The h Bar

@0celo7, okay. Then what you do is put the system of linear equations into matrix form $Ax=b$, where $A$ is the matrix of coeffiicients, $x$ is the vector with variables, and $b$ is the vector with constants.

0celo7

5:07 PM

it's not numbers

it's symbolic

the system is

Sanya

would not matter to mathematica&maple

0celo7

$A_1+A_1'=A_2+A_2'$, $k_1A_1-k_1A_1'=k_2A_2-k_2A_2'$, $A_2e^{i k_2 l}+A_2'e^{-i k_2l}=A_3e^{i k_1l}$, $A_2k_2e^{i k_2 l}-A_2'k_2e^{-i k_2l}=A_3k_1e^{i k_1l}$

user116211

@0celo7 Correct the commands T__T

heather

geesh

0celo7

@Sanya yes but that's cheating

@heather then what?

I took linear algebra

I've done all of this before

It left a scar on my soul

Jacobadtr

5:12 PM

I am a bit confused...

The semi empirical mass formula has the form of a quadratic $$B(Z,A) = aZ^2 + bZ + c$$, where $c$ depends on $A$ which is really a function of $Z$ too.

When differentiating $$B$$ wrt $$Z$$ for a given A, is it acceptable to treat A as a constant, or do I need to treat it as a function of Z?

I hope that makes sense

Sanya

@0celo7 sometimes, I'm really lost with you ....

3

0celo7

@Sanya Huh?

ok here comes the troll brigade starring anti-0celo7 sentiment

I'm sorry I have brain damage from crack abuse

nothing I can do about it

heather

Then you take $A^{-1}$ and multiply and get $x=A{^-1}b$

0celo7

If I could just """"take $A^{-1}$"""" I wouldn't be in this fuckign situtation

heather

To find inverse (searching for notes)

bolbteppa

5:14 PM

@0celo7 derive Cramer's rule using multilinear functionals then you can invert your matrices and think theoretically en.wikipedia.org/wiki/Cramer%27s_rule#Finding_inverse_matrix

user116211

@Jacobadtr for a given A...

Sanya

@0celo7 I just sometimes find it hard to help you or to communicate in a focused constructive manner - that does not mean that I do not like you or want to insult you or want to troll you or anything

user116211

@bolbteppa Wow! Only Cramer was left away from the party... and here he comes ;)

0celo7

I should have never taken you off ignore

Jacobadtr

@MAFIA36790 That is making me think I can just call A a constant, which would make life easy

Secret

5:16 PM

0celo7: First thing: You said there are 4 unknowns, which 4?

0celo7

@Secret there are four now

$A_1',A_2,A_2', A_3$

Fine I will do Gaussian elimination.

Sanya

looking at your equations, you can immediately eliminate A_3

0celo7

How??

Sanya

thus you only have a 3x3 for Gaußian elimination

$A_2k_2e^{i k_2 l}-A_2'k_2e^{-i k_2l}=A_3k_1e^{i k_1l}

is your last equation, right?

0celo7

bolbteppa, I'm sure you are a nice person. You're just so incredibly annoying that you cause me health issues. I'm literally shaking from rage

Please forgive me

No, don't forgive me

I want you to hate me too

bolbteppa

5:19 PM

haha

might be helpful

@heather Dude

I can compute an inverse matrix

The idea is for it to not take 24 hours

A brute-force cramer's rule or whatever would take LONG

bolbteppa

Dude just sit down with the hardest linear algebra book there is and treat it like a differential geometry book and you''ll get the best of all worlds

0celo7

Gauss might work

I'll try it

@Sanya yeah

heather

@0celo7, this isn't brute force, it is basically a formula! You plug in the numbers and go. A little slower but still.

Slereah

5:21 PM

Medieval Music - 'Hardcore' Party Mix

►

0celo7

@Sanya Oh I see.

i think

Sanya

you multiply the inverse of the exponential factor and you have A_3 in terms of the others

0celo7

yeah, multiply the other $A_3$ by $k_2$, then add

that was the first helpful comment today, thank you

Sanya

and sorry about talking of the inverse stuff, you're basically looking for the kernel of the matrix, so there is no inverse

I hadn't thought about that

0celo7

I'm looking for the kernel?

Sanya

5:25 PM

sure, basically if you reorder everything you have something like 0 = [Matrix]* (A1, A1', A2, A2',A3) in bad notation, right?

