt ¶2 + µ dy dt ¶2 ⇒ ds = sµdx dt ¶2 + µdy dt ¶2 dt Sum up those infinitesimal lengths ds to obtain the arc length s: The arc length s of a parametric curve x = x(t), y = y(t), a ≤t ≤b is: s = Zb a ds = Zb a sµ dx dt ¶2 + µdy dt ¶2 dt (8.5) Example 8.15 y x 1 0 1 x = cos3 t y = sin3 t Find the arc length of the parametric curve x = cos3 t, y = sin3 t, 0 ≤t ≤π/2. Solution: Since dx dt = −3cos2 t sint and dy dt = 3sin2 t cost, then the arc length s is: s = Zπ/2 0 sµ dx dt ¶2 + µ dy dt ¶2 dt = Zπ/2 0 q (−3 cos2 t sin t)2 +(3 sin2 t cos t)2 dt = Zπ/2 0 p 9 cos4 t sin2 t + 9 sin4 t cos2 t dt = Zπ/2 0 3 q sin2 t cos2 t (cos2 t + sin2 t) dt = Zπ/2 0 3 sin t cos t dt (since cos t ≥0 and sin t ≥0 for 0 ≤t ≤π/2) = 3 2 sin2 t ¯¯¯¯ π/2 0 = 3 2 The polar formula for arc length can be considered a special case of the parametric formula. A polar curve r = r(θ) for α ≤θ ≤β has Cartesian coordinates x = r(θ) cos θ and y = r(θ) sin θ, so that dx dθ = dr dθ cos θ −r sin θ and dy dθ = dr dθ sin θ + r cos θ . It is left as an exercise to show that putting these derivatives into formula (8.5)—using the parameter θ instead of t—yields the polar arc length formula: The arc length s of a polar curve r = r(θ) for α ≤θ ≤β is: s = Zβ α ds = Zβ α s r2 + µ dr dθ ¶2 dθ (8.6) | ElementaryCalculus_Page_276_Chunk3811 |
Arc Length and Curvature • Section 8.3 267 Example 8.16 Prove that the circumference of a circle of radius R is 2πR. Solution: Use the polar curve r = R for 0 ≤θ ≤2π. Then dr dθ = 0, so: s = Z2π 0 s r2 + µ dr dθ ¶2 dθ = Z2π 0 p R2 +02 dθ = Z2π 0 R dθ = R θ ¯¯¯¯ 2π 0 = 2πR ✓ Curvature y x 1 0 1 y = x2 Figure 8.3.2 In Chapter 4 you saw one simple measure of curvature: the sec- ond derivative. From Figure 8.3.2 it is clear that the parabola y = x2 is less curved at the point (1,1) than at the origin, yet d2 y dx2 = 2 at each point. So the second derivative—the rate of change of the slopes dy dx of the tangent lines—does not fully capture the curvature of a curve; more information is needed. It turns out that the rate of change of the angles of the tangent lines is the key to curvature. First consider a curve with arc length s between two points A and B on the curve. Let α be the angle between the tangent lines to the curve at A and B, as in Figure 8.3.3(a). α A B s (a) Small curvature α A B s (b) Larger curvature and α Figure 8.3.3 Curvature between A and B: same s, different α For the same arc length s but larger angle α as in Figure 8.3.3(b), the curvature appears greater. This suggests that curvature should measure α relative to s: The average curvature ¯κAB of a curve between two points A and B on the curve is ¯κAB = α s (8.7) where s is the arc length of the curve between A and B, and α is the angle between the tangent lines to the curve at A and B, called the angle of contingence. | ElementaryCalculus_Page_277_Chunk3812 |
268 Chapter 8 • Applications of Integrals §8.3 Similar to how the instantaneous rate of change of a function at a point is the average rate of change over an infinitesimal interval, the curvature of a curve at a point can be defined as the average curvature over an infinitesimal length of the curve: The curvature κ of a curve at a point A on the curve is κ = dφ ds (8.8) where s is the arc length function of the curve and φ is the angle that the tangent line to the curve at A makes with the positive x-axis. The idea is that moving an infinitesimal length ds along the curve induces an infinitesimal difference dφ in the angles of the tangent lines. Roughly, as B moves toward A: lim B→A ¯κAB = lim B→A α s = dφ ds = κ Curvature can be viewed as the instantaneous rate of change of direction of a curve, but with respect to arc length (i.e. distance traveled) instead of time. For a curve y = f (x), recall from formula (3.2) in Section 3.1 that φ = φ(x) = tan−1 f ′(x) at each point (x, f (x)) on the curve. Thus, by the Chain Rule: κ = dφ ds = d (tan−1 f ′(x)) ds = f ′′(x) dx 1+(f ′(x))2 ds So since ds = p 1+(f ′(x))2 dx by formula (8.4), then: The curvature κ of a curve y = f (x) is: κ = f ′′(x) (1 + (f ′(x))2)3/2 (8.9) The above formula makes κ a function of x. Note also that κ = 0 for a straight line, and that the curve y = f (x) is concave up if κ > 0 and concave down if κ < 0. Example 8.17 Find the curvature of the curve y = x2 for x = 0 and x = 1. Solution: For f (x) = x2, f ′(x) = 2x and f ′′(x) = 2, so for any x the curvature κ = κ(x) is: κ = f ′′(x) (1+(f ′(x))2)3/2 = 2 (1+4x2)3/2 In particular, κ(0) = 2 and κ(1) = 2 53/2 ≈0.1789. So y = x2 has more curvature at the origin than at (1,1). | ElementaryCalculus_Page_278_Chunk3813 |
Arc Length and Curvature • Section 8.3 269 For a parametric curve x = x(t), y = y(t), the curvature κ will be a function of the parameter t. Since dy dx = y′(t) x′(t) by formula (7.14) in Section 7.6, then by formula (7.15): d2y dx2 = d dt µ dy dx ¶ dx dt = d dt µ y′(t) x′(t) ¶ x′(t) = x′(t) y′′(t) −y′(t) x′′(t) (x′(t))3 So by formula (8.9): κ = d2y dx2 µ 1+ µ dy dx ¶2¶3/2 = x′(t) y′′(t) −y′(t) x′′(t) (x′(t))3 Ã 1 + µ y′(t) x′(t) ¶2!3/2 = x′(t) y′′(t) −y′(t) x′′(t) ¡ (x′(t))2¢3/2 Ã 1 + µ y′(t) x′(t) ¶2!3/2 Simplify the denominator to obtain the parametric curvature formula: The curvature κ of a parametric curve x = x(t), y = y(t) is: κ = x′(t) y′′(t) −y′(t) x′′(t) ¡ (x′(t))2 + (y′(t))2¢3/2 (8.10) The derivation of the curvature formula in polar coordinates is left as an exercise: The curvature κ of a polar curve r = r(θ) is: κ = (r(θ))2 + 2(r′(θ))2 −r(θ)r′′(θ) ¡ (r(θ))2 + (r′(θ))2¢3/2 (8.11) Example 8.18 Find the curvature of a circle of radius R. Solution: Use the polar curve r = r(θ) = R, so that r′(θ) = 0 = r′′(θ): κ = (r(θ))2 + 2(r′(θ))2 −r(θ)r′′(θ) ¡ (r(θ))2 + (r′(θ))2¢3/2 = R2 + 2·02 −R ·0 (R2 + 02)3/2 = R2 R3 = 1 R A circle thus has constant curvature, as you would expect by symmetry. It turns out that any planar curve with constant curvature is either a line or part of a circle.6 6See pp.62-63 in O’NEILL, B., Elementary Differential Geometry, New York: Academic Press, Inc., 1966. | ElementaryCalculus_Page_279_Chunk3814 |
270 Chapter 8 • Applications of Integrals §8.3 Exercises A For Exercises 1-10, find the arc length of the given curve over the given interval. 1. y = x3/2 ; 1 ≤x ≤4 2. y = x2 ; 0 ≤x ≤1 3. y = x2/3 ; 1 ≤x ≤8 4. y = x2 4 −ln x 2 ; 1 ≤x ≤2 5. y = x4 4 + 1 8x2 ; 1 ≤x ≤2 6. y = ln ex +1 ex −1 ; 1 ≤x ≤2 7. x = et cos t, y = et sin t ; 0 ≤t ≤π 8. x = cos t + t sin t, y = sin t −t cos t ; 0 ≤t ≤π 9. polar curve r = 1+cos θ ; 0 ≤θ ≤2π 10. polar curve r = eθ ; 0 ≤θ ≤2 11. Find the arc length of the curve in Example 8.15 for 0 ≤t ≤π. 12. Use formula (8.5) to find the circumference of the unit circle using two different parametrizations: (a) x = cos t, y = sin t, 0 ≤t ≤2π (b) x = 1−t2 1+ t2 , y = 2t 1+ t2 , −∞< t < ∞ For Exercises 13-18 find the curvature of the given curve at the indicated points. 13. y = sin x at x = 0 and x = π 2 14. y = ln x at x = 1 15. x2 a2 + y2 b2 = 1 at (a,0) and (0,b) 16. y = ex at x = 0 17. x2 −y2 = 1 at (1,0) 18. r = 1+cos θ at θ = 0 B 19. Prove formula (8.6). 20. Prove formula (8.11). 21. Let α and β be nonzero constants. Show that the arc length s of y = β sin x α over the interval [0,x0] can be put in terms of the elliptic integral E(k,φ): s = q α2 +β2 · E Ãs β2 α2 +β2 , x0 α ! 22. For −1 < k < 1 and −1 ≤x ≤1, define u(x) and K by u(x) = Zx 0 dt p 1−t2 p 1−k2 t2 and K = u(1) = Z1 0 dt p 1−t2 p 1−k2 t2 so that all the square roots are defined and positive. (a) Show that u is an increasing function of x and thus has an inverse function, call it x = sn u, with domain [−K,K] and range [−1,1]. (b) Define cn u = p 1−sn2 u and dn u = p 1−k2 sn2 u. Show that: d du (sn u) = cn u dn u , d du (cn u) = −sn u dn u , and d du (dn u) = −k2 sn u cn u The functions sn u, cn u, and dn u are called the Jacobian elliptic functions. (c) Suppose that sin φ = sn u. Show that E(k,φ) = Zu 0 dn2 v dv. 23. The ends of a 50 ft long catenary are fastened 40 ft apart. Use a numerical method to find how much the apex dips below the ends. (Hint: Solve for a in Example 8.14, then use symmetry.) | ElementaryCalculus_Page_280_Chunk3815 |
Surfaces and Solids of Revolution • Section 8.4 271 8.4 Surfaces and Solids of Revolution Long before calculus was invented the ancient Greeks (e.g. Archimedes) discovered the formu- las for the volume and surface area of familiar three-dimensional objects such as the sphere.7 Volumes and surface areas of arbitrary solids and surfaces can be found using multivariable calculus. However, single-variable calculus can be used in the special case of the objects pos- sessing symmetry about an axis, via methods that involve revolving a curve or region in the xy-plane around an axis. For example, revolve a curve y = f (x) ≥0 around the x-axis, for a ≤x ≤b. This produces a surface of revolution in three dimensions, as in Figure 8.4.1(a). y x a b y = f (x) (a) Revolve y = f (x) around x-axis y x a x b y = f (x) (b) dS over [x,x+ dx] f (x) f (x+ dx) y x ds dx (c) Side view of frustrum Figure 8.4.1 Surface of revolution over [a,b] with lateral surface area S r2 r1 l Figure 8.4.2 To find the total lateral surface area S, pick x in [a,b) then find the infinitesimal surface area dS swept out over the infinitesimal interval [x, x + dx], as in Figure 8.4.1(b). By the Microstraightness Property, the curve y = f (x) is a straight line segment of length ds over that interval, so that the infinitesimal surface is a frustrum—a right circular cone with the vertex chopped off by a plane parallel to the base circle. From geometry8 you might recall the formula for the lateral surface area of the frustrum in Figure 8.4.2: π(r1+r2)l. Use that formula with r1 = f (x), r2 = f (x+dx) = f (x)+dy, and l = ds = p 1+(f ′(x))2 dx (by formula (8.4) in Section 8.3) as in Figure 8.4.1(c), so that dS is dS = π(f (x)+(f (x)+ dy)) q 1+(f ′(x))2 dx = 2π f (x) q 1+(f ′(x))2 dx + π q 1+(f ′(x))2 dydx = 2π f (x) q 1+(f ′(x))2 dx + 0 since dydx = f ′(x)(dx)2 = 0. The surface area S is then the sum of all the areas dS: 7See Propositions 33 and 34 in On the Sphere and Cylinder, Book I, appearing in HEATH, T.L., The Works of Archimedes, Mineola, NY: Dover Publications, Inc., 2002. This work is also available at https://archive.org 8See pp.136-137 in WELCHONS A.M. AND W.R. KRICKENBERGER, Solid Geometry, Boston: Ginn & Co., 1936. | ElementaryCalculus_Page_281_Chunk3816 |
272 Chapter 8 • Applications of Integrals §8.4 The surface area S of the surface of revolution obtained by revolving the curve y = f (x) ≥0 around the x-axis for a ≤x ≤b is: S = Zb a dS = Zb a 2π f (x) q 1+(f ′(x))2 dx (8.12) For a general curve y = f (x), possibly negative in [a,b], the surface area S is: S = Zb a dS = Zb a 2π|y| ds = Zb a 2π ¯¯ f (x) ¯¯ q 1+(f ′(x))2 dx (8.13) y x a b y = ¯¯ f (x) ¯¯ y = f (x) Note that formula (8.13) holds by symmetry and formula (8.12). A curve y = f (x) < 0 and the curve y = ¯¯ f (x) ¯¯ = −f (x) are symmet- ric with respect to the x-axis, as in the figure on the right. Thus, both curves sweep out the same surface of revolution when revolved around the x-axis. This means formula (8.13) also holds if y = f (x) changes sign in [a,b]: similar to the area between two curves, you would split the integral over different subintervals depending on the sign. Example 8.19 Show that the surface area of a sphere of radius r is 4πr2. y x −r r y = p r2 −x2 Solution: Use the circle x2+y2 = r2. The upper half of that circle is the curve y = f (x) = p r2 −x2 over the interval [−r,r], as in the figure on the right. Revolving that curve around the x-axis produces a sphere of radius r, whose surface area S is: S = Zr −r 2π f (x) q 1+(f ′(x))2 dx = Zr −r 2π p r2 −x2 s 1+ µ −x p r2 −x2 ¶2 dx = Zr −r 2π✘✘✘✘ p r2 −x2 s r2 ✘✘✘ r2 −x2 dx = 2πrx ¯¯¯¯ r −r = 4πr2 ✓ A similar derivation using a frustrum yields the surface area S of the surface of revolution obtained by revolving a curve y = f (x) around the y-axis, for 0 ≤a ≤x ≤b: S = Zb a dS = Zb a 2π|x| ds = Zb a 2πx q 1+(f ′(x))2 dx (8.14) | ElementaryCalculus_Page_282_Chunk3817 |
Surfaces and Solids of Revolution • Section 8.4 273 Now suppose you revolve the region between a curve y = f (x) ≥0 and the x-axis around the x-axis, for a ≤x ≤b (see Figure 8.4.3(a)). This produces a solid of revolution in three dimensions, as in Figure 8.4.3(b). Notice that this solid consists of the surface of revolution as before along with its interior. y x a b y = f (x) (a) Revolve region around x-axis y x a x b y = f (x) (b) Solid f (x) f (x+ dx) y x ds dx dy (c) Infinitesimal strip Figure 8.4.3 Solid of revolution over [a,b] with volume area V The goal is to find the volume V of this solid. The idea is to divide the solid into slices, like a loaf of bread. First, the infinitesimal volume dV of the frustrum swept out by a strip of infinitesimal width dx at x in [a,b)—shown in Figure 8.4.3(c)—is needed. By the Micros- traightness Property the curve y = f (x) is a straight line of length ds over the interval [x, x+dx]. There is thus a right triangle at the top of the strip—unshaded in Figure 8.4.3(c)—whose area A is zero: A = 1 2(dy)(dx) = 1 2 f ′(x)(dx)2 = 0. f (x) dx Figure 8.4.4 That triangle thus contributes no volume when revolved around the x-axis: the volume dV swept out by that strip all comes from the shaded rectangle of height f (x) and width dx. That rectangle sweeps out a right circular cylinder of radius f (x) and height dx (see Figure 8.4.4). The volume of a right circular cylinder of radius r and height h is defined as the area of the base circle times the height: πr2h. Hence, dV = π(f (x))2 dx . The total volume V of the solid is then the sum of all those infinitesimal volumes dV: The volume V of the solid of revolution obtained by revolving the region between the curve y = f (x) and the x-axis around the x-axis for a ≤x ≤b is: V = Zb a dV = Zb a π(f (x))2 dx (8.15) This method for finding the volume is called the disc method, since the cylinder of volume dV resembles a disc. Think of the discs as being similar to infinitesimally thin slices of a loaf of bread. Notice that absolute values are not needed in formula (8.15) since f (x) is squared, so the formula holds even when f (x) is negative. | ElementaryCalculus_Page_283_Chunk3818 |
274 Chapter 8 • Applications of Integrals §8.4 Example 8.20 Show that the volume of a sphere of radius r is 4 3πr3. y x −r r y = p r2 −x2 Solution: Use the circle x2 + y2 = r2. Revolve the region between the upper half of the circle y = f (x) = p r2 −x2 and the x-axis around the x-axis over the interval [−r,r], as in the figure on the right. The solid of revolution swept out is a sphere of radius r, whose volume V is: V = Zr −r π(f (x))2 dx = Zr −r π(r2 −x2) dx = πr2x −1 3πx3 ¯¯¯¯ r −r = µ πr3 −1 3πr3 ¶ − µ −πr3 + 1 3πr3 ¶ = 4 3 πr3 ✓ Instead of memorizing formula (8.15), try to remember the more generic approach of revolv- ing an infinitesimal rectangular strip around an axis, which might not be the x-axis. The idea is to find the radius r and height h—typically dx or dy—of the disc swept out by that strip, so that the disc’s volume is dV = πr2h. Then integrate dV over the appropriate interval to find the volume V of the entire solid. Example 8.21 Suppose the region bounded by the curve y = x2 and the x-axis for 0 ≤x ≤1 is revolved around the line x = 1. Find the volume of the resulting solid of revolution. x = 1 y x 0 x 1 y = x2 r h x2 = y 1 Solution: The region is shaded in the figure on the right. Since the region is revolved around a vertical axis, the disc method will use discs with height dy, not dx. At a point x in [0,1] go up to the curve y = x2 and draw a horizontal rectangular strip to the line x = 1, as shown in the figure. Let h = dy and revolve that strip around the line x = 1, producing a disc of radius r −1 −x and height h = dy. Since y = x2 implies x = py, the volume dV of that disc is dV = πr2h = π(1−x)2 dy = π(1−py)2 dy = π(1−2py+ y) dy . The volume V of the entire solid is then the sum of those volumes dV along the y-axis for 0 ≤y ≤1: V = Z1 0 dV = Z1 0 π(1−2py+ y) dy = π µ y−4 3 y3/2 + 1 2 y2 ¶ ¯¯¯¯ 1 0 = π µ 1−4 3 + 1 2 ¶ = π 6 y x a −a 0 b −b y = f (x) The shell method can be used for finding the volume of a solid with a “hole” in the middle, as in the solid of revolution produced by revolving the shaded region in the figure on the right around the y-axis. The hole in the solid between x = −a and x = a is a result of the gap between the y-axis and the region. | ElementaryCalculus_Page_284_Chunk3819 |
