Serie de fourier

Páginas: 5 (1210 palabras) Publicado: 26 de enero de 2011
SERIE DE FOURIER

fx=x9-9, -π<&x<0xex, 0<&x<π

a) POR EULER
a0=12π-ππfxdx

a0=12π-π0X3-9dx+0πxexdx

a0=12π-π0x3dx+-π0-9dx+0πxexdx

a0=12π-π0x3dx-9-π0dx+0πxexdx

a0=12πx440-π-9x0-π+xex-exπ0

a0=12π044--π44+-90+9-π+πeπ-eπ-0e0-e0

a0=12π-π44-9π+πeπ-eπ+1

a0=-0.329238

PASO 3 OBTENCION DE LOS COEFICIENTES an y bn
OBTENEMOS anan=1π-ππf(x)cosnxdx
an=1π-π0x3-9cosnxdx+0πxexcosnxdx
an=1π-π0x3cosnxdx-9-π0cosnxdx+0πxexcosnxdx
an=1πx3sinnxn+3x2n2cosnx-6xn3sinnx-6n4cosnx0-π-91nsinnx0-π+xexcosnx1+n22+n2xexcosnx1+n22-excosnx1+n22+nxexsinnx1+n22+n3xexsinnx1+n22-2nexsinnx1+n22+n2excosnx1+n22π0an=1π03sin0nn+302cos0nn2-60sin0nn3-6cos0nn4--π3sin-πnn-3-π2cos-πnn2+6-πsin-πnn3+6cos-πnn4-9sin0nn+9sin-πnn+πeπcosπn1+n22+n2πeπcosπn1+n22-eπcosπn1+n22+nπeπsinπn1+n22+n3πeπsinπn1+n22-2neπsinπn1+n22+n2eπcosπn1+n22-0e0cos0n1+n22-n20e0cos0n1+n22+e0cos0n1+n22-n0e0sin0n1+n22-n30e0sin0n1+n22+2ne0sin0n1+n22-n2e0cos0n1+n22
LOS TERMINOS MULTIPLICADOS POR 0 SE CANCELANan=1π-6cos0n4+π3sin-πnn-3π2cos-πnn2-6πsin-πnn3+6cos-πnn4-9sinon+9sin-πnn+πeπcosπn1+n22+n2πeπcosπn1+n22-eπcosπn1+n22+nπeπsinπn1+n22+n3πeπsinπn1+n22-2neπsinπn1+n22+n2πeπcosπn1+n22+cos01+n22+2nsin01+n22-n2cos01+n22
COMO LOS TERMINOS cos0=1 ,sin0=0 ,sin-πn=-sinπn , cos-πn=cosπn
an=1π-6n4-π3sinπnn-3π2cosπnn2+6πsinπnn3+6cosπnn4-9sinπnn+πeπcosπn1+n22+n2πeπcosπn1+n22-eπcosπn1+n22+nπeπsinπn1+n22+n3πeπsinπn1+n22-2neπsinπn1+n22+n2eπcosπn1+n22+11+n22-n2cos01+n22
LOS TERMINOS sinπn=0an=1π-6n4-3π2cosπnn2+6cosπnn4+πeπcosπn1+n22+n2πeπcosπn1+n22-eπcosπn1+n22+n2eπcosπn1+n22+11+n22-n21+n22

