[  3  /  a  /  adv  /  an  /  c  /  cgl  /  ck  /  cm  /  co  /  diy  /  fa  /  fit  /  g  /  i  /  ic  /  jp  /  k  /  lit  /  m  /  mlp  /  mu  /  n  /  o  /  p  /  po  /  q  /  sci  /  sp  /  tg  /  toy  /  trv  /  tv  /  v  /  vg  /  vp  /  w  /  wg  /  wsg  /  x  ]

# /sci/ Science & Math

Warning: All the content of this page originally come from 4chan.org. This is only a partial archive made to avoid destruction. Some posts and images may be missing. All the messages below have been posted by anonymous users and we do not guarantee any truth of what they said.

Anonymous 2016-09-14 14:50:38 No.8344746

[Missing image file: ]

Can anyone explain how this equation was derived? It seemed out of nowhere to me

>>
Anonymous 2016-09-14 15:07:29 No.8344775
>>8344746
Not sure how they came up with the equations but they are two valid equations that have the properties needed to prove their claim. If you look at both equations, they each contain (p^2-2) which is the central idea to the theorem.

>>
Anonymous 2016-09-14 15:53:50 No.8344848
>>8344746
You're looking for a rational number $r$ such that $(p + r)^2 < 2$.
This can be rewritten as:[eqn] r(2p + r) < 2 - p^2 [/eqn]Let $\displaystyle r = \frac{2-p^2}{s}$. The hope is that this $s$ gets cancelled when you multiply $r$ by $2p + r$ so that only $2 - p^2$ remains.
So set $2p + r = s$ and solve for $s$. You'll indeed find that $s = p + 2$

>>
Anonymous 2016-09-14 15:57:25 No.8344862
How can I stop myself from trying to apply category theory to everything I encounter?

>>
Anonymous 2016-09-14 16:02:11 No.8344873
>>8344862
You don't, you embrace it.

>>
Anonymous 2016-09-14 16:05:18 No.8344877
>>8344873
fugg, am I inevitably going to become like one of those higher topos theoreticians, trying to unify logic, Hegelian dialectics, quantum mechanics and relativity in a single theory?

>>
Anonymous 2016-09-14 16:09:15 No.8344881
>>8344877
No bro you'll find something completely new to categorify and make n times more abstract.

>>
Anonymous 2016-09-14 16:53:52 No.8344997
how many stupid questions thread does /sci/ need

should we just kill /sci/ and make one permanent rolling sticky SQT

>>
Anonymous 2016-09-14 17:17:28 No.8345045
>>8344997
Sorry, I thought the previous one was dead ant also didn't the other one with 57+ replies

>>
Anonymous 2016-09-14 19:43:41 No.8345260
>>8345045
delete it

>>
Anonymous 2016-09-14 19:46:45 No.8345268
>>8344746
Can you post the entire thing in your image?

I am interested in the remark that says that for some reason whatever the author did shows that there are gaps in the rationals.

>>
Anonymous 2016-09-14 20:33:29 No.8345362
>>8345268

>>
Anonymous 2016-09-14 20:38:49 No.8345375
>>8345362
No, I meant everything before that.

The actual argument.

>>
Anonymous 2016-09-14 20:58:31 No.8345425
For k a natural number, if there exists an integer u such that $2^k + 1 = u^2$ then k must equal 3.

I can not for the life of me find how to prove this bullshit, what did I miss?

>>
Anonymous 2016-09-14 21:03:07 No.8345437
How do I integrate e^(x^2) ?

I tried wolfram alpha but it is giving me (1/2)(pi)^(1/2) erfi(x) + C .

I have no idea what "erfi(x)" is supposed to mean.

>>
Anonymous 2016-09-14 21:09:29 No.8345451
>>8345375

>>
Anonymous 2016-09-14 21:12:58 No.8345453
>>8345425
prove what?

>>
Anonymous 2016-09-14 21:16:14 No.8345460
>>8345453
That under those conditions k cannot have another value than 3.

>>
Anonymous 2016-09-14 21:31:00 No.8345499
>>8345437
exp(x2) is not an integrable function, i.e. you can't express its integral in terms of elementary functions. Wolfram Alpha is expressing it in terms of the error function, which you'll see is defined explicitly with this integral. However, if you wanted to compute a definite integral like the one that ranges from -? to ?, you could.

>>
Anonymous 2016-09-14 21:32:28 No.8345506
>>8345437
The problem is that e^(x2) is non-integrable, i.e. you can't express its integral in terms of elementary functions. Wolfram Alpha is outputting your "answer" in terms of the error function, which is really just a special functioned that's DEFINED in terms of this undoable integral. However, if you wanted to compute certain definite integrals like one that ranges from -? to ?, you could get a value.

>>
Anonymous 2016-09-14 21:34:53 No.8345510
>>8345506
>>8345499
don't mind my autism

>>
Anonymous 2016-09-14 21:37:00 No.8345512
Really dumb question, but I am unsure how to enter in the direction here?

The angle has to be below the x-axis, so I will put 36 degrees correct?

>>
G 2016-09-14 21:49:21 No.8345537
if science is accurate, then why falsify?

>>
Anonymous 2016-09-14 21:50:33 No.8345538
I know you're not supposed to put ice directly on skin because it causes frost bite, but what if your ice pack has a bunch of condensation on the outside, can that also cause frost bite?

>>
Anonymous 2016-09-14 21:57:24 No.8345550
>>8345512
anyone???

>>
Anonymous 2016-09-14 22:18:39 No.8345580
>>8345512
i guess

>>
Anonymous 2016-09-14 22:24:28 No.8345590
>>8345580
i don't have it :(, and i don't want to get it wrong. do you think it is 36 degrees? here is example from textbook

I am just not sure if it is negative because problem specifically asks for below x axis, while other one was just relative to x axis

>>
Anonymous 2016-09-14 22:32:14 No.8345609
>>8345590
just put it positive nigger, it has already specified that it's under the x axis

>>
Anonymous 2016-09-14 23:36:45 No.8345717
>>8345512
No, it would be -36 deg or 360-36=324 deg. I'd go with the - if I were you

>>
Anonymous 2016-09-14 23:37:05 No.8345719
>>8345510
Feynman's personality as I know it makes me think this might be an actual quote

>>
Anonymous 2016-09-14 23:40:32 No.8345723
>>8344746

actually this is a very common complaint with rudin

https://math.berkeley.edu/~gbergman/ug.hndts/m104_Rudin_notes.pdf

>>
Anonymous 2016-09-15 00:17:31 No.8345800
>>8345609
>>8345717
I put 36 and it accepted it, thanks anons!

>>
Anonymous 2016-09-15 00:19:25 No.8345803
>>8345499
>>8345506
Why is it not integrable? We can differentiate it but not integrate?

>>
Anonymous 2016-09-15 00:20:55 No.8345807
>>8345803
it's integrable. it just so happens that you can't express the integral with elementary functions

>>
Anonymous 2016-09-15 00:22:43 No.8345811
>>8345807
What does that mean?

>>
Anonymous 2016-09-15 00:26:32 No.8345817
>>8345811
it means that the equation
$dy = e^(x^2)$
has no solutions in the differential ring generated by $\mathbb{R}$, polynomials in x, rational functions in x, exponentials in x, logarithms in x.

>>
Anonymous 2016-09-15 01:35:02 No.8345940
wtf did I do in this problem? any anons know?

>>
Anonymous 2016-09-15 01:41:28 No.8345959
Anyone know how to evaluate a base 2 logarithm without a calculator? Heres the question

>>
Anonymous 2016-09-15 01:54:56 No.8345984
Suppose you have 2 points A and B that are a distance apart on an infinite plane. You have a small ruler that can only reach about 1/3 of the way there. How do you draw a line between the two points?

>>
Anonymous 2016-09-15 02:42:35 No.8346088
>>8345723
Honestly, I'm impressed by anyone that uses Rudin for their first introduction to analysis. I mean, I can read the bit in OP's pic and understand why he chose q how he did because I've already taken real analysis. But if you'd given that to me before I'd done any analysis, with no motivation at all, I'd be completely lost. It would seem as if he'd just pulled an equation out of his ass.
To me, Rudin's analysis textbooks seem to assume a level of mathematical maturity that usually hasn't developed in students studying the level of material they cover. Maybe things were different in his day. Maybe students who had to go through calculus when it was taught more rigorously wouldn't have found it unusual.

>>
Anonymous 2016-09-15 02:46:21 No.8346095
>>8345959
You don't need to evaluate it. Think about what the question is asking. What is the greatest integer you can raise 2 to and get an answer less than 22? What is the least integer you can raise 2 to and get an answer greater than 22?

>>
Anonymous 2016-09-15 02:58:51 No.8346127
Actual stupid question here. Been staring at it for an hour.

>>
Anonymous 2016-09-15 03:01:18 No.8346134
>>8346127
write as 2 integrals then try:
>substitution
>partial fraction decomposition

>>
Anonymous 2016-09-15 03:02:48 No.8346138
>>8346127
separate into $\int \frac{u^8}{u^2+1}$ and $\int \frac{u^5}{u^2+1}$ and do long division

>>
Anonymous 2016-09-15 03:06:55 No.8346145
>>8345425
2^k = (u-1)(u+1)
tells us that both u+1 and u-1 must be powers of 2. what powers of 2 differ by 2?