Surfaces and Solids of Revolution • Section 8.4 275 To find the volume V of that solid, at a point x in [a,b) form an infinitesimal strip of width dx from the x-axis up to the curve y = f (x), as in Figure 8.4.5(a). y x 0 a x b y = f (x) (a) Revolve region around y-axis f (x) f (x+ dx) y x ds dx dy (b) Infinitesimal strip f (x) x dx (c) Cylindrical shell Figure 8.4.5 Shell method Just like the strip in the disc method, the right triangle at the top of this strip—as in Figure 8.4.5(b)—has zero area and thus does not contribute to the volume dV of the right circular cylindrical shell swept out by the strip, shown in Figure 8.4.5(c). The volume of that shell is just the volume of the “outer” cylinder of radius x+dx minus the volume of the “inner” cylinder of radius x, both with height f (x): dV = π(x+ dx)2 f (x) −πx2 f (x) = ✘✘✘✘ πx2 f (x) + 2πx f (x)dx + π✟✟✟ ✯0 (dx)2 f (x) −✘✘✘✘ πx2 f (x) = 2πx f (x)dx The volume V of the entire solid is then the sum of those volumes dV, using an absolute value to handle any sign for f (x): The volume V of the solid of revolution obtained by revolving the region between the curve y = f (x) and the x-axis around the y-axis for 0 ≤a ≤x ≤b is: V = Zb a dV = Zb a 2πx ¯¯ f (x) ¯¯ dx (8.16) Example 8.22 Suppose the region bounded by the curve y = x2 and the x-axis for 0 ≤x ≤1 is revolved around the y-axis. Find the volume of the resulting solid of revolution. y x 0 x 1 y = x2 1 Solution: The region is shaded in the figure on the right. The vertical strip at x in [0,1) with infinitesimal width dx and height ¯¯f (x) ¯¯ = f (x) is shown in the figure. That strip produces the shell with volume dV in formula (8.16), so by the shell method the volume V of the solid of revolution is: V = Z1 0 dV = Z1 0 2πx ¯¯f (x) ¯¯ dx = Z1 0 2πx · x2 dx = π 2 x4 ¯¯¯¯ 1 0 = π 2 | ElementaryCalculus_Page_285_Chunk3820 |
276 Chapter 8 • Applications of Integrals §8.4 The volume dV in formula (8.16) can be generalized to dV = 2πrhw, where r is the distance from the axis of revolution to a generic vertical strip of infinitesimal width w in the region, and h is the height of the strip. Example 8.23 Suppose the region bounded by the curves y = x2 and y = x is revolved around the y-axis. Find the volume of the resulting solid of revolution. y x 0 x 1 y = x y = x2 1 Solution: The region is shaded in the figure on the right, along with a vertical strip with infinitesimal width w = dx at the distance r = x from the y-axis in the region and height h = x −x2. That strip produces the shell with volume dV = 2πrhw = 2πx(x−x2)dx, so that the volume V of the solid of revolution is: V = Z1 0 dV = Z1 0 2πx(x−x2) dx = 2π 3 x3 −π 2 x4 ¯¯¯¯ 1 0 = π 6 Exercises A For Exercises 1-3, find the surface area of the surface of revolution produced by revolving the given curve around the x-axis for the given interval. 1. y = p 4−x2 for 1 ≤x ≤2 2. y = cosh x for 0 ≤x ≤1 3. y = x3 6 + 1 2x for 1 ≤x ≤3 For Exercises 4-6, find the volume of the solid of revolution produced by revolving the region between the given curve and the x-axis around the x-axis for the given interval. 4. y = x3 for 0 ≤x ≤1 5. y = sin x for 0 ≤x ≤π 6. y = px for 0 ≤x ≤1 For Exercises 7-9, find the volume of the solid of revolution produced by revolving the region between the given curve and the x-axis around the y-axis for the given interval. 7. y = sin(x2) for 0 ≤x ≤pπ 8. y = sin x for 0 ≤x ≤π 9. y = x2 −x3 for 0 ≤x ≤1 10. Revolve the region in Example 8.23 around the line x = 1 and find the volume of the resulting solid. 11. Revolving the ellipse x2 a2 + y2 b2 = 1 around the x-axis produces an ellipsoid, for a > b > 0. Show that the surface area of the ellipsoid is 2πb2 ¡ 1+ a eb sin−1 e ¢ , where e is the eccentricity of the ellipse. 12. Show that the volume inside the ellipsoid from Exercise 11 is 4 3 πab2. 13. Find the surface area and volume of a right circular cone of radius r and height h. 14. Formulas (8.13), (8.15) and (8.16) can be extended to include regions over infinite intervals—the integrals in those formulas simply become improper integrals. Consider the region between the curve y = 1 x and the x-axis over the interval [1,∞). Revolve that region around the x-axis. (a) Show that the surface area of the resulting surface of revolution is infinite. (b) Show that the volume of the resulting solid of revolution is π. 15. For 0 < a < b, revolving the region inside the circle (x −b)2 + y2 = a2 around the y-axis produces a donut-shaped solid of revolution called a torus. Show that the volume of the torus is 2π2a2b. 16. Use formula (8.14) and symmetry to show that the torus from Exercise 15 has surface area 4π2ab. | ElementaryCalculus_Page_286_Chunk3821 |
Applications in Physics and Statistics • Section 8.5 277 8.5 Applications in Physics and Statistics This chapter concludes with a few applications showing how some familiar discrete sums can be replaced by integrals, which are essentially continuous sums. Center of Gravity Suppose a thin uniform rod has n > 1 masses m1,...,mn attached, with m1 and mn at the ends. The center of gravity of the masses is the point where—due to the Earth’s gravity—the rod would be balanced if a fulcrum were placed there (see Figure 8.5.1(a)). Imagine the rod as part of the x-axis and the weights as point masses—with each mass mk at xk—and let the center of gravity be at ¯x, as in Figure 8.5.1(b). m1 m2 ··· mn−1 mn (a) Rod balanced on fulcrum x x1 x2 ¯x xn−1 xn m1 m2 ··· mn−1 mn (b) Masses and ¯x on x-axis Figure 8.5.1 Center of gravity ¯x for masses m1,...,mn The rod is balanced if the masses do not rotate the rod, i.e. the total torque is zero. Torque is defined here as force times position relative to ¯x. Each mass mk applies a force mkg to the rod—where g is the (downward) acceleration due to the Earth’s gravity—at position (xx −¯x) relative to ¯x. The total torque is thus zero if (m1g)(x1 −¯x) + (m2g)(x2 −¯x) + ··· + (mng)(xn −¯x) = 0 so that solving for ¯x yields: ¯x = m1gx1 +···+ mngxn m1g +··· + mng = m1x1 +···+ mnxn m1 +···+ mn = Pn k=1 mkxk Pn k=1 mk (8.17) Each quantity mkxk is called the moment of the mass mk. Thus, ¯x is the sum of the moments divided by the total mass. This idea can be extended to regions in the xy-plane, using an integral of a continuum of moments instead of a finite sum. The center of gravity of a planar region is defined as the point such that any force along a line through that point produces no rotation of the region about that line.9 There should thus be zero torque in both the x and y directions, so the idea is to apply formula (8.17) in both directions to obtain the region’s center of gravity (¯x, ¯y). 9For a proof that such a point exists, see p.206 in BROWN, F.L., Engineering Mechanics, 2nd ed., New York: John Wiley & Sons, Inc., 1942. Some texts use the terms “center of mass” or “centroid” instead of “center of gravity,” and there are differences in the meanings. However, for the situation presented here, where the gravitational field is assumed to have constant magnitude and direction throughout the region, they all mean the same thing. | ElementaryCalculus_Page_287_Chunk3822 |
278 Chapter 8 • Applications of Integrals §8.5 A region can be thought of as a lamina—a thin plate with uniform density. Take the area of the region as its mass, which makes sense given the uniform density. For the region between two curves y = f1(x) and y = f2(x) over [a,b], with f1(x) ≥f2(x), take a vertical slice of width dx at some x, as in Figure 8.5.2(a). By the same arguments used in Section 8.4, all the area from that strip comes from the rectangle of height f1(x)−f2(x) and width dx (see the shaded rectangle in Figure 8.5.2(b)). y x 0 a x b y = f1(x) y = f2(x) (a) Region between y = f1(x) and y = f2(x) f1(x)−f2(x) y x ds dx dy (x+ 1 2dx, 1 2(f1(x)+ f2(x))) (b) Infinitesimal strip over [x,x+ dx] Figure 8.5.2 Center of gravity of a region By the assumption of uniform density, the center of gravity of that rectangle is clearly its geometric center, whose coordinates are ¡ x+ 1 2 dx, 1 2(f1(x)+ f2(x)) ¢ . The entire mass of the strip can be treated as if it is concentrated at that point. The moment mx of the strip about the x-axis is its mass times the position of its center of gravity relative to the x-axis (i.e. its y coordinate): mx = (f1(x)−f2(x))dx · ( 1 2 (f1(x)+ f2(x))) = 1 2 ((f1(x))2 −(f2(x))2)dx Similarly the moment my of the strip about the y-axis is its mass times the x coordinate of its center of gravity: my = (f1(x)−f2(x))dx · (x+ 1 2 dx) = x(f1(x)−f2(x))dx + 1 2(f1(x)−f2(x))(dx)2 = x(f1(x)−f2(x))dx The moments Mx and My of the entire region about the x-axis and y-axis, respectively, are defined as the sum of the respective moments mx and my of all strips over [a,b]: Mx = Zb a mx = Zb a 1 2 ((f1(x))2 −(f2(x))2) dx and My = Zb a my = Zb a x(f1(x)−f2(x)) dx Note in formula (8.17) that the denominator is the sum of all the masses in the system. For the region that total mass would simply be its area M: M = Zb a (f1(x)−f2(x)) dx Dividing the moments Mx and My by M yields the formula for the center of gravity: | ElementaryCalculus_Page_288_Chunk3823 |
Applications in Physics and Statistics • Section 8.5 279 The center of gravity (¯x, ¯y) of the region between the curves y = f1(x) and y = f2(x) over [a,b], with f1(x) ≥f2(x), is given by: ¯x = My M = Zb a x(f1(x)−f2(x)) dx Zb a (f1(x)−f2(x)) dx and ¯y = Mx M = Zb a 1 2 ((f1(x))2 −(f2(x))2) dx Zb a (f1(x)−f2(x)) dx (8.18) Example 8.24 Find the center of gravity of the region bounded by the curve y = x2 and the x-axis for 0 ≤x ≤1. y x 0 ¯x 1 y = x2 1 ¯y Solution: The region is shaded in the figure on the right. Using y = f1(x) = x2 and y = f2(x) = 0 in formula (8.18) yields Mx = Z1 0 1 2 (f1(x))2 dx = Z1 0 1 2 x4 dx = 1 10 x5 ¯¯¯¯ 1 0 = 1 10 My = Z1 0 x f1(x) dx = Z1 0 x3 dx = 1 4 x4 ¯¯¯¯ 1 0 = 1 4 M = Z1 0 f1(x) dx = Z1 0 x2 dx = 1 3 x3 ¯¯¯¯ 1 0 = 1 3 so that the center of gravity (¯x, ¯y) is: ¯x = My M = 1/4 1/3 = 3 4 and ¯y = Mx M = 1/10 1/3 = 3 10 Example 8.25 y x 0 ¯x 1 y = x y = x2 1 ¯y Find the center of gravity of the region bounded by the curves y = x and y = x2. Solution: The region is shaded in the figure on the right. Using y = f1(x) = x and y = f2(x) = x2 in formula (8.18) yields Mx = Z1 0 1 2 ((f1(x))2 −(f2(x))2) dx = Z1 0 1 2 (x2 −x4) dx = 1 6 x3 −1 10 x5 ¯¯¯¯ 1 0 = 1 15 My = Z1 0 x(f1(x)−f2(x)) dx = Z1 0 (x2 −x3) dx = 1 3 x3 −1 4 x4 ¯¯¯¯ 1 0 = 1 12 M = Z1 0 (f1(x)−f2(x)) dx = Z1 0 (x−x2) dx = 1 2 x2 −1 3 x3 ¯¯¯¯ 1 0 = 1 6 so that the center of gravity (¯x, ¯y) is: ¯x = My M = 1/12 1/6 = 1 2 and ¯y = Mx M = 1/15 1/6 = 2 5 | ElementaryCalculus_Page_289_Chunk3824 |
280 Chapter 8 • Applications of Integrals §8.5 Work Suppose that a constant force displaces an object along a line in the same direction in which the force is applied. The work done by the force is defined as the force times the displacement. For example, if the constant force F moves an object from position x = a to x = b on the x-axis, as in Figure 8.5.3(a), then the work W done by the force is: W = force × displacement = force × (final position −initial position) = F · (b −a) x a b F (a) Constant force F dx F(x) F(x+ dx) (b) Variable force F over [x,x+ dx] Figure 8.5.3 Work W as the effect of a force F displacing an object from x = a to x = b Suppose now that the force F is a function of position x over [a,b]: F = F(x). By the Micros- traightness Property, over an infinitesimal interval [x, x + dx] the curve y = F(x) is a straight line, as in Figure 8.5.3(b). How should the work dW performed by F over this infinitesimal in- terval be defined? After all, F is not constant over [x, x+dx]—it takes every value between F(x) and F(x+ dx). It is left as an exercise to show that any value in that range can be used—they all result in the same amount F(x)dx for the work performed.10 For example, suppose you use the value halfway between F(x) and F(x + dx) as the value of F: 1 2 (F(x)+ F(x+ dx)). Then the work dW as force times displacement is: dW = 1 2 (F(x)+ F(x+ dx)) dx = 1 2 (F(x)+ F(x)+ F′(x)dx) dx = F(x)dx + 1 2 F′(x)✟✟✟ ✯0 (dx)2 = F(x)dx Define the total work W over [a,b] as the sum of all the dW: The work performed by a force F(x) in displacing an object along the x-axis from x = a to x = b is: W = Zb a dW = Zb a F(x) dx (8.19) 10Note how this is different than claiming that F is “essentially constant over small intervals,” as most textbooks do. Instead, the additional infinitesimal force beyond F(x) contributes zero work over [x,x+ dx]. | ElementaryCalculus_Page_290_Chunk3825 |
Applications in Physics and Statistics • Section 8.5 281 Before continuing, some possible confusion needs to be cleared up. First, force is always a vector—it has both a magnitude and a direction. For the forces considered here, which act in a single dimension (e.g. along the x-axis), by convention the direction of the force is indicated by its sign: positive in the direction toward +∞, negative in the direction toward −∞. So a force of 3 N acts in the opposite direction as a force of −3 N, but they have the same magnitude |3| = 3. Second, work is not a vector—it is a scalar, meaning it has a magnitude but no direction. That magnitude can have any sign, though. Work is positive if the object is displaced in the same direction as the force, but is negative if the displacement is in the opposite direction of the force. For example, if you lift an object straight up from the ground, then you did positive work—the object moved in the same direction as the force you used. However, the force of gravity did negative work on the object as you lifted, since gravity works downward yet the object moved upward. F Fµ N −mg Last, zero work is done by a force if no displacement in its direction oc- curs. In particular, forces acting perpendicular to the line of displacement perform no work. For example, consider an object of mass m on a flat hori- zontal table top as in the figure on the right. If you push that object to the right with a force F (performing positive work), then both the downward force of gravity −mg and the upward normal force N exerted by the table perform zero work on the object. The force of friction Fµ from the table surface does negative work, as it opposes the force F. As an another example, no work is performed by holding a 100 lb object still and above the ground. Example 8.26 x 0 compress stretch Hooke’s law states that a coiled spring has an elastic restoring force F = −kx, where x is the displacement of the end of the spring from its equilibrium position as the spring is stretched or compressed, and k > 0 is the spring constant—or stiffness coefficient—specific to the spring. This force always tries to restore the spring to its equi- librium position, and the law holds only for a limited range of x. For a spring laid horizontally imagine it lies on the x-axis with the equilibrium position at x = 0, as in the figure on the right. (a) Find the spring constant k if a force of 2 N stretches the spring by 4 cm. (b) Use part (a) to find the work performed by compressing the spring 3 cm. Solution: (a) The force required to stretch the spring by an amount x is F = kx, since that force must counter the restoring force. Thus, k = F x = 2N 4cm = 2N 0.04m = 50 N/m. (b) By part (a) the force required to compress the string to position x is F(x) = kx = 50x, since again it must counter the restoring force. Thus, since 3 cm is 0.03 m, the work W performed is: W = Z−0.03 0 F(x) dx = Z−0.03 0 50x dx = 25x2 ¯¯¯¯ −0.03 0 = 25(−0.03)2 −0 = 0.0225 Nm | ElementaryCalculus_Page_291_Chunk3826 |
282 Chapter 8 • Applications of Integrals §8.5 Probability Suppose you flip two evenly balanced pennies and let X be the number of heads in the result. Then X is a discrete random variable—discrete because it can take only a discrete set of values (0, 1 and 2); random because its value is left to chance. The probability of a penny being flipped heads is 50% = 1 2, i.e. that is its theoretical likelihood since heads and tails are equally likely. The sample space S of all possible outcomes is the set S = {TT,TH,HT,HH}, where H is heads and T is tails (e.g. HT means the first penny came up heads and the second came up tails). Figure 8.5.4(a) shows a bar chart of the probabilities—as numbers between 0 and 1—with P(X = x) denoting the probability of the event that X equals the number x. Notice that the sum of the probabilities is 1, and P(X = x) = 0 if x is not 0, 1, or 2. P(X = x) x 0 1 2 1 4 1 2 probability (a) Discrete: P(X = x) f (x) x a b (b) Continuous: P(a < X < b) Figure 8.5.4 Probability: discrete vs continuous random variables The idea behind a continuous random variable X is to fill in those gaps between the bars in Figure 8.5.4(a), so that X would represent a continuous quantity, e.g. time, distance, temperature. Rather than finding P(X = x) you would find the probability that X is in a continuum such as an interval, e.g. P(a < X < b) (see Figure 8.5.4(b)). For a continuous random variable X define P(X = x) = 0 for all real x, and define the probability density function for X as a function f (x) ≥0 such that: (a) Z∞ −∞ f (x) dx = 1 (b) P(a < X < b) = Zb a f (x) dx for all a < b (including a = −∞and b = ∞) Notice that since P(X = a) = 0 then P(a ≤X < b) = P(a < X < b). In general, < and ≤are interchangeable for events involving continuous random variables (as well as > and ≥). In the remainder of this section it will be assumed that all random variables are continuous, for which the sample space is typically all of R or some interval, finite or infinite (e.g. (0,∞)). | ElementaryCalculus_Page_292_Chunk3827 |
