13.19. Prove that lim in VS .NET

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 13.19. Prove that lim
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f x sin nx dx lim
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1 a2 X 2 This follows at once from Problem 13.15, since if the series 0 a b2 is convergent, lim an n n!1 2 n 1 n lim bn 0.
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f x cos nx dx 0 if f x is piecewise continuous.
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The result is sometimes called Riemann s theorem.
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 13.20. Prove that lim
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We have 
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f x sin M 1 x dx 0 if f x is piecewise continuous. 2
f x sin M 1 x dx 2
f f x sin 1 xg cos Mx dx 2
f f x cos 1 xg sin Mx dx 2
Then the required result follows at once by using the result of Problem 13.19, with f x replaced by f x sin 1 x and f x cos 1 x respectively, which are piecewise continuous if f x is. 2 2 The result can also be proved when the integration limits are a and b instead of  and .
13.21. Assuming that L , i.e., that the Fourier series corresponding to f x has period 2L 2, show that M X sin M 1 t a 1  2 dt an cos nx bn sin nx f t x SM x 0   2 2 sin 1 t 2 n 1
Using the formulas for the Fourier coe cients with L , we have       1 1 an cos nx bn sin nx f u cos nu du cos nx f u sin nu du sin nx     1  f u cos nu cos nx sin nu sin nx du    1 f u cos n u x du   Also, a0 1 2 2 
f u du
CHAP. 13]
FOURIER SERIES
Then
SM x
M a0 X a cos nx bn sin nx 2 n 1 n M 1  1X  f u du f u cos n u x du 2   n 1  ( ) M 1  1 X f u cos n u x du   2 n 1  sin M 1 u x 1 2 du f u   2 sin 1 u x 2
using Problem 13.18.
Letting u x t, we have sin M 1 t 1  x 2 SM x dt f t x   x 2 sin 1 t 2
Since the integrand has period 2, we can replace the interval  x;  x by any other interval of length 2, in particular, ; . Thus, we obtain the required result.
13.22. Prove that  SM x
!  f x 0 f x 0 1 0 f t x f x 0 sin M 1 t dt 2 2   2 sin 1 t 2 ! 1  f t x f x 0 sin M 1 t dt 2  0 2 sin 1 t 2
From Problem 13.21, SM x 1  0
f t x
sin M 1 t 1 2 dt  2 sin 1 t 2
f t x
sin M 1 t 2 dt 2 sin 1 t 2
Multiplying the integrals of Problem 13.18(b) by f x 0 and f x 0 , respectively, sin M 1 t sin M 1 t f x 0 f x 0 1 0 1  2 2 dt dt f x 0 f x 0 1 2    0 2 sin 2 t 2 sin 1 t 2 Subtracting (2) from (1) yields the required result.
13.23. If f x and f 0 x are piecewise continuous in ;  , prove that
lim SM x
f x 0 f x 0 2
The function tinous. Also, lim
f t x f x 0 is piecewise continuous in 0 < t @  because f x is piecewise con2 sin 1 t 2
f t x f x 0 f t x f x 0 t f t x f x 0 lim lim exists, t!0 t t 2 sin 1 t 2 sin 1 t t!0 2 2 0 since by hypothesis f x is piecewise continuous so that the right-hand derivative of f x at each x exists.
t!0
Thus,
f t x f x 0 is piecewise continous in 0 @ t @ . 2 sin 1 t 2 f t x f x 0 is piecewise continous in  @ t @ 0. 2 sin 1 t 2
Similarly,
FOURIER SERIES
[CHAP. 13
Then from Problems 13.20 and 13.22, we have & ' f x 0 f x 0 0 or lim SM x M!1 2
lim SM x
f x 0 f x 0 2
BOUNDARY-VALUE PROBLEMS 13.24. Find a solution U x; t of the boundary-value problem @U @2 U 3 2 t > 0; 0 < x < 2 @t @x U 0; t 0; U 2; t 0 t > 0 U x; 0 x 0<x<2
A method commonly employed in practice is to assume the existence of a solution of the partial di erential equation having the particular form U x; t X x T t , where X x and T t are functions of x and t, respectively, which we shall try to determine. For this reason the method is often called the method of separation of variables. Substitution in the di erential equation yields 1 @ @2 XT 3 2 XT @t @x or 2 X dT d2X 3T 2 dt dx
where we have written X and T in place of X x and T t . Equation (2) can be written as 1 dT 1 d2X 3T dt X dx2 3
Since one side depends only on t and the other only on x, and since x and t are independent variables, it is clear that each side must be a constant c. In Problem 13.47 we see that if c A 0, a solution satisfying the given boundary conditions cannot exist. Let us thus assume that c is a negative constant which we write as 2 . Then from (3) we obtain two ordinary di erentiation equations dT 32 T 0; dt whose solutions are respectively T C1 e 3 t ;
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