/sqt/ Stupid Question Thread

[Anonymous No.8394032]

Post all of your stupid questions that don't don't deserve their own thread here.

I'll start: You think there will be another language other than Math that can translate the world where other people will be able to understand?

[Anonymous No.8394038]

Is the answer for A) 132J and the answer to B) .55 cents?

[Anonymous No.8394067]

Chris is trying to save Dana who is trapped in a room. The panel on the door of the room shows pic related.

To unlock the door, Chris must enter a password that consists of tracing a triangle with all three vertices among the points in the diagram. A password represented by triangle ABC does not depend on the order of vertices we choose to trace the triangle. So Triangles ABC, ACB, BAC,... all represent the same password. The following diagram shows some possible passwords:

Each tracing of a triangle takes one second. Chris is very talented and can easily keep track of the vertices he uses, which ensures he will not repeat any previously traced triangles. At the very moment Chris starts tracing his first triangle, a deadly gas is released inside the room. It takes exactly 9 minutes for the gas to render Dana unconscious and she cannot be saved once she is unconscious.

How many seconds will it take Chris to enter all possible passwords?

Is Chris guaranteed to save Dana?

[Anonymous No.8394125]

What is limiting light from going faster than c?

[Anonymous No.8394132]

>>8394125
the speed of light

[Anonymous No.8394156]

>>8394125

nothing is stopping it. It just doesn't go any faster.

You're thinking of the speed of light is bounded at c but that's not true. It's actually just the speed of light is c. Don't make it more complicated than it needs to be. It doesn't go any faster because that's not its speed.

[Anonymous No.8394158]

>>8394125
Because it does not feel like going faster

[Anonymous No.8394161]

>>8394125
It actually does go faster, but humans can't observe it going any faster than C.

[Anonymous No.8394194]

Can I get some help with logic here?

[Anonymous No.8394216]

How are midterms like at cal poly Pomona? I'm a transfer from a CC and one class is only 50 minutes long. Does that mean a midterm will only be 50 minutes to complete?

[Anonymous No.8394231]

My old question has not yet been answered:

>>8393612

>This is a question that google could not answer properly.
>Regarding a discussion about global warming, someone mentioned that the temperature has increased since the industrial revolution more rapidly than ever (let's say some several of thousands of years). I was looking at the charts (pic related) and I have the following questions: what is the accuracy of the geological methods for measurement of the earth temperature?
>Four important agency cannot agree with today's measurement (they agree on what's going on, but not on the precise amount.) If with today's tool there are difference in measurement+error, what is the error when measuring the temperature of the earth +1000 years ago?, as low as 0.1°C - 1°C?, or as high as +10°C?
>Also, could it be possible that we are comparing different data like temperature close to the sea vs temperature of the at 1000 masl?
>I couldn't find that on google.

[Anonymous No.8394239]

Where can I find

>G L Alexanderson and S Mac Lane, A Conversation with Saunders Mac Lane, The College Mathematics Journal 20 (1) (1989)

?

[Anonymous No.8394243]

>>8394067
He will not be able to do it. Take 3 (representing states A B C) to the 16th power (the number of points on the panel) to get every combination of ABC on the panel. But remember that ABC = ACB = BAC . . . For a total of 6 different combinations. So you have (3^16)/6 = # of combinations he can make. The # of triangles = the # of seconds it would take him. tl;dr He will not be able to save Diana, not even close.

[Anonymous No.8394249]

>>8394239
learn to google
http://www.jstor.org.sci-hub.bz/stable/2686814

[Anonymous No.8394251]

>>8394243
Thank you! I was not sure how to count them

[Anonymous No.8394255]

>>8394249
thx

[Anonymous No.8394323]

Why isn't de Maxwell–Boltzmann distribution normalized? It's just a facto of 1/2 so I don't get it why they put it that way.

[Anonymous No.8394367]

>>8394194
Woah, is this CSC236? I'm taking that course too.

[Anonymous No.8394399]

>>8394194
cmon at least explain your thoughts on the problem first.

[Anonymous No.8394425]

Can we use the Fast Fourier Transform to get a polynomial from a set of arbitrary coordinates? Or can this done with Lagrange polynomial interpolation?

[Anonymous No.8394460]

Hey /sci/, I'm doing dynamics homework and I'm having trouble solving a system of equations. I thought I was retarded at first (still might be) so I looked online for an answer but I haven't been able to come up with anything definite.

(3/5)x=4sin(a)+2
(4/5)x=4cos(a)

It seems that no matter which equation and variable I try to isolate it doesn't get me anywhere helpful. How retarded am I? Thanks for any help.

[Anonymous No.8394465]

>>8394425
any set of orthogonal functions will do
https://en.wikipedia.org/wiki/Generalized_Fourier_series

[Anonymous No.8394469]

>>8394460
you could try squaring both equations, then adding them since sin^2+cos^2 = 1 to eliminate the angle term

[Anonymous No.8394470]

>>8394469
Trying this now, will let you know how it goes. Either way thanks for the tip anon.

[Anonymous No.8394481]

>>8394469
>>8394470
Got me this far, similar point to previous but maybe more optons to explore than before? Maybe I can factor 4 out of the left side and continue from there...

[Anonymous No.8394487]

>>8394481
example
ax = sin(t)
bx = cos(t)

a^2x^2 = sin^2(t)
b^2x^2 = cos^2(t)

(a^2+b^2)x^2 - 1 = 0

it's just a quadratic equation

[Anonymous No.8394500]

>>8394487
Like this yeah?

[Anonymous No.8394505]

>>8394500
That was the thrust of my suggestion, yeah (though I didn't look in detail at your arithmetic).

[Anonymous No.8394516]

>>8394505
Okay cool, I'm not sure if it'll work but I'm glad to get another look and opinion on it anyway. If nothing else, I'll keep this trick in my head for when it may be relevant. Thanks a lot.

[Anonymous No.8394520]

>>8394460

(4/5)x = 4cos(a)

(x/5) = cos(a)

a = arccos(x/5)

(3/5)x = 4sin(a) + 2
(3/5)x = 4sin(arccos(x/5)) + 2

cos(a) = (x/5) ==> sin(a) = y/h = (5^2-x^2) / 5

(3/5)x = (4/5)(5^2-x^2)^(1/2) + 2

3x = 4(5^2-x^2)^(1/2) + 2

Square both sides and solve

[Anonymous No.8394523]

>>8394520

that last 2 should be a 10 but you get the point

[Anonymous No.8394544]

>>8394516
>>8394520
Okay various things. Firstly, these two methods give me the same answer, so that's good. The solution manual states x=4.86 but the answer I'm getting any way I look at it is 4.5, not much to stress about right now. Now, maybe my trig is just rusty as fuck, but I'm not entire sure on how you go from this

>(3/5)x = 4sin(arccos(x/5)) + 2

to this

>(3/5)x = (4/5)(5^2-x^2)^(1/2) + 2

I understand this part
>cos(a) = (x/5) ==> sin(a) = y/h = (5^2-x^2) / 5

But I'm confused as to how you got rid of the arccos(x/5). Is sin(arccos(a)) = sin(a)^1/2 or something? I currently feel that is not the case since the denominator of sin(a), 5, is not square rooted.

[Anonymous No.8394548]

>>8394544
Wait I think just re-reading it did it.

(5^2-x^2)^(1/2) = x isn't it

[Anonymous No.8394558]

>>8394544

My method gives you the x = 4.86 answer. So I don't know what it is you're doing.

To go from arccos(x/5) to the algebraic expression is literally just the definition

draw a triangle and put a in the corner

the adjacent side is going to be x

the hypotenuse 5

because cos(a) = x/5 = adjacent / hypotenuse

then you take the sin of that angle

which is just opposite / hypotenuse

use pythagorean theorem to find opposite, and you already 5 is the hypotenuse

[Anonymous No.8394572]

>>8394558
Hoooooly fuck okay my brain must be going full retard, where am I going wrong? Sorry for sideways pic.

[Anonymous No.8394582]

>>8394572

You're not squaring both sides.

C = (A + B)

C^2 = (A + B)^2 = A^2 + 2AB + C^2

= / = A^2 + B^2 (what you're doing)

[Anonymous No.8394602]

>>8394582
Duh, what a fucking idiot. Thank you so much.

[Anonymous No.8394626]

>>8394602

You're welcome.

Just a final tidbit, if it takes you more than 2 seconds to solve the equation after that step, then you probably didn't think to simplify your algebra a ton by isolating the radical before squaring.