Applications in Physics and Statistics • Section 8.5 283 Example 8.27 Let X be the lifetime—i.e. the time to failure—of an electronic component. If the average lifetime of the component is 700 days, then the probability density function f (x) for the random variable X is f (x) = ( λ e−λx if x ≥0, 0 if x < 0 (8.20) where λ = 1 700 and x is the number of days. In this case X is said to have the exponential distribution with parameter λ. Find the probability that the lifetime of the component is: (a) between 600 and 800 days (b) greater than 700 days Solution: (a) The probability is: P(600 < X < 800) = Z800 600 f (x) dx = Z800 600 1 700 e− x 700 dx = −e− x 700 ¯¯¯¯ 800 600 = −e−800 700 + e−600 700 ≈0.1055 Thus, there is about a 10.55% chance that the component’s lifetime will be between 600 and 800 days. (b) The probability is: P(X > 700) = Z∞ 700 f (x) dx = Z∞ 700 1 700 e− x 700 dx = −e− x 700 ¯¯¯¯ ∞ 700 = 0 + e−1 ≈0.3679 Exercises A For Exercises 1-3, find the center of gravity of the region bounded by the given curves over the given interval. 1. y = x3 and y = 0 ; 0 ≤x ≤1 2. y = −x+1 and y = 0 ; 0 ≤x ≤1 3. y = x2 and y = x3 ; 0 ≤x ≤1 4. Find the center of gravity of the region inside the circle x2 + y2 = r2 and above the x-axis. 5. Find the center of gravity of the region inside the circle x2 + y2 = r2 in the first quadrant. 6. Find the center of gravity of the region inside the ellipse x2 a2 + y2 b2 = 1 and above the x-axis. 7. Find the center of gravity of the region between the circle x2+ y2 = 4 and the ellipse x2 4 + y2 = 1 above the x-axis. 8. Would formula (8.18) for the center of gravity change if the mass of a region were proportional—but not equal—to its area, say, by a constant positive proportion δ ̸= 1? Explain. 9. If a spring requires 3 N of force to be compressed 5 cm, how much work would be performed in stretching the spring 8 cm? 10. The gravitational force F(x) exerted by the Earth on an object of mass m at a distance x from the center of the Earth is F(x) = −mgr2 e x2 where re is the radius of the Earth. If the object is released from rest at a distance ro from the center of the Earth, find the work performed by gravity in bringing the object to the Earth’s surface. | ElementaryCalculus_Page_293_Chunk3828 |
284 Chapter 8 • Applications of Integrals §8.5 11. Recall that the ideal gas law states that PV = RT, where R is a constant, P is the pressure, V is the volume, and T is the temperature. It can be shown that the work W done by an ideal gas in expanding the volume from Va to Vb is W = ZVb Va P dV . Calculate W. 12. Verify that Z∞ −∞ f (x) dx = 1 for the function f (x) in formula (8.20) in Example 8.27 for all λ > 0. 13. Find P(X < 300) in Example 8.27. 14. The distribution function F(x) for a random variable X is defined as F(x) = P(X ≤x) for all x. Show that F′(x) = f (x), where f (x) is the probability density function for X. B 15. Formula (8.18) can be extended to regions over an infinite interval, provided the area is finite. Use that fact to find the center of gravity of the region between y = e−x and the x-axis for 0 ≤x < ∞. 16. The expected value (or mean) E[X] of a random variable X with probability density function f (x) is E[X] = Z∞ −∞ x f (x) dx . Show that E[X] = 1 λ if X has the exponential distribution with parameter λ > 0. Note: The expected value can be thought of as the weighted average of all possible values of X, with weights determined by probability. It is analogous to the idea of a center of gravity. 17. A random variable X is said to have a normal distribution if its probability density function f (x) is f (x) = 1 σ p 2π e (x−µ)2 2σ2 for all x where σ > 0 and µ are constants. This is the famous “bell curve” in statistics. (a) Verify that Z∞ −∞ f (x) dx = 1. (Hint: Use Example 6.25 and a substitution.) (b) Show that E[X] = µ. (c) Use numerical integration to show that P(−1 < X < 1) ≈0.6827 when µ = 0 and σ = 1. 18. A random variable X has the beta distribution if its probability density function f (x) is f (x) = ( 1 B(a,b) xa−1(1−x)b−1 if 0 ≤x ≤1 0 elsewhere for positive constants a and b, where B(a,b) is the Beta function. Show that E[X] = a a+b. 19. Show that any value between F(x) and F(x+dx) for the force over [x,x+dx] gives the same formula dW = F(x) dx for the work performed over that interval. (Hint: Consider F(x+αdx) for 0 ≤α ≤1.) 20. A drop of water of mass M is released from rest at a height sufficient for the drop to evaporate completely, losing mass m each second (i.e. at a constant rate). Ignoring air resistance, show that the work performed by gravity on the drop up to complete evaporation is g2M2 6m2 . | ElementaryCalculus_Page_294_Chunk3829 |
Applications in Physics and Statistics • Section 8.5 285 21. This exercise is related to Einstein’s famous law E = mc2. The relativistic momentum p of a particle of mass m moving at a speed v along a straight line (say, the x-axis) is p = mv q 1−v2 c2 , where c is the speed of light. The relativistic force on the particle along that line is F = dp dt , which is the same formula as Newton’s Second Law of motion in classical mechanics. Assume that the particle starts at rest at position x1 and ends at position x2 along the x-axis. The work done by the force F on the particle is: W = Zx2 x1 F dx = Zx2 x1 dp dt dx (a) Show that dp dv = m ³ 1−v2 c2 ´3/2 . (b) Use the Chain Rule formula dp dt = dp dv dv dx dx dt to show that F dx = v dp dv dv . (c) Use parts (a) and (b) to show that W = Zv 0 dp dv v dv = Zv 0 mv ³ 1−v2 c2 ´3/2 dv . (d) Use part (c) to show that W = mc2 q 1−v2 c2 −mc2 . (e) Define the relativistic kinetic energy K of the particle to be K = W, and define the total energy E to be E = mc2 q 1−v2 c2 . So by part (d), K = E −mc2. Show that E2 = p2c2 + (mc2)2 . (Hint: Expand the right side of that equation.) (f) What is E when the particle is at rest? 22. A median of a triangle is a line segment from a vertex to the midpoint of the opposite side, and the three medians intersect at a common point. Show that this point is a triangle’s center of gravity. | ElementaryCalculus_Page_295_Chunk3830 |
CHAPTER 9 Infinite Sequences and Series 9.1 Sequences and Series In the 5th century B.C. the ancient Greek philosopher Zeno of Elea devised several paradoxes, the most famous of which—The Dichotomy—asserts that if space is infinitely divisible then motion is impossible. The argument goes like this: imagine a line segment of finite length, say 1 m, with a person at one end as in Figure 9.1.1. 0 ... 1 8 1 4 1 2 1 Figure 9.1.1 Zeno’s motion paradox Before traversing the entire distance the person would first need to travel one half the dis- tance. Before doing that, though, he would need to travel one fourth the distance, and before that one eighth the distance, and so on. There is thus no “first” distance for him to traverse, meaning his motion cannot even begin! Obviously motion is possible, or you would not be reading this. Does that mean Zeno’s reasoning is flawed? More on that later. In the meantime, notice a few things in Figure 9.1.1. First, the distance markers 1 2, 1 4, 1 8,... form an infinite sequence of numbers approaching 0. Second, the sum of the distances between successive markers is an infinite series which should equal the total segment length 1: 1 2 + 1 4 + 1 8 + ··· = 1 It will be shown shortly that the sum is indeed 1, which turns out to have no bearing on Zeno’s paradox. First some definitions are needed. 286 | ElementaryCalculus_Page_296_Chunk3831 |
Sequences and Series • Section 9.1 287 A sequence is an ordered list of objects, which in this book will always be real numbers. Sequences can be finite or infinite: finite if there is a last number in the list, infinite if every number in the list is followed by another number (i.e. a “successor”). Sequences should not be confused with sets—order matters in a sequence but not in a set, and numbers may repeat in a sequence but not in a set. For example, the sequences 〈1,2,3〉and 〈1,3,2〉are different, since order matters. However, the numbers in those sequences comprise the same set {1,2,3}. The simplest example of an infinite sequence is N, the set of natural numbers: 0,1,2,3,.... In fact, the numbers a0,a1,a2,... in any infinite sequence of real numbers can be written as the range of a function f mapping N into R: f (n) = an (9.1) Typically notation such as { an }∞ n=0 is used for representing an infinite sequence, or simply { an } when the initial value of the index n is understood (and n is always an integer). The intuitive notion of the limit of an infinite sequence can be stated formally: A real number L is the limit of an infinite sequence { an }, written as lim n→∞an = L or simply an →L , if for any given number ǫ > 0 there exists an integer N such that ¯¯an −L ¯¯ < ǫ for all n > N . (9.2) In this case the sequence { an } is said to converge to L and is called a convergent se- quence. A sequence that is not convergent is divergent. In other words, a sequence { an } converges to L if the terms an can be made arbitrarily close to L for n sufficiently large. In most cases the formal definition will not be needed, since by formula (9.1) the same rules and formulas from Chapters 1 and 3 for the limit of a function f (x) as x approaches ∞apply to sequences (e.g. sums and products of limits, L’Hôpital’s Rule). All you have to do is replace x by n. Example 9.1 For integers n ≥1 define an = 1 2n . Find lim n→∞an if it exists. Solution: Since lim x→∞ 1 2x = 0 then replacing x by n shows that lim n→∞an = lim n→∞ 1 2n = 0 . | ElementaryCalculus_Page_297_Chunk3832 |
288 Chapter 9 • Infinite Sequences and Series §9.1 Example 9.2 For integers n ≥0 define an = 2n+1 3n+2. Is { an } a convergent sequence? If so then find its limit. Solution: By L’Hôpital’s Rule, treating an integer n ≥0 as a real-valued variable x, lim n→∞an = lim n→∞ 2n+1 3n+2 = lim n→∞ 2 3 = 2 3 . Thus the sequence is convergent and its limit is 2 3. Example 9.3 For integers n ≥0 define an = en 3n+2. Is { an } a convergent sequence? If so then find its limit. Solution: By L’Hôpital’s Rule, treating an integer n ≥0 as a real-valued variable x, lim n→∞an = lim n→∞ en 3n+2 = lim n→∞ en 3 = ∞ Thus the sequence is divergent. Example 9.4 The famous Fibonacci sequence1 { Fn } starts with the numbers 0 and 1, then each successive term is the sum of the previous two terms: F0 = 0, F1 = 1, Fn = Fn−1 + Fn−2 for integers n ≥2 (9.3) Equation (9.3) is a recurrence relation. The first ten Fibonacci numbers are 0,1,1,2,3,5,8,13,21,34. Clearly { Fn } is a divergent sequence, since Fn →∞. For n ≥2 define an = Fn/Fn−1. The first few values are: a2 = F2 F1 = 1 1 = 1 , a3 = F3 F2 = 2 1 = 2 , a4 = F4 F3 = 3 2 = 1.5 , a5 = F5 F4 = 5 3 ≈1.667 Show that { an } is convergent. In other words, in the Fibonacci sequence the ratios of each term to the previous term converge to some number. Solution: There are many ways to prove this, perhaps the simplest being to assume the sequence is convergent and then find the limit (which would be impossible if the sequence were divergent). So assume that an →a for some real number a. Then divide both sides of formula (9.3) by Fn−1, so that Fn Fn−1 = 1 + Fn−2 Fn−1 ⇒ an = 1 + 1 an−1 for integers n ≥2 Now take the limit of both sides of the last equation as n →∞: lim n→∞an = lim n→∞ µ 1 + 1 an−1 ¶ ⇒ a = 1 + 1 a ⇒ a2 −a−1 = 0 ⇒ a = 1± p 5 2 Since a must be positive then a = 1+ p 5 2 . Thus, the sequence is convergent and converges to 1+ p 5 2 ≈1.618. Note: This number is the famous golden ratio, the subject of many claims regarding its appearance in nature and supposed aesthetic appeal as a ratio of sides of a rectangle.2 1Due to the Italian mathematician Leonardo Fibonacci (ca. 1170-1250). 2For example, see HUNTLEY, H.E., The Divine Proportion, New York: Dover Publications, Inc., 1970. | ElementaryCalculus_Page_298_Chunk3833 |
Sequences and Series • Section 9.1 289 An infinite series is the sum of an infinite sequence. If the infinite sequence is { an }∞ n=0 then the series can be written as ∞ X n=0 an = a0 + a1 + a2 +···+ an +··· or simply as Pan when the initial value of the index n is understood. There is a natural way to define the sum of such a series: The sum of an infinite series ∞ X n=0 an is ∞ X n=0 an = lim n→∞sn (9.4) where { sn }∞ n=0 is the sequence of partial sums of the series: sn = n X k=0 ak = a0 + a1 + a2 +···+ an for n ≥0 If the partial sums { sn } converge to a real number s then the series is convergent, and converges to s; if { sn } diverges then the series is divergent. One important convergent series is a geometric progression: a+ ar + ar2+ ar3 +···+ arn +··· (9.5) with a ̸= 0 and |r| < 1. Multiply the n-th partial sum sn by r: rsn = r(a+ ar + ar2+ ar3 +···+ arn−1 + arn) = ar + ar2+ ar3 + ar4 +···+ arn + arn+1 Now subtract rsn from sn: sn −rsn = a+ ar + ar2+ ar3 +···+ arn −(ar + ar2 + ar3 + ar4 +···+ arn + arn+1) sn −rsn = a−arn+1 so that lim n→∞sn = lim n→∞ a(1−rn+1) 1−r = a 1−r since rn+1 →0 as n →∞when |r| < 1. Thus: For a ̸= 0 and |r| < 1 the geometric progression ∞ X n=0 arn is convergent: ∞ X n=0 arn = a+ ar + ar2+ ar3 +···+ arn +··· = a 1−r (9.6) | ElementaryCalculus_Page_299_Chunk3834 |
290 Chapter 9 • Infinite Sequences and Series §9.1 Example 9.5 Show that ∞ X n=1 1 2n = 1. Solution: This is a geometric progression with a = 1 2 and r = 1 2. So by formula (9.6) the sum is: ∞ X n=1 1 2n = ∞ X n=0 µ1 2 ¶ · µ1 2 ¶n = a 1−r = 1 2 1−1 2 = 1 ✓ Example 9.6 Write the repeating decimal 0.17 = 0.17171717... as a rational number. Solution: This is a geometric progression with a = 0.17 = 17 100 and r = 0.01 = 1 100: 0.171717... = 0.17+0.0017+0.000017+··· = 0.17(0.01)0 +0.17(0.01)1 +0.17(0.01)2 +··· = ∞ X n=0 0.17(0.01)n = a 1−r = 0.17 1−0.01 = 0.17 0.99 = 17 99 Back to Zeno’s motion paradox, by Example 9.5 the sum of the infinite number of distances between successive markers in Figure 9.1.1 is 1, as expected. This fact is often mistaken as proof that Zeno was wrong, yet it does not actually address Zeno’s argument, since the person is attempting to begin motion at the tail end—the “infinite end”—of the geometric progression, not at the beginning (i.e. at n = 0). Zeno’s point remains that there is no “first step” at that tail end. In fact, even if you were to reverse the person’s position to start at the other end, so that he would first move a distance 1 2, then a distance 1 4, and so on, as in Figure 9.1.2, then a new problem is introduced: the person keeps moving no matter how close he is to the end point. There is now no “last step” and motion is thus still impossible. 0 ... 1 8 1 4 1 2 1 1 2 1 4 1 8 ··· Figure 9.1.2 Zeno’s motion paradox in reverse The convergence of the geometric progression has misled many people to argue that Zeno is wrong, by claiming it shows an infinite number of movements can be completed in a finite amount of time. But this line of argument fails on (at least) two counts. First, Zeno never argued about time—it is irrelevant to his paradox. The more fundamental flaw is that the introduction of time brings along the concept of speed, typically taken to be constant (though it need not be). Speed is distance over time, but it is precisely the distance part of that ratio that Zeno rejects—that distance can never be traveled. In other words, it is circular—and hence faulty—reasoning to “prove” that motion is possible by assuming it is possible. | ElementaryCalculus_Page_300_Chunk3835 |
Sequences and Series • Section 9.1 291 The geometric progression also doesn’t help when considering only the distances and ignor- ing time. Even assuming each of those distances could be traveled, the partial sums approach 1 but never actually reach it. The limit of a sequence is defined in terms of an inequality—you can get arbitrarily close to the limit, and that is all. The equality in formula (9.6) is merely a shorthand way of saying that. It is an abstraction based on properties of the real number system, not necessarily based on physical reality. Zeno’s paradox is not purely mathematical—it is about space and hence is physical, with a tinge of philosophy. Physicists have recognized this—and noted the flaw in the purely math- ematical line of attack—and devised new arguments against Zeno, some more sophisticated than others. However, they all invariably end up in some sort of circular reasoning. All this calls into question the original assumption: the infinite divisibility of space. If space had some smallest unit that could not be divided any further then there is no paradox—motion over a finite distance could always be decomposed into a large but finite number of irreducible steps.3 Exercises A For Exercises 1-8, determine if the given sequence is convergent. If so then find its limit. 1. { n e−n }∞ n=0 2. ½ n2 3n2 + 7n −2 ¾∞ n=1 3. ½ n2 3n3 + 7n −2 ¾∞ n=1 4. ½ n3 3n2 + 7n −2 ¾∞ n=1 5. n n ln n o∞ n=2 6. ½ n! (n+2)! ¾∞ n=0 7. n sin ³nπ 2 ´ o∞ n=0 8. { (−1)n cos nπ }∞ n=0 For Exercises 9-12 determine whether the given series is convergent. If so then find its sum. 9. ∞ X n=0 2 µ2 3 ¶n 10. ∞ X n=1 7−n 11. ∞ X n=0 3n + 5n+1 6n 12. ∞ X n=0 n For Exercises 13-16 use a geometric progression to write the given repeating decimal as a rational number. 13. 0.113 14. 0.9 15. 0.249 16. 0.017 B 17. In Example 9.4 define bn = Fn/Fn+1 for n ≥0 and show that { bn }∞ n=0 converges to p 5−1 2 . 18. Show that formula (4.1) for Newton’s method with f (x) = x2 −2 and x0 = 1 yields the sequence { xn }∞ n=0 with xn = 1 2 xn−1 + 1 xn−1 for n ≥1 , then assuming { xn } is convergent show that it must converge to p 2. (Hint: See Example 9.4.) 3As of this writing there is not yet a definitive answer as to whether space is continuous or discrete (quantized). A “smallest unit” would have to be below the Planck level—around 10−33 cm, well below current measurement capabilities. Some recent advances in the field of loop quantum gravity suggest the possibility of quantized space. See CHAMSEDDINE, A.H., CONNES, A. & MUKHANOV, V., “Geometry and the quantum: basics.” J. High Energy Phys. 2014, 98 (2014). https://doi.org/10.1007/JHEP12(2014)098 | ElementaryCalculus_Page_301_Chunk3836 |
