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# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
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22 n2n 2
16 nnx
sin-------]
n3n 3
2
-2
4 nn 2 n2n2 2
= (8/n2n2)cos(nn) = (-1) n8/n2n2 where the second line for an is found by integration by parts or a computer algebra system. Similarly,
bn = 1 J
2-
2 x2
sin— 2 -2 2 2
nnx 2 nnx
-----dx = 0, since x2sin------------ is an odd
28
function. Thus f(x) = — + —
2
(-1) n nnx
--------cos-------
n=1
n
2 8 m ( -1 ) n
21c. As in Eq. (27), we have sm(x) = — + —^^ -----------------------------2—
nnx
n=1
2
3
x
2
3
2
cos
2
21d. Observing the graphs we see that the Fourier series
converges quite rapidly, except, at x = -2 and x = 2, where there is a sharp "point" in the periodic function.
25.
Section 10.3
207
27a. First we have Ja+Tg(x)dx = Jag(s)ds by letting x = s + T
in the left integral. Now, if 0 < a < T, then from elementary calculus we know that
[a+Tg(x)dx = J Tg(x)dx + J a+Tg(x)dx = J Tg(x)dx + J ag(x)dx Ja Ja Jt Ja J0
using the equality derived above. This last sum is
I Tg(x)dx and thus we have the desired result.
0
Section 10.3, Page 562
2a. Substituting for f(x) in Eqs.(2) and (3) with L = n
yields a0 = (1/n)I^xdx = n/2;
0
am = (1/nn nxcosmxdx = (cosmn - 1)/nm2 = 0 for m even and
0
= -2/nm2 for m odd; and
bm = (1/n)I^xsinmxdx = -(ncosmn)/mn = (-1)m+1/m,
0
m = 1,2... . Substituting these values into Eq.(1) with
L = n yields the desired solution.
2b. The function to which the series converges is indicated in the figure and is periodic with period 2n. Note that
*-
-37T
the Fourier series converges to n/2 at x = —n, n, etc., even though the function is defined to be zero there. This value (n/2) represents the mean value of the left and right hand limits at those points. In (-Ï, 0), f(x) = 0 and f'(x) = 0 so both f and f' are continuous and have finite limits as x ^ -n from the right and as x ^ 0 from the left. In (0, n), f(x) = x and f'(x) = 1 and again both f and f' are continuous and have limits as x ^ 0 from the right and as x ^ n from the left. Since f and f' are piecewise continuous on [—n, n] the conditions of the Fourier theorem are satisfied.
4a. Substituting for f(x) in Eqs.(2) and (3), with L = 1 yields a0 = J1 (1-x2)dx = 4/3;
208
Section 10.3
4b.
7a.
7b.
7c.
12a.
= (1-x2)cos^xdx = (2/ïï ) 11xsinnïxdx
22 1 f1
= (-2/ï2ï2)[xcosnïx - cosnïxdx]
-1 J-1
= 4(-1) ï+1/ï2ï2 ; and bn = (1-x2)sinnïxdx = 0. Substituting these values
into Eq.(1) gives the desired series.
The function to which the series converges is shown in the figure and is periodic of fundamental period 2. In [-1,1] f(x) and f'(x) = -2x are both continuous and have finite limits as the endpoints of the interval are approached from within the interval.
As in Problem 15, Section 10.2, we have
2cos(2n-1)x (-1) n+1sinnx
f(x) =------+ > [------------- + --------------].
4 " ï (2n-1) 2 n
n=1
ï ^ 2cos(2k-1)x (—1) k+1sinkx
en(x) = f(x) +--------> [-------------- + ---------------].
4 , ï (2k—1) 2 k
k=1
Using a computer algebra system, we find that for n = 5, 10 and 2 0 the maximum error occurs at x = —ï in each case and is 1.6025, 1.5867 and 1.5787 respectively.
Note that the author's n values are 10, 20 and 40, since he has included the zero cosine coefficient terms and the sine terms are all zero at x = -ï.
It's not possible in this case, due to Gibb's phenomenon, to satisfy |en(x)I < 0.01 for all x.
a0 = 11 (x-x3)dx = 0 and an = 11 (x-x3)cos^xdx = 0 since
(x-x3) and (x-x3)cos^x are odd functions. bn = 11 (x-x3)sinnïxdx
x3 3x2 (ï2ï 2+6) (ï2ï 2+6)
[ — cos^x---------sinnïx--------------xcos^xH------------sinnïx] _1
2 2 33 44
ïï ï2ï 2 ï3ï 3 ï4ï 4
-12 12 ^ (-1) n
cosn^ so f(x) =----------> --------sinnïx.
ï3ï 3 ï3 ^ n3
n=1
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