>>
Anonymous 2016-09-15 03:08:59 No.8346150
>>8345506
never really understood this... what's an "elementary function"? why is erf() less elementary than cos()?

>>
Anonymous 2016-09-15 03:10:59 No.8346155
im so fucking dumb. i have a midterm on friday and i cant even do this. please help

>>
Anonymous 2016-09-15 03:11:06 No.8346156
>>8345940
you need an arccos() in there?

>>
Anonymous 2016-09-15 03:17:21 No.8346167
>>8346138
That's what I tried. Stuck on where to go with
$$\int \frac{u^8}{u^2+1}$$. Is there a way without doing long division?

>>
Anonymous 2016-09-15 03:20:45 No.8346172
>>8346167

>>
Anonymous 2016-09-15 03:21:16 No.8346173
>>8346167
write the integrand as

u^6-u^4+u^3+u^2-u-1 + (u+1)/(u^2+1)

the first terms are easy, the last looks pretty straightforward too.

>>
Anonymous 2016-09-15 03:22:07 No.8346174
>>8346172
forgot: b/c theres a remainder of one you need to make sure to add 1/(u^2+1

>>
Anonymous 2016-09-15 03:22:43 No.8346178
>>8346172
might as well keep the x^5 in there too, no need to split if you're going to use long division

>>
Anonymous 2016-09-15 03:29:44 No.8346192
>>8346138
>>8346127
Or just do long division directly
[eqn]\frac{u^8+u^5}{u^2+1} = x^{6}-x^{4}+x^{3}+x^{2}-x-1 + \frac{x+1}{x^2+1}[/eqn]

>>
Anonymous 2016-09-15 03:30:53 No.8346195
>>8346192
>>8346178
yeah, dont know why i didnt just say that. from there finding the integral is trivial

>>
Anonymous 2016-09-15 05:00:42 No.8346391
http://demonstrations.wolfram.com/AnEfficientTestForAPointToBeInAConvexPolygon/

>So the total costs for the test are just two additions (for the initial origin translation), two multiplications, one subtraction, and one "greater than zero" comparison for every vertex; finally an n-fold equality comparison if all the signs of the angles are equal.
>finally an n-fold equality comparison if all the signs of the angles are equal.

I don't get it, where does the last part come from? Is it not enough with just
X_(i + 1) * Y_(i) - X_(i) * Y_(i + 1) > 0

>>
Anonymous 2016-09-15 05:12:27 No.8346425
>>8344746
Someone explain why $(-a)(-b) = ab$. I understand that our algebra basically forces us to do so, but are there any explanations that make sense?

>>
Anonymous 2016-09-15 05:23:34 No.8346449
>>8346425
5 * 1 = 5
5 * -1 = -5
-5 * 1 = -5
-5 * -1 = 5

>>
Anonymous 2016-09-15 05:24:46 No.8346452
>>8346449
No shit?

>>
Anonymous 2016-09-15 05:47:04 No.8346492
>>8346452
$No shit?$

>>
Anonymous 2016-09-15 05:48:49 No.8346493
>>8346425
Because -(-a) = a and -a = (-1)a.

In more detail, we have (-a)(-b) = (-1)(a)(-1)(b) = (-1)(-1)(ab) = -(-1)(ab) = (1))(ab) = ab.

>>
Anonymous 2016-09-15 06:47:27 No.8346565
For calculating the error bounds of Trapezoidal Approximation, am I allowed to pick any arbitrary $K$ as long as $f(x) \leq K$? It seems like some of the solution manuals are picking wildly random values of $K$ for the calculation, unless I'm getting my local maximums wrong.

>>
Anonymous 2016-09-15 06:53:12 No.8346571
>>8344877
after a certain point, higher category theory starts to feel almost geometric in nature, and everything starts to make sense
the so-called "intuitive" parts of category theory finally actually become intuitive
any sort of connection between two things immediately sets of bells in your mind
you start to feel like reality is just the macroscopic manifestation of some braided, knotted, or twisted objects
you feel like you ~are~ the categories
it's like autism except worse
but possibly the worst part is that even though you've sacrificed your humanity for mathematics, there are still people who understand it better than you

>>
Anonymous 2016-09-15 07:06:32 No.8346588
>>8346565
if you make that |f''(x)|, sure.

>>
Anonymous 2016-09-15 07:09:50 No.8346592
>>8346391
I tried it without the equality and it works although I had to flip it around to X_(i) * Y_(i + 1) - X_(i + 1) * Y_(i) > 0 with counterclockwise vertices

>>
Anonymous 2016-09-15 07:46:41 No.8346629
how to solve 0.6125 = sin2x. using R = Vsin(2x)/g in physics and I'm supposed to find the angle.

>>
Anonymous 2016-09-15 08:00:51 No.8346637
>>8346629
arcsin

>>
Anonymous 2016-09-15 08:10:47 No.8346649
>>8344746
Is this text book by Rudin?

>>
Anonymous 2016-09-15 08:23:34 No.8346662
>>8346637
what happens to the 2 in the sin2x? arcsin(2*0.615)?

>>
Anonymous 2016-09-15 08:29:46 No.8346673
>>8346662
arcsin "cancels out" sin

apply to both sides:

arcsin(0.6125) = arcsin(sin(2x))

arcsin(0.6125) = 2x

arcsin(0.6125) / 2 = x

>>
Anonymous 2016-09-15 08:33:20 No.8346678
>>8346673
thanks!

>>
Anonymous 2016-09-15 08:54:35 No.8346695
How do I find a list of numbers that are NOT divisible by something in wolfram alpha?

Trying to find even numbers between 100-300 that are not divisible by 4. Wolfram alpha does not accept that phrasing. Best I can do is divisible by

>>
Anonymous 2016-09-15 09:08:43 No.8346705
>>8346695
never used it before (lol) but maybe try putting ! infront, in programming != means no equal so maybe that'll work

>>
Anonymous 2016-09-15 09:08:50 No.8346706
>>8346695
you could try doing it programmatically with some wolfram language syntax, like generate a list of even numbers and remove the ones that aren't divisible by 4

>>
Anonymous 2016-09-15 09:15:41 No.8346713
>>8344746
q = p - (p^2 - 2)/(p+2) = (p - sqrt(2))*(p+sqrt(2))/(p+2). If p^2 < 2, then the equation can be rewritten as q = p + (sqrt(2) - p)*(p + sqrt(2))/(p + 2). Clearly, the sqrt(2) - p term is meant to get p closer to sqrt(2), but the author doesn't want that, he wants to only cover a part of the distance. That is why he scales sqrt(2) - p down by (sqrt(2)+p)/(2+p).

>>
Anonymous 2016-09-15 09:32:11 No.8346728
How do I find the inverse (if it exists) for this kind of function?
f(a) = 1
f(b) = 2
f(c) = 4
f(d) = 3

>>
Anonymous 2016-09-15 09:47:16 No.8346743
Also, why is this not a function from R to R?
Goodness, if my university is going to introduce new concepts to me that are not easily researchable, it shouldn't be in the form of a question!

>>
Anonymous 2016-09-15 09:52:41 No.8346749
>>8346743
Because if R was the domain, then x=-2.9 would have to be in it, but that's rules out by the requirement x>1.
The domain and codomain of that functions are some open connected subsets of R.
(Besides, identifying a function with its model as a set of pairs is a little Plebeian an rough, but that's more a philosophical and educational point.

>>
Anonymous 2016-09-15 10:00:07 No.8346756
>>8346749
Why does -2.9 specifically have to be in the domain?

>>
Anonymous 2016-09-15 10:06:30 No.8346762
>>8346728
Is the function from {a,b,c,d} to {1,2,3,4}? Then if g is the inverse of f, you want g(f(x))=x for all x in {a,b,c,d} and f(g(y))=y for all y in {1,2,3,4}

So g defined as g(1)=a, g(2)=b, g(3)=d, g(4)=c is your inverse

>>
Anonymous 2016-09-15 10:45:39 No.8346798
>>8346756
You chose the wrong major

>>
Anonymous 2016-09-15 10:48:36 No.8346800
>>8346798
They seriously don't explain most of this at my university, it's a joke.
It's considered lucky if they can deliver exam papers without fucking something up.

>>
Anonymous 2016-09-15 10:52:37 No.8346804
>>8346756
-2.9 is a real number... it's part of R...

>>
Anonymous 2016-09-15 10:56:05 No.8346807
>>8346804
I'm aware, but why -2.9 specifically?

>>
Anonymous 2016-09-15 10:58:10 No.8346810
>>8346800
But you're also lazy.
You cab look up the definion of the domain and codomain of a function, the word I used.
You ask about why f doesn't go from R to R. The answer is that if you requre y="something of x" when x>1, then there is no information about what the value y=f(-6.353) is. Thus the function doesn't go from R somewhere. Much of R is not in its domain.

>>
Anonymous 2016-09-15 10:59:01 No.8346811
>>8346807
it was an example

>>
Anonymous 2016-09-15 10:59:55 No.8346812
>>8346807
It's a generic real. One over pi to the power of 5 also works as value that's not in the domain.

