/sqt/ - Stupid Question Thread

yes

are you lying?

no

Point set topology is awfully dry.

I've been reading and studying math daily for 3 months. I make sure I go through anywhere between 10% of a book to 50 pages of a book a day(which ever is greater). I now find myself dreaming in math and constantly counting things in my head. I get upset and feel old if I make a mistake. Am I actually learning or am I only deluding myself? I want to use my math to earn money either through the internet or by automating something like a food cart.

k thx

Well, can you go back to previous sections and solve problems? I would define that as having learned if so. I can't say how it necassarily relates to your goals though, perhaps a more CS orientated route may be more applicable than strictly math.

I feel like we're closer to solving all the millennium problems than we are to inventing a strong AI.

>automate every single job in the world, requiring either the abolishment of money altogether
this is why I can't wait for everyone to loose their jobs to automation

>I want to use my math to earn money
Stopped reading there.

Surface integrals of vector fields give you the flux of the vector field through the surface.

Surface integrals of f(x,y,z) = 1 give you the surface area of the bounded object you're integrating over.

What do surface integrals of z=f(x,y), a scalar function, give you? My book literally does not tell me what it quantity they calculate, and afaik neither does wikipedia. what the fuck

How can a process have a nonconstant mean?

If its distribution changed over time

>I make sure I go through anywhere between 10% of a book to 50 pages of a book a day(which ever is greater)
The fact that you can do this raises serious red flags about how much you're actually retaining. People studying full-time don't go through anything near 50 pages of material a day. Doing that in a week is a very solid pace.

What the other user said is a good idea though. If you can solve problems you studied a month ago, you probably actually learned that material.

How is quantum mechanics a random process? couldn't it be that we just don't know what the outcome will be, or that it's simply hard to predict?

do you not understand what "random process" means?

not that guy but I always hate the term "random process" because people never understand it

but I also feel like "Stochastic process" doesn't sound good either

Well that's why I asked?

How would I show the exponential of a diagonal matrix converges?

"random process" does not imply an absence of ability to predict

en.wikipedia.org/wiki/Stochastic_process

idk but probably just show each diagonal entry converges

are they playing magic the gathering? I quit playing recently and I kind of miss it already...

why don't we just use a symbol for 1/0 instead of just saying it's undefined?

for example, if we defined 1/0 as z then couldn't we say that 5/0 = 5z?

en.wikipedia.org/wiki/Wheel_theory

We do. 1/0=infinity = north pole of Riemann Sphere

o wow i see it now; it was a stupid question

Having trouble with a problem on ODEs

Uploaded the problem as a pdf, but here's the latex (Not sure if the formatting will get screwed up, but here goes nothing)
Given the following theorem for First order differential equations:

Given the following theorem for First order differential equations:

In an equation of the form
[math]
P(x,y)dx + Q(x,y) dy = 0[/math]

The solution is in the form:
[math]\\
\frac{1}{F} \frac{dF}{dx} = R
[\math]

where
[math]
R = \frac{1}{Q} * \big( \frac{\partial P}{\partial y} - \frac{\partial Q} {\partial x}\big)
[\math]

And the problem is:
[math]
\sin (x)\cos (y)dx + \cos (x)\sin (y) dy = 0
[\math]

So...
[math] R = \frac{1}{\cos x\sin y} \big( -\sin x\sin y + \sin x\sin y
\big)[/math]
which is zero,
\\Because [math]\frac{\partial P}{\partial y} = -\sin y \cos
x[/math] and [math]\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} [/math]\\
\\Am I making a mistake?

PS: The solution at the back of the book is given as:
[math]y
y = arccos(c/\cos (x))
[\math]
\end{document}

Given how they work, I'd say yes. The smallest I could find on Google is 0.01 kt.

A surface integral of some vector field dotted with the unit normal vector is flux. A surface integral of some scalar function z on some surface is the sum of the values of z on that surface. For instance if z represents the mass density of some surface as its x,y coordinates change then the surface integral of z on the surface is the total mass on the surface.

>Is it sufficient to say that for all x, there is some y such that y=x, so the condition that for all x, there exist (x,x) is clearly met, and it is therefore an equivalence relation?

Obviously not. What you would have needed to say is that because
1) every element of A maps onto some other element of A, so for all x in A, R contains (x,y) for some y of A
2) R is symmetric, so (x,y) in R implies (y,x) in R
3) R is transitive, so (x,y) in R and (y,x) in R imply (x,x) in R

i agree, but the powers that be probably dont just yet which is why the singularity probably wont happen for a long time, if ever.

