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## Calculus with Analytic Geometry II

by: Priscilla Thiel IV

8

0

6

# Calculus with Analytic Geometry II MAC 2312

Priscilla Thiel IV
University of Central Florida
GPA 3.95

David Rollins

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COURSE
PROF.
David Rollins
TYPE
Class Notes
PAGES
6
WORDS
KARMA
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## Popular in Calculus and Pre Calculus

This 6 page Class Notes was uploaded by Priscilla Thiel IV on Thursday October 22, 2015. The Class Notes belongs to MAC 2312 at University of Central Florida taught by David Rollins in Fall. Since its upload, it has received 8 views. For similar materials see /class/227436/mac-2312-university-of-central-florida in Calculus and Pre Calculus at University of Central Florida.

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Date Created: 10/22/15
MAC2312 Calculus 2 Spring 2009 Chapter 12 Review April 2 2010 Note Textbook section numbers are in parentheses 1 9 121 A sequence an is a set of numbers A sequence either converges or diverges by considering the limit lim an naoo 1f the limit exists and equals L then the sequence converges to L If the limit is in nite or does not exist then the series diverges Any of the standard methods L Hopitalls rule squeeze etc for evaluating limits can be use 122 A series is an infinite sum of terms written as 221 ak The partial sums of the rst N terms form a sequence SN 2161 ak 1f the sequence of partial sums converges then so does the series Divergence Test 1f lim an y 0 then the series diverges You should always check this as if the terms naoo in the series donlt go to zero in the limit then the series diverges However note that the terms going to zero in the limit does not guarantee the series converges Geometric series can be summed a 00 16 2 1 7 akEOT a TT 1 T provided M lt 1 The geometric series diverges for W 2 1 Telescoping series are of the form 00 a Do A B 7 7 kZ1k2bkC 161k701 1676 First write using partial fraction Then write out the rst few terms and the last couple in the partial sums SN to see what cancels and what stays Then evaluate Nlim N H00 123 Integral Test For a series of the form 00 00 Z ak 2 f 1 161 161 where is a continuous positive decreasing function on 1 Evaluate the improper integral 1 00 mm and if the integral converges so does the series If the integral diverges then so the series This test works best if when it is straight forward to nd an antiderivative for 1f the series sum is approximated by the sum of the rst N terms 00 N Zak m Zak 161 161 g 5 on 8 then the maximum error made is N mm 124 Comparison test In this test you will compare the convergencedivergence property of a known series with an unknown series The known series will either be the geometric series 220 Tk or the pseries 2211161 The pseries converges if p gt 1 and diverges if p g 1 To decide which series to compare with a good strategy is examine what the unknown series is like for large k Inequality version If the series 2 ak and E bk have positive terms and ak S bk 1f Zak diverges then so does 212k 1f 212k converges then so does Zak You will have to verify that the inequality is in fact true Limit version Evaluate the limit lim k C kaoo bk with O f 0 or O f 00 Then either both series converge or both diverge 125 Alternating series This applies to series of the form or something similar iPNak 160 where the ak gt 0 This series converges if the following two conditions are met 1 lim ak 0 and 2 naoo ak1 g ak for all Is To check 2 this can be done directly by verifying this inequality or if ak then show f k S 0 126 Absolute convergence Given the series 220 ak with some ak that are negative then we may instead consider the related series 2201am which has all positive terms This is useful since if 220 lakl converges then so does 220 ak Such a series is said to be absolutely convergent 1f 0 ak converges but 220 lakl diverges then the series is said to be conditionally convergent Ratio Test Evaluate the following limit m ak1 LA kaoo ak 1f L lt 1 the series converges L gt 1 the series diverges and if L 1 then there is no conclusion In this last case a different test should be used The ratio test works well with factorials exponentials and powers Root Test Evaluate the following limit lim 5 lakl L kaoo 1f L lt 1 the series converges L gt 1 the series diverges and if L 1 then there is no conclusion In this last case a different test should be used The root test works best if ak k Note that the root and ratio test can also be applied to series with only positive terms 12 A good strategy is to rst notice if the