Exam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH1105 Instructor: Attila Máté 1


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1 Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixedpoint itertion to converge when solving the eqution x = f(x). Solution. The condition is described be the following Theorem. Assume x = c is solution of the eqution x = f(x). Assume further tht there re numbers r > nd q with q < 1 such tht f (x) q for ll x with c r < x < c+r. Then strting with ny vlue x 1 (c r,c+r), the sequence { } n=1 defined by = f( 1 ) for n > converges to c. b) Define the sequence { } n=1 by x 1 = 1 nd +1 = +. Explin why converges to. Solution. Writing f(x) = x +x we hve +1 = f( ), tht is, we re deling with n exmple of fixedpoint itertion. Further, we hve f (x) = x +1, = x x = 1 1 x > 4 whenever 1/x < 5/4, or else x > / 5. On the other hnd, f (x) < 1/ for ll x. So, we hve f (x) < /4 for ll x in the intervl ( ) ( ( ) 5,+, ( )) Note tht x 1 = 1 is in the intervl on the righthnd side. Hence converges to ccording to the Theorem mentioned in the solution to Prt ) (tke c =, r = / 5 nd q = /4 in the Theorem). Note. The following more generl sttement is true: Let > be rel number, nd define the sequence { } n=1 by x 1 = 1 nd +1 = + Then converges to. To see tht this sttement is true, observe tht for every n > 1. Indeed, for n 1 we hve +1 = + =, where the inequlity holds since the rithmetic men is lwys greter thn or equl to the geometric men. Now, writing x f(x) = +x, 1 All computer processing for this mnuscript ws done under Debin Linux. AMSTEX ws used for typesetting. 1.
2 we hve f (x) = x +1 = x x So, noting tht +1 = f( ) nd = f( ), for n > 1 we hve = 1 x < = f( ) f() = f (ξ)( ) with some ξ (, ) ccording to the MenVlue Theorem of differentition. As f (ξ) < 1/, this implies tht +1 (1/)( ) for ll n. Therefore (1/) n ( ) (n ) follows by induction on n. Hence it follows tht tends to. Observe tht this proof simply repets the rguments used in estblishing the theorem bout the convergence of fixed point itertion, while certin technicl detils re different (since c for every n 1 with c = in the present cse). Further, observe tht when solving the eqution g(x) = with g(x) = x by Newton s method, we hve +1 = g() g ( ) = = + I.e., the itertion described bove cn be viewed s solving the eqution x = by Newton s method... We wnt to evlute dx lnx using the composite trpezoidl rule with five deciml precision, i.e., with n error not exceeding Wht vlue of n should one use when dividing the intervl [,] into n prts? Solution. The error term in the composite Simpson formul when integrting f on the intervl [, b] nd dividing the intervl into n prts is (b ) 1n f (ξ). We wnt to use this with =, b = 1, nd f(x) = 1/lnx. We hve f (x) = 1 ( 1 x (lnx) + ) (lnx). It is cler tht f (x) is decresing on the intervl (1,+ ), so it ssumes its mximum of the intervl [,] t x =. We hve f ().17. So, noting tht = nd b =, the bsolute vlue of the error is (b ) 1n f (ξ) = 1 1n f (ξ) n n. In order to ensure tht this error is less thn 5 1 6, we need to hve /n < 5 1 6, i.e., n > = Recll tht. In cse =, we cnnot require ξ (,), since then this intervl is empty. In this cse, tke ξ =. It is esy to show tht = cn hppen only in cse = 1, but it is not worth doing so for the completion of the present proof. Equlity below cn hold only in cse =.
3 So one needs to mke sure tht n Thus one needs to divide the intervl [,] into (t lest) 184 prts in order to get the result with 4 deciml precision while using the trpezoidl rule..) Describe how to del with the singulrity in the integrl x / (e x 1)dx if one wnts to evlute this integrl using Simpson s rule. Solution. One cn subtrct the singulrity by tking n initil segment of the Tylor series t x = of e x. We hve e x = n! ; this Tylor series is convergent on the whole rel line. Therefore, we hve where the symbol O( ) is ment s x. Hence n= e x = 1 x + x4 +O(x6 ), ( ) x / e x 1+x x4 = O(x 6 / ) = O(x 9/ ). The fourth derivtive of this ner x = is O(x 1/ ); tht is, the fourth derivtive tends to zero when x ց. So the fourth derivtive is bounded on (, 1), the intervl of integrtion; therefore, Simpson s rule cn be used to clculte the integrl. To sum up, we hve x / (e x 1)dx = ( ) x / e x 1+x x4 dx+ ) ( x 1/ + x5/ dx. The first integrl on the righthnd side cn be clculted by Simpson s rule (t x = tke the integrnd to be ), nd the second integrl cn be evluted directly, by clculting the integrl explicitly. b) We wnt to evlute the integrl x sin 1 x dx (tke the integrnd to be for x = ) using some numericl method, such s the trpezoidl rule, Simpson s rule, dptive integrtion, or Romberg integrtion. Which method would be best suited for evluting the integrl? Give resons for your nswer. Solution. The integrnd behves unevenly on the intervl of integrtion; it behves bdly ner zero (so the intervl of integrtion needs to be divided into mny prts ner zero), while it behves well wy from zero (so, wy from zero we do not need to divide the intervl of integrtion into very smll prts). Hence dptive integrtion needs to be used. Adptive integrtion with the trpezoidl rule works better thn dptive integrtion with Simpson s rule. This is becuse the fourth derivtive of the integrnd is x 6 sin 1 x 4x 5 cos 1 x ; this fetures in the error term of Simpson s rule, while the second derivtive, which fetures in the error term of the trpezoidl rule, is x sin 1 x x 1 cos 1 x +sin 1 x.