Again, take a notebook and write down some definitons you can come back to. Do it.

>>
Anonymous 2016-09-15 11:04:39 No.8346821
>>8346810
Ah, I suspected that was how it worked but it seemed overly trivial.
Kinda cheeky, making all the assignments hard then giving one where every single answer is trivial.

>>
Anonymous 2016-09-15 11:07:56 No.8346824
>>8346810
Wait, actually, if y is the codomain why does it matter that that there is no corresponding f(x)?
Or is the problem that f(x) is supposed to include even explicitly ruled out values of x?

>>
Anonymous 2016-09-15 11:29:55 No.8346836
>>8346695
it is well past the time for you to learn how to program

>>
Anonymous 2016-09-15 12:00:36 No.8346858
If f and g are functions, what does
g o f mean?
I'm using o in place of a circle symbol, it's not the letter o.

>>8346824
Darn, I guess that anon is gone.

>>
Anonymous 2016-09-15 12:03:58 No.8346862
>>8346858
https://en.wikipedia.org/wiki/Function_composition

>>
Anonymous 2016-09-15 12:42:10 No.8346888
Given 2 particles "A" and "B", does observing "A" collapse the wave function of "B" if B itself has not been observed yet?

I'm asking in regards to "quantum communication", with the following scenario:

If we have two "stations" an arbitrary distance apart (let's just say 100 miles for arguments sake) and place a photon gun right in the middle at the 50 mile point. It creates sets of entangled photons and sends one to "Station A" and one to "Station B". It sends as many photons in either direction as needed to make a decent enough interference pattern (let's just say 1000 entangled photon pairs, but whatever).

"Station A" would have an on/off "observer switch" that could choose to either collapse the wave function of the incoming 1000 photons, or not observe them and keep the wave function in tact. "Station B" would then see the intended result of Station A when it receives its photon and is sent through a double slit. If Station B observes an interference pattern, it means Station A had their "observer switch" off and this could be interpreted as a binary "bit 0". If there's no interference pattern, it means it was observed at station A and this could be interpreted as binary "bit 1". After a pre-set time at both stations and photon gun, the bit is recorded and then another 1000 entangled photon pairs are sent.

Is there a reason why this wouldn't work, or why this wouldn't be classified as "Quantum communication"?

>>
Anonymous 2016-09-15 13:01:39 No.8346898
>>8346888
Yes, observing A collapses B.

Look at it this way:
If you split apart a positive and negative particle, sending one to A and one to B, until observed you could have either the positive particle or the negative particle arriving at B.
But if you receive a positive particle at A, there's zero chance that the particle at B will also be positive, no matter how long you wait before observing B.

As for your proposed quantum communication however, that's a lot more complicated.
Observing particle A will not make the location of B 100% certain as there's lots of ways for quantum uncertainty to creep in, but the question is if it can be even a smidgen more certain than it would be otherwise, since one could make a double slit which favours particles that have especially uncertain location.

>>
Anonymous 2016-09-15 13:07:42 No.8346903
>>8346898
Thanks for your help anon, it was nice to feel like I was onto something for a few moments kek.

>>
Anonymous 2016-09-15 13:07:55 No.8346904
Some pretty basic number theory.

Prove that no matter what value of n you start with you will eventually reach 1 with successive applications of the following procedure:

Given any positive integer n, if n is even divide it by 2, if n is odd multiply it by 3 and add 1.

>>
Anonymous 2016-09-15 13:28:33 No.8346936
>>8346904
This is too trivial, even for this thread :^)

>>
Anonymous 2016-09-15 13:36:33 No.8346947
>>8346904
Prove that hailstone numbers always go down to 1?
Pretty sure that there's only empirical evidence for that so far.

>>
Anonymous 2016-09-15 13:39:43 No.8346953
>>8346904
stop wasting our time with easy questions

>>
Anonymous 2016-09-15 13:41:03 No.8346955
>>8346127
do polynominal division to simplify the expression into P(u) + (Au + B)/(u^2 + 1), then break (Au+B)/(u^2 + 1) into (A/2)*(2u)/(u^2 + 1) + B/(u^2 + 1) (because 2udu = d(u^2 + 1)). Clearly, these three integrals can be solved with the power rules, a natural logarithm, and an arctangent.

>>
Anonymous 2016-09-15 13:44:34 No.8346961
>>8346947
seems like it would be fairly easy to prove though?

>>
Anonymous 2016-09-15 14:17:56 No.8347022
>>8346961 (Me)
so it's an unsolved problem, too bad it isn't a $1 million millennium prize problem lmfao what do i get if i prove it? >> Anonymous 2016-09-15 14:36:17 No.8347042 Maybe more home electrical than science, but would drawing 46 amps on a 50 amp circuit be fine? I know you're supposed to leave some leeway in there, not sure is 8% is sufficient, though. I'm setting up an electric brewery in my garage and don't really want to burn my house down. >> Anonymous 2016-09-15 14:42:00 No.8347053 >>8347042 46 amp could be the "average" draw while running, if you're pulling more than that even if it's transient (like when you're turning it on or something) that might trigger the circuit breaker also >>>/diy/ >> Anonymous 2016-09-15 15:42:04 No.8347156 >>8347022 >what do i get if i prove it? You will literally become famous. Also I'm sure Erdös has promised a price for the solution, maybe$1000 or something similar.

>>
Anonymous 2016-09-15 15:48:16 No.8347178
>>8346145
Damn! I wrote that down but was clearly too tired to see the implication, then u-1=2 and u+1=4 therefore u=3, 2^k+1 = 3^2 =9 and k must be 3. Thanks.

>>
Anonymous 2016-09-15 15:51:32 No.8347198
Just had a test for one if my cs classes. Ine of the problems was to create a proposition for the statement O(n)= "Is an idd number" where the domain is all Natural numbers. I got hung up on an earlier problem and had to rush through this part without a while lot of thought.

I put "for all n((n-1)/(2(n/2)=1)" immediately after the test I realized "for all n there is an x (n=2x+1)" was a better answer. But was my answer wrong? From my understanding deviding and odd natural number by 2 would round down (3/2=1) so my answer would be ((3-1)/(2 (3/2)) = (2/(2(1)) =1

Or am i too sleep deprived to think clearly

>>
Anonymous 2016-09-15 15:58:06 No.8347228
>>8347022
Good luck inventing the entire area of math you will need first. Learn some category theory.

>>
Anonymous 2016-09-15 16:20:23 No.8347325
>>8347228
>>8347156
for 1 million dollars i would (try to) do it... don't think i'll put any serious amount of effort into it if it's just for fame

>>
Anonymous 2016-09-15 16:26:03 No.8347353
>>8347325
try the Hodge conjecture then

>>
Anonymous 2016-09-15 16:45:12 No.8347434
Is there a good website for learning Calc III?

None of this shit makes any sense

>>
Anonymous 2016-09-15 16:48:03 No.8347446
How do you get the differential of an integral when substituting two of the previous coordinates with one?

>>
Anonymous 2016-09-15 17:03:34 No.8347507
>>8347434
No. Only textbooks

>>
Anonymous 2016-09-15 17:13:41 No.8347553
why is this not the answer

side note: fuck online homework

>>
Anonymous 2016-09-15 17:43:10 No.8347655
No idea to make new thread so I ask here instead. What are some really good linear algebra books?

>>
Anonymous 2016-09-15 17:46:29 No.8347666
>>8347553
They don't bite, usually.

>>
Anonymous 2016-09-15 17:47:30 No.8347671
>>8347655
Axler.
Check the wiki next time.

>>
Anonymous 2016-09-15 17:49:13 No.8347673
>>8347671
Thanks. Will do that.

>>
Anonymous 2016-09-15 18:06:54 No.8347721
>>8347553
Is there a "try ungraded version"? Looks like webassign

>>
Anonymous 2016-09-15 18:11:52 No.8347735
>>8347198
this problem depends on many things

for instance, if it is a binary machine and you have direct access to testing bits, then you can just test whether the units bit is 1

>>
Anonymous 2016-09-15 19:12:28 No.8347889
If you have an 18 square inch peice of cardboard and you cut into an open top square box out if it. What would its volume be? I thought it was (18-x)cubed because lwh and since its a cube all sides are equal length.

>>
Anonymous 2016-09-15 19:27:55 No.8347916
>>8347889
If your square box is a cube...
Try drawing a map of an open cube.
You end up with a cross 3 squares tall and 3 squares wide
Thus each side of your cube is 18/3=6 and your volume is 6^2=36

>>
Anonymous 2016-09-15 19:28:01 No.8347917
>>8347889
You cant do it with a square. You cant get 6 squares out of a square. The cardboard would need to be 2x by 3x as a cube has 6 sides. Each side being x, volume is x^3.

>>
Anonymous 2016-09-15 19:58:55 No.8347990
>>8345451
>>8345723
>>8346088

It's funny because I just started reading rudin too, and thought to myself "Nigga just use proof by contradiction" I know he had a purpose with his argument but why the fuck is the first two pages so blah. I might jump ship and look for another analysis book. I just finished pic related.