3x - 10 = 4sqrt(25-x^2)
(3x-10)^2 = (4sqrt(25-x^2))^2

is much preferable to the messier alternative of

3x = 4sqrt(25-x^2) + 10
(3x)^2 = (4sqrt(25-x^2) + 10)^2

[Anonymous No.8394633]

>>8394626
Yuuuuuuuup I see that......

[Anonymous No.8394647]

NUCLEAR CHEMISTRY

So I went over a few concepts about nuclear decay at my uni and need some clarification

When an atom undergoes either Alpha or Beta emission, does it really become a completely different atom

Say, a Thalium emits an electron does that mean it has the properties of a Mercury?

Help a chem fag out

[Anonymous No.8394690]

Hey what is a determinant? I've read Lightstone's Linear Algebra and I don't remember him explaining exactly why a determinant of a matrix 2x2 matrix = ad - bc, like why not ac - bd??

I think he used induction like, given a matrix M;

if M_(1,1), then |M| = m_1,1

else if M_(n,n), then |M| = some row, then the matrix of that row, you know the deal

but he never explained WHY ad-bc, like why do we take the a row or a column than ignore them then take the determinant. It seems to me that mathematician define ad-bc as a matrix without a particular reason why other than it is somehow useful.

[Anonymous No.8394727]

>>8394690
Here's one way to look at the determinant of a 2x2 matrix.

[Anonymous No.8394765]

>>8394194
ill help with the first, but I'm not gonna do your homework for you because you are to lazy to do it yourself
∃(E(s,c) Λ D(s))
We should use "there exists" because we don't know if all students who are enrolled in a comp sci course has a dog, but there exists at least one. The conjunction connects the two predicates in the most logical way as well, basically making the saying "There exists a student s enrolled in a comp sci course c, and student s has a dog" if written correctly.

[Anonymous No.8394810]

>>8394690
> Hey what is a determinant?
It's the scale factor. Transforming a region of R^2 by a 2x2 matrix scales its area by the determinant. Transforming a region of R^3 by a 3x3 matrix scales its volume by the determinant. Etc.

If the determinant is zero, then the rank of the matrix is less than the number of columns, i.e. it has a null space.

A square matrix has an inverse if and only if the determinant is non-zero.

A negative determinant means that the matrix will change a right-handed coordinate system to left-handed and vice versa.

[Anonymous No.8395263]

I have no clue how to separate this.

[Anonymous No.8395268]

>>8395263
k and h are integers right?

[Anonymous No.8395273]

>>8395268
No, but I didn't think integers mattered for algebra.

[Anonymous No.8395281]

>>8395273
they don't, i'm just retarded.
basically you dont need to separate anything. there are no values of t to worry about. are you trying to get it in terms of y and t?

[Anonymous No.8395288]

Can anybody explain "If you throw a ball directly up in the air at ____m/s, how high does it go, how long is it in the air, etc."

Please explain it like I'm five, sock puppets may be required

[Anonymous No.8395294]

>>8395263
You can literally bring the whole thing with y's and h's to the left; the h is just a constant.

To integrate it, you'll probably have multiply it out and factor (so you have two linear factors with y's and constants), then do some partial fractions shit.

[Anonymous No.8395297]

>>8395281
I'm trying to get y in terms of t.
t is created by moving dt to the right side and then integrating presumably.
What I struggle to do is move all occurences of y onto the left.

>>8395294
The problem is I lack the skill for that.

[Anonymous No.8395301]

>>8395297
senpai its literally this

[Anonymous No.8395303]

>>8395288
The earth is pulling on everything trying to increase their velocity towards it. When you throw something away from the earth it has a negative velocity because it's moving away from earth. The earth causes it it to slow down and eventually start moving towards it. This happens at about a rate of 10m/s per second, so if you threw something with that speed, it would turn around after one second. How far it goes is a bit more complicated and will require some math. a=dv/dt so v=at+v_0 and v=ds/dt so s=at^2/2+v_0t+s_0
For t=|v/a| you get
s=v_0^2/2/a+sign(v_0)v_0^2/a+0
which in the case of v_0=-10m/s gives
s=100/2/10-100/10=-100/10/2=-5
so it rises 5 meters before starting to fall.

[Anonymous No.8395304]

>>8395297
>The problem is I lack the skill for that.
I worked it out a bit; after multiplying it out, you need the quadratic formula to factor the denominator.

But then, you should be able to use partial fractions.

[Anonymous No.8395306]

>>8395304
Alright, how would you use partial fractions for this?

[Anonymous No.8395312]

>>8395288
d2x/dt2=-g (acceleration)
dx/dt=u-gt (velocity)
x=ut-gt2/2 (position)

x=0 => t=0 or t=2u/g
dx/dt=0 => t=u/g => x=u2/2g

[Anonymous No.8395318]

>>8394647
not quite, but yeah. in this case, the path would be beta decay which means the neutron turns into a proton and an electron, making the thalium atom a lead atom.

[Anonymous No.8395355]

>>8395306
You can use partial fractions to turn a product into a sum:
a/p+b/q = (a*q+b*p)/(p*q)

So choose a,b such that a*q+b*p=1.

Having factored the denominator into (y-r1)*(y-r2), you end up with a/(y-r1)+b/(y-r2), and can integrate each term separately to a*log(y-r1) and b*log(y-r2).

Note that the sign of k-4*h determines whether the roots of the quadratic are real or complex. If they're complex, then the logs turn into arctan.

[Anonymous No.8395360]

>>8395355
I can't see at all how this applies to the problem though.

[Anonymous No.8395380]

I had problems in school, and didn't study math, is there a good place to learn what I didn't learn? I finished school, and I don't work, so I have time and need to learn it. I know aritmhetic, but I don't know about geometry and stuff.

[Anonymous No.8395393]

>>8395360
Are you expecting us to just do the entire problem for you?

>>8395301
shows what you're trying to integrate.

Factor the denominator, then use partial fractions to convert the product to a sum, then integrate each term.

If that's too hard, then give up math and switch to sociology instead.

[Anonymous No.8395404]

>>8395393
No, but I certainly don't know how to rearrange it into an integratable form despite looking up lots of stuff.

Which really sucks since the assignment is due in like, 9 hours, and it's the first part of the first question that's got me stumped.

[Anonymous No.8395468]

>>8395301
>>8395393
i fucked it up its actually this

[Anonymous No.8395544]

I reviewed Riemann Sum and notice something strange to me, pic related.

How is the area under a graph that is bijective to R equal to the sum partitions which are bijective to the natural number? If I zoom in into the graph and adjust the axis a bit, I will get a interval of the graph where the partition will appear to be too big and is only an approximation.

[Anonymous No.8395553]

>>8395544
>How is the area under a graph that is bijective to R equal to the sum partitions which are bijective to the natural number?
Because real numbers are limits of rationals in some way or another depending on the particular construction.

[Anonymous No.8395562]

Will computer science cease to be a meme degree in 4 years?

I'm going to college and I don't want to be without a job. Thinking either Comp-sci or Business (Logistics)

[Anonymous No.8395576]

>>8395553

So what? For any paritition k there exist an interval x_k-1 < x < x_k and there are infinitely many reals here, so if I am to take the rieman sum of this partition with n = 1, i will get an approximation, not an equality.

[Anonymous No.8395580]

>>8395576
But the image you posted doesn't set n=1. It is a limit.

[Anonymous No.8395582]

>>8395562

no it's going to be even more worthless

compsci turned to complete shit

[Anonymous No.8395585]

>>8395576

cont.

what I am trying to say is that if you want to find the exact true area under the graph you have to find the rieman sum of the rieman sum of the rieman ....

[Anonymous No.8395589]

>>8395582
Explain. I know Compsci used to be just a math degree, but now they also teach programming classes. Why will it be more worthless when the BLS says it will grow by 22% in 2020? Is the government Shilling?

[Anonymous No.8395591]

>>8395580

No if I somehow zoom in into any partition and since the width of the partition is not zero, then there must be some length. So this single partition itself will have curves, despite very small.

[Anonymous No.8395595]

>>8394216
Yes, it's not bad though. Just practice speed a bit

t.UCLA

[Anonymous No.8395840]

>>8395591
The limit of the width of the partition is 0.

[Anonymous No.8395877]

What is this kind of differential equation called?

I have to solve using the method of characteristics.

[Anonymous No.8395885]

>>8395877
nevermind, had to tweak wolfram input a bit
http://www.wolframalpha.com/input/?i=(d(u(x,y))%2Fdx)(d(u(x,y))%2Fdy)+%3D+xy

[Anonymous No.8395918]

>>8395263
Move dy to the side with all the y's and everything else. Then solve for dt. Then integrate both sides.

