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Note.tex 2008/09/19( )

2000年度『数学展望 I』講義録

数学の基礎訓練I

Transcription:

17 3 31 1

1 3 2 5 3 9 4 10 5 15 6 21 7 29 8 31 9 35 10 38 11 41 12 43 13 46 14 48 2

15 Radon CT 49 16 50 17 53 A 55 1 (oscillation phenomena) e iθ = cos θ + i sin θ, cos θ = eiθ + e iθ 2, sin θ = eiθ e iθ. 2i f(t) = ce iωt, d 2 f dt 2 + ω2 f = 0, θ = ωt. (periodic function (period) f(t + T ) = f(t). e iωt T = /ω T ω (frequency) f = 1/T 1. e iωt T > 0 ω [0, T ) x ( π < x < π) x ( π < x < π) θ = t/t 3

(periodical integration) T a+t f(t)dt = f(t)dt = f(t)dt. 0 a T t f(t) f(t) = g(t) + ih(t) b a b a f(t) dt = b a f(t) dt = lim n g(t) dt + i b a h(t) dt n f(τ j )(t j t j 1 ) j=1 b a f(t) dt = F (b) F (a), F (t) = f(t) b b f(t) dt f(t) dt (a b) a a (e iωt ) = iωe iωt n Z { π if n = 0, e int dt = π 0 otherwise 2. e (a+ib)t e at cos(bt) dt, e at sin(bt) dt 3. c n t n e ct dt n = 1, 2 4

2 T f(t) 2 dt < + (square integrable) T H T H T = { f; f(t + T ) = f(t), T } f(t) 2 dt < +. H T L 2 (0, T ) L 2 ( T/2, T/2) f(t) + g(t) 2 2( f(t) 2 + g(t) 2 ) f, g H T = αf + βg H T H T (f g) = T 2 f(t)g(t) f(t) 2 + g(t) 2 f(t)g(t)dt = T 0 f(t)g(t)dt = 4. z, w T/2 T/2 z + w 2 2( z 2 + w 2 ), 2 zw z 2 + w 2 (i) (f g 1 + g 2 ) = (f g 1 ) + (f g 2 ), (f βg) = β(f g). f(t)g(t)dt 5

(ii) (f 1 + f 2 g) = (f 1 g) + (f g 2 ), (αf g) = α(f g). (iii) (f g) = (g f). (iv) (f f) 0. 5. (f f) = 0 f 0 (f g) H T [] (Cauchy-Schwarz inequality) (f g) (f f)(g g) f(t)g(t) dt f(t) 2 dt g(t) 2 dt. T T [] f (Hermann Schwarz) b b b f(t) dt 1 dt f(t) 2 dt a 6. [a, b] f b a f(t) dt < +, a b a a T f(t) 2 dt = + f = (f f), f + g f + g, αf = α f. OrthoNormal System 7. e at (0 t ) a + 2.1. {e int / } n Z L 2 (0, ) {cos(nt)/ π} n=1,2,... {sin(nt)/ π} n=1,2,... 1/ L 2 (0, ) 6

8. T 9. 2.2 ( ). H {e n } n 1 f f = f n (e n f)e n {z n } f n z n e n 2 = f 2 + n z n (e n f) 2 (f e n ) = 0, n = 1, 2,... f (e n f)e n f z n e n n n best approximation (e n f) 2 (f f) = f 2 n 1 f H Bessel (Bessel s inequality) 2.3 (). [a, b] f(t) lim n ± b a f(t)e int dt = 0. Proof. [a, b] [ π, π] f t [ π, π] \ [a, b] 0 b a f(t)e int dt = (e n f) 0 (n ± ) 7

[a, b] [ π, π] [a, b] [ π + k, π + k] (k Z) π+k π+k f(t)e int dt = = π π π π f(s + k)e in(s+k) ds f(s + k)e ins ds 10. f(t) = 1, f(t) = t Remark. b b f(t)e int dt f(t)e int dt a f(t) = t 1/2 (0 < t 1) 1 0 a f(t) 2 dt = + 0 < t δ, δ t 1 1 lim n 0 f(t)e int dt = 0 lim f(t) cos(nt) dt = 0, n f(t) sin(nt) dt = 0 lim n cos(nt) sin(nt) f (amplitude modulation) f f cos(nt), sin(nt) /n 0 8