Secret

[Type too slow]
Using $A_2e^{i k_2 l}+A_2'e^{-i k_2l}=A_3e^{i k_1l}$ you should get:
$$A_1+A_1'=A_1+A_2'$$
$$k_1A_1-k_1A_1'=k_2A_2-k_2A_2'$$
$$A_2k_2e^{ik_2l}-A_2'e^{-ik_2 l}=k_1(A_2e^{ik_2l}+A_2'e^{ik_2l})$$
These can be rearranged into...

0celo7

For the record, liner algebra and number theory (equations mod n specifically) are the hardest branches of mathematics.

bolbteppa

Number theory and probability are

Sanya

so you are looking for the vectors which are mapped onto zero

if I see it correctly

0celo7

no

$A_1$ is a constant

nonzero constant

Sanya

5:26 PM

ah ok, didn't get that

0celo7

oh also fucking poker hand probabilities

bolbteppa

I literally can't do Euclid's algorithm, I have to write it out with matrices, I just f**king hate it so much

0celo7

I take everything back

Sanya

then the inversion works

0celo7

computing poker hand probabilities is the hardest math in existence

bolbteppa

5:26 PM

Euclidean algorithm

In mathematics, the Euclidean algorithm, or Euclid's algorithm, is an efficient method for computing the greatest common divisor (GCD) of two numbers, the largest number that divides both of them without leaving a remainder. It is named after the ancient Greek mathematician Euclid, who first described it in Euclid's Elements (c. 300 BC). It is an example of an algorithm, a step-by-step procedure for performing a calculation according to well-defined rules, and is one of the oldest algorithms in common use. It can be used to reduce fractions to their simplest form, and is a part of many other number...

Sanya

well, anyway, I'm positive you will solve it with time :)

0celo7

I'm not...

user218912

stop acting like this @0celo7

bolbteppa

Also the Jordan Normal Form, literally cannot make sense of it, have to derive it using holomorphic calculus

user218912

we all know you're good at math so deal with it...

0celo7

5:28 PM

@bloo If I were good at math I would be able to do linear algebra

I truly believe there is something wrong with my brain

bolbteppa

No linear algebra takes a solid year or two to make sense of, no excuses

Hoffman/Kunze is one of the hardest books on it, not an easy one

user218912

@0celo7 yes that it can't understand basic stuff.

0celo7

I understand basic stuff just fine

I don't understand linear algebra specifically

user218912

@0celo7 maybe the part of your brain that deals with linear algebra is missing.

0celo7

Right

it's a disease

user218912

5:31 PM

xD

0celo7

oh god I have to do this for the other energy too

I probably won't finish this assignment

user218912

what do you have in QM right now?

0celo7

99

I was considering asking the prof for a job, but I'll forget that now

user218912

you better forget it, 99 is for noobs.

user218912

you don't even have 100.

0celo7

5:34 PM

I know

Secret

After rearrange you should get
$$-A_1'+A_2+A_2'=A_1$$
$$k_1A_1'+k_2A_2-k_2A_2'=k_1A_1$$
$$0+(k_2e^{ik_2l}-k_1e^{-ik_2l})A_2-(k_2e^{ik_2l}+k_1e^{-ik_2l})A_2'=0$$

Since All the unknowns $A_1',A_2,A_2'$ are on the LHS and all the knowns are on the RHS, we can then rewrite the above into a matrix equation:

$$\begin{pmatrix}-1 & 1 & 1 \\ k_1 & k_2 & -k_2 \\ 0 & (k_2e^{ik_2l}-k_1e^{-ik_2l}) & -(k_2e^{ik_2l}+k_1e^{-ik_2l})\end{pmatrix}\begin{pmatrix}A_1' \\ A_2 \\ A_2'\end{pmatrix}=\begin{pmatrix}A_1 \\ k_1A_1 \\ 0\end{pmatrix}$$

0celo7

$$\begin{pmatrix}
1 & 1 & 1\\
-k_1 & -k_2 & k_2\\
0 & (k_2-k_1)\ee^{\ii k_2 l} & -(k_1+k_2)\ee^{-\ii k_2 l}
\end{pmatrix}
\begin{pmatrix}
A_1'\\
A_2\\
A_2'
\end{pmatrix}=
\begin{pmatrix}
-A_1\\
-k_1A_1\\
0
\end{pmatrix}$$

it works on my TeX

Secret

{pmatrix : missing "}"

user218912

you're missing a curly brace at the end.

