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/sci/ - Science & Math

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/sqt/ - stupid questions thread or QTDDTOT

For book recommendations, check the sticky and/or the /sci/ wiki

For learning how to use the inboard latex, check the sticky. You can also test your latex before you post by clicking the "TEX" button in your reply box.

If your latex isn't working, it's because your adblocker is blocking it.

If you ask any question, remember that there is almost no universal notation:
>what constitutes a BAD question
If p divides |G|, show that there exists an element of order p.
>what constitutes a GOOD question
Suppose p is a prime that divides the order of a finite group G. Show that there exists an element of order p.
>>
The Radius r and height h of a right circular cylinder are measured with possible errors of 5% and 3% respectively. Approximate the maximum possible percent error in measuring the volume.

dr = .05
dh = .03
V = pi*r^2*h
Here I took the partial of V with respect to r times dr + partial of V with respect to h times dh
dV = 2pi*r*dr + pi*r^2*dh

dV/V = 2/r * dr + 1/h * dh

What do I do now? There are no values of R or H to work with. Any ideas?
>>
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>The negation of the bijection (double arrow) operator is the same as NOT XOR(A,B)
>mfw
This shit blew my mind in lecture today... Why is this?
>>
>>10058316
Think about it this way: A XOR B is true if and only if A and B are different.

Another way to think of it: XOR is the same as addition modulo 2, but modulo 2, we have 1 = -1, so it's the same as subtraction. Thus you can think of XOR as taking the difference of two bits.
>>
#!/bin/bash
while ! foo
do
sleep 1
echo "Restarting program..."
done

Why does the above script open foo in the first place? How does it start foo again if it fails? Shouldn't it just check if foo is not running and echo a message?
>>
ok, peak stupid question here

light. How the fuck do we create black and white colored things

Black is supposed to be the absorption of all light, white is every visible color on the spectrum. But for something to be a color,m it has to reflect that color into the eye odf the person looking at it. How do we make threads black and white
>>
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>>10059060
$\text{pressure} = P = \rho g h$
$\text{vert force} = \int P cos(\theta) dA$

in this case, i'd use $dA = L R d \theta$ and $h = R \cos \theta$ and integrate from 0 to pi/2.
>>
>>10059068
It's the wrong answer, it's supposed to be ~641 kN
>>
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>>10059068
(rewrote this post, missed a cosine)
to make it explicit,

$\text{vert force} = \int_{0}^{\pi / 2} (\rho g (H - R \cos \theta))(\cos \theta)(L R d \theta)$

where H = total height of container (7 m), L = width (6 m), and theta is the angle on the curved surface from the vertical face
>>
>>10059081
>>10059085
to be extra sure :p
http://www.wolframalpha.com/input/?i=int+from+0..pi%2F2+of+(997+*+9.81+(7+-+2+*+cos(theta)))+cos(theta)+*+6+*+2+*+dtheta
>>
>>10059068
and because i'm annoyed at small mistakes

i should have said $h = H - R \cos \theta$, not just $h = R \cos \theta$
>>
This is probably a really stupid question, but when doing a coordinate transform from Cartesian to polar (or in general really), if i have
$dx = \cos \varphi dr - r \sin \varphi d\varphi$
Does that mean that transforming the differential operator is as simple as
$\frac{\partial}{\partial x} &= \frac{1}{\cos \varphi}\frac{\partial}{\partial r} - \frac{1}{r \sin \varphi} \frac{\partial}{\partial \varphi}$
>>
>>10059108
fixed 2nd expression:
$\frac{\partial}{\partial x} = \frac{1}{\cos \varphi}\frac{\partial}{\partial r} - \frac{1}{r \sin \varphi} \frac{\partial}{\partial \varphi}$
>>
>>10059060
Aren't there formulas for this type of stuff in fluid mechanics?

Look inside your book dude, seriously.
>>
>>10059110
nu,

$\frac{\partial}{\partial x} = \frac{\partial r}{\partial x}\frac{\partial}{\partial r} + \frac{\partial \phi}{\partial x}\frac{\partial}{\partial \phi}$
>>
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Let a = -rx and b = 1 + x^2.

How do I tell which function affects the flow of x(dot)? Why does b > a mean the flow I don't understand their explanation.
>>
data is sent through a channel, with error free probability of p=0.9, find the probability that at least 3 of the first 5 packets are error free.

Is this a binomial, or geometric PMF?
>>
How should I go about finding the area of the triangle in the centre? (The big triangle is equilateral)
>>
>>10058465
programs generally return 0 if they were successfull or nonzero if not successfull.

the first line of that code says:
while program "foo" does not return 0, loop (done=loop)
>>
>>10058316
If you draw them as Venn diagrams, it's obvious.
>>
>>10058290
Never really gave this some thought since it's basically second nature to me but should I consider using banker's rounding or should I just continue using what I've learned from grade school and just round half away from 0?
>>
This is a real brainlet question, but how would I represent 560000 in the form x = ±b * 10^e where 0.1 ≤ b ≤ 1?
>>
>>10059298
>This is a real brainlet question
Yes, it is. How often do you need to divide 560000 by 10 until the result (call it b) is less than 1? That's your e.
>>
>>10059305
Oh, I knew how to do that, I'm just wondering how to actually represent it; 0.56 * 10^6?
>>
>>10059311
yep, that's right.
people generally dont use e as a free variable as that often gets people confused with euler's number
>>
>>10059321
Thank you! It's just supposed to represent the exponent according to the assignment so it's fine within the scope of the module.

What I don't know is how to represent a negative number in the form though; how would you suggest I do go about that?
>>
>>10058509
Use a dye that absorbs most of the spectrum for black, and reflects most of it for white... Whaddya missing here?
>>
>>10059202
The values of $\dot{x}$ dictate flow in the sense that as we evolve with time $\dot{x}$ tells us how quickly our solution $x(t)$ grows or shrinks. You let time evolve and your solution $x(t)$ 'flows', perhaps growing and growing towards some fixed point.

You can get a quick, qualitative understanding of this flow by checking when $\dot{x}$ is positive, negative, or zero. If, say, $\dot{x} > 0$ then the value of $x(t)$ grows with time. Just as you are used to with derivatives.

So lets say we want to find when $\dot{x} = 1 + rx + x^2 > 0$. This happens when $1+x^2 > -rx$. So if we plot $y = 1 + x^2$ and $y = -rx$ we know $\dot{x} > 0$ when the curve of $y = 1 + x^2$ is above the curve $y = -rx$.
>>
>>10059330

how do you make dyes?
>>
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>>10059486
That's very easy -- see pic related
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>>10059486
trial and error mostly, with chemicals that you know what colour will turn into. In chemistry it is almost impossible to guess what colour a compound will turn out to be before you observe it
>>
Are the chances of becoming a tenure track math professor still really slim if you are applying internationally? I will teach mathematics in Rwanda if it's a tenure track position.
>>
>>10059460
Is there an explanation as to why we subtract - rx rather than the other part of the equation? Is it obvious to someone who has a decent background in calculus? Does it have to do with the growth of rx vs x^2?
>>
Anyone know anything about logic design?
So a ripple carry adder is slower than a carry look agead adder when there are many inputs.
So why would you choose a ripple carry over a carry look ahead? Price?
>>
Guys, would (121 + 0.327) - 119 and (121 - 119) + 0.327 when rounded to three significant figures not have the exact same absolute and relative errors?
>>
>>10059765
Oh wait, nevermind. The specification says to round to three significant figures after each operation.
>>
>>10059591
Tenure at all outside of North America is uncommon. A few Euro countries have it in some form, but it's generally even more absurd to get into than in America. I think I heard a number once that something like <10% of German professors are tenure-track or tenured.
However if you are trying for a US or Canadian school as an international PhD your chances are probably only marginally more tiny than domestic applicants. I've met plenty of tenure-track or tenured professors in the US with PhDs from England, Russia, Germany, Australia, etc.
>>
>>10059765
Shitty lesson to never round until the end?
>>
>>10059766
are you sure? I dont think you're supposed to round till the end

>>10059325
let b be negative
>>
>>10059755
I'm not certain but it probably doesn't matter. Choosing certain representations may lead to an improved understanding of the problem.

