# 1 Numerical Solution to Quadratic Equations

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1 cs42: introduction to numericl nlysis 09/4/0 Lecture 2: Introduction Prt II nd Solving Equtions Instructor: Professor Amos Ron Scribes: Yunpeng Li, Mrk Cowlishw Numericl Solution to Qudrtic Equtions Recll from lst lecture tht we wnted to find numericl solution to qudrtic eqution of the form x 2 + bx = c. () One obvious method for solving the eqution is to use the fmilir qudrtic formul: x,2 = b ± b 2 + 4c. (2) 2 Now, the formul does not provide us with solution tht is written in deciml form; to get such solution we need to evlute the bove expression. To this end, since most computers perform dditions, subtrctions, multiplictions, nd even divisions very fst, we should focus on the need to tke squreroot. In short, the explicit solution for the qudrtic eqution ctully reduces our problem to finding n lgorithm for computing squreroots. Even if we hve such n lgorithm, we hve to sk if using the qudrtic formul is the most efficient wy to solve this eqution; in other words, whether the best lgorithm for finding squreroots is superior to the best lgorithm tht solves the qudrtic eqution without using the formul. As we shll show now, we cn extend the powerful squre root lgorithm we proposed in the lst lecture so tht it solves generl qudrtic equtions, mking the use of the qudrtic formul (2) unnecessry (nd, in fct, inefficient). 2 Finding Squre Roots nd Solving Qudrtic Equtions 2. Finding Squre Roots As we discussed lst time, there is simple scheme for pproximting squre roots to ny given precision. More formlly, we cn find nonnegtive solution to the qudrtic eqution: x 2 = c, x 0, c 0 (3) using n itertive method tht llows us to control the precision of the solution. The ide is tht we find function tht, given n pproximtion of the solution s input (x old ), outputs more precise pproximtion (x new ). If we use this function itertively by recycling the vlues it produces on ech itertion, we will rrive t better nd better solutions. We cn continue this process until we rech ny level of precision we wnt. Recll tht the itertion we used ws: x new = x2 old + c 2x old (4)

2 An interesting property of this lgorithm is tht, roughly speking, the precision of the output (x new ) is doubled with ech intertion. As we shll see lter, this speed of convergence of the lgorithm will be n importnt fctor in evluting numeric lgorithms. 2.2 Solving Qudrtic Equtions Now tht we hve scheme for solving restricted kind of qudrtic eqution, cn we use the scheme to solve our originl problem? The nswer is yes. To solve equtions of the form x 2 + bx = c (5) We simply need to dd nother term to the denomintor of the formul: x new = x2 old + c 2x old + b We cn use this new formul itertively to rrive t numericl solutions of the qudrtic eqution tht re rbitrrily precise. Ech itertion will tke only 5 opertions (2 multiplictions, 2 dditions, nd division). Conclusion: the explicit qudrtic eqution reduced our problem from iterting with (6) to iterting with (4). While pedntic student my rgue (correctly!) tht the ltter itertion involves fewer opertions, we should ll gree tht the thought-to-be-wonderful qudrtic formul hs ctully only fringe vlue (t most). So, it should now be cler tht lgorithms tht re good for finding numericl solutions re often quite different from the methods we use to find nlytic solutions. 3 Clculting Definite Integrls Another good exmple of the difference between numericl computtion nd nlytic solutions is in clculting definite integrls. Recll tht the definition of the definite integrl of continuous (nd we will ssume positive) function f(t) over the intervl [, b], b f(t)dt is the re under the curve, s shown in Figure 3. (6) b Figure : The Definite Integrl of f(t) over [, b] 2

