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{sin(klx), cos(klx)} oinarria: Fourier-en serieak
(a,b) tarteko karratu-batugarria duten funtzioen Hilbert-en bektore-espazioaren oinarri ortogonal bat bilatuko dugu hurrengo funtzioetatikabiatuz:
€
gl (x) = sin(klx) ; l =1,2,3,Lf l (x) = cos(klx) ; l = 0,1,2,3,L
non l indizeak balore arruntak har ditzakeen eta k zehaztu behar den funtzioen ortogonaltasuna baieztatu dadin:
(gl(x), f j(x))=0(gl(x),gj (x))=δlj gl (x)
2
( fl (x), fj(x))=δlj fl (x) 2
![Page 2: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/2.jpg)
(gl(x), f j(x))=0
€
sin(klx)cos(kjx)a
b
∫ dx = 0
€
12
sin[k(l + j)x] + sin[k(l − j]x)[ ]a
b
∫ dx = 0
−12
cos[k(l+j)x]k(l+j) +cos[k(l −j)x]
k(l −j)⎡ ⎣ ⎢ ⎢
⎤ ⎦ ⎥ ⎥ a
b
=0
l−j ≠0
![Page 3: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/3.jpg)
(gl(x), f j(x))=0
€
sin(klx)cos(kjx)a
b
∫ dx = 0
€
12
sin[k(l + j)x] + sin[k(l − j)x[ ]a
b
∫ dx = 0
−12
cos[k(l+j)x]k(l+j) +cos[k(l −j)x]
k(l −j)⎡ ⎣ ⎢ ⎢
⎤ ⎦ ⎥ ⎥ a
b
=0
−12
cos[k(l+j)b]−cos[k(l+j)a]k(l+j) +cos[k(l−j )b]−cos[k(l−j)a]
k(l−j)⎡ ⎣ ⎢ ⎢
⎤ ⎦ ⎥ ⎥ =0
![Page 4: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/4.jpg)
(gl(x), fl (x))=0
€
sin(klx)cos(klx)a
b
∫ dx = 0
€
sin(2klx)2a
b
∫ dx = 0
−cos(2klx)2kl
⎡ ⎣ ⎢ ⎤
⎦ ⎥ a
b
=0
− cos(2klb)−cos(2kla)2kl
⎡ ⎣ ⎢ ⎤
⎦ ⎥ =0
![Page 5: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/5.jpg)
(gl(x),gj (x))=0 ; l ≠j
€
sin(klx)sin(kjx)a
b
∫ dx = 0
12 cos[k(l −j)x]−cos[k(l+j)x[ ]
a
b
∫ dx=0
€
12
sin[k(l − j)x]k(l − j)
− sin[k(l + j)x]k(l + j)
⎡ ⎣ ⎢
⎤ ⎦ ⎥a
b
= 0
€
12
sin[k(l − j)b]− sin[k(l − j)a]k(l − j)
− sin[k(l + j)b] − sin[k(l + j)a]k(l + j)
⎡ ⎣ ⎢
⎤ ⎦ ⎥= 0
![Page 6: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/6.jpg)
( fl (x), fj(x))=0 ; l ≠j
cos(klx)cos(kjx) a
b
∫ dx=0
12 cos[k(l +j)x]+cos[k(l−j)x[ ]
a
b
∫ dx=0
€
12
sin[k(l + j)x]k(l + j)
+ sin[k(l − j)x]k(l − j)
⎡ ⎣ ⎢
⎤ ⎦ ⎥a
b
= 0
€
12
sin[k(l + j)b]− sin[k(l + j)a]k(l + j)
+ sin[k(l − j)b] − sin[k(l − j)a]k(l − j)
⎡ ⎣ ⎢
⎤ ⎦ ⎥= 0
![Page 7: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/7.jpg)