One thing to note is that we could have also plotted y = 1 + rx + x^2 and checked where 1 + rx + x^2 > 0 for certain r values.
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Someone explain me pic related please.
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>>10059962
Write $a = e^{lna}$, since e and log are inverses to each other.
then raise both sides to b.
>>
I know f-tests are used to determine whether the coefficients in a regression are jointly zero (or rather, whether the null hypothesis that they are can be rejected)

I'm just curious why this is the case. How does the f-distribution happen to enable this?
>>
>>10059962

if it makes it easier just put the $b$ back into the log term to get $e^{ln(a^{b})}$ and then do the reverse step
>>
>>10059962
$(x^y)^z = x^(y z)$ so $e^(b \ln a) = (e^{\ln a})^b = a^b$

also, $(x^y)^z = e^(x y)$ since you take $x$, multiply it by itself $y$ times, then take all that and multiply it by itself $z$ times, so it's like just multiplying by itself $y*z$ times in total
>>
>>10059985
>>10060124
>>10060131
I understand it now.
Thank you, anons!
>>
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How can I avoid the Dunning-Kruger Effect?
>>
Meant to post >>10060216 here. Any takers?
>>
How hard is it to get into a grad program in a first world country, assuming I have two majors with 4.0 in a top 100 college?
>>
>>10060269
What color is your skin?
>>
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Whenever I learn something or try to solve a problem I always feel as if some part of my brain is asleep and therefore unable to work at full potential. How can I either switch on that part or get rid of this feeling?
>>
>>10060302
White, at least compared with most of the third world.
>>
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how do I show $A\subseteq \mathbb{R}$ is Lebesgue measurable if and only if $A+y$ is Lebesgue measurable, for $y\in \mathbb{R}$. I know that the outer measure is translation invariant. For instance, $m(A+y) = m^{*}(A+y) = m^{*}(A)$, but I can't just conclude from this that $m(A)$ follows, right?
>>
>>10060219
First of all note that $\bigcap^{\infty}_{n=1} \bigcup^{\infty}_{k=n} A_{k}$ is just $lim{sup{A_{n}}}$. If you understand what the concept behind $lim{sup{A_{n}}}$ is, then it's gonna be way easier.

Suppose $m(\bigcap^{\infty}_{n=1} \bigcup^{\infty}_{k=n} A_{k}) \neq 0$. Then, since $m$ is a measure, it can't be lower than zero, so the only possibility is that $m(\bigcap^{\infty}_{n=1} \bigcup^{\infty}_{k=n} A_{k}) > 0$.

In that case, what does the $m(lim { }sup{A_{n}})$ tell you about each $A_{k}$ (at least about an infinite amount of them)? When you figure that out, analyze the sequence of real numbers $m(A_{k})$ $\forall k \in \mathbb{N}$. Can you say anything about its convergence? Can you tell now where is the contradiction, when evaluating the series sum?
>>
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>>10060336
>>10060219
>that one guy who has been really struggling with his measure theory/analysis course and is in these threads daily
I feel for you senpai
>>
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>>10060392
>>
Retard here:
Why do dead things decompose in a week? What's going on when you're alive that stops bacteria from decomposing you? Or animals, trees, whichever.
>>
>>10060421
Cells are still alive and being fed, which makes them harder to break down, in addition to various living defense systems. Once the food pipeline gets cut and they die, they kinda go to mush and leak their sweet juicy bits all over for anything to come along and pick up, and nothing keeps the bacteria and other life that was previously struggling to survive inside the organism from suddenly running rampant.

Plants generally take a lot longer to decompose, in part due to their cell walls and being less dependant on central systems (in some cases to the degree where they can re-root and make a new plant).
>>
>>10059210
Dunno. I got 12 using coordinate geometry (interpolant=1/3), but I'm guessing that there's a much simpler geometric approach.
>>
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>>10060336
bump
>>
>>10060336
>>10060461
Give the definition of Lebesgue measurable set you were using earlier on.
>>
>>10060451
Well it's a math olympiad so yes, there has to be
>>
>>10060502
A is Lebesgue measurable if $\forall \epsilon>0$ there exists an open set $G$ with $A\subseteq G$ and $m^{*}(G\setminus A) < \epsilon$
>>
>>10060521
If G is an open set of R with the standard topology, wouldn't a translation of G be an open set as well, so couldn't you run the same argument to show A+y is Lebesgue measurable?
>>
>>10060521
1) If $A \subseteq \mathbb{R}$ is Lebesgue measurable, then $A+y$ is Lebesgue measurable $\forall y \in \mathbb{R}$
$A$ is Lebesgue measurable, then $\forall \varepsilon > 0$ $\exists G$ such that $A \subseteq G$ and $m^{*}(G \setminus A) < \varepsilon$.
Consider the set $G+y$, which must also be open $\forall y \in \mathbb{R}$ since $G$ is open (don't know if you have proved it but it's basic topology of the real line, so it's simple to prove). We have to prove that $\forall \varepsilon > 0$ we can use $G+y$ such that $m^{*}(G+y \setminus A+y) < \varepsilon$.

$m^{*}(G \setminus A) < \varepsilon$ means that $(G \setminus A)$ can be covered with a (countably) infinite bunch of closed intervals of arbitrary length. The important thing to note is that now that we reduced the work to just intervals, and it is really easy to proof that the length of closed intervals is translation invariant. Since the length of an interval $\cal{l} ([a,b])=b-a$, you can proof that the interval $[a+y,b+y]$ has the same length, for every $y \in \mathbb{R}$.

Now, for each $\varepsilon > 0$, you can find a set $G$ such that an outer cover of $(G \setminus A)$ has an outer measure lower than $\varepsilon$. That cover is made of intervals, so if it is given by $\bigcup_{k=1}^{\infty} I_{k}$, consider the cover $\bigcup_{k=1}^{\infty} I_{k}+y$. It still retains the same total length, but now it covers $(G+y \setminus A+y)$ instead. With that, you should be able to proof that $m^{*}(G+y \setminus A+y) < \varepsilon$, and thus you found the open set that lets you state that $A+y$ is Lebesgue measurable.
>>
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>>10060582
this makes perfect sense...
>>
>>10060605
this too >>10060584
>>
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Can someone help me with this derivative
>>
>>10060521
Run out of space in >>10060584 so I'll just leave some other stuff here. The second part of the proof (to prove that it is "if and only if") should be:
2) If $A+y$ is Lebesgue measurable, for $y \in \mathbb{R}$, then A is Lebesgue measurable
This part is trivial because if it is $A+y$ is always Lebesgue measurable for any value of $y$, then, in particular for $y=0$, it must also hold true.

Also I just noticed that the first part can be done much faster if you already know that the Lebesgue outer measure is translation invariant (which you do). If $m^{*}(G \setminus A) < \varepsilon$, since you know that the outer measure is translation invariant, you can immediately state that $m^{*}((G \setminus A)+y) < \varepsilon$. Then, since $G+y$ is open (from what I told earlier), and $(G \setminus A)+y=G+y \setminus A+y$, you already get the result that $m^{*}(G+y \setminus A+y) < \varepsilon$, and that's basically it.
>>
>>10060608
Try factoring out from the top instead of trying to multiply things together.
>>
>>10060615
wat, how do you factor x^2 + 4?
>>
>>10060619
I mean taking out $x^{2}-4$ 's from the numerator to cancel with the denominator.
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>>10060608
Distinguish between y and 4.
>>
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>>10060611
Using the translation invariance of the outer measure was my original plan, but I failed to realize to use the definition I was given to apply it. Thanks again.
>>
How would I find the shortest possible path that intersects all possible tangent lines of the unit circle?
>>
>>10060691
>>
How can I combat the symptoms of burnout so I can get through the month? Exercise is about the only thing that makes me less pissed off at everything, and that doesn't exactly help me learn shit.

Caffeine just pisses me off and distracts me. Literally punched a brick wall earlier because my useless piece of shit lecturer fucked up the video recording of her lecture.
>>
>>10060790
very light cbd doses and sex is the only thing that will help, cocaine and speed will just make you angrier
>>
>>10060798
>very light cbd doses and sex is the only thing that will help
Yes the kind of person posting on this board has ready access to both of those.
>>
how to do e-d proof for limit of x/(x^2+a) as x approaches 0?
>>
guys
what do if my iq was sketchily tested at ~115
should i off myself
>>
>>10060863
>how to do e-d proof for limit of x/(x^2+a) as x approaches 0?
What have you tried?
>>
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How can I calculate power consumption here?
I figure that the power factor means the current and voltage are 45 degrees out of phase, but that's all I'm able to figure out.
>>
How would I find the average out of two percentages?
E.g year 1 of school = x%, year 2 = y%. How would I get the average for years 1 and 2?
>>
>>10061274
Well you can represent any percentage as a fraction of something over 100. Add those two fractions and divide them by two, just like you would with any other number.
>>
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>>10059210
bump
>>
My professor told me to use taylor series for these. After I use taylor series, can't I just use change of variables to make them look like whichever type of bifurcation occurs?
>>
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>>10058290
can anyone help me with a high school physics problem?
So a golf ball with a radius of 2.1 cm ends up in the hole if it falls at least its radius before it hits the other side of the hole. The diameter of the hole is 10.8 cm. I'm supposed to calculate the highest velocity the ball can have in point A (refer to pic) if the ball were to end up in the hole.
The answer is apparently 1.3 m/s or at least that's what the book says but i'm getting other results.
>>
>>10061440
Under what condition (in terms of its center) do you suppose the ball misses the hole?
>>
I took my linear algebra exam earlier today. One question started with $P_1(\mathbb{R})\to P_2(\mathbb{R})$ given by
[eqn]T(a_0+a_1x) = 8+a_0x+\frac{1}{2}a_1x^2[/eqn]
First thing we had to do was show it's linear. It's not thought right? Or am I crazy? Because $T(0+0) = T(0) = 8 \neq 16 = T(0)+T(0)$. However, we later had to make a transformation matrix for T but it needs to be linear for that. Am I going about this wrong?
>>
>>10061440
You're probably assuming (high school) that it will go in if it falls at least half the ball's diameter over the distance to the other side of the hole. This isn't actually true as it can hit higher and still fall in, but for simplicity and avoiding the combination of collisions and a second projectile motion problem that is what you should probably assume. Then it's a 2D kinematic problem where delta y is the radius of the golf ball (be mindful of sign), delta x is the diameter of the hole, and of course you know acceleration.
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>>10061637
I think delta x would be the diameter of the hole minus the ball radius
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>>10061623
>8≠16
But you are in projective space.
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>>10061645
Yes, technically true. Not sure if their answer would include that detail though, but yes.
>>
>>10061653
Could you elaborate? We haven't covered that at all.
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>>10058290
what would it feel like to jump into a pool of mercury?
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>>10061714
>We haven't covered that at all.
I very much doubt this is actually true if you're being tested on properties of projective spaces. Roughly speaking in projective space things that are multiples of each other as the same. All constant quadratics which look like p+0x+0x^2 are multiples of each other.
>>
>>10058290
Is it possible to get a normal vector to any theoretical velocity of some position r(t), regardless of whether v is constant or not?
>>
>>10061637
Yeah i'm in highschool and the book itself says it's simplified. And it's actually a projectile motion problem.

