/sqt/ - Stupid Questions Thread

Why is learning math/science history important? Why do I need to know who discovered what by doing something in the year X?
When was the last time anyone referred to a historical event when they were solving a problem? Is learning the concepts not enough?

IF the LD50 of caffeine anhydrous is 200mg/kg in humans, and I, a human, weigh 70kg, while the bioavailability of powdered caffeine anhydrous is 90%...

The highest likelihood of fatal caffeine toxicity would come from ingestion of 30800mg / 30.8g or greater, right? Is this math correct?

It gives you a deeper understanding of how the problem came about and how it was originally solved, while giving you proper context for its importance.

How to increase IQ?
Should I just spend an entire day studying advanced academics and then get into the habit of learning new things?

Units for moment of inertia is distance^4 (mm^4, ft^4, m^4, etc.)

What is the physical representation of distance^4?

An IQ test measures your ability to detect patterns, and if you include verbal IQ, ability to infer text. So your best idea would to just get really good at pattern detection.

What's the best grad degree to get if I'm studying Mathematics in University right now and I want to be set for life?

Anyone in academia?

Does 'pubish or perish' get in the way of productive research?

Is there something like 'The Journal of Negative Results' where papers with no methodology section or results that cannot be reproduced are refuted?

Define a sequence of functions [math](f_{n})_{n \in \mathbb{N}}[/math] for [math] f_{n} : \mathbb{R} \rightarrow \mathbb{R}, \; f_{n}: x \mapsto \frac{x}{n} [/math]. Does [math] f_{n} [/math] converge uniformly to some [math] f [/math]?

I think it does not, but I am not doing well a proving it.

dude 4-dimensions lmao

Isn't the 4th dimension time? How does that relate to inertia?

this kind of depends on whether you abuse notation like most people do by disregarding the collection of sets

Well (fn) converges pointwise to the function which is zero everywhere. So if you suppose (fn)->f uniformly, you must have f=0; you can look for a contradiction from here.

I need help proving that [math](1-1/2^a)*...*(1-1/n^a)[/math] converges to [math]0[/math] if [math]a < 1, a = 1[/math] or to a positive real if [math]a > 1[/math]

My reasoning so far.
->If a < 1, let a = 1/b with b superior or equal to 1.
We have : [math](1 - 1/n^\frac{1}{b}) = (n - n^\frac{1}{b})/n[/math], so if you make the product, you've got [math]\frac{1}{n!}*(1-2^\frac{1}{b})*...*(1-n^\frac{1}{b})[/math], and of course, the factorial win this one.

For the second part, I have the reasoning, but I need it more rigorously.

Basically I transform the thing on [math](n^a - 1)/(n^a)[/math] and since all terms are positives, and the general term of the product converges to one, there's "no way" it can go to zero or diverges, but I don't feel like putting it like that. How can I explain it ?

Your series of functions can be factorised with respect to x : x (1 + 1/2 + 1/3 + ... + 1/n)

(1 + 1/2 + ... + 1/n) is a well-known divergent series.

Whoops, I've read series of functions instead of sequence of functions. My bad.

When a=1, you have (1-1/2)...(1-1/n)=(1/2)(2/3)...(n-1/n)=1/n, so this of course goes to zero

Now when a>1, the sequence converges since it is decreasing and bounded below by 0. Also we have (1-1/2^a)...(1-1/n^a)>=(1-1/n^a)^n, and this converges to e^(-1/n^(a-1)), hence the limit must be strictly positive.

My bad, that second bit is wrong

I was reading this
world-nuclear.org/information-library/nuclear-fuel-cycle/conversion-enrichment-and-fabrication/uranium-enrichment.aspx

to better understand the implications of 20% of the US Reserve of U3O8 being sold to a Russian company but I am an uneducated plebeian and I'm having trouble fully understanding the meaning of the sentence

>Natural uranium contains 0.7% of the U-235 isotope. The remaining 99.3% is mostly the U-238 isotope

so does this mean that 99.3% of all Natural Uranium is the U-238 isotope and 0.7% of all Natural Uranium is U-235, or that all Natural Uranium is made of 99.3% U-238 /and/ 0.7% U-235?

basically: Can two isotopes exist within the same unit of an element?

>since it is decreasing and bounded below by 0.
Yeah, but does the product of every member of the sequence converges ?

Also, is my reasoning with the exponential good or bad ?

In every mole of natural Uranium, there is 0.993 mole of U-238 and 0.007 mole of U-235.
It's like, natural uranium is a mix of both, but it's not an element as itself.