Aurreko ortogonaltasun-baldintzak bete daitezen:
−12
cos[k(l+j)b]−cos[k(l+j)a]k(l+j) +cos[k(l−j )b]−cos[k(l−j)a]
k(l−j)⎡ ⎣ ⎢ ⎢
⎤ ⎦ ⎥ ⎥ =0
− cos(2klb)−cos(2kla)2kl
⎡ ⎣ ⎢ ⎤
⎦ ⎥ =0
€
12
sin[k(l − j)b] − sin[k(l − j)a]k(l − j)
− sin[k(l + j)b]− sin[k(l + j)a]k(l + j)
⎡ ⎣ ⎢
⎤ ⎦ ⎥= 0
€
12
sin[k(l + j)b]− sin[k(l + j)a]k(l + j)
+ sin[k(l − j)b] − sin[k(l − j)a]k(l − j)
⎡ ⎣ ⎢
⎤ ⎦ ⎥= 0
Egiaztatu behar dena hauxe da:
€
kb = ka + 2πn ; n ∈ Ζ
![Page 8: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/8.jpg)
Hortaz hurrengo funtzioek osatzen dute oinarri ortogonal bat:
€
gl (x) = sin( 2πb − a
lx) ; l =1,2,3,L
f l (x) = cos( 2πb − a
lx) ; l = 0,1,2,3,L
eta beraien normak hurrengo hauek dira:
€
(gl (x),gl (x)) = sin(klx)sin(klx)a
b
∫ dx
€
= sin2(klx)a
b
∫ dx
(gl(x),gl(x))= 1−cos(2klx)2a
b
∫ dx
€
=x2
− sin(2klx)4kl
⎡ ⎣ ⎢
⎤ ⎦ ⎥a
b
€
=b − a
2
![Page 9: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/9.jpg)
( fl (x), fl(x))= cos(klx)cos(klx)a
b
∫ dx= cos2(klx)a
b
∫ dx
( fl (x), fl(x))= 1+cos(2klx)2a
b
∫ dx
€
=x2
+ sin(2klx)4kl
⎡ ⎣ ⎢
⎤ ⎦ ⎥a
b
l ≠0( f0(x), f0(x))=
a
b
∫dx =b−a =L
=b−a2
![Page 10: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/10.jpg)
€
gl (x) = sin(2πL
lx) ; l =1,2,3,L
f l (x) = cos(2πL
lx) ; l = 0,1,2,3,L
(gl(x), f j(x))=0
(gl(x),gj (x))=δljL2
( fl (x), fj(x))=L2(1+δl0)δlj
![Page 11: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/11.jpg)
Beraz, (a,b) tarteko karratu-integragarridun funtzioa, f(x), oinarri ortogonal honetan adieraz daiteke. Adierazpen edo garapen honi f(x)funtzioaren “Fourier-en seriea” deitzen zaio:
f (x) = αl fl(x)l=0
∞∑ + βlgl(x)l=1
∞∑
f (x) =α0 + αl cos(2πLlx)
l=1
∞∑ + βl sen(2πLlx)
l=1
∞∑ (a,b) tartearen barnean f(x) eta beraren Fourier-en seriearen garapena elkarren berdinak dira, baina (a,b) tartetik kanpo seriearen garapenak badu periodizitate-propietatea, oinarri ortogonalaren funtzioak periodikoak direlako; aitzitik f(x) ez du zertan periodikoa izan behar eta,orokorrean, (a,b) tartetik at, f(x) eta seriearen garapenak desberdinakizango dira.
