Calculus MATH 1B
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Math 1B Discussion Exercises GSl Theo Johnson Freyd http math berkeley edutheojfOQSpring1B Find two or three classmates and a few feet of chalkboard Introduce yourself to your new friends and write all of your names at the top of the chalkboard As a group try your hand at the following exercises Be sure to discuss how to solve the exercises 7 how you get the solution is much more important than whether you get the solution If as a group you agree that you all understand a certain type of exercise move on to later problems You are not expected to solve all the exercises some are very hard Many of the exercises are from Single Variable Calculus Early Transcendentals for UC Berkeley by James Stewart these are marked with an Others are my own or are independently marked Always draw pictures Integration by Parts ln 1A you learned the product rule for di erentiation d d d xww Ema gltzgt M Ereoi This is often abbreviated fg f g 9 or dfg 1 lg golf By the fundamental theorem of calculus the integral of the derivative of a function is the original function Thus fg C fdfg f 1 lg golf ffdg fgdf Rearranging the equality gives the integration by parts formula which is a kind of product rule for inde nite integrals fzg zdz fzgz e anova What about de nite integrals Then fjfdg ffgdf f dfg fglg fbgb 7 faga b b b l fltzgtg ltzgtdzifltzgtgltzgtie govom 1 a Use the integration by parts formula to integrate memdm Hint let x z and g m em What is gz b Based on your answer to part a nd fzz e1 la Based on that nd fzg em dz and f m4 em dx c Guess the pattern from part Prove your pattern use integration by parts to write f z e1 dz in terms of fan 1 e1 la This is an example of a reduction formula 2 a Use the integration by parts formula to integrate lnmdz Hint let x lnz and g m 1 03 F U 00 to Let n 2k 1 be an odd integer Calculate f b Use the integration by parts formula to integrate lnz2 dz Hint let fz lnz2 and g z 1 c Use the integration by parts formula to integrate ln z dz Find a formula for f fz g z dz by applying the integration by parts formula twice Sup pose that f1 2 f4 7 f 1 5 and f 4 3 and that f z is continuous on 14 What is f14zf zdz7 By integrating by parts twice and rearranging nd fem cosz dz What is fem sinzdz How about f a coszdz where a is a constant How about fzcm coszdz Use integration by parts to evaluate the following integrals For some you will rst need to make a substitution a tsin2tdt d 325ds g 07rz3coszdz j c arcsinzdz f 911175dy i 0 as sint 7 3 d3 b z2 sin 7rz dz 1 e 0 z2 1e m dz xg h 1 arctan1z dz 03 cos02 d0 Am 1 sinlnz dz 7r k 5C0 sin 2t dt 0 What s wrong with the following proof that 0 1 lnzldzlzi zdz1ldzllnz z z z z Why can we forget to add an arbitrary constant during the intermediate steps when integrat ing by parts lntegrate fznemdz completely honestly let u z and dv emdz but this time let u e1 C 7r2 0 m cos z dz in two different ways a Using a reduction formula What happens to the boundary terms the M in fudv M 7 f1 du Why does it matter that n is odd The formula is different for even b Using a u substitution Hint cos2 z 1 7 sin2 z and sinz cos For any given k you could then expand out and evaluate the integral For a general k you can use integration by parts to get a reduction formula What is fOWZ dz Use the reduction formula from the previous problem to evaluate OWZ cos z dz for 71 even Math 1B Discussion Exercises GSl Theo Johnson Freyd http math berkeley edutheojfOQSummer1B ay by Cy 905 Recall that if yl and yg are solutions to the inhomogeneous equation ay by 0y gt then yl 7 yg is a solution to the homogeneous equation ay by 0y 0 Therefore the general solution to ay by 0y gt is y 9p 0191 Czyz where yl and yg are linearly independent solutions to the homogeneous equation and yp is any particular solution to the inhomogeneous equation Thus we can solve inhomogeneous linear differential equations provided we have some methods to guessing particular solutions to them The method we will outline is called the method of undetermined coefficients lt generalizes well to higher order differential equations but it does not work for all functions gt The idea is as follows Certain special functions 7 namely polynomials exponentials Sines and cosines and sums and product of these 7 have the special property that upon repeated di erentiation the functions cycle through77 a nite list For example if we repeatedly differentiate 5 sin 25 we get etsint cost et2 cost et2 cost 7 2 sin 25 etc all of which are linear combinations of the functions etsint and 5 cost So by cycle through a nite list77 I mean that a function ft is special in this sense if there is a nite set of functions f1t fnt so that all derivatives of ft including 1 itself are linear combinations of elements of the set In particular let s say that ft is a special function and 1305 fnt the corresponding nite set of functions Then af t Inct cft is de nitely a linear