3 f(x) f(x) 2 dx < + f(x) = n Z f n e inx, f n C (Fourier series) f(x) f n = 1 f(x)e inx dx, n = 0, ±1, ±2,... f n f f (Fourier coefficient) {e n (x) = e inx / } (e n f)e n (x) n Z D. Bernoulli (1700 1782), L. Euler (1707 1783) J. Fourier (1768 1830) ( ) Fourier P. Dirichlet (1805 1859) 3.1. { 1 if 0 x < π, f(x) = 0 if π x < 0. 9

f 0 = 1 2, f n = 1 ( 1)n in f n e inx n Z Remark. 11. cos(mx), sin(mx) 12. m x m + = { x m if x 0, 0 otherwise [ π, π] 13. f 4 (regularization) Fejer Poisson lim f n = 0 n 0 < r < 1 f n r n e inx n 10

r 1 f n f 1 π π P r (y) P r (y) = n Z r n e iny = = f(y)p r (x y) dy (re iy ) n + n=0 (re iy ) n n=1 1 re iy + 1 reiy 1 re = 1 r 2 iy 1 2r cos y + r 2 Poisson (Poisson kernel) 4.1 (Poisson ). (i) P r (y) 0 ( 1+r P 1 r r(y) 1 r ) y ( y 1+r (ii) (iii) 1 π P r (y)dy = 1, π lim P r(y) = 0 r 1 0 for y 0. More precisely, δ > 0, ɛ > 0, r < 1, P r (y) ɛ for y δ and r r < 1. 14. P r (y) Poisson P r (x) = 1 r 2 (1 r) 2 + 4r sin 2 x 2 P r 4.2. f(x) {f n } f(x) = lim f n r n e inx r 1 0 n Z x 11

Proof. ɛ > 0 f(x) f(y) ɛ for x y δ δ > 0 P r (x y) ɛ if x y δ r < 1 1 π f(x) π f(y)p r (x y) dy = (f(x) f(y))p r (x y) dy π = ɛ π x y δ π π π f(x) f(y) P r (x y) dy f(x) f(y) P r (x y) dy + π ɛ + 4Mπɛ P r (x y) dy + ɛ π π x y δ f(x) f(y) dy (M = f = sup{ f(x) ; x R}) 15. f n r n e inz n Z f(x) f(y) P r (x y) dy Iz < log r z 4.3 ( ). ɛ > 0, N, {a n } N n= N N f N a n e n = sup f(x) a n e inx ɛ. N f N f 2 (e n f)e n f n= N x R N n= N n= N a n e n 2 f a n e n 2 0 f g f g N f (e n f)e n f g (e n f g)e n + g (e n g)e n n= N f g + g (e n g)e n 12

f(x) 2 dx < + 4.4. f(x), g(x) π π f(x)g(x)dx = n Z (f g) = (f e n )(e n g) f n g n, f n = 1 π e inx f(x) dx. π π π f(x) 2 dx = n Z f n 2, (f f) = (e n f) 2 n Z H {e n } f (f f) = (e n f) 2 n (complete) Parseval (Parseval s equality) f = n (e n f)e n lim n n f (e k f)e k = 0 k=1 (Cauchy-Schwarz ) (f g) = (f e n )(e n g) n 13

I = n e n )(e n (Dirac ) 16. n f (e k f)e k 2 = (f f) k=1 n (e k f) 2. 17. L > 0 f(x) = F (Lx/). y R x k e iyx dx e iyx dx = i y e iyx k=1 y xe iyx dx = ix y e iyx + 1 y 2 e iyx x 2 e iyx dx = i x2 y e iyx + 2x y 2 e iyx 2i y 3 e iyx x, x 2 ( π < x < π) i 2 n ( 1)n (n 0), n 2 ( 1)n (n 0) Parseval 1 ζ(2) = n = π2 2 6 n=1 1 ζ(4) = n = π4 4 90. n=1 14

18. x 2 x 3 x ζ(x) = 1 + 1 2 x + 1 3 x + 1 4 x +... L. Euler (infinite product formula) ζ(x) = p:prime (1 1p x ) 1 ζ(2n) B. Riemann ζ(z) (i) z = 1 (ii) z = 2, 4,... (iii) 0 < Rz < 1 (iii) Rz = 1/2 5 (smooth) (piecewise smooth) 5.1. f f (x) 2 dx < + f f n < +. n Z 15

Proof. f f n f f n 2 < + n f f n = inf n f n = n n ( ) 1/2 ( ) 1/2 1 n f n 1 f n n 2 < + 2 n n 19. f(x) f(x) = x ( π < x < π) 20. 21. f(x) = x sin(1/x) ( 2/π x 2/π) (i) (ii) 5.2 ( ). f(x) = n Z f n e inx x Proof. f(x) = lim f n r n e inx r 1 0 n Z f(x) f n e inx n Z f(x) f n r n e inx n Z + f n (1 r n ) n Z 16