OBTENEMOS bn
bn=1π-ππfxsinnxdx
bn=1π-π0x3-9sinnxdx
bn=1π-π0x3cosnxdx-9-π0sinnxdx+0πxexcosnxdx
bn=1π-x3cosnxn+3x2sin(nx)n2+6xcosnxn3-6sinnxn40-π+9cosnxn0-π+xexsinnx1+n22+n2xexsinnx1+n22-exsinnx1+n22-nxexcosnx1+n22-n3xexcosnx1+n22+2nexcosnx1+n22+n2exsinnx1+n22bn=1π-03cos0nn+302sin(0n)n2+60cos0nn3-6sin0nn4+-π3cos-πnn-3-π2sin(-πn)n2-6-πcos-πnn3+6sin-πnn4+9cos0nn-9cos-πnn+πeπsinπn1+n22+n2πeπsinπn1+n22-eπsinπn1+n22-nπeπcosπn1+n22-n3πeπcosπn1+n22+2neπcosπn1+n22+n2eπsinπn1+n22-0e0sin0n1+n22+n20e0sin0n1+n22-e0sin0n1+n22-n0eocoson1+n22-n30e0cos0n1+n22+2ne0cos0n1+n22+n2e0sin0n1+n22
LOS TERMINOS MULTIPLICADOS POR 0 SE CANCELAN Y sin0n=0bn=1π-π3cos-πnn-3π2sin(-πn)n2+6πcos-πnn3+6sin-πnn4+9cos0nn-9cos-πnn+πeπsinπn1+n22+n2πeπsinπn1+n22-eπsinπn1+n22-nπeπcosπn1+n22-n3πeπcosπn1+n22+2neπcosπn1+n22+n2eπsinπn1+n22-2ncos01+n22
COMO LOS TERMINOS cos0=1 ,sin0=0 ,sin-πn=-sinπn , cos-πn=cosπn
bn=1π-π3cosπnn+6πcosπnn3+9n-9cosπnn-nπeπcosπn1+n22-n3πeπcosπn1+n22+2neπcosπn1+n22-2n1+n22
PASO 4 DAR VALORES A a1………………………………………a10 y b1 …………………………..b10
an=1π-6n4-3π2cosπnn2+6cosπnn4+πeπcosπn1+n22+n2πeπcosπn1+n22-eπcosπn1+n22+n2eπcosπn1+n22+11+n22-n21+n22an=1π1n4-6-3n2πcosnx+6cosnx+11+n22πeπcosnx+n2πeπcosnπ-eπcosnπ+n2eπcosπn+1-n2
a1=1π11-6+3π-6+14-πeπ-πeπ+eπ-eπ+1-1=-10.5485
a2=1π116-6-12π+6+125πeπ+4πeπ-eπ+4eπ+1-2=4.7493
a3=1π181-6+27π-6+1100-πeπ-9πeπ+eπ-9eπ+1-9=-2.6426
a4=1π1256-6-48π+6+1289πeπ+16πeπ-eπ+16eπ+1-17=1.5395
a5=1π1625-6+75π-6+1676-πeπ-25πeπ+eπ-25eπ+1-25=-1.0489
a6=1π11296-6-108π+6+11369πeπ+36πeπ-eπ+36eπ+1-36=0.7222

a7=1π12401-6+147π-6+12500-πeπ-49πeπ+eπ-49eπ+1-49=-0.5507a8=1π14096-6-192π+6+14225πeπ+64πeπ-eπ+64eπ+1-64=0.4142
a9=1π16561-6+243π-6+16724-πeπ-81πeπ+eπ-81eπ+1-81=-0.3371
a10=1π11000-6-300π+6+110201πeπ+100πeπ-eπ+100eπ+1-100=-0.00248

bn=1π-π3cosπnn+6πcosπnn3+9n-9cosπnn-nπeπcosπn1+n22-n3πeπcosπn1+n22+2neπcosπn1+n22-2n1+n22

bn=1π1n3-n2πcosnx+6πcosnx+9n2-9n2cosπn+11+n22-nπeπcosnπ-n3πeπcosπn+2neπcosπn-2nb1=1π11π-6π+9+9+14πeπ+πeπ-2eπ-2=8.4578

b2=1π18-4π+6π+36-36+125-2πeπ-8πeπ+4eπ-4= -7.8786

b3=1π1279π-6π+81+81+11003πeπ+27πeπ-6eπ-6=8.5021

b4=1π164-16π+6π+144-144+1289-4πeπ-64πeπ+8eπ-8=-5.4060

b5=1π112525π-6π+225+225+16765πeπ+125πeπ-10eπ-10=5.6152

b6=1π1216-36π+6π+324-324+11369-6πeπ-216πeπ+12eπ-12=-3.8229...
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