[Anonymous No.8396038]

Should I minor in computer science? My major is computer engineering.

[Anonymous No.8396051]

>>8395544
>>8395576
>>8395585
>riemannian integration
lebesgue laughs in his grave at you all

[Anonymous No.8396054]

>>8394032
Is any finite field commutative?

[Anonymous No.8396055]

Prove that if a probability density function is symmetric about x=a then the expected value of x is a

It's really damn obvious but I can't into proofs

[Anonymous No.8396064]

When I plug -2 in for f(x) on the table function on my calculator I get 0
How ever when I solve it myself I get 2.828

What is going on here?
pic related

[Anonymous No.8396072]

>>8396064
Lol bait

[Anonymous No.8396074]

>>8396072
no I'm serious

[Anonymous No.8396080]

>>8396072
nevermind I figured it out

[Anonymous No.8396162]

>>8395263
Bump

[Anonymous No.8396252]

[eqn]\mathbb{E}(X-a) = \int_\mathbb{R} (x-a)\,p(x)\,dx = \int_\mathbb{R} z\,q(z)\,dz,[/eqn] where [math]z := x-a[/math] and [math]q[/math] is a function such that [math]q(x) = p(x-a)[/math], i.e. shifted by [math]-a[/math]. Then [math]q[/math] is clearly symmetric about [math]x=0[/math] and so [eqn]-\int_{-\infty}^0 z\,q(z)\,dz = \int_0^{\infty} z\,q(z)\,dz \implies 0 = \int_{-\infty}^0 z\,q(z)\,dz + \int_0^{\infty} z\,q(z)\,dz = \int_\mathbb{R} z\,q(z)\,dz = \mathbb{E}(X-a).[/eqn]
Then by linearity of the expected value operator, [eqn]\mathbb{E}(X-a) = \mathbb{E}(X) - \mathbb{E}(a)[/eqn] and so you have [eqn]\mathbb{E}(X) = \mathbb{E}(a) = a.[/eqn]

[Anonymous No.8396266]

>>8394323
But it is normalised

[Anonymous No.8396273]

>>8396054
Refer to the definition of a field, i.e. what criteria must be satisfied to make something a field. This will quickly answer your question.

[Anonymous No.8396280]

>>8396252
but thats wrong
we have
[math]\mathbb{E}(X-a) = \mathbb{E}(X) - a [/math]

[Anonymous No.8396288]

What degree won't get me laughed at and shitposted to death on /sci/?

[Anonymous No.8396344]

Say you have three As and four Bs in a sequence, like this for example:

AAABBBB

How many unique combinations can you get? AABABBB, AABBABB, etc. Obviously it's a multiple of 7, and I'm pretty sure it's 35 but I'm wondering if I'm missing some, or if there's some way to calculate it without writing them all out.

[Anonymous No.8396362]

>>8394194
>CS students aren't logical
xDDDDDDDDDDD

[Anonymous No.8396402]

>>8396344
You're correct. To calculate it, imagine you have 7 empty slots and 3 A's. How many possible positions can you place these 3 A's into the 7 slots? There are "7 choose 3" positions:
[eqn]C(7, 3) = \frac{7!}{3!(7-3)!}=35.[/eqn]
(notice that we don't care about the B's, since they obviously just fill the other 4 slots)

Here's another way of thinking about the above calculation. Let's say we have a set of size 7: S = {a1, a2, a3, ..., a7}. We can think of the a1...a7 as the empty "slots." Then each subset of S of size 3 is a possible position for the A's. To count the number of possible subsets of a certain size of a set, we use the same C(n, k) formula.

[Anonymous No.8396424]

>>8396402
Brilliant, thanks a lot, it's been years since I last looked at permutations/combinations

[Anonymous No.8396452]

Just wondering, why isn't the right interval pi? Doesn't it need to include 1 or am I going full retard?

[Anonymous No.8396475]

>>8396452
>Doesn't it need to include 1
pi/2 = 1.57

[Anonymous No.8396490]

>>8396475
For some reason I was seeing it as sin(pi/2), but why wouldn't (0,pi) work then?

[Anonymous No.8396503]

>>8396490
Because your existence theorem requires continuity on the interval
tan(x) is not continuous at pi/2

[Anonymous No.8396508]

>>8396503
Oh, I'll keep that in mind.

[Anonymous No.8396584]

[math]\mathcal{O}(x^3)+x\cdot\mathcal{O}(x^2)\approx \mathcal{O}(x^3)[/math]

Correct?

[Anonymous No.8396684]

does this mean f, f', and f'' are continuous on [a, b]?

[Anonymous No.8396692]

>>8396584
ya, why were you having second thoughts

[Anonymous No.8396693]

Can someone explain this proof to me?

[Anonymous No.8396743]

>>8396693
what are you confused by? the restriction step is because a^6=(-a)^6

you choose a nontrivial triple (a,b,c) with max(a,b,c) > 0 as small as possible (this exists by applying the well ordering principle to the set { max(a,b,c) : a^6+2b^6=4c^6, max(a,b,c)>0 }

you look at a^6+2b^6=4c^6 and see that a has to be even since a^6=4c^6-2b^6 is even.

so you can write a=2(a_1), and substitute that into the original equation to get
64(a_1)^6+2b^6=4c^6.

Then dividing by two gives 32(a_1)^6+b^6=2c^6, which rearranges to show you that b is even:
b^6=2c^6-32(a_1)^6

substitution b=2b_1 gives:
32(a_1)^6+(2b_1)^6=2c^6
which implies
32(a_1)^6+64(b_1)^6=2c^6
divide by two:
16(a_1)^6+32(b_1)^6=c^6
and so c is even, write c=2c_1

then
16(a_1)^6+32(b_1)^6=(2c_1)^6=64(c_1)^6
which implies
(a_1)^6+2(b_1)^6=4(c_1)^6

but now we have a new nonnegative solution to our original equation, and we also have max(a_1,b_1,c_1)< max(a,b,c) (since at least one of the a,b,c are nonzero, half of that value has to be strictly smaller than its original value). but this contradicts the minimality of max(a,b,c), and so the set { max(a,b,c) : a^6+2b^6=4c^6, max(a,b,c)>0 } we originally considered was empty

[Anonymous No.8396746]

>>8396684
yes

[Anonymous No.8396756]

>>8396693
1. If a,b,c is a solution, then a,b,c must all be even, i.e. a/2,b/2,c/2 are all integers.
2. If a,b,c is a solution, then so is a/2,b/2,c/2.
3. So a/2,b/2,c/2 must also be even.
4. Repeat ad infinitum. I.e. it must be possible to divide a,b,c by 2 infinitely many times without ending up with an odd number. I.e. they must all be zero.

[Anonymous No.8396840]

Do black holes exist?

[Anonymous No.8396854]

>>8396840
Primordial black holes exist?

[Anonymous No.8396875]

>>8394243
>>8394251
This guy was probably trolling but this answer seems very wrong, you should be using the nCr function not just exponents. Also you need to weed out the combinations where all three points are colinear and thus don't form a triangle.

[Anonymous No.8396901]

>>8396693
>>8396743
>>8396756
I think I'm misunderstanding what 'max' means.

[Anonymous No.8396921]

Is negative one smaller or bigger than zero?
I mean, teachers explain it using debt, and having negative money is the same as zero money. And negative one has bigger magnitude. Like whoah

[Anonymous No.8396924]

>>8396921
-1 < 0
but |-1| > |0|

>>8396901
max(a,b,c) is equal to the largest of these 3 nonnegative integers

[Anonymous No.8396929]

>>8394161
[Citation required]
In what thought experiment was this bullshit thought up?

I know Tachyons are an entire other meme but I've literally never heard of someone suggest mathematically that photons can exist at speeds higher than c.

[Anonymous No.8396930]

>>8396924
wait, but aren't they saying:

choose:

1. a triplet of nonnegative integers a,b,c that satisfy the equation
2. with max(a,b,c) > 0 as small as possible

how can you choose the largest of the 3 nonnegative integers and have it be as small as possible?

[Anonymous No.8396932]

>>8396930
youre choosing the maximum of a given solution (a,b,c) to be as small as possible

for example suppose you had solutions

(1,4,7)
(6,4,10)
(0,20,2)

then max(1,4,7)=7
max(6,4,10)=10
and max(0,20,2)=20
and the smallest max(a,b,c) is 7

[Anonymous No.8396934]

so every time I ask what a sine wave is, I always get the answer: "well, it's what sin(x) spits out"

that's not good enough

How is sin() actually computed?
any thing I can read about this?

