Question

Paserval Identiy

Answer 1

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

Answer 2

line 3 to 4

Answer 3

i think he integrated to get to line 4

Answer 4

but it doesnt make sense to me

Answer 5

bumpity bump! HALP GUISE <3

Answer 6

( ͡° ͜ʖ ͡°)

Answer 7

i think there might be something missing in your notes

have a look at this.

Answer 8

whilst this recognises the turgidity of the problem but takes it head on

https://www.youtube.com/watch?v=_Q04yoKYJn8

Answer 9

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

Answer 10

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 \(L^2\))

Answer 11

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

Answer 12

you're amazing, thanks! <3