5.3. f(x) = x ( π x π). xe inx dx = i n xe inx + 1 n 2 e inx f 0 = π/2 x = π 2 2 π n:odd 1 π x e inx dx = ( 1)n 1 (n 0) π πn 2 1 n 2 einx = π 2 4 π (cos x + 13 2 cos(3x) + 15 2 cos(5x) +... ). 5.4. f(x) = x α ( π x π) α > 0 { 2α 2 π 2α 1 f (x) 2 if 2α 1 > 0, 2α 1 dx = + otherwise. Dirichlet n n f k e ikx = k= n = k= n k= n = 1 Dirichlet D n (y) D n (y) = n k= n 1 f(y)e ik(x y) dy n 1 f(x y)e iky dy f(x y)d n (y)dy. e iky = e iny 2n k=0 e iky = sin(n + 1 2 )y sin y 2 22. n Dirichlet 5.5 (Dirichlet ). (i) f(x) δ > 0 lim f(x)d n (x) dx = 0. n x δ 17

(ii) n 1 D n (x) dx =. Proof. (i) Dirichlet sin(n + 1/2)x e (n+1/2)x e i(n+1/2)x = 2i sin(n + 1/2)x x δ D n sin(x/2) 0 (ii) {e ikx } n D n (x) dx = e ikx dx =. k= n 5.6 ( (principle of localization)). f(x) x = a n f(x) = lim f k e ikx n k= n f x = a f(x), g(x) f(x) = g(x) ( x a 2δ) n lim (f k g k )e ika = 0 n k= n x a δ Proof. h(x) = f(x) g(x) h(x) = 0 ( x a 2δ) (h(x y) h(x))d n (y)dy = 0 uniformly for x a δ lim n (h(x y) h(x))d n (y)dy + y δ y δ (h(x y) h(x))d n (y)dy 0 Dirichlet n x a δ 0 18

5.7. f(x) = x ( π < x < π) 1 π π xe inx dx = i( 1)n, n 0 n x = i( 1) n e inx ( 1) n 1 = 2 sin(nx), n n n 0 n=1 π < x < π x = π/2 x = ±π π 4 = 1 1 3 + 1 5 1 7 +... lim n n k= n f k e ikx = 0 5.8 (Dirichlet). f x = a x = a x = a lim n n k= n f k e ika = f(a + 0) + f(a 0) 2 f(a) Proof. x = a f x = a a = ±π f(a ± 0) = f( π) g(x) = f(x) g( π) = g(π) f(π) + f( π) 2 h(x) = g(x) g(π)x = g(x) + g( π)x 19

x = ±π h f h x = ±π x = ±π h x = ±π h(±π) = 0 Ax + B x = ±π B f x = ±π f(π) + f( π) 2 Remark. Gibbs 5.9 (Kolmogorov). [ π, π] f(t) π π f(t) dt < + f t 5.10 (Carleson). f(x) N f(x) = lim f n e inx N n= N x 5.11. f(x + ) = f(x) {f n } f m 1 f (m 1) (x) x f (m) n= n 2m f n 2 < + 20

0 k < m f (k) (x) = (in) k f n e inx x n= 5.12. (i) f m f (m) ( ) 1 f n = o n m (ii) f f n ( ) 1 f n = O n m+2 f m f (m) 23. 6 ξ f(ξ) = + f(x)e ixξ dx f 2L f(x) [ L, L] 0 f (support) [ L, L] F ( ) L F (x) = f π x, π x π 21

F n = 1 = 1 2L = 1 2L π π L L F (x)e inx dx f(y)e iπny/l dy f(x)e iπnx/l dx L F n 2LF n n ξ = πn/l f 6.1. f f f (m) (x) dx < + f(ξ) = O(1/ ξ m ). Proof. [a, b] f(ξ) = ( i) n b ξ n x n f(x) dx n! n 0 a ξ f (m) (x)e ixξ dx = (iξ) m f(x)e ixξ dx ξ m f(ξ) f (m) (x) dx 24. 22

f 2L > 0 f(x) = 0 for x L f [ L,L] 2L > 0 f(x) = 1 L e iπn(x y)/l f(y) dy = 1 π 2L L L eiπnx/l f(πn/l) n Z n Z f f(ξ) = O(1/ ξ 2 ) L 1 Parseval e ixξ f(ξ) dξ f(x) 2 dx = 1 π L f(πn/l) 2 n Z L 1 f(ξ) 2 dξ f(ξ) 25. f(x) f(ξ) f(x) (integrable) f(ξ) = f(x) dx < + f(x)e ixξ dx 23