Since it's a horizontal launch i figured these equations apply
X=V(initial)*t
Y=-0.5g*t^2
The problem now boils down to finding V(initial) right? After some rearranging and plugging in, i got V(initial) that is off by 0.2 from the correct answer. It might be the case that the book is wrong but i'm not sure. I would really be thankful if someone could show me the entire step to the correct answer.
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>>10061440
-0.021=-9.81/2*t^2
t=0.065 sec
0.087=v*0.065
v=1.3 m/s
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>>10061769
We aren't being tested on properties of projective spaces. Just linear transformations.
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>>10061843
Why do we have to subtract the diameter of hole with the radius of the ball?
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>>10061841
I'm getting 1.33 m/s. Account for delta x being the diameter minus radius of ball since it only starts falling after the bottom of the ball is over the hole and the right side is what hits the edge. Also, don't round until the end. Do the algebra before whole expression into your calculator and make sure to use parentheses.
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>>10061623
Something seems wrong with that map. A linear map takes the zero vector of one vector space to the corresponding zero vector of the other space. Are you sure there wasn't an "x" on that 8 term?
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>>10061942
100% sure
I think my professor fucked up the exam. That's all I can figure from it.
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d) is just a pascal distribution right?
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>>10061035
Bump
>>
Do floating point and fixed-point numbers have the same hex value?
I need to convert -25.5625 to hexadecimal.
I have the binary format which is 11001.101
Typically I see Hex as something like CF6, but I'm looking at my notes on how to find floating point to hexadecmial and we would get something like 0x43640000. Obviously these are hex values pulled from my ass.
>>
>>10058295
>using calculus to solve this
Why?
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>>10059210
Sure would help if I knew exactly which sides were measured 4 and 20.
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>>10062336
I'm pretty sure those are not measurements for the sides.
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>>10062345
Not what I meant. Essentially, go from this and apply Thales fifteen times.
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>>10062351
What I mean is that the 4s and 20s don't seem like length measurements in general. They seem like the area measurements of the regions they are in.
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>>10062369
Right, that.
The side measurements can be obtained by triangle similarities.
Once we have the two components of the base, we know that the areas corresponding to the triangle are obtained by multiplying them by the height and dividing by two, but they have the same height and you have the area of the left triangle, so you obtain the area on the right by proportion and subtract the rest until only the triangle in the middle's left.
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>>10062415
Fuck it's root of six.
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>>10062016
can someone respond?
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>>10062460
Pretty sure it's binomial.
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>>10062461
How? Binomial measures the probability given a distribution of trials.
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>>10062293
> Do floating point and fixed-point numbers have the same hex value?
Are you referring to the machine representations?

Floating-point stores an exponent (and typically doesn't store the leading digit, because it will always be 1). Fixed point is just an integer with an implied power-of-two scale factor.

> I need to convert -25.5625 to hexadecimal.
In hex, that would be -19.9 (i.e. -(16+9+9/16))

> Typically I see Hex as something like CF6, but I'm looking at my notes on how to find floating point to hexadecmial and we would get something like 0x43640000. Obviously these are hex values pulled from my ass.
In IEEE-754 machine representation, it would first be normalised to -1.59765625*2^4. The leading 1 would be discarded leaving the fraction 0.59765625. The exponent is stored with a bias (127 for single precision, 1023 for double precision).

So for single precision you'd have a sign bit of 1, an exponent of 127+4=131 (10000011), and a fraction of 0.59765625 (0.10011001):
11000001110011001000000000000000
= C1CC8000
>>
>>10062495
each trial is each packet
PS a Poisson distribution is the limit of a binomial distribution when the number of trials goes to infinity
>>
I can't find a haskell equivalent of toString. Is there an easier way to print out primitives I can't figure this out.
>>
Write the equations for the plane tangent to the surface z = ysinx - y^2 + 1 at the point (pi/2 , 3)

My question is what is zo for the equation of tangent plane? 0?
>>
>>10062807
"show" converts any instance of Show (which is most built-in types, except functions) to a String. You can implement Show for most user-defined types using "deriving".

There's also "shows", which returns a function which concatenates the string representation to an existing string. This allows you to concatenate using function composition, which is more efficient that string concatenation.
>>
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Why does the radius of a circle inscribe a hexagon and not some other regular polygon?
>>
My initial thought is to take the the partial derivative of x1 and x2.
Then where those derivatives are equal to 0. Assuming they have a max, I take the value that makes it 0 then plug it back in the original function?
>>
>>10062832
Thanks I'm not so worried with efficiency at the moment. I'm mainly learning haskell as a self indulgent thing.

>>10062868
A maximum necessarily lies at a place where the gradient is zero. However just because the gradient is zero does not necessarily mean you've reached a maximum. This link should explain it a bit better than I can.

http://tutorial.math.lamar.edu/Classes/CalcIII/AbsoluteExtrema.aspx
>>
>>10062830
z(pi/2, 3)?
>>
>>10062868
kek that book...
>$y_1 = x_2$

also, yes, but
>I take the value that makes it 0 then plug it back in the original function
no need, it just asks for x1 and x2
>>
>>10062883
Holy fuck i’m stupid.
>>
>>10061913
You're right i didn't pay too much attention to the diagram on the book that was my mistake cause the ball is supposed to be on the edge of the hole from the balls center on point A.
Thanks
>>
>>10062875
Wouldnt finding an absolute extremum require a closed region? I remember doing that in class and we would need some boundaries to find points of intersection.
>>
>>10062893
:3
>>
>>10062902
it's a second order polynomial, so this is probably more complicated than it needs to be

there will only be one extremum, and if a maximum exists it'll be that one (you'll know this when you solve and find only one solution)
>>
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>>10062854
>>
>off topic, but I don't think I can find a better place to discuss this
What the fuck is a smart campus?
Can technology possibly augment our universities anymore? Should it? What do you see universities will look like in 50 years. Or, are they dying a slow, painful death?

>Should I rather create a different thread to discuss this?
>>
>>10062854
It does for literally every regular polygon
>>
>>10042854

Equilateral triangles, no other regular polygon's side length = distance from corner to center. (center of circumscribed circle)
>>
>>10062969
What? Are you saying other regular polygons can be inscribed with the radius?
>>
>>10062982
Yes, if you go from the center of the polygon to the vertex of two edges as you have with the hexagon...
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>>10061623
stupid question but why do some anons write in this awful notation and others use actual mathematical symbols? Is that the input for some program i'm too green to know about and the other is unicode or something?
>>
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>>10062982
In case you're somehow not a troll...
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I'm trying to crack the recursion 2An+An-1+An-2. I want to use partial fraction decomposition on pic related to yield some sort of geometric sum then solve for A and B of the decomposition. Im completely stuck. Help?
>>
>>10062990
ah i figured it out, thank you for not replying
>>
>>10063022
To clarify, I need that recursion just in terms of some power of n.
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>>10063022
What are the first three terms in the sequence and what does this expression equal? Writing out some terms may help you to recognize a separate pattern.
>>
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>>10063029
This is what I've done so far. The box on the right are the terms in the sequence and z=1/2(-r-t) is how the sequence is defined where r is 0.25 and t is 0.125. This is the same as saying An+2 =1/2(-An-An+1)
>>
>>10063040
The recursion that I'm trying to solve specifically is trying to find the coeff in front of r (i'll be doing t after) The sequence starts 1r + 0t then 0r+1t then 1/2(-r-t). These starting terms are given. Its probably super confusing, but I mainly need help with proceeding with the actual equation as x^+x+2 is unfactorable.
>>
Learning combinatorics and hit a stumbling block:

How do I expand $\sum_{k=1}^{n} (k*{n \choose k - 1})$ into a closed form in terms of n?