>>
Anonymous 2016-09-15 20:40:52 No.8348069
>>8347917
3^2 = 1^2 + 1^2 +1^2 + 1^2 + 1^2 + 2^2

>>
Anonymous 2016-09-16 00:36:43 No.8348524
How many objects of space debris where left in earths orbit during an apollo moon mission?

>>
Anonymous 2016-09-16 00:44:01 No.8348541
Can someone explain this triangle inequality? Doesn't really matter what D or Hn are. I guess just prove

$| \sum_{i=0}^n (x_i) - \sum_{i=0}^n (y_i) | \leq \sum_{i=0}^n |x_i - y_i|$

>>
Anonymous 2016-09-16 00:45:45 No.8348546
>>8344746
>Can anyone explain how this equation was derived? It seemed out of nowhere to me

Welcome to baby Rudin

http://math.stackexchange.com/questions/141774/choice-of-q-in-baby-rudins-example-1-1
http://math.stackexchange.com/questions/14970/no-maximumminimum-of-rationals-whose-square-is-lessergreater-than-2

>>
Anonymous 2016-09-16 00:48:27 No.8348553
>>8348541
nvm got it, I just needed to rearrange the terms on the left side to get $(x_i-y_i)$ which turns it into a sum instead of a difference, then use absolute value form of triangle inequality

>>
Anonymous 2016-09-16 00:53:28 No.8348570
>>8347042
The fuck do you need 46 Amps for?

>>
Anonymous 2016-09-16 00:57:41 No.8348580
Notation question, if I have ( \frac{dy}{dx} )^{2} that's a first order derivative right? What is it supposed to look like when solved? Assuming y' equaled something like e^{x} would ( \frac{dy}{dx} )^{2} be (e^{x})(e^{x})?

>>
Anonymous 2016-09-16 01:00:22 No.8348593
>>8348580
Yeah, that's just the derivative squared.

By contrast,
[eqn]\frac{d^2y}{dx^2}[/eqn]
is the second order derivative.

>>
Anonymous 2016-09-16 01:02:11 No.8348600
>>8348342
you missed a minus sign in your last double series, there should be an overall minus before the two summations.

You want to convert your double series to a power series, so look at what terms contribute for a given power of z, say n.

If n=0, set n=k-2j=0 and sum over all values of k,j which satisfy this constraint in the double series --> k=2j and sum over j gives you factor of -e, since it's just a sum over the factorial factors which is the taylor series for -exp[x] at x=1.

If n>0, keep doing this and realize you get n=k-2j--->k=n+2j does nothing except pop out a z^n factor and you still do a summation over the inverted factorials (index by j) which gives the factor -e.

If n<0, then you can only include positive values of j in setting k=n+2j. Because now you must always start at some positive value of j, you won't get the full series for -e--instead you'll be missing the first few terms. That's what these positive factors are. If n=-1, you start at j=1, and so you miss the 1=1/0!) term, but keep the rest of the series for -e starting from -1/(1!) onward. If n=-2, you still start at j=1, but now you k value starts at 0 instead of 1. This continues onward, since for odd powers of n<0 k will start at k=1 but even powers k will start at k=0--so the initial starting point in the j series will be unchanged for the constant factor in consecutive odd, even powers. Every even negative power will lose another term in the series expansion for -e, and so must be added to compensate.

This is a rough outline, you can be more precise for the negative terms if you really want but this is how you resum series to get the desired power series--fix the power of the term you want to find and sum over all values that produce such a term.

>>
Anonymous 2016-09-16 01:13:14 No.8348627
>>8348593
Thanks a ton!

>>
Anonymous 2016-09-16 01:36:40 No.8348665
>>8348600
Thanks a lot for this. I can actually go to sleep now without this bugging me.

>>
Anonymous 2016-09-16 01:43:39 No.8348673
I don't have a question just wanted to say thanks high school children for helping me with my math

>>
Anonymous 2016-09-16 02:01:20 No.8348687
We select from a deck of cards the four kings and the four queens. From these eight cards we draw one
card at a time, without replacement, until all eight cards are drawn. Find the probability that
a) All kings are drawn before the queen of spades.
b) There is at least one queen that is drawn after all the kings.
c) Each queen is drawn before each of the kings.
d) The last king to be drawn is the sixth card to be drawn.
e) Each queen is drawn before the king of the same suit.

The answers were 1/5, 1/2, 1/70, 1/16, 1/7

I've been mad at statistics for a week now

>>
Anonymous 2016-09-16 02:05:17 No.8348693
>>8348673
man the harpoons

>>
Anonymous 2016-09-16 03:23:46 No.8348833
>>8348342
>>8348665
>>8348600

Self study dumbie me is back with another one.

Laurent series for $\frac{1}{z^2-4} \text{ about } z = 2.$

I decompose $\frac{1}{(z-2)^{2}-4} = \frac{1}{z(z-4)}$ into partial fractions and get a strictly negative series even though it should be alternating apparently. The book's answer is
$\sum\limits_{k=-1}^{\infty}{\frac{-1^{k+1}(z-2)^{k}}{4^{k+2}}}$ I just don't see how it's alternating. Does anyone recommend any practice resources also to refine my skills with series manipulations? Seems like I lost a bit of it if I need to keep asking this shit.

>>
Anonymous 2016-09-16 03:27:38 No.8348850
It's alternating due to the k=-1 my brainlet friend

>>
Anonymous 2016-09-16 03:54:11 No.8348919
>>8348687
i fucking hate combinatorics too

>>
Anonymous 2016-09-16 03:57:34 No.8348923
>>8348833
You need to practice these on your own my friend, this is the last one I'll help with. You clearly haven't got the hang of these, do like 10 more that look difficult to you/not immediately obvious.

$\frac{1}{z^2-4} =\frac{1}{z-2} \frac{1}{z+2} = \frac{1}{z-2}\frac{1}{(z-2)+4} = \frac{1}{z-2}\left(\frac{1}{4}\right)\frac{1}{1-[-(z-2)/4]} \\ = \frac{1}{4(z-2)}\sum_{k=0}^\infty\frac{(-1)^k (z-2)^k}{4^k} = \sum_{k=0}^\infty\frac{(-1)^k (z-2)^{k-1}}{4^{k+1}} \\ = \sum_{k=-1}^\infty\frac{(-1)^{k+1}(z-2)^k}{4^{k+2}}$

>>
Anonymous 2016-09-16 03:59:09 No.8348924
>>8348833
as far as a good resource, crack open any calc 2 sequence text and work through some of the series problems if the Laurent ones are too tricky.

>>
Anonymous 2016-09-16 07:08:18 No.8349201
Regarding maximum data rate of a channel, I used Nyquist and Shannon theorems. Which one do I choose? The Shannon one is higher.

>>
Anonymous 2016-09-16 09:46:24 No.8349411
How can $\frac{3}{10} ~<~ \frac{2}{5}$ if both numbers of 3/10 are larger than in 2/5????

>>
Anonymous 2016-09-16 09:50:12 No.8349415
>>8349411
2/5 = 4/10

>>
Anonymous 2016-09-16 09:51:07 No.8349416
>>8349415
That's cheating, you're increasing both numbers, of course it's bigger then.

>>
Anonymous 2016-09-16 09:55:50 No.8349421
>>8349416
1/1 = 1
2/2 = 1

>>
Anonymous 2016-09-16 09:56:53 No.8349422
>>8349421
BUT WHY
Logically 1/1 should be smaller than 2/2 because both numbers are bigger!

>>
Anonymous 2016-09-16 09:58:26 No.8349424
>>8349422
Because that isnt how math works

>>
Anonymous 2016-09-16 09:59:47 No.8349428
>>8349424
Can you please explain logically why it wouldn't work instead of resorting to every primary school math teacher's favorite argument "because I say so"?

>>
Anonymous 2016-09-16 10:00:24 No.8349429
>>8349422
If you arent trolling, think of it more as a proportion

If you cut a pizza in quarters and take 2 quaters (2 / 4), you have half the pizza (1 / 2)

>>
Anonymous 2016-09-16 10:01:15 No.8349430
>>8349428
Because 1/1 and 2/2 mean 1 divided by 1 and 2 divided by 2, both of which equal

>>
Anonymous 2016-09-16 10:02:17 No.8349432
>>8349430
equal 1*

>>
Anonymous 2016-09-16 10:02:37 No.8349433
>>8349429
Wrong, I have two quarters, plus a pizza slicer, minus the money I could have earned in the time I spent to cut it.
Also two quarters of a pizza last longer than one half because you're eating two pieces instead of one, which is just proving my point.

>>8349430
> 1 = 2 because 1/1 = 2/2

>>
Anonymous 2016-09-16 10:02:50 No.8349434
>>8346571

>>
Anonymous 2016-09-16 10:04:03 No.8349437
>>8349433
1/1 and 2/2 both equal 1. 1 does not equal 2

>>
Anonymous 2016-09-16 10:04:12 No.8349439
>>8349433
plug it into a calculator

2 divided by 2 is...?

>>
Anonymous 2016-09-16 10:06:31 No.8349442
>>8349437
They can't both equal the same number because 1/1 is smaller than 2/2 because both numbers in 1/1 are smaller than the numbers in 2/2.
If 1/1 = 2/2, then logically it must follow that 1 = 2.

