MAE305-MAT301 Problem Set # 0, Self-Test: Review of Basic Ideas

MAE305-MAT301 Problem Set # 0, Self-test: Review of basic ideas

Assigned:14September2016 Due:Never This set of exercises is for review of elementary operations and ideas from calculus. It is not to be turned in. The exercises also provide a few opportunities to gain familiarity with elementary mathematical operations (including plotting) with Mathematica.

There are several topics from calculus that you should feel comfortable with before begin- ning this course. We will now provide a brief review of some of the more important topics that will be of use either shortly or later in the course. These topics include changes of variables and the chain-rule, standard integrals, integration by parts, partial di↵erentiation, Taylor series, and elementary operations involving complex numbers. It may sound obvious, but is nevertheless worth stating, that it is wise to feel comfortable and understand each of the operations, techniques and ideas discussed below. You should not feel like you are being asked to memorize a menagerie of unrelated topics. All of these themes are part of the toolbox of calculus that will appear in this course as well as other quantitative courses you will take. In addition, I suggest that you take this opportunity to learn a few basic commands using Mathematica. On the course website we have provided a separate document with more details including getting your own copy of the student version of Mathematica for your computer. Here we only give brief instructions. To get started, it should be as easy as

(i) Open Mathematica (ii) Elementary mathematical operations are +, , ,/ and powers are indicated with ˆ, ⇤ e.g. 2 3,x/6 and xˆ2. Also, Mathematica commands have a capitalized ﬁrst letter ⇤ and arguments are contained within brackets, e.g. Cos[x], Sin[x], Exp[x2]. (iii) Important: To run a Mathematica command requires you to type “shift-return” at the same time. (iv) You integrate the function f(x)=x2 with the command Integrate[x2,x] and the deﬁnite integral 1 x2 dx is obtained by typing Integrate[x2, x, 0, 1 ]. 0 { } 2 x (v) Similarly, you plotR the function f(x)=x e , from x = 0 to 2 with the command Plot[x2e x, x,0,2 ]. If you want to plot more than one function, then { } you simply place the list of functions within curly brackets, as in Plot[ x2e x, { Sqrt[x] , x,0,2 ]. } { } (vi) Try each of the above and then explore some of the Mathematica help pages.

0. Basic mathematical facts:

(a) ex+y = exey = ex + ey;andforanyconstanta,wehaveax+y = axay = ax + ay. 6 6 g ln g ln g 1 1 (b) ln e =1;ln(e )=g;also,e = g;moreover,e = g = = g. g 6

1 (c) Show that lim t ln t 0. Hint: Use L’Hopital’s rule. t 0 ! ! Explain why we then say that “t goes to zero faster than ln t goes to inﬁnity”. Solution: We can write ln t t 1 lim t ln t = = = t (1) t 0 t 1 t 2 ! after using L’Hopital’s rule. Thus, lim t ln t = t 0. t 0 ! ! t t (d) Show that lim te 0. Hint: Use the change of variables u = e to reduce t ! the given statement!1 to item (c). t Explain why we then say that “e goes to zero faster than t goes to inﬁnity”. (e) An integral from calculus (that comes up later in the course): Use the identity 2 1 cos(2x) sin x = 2 (familiar from calculus or high school) to show ⇡/2 ⇡ sin2 x dx = . (2) Z0 4 Solution: Given the hint we write ⇡/2 1 ⇡/2 ⇡ sin2 x dx = (1 cos(2x)) dx = , (3) Z0 2 Z0 4 since the second term in parentheses integrates to sin(2x), which vanishes at x =0,⇡/2. If you want to test your skill further, then show ⇡/2 3⇡ sin4 x dx = . (4) Z0 16 Solution: We write ⇡/2 1 ⇡/2 1 ⇡/2 sin4 x dx = (1 cos(2x))2 dx = 1 2cos(2x)+cos2(2x) dx, 0 4 0 4 0 Z Z Z ⇣ (5)⌘ which leads to (only giving the non-zero terms) ⇡/2 1 ⇡ ⇡ 3⇡ sin4 x dx = + = . (6) Z0 4 ✓ 2 4 ◆ 16 If you feel energetic, show ⇡/2 5⇡ sin6 x dx = . (7) Z0 32 For perspective, these integrals were known already in 1792 by David Ritten- house (1732-1796), a very early American intellectual, who apparently had no