3 Wht is the best wy to clculte the definite integrl? Our experience in mth clsses suggests tht we might use the nti-derivtive pproch. We find function F such tht F = f, nd then b But this brings up two importnt questions:. How do we determine n nti-derivtive function? f(t)dt = F(b) F(). (7) 2. How difficult is it to evlute the nti-derivtive function t nd b? Sometimes the nswer to both questions is it s esy. For exmple, () consider 0 t3 dt. We know how to compute n nti-derivtive, nmely 4 t4, nd since this nti-derivtive is polynomil, we know how to evlute it efficiently using few rithmetic opertions. So we evlute 4 t4 0 = 4 0 = 4. Unfortuntely, most exmples re not so nice. For exmple, (b) consider x tdt. We know how to compute n ntiderivtive, nmely log e t, so we cn rewrite the integrl s x t dt = log e x log e = log e x. (8) To get numeric solution, we now need to evlute log e t x. Unfortuntely this is difficult - much more complicted thn evluting t. Any lgorithm for evluting log will be more complicted thn one for evluting t, while the re tht we re fter is completely determined by t. Some exmples re even worse. (c) Consider b e t2 dt, n integrl tht is useful in sttistics. The integrnd does hve n nti-derivtive, sy F (t) = e t2. But it is not esy to compute the nti-derivtive, or even to write down simple closed-form. In fct, n efficient wy for evluting F t x is to compute the integrl x 0 f(t)dt. Finlly, (d) consider the integrl of the step function shown in Figure 3. Computing this integrl Figure 2: Approximting the Definite Integrl Using Subintervls is trivil. But no nti-derivtive exists. The bove discussion highlights fundmentl issue: the re we need to compute depends on the function f. Our ttempt to use the nti-derivtive pproch brings to the scene new function F whose properties my not be in tndem with those of the originl f. So is there method for clculting the definite integrl tht uses only informtion bout the function f? The nswer is yes, we cn use the definition of the definite integrl to pproximte its vlue, s shown in Figure 3. We strt by splitting up the intervl into subintervls of equl size h, then estimting the definite 3

4 integrl over ech subintervl, then dding them ll together. If we split the intervl [, b] into N subintervls of equl width h = b N, we cn clculte the definite integrl s: b f(t)dt = N j= +jh +(j )h f(t)dt. (9) + 2h + h b Figure 3: Approximting the Definite Integrl Using Subintervls We hve reduced the problem to estimting the definite integrl over smller subintervls. Here re two possible strtegies for estimting the smller definite integrls: Rectngle Rule +h f(t)dt f()h (0) In the rectngle rule, we clculte the re of the rectngle with width h nd height determined by evluting the function t the left endpoint of the intervl. Midpoint Rule +h f(t)dt f( + h )h () 2 In the midpoint rule, we clculte the re of the rectngle with width h nd height determined by evluting the function t the midpoint of the intervl. These two methods re depicted in Figure 3. Using these methods, we cn clculte the definite integrl by simply evluting the originl function t definite points, which we should lwys be ble to do (the previling ssumption is tht we hve n lgorithm for evluting f. In the bsence of such lgorithm, we ctully do not know wht f is). We cn improve the precision in ech cse by chnging the size of the subintervls h - the smller the subintervl, the greter the precision. The next question is, how cn we decide which of these lgorithms to use? Intuitively, it seems tht the midpoint rule would be better choice, but how cn we quntify this intuition? 4

5 Rectngle Rule Midpoint Rule Figure 4: The Rectngle nd Midpoint Rules 3. Evluting Numeric Algorithms In generl, we evlute numeric lgorithms using 3 mjor criteri: Cost - The mount of time (usully clculted s the number of opertions) the computtion will tke. Frequently we cn trdeoff cost versus ccurcy by chnging prmeter like the number of subintervls. Accurcy - How quickly the lgorithm pproches our desired precision. If we define the error s the difference between the ctul nswer nd the nswer returned by the numeric lgorithm, we often mesure the rte t which the error pproches 0 s cost is incresed. Robustness - How the correctness of the lgorithm is ffected by different types of functions nd different types of inputs. The lgorithm for clculting squre roots turns out to be extremely robust: it will work for ny squre root, even if we give it terrible initil vlue. 3.2 Evlution of Algorithms for Definite Integrls Applying these criteri to the midpoint nd rectngle rules: Cost Ech lgorithm performs single function evlution to estimte the definite integrl over subintervl, so the cost is identicl. The totl cost over the entire computtion grows linerly with the number of intervls N in ech cse, mening tht there is some fixed constnt c, not dependent on the number of intervls, such tht the totl cost is c N = O(N). For exmple, if the number of intervls doubles, the cost will double s well. Note tht we cn trdeoff between cost nd ccurcy by chnging the number of intervls N. Accurcy Clerly s the size of ech subintervl h gets smller, the error gets smller s well. Is there ny difference between the rte t which the rectngle nd midpoint rules error pproches 0? Rectngle Rule - For the rectngle rule, the error pproches 0 in direct proportion to the rte t which h pproches 0. In other words, the error ǫ = c h for some fixed constnt c, or ǫ = O(h). For exmple, hlving the size of h will hlve the mount of error. Midpoint Rule - For the midpoint rule, the error pproches 0 in proportion to the rte t which the squre of h pproches 0. Tht is the error ǫ = c 2 h 2 for some fixed constnt c 2, or ǫ = O(h 2 ). For exmple, hlving the size of h will divide the error by 4. 5