292 Chapter 9 • Infinite Sequences and Series §9.1 19. In this exercise you will prove a formula for the general number Fn in the Fibonacci sequence. Denote the positive and negative solutions to the equation x2−x−1 = 0 by φ+ = 1+ p 5 2 and φ−= 1− p 5 2 , respectively. Note that φ+ is the golden ratio mentioned in Example 9.4. (a) Use the equation x2 −x−1 = 0 to show that φn+1 + = φn + + φn−1 + and φn+1 − = φn −+ φn−1 − . (b) Use part (a) and induction to show that Fn = φn + −φn − p 5 for n ≥0. 20. A ball is dropped from a height of 4 ft above the ground, and upon each bounce off the ground the ball bounces straight up to a height equal to 65% of its previous height. Find the theoretical total distance the ball could travel if it could bounce indefinitely. Why is this physically unrealistic? 21. A continued fraction is a type of infinite sum involving a fraction with a denominator that continues indefinitely: a = a0 + 1 a1 + 1 a2 + 1 a3 + ··· (a) Show that the golden ratio φ+ = 1+ p 5 2 (a solution of x2 −x−1 = 0) can be written as φ+ = 1 + 1 1 + 1 1 + 1 1 + ··· . (Hint: Look for a recurrence relation in the fraction.) (b) Show that p 2 = 1 + 1 2 + 1 2 + 1 2 + ··· . 22. Show that the golden ratio φ+ = 1+ p 5 2 can be written as φ+ = q 1+ p 1+ p 1+···. 23. Write a computer program to approximate the result in Exercise 22 using 100 terms. After how many iterations do the approximate values start repeating? 24. Would the existence of infinitesimals as a measure of space resolve Zeno’s paradox? Explain. | ElementaryCalculus_Page_302_Chunk3837 |
Tests for Convergence • Section 9.2 293 9.2 Tests for Convergence There are many ways to determine if a sequence converges—two are listed below. In all cases changing or removing a finite number of terms in a sequence does not affect its convergence or divergence: 1. Monotone Bounded Test: A sequence that is bounded and monotone—i.e. either always increasing or always decreasing—is convergent. 2. Comparison Test: If { bn } diverges to ∞and if { an } is a sequence such that an ≥bn for all n > N for some N, then { an } also diverges to ∞. Likewise, if { cn } diverges to −∞and an ≤cn for all n > N for some N, then { an } also diverges to −∞. x a1 a2 a3 a4 ··· →M Figure 9.2.1 The Comparison Test makes sense intuitively, since something larger than a quantity going to infinity must also go to infinity. The Monotone Bounded Test can be understood by thinking of a bound on a sequence as a wall that the sequence can never pass, as in Figure 9.2.1. The increasing sequence { an } in the figure moves toward M but can never pass it. The sequence thus cannot diverge to ∞, and it cannot fluctuate back and forth since it always increases. Thus it must converge somewhere before or at M.4 Notice that the Monotone Bounded Test tells you only that the sequence converges, not what it converges to. Example 9.7 Show that the sequence { an }∞ n=1 defined for n ≥1 by an = 1 · 3 · 5 ··· (2n−1) 2 · 4 · 6 ··· (2n) is convergent. Solution: Notice that { an } is always decreasing, since an+1 = 1 · 3 · 5 ··· (2n−1) · (2n+1) 2 · 4 · 6 ··· (2n) · (2n+2) = an · 2n+1 2n+2 < an · (1) = an for n ≥1. The sequence is also bounded, since 0 < an and an = 1 2 · 3 4 · 5 6 ··· 2n−1 2n < 1 for n ≥1 since each fraction in the above product is less than 1. Thus, by the Monotone Bounded Test the sequence is convergent. Note that for a decreasing sequence only the lower bound is needed for the Monotone Bounded Test, not the upper bound. Similarly, for an increasing sequence only the upper bound matters. 4For a proof see pp.48-49 in BUCK, R.C., Advanced Calculus, 2nd ed., New York: McGraw-Hill Book Co., 1965. | ElementaryCalculus_Page_303_Chunk3838 |
294 Chapter 9 • Infinite Sequences and Series §9.2 Some tests for convergence of a series are listed below: 1. n-th Term Test: If Pan converges then lim n→∞an = 0. 2. Ratio Test: For a series Pan of positive terms let R = lim n→∞ an+1 an . Then (a) if R < 1 then the series converges, (b) if R > 1 (including R = ∞) then the series diverges, (c) if R = 1 then the test fails. 3. Integral Test: For a series ∞ X n=1 an of positive terms let f (x) be a decreasing function on [1,∞) such that f (n) = an for all integers n ≥1. Then ∞ X n=1 an and Z∞ 1 f (x) dx either both converge or both diverge. 4. p-series Test: The series ∞ X n=1 1 np converges for p > 1, and diverges for p ≤1. 5. Comparison Test: If 0 ≤an ≤bn for n > N for some N, and if Pbn is convergent then Pan is convergent. Similarly, if 0 ≤bn ≤an for n > N for some N, and if Pbn is divergent then Pan is divergent. 6. Limit Comparison Test: For two series Pan and Pbn of positive terms let L = lim n→∞ an bn . Then (a) if 0 < L < ∞then Pan and Pbn either both converge or both diverge, (b) if L = 0 and Pbn converges then Pan converges, (c) if L = ∞and Pbn diverges then Pan diverges. 7. Telescoping Series Test: Suppose Pan = P(bn −bn+1) for some sequence { bn }. Then Pan converges if and only if bn →L, in which case Pan = b1 −L. | ElementaryCalculus_Page_304_Chunk3839 |
Tests for Convergence • Section 9.2 295 Most of the above tests have fairly short proofs or at least intuitive explanations. For exam- ple, the n-th Term Test follows from the definition of convergence of a series: if Pan converges to a number L then since each term an = sn −sn−1 is the difference of successive partial sums, taking the limit yields lim n→∞an = lim n→∞(sn −sn−1) = L −L = 0 by definition of the convergence of a series. ✓ Since the n-th Term Test can never be used to prove convergence of a series, it is often stated in the following logically equivalent manner: n-th Term Test: If lim n→∞an ̸= 0 then Pan diverges. Example 9.8 Show that ∞ X n=1 n 2n+1 = 1 3 + 2 5 + 3 7 +··· is divergent. Solution: Since lim n→∞ n 2n+1 = 1 2 ̸= 0 then by the n-th Term Test the series diverges. The Ratio Test takes a bit more effort to prove.5 When the ratio R in the Ratio Test is larger than 1 then that means the terms in the series do not approach 0, and thus the series diverges by the n-th Term Test. When R = 1 the test fails, meaning it is inconclusive—another test would need to be used. When the test shows convergence it does not tell you what the series converges to, merely that it converges. Example 9.9 Determine if ∞ X n=1 n 2n is convergent. Solution: For the series general term an = n 2n , R = lim n→∞ an+1 an = lim n→∞ n+1 2n+1 n 2n = lim n→∞ n+1 2n = 1 2 < 1 , so by the Ratio Test the series converges. 5See pp.612-613 in TAYLOR, A.E. AND W.R. MANN, Advanced Calculus, 2nd ed., New York: John Wiley & Sons, Inc., 1972. | ElementaryCalculus_Page_305_Chunk3840 |
296 Chapter 9 • Infinite Sequences and Series §9.2 Figure 9.2.2 shows why the Integral Test works. a1 a2 a3 a4 a5 y x 0 1 2 3 4 5 y = f (x) (a) Z∞ 1 f (x) dx > ∞ X n=2 an a1 a2 a3 a4 y x 0 1 2 3 4 5 y = f (x) (b) Z∞ 1 f (x) dx < ∞ X n=1 an Figure 9.2.2 Integral Test In Figure 9.2.2(a) the area R∞ 1 f (x)dx is greater than the total area S of all the rectangles under the curve. Since each rectangle has height an and width 1, then S = P∞ 2 an. Thus, since removing the single term a1 from the series does not affect the convergence or divergence of the series, the series converges if the improper integral converges, and conversely the integral diverges if the series diverges. Similarly, in Figure 9.2.2(b) the area R∞ 1 f (x)dx is less than the total area S = P∞ 1 an of all the rectangles, so the integral converges if the series converges, and the series diverges if the integral diverges. Notice how in both graphs the rectangles are either all below the curve or all protrude above the curve due to f (x) being a decreasing function. Example 9.10 Show that the p-series ∞ X n=1 1 np converges for p > 1 and diverges for p = 1. Solution: For p ≥1 let f (x) = 1 xp on [1,∞). Then f (x) is decreasing, and f (n) = an = 1 np for all integers n ≥1. For p > 1, Z∞ 1 f (x) dx = Z∞ 1 dx xp = −1 (p−1)xp−1 ¯¯¯¯ ∞ 1 = 0 − µ −1 (p−1)(1)p−1 ¶ = 1 p−1 < ∞ so the integral converges. Thus, by the Integral Test, the series converges for p > 1. For p = 1, Z∞ 1 f (x) dx = Z∞ 1 dx x = ln x ¯¯¯¯ ∞ 1 = ∞ so the integral diverges. So by the Integral Test, the series diverges for p = 1. The harmonic series ∞ X n=1 1 n = 1 + 1 2 + 1 3 + 1 4 + ··· thus diverges even though an = 1 n →0 (which is hence not a sufficient condition for a series to converge). Note that this example partly proves the p-series Test. The remaining case (p < 1) is left as an exercise. | ElementaryCalculus_Page_306_Chunk3841 |
Tests for Convergence • Section 9.2 297 The divergence part of the Comparison Test is clear enough to understand, but for the con- vergence part with 0 ≤an ≤bn for all n larger than some N, ignore the (finite) number of terms before aN and bN. Since Pbn converges then its partial sums must be bounded. The partial sums for Pan then must also be bounded, since 0 ≤an ≤bn for n > N. So since an ≥0 means that the partial sums for Pan are increasing, by the Monotone Bounded Test the partial sums for Pan must converge, i.e. Pan is convergent. Example 9.11 Determine if ∞ X n=1 1 nn is convergent. Solution: Since nn ≥n2 > 0 for n > 2, then 0 < 1 nn ≤ 1 n2 for n > 2. Thus, since P∞ n=1 1 n2 converges (by the p-series Test with p = 2 > 1, as in Example 9.10), the series P∞ n=1 1 nn converges by the Comparison Test. For the Limit Comparison Test with an bn →L < ∞and L > 0, by definition of the limit of a sequence, an bn can be made arbitrarily close to L. In particular there is an integer N such that L 2 < an bn < 3L 2 for all n > N. Then 0 < an < 3L 2 bn and X bn converges ⇒ X an converges by the Comparison test. Likewise, 0 < L 2 bn < an and X bn diverges ⇒ X an diverges by the Comparison Test again. The cases L = 0 and L = ∞are handled similarly. Example 9.12 Determine if ∞ X n=1 n+3 n · 2n is convergent. Solution: Since P∞ n=1 1 2n is convergent (as part of a geometric progression) and lim n→∞ (n+3)/(n · 2n) 1/2n = lim n→∞ n+3 n = 1 then by the Limit Comparison Test P∞ n=1 n+3 n·2n is convergent.. | ElementaryCalculus_Page_307_Chunk3842 |
298 Chapter 9 • Infinite Sequences and Series §9.2 A series Pan is telescoping if an = bn −bn+1 for some sequence { bn }. Assume the series Pan and sequence { bn } both start at n = 1. Then the partial sum sn for Pan is sn = a1 + a2 + ··· + an = (b1 −b2) + (b2 −b3) + ··· + (bn −bn+1) = b1 −bn+1 for n ≥1. Thus, since b1 is a fixed number, limn→∞sn exists if and only if limn→∞bn+1 exists, i.e. Pan converges if and only if { bn } converges. So if bn →L then sn →b −L, i.e. Pan converges to L, which proves the Telescoping Series Test. Note that the number b1, as the first number in the sequence { bn }, could be replaced by whatever the first number is, in case the index n starts at a number different from 1. Example 9.13 Determine if ∞ X n=1 1 n(n+1) is convergent. If it converges then can you find its sum? Solution: For the sequence { bn } with bn = 1 n for n ≥1, each term in the series can be written as 1 n(n+1) = 1 n − 1 n+1 = bn −bn+1 Thus, since { bn } converges to 0, by the Telescoping Series Test the series also converges, to b1 −0 = 1. Convergent series have the following properties (based on similar properties of limits): Let Pan and Pbn be convergent series, and let c be a number. Then: (a) P(an ± bn) is convergent, with P(an ± bn) = Pan ±Pbn (b) P can is convergent, with P can = c · Pan Exercises A For Exercises 1-5 show that the given sequence { an }∞ n=1 is convergent. 1. an = 2 · 4 · 6 ··· (2n) 3 · 5 · 7 ··· (2n+1) 2. an = 1 −2n n! 3. an = 2 · 4 · 6 ··· (2n) 1 · 3 · 5 ··· (2n−1) · 1 2n+2 4. an = 1 n µ 2 · 4 · 6 ··· (2n) 1 · 3 · 5 ··· (2n−1) ¶2 5. an = 1 2 · 3 2 · 3 4 · 5 4 · 5 6 · 7 6 ··· 2n−1 2n · 2n+1 2n For Exercises 6-17 determine whether the given series is convergent. 6. ∞ X n=0 sin ³nπ 2 ´ 7. ∞ X n=1 1 n(n+2) 8. ∞ X n=1 1 2n 9. ∞ X n=1 n (n+1)2n | ElementaryCalculus_Page_308_Chunk3843 |
Tests for Convergence • Section 9.2 299 10. ∞ X n=2 1 n p ln n 11. ∞ X n=1 n! (2n)! 12. ∞ X n=1 n en 13. ∞ X n=1 1 cosh2 n 14. ∞ X n=1 n! 2n 15. ∞ X n=1 1 pn 16. ∞ X n=1 1 n(2n−1) 17. ∞ X n=1 ln(n+1) n2 For Exercises 18-21 determine whether the given series is convergent. If convergent then find its sum. 18. ∞ X n=1 1 (2n+1)(2n+3) 19. ∞ X n=1 1 (2n+3)(2n+5) 20. ∞ X n=1 2 (3n+1)(3n+4) 21. ∞ X n=1 1 4n2 −1 B 22. Continue Example 9.10 with a proof of the p-series Test for p < 1. 23. Show that { an }∞ n=1 is convergent, where an = 1 1! + 1 2! + 1 3! + 1 4! + ··· + 1 n! for n ≥1. (Hint: Use the Monotone Bounded test by using a bound on 1 n! for n > 2.) 24. Consider the series ∞ X n=1 1 2n−1 = 1 + 1 3 + 1 5 + 1 7 + ··· . (a) Show that the series is divergent. (b) The textbook Applied Mathematics for Physical Chemistry (3rd ed.), J. Barrante, provides the following argument that the above series converges: Since 1 + 1 4 + 1 9 + 1 16 + ··· < 1 + 1 3 + 1 5 + 1 7 + ··· < 1 + 1 2 + 1 3 + 1 4 + ··· where the series on the left converges (by the p-series Test with p = 2) and the series on the right diverges (by the p-series Test with p = 1), and since each term in the middle series is between its corresponding terms in the left series and right series, then there must be a p-series for some value 1 < p < 2 such that each term in the middle series is less than the corresponding term in that p-series. That is, 1 + 1 4 + 1 9 + 1 16 + ··· < 1 + 1 3 + 1 5 + 1 7 + ··· < 1 + 1 2p + 1 3p + 1 4p + ··· for that value of p between 1 and 2. But p > 1 for that p-series on the right, so it converges, which means that the middle series converges! Find and explain the flaw in this argument. 25. Wallis’ formula6 for π is given by the infinite product π 2 = 2 1 · 2 3 · 4 3 · 4 5 · 6 5 · 6 7 ··· 2n 2n−1 · 2n 2n+1 · ··· . Notice that this is the limit of the reciprocal of the sequence in Exercise 5. Write a computer program to approximate the limit using 1 million iterations. How close is your approximation to π 2 ? 6Due to the English mathematician and theologian John Wallis (1616-1703). For a proof of the formula see pp.738- 739 in TAYLOR, A.E. AND W.R. MANN, Advanced Calculus, 2nd ed., New York: John Wiley & Sons, Inc., 1972. | ElementaryCalculus_Page_309_Chunk3844 |
300 Chapter 9 • Infinite Sequences and Series §9.3 9.3 Alternating Series In the last section the harmonic series ∞ X n=1 1 n = 1 + 1 2 + 1 3 + 1 4 + 1 5 + ··· was shown to diverge. If you were to alternate the signs of successive terms, as in ∞ X n=1 (−1)n−1 n = 1 −1 2 + 1 3 −1 4 + 1 5 −··· (9.7) then it turns out that this new series—called an alternating series—converges, due to the following test: Alternating Series Test: If Pan is an alternating series—i.e. the signs of the terms an alternate between positive and negative—such that the absolute values of the terms are decreasing to 0, then the series converges. The condition for the test means that |an+1| ≤|an| for all n and |an| →0 as n →∞. To see why the test works, consider the alternating series given above by formula (9.7), with an = −1n−1 n . The odd-numbered partial sums s1, s3, s5, ..., can be written as s1 = 1 , s3 = 1 − µ1 2 −1 3 ¶ | {z } > 0 , s5 = 1 − µ1 2 −1 3 ¶ | {z } > 0 − µ1 4 −1 5 ¶ | {z } > 0 , ... while the even-numbered partial sums s2, s4, s6, ..., can be written as s2 = 1 −1 2 , s4 = 1 −1 2 + µ1 3 −1 4 ¶ | {z } > 0 , s6 = 1 −1 2 + µ1 3 −1 4 ¶ | {z } > 0 + µ1 5 −1 6 ¶ | {z } > 0 , ... Thus the odd-numbered partial sums { s2n−1 } are decreasing from s1 = 1, and the even-numbered ones { s2n } are increasing from s2 = 1−1 2 = 1 2, with 1 2 < sn < 1 for all n, i.e. the partial sums are bounded. So by the Monotone Bounded Test, both sequences { s2n−1 } and { s2n } must converge. Since s2n −s2n−1 = a2n = −1 2n for all n ≥1, then lim n→∞(s2n −s2n−1) = lim n→∞ −1 2n = 0 ⇒ lim n→∞s2n = lim n→∞s2n+1 Thus, the partial sums sn have a common limit, so the series converges. Notice that the key to the convergence was having the terms decreasing in absolute value to zero. | ElementaryCalculus_Page_310_Chunk3845 |
Alternating Series • Section 9.3 301 Example 9.14 Determine if ∞ X n=2 (−1)n ln n is convergent. Solution: For the general term an = (−1)n ln n , since ln(n+1) > ln n for n ≥2 and ln n →∞as n →∞, then |an| decreases to 0 as n →∞. Thus, by the Alternating Series Test the series converges. The series P∞ n=1 (−1)n−1 n converges by the Alternating Series Test, though the series P∞ n=1 1 n diverges. This makes P∞ n=1 (−1)n−1 n an example of a conditionally convergent series: A series Pan is conditionally convergent if Pan converges but P|an| diverges. If P|an| converges then Pan is absolutely convergent. For example, P∞ n=1 (−1)n−1 n is not absolutely convergent, since P∞ n=1 1 n diverges. Example 9.15 Is ∞ X n=1 (−1)n−1 n2 conditionally convergent or absolutely convergent? Solution: Since P∞ n=1 1 n2 converges (by the p-series Test) then P∞ n=1 (−1)n−1 n2 is absolutely convergent. It turns out that absolute convergence implies ordinary convergence: Absolute Convergence Test: If P|an| converges then Pan converges. The test is obvious if the terms an are all positive, so assume the series has both positive terms (denoted by the sequence © apos ª ) and negative terms (denoted by © aneg ª ). Then the series can be decomposed into the difference of two series: X an = X apos − X |aneg| Since each of the sums on the right side of the equation is part of the convergent series P|an|, then each sum itself converges (being part of a finite sum). Thus their difference, namely Pan, is finite, i.e. Pan converges. The test can be stated in the following logically equivalent manner: Absolute Convergence Test: If Pan diverges then P|an| diverges. | ElementaryCalculus_Page_311_Chunk3846 |