combination of the same nite list of functions Conversely let s say that gt is a special function and 91 gn its nite list Then for any constants A1 An the function ft A191t Angnt is special and its reasonable to hope that we nd constants A1 An so that af tbf tcft gt since nding such constants requires only that we solve a system of 71 linear equations in n unknowns Of course not every system of 71 linear equations in n unknowns has a solution It turns out that the only way we could fail to nd a solution in the above paragraph is if there are some constants B1 B so that B19105 Bngnt is a solution to the homogeneous equation ay by 0 0 In this case we have to expand our set 91 gn We can use the fact though that tg g tg and expand the set by multiplying each member by t For a second order linear differential equation you may have to do this at most twice for an nth order differential equation 71 times All in all we get the following rules for guessing the form of particular solutions to differential equations lf gt ektpt where p is a polynomial of degree n guess ypt ektqt where q is a degreen polynomial with undetermined coef cients If gt ektpt cos mt or ektpt cos mt guess ypt ektqt cos mt ektrt sin mt lfgt 91t 9225 it might be simpler to solve each equation ay by 0y 91t and ay by 0y 92t separately and add the answers If any yt of the form of the guess ypt is itself a solution to the complementary equation ay by 0y 0 you may have to multiply those terms by t or 252 1 to 03 F U a 1 00 Exerc Write a trial solution for the following differential equations Do not determine the coef cients b y 317 4y 903 006 d y 1 4y 531 zsin 2x a y 93 1 megm c y 1 231 10y mze m cos 3x Find the general solution for the following differential equations a y 7 42 5y 6 b yquot 22 y 906 Solve the initial value problem 24quot 2 i 2y z 81112957 y017 20 0 Use the method of undetermined coef cients to nd the general solution to the following rst order linear differential equation 3 a sin bx where a b are constants How would you normally solve this differential equation Which method do you prefer Use the method of undetermined coef cients to nd the general solution to the following rst order linear differential equation y mneam where a is a constant and n is a positive integer How would you normally solve this differential equation Which method do you prefer A spring with spring constant k mass m and damping constant c is hung vertically so that it experiences a constant downward force mg where g is the acceleration due to gravity Find the equilibrium solution ie nd the position at which the spring will hang without moving Then nd the general solution Explain how why for the purposes of solving problems with springs we can ignore gravity if we measure the displacement from the equilibrium solution rather than from the location in which the spring doesn t apply any force A series circuit contains a resistor with R 40 9 an inductor with L 2 H and a capacitor with C 00025 F Let s assume that the initial charge on the capacitor is 0 and that the initial current is 0 a If we add a battery to the circuit that applies a constant potential of 12 V how will the circuit respond b How will the circuit respond if we instead use a power source that applies an alternating potential of 12 sin50tsV Let f and g be special functions so that all the derivatives of f are linear combinations of f1 fm and all derivatives of g are linear combinations of 91 gn a Prove that the sum 1 9 is a special function by nding a nite list of functions so that all derivatives of f g are linear combinations of members of the list b Prove that the product fg is a special function by nding a nite list of functions so that all derivatives of fg are linear combinations of members of the list Hint think about the product rule ises marked with an are from Single Variable Calculus Early T musaeudeumls for UC Berkeley by James Stewart Math 1B Discussion Exercises GSl Theo Johnson Freyd http math berkeley edutheojfOQSummer1B Find two or three classmates and a few feet of chalkboard As a group try your hand at the following exercises Be sure to discuss how to solve the exercises 7 how you get the solution is much more important than whether you get the solution If as a group you agree that you all understand a certain type of exercise move on to later problems You are not expected to solve all the exercises some are very hard Exercises marked with an are from Single Variable Calculus Early Transcendentals for UC Berkeley by James Stewart Others are my own or are independently marked In nite Series A series is an in nite sum of numbers There are two sequences associated to each series First of all there s the sequence of terms in the sum aka summands 1111 1111 2 4 8 727478739 Second there s the sequence of partial sums 1lll 2 4 8 727478739 We normally write the former explicitly 1 1 1 1 1 7 7 7 7 2 4 8 Z 2 ri0 On the other hand a series converges exactly if the sequence of