lmao close

wtf

@Danu PLEASE

freaking LaTeX

A bloody space

Caption: summary of the two susskind theoretical minumum books

@ACuriousMind then - would you do it?

I have no idea @bernard gl..

Okay

So

Continuing:$$\sum\limits_{k=0}^{n+1}\frac{k}{(k+1)!}=\sum\limits_{k=0}^{n} \frac{k}{(k+1)!}+\frac{(n‌​+1)}{(n+2)!}=1-\frac{1}{(n+1)!}+\frac{n+1}{(n+2)!}=1-‌​\frac{1}{(n+2)}$$

(Thanks @Danu!)

My physics friend comment how one way to think about the relation between classic and quantum is that classic does not allow superposition, such as points in state spaces have only one value for their coordinates and not a superposition of them, thus in effect all points in classical state space are pointer states

Okay... You take it from here?

This was exactly what was on the board

What the heck is that

Oh, scratch that. You messed up the upper limit for the FIRST.

It should be $n+1$

He's using the case $n$ to prove the case $n+1$

@Sanya You must have missed the part where I'm one of the users who said they'd only do it if no other qualified candidate runs. I'm increasingly thinking of running to end this a bit ridiculous standoff, though.

I messed up on the text

@JohnDuffield Well, your last sentence is an accurate recount of the events that happened, as far as I know, but none of us know why he left.

it was right on my book

There

@ACuriousMind Perfect

that's exactly as was on the board & on my notes

@BernardMeurer Do you see now why it's true?

@BernardMeurer Can you be a bit more specific?

Not really, I don't understand that proof

You take the $n+1$ case, split it into $n$-case plus one more term. Then use the inductive step to substitute the result for the $n$-case

I'm guessing you are confused because there's no indication why the second equality is true?

@DavidZ : noted. These things happen I suppose.

@Danu The lack of an explicit "inductive step" on the board makes me think that might not tell Bernard much

Where did that second sum come from

@BernardMeurer It's the full first sum, minus the last term

Then he adds the last term separately

@ACuriousMind sorry, missed it indeed - wasn't around too much these days

He just "split off" the last term.

@0celo7 do you think "a clear and complete discussion of the theory behind the experiment" means only the theory discussed in the experiment? For ex: I'm doing a report on $Q = mc\Delta T$ essentially and it's extremely basic theory. Idk if I should go into derivations for the specific heat of solids using debye or einstein's models.

@BernardMeurer Do you agree with $\sum_{k=0}^n f(k) = \sum_{k=0}^{n-1}f(k) + f(n)$?

lmao

it's happening to ACM too.. oh lord

nobody can tex anymore

Phew, I'm not the only one screwing up

Man, your TeX failure is infectious :P

STD: Sexual TeX Disease

Wait, let me think about that

Yes, I agree with that

@BernardMeurer Then that's exactly what happened here

@Sanya No need to feel sorry

Thinking

How does the last step happen?

$$1-\frac{1}{(n+1)!}+\frac{n+1}{(n+2)!}=1-‌​\frac{1}{(n+2)!}$$

Ah

Just write out $(n+2)!$

Then group the two fractions together

Then simplify the fraction

$n!*n+1*n+2$?

That right?

Wait

Sure, but $(n+2)\cdot(n+1)!$ is more useful

@Secret oh my god

is that yours?

@BernardMeurer I'm assuming you intended to have parentheses about the sums (else it's wrong)

@0celo7 Yup, wrote today when I was hangning out with my physics friends in the physics society room

Which parentheses?

$(n+2)\cdot(n+1)\cdot n!$

note the parentheses

AH

Yes, yes

Forgot to put them

In any case, I'm seeing that it's actually more useful to rewrite $-1/((n+1)!)$ instead

It doesn't really matter, but that'll be a bit faster

The idea is always to make sure that the denominator is the same

Then put them into a single fraction and simplify

@Obliv that one fell victim to this

The example works fine, lol

painingest thing in the world to debug

@Danu Thinking

@emilio wow thanks. Would have never known.

since i don't use 80+ char tex strings

So wait $$1-\frac{1}{(n+1)!}+\frac{n+1}{(n+2)(n+1)!}$$

@BernardMeurer Write it out

@BernardMeurer That's true. That's one way to rewrite the thing

This doesn't help me much though, does it? The denominator is still different

@BernardMeurer $$1-\frac{1}{(n+1)!}+\frac{n+1}{(n+2)!}=1-\frac{n+2}{(n+2)\cdot (n+1)!}+\frac{n+1}{(n+2)!}$$

OH

FUCk

@BernardMeurer It works

JEEZ

The other way works too

WTF

Sigh, I need beer

Obvious

Just put $\frac{n+1}{n+2}$ as the numerator of the one

But yeah, what I said earlier about rewriting the other one was hinting at what Emilio said.

It's much faster

@Danu way faster

if I do say so myself

Well... Depends mostly what you're measuring by

Wait

Hmmm

Both take less than... 10 seconds in my head :P

Oh, okay

But the other way ends up using the same step as the one way anyways

Understood

Which is a waste

So @BernardMeurer If you're trying to prove equality of some sums by induction, this should be your go-to method

Okay I fucked up

Split the last term off

@Obliv Not an easy thing to troubleshoot. Just look at all the linked dupes on that meta.math.se thread.

@BernardMeurer signs

Oh god

WHAT IS HAPPENING

Deleted for stupidity

lmao

Relax man

yeah, chill

you just learn to see these ones

nothing but practice

Okay, I get this 100% now

makes perfect sense

thanks a lot guys!

@Obliv how many words are your lab reports

@JohnDuffield How do you feel about that?

@0celo7 I need to start making these

and they are a very significant portion of my grades

Experimental plebs :))))

::hides::

@DanielSank : a little sad that he felt the need to call for downvote collusion and banning, and that he then felt the need to leave.

Aw come on guys, do you have to keep on this?

@BernardMeurer Like 30% of mine

@DavidZ How do you feel about that? ;)

:-P

@Danu is $4^\sqrt{2}\in\mathbb{P}$ true?

$\mathbb{P} = \mathbb{R}\setminus\mathbb{Q}$

@BernardMeurer What is $\Bbb P$ supposed to mean? To me it usually means complex projective space

@BernardMeurer Just ask if it is irrational :P

Irrationals

@Danu There must be rigor :p

That's perfectly rigorous.

What you were doing was using needlessly complicated notation :P

Leave me and my symbols alone :p

But before you leave us answer the question :D

$\setminus$ is needlessly complicated

@Danu Any idea?

@BernardMeurer What have you been shown in class?

$\sqrt 2,\sqrt 3$ are irrational?

@0celo7 Yes

we didn't prove why though

we stated that they are, which I already knew

But you're supposed to prove this now?

Wait

It seems nontrivial.

I'm not sure how to prove it for hte record

Stupid me

@BernardMeurer What are you doing?

Nothing right now

I want to prove $4^\sqrt{2} \in \mathbb{R}\setminus\mathbb{Q}$

@BernardMeurer Proving that any non reducible square root is irrational is pretty straightforward. A concise and simple proof of \sqrt{2} is given here: math.utah.edu/%7Epa/math/q1.html

@QuantumBrick I think his current problem is much harder.

@BernardMeurer I have a question.

What does $4^{\sqrt 2}$ mean?