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.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full 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+2:17rzsm2n7r1 = 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=—lJ 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]_%[ﬂzzsmnt]+% = 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 n27r22 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
 Math

Click to edit the document details