[Anonymous No.8396938]

>>8396840
>>8396854
No one will answer my questions ...

[Anonymous No.8396940]

>>8396934
you can do the taylor series definition to compute sin(x)

or look at a unit circle and compute sin(theta) as the b coordinate of the point (a,b) which has angle theta (taken counterclockwise) with the x-axis

[Anonymous No.8396959]

>>8396940
so I did some research and it seems Sun Microsystems uses the taylor series with thirteen terms

https://git.uclibc.org/uClibc/tree/libm/k_sin.c

[Anonymous No.8396968]

>>8396932
I understand now, thank you very much.

[Anonymous No.8396972]

>>8396968

np

[Anonymous No.8397009]

>>8396934
> so every time I ask what a sine wave is, I always get the answer: "well, it's what sin(x) spits out"
If you're talking about a sine *wave* as opposed to the sine function, the solution to d2x/dt2+ω2x=0 is a*sin(ω*t+φ), i.e. a sine wave with frequency ω/2π, amplitude a and phase φ. Amplitude and phase are determined by initial conditions.

> How is sin() actually computed?
However you like.

Taylor series, CORDIC, lookup table of approximating polynomials, repeated application of half-angle formulae or double-angle formulae.

It depends upon how much accuracy you need and what kind of hardware you have.

CORDIC is efficient on simple hardware (shift, add/subtract, compare), Taylor series is faster if you have fast multiplication, and can be used for arbitrary precision. Lookup tables allow the use of lower-degree polynomials.

[Anonymous No.8397011]

I'm probably going to get downboated or told to fuck off for this question, but I'll ask anyway;

Out of 600% of a total value, 512.548% of that value is 26,618.The problem is that I'm trying to find the value of the remaining 87.452%. Please great math wizards, help this plebeian with this problem, as it pertains to an outstanding loan. Any help is greatly appreciated.

[Anonymous No.8397027]

>>8397011
This isn't Reddit senpai. That being said, here is the solution to your problem. Someone correct me if I'm wrong, even though I am pretty sure I'm not.

512.548/600 * x = 26618

Solving for x, we get x = 31159.62 (2 decimal places)

Now x is the total value you were talking about in your question. To get the remaining 87.452% you do:

31159.62 - 26618 = 4541.62

4541.62 (approximately) should be the value of the remaining 87.452%.

[Anonymous No.8397033]

what the FUCK is a tensor?

and how do I into it?

[Anonymous No.8397102]

>>8397033
>what the FUCK
hey, there's no need to be so tense :^)

[Anonymous No.8397117]

currently reading through a math paper on primes and it has a notation i dont understand

"Consider a prime [math] q [/math] that is a factor of [math] n [/math] and let [math] q^k || n [/math] ."

What does [math] q^k || n [/math] mean?

[Anonymous No.8397119]

>>8397117
exactly divides

as in its the highest power of q dividing n

i.e. 2^3 || 8 but you wouldnt write 2^2 || 8

[Anonymous No.8397121]

>>8397119
equivalently q^k || n iff gcd(q^k, n/q^k)=1

[Anonymous No.8397175]

>>8396280
That doesn't contradict what he said, in fact what he said implies that.

[Anonymous No.8397185]

>>8396921
>having negative money is the same as zero money

Being in debt and having no money isn't the same as not being in debt and having no money though.

[Anonymous No.8397191]

>>8397033
An element of the tensor product of spaces.

Basically you pretend that you can multiply vectors together, like [math]v \otimes w[/math], and then you add these together to get the most general kind of tensor which is [math]\sum v_i \otimes w_i[/math].

[Anonymous No.8397199]

Guys quick should i sleep for 1.5 hours or 3 hours for maximum rest gains for exam today

[Anonymous No.8397214]

Doing a paper on (special) relativity, need to have subquestions for a deadline today. Anyone know any?

[Anonymous No.8397239]

>>8397199
as much sleep as you can get w/o being woken up in the middle of a REM sequence

Also, I reccomend taking a strong coffee before a short nap so you wake up extra refreshed

[Anonymous No.8397398]

Can someone recommend a book that would teach me how to prove? I can't do it for shit

[Anonymous No.8397423]

>>8397398

brah

how to prove it by velleman, i heard it is quite well received

[Anonymous No.8397592]

>>8394032
Starting Chemistry this year. Not sure which direction to specialize in. Anyone got any tips to which is personally more fun, or which makes more cash?

[Anonymous No.8397654]

In pic related, when it says k=1, is that designating the starting value, or the interval?

And whichever one it is, how is the other expressed?

Sorry, I know this is painfully basic. I've only completed up to Calc 1 and just encountered summation for the first time in a non-math class.

[Anonymous No.8397656]

>>8396584
You don't need the squiggle-equals-sign there.

O(x^3) + x*O(x^2) literally is "big-O-of x^3"

[Anonymous No.8397660]

>>8396684
Yes, but saying f(x) instead of f in this context is triggering me.

[Anonymous No.8397758]

>>8397654
k=1 is the starting value.

The index interval in a summation is always 1, if you want intervals of 2, for example, you'd just replace k with 2k.

[Anonymous No.8397836]

>>8394067
>>8394243
>>8396875
I'll give this a shot.

In order to draw a (non-degenerate) triangle, we require each of its three vertices to be unique, that is, no vertex can be picked more than once. Since we don't care about the order in which the vertices are picked, the number of ways to choose 3 elements out of a possible 16 is given by the binomial coefficient
[eqn]{{16}\choose{3}} = \frac{16!}{3!(16-3)!} = \frac{14\cdot15\cdot16}{2\cdot3} = 560[/eqn]
so there are 560 possible distinct combinations to try. To guarantee success, we'll have to try all of them.

In nine minutes, there are [math]9\cdot60 = 540[/math] seconds. Since Chris can draw one triangle per second, he'll be able to try 540 out of the 560 combinations, which means he's not guaranteed to be able to save Dana. He is fairly likely to do so, though.

Fun sidenotes: The number of ways to choose 3 vertices from the 3 "correct" dots and 0 vertices from the 16-3 = 13 "incorrect" ones is given by
[eqn]{{3}\choose{3}} {{13}\choose{0}} = 1[/eqn]
Thus, the probability of succeeding on any one try is [math]\frac{1}{560}[/math], as is to be expected. However, this method of determining the probability shows that this exercise can be described by a hypergeometric distribution of the form
[eqn]p_x(k) = \frac{ {{v}\choose{k}} {{s}\choose{n-k}} }{{{v+s}\choose{n}}}[/eqn]

[Anonymous No.8398120]

What does a|b = c mean? I thought a|b was either true or false.

[Anonymous No.8398164]

>>8397836
you picked 3 points to form a triangle in the correct way but still forget to get rid of the collinear ones (i.e. picking three points on the top or bottom edge, or three points on the left or right edges). there are (5 choose 3) ways to each of these, so you need to subtract 4(5 choose 3)=40 degenerate cases from 560, giving 520, and so chris always saves dana

[Anonymous No.8398223]

>>8398164
Ah, good point. I did indeed forget about the collinear cases. Your correction is now the official answer to the problem.

[Anonymous No.8398228]

>>8398223
i won two movies passes with it, pretty worth doing some baby combinatorics

[Anonymous No.8398230]

Somebody knows how to solve for x:
x^t + x(b-1) + b= 0, t is a natural number and b is a real number

[Anonymous No.8398241]

>>8398230
is there any more context for this? if t=1 or 2 its easy but since you're always going to have t solutions it seems like a very general problem

[Anonymous No.8398264]

>>8398241
>>8398230
at least if t is odd you have
(-1)^t+(-1)(b-1)+b=-1-b+1+b=0
so that's one root

so you can factor out as

(x^t+x(b-1)+b)=(x+1)(x^[t-1]-x^[t-2]+x^[t-3]-...+x^2-x+b)

[Anonymous No.8398340]

>>8397758
Thanks!

