Mimi_x3
  • Mimi_x3
Paserval Identiy
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Mimi_x3
  • Mimi_x3
can someone explain how my prof got from https://gyazo.com/0abbc1f224a0c50446069bc955fb3b99
Mimi_x3
  • Mimi_x3
line 3 to 4
Mimi_x3
  • Mimi_x3
i think he integrated to get to line 4

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Mimi_x3
  • Mimi_x3
but it doesnt make sense to me
Mimi_x3
  • Mimi_x3
bumpity bump! HALP GUISE <3
imqwerty
  • imqwerty
( ͡° ͜ʖ ͡°)
IrishBoy123
  • IrishBoy123
i think there might be something missing in your notes have a look at this.
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IrishBoy123
  • IrishBoy123
whilst this recognises the turgidity of the problem but takes it head on https://www.youtube.com/watch?v=_Q04yoKYJn8
anonymous
  • anonymous
The L2 norm and corresponding inner product (its Hilberrt space) are bilinear, so we can distribute it over sums -- in this case: $$\|f\|=\langle f,f\rangle=\int_{-L}^L ff^*\, dx$$
anonymous
  • anonymous
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 \(L^2\))
anonymous
  • anonymous
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 $$\frac1{L}\left\|\frac{a_0}2\psi_0+\sum a_n\psi_n+\sum b_n\phi_n\right\|=\frac1L\left(\frac{a_0^2}4(2L)+L\sum a_n^2+L\sum b_n^2\right)$$
Mimi_x3
  • Mimi_x3
you're amazing, thanks! <3

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