Math thread! Come in and have some pi...

[Izrael]

First rewrite








In contrast to what we started with, the resulting expression is continuous at t=0. So the answer is -5/2.

or much quicker for non noobs (i develop a lot) (i'm not calling you a noob of course, i just give an other method)

1/sqrt(1+t)=1-t/2+o(t) at t=0

so

1/sqrt(1+t)-1~-t/2 at t=0

(5/t)[1/sqrt(1+t)-1]~-5/2 at t=0

as a consequence (5/t)[1/sqrt(1+t)-1] -> -5/2 at t=0

[Izrael]

you're right

After years using it (since first year in college) i forgot that it is not taught in high school.

[Shutuponions]

First rewrite








In contrast to what we started with, the resulting expression is continuous at t=0. So the answer is -5/2.

Hi thanks for the response, I like how you pulled out the 5/t in the first step. Then at the 2nd step you find a common denominator so you got sqrt(1+t). Then what did you do next?

[ctownballer04]

Hi thanks for the response, I like how you pulled out the 5/t in the first step. Then at the 2nd step you find a common denominator so you got sqrt(1+t). Then what did you do next?

3rd step: He pulled out 1/sqrt(1+t) so (5/t)*1/sqrt(1+t)= 5/t*sqrt(1+t). This is how we get 5/(t*sqrt(1+t))*(1-sqrt(1+t))
4th step: He multiplied by the conjugate of 1-sqrt(1+t), which is 1+sqrt(1+t).

[Izrael]

This exercise is useful to learn some basic technics to beat indeterminate forms:

-Always find a common denominator
-Always try to multiply by the conjugate when square roots

There are more powerfull tools (limited development or L'Hôpital's rule) but these reflexes are to be mastered.

[MrJensenn]

I agree. Might just write something like this

Since any number N can be expressed by



Where each a_i belongs to {0,1}. It is self evident that a right-shift by k places results in (assuming k<n)

If I am not 100% idiot this is correct, I think:


Just a little change by doing a_0*2^0. Idk if that should be or not.

[NephilimRising]

http://web.iitd.ac.in/~pmvs/courses/...x-notation.pdf

can anyone tell me why on the top of page 5 there is (xj xj ) in the denominator? Why not use the same index as at the top, namely i?

[CountryPunk]

MiscMathematician what are you teaching this semester

[CountryPunk]

I'm not unfortunately. Next semester I'll probably get a couple gigs

Aw ****, what are you gonna do for money then? Tutor? NSA?

[charity4thepoor]

Aw ****, what are you gonna do for money then? Tutor? NSA?

g4p

[CountryPunk]

g4p

what is g4p, is it some sort of MMA fighting association

[CountryPunk]

Yes tutoring if I cant find anything else, but it pays chit. I may have to look at private tutoring.

ehehehehehe "private" tutoring ehehehehehe

[Plyoz]

As much as I dislike straight math, as soon as we start talking about boring subjective chit in philosophy class I wish I was doing Cal 2 homework. Starting to love working problems way more than reading and discussing even if it is way harder

[CountryPunk]

MiscMathematician what is the hardest grad course you've ever taken

[CRyan64]

As much as I dislike straight math, as soon as we start talking about boring subjective chit in philosophy class I wish I was doing Cal 2 homework. Starting to love working problems way more than reading and discussing even if it is way harder

lolz, what are yall studying in your phil class?

[NephilimRising]

So my "Mathematical methods for physicists class" ****ing sucks. Very superficial treatment.

Also there is this fukking Russian Boy Genius who already knows everything and who won't stfu. He makes it impossible for anyone else to learn or engage.

I want to tie a knife to my dick and **** him in his face. I hope he gets aids.

[juiceboxx03]

Also there is this fukking Russian Boy Genius who already knows everything and who won't stfu. He makes it impossible for anyone else to learn or engage.

Had a guy like that in multivar calc, brb can't progress in lecture because he keeps asking questions that weren't really even questions, just loudly confirming that he already knew it.

Diff eq is going well so far, linear algebra not so much

[CountryPunk]

I hate the kids in my class who failed the first time and are re-taking it and always speak up like they know everything already

[Plyoz]

lolz, what are yall studying in your phil class?

It's like Environmental Ehtics/Philosophy

Yesterday we talked about how Christians were to blame for ruining the environment. Needless to say things heated up pretty fast and I had some lols

[NephilimRising]

Had a guy like that in multivar calc, brb can't progress in lecture because he keeps asking questions that weren't really even questions, just loudly confirming that he already knew it.

Diff eq is going well so far, linear algebra not so much

Let me shed some linear algebra insight on you.

Linear Algebra is about shedding geometrical intuition and abstracting the idea of a vector space. Linear Transformations are functions that map from one vector space to another. It's not really a class about matrices, but matrices are an immensely useful tool to approach the study of vector spaces with.

I hate the kids in my class who failed the first time and are re-taking it and always speak up like they know everything already

This kid is not one of those kids. He's a boy genius who already knows most of the stuff since he's somewhat of a prodigy. No one else can get a word in, he's a kunt.