6.2. f(x) f(ξ) ξ lim f(ξ) = 0 ξ ± Proof. a < b b a f(x)e ixξ dx ξ ξ ± 0 a f(x)e ixξ dx, b f(x)e ixξ dx f(x) Remark. 26. b a f(x) 2 dx < + b f(x) dx < + a 27. ɛ-δ ɛ-δ ɛ > 0, a > 0, f(x) dx ɛ. x a ξ η e ixξ e ixη ɛ ( x a) f(ξ) f(η) a ( ) f(x)(e ixξ e ixη dx + 2ɛ ɛ f(x) dx + 2 a 0 Parseval 24

6.3. f(x) [a, b] 0 b a f(x) 2 dx < + f 6.4. f(x) f(x) 2 dx = 1 f(ξ) 2 dξ f(x) dx < +, f(x) 2 dx < + f f(ξ) 2 dξ < + f f/ L 2 (R) L 2 (R) f(x)g(x)dx = 1 f(ξ)ĝ(ξ)dξ f(x) = 1 f(ξ)e ixξ dξ Lebesgue x Proof. f(x) { f(x) if x a, f a (x) = 0 otherwise f a f a (ξ) = a a f(x)e ixξ dx 25

f(ξ) f a (ξ) f(ξ) f a (ξ) = f(x)e ixξ dx f(x) dx 0 (a + ) x a ξ f a (ξ) 2 dξ = a b f b (ξ) f a (ξ) = g(x) = 1 f b (ξ) f a (ξ) 2 dξ = x a a a x b a f(x) 2 dx f(x)e ixξ dx { f(x) if a x b, 0 otherwise a x b f(x) 2 dx x a f(x) 2 dx b + a + b + c f b (ξ) f a (ξ) 2 dξ f(x) 2 dx c x a b + f b (ξ) f(ξ) c c f(ξ) f a (ξ) 2 dξ f(x) 2 dx x a c + f(ξ) f a (ξ) 2 dξ 26 x a f(x) 2 dx

a + Parseval f(ξ) 2 dξ = lim a + f a (ξ) 2 dξ a = lim f(x) 2 dx a + = a f(x) 2 dx. Parseval 28. f f(ξ) = f( ξ). f f f(ξ) (ξ 0) f(ξ) 2 ξ > 0 power spectrum 29. h(ξ) c h = h(ξ) 2 dξ c f f a f f b + f b f a f f a 2 f(x) 2 dx x a f a f f a f f a + f f a lim f a = f a + 27

Parseval f L 1 (R) f f(ξ) ξ + f(ξ)e ixξ dξ = N lim N + N f(ξ)e ixξ dξ N N f(ξ)e ixξ dξ = + N (x) = N f(y) e i(x y)ξ dξ dy = N N N + e ixξ dξ = 2 sin(nx). x f(y) N (x y) dy Dirichlet 30. N N (x) D n (x) 6.5 ( ). f L 1 (R) L 2 (R) x = a 1 N f(x) = lim f(ξ)e ixξ dξ N N f(x) x = a f, g L 1 (R) L 2 (R) f(x) = g(x) ( x a 2δ) N lim ( f(ξ) ĝ(ξ))e ixξ dξ = 0 N N x a δ 31. 28

6.6 (Dirichlet). f L 2 (R) L 2 (R) x = a f(a + 0) + f(a 0) 2 N = lim N N f(ξ)e iaξ dξ 7 7.1. e ax2 ixξ dx = π a e ξ2 /4a. ξ = 0 (Gaussian integral) ξ F (ξ) df dξ = ξ 2a F Cauchy ξ = 0 32. π e ax2 dx = a. y = x + ξ 33. a > f(x) = (x )e ax2 f(x) dx < + 29

34. a > 0 b π e ax2 +bx /(4a) = a eb2 ( π x n e ax2 iξ = i ) n e ξ2 /(4a). a ξ { e λx if x > 0, f(x) = 0 otherwise (λ > 0 ) e λx ixξ dx = 1 λ + iξ 1 + 0 e ixξ λ + iξ dξ = { e λx if x > 0, 0 if x < 0 e ixξ /(λ + iξ) 35. λ 1 e ixξ iξ λ dξ = { 0 if x > 0, e λx if x < 0 36. Cauchy x x = 0 1/2 + 1 z iλ dz = log(z iλ) z=+ = πi. z= 30