Basically i started studying CS even though im a complete noob.
We doing Haskell now.
I have to write a funktion "berechneNote k b" that sums up two Floats k = "Klausurpunkte" and b = "Bonuspunkte" and gives out a Float. Let the german words not confuse you.
So far i managed to do all that since it's the basics i remember from playing with Haskell for 2 days a year ago.
However if the value of k > 100 it should give you an error message like "more than 100 is imposible" and if b > 20 a different error message. This i cannot do anymore since i just dont have an example for this eccact problem.
I just don't understand where to put "if k > 100" etc...

If possible give a solution with explenation on the syntax.

Why does he pose with the buttcrack?

F isn't really a solution to the differential equation. It's an "integrating factor" - a function which can be used as a scaling factor for the original differential equation so that it becomes exact: [math] \displaystyle \frac{ \partial FP }{ \partial y} = \frac{ \partial FQ }{ \partial x} [/math].

cliffsnotes.com/study-guides/differential-equations/first-order-equations/integrating-factors

For all x there is a y such that xRy, but R is symmetric, so yRx. Use transitivity to conclude that xRx.

Just started an intro to probability book and I'm a little confused about notation and i want to make sure I have all of the terminology and concepts ironed out properly.

So if we have a sample space S, a random variable, say X:S->R (call the image X(S)), and a pdf, call it p, it makes sense that p:X(S) -> [0, 1] and when we want to talk about the probability X assumes a particular value we write p(x). Then my book starts using capital P(X = x) out of no where and doesn't really explain what P is. So what is the difference between p and P here?

I still don't really understand how to solve it though... Shouldn't the integrating factor be zero because R is zero?

At least in my stats class, lowercase probabilities were PDFs and Upper cases were CDFs (Cummulative Distribution Function)

ie: In a fair dice roll, f(3) = 1/6
F(3) = f(1) + f(2) + f(3) = 1/2

[eqn]p(x) = P(X = x) = P(X^{-1}(\{x\})) = P\left( \left\{ \omega \in S \middle| X(\omega) = x \right\} \right) [/eqn]

>pdf
Are you working with discrete random variables? If so then the terminology "probability mass function" for p is more accurate.
Most intro books won't be so pedantic, but when you get into continuous random variables (e.g., height or weight, which can take values to arbitrary decimal precision) the analogous "probability density function" -- which is usually then written as f(x) rather than p(x) -- can take values greater than 1. This doesn't violate the laws of probability, but I'll leave it to your book to explain why.

Otherwise, so long as your random variable X is discrete, P(X=x) is basically shorthand for p(x) and they have the same meaning. But when X is continuous you'll need to use something like [math]P(X\leq x) = \int_{t \leq x} p(t) \,\mathrm{d}t [/math] instead.

Sidenote: Even when X is discrete, you can choose to specify its distribution in this "cumulative" form instead, since you can convert between the two using [math]P(X\leq x) = \sum_{t \leq x}p(t)[/math] and [math]P(X=x) = P(X\leq x) - P(X < x)[/math], but when X is continuous the cumulative form is the only one you can use.

Is there an easy way to remove calcium from a solution I need to drink? Calcium ascorbate + broc powder + water and I dont want the calcium.

Does anyone have a reference for a calculus 3 practice set of problems that is just about functions from [math] \mathbb{R} [/math] to [math] \mathbb{R}^n [/math] with solutions but preferably full procedures at the back?

I have a test about these functions next week and I'd like to practice while being able to compare my procedure with the answers. I already exhausted Paul's Math Notes.

h-hi
just got into series for the first time in calc 2
does pic related converge or diverge and help me solve it pls?
I know comparison test, lim comp test, integral test, root test and ratio test
idk anything about absolutely conv or div yet
I tried using comp test with 1/k (as a p series would be easy if it worked) but it was one of those cases that didnt conclude anything

[math] \sum_{n=1}^{\infty} \frac{2 + cosk}{k} = \sum_{n=1}^{\infty} \frac{2}{k}
+ \sum_{n=1}^{\infty} \frac{cosk}{k} = 2\sum_{n=1}^{\infty} \frac{1}{k}
+ \sum_{n=1}^{\infty} \frac{cosk}{k}[/math]

So this is a series that has all of its partial sums bigger than the equivalent partial sum for the harmonic series.