![Page 12: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/12.jpg)
€
gl (x) = sin(2πL
lx) ; l =1,2,3,L
f l (x) = cos(2πL
lx) ; l = 0,1,2,3,L
gl(x+nL) =gl(x) ; n∈Ζfl (x+nL) = fl(x) ; n∈Ζ
€
sin 2πL
(x + nL) ⎡ ⎣ ⎢
⎤ ⎦ ⎥= sin 2π
Lx + 2πn
⎛ ⎝ ⎜
⎞ ⎠ ⎟= sin(2π
Lx); n ∈ Ζ
cos 2πL
(x + nL) ⎡ ⎣ ⎢
⎤ ⎦ ⎥= cos 2π
Lx + 2πn
⎛ ⎝ ⎜
⎞ ⎠ ⎟= cos(2π
Lx); n ∈ Ζ
![Page 13: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/13.jpg)
Kalkulatu hurrengo funtzioaren (Heaviside-ren funtzioaren, alegia) Fourier-en seriea (-1,1) tartean:
h(x) = −1 , −1≤x<01 , 0≤x<1
⎧ ⎨ ⎩
h(x) = α l fl (x)l=0
∞∑ + βlgl(x)l=1
∞∑Hurrengo seriearen l eta l kalkulatu behar ditugu:
€
gl (x) = sin(2πL
lx) ; f l (x) = cos(2πL
lx)
Tartea (-1,1) denez:
€
L = b − a =1− (−1) = 2gl (x) = sin(πlx) ; f l (x) = cos(πlx)
![Page 14: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/14.jpg)
€
h(x) = α l cos(πlx)l= 0
∞
∑ + β l sin(πlx)l=1
∞
∑
αn =(cos(πnx),h(x))cos(πnx) 2
αn = 1cos(πnx) 2 cos(πnx)h(x)dx
−1
1
∫ =0
αn =0 ; ∀n
![Page 15: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/15.jpg)
€
=2 −cos(πnx)πn
⎡ ⎣ ⎢
⎤ ⎦ ⎥0
1
€
=2
πn[−cos(πn) +1]
€
=2
πn[−(−1)n +1]
€
=0 , n bikoitia bada4
πn , n bakoitia bada
⎧ ⎨ ⎪
⎩ ⎪
€
h(x) = α l cos(πlx)l= 0
∞
∑ + β l sin(πlx)l=1
∞
∑
€
n = (sin(πnx),h(x))sin(πnx) 2
€
n = 1sin(πnx) 2 sin(πnx)h(x)dx
−1
1
∫
€
n = 2 sin(πnx)h(x)dx0
1
∫
€
=2 sin(πnx)dx0
1
∫
![Page 16: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/16.jpg)
€
h(x) = α l cos(πlx)l= 0
∞
∑ + β l sin(πlx)l=1
∞
∑
€
l =0 , l bikoitia bada4πl
, l bakoitia bada
⎧ ⎨ ⎪
⎩ ⎪αl =0
€
h(x) = 4π
sin(πlx)ll=1
∞
∑
€
l bakoitia
![Page 17: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/17.jpg)
-1.5
-1
-0.5
0
0.5
1
1.5
-4 -3 -2 -1 0 1 2 3 4
f(x)1357
x
f ( x ) =4
π
sen[ π ( 2 k + 1 ) x ]
2 k + 1k = 0
∞
∑
€
f (x) = 4π
sin[π (2l +1)x]2l +1l= 0
∞
∑
![Page 18: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/18.jpg)
{exp(iklx)} oinarria:
Euler-en formula kontutan hartuz:
€
ixe = cos(x) + isin(x)
sinuen eta cosinuen oinarri ortogonalaren elementuak esponentzial konplexuen funtzioez idatz daitezke:
€
cos(klx) = 12
iklxe + −iklxe( )
sin(klx) = 12i
iklxe − −iklxe( )
![Page 19: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/19.jpg)
Hori ez ezik, egiaztatu daitekeen bezala, esponentzial konplexu hauek betetzen dute ortogonaltasunaren baldintza:
iklxe , ikjxe( ) =0 ; l ≠j
Beraz, benetan:
k= 2πb−a
⇒ iklxe , ikjxe( )=0 ; l≠ j
iklxe , ikjxe( ) = [ iklxe ]*
a
b
∫ ikjxe dx
iklxe , ikjxe( ) = −iklxea
b
∫ ikjxe dx = −ik(l−j)xea
b
∫ dx=−ik(l−j )xe
−ik(l−j)⎡ ⎣ ⎢ ⎢
⎤ ⎦ ⎥ ⎥ a
b
![Page 20: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/20.jpg)
Bestaldetik, esponentzial konplexu baten eta bere buruaren artekobiderkaketa, hau da, bere normaren karratua hurrengo hau da:
iklxe , ikjxe( ) =Lδlj ; k=2πL
Beraz, esponentzial konplexuen multzo hau {exp(iknx)}, nonk=2π/(b-a) den eta n zenbaki osoa den, da (a,b) tarteko karratu batugarrien funtzioen Hilbert-en espazio bektorialaren oinarri ortogonalbat da :
iklxe , iklxe( ) = [ iklxe ]*
a
b
∫ iklxe dx
iklxe , iklxe( ) = −iklxea
b
∫ iklxe dx =a
b
∫dx =b−a =L
![Page 21: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/21.jpg)
Ondorioz,(a,b) tartean karratu bateragarria duen funtzio bat, f(x), oinarri ortogonal honekiko garatu daiteke :
f (x) = all=−∞
∞∑ iklxe
Kontutan hartu behar da, oraingo honetan, garapenaren eskalareak, hau da, al eskalareak, orokorrean zenbaki konplexuak izango direla.