>>8349439
> it works because the calculator says so

>>
Anonymous 2016-09-16 10:09:14 No.8349445
>>8349442
x/y means that of y possible parts you have x of them. So 1/4 means that of 4 parts you have 1, or 25%, or 0.25. 2/2 means that of 2 parts you have 2, or 100%, or 1. 1/1 means that of 1 part you have 1, or 100% or 1

I mean I know you're trolling but this is vaguely amusing for some reason

>>
Anonymous 2016-09-16 10:13:50 No.8349452
>>8349445
I know what that means but how is it possible that when you have two fractions and the numbers in one fraction are larger than those in the other fraction, they still evaluate to the same number, or the fraction with the smaller numbers like 1/2 is actually larger than the fraction with the bigger numbers like 5/20?

>>
Anonymous 2016-09-16 10:16:21 No.8349453
>>8349452
Its possible because 1 divided by 2 is bigger than 5 divided by 20. Its the relationship between the numbers in the fraction that matter not the numbers themselves

>>
Anonymous 2016-09-16 10:17:13 No.8349455
>>8349453
> it's possible because that's how it works
circular logic

>>
Anonymous 2016-09-16 10:18:23 No.8349456
>>8349455
All of math is circular logic. Its true because thats how we define it to be

>>
Anonymous 2016-09-16 10:20:10 No.8349457
>>8349456
Whoa man, that's like totally deep.

>>
Anonymous 2016-09-16 10:21:08 No.8349458
>>8349457
No its not, its a basic fact of mathematics

>>
Anonymous 2016-09-16 10:34:47 No.8349472
>>8349452
Beacause its a new number. A fraction is a percentage. Think of it like 50% = 1/2 or 2/4 rather than 1/2 = 2/4 and it makes more sense. The specific numbers dont matter as there are infinite solutions to x/2x=50%. Division works that way because we say it works that way.

You could come up with your own operation (lets call it +÷) that does what you are describing.
x+÷y=z_n,d and z_n-1,d-1 < z_n,d
Where subscripts denote the num and denom.

So

1+÷2=0.5_1,2 and 2+÷4=0.5_2,4 and 0.5_1,2<0.5_2,4

But this would get stupid confusing after you perform this operator on the same element a few times. The practical applications of it dont seem to warrant its use.

>>
Anonymous 2016-09-16 10:40:32 No.8349477
>>8349472
>Think of it like 50% = 1/2 or 2/4
Again you are assuming that 1/2 = 2/4 when really that shouldn't be the case as they are comprised of different numbers.
>The specific numbers dont matter
Then why use them?
>infinite solutions to x/2x=50%
Only if you assume that which I am trying to get you to prove. PROVE that x/2x = 50% for all x.

This started out as trolling but now I'm genuinely interested in seeing a proof for something this "basic". I mean, everyone understands it intuitively but explaining it to other people formally still seems to be a challenge.

>>
Anonymous 2016-09-16 10:44:56 No.8349482
>>8349477
>Again you are assuming that 1/2 = 2/4 when really that shouldn't be the case as they are comprised of different numbers
Fractions are an expression of a relationship between values. The relation between 2 and 4 is the same as the relationship between 1 and 2

>Then why use them?
Have to use something

>PROVE that x/2x = 50% for all x
2 lots of something is twice as much as 1 lot of something by tautological definition. Its true because thats the definition of what those words mean

>>
Anonymous 2016-09-16 10:48:40 No.8349487
>>8349482
>The relation between 2 and 4 is the same as the relationship between 1 and 2
Wrong, 4 is 2 larger than 2 whereas 2 is only 1 larger than 1.
So $\frac{2}{4} = \frac{2}{2+2} = \frac{1+1}{1+1+1+1}$ and $\frac{1}{2} = \frac{1}{1+1}$. Note that there are twice as many 1s in 2/4 as there are in 1/2, so it's obviously a larger number.

What you posted was no proof. I want a formal proof, using axioms.

>>
Anonymous 2016-09-16 10:49:25 No.8349489
>>8349477
You can't prove a definition you shitlord. Its defined. Division is an operator that is defined. Addition is defined. Subtraction is defined. They are human made rules. They exist because we say they do, literally. Thats the reason. Some guy said + means you combine 2 elements.
I can say a crude drawing of a penis means multiply the reciprocal if 2 numbers. You can make up any fucking operator you want and you dont need to prove it. I did it in my reply. You learn this in algebra 2 or some shit.

>>
Anonymous 2016-09-16 10:50:01 No.8349490
>>8348924
It's just rust and I know I need practice as I acknowledged. Strange thing is the other ones I was doing came out as silk then I just crashed. Maybe should have just went to bed. I guess I'll just see what power series stuff I can find but things like converting a double sum I can't find much of. I'll look. I also have like 4 more Complex texts I can do Laurent problems.

>>
Anonymous 2016-09-16 10:52:31 No.8349493
>>8348850
It was more of how they get to that point to begin with. Not why it wasn't alternating from the answer.

>>
Anonymous 2016-09-16 10:52:54 No.8349494
>>8349487
There are as many 2's in 4 as there are 1's in 2. That is the relationship being express by 1/2=2/4

These are axioms

>>
Anonymous 2016-09-16 10:55:18 No.8349498
>>8349489
Still not seeing a formal proof.
There is no definition that says "1/2 = 2/4", that's an assumption that I want you to prove using well-defined axioms.
> Division is an operator that is defined
So are the comparison operators, and look how well they work.
> if one number is larger than the other we use the greater than operator
> 4 is larger than 2, so 4 > 2
> 2 is larger than 1, so 2 > 1
$\frac{4 > 2}{2 > 1} \rightarrow \frac{4}{2} > \frac{2}{1}$

>>8349494
> There are as many 2's in 4 as there are 1's in 2
You're comparing apples to oranges here.

>>
Anonymous 2016-09-16 10:58:56 No.8349503
>>8349498
>You're comparing apples to oranges here
No i'm comparing values. In this case the same value is expressed in different ways

>>
Anonymous 2016-09-16 11:00:20 No.8349505
>>8349503
If there are as many 2's in 4 as 1's in 2, and 2>1, logically 4>2 right?

>>
Anonymous 2016-09-16 11:00:50 No.8349507
>>8349489
In addittion the natural language definition for division is domething like "for x,y,z: x devided by y, where y is not zero, z is the number or ys contained in x. In addition to this z times y is x." ×/y=z
So 1/2=.5 "there is one half 2s in 1."

2/4=.5 "there is one half 4s in 2."

6/3=2 "there are 2 3s in 6.

>>
Anonymous 2016-09-16 11:01:26 No.8349508
>>8349498
There are as man "I" in "II" as "II" in "IIII". You can see this just by looking, no math involved

>>
Anonymous 2016-09-16 11:01:52 No.8349509
>>8349498

>>
Anonymous 2016-09-16 11:02:32 No.8349510
>>8349487
What if 5/4th's of something never exists, and all fractions that exist outside of just being a relationship, are an expression of that which is less than 1/1?

>>
Anonymous 2016-09-16 11:02:35 No.8349511
>>8349505
Yes, 4>2 in exactly the same ratio as 2>1

>>
Anonymous 2016-09-16 11:03:40 No.8349513
>>8349511
So if 4>2 and 2>1, 4/2 > 2/1. See >>8349498

>>
Anonymous 2016-09-16 11:05:05 No.8349516
>>8349498
Prove that 4>2 faggot i dont believe you

>>
Anonymous 2016-09-16 11:05:43 No.8349517
>>8349411
If you are not trolling:
You probably know addition right? If not you should learn that first.

If you know addition you should also know subtraction:
Saying 7-10=x is that same as searching for the answer to the Question:
Which whole number do I need to choose so that 7=10+x.
You might argue that no one knows that that is true and the result should not be -3 but because the natural numbers (1,2,3,...) and the whole numbers (...,-1pro,0,1,...) are defined a certain way(based on Axioms, that means things that are not logically derived) it is true.

You probably also now multiplication. Else you need to learn that first.

Now consider a similar example to subtraction:
Saying 7/10 = x means really that you are searching for a whole number so that 7=x*10 is true. That is the way the ration numbers (all fractions) are defined.
Again there is little arguing here because mathematics is based on axioms.

Why are 2/2 and 1/1 the same.

2/2 is the solution to 2=2*x, you see here that x has to be 1.
1/1 on the other hand is the solution to 1=1*x, again that has to 1.

This will work for all fractions like 1/1,2/2,3/3,... they are all equal to 1.
Secondly:
Why are 1/2 and 2/4 the same?

Because it is defined that way. There is really no better explanation except that it is completely consistent with reality.
The exact definition is: a/b=c/d if and only if b*c=a*d.

Thirdly:
why is 3/10 < 2/5.
That follows logically from "secondly".

2=5*x has by definition the same solution as 4=10*x
It will logically follow that the solution to 4=10*x is bigger then the solution to 3=10*x.
Im sorry if you didnt understand, but foundation of mathematics is a very complicated topic and many Analysis courses will start with precisely defining what the numbers are, but that usually takes up a couple of lectures.

The way the numbers are constructed from the ground up will explain many question like yours and many more like, 0.9999...=1?.

