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フーリエ級数展開は ベクトルの分解ですよね!?

horiem
March 29, 2017
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フーリエ級数展開は ベクトルの分解ですよね!?

2017年3月29日@プログラマのための数学LT会
詳細: http://blog.physips.com/entry/fourier_orthogonal

horiem

March 29, 2017
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  1. ϑʔϦΤڃ਺ల։ͱ͸ • 2 π पظؔ਺Λ sin ͱ cos ʹ෼ղʢʹల։ʣ •

    Ի੠ॲཧ΍ɺը૾ॲཧʹར༻ • ϑʔϦΤม׵͸ɺϑʔϦΤڃ਺ల։ͷ֦ு • पظؔ਺Ͱͳͯ͘΋ల։Մೳ f(x) = a0 2 + 1 X k=1 (ak cos kx + bk sin kx) ak = 1 ⇡ Z ⇡ ⇡ f(x) cos kx dx bk = 1 ⇡ Z ⇡ ⇡ f(x) sin kx dx
  2. f(x) = a0 2 + a1 cos 1x + b1

    sin 1x + a2 cos 2x + b2 sin 2x
  3. f(x) = a0 2 + a1 cos 1x + b1

    sin 1x + a2 cos 2x + b2 sin 2x + a3 cos 3x + b3 sin 3x
  4. f(x) = a0 2 + a1 cos 1x + b1

    sin 1x + a2 cos 2x + b2 sin 2x + a3 cos 3x + b3 sin 3x + a4 cos 4x + b4 sin 4x + a5 cos 5x + b5 sin 5x
  5. f(x) = a0 2 + a1 cos 1x + b1

    sin 1x + a2 cos 2x + b2 sin 2x + a3 cos 3x + b3 sin 3x + a4 cos 4x + b4 sin 4x + a5 cos 5x + b5 sin 5x + a6 cos 6x + b6 sin 6x + a7 cos 7x + b7 sin 7x + a7 cos 7x + b7 sin 7x
  6. ܎਺ͷٻΊํෳࡶ͗͢໰୊ • ͳͥ͜Μͳܗͳͷ͔ʁ ➡ ϕΫτϧ͔ΒͷྨਪͰཧղՄೳʂ f(x) = a0 2 +

    1 X k=1 (ak cos kx + bk sin kx) ak = 1 ⇡ Z ⇡ ⇡ f(x) cos kx dx bk = 1 ⇡ Z ⇡ ⇡ f(x) sin kx dx
  7. ؔ਺͸ϕΫτϧͰ͋Δ ੒෼൪߸ ੒෼ v = 0 B B B @

    v1 v2 . . . vn 1 C C C A • ؔ਺͸ແ਺ͷ఺ͷू·Γ • ϕΫτϧ΋੒෼ͷू·Γ ➡ ؔ਺͸ແݶݸͷ੒෼Λ΋ͬͨϕΫτϧ
  8. ؔ਺ͷ಺ੵ fk fk 1 fk+1 fk+2 gk+2 gk+1 gk 1

    gk < f, g > = lim x !0 [ f1g1 x + f2g2 x + · · · + fngn x ] = Z f ( x ) g ( x ) dx
  9. ؔ਺ͷ಺ੵ fk fk 1 fk+1 fk+2 gk+2 gk+1 gk 1

    gk ͜ͷ΁Μ ͍͍ײ͡ʹ ܾΊΕ͹OK < f, g > = lim x !0 [ f1g1 x + f2g2 x + · · · + fngn x ] = Z f ( x ) g ( x ) dx
  10. ϕΫτϧͷܭࢉ • ϕΫτϧͷ௕͞ • ϕΫτϧͷ௚ަ • ਖ਼ن௚ަجఈ • ϕΫτϧͷ෼ղ |U|

    = p U · U U · V = 0 ei · ej = ij U = (U · e1)e1 + (U · e2)e2 + . . .
  11. ؔ਺ͷܭࢉ • ؔ਺ͷ௕͞ʢϊϧϜʣ • ؔ਺ͷ௚ަ • ਖ਼ن௚ަجఈ • ؔ਺ͷ෼ղ ||f||

    = p < f, f > < f, g >= 0 < hi, hj >= ij f =< f, h1 > h1+ < f, h2 > h2 + . . .
  12. ؔ਺Λࡾ֯ؔ਺Ͱ෼ղ • ͜Ε͕ϑʔϦΤڃ਺ల։ • ಺ੵΛ͍͍ײ͡ʹܾΊ͍ͨ ➡ ࡾ֯ؔ਺͕ਖ਼ن௚ަجఈͱͳΔΑ͏ʹ಺ੵΛఆٛ f = a0

    2 + < f, cos 1x > cos 1x+ < f, cos 2x > cos 2x + . . . + < f, sin 1x > sin 1x+ < f, sin 2x > sin 2x + . . .
  13. ϑʔϦΤڃ਺ల։ͷͨΊͷ಺ੵ • ಺ੵΛ͜ͷΑ͏ʹఆٛ͢Ε͹ɺҎԼ͕ຬͨ͞ΕΔɿ < f, g > = 1 ⇡

    Z ⇡ ⇡ fg dx < cos ix, cos jx > = ij < sin ix, sin jx > = ij < cos ix, sin jx > = 0
  14. f(x) = a0 2 + 1 X k=1 (ak cos

    kx + bk sin kx) ak = 1 ⇡ Z ⇡ ⇡ f(x) cos kx dx bk = 1 ⇡ Z ⇡ ⇡ f(x) sin kx dx
  15. f(x) = a0 2 + 1 X k=1 (ak cos

    kx + bk sin kx) ak = 1 ⇡ Z ⇡ ⇡ f(x) cos kx dx bk = 1 ⇡ Z ⇡ ⇡ f(x) sin kx dx =< f, cos kx > =< f, sin kx > =< f, cos kx > =< f, sin kx >