# Review of basic calculus

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1 Review of bsic clculus This brief review reclls some of the most importnt concepts, definitions, nd theorems from bsic clculus. It is not intended to tech bsic clculus from scrtch. If ny of the items below is not fmilir to you, plese consult clculus book nd review the topic in more detil. Limits nd continuity Clculus begins with the definition of limit. In fct, there re two, relted definitions. We sy tht sequence of numbers n pproches the limit L in cse for ll positive ɛ there exists rel number N such tht n L < ɛ whenever n > N. This is the definition of sequentil limit. We write lim n L n s shorthnd nottion to men tht the sequence n pproches the limit L. A second nottion for this is to write n L s n. The second, relted definition is functionl limit: we sy tht f(x) pproches L s x pproches in cse for ll positive ɛ there exists positive δ such tht f(x) L < ɛ whenever 0 < x < δ. As bove, we use the nottion lim f(x) L nd lso the nottion f(x) L s x. x One cn lso define one-sided limits: for the left-hnd limit we chnge the definition to require tht f(x) L < ɛ when δ < x < ; for the right-hnd limit the requirement holds when < x < + δ. The left- nd right-hnd limits re denoted lim f(x) nd lim f(x), respectively. x x + It is theorem tht lim f(x) L if nd only if whenever n we hve tht f( n ) L. It is lso x theorem tht f hs limit t if nd only if f hs both left- nd right-hnd limit nd these limits re equl. lim x The notion of limit is the bsis for the definition of continuity: we sy tht f is continuous t if f(x) f(). Sid nother wy, f is continuous t if the following three conditions re stisfied: f is defined t. f hs limit t. The limit equls f(). We cn interpret limits in the context of pproximtions. To sy tht f(x) L s x is to sy tht f(x) L when x. More precisely, to sy tht f(x) L < ɛ is to sy tht L ɛ < f(x) < L+ɛ, which is to sy tht f(x) L with error less thn ɛ. There re mny wys function cn be discontinuous. One wy the most importnt for us is to hve removble discontinuity. This hppens when f hs limit L t point, but either f is not defined t or else f() L. In this cse, if we redefine f() to be L, then the new definition gives function which is continuous t. If both the left- nd right-hnd limits exits, but re not equl to one nother, then we sy tht f hs jump discontinuity. Some of the most importnt exmples of functions with jump discontinuity re the unit step functions sometimes clled the Heviside functions which re defined by the rule { 0, if x < c u c (x), if x c A third type of discontinuity is verticl symptote. This hppens when for ll M there is positive δ such tht f(x) > M whenever 0 < x < δ. We sometimes use the nottion lim f(x) or x f(x) s x. There re mny, mny other kinds of discontinuities. Just to give yourself n inkling of how bd discontinuity cn be, look t grph of the function sin(/x) ner 0. One of the min properties of continuous function is the Intermedite Vlue Theorem, which sys tht if f is continuous on intervl contining points nd b, nd if z is ny vlue between f() nd f(b), then there is point c between nd b such tht f(c) z. This is often used in the following specil cse: if f is continuous on some intervl, nd if f tkes positive vlue somewhere on the intervl nd negtive vlue t nother, then somewhere in between f must equl 0.