series is a known series geometric or p series Next check if the terms in the series go to zero in the limit as n A 00 1f the terms are all positive then the following tests can be used integral comparison ratio and root If some terms in the series are negative the following tests can be used alternating absolute convergence approach using any of the tests for positive term series including the root and ratio tests 128 A power series is of the form 2 ak I 7 116 160 H O H where a is a constant and the ak are speci ed A power series converges for lz 7 al lt R where R is called the radius of convergence This is equivalent to a 7 R lt z lt a R It is possible that the series may converge at both one or none of the endpoints Convergence is determined by using the ratio test This will also determine R Convergence at the endpoints is determined by other tests as the ratio test is inconclusive at these points If R 00 the series converges for all I If R 0 the series converges for only I a 129 A new power series can be found from a known power series by a the substitution property for functions b algebraic 391 39 c quotm 39 39 and d 39 t 39 1n the latter two cases a power series can be differentiated or integrated term by term and the radius of convergence remains the same 129 Taylor and MacLaurin a 0 series are power series with fka T so the Taylor series expansion about I a is fk a M 7 ZTW 7 a k0 where the function is equal inside the interval of convergence To save time it is good to memorize the series for excoszsinz and 11 7 z geometric series For other functions you should be able to nd the series given and a Note all the ideas about constructing series from 129 apply to Taylor series Taylor polynomials Tn are just a truncation of the series and are nth degree polynomials fk a mac Z ltz 7 a 160 The error Rn made when approximating a function by a Taylor polynomial has the property that lim Rnz 0 n7gtoo for a given 1 within the interval of convergence Taylorls inequality gives a way of estimating this error anIl S Iia n1 M n 1 where lf 1 g M for lz 7 al g d This inequality can be used in two ways i determine n with d given so that the error requirements are met and ii determine d given n so that the error requirements are met In either case M must be calculated Note that if the series happens to be alternating then the error term is simply the next term neglected and the error analysis is simpler 1211 The Taylor expansion of 1 zk about I 0 is given by k kk71 kk71 k72 11k1wx2xsm where k can be any real number This binomial series is often written with the following shorthand 1 zk I Where In nl TL lt97 H This series converges for lt 1 The binomial series is used for functions like for example7 x1 7 12 or 1 z3Qi 1 1212 Taylor series and polynomials can be used to approximately evaluate functions7 nd limits7 evaluate de nite integrals They can also be used to approximate expressions by using an approximating polynomial ydz z dzdt 533 7 00 with concavity determined by the sign of the second derivative de ned curve is given by The arclength of a parametrically b L WW MW dt where z and y gti The same comments in 91 apply for this type of arclength calculation 1113 Polar coordinates provide an alternative approach to locating points in the plane The Cartesian coordinates 17y are related to T 9 by the relations I Tcost y Tsint9 and 7 2 12 y2 tant9 yz Take care when nding 9 for an angle in the second or third quadranti When plotting 7 ft97 it is possible that 7 lt 0 Plotting such a point means by convention that we plot lTl9 N You should be able to plot such a graph in simple cases The slope of the tangent line is found using the same formula as in 1112 with t replaced by 9 and z cost9 and y sin 9 then dy 7 dydt E dagd6 1114 Area in polar coordinates is given by 1 3 A i f 92d6 and the arclength is L f62 f62 49 The limits of integration are often found by seeing when 0 or when two curves intersect ft9 99 Be sure to have a sketch of the curves and identify the area to be found Note that when nding the area between curves assume f gt g the integral takes the form A g 166 7glt6gt2d61 0 i 1115 The basic equation of a parabola with focus 017 and directrix y 7p is 12 4pyi The basic equation of an ellipse with foci i5 0 and vertices ia7 0 is where c2 a2 7 122 A circle corresponds to the case a 12 The basic equation of a hyperbola with foci i5 0 and vertices ia7 0 is where c2 a2 122 Its asymptotes are y ibazi For any of the three conic sections7 replacing I by z 7 h and y by y 7 k shifts the curve by h units horizontally and k units verticallyi

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