>>
Anonymous 2016-09-16 11:05:44 No.8349518
>>8349513
No, because 4/2 =/= 4

>>
Anonymous 2016-09-16 11:06:07 No.8349519
>>8349509
Oh no I'm having a discussion about mathematics on the science and mathematics board on 4chan, better call the mods.

>>
Anonymous 2016-09-16 11:07:42 No.8349522
>>8349516
By the definition of the natural numbers that is wrong.

4=n(n(2)) therefore 4 is the successor of the successor of 2 and therefore bigger.

>>
Anonymous 2016-09-16 11:08:41 No.8349524
>>8349519
kill yourself you admitted that you were trolling and now you're just high-jacking the thread with your retarded off-topic "discussion"

>>
Anonymous 2016-09-16 11:15:04 No.8349537
>>8349530
Agree. As much as youve triggered my autism and kept me up til 6:30 this thread it was a stupid question and this is the stupid questions thread.

>>
Anonymous 2016-09-16 11:17:58 No.8349541
>>8349530
>You're missing a couple steps and didn't specifically write down any of the axioms
Yeah, i was at the 2000 character limit.
But foundations of mathematics is really interesting especially the construction of the real numbers that would give people a lot of understanding why certain things are the way they are and why certain infinite series can be equal to certain numbers.

>>
Anonymous 2016-09-16 11:18:38 No.8349544
>>8349530
Believe it or not this isn't challenging in any way.
It's just remarkable to imagine that somehow who can write down the words "formal proof" doesn't about how to compare fractions. Something made a hole in your head.

>>
Anonymous 2016-09-16 11:23:31 No.8349555
>>8349544
So instead of simply posting a consistent formal proof that will satisfy me (shouldn't take you long, after all "this isn't challenging in any way") and get this topic over with you decided to spend your time writing a post crying about how I should make my own thread (thereby pruning a possibly more interesting or valuable thread) and kill myself? GJ, you really improved the board quality. Why don't you go and suck some Hiro dick while you're at it, maybe he'll make you a janitor one day.

>>
Anonymous 2016-09-16 11:26:26 No.8349565
>>8349555
>shouldn't take you long
Just because something is trivial doesn't mean it's short to write. This was also my first post.
I mean it man, it's really fucking weird. Like how can you reach a level of education where you understand what a proof is and not know how to compare fractions?

>>
Anonymous 2016-09-16 11:28:27 No.8349569
>>8349555
Once again. "I' goes into "II" the same number of times as 'II" goes into "IIII". Proven without even using any math

>>
Anonymous 2016-09-16 11:29:40 No.8349570
>>8349565
> not know how to compare fractions?
I literally said I was trolling at the beginning.
Do you figure your elementary school teacher didn't know how to multiply numbers when he asked you to do it on a test or as homework? I wanted to see what that proof would look like, wanted to see if you could do it and wanted to motivate you to think about "trivial" things instead of accepting them as fact.
And still you're arguing in a meta-discussion about whether this discussion is pointless or not instead of contributing to the thread.

>>8349569
where are the axioms

>>
Anonymous 2016-09-16 11:35:23 No.8349575
>>8349570
>and wanted to motivate you to think about "trivial" things instead of accepting them as fact.
>hurr durr what if you, you know, built math on foundations
woaaaaw anoooon, you might be the first person eeeeeveeeer to have fought of that!
literally anyone who has studied math does that in their first year

>I literally said I was trolling at the beginning.
Then why are you surprised at the annoyance? When answering the question you realize we have to adapt it to who we are answering to? If you display the knowledge of a middle schooler, expect an answer for a middle-schooler. If you wanted the formal foundations of arithmetic you could have just asked so. And we would still have told you to go fuck yourself, because no way we're gonna LaTeX all that shit for your pleasure and you can find it anywhere.

>>
Anonymous 2016-09-16 11:35:54 No.8349576
>>8349570
Never fucking reply to me again unless you are contributing to the thread.

>>
Anonymous 2016-09-16 11:36:43 No.8349579
>>8349570
>where are the axioms
That is the axiom

>>
Anonymous 2016-09-16 11:39:56 No.8349584
>>8349575
> still complaining
> still not proving
I've been asking for a formal proof for literally *checks* close to an hour. See >>8349477

>>8349576
Here's a free (You).

>>8349579
Doesn't look like much of an axiom to me.
https://en.wikipedia.org/wiki/Axiom#Mathematical_logic

>>
Anonymous 2016-09-16 11:41:07 No.8349585
>>8349584
>I've been asking for a formal proof for literally *checks* close to an hour.
and you're not getting what you want? Oh my God poor you, whatever will you do?

>>
Anonymous 2016-09-16 11:42:02 No.8349587
Hey guys, I'm really stumped with this differential equation..
$\frac{d^2y}{dx^2}+ 4y=xsin(2x)$

I can get the complimentary function which is
$y_c=c_1cos(2x)+c_2sin(2x)$

I'm really struggling to find a guess for the particular solution. I want to solve this problem by the method of undetermined coeffecients.
Could someone just give me a hint as to what my guess should contain please! Thank you in advance.

>>
Anonymous 2016-09-16 11:42:04 No.8349588
>>8349584
I have no clue how to formulate a mathematical proof but the relationship has been clearly demonstrated

>>
Anonymous 2016-09-16 11:42:07 No.8349589
>>8349585
> whatever will you do

>>
Anonymous 2016-09-16 11:49:36 No.8349594
>>8349587
Guess a solution of the form $a x \cos x + b x \sin x + c \cos x + d \sin x$

>>
Anonymous 2016-09-16 11:51:32 No.8349598
>>8346571
That...that was beautiful...

>>
Anonymous 2016-09-16 11:53:54 No.8349602
>>8349594
Nevermind that, didn't see the 2 in there
guess:
$a x^2 \cos 2x + b x^2 \sin 2x + c x \cos 2x + d x \sin 2x$

>>
Anonymous 2016-09-16 12:05:40 No.8349614
>>8349602
Wow thanks for clarifying this. Turns out my simplifications are horrible af and thats why I wasnt getting the answer! Thanks a lot for the help anyways!

>>
Anonymous 2016-09-16 12:09:10 No.8349617
>>8349614
If you want the general method for that sort of differential equation see page 14 of this:
http://www.math.psu.edu/tseng/class/Math251/Notes-2nd%20order%20ODE%20pt2.pdf

In that case since the constant term is a product of a degree 2 polynomial and a trig function you should look for a solution of the same form (product of degree 2 polynomial and both sin and cos)

>>
Anonymous 2016-09-16 12:37:21 No.8349639
>>8349617
Wow thanks so much for the help! But the reason I asked that question is because I had the right guess but I lost track of all the various terms and made a mistake like over 4-5 times!

>>
Anonymous 2016-09-16 15:05:04 No.8349833
Harro anons,

does an abelian group satisfy distributive property? is it not an axiom?

>>
Anonymous 2016-09-16 15:15:02 No.8349850
>>8349833
For there to be a distributive property, you'd need a second operation - which you don't necessarily have in an abelian group. However, you can take an abelian group, with a second operation and a multiplicative identity - without division. Then you'd have a Ring, which does satisfy the distributive property.

>>
Anonymous 2016-09-16 15:18:08 No.8349855
>you'd need a second operation

what is this sneaky second operation that you refuse to name. Also ring is algebraic? cool.

>>
Anonymous 2016-09-16 15:23:23 No.8349861
>>8349855
There are lots of examples of rings, like Matrix algebra. Matrix Algebra, over addition forms an abelian group. With matrix multiplication, it forms a ring. It's important to note that AB isn't necessarily equal to BA. Furthermore, a matrix A isn't necessarily invertible; but, A(B+C) = AB + AC whenever you have matrices that are "conformable" to multiplication.

>>
Anonymous 2016-09-16 15:25:28 No.8349865
>>8349833
>>8349855
Distributive property requires two operations; otherwise there's nothing to distribute.

It's taken as an axiom for things that have two or more operations (e.g. rings) to make them compatible in some sense.

>>
Anonymous 2016-09-16 15:27:18 No.8349871
>>8349855
Another example of a Ring, is the Ring of integers. Clearly, the integers with addition & 0 form an abelian group. However, if you throw in regular multiplication, then your multiplicative Identity is 1. Furthermore ab = ba and If you divide two integers you don't necessarily get an integer back, you could get a rational! Hence, the Integers under multiplication & addition form a Ring. Note: the integers clearly have your beloved distrubution property, i.e. a(b+c) = ab + ac for all integers a, b, c.

>>
Anonymous 2016-09-16 15:31:51 No.8349878
>>8349871

>he integers clearly have your beloved distrubution property

which is an axiom, right?

Also thanks for the cool introduction to rings, I have always to study them. being in an engineering program sucks because ur missing out on the kool stuff

>>
Anonymous 2016-09-16 15:45:59 No.8349895
>>8349878
You can actually construct the integers from the naturals, and show it satisfies the properties that you use all the time.

>>
Anonymous 2016-09-16 15:58:26 No.8349915
>>8349456
Making definitions isn't circular.

>>
Anonymous 2016-09-16 16:31:21 No.8349975
Ola senorita /sci/, blighstone here is back with another question:

>show that a 1x1 matrix A is triangular.