>all Natural Uranium is made of 99.3% U-238 /and/ 0.7% U-235?
I think it's this one. It's basically the reason Oxygen's molar mass is not 16 but 15,9994 +/- 0.0004 g/mole.

Thank you, I thought so but I wasn't sure and it was making me feel incredibly stupid.

you said - since all terms are positives, and the general term of the product converges to one, there's "no way" it can go to zero or diverges

this reasoning is not correct, take your example when a=1

now i've figured out what to do when a>1

we have log( (1-1/2^a) ... (1-1/n^a) ) = log(1-1/2^a) + ... + log(1-1/n^a)

now in the range [0,1/2] we have log(1-x)>=-x, so using this inequality we get

log( (1-1/2^a) ... (1-1/n^a) ) >= -1/2^a - ... -1/n^a > -1 * (sum from k=2 to inf 1/k^a), and since a>1 this sum converges, say to C

then (1-1/2^a) ... (1-1/n^a) >e^-C and you're done

Does having a bigger brain actually make you smart?

a = 1 is excluded, I said that it goes to 0 for a < 1 and a = 1, but to a positive real if a > 1.

If a = 1, we have (2-1)/2 * (3-1)/3 * ... * (n-1)/n
Which is 1*2*3*(n-1) / 2*3*...*n = 1/n which converges to 0.

Thanks for the rest.

> Can two isotopes exist within the same unit of an element?
Define "unit".

Any given atom of uranium is of a specific isotope. If you mine uranium ore, 0.7% of the uranium atoms will be U-235, the other 99.3% U-238. These proportions will remain unchanged by chemical processes such as smelting.

The only viable process for changing the relative proportions of the isotopes is isotopic separation (enrichment) of UF6 using gas centrifuges. Because U238 is heavier than U235, gas drawn from the outside of the centrifuge will contain a slightly lower proportion of U235 while gas drawn from the centre will contain a slightly higher proportion.

The difference in concentrations is fairly slight, so practical enrichment requires a "cascade" consisting of a chain of centrifuges, with the outputs from each feeding the inputs of its neighbours. Thus, U235 concentration increases from one centrifuge to the next.

reposting from last thread, would really appreciate any help

I just failed an exam in a basic chemistry class. It's the only chemistry I need to take and I need to pass it this next exam or else I will need to re-do this whole year.
How do I best teach myself chemistry? I was pretty far away from the passing mark(I got 22 points, needed 40 for a passing grade).
Do I just keep on studying through the book or are there any simple tips for learning basic chemistry? This is some basic shit I just failed it, no excuses honestly.
The exam is in the end of Jan.

How do I know y if i have a system of this form:

y''+y=1+f(t)+g(t)+h(t)+...

y'+y=2+f(t)+g(t)+h(t)+...

Can it even be done?

You have two separate equations. Each can be solved individually, but there's no reason to suspect that they will have the same solution.

just starting with derivatives, how did this guy get from the second step to the third one?

or rather: from the first step to the second

just fucking study holy shit

how can anyone be this helpless

I actually did find that the sequence converges to the constant zero function, I conveniently forgot uniform convergence implies pointwise convergence. I can take it from here, thanks!

Thats alright dude, thanks for replying anyway.

converges pointswise to the constant zero function *

rootX(x) = x^(1/x)

and using the definition of logarithms ( the fact that e^ln(n) = n )

x^(1/x) = (e^ln(x))^(1/x)

thanks, and the next part? after the =

This came from a DE system where I already solved for x. Then I used x in the original equations and that's where the functions of t come from. The y solution should be such that it works for both of those equations right? But I don't know how to solve them. I thought I could do:

y''=-y+1+f(t)+g(t)+h(t)+...

y'=-y+2+f(t)+g(t)+h(t)+...
y''=-y'+f'(t)+g'(t)+h'(t)+...

-y+1+f(t)+g(t)+h(t)+... = -y'+f'(t)+g'(t)+h'(t)+...
-y+y' = stuff

But I still have y and y' and I don't know how to solve mixed stuff like that. What do?

well then you differentiate (e^ln(x))^(1/x)

derivative of e^f(x) = e^f(x) * f`(x)

two months

just study

Polite bump

Help me Veeky Forums, I am complete idiot. How can I prove that [math]sinh(x) -sin(x)\ge 0[/math]?

I tried to estimate that the derivative is positive so that the function is increasing and then show its zero when x = 0. And that proves it. But I'm having difficulties with showing that [math]f'(x) = 1/2(e^x + e^{-x}) - cos(x) \ge 0[/math]

oh okay, thanks, I kept looking at (e^x)' = e^x

Keep in mind that sinh(x) = (e^x-e^-x)/2
Shouldn't be that hard to prove that this is always higher than 1, and therefore sin(x). (by the way, your equation is only true in R+.

