SQT Stupid Questions Thread

Can you post the entire thing in your image?

I am interested in the remark that says that for some reason whatever the author did shows that there are gaps in the rationals.

...

No, I meant everything before that.

The actual argument.

For k a natural number, if there exists an integer u such that [math]2^k + 1 = u^2[/math] then k must equal 3.

I can not for the life of me find how to prove this bullshit, what did I miss?

How do I integrate e^(x^2) ?

I tried wolfram alpha but it is giving me (1/2)(pi)^(1/2) erfi(x) + C .

I have no idea what "erfi(x)" is supposed to mean.

...

prove what?

That under those conditions k cannot have another value than 3.

exp(x2) is not an integrable function, i.e. you can't express its integral in terms of elementary functions. Wolfram Alpha is expressing it in terms of the error function, which you'll see is defined explicitly with this integral. However, if you wanted to compute a definite integral like the one that ranges from -∞ to ∞, you could.

The problem is that e^(x2) is non-integrable, i.e. you can't express its integral in terms of elementary functions. Wolfram Alpha is outputting your "answer" in terms of the error function, which is really just a special functioned that's DEFINED in terms of this undoable integral. However, if you wanted to compute certain definite integrals like one that ranges from -∞ to ∞, you could get a value.

don't mind my autism

Really dumb question, but I am unsure how to enter in the direction here?

The angle has to be below the x-axis, so I will put 36 degrees correct?

if science is accurate, then why falsify?

I know you're not supposed to put ice directly on skin because it causes frost bite, but what if your ice pack has a bunch of condensation on the outside, can that also cause frost bite?

anyone???

i guess

just download the solutions manual and check

i don't have it :(, and i don't want to get it wrong. do you think it is 36 degrees? here is example from textbook

I am just not sure if it is negative because problem specifically asks for below x axis, while other one was just relative to x axis

just put it positive nigger, it has already specified that it's under the x axis

No, it would be -36 deg or 360-36=324 deg. I'd go with the - if I were you

Feynman's personality as I know it makes me think this might be an actual quote

actually this is a very common complaint with rudin

math.berkeley.edu/~gbergman/ug.hndts/m104_Rudin_notes.pdf

I put 36 and it accepted it, thanks anons!

Why is it not integrable? We can differentiate it but not integrate?

it's integrable. it just so happens that you can't express the integral with elementary functions

What does that mean?

it means that the equation
$ dy = e^(x^2)$
has no solutions in the differential ring generated by $\mathbb{R}$, polynomials in x, rational functions in x, exponentials in x, logarithms in x.

wtf did I do in this problem? any anons know?

Anyone know how to evaluate a base 2 logarithm without a calculator? Heres the question

Suppose you have 2 points A and B that are a distance apart on an infinite plane. You have a small ruler that can only reach about 1/3 of the way there. How do you draw a line between the two points?

Honestly, I'm impressed by anyone that uses Rudin for their first introduction to analysis. I mean, I can read the bit in OP's pic and understand why he chose q how he did because I've already taken real analysis. But if you'd given that to me before I'd done any analysis, with no motivation at all, I'd be completely lost. It would seem as if he'd just pulled an equation out of his ass.
To me, Rudin's analysis textbooks seem to assume a level of mathematical maturity that usually hasn't developed in students studying the level of material they cover. Maybe things were different in his day. Maybe students who had to go through calculus when it was taught more rigorously wouldn't have found it unusual.

You don't need to evaluate it. Think about what the question is asking. What is the greatest integer you can raise 2 to and get an answer less than 22? What is the least integer you can raise 2 to and get an answer greater than 22?

Actual stupid question here. Been staring at it for an hour.

write as 2 integrals then try:
>substitution
>partial fraction decomposition

separate into [math]\int \frac{u^8}{u^2+1}[/math] and [math]\int \frac{u^5}{u^2+1}[/math] and do long division

2^k = (u-1)(u+1)
tells us that both u+1 and u-1 must be powers of 2. what powers of 2 differ by 2?

never really understood this... what's an "elementary function"? why is erf() less elementary than cos()?

im so fucking dumb. i have a midterm on friday and i cant even do this. please help

you need an arccos() in there?

That's what I tried. Stuck on where to go with
[tex]\int \frac{u^8}{u^2+1}[/tex]. Is there a way without doing long division?