0celo7

lol commands

wait what did I do wrong

user218912

5:37 PM

\e?

0celo7

no

Secret has different equations

ugh

I copied my own equations wrong

@Secret I think your third eq is wrong

$A_2$ should only ever be multiplying a right moving wave.

bolbteppa

Given $T(\mathbf{u}) = \mathbf{v}$, perturb $T$ so that the new operator becomes invertible, $(T - \lambda I)(\mathbf{u}) = \mathbf{v}$, thus $\mathbf{u} = ( T - \lambda I)^{-1}(\mathbf{v}) = R(\lambda)(\mathbf{v})$. Expand $R(\lambda)$ in a Laurent expansion (!) $$R(\lambda) = \frac{-1}{\lambda}\frac{1}{1-T/ \lambda} = \frac{-1}{\lambda} \sum (T/\lambda)^n = \frac{-1}{\lambda} + \frac{-T}{\lambda^2} + \dots$$ so that $T$ is a residue of $ \lambda R(\lambda)$, then

$$T = \frac{-1}{2 \pi i}\sum_i \oint_{\gamma_i} \lambda R(\lambda) d \lambda = \frac{-1}{2 \pi i}\sum_i \oint_{\gamma_i} ( \lambda - \lambda_i + \lambda_i) R(\lambda) d \lambda = \sum_i (\lambda_i P_i + D_i)$$ is the Jordan Normal Form decomposition of $T$ :\

0celo7

$\newcommand{\ee}{\mathrm{e}}
\newcommand{\ii}{\mathrm{i}}$

$$\begin{pmatrix}
-1 & 1 & 1\\
-k_1 & -k_2 & k_2\\
0 & (k_2-k_1)\ee^{\ii k_2 l} & -(k_1+k_2)\ee^{-\ii k_2 l}
\end{pmatrix}\begin{pmatrix}
A_1'\\
A_2\\
A_2'
\end{pmatrix}=
\begin{pmatrix}
-A_1\\
-k_1A_1\\
0
\end{pmatrix}$$

ok I fixed chat

ok, how do I solve this?

no trolling

bolbteppa

of $- \lambda R(\lambda)$

That is mental, but it gives you the JNF

0celo7

@Secret any ideas?

bolbteppa

5:48 PM

Is that even linear algebra?

Secret

typing...

0celo7

@Secret well, did you check your matrix?

I think what you wrote is wrong, but maybe I did something wrong when I gave you the original equations

@Secret this is correct

so how did you get $e^{-ik_2l}$ multiplying $A_2$?

Secret

Ok I got this for the 3rd row:
$$0 , (k_2-k_1) e^{i k_2 l} , -(k_1e^{i k_2 l}+k_2e^{-i k_2 l})$$

0celo7

yes!

That's what I have, see above

I think I've figured this thing out

@Secret @Sanya @heather I have it

Secret

(It seems this matrix will be rather time consuming to find its eigenvalues because of those exp terms. Also given that zero entry there, gaussian elimination might be fast enough)

Begin by rewritting the matrix as an augmented matrix

$$\begin{pmatrix} -1 & 1 & 1 & | -A_1\\ -k_1 & -k_2 & k_2 &| -k_1A_1\\ 0 & (k_2-k_1) e^{i k_2 l} & -(k_1e^{i k_2 l}+k_2e^{-i k_2 l}) &| 0 \end{pmatrix}$$

Next (still typing...)

0celo7

6:02 PM

PhD linear algebra

Secret

Next replace row 2 by row 2 -k_2 (row 1) to get

$$\begin{pmatrix} -1 & 1 & 1 & | -A_1\\ 0 & -k_2-k_1 & k_2-k_1 &| -k_1A_1-k_1\\ 0 & (k_2-k_1) e^{i k_2 l} & -(k_1e^{i k_2 l}+k_2e^{-i k_2 l}) &| 0 \end{pmatrix}$$

Then

0celo7

how to TeX an augmented matrix?