302 Chapter 9 • Infinite Sequences and Series §9.3 One unusual feature of a conditionally convergent series is that its terms can be rearranged to converge to any number, a result known as Riemann’s Rearrangement Theorem. For example, the alternating harmonic series 1 −1 2 + 1 3 −1 4 + 1 5 −··· consists of one divergent series of positive terms subtracted from another series of positive terms, namely: 1 −1 2 + 1 3 −1 4 + 1 5 −··· = µ 1 + 1 3 + 1 5 + 1 7 + ··· ¶ − µ1 2 + 1 4 + 1 6 + ··· ¶ The idea is that since the first series on the right diverges then that series has some partial sum that could be made just larger than any positive number A. Likewise, since the second series on the right that is being subtracted also diverges, it has some partial sum that when subtracted from the first partial sum results in a number just less than A. Continue like this indefinitely, first adding a partial sum to get a number just bigger than A then subtracting another partial sum to get just less than A. Since the terms in each series approach zero, the overall series can be made to converge to A! It also turns out that an absolutely convergent series does not have this feature—any rearrangement of terms results in the same sum.7 Exercises A For Exercises 1-5 determine whether the given alternating series is convergent. If convergent, then determine if it is conditionally convergent or absolutely convergent. 1. ∞ X n=1 (−1)n−1 pn 2. ∞ X n=1 (−1)n−1 n! 2n 3. ∞ X n=1 (−1)n−1 2n−1 4. ∞ X n=2 (−1)n n ln n 5. ∞ X n=1 (−1)n−1 n! 6. Rearrangements of a divergent alternating series can make it appear to converge to different num- bers. For example, find different rearrangements of the terms in the divergent series ∞ X n=0 (−1)n = 1 −1 + 1 −1 + ··· so that the series appears to converge to 0, 1, -1, and 2. 7. A guest arrives at the Aleph Null Hotel,8 which has an infinite number of rooms, numbered Room 0, Room 1, Room 2, and so on. The hotel manager says all the rooms are taken, but he can still give the guest his own room. How is that possible? Would it still be possible if an infinite number of guests showed up, each wanting his own room? Explain your answers. 7For formal proofs of these statements, see pp.442-444 in KLAMBAUER, G., Aspects of Calculus, New York: Springer-Verlag, 1986. 8The symbol ℵ0 is called aleph null and represents the cardinality of N, i.e. its size. ℵ0 is the smallest infinity. | ElementaryCalculus_Page_312_Chunk3847 |
Power Series • Section 9.4 303 9.4 Power Series A power series is an infinite series whose terms involve constants an and powers of x −c, where x is a variable and c is a constant: P an (x−c)n. In many cases c will be 0. For example, the geometric progression ∞ X n=0 rn = 1 + r + r2 + r3 + ··· = 1 1−r converges when |r| < 1, i.e. for −1 < r < 1, as shown in Section 9.1. Replacing the constant r by a variable x yields the power series ∞ X n=0 xn = 1 + x + x2 + x3 + ··· = 1 1−x (9.8) that converges to 1 1−x when −1 < x < 1. Note that the series diverges for |x| ≥1, by the n-th Term Test. In general a power series of the form P fn(x), where fn(x) = an(x −c)n is a sequence of func- tions, has an interval of convergence defined as the set of all x such that the series con- verges. The interval can be any combination of open or closed, as well as the extreme cases of a single point or all real numbers. On its interval of convergence the power series is thus a function of x. The radius of convergence R of a power series is defined as half the length of the interval of convergence. In the case where the interval of convergence is all of R you would say R = ∞. For example, for the above power series P∞ n=0 fn(x), where fn(x) = xn for n ≥0, the interval of convergence is −1 < x < 1, so the radius of convergence is R = 1. Notice that f (x) = ∞ X n=0 xn for −1 < x < 1 is thus a well-defined function on the interval (−1,1), where it happens to equal 1 1−x. This power series can be thought of as a polynomial of infinite degree. To find the interval of convergence of a power series P fn(x), you typically would use the Ratio Test on the absolute values of the terms (since the Ratio Test requires positive terms): r(x) = lim n→∞ ¯¯¯¯ fn+1(x) fn(x) ¯¯¯¯ (9.9) Note that the limit r(x) in this case is a function of x. When taking the limit, though, treat x as fixed. By the Ratio Test the power series will then converge for all x such that r(x) < 1, and diverge when r(x) > 1. When r(x) = 1 the test is inconclusive, so you would have to check those cases individually to see if those values of x should be added to the interval of convergence (along with the points where r(x) < 1). | ElementaryCalculus_Page_313_Chunk3848 |
304 Chapter 9 • Infinite Sequences and Series §9.4 Example 9.16 Find the interval of convergence of the power series ∞ X n=0 xn n! . Solution: For fn(x) = xn n! , r(x) = lim n→∞ ¯¯¯¯ fn+1(x) fn(x) ¯¯¯¯ = lim n→∞ ¯¯¯¯ xn+1/(n+1)! xn/n! ¯¯¯¯ = |x| · lim n→∞ ¯¯¯¯ 1 n+1 ¯¯¯¯ = |x| · 0 = 0 for any fixed x. Thus, r(x) = 0 < 1 for all x, so the interval of convergence is all of R, i.e. −∞< x < ∞. Example 9.17 Find the interval of convergence of the power series ∞ X n=1 xn n . Solution: For fn(x) = xn n , r(x) = lim n→∞ ¯¯¯¯ fn+1(x) fn(x) ¯¯¯¯ = lim n→∞ ¯¯¯¯ xn+1/(n+1) xn/n ¯¯¯¯ = lim n→∞ ¯¯¯¯ nx n+1 ¯¯¯¯ = |x| · lim n→∞ ¯¯¯¯ n n+1 ¯¯¯¯ = |x| · 1 = |x| for any fixed x. Thus, the series converges when r(x) = |x| < 1 and diverges when r(x) = |x| > 1. The cases r(x) = |x| = 1 need to be checked individually. For x = 1 the series is P∞ n=1 1 n, which diverges. For x = −1 the series is P∞ n=1 (−1)n−1 n , which converges. Thus, the interval of convergence is −1 ≤x < 1. Example 9.18 Find the interval of convergence of the power series ∞ X n=0 n! xn . Solution: For fn(x) = n! xn, r(x) = lim n→∞ ¯¯¯¯ fn+1(x) fn(x) ¯¯¯¯ = lim n→∞ ¯¯¯¯ (n+1)! xn+1 n! xn ¯¯¯¯ = |x| · lim n→∞|n+1| = ( 0 if x = 0, ∞ if x ̸= 0. Thus, r(x) = ∞> 1 for all x ̸= 0, so the series diverges for x ̸= 0. So since r(x) = 0 < 1 only for x = 0, the interval of convergence is the single point x = 0. It turns out that power series can be both differentiated and integrated term by term:9 For a power series f (x) = ∞ X n=0 an (x−c)n that converges for |x−c| < R, both f ′(x) = ∞ X n=1 nan (x−c)n−1 and Z f (x) dx = C + ∞ X n=0 an n+1 (x−c)n+1 (9.10) converge for |x−c| < R. 9The proofs require uniform convergence, a stronger condition than ordinary convergence. See pp.129-134 in BROMWICH, T.J., An Introduction to the Theory of Infinite Series, 2nd ed., London: Macmillan & Co. Ltd., 1955. | ElementaryCalculus_Page_314_Chunk3849 |
Power Series • Section 9.4 305 Notice that the above statement says nothing about the convergence of f ′(x) or R f (x)dx at the endpoints of the interval |x −c| < R. In each case convergence at the endpoints can be checked individually. Example 9.19 Write the power series form of the derivative of f (x) = ∞ X n=0 xn and find its interval of convergence. Can f ′(x) be written in a non-series form?. Solution: Differentiate f (x) term by term: f ′(x) = d dx à ∞ X n=0 xn ! = d dx (1 + x + x2 + x3 + ···) = 0 + 1 + 2x + 3x2 + ··· = ∞ X n=1 nxn−1 Since f (x) converges for −1 < x < 1 then so does f ′(x). Checking the endpoints, at x = 1 and x = −1 the series for f ′(x) are P∞ n=1 n and P∞ n=1(−1)n−1 n, respectively, both of which diverge by the n-th Term Test. Thus, the interval of convergence for f ′(x) is−1 < x < 1. Since f (x) = 1 1−x for −1 < x < 1 then f ′(x) = 1 (1−x)2 for −1 < x < 1. Thus, ∞ X n=1 nxn−1 = 1 + 2x + 3x2 + ··· = 1 (1−x)2 for −1 < x < 1. Bessel Functions Many applications in engineering and physics—especially those involving oscillations or me- chanical vibrations—require solving differential equations of the form d2y dx2 + 1 x dy dx + y = 0 , (9.11) known as Bessel’s equation. This equation has a solution J0(x), known as Bessel’s function of order zero,10 defined in terms of a power series: J0(x) = ∞ X n=0 (−1)n x2n (n!)2 · 22n = 1 −x2 22 + x4 22 · 42 − x6 22 · 42 · 62 + ··· (9.12) The Ratio Test shows that J0(x) converges for all x, since for any fixed x, r(x) = lim n→∞ ¯¯¯¯¯¯¯¯¯ (−1)n+1 x2n+2 ((n+1)!)2 · 22n+2 (−1)n x2n (n!)2 · 22n ¯¯¯¯¯¯¯¯¯ = x2 · lim n→∞ ¯¯¯¯ 1 4(n+1)2 ¯¯¯¯ = 0 < 1 . 10Due to the German astronomer Friedrich Wilhelm Bessel (1784-1846), from a study of elliptic planetary motion. For some historical background and examples of how Bessel’s equation arises in applications, see § I.2 and § I.8 in MCLACHLAN, N.W., Bessel Functions for Engineers, London: Oxford University Press, 1934. | ElementaryCalculus_Page_315_Chunk3850 |
306 Chapter 9 • Infinite Sequences and Series §9.4 The general Bessel equation of order m, d2y dx2 + 1 x dy dx + µ 1−m2 x2 ¶ y = 0 (9.13) for m = 0,1,2,..., has a solution Jm(x), called Bessel’s function of order m: Jm(x) = ∞ X n=0 (−1)n 1 n! · (n+ m)! ³ x 2 ´2n+m (9.14) For example, the Bessel function J1(x) of order 1 is J1(x) = ∞ X n=0 (−1)n 1 n! · (n+1)! ³ x 2 ´2n+1 = x 2 − x3 22 · 4 + x5 22 · 42 · 6 − x7 22 · 42 · 62 · 8 + ··· Term by term differentiation shows that J′ 0(x) = −J1(x): d dx (J0(x)) = d dx µ 1 −x2 22 + x4 22 · 42 − x6 22 · 42 · 62 + x8 22 · 42 · 62 · 82 −··· ¶ = −x 2 + x3 22 · 4 − x5 22 · 42 · 6 + x7 22 · 42 · 62 · 8 −··· = −J1(x) Graphs of J0(x) and J1(x) are shown in Figure 9.4.1 below. As you can see, J0(x) and J1(x) behave as sort of “poor man’s” cosine and sine functions, respectively. 0 5 10 15 20 25 -1 -0.5 0 0.5 1 x y J0(x) J1(x) Figure 9.4.1 Bessel functions J0(x) and J1(x) | ElementaryCalculus_Page_316_Chunk3851 |
Power Series • Section 9.4 307 Exercises A For Exercises 1-8 find the interval of convergence of the given power series. 1. ∞ X n=1 nxn (n+1)2 2. ∞ X n=1 nxn 2n 3. ∞ X n=1 n2 (x−2)n 4. ∞ X n=0 (x+4)n 2n 5. ∞ X n=1 (x+1)n nn 6. ∞ X n=1 nn xn 7. ∞ X n=0 (−1)n xn 8. ∞ X n=1 nxn n+1 9. Note that power series of the form P∞ n=0 anxn have an issue at x = 0 when n = 0: 00 is an indeter- minate form—it can equal anything (or nothing). What value has it implicitly been assigned so far? What would be the technically correct way to write the series P∞ n=0 anxn so that this issue goes away? 10. Show that ∞ X n=1 nxn = x (1−x)2 for −1 < x < 1. 11. Write the following infinite series as a rational number: 1 10 + 2 100 + 3 1000 + 4 10000 + ··· B 12. Differentiating term by term, verify that the Bessel function J0(x) satisfies Bessel’s equation (see equation (9.11)). 13. Show that for all m ≥1 the Bessel functions Jm(x) converge for all x. 14. For all m ≥1 verify that the Bessel functions Jm(x) satisfy the general Bessel equation of order m (see equation(9.13)). 15. For the Bessel functions J0(x) and J1(x) show that: (a) d dx (x J1(x)) = x J0(x) (b) R J0(x) J1(x) dx = −1 2 J2 0(x) (c) R x J0(x) J1(x) dx = −1 2 x J2 0(x) + 1 2 R J2 0(x) dx (d) For all integers n ≥2, Z xn J0(x) dx = xn J1(x) + (n−1)xn−1 J0(x) −(n−1)2 Z xn−2 J0(x) dx . (Hint: Use part (a) and integration by parts twice.) 16. For all integers m ≥2 show that the Bessel functions Jm(x) satisfy the relations: (a) m Jm(x) + x J′ m(x) = x Jm−1(x) (b) Jm−1(x) −Jm+1(x) = 2J′ m(x) 17. Use long division to obtain the first three terms of 1 xJ2 0(x), then integrate term by term to show that J0(x) Z dx xJ2 0(x) = J0(x) · ln x + x2 4 −3x4 128 + ··· . This function is a Bessel function of the second kind and is another solution of Bessel’s equation. | ElementaryCalculus_Page_317_Chunk3852 |
308 Chapter 9 • Infinite Sequences and Series §9.5 9.5 Taylor’s Series In the previous section a few functions, e.g. f (x) = 1 1−x, turned out to be the sum of a power series. This section will discuss a general method for representing a function as a power series, called a Taylor’s series.11 Suppose that a function f (x) can be written as f (x) = ∞ X n=0 an (x−c)n either for all x or for |x −c| < R, for some R > 0. Then f (c) = a0, and differentiating term by term yields f ′(x) = ∞ X n=1 nan(x−c)n−1 ⇒ f ′(c) = 1 · a1 f ′′(x) = ∞ X n=2 n(n−1)an(x−c)n−2 ⇒ f ′′(c) = 2 · 1 · a2 f ′′′(x) = ∞ X n=3 n(n−1)(n−2)an(x−c)n−3 ⇒ f ′′′(c) = 3 · 2 · 1 · a3 ··· ··· f (k)(x) = ∞ X n=k n(n−1)(n−2) ··· (n−k +1)an (x−c)n−k ⇒ f (k)(c) = k! ak so that in general (since f (0)(x) = f (x) and 0! = 1): an = f (n)(c) n! for n ≥0 (9.15) These { an } are the Taylor’s series coefficients of f (x) at x = c. The full power series repre- sentation of f (x) can now be stated: Taylor’s formula: If f (x) has a power series representation in powers of x−c, where x = c is inside the interval of convergence, then that representation is unique in that interval and is given by f (x) = ∞ X n=0 f (n)(c) n! (x−c)n (9.16) for all x in the interval. This is the Taylor’s series for f (x) about x = c. 11Named after English mathematician Brook Taylor (1685-1731), though such series were known to others (e.g. James Gregory, Johann Bernoulli) before Taylor. | ElementaryCalculus_Page_318_Chunk3853 |
Taylor’s Series • Section 9.5 309 Example 9.20 Find the Taylor’s series for f (x) = ex about x = 0. Solution: Since d dx (ex) = ex, then for all n ≥0, f (n)(x) = ex ⇒ f (n)(0) = e0 = 1 . Thus, by Taylor’s formula with c = 0:12 ex = ∞ X n=0 f (n)(0) n! xn = ∞ X n=0 xn n! = 1 + x + x2 2! + x3 3! + x4 4! + x5 5! + ··· For what x is this Taylor’s series valid? Recall that Example 9.16 showed the interval of convergence is all of R. Thus the above Taylor’s series holds for all x. Before continuing, you might be wondering why you should even bother with finding the Taylor’s series—after all, in the above example why replace a simple function like ex by a far more complicated expression? One reason is that it often helps simplify some computations, especially in integrals. The idea is to use only a few terms in the series, i.e. a polynomial, as an approximation, since polynomials are generally easier to work with. Perhaps surprisingly, in many practical applications no more than two terms are needed, and often only one. For example, using only the first two terms of the Taylor’s series for ex in Example 9.20, ex ≈1+ x is a good approximation when x is close to 0 (i.e. |x| ≪1). Using more terms does not necessarily help—for |x| ≪1 and n > 1, xn will be effectively 0. So the added complexity would not make the approximation significantly better. Example 9.21 The energy density E of electromagnetic radiation at wavelength λ from a black-body at temperature T degrees Kelvin is given by Planck’s Law of black-body radiation, E(λ) = 8πhc λ5(ehc/λkT −1) where h is Planck’s constant, c is the speed of light, and k is Boltzmann’s constant. Show that for λ ≫1: E(λ) ≈8πkT λ4 Solution: Since ex ≈1+ x for |x| ≪1 by the Taylor series for ex, let x = hc/λkT. Then x ≪1 and so E(λ) = 8πhc λ5(ehc/λkT −1) ≈ 8πhc λ5 ³³ 1+ hc λkT ´ −1 ´ ≈ 8πhc λ5 hc λkT ≈8πkT λ4 12The special case of c = 0 in Taylor’s formula yields what is sometimes called the Maclaurin’s series for f (x), though that terminology is typically not used in fields of study outside mathematics. | ElementaryCalculus_Page_319_Chunk3854 |
310 Chapter 9 • Infinite Sequences and Series §9.5 Example 9.22 Find the Taylor’s series for f (x) = sin x about x = 0. Solution: The derivatives of f (x) = sin x repeat every four derivatives: f (x) = sin x , f ′(x) = cos x , f ′′(x) = −sin x , f ′′′(x) = −cos x , f (4)(x) = sin x So at x = 0: f (0) = 0 , f ′(x) = 1 , f ′′(x) = 0 , f ′′′(x) = −1 , f (4)(x) = 0 So for n ≥0, f (n)(0) = 0 if n is even, 1 if n = 1,5,9,..., −1 if n = 3,7,11,.... Thus, by Taylor’s formula with c = 0 sin x = ∞ X n=0 f (n)(0) n! xn = x −x3 3! + x5 5! −x7 7! + x9 9! −··· = ∞ X n=0 (−1)n x2n+1 (2n+1)! By the Ratio Test this series converges for all x, since for any fixed x, r(x) = lim n→∞ ¯¯¯¯¯¯¯¯¯ (−1)n+1 x2n+3 (2n+3)! (−1)n x2n+1 (2n+1)! ¯¯¯¯¯¯¯¯¯ = x2 · lim n→∞ ¯¯¯¯ 1 (2n+3)(2n+2) ¯¯¯¯ = 0 < 1 . Example 9.23 Find the Taylor’s series for f (x) = cos x about x = 0. Solution: The Taylor’s series can be found using the same procedure as in Example 9.22, but it is simpler to just differentiate the Taylor’s series for sin x term by term for all x: cos x = d dx (sin x) = d dx µ x −x3 3! + x5 5! −x7 7! + x9 9! −··· ¶ = 1 −x2 2! + x4 4! −x6 6! + x8 8! −··· = ∞ X n=0 (−1)n x2n (2n)! Since the Taylor’s series for sin x converges for all x then so does its derivative. Thus the Taylor’s series for cos x converges for all x. Notice that the Taylor’s series for cos x has only even powers of x, while the series for sin x has only odd powers of x. This makes sense since cos x and sin x are even and odd functions, respectively. | ElementaryCalculus_Page_320_Chunk3855 |