{ x λ e ax if x > 0, f(x) = 0 otherwise. a > λ x λ = e λ log x, x > 0. f(ξ) = + 0 x λ e ax ixξ dx x z = (a + iξ)x f(ξ) = (a + iξ) λ 1 z λ e z dz L a + iξ e z + f(ξ) = (a + iξ) λ 1 x λ e x dx = (a + iξ) λ 1 Γ(λ + 1) 0 { x λ (a + iξ) λ 1 e ixξ e ax /Γ(λ + 1) if x > 0, dξ = 0 otherwise L 8 vs. 31

f(x) f(x a) T a (translation) lim T yf T a f = 0. y a f(x), g(x) (convolution) (f g)(x) = f(x y)g(y)dy f g = g f, (f g) h = f (g h). f(x) f(rx) (0 r R) f(x) rf(rx) f(x + a) f (n) (x) f(ξ) f(ξ/r) e iaξ f(ξ) (iξ) n f(ξ) e iαx f(x) f(ξ + α) f(x)g(x) f g f ĝ f(ξ)ĝ(ξ) parameter a Taylor 37. 8.1 (Riemann-Lebesgue). f(x) f(x) dx < + f(ξ) ξ lim f(ξ) = 0. ξ ± 32

Proof. f g f(x) g(x) dx f(ξ) f(ξ) ĝ(ξ) + ĝ(ξ) 6.1 ( ĝ(ξ) 0 (ξ ) f g ξ 8.2. (i) f(x) m lim f (l) (x) = 0(0 l m 1), x ± f (m) (x) dx < + f(ξ) f(ξ) = o ( ) 1 ξ m lim ξ ± ξm f(ξ) = 0 (ii) f(ξ) f(ξ) = O(1/ ξ m+2 ) f m f (m) lim f (l) (x) = 0(0 l m), x ± Proof. (i) f (l) f (m) (x)e ixξ dx = (iξ) m f(x)e ixξ dx (iξ) m f(ξ) 33

(ii) f f(ξ) = O(1/ ξ 2 ) Riemann-Lebesgue f(x) = 1 f(ξ)dξ f(ξ) = O(1/ ξ 3 ) f f (x) = 1 (iξ) f(ξ)e ixξ dξ iξ f(ξ) Riemann-Lebesgue f (x) f L 2 (R) lim (T hf f)/h = g h 0 g L 2 (R) lim (T hf f)/h g = 0 h 0 f L 2 - g = f f C 1 f L 2 (R) f L 2 - L 2 - f f 8.3. f L 2 (R) (i) f (n) L 2 ξ n f(ξ) L 2 (R) (ii) ξ n f(ξ) L 2 (R) f C n 1 Proof. ξ n f(ξ) L 2 (R) ξ m f(ξ) L 2 (R) ξ 1, ξ 1 ξ 1 Cauchy-Schwartz 38. n λ > 0 f n (x) = 1 1 (λ + iξ) n eixξ dξ f n+1 (x) = x n f n(x) (λ+iξ) n f n n 34

9 f(x) = 1 dξ dyf(y)e i(x y)ξ x ξ f(x) = f(y)δ(x y)dy = δ(x) = 1 dξe ixξ f(x y)δ(y)dy, g(y) = f( y) g(y)δ(y)dy = g(0) g(y) δ(y) g(y) y = a 0 a+ɛ a ɛ g(y)δ(y) = 0 = g(0) δ(a) = 0 (a 0) g(y) = 1 ( y ɛ) δ(y)dy = g(y)δ(y)dy = g(0) = 1 δ(y) y = 0 1 1 e aξ2 +ixξ dξ = 1 4πa e x2 /4a 35

a +0 1 a a dξe ixξ = sin(ax) πx a + wild δ(x) ϕ a (x) δ(x) = lim a 0 ϕ a (x) {ϕ a } (i) ϕ a (x) (ii) f(x) lim a 0 f(x)ϕ a (x) dx = f(0) f f(x) g g(x)f(x)dx f g g(x)f(y)dx δ g g(0) 36

ϕ a ϕ a (g) = g(x)ϕ a (x)dx δ = lim a +0 ϕ a regularization 0 e ±ixξ dξ 0 e ±i(x±iɛ)ξ dξ, ɛ > 0 ɛ +0 0 e ±ixξ 1 dξ = lim ɛ +0 ɛ ix = ±i x ± i0. + 1 f(x) dx = x ± i0 = = + + + 1 (f(x) f(0) + f(0)) dx x ± i0 f(x) f(0) x f(x) f(0) x f(x) f(0) x δ(x) = dx + f(0) lim dx πif(0). = f (x) x=0 i x + i0 i x i0 lim ɛ +0 r + 37 r r 1 x ± iɛ dx