[Anonymous No.8398474]

Let A =
[2 -1]
[-6 3]

and b =
[b1]
[b2]

Show that the equation Ax=b doesn't have a solution for all possible p and describe the set of b where it does have a solution.
~~~~~
They never said what x is but i assume it's just [x1, x2, x3]t.
Is putting this into an augmented matrix form
[ 2 -1 b1]
[ 0 0 b1+b2/3]

and using the rule that you can't have a pivot in the last column for it to be consistent, so b1+b2/3 has to = 0 a valid way to solve it? I'm reading the chapter and the way it tells us to do it is different from what the professor has been teaching in class. The way the book wants us to do it is to multiply matrix A by [x1 x2 x3]t and have it be equal to b. are these equivalent or totally different?

[Anonymous No.8398563]

How do I add these two numbers? I can't brain today
4.35x10^17 + 3.15x10^111

[Anonymous No.8398578]

>>8398563
>4.35x10^17 + 3.15x10^111

10^111 (0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000435 + 3.15)

= 3.15000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000435 * 10^111

[Anonymous No.8398635]

>>8394032

If a thread with X posts starts on post number Y and ends with post number Z, how many posts should we expect to get dubs, trips and quads.

[Anonymous No.8398697]

>>8398474
>They never said what x is but i assume it's just [x1, x2, x3]
Why would you assume it has three elements? You can't multiply a 2x2 matrix by a 3x1 matrix to yield a 2x1 matrix.

[Anonymous No.8398760]

>>8398635
1/10 posts is dubs
1/100 posts is trips
1/1000 posts is quads.
However. This theory doesn't allow for the variable of a post ending in 98888 to be counted as only one number. It is considered a get as dubs, trips, and quads depending on how many of the digits you observe. Therefore blah blah blah blah math stuff.... Figure the rest out yourself.

[Anonymous No.8398762]

Quads

[Anonymous No.8398764]

>>8398762
Oh. I was using the expected posting time of /b/ and forgot /sci/ operates much slower.

[Anonymous No.8398776]

derivative/function, is there any way tell which goes with which for c and d?

why does c go with i?

why does d go with iii?

[Anonymous No.8398817]

>>8398776
The the gradient at a point on the top graph is the Y value on the bottom graph.

C has only one point where the gradient is 0 so its corresponding bottom graph will only cross the X axis in one place.

D has three points where the gradient is 0 so its corresponding bottom graph will only cross the X axis in three places.

[Anonymous No.8398875]

Questionable quads?

[Anonymous No.8398877]

>>8398875
I was so close. But there is still time to get given that /sci/ is so slow tonight

[Anonymous No.8398879]

Reality is an illusion.

[Anonymous No.8398880]

>>8398879
Goddamn why is /sci/ even slower now.
8398881 get

[Anonymous No.8398882]

If space itself expands faster than light, and this destructive wave can only travel light speed, doesn't that mean we can just "outrun" the wave?

[Anonymous No.8398883]

>>8398882
Only if we can achieve/exceed the speed of light. Highly unlikely.

[Anonymous No.8398886]

Philosophy is the science of quads

[Anonymous No.8398887]

Science dictates that jet fuel can't melt steal beams
Quads confirm

[Anonymous No.8398888]

>>8398883
Either way this "wave" would kill us?
(Assuming that our galaxy continue increasing its speed)

[Anonymous No.8398890]

>>8398888
FUCK YOU MOTHER FUCKER THOSE QUADS ARE MINE

[Anonymous No.8398891]

>>8398888
But yes. Death is coming faster than we can outrun.

[Anonymous No.8398898]

Cursed false vacuum state...

[Anonymous No.8398905]

>>8398898
>>8398891
Oh shit, I remember that, we will die at any time.

[Anonymous No.8398913]

>>8394216
Are you tranfering from mt San Antonio?

[Anonymous No.8398916]

I'm currently covering real integration techniques in one of my classes, and how the shit is this stuff intuitive? Is there any way to really recognize which technique is required to solve these things, or is it more of a "solve a thousand problems and it'll just click" type of deal.

[Anonymous No.8398929]

>>8398916
intuition comes from experience

[Anonymous No.8398960]

What portion of sunlight produces heat?

visible light itself produces no heat, right?

[Anonymous No.8398965]

How do birds navigate towards their destination during migration?

[Anonymous No.8398971]

>>8398965
Magnets
>Not even a meme

[Anonymous No.8398974]

>>8398965
GPS

[Anonymous No.8398981]

Disprove horoscopes and astrology for me

[Anonymous No.8398994]

>>8398981
use your brain

[Anonymous No.8399046]

>>8398776
The derivative indicates the slope of the function.

So (x, y) on the derivative graph will correspond with the slope of the function at x.

Ex: If the y value of the derivative graph at x = 1 is 4, then the slope of the function (not the derivative of the function) will be 4.

If you understand why B corresponds to IV, you should be able to understand how the others relate without much issue.

Have fun with calc, it's when math gets cool.

[Anonymous No.8399047]

>>8398960
> visible light itself produces no heat, right?
Wrong.

Electromagnetic radiation falls on the surface of an object. Typically (i.e. in the absence of less common mechanisms such as photoluminescence), it either gets reflected (at its original wavelength) or absorbed. If it's absorbed, then it ends up as heat, i.e. increases the temperature of the object.

Heat energy is then emitted as black-body radiation, i.e. the distribution of wavelengths depends upon the objects' temperature, and is unrelated to the wavelengths of the radiation which heated it to that temperature in the first place.

[Anonymous No.8399086]

Is writing g mol^-1 the same as writing g/mol?
i.e the ^-1 is acting as dividing

[Anonymous No.8399104]

>>8399086
No it's written different.

[Anonymous No.8399111]

>>8398965
In winter it gets cold in North and warm in south. Birds feel the warm and fly to it.
In summer it gets warm in North and cold n south. Again they move towards the warm.

[Anonymous No.8399135]

How do you read the p with the tilde hat? Would you just say p tilde?

[Anonymous No.8399139]

>>8399086
Yes.

[Anonymous No.8399141]

>>8399135
yes

[Anonymous No.8399177]

>>8399086
Yes

[Anonymous No.8399216]

https://math.stackexchange.com/questions/1161244/find-derivative-of-fraction-using-first-principles


Can someone give me a more in depth working out of this? My worksheets (or maybe its me) are hopeless

[Anonymous No.8399349]

>>8398760
>>8398635

The general formula is then 1/10^n - 1/10^(n+1) = 9/10^(n+1). So

dubs: 0.09 (9%)
trips: 0.009 (10%)

It will depend on Y and Z, but if you know Y and Z then you know exactly which ones occur.

[Anonymous No.8399350]

>>8399349
*not 10%, 0.9%
etc.

[Anonymous No.8399354]

>>8399216
f(x)=(3x-1)/(x+2)
f(x+h)=(3x+3h-1)/(x+h+2)
f(x+h)-f(x)=(3x+3h-1)/(x+h+2)-(3x-1)/(x+2)
=((3x+3h-1)(x+2)-(3x-1)(x+h+2))/((x+2)(x+h+2))
=((3x^2+6x+3xh+6h-x-2) - (3x^2+3xh+6x-x-h-2))/((x+2)(x+h+2))
=7h/((x+2)(x+h+2))
(f(x+h)-f(x))/h=7/((x+2)(x+h+2))
lim[h->0]((f(x+h)-f(x))/h) = 7/((x+2)(x+2))
=7/(x^2+4x+4)

[Anonymous No.8399359]

How do I find the Laurent series expansions about [math]z=2[/math] (and their radii of convergence) for:

[math](4-z^2)^{3/2}[/math]
[math](z^2+1)^{1/4}[/math]
[math](z^2+1)^{1/2}[/math]

The textbook doesn't mention any way of finding Laurent series for fractional powers and I can't find any way to do it online, and to make things worse, the first of those functions has a branch point at [math]z=2[/math]. Wat do?

[Anonymous No.8399364]

>>8398120
Context? It may not mean divides. Or it could just mean that that's the quotient.

[Anonymous No.8399388]

Can someone guide me on the difference between b and c, preferably with an example? It's my first time dealing with floor functions

[Anonymous No.8399458]

>>8399388
c) is just g(0). For x in [0,1), floor(x)=x. So as x approaches 0 from above, floor(x) is just constant 0, g(floor(x)) is g(0), and the limit is g(0).

b) is ... well, if floor(g(x))=floor(g(0)) for x in some continuous interval [0,epsilon], then it's just floor(g(0)). If they're equal over (0,epsilon] but there's a discontinuity at x=0, then it's floor(g(epsilon)).

[Anonymous No.8399466]

I cannot understand ii) for the life of me; I simply can't see how P(A) - P(A ^ B) = P(A|B)
It makes no sense.. am I reading the notation wrong?

[Anonymous No.8399513]

>>8399047
>i.e.i.e.i.e.i.e
take it easy nigga

[Anonymous No.8399537]

>>8399466
You're reading the notation wrong. It means p(a not b).

[Anonymous No.8399635]

Can integral transforms only be used for linear differential equations?

[Anonymous No.8399724]

