/sqt/ Stupid Questions Thread

Well then what does 62*3-2*3^2-(62*2-2*2^2) have to do with anything? It's just f(3)-f(2)?

For some reason I thought it should be like the derivative.

>For some reason I thought it should be like the derivative.
It's not.

Hypothetically, what is the legality of writing a research paper about CP, and including links to actual CP in the paper? Strictly hypothetically of course.

thanks

cot(πz) has a pole of order 1 at 0, since it is cos/sin and 1/sin can easily be seen to have pole of order 1 at 0 (see its series expansion).
The denominator makes the order of the whole thing 2.
Therefore the residue at 0 is the limit of the derivative of z^2 f(z) as z goes to 0.

Now for the pole at -1. It's obvious that the pole of order 1, therefore the residue is the limit of (z+1) f(z) as z goes to -1.

actually my bad, the pole at -1 is of order 2, so you have to do the same as for 0

What are the implications of a vector space not necessarily being naturally isomorphic to its double dual in the infinite dimensional case?

Given that x,a,b are between -1 and 1, how do I find the maximum of |(x-a)(x-b)| ?
I have absolutely no clue.

x=1,a=b=-1
>how do I
split the terms and maximise both (semi)independently

|(x-a)(x-b)| =
|(x-a)||(x-b)|

Is their an approach for finding a bijection between [math]\mathbb{N}[/math] and some set [math]A[/math]? For instance, for [math]a,b \in \mathbb{R}[/math], [math]a

>Is their an approach for finding a bijection between N and some set A?
depends on the A

> For instance, for a,b∈R, a

if you don't deal with unbounded sets you can just use a linear approach
lets say f(x) = mx+n and you want f(0) = a and f(1) = b, you'll get a set of 2 equations.
0*m+n = a
1*m+n = b
which you can solve and get n=a, m = b-a, so
f(x) = (b-a)x + a
has the properties you want.
you only have to involve more complex functions, when you want to find a bijection from a subset of the reals that includes infinity

Ah, I see. Thank you.

I prefer looking at that formula this way:
α + (b-α)x
you start at α and add a scaled version of x by he length of (a,b) so that it can manage to reach b in time.

I'm trying to remember some proof from a while ago

Consider a hilbert space [math] \mathcal{H} [/math] and a linear operator [math]A [/math] on [math] \mathcal{H} [/math]. Lets say for some [math] b\in \mathcal{H} [/math] there exists an [math] x\in \mathcal{H} [/math] such that [math] \left \| Ax-b \right \|_\mathcal{H} [/math] is minimised.

I want to prove that [math] (Ax-b)~ \bot ~im(A)[/math] or [math] \left \langle Ax-b,Ay \right \rangle = 0~~ \forall y\in\mathcal{H}[/math]

I want to prove least squares on general hilbert spaces, so I cant use that [math] A^* (Ax-b) = 0 [/math]

Best Calculus textbook? I'm looking at Thomas Early Transcendentals.

Haven't studied functional analysis. but if the theorem in pic works for infinite dimensional Hillbert spaces, then it proves what you want, since the minimum would be achieved by the projection of [math] b [/math] to the subspace [math] Im(A) [/math] and [math] Ax-b [/math] .

>and [math] Ax-b [/math]
ignore that

|| A(x + t y) -b||^2 is quadratic in t and must vanish at t=0 for all y because it has a minimum there. This gives =0.