2 The Intermedite Vlue Theorem leds to method for solving equtions defined by continuous functions. If f is continuous on the intervl [, b] nd if f chnges sign between nd b tht is, f() nd f(b) re of opposite sign then either f chnges sign somewhere between nd the midpoint ( + b)/2, or else between this midpoint nd b. In the former cse we replce b by the midpoint; in the ltter we replce by the midpoint. In either cse we hve replced the intervl by one hlf s long, nd so hve nrrowed by hlf the possible error in knowing where f(x) 0. This is clled the bisection method. It is very slow, but hs the dvntge tht it includes esily predictble error estimte. One of the min theorems is the continuity of lgebr, which sys tht ny function defined by lgebric opertions ddition, subtrction, multipliction, nd division is continuous t every point of the domin (tht is, t ny point where you don t try to divide by 0). In prticulr, ll polynomil functions re continuous everywhere. It is lso theorem tht if g is the inverse function of one-to-one function f, nd if f is continuous t, then g is continuous t f(). In prticulr, the frctionl powers x /n re ll continuous wherever they re defined. Slope nd derivtive A nonverticl line hs constnt slope. In other words, if points (, b) is ny point of the line nd if ( + x, b + y) is ny second point, then the rtio y/ x lwys yields the sme vlue, no mtter wht two points we choose. (Of course, when we chnge the line then the vlue of this slope chnges.) The slope is the most importnt chrcteristic of the line. A positive slope mens the line represents n incresing function; negtive slope corresponds to decresing function; nd constnt function hs slope 0. Lines cn be defined by mny types of equtions. For exmple 2x 3y 7, x 9, y 8 5(x + 4) re ll equtions which define lines. The lst is n exmple of point-slope formul: if the line psses through the point (, b) nd hs slope m then one eqution for the line is y b m(x ). Every nonverticl line hs well-defined slope, nd hence every point on it yields n eqution in point-slope form. For us this will be the most useful formul for lines. The derivtive of f t point is defined to be y lim x 0 x lim x 0 f( + x) f(), x provided the limit exists. If it does we use either of the nottions f () or df () to denote this limit, nd refer to it s the instntneous slope of f t. When the derivtive exists we sy tht the function is differentible. Another wy to think bout the derivtive is to consider the slope function m( x), which mesures the slope of the secnt line thru (, f()) nd ( + x, f( + x)). This function is not defined when x 0, but when it hs removble discontinuity there the vlue we define when x 0 is the derivtive of f t. A differentible function cn be pproximted by liner function on sufficiently smll intervls. So, if you grph differentible function on grphing clcultor nd then zoom in on smll intervls then wht you see is indistinguishble from stright line. This is becuse if y/ x f (), then for ny positive ɛ we hve tht y f () x, with error less thn ɛ x, whenever x is sufficiently smll. One of the most importnt consequences of this interprettion is the theorem tht if f is differentible then f is continuous. When we hve function of more thn one vrible, f(x, y,...) nd if we hold ll but one vrible constnt, nd differentite with respect to the remining vrible, we obtin the prtil derivtive of f. For exmple, if f is function of x nd y, then it hs two prtil derivtives, which we denote f/ x nd f/ y. We will mke only slight use of prtil derivtives this semester. The higher order derivtives of function re defined by iterting the opertion of the differentition. So, the second derivtive is the derivtive of the derivtive, nd is denoted either f or d2 f. The n-th 2 derivtive of f is denoted either f (n) or dn f n. 2

3 The Men Vlue Theorem If f is continuous on closed intervl [, b] then there is point c in tht intervl such tht f(c) f(x) for every x in [, b]. Such vlue f(c) is clled the mximum of f on [, b]. Note tht there my be severl points t which f tkes its mximum vlue. Similrly, if f is continuous on [, b] the f tkes minimum vlue somewhere in [, b]. Fermt s Theorem is the observtion tht if the f tkes on mximum (or minimum) vlue t some point c in the interior (, b), nd if f is differentible t c, then f (c) 0. When f (c) 0 we sy tht c is criticl point for f. So, if f hs mximum (or minimum) t c then either c is criticl point, n endpoint, or point where f is not differentible (sometimes clled singulrity). Using Fermt s Theorem we cn prove the Men Vlue Theorem for Derivtives, which sys tht if f is continuous on [, b], nd differentible on the interior (, b), then there is point c in the interior such tht f f(b) f() (c). b The right-hnd side of this eqution is the verge slope of f on [, b], whence the nme of the theorem. We will stte generliztion of this theorem to higher order derivtives when we come to power series. The Men Vlue Theorem hs n importnt consequence for the study of differentil equtions the uniqueness of solutions to differentil equtions. The simplest form of this sys tht of f nd g re differentible on (, b), nd if f (x) g (x) for ll x in (, b), then in fct f(x) g(x) is constnt on the intervl. In geometric lnguge this sys tht if the grphs of f nd g hve the sme (instntneous) slope t every point, then the grph of one cn be shifted up or down to mtch the grph of the other. A third wy to stte this theorem is to sy tht if we re given the vlue of the derivtive of f t every point, nd the vlue of f t one point, then there is t most one possibility for f. When we hve covered integrtion we will see in fct tht there is lwys exctly one possibility for f, ltho it my be difficult or even impossible to find net formul for it. Antidifferentition From the Men Vlue Theorem for Derivtives we lerned tht if we re given function g(x) is defined on [, b] nd rel number C then there is t most one differentible function F(x) such tht F (x) g(x) nd F() C. Such solution F(x) is clled n ntiderivtive of g(x), nd the process of finding the ntiderivtive is clled ntidifferentition. Is there lwys n ntiderivtive? How would we find it? In bsic clculus we nswered these questions with Riemnn integrls nd the Fundmentl Theorem of Clculus: if g(x) is continuous then stisfies the given initil vlue problem. Riemnn integrls F(x) C + g(t) dt Recll tht derivtive, slope, velocity, nd rte of chnge re ll essentilly synonyms. We use different words in different contexts, but mthemticlly they re ll the sme concept. We cn use this to motivte our solution the simple initil problem F (x) g(x), F() C. Here, g(x) nd C re given; F(x) is the unknown. The ide is to think of F(x) s the position of prticle moving long line with velocity g(x) t time x. In this interprettion C represents the position t time. If the velocity g(x) is continuous then on very smll time intervl, sy from t to t 2, the velocity is pproximtely constnt, roughly equl to g(t ), nd so we expect tht on the intervl [t, t 2 ] we would compute the chnge in F s follows: F g(t ) t where t t 2 t. This is the fmilir distnce rte time formultion, which works when the rte is constnt. Now if we divide the intervl [, x] into series of smll subintervls [t, t 2 ], [t 2, t 3 ],... 3