(i) regularization (ii) regularization L. Schwartz distribution, I.M. Gelfand generalized function, hyperfunction 39. π e ax2 ixξ /4a dx = a e ξ2 Fresnel + e itx2 ixξ dx = it + 0 π it + 0 e iξ2 /4t t e πi/4 if t > 0, it + 0 = 0 if t = 0, t e πi/4 if t < 0 t R (t 0) h(x) = h = δ { 1 if x 0, 0 otherwise h Heaviside function 10 (wave equation) 2 u t 2 2 u x 2 = 0 38

u(0, x) = f(x), u (0, x) = g(x) t u(t, x) = 1 v(t, ξ)e ixξ dξ v(0, ξ) = v(0, ξ) = f(x)e ixξ dx, g(x)e ixξ dx, 2 v t 2 (t, ξ) = ξ2 v(t, ξ) v(t, ξ) = iξ f(ξ) + ĝ(ξ) 2iξ u(t, x) = (heat equation) f(x + t) + f(x t) 2 e itξ + iξ f(ξ) ĝ(ξ) e itξ 2iξ + 1 2 u t = D 2 u x 2 (D > 0) x+t x t g(y) dy. u(0, x) = f(x) u(t, x) = 1 v(0, ξ) = 39 e ixξ v(t, ξ)dξ f(y)e iyξ dy.

u(t, x) v t (t, ξ) = Dξ2 v(t, ξ) v(t, ξ) = e Dξ2t v(0, ξ) = e Dξ2 t f(y)e iyξ dy. u(t, x) = 1 dy f(y) dξe Dtξ2 +i(x y)ξ dξ 1 = e (x y)2 /(4Dt) f(y) dy 4πDt f(x) = δ(x) x = 0 u(t, x) = 1 4πDt e x2 /(4Dt) 40. Dirichlet y 0 (Laplace equation) f(x, 0) = h(x), f(x, y) = 2 f x 2 + 2 f y 2 = 0 lim f(x, y) = 0 y + F (ξ, y)e ixξ dξ (iξ) 2 F + 2 F y 2 = 0 40

F (ξ, y) = A(ξ)e ξy + B(ξ)e ξy f(x, y) = 1 = y π F (ξ, y) = 1 ĥ(ξ)e y ξ ĥ(ξ)e ξ y e ixξ dξ = 1 h(t) (x t) 2 + y dt 2 dtdξ h(t)e y ξ e i(x t)ξ 11 f(x + L) = f(x) f(x) = n Z f n e inx f(ξ) = dx f n e ix(ξ n) n = f n dxe ix(ξ n) n = f n δ(ξ n) n 1 dξe ixξ f(ξ) = f n e inx 41 n

f(x) g(x) = n Z f(x + n) g g(x)e inx dx = f(x+k)e in(x+k) dx = f(x)e inx dx = f(n) k 11.1 (Poisson s summation formula). f(x + n) = 1 f(n)e inx. n f(x) = δ(x) 1 e inx = δ(x + n). n Z n Z f(x) f(ξ) n f(ξ) = 0 for ξ > α f(ξ) ( ξ α) 2α g g(ξ) = 1 α e iπnξ/α f(η)e iπnη/α dη 2α n α = π e iπnξ/α f(πn/α) α n f(x) = 1 α dξ π α α = n f(πn/α) n f(πn/α)e iξ(x πn/α) sin(αx n) (αx n) f(x) f(πn/α) sampling theorem α α f(ξ) dξ < + f(x) x 42

12 P (a X b, c Y d) = P (a X b)p (c Y d). X + Y : P (a X + Y b) = = = a x+y b a u b b a ρ X (x)ρ Y (y) dxdy ρ X (v)ρ Y (u v) dudv ρ X ρ Y (u) du. uρ X ρ Y (u) du = u ρ X (v)ρ Y (u v) dv du = (x + y)ρ X (x)ρ Y (y) dxdy = xρ X (x) dx + yρ Y (y) dy = µ X + µ Y. f g f g = g f, (f g) h = f (g h) 41. 12.1. ρ j X j (1 j n) X 1 + + X n ρ 1 ρ 2 ρ n µ σ µ = µ 1 + + µ n, σ 2 = σ 2 1 + + σ 2 n. X Z = X µ σ f X ρ Z f(z) = σρ(σz + µ) 43