I have no idea how to apply the binomial theorem here.
>>
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Doing homework and there's only one task left that I am stuck on

Say you have an ordinary coordinate system K with coordinates x and y, with the normed basis vectors ex and ey
A vector R that we assume is in the first quadrant can be written as x*ex+y*ey
A new coordinate system is introduced, called K'. This has coordinates x' and y'. Origin in this system located at the origin of K so we can say that it's almost like K just rotated θ degrees in positive direction relative to K

R can be decomposed in K' as
$x'\vec{e}_{x'}+y'\vec{e}_{y'}$
The basis vectors in the coordinate directions x' and y' are normed
>>Find the coordinate position (x', y') relative to K' by the coordinate position (x, y) relative to K

At first I thought I would just do a coordinate changing using an inverse matrix of K but of course, that won't work in this case
>>
>>10063202
fucking hell
x'*ex'+y'*ey'
>>
Give me a short guide how to proof any limit using cauchy's definition. Can we say that function has limit only of |f(x)-A|=a*|x-x0|, where a is any possible factor?
>>
>>10062993
Right, but they're not inscribed using the radius.
>>
>>10063651
Just to clarify, you want to know if there exists any other regular polygons you can inscribe in a circle where the side lengths of that polygon are integer multiples of the circle's radius?
>>
>>10063651
What do you mean "using the radius"?
>>
>>10063050
There is a cute trick you can use for sums that look like this. The binomial theorem gives you $(1+x)^n = \sum_{i=1}^n{{n \choose k}x^k}$, and then you take a derivative to get the k to come down in front. Once you're done your manipulations, plug in x = 1.
Yours isn't exactly like this but there are a couple simple things you could do to wrestle it into a form where this will work.
>>
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Im trying to graph f(x)=Cos(x) with the domain {0<= x <= 2pi}.
Im trying to get the Range(y) values using my calculator ( pic related ).
But I know what a Cos function looks like and its giving me back values
between 1 and 0.9945. If i sketch would look like a straight line. I think its
interpreting my values as degrees. Is there a certain syntax to get what Im after?
I know this going to come across as a stupid question but google just keeps giving me back American
calculators that graph it on screen.
>>
>>10063763
If you can't find a radian mode, convert in your function.
>>
>>10063763
Why the fuck are you graphing with a calculator in the year of our Lord Jesus H. Christ 2018?
>>
>>10063779
Thank you, ive never understood trig, nows time to start
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>>10058295
shouldn't it just be (1.05^2)*(1.03)?
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>>10063796
Im just getting the range, and more so wanted a better understanding of solving trig functions with my calculator, as I knew it wasnt giving me what I wanted. Just had to switch to Radian thanks to >>10063779
>>
How can i find all crit values for
8x^3+y^3-24xy+2?
I found one at (2,4) but I don’t know if I found them all.
I found the first one by taking both partial derivative. Fx = x^2 = y then i substituted y into Fy.
>>
>>10062684
Yeah, but a binomial random variable measures the number of successes in N trials, while a Pascal measures the number of trials given K-th success. So if I wanted to know the probability that the 4th trial will yield my 2nd success is that not a Pascal?
>>
What do you do when, after you get your physics degree, you realize you have no talent or passion for physics, engineering or mathematics?
>>
>>10064125
an hero
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Does anyone know how they got it and -it?
>>
need a linearly independent solution y2(x) to the ode
>>
why do we call it 'oxidation' if it's about losing electrons? is it because oxygen is such an electronegative slut and we just associate the two?
>>
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>>10064125
Become pic related.
>>
Any CS majors here? CS pleb here, I'm stuck in this question:
Show that $\log(n^n + n)$ is $O(nlogn)$
$\log(n^n + n)$ is $O(nlogn)$
First I did this $\log(n^n + n) = log(n) + log(n^{n-1}+1)$
But then I got stuck. Much appreciated bois.
>>
>>10064360
It is probably easier for you to exponentiate first and try to show that (equivalently) $n^n + n ~ O(n^n)$
>>
>>10064443
What if I use this $log(a + b) = log(a) + log(1 + b/a)$ to get
$log(n^n + n) = log(n^n) + log(1 + n/n^n)$
>>
>>10064341
>is it because oxygen is such an electronegative slut and we just associate the two?
Basically yes. The first discovered oxidizing agent was O2 and the whole thing basically meant "to react with oxygen and form an oxide", so the name stuck since then.
>>
I have 2 matrices

|5cos(x)+3cos(x) |
|3cos(x)-5sin(x) |

and

|cos(x)+sin(x)|
|cos(x)-sin(x)|

Multiplying them together would look like this:
(5cos(x)+3sin(x))(cos(x)+sin(x))+(3cos(x)-5sin(x))(cos(x)-sin(x))

Now the answer to this should be 8. How is that possible?
>>
>>10064550
they ought to have tought you sometime in high school that $cos^2(x)+sin^2(x) = 1$. The terms involving sinxcosx just cancel each other.
>>
>>10064360
>>10064517
Both ways work, just choose an appropriate n_0.

For example, $log(n) + log(n^{n-1}+1) < 2log(n^n)$ for $n>1$.
How were you taught to show that $f \in O(g)$ ?
>>
>>10064571
I was taught to do the following:
$f(n) <= c$ then solve for c, but I suck at this.
Your way is much better and easier, thanks a lot man!
>>
>>10064550
Yes I know that but cos^2+sin^2 with different coefficients don't? Here the coefficients are 5 and 3
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>>10064559
And I knew this would have something to do with cos^2x+sin^2x
I probably asked the question poorly because it's late. All I wonder is what do I do if the term contains differing coefficients
>>
>>10064595
But you get a pair of each multiplying out. The first two parentheses give you a 5cos^2(x) and the second two parentheses give you a 5sin^2(x). Pair these up to get 5. Do a similar thing with the 3sin^2(x) and 3cos^2(x).
>>
>>10064595
>term contains differing coefficients
You're misunderstanding. Multiply everything out, group the terms by coefficient. You'll see what was meant.
>>
>>10064603
Oh yeah, you're right
Sorry, been doing this for 12 hours. A fuckhuge matrix assignment that I've been doing since 12pm
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>>10064292
The trivial one?
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>>10064604
>>10064603
thanks alot. I have been sitting on this for an hour because I'm so damn sleepy and couldn't figure out that there would be 2 groups of cos^2x+sin^2x, each with their own coefficient. I thought I saw it was mixed with 5 and 3
>>
>>10064630
Better to get at least some sleep usually. Wake up earlier and finish it.
>>
>>10064140
bump
>>
For a 2nd order non-homogeneous ODE, if the free term equals bxcos(ax), how can I find the particular integral?

is it (cx+d)(mcos(x)+nsin(x))?
>>
>>10064669
The values of it and -it aren't relevant, the idea is that there is a discontinuity because you are just picking the principal branch. Read again the second and third lines. The idea is that there is a jump between the values in the positive and negative imaginary axis. it just represents an arbitrary number in the upper part of the axis and -it the equivalent in the lower part of the axis. There is a sudden jump from the former to the latter because of said discontinuity, but they aren't any specific numbers, just a general placeholder.
>>
Complex topology help

It says "Show A = {z in C (complex) : Re(z) > 0) is open".

I picked an arbitrary w in A, so that Re(w) > 0.

Then, I chose r = Re(w) and defined an open disk such that

Dr(w) = {x in C : |x - w| < r}

Then, I chose an arbitrary point in the disk, y (y in Dr(w)), and now I'm struggling to show that Re(y) > 0.

I know that |Re(y) - Re(w)| <= |y - w| < Re(w), but not sure what else to do.
>>
>>10064759
Sorry, I just don't understand why there's a discontinuity (i.e., how did they find that discontinuity algebraically). I'm probably missing something really stupid right now or overthinking
>>
wtf lads, how is that hint even possible? My idea is that any line must contain two points that have the same coordinate being zero (eg: (0,1,1,1) and (0,1,2,3) ), and hence you can eliminate a variable from a the line equation
>>
>>10064762
I'm not sure exactly how much you know about complex analysis, but if you know what mobius transformations are, then you should understand easily:

Let f be the mobius transformation that takes the open disk to the upper half plane (i think it's called the cayley map, or something along the lines of (z-i)/(z+i) ), and g the rotation by 90º clockwise (ie: multiplication by e^-i pi /2 ). Then their composition h = g o f is biholomorphic, in particular, its inverse h^-1 is continuous, and h^-1 maps your region A to the open disk around 0. The open disk is clearly open, and the preimage under h^-1 is open since h^-1 is continuous. In particular, A is open.
>>
>>10064762
>>10064773
Also, here's another explanation for that mapping: https://math.stackexchange.com/questions/114733/mapping-half-plane-to-unit-disk
>>
>>10064773
I have no idea about any of that, but I appreciate the help (I only have undergrad knowledge)

What I did was q = y - w so Re(y) = Re(q) + Re(w), and then |Re(q)| < Re(w), so -Re(w) < Re(q) < Re(w), which means 0 < Re(z) < 2Re(w)

This means z is in the set
>>
>>10064763
The principal branches of the complex square root and logarithm are defined from (-π,π] (usually at least, from your text it seems to be the same). This is because the square root of any complex number is technically different if you have arg(z)=k or arg(z)=k+2πn for some integer, non-zero values of n. Check https://math.stackexchange.com/questions/2536235/what-are-the-branches-of-square-root-function for a bit more info on that, if you don't know it already.