Taylor’s Series • Section 9.5 311 Example 9.24 The function ln x is not defined at x = 0 and hence has no Taylor’s series about x = 0. Instead, find the Taylor’s series for f (x) = ln(1+ x) about x = 0. Solution: Take successive derivatives: f (x) = ln(1+ x) , f ′(x) = 1 1+ x , f ′′(x) = − 1 (1+ x)2 , f ′′′(x) = 1 · 2 (1+ x)3 , f (4)(x) = −1 · 2 · 3 (1+ x)4 So f (0) = 0 and for n ≥1: f (n)(x) = (−1)n−1 (n−1)! (1+ x)n ⇒ f (n)(0) = (−1)n−1(n−1)! Thus, by Taylor’s formula, ln(1+ x) = ∞ X n=0 f (n)(0) n! xn = ∞ X n=1 (−1)n−1 (n−1)!xn n! = ∞ X n=1 (−1)n−1 xn n = x −x2 2 + x3 3 −x4 4 + x5 5 −··· Use the Ratio Test to find the interval of convergence: r(x) = lim n→∞ ¯¯¯¯¯¯¯¯ (−1)n xn+1 n+1 (−1)n−1 xn n ¯¯¯¯¯¯¯¯ = |x| · lim n→∞ ¯¯¯¯ n n+1 ¯¯¯¯ = |x| · 1 = |x| So the series converges when |x| < 1. Check the cases r(x) = |x| = 1 individually. For x = 1 the series is the alternating harmonic series P n=1 (−1)n−1 n , which converges. For x = −1 the series is −P n=1 1 n, the negative of the harmonic series, which diverges. Thus, the series converges for −1 < x ≤1. Example 9.25 Find the Taylor’s series for f (x) = ex2 about x = 0. Solution: The Taylor’s series can be found using the same procedure as in Example 9.20, but it is simpler to just replace each occurrence of x in the Taylor’s series for ex by x2 (since the series for ex converges for all x). In other words, make the substitution u = x2 in the Taylor’s series for eu about u = 0: eu = ∞ X n=0 un n! ex2 = ∞ X n=0 (x2)n n! = ∞ X n=0 x2n n! = 1 + x2 + x4 2! + x6 3! + x8 4! + x10 5! + ··· | ElementaryCalculus_Page_321_Chunk3856 |
312 Chapter 9 • Infinite Sequences and Series §9.5 Define the n-th degree Taylor polynomial Pn(x) for a function f (x) about x = c by Pn(x) = nX k=0 f (k)(c) k! (x−c)k = f (c) + f ′(c) 1! (x−c) + f ′′(c) 2! (x−c)2 + ··· + f (n)(c) n! (x−c)n for x in the interval of convergence for the full Taylor’s series. In other words, Pn(x) is the n-th partial sum of the Taylor’s series. Since some of the coefficients could be zero, Pn(x) is a polynomial of degree at most n. Thus, Pn(x) = O(xn). For that reason Pn(x) is sometimes called the O(xn) approximation to f (x). Figure 9.5.1 shows a comparison of sin x with a few of its approximations: −8 −6 −4 −2 0 2 4 6 8 −1 −0.5 0 0.5 1 x y O(x7) approx. O(x11) approx. O(x15) approx. sin(x) Figure 9.5.1 sin x and Taylor’s series approximations As you can see, the Taylor polynomials of degree 7, 11 and 15 are all good approximations over the interval [−2,2], with the O(x15) approximation still being fairly good over [−6,6]. Clearly those approximations all become poor quite quickly for |x| > 6; they approach ±∞, unlike sin x. The following theorem shows how to measure the accuracy of the approximations: | ElementaryCalculus_Page_322_Chunk3857 |
Taylor’s Series • Section 9.5 313 Remainder Theorem:13If Pn(x) is the n-th degree Taylor polynomial about x = c for a function f (x) in some interval containing x = c, then for all x in that interval, f (x) = Pn(x) + Rn(x) , (9.17) where Rn(x) = f (n+1)(c +θ(x−c)) (n+1)! (x−c)n+1 (9.18) for some number θ between 0 and 1. Alternatively, Rn(x) = 1 n! Zx c (x−t)n f (n+1)(t) dt . (9.19) Since the number θ is unknown in equation (9.18), usually only an upper bound on the remainder Rn(x) can be found by that formula. For practical purposes formula (9.19) for Rn(x) might be easier to use (via numerical integration). A common misconception is that hand-held calculators use Taylor’s series to compute values of functions like sin x, cos x, ex, etc. However, that is typically well beyond their capability, especially for large values of x—far too many terms would be required. Instead, many calcula- tors use an algorithm called CORDIC14 (Coordinate Rotation Digital Computer), and—perhaps surprisingly—lookup tables. CORDIC uses the computationally inexpensive operation of bit- shifting to translate large input values into a smaller range, then uses tables stored in memory for values in that range, along with interpolation for numbers between those in the tables.15 Exercises A For Exercises 1-9 write out the first three nonzero terms in the Taylor’s series for the given function f (x) about the given value c. You may use any method you like. 1. f (x) = sin x ; c = π/2 2. f (x) = sinh x ; c = 0 3. f (x) = cosh x ; c = 0 4. f (x) = tan x ; c = 0 5. f (x) = tanh x ; c = 0 6. f (x) = sec x ; c = 0 7. f (x) = 1 1+ x2 ; c = 0 8. f (x) = 1 1+ x2 ; c = 1 9. f (x) = p 1+ x2 ; c = 0 10. Use Example 9.19 from Section 9.4 to write out the first three nonzero terms in the Taylor’s series for f (x) = 1 (1−x)3 about x = 0. 11. Use the derivative of the Taylor’s series for p 1+ x2 from Exercise 9 to write out the first three nonzero terms in the Taylor’s series for f (x) = x2 p 1+x2 about x = 0. 13For a proof see pp.171-172 in KLAMBAUER, G., Aspects of Calculus, New York: Springer-Verlag, 1986. 14See Ch.7 in SCHMID, H., Decimal Computation, New York: John Wiley & Sons, Inc., 1974. 15To see how inaccurate calculators can be, compute tan(355/226) in radian mode on a calculator. The true value to 3 decimal places is −7497258.185, but few calculators produce an answer close to that. | ElementaryCalculus_Page_323_Chunk3858 |
314 Chapter 9 • Infinite Sequences and Series §9.5 For Exercises 12-15 replace the function f (x) by its Taylor’s series about x = 0 to evaluate the given indefinite integral R f (x) dx (up to the first three nonzero terms in the series). 12. Z sin x x dx 13. Z cos(x2) dx 14. Z e−x2 dx 15. Zp 1+ x6 dx 16. Use d dx (tan−1 x) = 1 1+x2 along with Exercise 7 to find the Taylor’s series for f (x) = tan−1 x about x = 0, along with its interval of convergence. 17. Use Exercise 16 to show that π = 4 µ 1 −1 3 + 1 5 −1 7 + ··· ¶ . 18. Use the first three nonzero terms in the Taylor’s series about x = 0 for e−x2/2 to evaluate the definite integral Z1 −1 1 p 2π e−x2/2 dx . Note: The actual value (rounded to 4 decimal places) is 0.6826. 19. Recall that the surface area S of the solid obtained by revolving the curve y = x2 around the x-axis between x = 0 and x = 2 is given by the integral S = Z2 0 2πx2p 1 + 4x2 dx . The exact value of the integral rounded to 3 decimal places is S = 53.226. Use the first two nonzero terms in the Taylor’s series for p 1+4x2 about x = 0 to approximate the integral. How close is this approximation to the actual value? Does the approximation become better if you use the first three nonzero terms in the Taylor’s series? Justify your answer. 20. The tangential component of a space shuttle’s velocity during reentry is approximately v(t) = vc tanh µ g vc t + tanh−1 µ v0 vc ¶¶ where v0 is the velocity at time t = 0 and vc is the terminal velocity. If tanh−1³ v0 vc ´ = 1 2 then show that v(t) ≈gt+ 1 2vc. 21. The velocity of a water wave of length L in water of depth h satisfies the equation v2 = gL 2π tanh ³ 2πh L ´ . Show that v ≈ p gh. 22. A disk of radius a has a charge of constant density σ. A point P lies at a distance r directly above the disk. The electrical potential V at point P is given by V = 2πσ( p r2 + a2 −r). Show that V ≈πa2σ r for large r (i.e. r ≫1). 23. The fifth-degree Padé approximation uses rational functions to approximate tanh x: tanh x ≈ x5 +105x3 +945x 15x4 +420x2 +945 Compare the values of the Padé approximation and the fifth-degree Taylor’s series approximation from Exercise 5, evaluated at x = 1. Which is better? The actual value of tanh(1) is 0.7615941559558. How do the two approximations compare at x = 2? | ElementaryCalculus_Page_324_Chunk3859 |
APPENDIX A Answers and Hints to Selected Exercises Chapter 1 Section 1.1 (p. 6) 1. 2t 2. 19.6t 3. −32t+2 4. 3t2 Section 1.2 (p. 14) 1. 0 3. 2x+2 5. − 1 (x+1)2 7. −2 x3 9. 1 2 p x+1 11. 2x+3 2 p x2+3x+4 Section 1.3 (p. 20) 5. Hint: Use the sine double-angle formula. 7. Hint: Use Exercise 5 and the sine addition formula. Section 1.4 (p. 26) 1. 2x−1 3. 4x5 + 9 x7 5. x cosx+sin x 7. x cos x−sinx x2 9. 2−2t2 (1+t2)2 11. ad−bc (cx+d)2 13. 2πr Section 1.5 (p. 30) 1. −20(1−5x)3 3. − 1 p 1−2x 5. 1−x 2px(x+1)2 7. −8(1−t)3 (1+t)5 9. 2sinx cos x 11. 15sec2(5x) 13. 2xsec(x2) tan(x2) 15. β(1−β2)−3/2 17. sin(cos x) sinx 21. Hint for part(b): Use part(a) and the Chain Rule to find Sp, then re- call how to convert from radians per second to revolutions per minute. Section 1.6 (p. 35) 1. 6x+2 3. −9cos3x 5. 2 x3 Chapter 2 Section 2.1 (p. 40) 1. f −1(x) = x, ¡ f −1¢′ (x) = 1 3. f −1(x) = px, ¡ f −1¢′ (x) = 1 2px 5. f −1(x) = 1 x, ¡ f −1¢′ (x) = −1 x2 7. f −1(x) = 1 px, ¡ f −1¢′(x) = −1 2 x−3/2 Section 2.2 (p. 44) 1. 6sec2 3x tan3x 3. −3csc2 3x 5. 3 9+x2 7. − 3 1+9x2 9. 1 1+x2 11. 6sin−1 3x p 1−9x2 13. 1 1+x2 15. cot−1 x− x 1+x2 Section 2.3 (p. 52) 1. 2e2x 3. −e−x −ex 5. 2ex (1−ex)2 7. eex ex 9. 1 x 11. 6x(ln(tanx2)) 2 sec2 x2 tanx2 315 | ElementaryCalculus_Page_325_Chunk3860 |
316 Appendix A: Answers and Hints to Selected Exercises 15. xx2(x+2xlnx) 17. xsinx ³ cos x lnx+ sin x x ´ 19. 15.5 hours 21. 12 hours Section 2.4 (p. 55) 1. ln3(3x−3−x) 2 3. (ln2)2 22x 2x 5. 2x (ln2)(x2+1) 7. cos(log2 πx) xln2 9. 3x2 Chapter 3 Section 3.1 (p. 61) 1. y = 4x−3 3. y = −6x+10 5. y = 4x 7. y = x+3 9. y = 240x+176 11. y = 2x 13. y = 3x+ 31 27, y = 3x+1 15. 75.96◦ 17. 0◦ 19. 116.6◦ 21. 5.71◦ 23. y = −1 4 x+ 11 2 25. y = −1 4 x−81 4 , y = −1 4 x+ 1159 108 Section 3.2 (p. 72) 1. 7 3 3. 0 5. −1 7. 0 9. 2 11. 0 14. 1 2 15. 0 17. 0 Section 3.3 (p. 78) 1. continuous 3. discontinuous 5. discontinuous 7. discontinuous 9. continuous 11. continuous 13. continuous 15. discontinuous 17. continuous 19. 1 21. e−1 25. Hint: Use the Intermediate Value Theo- rem. Section 3.4 (p. 81) 1. −3x2y+4y2+2x x3−8xy−1 3. 2(x−y+1)−3(x+y)2 2(x−y+1)+3(x+y)2 5. 2x(1−(x2−y2)) y(2(y2−x2)−1) 7. −y x 9. −−2x−y+3x2y2esin(xy)+x3y3esin(xy) cos(xy) x4y2esin(xy) cos(xy)+2x3 yesin(xy)−3y2−x 13. −x2+y2 y3 Section 3.5 (p. 83) 1. 80π ft/s 3. 2.4 ft/s 5. 10 ft/s 7. −76π cm3/min 9. 45.14 mph 11. 155.8 ft/min Section 3.6 (p. 88) 1. (2x−2)dx 2. 4x sin(x2) cos(x2)dx 11. Hint: Mimic Example 3.35. Chapter 4 Section 4.1 (p. 98) 1. (1,1) 3. 125,000 sq yd 5. U = V 2 7. R = r 9. 2ab 13. Q = q 2DP I+W 15. r = q r2 0 + x2 0 17. 380.62 minutes 19. 12π p 3 21. 12.8 ft 22. Hint: Place the right angle of the triangle at the origin in the xy-plane. 25. x = q RN r 27. x = a p 2 33. (a2/3 + b2/3)3/2 34. Hint: You can leave your answer in terms of R and an angle satisfying a certain equa- tion. Section 4.2 (p. 108) 2. local maximum at x = 0, local minimum at x = 2, inflection pt at x = 1, increasing for x < 0 and x > 2, decreasing for 0 < x < 2, concave up for x > 1, concave down for x < 1 3. local maximum at x = 1, inflection pt at x = 2, increasing for x < 1, decreasing for x > 1, concave up for x > 2, concave down for x < 2, horizontal asymptote: y = 0 5. local maximum at x = 0, inflection pts at x = ± 1 p 3, increasing for x < 0, decreasing for x > 0, concave up for x < −1 p 3 and x > 1 p 3, concave down for −1 p 3 < x < 1 p 3, horizontal asymptote: y = 0 | ElementaryCalculus_Page_326_Chunk3861 |
Appendix A: Answers and Hints to Selected Exercises 317 7. local maximum at x = ln2, inflection pt at x = ln4, increasing for x < ln2, decreasing for x > ln2, concave up for x > ln4, concave down for x < ln4, horizontal asymptote: y = 0 Section 4.3 (p. 117) 1. x = 0.450184 3. x = 0.567143 5. x = 1.414213 11. global maximum at x = 2.8214 14. 50 Section 4.4 (p. 122) 1. No 3. Yes 6. No 18. Hint: Calculate f ′(x) and use the cosine addition formula. Chapter 5 Section 5.1 (p. 131) 1. x3 3 + 5x2 2 −3x+C 3. 4ex +C 5. −5cos x+C 7. 6ln|x|+C 9. −4 3 x3/2 +C 11. x2 2 + 3 7 x7/3 +C 13. 3secx+C 15. −7cotx+C Section 5.2 (p. 139) 3. 1 2 4. 1 3 5. 1 6. 1 4 Section 5.3 (p. 145) 1. 1 3 3. 1 4 5. 1 2 7. 1 9. 2e −2e−1 11. 16 3 Section 5.4 (p. 150) 1. 3sin5x−4cos5x 5 +C 3. −1 2 e−x2 + 1 3 sinx3 + C 5. ln(1+ ex)+C 7. 2 5(x+4)5/2 −8 3(x+4)3/2 +C 9. tan x−x+C 11. 3 10 tan−1¡ 5x 2 ¢ +C 13. 10 15. 1192 15 17. 1 19. −1 48 21. π 6 23. 1 2 Section 5.5 (p. 158) 1. 1 2 3. 1 5. divergent 7. 1 ln2 9. divergent 11. 6 13. divergent 15. π 2 19. Yes 20. No Chapter 6 Section 6.1 (p. 165) 1. x2 ln x 2 −x2 4 +C 2. (x2 −2x+2)ex +C 3. xsinx+cos x+C 5. x2ax lna −2xax ln2 a + 2ax ln3 a +C 7. xlnx2 −2x+C 9. Hint: Use a double-angle identity. 11. xsin−1 x+ p 1−x2 +C 13. xtan−1 3x−1 6 ln(1+9x2)+C 15. −3 8 sinxcos3x+ 1 8 cos xsin3x+C 17. 1 4 x4ln2 x−1 8 x4ln x+ 1 32 x4 +C 19. 16 3 20. 2 p 2+2 15 21. xsin(ln x) 2 −xcos(ln x) 2 +C 23. x2 tan−1 x 2 −x 2 + tan−1 x 2 +C 24. xcot−1px+px+cot−1 px+C 25. Hint: Try the substitution t = px. Section 6.2 (p. 171) 1. −1 14 cos 7x+ 1 6 cos 3x+C 3. −1 14 sin 7x+ 1 6 sin 3x+C 5. −2 5 cos5/2 x+ 2 9 cos9/2 x+C 7. −1 4 sin 2x+ 1 48 sin2 2x+ 5 16 x+ 3 64 sin 4x+C 9. 1 3 tan3 x+tan x+C 11. 1 4 sin4 x+C Section 6.3 (p. 177) 1. 1 2 x p 9+4x2 + 9 4 ln|2x+ p 9+4x2|+C 3. 1 2 x p 4x2 −9−9 4 ln|2x+ p 4x2 −9|+C 5. −sin−1 x − p 1−x2 x +C 7. ln|x|−ln|1+ p 1+ x2|+C 9. 1 3(x2 +4)3/2 −4 p x2 +4+C 11. 1 108 tan−1 ¡2x 3 ¢ + x 18(9+4x2) +C | ElementaryCalculus_Page_327_Chunk3862 |
318 Appendix A: Answers and Hints to Selected Exercises 13. −9 p 9−x2 + 1 3(9−x2)3/2 +C 15. −1 9 p −9x2 +36x−32−2 3 sin−1¡3x−6 2 ¢ +C Section 6.4 (p. 184) 1. −ln|x|+ln|x−1|+C 3. 1 5 ln|2x−1|−1 5 ln|x+2|+C 5. 1 x + 1 2 ln|x−1|−1 2 ln|x+1|+C 7. 2ln|x|+ 1 x −2ln|x+1|+C 9. −3ln|x|+ 2 x +3ln|x−1|+ 1 x−1 +C 11. 1 3 tan−1 x−1 6 tan−1¡ x 2 ¢ +C Section 6.5 (p. 193) 1. 2 ln ¯¯sec 1 2θ ¯¯−ln|sin θ|+C 3. 2 p 3 tan−1 à 2tan 1 2 θ−1 p 3 ! +C 5. 4 p 3 tan−1 à 2tan 1 2 θ−1 p 3 ! −θ +C 7. ln ¯¯tan 1 2θ ¯¯−ln ¯¯tan 1 2θ +1 ¯¯+C 9. ln ¯¯tan 1 2θ ¯¯−2ln ¯¯tan 1 2θ +1 ¯¯+C 11. p 2π 23. 3 2Γ( 2 3) x2/3 Section 6.6 (p. 201) 1. The true value is P ≈7.4163 p l/g (i.e. the integral is ≈1.8541) 2. 7.416331870724302 p l/g 3. 0.8948311310564181 7. 119.9785845899309 9. 0.5967390281992041 (The true value is 0.5963473623231939) Chapter 7 Section 7.1 (p. 209) 2. Foci: (±3,0), vertexes: (±5,0), e = 3 5 3. Foci: (0,± p 5), vertexes: (0,±3), e = p 5 3 5. Foci: ³ ± p 3 2 ,0 ´ , vertexes: (±1,0), e = p 3 2 9. ³ ± ab p a2+b2 ,± ab p a2+b2 ´ 13. Hint: Use the two points you know for certain are on the ellipse to find the location of the directrix. 16. Hint: Use Exercise 15 and formula (7.3). Section 7.2 (p. 215) 2. Focus: (0,2), vertex: (0,0), directrix: y = −2 3. Focus: ¡ 0, 1 32 ¢ , vertex: (0,0), directrix: y = −1 32 4. Focus: ¡1 4,0 ¢ , vertex: (0,0), directrix: x = −1 4 5. Focus: ¡ −1 12 ,0 ¢ , vertex: (0,0), directrix: x = 1 12 7. (0,0) and (4p,4p); y = x 9. |4p| 11. (3p,±2 p 3p) 16. Focus: ³ −b 2a , 4ac−b2+1 4a ´ , vertex: ³ −b 2a , 4ac−b2 4a ´ , directrix: y = 4ac−b2−1 4a Section 7.3 (p. 223) 2. Foci: (±5,0), vertexes: (±4,0), directrices: x = ± 16 5 , asymptotes: y = ± 3 4 x, e = 5 4 3. Foci: (± p 23,0), vertexes: (±2 p 2,0), direc- trices: x = ± 8 p 23, asymptotes: y = ± p 15 2 p 2 x, e = p 23 2 p 2 4. Foci: (± p 41 2 ,0), vertexes: (± 5 2,0), directri- ces: x = ± 25 2 p 41, asymptotes: y = ± 4 5 x, e = p 41 5 5. Foci: (± p 5 2 ,0), vertexes: (±1,0), directrices: x = ± 2 p 5, asymptotes: y = ± 1 2 x, e = p 5 2 6. Foci: (0,± p 34), vertexes: (0,±3), directri- ces: y = ± 9 p 34, asymptotes: y = ± 3 5 x, e = p 34 3 7. x2 9 −y2 16 = 1 17. x2 302500 − y2 697500 = 1 Section 7.4 (p. 229) 1. Foci: (0,2) and (6,2), vertexes: (−2,2) and (8,2) 3. Foci: (−3,1 ± 2 p 3), vertexes: (−3,−3) and (−3,5) | ElementaryCalculus_Page_328_Chunk3863 |
Appendix A: Answers and Hints to Selected Exercises 319 5. Focus: (−3,−239 16 ), vertex: (−3,−15), direc- trix: y = −241 16 7. Focus: ¡1 2, 5 4 ¢ , vertex: ¡1 2, 3 2 ¢ , directrix: y = 7 4 9. Foci: (−1± p 13,−3), vertexes: (−4,−3) and (2,−3), directrices: x = −1 ± 9 p 13, asymptotes: y = ± 2 3(x+1)−3 11. Foci: ( p 2, p 2) and (− p 2,− p 2), vertexes: (1,1) and (−1,−1), directices: y = −x ± p 2, asymptotes: x = 0 and y = 0 15. hyperbola Section 7.5 (p. 236) 12. Hint: Use Exercise 11. 24. local maximum at x = ln p 3, inflection pt at x = ln3, horizontal asymptote: y = 0 26. (b) Hint: See Exercise 10. 27. (b) k1 = c1 + c2, k2 = c1 −c2 30. s0 ≈1.006237835313385, eπ/s0 = 22.69438187638412 33. (a) x = cx+cy 2 −y−x 2c , y = cx+cy 2 + y−x 2c (b) c = cosh a+sinh a (c) Hint: Use Example 7.12. Section 7.6 (p. 243) 5. Yes 6. (a) Hint: Solve for t in terms of x then substitute into y. (b) Hint: Use the distance formula. 8. Hint: Does BP = AB? 9. (a) − 7 2(t+1)3 (c) ¡ 49 25, 14 5 ¢ 12. x = t2 −1, y = t(t2 −1) Section 7.7 (p. 250) 1. r = 6 cos θ 3. r2 = sec 2θ 5. θ = 3π 4 7. r = sec θ tan θ 9. x4 +2x2y2 + y4 = 4x2 −4y2 11. y−1 = −3 p 3 5 (x+ p 3) 13. local maxima at ¡ 2, π 2 ¢ and ¡ 0, 3π 2 ¢ , local minima at ¡1 2, 7π 6 ¢ and ¡1 2, 11π 6 ¢ 15. local maxima at ³ 2 p 2 3 ,α ´ and ³ −2 p 2 3 ,2π−α ´ , local minima at ³ −2 p 2 3 ,π−α ´ and ³ 2 p 2 3 ,π+α ´ , where α = tan−1p 2 17. π 2 Chapter 8 Section 8.1 (p. 256) 1. 32 3 2. 4 3 3. 1 12 5. 9 2 7. 19 3 9. 1 12 11. 3π 13. 25 ¡ π 3 +1− p 3 ¢ 17. 6250π3 3 sq ft Section 8.2 (p. 261) 1. 1 3. 4 3 5. 2 π 7. 0 9. 1 2 ln3 Section 8.3 (p. 270) 1. 8 27(103/2)−1 27(133/2) ≈7.634 2. p 5 2 + 1 4 sinh−12 ≈1.479 3. 8 27(103/2)−1 27(133/2) ≈7.634 4. 3 4 +ln p 2 ≈1.097 7. p 2(eπ −1) ≈31.312 9. 8 11. 3 13. κ(0) = 0, κ( π 2 ) = −1 15. κ(0) = a b2 at (a,0), κ( π 2 ) = b a2 at (0,b) 17. −1 23. 13.27 ft Section 8.4 (p. 276) 1. 4π 2. π 2 (2+sinh2) 3. 208π 9 5. π2 2 7. 2π 9. π 10 10. π 6 13. S = πr p r2 + h2, V = 1 3πr2h Section 8.5 (p. 283) 1. ¡4 5, 2 7 ¢ 3. ¡ 3 5, 12 35 ¢ 5. ¡ 4r 3π, 4r 3π ¢ 7. ¡ 0, 11 4π ¢ 9. 0.192 Nm 11. RT log ³ Vb Va ´ 13. 0.3486 15. ¡ 1, 1 4 ¢ 18. Hint: Use Exercise 28 in Section 6.1. 20. Hint: Use the equation from Section 5.1 for free fall motion to write time as a function of height. | ElementaryCalculus_Page_329_Chunk3864 |
