1 Approximating Integrals using Taylor Polynomials

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1 Seughee Ye Ma 8: Week 7 Nov Week 7 Summary This week, we will lear how we ca approximate itegrals usig Taylor series ad umerical methods. Topics Page Approximatig Itegrals usig Taylor Polyomials. Defiitios Examples Approximatig Itegrals Numerical Itegratio 5 Approximatig Itegrals usig Taylor Polyomials. Defiitios Whe we first defied the derivative, recall that it was supposed to be the istataeous rate of chage of a fuctio f(x at a give poit c. I other words, f gives us a liear approximatio of f(x ear c: for small values of ε R, we have f(c + ε f(c + εf (c But if f(x has higher order derivatives, why stop with a liear approximatio? Taylor series take this idea of liear approximatio ad exteds it to higher order derivatives, givig us a better approximatio of f(x ear c. Defiitio (Taylor Polyomial ad Taylor Series Let f(x be a C fuctio i.e. f is -times cotiuously differetiable. The, the -th order Taylor polyomial of f(x about c is: f (k (c T (f(x (x c k The -th order remaider of f(x is: k R (f(x f(x T (f(x If f(x is C, the the Taylor series of f(x about c is: T (f(x k f (k (c (x c k Note that the first order Taylor polyomial of f(x is precisely the liear approximatio we wrote dow i the begiig. Now that we defied Taylor polyomials as higher order extesios of the liear approximatio, we have to justify our claim that Taylor polyomials are ideed approximatios. So what does it mea for a Taylor polyomial T (f(x to be a good approximatio of f(x? It meas that T (f(x should be close to the true value of f(x. I other words, we wat f(x T (f(x to be close to. But we defied this differece to be somethig... Theorem.. Let R (f(x be the -th order remaider of f(x. The, R (f(x is o((x c. Page of

2 Seughee Ye Ma 8: Week 7 Nov Theorem. is sayig precisely that T (f(x is very close to the real value of f(x whe x is ear c. Hece, we have our justificatio for callig Taylor polyomials higher order approximatios of f(x. Now, we look at aother very useful theorem, which will actually let us compute R (f(x. Theorem.. Suppose f(x is ( + -times cotiuously differetiable. The, R (f(x c f (+ (y (x y dy! With Theorem., we will be able to kow exactly by how much T (f(x is off the true value of f(x.. Examples Example.. Compute the Taylor series for f(x e x about. Solutio. Recall that the Taylor series of f(x is simply k f (k ( x k However, for all k, f (k (x e x. Hece, for all k, f (k ( e. Therefore, as we already kow. e x k f (k ( x k Here are the Taylor series about for some of the fuctios that we have come across several times. Try to do a couple of them as a exercise! si x x x! + x5 5! ( k x k+ (k +! k cos x x! + x4 4! x k x k k ( k xk (k! Let s look closely at the Taylor series for si x ad cos x. It looks like we ve split up the Taylor series of e x + x + x! + ito two ad alterated sigs. So ca we fid ay relatio betwee these three Taylor series? The aswer is yes ad i fact, we will see somethig amazig come out of the ispectio. Let i be the imagiary umber. If you have ever see i before, it s just some umber (ot real with the property that i. As bafflig as it might be to raise a umber to a complex power, let s take a leap of faith ad k x k Page of

3 Seughee Ye Ma 8: Week 7 Nov raise e to ix power ad see what we get. Pluggig i ix i the Taylor series for e x we get: e ix + (ix + (ix! + (ix! + (ix4 4! + (ix5 5! + ( + ix x! ix + x4! 4! + ix5 ( 5! ( x + x4 4! x! + + (ix ix + ix5 ix7 + (! 5! 7! ( x + x4 4! x! + + i (x x! + x5 5! x7 7! + (4 ( ( ( k xk + i ( k x k+ (5 (k! (k +! k k cos x + i si x ( So there we have it! e ix cos x + i si x If you are still ot coviced that this formula is the most beautiful thig you ve see i math, try pluggig i x π. If we let x π, what do we get? e iπ cos π + i si π e iπ + This equatio is called the Euler s formula ad dubbed the most beautiful equatio i mathematics.,, π, e, ad i are arguably the five most importat umbers i all of math ad to see them appear i oe equatio is ideed quite amazig!. Approximatig Itegrals Now, we will see how Taylor polyomials ca help us approximate itegrals. For example, cosider the Gaussia itegral e x dx called the Gaussia for short. The Gaussia is a very importat itegral, oe of the properties beig that it is the curve that represets the ormal distributio a.k.a. the bell curve. We would like to evaluate the Gaussia but there is oe problem: there is o elemetary atiderivative of e x. This meas that we caot rely o the Fudametal Theorem of Calculus to evaluate the itegral. But usig Taylor series, we ca approximate the value of this itegral. Example.. Approximate e x dx to withi of its actual value. Solutio. To simplify otatio, we will write T (x ad R (x for T (e x (x ad R (e x (x, respectively. For ay, we have e x T (x + R (x. By itegratig both sides, we obtai e x dx Now, T (x is just a polyomial. Therefore, T (xdx + R (xdx T (xdx is a itegral that we ca explicitly compute. O the other had, we kow that R (x goes to as icreases. So the idea is to make R (xdx small by icreasig : I our case, we wat to fid such that / R (xdx <. By Theorem., we have R (e x (x ( + e y! (x y dy Page of