2 formal education and never earned a degree (his method for evaluating the integrals was also di↵erent).1 In case you think that it is easy to spot a pattern, ⇡/2 8 35⇡ note that 0 sin x dx = 256 . 1. Changes of variablesR and the chain-rule: Recall the chain-rule df (x(t)) dx(t) df (x) = . (8) dt dt dx df (a) Given f(x)=x2e x.Letx = t2/3.Evaluate as a function of t. dt 2 x 2/3 Solution: Given f(x)=x e with x = t ,then

df (x) x 2 x =2xe x e , (9a) dx dx 2 1/3 = t . (9b) dt 3 Therefore,

df (x(t)) dx df (x) 2 x 2 1/3 2 1/3 t2/3 = =(2x x )e t = (2t t)e . (10) dt dt dx 3 3

2 x (b) Where does the function f(x)=x e have a maximum? Use your knowledge of calculus, then plot the function and verify your answer. df (x) x 2 Solution: We calculate dx = e (2x x )=0wherex =0orx =2.You can plot the function to verify the location of the maximum.

2. How ideas with simple changes of variables often arise in di↵erential equations:

Consider the function y(t), which satisﬁes the equation dy = ay2 with y(0) = y . (11) dt 0

Here a, y0 and t0 are considered known numbers (constants). We will likely discuss this equation the ﬁrst day of class. Rescale the equation with the change of variables: t y T = ,Y= , (12) tc y0

where the function Y (T )=y(t)andtc is a constant you are to identify. The key is to choose tc to simplify the di↵erential equation to have all of the remaining constants equal to unity.

1See D.E. Zitarelli, “David Rittenhouse: Modern Mathematician”, Notices of the AMS, p. 11-14, January 2015.

3 In the di↵erential equation, make the change of variables indicated. This question tests your understanding of making a change of variables.

Show that Y (T )satisﬁesthedi↵erentialequation dY = Y 2 , with Y (0) = 1. (13) dT

What did you select for tc? Remark: Notice how equation (13) is “cleaner” (or we would say “nicer”) than the original equation (11). The two equations are really structurally the same, but it is easier to think about the equation that does not have extra algebraic quantities that are cumbersome and do not influence how we think about the solution to the equation. Moreover, in an actual physical problem, the quantity tc represents the time scale over which significant changes occur, so with this identification you have learned something before solving the problem. Solution: We begin with the indicated change of variables. We write

dy d dT dY y0 dY = (y0Y (T )) = = , (14) dt dt dt dT tc dT where we have used the fact that y is a constant and dT = 1 .TheODEcanthen 0 dt tc be written y0 dY 2 2 = ay0Y , with Y (0) = 1. (15) tc dT

Choose the constant tc to eliminate the other constants in the equation. Thus, we choose t = 1 .TheODEthenreducestoequation(13). c ay0 3. Integrals that yield logarithms:2 d ln x d ln( x) Reminder: Since =1/x for x>0and =1/x for x<0, we use dx dx the absolute value symbol to write x ds =ln x +constant. (16) Z s | | 2 dx (a) Evaluate . Are there any restrictions on the values of a? Z0 x + a Solution:

2 dx 2 1 =[ln x + a ]0 =ln 2+a ln a =ln 1+2a . (17) Z0 x + a | | | | | | | | 2In calculus you should beware that log x and ln x are often used interchangeably, as we do in this document.

4 Notice that a =0anda = 2fortheinterpretationbasedontheargumentsof the logarithm,6 which is discussed6 in advanced books as the meaning in terms of the principal value. Otherwise, the integrand has a zero in the denominator and is ill-deﬁned for 2 a 0.   ⇡/2 cos x (b) Evaluate dx. Z⇡/6 sin x Solution: Recognize that the integrand is the derivative of ln sin x so that | | ⇡/2 ⇡/2 cos x d ln sin x ⇡/2 dx = | |dx =[ln sin x ]⇡/6 =ln(1) ln(1/2) = ln 2. Z⇡/6 sin x Z⇡/6 dx | | (18) 2 dx (c) Evaluate 2 2 Z1 x(a + x ) Solution: Expand to obtain two straightforward integrals.