If you cross the non-positive real half-line, what you are basically doing is jumping from the value of arg(z)=π straight into values of arg(z)> -π, since the principal branch of the square root isn't defined for values higher than π. If you evaluate your square roots, you'll notice that this sudden change in the value arg(z) messes up the result of your square root, the imaginary part goes from positive to negative suddenly - a discontinuity.
>>
>>10064787
>>10064762
>complex topology
>I only have undergrad knowledge
>>
>>10064793
sorry I guess "complex topology" is a bit over the top for what that problem was, when I was googling for help someone called it that and it sounded cool
>>
>>10064787
nigga i did complex analysis before topology, literally learnt that shit in second year.
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>>10064792
I understand what the principal branch is and why the value jumps between -pi and pi, what I was a little unclear on is how to find exactly where the discontinuities occur for a function sqrt(1 + z^2), or even sqrt(1 - z^2) - like how do I find where the function has discontinuities/what values is z around when it jumps/ etc.

The example provided is for sqrt(1+z^2) and there's an exercise that asks a similar question (where do the jumps occur) for sqrt(1 - z^2). I assumed that they wouldn't both have the same answer, or both be (-pi, pi], that just seemed too obvious but I could be wrong
>>
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>>10064627
its ok thank you, I've solved it
>>
A machine can be used to produce five types of parts. We need to produce 12 parts of type 1, 18 of type 2, 15 of type 3, 20 of type 4, and 10 of type 5. How many different production schedules ( the sequence in which the parts are produced) can be constructed if the parts can be produced in any order?

I've been stuck on this for like an hour
>>
>>10064835
>I've been stuck on this for like an hour
What have you tried?
>>
>>10064835
That's just permutation with repeated letters.
>>
>>10064848
Might as well drop a link.
https://en.wikipedia.org/wiki/Permutation#Permutations_with_repetition
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>>10064848
Would it then just be (75!) / (12!*18!*15!*20!*10!) ?

So it's analogous to the different distinct strings of "AAABBCCCCEEFFF"?
>>
>>10064869
Ya.
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>>10064815
Think about sqrt(1+z^2) as a composition of functions. You already know that sqrt(z) has the discontinuity whenever arg(z) goes past π in the counterclockwise direction. But you are not evaluating z in the square root, you are evaluating 1+z^2. So you need to find the values for z such that arg(1+z^2) goes past π. Speaking more geometrically, as the text does, you need to cross the non-negative real halfline. Which means you are looking for values where Im(1+z^2)=0 and Re(1+z^2)≤0 as the values that will cause the discontinuity, you can find the direction afterwards.

If z=a+bi, z^2= a^2 -b^2 +2abi, and 1+z^2=(1+a^2-b^2)+2abi.
Because of the 2 conditions we had for the real and imaginary parts, we know that 2abi=0 (so either a or b is 0), and (1+a^2-b^2)≤0. b can't be zero, because otherwise 1+a^2 would always be positive. Then, a is zero, and the only condition we have left is that 1-b^2≤0. Then, 1≤b^2, which implies that either b≤-1 or 1≤b. So the values you are looking for are z=bi, where b≤-1 or b≥1. That is, the complex number z is on the imaginary axis (a=0), and on the half lines that are mentioned in your book.

That's how you find where are the jumps, as an example. You need to see the sqrt as the composition of two functions. See if you can proof in which direction you'd need to cross the line.

In short just see which values make the function inside of the sqrt match the conditions necessary for the principal branch.
>>
>>10064971
>it's the same answer, which makes sense, but it's psyching me out.
You mean the velocities? It's right. Think of it as (explain it like) putting energy into a frictionless system. What goes in must come out.
>>
>>10064995
Yeah the velocities, and then the time was just doubled, ie, the time up is the same as the time down.
I even solved it another way, just couldn't explain it well and was questioning myself because like I said, been years and thought I was maybe forgetting something stupidly simple.
Thanks for the confirmation.
>>
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please for the love of god somebody explain to me how the x component of gravity (aligned with the ramp) is equal to g*sin@ and not g/sin@, this genuinely doesnt make sense to me
FUCK phydicks, i just wanna do calculus
>>
>>10065045
opp/hyp = g_x/g
>>
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TURBO brainlet here, but how the FUCK do I find the five key points of a trigonometric graph?

I understand using default values like 0, pi/2, pi, 2pi/3 and then 2pi but for stuff like

y=sin(2x) where the period changes obviously to just pi since period = 2pi/B

pic related are the values that were worked in the book for the above equation

how did they get them? I can understand what do kind of I just need to know what they did mathematically to get those points!!!
>>
>>10065053
So sin(x) has key points at 0, pi/2, pi,... To find the key points of sin(2x) for example you need to find when 2x equals 0,pi/2,pi... For example 2x = pi/2 when x = pi/4.
>>
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mathlet working his way through the apostol meme
help

>>10065045
tilt the axis so the normal force is perpindicular to the surface plane, gravity will make a triangle
>>
>>10065064
>help
What have you tried?
>>
>>10065064
What is your problem exactly? Have you tried setting p to a polynomial with variable coefficients and then plugging it in?
It might also help you to transform the coordinates so you see the symmetry more easily.
>>
>>10065061
I still don't get it. On that graph where the fuck did they pull pi/4 , 3pi/4 from etc.
>>
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>>10065068
the last problem i could reduce to a system of linear equations to solve for the polynomial, but it obviously won't work here
>>
>>10065077
"Five key points" sounds like an abandoned scorsese movie title..
What you're asked to find is just the values of x for the three zero crossings and two points of local maximum in one period.
>>
>>10065050
but my diagram is showing the opposite of what youre telling me
I hope im not drawing it wrong or anything
>>10065064
my normal force is already perpendicular to the ramp
>>
>>10065086
oh it is durr
tilt the plane so that it's flat, is what i meant to say
>>
>>10065080
It is probably easier to just directly solve ax^2+bx + c = a(1-x)^2+b(1-x)+c.
If you expand the right you'll get ax^2+bx+c = a-2ax+ax^2+b-bx+c. Cancel some shit around and you end up with 2ax+2bx - a - b = 0.
The key idea is that you can group by degree of x, so (2a+2b)x - (a+b) = 0 for all x. So you just need a = -b for the first one. The other ones will be similar.
>>
>>10064877
>the non-negative real halfline
Do you mean negative real halfline?

Also, thank you! This cleared it up a ton.
>>
>>10065077
>>10065077
Try solving for 2x = 0, 2x = pi/2, 2x = pi,... and see how those points correspond to 0, pi/2, pi,.. in the plot of sin(2x) and sin(x).

sin(2x) behaves similarly to sin(x), it's just the argument of sin is different. The 'key' values for sin itself always occur at 0, pi/2, pi,... When you screw with the argument you need to find when the ARGUMENT is equal to these values 0,pi/2,pi,... Say you had sin(10x+2.54321). You need to find when 10x+2.54321 = 0,pi/2,pi...
>>
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>>10065086
>but my diagram is showing the opposite of what youre telling me

You're just reading it wrong. See candy colored pic related.
>>
>>10065098
Sorry, I messed up there. I actually meant non-positive mainly because I didn't really analyze the point 0 so I wanted to play it safe, but after reading your text and thinking about it, negative is just fine, so yes.
>>
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>>10065099
!!!!!!!!!!!!!!!!!!!

This made sense thank you now I don't have to fucking kill myself

getting all the proper values now that it says in the book for the worked examples if I run through them myself
>>
>>10065105
thank you so much
>>
>>10064360
log(n^n+n)<log(2n^n)
>>
>>10064360
For any k>1, for sufficiently large n
log(n^n)<log(n^n+n)<log(n^(k*n))
log(n^n)=n*log(n)
log(n^(k*n))=k*n*log(n)
>>
Prove that if f is a continuous function defined on an open subset U of C, then sets of the form {z∈U:|f(z)|<r} and {z∈U: Re(f(z))<r} are open.

Very lost on this - in a complex analysis class. I said let x be in U and in the first set, then |f(x)| < r, and then I was going to define a disk around x, but I'm not sure anymore.
>>
So I have this real analysis problem:
Let $f : [a,b] \longrightarrow \mathbb{R}^n$ be a differentiable path such that $f(a) = f(b) = 0$. Show that $\exists c \in (a,b)$ such that $\langle f(c),f'(c) \rangle =0$.