4 Seughee Ye Ma 8: Week 7 Nov However, ote that ( ( x k R (x e x T (x e x R (e x (x k Hece, we see that R (x + e y (! (x y dy Ufortuately, this is ot somethig we ca easily itegrate. However, we are ot iterested i the actual value of the itegral. We are oly iterested i makig this itegral close to. How do we boud R (x? First, ote that for ay y [, x ], e y e ; ad (x y (x x. Note also that for all y [, x ], we have y! (x y. This gives us I our case, we wat R (x R (xdx [ x! y x+! Therefore, we eed to fid a value of for which + e y (! (x y dy (7 e y! (x y dy (8! x dy (9 x +! ( +! ] x [ dx x + ( +! ( + < ] < Checkig small values of,... we see that whe, the iequality is satisfied. This meas that T (xdx is withi of the true value of which is easy sice we already kow that Taylor series of e x : T (x x + x4 x! We ca ow itegrate T (x over [, ] to approximate the Gaussia: ( ( e x dx. First, let s compute T (x T (xdx x + x4 x dx ( ] [x x + x5 x7 ( (4 Page 4 of

5 Seughee Ye Ma 8: Week 7 Nov Therefore, we coclude that e x dx ± This example shows that Taylor polyomials ca be used effectively to approximate itegrals. However, we should ote that approximatig with Taylor polyomials works well ear the poit about which we are writig the Taylor polyomial. For example, if we were to approximate value usig Taylor polyomials, we would eed to compute T (xdx. e x dx to withi of its true I our example, the third order Taylor polyomial was good eough to approximate the itegral to withi. However, as we get farther away from (for us from to, we eed the eleveth order Taylor polyomial just to get a value that is withi of the true value. Numerical Itegratio I Sectio, we saw how we ca use Taylor polyomials to approximate itegrals. Now, we see a few other umerical methods we ca use to approximate itegrals. Recall that the itegral was defied as the greatest lower boud of all the upper Riema sums. Equivaletly, we saw that it was the limit of the upper Riema sums associated to the uiform partitios as. Numerical itegratio tries to approximate the itegral by a sum that resembles the Riema sums. Midpoit rule Let f be a fuctio o [a, b] ad cosider the uiform partitio P of [a, b]. The, defie m j a + (j (b a I other words, we defie m j to be the midpoit of the subiterval (t j, t j. Now, defie J ( b a f(m j j Note that J looks like the Riema sum except we have replaced sup f(x with f(m j. It turs out that J provides a decet approximatio of b f(xdx. I fact, we have a b ( J f(xdx O a Istead of provig this theorem, let s look at a more advaced umerical itegratio method: Simpso s rule Agai, we start with a fucito f(x defied o [a, b] ad let P be a uiform partitio of [a, b]. Hece, we have j(b a x j a + Agai, let m j be the midpoit of [x j, x j ] i.e. m j x j, x j Page 5 of

6 Seughee Ye Ma 8: Week 7 Nov Now, we cosider J j f(x j + 4f(m j + f(x j ( b a Note that this time, we chose a weighted average of f(x j, f(m j ad f(x j istead of choosig a particular poit. Why is this a good idea? Let s look at the followig lemma. Lemma.. Let I [a, b] be ay iterval ad let q(x be a quadratic polyomial o I. Let m b a be the midpoit of I. The, b q(xdx b a (q(a + 4q(m + q(b a Before we prove the lemma, ote that this lemma says that if f is a quadratic polyomial, the J is i fact equal to f(xdx. Proof. Suppose q(x is quadratic. By cosiderig p(x q(x m, we ca assume without loss of geerality that the iterval [a, b] is of the form [ a, a]. Suppse p(x c + c x + c x. The, we get a a c + c x + c x dx [c x + c x c a + c a + c x ] a a Computig the right had side, we see that the equality is ideed satisfied. (5 ( Now, suppose f(x is C. The, we ca cosider the secod order Taylor polyomial of f(x: T (f(x f(m j + f (m j (x m j + f (m j (x m j By Theorem., we kow that R (f(x is o((x m j. I fact, R (f(x is O((x m j. Sice T (f(x is a quadratic polyomial, by Lemma., we kow that j T (f(xdx T (x j + 4T (f(m j + T (f(x j x j (x j x j j I particular, if we cosider the differece betwee f(xdx ad the j-th term i J, the differece is x j b a O ((. Hece, the differece betwee f(xdx ad J is O ((. Page of

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