4 then the new position t time x is pproximtely equl to the strting position C plus the sum of the chnges F over ll of the subintervls: F(x) C + n g(t i ) t. Riemnn s Theorem is tht the error in this pproximtion vnishes s the length of ll the subintervls shrinks to 0. More precisely, if g(x) is continuous then the limit g(t)dt i lim mx t 0 i n g(t i ) t does not depend on how you divide up the intervl [, x] into subintervls now how you choose the representtive point t i from the i-th subintervl. No mtter how you mke these choices, the limit lwys exists, nd lwys gives the sme result, provided only tht the lengths of ll these subintervls shrink to 0. This limit is clled the Riemnn integrl of g on the intervl [.x]. There re mny wys to interpret the Riemnn integrl, nd hence mny pplictions. If we drw grph of g then we see tht its integrl lso mesures the re between the grph nd the horizontl xis, t lest where g(t) > 0: if g(t) > 0 then g(t) t is the re of rectngle of height g(t) nd width t. If g(t) < 0 then this differentil mesures the negtive of the re. The Men Vlue Theorem for Integrls Another importnt interprettion of Riemnn integrl is s n verging process. It is simplest to explin this when we divide up the intervl [, b] into n equl-size subintervls, hence of size (b )/n. Since the common fctor (b ) is constnt, we cn fctor it out of the Riemnn sum nd lso out of the limit: n g (b ) lim g(t i ). n i Since the sum inside the limit is simply the men (verge) vlue of the g(t i ), it is resonble to interpret b g men vlue of g on [, b]. The men vlue of the g(t i ) lies between the minimum nd mximum vlues of g on [, b]. By the Intermedite Vlue Theorem we deduce tht if g is continuous then g tkes its men vlue somewhere in [, b]. Tht is, if g is continuous on [, b] then for some c in [, b] we hve tht g(c) b The Fundmentl Theorem of Clculus We now hve ll of the tools we need to see tht the Riemnn integrl does indeed solve our simple initil vlue problem tht is, tht g is n ntiderivtive of g. To see this we compute the difference quotient nd pply the Men Vlue Theorem for Integrls: Fundmentl Theorem of Clculus If g is continuous nd we define F(x) C + g then proof. df df lim F x 0 x lim x 0 x g. g(x), F() C. + x x g(t)dt lim g(c) g(x). x 0 This lst equlity is becuse of the Squeeze Lw: since c is between x nd x + x nd since x 0 we conclude tht c x. 4

