Power series and Taylor series - University of Pennsylvania

1 Power series and Taylor seriesD. DeTurckUniversity of PennsylvaniaMarch 29, 2018D. DeTurckMath 104 002 2018A: Series1 / 42 SeriesFirst .. a review of what we have done so far:1We examined series of constants and learned that we can sayeverything there is to say about geometric and developed tests for convergence of series of considered Power series , derived formulas and other tricksfor finding them, and know them for a few DeTurckMath 104 002 2018A: Series2 / 421. geometric and telescoping seriesThe geometric series is n=0anrn=a+ar+ar2+ar3+ =a1 rprovided|r|<1 (when|r| 1 the series diverges).We often use partial fractions to detect telescoping series , forwhich we can calculate explicitly thepartial sums DeTurckMath 104 002 2018A: Series3 / 422.

2 Tests for convergence of series of constants1 Fundamental divergence test (nth term must go to zero forconvergence to be possible)2 Integral test3 Comparison and limit comparison tests4 Ratio test5 Root test6 Alternating series testD. DeTurckMath 104 002 2018A: Series4 / 423. Power seriesf(x) =a0+a1x+a2x2+a3x3+ wherean=f(n)(0)n!orf(x) = n=0f(n)(0)n!xnand we know the series forex, sinxand11 DeTurckMath 104 002 2018A: Series5 / 42 Convergence of Power seriesBefore we get too excited about finding series , let s make surethat, at the very least, the series , we ll deal with the question of whether they converge to thefunction we expect. But for now, we ll assume that if theyconverge, they converge to the function they came from.

3 (Strictly speaking, this is not always true but it is true for alarge class of functions, which includes nearly all the onesencountered in basic science and mathematics. This fact was notfully appreciated until the early part of the twentieth century.)Fortunately, most of the question of whether Power series convergeis answered fairly directly by theratio DeTurckMath 104 002 2018A: Series6 / 42 Ratio test reviewRecall that for a series of constants n=0bn, we have that the seriesconverges (absolutely) iflimn bn+1bn is less than one, diverges if the limit is greater than one, and thetest is indeterminate if the limit equals use the ratio test on Power series , just leave thexthere andcalculate the limit for each value ofx.

4 This will give an inequalitythatxmust satisfy in order for the series to DeTurckMath 104 002 2018A: Series7 / 42 For the exponential functionThe Power series forexisex= 1 +x+x22!+x33!+x44!+ = n=0xnn!Thereforelimn bn+1bn = limn xn+1(n+ 1)!n!xn = limn |x|n+ 1= matter whatxis, the limit is 0, which is less than 1. Sotheseries for the exponential function converges for all values of turn!For which values ofxdoes the series forf(x) = sinxconverge?D. DeTurckMath 104 002 2018A: Series8 / 42A more interesting example:For the seriesf(x) = n=1xnn, This time the ratio test giveslimn xn+1n+ 1nxn = limn |x|nn+ 1=|x|.So the series converges if|x|<1 and diverges if|x|>1(reminiscent of the geometric series ).

5 It remains to check the endpointsx= 1 andx= 1 Forx= 1 the series is n=11n, the (divergent) harmonic 1 the series is n=1( 1)nn, the alternating harmonicseries, which we know to be (conditionally) n=1xnnconverges if 1 x<1and diverges DeTurckMath 104 002 2018A: Series9 / 42OK, your turn..For which values ofxdoes the series n=1(2x)nn2converge?A. 1<x<1B. 2<x<2C. 12 x<12D. 2 x 2E. 12 x 12 For which values ofxdoes the series n=1nxnconverge?A. 1<x<1B. 1 x<1C. 1<x 1D. 1 x 1E. 0 x<1D. DeTurckMath 104 002 2018A: Series10 / 42 From these examples.. it should be apparent that Power series converge for values ofxin an interval that is centered at zero, , an interval of the form[ a,a], ( a,a], [ a,a) or ( a,a) (whereamight be either zero orinfinity).

6 The interval is called theinterval of convergenceand the numberais called theradius of DeTurckMath 104 002 2018A: Series11 / 42 Let s go back to finding series for functionsThere are two ways:The standard way:Use the formulaan=f(n)(0)n!to find the coefficients. We ve foundseries forexand sinxthis way:ex= 1 +x+x22!+x33!+x44!+ = n=0xnn!andsinx=x x33!+x55! x77!+ = n=0( 1)nx2n+1(2n+ 1)!We could calculate other series this way (and sometimes we dohave to resort to this), but the other way is more fun:D. DeTurckMath 104 002 2018A: Series12 / 42 The other wayMagic tricks:Start from known series and use algebraic and/oranalytic manipulation to get others:Substitutex2forxeverywhere in the series forexto get:ex2= 1 + [x2] +[x2]22!