(a,b) tartearen barnean f(x) eta beraren garapena esponentzialen bidezelkarren berdinak dira, baina (a,b) tartetik kanpo seriearen garapenak badu periodizitate-propietatea, oinarri ortogonalaren funtzioak periodikoak direlako; aitzitik f(x) ez du zertan periodikoa izan behar eta,orokorrean, (a,b) tartetik at, f(x) eta seriearen garapenak desberdinakizango dira.
![Page 22: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/22.jpg)
f(x) funtzio bati dagozkion garapenaren al eskalareak kalkulatu nahi baditugu, hurrengo erara egin dezakegu ortogonaltasuna baliatuz :
f (x) = all=−∞
∞∑ iklxe
( ikjxe , f(x))=( ikjxe , all=−∞
∞∑ iklxe ) = all=−∞
∞∑ ( ikjxe , iklxe )
( ikjxe , f(x))= all=−∞
∞∑ δljL =ajL
aj = 1L
( ikjxe , f (x))
![Page 23: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/23.jpg)
f(x) funtzioa erreala bada ondorengo hau egiaztatzen da:
( f (x), ikjxe ) = f(x)* ikjxe dxa
b
∫ = f (x) ikjxe dxa
b
∫
( ikjxe , f(x))= ikjx[e ]* f(x)dxa
b
∫ = −ikjxe f (x)dxa
b
∫
aj = 1L
( ikjxe , f (x))
aj = 1L
−ikjxe f (x)dxa
b
∫
= ikjxe f(x)dxa
b
∫ = ik(−j )x[e ]* f(x)dxa
b
∫=( −ikjxe , f(x))
![Page 24: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/24.jpg)
Beraz, f(x) funtzioa erreala baldin bada:
( f (x), ikjxe ) =( −ikjxe , f (x))
bestaldetik, funtzio ororako:
( f (x), ikjxe ) = ( ikjxe , f (x))[ ]*
=a−jL
= aj L[ ]*
Ondorioz, f(x) funtzioa erreala bada, hurrengo hau betetzen da:
aj* =a−j
![Page 25: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/25.jpg)
Aurrekoaz baliatuz, ikus dezagun nola aurki dezakegun esponentzial konplexuen oinarri ortogonalaren eta Fourier-en seriearen (hau da, sinuen eta kosinuen oinarri ortogonalaren) koefizienteen arteko erlazioa funtzio erreal baterako:
f (x) = all=−∞
∞∑ iklxe =a0 + (a−l−iklxe +al iklxe )
l=1
∞∑f(x) funtzioa erreala bada: a−l =al
*
f (x) =a0 + [al* iklxe( )* +al iklxe ]
l=1
∞∑f (x) =a0 + 2Re[al iklxe ]
l=1
∞∑
![Page 26: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/26.jpg)
al koefizienteak hurrengo erara idazten baditugu:
al =aliθle
f (x) =a0 + 2Re[al iklxe ]l=1
∞∑ =a0 + 2Re[al i(klx+θ l )e ]l=1
∞∑
f (x) =a0 + 2all=1
∞∑ cos(klx+θl )
€
f (x) = a0 + 2 all=1
∞
∑ [cos(klx)cos(θ l ) − sin(klx)sin(θ l )]
€
f (x) = α 0 + [α ll=1
∞
∑ cos(klx) + β l sin(klx)]
f(x) funtzio errealaren Fourier-en seriearekin alderatuz:
![Page 27: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/27.jpg)
hurrengo erlazioak erdiesten dira:
€
0 = a0
α l = 2 al cos(θ l )
β l = −2 al sin(θ l )
⎫ ⎬ ⎪ ⎭ ⎪l ≠ 0
edo baliokideak direnak:
a0 =α0
al =12 αl
2 +βl2 iarctg−βlαle
![Page 28: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/28.jpg)
f x( ) =x−12 ; x∈(0,1)
€
oinarri ortogonala ⏐ → ⏐ ⏐ ⏐ ⏐ ⏐ i2πjxe{ } ; j ∈ Ζ
f x( ) = aj i2πjxej=−∞
∞∑ → al = −i2πlxe f(x)dx0
1
∫al = −i2πlxe (x−1
2)dx0
1
∫ = −i2πlxe x dx0
1
∫ −12
−i2πlxe dx0
1
∫
al = −i2πlxe x dx0