NeuroFuzzy

"PhD linear algebra" xD

0celo7

apparently it's not built in

well this is probably wrong

\[\begin{pmatrix}
1 & -1 & -1\\
0 & (k_1-k_2)(k_1+k_2) & (k_2-k_1)^2\\
0 & 0 & (k_2-k_1)^2\ee^{\ii k_2l}-(k_1+k_2)^2\ee^{-\ii k_2l}
\end{pmatrix}\begin{pmatrix}
A_1'\\
A_2\\
A_2'
\end{pmatrix}=
\begin{pmatrix}
A_1\\
0\\
0
\end{pmatrix}\]

that would mean $A_2'=0$

which I don't believe

UGH

PhD too powerful

I give up

Secret

Is A2' correspond to waves moving from the right hand side back to the square potential?

0celo7

6:14 PM

yeah

the back scattered wave on the right of the potential

or is that correct?

I'm done with this

using a CAS to solve this shit

I can't be bothered

Secret

So you have A1 heading towards the potential, A2 is transmitted and A1' reflected? If that's the case, then we don't expect there is A2' coming form the otther side of the potential

because your incoming wave is from the left side

0celo7

holy shit

I was RIGHT

DID I SOLVE A LINEAR SYSTEM

I SOLVED A LINEAR SYSTEM

OH MY GOD I SOLVED A LINEAR SYSTEM

Hmm

Secret

You do solved a linear system, note that you can read form row 3 to obtain A2', followed by A2 and thne A1'

0celo7

It's wrong

I solved that system correctly but I don't think it was the correct system.

:/

Secret

Well that's a good start, at least you now know how to solve the system now via gaussian elimination

I storngly suggested based on the equations you gave me, A2' has to be zero

0celo7

6:18 PM

yeah but I also got $A_2=0$

which gives $A_3=0$ from earlier

but that means nothing is transmitted through the potential

which is probably wrong

@Secret hmm

But you'd expect $A_2$ to be nonzero, right?

Because part of the wave makes it into the rectangle

Secret

Well A2 I got nonzero because the -k_1A_1-k_1 stuff on the right of row 2 will onyl get more complicated nonzero when row 3 is done with a row reduction

0celo7

I'll solve the original 4x4 system to confirm.

yikes

ok this is positively awful

it "simplified" everything

@Secret Are we both solving this system?

user116211

Hey, @0celo7, you know the proof of Cartesian Product being countably infinite?

Secret

I have exp(+ve) with k1 and exp(-ve) with k2

so I don't have exp being pulled out for the 3rd column 3rd row term

user116211

0celo7

6:27 PM

@Secret How did that happen?

$$A_2\ee^{\ii k_2 l}+A_2'\ee^{-\ii k_2l}=A_3\ee^{\ii k_1l}\quad\text{and}\quad A_2k_2\ee^{\ii k_2 l}-A_2'k_2\ee^{-\ii k_2l}=A_3k_1\ee^{\ii k_1l}.$$

that's what I derived from QM

Multiply the first one by $k_1$

Then subtract

Right?

Secret

sorry that was my mistake, your matrix is correct

0celo7

@Secret Ok, but $(A_1,0,0)$ is a solution :/

user116211

$$|m+n +m^\prime +n^\prime|\cdot |m +n -m^\prime -n^\prime| = |n^\prime -n|·$$

0celo7

but it makes zero sense

@MAFIA36790 I have no clue what you're trying to prove here

Secret

No it's not ,the 2nd 0 in A1 0 0 is not zero. Give me a sec when I type this up:

btw, because the exp all bubble out we can multiply the 3rd row by exp(+ve) to save outselfves some trouble

user116211

6:31 PM

> If $ n \ne n^\prime,$ then the natural number $|m + n + m^\prime + n^\prime|$ both divides and is greater than the natural number $|n^\prime - n |,$ which is impossible.

0celo7

@Secret Why not? If I plug in $(A_1,0,0)$ is that not a solution?

user116211

@0celo7 I'm not getting the quoted statement above $\uparrow$ :(

0celo7

@ACuriousMind If you're there, I really need some help...

user116211

Why can't $|m + n + m^\prime + n^\prime|$ both divide and be greater than the natural number $|n^\prime -n|\,?$

user116211

._.

user116211

6:34 PM

I GOT it!!!

user116211

I'm dumb and absent-minded!!!!

user116211

Damn ;/

user116211

What an idiot ;/

peterh

In the "New feed items", I get regularly worldbuilding questions... how?

user116211

@peterh check the feeds of the room.

0celo7

6:35 PM

ok my calculator agrees with wiki now

so what the hell are we doing wrong

user116211

^^ @peterh.