320 Appendix A: Answers and Hints to Selected Exercises Chapter 9 Section 9.1 (p. 291) 1. Converges to 0 2. Converges to 1 3 3. Converges to 0 5. Divergent 7. Divergent 9. 6 11. 32 13. 113 999 14. 1 15. 1 4 20. 132 7 ft 24. No Section 9.2 (p. 298) 6. Divergent 7. Convergent 8. Divergent 9. Convergent 10. Divergent 11. Convergent 12. Convergent 13. Convergent 14. Divergent 15. Divergent 16. Convergent 17. Convergent 18. 1 6 19. 1 10 20. 1 6 21. 1 2 Section 9.3 (p. 302) 1. Conditionally convergent 3. Conditionally convergent 5. Absolutely convergent 7. Answer to second question: Yes Section 9.4 (p. 307) 1. −1 ≤x < 1 2. −2 < x < 2 3. 1 < x < 3 4. −6 < x < −2 5. −∞< x < ∞ 6. x = 0 7. −1 < x < 1 10. Hint: Use Example 9.19 Section 9.5 (p. 313) 1. 1− (x−π 2 )2 2! + (x−π 2 )4 4! −··· 2. x+ x3 3! + x5 5! +··· 3. 1+ x2 2! + x4 4! +··· 4. x+ x3 3 + 2x5 15 +··· 5. x−x3 3 + 2x5 15 −··· 6. 1+ x2 2 + 5x4 24 +··· 7. 1−x2 + x4 −··· 8. 1 2 −(x−1) 2 + (x−1)2 4 −··· 9. 1+ x2 2 −x4 8 +··· 11. x2 −x4 2 + 3x6 8 −··· 13. x−x5 10 + x9 216 −··· 15. x+ x7 14 −x13 104 +··· 18. 0.68485 19. 97.18; −132.605 | ElementaryCalculus_Page_330_Chunk3865 |
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History This section contains the revision history of the book. For persons making modifications to the book, please record the pertinent information here, following the format in the first item below. 1. VERSION: 0.1 Date: 2016-01-24 Author(s): Michael Corral Title: Elementary Calculus Modification(s): Initial version 2. VERSION: 0.5 Date: 2020-05-27 Author(s): Michael Corral Title: Elementary Calculus Modification(s): Numerous corrections of typos and errors, as well as enhancements of some sections and more exercises. 3. VERSION: 0.5a Date: 2020-06-03 Author(s): Michael Corral Title: Elementary Calculus Modification(s): Replaced the old Section 5.5 (Average Value of a Function) with the new Section 5.5 (Improper Integrals). Average Value of a Function will be moved to Chapter 8 in the second half of the book. 4. VERSION: 1.0 Date: 2020-12-31 Author(s): Michael Corral Title: Elementary Calculus Modification(s): Full version. 329 | ElementaryCalculus_Page_339_Chunk3874 |
Index Symbols ∆x. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .8 〈f 〉. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .258 ˙f . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 dx. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .15 d2 y dx2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .32 dn dxn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 ≫. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .238 ≪. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 N . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Q . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4 R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 f ′ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 f (n) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 n-th Term Test . . . . . . . . . . . . . . . . . . . . . . . . . 294 A absolute convergence. . . . . . . . . . . . . . . . . . .301 absolute value | ElementaryCalculus_Page_340_Chunk3875 |
. . . . . . . . . . . . . . . . . . . . . . . . . . . 13 accelerating.. . . . . . . . . . . . . . . . . . . . . . . . . . . . .34 acceleration.. . . . . . . . . . . . . . . . . . . . . . . . . . . . .33 algebraic curve . . . . . . . . . . . . . . . . . . . . . . . . . . 80 alternating series . . . . . . . . . . . . . . . . . . . . . . 300 angle of contingence.. . . . . . . . . . . . . . . . . . .267 angle of tangent line.. . . . . . . . . . . . . . . . . . . .59 antiderivative .. . . . . . . . . . . . . . . . . . . . . . . . . 124 arc .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 arc length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 parametric .. . . . . . . . . . . . . . . . . . . . . . . . 266 polar .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266 area between curves. . . . . . . . . . . . . . . . . . . .252 area function .. . . . . . . . . . . . . . . . . . . . . . . . . . 140 area under a curve .. . . . . . . . . . . . . . . . . . . . 133 asymptote . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 horizontal .. . . . . . . . . . . . . . . . . . . . . . . . . . 67 vertical . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 average value . . . . . . . . . . . . . . . . . . . . . . . . . . 258 axes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 of an ellipse . . . . . . . . . . . . . . . . . . . . . . . . 203 B Bézier curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241 base a logarithm.. . . . . . . . . . . . . . . . . . . . . . . .54 Bessel Functions .. . . . . . . . . . . . . . . . . | ElementaryCalculus_Page_340_Chunk3876 |
. . . . . 305 Bessel functions order m . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306 order zero . . . . . . . . . . . . . . . . . . . . . . . . . . 305 Bessel’s equation .. . . . . . . . . . . . . . . . . . . . . . 305 order m . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306 Beta function.. . . . . . . . . . . . . . . . . . . . . . . . . .188 Big O notation .. . . . . . . . . . . . . . . . . . . . . . . . . . 68 bipolar coordinates .. . . . . . . . . . . . . . . . . . . . 251 bisection method . . . . . . . . . . . . . . . . . . . . . . . 109 Bohr radius . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 brachistochrone .. . . . . . . . . . . . . . . . . . . . . . . 243 C catenary. . . . . . . . . . . . . . . . . . . . . . . . . . .231, 264 ceiling function .. . . . . . . . . . . . . . . . . . . . . . . . . 74 center of gravity. . . . . . . . . . . . . . . . . . . . . . . .277 Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 Chebyshev polynomials.. . . . . . . . . . . . . . . . .44 330 | ElementaryCalculus_Page_340_Chunk3877 |
Index 331 circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202 definition . . . . . . . . . . . . . . . . . . . . . . . . . . 202 involute . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244 circular functions.. . . . . . . . . . . . . . . . . . . . . .230 closed interval . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 Comparison Test . . . . . . . . . . . . . . . . . . . . . . . 293 concave down. . . . . . . . . . . . . . . . . . . . . . . . . . .102 concave up.. . . . . . . . . . . . . . . . . . . . . . . . . . . . .102 concavity .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 conditional convergence . . . . . . . . . . . . . . . . 301 conic sections. . . . . . . . . . . . . . . . . . . . . . . . . . .221 conjugate axis . . . . . . . . . . . . . . . . . . . . . . . . . . 217 constant function .. . . . . . . . . . . . . . . . . . . . . . . . 9 Constant Multiple Rule . . . . . . . . . . . . . . . . . 21 continued fraction .. . . . . . . . . . . . . . . . . . . . . 292 continuity.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .73 continuous function . . . . . . . . . . . . . . . . . . . . . 73 convergent integral . . . . . . . . . . . . . . . 152, 155 coordinate transformations . . . . . . . . . . . . 224 CORDIC. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .313 countable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 crackle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .34 critical point .. . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 curvature.. . . . . . . . . . . . . . . . . . . . . . . . .267, 268 average . . . . . . . . . . . . . . . . . . . . | ElementaryCalculus_Page_341_Chunk3878 |
. . . . . . . . 267 parametric .. . . . . . . . . . . . . . . . . . . . . . . . 269 polar .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 cycloid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242 D Darboux’s Theorem. . . . . . . . . . . . . . . . . . . . .122 decelerating . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 definite integral .. . . . . . . . . . . . . . . . . . . . . . . 132 derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3, 8 n-th . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . 27 Constant Multiple Rule . . . . . . . . . . . . . 21 Difference Rule . . . . . . . . . . . . . . . . . . . . . 21 exponential function .. . . . . . . . . . . . . . . 46 first . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 fractional . . . . . . . . . . . . . . . . . . . . . . . . . . 189 higher order.. . . . . . . . . . . . . . . . . . . . . . . .32 natural logarithm function.. . . . . . . . .48 polar coordinates . . . . . . . . . . . . . . . . . . 248 Power Rule .. . . . . . . . . . . . . . . . . . . . . . . . .24 Product Rule .. . . . . . . . . . . . . . . . . . . . . . . 21 Quotient Rule . . . . . . . . . . . . . . . . . . . . . . . 21 second. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .32 Sum Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 trigonometric functions .. . . . . . . . . . . . 23 with respect to x . . . . . . . . . . . . . . . . . . . . 15 zero-th . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 Difference Rule . . . . . . . . . . | ElementaryCalculus_Page_341_Chunk3879 |
. . . . . . . . . . . . . . . . 21 differentiable .. . . . . . . . . . . . . . . . . . . . . . . . . . . 14 differential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 differential equation . . . . . . . . . . . . . . . . . 36, 47 differentiation .. . . . . . . . . . . . . . . . . . . . . . . . . . 14 implicit.. . . . . . . . . . . . . . . . . . . . . . . . . . . . .79 under the integral sign . . . . . . . . . . . . 185 Dirac delta function .. . . . . . . . . . . . . . . . . . . 152 directrix .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 disc method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273 discontinuous.. . . . . . . . . . . . . . . . . . . . . . . . . . .73 distribution function .. . . . . . . . . . . . . . . . . . 284 divergent integral .. . . . . . . . . . . . . . . . 152, 155 domain .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8, 37 dot notation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .15 E e .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 eccentric angle . . . . . . . . . . . . . . . . . . . . . . . . . 243 Efimov trimer .. . . . . . . . . . . . . . . . . . . . . . . . . 237 eigenfunction .. . . . . . . . . . . . . . . . . . . . . . . . . . . 36 eigenvalue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 elasticity. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .88 ellipse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202 area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206 center .. . . . . . . . . . . . . . . . . . . . . . . . . | ElementaryCalculus_Page_341_Chunk3880 |
. . . . 203 construction.. . . . . . . . . . . . . . . . . . . . . . .203 definition . . . . . . . . . . . . . . . . . . . . . . . . . . 202 diameter . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 directrix.. . . . . . . . . . . . . . . . . . . . . . . . . . .209 eccentric angle . . . . . . . . . . . . . . . . . . . . . 243 foci . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202 | ElementaryCalculus_Page_341_Chunk3881 |
332 Index latus rectum .. . . . . . . . . . . . . . . . . . . . . . 209 major auxiliary circle .. . . . . . . . . . . . . 243 major axis. . . . . . . . . . . . . . . . . . . . . . . . . .203 minor auxiliary circle .. . . . . . . . . . . . . 243 minor axis . . . . . . . . . . . . . . . . . . . . . . . . . 203 principal axis . . . . . . . . . . . . . . . . . . . . . . 203 reflection property .. . . . . . . . . . . . . . . . 206 second definition .. . . . . . . . . . . . . . . . . .209 semi-major axis . . . . . . . . . . . . . . . . . . . . 203 semi-minor axis. . . . . . . . . . . . . . . . . . . .203 vertex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 ellipsograph .. . . . . . . . . . . . . . . . . . . . . . . . . . . 244 ellipsoid .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276 elliptic curve. . . . . . . . . . . . . . . . . . . . . . . . . . . . .79 elliptic function . . . . . . . . . . . . . . . . . . . . . . . . 270 elliptic integral . . . . . . . . . . . . . . . . . . . . . . . . . 264 enthalpy .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 entropy .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 escape velocity . . . . . . . . . . . . . . . . . . . . . . . . . 216 Eudoxus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Euler number.. . . . . . . . . . . . . . . . . . . . . . . . . . .45 even function .. . . . . . . . . . . . . . . . . . . . . . . . . . . 13 event . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282 expected value. . . . . . . . . . . . . . . . . . . . . . . . . .284 exponential decay . . . . . . . . . . . . . . . . . . . . . . . | ElementaryCalculus_Page_342_Chunk3882 |
50 exponential distribution.. . . . . . . . . . . . . . .283 exponential functions.. . . . . . . . . . . . . . . . . . .45 exponential growth .. . . . . . . . . . . . . . . . . . . . . 51 Extended Mean Value Theorem . . . . . . . . 121 F factorial.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .34 Fermat’s Principle .. . . . . . . . . . . . . . . . . 95, 101 Fibonacci sequence . . . . . . . . . . . . . . . .288, 292 First Derivative Test . . . . . . . . . . . . . . . . . . . 105 floor function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 focal radius.. . . . . . . . . . . . . . . . . . . . . . .212, 223 fractional derivative. . . . . . . . . . . . . . . . . . . .189 free fall formulas . . . . . . . . . . . . . . . . . . 129, 130 frustrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271 function.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .37 absolute value. . . . . . . . . . . . . . . . . . . . . . .13 area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 average value . . . . . . . . . . . . . . . . . . . . . . 258 ceiling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 circular . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230 composition . . . . . . . . . . . . . . . . . . . . . . . . . 27 constant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 continuous . . . . . . . . . . . . . . . . . . . . . . . . . . 73 differentiable.. . . . . . . . . . . . . . . . . . . . . . .14 elliptic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270 even . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 exponential. . . . . . . . . . . . . . . . . . . . . . . . | ElementaryCalculus_Page_342_Chunk3883 |
. .45 floor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 hyperbolic.. . . . . . . . . . . . . . . . . . . . . . . . .230 infinitesimal . . . . . . . . . . . . . . . . . . . . . . . . 72 inverse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 inverse hyperbolic . . . . . . . . . . . . . . . . . 234 inverse trigonometric .. . . . . . . . . . . . . . 41 Langevin. . . . . . . . . . . . . . . . . . . . . . . . . . .238 linear .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .9 natural logarithm . . . . . . . . . . . . . . . . . . . 47 objective. . . . . . . . . . . . . . . . . . . . . . . . . . . . .90 odd . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13, 143 rational . . . . . . . . . . . . . . . . . . . . . . . . 70, 178 root . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 sigmoid neuron.. . . . . . . . . . . . . . . . . . . . .52 step . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 trigonometric.. . . . . . . . . . . . . . . . . . . . . . .41 Fundamental Theorem of Calculus . . . . 140 G Gamma function . . . . . . . . . . . . . . . . . . 159, 188 Gaussian quadrature.. . . . . . . . . . . . . . . . . .199 Gibbs energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 golden ratio . . . . . . . . . . . . . . . . . . . . . . . 288, 292 H half-angle substitution .. . . . . . . . . . . . . . . . 190 half-life . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 harmonic series . . . . . . . . . . . . . . . . . . . . . . . . 296 height function .. . . . . . . . . . . . . . . . . . . . . . . . 252 Hooke’s Law. . . . . . . . . . . . . . . . . . . . . . | ElementaryCalculus_Page_342_Chunk3884 |
. . . . . .281 horizontal asymptote . . . . . . . . . . . . . . . . . . . . 67 hyperbola. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .216 | ElementaryCalculus_Page_342_Chunk3885 |
Index 333 asymptotes. . . . . . . . . . . . . . . . . . . . . . . . .217 center .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 conjugate axis. . . . . . . . . . . . . . . . . . . . . .217 construction.. . . . . . . . . . . . . . . . . . . . . . .219 definition . . . . . . . . . . . . . . . . . . . . . . . . . . 216 directrix.. . . . . . . . . . . . . . . . . . . . . . . . . . .216 eccentricity . . . . . . . . . . . . . . . . . . . . . . . . 216 focal radius . . . . . . . . . . . . . . . . . . . . . . . . 223 focus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216 latus rectum .. . . . . . . . . . . . . . . . . . . . . . 223 reflection property .. . . . . . . . . . . . . . . . 220 sector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230 transverse axis . . . . . . . . . . . . . . . . . . . . 217 vertex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 hyperbolic angle. . . . . . . . . . . . . . . . . . . . . . . .230 hyperbolic cosine .. . . . . . . . . . . . . . . . . . . . . . 230 hyperbolic functions.. . . . . . . . . . . . . .230, 231 inverses . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234 hyperbolic rotation . . . . . . . . . . . . . . . . . . . . . 238 hyperbolic sector . . . . . . . . . . . . . . . . . . . . . . . 230 hyperbolic sine .. . . . . . . . . . . . . . . . . . . . . . . . 231 I ideal gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 implicit differentiation . . . . . . . . . . . . . . . . . . 79 improper integral . . . . . . . . . . . . . . . . . . . . . . 152 indefinite integral .. . . . . . . . . . . . . . . . . . . . . 125 indeterminate form . . . . . . . . . . . . . . . . . . 69, 75 induction . . . . | ElementaryCalculus_Page_343_Chunk3886 |