partial sums converges and the limit of the sequence of partial sums is the value of the series Standard notation the sequence of summands is an the sequence of partial sums is 3 220 ak and if 3 converges then 220 an limyH00 3 Here are some facts about series 0 A series cannot converge unless the sequence of summands tends to 0 But the sequence of summands can go to 0 without the series converging o The geometric series a ar arz ar3 220 ar converges if and only if lrl lt 1 If it converges then it converges to a17 r A proof and generalization of this is in exercise 3 o If 2a A and 2b B then 2an b A B Let 0 220 akbn7k Then 2 on AB This is called the Cauchy product77 or discrete convolution of the two series 0 This one isn t so much a fact as a technique If we can write each an as a difference an 1 7 bnirl then 3 b0 7 bn1 so the sum 2a converges if the sequence bn converges Partial fractions are a good way to nd such decompositions of terms as differences 1 What is the sum i1ni1111 2 2 4 8 16 ri1 to Howabout gtj1ni1111 1 3 73 9 27 81 If k is some number strictly greater than 1 what is 1 1 1 1 1 i 77777 Z k k k2 k3 k4 n1 Determine whether the following series are convergent or divergent lf convergent nd the sums 00 00 00 73n71 1 W a E b X i c 2 71 1 4n 0 1 3n 1 3n co d 2 2 e Z 7 f Zcos1 n0 n0 n0 00 3 00 00 n g g m h 2 arctan n 1 2 1H m Remember how to evaluate the geometric series if lrl lt 1 then we can evaluate S 220 ar a ar arz a7 3 by mutiplying by r and subtracting S aara72a73 7 rS arar2a73 SirS a a th S 7 us 17 Use this method to compute the following sums 12 3i b1 gg2i5 a 3 9 27 81 243 2 4 8 32 64 When money is spent on goods and services those who receive the money also spend some of it The people receiving some of the twicespent money will spend some of that and so on this chain reaction is called the multiplier e ect In a hypothetical isolated community the local government begins the process by spending D dollars Suppose that each recipient of spent money spends 10OC and saves 1008 of the money she or he receives The values 0 and s are called the marginal propensity to consume and the marginal propensity to save and of course 0 s 1 a Let Sn be the total spending that has been generated after it transactions Find an equation for Sn b Show that limH00 Sn kD there k 15 is the multiplier What is the multiplier if the marginal propensity to consume is 80 c In fact the marginal propensities to save and to consume vary by socioeconomic class among other things poor people spend a larger proportion of the money they receive and rich people save a larger proportion If the government is trying to stimulate the economy to whom should it give its money Math 1B Discussion Exercises GSl Theo Johnson Freyd http math berkeley edutheojfOQSummer1B Find two or three classmates and a few feet of chalkboard As a group try your hand at the following exercises Be sure to discuss how to solve the exercises 7 how you get the solution is much more important than whether you get the solution If as a group you agree that you all understand a certain type of exercise move on to later problems You are not expected to solve all the exercises some are very hard Exercises marked with an are from Single Variable Calculus Early Transcendentals for UC Berkeley by James Stewart Others are my own or are independently marked Alternating Series and Absolute Convergence Let on be a positive decreasing sequence on 2 by 2 0 for every n Then 271 bn converges Moreover if we truncate the series after then Nth term and estimate 23071 bn by 3N Z v71 bn then the error of the estimate is at most le1l A series Zan converges absolutely if 2 lanl converges It is a theorem that if Zan is abso lutely convergent then it is convergent But many series are convergent without being absolutely convergent For example the Alternating Harmonic series 2071W 1n converges by the alter nating series test but l71 1nl 1n is the divergent Harmonic series A series that converges but does not absolutely converge is said to converge conditionally 0 71n71 1 For what values of p does 2 T n a converge absolutely b converge conditionally c diverge 00 2 For what values of r does 2 r n1 a converge absolutely A b converge conditionally c diverge 3 Determine whether the following series are absolutely convergent conditionally convergent or divergent 4 U 1 00 The Riemann C function is de ned to be the analytic continuation of77 5 How many terms of the series would you need to add in order to nd the sum to the indicated accuracy a El inn 0 0 10 lerrorl lt 0000005 10 271 71ne lerrorl lt 00 n1 Show that the series 271 1bn where bn 1n if n is odd and bn 1n2 if n is even is divergent Why does the alternating series test not apply a Find a sequence an so that 221 an diverges but 221an2 converges 10 Find a sequence an so that 221 an converges but 21an2 diverges 221 a For what 3 does the above de nition of 5 converge le what is the domain of the right hand side b Prove that