0celo7

wait

there's a typo

lol

I hate my life

user116211

@0celo7 :(

0celo7

welp, everything is wrong

\[\begin{pmatrix}
-1 & 1 & 1\\
-k_1 & -k_2 & k_2\\
0 & (k_2-k_1)\ee^{\ii k_2 l} & -(k_1+k_2)\ee^{-\ii k_2 l}
\end{pmatrix}\begin{pmatrix}
A_1'\\
A_2\\
A_2'
\end{pmatrix}=
\begin{pmatrix}
-A_1\\
-k_1A_1\\
0
\end{pmatrix}\]

should be

\[\begin{pmatrix}
-1 & 1 & 1\\
-k_1 & -k_2 & k_2\\
0 & (k_2-k_1)\ee^{\ii k_2 l} & -(k_1+k_2)\ee^{-\ii k_2 l}
\end{pmatrix}\begin{pmatrix}
A_1'\\
A_2\\
A_2'
\end{pmatrix}=
\begin{pmatrix}
A_1\\
-k_1A_1\\
0
\end{pmatrix}\]

er

no that's still wrong...

Secret

6:40 PM

O it's not all zero:

$$\begin{pmatrix} -1 & 1 & 1 & | -A_1\\ -k_1 & -k_2 & k_2 &| -k_1A_1\\ 0 & (k_2-k_1) e^{i k_2 l} & -(k_1+k_2)e^{-i k_2 l} &| 0 \end{pmatrix}$$

$$\begin{pmatrix} -1 & 1 & 1 & | -A_1\\ 0 & -k_2-k_1 & k_2-k_1 &| -k_1(A_1+1)\\ 0 & (k_2-k_1) e^{i k_2 l} & -(k_1+k_2)e^{-i k_2 l} &| 0 \end{pmatrix}$$

$$\begin{pmatrix} -1 & 1 & 1 & | -A_1\\ 0 & -k_2-k_1 & k_2-k_1 &| -k_1(A_1+1)\\ 0 & (k_2-k_1) & -(k_1+k_2)e^{-2i k_2 l} &| 0 \end{pmatrix}$$

$$\begin{pmatrix} -1 & 1 & 1 & | -A_1\\ 0 & 1 & -\frac{k_2-k_1}{k_2+k_1} &| \frac{k_1(A_1+1)}{k_2+k_1}\\ 0 & (k_2-k_1) …

0celo7

\[\begin{pmatrix}
-1 & 1 & 1\\
k_1 & k_2 & -k_2\\
0 & (k_2-k_1)\ee^{\ii k_2 l} & -(k_1+k_2)\ee^{-\ii k_2 l}
\end{pmatrix}\begin{pmatrix}
A_1'\\
A_2\\
A_2'
\end{pmatrix}=
\begin{pmatrix}
A_1\\
k_1A_1\\
0
\end{pmatrix}\]

there

@Secret Dude the original matrix was wrong

we both somehow copied it wrong

peterh

@MAFIA36790 Thanks! But I am not sure ordinary mortals can see this function. Although here on the chat, as I can see, my SE-wide combined reputation counts (18k), I can't see this anywhere.

Secret

Ok should be working now

$$\begin{pmatrix} -1 & 1 & 1 & | A_1\\ k_1 & k_2 & -k_2 & | k_1A_1\\ 0 & (k_2-k_1)e^{i k_2 l} & -(k_1+k_2)e^{-i k_2 l} & | 0 \end{pmatrix}$$

$$\begin{pmatrix} -1 & 1 & 1 & | A_1\\ k_1 & k_2 & -k_2 & | k_1A_1\\ 0 & (k_2-k_1) & -(k_1+k_2)e^{-2i k_2 l} & | 0 \end{pmatrix}$$

$$\begin{pmatrix} -1 & 1 & 1 & | A_1\\ 0 & k_2+k_1 & -k_2+k_1 & | k_1(A_1+k_1)\\ 0 & (k_2-k_1) & -(k_1+k_2)e^{-2i k_2 l} & | 0 \end{pmatrix}$$

$$\begin{pmatrix} -1 & 1 & 1 & | -A_1\\ 0 & 1 & -\frac{k_2-k_1}{k_2+k_1} &| \frac{k_1(A_1+1)}{k_2+k_1}\\ 0 & (k_2-k_1) & -(k_1+k_2)e^{-2i k_2 l} &| 0 …

This stuff is insanely messy, but it is clear that A2, A2' A1' are all nonzero

0celo7

Uh did you fix the typo?