X 1 + + X n nµ, nσ Z n = X 1 + + X n nµ nσ Y j = X j µ g(y) Y 1 + + Y n g n = g g Z n h n (z) = nσg n ( nσz) nσ (g g)( nσ)e ixξ dx = ĝ = ĝ (g g)(y)e iyξ/( nσ) dy ( ξ nσ ) n ĝ(t) = 1 σ2 2 t2 +... ( ) n ) n ξ ĝ = (1 ξ2 nσ 2n +... e ξ2 /2 Z n n + 1 + e ξ2 /2 e ixξ dξ = 1 e x2 /2 12.2 ( (central limit theorem)). [a, b] lim P (a Z n b) = n 12.3. t = 0 b ĝ(t) = 1 σ2 2 t2 + o(t 2 ). 44 a e ξ2 /2 dξ.

Proof. y y 2 ( y 1) y g(y) dy 1 + y 2 g(y) dy < + ĝ ĝ(0) = 1, ĝ (0) = 0, ĝ (0) = σ 2 Taylor ĝ(t) = 1 + t 0 = 1 σ2 2 t2 + ĝ (s)(t s) ds t = 1 σ2 2 t2 + o(t 2 ) 0 (ĝ (s) + σ 2 )(t s) ds 12.4. [α, β] Proof. n log ĝ lim ĥ n (ξ) = lim ĝ n n ( ξ nσ ) n = e ξ2 /2 ( ) ) ) ξ = n log (1 ξ2 nσ 2n + o(ξ2 /n) = n ( ξ2 2n + o(ξ2 /n) = ξ2 2 +no(ξ2 /n) n ξ [α, β] ξ 2 /2 12.5. f, g f L 1 (R), g L 1 (R) g(x)f(x) dx = 1 ĝ(ξ) f(ξ) dξ f, ĝ Proof. ( f ĝ) f 1 ĝ f 1 g 1 g f L 1 (R) L 1 (R) L 1 L 2 45

12.6. ĥn(ξ) 1 f L 1 (R) lim h n (x)f(x) = 1 e ξ2 /2 f(ξ) dξ n h(x) = e x2 /2 / [a, b] f 0 f ± n (x) 1 f n (x) f(x) f + n (x) f m ± L 1 1 ĥ n f m dξ = h n (x)fm(x) dx h n (x)f(x) dx h n (x)f m(x) + dx = 1 + ĥ n f m dξ h(x)fm(x) dx lim inf h n (x)f(x) dx lim sup h n (x)f(x) dx h(x)f + m(x) dx m b h(x) dx lim inf h n (x)f(x) dx lim sup h n (x)f(x) dx a a b h(x) dx 13 µ f = x f(x) 2 dx, σ f = 46 (x µ f ) 2 f(x) 2 dx.

(Quantum Probability), A = A, B = B A 0 = A (f Af), B 0 = B (f Bf) 0 (f (A 0 + itb 0 ) (A 0 + itb 0 )f) = (f A 2 0f) + t 2 (f B 2 0f) + it(f [A 0, B 0 ]f) σ A = 1 4 (f [A, B]f) σ Aσ B. (f A 2 0f), σ B = (f B0f) 2 A = x, B = i d dx [A, B] = i 1 4 (f f) σ Aσ B µ B = (f i d dx f) = ( f ξ f) = µ bf, σ 2 B = ( f (ξ µ bf ) 2 f) = σ 2 bf 13.1 ( (uncertainty principle)). f L 2 (R) σ f σ bf 1 4 f(x) = Ce tx2 +cx, t > 0, c, C C (B 0 + ita 0 )f = 0 f ( ) d dx tx f = cf c C f(x) = Ce tx2 /2+cx 47

t > 0 c = a + ib f(x) = ( ) 1/4 t e tx2 /2+cx a 2 /(2t)+iθ π σ f = 1 2t, σ b f = t 2 42. 43. 14 20Hz 20kHz localize test function (750nm) RGB (350nm) Digital Audio sampling frequency 8kHz, CD, 44.1kHz) 48