7 +[x2]33!+[x2]44!+ = 1 +x2+x42!+x63!+x84!+ = n=0x2nn!D. DeTurckMath 104 002 2018A: Series13 / 42 Another exampleTake the derivative of the series for sinxto get:cosx=ddxsinx=ddx(x x33!+x55! x77!+ )= 1 3x23!+5x45! 7x67!+ = 1 x22!+x44! x66!+ = n=0( 1)nx2n(2n)!D. DeTurckMath 104 002 2018A: Series14 / 42 Yet another exampleIntegrate both sides of the geometric series from 0 toxto get: x011 tdt= x0(1 +t+t2+t3+ )dt ln(1 x) =x+x22+x33+x44+ Negate both sides and replacexby xeverywhere to get:ln(1 +x) =x x22+x33 x44+ = n=1( 1)n+1xnnPutx= 1 to learn that the fourth series from before sums to ln DeTurckMath 104 002 2018A: Series15 / 42 Start from the geometric series again:and substitute x2forxeverywhere it appears to get:11 +x2= 1 x2+x4 x6+x8 x10+ = n=0( 1)nx2nNow integrate both sides from 0 toxto get.

8 Arctanx=x x33+x55 x77+x99 x1111+ = n=0( 1)nx2n+12n+ 1 to show that the first series from before sums to still have the challenge of showing that n=11n2= DeTurckMath 104 002 2018A: Series16 / 42 Applications of series I: Limits atx= know for other reasons that limx 0sinxx= we could prove this using series :limx 0sinxx= limx 01x n=0( 1)nx2n+1(2n+ 1)!= limx 0 n=0( 1)nx2n(2n+ 1)!= limx 0(1 x23!+x45! x67!+ )= 1 + 0 + 0 + = 1 You can do this for complicated limits at 0 substitute the seriesfor the functions and do the limit: limx 0x sinx1 e x3A. does not existD. DeTurckMath 104 002 2018A: Series17 / 42 Applications of series II: Approximate evaluation ofintegralsMany integrals that cannot be evaluated in closed form ( , forwhich no elementary anti-derivative exists) can be approximatedusing series (and we can even estimate how far off theapproximations are).

9 Calculate 10e x2dxto the nearest begin by substituting x2forxin the known series forex, andthen integrating it. This will give us a numerical series thatconverges to the answer: 10e x2dx= 101 x2+x42! x63!+ dx= 1 13+15 2! 17 3!+ D. DeTurckMath 104 002 2018A: Series18 / 42 Error estimatesSo far we have 10e x2dx= 1 13+15 2! 17 3!+ The series on the right is alternating, so if we want the error to beless than =11000, we need to take all the terms before thefirst one that is less than that. In other words, which is the firstterm with denominator greater than 1000?Well, 9 4! = 216 and 11 5! = 1320, so the term111 5!should bethe first omitted term. So we write 10e x2dx 1 13+15 2!

10 17 3!+19 4!=56517560 (according to Maple, the answer to 5 decimal places is ).D. DeTurckMath 104 002 2018A: Series19 / 42 Try this:Which of the following is closest to 10cos x dx?A. DeTurckMath 104 002 2018A: Series20 / 42 How good are our approximations?We ve been associating series with functions and using them toevaluate limits, integrals and the integrals we ve estimated, we ve been fortunate that theresulting numerical series were alternating series so the error is easyto estimate. What happens when the series are not alternating?We ll continue to concentrate on questions like:1If I use only the first three terms of the series , how big is theerror?2 How many terms do I need to get the error smaller DeTurckMath 104 002 2018A: Series21 / 42To get error estimates:Use a generalization of the Mean Value Theorem for Mean Value Theorem approach:Recall the mean-value theorem:f (somewhere betweenaandb) =f(b) f(a)b aSeta= 0 andb=xand get thatf (somewhere between 0 andx) =f(x) f(0)xRearrange this to getf(x) =f(0) +f (somewhere) DeTurckMath 104 002 2018A: Series22 / 42 Turn this into an error estimateStart fromf(x) =f(0) +f (somewhere) that if you know that the absolute value of the derivativeoffis always less thanM, then you know that|f(x) f(0)|<M|x|.