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 infinite series . . . . . . . . . . . . . . . . . . . . . . . . 4, 289 convergent . . . . . . . . . . . . . . . . . . . . . . . . . 289 divergent. . . . . . . . . . . . . . . . . . . . . . . . . . .289 partial sum . . . . . . . . . . . . . . . . . . . . . . . . 289 sum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289 infinitesimal .. . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 infinitesimal function. . . . . . . . . . . . . . . . . . . .72 inflection point .. . . . . . . . . . . . . . . . . . . . . . . . 103 instantaneous rate of change.. . . . . . . . . .3, 8 integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 integral .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 convergent . . . . . . . . . . . . . . . . . . . . 152, 155 definite. . . . . . . . . . . . . . . . . . . . . . . . . . . . .132 divergent . . . . . . . . . . . . . . . . . . . . . 152, 155 exponential function .. . . . . . . . . . . . . . 128 improper .. . . . . . . . . . . . . . . . . . . . . . . . . . 152 indefinite . . . . . . . . . . . . . . . . . . . . . . . . . . 125 trigonometric .. . . . . . . . . . . . . . . . 128, 167 integral sign. . . . . . . . . . . . . . . . . . . . . . . . . . . .125 Integral Test . . . . . . . . . . . . . . . . . . . . . . . . . . . 294 integrand. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .139 integration .. . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 by parts .. . . . . . . . . . . . . . . . . . . . . . . . . . . 159 Monte Carlo . . . . . . . . . . . . . . . . . . . . . . . 255 numerical .. | ElementaryCalculus_Page_343_Chunk3887 |
. . . . . . . . . . . . . . . . . . . . . . . . 195 tabular method .. . . . . . . . . . . . . . . . . . . 163 interval of convergence. . . . . . . . . . . . . . . . .303 inverse function . . . . . . . . . . . . . . . . . . . . . . . . . 38 inverse hyperbolic functions . . . . . . . . . . . 234 inverse trigonometric functions . . . . . . . . . 41 involute of circle.. . . . . . . . . . . . . . . . . . . . . . .244 irrational numbers.. . . . . . . . . . . . . . . . . . . . . . .5 J Jacobian elliptic functions.. . . . . . . . . . . . .270 jerk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 jump discontinuity . . . . . . . . . . . . . . . . . . . . . . 73 L L’Hôpital’s Rule. . . . . . . . . . . . . . . . . . . . . . . . . .69 Laguerre polynomials .. . . . . . . . . . . . . . . . . 201 lamina. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .278 Langevin function .. . . . . . . . . . . . . . . . . . . . . 238 Laplace transform.. . . . . . . . . . . . . . . . . . . . .237 latus rectum.. . . . . . . . . . . . . . . . . . . . . .209, 223 of ellipse .. . . . . . . . . . . . . . . . . . . . . . . . . . 209 of parabola. . . . . . . . . . . . . . . . . . . . . . . . .215 left continuous.. . . . . . . . . . . . . . . . . . . . . . . . . .73 Legendre polynomials . . . . . . . . . . . . . . . . . . 201 Leibniz integral rule . . . . . . . . . . . . . . . . . . . 185 Leibniz notation . . . . . . . . . . . . . . . . . . . . . . . . . 15 limit. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3, 62 infinite .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 left . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 one-sided . . . . . . . . . . . . . . . . . . . . . | ElementaryCalculus_Page_343_Chunk3888 |
. . . . . . . 64 right. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .65 | ElementaryCalculus_Page_343_Chunk3889 |
334 Index Limit Comparison Test. . . . . . . . . . . . . . . . .294 limits of integration .. . . . . . . . . . . . . . . . . . . 139 line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 normal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 tangent. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .56 linear combination . . . . . . . . . . . . . . . . . . . . . . 24 linear function .. . . . . . . . . . . . . . . . . . . . . . . . . . . 9 linear operator. . . . . . . . . . . . . . . . . . . . . . . . . . .24 logarithmic differentiation.. . . . . . . . . .49, 86 Lorentz factor . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 Lorentz transformation . . . . . . . . . . . . . . . . 237 M Mach number. . . . . . . . . . . . . . . . . . . . . . . . . . . .89 Maclaurin’s series . . . . . . . . . . . . . . . . . . . . . . 309 magnitude . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 major auxiliary circle.. . . . . . . . . . . . . . . . . .243 maximum. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .90 Maxwell speed distribution . . . . . . . . . . . . 166 Mean Value Theorem. . . . . . . . . . . . . . . . . . .118 alternative form . . . . . . . . . . . . . . . . . . . 121 Extended. . . . . . . . . . . . . . . . . . . . . . . . . . .121 median of triangle.. . . . . . . . . . . . . . . . . . . . .285 method of exhaustion . . . . . . . . . . . . . . . . . . . . . 6 Microstraightness Property . . . . . . . . . . . . . 17 minimum .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 minor auxiliary circle . . . . . . . . . . . . . . . . . . 243 modulated wave . . . . . . . . . . . . . . . . . . . . . . . . 167 moment . . . . . . . . . . . . . . . . . . . . | ElementaryCalculus_Page_344_Chunk3890 |
. . . . . . . 277, 278 Monotone Bounded Test. . . . . . . . . . . . . . . .293 Monte Carlo integration.. . . . . . . . . . . . . . .255 Monte Carlo method . . . . . . . . . . . . . . . . . . . 260 N nappe.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .221 natural logarithm . . . . . . . . . . . . . . . . . . . . . . . 47 natural numbers. . . . . . . . . . . . . . . . . . . . . . . . . .4 Newton’s method .. . . . . . . . . . . . . . . . . . . . . . 109 norm of a partition .. . . . . . . . . . . . . . . . . . . . 135 normal distribution . . . . . . . . . . . . . . . . . . . . 284 normal line .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 Nth Derivative Test. . . . . . . . . . . . . . . . . . . . .106 numerical integration .. . . . . . . . . . . . . . . . . 195 O O notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 objective function.. . . . . . . . . . . . . . . . . . . . . . .90 odd function .. . . . . . . . . . . . . . . . . . . . . . . 13, 143 one-to-one . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 open interval . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 optimization .. . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 P p-series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .296 p-series Test. . . . . . . . . . . . . . . . . . . . . . . . . . . .294 Padé approximation .. . . . . . . . . . . . . . . . . . . 314 parabola. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .210 axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210 construction.. . . . . . . . . . . . . . . . . . . . . . .210 definition . . . . . . . . . . . . . . . . . . . . . . . . | ElementaryCalculus_Page_344_Chunk3891 |
. . 210 directrix.. . . . . . . . . . . . . . . . . . . . . . . . . . .210 eccentricity . . . . . . . . . . . . . . . . . . . . . . . . 210 focal radius . . . . . . . . . . . . . . . . . . . . . . . . 212 focus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210 latus rectum .. . . . . . . . . . . . . . . . . . . . . . 215 reflection property .. . . . . . . . . . . . . . . . 212 vertex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210 paraboloid .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212 parameter .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239 parametric equations. . . . . . . . . . . . . . . . . . .239 parametrization of a curve . . . . . . . . . . . . . 239 partial fractions .. . . . . . . . . . . . . . . . . . . . . . . 178 partition.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .134 norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 periodic waveform. . . . . . . . . . . . . . . . . . . . . .261 Planck’s constant .. . . . . . . . . . . . . . . . . . . . . . . 36 plane curve. . . . . . . . . . . . . . . . . . . . . . . . . . . . .239 point of tangency . . . . . . . . . . . . . . . . . . . . . . . . 56 polar axis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .245 polar coordinates .. . . . . . . . . . . . . . . . . . . . . . 245 derivatives . . . . . . . . . . . . . . . . . . . . . . . . . 248 pop. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .34 Power Formula . . . . . . . . . . . . . . . . . . . . . . . . . 127 Power Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 power series . . . . . . . . | ElementaryCalculus_Page_344_Chunk3892 |
. . . . . . . . . . . . . . . . . 4, 303 differentiation . . . . . . . . . . . . . . . . . . . . . 304 | ElementaryCalculus_Page_344_Chunk3893 |
Index 335 integration.. . . . . . . . . . . . . . . . . . . . . . . .304 interval of convergence . . . . . . . . . . . . 303 radius of convergence . . . . . . . . . . . . . . 303 principal axis. . . . . . . . . . . . . . . . . . . . . . . . . . .203 of an ellipse . . . . . . . . . . . . . . . . . . . . . . . . 203 probability.. . . . . . . . . . . . . . . . . . . . . . . . . . . . .282 probability density function .. . . . . . . . . . . 282 Product Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 product-to-sum formulas . . . . . . . . . . . . . . . 167 proof by induction .. . . . . . . . . . . . . . . . . . . . . . 24 Q q-derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 q-differential . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 quantum calculus . . . . . . . . . . . . . . . . . . . . . . . 89 quicksort algorithm .. . . . . . . . . . . . . . . . . . . 139 Quotient Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 R radioactive decay . . . . . . . . . . . . . . . . . . . . . . . . 50 radius of convergence . . . . . . . . . . . . . . . . . . 303 random variable.. . . . . . . . . . . . . . . . . . . . . . .282 continuous .. . . . . . . . . . . . . . . . . . . . . . . . 282 discrete . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282 range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 Ratio Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294 rational curve . . . . . . . . . . . . . . . . . . . . . . . . . . 244 rational function.. . . . . . . . . . . . . . . . . . .70, 178 rational numbers . . . . . . . . . . . . . . . . . . . . . . . . . 4 real numbers. . . . . . . . . . . . . . . . . . . . . . . . . . . . | ElementaryCalculus_Page_345_Chunk3894 |
. .4 recurrence relation . . . . . . . . . . . . . . . . . . . . . 288 reflection property.. . . . . . . . . . . . . . . . . . . . .206 ellipse.. . . . . . . . . . . . . . . . . . . . . . . . . . . . .206 hyperbola . . . . . . . . . . . . . . . . . . . . . . . . . . 220 parabola . . . . . . . . . . . . . . . . . . . . . . . . . . . 212 regular polygon . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 related rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 relativistic momentum . . . . . . . . . . . . . . . . . 285 relativistic velocity addition law . . . . . . . 123 reliability.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .158 Riemann sum . . . . . . . . . . . . . . . . . . . . . . . . . . 134 Riemann’s Rearrangement Theorem . . . 302 right continuous .. . . . . . . . . . . . . . . . . . . . . . . . 73 Rolle’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . 119 root mean square. . . . . . . . . . . . . . . . . . . . . . .261 root of function .. . . . . . . . . . . . . . . . . . . . . . . . 109 rotation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .226 rotational velocity . . . . . . . . . . . . . . . . . . . . . . . 31 S scalar. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .281 Schrödinger equation. . . . . . . . . . . . . . . . . . . .36 Schwarzchild radius.. . . . . . . . . . . . . . . . . . .101 secant line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 secant method. . . . . . . . . . . . . . . . . . . . . . . . . .112 Second Derivative Test . . . . . . . . . . . . . . . . . . 91 second-degree equation . . . . . . . . . . . . . . . . 227 sequence . . . . . . . . . . . . . . . . . . . . . . . . 4, 286, 287 Comparison Test. . . . . . . . . . . . . . . | ElementaryCalculus_Page_345_Chunk3895 |
. . . .293 convergent . . . . . . . . . . . . . . . . . . . . . . . . . 287 divergent. . . . . . . . . . . . . . . . . . . . . . . . . . .287 Fibonacci.. . . . . . . . . . . . . . . . . . . . . . . . . .288 limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287 Monotone Bounded Test . . . . . . . . . . . 293 series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4, 286 absolutely convergent. . . . . . . . . . . . . .301 alternating.. . . . . . . . . . . . . . . . . . . . . . . .300 conditional convergent. . . . . . . . . . . . .301 harmonic.. . . . . . . . . . . . . . . . . . . . .296, 300 infinite .. . . . . . . . . . . . . . . . . . . . . . . . . . . . 289 Integral Test . . . . . . . . . . . . . . . . . . . . . . . 294 Limit Comparison Test . . . . . . . . . . . . 294 n-th Term Test . . . . . . . . . . . . . . . . . . . . .294 p-series Test . . . . . . . . . . . . . . . . . . . . . . . 294 Ratio Test . . . . . . . . . . . . . . . . . . . . . . . . . .294 rearrangement. . . . . . . . . . . . . . . . . . . . .302 telescoping. . . . . . . . . . . . . . . . . . . . . . . . .298 Telescoping Series Test . . . . . . . . . . . . 294 shell method . . . . . . . . . . . . . . . . . . . . . . . . . . . 274 sigma notation . . . . . . . . . . . . . . . . . . . . . . . . . 135 sigmoid neuron function.. . . . . . . . . . . . . . . .52 Simpson’s rule.. . . . . . . . . . . . . . . . . . . . . . . . .197 slope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 slope of a curve . . . . . . . . . . . . . . . . . . . . . . . . . . 58 smooth curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 | ElementaryCalculus_Page_345_Chunk3896 |
336 Index snap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 Snell’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 solid of revolution .. . . . . . . . . . . . . . . . . . . . . 273 disc method . . . . . . . . . . . . . . . . . . . . . . . . 273 speed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2, 34 Squeeze Theorem. . . . . . . . . . . . . . . . . . . . . . . .71 stationary point . . . . . . . . . . . . . . . . . . . . . . . . . 91 step function.. . . . . . . . . . . . . . . . . . . . . . . . . . . .74 straight line motion. . . . . . . . . . . . . .1, 33, 129 substitution .. . . . . . . . . . . . . . . . . . . . . . . . . . . 146 half-angle . . . . . . . . . . . . . . . . . . . . . . . . . . 190 trigonometric . . . . . . . . . . . . . . . . . . . . . . 172 substitution method. . . . . . . . . . . . . . . . . . . .146 Sum Rule .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 surface of revolution . . . . . . . . . . . . . . . . . . . 271 T tabular integration .. . . . . . . . . . . . . . . . . . . . 163 tangent line. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .56 angle of. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .59 Taylor polynomial . . . . . . . . . . . . . . . . . . . . . . 312 Taylor’s formula . . . . . . . . . . . . . . . . . . . . . . . . 308 Taylor’s series . . . . . . . . . . . . . . . . . . . . . . . . . . 308 Remainder Theorem . . . . . . . . . . . . . . . 313 telescoping series. . . . . . . . . . . . . . . . . . . . . . .298 Telescoping Series Test. . . . . . . . . . . . . . . . .294 Thales’ Theorem. . . . . . . . . . . . . . . . . . . . . . . . .18 time dilation.. . . . . . . . | ElementaryCalculus_Page_346_Chunk3897 |
. . . . . . . . . . . . . . . . . . . .72 torque .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277 torus.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .276 trajectory envelope . . . . . . . . . . . . . . . . . . . . . 213 translation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224 transverse axis . . . . . . . . . . . . . . . . . . . . . . . . . 217 trapezoid rule . . . . . . . . . . . . . . . . . . . . . . . . . . 197 trigonometric functions .. . . . . . . . . . . . . 23, 41 inverses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 trigonometric integrals. . . . . . . . . . . .128, 167 trigonometric substitution . . . . . . . . . . . . . 172 U unit hyperbola. . . . . . . . . . . . . . . . . . . . . . . . . .230 V vector. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .281 velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 average . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2 instantaneous . . . . . . . . . . . . . . . . . . . . . . . . 2 vertical asymptote. . . . . . . . . . . . . . . . . . . . . . .66 W Wallis’ formula .. . . . . . . . . . . . . . . . . . . . . . . . 299 work. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .280 Z Zeno’s motion paradox . . . . . . . . . . . . 286, 290 | ElementaryCalculus_Page_346_Chunk3898 |
Michael Biehl The Shallow and the Deep A biased introduction to neural networks and old school machine learning | The+Shallow+and+the+Deep_Page_1_Chunk3899 |
The Shallow and the Deep | The+Shallow+and+the+Deep_Page_3_Chunk3900 |
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