I'm not really sure on how to start. This is basically saying that in at least one point, the vector is perpendicular to the tangent vector at that point, right? I can kinda see that working geometrically, but I'm not sure how to prove it rigorously.
>>
>>10065347
Isn't this basically the mean value theorem?
>>
>>10065335
An equivalent definition of continuity on a metric space is that the preimage of a continuous function on an open set is open.

See https://en.wikipedia.org/wiki/Continuous_function#Continuous_functions_between_topological_spaces
>>
>>10065363
Oh boy, we haven't learned about this at all. Not completely sure if I even know what the inverse image is in this sense. Any more tips?
>>
>>10065377
If we have an open set $A \subset X$ then the preimage of $A$ under a function $f$ is defined as $f^{-1}(A) = \{x \in X : f(x) \in A\}$.

You should try proving that the preimage of an open set under a continuous function is open. Use your definition of open sets. Prove the other direction for equivalence.
>>
>>10065360
I assume so because the problem is stated shortly after another problem that seems to be related to the mean value theorem, but the way it's worded in that problem is that, under the conditions that were mentioned earlier (assuming $f = (f_{1},f_{2},...,f_{n})$ of course), $\exists c_{1},c_{2},...,c_{n} \in (a,b)$ such that
$\frac{f(b)-f(a)}{b-a} = (f_{1}'(c_{1}), f_{2}'(c_{2}),...,f_{n}'(c_{n}))$. Since each of the component functions may cancel out for different values of $c_{i}$ $\forall i \in {1,2,...,n}$, I think I can't just pick a single $c$ for which $f(c)$ cancels out completely on all of its components. Or do I need to prove something else as well?
>>
>>10065403
Cool, don't know how I fucked up the TeX. What I mean is that I don't really know how to find a single point $c$ out of all those $c_{1},c_{2},...,c_{n}$ that takes me from that previous result to the scalar product result I'm looking for.
>>
>>10065420
because chink moot can't code the site can't handle more than one set of math tags per line
>>
>>10065420
Yes, it won't work on the vector value .. but what if you took another ... measure? Hint, hint.
>>
>>10065435
I can only imagine you are referring to the norm of the vector? But isn't that also only zero iff the vector is the zero vector? Which would take me back to a value c such that $f'(c)=(0,0,...,0)$, wouldn't it?
>>
>>10065452
Try computing the derivative of $\frac{1}{2}\left\|f\right\|^2$ and see if that helps you out in some way.
>>
>>10065458
Ok let me see if I got this clear. I can define a function $g:[a,b] \longrightarrow [0, \infty)$ such that $g(x)= ||f(x)||$, which is differentiable in that interval because the norm is the square root of the sum of the squares of differentiable functions, and the operations of those functions in the interval is differentiable.
Then, since it's a real-valued function, I can apply the mean value theorem, and find a value c such that $g'(c)=\frac{g(b)-g(a)}{b-a}$, but the numerator is 0 because it's the difference of the norms of the zero vector, and hence $g'(c)=0$.
But the derivative of g is, because of the chain rule, $g'(x)=\frac{\langle f(x) , f'(x) \rangle}{||f(x)||}$, and if $g'(c)=0$, that means that at that same point $\frac{\langle f(c) , f'(c) \rangle}{||f(c)||} = 0$, so the scalar product must be 0.
Is that all correct? Thanks for the guidance by the way.
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>>10065523
Yeah, except for one thing. If you define g to be the square of the norm of f, the whole thing becomes much clearer:$\frac{d}{dx}\left\|f(x)\right\|^2 = \frac{d}{dx} \left<f(x),f(x)\right> = 2 \left<f(x),f'(x)\right>$
Well done.
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I have a question:
Let a, b be integers, if $a \equiv b (mod 5)$ and $x \equiv y (mod 5)$, prove that $a+x \equiv b+y (mod 5)$.
All I have to do is to prove that a - b, x - y, and (a + x) - (b + y) are multiples of 5 isn't it?
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>>10065546
>All I have to do is to prove that a - b, x - y, and (a + x) - (b + y) are multiples of 5 isn't it?
Yes, you need to use the first two (which are given) to show the last one.
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Can someone teach me how to prove stuff? I don't mean I lack the prerequisites or anything. I just can't think out of the box. Maybe I'm just stupid.
For instance, proving that the square root of 2 is irrational: I already know how to prove even/odd numbers, etc. but without someone telling me to use those proves, I would never solve it.
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>>10065588
To learn how to prove stuff, read more proofs. If you can't think out of the box, as you put it, get a bigger box. Look for similarities in structure, expand your repertoire. There is no easy way to "just learn" it.
The ancient greeks threw people off their boats for insinuating that there even might be such a thing as an irrational number, so if it takes you a couple of days to grok something like a reductio ad absurdum proof, that's not too bad.

I don't know what you mean by "I know how to prove even/odd numbers".
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>>10065607
Thanks.

I'm referring to the proof by contradiction shown here: https://math.stackexchange.com/questions/1415235/prove-the-existence-of-the-square-root-of-2.

Basically what I'm trying to say is that I know the basic algebra in required to prove that theorem. But if no one tells me how to use those basic algebra, I can't do shit.
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>Government want to promote the recycling of no longer useful electrical waste
>Look at the current status quo: There are some street recyclers that buy and collect those electrical waste and then sell them to recycling business
>Government: NO, that's harmful, they must atop doing so. We should instead hire some dedicated recyclers to collect those electric waste.
>In order to ensure the recycling program to be successful, a built in recycling cost will be added to product price tag, which would pay for professional recycling company to help take those electric wastes away. Oh, because the demand is so huge, they would need to sit there for months before someone can come and take them away. You said why not hire those street recyclers? Simple, who knows where will they drop those electric waste, they're banned, you should use service from licensed company. Also they would damage the environment.
Is the move by the government making any actual sense?
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>>10065614
> But if no one tells me how to use those basic algebra, I can't do shit.
Well, that's math for you. Understanding a method or proof which someone else figured out is a lot easier than being the first person to discover it.

The main things are to have a decent knowledge base so that you can recognise plausible lines of enquiry, and be fluent in the fundamentals so that you can easily spot places where it might be useful to apply specific knowledge.
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I'm fresh into college taking an intro to compsci course with Java.
This has been my first real venture into coding.
So here's the question, am I supposed to fucking hate it? Every second I spend learning something new is like pulling teeth. It just feels like a giant mish-mash of rules and jargon, and this is coming from a guy who loves math.
Am I a brainlet? Is it just a bad teacher? The "lecture" is her reading off a powerpoint that I could have just read at home.
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>>10065395
>>10065395
>>10065335
I'm still super lost on this and have spent over 2 hours on it. Can anyone walk me through? I know 0 about topology.
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>>10065698
>>10065697
Things will get easier once you understand how expressions are parsed and evaluated.
If all your lecturer does is reiterate the points found in lecture notes, I suggest you skip your lectures and practice programming yourself, or even better, write a program that you will use daily.
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>>10065064
for a
for degree 0, obviously all polynomials satisfy it
for degree 1, no polynomial satisfies it (just plug in and see)
for degree 2, all polynomials than can be factored with x(1-x)
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How do I solve this:

Prove that there is no positive random number such that $a^2 = 3$.

Pic related what I got so far
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I'm about to choose my masters. I'm thinking about pure math, theoretical physics or subatomic physics. Which of these gives the best job opportunities for working with mathematical physics? Which is "best" overall in the job market?

My program is engineering physics if that matters.
>>
Brainlet studying chemistry from the ground up. How complex does the math in chemistry get the further you go? I imagine it's not always gonna be a matter of simple formulae and numbers with lots of decimals.
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So I've been reading about basics of integration, because I think I had a bunch of flaws in my fundamentals. Can someone please tell me if I got it right now?

As I understand, if a function is continuous in an interval, the first Fundamental Theorem of Calculus guarantees that it has an antiderivative. However, it doesn't necessarily guarantee that said antiderivative can be expressed in terms of elementary functions (like the antiderivative of $x^{x}$).

However, not being continuous doesn't mean that a function can't be integrated, like for example the indicator function for the rationals, which can be integrated through the Lebesgue integral.

In case the function DOES have an antiderivative, it can be easily evaluated through the second Fundamental Theorem of Calculus. However here is another catch I have right now. According to the wiki page on the FTC, the second part doesn't need the condition that a function f needs to be continuous in order to evaluate the definite integral in terms of f's antiderivative. However, the first part of the TFC only guarantees that the antiderivative exists when the function f is continuous. So can you explicitly find the antiderivative of a function that isn't continuous? Or does it just mean that f doesn't need to be continuous, but at least piecewise continuous?
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Find the charge on the capacitor as a function of time
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i cant figure this out FUCK
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>>10066197
I'll do the first inclusion
Let a be an element in the intersection of X and the inverse image of Y. Then, by definition of intersection, a is an element of X, and a is also an element of the inverse image of Y, which means also by definition that f(a) is an element of Y. Also, if a is in X, f(a) is an element of f(X). And now we know that f(a) is in both Y and f(X). Which means f(a) is in the intersection of f(X) and Y.