5 There re two wys (t lest!) tht we pply the Fundmentl Theorem: to define functions s solutions to initil vlue problems, nd to evlute integrls when we know n ntiderivtive by some other mens (usully rote memory). Let s look t exmples of ech of these pplictions. None of the bsic rules for derivtives Power Rule, Chin Rule, Product Rule,... combine to give formul for the ntiderivtive of /x. But /x is continuous, t lest when x 0, nd so the Fundmentl Theorem gurntees tht it hs n ntiderivtive. Since this ntiderivtive is importnt in pplictions, we give it nme: logrithm. (The word nd the definition re due to John Npier, who lived before Clculus ws invented!) More precisely, dt log(x) ln(x) t. We use the nottion log(x) nd ln(x) interchngebly. The nturl logrithm is in essence the only logrithm: ll other logrithms re sclr multiples of it. So, whenever we sy logrithm in this clss we men the nturl logrithm, unless we explicitly sy otherwise. So, the nturl logrithm is the unique solution to the initil vlue problem d log(x), log() 0. x Hence it is n incresing function on (0, + ). Therefore it hs n inverse, which we cll the exponentil function: exp(log(x)) x, log(exp(x)) x. If we pply the Chin Rule to this second reltion then we obtin the derivtive of the exponentil function: exp(x) d exp(x), whence d exp(x) exp(x). We sometimes use the nottion e x for exp(x). This is becuse it obeys the usul lws for exponents. To see this, we strt with the identities log(b) log() + log(b), log( n ) n log(). The first cn be proved by replcing b by x nd then observing tht both sides hve the sme derivtive nd the sme vlue t. By the Men Vlue Theorem for Derivtives they must be the sme everywhere. The second reltion (for positive integers n) is simply repeted ppliction of the first. If we trnslte these reltions using the definition of exp(x) s the inverse function for log(x) we find tht e x e y e x+y, e nx (e x ) n. More generlly, we define logrithms nd exponentils to ny positive bse, other thn, by the rules log b (x) log(x)/ log(b), b x exp(x log(b)). Note tht log(2 n ) n log(2) + s n +. Since log(x) is strictly incresing we conclude tht log(x) + s x +. Similrly log(x) s x 0 + ; e x + s x + ; nd e x 0 s x. Sometimes we know the ntiderivtive lredy, nd wnt to evlute the integrl. The Fundmentl Theorem tells us how to do this: if F (x) g(x) then For exmple, if n then g F(b) F(). x n bn+ n+. n + 5

6 Power series We cn pply this to find power series representtion for log(x). If r < then r + r + r2 + r 3 + r k. This is the geometric series. If we tke r t in the definition of log(x) we find tht log(x) dt t dt ( t) ( t) k dt ( ) k (t ) k dt ( ) k (x )k+ 0 k+ k + (x ) 2 (x )2 + 3 (x )3 4 (x )4 + In prticulr, ln( 0 2 ) (k + ) 2k+ Homework problems: due Tuesdy, 27 June Compute the slope function m( x) for the function x 2 3x t the point (, 4), nd sketch its grph. Use the grph of m( x) to determine the vlue of the derivtive of x 2 3x t. Explin your resoning! 2 Compute the slope function m( x) for the function x+ x t the point (0, 0), nd sketch its grph. Use this grph to explin why x + x is not differentible t 0. Wht kind of discontinuity does the grph of m( x) hve? 3 Use the definition of derivtive to compute f (), for ech of the following. f(x) /(x 3), 2. b f(x) x, 3. c f(x) (2x + ) 7,. 4 Use the Bisection Method to pproximte the root of the eqution x 5 + x 3 0, ccurte to within Let f(x) (x+)x 2 +(x 3)x. Find ll points c in the intervl [, 3] t which the instntneous slope equls the verge slope over the intervl. 6 Use Fermt s Theorem to find the mximum nd minimum vlues of x 3 x on [.5, 0.5]. 7 Drw the grphs of the following integrnds nd evlute the given integrls by interpreting the integrl s re. (Do not find ntiderivtives!) 6 3 (2t 5)dt 6

7 b c d (2t 5)dt (3 t )dt 4 t2 dt 8 Drw grph of f(x) cos(x/2) on [0, 4π], nd use this to find the men vlue of f on this intervl. (Do not find ntiderivtives!) 9 Find power series representtion of f(x) rctn(x) round the point x 0, nd use this to pproximte π/4. 0 Find power series representtion of e x ner x 0, nd use this to pproximte e. 7

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