Regarding these. Can I just assume the terms in common are y? For example:

y''+y=1+f(t)+g(t)+h(t)+...

y'+y=2+e(t)+g(t)+h(t)+...

Then y = g(t)+h(t), since those are the terms in common in both equations. Would this work?

You can solve each of those equations using the Laplace transform.

E.g.
y''=-y+1+x(t)
=> Y(s).s^2 - y(0).s - y'(0) = -Y(s) + 1/s + X(s)
=> Y(s).(s^2+1) = X(s) + y(0).s + y'(0) + 1/s
=> Y(s) = X(s)/(s^2+1) + y(0).s/(s^2+1) + y'(0)/(s^2+1) + 1/s(s^2+1)
=> y(t) = x(t)*sin(t) + (y(0)-1).cos(t) + y'(0).sin(t)
where * indicates convolution. The convolution x(t)*sin(w*t) is simply the frequency component of x(t) having angular frequency w, phase-shifted by 90 degrees.

IOW, it's a system which oscillates at a frequency of one radian per second, the phase and amplitude varying according to initial conditions and resonance with the driving signal x(t).

Thanks lad

I doubt it.

How exactly did you end up with a single parameter being described by multiple equations in the first place?

In particular, a system with y proportional to -y'' tends to be sinusoidal, whereas y proportional to -y' tends to be exponential, while a damped oscillator has y'' proportional to both -y and -y'.

But (e^x-e^-x)/2 isn't always bigger than 1, even for x => 0? You can even see that on your graph? Did you mean 1/2(e^x + e^-x) => 1? Thats excatly where I'm having trouble :D

Whats the integral of? Do I used trigonometic sub or what?

[eqn]\int_{0}^{2\pi}\sqrt(4cos^2\theta)d\theta[/eqn]

When finding zeros of a polynomial, how do you know what numbers to try next with the intermediate value theorem?
People keep saying how intuitive it is, but I still don't fucking get it.
I end up just trying a shit ton of different numbers and it takes me forever to factor out a polynomial of degree 4 or greater.
It's embarrassing.

Can science one day solve moral problems?

Learn conversions. Learn how stoicheometry works.
PV=NRT
M1V1=M2V2.
Work them examples dude.

That's just 2*abs(cos(theta)) under the integral m8...

Just keep trying with it. Usually it's pretty obvious. Also cheat and take the derivative to see where the function is increasing and decreasing. Try the obvious stuff, -1, 0, 1, 100 etc.

I actually completely forgot about that, damn.
Thanks

Is it possible to find closed form solutions to
[eqn]\zeta (x)=x[/eqn]

do you know that there are any?

If I have to solve a Laplace problem with initial conditions, there comes a time where y(0) appears and I plug the initial condition there. But what happens if there are no initial conditions? For example, I know that

L{y'} = sL{y} - y(0)

If my initial conditions said that y(0)=7, then I would write

L{y'} = sL{y} - 7

And proceed to do a bunch of algebra and finally the inverse Laplace to know y (Am I right?), but if I don't have any initial conditions, What should I do with the y(0) term? All I know is from watching videos and solving stuff with tables, so please don't mention any buzzwords about frequency and domainwhatever.

how did he get to the last line?

cant solve a DE by laplace without initial conditions, that is its downside
different methods work for different cases

Have most optimistic people experienced trauma in their lives, in which they rather feel happy all the time than not?

I guess z=sqrt{a(x-r_1)/(x-r_2)}

>cant solve a DE by laplace without initial conditions

The funny thing is that I've got some excercises of Laplace without initial conditions. For example:

x' = x-2y

y' = 5x-y

And the alleged answers are:

x = -Cos(3t)-(5/3)Sen(3t)

y = 2Cos(3t)-(7/3)Sen(3t)

Did the book forget to print the initial conditions or is there something to be done?

Not necessarily. Its mainly about how wrinkled your brain is, which increases the surface area.

Would it be correct to say that tidal locking is the result of gravity neutralizing angular momentum? or matching angular momentum?

Are shadows two dimensional?

If you don't know the initial conditions then you just leave y(0), y'(0) etc as constants, which will end up as coefficients in the resulting time-domain equation.

E.g. given the DE
y''(t) = -w^2.y(t)

the solution is
y(t) = (y'(0)/w)*sin(w*t) + y(0)*cos(w*t)

This should be reasonably intuitive.

