homework 11 solutions

# homework 11 solutions - Section 3.8 7(a If xi = 0 and y(x =...

• Homework Help
• davidvictor
• 7

This preview shows pages 1–7. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

This preview has intentionally blurred sections. Sign up to view the full version.

This preview has intentionally blurred sections. Sign up to view the full version.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Section 3.8 7. (a) If xi = 0 and y(x) = A +Bx, then y(0) = A = 0, so y(x) = Bx. But then y(1)+y'(l) = 23 = 0, so A = B = 0 and A = O is notaneigenvalue. (b) If a = a2>0 and y(x) = Acosax+Bsinaoc, then y(0) = A = 0 so y(x) = B sin ax. Hence y(1)+y’(1) = B(sin 01+ acos a) = 0. so a must be a positive root of the equation tan a = —a, and hence the abscissa of a point of intersection of the lines y = tan z and y = —z. We see from the figure below that an lies just to the right of the vertical line z = (2n — l)7r/ 2, and lies closer and closer to this line as It gets larger and larger. Y 13. (a) With 1 = 1, the general solution of y”+2y’+ y = 0 is y(x) = Ae‘x + Bxe‘x. Butthen y(0) = A = 0 and y(1) = e'1(A +B) = 0. Hence 1 = 1 is notan eigenvalue. (b) If A < 1, then the equation y" + 2y' + ky == 0 has characteristic equation r2 + 2r + A = 0. This equation has the two distinct real roots (—2 1 J4 — 42. )/ 2 ; call them r and s. Then the general solution is y(x) = Aer" + Be“, and the conditions y(0) = y(1) = 0 yield the equations A+B = O, Ae’+Be‘ = 0. If A,B ¢ 0, then it follows that e’ = es. But r at s, so there is no eigenvalue x1< 1. (c) If A>1 let ’1 — 1 = a}, so /1 = 1+ d2. Then the characteristic equation r2+2r+11 = (r+1)2+arz = 0 has roots —1:l:ai, so y(x) = e‘x(A cos ax+Bsin ax). Now y(0) = A = 0, so y(x) = Ae’xsin ax. Next, y(1) = Ae‘lsin a = 0, so a = mt with n aninteger. Thus the nth positive eigenvalue is A“ = n2722+1. Because 2. = d2 +1, the eigenfunction associated with 2,, is yn(x) = e‘xsin max. 16. (a) The endpoint conditions are y(0) = y'(0) = 0 and y(L) = y'(L) = 0- (b) The derivative y’(x) = k(4x3 — 6sz + 2L2x) = 2loc(2x — L)(x - L) vanishes at x = 0, L/2, L. Because y(0) = y(L) = 0, the argument of Problem 15(b) implies that ymax = y(L/2). Section 9.1 26. a0 = if(sint)dt = 3— 7f 72' 1 . al —-—J:smtcostdt = 0 7r —2/ 2—1 f an = _1_E(Sint) cosmdt = 1+cosn7r _ { Jr(n ) or n even 7: 7r(1—n2) — 0 for n>l odd 1 1 b = — sinztdt = — ‘ 72' I 2 1 sinnrr b=- s‘tsintdt=——-—————=0forn>1 n ﬂﬂhn) n ﬂamz) 1 1 . 2 cosZt cos4t cos6t cosSt f(t) ~ —+—smt—— + + + +--- 7r 2 7r 1 15 35 63 27, 28. We can handle both of these at once with a less tedious method than that suggested in the book. This requires no 6”" or obscure trig identities. Assume that either f(i7r) = g(:i:7r) : 0 or f’(:l:7r) = g’(:i:7r) : 0. By two integrations by parts, *"f”(t)g(t)dt=f’(t)g(t) - _"f'(t>g'(t>dt =f’(t)g(t) —f(t)g’(t) + _"f(t)g"(t>dt = _”f(t>g"<t>dt. To verify Equation (9), take f(t) = cos(mt) and g(t) = cos(nt). Then, by the above observation, if m and n are integers then 1|' 7|" —m2 cos(mt) cos(nt) dt = —n2 / cos(mt) cos(nt) dt, —7'|' -7? so (112 — m2)/ cos(mt) cos(nt) dt = 0. If m and n are 2 0 and m ¢ n then 722 v m2 75 0, which forces / cos(mt) cos(nt) dt = 0. To verify Equation (10), take f(t) = sin(mt) and g(t) = sin(nt). Then, by the above observation, if m and n are integers then ‘K 7? ——m2 sin(mt) sin(nt) dt 2 —n2/ si11(mt) sin(nt) dt, SO 1r (722 ~ mz) sin(mt) sin(nt) dt = 0. 2 If m and n are 2 0 and m 75 n then H — m2 # 0, which forces f” sin(mt) sin(nt) dt = 0. —‘7'l' 29. 