2 2 2 dx 1 1 x 1 1 2 2 2 2 = 2 2 2 dx = 2 ln x ln(a + x ) Z1 x(a + x ) a Z1 ✓x a + x ◆ a  2 1 1 x2 2 1 4(1 + a2) = 2 ln 2 2 = 2 ln 2 . (19) 2a " a + x #1 2a 4+a

(d) Use a partial fraction expansion to evaluate

1 dx . (20) Z0 (x + a)(x + b) Solution: Expanding the integrand we have 1 dx 1 1 1 1 = dx 0 (x + a)(x + b) b a 0 x + a x + b Z Z ✓ ◆ 1 1 (1 + a)b = [ln x + a ln x + b ]1 = ln (21). b a | | | | 0 b a (1 + b)a Note: For this problem, other problems here, and indeed standard problems in calculus, this course, and other mathematics courses involving questions whose answers require the use of an algorithm, you can check your answers by using Maple or Mathematica,whichareavailabletostudentsatPrinceton,orviathe Wolfram Alpha website, www.wolframalpha.com. We encourage you to use these tools to aid in your learning.

4. Integration by parts:

5 Review the method of integration by parts:3

b b dv(x) b du(x) u(x) dx = u(x)v(x) a v(x) dx. (23) Za dx | Za dx 1 (a) Evaluate x log xdx. Can you a priori determine whether the integral is 0 positive or negative?Z Solution: One application of integration by parts is necessary.

1 x2 1 1 1 1 x ln xdx = ln x x2 dx Z0 " 2 #0 2 Z0 x 1 x2 1 1 = = . (24) 2 " 2 #0 4

2 (Reminder: Use l’Hopital’s rule to show that limx 0 x ln x = 0.) Note that ! the integral is clearly negative since on the interval 0

1 sin x 1 2 1 x2 cos x dx = x2 x sin x dx Z0  0 Z0 sin 2 cos x 1 1 1 = x + cos x dx " ✓ ◆0 Z0 # sin 2 cos 2 sin 2 cos 2sin = + [sin x]1 = + (25). 2 3 0 2 3

5. Partial di↵erentiation: @f @f (a) Given a function f(x, y), state the meaning of versus . @x @y Solution: In evaluating the partial derivative (@f/@x), the variable y is ﬁxed, i.e. it is regarded as a “constant” (independent of x). Similarly, the variable x is regarded as a “constant” (independent of y)inevaluatingthepartialderivative (@f/@y).

3This equation is remembered easily by recalling the product rule for di↵erentiation

d du(x) dv(x) (u(x)v(x)) = v(x)+u(x) , (22) dx dx dx and then integrating both sides.The result was obtained ﬁrst by the German philosopher and mathematician Gottfried Wilhelm Leibniz in 1675.

6 @f @f (b) Given f(x, y)=x2exy log y,evaluate and . @x @y Solution: @f =(2xexy + yx2exy)logy, (26a) @x @f exy = x2 xexy log y + . (26b) @y y ! (c) Suppose that u(x, y)isaspeciﬁedfunctionwithx(t)andy(t). Write down an equation to determine the derivative of u with respect to t. Solution: Using the chain rule, du(x(t),y(t)) dx @u dy @u = + . (27) dt dt @x dt @y 6. Changes of variables for special functions: Later in the course we will learn a variety of special functions, which are deﬁned in terms of integrals. (a) Given the identify 1 2 t t e dt =2, (28) Z0 1 2 2t evaluate t e dt. Z0 Solution: Start with equation (28) and make the change of variables u =2t. Then, we have 2 1 2t 2 1 u u 1 1 1 2 u 1 e t dt = e du = u e du = . (29) Z0 Z0 4 2 8 Z0 4 =2 (b) Given the identify | {z } sin x ⇡ 1 dx = , (30) Z0 x 2 sin(4x) evaluate 1 dx. Z0 x Solution: Start with equation (30) and make the change of variables u =4x. Then, as du =4dx,wehave sin(4x) sin u ⇡ 1 dx = 1 du = . (31) Z0 x Z0 u 2 7. Some remarks about Taylor series:4 This exercise is more of a discussion then a set of problems, but you may wish to write out some of the details to gain experience with the ideas of a Taylor series. 4Introduced by the English mathematician Brooke Taylor (1685-1731), though the idea was apparently known earlier. Taylor series are one of the most useful ideas in calculus!

7 Why do we introduce the Taylor series? What does it tell you about functions? e.g. the first term in the Taylor series? the second term? the third term? Draw sketch of a typical smooth function and indicate the first few terms of a Taylor series of the function near some point. If you have trouble thinking about these problems than you did not pick one of the key intellectual features of calculus, which asks you to think qualitatively about functions, as well as qualitatively about functions, such as how they change, the rate of change, etc. The Taylor (or MacLaurin) series of a function f(x)aboutthepointx =0isdefined as (n) n 1 f (0) d f f(x) xn where f (n) . (32) ⌘ n! ⌘ dxn nX=0 Typically there is a range of x values, called the interval of convergence, for which the series on the right-hand side converges to the value of f(x). The generalization of the Taylor series for an expansion about the point x = a is written (n) 1 f (a) f(x) (x a)n . (33) ⌘ n! nX=0 Verify each of the Taylor series shown below (expansions about x =0):

2 4 2n x x 1 x cos x =1 + + = ( 1)n (34a) 2! 4! ··· (2n)! nX=0 3 5 2n+1 x x 1 x sin x = x + + = ( 1)n (34b) 3! 5! ··· (2n +1)! nX=0 2 3 n x x 1 x ex =1+x + + + = (34c) 2! 3! ··· n! nX=0 1 n(n +1) n(n +1)(n +2) =1 nx + x2 x3 + for x < 1(34d) (1 x)n ⌥ 2 ⌥ 3! ··· | | ± n+1 1 1 1 1 ( 1) ln(1 + x)=x x2 + x3 x4 + = xn for 1

n n n 1 x d cos x x cos x = = ( 1)n/2 (36a) n! dxn ! n! nX=0 x=0 n=Xeven 8 2m 1 x = ( 1)m , (n =2m). (36b) (2m)! mX=0 To see how well this idea works, plot the first few terms of the series and compare with cos x. We may follow a similar procedure as above to obtain the Taylor series for sin x about x =0.Or,bytheidentity d cos x = sin x, dx we obtain 2m d cos x d 1 x sin x = = ( 1)m (37a) dx dx (2m)! mX=0 2m 1 2n+1 1 x 1 x = ( 1)m = ( 1)n , (m = n +1) (37b) (2m 1)! (2n +1)! mX=1 nX=0 Since n x d e x n =[e ]x=0 =1, " dx #x=0 then using the Taylor series formula we find n n x n 1 x d e 1 x ex = = (38) n! " dxn # n! nX=0 x=0 nX=0 Similarly, d 1 n = , (39) dx (1 x)n ⌥(1 x)n+1 ± ± then using the Taylor series formula we find 1 1 x d 1 x2 d2 1 = + + +(40a) (1 x)n "(1 x)n # 1! "dx (1 x)n # 2! "dx2 (1 x)n # ··· ± ± x=0 ± x=0 ± x=0 x2 x3 =1 nx + n(n +1) n(n +1)(n +2)+ (40b) ⌥ 2 ⌥ 3! ··· Since d 1 ln(1 + x)= , (41) dx (1 + x) then by the expansion in equation (40) (n =1),weobtain

