SQT

In general relativity the position of energy is kind of fucked up.The gravity due to the kinetic energy is not directed towards the galaxy like if it was mass, since the momentum causes some dragging effects, this wont cause the galaxy to collapse since the kinetic energy isn't 'in' the galaxy. On the other hand, if some 3rd person saw you and the galaxy moving towards each other and combined you 2 would co;apse due to the kinetic energy, then you 2 will.

Check lines 2 and 3

Is [eqn]\frac{1}{2}||f(x)||^2[/eqn] continuously differentiable? I thought it was not differentiable at 0 but I'm not sure anymore

what conditions do you have on f?

No explicit conditions, but I think it's continuously differentiable

why do you think its not differentiable at 0 if there's no given condition on it at 0?

Cause the absolute function is not differentiable at 0 and I thought there's some relationship

Why is [math]\{0\}[/math] a linearly dependent set?

>In the theory of vector spaces, a set of vectors is said to be linearly dependent if one of the vectors in the set can be defined as a linear combination of the other

There are no other vectors in the set that could define 0

dont use definitions from the start of a wikipedia page

the standard definition is the one here
en.wikipedia.org/wiki/Linear_independence#Definition

and you have c*0 = 0 for any scalar c, so the set is linearly dependent

yes but [math]\|x\|^2[/math] is differentiable, because [math]\|x\|^2=x^2[/math]

select a hash function, implement it, and use it to determine the index. youre confined to an array so youll obv need to use some probing technique since you cant just chain in the event of collisions. may as well just use linear probing.

Why do people get sweaty at temperatures at or above 80 F when our core temperatures are 98.7 F?

the body is trying to maintain a temperature of 98.7F. it also naturally produces heat. hence, it needs an ambient temperature below the core temperature to maintain a temp of 98.7F.

The rate of heat transfer due to conduction is proportional to the temperature difference. As the air temperature rises, the temperature difference drops and heat transfer by conduction alone becomes insufficient, so evaporation (i.e. sweating) is used.

What kind of mental deficiency causes people to think that Beatles are a good band?

120=5*3*2^3
From you factorization, it's clear that at least one term is divisible by 5, and one is divisible by 3. One must be divisible by 4 as well, and then a different one by 2. That's all you need.

||f||^2 is a sum of squares. It's at least as smooth as f is.

If I'm travelling at the speed of light in one direction and I fire off some electromagnetic radiation in the opposite direction what is its velocity? How can it travel at the speed of light if its source is on an opposing vector also at the speed of light? it would have to have a subjective velocity double that of light?

You made me do this.

Ahem,
"The fact that so many books still name the Beatles as “the greatest or most significant or most influential” rock band ever only tells you how far rock music still is from becoming a serious art. Jazz critics have long recognized that the greatest jazz musicians of all times are Duke Ellington and John Coltrane, who were not the most famous or richest or best sellers of their times, let alone of all times. Classical critics rank the highly controversial Beethoven over classical musicians who were highly popular in courts around Europe. Rock critics are still blinded by commercial success. The Beatles sold more than anyone else (not true, by the way), therefore they must have been the greatest. Jazz critics grow up listening to a lot of jazz music of the past, classical critics grow up listening to a lot of classical music of the past. Rock critics are often totally ignorant of the rock music of the past, they barely know the best sellers. No wonder they will think that the Beatles did anything worthy of being saved."

When will science artificially create women who are not dumb cunts?

Haha how are imaginary numbers ever real my man just close your eyes nigga haha.

Anyone know what this file is?
pastebin.com/bebTpGKN
it appears to be a key or encrypted data on a stack.

Basic statistical analysis on the data (excluding the first and last 16 bytes) indicates that it's quite close to random

Shannon entropy = 0.99016, distribution is quite close to binomial.

So either random, pseudo-random or compressed.

It would be traveling at c with regards to your reference frame.

Also smartalec answer; an object that could 'give off' radiation wouldn't be traveling at c anyways.

I thought the Borel sigma algebra on R^n was just the intersection of all the sigma algebras on R^n that contained the standard topology.

How the fuck do I solve this shit?

Draw a line through D, perpendicular to AB. Let P be the point of intersection with AB.

The area of the rectangle PBCD is a^2*sin(30)*cos(30)=a^2*sqrt(3)/4.