I mean the derivative must vanish.

that should do the trick, thanks.
I remember doing something like that in numerics, but isn't
[eqn]\left [\frac{\mathrm{d} }{\mathrm{d} t}|| A(x + t y) -b||^2 \right ]_{t=0}=2\cdot\mathfrak{Re}(\left \langle Ay,Ax-b \right \rangle) [/eqn]
which doesn't exactly give the result in complex hilbert spaces

>that should do the trick, thanks.
Are you sure though? Cause I googled and found this:
math.stackexchange.com/questions/275391/orthogonal-projection-on-the-hilbert-space
and an operator's image isn't always closed

ARE YOU SMART ENOUGH TO FIGURE OUT THE GREATEST PUZZLE OF ALL TIME?

oh you're right, the orthogonal projection doesn't exist if the subspace isn't closed.
I guess my question only really makes sense for operators with closed image, because you couldn't otherwise guarantee a minimum

I've got a BS in biology and 4 years experience as a technician in academia. Can I get a decent paying job in industry without any further education/certifcation bullshit? I really don't want to get a PhD, and I want to get a Master's even less since I feel like a lot of Master's degree programs are horseshit money grabs.

> 4 years experience as a technician in academia

Whytho

shit, fixed.

maximum domain of convergence for this laurent series, without using any convergence tests from calculus?

thought i was gonna go to grad school, got to last year of undergrad and decided fuck school. lab i was in offered me a job, so i took it. then another lab was moving and offered me another job for more money so i took that.

IMPORTANT SCIENTIFIC QUESTION:

How far out were they? Assume the following:

-The galaxy depicted is their home galaxy - the one far, far away.

-This spiral galaxy is equal in size to our own spiral galaxy, the Milky Way.

-The humanoids depicted in the films are 1-1 scale with real human beings (specifically, the actors who portray them), and aren't self-contained in sci-fi-fantasy logic where everything is really twice as big or five times as small as its earth-counterparts. So both Luke and Leia are approximately 170cm tall, give-and-take their true heights (that is, the actors' heights).

-Thus, that window is about the size of a big picture window in a nice living room on Earth, say roughly 3M x 5M.

>correlation isn't causation

How else do we prove causation other than doing a bunch of experiments and saying "well the two events correlated a whole bunch"?

natural science doesn't provide proofs

So I'm an idiot that never did much math in high school. I'm trying to learn some math now on my own time, and I have started to use Khan Academy but I've been quite confused.

There seems to be no way that I can see to do the World of Math mission without trudging through gradeschool math. I know I'm dumb, but I'm not exactly "can't count to 10" dumb.

How do I skip/complete that shit quickly?

Related to this, I'm trying to go through it with Mastery Challenge, and it'll give me like 90 gradeschool math problems then drop a quadratic equation in between "count the pictures of beetles" and "name this angle".

tl;dr how khan difficulty curve

Yes. Didn't think about complex spaces. If the inner product has no real part, it's pure imaginary, and you can replace y with i*y and get pure real, and so zero. Is that right?

Find all funcs G(t) such that u(x,y) = G(x^2+y^2) is harmonic, shid pls help :-DDDDDDD

Calculate the laplacian of u

What's the most recommended book on artificial intelligence, according to Veeky Forums? I am reading artificial intelligence a modern approach, but I also want to try my skills with Deep learning by Ian Goodfellow.

What do you think Veeky Forums? P.S. I would ask /g/, but they're retards.

are there any journals/textbooks for learning more about current thought about abiogenesis?

Basic statistics question:
I've got a sample exam, with 12 questions on it. 8 of these exact questions will be on the real exam. on the real exam, i have to answer 5 out of the 8. how many questions, of the 12 on the sample exam, can i choose to ignore?

Will AIs ever replace scientists?

is the real exam also 12 questions?

the real exam is 8 questions

QUICK

is this right

i'm too tired to think

nigga is you high

yyyyyyyyyyerteyru tert i'm gonna sleep thanks my dude

You can ignore 3. Then if you get the three you ignored, you can answer the other 5.
If you ignore more than 3, you might get all the questions you ignored on the exam.

If you want only a high probability of only getting questions you studied for you might be able to ignore more.

I'd say they were pretty far out, bro

For example if you only study 8 and neglect 4 then you have a 86% chance of being able to choose 5 questions you know.

what are all the domains where f(z) = |x^2-y^2| +2i|xy| is analytic? z = x+iy, ofc

Is the hairline pictured there normal for a 27 year old?

u(x,y) = |x^2-y^2| , v(x,y) = 2|xy|
check cauchy riemann conditions

that works, yes

let's check for absolute convergence.
The series from -infinity to 0 behaves differently then the series from 0 to infinity, so we split them up.

case n>0: We have
[math] \left | \frac{z^n}{3^n+1} \right |

Not really a stupid question, but don't feel like it deserves its own thread.

Does anyone know of any online accredited college courses on artificial intelligence? Not a whole major, just a single introductory class.

If I have a limit where the denominator is not going to be 0, can I do this?
[math] \lim_{x\rightarrow 0} \frac{f(x)}{x+1} =\frac{\lim_{x\rightarrow 0}f(x)}{\lim_{x\rightarrow 0}x+1} = \lim_{x\rightarrow 0} f(x)[/math]

en.wikipedia.org/wiki/Limit_of_a_function#Properties

yup
if [math] \lim_n a_n = a [/math] and [math] \lim_n b_n = b [/math] exist, [math] \lim_n \frac{a_n}{b_n} [/math] also exists and is equal to [math] \frac{a}{b} [/math]

Am I correct in saying that, given [math]A=\{1,2,3 \}, B = \{2,4,6 \}[/math], the subset of [math]A\times B[/math] [math]: C = \{ (1,2),(2,4) \}[/math], is not a function, since f(3) isn't defined?

I'd also like to know. Really bad at derivatives

it is a function, but it's a not a function from A to B

Go work for a magazine if you want that

In the context though, is it not obvious that the function's domain/codomain is A and B, since a function is defined as a subset of the Cartesian product.

> is it not obvious that the function's domain/codomain is A and B, since a function is defined as a subset of the Cartesian product.
the issue is that a set can be a function for different codomains, i.e. {(1,2),(2,4)} is a function from {1,2} to {2,4}, it's also a function from {1,2} to {2,4,6}, and in general it's a function from {1,2} to any set containing {2,4}. the set making up the function makes no reference to where the elements 'came from'

How does one show that the integral [math] \int_{1}^{\infty} \frac{1+e^{-x}}{x} [/math] diverges?

In probability, how do 3+ mutually exclusive events work?
From the description I've read, if it's for every event A1, A2 ... An, every two events cannot happen at the same time.
So if you have A1 A2 and A3, does that mean they're mutually exclusive if A1^A2, A1^A3 and A2^A3 = O?

Hi Veeky Forums, recently an interesting question come up while im studying combination and permutation. I can't get the right answer so I hope we can discuss about this.

There are 5 big beads, each of them is different and unlimited amount of 5 different types small beads.

How many ways can you arrange them into a necklace so that
-there are 5 any type of small beads between every big beads
-all big beads is used

So far, I've only find out that there must be more than 4! ways of arranging them since the 5 big beads is a circular permutation (5-1)!
Pic related, I will try to clarify some things if the question is not clear enough. I'm thinking of making a thread about this but this might be just a stupid question.

Thank you

Anyone familiar with any research groups focussed on finding higher-bandwidth methods of getting information to a brain?

I'm thinking the answer would be the number of way the big beads can be arranged in circular permutation multiplied by the number of ways the 5 different types of small beads to 25 places.
Any thought?

I think including a link to an actual CP content or forum or somekind is a form of sharing it so it may be against the law.

Man, you are so lazy. I immediately graphed your curve and noticed it looks like [math] \frac{1}{x} [/math] but bigger. So all you have to do now is prove that [math] \frac{1 + e^{-x}}{x} > \frac{1}{x} [/math] which is trivial because [math] 1 + e^{-x} > 1 [/math]. Then remember that integrals preserve inequalities so:[math] \int_{1}^{N} \frac{1 + e^{-x}}{x}dx > \int_{1}^{N} \frac{1}{x}dx [/math] and by comparison it diverges.

I hope you take this as a lesson. The difference between a good analyst and a bad analyst is that the good analyst has memorized hundreds of curves and knows immediately when to compare a curve with another. I recommend trying a similar problem now. Train your intuition. Rise above good man. Analysis is better than algebra or topology, everyone knows that.

yes

If [math]f^{-1}(f(x))=x[/math] does is follow that [math]f(f^{-1}(x))=x[/math]?

yes, by definition of an inverse function

Hi, I'd like to go to university and wonder which STEM field(s) to choose.
I'm OK with math, not particularly good but obviously willing to make an effort and learn.
I'm somewhat proficient in Biology and also interested, but the other fields are also cool.

Which field has the best job opportunities and what can I expect from each field?

Thanks in advance.

>STEM meme
unless you really enjoy it with a passion forget it. Ching zing Tang is already studing 25 hours a day and will out class you in everything.

Are you being serious? What am I supposed to do else, history? I don't want to do an apprenticeship to be honest, I did very well in school and want to learn more.

thanks!

from personal experience you have the chinks, poos, and literal 500IQ autists who will make you work like a fucking horse to keep up to the top echelon.

You can go into STEM but aim for getting lab time. If you can get involved in some research projects and have some experience in you CV by the time you finish your undergrad you will be just as viable as the academically top students, if not more so (hell, I got two separate offers for post-grad placements in my third year of undergrad from groups that I did some work with).

>Analysis is better than algebra or topology

scienceforums.net/topic/59514-calculating-the-luminance-of-the-sun/
>Sun subtends about 0.5 degree on earth surface solid angle Omega becomes = 2pi (1- cos (0.5/2)) = 5,98e-5 steradians
>At bright day light and at 90 degree normal incident, illuminance of sun at earth's surface is around 10.000 lux = lm/m2
>Luminance = illuminance / Omega = 10.000/5,89e-5 =1.67e9 cd/m2
how does this calculation make any sense??

how do you explain the wild discrepancy vs e.g.
>The average luminance B of the sun, observed outside the atmosphere, is a fixed quantity having a value of about 200 000 candles/cm2 . The values of B found in the literature are usually those calculated from the measurements of Bo, the only reference to its direct measurement being that by Teele, who compares his value of 190 000 candles/ cm2 calculated from the normal insolation observed during the 1935 stratosphere flight to "a directly measured value of 200 000 candles/cm2 reported by Worthing"
webcache.googleusercontent.com/search?q=cache:GRKFNJ2KT4MJ:https://www.osapublishing.org/viewmedia.cfm?id=51094&seq=0

fucking retarded sites like wikipedia and others just parrot the same 1.6 Gcd/m^2 without any explanation

schorsch.com/en/kbase/glossary/luminance.html

Is there a typo here or something? Why is the answer in terms of "N" if x is approaching positive infinity? I know the limit itself equals a negative value, but isn't that irrelevant? The definitions seem to imply that the behavior along the x-axis is what determines the answer.

Top portion of the picture is the definitions, middle is the problem, and bottom is the book's answer. Sorry about the shitty lighting.

A bit baffled by this one. It's derivates-related.

if x is bigger than N, then the difference between f(x) and it's limit at infinity is smaller than epsilon
basic definition of convergence

r'(x) = g'(x)*f'(g(x))
r'(1) = g'(1)*f'(g(1)) = g'(1)*f'(4) = 0 *5/4 = 0
now try it for s

>Why is the answer in terms of "N" if x is approaching positive infinity?
N is the number that x has to be greater than to guarantee that f(x) falls within epsilon of -3 man
Think about it, if f(x) has a limit at x going to positive infinity, you can always find some interval for x where f(x) is within an arbitrary fixed value of the limit. This problem is about finding N (and thus a satisfactory interval for x) as a function of the distance, epsilon, from -3, thus showing there's a satisfactory N for *every* distance from -3, proving the limit

g(1)=4
f(g(1))=f(4)
f'(4)=?

It's 5/4 because that's the slope. I'll solve s'(4) then. Thanks!