[turkey_server]

This kid is not one of those kids. He's a boy genius who already knows most of the stuff since he's somewhat of a prodigy. No one else can get a word in, he's a kunt.

Doubt it. The real prodigies are well known by their professors, and get told from day 1 not to come to class, and just to stay home and read a book. Srs.

[CountryPunk]

Doubt it. The real prodigies are well known by their professors, and get told from day 1 not to come to class, and just to stay home and read a book. Srs.

I'm sure "prodigy" might've been a slight exaggeration on his part (the undergrad "geniuses" at my school are like 15 years old and taking only graduate-level courses, etc.). The kid probably just studied ahead or learns rather quickly at an above-average level.

[CRyan64]

It's like Environmental Ehtics/Philosophy

Yesterday we talked about how Christians were to blame for ruining the environment. Needless to say things heated up pretty fast and I had some lols

lolz. I like philosophy...but ethics ain't a branch that I care for.

[BigDardHick]

Can someone help me with these two review problems from my calc homework?


These two are the only ones that I haven't been able to do. Measlies for help

[BigDardHick]

^^ I did not read (1.B) correctly. He carries the water-filled bucket slower now.

Noticce that until he reaches the river,

time(to river) = distance/rate = sqrt(x^2+50^2)/5

time(from river) = distance/rate = sqrt([500-x]^2 + 200^2)/3

minimize time t(x) = time1+time2 = sqrt(x^2+50^2)/5 + sqrt([500-x]^2 + 200^2)/3

you can do it this same way for part A

Thanks man, I guess I should've mentioned that I was struggling with the setup for both of them. I got them now with your help.

[CountryPunk]

Thanks man, I guess I should've mentioned that I was struggling with the setup for both of them. I got them now with your help.

You have a great username btw

[hitah0]

any brahs wanna help with these pde problems?




I am not sure how to start, any input would be great

[2445ricardo]

Abstract algebra brahs, I need some help with a couple of questions.

1.) Let f be a function from S to T, A contained in S, B contained in T, and {Bi}I is a collection of subsets of T. Show that:
a.) B is contained in f(f^-1(B)), and B = f(f^-1(B)) if and only if B is contained in the image of f
b.) f^-1(UI Bi) = UI f^-1(Bi)

Note: UI Bi = Union of the sets Bi in {Bi}I (didn't know what other way to type it.)

2.) Let f be a function from S to T and let g be a function from T to S so that g composed with f is injective and f composed with g is surjective. Show that both compositions are bijections.


Pls help.

[2445ricardo]

Let S={1,2}, B=T={a,b}, f(1) = f(2) = a. Then f(f^{-1}(B)) = {a} and B certainly is not contained inside it. Question is flawed

I'm sorry, the question isn't flawed, I just typed it incorrectly. I meant to say that f(f^-{1}(B)) is contained in B, and B = f(f^-1(B)) if and only if B is contained in the image of f.

[2445ricardo]

Let P = gof, Q=fog. If P is injective, then we also have that f is injective, since f(a)=f(b) => P(a)=P(b)=> a=b. By a similar argument (definitions), if Q is surjective, then so is f.

Now notice this means f is a bijection between S and T. Hence Q(x)=Q(y) => g(x)=g(y). Since x and y are both in T, there exist unique a and b such that f(a)=x and f(b)=y. So g(x)=g(y) => g(f(a)) = g(f(b)) => P(a) = P(b) => a=b. Hence x=y. So Q is also injective (therefore bijective).

Can you show it for P? Note since Q is injective, so is g.

Well since we know now that g is injective, I'm assuming that if I can show that g is surjective, I can just use a similar argument to the one you used for Q?

Okay, but first, applying a function always results in (at best) retaining information. Functions will usually lose information. Functions which lose no assignment information are called injective (knowing the gun and the target, you know the bullet that hit it). Functions which are surjective mean you can pick any target and find at least one bullet that hit it. The best functions are both, every target is hit by one and only one bullet. That is why they are special names. Applying f^{-1} to an element may result in nothing (empty set), since maybe f never sends something to that element. That is the point of this exercise.

To make it easier, lets let A=f^{-1}(B). If y belongs to f(A) then there is something in A, say x, such that f(x)=y. That is, x belongs to A=f^{-1}(B). By definition, this means f(x) belongs to B. But f(x)=y. So y belongs to B.

So what if they were equal? Well, since f^{-1}(B) is a subset of S, B=f(f^{-1}(B)) is a subset of f(S), which is of course the image of f.

This makes a lot more sense now.

Conversely, if they were not equal, then there must exist some element y in B that does not belong to f(f^{-1}(B)). Suppose f^{-1}(y) was not empty, there exists some x in S with f(x)=y. But since y belongs to B, x belongs to f^{-1}(B) by definition. Hence y=f(x) belongs to f(f^{-1}(B)), contradiction. Therefore f^{-1}(y) is empty, so y does not belong to f(S).

But how does this imply that the not equal case holds iff B is a subset of the Image of F?

Once again, thanks for all your help.