Question

A-9 Derivation/Proof

Answer 1

@sillybilly123

Answer 2

@Hero

Answer 3

I actually figured it out so I'm gonna close dis

Answer 4

Oh, nice :)

Answer 5

Post the solution? :P

Answer 6

yeah I could scan it in hold on

Answer 7

Messy handwrting but

Answer 8

Very nice :)

I wouldn't bother with DeM. And use the coolest ID ever: \(e^{i \pi} + 1 = 0 \implies e^{2 \pi }= 1\), the last one coming in useful in the integration

\(\square = \int\limits_0^{2 \pi} ~ d \varphi \dfrac{1}{\sqrt{2 \pi}} e ^{ i m \varphi}\)

For \(m = 0\):

\(\square = \int\limits_0^{2 \pi} ~ d \varphi \dfrac{1}{\sqrt{2 \pi}} = \sqrt{2 \pi}\)

For \(m \ne 0\):

\(\square = \dfrac{1}{\sqrt{2 \pi}} \left( \dfrac{1}{i m} e^{ i m \varphi} \right)_0^{2 \pi} \)


\(= - \dfrac{i}{m \sqrt{2 \pi}} \left( e^{ i m 2 \pi} - e^{ i 0} \right) \)

\(= - \dfrac{i}{m \sqrt{2 \pi}} \left( (e^{ i 2 \pi})^m - e^{ i 0} \right) \)

\(= - \dfrac{i}{m \sqrt{2 \pi}} \left( (1)^m - 1 \right) \)


For the next bit you have:

\( \int\limits_0^{2 \pi} ~ d \varphi ~ \dfrac{1}{\sqrt{2 \pi}} e ^{- i m \varphi} \dfrac{1}{\sqrt{2 \pi}} e ^{ i n \varphi}\)

\( = \dfrac{1}{2 \pi} \int\limits_0^{2 \pi} ~ d \varphi ~ e ^{- i (m + n) \varphi} \)

So you can say let \(k = - m + n\) and then pattern match back into the previous problem. eg \(m = n \equiv k = 0\)

Answer 9

Question looks out of place on that page. Presume that's what attracted you to it.

Answer 10

typo last line

\( = \dfrac{1}{2 \pi} \int\limits_0^{2 \pi} ~ d \varphi e ^{ i (\color{red}{- m} + n) \varphi} \)