30. As for the m = n # 0 case, ﬁrst notice that cos(:c —— 7r/2) = sin(z), and cos(mt — 271') = cos(mt), and therefore 7r 7r 1r/2 / sin2(1nt) dt = / cosz(7nt - 7r/2) (It 2 / c0s2(m,t) dt : f ._-,r ~7r —37r/2 —7r 7f c0s2(mt) dt. Then, since 7r ldt = 2n“, /W [0052(mt) + sin2(mt)] dt = / _.,, —-rr we conclude that / cosz(mt) dt :/ sin2(mt) dt = 7r. —7|' --7r The m = n = 0 case (which the book doesn’t address) is trivial: 7r / cosz(0 ~ t) dt 2 271' —7I' and / sin2(0 - t)dt = 0. The fact that you must divide (10 by 2 when writing down Fourier series is directly related to the fact that If" c0s2(mt) dt is just 71' for m 75 0, but is 271' for m = 0. Notice that c0s(mt) is even and siu(nt) is odd, so their product is odd. Since the integral of an odd integrand over a symmetric interval is zero, we have that / cos(mt) sin(nt) dt = 0. Do m and n really need to be integers to make this conclusion? By the change of variables theorem, with u = t — P, /a+Pf(t)dt = f(u)du. P This works no matter what a is. Remember that the variable of integration is immaterial. That is, foaf(u)du=/Oaf(t)dt=/Oaf- Using the theorem on breaking up the range of integration twice, a+P P a+P P a P / m) dt = / f(t) dt +/ f(t) dt 2/ f(t) dt +/ f(t) dt 2/ m) dt. a . a P a 0 0 Again this works no matter what a is. I see no point in doing part (b)7 as we have already proved the desired result. Section 9.2 2 17. (a) ao=Ltdt=2 a" = ftcosnmdt = c052n7r+2:17rzsm2n7r-1 = 0 n7r b" = L2 tsinnmdt = sm2mr ~n3:72r0052nrr = __nZ_ﬂ f0) _ 1_£isinnm 7" 11:] n (b) Substitution of t = 1/2, f (t)=1/ 2in this series gives 1 2 1 1 1 1 1 1 7r ——= —— —— -———+--- , so 1——+———+---=—. 2 1 #0 3+5 7 j 3 5 7 4 2 1 20- First we derive the the Fourier series of the function f(t) of period 27r deﬁned for 0<t<27rbyf(t)=t. 1 27: a0 = —— tdt = 27: 7t 1 2» cos 2m: + 2m: sin 2m: —1 an 2 — tcosntdt = ———-———2——~— = 0 7! n 72' 1 2n . sin 2m: - 2n” cos 2n7r 2 b” = —— tsmntdt = —-——————-—2—— = __ 7: n 7r n °° sinnt ﬂ!) = n—2Z n "=1 First we derive the the Fourier series of the function g(t) of period 2n deﬁned for 0 <t<2nbyg(t)=t2. 2n 2 a0=—l-J tzdt=§£— no 3 12 4n7rcosZmr+2 2 27r2—-1 ’ 2 4 a" = _ ” tzcosmdt = = -2. 7r n 7r n b" = _1_ 27: tzsinntdt = (2—4n27r2)0032n::+4n7rsin2n7r—2 = _i75 n n 7r n 2 :1) co ' g“) = 472' +42 cosnt_4ﬂz smnt 2 3 "-1 n ":1 n for 0 < t < Zn we have 2 3t2——671't+27zr2 1 72' 72' —— = _ t __ t +__.. 4g() 2f() 6 12 2 2 no a) ' m ' no =_i4_[4731 +4Zcosznt_4”Zsmnt]_%[ﬂ-zzsmnt]+% = Zcosznn n=l n=l n n=1 n n n 1 n 25. Now we want to sum the alternating series 1 1 1 1 “Ts—37+?“ of reciprocals of odd cubes. Having used a Fourier series of t4 in Problem 24 to evaluate 2(1/n4), it is natural to look at a Fourier series of t3 . Let ﬁt) be the period 27: function with f(t) = t3 if —7r < t < 7:. We calculate the Fourier coefﬁcients of ﬁt), and get _ 1 3 _ _ 1 3 _ a0 — ;J:t dt — 0, a" — ;J: t cosntdt — 0 2m: n27z'2—6 COSﬂﬂ—6 n27r2-2 sinmr 2 7r Ir n7r n n °° ,,+ sinnt °° M sinnt ’3 = Zﬂzz (‘1) l n ‘122(—1) 1 n3 - (ﬁgure below) n=l "=1 e t. 37T 57r If we substitute t = 72/2 and use Leibniz's series 2(—1)"+1/n = 72/4 of Problem 17 we ﬁnd that 1_.1_+L_-1_+i+... —- ﬂ_3 33 53 73 93 32' There is no value of t whose substitution in the Fourier series of ﬁt) = 13 yields the series 2(1/n3) containing the reciprocal cubes of both the odd and even integers. Indeed, the summation in "closed form" of the series 00 .1... —1+_.1._+.1_+__l_.+_!..+ ,,=, n3 23 33 43 53 is a problem that has challenged many ﬁne mathematicians since the time of Euler. Only in modern times (by R. Apery in 1978) has it been shown that this sum is an irrational number. For a delightful account of this work, see the article "A Proof that Euler Missed . . . An Informal Report" by Alfred van der Poorten in the The Mathematical Intelligencer, Volume 1 (1979), pages 195—203. ...
View Full Document

• Spring '07
• TERRELL,R

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern