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

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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↵er- entiation, 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 com- mand 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 in- tegrals 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).
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