low-pass filter sampling () PCM (pulse code modulation) AM (amplitude modulation) FM (frequency modulation) f(t) ω AM FM f(t)e iωt Ae i(ω+f(t))t 44. 15 Radon CT CT (computer tomography) tomo (cut), anatomy Johann Radon (1887 1956) f(ξ, η) = f(x, y)e ixξ iyη dxdy, R 2 f(x, y) = 1 () 2 = 1 () 2 R 2 0 f(ξ, η)e ixξ+iyη dξdη 0 F (r, θ)e ir(x cos θ+y sin θ) rdrdθ. F (r, θ) = f(x, y )e ir(x cos θ+y sin θ) dx dy R 2 + = du f(x, y )e iru. l u,θ l ρ,θ f(x, y )e irρ = + f(u cos θ v sin θ, u sin θ + v cos θ)dv 49

f(x, y) x cos θ + y sin θ = ρ (u, θ) f Radon R f (u, θ) F (r, θ) = + R f (u, θ)e iru du f(x, y) R f (u, θ) R f (u, θ+ π) = R f ( u, θ) A. Cormack G. Hounsfeld 16 N F (n + N) = F (n) (n Z) l N F C N = {z T; z N = 1} ζ n F (n) C ζ = e i/n 1 N f(θ + ) = f(θ) F (n) = f(n/n) (f g) = 0 f(θ)g(θ)dθ (F G) = N n=1 F (n)g(n) N 50

F (m) = n=1 f m = 0 f(θ)e imθ dθ N ( ) n imn/n f e N N = N N F (n)e imn/n F (m) F (m + N) = F (m) l N (discrete Fourier transform) N 1 N /N F m N N 1 /N 2N m F 2N 2 N N Cooley Tukey (Fast Fourier Transform) n=1 51

C N = {ζ C; ζ N = 1} N e ik/n F (k) F l N N 16.1. l N M F l 2N G Ĝ 2M + 6N Proof. G l 2N G 0 (k) = G(2k), G 1 (k) = G(2k + 1) (1 k N) G j l N Ĝ0, Ĝ1 2M 2N Ĝ(m) = G(k)e imk/(2n) 2N k=1 ( = π N ) N G(2k)e imk/n + G(2k + 1)e imk/n e πim/n N k=1 k=1 = 1 ) (Ĝ0 (m) + 2 Ĝ1(m)e πim/n 1 m 2N 3 2N = 6N N = 1 = 2 0 M(0) = 0 N = 2 = 2 1 M(1) = 2M(0) + 3 2 1 N = 2 n l N M(n) M(n) M(n) = 2M(n 1) + 32 n, n = 1, 2,... 52

M(n) M(n 1) = + 3 2 n 2 n 1 M(n)/2 n M(n)/2 n = 3n M(n) = 3n2 n = 3N log 2 N 2N 2 n = 10, N = 2 10 = 1024 3N log 2 N 2N 2 = 3n 2 n+1 = 1 51.2 45. N 46. 2 N = d n (d 2) M d (n) M 2 (n) = 3n2 n 17 (isoperimetric problem) (isoperimetric inequality) A. Hurwitz 4πA l 2. C z(s) (0 s 1), z(0) = z(1) C 1 l = dz ds ds s t = s dz l ds ds 0 0 53

0 t dz dt = l l 2 = 0 dz dt C A = x dy = 0 z + z 2 2 d dt dt ( ) z z dt 2i z(0) = z() A = 1 (zdz zdz) = 1 4i 2 I z dz dt dt l 2 2A = z(t) 0 ( dz dz dt dt I z dz ) 0 dt dt 0 z(t) = n c n e int l 2 2A = n (n 2 n) c n 2 n 2 n 0 (n Z) n 2 n 0 n c n = 0 z(t) = c 0 + c 1 e int C ( c 0 c 1 ) 54

A X P (a X b) P (a X b) = b a ρ(x) dx = 1. ρ ρ(x) 0 ρ(x) dx = 1 µ = X = f(x) = P (a αx + β b) = f(x)ρ(x) dx. xρ(x) dx, σ 2 = (X µ) 2 = a αx+β b ρ(x) dx = b a (x µ) 2 ρ(x) dx. ( ) y β 1 ρ α α dy. X, Y P (a X a, b Y b ) = ρ(x, y) dxdy. a x a,b y b ρ X (x) = ρ Y (y) = 55 ρ(x, y) dy, ρ(x, y) dx.

X = ax + by, Y = cx + dy. ρ (x, y )dx dy = ρ (ax+by, cx+dy)(ad bc)dxdy = D D D ρ(x, y)dxdy (ad bc)ρ (ax + by, cx + dy) = ρ(x, y) ax + by = x ρ (x, y )dx dy = (ax + by)ρ (ax + by, cx + dy)(ad bc)dxdy = (ax + by)ρ(x, y)dxdy = a X + b Y µ X+Y = µ X + µ Y 56