The angle ADP is 45, so the area of ADP is (1/2)*a^2*sin(30)^2=a^2/8.

Total area is a^2*(1/8+sqrt(3)/4) = a^2*(1+2*sqrt(3))/8.

I cant see where I fucked up, any more tips?

Prove that all natural numbers with the form of n^3 + 17n are divisible by 6.
I have a tip to express 17n as 18n - n. But how does that help?
Help?

0 is a nontrivial combination of itself, namely [math]0 = 1\cdot 0[/math]

n^3+17n = n^3-n+18n
= n(n^2-1)+18n
= n(n-1)(n+1)+18n
= (n-1)n(n+1)+18n

What follows may seem familiar from the answer to your previous question.

(n-1)n(n+1) is the product of 3 consecutive numbers, n-1, n and n+1.

At least one of those will be a multiple of 2, and exactly one of them will be a multiple of 3. So their product must be a multiple of 6.

And 18n=6*3n, which is clearly a multiple of 6.

Did you never wonder why the binomial coefficient C(n,r)=n!/(r!(n-r)!) (for r

Does the analytification of the hilbert scheme coincide with the douady space of the analytification?

Thank you, divisibility is hard for me for some reason but I'm almost done with it, just 2 more problems.

(n^3 - n)+ 18n

n^3 - n is divisible by 3 by Fermat's theorem, and also is even, thus is divisible by 6

Is this a valid way to argue pic related, or is a bunch of words a babby-tier way to prove a proposition such as this?
Translation:
"G is a simple undirected graph containing more than two vertices. Show that G contains a vertex v such that (the inequality) holds."

My approach would be the following:
Note that the RHS is equivalent to the mean degree of the vertices in G (seems fairly obvious to me in this setting). The LHS is equivalent to the mean degree of the nodes adjacent to v, as it sums the degrees of the neighbours and divides by the degree of v, the latter of which is simply the number of neighbours.
Thus, if the inequality didn't hold, the mean degree of the neighbours of any v would be strictly less than the mean degree of the graph, a contradiction. Hence, the inequality holds.

sorry, forgot pic.

Would it be feasible for a layman to understand this?

oeis.org/A000262

>integrals of parabaloids in spherical coordinates

wew lad, how the fuck is overcomplicated bullshit like this allowed?

Is Engineering a branch of Science?

No, Science is a branch of Engineering.

True or false: let R be an arbitrary ring, and x [math]\epsilon[/math] R with x [math]\neq[/math] [math]0_R[/math] then x is either a unit or a zero divisor. The answer if false, but I don't know why.

Is it not obvious? Just consider the integers.

Can you propel yourself in space with a lantern

Guys did I fuck up?
>didnt get a good enough rank for top 5 IIT in core branch
>Could've taken mining but its not good
>applied abroad and go into uni of sydney and got in easily
Did I do good considering I want to apply to top american unis for masters?

I have a transformation [math] x_ { \mu } \to \alpha x_ { \mu } [/math] and I need to deduce that it can be generated by [math] D = i x^{ \mu } \partial _{ \mu } [/math] I don't know how to go about doing this, other than I should expand near the identity so [eqn] x_ { \mu } \approx I + i \beta \partial_{ \mu } x_ { \mu } [/eqn] Which has a passing familiarity to what I'm supposed to show.

How would you differentiate a function like [eqn] \frac { \partial } { \partial x^{ \nu } } \frac { 1 } { x^2 } [/eqn] The way I've been going about it is to write [math] x^{-2} [/math] as [math] x^{ -2 } = \left ( x^{ \nu } x_{ \nu } \right )^{ -1 } [/math] where the index was chosen so that it coincided with the index on the derivative. The just apply the chain rule and product rule, giving [eqn] \frac { \partial } { \partial x^ { \nu } } \frac { 1 } { x^2 } = \frac { \partial } { \partial x^ { \nu } } (x^{ \nu } x_{ \nu })^{-1} = - \frac { x_{ \nu } } { (x^{ \nu } x_{ \nu })^2 } [/eqn] Which seems right.

Can someone help me here?
In 3, how did the 2 went to the denominator of fraction below?
In 4, where he came up with the ax on the denominator of the fraction above and the bx on the denominator of the fraction below?