>>
Anonymous 2016-09-16 17:05:26 No.8350036
How do asymptotes exist, doesn't 0.999…=1 mean an asymptote would be 0?

>>
Anonymous 2016-09-16 17:06:13 No.8350040
>>8349850

so to show that an abelian group is distributive you we have that group A is closed, so (a + b)c = (d)c and bc + ac = dc somehow???

wtf

>>
Anonymous 2016-09-16 17:07:33 No.8350043
>>8349975
List out all the entries not on the main diagonal and exhaustively prove that they are zero.

>>
Anonymous 2016-09-16 17:49:42 No.8350123
>>8350043

Erm what? Theres nothing outside of the diagonal, they are not even numbers. I am thinking of proof by case or using induction by showing that it satisfy all triangular properties.

>>
Anonymous 2016-09-16 18:02:51 No.8350150
how do you write (1+i)^{n}-(1-i)^{n} using exponentials?

>>
Anonymous 2016-09-16 18:15:18 No.8350180
>>8350040
You don't show that an "Abelian group" is distributive, for that you'd need a second operation. What you can do is realize that an abelian group is embedded in a larger Set with at least two operations, then realize that this larger set has the distributive property.

>>
Anonymous 2016-09-16 18:22:21 No.8350190
>>8350180

...and what do you mean by "second operation"..?

God i might actually be a retard

>>
Anonymous 2016-09-16 18:30:47 No.8350203
>>8350190
there is no second operation in a group, only one, i.e. the integers under addition

there are two operations in rings which you might be confusing a group with, i.e. the integers with addition and multiplication

>>8350150
(1+i)^n-(1-i)^n
= sqrt(2)(e^{n*i*pi/4}-e^{n*i*(-pi/4)})

>>
Anonymous 2016-09-16 18:39:45 No.8350225
>>8350203
whoops:
(1+i)^n-(1-i)^n
= sqrt(2)^n(e^{n*i*pi/4}-e^{n*i*(-pi/4)})

>>
Anonymous 2016-09-16 19:01:14 No.8350280
>>8350225
ok, this is what I found too.
I was trying to find a way to write it using a single exponential but I don't think it can be done.

>>
Anonymous 2016-09-16 19:06:40 No.8350289
>>8350280
its also equal to
sqrt(2)^n(e^{n*i*pi/4}-e^{n*i*(-pi/4)})
=sqrt(2)^n(2*i*sin(n*pi/4})

since the cosines cancel out

>>
Anonymous 2016-09-16 19:56:35 No.8350356
>>8344746

So I just saw a video of an alleged ladyboy harvesting piss from a goat and then injecting the piss in his pecs to make his "man boobs" swell. My question is: can he get an infection by doing this?

>>
Anonymous 2016-09-16 19:59:07 No.8350360
Within bending moments what does the x in this equation mean? I wondered if it was the axis of the cross section but then I don't know how this would have a integer

>>
Anonymous 2016-09-16 20:12:24 No.8350377
What are equations of the form: aq+bp=1, where a,b are known integers, called? Need the name since I want to check why they only have a solution if a and b are relatively prime. Just started an abstract algebra course after a year of no math

>>
Anonymous 2016-09-16 20:17:23 No.8350389
How do you justify spending much of your life on science and math?

>>
Anonymous 2016-09-16 20:19:13 No.8350391
>>8350389
it's useful.

>>
Anonymous 2016-09-16 20:22:04 No.8350395
>>8350389
For myself? Fun things are fun
For others? Math is useful for comp.sci.

>>
Anonymous 2016-09-16 20:22:48 No.8350396
>>8350389
You mean the align? I don't know the shortcut.

>>
Anonymous 2016-09-16 20:32:36 No.8350410
>>8350377
there's probably no name for it, but in general it's aq+bp=d where d is the greatest common divisor of a and b, and you use the euclidean algorithm to find such q and p

>>
Anonymous 2016-09-16 20:34:48 No.8350412
>>8350410
i take it back, i guess you could say it's a linear diophantine equation, and you specifically want the extended algorithm
https://en.wikipedia.org/wiki/Extended_Euclidean_algorithm

>>
Anonymous 2016-09-16 20:37:20 No.8350416
>>8350389

I am taking engineering course, planning to make a minor in math. Why? They are fun, how fun? I automatically grin when I understand a proof/ see cool observations in math.

>>
Anonymous 2016-09-17 00:55:53 No.8350828
What would the conjugate base be for this? Last question and it has me stumped.

>>
Anonymous 2016-09-17 02:07:05 No.8350927
>>8350377
There's no name for that specific equation, although the more general case is a linear diophantine equation.

Also, see Bezout's identity: a*x+b*y=d where d is the greatest common divisor of a and b.

Clearly, if a and b are both multiples of d, so are a*x and b*y, and therefore so is a*x+b*y.

So a*x+b*y=1 only has a solution if a and b have no common divisors (other than one), i.e. they are coprime.

>>
Anonymous 2016-09-17 02:30:21 No.8350970
>>8350828
strong base converts ketones to an enolate

>>
Anonymous 2016-09-17 07:01:08 No.8351366
>>8350389
graphics programming

you need math to get the coolest visual effects without taking a dump on performance

>>
Anonymous 2016-09-17 10:06:48 No.8351538
>>8344746
What really makes one (male/female) beautiful?

>>
Anonymous 2016-09-17 10:11:32 No.8351541
bump

>>
Anonymous 2016-09-17 10:15:30 No.8351548
>>8351538
beauty is in the eye of the beholder

and for example men tend to like relatively wide hips in women because those women tend to be more successful in producing fertile offspring, it's evolution

>>
Anonymous 2016-09-17 10:22:22 No.8351557
>>8348546
>Möbius transformation
nice

>>
Anonymous 2016-09-17 14:44:39 No.8351874
I have 2 stupid, basic physics questions which are surprisingly hard to find an answer to.

1) I have a table on the ground, with a cup on top, and the gorund vibrates. The vibrations must travel up the table legs to shake the coffee cup on top of the table. What cartesian axis are the vibrations travelling in? Is it z (up the leg) only? Is it planar vibration in x and y? Is it the same as the gorund vibration? If I stomp on the ground, what direction are the vibrations travelling in?

2) If I have a coil of wires (solenoid) with an iron core, and I pass a current through, what is the time it takes to establish a magnetic field in my core (from 0 current)? Is it the same speed as the current travelling through my wires, i.e. close to light speed? Does reluctance not inhibit this speed in some way?

>>
Anonymous 2016-09-17 14:59:14 No.8351893
>>8351874
The inductance will limit the rate of change of current, but it won't cause the current or field to remain at zero for some time interval.

>>
Anonymous 2016-09-17 15:02:39 No.8351897
>>8351538
Literally money.

>>
Anonymous 2016-09-17 15:23:41 No.8351921
>>8351874
1.Vibrations(Phonons) travel in all possible directions.

2.Depends on the distance of the iron from the coil(how long light travels from the coil to the iron) and the orientations of the magnetic domains

>>
Anonymous 2016-09-17 16:47:57 No.8352026
Best beginner textbook on discreet math?

>>
Anonymous 2016-09-17 17:48:53 No.8352135
Is there a way to get the modulus of a complex number in polar form and shove it up into the exponent of the Euler formula of that polar form?

>>
Anonymous 2016-09-17 18:01:28 No.8352150
>>8352135
The motivation is that I want to simplify calculating the power of a complex number somehow, because...

$(re^{i\theta})^n\\\text{Fuck you, buddy}\rightarrow r$

>>
Anonymous 2016-09-17 18:17:10 No.8352171
>>8352135
>>8352150
Nevermind, as expected I'm retarded.
$(re^{i\theta})^n = r^ne^{i\theta n}$

>>
Anonymous 2016-09-17 19:09:57 No.8352262
Is the gas constant R measured as energy?

>>
Anonymous 2016-09-17 19:35:49 No.8352317
If combustion is an exothermic process then why do you need heat to trigger it?

>>
Anonymous 2016-09-17 19:35:55 No.8352319
>>8352262
energy per temperature increment per mole

>>
Anonymous 2016-09-17 20:23:34 No.8352387
I know it takes a lot of concrete to shield against gamma rays, but perhaps it would be better to route it somehow if it's possible?

Since this seems fairly obvious, why can't you direct/route radiation?

>>
Anonymous 2016-09-17 20:24:46 No.8352390
Brainlet here.
What's a good resource for learning math notation?

>>
Anonymous 2016-09-17 20:27:51 No.8352393
>>8352390
Don't know about resources for math notation, but the best way to learn it is to do math, so git gud.

>>
Anonymous 2016-09-17 20:35:39 No.8352407
>>8352390
I don't know of any specific resource. Pretty much everybody picks up the notation as they're learning it. The reason being is that it's not fully standardized, and it can change drastically from subject to subject or even from professor to professor or textbook to textbook. More important is to become familiar with the concepts, and learn what notation each textbook or professor uses for those concepts by reading textbooks, watching lectures, reading Wikipedia articles, etc.

>>
Anonymous 2016-09-17 20:42:24 No.8352424

>>
Anonymous 2016-09-17 20:56:52 No.8352459
>>8352424
i was having trouble with same problem lmao

>>
Anonymous 2016-09-17 20:57:14 No.8352461
Hey guys, I was hoping for help with a probability / economics question.