4 The second fourth derivtive behves much worse (is often much lrger) thn the second derivtive ner. In n ctul clcultion, one would write the integrl s x sin 1 4 x dx = The bsolute vlue of the first integrl is less thn 4 x dx = x 1 4 x= x sin 1 x dx+ 1 4 x sin 1 x dx. = < The second integrl cn be evluted s.86,59,55,596 with n error less by dptive integrtion using the trpezoidl rule. So, this result gives the vlue of the integrl on the left with n error of < ; tht is, the integrl of the lefthnd side is.86,59,55,596 with 11 deciml digit precision. Trying to evlute the integrl on the lefthnd side directly, with dptive integrtion on the whole intervl [,1], will fil becuse of the bd behvior of the integrl ner. 4.) Given certin function f, we re using the formul to pproximte its derivtive. We hve f(x,h) = f(x+h) f(x h) h f(,1/8).84,5 nd f(,1/16).871,7. Using Richrdson extrpoltion, find better pproximtion for f (). Solution. We hve f (x) = f(x,h)+c 1 h +c h 4... f (x) = f(x,h)+c 1 (h) +c (h) 4... with some c 1, c,... Multiplying the first eqution by 4 nd subtrcting the second one, we obtin Tht is, with x = nd h = 1/16 we hve f (x) = 4 f(x,h) f(x,h)+1c h f (x) 4 f(x,h) f(x,h) 4.871,7.84,5 =.88,85 The function in the exmple is f(x) = sin 1 x, nd f ().88,511. b) In estimting the error of the trpezoidl rule nd Simpson s rule, the following lemm ws used. Lemm. Let g(x) for ll x [,b]. Assume f is differentible on (,b), nd for ech x [,b] we hve ξ x (,b). Then there is n η (,b) such tht f (ξ x )g(x)dx = f (η) provided tht the integrls on both sides of this eqution exist. g(x) dx, Briefly outline the proof of this lemm. 4
5 Solution. We my ssume tht g. Indeed, if g = then both integrls in the bove eqution re zero, so the eqution holds with ny η (,b). 4 Assume m < f (x) < M for ll x (,b), where we llow m = nd M = +, if f is not bounded. Then or else m g(x)dx < f (ξ x )g(x)dx < M f (ξ x )g(x)dx = H with some H with m < H < M. In fct, one only needs to tke H = f (ξ x )g(x)dx g(x)dx. g(x) dx, g(x) dx, Since the derivtive stisfies the intermedite vlue property, 5 it is esy to conclude tht we hve H = f (η) for some η (,b). Hence bove Lemm follows. 5.) Write third order Tylor pproximtion for the solution y(x) t x = 1 + h of the differentil eqution y = x + y with initil condition y(1) = (i.e., the error term in expressing y(1 + h) should be O(h 4 )). Solution. We hve y(1) =, y (1) = x+y = 5; the righthnd side ws obtined by substituting x = 1 nd y =. Differentiting, then using the eqution y = x+y, nd gin substituting x = 1 nd y =, we obtin y (x) = (x+y ) = 1+yy = 1+y(x+y ) = 1+xy +y = 1. Differentiting, then using the eqution y = x+y, nd gin substituting x = 1 nd y =, we obtin y (x) = (1+xy +y ) = y +xy +6y y = y +(x+y )y = y +(x+y )(x+y ) = 14. for h ner. y(1+h) = y(1)+y (1)h+y (1) h +y (1) h 6 +O(h4 ) = +5h+1 h +14h 6 +O(h4 ) = +5h+ 1 h + 67 h +O(h 4 ) b) Consider the differentil opertors D 1 = h x +k y nd D = x +f y, where h nd k re fixed numbers, nd f is function of the vribles d y ll whose prtil derivtives (of ny order) re continuous. These opertors re to be pplied to functions g of the vribles d y ll 4 To show tht f (ξ x)g(x)dx = ssuming g(x)dx = is little messy when one works with the Riemnn integrl, but quite esy if one works with the Lebesgue integrl, the ltter being generliztion of the Riemnn integrl, in common use in modern mthemtics. We cnnot go into detils here. 5 Tht is, if for c,d (,b), we hve f (c) H f (d), then there is n η (,b) (in fct, n η between c nd d cn be found) such tht f (η) = H. 5
6 whose prtil derivtives (of ny order) re continuous. Explin why one cn use the Binomil Theorem to clculte D 1 n, where n is positive integer, while one cnnot pply the Binomil Theorem to clculte D n. Do not do detiled clcultion, just clerly describe the min reson. Solution. The proof of Binomil Theorem sying tht (A+B) n = n k= ( ) n A n k B n k fortheelementsandb ofring 6 mkesuseofthessumptionthtandb commute(i.e., thtab = BA). The differentil opertors h d k y forming D 1 commute, so the Binomil Theorem is pplicble with D 1, while the differentil opertors d f y do not commute, so the Binomil Theorem is not pplicble with D. 6 The differentil opertors described generte ring.
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