1
∫ = ix2πl
−i2πlxe⎡ ⎣ ⎢ ⎤
⎦ ⎥ 0
1− i
2πl−i2πlxe dx
0
1
∫ = i2πl
u=x ; du=dx
dv= −i2πlxe dx ; v=−i2πlxe
−i2πl =i −i2πlxe2πl l ≠ 0
![Page 29: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/29.jpg)
l =0 → a0 = (x−12)dx
0
1
∫ =0
f (x) = i2πj
i2πjxe0≠j=−∞
∞∑ =2 Re i2πj
i2πjxe⎡ ⎣ ⎢ ⎢
⎤ ⎦ ⎥ ⎥ j>0
∞∑
€
f (x) = −1π
sin(2πjx)jj>0
∞
∑€
=−1π
sin(2πjx)jj>0
∞
∑
![Page 30: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/30.jpg)
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0
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-1.5 -1 -0.5 0 0.5 1 1.5 2 2.5
x-1/2
x
-1/π (sin(2πx))
![Page 31: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/31.jpg)
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0
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-1.5 -1 -0.5 0 0.5 1 1.5 2 2.5
x-1/2
x
-1/π (sin(2πx)+sin(4πx)/2)
![Page 32: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/32.jpg)
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-1.5 -1 -0.5 0 0.5 1 1.5 2 2.5
x-1/2
x
-1/π (sin(2πx)+sin(4πx)/2+sin(6πx)/3)
![Page 33: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/33.jpg)
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0
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-1.5 -1 -0.5 0 0.5 1 1.5 2 2.5
x-1/2
x
-1/π (sin(2πx)+sin(4πx)/2+sin(6πx)/3+sin(8πx)/4)
![Page 34: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/34.jpg)
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0
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-1.5 -1 -0.5 0 0.5 1 1.5 2 2.5
x-1/2
x
-1/π (sin(2πx)+sin(4πx)/2+sin(6πx)/3+sin(8πx)/4+sin(10πx)/5 )
![Page 35: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/35.jpg)
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0
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x-1/2
x
-1/π (sin(2πx)+sin(4πx)/2+sin(6πx)/3+sin(8πx)/4+sin(10πx)/5 +sin(12πx)/6)
![Page 36: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/36.jpg)
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0
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x-1/2
x
-1/π (sin(2πx)+sin(4πx)/2+sin(6πx)/3+sin(8πx)/4+sin(10πx)/5 +sin(12πx)/6+sin(14πx)/7)
![Page 37: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/37.jpg)
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0
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x-1/2
x
-1/π (sin(2πx)+sin(4πx)/2+sin(6πx)/3+sin(8πx)/4+sin(10πx)/5 +sin(12πx)/6+sin(14πx)/7+sin(16πx)/8)
![Page 38: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/38.jpg)
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0
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x-1/2
x
-1/π (sin(2πx)+sin(4πx)/2+sin(6πx)/3+sin(8πx)/4+sin(10πx)/5 +sin(12πx)/6+sin(14πx)/7+sin(16πx)/8+sin(18πx)/9)
![Page 39: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/39.jpg)
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0