Now comparison test and you are done.

Wait no no. I am fucking retarded.

Notice that cosine of k always varies inside [1,-1]

So at best the numerator becomes 2 + 1 = 3
and at worst it becomes 2 - 1 = 1

So this sum's partial sums are always larger than or equal to the equivalent partial sum of the harmonic series.

Say I have the usual ring axioms, except the axiom about inverses only mentions the right inverse. Is it still possible to prove the cancellation law?

Not Veeky Forums here
Where should be one's eye located to see a real image
youtube.com/watch?v=h6NqaDxi2KY

It's possible but it never says where one's eyes should be

if the harmonic series is bigger and we know it diverges doesnt that mean we cant use the comp test since the comp test says if the biggest converges so does the smallest and if the smallest diverges so does the biggest?

I proved the harmonic series is smaller than this series.

If the smallest diverges so does the biggest.

When finding the derivative using the definition of a derivative, is it ok if there is an "h" left over in the end?

no because h disappears when you take the limit

o-ok thanks senpai

The problem didn't have a limit though
It just asked to find the derivative

the definition of derivative is a limit

Yeah because the h vanishes.

If we have a random process X we know to be wide sense stationary, is |X| WSS as well?

I know the mean would still be constant but I'm not sure about the autocorrelation

Depends on the item and manufacturer. I know that at Scania here in the Netherlands, were the taktime is a few minutes (every few minutes a fully functional truck is made), they do test every single one. Probably not every little aspect but they sure as hell check is they drive because they are often driven to the customer after.

Why is the answer u(t-1)e^(-t+1) - u(t-3)e^(-t+3)
and not u(t-1)[1 - e^(-t+1)] - u(t-3)[1- e^(-t+3)]?
If I use convolution, my bounds are from 1 to t and from 3 to t
When I plug in t in the integral I get e^0 which gives 1.

what did you get instead of y(t)=s(t-1)-s(t-3)?

Why are you doing an integral? You're given the step response and step inputs

Oh so the h = 0

I understand the answer is easy to come to just by looking at it but shouldn't I be able to use convolution to solve it too? I guess by step response they don't mean impulse response

No they obviously don't

Step response is the result of the convolution of u(t) and the impulse response

suppose [math]\mathcal{R}[/math] is an equivalence relation on a set X. then [math]\mathcal{R}[/math] is an equivalence relation on [math]X'\subseteq X[/math] right? or is there a counterexample

plz go see a real doctor and have your blood tested, it costs less than 100 dollar

it's only cropped like that

no. also you can't overdose on calcium.

my blood's tested every 2 months when i donate blood so i know i have enough iron but i only know that once they prick my finger to test

are there really none that can be tested without 100 dollars? it's something i'd want to be able to check sort of regularly, i'm just a grad student who gets into shit diets sometimes so it'd be nice to be able to know what i'm missing

>suppose R is an equivalence relation on a set X. then R is an equivalence relation on X′⊆X right? or is there a counterexample
just think about it

stop donating blood. it's enough when you do once per five years and only 5 times in your life.
how shitty is your diet? i'm almost sure you dont miss anything.

How do I do GR?

I learn how to calculate things like covariant derivatives and play around with index notation and metrics but none of it means anything to me.

I have no idea what I am doing

i assumed it was true because the [math]\forall[/math] conditions are inherited by the subset. is that right?

If [math]\mathcal{R}[/math] is an equivalence relation on a set [math]X[/math], then the sets [math][x]_\mathcal{R}=\{ y\in X\ |\ x\mathcal{R}y\}[/math] partition [math]X[/math]. Now, restricting to a set [math]X'\subset X[/math], one gets a partition of [math]X'[/math] with sets [math][x]_\mathcal{R}\cap X'[/math], defining an equivalence relation [math]\mathcal{R}'[/math] on [math]X'[/math] by [math]x\mathcal{R}'y \Leftrightarrow (x\mathcal{R}y)\land (x, y\in X')[/math].

