Z b. f(x)dx. Yet in the above two cases we know what f(x) is. Sometimes, engineers want to calculate an area by computing I, but...


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1 Chpter 7 Numericl Methods 7. Introduction In mny cses the integrl f(x)dx cn be found by finding function F (x) such tht F 0 (x) =f(x), nd using f(x)dx = F (b) F () which is known s the nlyticl (exct) solution. But consider Z p +x 3 dx, 0 nd Z 0 e x2 dx. Neither of the bove integrls cn be expressed in terms of functions tht we know. However both of these integrls do exist, since they both represent the re below the curves p +x 3 nd e x2 between x = 0 nd x = (nd both curves re wellbehved). Yet in the bove two cses we know wht f(x) is. Sometimes, engineers wnt to clculte n re by computing I, but... They don t know the eqution for f(x). There might be no formul for f(x) t ll! Thnkfully, there re some prcticl methods out there for clculting res under grphs, e.g. counting squres. But this is timeconsuming nd boring! Besides, there re other methods of clculting res which re much more ccurte, even though they re still only pproximtions. 7
2 CHAPTER 7. NUMERICAL METHODS The Rectngulr Rule The rectngulr rule (lso clled the midpoint rule) is perhps the simplest of the three methods for estimting n integrl you will see in the course. Integrte over n intervl pple x pple b. Divide this intervl up into n equl subintervls of length h =(b )/n. is the midpoint of the subin Approximte f in ech subintervl by f(x j ), where x j tervl. Are of ech rectngle: f(x )h, f(x 2 )h,..., f(x n)h. ) f(x)dx h [f(x )+f(x 2)+ + f(x n)]. The pproximtion on the RHS becomes more ccurte s more rectngles re used. In fct, f(x)dx =lim h!0 {h [f(x )+f(x 2)+ + f(x n)]} Note: As h! 0,n!,sinceh = b nd (b n ) is fixed. Remrk 7.. Actully, there re severl di erent versions of the rectngulr rule out there. If you re interested, these re mentioned in Sections 5. nd 5.2 of Thoms Clculus ( th edition). 7.3 The Trpezium Rule Another method of clculting n integrl pproximtely is the trpezoidl (trpezium) rule. The procedure is s follows... Agin, divide the intervl pple x pple b into n equl subintervls, i.e. = x 0 <x <x 2 <...<x n <x n = b, ech with length h =(b )/n. Then the re under the curve is pproximted by sum of n trpezi, insted of rectngles. Are of first trpezium: A = re of rectngle + re of tringle = f()h + 2 (f(x ) f()) = 2 h [f()+f(x )].
3 CHAPTER 7. NUMERICAL METHODS 73 Are of next trpezium:a 2 = 2 h [f(x )+f(x 2 )] Are of penultimte trpezium:a n = 2 h [f(x n 2)+f(x n )]. Are of lst trpezium:a n = 2 h [f(x n )+f(b)] Then f(x)dx Sum of ll n trpezi = 2 h {f()+f(x )+f(x )+f(x 2 )+f(x 2 )+ +f(x n 2 )+f(x n 2 )+f(x n )+f(x n )+f(b)}, i.e. I h 2 {f()+f(b)+2[f(x )+f(x 2 )+ + f(x n )]}. where Exmple 7.. Estimte h = b n using the trpezium rule with n = 5. x i = + ih, i =, 2, 3,...,n. Note tht we hve b =2,= nd n = 5. x dx So nd ) h = b n = 2 = 5 5 =0.2. =, x =.2, x 2 =.4, x 3 =.6, x 4 =.8, b =2, I {f()+f(b)+2[f(x )+f(x 2 )+f(x 3 )+f(x 4 )]} =0. {f() + f(2) + 2 [f(.2) + f(.4) + f(.6) + f(.8)]} = (4 d.p) pple Notes: In the previous exmple, the nlyticl vlue is given by x dx =[lnx]2 =ln2 ln = ln 2 = (4.d.p).
