ITT: Veeky Forums conducts a scientific study. We shall find the link between the scores on these three tests...

Show me a picture of what you're talking about, and don't use the R word, it's not 2003 anymore.

>rational

Fuck off.

Sigh, why are you so aggressive?

bump. give me your data.

>guaranteed _replies_(Veeky Forums).jpg

>MBTI

Top kek, you may as well compare horoscopes.

Right off the website you retard.

I'm sorry, autistic.

Here you go.

Pfft

It's pronuncd "debator".

my hypothesis is those that take these tests are faggots

how can we test it experimentally?

So describing the MBTI => the site is MBTI? Because that's definitely not true. And please don't use the R word.

so because it uses acronyms, it's MBTI? That's definitely not true. OCEAN/CANOE (Also known as the Big 5) isn't MBTI.

Dammit dude you're fucking stupid
MBTI has four categories
E vs I
S vs N
F vs T
P vs J

Are you really, seriously, honestly fucking retarded enough to think "ENTP" would seriously stand for some other arbitrary psychological topology based on a series of 4 binary categories and not the MBTI category "Extroverted, Intuitive, Thinking, Perceiving"?

You've pretty much proven my first post right, you seem intolerable to be around

I do think that, and if you actually read about their theory, so might you. This is completely different than the MBTI.

Jesus fucking christ, kill yourself

>reads the theory
>knows he's dead wrong
>k-k-kill y-y-yourself. Haha lmfao XD.

so close!

god tier

i know right

also got INTJ and skeptic

samefag, nobody takes online iq tests seriously

Filename says it all. Some useless, but interesting formula
[eqn] 3^4 + 4^4 + 5^4 = 6^4 [/eqn]
[eqn] \sum_{k=1}^n k^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30} [/eqn]
[eqn] \sum_{A_{\ k}}^n \ \ \ \sum_{A_{\ k-1}}^{A_{\ k}} \ \ \ \sum_{A_{\ k-2}}^{A_{\ k-1}} \ \cdots \ \sum_{A_{\ 2}}^{A_{\ 3}} \ \ \ \sum_{A_{\ 1}}^{A_{\
2}} A_{\ 1} = \frac{n(n+1)(n+2)(n+3)\cdots(n+k)}{1 \cdot 2 \cdot 3 \cdots (k+1)} [/eqn]
where in the last term all sums begin at 1 and start from the rightmost sum and go in order to the leftmost

>start test
>first question
>which vowel
>close tab

op, fuck off

I'm really not sure what to make of this

Some amusing, but useless elementary equations I found messing around back in high school
[eqn] 3^3 + 4^3 + 5^3 = 6^3 [/eqn]
[eqn] \sum_{k=1}^n k^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30} [/eqn]
[eqn] \sum_{A_{\ k}}^n \ \ \ \sum_{A_{\ k-1}}^{A_{\ k}} \ \ \ \sum_{A_{\ k-2}}^{A_{\ k-1}} \ \cdots \ \sum_{A_{\ 2}}^{A_{\ 3}} \ \ \ \sum_{A_{\ 1}}^{A_{\ 2}} A_{\ 1} = \frac{n(n+1)(n+2)(n+3)\cdots(n+k)}{1 \cdot 2 \cdot 3 \cdots (k+1)} [/eqn]
where in the last term all sums begin at 1 and start from the rightmost sum and go in order to the leftmost

...

You're intelligent but irrational

>"how rational are you?"
>first question
>you're watching a shitty movie, do you leave half way when it obviously won't get better

Obviously they're checking for the Sunk Cost Fallacy, but I would stick with it so I can talk about the movie with other people.

cute formulas user, especially the last one. the answer is a binomial coefficient, which got me thinking that it probably has a combinatorial proof.

let Z denote the set of all partitions of n of size k+1. In other words, Z is the set of all k+1-tuples of non-negative integers which sum to n. by the stars-and-bars argument, Z has size[math]
{n+k\choose k}[/math]. we may consider Z to be a probability space by endowing it with the uniform measure.

define the random variables [math] X_j: Z\to N[/math] by
[math] X_j(p_1, \dots,p_{k+1})=p_j[/math]

we compute EX_1 in two different ways.

on the one hand, we may simply sum over all elements of Z directly, weighting each one with a factor of [math] 1/{n+k\choose k}[/math] to get

[math] EX_1=[/math](the sum on the left side of your formula)[math]/{n+k\choose k}[/math]

where we interpret A_j to be the sum of the first j elements of a given partition

on the other, clearly [math] \sum_{j=1}^{k+1} X_j=n[/math], which implies that [math] \sum_{j=1}^{k+1} EX_j=n[/math]

I claim that all X_j's have the same distribution. this is because we can switch any two entries in a given partition to get another partition. this gives a bijection between the set of all partitions with p_1=i and p_j=i, for any i and j.
in particular EX_j=EX_i for every j. so
[math] EX_1=n/k+1[/math]

equating the two expressions for the mean gives the formula

When I was in high school I found formulas for the sum of k^r, for r=4, 5, 6, and I think 7.

Basically you assume that the formula is a polynomial of degree one bigger than the power you're summing. Then you calculate several specific sums so that you can solve for the coefficients of the unknown polynomial. Then you finally prove that the formula works for all n by induction.

Not very elegant but I thought it was cool as fuck when I was 16.

that's pretty cool user. there's a general formula for any r called faulhaber's formula.

I appreciate this proof. Rarely do people solve problems for fun because they're so lazy. I have a proof of my own, but I forgot it. It had something to do with the analogy between this sum and a multiple integral.

Are you a graduate? I'm dropped out now, but I won a state math contest in high school and wouldn't mind discussing math with you in the future.

You probably don't have time, but lord knows it'd be an upgrade for me from this place. Most of the time I end up scouring Veeky Forums for hours just looking for interesting problems and proofs

if you're looking for interesting problems and proofs, there are lots of other sites way better than this one, like math stackexchange, quora, or mathoverflow

98% mensa
skeptic
INTP

this is retarded

Thanks, you probably changed my life. As a teenager I catapulted to advanced math classes and and almost graduated, but my life spiraled out of control because I'm bipolar.

At least this way I can start doing math again