If you have a path connected subspace [math]A\subset X[/math] such that the inclusion induces a surjection on fundamental groups, then any path in [math]X[/math] with endpoints in [math]A[/math] is homotopic to a path solely in [math]A[/math]. It's apparent that the idea is to take a path between the endpoints of your path to get a loop which you can homotope into [math]A[/math], but the other endpoint will likely move under this homotopy. This can be saved as long as you make sure the endpoint doesn't leave [math]A[/math] at any point in the homotopy, but why should this be possible?

[Anonymous No.8399740]

I've been programming since I was a kid but only coding not getting into theoretical or mathematical computer science. I see when people talk in plain English about the cool interesting things they do they describe it with buzwords like data mining, supervised and unsupervised machine learning, optimization, clustering, classifiers, support vector machines, similarity search, recommender systems, artificial neural networks, etc etc. I look up what these things are and all see is math which is like gibberish. So I look up what prerequisites its things like linear algebra, statistics, calculus, etc. I've long forgotten most of highschool algebra. How can I ever hope to get to this level. It seems like a goal which will be impossible to reach from where I currently am. Will I ever be able to get to where I want to go from where I'm at?

[Anonymous No.8399770]

>>8399740
Sure, provided that you put in the effort and you aren't allergic to math. Strong AI is mostly math with a bit of programming.

[Anonymous No.8399785]

we are moving about 1/9th of the speed of light in space, do we account for that when calculating for instance time dilation? when my textbook says someone is moving with 0.8c on earth, does it take into account that we are already moving about 1/9th of the speed of light relative to (not sure about this-->) "space"

[Anonymous No.8399859]

>>8399635
mellin transform is an example of something different

[Anonymous No.8399873]

>>8399635
>>8399859
https://duckduckgo.com/?q=mellin+transform+fractional+differential+equation

[Anonymous No.8399884]

>>8399785

That's meaningless.

You can't create the notion of an absolute speed by using 'space" as a reference point

[Anonymous No.8399900]

>>8399724
You've made your loop using some path contained solely in [math]A[/math] between the same end-points, ok?

Just concatenate this exact same path to your whole loop homotopy and it will give you the needed path homotopy.

pic related

[Anonymous No.8400171]

>>8399364
It in a proof of gcd(a,b)=gcd(b,r) I think. It was definitely divides, though I think it's an abbreviation for something. Maybe that a|b with remainder c? I was hoping someone may have seen it before.

[Anonymous No.8400174]

What is the best institute on anti aeging that will make good use of my money?

[Anonymous No.8400204]

I'm sorry for bosting homeworks, and don't want an answer, just someone to explain what the fuck is going on with the value of Cv,m here - why has the lecturer attached operations instead of just giving the value?

[Anonymous No.8400207]

>>8400204
Cv,m is just a function of T, bro. Plug it in and integrate.

[Anonymous No.8400214]

>>8400207
Danks, man.

[Anonymous No.8400236]

Hey guys mathbaby here
I need to learn functions (linear/trig/etc) -> limits -> derivatives in 10 days, what do you recommend?

[Anonymous No.8400301]

>>8400236
Depends on how extensive your knowledge needs to be. Your best bet is Khan Academy.

I don't really know why you need to learn this in 10 days, but I'll assume you're about to get into a Calc course or something. So here's the run-down of what you need to know (I'll probably miss some things):

Functions:
Know the definitions of the domain and range of a function.
Learn how to determine the domains and ranges of functions.
Know the domains and ranges of the following functions:
f(x)=x
f(x)=x^2
f(x)=x^3
f(x)=1/x
f(x)=sin(x)
f(x)=cos(x)
f(x)=tan(x)
f(x)=ln(x)
f(x)=e^x
Know how to evaluate all of the trigonometric functions at the points on the unit circle.
Know how to find the composite of two functions.
Know what a one-to-one function is.
Know how to find inverses of one-to-one functions.

Limits:
Limits are pretty straight-forward, but you need to be very fluent in algebra, i.e. you need to remember several different tricks. Learn how to evaluate limits both at specific points and at positive and negative infinity.

Derivatives:
Know the definition of the derivative. Many calc noobs are scared of the definition, but you should learn it because your calc class might need you to prove some of the various differentiation rules using the definition.
Know what a derivative is, intuitively.
Know the power rule, product rule, quotient rule, and chain rule.
Know the derivatives of the trigonometric and logarithmic functions.

Good luck. I can't really give suggestions on how to learn this stuff in 10 days. It's not really that hard; it all really amounts to lots of memorization and doing a lot of exercises. Khan Academy is a good resource for exercises. Look the concepts up online. There are a shit-ton of resources on this stuff a Google search away.

[Anonymous No.8400303]

I'm still struggling to apply orthogonality arguments when solving for certain things in my intro to QM class.

mainly expectations values in the infinite square well. I'm convinced that explicitly solving complex integrals for lets say the expectation value of x^2 isnt necessary, but i've yet to fully understand how to apply the orthonormality of psi to make the problem less tedious.

[Anonymous No.8400309]

>>8400301
wow, thank you for your detailed answer!
Do you think I should do stewart calculus exercises as well?

[Anonymous No.8400323]

I have a virology question.

To my understanding when measuring the pro-viral load of a blood specimen you are taking into consideration the viral genome integrated into the blood cells right?

Why then is it necessary not to centrifuge the plasma (assuming using an EDTA blood specimen tube) from the blood? Why can't you still spin it and then just measure the blood cells minus plasma?

[Anonymous No.8400325]

>>8400309
>>8400309
>Do you think I should do stewart calculus exercises as well?
There's no such thing as doing too few exercises. If you think you have the time in these 10 days to do those exercises, go for it. And Stewart will probably have better explanations of the concepts than any half-assed online explanation would (though I haven't read the book myself).

My list is in no ways exhaustive, by the way. I would definitely spend some time focusing on limits, because they tend to really fuck people up. Brush up on your algebra, know what you can and cannot do. Then learn all the specific tricks there are to evaluating limits. Differentiation is important, but it can come later (again, I don't know what your situation is; I'm assuming you're preparing for a calc class).

[Anonymous No.8400335]

>>8394032
How do i get a job at Roskosmos,im not Russian

[Anonymous No.8400336]

>>8400325
You're right (prepping for test).
I'm not in a bad shape though, I kinda understand the concepts of limits and derivatives , the only thing I might really need to get gud at are the e^x indetermination or something like that.

But anyways, thank you for your time!

[Anonymous No.8400337]

some kid in 5th grade with aspergers is saying that math is "illogical" to him and having a hard time with numbers
is there any way to help him?

[Anonymous No.8400406]

Why can't we take a page from cancer for life extensions studies?

[Anonymous No.8400517]

>>8399359
Does nobody know? ;_;

[Anonymous No.8400583]

Is there a degree where I don't have to do labs?

[Anonymous No.8400617]

>>8400583
Sociology

[Anonymous No.8400669]

>wanted to become an engineer
>terrible at labs
>get carried by lab partner
>I know he hates me

What should I do? I'm talking about all the labs. Even in Physics.

[Anonymous No.8400687]

>>8400583
math

[Anonymous No.8400690]

Where does /sci/ get their science related news?
I'm interested in reading about new studies as they're released.
Preferably no sites with sensationalized articles.

[Anonymous No.8400706]

Could someone help me a little bit here? I'm reading this proof about polynomial congruences but I'm not a native English speaker so I'm having some trouble understanding what this one part means.

"If [math]s[/math] is a solution of [math]x^{2} \equiv a \ (mod \ 2^{k})[/math], then squaring [math]-s[/math] and [math]\pm s + 2^{k-1}[/math] modulo [math]2^{k}[/math] shows that these are also solutions; by taking least positive residues, we may suppose that all of the solutions are positive and less than [math]2^{k}[/math]."

What does the "by taking least positives residues" part mean? Does "least positive" just mean "smallest" or what? I'm having a bit of trouble understanding what we're taking here.

