barcode in ssrs report 14.4. Solve the integral equation in .NET framework

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1 14.4. Solve the integral equation
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& f x cos x dx
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FOURIER INTEGRALS r 1 & p 2 2= 1 f x cos x dx F and choose F 0  0 0@ @1 . >1
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[CHAP. 14
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Let 14.3,
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r 1 r 1 r 2 2 2 F cos x d f x 1 cos x d  0  0  1 2 2 1 cos x 1 cos x d  0 x2
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1 14.5. Use Problem 14.4 to show that
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sin2 u  du . 2 u2
As obtained in Problem 14.4, & 2 1 1 cos x 1 0@ @1 cos x dx 0 >1  0 x2 Taking the limit as ! 0 , we nd 1
1 cos x  dx 2 x2 1
1 But this integral can be written as
2 sin2 x=2 dx which becomes x2
sin2 u du on letting x 2u, so that u2
the required result follows.
1 14.6. Show that
cos x  d e x ; x A 0. 2 1 2
Let f x e x in the Fourier integral theorem 1 2 1 f x cos x d f u cos u du  0 0 Then 2  1 cos x d
e u cos u du e x 1 . 2 1
1 But by Problem 12.22, 12, we have 2  1
e u cos u du 1 or
Then
cos x d e x 2 1
cos x  d e x 2 2 1
PARSEVAL S IDENTITY 14.7. Verify Parseval s identity for Fourier integrals for the Fourier transforms of Problem 14.1.
We must show that 1 f f x g2 dx
1 1
1 fF g2 d
( where f x
1 jxj < a 0 jxj > a
and F
r 2 sin a : 
CHAP. 14]
FOURIER INTEGRALS
This is equivalent to a
1 1 2 dx
2 sin2 a d 2 1  1
1 or
sin2 a d 2 2 1 1
sin2 a d a 2
i.e.,
sin2 a a d 2 2
By letting a u and using Problem 14.5, it is seen that this is correct. The method can also be used to 1 2 sin u nd du directly. u2 0
PROOF OF THE FOURIER INTEGRAL THEOREM 14.8. Present a heuristic demonstration of Fourier s integral theorem by use of a limiting form of Fourier series.
Let f x L
L 1 a0 X nx nx bn sin an cos L L 2 n 1
where an
f u cos
nu du and L
bn
f u sin
nu du: L
Then by substitution (see Problem 13.21, 13), 1 1 L 1X L n u x du 2 f x f u du f u cos 2L L L n 1 L L 1 If we assume that j f u j du converges, the rst term on the right of (2) approaches zero as L ! 1,
while the remaining part appears to approach 1 1X 1 n lim u x du f u cos L!1 L L n 1 1 This last step is not rigorous and makes the demonstration heuristic. Calling =L, (3) can be written f x lim where we have written F But the limit (4) is equal to f x
0 1 X n 1 !0
F n
f u cos u x du 1 d
0 1
1 F d
f u cos u x du
which is Fourier s integral formula. This demonstration serves only to provide a possible result. To be rigorous, we start with the integral 1 1 1 d f u cos u x dx  0 1 and examine the convergence. This method is considered in Problems 14.9 through 14.12.
370 L 14.9. Prove that: (a) lim
!1 0
FOURIER INTEGRALS
[CHAP. 14
sin v  dv ; v 2
L sin v dv lim !1 v 0
b lim
L
!1 L
(a) Let v y. Then lim
!1 0
sin y dy y L
sin v  dv . v 2
sin y  dy by Problem 12.29, Chap. 12. y 2
Let v y.
Then lim
!1 L
sin v dv lim !1 v
sin y  dy : y 2
14.10. Riemann s theorem states that if F x is piecewise continuous in a; b , then b F x sin x dx 0 lim
!1 a
with a similar result for the cosine (see Problem 14.32). L sin v  dv f x 0 a lim f x v !1 0 v 2 b lim 0 f x v sin v  dv f x 0 v 2
Use this to prove that
!1 L
where f x and f 0 x are assumed piecewise continuous in 0; L and L; 0 respectively.
(a) Using Problem 9(a), it is seen that a proof of the given result amounts to proving that L sin v dv 0 lim f f x v f x 0 g !1 0 v f x v f x 0 This follows at once from Riemann s theorem, because F v is piecewise continv uous in 0; L since lim F v exists and f x is piecewise continuous.
n!0
(b) A proof of this is analogous to that in part (a) if we make use of Problem 14.9(b).
1 14.11. If f x satis es the additional condition that 1 sin v  dv f x 0 ; a lim f x v !1 0 v 2
We have 1 f x v 1 f x 0
j f x j dx converges, prove that 0 sin v  b lim dv f x 0 : f x v !1 1 v 2
1 f x v 1 f x 0
sin v dv v sin v dv v
L L
f x v f x 0
sin v dv v sin v dv v
sin v dv v sin v dv v
1 2
Subtracting, 1 f f x v f x 0 g
sin v dv v
sin v f f x v f x 0 g dv v 1 1 sin v sin v f x v f x 0 dv dv v v L L 4
Denoting the integrals in (3) by I; I1 ; I2 , and I3 , respectively, we have I I1 I2 I3 so that jIj @ jI1 j jI2 j jI3 j Now jI2 j @  1 1    f x v sin v dv @ 1 j f x v j dv  v  L L L
CHAP. 14]
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