## Mimi_x3 one year ago Paserval Identiy

1. Mimi_x3

can someone explain how my prof got from https://gyazo.com/0abbc1f224a0c50446069bc955fb3b99

2. Mimi_x3

line 3 to 4

3. Mimi_x3

i think he integrated to get to line 4

4. Mimi_x3

but it doesnt make sense to me

5. Mimi_x3

bumpity bump! HALP GUISE <3

6. imqwerty

( ͡° ͜ʖ ͡°)

7. IrishBoy123

i think there might be something missing in your notes have a look at this.

8. IrishBoy123

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

9. anonymous

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

10. 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 $$L^2$$)

11. 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 $$\frac1{L}\left\|\frac{a_0}2\psi_0+\sum a_n\psi_n+\sum b_n\phi_n\right\|=\frac1L\left(\frac{a_0^2}4(2L)+L\sum a_n^2+L\sum b_n^2\right)$$

12. Mimi_x3

you're amazing, thanks! <3