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Mimi_x3

  • one year ago

Paserval Identiy

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  1. Mimi_x3
    • one year ago
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    can someone explain how my prof got from https://gyazo.com/0abbc1f224a0c50446069bc955fb3b99

  2. Mimi_x3
    • one year ago
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    line 3 to 4

  3. Mimi_x3
    • one year ago
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    i think he integrated to get to line 4

  4. Mimi_x3
    • one year ago
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    but it doesnt make sense to me

  5. Mimi_x3
    • one year ago
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    bumpity bump! HALP GUISE <3

  6. imqwerty
    • one year ago
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    ( ͡° ͜ʖ ͡°)

  7. IrishBoy123
    • one year ago
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    i think there might be something missing in your notes have a look at this.

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  8. IrishBoy123
    • one year ago
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    whilst this recognises the turgidity of the problem but takes it head on https://www.youtube.com/watch?v=_Q04yoKYJn8

  9. anonymous
    • one year ago
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    The L2 norm and corresponding inner product (its Hilberrt space) are bilinear, so we can distribute it over sums -- in this case: $$\|f\|=\langle f,f\rangle=\int_{-L}^L ff^*\, dx$$

  10. anonymous
    • one year ago
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    observe: $$\left(\frac{a_0}2\psi_0+\sum a_n\psi_n+\sum b_n\phi_n\right)\left(\frac{a_0}2\psi_0+\sum a_m\psi_m+\sum b_m\phi_m\right)\\\quad =\frac{a_0}2\left(\frac{a_0}2\psi_0^2+\sum a_m\psi_m\psi_0+\sum b_m\phi_m\psi_0\right)\\\begin{align*}\qquad&+\sum a_n(\frac{a_0}2\psi_0\psi_n+\sum a_m\psi_m\psi_n+\sum b_m\phi_m\psi_n)\\&+\sum b_n(\frac{a_0}2\psi_0\phi_n+\sum a_m\psi_m\phi_n+\sum b_m\phi_m\phi_n)\end{align*}$$now integrate and 'distribute' the integral into the sums (formally this requires interchange of limits, but its 'free' with the structure we impose on \(L^2\))

  11. anonymous
    • one year ago
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    point is, the only terms in the sums that survive in the integral for orthogonality are where teh indices match. that means our first part of the expression reduces to $$\frac{a_0}2\left(\frac{a_0}2\int_{-L}^L\psi_0^2\, dx+\sum a_m\int_{-L}^L\psi_m\psi_0\,dx+\sum b_m\int_{-L}^L\phi_m\psi_0\, dx\right)\\\quad=\frac{a_0^2}4\cdot 2L+\sum a_m\cdot 0+\sum b_m\cdot 0\\\quad=\frac{a_0^2}4(2L)$$ similarly for our second part we get $$\sum a_n\left(\frac{a_0}2\int_{-L}^L\psi_0\psi_n\, dx+\sum a_m\int_{-L}^L\psi_m\psi_n\,dx+\sum b_m\int_{-L}^L\phi_m\psi_n\, dx\right)\\\quad=\sum a_n\left(\frac{a_0}2\cdot0+\sum a_m\cdot L\delta_{m,n}+\sum b_m\cdot 0\right)\\\quad=\sum a_n\sum a_m(L\delta_{m,n})\\\quad=\sum a_n \cdot La_n\\\quad=L\sum a_n^2$$ and the third part: $$\sum b_n\left(\frac{a_0}2\int_{-L}^L\psi_0\phi_n\, dx+\sum a_m\int_{-L}^L\psi_m\phi_n\,dx+\sum b_m\int_{-L}^L\phi_m\phi_n\, dx\right)\\\quad=\sum b_n\left(\frac{a_0}2\cdot0+\sum a_m\cdot0+\sum b_m\cdot(L\delta_{m,n})\right)\\\quad=\sum b_n\sum b_m(L\delta_{m,n})\\\quad=\sum b_n \cdot L b_n\\\quad=L\sum b_n^2$$ so overall it simplifies to $$\frac1{L}\left\|\frac{a_0}2\psi_0+\sum a_n\psi_n+\sum b_n\phi_n\right\|=\frac1L\left(\frac{a_0^2}4(2L)+L\sum a_n^2+L\sum b_n^2\right)$$

  12. Mimi_x3
    • one year ago
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    you're amazing, thanks! <3

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