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x-1/2
x
-1/π (sin(2πx)+sin(4πx)/2+sin(6πx)/3+sin(8πx)/4+sin(10πx)/5 +sin(12πx)/6+sin(14πx)/7+sin(16πx)/8+sin(18πx)/9+sin(20πx)/10)
![Page 40: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/40.jpg)
H(x) = 0 , −1≤x<01 , 0≤x<1
⎧ ⎨ ⎩
Kalkulatu (-1,1) tartean hurrengo funtzioen Fourier-en serieak:
f (x) =x
f (x) = xe
f (x) =δ(x)
![Page 41: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/41.jpg)
H(x) = 0 , −1≤x<01 , 0≤x<1
⎧ ⎨ ⎩
€
oinarri ortogonala ⏐ → ⏐ ⏐ ⏐ ⏐ ⏐ iπjxe{ } ; j ∈ Ζ
H x( ) = aj iπjxej=−∞
∞∑ → al =12
−iπlxe H(x)dx−1
1
∫al =1
2−iπlxe H(x)dx
−1
1
∫ =12
−iπlxe dx0
1
∫ =12
1−iπl
−iπlxe⎡ ⎣ ⎢ ⎤
⎦ ⎥ 0
1
al =12iπl
−iπle −1[ ]
€
=i
2πlcos(πl) − isin(πl) −1[ ]= i
2πl cos(πl)−1[ ]
€
=0 ; l bikoitia bada−iπl
; l bakoitia bada
⎧ ⎨ ⎪
⎩ ⎪ l ≠ 0
Kalkulatu Heaviside-ren funtzioaren Fourier-en seriea:
![Page 42: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/42.jpg)
al =12
−iπlxe H(x)dx−1
1
∫
H(x) = 0 , −1≤x<01 , 0≤x<1
⎧ ⎨ ⎩
€
oinarri ortogonala ⏐ → ⏐ ⏐ ⏐ ⏐ ⏐ iπjxe{ } ; j ∈ Ζ
H x( ) = aj iπjxej=−∞
∞∑
0-iπ0x =1
2 dx0
1
∫ =12
€
al =0 ; l bikoitia bada
−iπl
; l bakoitia bada
⎧ ⎨ ⎪
⎩ ⎪a0 =1
2 ;
=12 + −i
πjiπjxe
0≠j=−∞
∞∑j bakoitia
Kalkulatu Heaviside-ren funtzioaren Fourier-seriea:
![Page 43: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/43.jpg)
=12 + 2Re −i
πjiπjxe
⎡ ⎣ ⎢ ⎢
⎤ ⎦ ⎥ ⎥ j>0
∑j bakoitia
=12 + −i
πjiπjxe
0≠j=−∞
∞∑j bakoitia
H(x) = 0 , −1≤x<01 , 0≤x<1
⎧ ⎨ ⎩
€
oinarri ortogonala ⏐ → ⏐ ⏐ ⏐ ⏐ ⏐ iπjxe{ } ; j ∈ Ζ
H x( ) = aj iπjxej=−∞
∞∑
€
H x( ) = 12
+ 2π
Re−i cos(πjx) + isin(πjx)( )
j
⎡ ⎣ ⎢
⎤ ⎦ ⎥
j>0∑
j bakoitia
€
H x( ) = 12
+ 2π
sin(πjx)jj>0
∑j bakoitia
Kalkulatu Heaviside-ren funtzioaren Fourier-seriea:
![Page 44: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/44.jpg)
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0
0.2
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0.8
1
1.2
-4 -3 -2 -1 0 1 2 3 4
x
![Page 45: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/45.jpg)
-0.2
0
0.2
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0.8
1
1.2
-4 -3 -2 -1 0 1 2 3 4
x
![Page 46: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/46.jpg)
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![Page 47: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/47.jpg)
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![Page 48: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/48.jpg)
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x
![Page 49: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/49.jpg)
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![Page 50: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/50.jpg)
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![Page 51: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/51.jpg)
