A community for students.
Here's the question you clicked on:
 0 viewing
Mimi_x3
 one year ago
Paserval Identiy
Mimi_x3
 one year ago
Paserval Identiy

This Question is Closed

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0can someone explain how my prof got from https://gyazo.com/0abbc1f224a0c50446069bc955fb3b99

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0i think he integrated to get to line 4

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0but it doesnt make sense to me

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0bumpity bump! HALP GUISE <3

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1i think there might be something missing in your notes have a look at this.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1whilst this recognises the turgidity of the problem but takes it head on https://www.youtube.com/watch?v=_Q04yoKYJn8

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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
 one year ago
Best ResponseYou've already chosen the best response.0observe: $$\left(\frac{a_0}2\psi_0+\sum a_n\psi_n+\sum b_n\phi_n\right)\left(\frac{a_0}2\psi_0+\sum a_m\psi_m+\sum b_m\phi_m\right)\\\quad =\frac{a_0}2\left(\frac{a_0}2\psi_0^2+\sum a_m\psi_m\psi_0+\sum b_m\phi_m\psi_0\right)\\\begin{align*}\qquad&+\sum a_n(\frac{a_0}2\psi_0\psi_n+\sum a_m\psi_m\psi_n+\sum b_m\phi_m\psi_n)\\&+\sum b_n(\frac{a_0}2\psi_0\phi_n+\sum a_m\psi_m\phi_n+\sum b_m\phi_m\phi_n)\end{align*}$$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
 one year ago
Best ResponseYou've already chosen the best response.0point 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
 one year ago
Best ResponseYou've already chosen the best response.0you're amazing, thanks! <3
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.