>stop donating blood.
but it helps people, it's too easy to save lives to not do it

>how shitty is your diet?
sometimes i'll eat instant noodles exclusively for 2-3 days, and i usually eat vegetables once a month

yes
as said equivalence relations are in bijection with partitions, so deleting the representatives of each equivalence class that aren't in X' gives a partition of X', and so gives an equivalence relation on X'

>but it helps people, it's too easy to save lives to not do it
it hurts your own life if you overdo it. stop it for minimum three years if you did it more than twice in one year.
>instant noodles exclusively for 2-3 days, and i usually eat vegetables once a month
that's alright for a period of a year or even two

btw if your diet is so shitty, I hope they throw away your blood. i wouldnt want to receive your shitty blood if i were ill.

>it hurts your own life if you overdo it.
wouldn't they recommend a longer wait time between donations if so?

> stop it for minimum three years if you did it more than twice in one year.
i've been trying to do it 6 times per year for the past 4 years or so

>btw if your diet is so shitty, I hope they throw away your blood. i wouldnt want to receive your shitty blood if i were ill.
they tell you if they throw it away and they've never told me so I guess it must not be that bad

I've done it 5 times the past years, what's so bad about it?

past year*

Why would you remove calcium?

First, filter the powder out, coffee filters work fine just as any porous fibre.
Then precipitate calcium either as sulfate or carbonate (pick whichever is easier for you to get at food-grade purity):
a) sulfate route
Add a solution of MgSO4 (Epsom salt) or Na2SO4 (Glauber's salt), the difference is that with Epsom you're adding magnesium to your solution and with Glauber's it's sodium.
b) carbonate route
Add aqueous Na2CO3 to the drink. Na2CO3 can be prepared by heating NaHCO3 (baking soda) to 50-100 degrees Celsius until it stops producing CO2.

Filter out the calcium sulfate/carbonate. Finally add the broc powder again.
Check out the alternative names for these compounds.
Alternatively you could tinker with phosphates, stearates, or whatever.

dude no this will make mustard gas

A question on MATH progression.

I've never had a formal treatment on algebra topics, should I get my feet wet on those before starting Spivaks? Or can I get to that when I am done?

IF I indeed need to understand everything related to Algebra, should I give Artin a try?

i'm assuming you mean spivak's calculus

i've never read it but i don't think you need any abstract algebra for any calc 1-2 material

Apparently Spivaks Calculus is more like a intro to analysis book, so people say.

Do I need Algebra for that, what do you think?

probably not

but it's worth learning algebra anyway

the broc powder is full with calcium, too

Recall that the differential of f(x,y) is

df = (df/dx)dx + (df/dy)dy

Now look at the ODE you were given. It matches this form. This means that there could be a function f whose differential is the given ODE and whose value df = 0.

If this is true, then P = df/dx, Q = df/dy.

We know that the mixed partial derivatives of a function equal eachother, so an easy way to check if f exists is to take the derivative of P wrt y and the derivative of Q wrt x and see if they equal eachother.

In your case, they both equal -sin(y)cos(x) and so we know f exists. We call this ODE exact.

Now the task is to find f. You use the clues available to you.

Take the integral of P wrt x:

f(x,y) = -cos(x)cos(y) + C

We know C cannot be a function of x, since it disappears upon differentiation wrt x, so this becomes

f(x,y) = -cos(x)cos(y) + h(y)

Now take df/dy and see how it compares with Q.

df/dy = -cos(x)sin(y) + h'(y) = cos(x)sin(y)

h'(y) = 2cos(x)sin(y)

Integrate

h(y) = -2cos(x)cos(y) + C

Now we know what f is.

f(x,y) = -cos(x)cos(y) - 2cos(x)cos(y) + C

= -3cos(x)cos(y) + C

Since the differential of f is 0, we can say that f = D, some constant, since the differential of a constant is always 0.

D = -3cos(x)cos(y) + C

Combining constants and solving for y gives

y = arccos(C/cos(x))

Note: The constant of integration in h is usually ignored, since it always combines later.

I have another question, friend, to you or anybody who has any idea.

I am studying elementary geometry and I also read some of the Geometry chapters in Langs Basic Math, or what he calls ''Transformations'' (Mapping, Isometries, Congruences, etc)

After I was done, I wanted to move to a book called ''Introduction to Geometry'' by Coxeter which is another book which deals with geometry through transformations or algebra.

I suppose a treatment of abstract algebra would be preferable than any other training in calc?

> Studying CS
> Actually using a good language
Im fucking envious
I also see u are german, would you mind telling me where u study?

Drink water instead.
Again, why do you want to get rid of calcium?

Can anyone help me out with pic related please?