when both sides converge we have With 128 Proofs without Words 39 Make sens by Roger I The Alternating Harmonic Series y 1 1 1 1 1 1n2153 Zm l sin2n16 1 y z 1 1 2 I A Ax 39 V1 1 2 z 2 1 j 4 2 4 1 3 1 ll 1 2 3 4 3 4 l 11 A 14 l 1l 1 45 6395g39 478 73 1 quot 2quot 1 1 k 12 2H 2quot 2quot2k 1 2quot2k 2quot2k 1 2quot2k n 12 R fzdr 1 l1fl i 1 l 71 33W sin2n 19 Sine Math 113 Fall 2008 Recursive Sequence Example De ne the sequence an by a1 1 3an6i We will prove that an converges and nd the limit This kind of sequence where an1 is de ned in terms of an is called recursively de ned Welve done a couple of problems where we could only nd the limit of a recursively de ned sequence after we already knew that the sequence converged for example see problem 3 on Section 108 Quiz 4 Solutions So here is the outline we will follow an1 l Prove that an converges via ab and c a Prove that an is bounded b Prove that an is monotonici c Use the Monotonic Sequence Theoremi 2 Find limnn00 a Start with 1a We will prove an is bounded by induction In particular we will show that 0 S an S 2 for all n 2 1 In a problem you will most likely be given these bounds 0 and 2 If you are not then you7ll have to try to guess themi Here is the proof that 0 S an S 2 for n 2 1 First the base case n 1 Since a1 l we have 0 S a1 S 2 Next the inductive step k gt k 1 So assume 0 S ak S 2 and we will prove 0 S ak1 S 2 Since ak is positive ak1 fan 6 must also be positive So ak1 2 0 Also xgak 6 S 32 6 because ak S 2 and s is increasing 2i ak1 So ak1 S 2 We have shown 0 S ak1 S 2 which nishes the inductive proof thatOSanS2foralln21i Now we do 1b and show that an is monotonici First some general thoughts on this part of the problemi De ne 31 6 so an an1i As a rule if f is increasing on the interval of possible values of an 0 S I S 2 in this case then the sequence an is monotonici It is tempting to think that f increasing means that an is increasing but that is not the case If f is in creasing then to see if the sequence is increasing or decreasing just see if you have a1 lt a or al gt agi In this case the sequence should be increasing We now prove that an lt an1 for all n 2 l by induction First the base case n 1 We can nd a2 W gt 1 a1 so a1 lt agi Now the inductive step k gt k 1 So we assume that ak lt ak1 and prove that ak1 lt ak2i By our de nition of f above ak ak1 and fak1 ak2i We now check that f is increasing for 0 S I S 2 the possible values of an by step lai To do this differentiate to get 1 Hz gm 6W3 2 0 for 0 S I S 2 Since f is increasing and ak lt ak1 we get ak lt fak1 the idea is that if f is increasing then bigger inputs give bigger outputs But t is means ak1 lt ak2 which nishes the proof by induction that an is increasing We have shown that an is bounded and increasing so by the Monotone Sequence Theorem an convergesi This was 1C We now turn to step 2 which is nding the limit of an Let L limnn00 ani Then we also have L limnn00 an1 essentially all we have done is throw out the rst element of the sequence So starting wit an1 3 an 6 and taking limits of both sides gives L 71320 aw lim 3 an 6 naoo 3 lim an 6 limit rulescontinuity naoo fL 6 So L xgL 6 Cubing both sides rearranging and then factoring gives 0L37L76 L72L22L3A So L is root of the polynomial z 7 2 12 2x 3 Since 12 2x 3 has no real roots we must have L 2 In conclusion lim an 2i naoo So this is a pretty long example and you wouldnlt have to write so much if you did it But here7s a checklist of important steps 0 Prove by induction that an is bounded p S an S q for all n 2 1 You may have to use some algebra to get this to work depending on the problemi 0 De ne so that an an1i 0 Make sure is increasing iiei 2 0 for p S I S q o Prove that an is increasing or decreasing by induction To choose between increasing or decreasing just check whether a1 lt a2 or vice versa This proof will always be pretty much the same as the one in the example above 0 Conclude that an converges and let L limnn00 ani Set L and solve for L If there is more than one such L use your knowledge of the sequence to eliminate all but one of themi For example if an is always positive you can eliminate negative choicesi Similarly if a1 1 and an is increasing the limit cannot be 12 Rob Bayer Worksheet February 127 2009 The Limit Comparison Test 1 Determine whether each of the following integrals converges or diverges I4 7 312 10 sine d x a 10 I5 31 7 2 I d 2 lnz 1 7 lnzdz b I2M W41 ltegt 123 12 7 Is dz ztanilz l m21231 39 7 d f 7d c1 zs1nltx2gt z W 1 sin2 I we oo 2 Determine whether the integral converges or diverges by using 1 a The Limit Comparison Test b The regular Comparison Test c Which is easier for problems like this Integration Practice 1 Find each of the following integralsi Be sure to work as a group so everyone knows how to do all these problems a exezdz h ltbgt 75601 w 1 Cgtmd1 d iln1dz j e Site k f cos4t7sin4 tdt 1 g cos3 21 sin Zrdz m lnsec 9 sec2 Odt
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