It's a trig identity [math] 1- \cos (x) = 2 \sin ^2 ( x/2 ) [/math] then it looks like they re-wrote it, since in general [math] 1/a/b = b/a [/math] so you could think of it as [eqn] \frac { 1 } { \frac { x } { 2 } } \sin ^2 \left ( \frac { x } { 2 } \right ) = \frac { 2 \sin ^2 \left ( \frac { x } { 2 } \right ) } {x }[/eqn]

Similar idea with 4), he re-writes so you can use the previous result. Notice that [eqn] \frac { \sin ( \alpha x ) } { \frac { \alpha x } { \frac { \sin ( \beta x } { \beta x } } } = \frac { \sin ( \alpha x ) } { \alpha x } \times \left ( \frac { \sin ( \beta x ) } { \beta x } \right ) ^{-1} = \frac { \sin ( \alpha x ) } { \alpha x } \times \frac { \beta x } { \sin ( \beta x ) } = \frac { \beta \sin ( \alpha x ) } { \alpha \sin ( \beta x ) } [/eqn] So now if you multiply it by [math] \alpha / \beta [/math] then you'll get your original limit.

>inb4 jsmath fail.

Thank you, user.

>Let X, Y and Z be sets. Draw Venn diagrams to illustrate the sets (X u Y) n Z' and blah blah blah...

I get how to answer the question, but how do I actually draw Venn diagrams on a computer? Every online editor wants data sets, and you can't shade intersections in WordArt. Right now I'm seriously leaning towards just using mspaint, but surely there's gotta be a better option.

why would you choose the index to coincide with the derivative's index, isn't that a divergence? moreover, it isn't clear what the x^2 term is, its ambiguous.

Yes. Quite remarkable from just looking at the formula.

Isn't that what the text is saying?

That's a good point, so suppose that I introduce a new index [eqn] \frac { \partial } { \partial x^ { \nu } } \frac { 1 } { x^2 } = \frac { \partial } { \partial x^ { \nu } } (x^{ \rho } x_{ \rho })^{-1} = - \frac { \eta ^{ \rho } _{\nu} x_{ \rho } } { (x^{ \rho } x_{ \rho })^2 } = - \frac { x_{ \nu } } { (x^{ \rho } x_{ \rho })^2 } [/eqn] Which would explain how I managed to end up with the right answer with the wrong method, thanks. Although I'm annoyed that it didn't occur to me sooner.

How do I prove that this system is linear and not time-invariant?

y(t)=3tx(t)

Seven digit number 6pqpqpq is divisible by 18.
Five digit number pqpqp is divisible by 6.
What is p?

No matter what I try I can't solve it.

I dunno, it's late where I am and I've been up all night, so I might be wrong here, but when I did it I got an extra factor of 2.

[eqn]\partial_{\alpha}[(x^{\mu}x_{\mu})^{-1}]=-x^{-2}\partial_{\alpha}[x^{\mu}x_{\mu}]=-x^{-4}(x_{\mu}\partial_{\alpha}x^{\mu}+x^{\mu}\partial_{\alpha}x_{\mu})=x^{-4}(x_{\mu}\partial_{\alpha}x^{\mu}+x^{\mu}\eta_{\mu\sigma}\partial_{\alpha}x^{\sigma})=...=-x^{-4}(2x_{\alpha})[/eqn]

heh, fugg

[math]\partial_{\alpha}[(x^{\mu}x_{\mu})^{-1}]=-x^{-2}\partial_{\alpha}[x^{\mu}x_{\mu}]=-x^{-4}(x_{\mu}\partial_{\alpha}x^{\mu}+x^{\mu}\partial_{\alpha}x_{\mu})=x^{-4}(x_{\mu}\partial_{\alpha}x^{\mu}+x^{\mu}\eta_{\mu\sigma}\partial_{\alpha}x^{\sigma})=...=-x^{-4}(2x_{\alpha})[/math]

0 (mod 3)

Nevermind. I really misunderstood that question.

That's it, I missed the factor of two in the example. Thanks a lot user, that looks right to me. Also when using tex on Veeky Forums it's best to just throw in whitespace everywhere.

If you do a melting point test in a pan over a candle using an ir thermometer will your results be anywhere near accurate?

I really need to know before I take these drugs

Brehs...