...

write the integrand as

u^6-u^4+u^3+u^2-u-1 + (u+1)/(u^2+1)

the first terms are easy, the last looks pretty straightforward too.

forgot: b/c theres a remainder of one you need to make sure to add 1/(u^2+1

might as well keep the x^5 in there too, no need to split if you're going to use long division

Or just do long division directly
[eqn]\frac{u^8+u^5}{u^2+1} = x^{6}-x^{4}+x^{3}+x^{2}-x-1 + \frac{x+1}{x^2+1}[/eqn]

yeah, dont know why i didnt just say that. from there finding the integral is trivial

demonstrations.wolfram.com/AnEfficientTestForAPointToBeInAConvexPolygon/

>So the total costs for the test are just two additions (for the initial origin translation), two multiplications, one subtraction, and one "greater than zero" comparison for every vertex; finally an n-fold equality comparison if all the signs of the angles are equal.
>finally an n-fold equality comparison if all the signs of the angles are equal.

I don't get it, where does the last part come from? Is it not enough with just
X_(i + 1) * Y_(i) - X_(i) * Y_(i + 1) > 0

Someone explain why [math](-a)(-b) = ab[/math]. I understand that our algebra basically forces us to do so, but are there any explanations that make sense?

5 * 1 = 5
5 * -1 = -5
-5 * 1 = -5
-5 * -1 = 5

No shit?

[math]No shit?[/math]

Because -(-a) = a and -a = (-1)a.

In more detail, we have (-a)(-b) = (-1)(a)(-1)(b) = (-1)(-1)(ab) = -(-1)(ab) = (1))(ab) = ab.

For calculating the error bounds of Trapezoidal Approximation, am I allowed to pick any arbitrary [math]K[/math] as long as [math]f``(x) \leq K[/math]? It seems like some of the solution manuals are picking wildly random values of [math]K[/math] for the calculation, unless I'm getting my local maximums wrong.

after a certain point, higher category theory starts to feel almost geometric in nature, and everything starts to make sense
the so-called "intuitive" parts of category theory finally actually become intuitive
any sort of connection between two things immediately sets of bells in your mind
you start to feel like reality is just the macroscopic manifestation of some braided, knotted, or twisted objects
you feel like you ~are~ the categories
it's like autism except worse
but possibly the worst part is that even though you've sacrificed your humanity for mathematics, there are still people who understand it better than you

if you make that |f''(x)|, sure.

I tried it without the equality and it works although I had to flip it around to X_(i) * Y_(i + 1) - X_(i + 1) * Y_(i) > 0 with counterclockwise vertices

how to solve 0.6125 = sin2x. using R = Vsin(2x)/g in physics and I'm supposed to find the angle.

arcsin

Is this text book by Rudin?

what happens to the 2 in the sin2x? arcsin(2*0.615)?

arcsin "cancels out" sin

apply to both sides:

arcsin(0.6125) = arcsin(sin(2x))

arcsin(0.6125) = 2x

arcsin(0.6125) / 2 = x

thanks!

How do I find a list of numbers that are NOT divisible by something in wolfram alpha?

Trying to find even numbers between 100-300 that are not divisible by 4. Wolfram alpha does not accept that phrasing. Best I can do is divisible by

never used it before (lol) but maybe try putting ! infront, in programming != means no equal so maybe that'll work

you could try doing it programmatically with some wolfram language syntax, like generate a list of even numbers and remove the ones that aren't divisible by 4

q = p - (p^2 - 2)/(p+2) = (p - sqrt(2))*(p+sqrt(2))/(p+2). If p^2 < 2, then the equation can be rewritten as q = p + (sqrt(2) - p)*(p + sqrt(2))/(p + 2). Clearly, the sqrt(2) - p term is meant to get p closer to sqrt(2), but the author doesn't want that, he wants to only cover a part of the distance. That is why he scales sqrt(2) - p down by (sqrt(2)+p)/(2+p).

How do I find the inverse (if it exists) for this kind of function?
f(a) = 1
f(b) = 2
f(c) = 4
f(d) = 3

Also, why is this not a function from R to R?
Goodness, if my university is going to introduce new concepts to me that are not easily researchable, it shouldn't be in the form of a question!