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![Page 52: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/52.jpg)
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![Page 53: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/53.jpg)
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![Page 54: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/54.jpg)
f x( ) =x ; x∈(−1,1)
€
oinarri ortogonala ⏐ → ⏐ ⏐ ⏐ ⏐ ⏐ iπjxe{ } ; j ∈ Ζ
f x( ) = aj iπjxej=−∞
∞∑ → al =12
−iπlxe f(x)dx−1
1
∫al =1
2−iπlxe xdx
−1
1
∫ u=x ; du=dx
dv= −iπlxe dx ; v=−iπlxe
−iπl =i −iπlxeπl l ≠ 0
€
al = 12
−iπlxe x dx−1
1
∫ = 12
ixπl
−iπlxe ⎡ ⎣ ⎢
⎤ ⎦ ⎥
−1
1
− 12
iπl
−iπlxe dx−1
1
∫
![Page 55: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/55.jpg)
al =12
−iπlxe x dx−1
1
∫ = iπl
cos(πl) = iπl
(−1)l
al =12
−iπlxe x dx−1
1
∫ =12
iπl
−iπle + iπle( )⎡ ⎣ ⎢ ⎤
⎦ ⎥
al =12
−iπlxe x dx−1
1
∫ =12
iπl
−iπle + iπle( )+ 1(πl)2
−iπle − iπle( )⎡ ⎣ ⎢ ⎤
⎦ ⎥
€
al = 12
−iπlxe x dx−1
1
∫ = 12
iπl
− iπle + iπle( ) ⎡ ⎣ ⎢
⎤ ⎦ ⎥+ 1
21
(πl)2−iπlxe
⎡ ⎣ ⎢
⎤ ⎦ ⎥
−1
1
€
al = 12
−iπlxe x dx−1
1
∫ = 12
iπl
−iπle + iπle( ) ⎡ ⎣ ⎢
⎤ ⎦ ⎥ − 1
2i
πl−iπlxe dx
−1
1
∫
![Page 56: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/56.jpg)
f x( ) =x= aj iπjxej=−∞
∞∑aj = i
πl(−1)j
= i(−1)jπj
iπjxej=−∞
∞∑
x= 2Re i(−1)jπj
iπjxe⎡ ⎣ ⎢ ⎢
⎤ ⎦ ⎥ ⎥ j>0
∞∑
€
= 2 (−1) j +1
πjsin(πjx)
j>0∑
€
x = 2π
(−1) j +1
jsin(πjx)
j>0∑
![Page 57: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/57.jpg)
f(x)δ(x−xo)dx−1
1
∫ = f(xo)
xo ∈ (−1,1)
⎫ ⎬ ⎪ ⎪
⎭ ⎪ ⎪
Kalkulatu (x)-en Fourier-en seriea:
€
oinarri ortogonala ⏐ → ⏐ ⏐ ⏐ ⏐ ⏐ iπjxe{ } ; j ∈ Ζ
δ x( )= aj iπjxej=−∞
∞∑ → al =12
−iπlxe δ(x)dx−1
1
∫al =1
2−iπlxe δ(x)dx
−1
1
∫ =12
δ x( )=12
iπjxej=−∞
∞∑ =12 +1
2 iπjxe + iπjxej>0∑
j<0∑⎛
⎝ ⎜ ⎜
⎞ ⎠ ⎟ ⎟ =1
2 +12 −iπjxe + iπjxe
j>0∑
j>0∑⎛
⎝ ⎜ ⎜
⎞ ⎠ ⎟ ⎟ =1
2 +12 ( −iπjxe + iπjxe
j>0∑ )δ x( )=1
2 + cos(πjx)j>0∑=12 + cos(πjx)
j>0∑
![Page 58: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/58.jpg)
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2
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x
![Page 59: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/59.jpg)
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-4
-2
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-4 -3 -2 -1 0 1 2 3 4
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![Page 68: {sin(klx), cos(klx)} oinarria: Fourier-en serieak](https://reader037.vdocuments.pub/reader037/viewer/2022102619/56815652550346895dc3f255/html5/thumbnails/68.jpg)
Δ x( ) =iπ2
iπjxje
j=−∞
∞∑
δ(x) =12 + cos(πjx)
j>0∑
=iπ2 j iπjxe
0≠j=−∞
∞∑=−π jsen(πjx)j>0∑=π
2 2Re j i iπjxe[ ]j>0∑
f(x)Δ(x−a)dx−1
1
∫ =−f' (a)
a∈ (−1,1)
⎫ ⎬ ⎪ ⎪
⎭ ⎪ ⎪
Kalkulatu (x)-en Fourier-en seriea:
€
oinarri ortogonala ⏐ → ⏐ ⏐ ⏐ ⏐ ⏐ iπjxe{ } ; j ∈ Ζ
Δ x( ) = aj iπjxej=−∞
∞∑ → al =12
−iπlxe Δ(x)dx−1
1
∫al =1
2−iπlxe Δ(x)dx
−1
1
∫ =iπl2
€
=−π j sin(πjx)j>0∑Δ x( )