I'm on mobile, can't tell

Secret

Another Small typo at the final ans:
$$\begin{pmatrix} -1 & 1 & 1 & | A_1\\ 0 & 1 & -\frac{k_2-k_1}{k_2+k_1} &| \frac{k_1(A_1+1)}{k_2+k_1}\\ 0 & 0 & -(k_1+k_2)e^{-2i k_2 l}+\frac{(k_2-k_1)^2}{k_2+k_1} &| -\frac{k_1(k_2-k_1)(A_1+1)}{k_2+k_1} \end{pmatrix}$$

should be all fixed now

0celo7

6:51 PM

Jesus Christ

Secret

Ack, I hate careless mistakes: the k1(A1+1) should be 2k1A1
$$\begin{pmatrix} -1 & 1 & 1 & | A_1\\ 0 & 1 & -\frac{k_2-k_1}{k_2+k_1} &| \frac{2k_1A_1}{k_2+k_1}\\ 0 & 0 & -(k_1+k_2)e^{-2i k_2 l}+\frac{(k_2-k_1)^2}{k_2+k_1} &| -\frac{2k_1(k_2-k_1)A_1}{k_2+k_1} \end{pmatrix}$$

now it makes a lot more sense

0celo7

7:08 PM

@Secret do you think that is correct?

user116211

@peterh Go to info; you will see the feeds option.

user116211

peterh

@MAFIA36790 Thanks!

Secret

It does have the expected $k_1+k_2$ denominator which is related to the transmission and reflection coefficients

0celo7

yeah

Secret

7:09 PM

the exp(-2ik2l) term looks strange for an amplitude through...

but I do remeber square potential problems also have that as part of the constraints of the amplitudes

0celo7

no, that part is actually there

Ok, so in this problem, with three parts, what the heck are the TR coeffs actually

like

user116211

Anyways, get you linear algebra job done @0celo7.

0celo7

$$R=\frac{j_1^r}{j_1^i}$$

and

user116211

I'm off now early; night comrades.

0celo7

$$T=\frac{j^t_3}{j_1^i}$$

but probability is conserved...

so I can just compute one of them

...right?

@ACuriousMind Please take pity on me...

Secret

7:13 PM

Transmission coefficient is amplitude of the transmitted wave against the amplitude of the incoming wave. similarly for reflection coefficient

T+R=1 always

0celo7

yes

but is it easier to compute the reflection coefficient

then subtract from 1

Secret

You can of course compute it that way. For me I tend to use T+R=1 as a sanity check to see if I do compute all the amplitudes correctly

0celo7

hmm, ok

@Secret thanks for the help!

I think I get this PhD stuff now

@Secret hmm

what is the final step here?

Secret

row 3= row 3 - (k2-k1) row 2

0celo7

k

@Secret Hmm, in step one, you're multiplying the top row by $k_1$ then adding to the second?

er

second step

shouldn't you end up with like $2k_1A_1$ on the right?

Secret

7:25 PM

Refer to the newest image I posted, the k1(A1+1) one is a careless mistake typo

0celo7

ok

Secret

which has the 2k1A1 you mentioned

0celo7

@Secret OK, we have the same equations

Now time to compute the transmission/reflection coefficients

lol

these equations

Secret

don't forget after solving for the A1' A2 A2', you need to plug it back to solve A3

0celo7

@Secret yeah

but $A_2'$ is horrible!

wonder if I can simplify

aha

Secret

7:33 PM

and yes they do become messy, which is the reason we use computers. But solving them by hand help you understand what is happening so you can build intuition for more complicated cases

0celo7

$$A_2'=\frac{2k_1 A_1(k_1-k_2)}{(k_1-k_2)^2-(k_1+k_2)^2 e^{-2i k_2l}}$$

Seems reasonable?

@Secret Is that what you get?

Secret

yup

0celo7

Then $$A_2=\frac{2k_1 A_1}{k_1+k_2}-\frac{k_1-k_2}{k_1+k_2}A_2'?$$

$$\frac{A_2}{A_1}=\frac{2k_1}{k_1+k_2}-\frac{2k_1 (k_1-k_2)^2}{(k_1+k_2)[(k_1-k_2)^2-(k_1+k_2)^2 e^{-2i k_2l}]}$$

ugh

Secret

It's nearly 7:00 now, I have to go to sleep else I will be terribly exhausted