Because if R was the domain, then x=-2.9 would have to be in it, but that's rules out by the requirement x>1.
The domain and codomain of that functions are some open connected subsets of R.
(Besides, identifying a function with its model as a set of pairs is a little Plebeian an rough, but that's more a philosophical and educational point.

Why does -2.9 specifically have to be in the domain?

Is the function from {a,b,c,d} to {1,2,3,4}? Then if g is the inverse of f, you want g(f(x))=x for all x in {a,b,c,d} and f(g(y))=y for all y in {1,2,3,4}

So g defined as g(1)=a, g(2)=b, g(3)=d, g(4)=c is your inverse

You chose the wrong major

They seriously don't explain most of this at my university, it's a joke.
It's considered lucky if they can deliver exam papers without fucking something up.

-2.9 is a real number... it's part of R...

I'm aware, but why -2.9 specifically?

But you're also lazy.
You cab look up the definion of the domain and codomain of a function, the word I used.
You ask about why f doesn't go from R to R. The answer is that if you requre y="something of x" when x>1, then there is no information about what the value y=f(-6.353) is. Thus the function doesn't go from R somewhere. Much of R is not in its domain.

it was an example

It's a generic real. One over pi to the power of 5 also works as value that's not in the domain.

Again, take a notebook and write down some definitons you can come back to. Do it.

Ah, I suspected that was how it worked but it seemed overly trivial.
Kinda cheeky, making all the assignments hard then giving one where every single answer is trivial.

Wait, actually, if y is the codomain why does it matter that that there is no corresponding f(x)?
Or is the problem that f(x) is supposed to include even explicitly ruled out values of x?

it is well past the time for you to learn how to program

If f and g are functions, what does
g o f mean?
I'm using o in place of a circle symbol, it's not the letter o.

Darn, I guess that user is gone.

en.wikipedia.org/wiki/Function_composition

Given 2 particles "A" and "B", does observing "A" collapse the wave function of "B" if B itself has not been observed yet?

I'm asking in regards to "quantum communication", with the following scenario:

If we have two "stations" an arbitrary distance apart (let's just say 100 miles for arguments sake) and place a photon gun right in the middle at the 50 mile point. It creates sets of entangled photons and sends one to "Station A" and one to "Station B". It sends as many photons in either direction as needed to make a decent enough interference pattern (let's just say 1000 entangled photon pairs, but whatever).

"Station A" would have an on/off "observer switch" that could choose to either collapse the wave function of the incoming 1000 photons, or not observe them and keep the wave function in tact. "Station B" would then see the intended result of Station A when it receives its photon and is sent through a double slit. If Station B observes an interference pattern, it means Station A had their "observer switch" off and this could be interpreted as a binary "bit 0". If there's no interference pattern, it means it was observed at station A and this could be interpreted as binary "bit 1". After a pre-set time at both stations and photon gun, the bit is recorded and then another 1000 entangled photon pairs are sent.

Is there a reason why this wouldn't work, or why this wouldn't be classified as "Quantum communication"?

Yes, observing A collapses B.

Look at it this way:
If you split apart a positive and negative particle, sending one to A and one to B, until observed you could have either the positive particle or the negative particle arriving at B.
But if you receive a positive particle at A, there's zero chance that the particle at B will also be positive, no matter how long you wait before observing B.

As for your proposed quantum communication however, that's a lot more complicated.
Observing particle A will not make the location of B 100% certain as there's lots of ways for quantum uncertainty to creep in, but the question is if it can be even a smidgen more certain than it would be otherwise, since one could make a double slit which favours particles that have especially uncertain location.

Thanks for your help user, it was nice to feel like I was onto something for a few moments kek.

Some pretty basic number theory.

Prove that no matter what value of n you start with you will eventually reach 1 with successive applications of the following procedure:

Given any positive integer n, if n is even divide it by 2, if n is odd multiply it by 3 and add 1.

This is too trivial, even for this thread :^)

Prove that hailstone numbers always go down to 1?
Pretty sure that there's only empirical evidence for that so far.

stop wasting our time with easy questions

do polynominal division to simplify the expression into P(u) + (Au + B)/(u^2 + 1), then break (Au+B)/(u^2 + 1) into (A/2)*(2u)/(u^2 + 1) + B/(u^2 + 1) (because 2udu = d(u^2 + 1)). Clearly, these three integrals can be solved with the power rules, a natural logarithm, and an arctangent.