4 CHAPTER 7. NUMERICAL METHODS 74 If we used n = 0, we would hve I , which is even more ccurte thn using n = 5. Error in using the Trpezuim Rule Let ˆ I be the trpezium pproximtion to I, then we define the error " T s Then it turns out tht if then " T = ˆ I I. f 00 (x) pple M for ll x with pple x pple b, (b )3 " T pplem 2n 2. Exmple 7.2. Wht is the smllest n such tht hs mximum error of? 0 e x2 dx We must choose n lrge enough such tht " T pple. Note tht f(x) =e x2 ) f 00 (x) = 2+4x 2 e x2 We re interested in 0 pple x pple 2; on this intervl the mximum vlue of f 00 (x) occurs t x = 2, thus M = f 00 (2) 983 (rounded up). So (b )3 " T pplem 2n 2 pple n n 2 i.e we need 655 n 2 pple ) n The smllest such n tht stisfies this is n = Simpson s Rule Simpson s Rule is yet nother method of numericl integrtion. It is credited to Thoms Simpson (7076), n English mthemticin, though there is evidence tht similr methods were used 00 yers prior to him. So fr, we looked t two methods for numericl integrtion: Piecewise constnt pproximtion =) Rectngulr Rule
5 CHAPTER 7. NUMERICAL METHODS 75 Piecewise liner pproximtion =) Trpezium Rule Piecewise qudrtic pproximtion =) Simpson s Rule For Simpson s rule we divide pple x pple b into n even number of subintervls n of length h =(b )/n with endpoints = x 0 <x <x 2 <...<x n <x n = b, Min ide: Suppose typicl prbol (i.e. x 2 + bx + c) psses through three consecutive points (x i,y i ), (x i,y i ), (x i+,y i+ ) We will not go through the derivtion, but I cn tell you tht Simpson s formul turns out to be... h 3 (S 0 +4S +2S 2 ), where S 0 = f()+f(b), (7.) S = f(x )+f(x 3 )+...+ f(x n ), (7.2) S 2 = f(x 2 )+f(x 4 )+...+ f(x n 2 ). (7.3) (7.4) Observe tht for ll the indices tht pper in S, re odd, while those for S 2 re even (remember tht s n must be even, we hve tht (n ) is odd whilst (n 2) is even). Menwhile it cn be shown for Simpson s rule tht if then Exmple 7.3. Evlute f (4) (x) pplem for ll x with pple x pple b, " S pple M(b )5 80n 4. using Simpson s rule with n = 0,=,b= 2. x dx Note tht h = 2 0 = 0 =0., nd keep trck of ll the vlues of x i nd f(x i ) s follows...
6 CHAPTER 7. NUMERICAL METHODS 76 i x i f(x i )=/x i / 2.2 5/ / / / / / / / /2 Sums i.e. nd therefore Compre with the exct vlue S 0 = S = S 2 = ˆ h 3 (S 0 +4S +2S 2 )= dx x =ln2= , hence this is correct to FIVE d.p. (Trpezium Rule ws correct to d.p.) 7.5 Newton s Method for RootFinding In engineering, it is often required to find x such tht f(x) =0. (7.5) These vlues of x re known s roots of f(x). Exmples: ) x 2 3x +2=0 2) sin x = 2 x 3) cosh x cos x = All of these cn be written in the form (7.5). In this course, I will introduce you to one of the fstest methods for finding roots of f(x)...newton s Method (.k.. NewtonRphson Method).
7 CHAPTER 7. NUMERICAL METHODS 77 How the method works: Let our first (initil) guess to the root be x 0.Thenx is the point where the tngent to the curve f t x 0 intersects the xxis. i.e. tn = f 0 (x 0 )= f(x 0) x 0 x, x = x 0 f(x 0 ) f 0 (x 0 ). Now x is our new guess for the root of f(x).but we might wnt better guess; cll this x 2. It turns out the next itertion is nd we cn repet the procedure yet gin: x 2 = x f(x ) f 0 (x ), x 3 = x 2 f(x 2 ) f 0 (x 2 ), nd so on. We cn keep iterting until we get the desired ccurcy, using the formul: Exmple 7.4. Find the positive solution of x n+ = x n f(x n ) f 0 (x n ). 2sinx = x. Figure 7.: A plot of y =2sin x nd y = x. The root we re fter is the positive xvlue t the point where the two functions intersect. First, get the originl eqution into the form f(x) = 0: f(x) =x 2sinx ) f 0 (x) = 2 cos x.
8 CHAPTER 7. NUMERICAL METHODS 78 Then, by Newton s Method, We need n initil guess, e.g. x 0 = 2. x n+ = x n x n 2sinx n 2 cos x n = 2(sin x n x n cos x n ) 2 cos x n = N n D n. n x n N n D n x n+ = N n /D n The ctul solution to 4 d.p is Advntges of Newton s method: Converges very fst! You only need to give one initil guess (some methods require TWO). Disdvntges: You need to clculte the derivtive of f(x). Sometimes the method doesn t converge to root t ll! The method is useless if your first guess is sttionry point of f(x) (becuse you get division by zero).
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