6 This difference is substntil. For exmple, if we wnt to compute the integrl of function between 0 nd, nd we use N = 000 subintervls, ech subintervl will hve size h = 000, so the rectngle rule s error will be on the order of 000, while the midpoint rule s error will be on the order of 0 6. Robustness It turns out tht for functions tht re resonbly well-behved, both the endpoint rule nd the midpoint rule re quite robust. There do exist situtions, however, in which it is possible to evlute function t the left endpoint of n intervl, nd not t the midpoint. Through this evlution, we hve shown tht, in generl, the midpoint rule is fr better choice thn the rectngle rule for pproximting the definite integrl, since, for the sme cost, it will give us much more precision. 4 Loss of Significnce One finl wy tht numericl computtion is different thn the wy you hve done mthemtics to this point is the potentil loss of significnce. Consider two rel numbers stored with 0 digits of precision beyond the deciml point: π = b = The ctul numbers π nd b contin dditionl digits tht we hve not stored. So our stored numbers re ctully good, cceptble pproximtions of the true π, b. Now, we wnt to compute π b, nd wnt to hve similrly good pproximte representtion: 0- significnt digits, i.e., once the zeros end nd the number begins. However, ll we cn do is subtrct the given pproximtion to obtin : only two significnt digits! So, we hve lost most of the precision of the two originl numbers. One might try to solve this problem by incresing the precision of the originl numbers, but this is not solution: For ny finite precision storge, numbers tht re close enough will be indistinguishble. There is no universl wy to void loss of precision! The only generl solution is to void subtrcting numbers tht re lmost equl. An exmple of the effect of loss of significnce occurs in clculting derivtives. Exmple 4. (Clculting Derivtives). Given the function f(t) = sint, wht is f (2)? We ll know from clculus tht the derivtive of sint is cos t, so one could clculte cos(2) in this cse. However, just s with clculting definite integrls, it is not lwys possible, or efficient, to evlute the derivtive of function. Insted, recll tht the vlue of the derivtive of function f t fixed point t 0 is the slope of the line tngent to f t t 0. Recll, lso, tht we cn pproximte this slope by clculting the slope of secnt line tht intersects f t t 0 nd t point very close to t 0, sy t 0 + h, s shown in Figure 4. We lerned in clculus, tht s h 0, the slope of the secnt line pproches the slope of the tngent line, or: f f(t 0 + h) f(t 0 ) (t 0 ) = lim. (2) h 0 h 6

7 Tngent line to sin(2) Secnt line through sin(2), sin(2+h) h Figure 5: Tngent nd Secnt Lines on f(t) = sint Just s with the definite integrl, we could pproximte the derivtive t t 0 by performing this clcultion for smll vlues of h. However, notice tht s h gets very smll, the numertor of Eqution 2 will become 0, due to loss of significnce, so tht we will erroneously clculte ll derivtives s 0. Preventing loss of significnce in our clcultions will be n importnt prt of this course. 5 Topic I: Solving Equtions Now tht we hve some ide of wht lgorithms for computtion re, we will discuss lgorithms for performing one of the most bsic numericl tsks, solving equtions in one vrible. sin(x) x Figure 6: Grphicl Solution for x 3 = sinx Consider the following eqution, with solution depicted in Figure 5. x 3 = sinx. (3) There is no nlytic solution to this eqution. Insted, we will present n itertive method for finding numericl solution. 5. Itertive Method As in our solution to finding squre roots, we would like to find function g, such tht if we input n initil guess t solution, g will output better pproximtion of the ctul solution. More 7

8 formlly, we would like to define sequence x 0, x, x 2,... with x n+ g(x n ) such tht (x n ) 0 converges to the solution. We cn continue clculting vlues x i until we rech two vlues with x j = x j+. Clerly, one property of g is tht g(r) = r, where r is the rel solution to Eqution 3. A trnsformtion of Eqution 3 (e.g. x = sinx x 3 + x) will stisfy this property, the trick is to choose the trnsformtion tht produces converging series of vlues x, x 2,.... 8

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