[Anonymous No.8400754]

>>8399354
tyvm anon

[Anonymous No.8400804]

>>8400706
the phrase states that if s is a solution,
i.e. s^2=a mod 2^k,
then
(-s)^2=s^2=a mod 2^k
and
(plus minus s + 2^(k-1))^2
= s^2 plus minus 2*s*2^(k-1)+2^[2(k-1)]
= a plus minus s*2^k+(2^k)*(2^[k-2])
=a mod 2^k


one of these four solutions is positive and closest to 0, that is the least positive residue

for example, if s>2^(k-1) then:
-s<-2^k,
s+2^(k-1)>2^k,
and
-s+2^(k-1) < 0.

in which case taking the least positive residue would be taking s

[Anonymous No.8400813]

>>8400706
the least positive residue is the number b, with 0<=b<= 2^k - 1, such that b is congruent to whatever number you're taking.

Every negative solution or solution bigger than 2^k will be congruent to one of these

[Anonymous No.8400821]

>>8400813
there might be multiple numbers in the [0, 2^k-1] interval, you need to specifically choose the smalest

[Anonymous No.8400837]

>>8400804
Okay, so it simply means choosing the smallest positive solution. Thanks for clearing that up.

[Anonymous No.8400942]

how is the final image not 20.44cm?
Treating the image of lens 1 as the object for lens 2, it seems like this should be right....

[Anonymous No.8400948]

>>8400942
also, that interactive learning ware shit doesnt work

[Anonymous No.8400949]

Is it possible to make something like a solar forge with a shit ton of LED lights?

Like if you have a massive pile of 100W LED lights
had them in an array of some sort
using dozens of lens to combine them, and focus them all on a pinprick

Could you essentially make that pinprick at a designated range, based on the final lens, able to do things list instantly burn through wood.


Not asking if it's practical, a laser can obviously do it better, just want to know if it's possible to do with flashlights.

[Anonymous No.8400956]

any good circuit analysis youtube channels?

[Anonymous No.8400963]

>>8400949
It's less efficient then just using the electricity directly. But it's theoretically possible.

[Anonymous No.8400980]

>>8400963
I forgot to mention I'm not talking about actually making a forge, I was just using it as an example of the concept.

[Anonymous No.8401368]

>>8398882
Anyone know what "destructive wave" this anon is referring to?

[Anonymous No.8401849]

When are we going to get China's results on the first lung cancer patient treated with CRISPR?

[Anonymous No.8401865]

[math](x^2 -4x) * (x^2 +2x -3) = 0 [/math]
I'm supposed to solve this using the zero rule. It works well if I say x=0, I think, but I threw it into a calculator and got four answers. Like
[math]x-4=0 [/math] so [math] x = 4 [/math]
[math]x+3=0 [/math] so [math] x = -3 [/math]
[math]x-1=0 [/math] so [math] x = 1 [/math]
I really don't get it.

[Anonymous No.8401884]

>>8401865
You need to fully factorize the equation.

[Anonymous No.8401889]

>>8400171
Not enough context, post the full sentence it came from

[Anonymous No.8401894]

>>8401865
Yeah, 0 is one answer, but there are other possible values for x.

First, factor out some stuff. You can factor out an x from [math]x^2 - 4x[/math], so [math]x^2 - 4x = x(x - 4)[/math]. Also, [math]x^2 + 2x - 3 = (x + 3)(x - 1)[/math].
So we have
[math]x(x - 4)(x + 3)(x - 1) = 0[/math].

Divide this by [math](x - 4)(x + 3)(x - 1)[/math] to get [math]x = 0[/math].
Divide this by [math]x(x + 3)(x - 1)[math] to get [math]x - 4 = 0[/math].
Divide this by [math]x(x - 4)(x -1)[/math] to get [math]x + 3 = 0[/math].
Divide this by [math]x(x - 4)(x + 3)[/math] to get [math]x - 1 = 0[/math].

From there, you can find the four possible values of x.

[Anonymous No.8401914]

>>8401884
Thanks. I'm researching how to do that, however I'm stuck at something.
>>8401894
>x2−4x=x(x−4)
Bro, thank you, but you lost me here. If it's [math]x*x[/math] and [math] -4*x[/math] and I can see that the way you factored it is the same but why can't I say [math]x*x-4*x[/math]?

[Anonymous No.8401934]

>>8401914
I'm not really sure what you're asking, but [math]x*x - 4*x[/math] is not fully factored. In fact, when you go from [math]x^2 - 4x[/math] to [math]x*x - 4*x[/math], you didn't factor at all. You want to "pull out" the x and get [math]x(x - 4)[/math] because x is a common factor of x^2 and -4x.

[Anonymous No.8401950]

Do I have to work with people when taking Computer Science classes? Can I work by myself?

[Anonymous No.8401956]

>>8401934
Oh shit! Yeah I see. Thank you!

[Anonymous No.8402065]

How do I even do this?

[Anonymous No.8402073]

>>8402065
10 volts across six ohms is how i current.

In the second loop there's 3i current through 5 ohms.

[Anonymous No.8402100]

What concrete proof do we have that Earth is 4.6 billions years old and not 13.8billions years old apart from radiometric dating? What if we find a rock that is 5billions years old on Earth.

[Anonymous No.8402187]

I need to find the arc length of a curve given in parameter form

x= t ^5
y= t^2
T goes from 1 to 3.
I have no idea of how to solve the integral.

[Anonymous No.8402195]

I dont understand vector calculus. In the current PDE's class, they're doing stuff in Rn and I'm not sure how they play around with everything so fast. I never took physics courses, so is there a short book that will let me git gud?

[Anonymous No.8402198]

>>8402187
whats the integral?

[Anonymous No.8402206]

>>8402198
Ah sorry. uh to calculate the arc length in parametric form you need to solve the integral of the sqrt((derivative of x)^2 +(derivative of y)^2) both in relation to t. I really wish to print pretty math in /sci/, how do I do that?

[Anonymous No.8402228]

>>8402206
you sure the exponents are right? seems pretty non-trivial

[Anonymous No.8402241]

>>8402228
Yup. And she isn't a bad teacher or gives this kind of shit normally so I thought I was being stupid. These two are quite hairy.
The answers are (according to her)
>a) 1/27 (85*sqr(85)- 13 * sqrt (13))
>b) 8

And I might be stupid and some anon finds a easy way of solving them.

[Anonymous No.8402250]

>>8402241
well
>a) 1/27 (85*sqr(85)- 13 * sqrt (13))
is about 27 according to wolfram

while the integral is about 242:
http://www.wolframalpha.com/input/?i=integrate+((5t%5E4)%5E2%2B(2t)%5E2)%5E(1%2F2)+from+1+to+3

and in the graph, looks much longer than 27:
http://www.wolframalpha.com/input/?i=(t%5E5,+t%5E2)+from+t%3D1+to+3

[Anonymous No.8402255]

>>8402250
Well, I'll see her monday. Thanks anon.

[Anonymous No.8402257]

>>8402250
and
(t^3, t^2) from t=1 to 3 seems to be the curve that has length 1/27(85sqrt(85)-13sqrt(13)):
http://www.wolframalpha.com/input/?i=(t%5E3,+t%5E2)+from+t%3D1+to+3

[Anonymous No.8402258]

>>8402257
>>8402255
(which is much easier to integrate of course)

[Anonymous No.8402259]

>>8402255
Probably a typo that made me lose 30 minutes and lots of paper. Thanks again.

[Anonymous No.8402266]

How fast would medicine improve if we said "fuck it, let's experiment on humans?"

[Anonymous No.8402276]

>>8402259
looks like you want a double angle formula for question c (you should get a 1-cos(t) in a square root)

http://www.sosmath.com/trig/douangl/douangl.html

[Anonymous No.8402350]

Hello bros, can u please give examples of E1, E2 and E1CB reaction mechanisms with biomolecules.

[Anonymous No.8402429]

doing a paper on traveling near light speed, what you would observe, feel, and how others would observe you, so mostly relativity.

Done with phase1, now need loads of sources, would appreciate if you could suggest articles, books, etc. thanks

[Anonymous No.8402511]

Is "interstellar speeds" okay way to call speeds that near the speed of light

[Anonymous No.8402538]

How come Engineering is the only discipline where the professors essentially tell you to quit if you don't know how to do something?

[Anonymous No.8402543]

How ends the universe? (Your opinion)

[Anonymous No.8402607]

>>8402065

Is this correct? Also, I still don't get how 3is is positive. Am I correct to say that the Io current going to the resisters will be negative and going away would make the current positive, right? Is 3Is positive because it's dependent to Vo and since the current is going from left to right then Vo is positive which makes 3Is positive?

[Anonymous No.8402640]

>>8402543
1) With big me being the universe itself, which by the way will be totally self-descripted quine in its happy unchanging state.
2) The same but it doesn't end, only converges to this state

[Anonymous No.8402658]

>>8397592
It's all about tastes, but I think that organic chemistry is one of the most interesting, although difficult for most (and can't figure out why). Organic is also probably one of the most useful work-wise, e.g. polymers or pharmaceutical

[Anonymous No.8402669]

>>8402658
>organic chemistry is one of the most interesting, although difficult for most (and can't figure out why).