2 3 4 n x dt x x x 1 x ln(1 + x)= = x + + = . (42a) (1 + t) 2 3 4 ··· n Z nX=1

9 8. Comparing functions when variables are small: In many practical problems we have to compare the typical magnitude of two di↵erent terms. In a typical calculus course, you learn the idea of a limit and so, for example, can evaluate (you might wish to consult the Taylor series above for sin x) sin2(2x) (2x)2 lim sin x x 0orlim = =4. (43) x 0 x 0 2 2 ! ⇡ ! ! x x It is helpful to think about these limits as being taken for x 1. Hence, it is gener- ally most useful to write lim sin x x and to think about this⌧ as the approximation x 1 ⇡ of sin x when the argument⌧ is small compared to 1. You do not simply substitute x =0intheformulasinceweseekanapproximation to the original function. Using the idea of a Taylor series, approximate the following ratios when an argument is small. Report your answer as a function of x. sin(4x) (a) Approximate for x 1. Is this ratio much smaller or much larger x3/2 ⌧ than unity? Solution: We have sin(4x) 4x 4 lim = = 1. (44) x 1 3/2 3/2 1/2 ⌧ x x x For x 1theratioisclearlyalargenumber. ⌧ 2x2 1 e (b) Approximate x2 for x 1. ⌧ u Solution: For small arguments, e 1 u,thenwehave ⇡ 2x2 2 1 e 2x = =2. (45) x2 x2 9. Elementary properties of complex numbers: (there are no exercises here) We only require a few elementary operations in this course but the subject is studied in more detail in MAE306. (a) We begin with the deﬁnition i2 = 1ori = p 1. A complex number is written c = a + ib where the pair (a, b)arerealnumbers.Wereferto a as the real part of the complex number, a =Rec,andb as the imaginary part of the complex number, b =Imc. (b) The usual rules of algebra apply, with the caveat that i2 = 1. Thus, (a + ib)(c + id)=(ac bd)+i(ad + bc). (c) Euler’s formula:5 (This is one of many formulas associated with Euler’s name.) eix =cosx + i sin x (49)

5We note that if z is a complex number, then we deﬁne ez according to equation (3) so that z2 z3 z4 ez =1+z + + + + . (46) 2! 3! 4! ···

10 (d) imaginary powers: Recall that y = x↵ may be written as y = e↵ ln x where we have used the identity c = eln c.Thus,weseethatifthepowerisacomplex number, we may write

xa+ib = xaxib = xaeib ln x = xa [cos(b ln x)+i sin (b ln x)]. (50)

Note: This idea will reappear early in the course when we study the Euler equidi- mensional di↵erential equation (remember we said Euler’s name appears a lot).

Thus, letting z = ix we have x2 x3 x4 eix =1+ix i + + , (47) 2! 3! 4! ··· which may be written by grouping real and imaginary parts as

x2 x4 x3 x5 eix =1 + + + i x + + = cos x + i sin x. (48) 2! 4! ··· 3! 5! ··· ✓ ◆ The ﬁnal identity follows by recognizing the Taylor series for cos x and sin x.

11 Figure 1: Reference: D. J. Struik, A Concise History of Mathematics,Dover(1948).

Historical note: Leonard Euler was one of the great mathematicians of all time. Born and raised in Basle Switzerland, Euler studied under Johann Bernoulli, one of the Bernoulli clan of mathematicians that spanned more than four generations (Euler’s father, a clergy- man, in fact studied under Jakob Bernoulli who was Johann’s older brother). Euler moved to the St. Petersburg Academy in Russia in 1727, the academy having been founded some years earlier by Peter the Great. In 1741 Euler moved to the Berlin Academy only to return to St. Petersburg in 1766. Euler lost one eye in 1735 and the other eye in 1766 but somehow this did not deter the enormous breadth and productivity of his mathematics. Euler’s notation is often the notation we use today. For example, notation in trigonometry, the exponential e,andthedefinitiveuseof⇡ for the ratio of the cir- cumference to the diameter of a circle are largely due to Euler, as are the use of i for p 1 and ⌃to designate summation. Euler introduced the Gamma and Beta functions, which we will see later in this course, as well as the Euler constant defined as 1 1 1 lim ( + + ...+ ln n)=0.577216 ... .InoneofEuler’smanytextbookswe n 1 2 n ⌘ find!1 a section on di↵erential equations that distinguishes linear, exact and homogeneous equations and serves as a model for textbooks today (hence the first week of this class). References: D. J. Struik, A Concise History of Mathematics, Dover (1948); C. B. Boyer, A History of Mathematics, as revised by U. C. Merzbach, John Wiley & Sons (1991).