The second inclusion is more of the same. But now your starting element is in the codomain instead.
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>>10066236
wait didnt you just proof everything now?
or do i have to do the other way round aswell to show it really is EQUAL and not a subset of the codomain?
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>>10058290
How can i decide if a b and c of a parabola are positive or negative just from a drawing? For example pic related
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>>10066273
a: parabola is open upwards (+) or downwards (-)
b: dunno, but if the maximun/minimum is on the y-axis its 0
c: at x=0, thats your c
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>>10066272
You need to proof the other way around too. You got it right, the equality of 2 sets means that they are mutually included in each other.
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>>10066273
>>10066276
a is curvature.
b is slope at x=0.
c is y-intercept.
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>>10065899
>random
you mean rational?

if so, then suppose 3 = (m/n)^2 where m,n are integers with no common factors
then 3n^2=m^2. But 3 divides m^2, so in particular in must divide m, so write m=3r. Then 3n^2=(3r)^2 which implies n^2=3r^2. But this implies 3 divides n^2, in particular n. But this is in contradiction with m, n having common factors
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>>10066288
blessed post
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First time using simulink here. I want to multiply these constants and obtain a constant, but for some reason I obtain a matrix with the constant result associated with time. Where am I fucking up?
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>>10066287
ok so:
Let b be an element of f(X) intersecting Y. Therefore there exists an x in X such that f(x) = b, and that b in an element of Y. Because b is an element of Y, the reversed image of b is an element of the reversed image of Y.
But whats up with the brackets now? Is there even a difference between f(X [intersec] f^-1(Y)) and f(X) [intersec] f^-1(Y)?
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>>10066328
The first part is looking good.
>Is there even a difference between f(X [intersec] f^-1(Y)) and f(X) [intersec] f^-1(Y)?
Be careful with where the sets you are evaluating are. f(X) is a subset of the codomain, whereas f^-1(Y) is a subset of the domain. The intersection could always be empty, for all we know. For example, if the domain and codomain are different, disjointed spaces.

The brackets just tell you that you are evaluating a single set, defined by the intersection of 2 others. You are evaluating two sets. The difference is on where the intersection is actually happening, in the domain or codomain.

Note that you defined (small) x as the inverse image of b. That means that x should also be in the inverse image of Y, right?
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>>10066386
>You are evaluating two sets
Meant "You are not evaluating two sets" here.

And also
>Note that you defined (small) x as the inverse image of b
x is one element of the inverse image, not necessarily the whole inverse image
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>>10065727
Let's work in $\bf{C}$. A set $A \subset \bf{C}$ is open when for all $x \in A$ there exists real $r > 0$ such that for all $y \in \bf{C}$ if $|x - y| < r$ then $y \in A$. One way to interpret this is a set is open if for any point in the set you can find a ball centered at the point such that the ball is entirely contained within the set. For example $\{z \in \bf{C} : |z| < 1\}$ is open in $\bf{C}$.

According to one definition a set is continous at a point $z \in \bf{C}$ if for all real $\epsilon > 0$ there exists a real $\delta > 0$ such that, for all $w \in \bf{C}$, if $0 < |x-w| < \delta$ then $|f(z)-f(w)| < \epsilon$.

Use these definitions to prove the following statement: If $f: \bf{C} \to \bf{C}$ is continuous then for any open set $A \subset \bf{C}$ the preimage of $A$ under $f$ is open.

The goal of the proof is to show the preimage is open. So choose a point in the preimage and try to show there exists a contained ball centered at that point. Be sure to make use of continuity.

Once you have proven this statement you should be able to prove those two sets in your original problem are open using the statement.
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>>10066386
>>10066405

i dont get it, someone help me
atm im at:
>b in (f(X) insc Y
>thus b in f(X) AND b in Y
>thus x in f^-1(b) AND f^-1(b) in f^-1(Y)
>thus x in f^-1(Y)
>f(X insc f^-1(Y)

sorry for not LaTeX, dont know how to use that
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>>10066909
We are analyzing the intersection of f(X) and Y. Since f(X) intersected with Y is a subset of f(X), for every element b in said intersection there has to be an element of X (which will be the small x) such that f(x) will be in the intersection of f(X) and Y. If that is the case, then there has to be at least one value of x in X that maps to b (otherwise it couldn't be in the image of X) and it also has to be in the inverse image of Y, by definition of inverse image. Then, x has to be in both X and f^-1(Y).

In other words
Let $b$ be an element of $f(X) \cap Y$
Because of an intersection property, $f(X) \cap Y \subseteq f(X)$
Thus, there has to be at least one element of $X$ that maps to the element b in that intersection. Let's call that element x.
We picked an element of X, so of course we have immediately $x \in X$. However, we also know that $b \in Y$.
Thus, by definition of preimage, $x \in f^{-1}(Y)$.
Now, you have $x \in X \cap f^{-1}(Y)$. You can finish it now after evaluating x.

Always keep track on where each element you pick belongs to. You were already close enough but are missing the relations between some subsets. For example, x being in f(X) and also being in the preimage of b already tells you that you can always find an element of X that is in both, for every value of b.
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Thank you dear anon

>You were already close enough but are missing the relations between some subsets. For example, x being in f(X)

after reading through your line of argument, I dont see where I made a mistake. It both sounds correct to me, do you mind to tell me where exactly my mistake is?

t. brainlet math undergrad
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>>10067051
I don't know if I'm misunderstanding what you meant to say, but I expected to see the result $x \in X$ somewhere.
You already showed that $x \in f^{-1}(b)$, but keep in mind that just being in the preimage of b and having b be an element of (X) doesn't really guarantee that any x in the preimage of b will be in X. For all we know you could be picking an element z in $f(X) \setminus Y$ such that it also gets mapped to b.

The reason I say this even though it seems pretty obvious is that if you don't bother showing why that same value of x can be found in the set X, all you show is that $b \in f(X) \wedge b=f(x) \in f(f^{-1}(Y))$. Which implies that $b \in f(X) \cap f(f^{-1}(Y))$.

However, if A and B are arbitrary sets, the equality $f(A \cap B) = f(A) \cap f(B)$ doesn't necessarily hold, and you would have just proved that b is contained in the latter.

I guess it's more of a nitpick because I expected that you would explicitly show that the element x could be simultaneously in $X \wedge f^{-1}(Y)$, but if you already had that in mind and just didn't write it then it's alright. At least in this more informal setting.
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>>10067124
again, thanks for taking the time out of your day