Okay so I looked at the answer because after an hour I still couldn't do it and p = 4. So I think q is 6, since 6464646 (6pqpqpq) is divisible by 18 and 46464 (pqpqp) is divisible by 6. But I still have no idea how to solve it. Help?

x'(t)=k*x(t) => y'(t)=3*t*x'(t) = 3*t*k*x(t) = k*y(t)
x'(t)=xa(t)+xb(t) => y'(t)=3*t*(xa(t)+xa(t))=3*t*xa(t)+3*t*xa(t)=ya(t)+ya(t)

y(t+dt)=3*(t+dt)*x(t+dt)
x'(t)=x(t+dt) => y'(t)=3*t*x(t+dt) =/= y(t+dt)

Dammit it also works if q = 0. Now I'm lost.

> Seven digit number 6pqpqpq is divisible by 18.
6000000 + 101010*p + 10101*q = 0 (mod 18)
> Five digit number pqpqp is divisible by 6.
10101*p + 1010*q = 0 (mod 6)

I tried something like this, but I don't know how to go from there.

is being a geneticists worth it

6000000=6, 101010=12, 10101=3 (mod 18)
10101=3, 1010=2 (mod 6)

6 + 12*p + 3*q = 0 (mod 18)
=> 12*p + 3*q = 12 (mod 18)
=> 4*p + q = 4 (mod 6)

3*p + 2*q = 0 (mod 6)

[4 1][p]=[4]
[3 2][q]=[0]

Note that matrix is its own inverse:
4*4+1*3=19=1 (mod 6)
4*1+1*2=6=0 (mod 6)
3*4+2*3=18=0 (mod 6)
3*1+2*2=7=1 (mod 6)

[p]=[4 1][4]
[q]=[3 2][0]

=> p=4,q=0

Dammit dude, I just solved it haha. But thank you, anyway.

Also, because mod 6, q=6 is also a solution.

Verify that the set of complex numbers of the form [math]x+y sqrt{2}[/math]
where x and y are rational, is a subfield of the field of complex numbers.

Note: I've found that the solutions online haven't been very useful so please give me a new explanation and thorough solution.

Also, Why is this a set of complex numbers? Aren't complex numbers only x+yi?

Correction: [math]x+y \sqrt{2}[/math]

>Aren't complex numbers only x+yi?
yes, every complex number can be written as a+bi for some real numbers a,b.

in this case you have a=(x+ysqrt2) and b=0

to see that its a subfield you just need to check that products (a+bsqrt2)(c+dsqrt2), quotients (a+bsqrt2)/(c+dsqrt2), sums a+bsqrt2 - c+dsqrt2 can be written as e+fsqrt2 for some e,f

> Why is this a set of complex numbers?
It's a set of real numbers, and the real numbers are a subset of the complex numbers.

What is this 'Co' operator?

Ah yes, like all naturals are also complex numbers. right?

yes but not a subfield or even a subring of course

I think I figured it out as being a convex combination of the set.

How would I do ii) here? I've done i) easily enough, but not too sure where to go with ii)

I know the area of a tetrahedron is 3^1/2 a^2.

How do I find the beginning and endpoints ?

I think that the easiest way to do is:
1.Because the Gram matrix is symmetric, it can be diagonalized, and using the properties of the determinants det(AB)=det(A)*det(B) the determinant of the diagonal matrix is the same as the determinant of the original one. Therefore we only have to prove the statement for an (non-normalized) orthogonal basis.
2. And it's easy to prove that the Gram matrix for an orthogonal basis (not necessarily normalized) has det>0, because the norm is positively defined.

No takers?

Let's take a function [math]f(x^\mu)[/math], if we apply that transformation what we have is [math]f(\alpha x^\mu)[/math] but if we expand it near [math]\alpha\approx 1[/math] we get
[math]f(\alpha x^\mu)\approx f( x^\mu)+(\alpha -1) x^\nu \partial_\nu f(x^\mu) [/math]
Therefore the generator of the dilatations is x^\nu \partial_\nu, and we add an extra i because we are physicist and not mathematicians.

Thanks user.

If V is the matrix you get by putting the vectors into columns of a matrix, then G = V^t V.

If the vectors are linearly dependent, there is a non-zero x such that Vx =0. Then G x = V^t V x =0, so G is singular and so has zero determinant.

If the vectors are linearly independent, then for any non-zero x, V x is not zero, so x G x = || Vx||^2 >0 , so G is positive definite and so has positive determinant.

How would I go about starting this?