Organic chem is a series of "tricks" and "tips" that "work" "most times."

[Anonymous No.8402779]

>>8402669
I know, and yet when I did the first course few years ago, at least 80% of the class failed it and a good 30% failed it more than once...

[Anonymous No.8402795]

>>8401950
There might be some group programming assignments. Ask the professors.

[Anonymous No.8402797]

https://www.youtube.com/watch?v=kjupT6snXH0

[Anonymous No.8402800]

>>8402429
Mr. Tompkins

[Anonymous No.8402878]

Does Fermat's Last Theorem hold for [math]a, b, c \in \mathbb{R}[/math]?

[Anonymous No.8403057]

>>8402607
Look at the polarity markings on the voltage and current sources.

[Anonymous No.8403059]

>>8402878
No. There is always a solution in R:
c=(a^n+b^n)^(1/n).

[Anonymous No.8403117]

>>8394032
already made a thread on this but I think it's better suited here:
I have no calculus experience whatsoever (furthest I've been is trig) and I have recently been exposed to the idea of PID (proportional, integral, derivative) loop control. For the sake of helping myself better understand it, I decided to sketch graphs of each one independently. I would assume the correction had a 1 to 1 correlation with the adjustment made and declared a setpoint from which the error was measured.
P and D were easy. As long as P is less than 1 (in my particular circumstance), the process approaches but never reaches the setpoint ala Zeno's paradox. As long as D is less than 1 (in my particular circumstance) any disturbance will immediately flatten out and never making an attempt to move to the setpoint. "I" is the one I have a question about though.

No matter what "I" value I chose, it resulted in unending oscillation that never shrinks. If the "I" value is greater than 3, it continuously grows. The thing that interests me, though, is what happens when I round the numbers. with a value of 0.5, I get pic related. It's an endless oscillation like usual, every 0.5 was rounded up to 1. But it follows a pattern. Since each apex is the point where the sum of the opposite side becomes less than that on its side, each wave can be cut in half, halves with like sums are highlighted with like colors.

To be clear on how my graph works: red line is the set point, essentially 0. Yellow lines are the error. The total error is summed up and multiplied by my "I" value (in this case, 0.5) to get a correction. The next point on the graph is the current position, - (the sum of the all errors * "I" value).

The pink sums are 6, grey is 10, yellow is 3, reddish-brown is 8 and green is 5.
This pattern continues after my graph in the opposite direction.

So... anything special about this pattern? Or about what I am doing here?

[Anonymous No.8403143]

>>8402511
not really, "interstellar speed" implies it's a reasonable speed to travel from one star to another. The nearest star to us (proxima centauri) is 4.37 light years away. Unless you consider that a reasonable amount of time, it's not really "interstellar speed"

[Anonymous No.8404293]

I have a question.
[math]y=-x^2+8x-13[/math]
What is a? How do I write it in, for instance, [math]D=b^2-4*a*c[/math]

[Anonymous No.8404312]

>>8404293
The coefficient of x^2, in this case -1.

[Anonymous No.8404320]

>>8404312
If a isn't there then is it always -1 or 1 then? For instance if it was just [math]x^2[math] it would be 1?

[Anonymous No.8404339]

>>8404320
a is there; it's -1.

>if it was just x^2 it would be 1
Yes.

[Anonymous No.8404345]

>>8404339
> is there; it's -1
Sorry it slipped.
And thank you.

[Anonymous No.8404346]

>>8404293
a is the coefficient attached to the x^2.

[Anonymous No.8404365]

Show that if P and Q are non zero polynomials, then the degree of (PofQ) is equal to the product of the degree of P with the degree of Q

[Anonymous No.8404397]

>>8404365
by the fundamental theorem of algebra, every polynomial of degree n > 1 can be factored into n terms of degree 1. Substituting in a polynomial of degree m, this factored form will create n products of degree m polynomials. Specifically, the leading term of each polynomial will be on the order of [math]x^m[/math] and there will be n of these products, so [math](x^m)^n[/math] which by the law of exponents is just [math]x^{m \cdot n}[/math]

[Anonymous No.8404402]

[Anonymous No.8404410]

what "functions"? I don't get it where are [math]f_{i}[/math] coming from? I thought it's just one function that takes a vector and outputs a vector?

[Anonymous No.8404411]

How ends the universe? Your opinion.

[Anonymous No.8404417]

>>8404410
Yes, they mean [math]f[/math]'s coordinates in [math]\mathbb{R}^m[/math]

[Anonymous No.8404418]

>>8404410
I was going to persercute you but then I realized this is SQG.
We have
[eqn]
f(x_1, x_2, \hdots, x_n) = (f_1(x_1, x_2, \hdots, x_n), \hdots, f_m(x_1, x_2, \hdots, x_n))
[/eqn]
General advice: try reading a little more closely next time.

[Anonymous No.8404446]

>>8394067
We'll be using the "choose" function here, i.e. a binomial coefficient, since order does not matter and since there is no repetition.
Note that each unique 3-set of dots corresponds to a unique triangle, so we are not worried about double counting.
There are 16C3 possible ways to select a 3-set of dots. However, we can be more restrictive by not considering straight lines as triangles. We only obtain a straight line when all three dots in our 3-set lie on the same line. There are 4C1 * 5C3 ways to select a straight line (first term selects the line on which the dots lie, second term selects the dots themselves.
Therefore there are 16C3 - 4C1 * 5C3 ways of selecting triangles, or 520 ways select a 3-set of dots that form a unique triangle.

Chris can draw all of these triangles in 8 minutes and 40 seconds, so he will be able to save her if he can exclude straight lines.

[Anonymous No.8404466]

I believe this problem is impossible.

I could find a solution if it was asking to make them all upside up (by flipping DDDU 15times and using the remaining 5 Down cups to flip these remaining downs)

Pretty sure this is impossible, but how can it be shown?

[Anonymous No.8404470]

>>8404411

Checked.

Space loops round on itself, why not time? The Universe is a mobius strip.

[Anonymous No.8404471]

>>8404466
>Pretty sure this is impossible, but how can it be shown?
find an invariant

[Anonymous No.8404481]

>>8404470
False, the universe is a klein bottle.

[Anonymous No.8404491]

Why does \begin {align*} blocks align all my equations to the right?

[Anonymous No.8404496]

>>8404466
It's impossible because you can only change the number of "corrected" cups by a multiple of two:
Correct, correct, correct, correct (+4)
Correct, correct, current, incorrect (+2)
Correct, correct, incorrect, incorrect (0)
Correct, incorrect, incorrect, incorrect (-2)
Incorrect, incorrect, incorrect, incorrect (-4)

Since 51 is not divisible by two, it's impossible.

[Anonymous No.8404540]

>>8404491
Post a minimal example code on pastebin.
In case its something really silly...are you ending lines with ' &\\'?

[Anonymous No.8404546]

>>8404540
Oh.. maybe because I'm ending lines with \\? How should I do it instead?

http://pastebin.com/gDJ1EKrK

[Anonymous No.8404549]

>>8404466
At any given time you can change the number of upside down cups by -4, +4, -2, +2, or 0. Can you find a solution (q, r, s, t) to the equation -4q + 4r - 2s + 2t + 50 = 51 + 4q - 4r + 2s - 2t?

[Anonymous No.8404550]

If a conditional statement is opinionative, what conclusions can you make on its truth value?
>if food is from KFC, then it is finger licking good

Thanks.

[Anonymous No.8404554]

>>8404546
You have to indicate where you want your equation to align by using &, e.g.
f(x) &= x^2 +1 \\
&= (x + 1)(x - 1)

[Anonymous No.8404565]

>>8404554
I just want them to be one after the other like multiple lines.. I don't have to make an align block for each of them, do i? The different lines dont continue the previous lines

[Anonymous No.8404568]

>>8404481
>False, the universe is a klein bottle
Theory? links?

[Anonymous No.8404576]

>>8404565
Then you just want normal alignment? Try using the equation environment instead of align* instead, and see if that tickles your fancy. If you're actually looking for some sort of intelligent alignment, you'll have to figure that out for yourself by actually reading about how the environment works. Maybe try putting '&' before 'Optimisation', 'Objective', and 'Constraints'

[Anonymous No.8404599]

>>8404397
That has to be the worst case of overkill I've ever seen

[Anonymous No.8404600]

>>8404576
>Then you just want normal alignment?
Yes please, but I don't know how to get it. usint \equation{} numbers them though, and I wanted not to number everything

[Anonymous No.8404632]

>>8404600
Try
\[
latex here
\]
instead.