added an extra $x \in X$
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What do you do when you realize you’re entering your Junior year as an EE student and realize you don’t want to do EE anymore?
In Community college, I went from EE to compe to EE, and now Im at a CSU and realize I just want to do software, games, and music. Anything in EE that will let me do this or do i just grind out being an EE and just try and find a job.
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>>10066194
Did you even try and attempt this problem? This is so easy.
t. intro to circuits student
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>>10061781
Yes. For any orientable manifold $M$ the normal space $N_xM$ at $x\in M$ can be defined as the spaces of normal vectors perpendicular to every vector in the tangent space $T_xM$, where the notion of "perpendicular" comes from the orientation on $M$. In fact, orientability is equivalent to the existence of a globally defined normal bundle $NM$. The vector normal to a trajectory $t\mapsto x(t)$ is just $n \in N_{x(t)}M$ such that $n \perp X_f$, where $X_f$ is the Hamiltonian vector field generating the trajectory.
>>10063202
Suppose $B = \{e_1,\dots,e_n\}$ spans the vector space $V$ over a feld $k$, then there is an isomorphism $\operatorname{End}(V) \cong V^* \otimes V$ between linear self-maps of $V$ with the matrix representation, where $L \in V^*\otimes V$. Endow $V$ with an inner product $(\cdot,\cdot):V\times V\rightarrow k$, then $L\in \operatorname{End}(V)$ is a coordinate transform iff $(Lx,Ly) = (x,y) = (x,L^*Ly)$ for every $x,y\in V$, hence $L \in O(n)$ necessarily is an element of the orthogonal group. Since $O(n)\hookrightarrow \operatorname{End}(V)$ embeds into the group of self-maps, the above isomorphism descends to a representation $\rho:O(n)\rightarrow SL_k(n) \subset V^*\otimes V$. It is always possible to express coordinate transforms by a matrix of unit determinant. You probably did something wrong.
>>10064773
>Then their composition h = g o f is biholomorphic
Only on the Riemann sphere $\mathbb{C}\cup \{\infty\}$. Remember, Mobius transforms generates the group of conformal transformations in $\mathbb{C}$, the element $\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}$ of which has no inverse unless you append the point at infinity.
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>>10064765
It's talking about a line in the projective space. The homogeneous coordinates parameterizes every point in the projective space with lines in the total space $P \rightarrow \mathbb{P}$. Now he's talking about lines $in$ the projective space itself, which is defined with an equation in projective coordinates.
Combine this with the equation you get from the homogeneous coordinates gets you 2 equations.
>>
which level of chemistry/physic i must have to understand properly some subjects of medicine like biochemistry, pharmacology, physiology, etc?
i must dominate org chem? (level student of chemistry)
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>>10065347
First let us rescale the compact interval so that $[a,b] \cong[0,1]$. Since $f(0) = f(1)$, it descends to a smooth map $\tilde{f}:S^1\rightarrow \mathbb{R}^n$ under the identification $[0,1] \rightarrow [0,1]/\sim ~ \simeq S^1$ of the endpoints $t\mapsto e^{i2\pi t} \equiv e^{i\phi}$. The derivative $X = \partial_t$ as a tangent vector $X\in T_t[0,1]$ then also descends along with the quotient to $X' = \partial_\phi \in T_\phi S^1$. The condition [eqn]\langle f(c),f'(c)\rangle = 0[/eqn] is then equivalent to the tangent vector $X' \in T_{\tilde{c}}S^1$ being singular as a linear map on $\mathbb{R}^n$, where $\tilde{c}\in S^1$ is wherever the point $c \in [0,1]$ descends to in $S^1$.
Now let us define a Morse function $g(\cdot) = \langle f(\cdot),X' f(\cdot) \rangle: S^1 \rightarrow \mathbb{R}$, then from the fact that $S^1$ is homotopically non-trivial, $g$ must acquire a critical point in the sense that $X'g(\tilde{c}) = 0$ at some $\tilde{c}\in S^1$.
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>>10067810
Sorry, the Morse function is $g(\cdot) = |f(\cdot)|^2$.
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>>10067769
pls rspnd
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What kind of function will give me a line like this?
>>
I don't really see how you make the jump from (7) to the highlighted line.
This is an example from Complex Analysis by Bak and Newman, chapter 11.2
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>>10067861
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>>10067863
Geometric series, retard
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>>10067870
why the homophobia?
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>>10067870
Unironically thanks, my dude
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>>10067858
>>10067858
>>10067858
>>10067858
>>10067858
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>>10067933
Faggot
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>>10067858
[eqn] -e^{-(x-a)/b}+c [/eqn]
this somewhat looks like your drawing, adjust a,b and c as needed
>>
Here is my question, I was not taught how to isolate a variable like this

An analysis of temperature record for a city indicates that the average daily temperature T in C° during the year is approximately

T = 6 - 15cos[π/6(m - 0.5)]

where m is measured in months. Solve for m such that T is minimized. (Hint: what is the value produced by the trigonometric function so that the difference is largest?)

Would I use a sum or difference identity? any help would be fantastic.
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>>10058316
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>>10068007
T is minimised when cos(π/6(m - 0.5))=1 => π/6(m - 0.5)=0 => m=0.5
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>>10068007
consider the range of the cosine function, determine which output value in that range meets the specification of your problem, then find the input values in the domain of the cosine function that map to that particular output value. From there it is a simple algebraic equation to solve for m no identities required
>>
How do I evaluate whether a change in a variable leads to an improvement in data?
Are there statistical techniques for this?
>>
>Let X be a set with exactly 7 elements. How many equivalence relations with exactly four equivalence classes are there on X?

I don't understand equivalence relations well so I don't know if my answer is right or even close:

(7choose4)+(7choose3)(4choose2)+(7choose2)(5choose2)(3choose2)=875

I'm just adding up the number of ways you can partition 7 elements (I think). Is this the right approach?
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>>10068412
You should be indeed adding the number of ways you can partition 7 elements into 4 sets.

But you start with (7 choose 4). I'm trying to figure out which set of these partitions you're trying to count with (7 choose 4) but I can't. What do the terms of your sum represent?
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>>10068421
e.g. with (7 choose 4) I'm choosing a set of 4 from the 7 which leaves 3 elements where the only remaining choice is for each to be its own set

like {1,2,3,4,5,6,7} -> {{1,2,3,4},{5},{6},{7}} is one configuration

likewise with (7choose3)(4choose2), {{1,2,3},{4,5},{6},{7}} would be one configuration

sorry in advance if I'm retarded, I'm having trouble understanding these
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>>10068427
Okay, I see, so 1st term is the partitions with subsets of cardinalities 4,1,1,1, 2nd term is for 3,2,1,1, and 3rd term is for 2,2,2,1, and these are the only options.

I think the first 2 terms are correct. However for the 3rd term, you counted {1,2},{3,4,},{5,6},{7} and also counted {3,4},{1,2,},{5,6},{7}, and others with permutations of the three sets of cardinality 2. So you counted each term 6 times if I'm not being stupid, and should divide that last term by 6.
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Hey guys, pretty stupid question.
Can someone help me prove that theRiemann zeta functionhas itszerosonly at the negative even integers andcomplex numberswithreal part1/2. Just need it for some homework haha.
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>>10068456
> Goes for a low-effort troll
> Can't even copy-paste properly
Dude...
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>>10058290
Light is travelling in smaller ball. In a second I measured in bigger ball, it passed 320_000KM/s.

Are my measures wrong or it goes over C?

To which zone around travelling light this limits apply?
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>>10068461
How can you be certain this wasn't a form of meta-trolling? In that he merely pretended to not know how to copy-paste in order to garner attention?
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>>10068467
How can you be certain this wasn't a form of meta-trolling? In that he merely pretended to not know that it could be someone pretending to not know to copy-paste in order to garner attention in order to gamer attention?
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>>10068471
>gamer
Are you meta-meta-meta-trolling me pal?
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>>10067769
you need good physic fren
>>
>>10066159
Your statements are correct. As for your question, there are a lot of subtle conditional statements going on here.
The second FTC can be seen as saying that IF a function f is Riemann integrable on [a,b] and IF it has an antiderivative F, THEN the antiderivative is given by F(x) = int{_a}{^x} f(z) dz + [constant]. It's not very illuminating, but this IS explicitly the antiderivative even when f is not continuous. Finding a nicer expression is generally not possible (you can't even do it for x^x, which is continuous). But beware both conditions I stated must be satisfied for the theorem to hold: there are functions that have an antiderivative but are nevertheless not Riemann integrable, and there are also functions that are Riemann integrable but do not have an antiderivative (for example Thomae's function).
Note that the FTC says nothing about WHEN a discontinuous function has an antiderivative. The easiest way to tell if a given function does not have an antiderivative is by Darboux's theorem: if the function doesn't have the intermediate value property, then it doesn't have an antiderivative. But functions that have antiderivatives can be very badly behaved - for example, they can be discontinuous on a dense set, so they definitely don't have to be piecewise continuous. As far as I know, finding a necessary and sufficient condition for a function to have an antiderivative is still an open problem.
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>>10068454
That make sense. Am I not overcounting twice with (7choose3)(4choose2) as well? meaning the answer is

(7choose4)+(7choose3)(4choose2)/2!+(7choose2)(5choose2)(3choose2)/3!=245?
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>>10068521
Looks good, I think. Dunno why I didn't realize the 2nd term had to be divided by 2 as well.
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>>10065966
It depends what you mean by mathematical physics. If you mean the actual field of mathematical physics, I would recommend going into whatever department the mathematical physicists at your institution are in. If you just mean physics that has a lot of math, go into physics instead. In terms of the academic job market, your listed options are all about the same and very competitive. Given your background, you might also want to consider materials science depending on how worried you are about job prospects. But pure math and theoretical physics grads do quite well in the finance sector if that interests you.
>>
What is the universe expanding into
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>>10068626
Ok I googled it but I still do not understand it
Theexpansion of the universeis the increase of thedistancebetween twodistant parts of the universewithtime.[1]It is anintrinsicexpansion wherebythe scale of space itself changes. The universe does not expand "into" anything and does not require space to exist "outside" it. Technically, neither space nor objects in space move. Instead it is themetricgoverning the size and geometry ofspacetimeitself that changes in scale. Although light and objects within spacetime cannot travel faster than thespeed of light, this limitation does not restrict the metric itself. To an observer it appears that space is expanding and all butthe nearest galaxiesare receding into the distance
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I have done a Fourier transform and I have no idea how to plot this in the frequency domain. Please help a brainlet out. This is how it looks like:
H(ω)=(3i/2)[δ(ω+ 5)−δ(ω−5)] +ω[δ(ω+ 4)−δ(ω−4)]

I assume that I need to set ω as values that makes each of the delta functions into 0.
Thus I would get these points:
H(−5) =3i/2
H(5) =−3i2
H(−4)=−4
H(4) = −4
How can I plot the function from these?
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>>10068675
>I assume that I need to set ω as values that makes each of the delta functions into 0.
I mean give one of them value zero and the rest non-zero as you know...δ(0)=1
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>>10068491
Thanks, this helped clear up a bunch of misconceptions.
>>