If y'(0)=0 (e.g. moving the mass in a spring-mass system to a fixed offset and releasing it at t=0, or charging a capacitor in an L-C circuit to a fixed voltage and connecting it to an inductor at t=0), then you get a cosine wave whose amplitude is equal to the initial value.

Conversely, if y(0)=0 (e.g. imparting an impulse at t=0 to a mass which is at the equilibrium point, or starting an L-C circuit with zero capacitor voltage but non-zero inductor current), then you get a sine wave whose derivative at the zero-crossing is its derivative at t=0.

And if both y(0)=0 and y'(0)=0, you get nothing; no energy = no motion/tension/voltage/current = zero amplitude.

The thing is, any time-varying system ... varies with time, duh. If you change your zero reference point, you change the equation describing the system.

So a system which is described by the equation y(t)=a*sin(w*t) is just as well described by y(t')=a*cos(w*t') where t'=t+pi/2w.

IOW, the only system whose behaviour is independent of its initial conditions is one which never changes.

this slow ass thread mang

yahoo answers is probably faster

Yahoo Answers! is unironically very fast.

Not really a science or math realted question but, I didn't want to make a thread about it.

Do schools charge more if you pick a dual degree program? I'm thinking about going for an MD/MPH program when I graduate soon.

why dont you look at your school's website brainlet?

> x' = x-2y
> y' = 5x-y
x = ((x(0)-2*y(0))/3)*sin(3t)+x(0)*cos(3t)
y = ((5*x(0)-y(0))/3)*sin(3t)+y(0)*cos(3t)

> x = -Cos(3t)-(5/3)Sen(3t)
> y = 2Cos(3t)-(7/3)Sen(3t)
=> x(0)=-1, y(0)=2

But suppose you had x(0)=y(0)=1, i.e.
x = (-1/3)*sin(3t)+cos(3t)
y = (4/3)*sin(3t)+cos(3t)
so:
x' = -cos(3t)-3*sin(3t)
y' = 4*cos(3t)-3*sin(3t)
x-2y = (-1/3)*sin(3t)+cos(3t) - (8/3)*sin(3t)-2*cos(3t)
= -3*sin(3t) - cos(3t)
= x'
5x-y = (-5/3)*sin(3t)+5*cos(3t) - (4/3)*sin(3t)-cos(3t)
= -3*sin(3t) + 4*cos(3t)
= y'

It should be self-evident that, while a system of differential equations can describe the nature of a system, the exact behaviour depends upon its inputs.

In particular, if a system of equations is linear and time-invariant (i.e. it consists only of the unknown functions and their derivatives), scaling all of the functions by the same factor scales all of their derivatives by the same factor, which (as there's nothing else in the equations) scales both sides of every equation by that factor. IOW, such a system always has an "amplitude" parameter.

Even more particularly, anything that models a physical system will have a "zero energy" state where it just sits there with nothing changing.

>Even more particularly, anything that models a physical system will have a "zero energy" state where it just sits there with nothing changing.

kind of like your life

I did but, no answer. That's why I asked here.

the two rotations thing is a consequence of being a spinor. the rotation operator ends up having factors of like, h/2 or something so that you need 4 pi to cancel with the 1/2 to make 2pi (a full rotation)

first of all, moment of inertia has units kg*m^2, not m^4. secondly, that guy above is trolling you

Is physics actually useful?

it's arguably the most useful field in existence

depends on your occupation, but in general, a familiarity with the essential concepts contributes immensely to general understanding of nature and the world

I rate physics among the noblest branches of human knowledge. I think it would be a scandal for anyone who considers themselves a philosopher to lack at least a rudimentary understanding of modern physics, including classical mechanics, quantum mechanics, general relativity and cosmology.

Speaking solely of tangible benefits I would say agriculture is more useful. Yeah you can do neat things with physics, but it's not gonna do you a lot of good if you can't eat.

Is there a minimum unit to empty space? Or can it be divided infinitely?

>implying there is a such thing as 'empty space'

that being side the limit of divisibility is probably close the planck length

being said*

If the universe is expanding then doesn't that imply that there is something for it to expand into? That space should be empty since the universe has not expanded into it yet.

What would you do if you had

a) 10 minutes to live?
b) one week to live?
c) one year to live?

More importantly, what wouldn't you do in the shorter timespan that you would do in the longer span? How to prioritize?

that's a common fallacy, akin to arguing there must be a beginning to time. all beginnings are in time; in just the same way, all limits are in space. it is nonsensical to speak of a limit to space. material space is not expanding into an infinite void, any more than time is expanding into an infinite future

So you got the initial values based on the answers right? But it would've been impossible to solve without previously knowing the expected answer. Am I getting it right?