From 'Probability and Statistical Inference', (Hogg et al.) pg 104 Q 3.2-19:

A bakery sells rolls in units of a dozen. The demand X (in 1000 units) for rolls has a gamma distribution with paramaters a = 3, B = 0.5, where B is in units of days per 1000 units of rolls. It costs $2 to make a unit that sells for$5 on the first day when the rolls are fresh. Any leftover units are sold on the second day for \$1. How many units should be made to maximize the expected value of the profit?

Solution in the back of the book is 1.96 units. I have not had any success with this question :(

>>
Anonymous 2016-09-17 21:42:01 No.8352550
>>8351897
Have to disagree. There are lots of rich people who are anything but beautiful.

>>
Anonymous 2016-09-17 21:42:56 No.8352554
>>8352424
Here is proof of my attempt in solving the problem, so you guys don't think I'm just looking for someone to do my homework and I just do nothing.

>>
Anonymous 2016-09-17 21:50:02 No.8352577
>>8352424

>>
Anonymous 2016-09-17 21:52:02 No.8352582
>>8352577
Like the axis it is around? It's a washer is it not?

>>
Anonymous 2016-09-17 21:53:07 No.8352587
>>8352577
It's a solid of revolution. Take the area bounded by the curves and revolve it around the line x = 3. He's looking for the volume.

>>
Anonymous 2016-09-17 21:54:06 No.8352591
>>8352582
But why would you need an axis when you just have to calculate the area between two curves?

>>
Anonymous 2016-09-17 21:55:08 No.8352594
>>8352582
>>8352587
Ah, thank you, I see

>>
Anonymous 2016-09-17 22:03:37 No.8352617
>>8352587
>>8352582
>>8352577
Do you guys have any idea where I'm going wrong?

>>
Anonymous 2016-09-17 22:05:18 No.8352622
>>8352554
Your figure revolves around axis (x) that's not the same as the argument (y).

So the formula should be https://en.wikipedia.org/wiki/Solid_of_revolution#Cylinder_method instead

>>
Anonymous 2016-09-17 22:14:51 No.8352655
>>8352554

In your first line, you should be subtracting 3 from your 'radius functions', rather than how you have it.

I.e. 3 - y^2 should be y^2 - 3, and 3 - (1 -y^2) should be -y^2 - 2

>>
Anonymous 2016-09-17 23:23:10 No.8352814
>>8352655
>>8352622
ANONS

I'm such a brainlet :(

Where am I messing up now?

>>
Anonymous 2016-09-17 23:31:37 No.8352838
Mineralogy question.

How do I find the edge lengths of a unit cell? Is there a formula or am I supposed to just google it/get it from a chart somewhere?
Same thing for the angles

I've got six to do but I can't find any information on how to find those two things.
For example, how would I find KCl's unit cell edge length?
I know the formula, and that it's Isometric and the Radius of K+ in this is 1.51Angstroms

>>
Anonymous 2016-09-17 23:50:32 No.8352895
Group Theory question, what do they mean by G/N where G is a group and N is a normal subgroup of G?

>>
Anonymous 2016-09-17 23:53:31 No.8352907
>>8352895
It's something called the quotient group. It can be realized as the set of cosets of N, {gN : g in G}, and It has the property that any homomorphism out of G that sends N to the identity factors through the quotient map G -> G/N.

>>
Anonymous 2016-09-17 23:56:36 No.8352919
>>8352907
Ah, I see, thanks a bunch

>>
Anonymous 2016-09-18 00:04:03 No.8352938
Can I ask programming questions here? Its beyond basic. I just cant find what I'm looking for in google.

In Java can I not initialize variables like
int a,b = 0; ?
You can do it in C but java is saying a is not initialized, and b is.

>>
Anonymous 2016-09-18 00:20:53 No.8352981
>>8352938
You are mistaken about what that line does in C.

In C (and in Java and C++ for that matter) that line reads "create the variable 'a', then create the variable 'b' and initialize it to 0." In C, if an int variable is not explicitly given a value, it is initialized to 0 by default (unless I'm mistaken). To see what I mean, try the line
int a, b = 1;
in C and look at the values of a and b. The statement is just a list of variable declarations. You could easily extend it:
int a, b = 1, c = 3, d, e, f = 5;

Java doesn't initialize int variables to 0 by default, so you will just get an error if you try to use 'a' without giving it a value.

>>
Anonymous 2016-09-18 00:28:29 No.8352996
>>8352981
Ahhh that makes sense, I did find out
int a=0, b = 0;
worked properly, as a result I figured that part out.
But In C our teacher taught us that
int a,b,c,... = whatever;
so 0 would set them all to 0. At least that's what I remember. And technically it did I guess but it essentially assigns the last one separately or redundantly(?) or whatever you call it since
int a,b,c,...;
would set them all to 0 as well. Right?
Thanks for clarifying anon.

>>
Anonymous 2016-09-18 00:28:34 No.8352997
>>8352814
Again in the first line, you forgot to square your inner and out radii.

It should read (y^2 - 3)^2 - ( -y^2 - 2)^2

>>
Anonymous 2016-09-18 00:40:28 No.8353028
>>8352424
>>8352814
>>8352554

You were on the right track in >>8352554 , but you screwed up a minus sign in expanding the top line.

pic related is a solution

>>
Anonymous 2016-09-18 00:48:19 No.8353064
>>8352996
>Right?
Yup. Though if you really do want all of those variable initialized to 0, it's good practice to state that explicitly:
int a = 0, b = 0, c = 0, ...;
Because 1) it's easier for other people to see your intention and 2) god know's if that other code will compile correctly with all compilers on all systems.

>>
Anonymous 2016-09-18 00:54:24 No.8353076
>>8352981
In C, if an int variable is not explicitly given a value, it is initialized to 0 by default (unless I'm mistaken).
That's only true for variables with static storage duration (global variables, and local variables with the "static" qualifier).

Variables with automatic storage duration (i.e. local variables without the "static" qualifier) are left uninitialised (i.e. their initial value is indeterminate) if not explicitly initialised.

>>
Anonymous 2016-09-18 01:26:57 No.8353133
>>8352387

(1/2)

Short answer: It's tough, but doable. However, it's much simpler to shield

Long answer: It's a topic of some interest for particle physicists. Assuming you're talking about reflecting the radiation, it is certainly possible. However, the refractive index of a material is (usually) a function of the incident photon energy. In turn, the refractive index tells us something about how much radiation will be reflected (c.f. reflectivity, total internal reflection).

For gamma rays, the incident photon energy is incredibly high, so the corresponding critical angle for total internal reflection is tiny i.e. you can only reflect it a bit. Here's an image to give you an idea: http://imagine.gsfc.nasa.gov/Images/science/xray_telescope_1mirror_full.jpg. There are X-ray mirrors/telescopes that do it for X-rays, so it is sort of possible (http://universe.gsfc.nasa.gov/xrays/MirrorLab/xoptics.html).

>>
Anonymous 2016-09-18 01:28:01 No.8353136
>>8352387

(2/2)

The problem is that this isn't what you want. It would be nice to surround a radiation hazard with nice 45 degree mirrors that reflect it out into outer space because the 'mirror' will just absorb the radiation since the angle is too large. You'd have to an incredibly low angle, maybe a succession of mirrors (which incidentally would have to be very finely constructed). With a small angular deflection ~ 1 degree, then you'll end up needing to leave a lot of space to ensure that you're not just beaming gamma rays at the nearest skyscraper (much more than the few metres of shielding*). Then you have to maintain the mirrors. And your reflection probably won't be 100% effective anyway. so you still probably need shielding. And you'd be collimating a beam of gamma rays into space which could be a nightmare for passing satellites. In short, lumps of concrete are a much simpler solution for containment/shielding.

That doesn't preclude people trying, because collimated gamma rays are of interest to many types of experimental physicist. It's just not a viable containment method.

*Say you have a deviation of 1 degree = 0.017 radians. For a 100m tall building, you'd need to leave 5700m to allow the beam to clear it. (separation required = building height / tan(angle)).That is, you can't have any buildings over 100m within 6km of it. Now, most people don't want to live near radiation hazards, but if we're talking labs shielding themselves or nuclear power stations etc. then it becomes a problem. Plus, the source isn't contained, so if anyone in a lab wanted to work at it, they couldn't get closer than about 100m without being exposed for the same reason as the building.. You could use lots of consecutive mirrors I suppose, but it would be much more tricky and expensive than just using concrete.

>>
Anonymous 2016-09-18 02:11:06 No.8353239
>>8352550
Then why do they have any pussy they want?

>>
Anonymous 2016-09-18 02:13:04 No.8353242
>>8353028
Thanks Anon!

>>
Anonymous 2016-09-18 02:45:24 No.8353309
>>8351538

Symmetry.

>>8352026

A Transition to Advanced Mathematics. Read that first, will teach basic logic, set theory and proof-writing. After that, pick a book out of here:

http://hbpms.blogspot.com/2008/05/stage-1-introductory-discrete.html

>>
Anonymous 2016-09-18 02:46:57 No.8353312
>>8352390

Do math and learn set theory

>>
Anonymous 2016-09-18 02:54:09 No.8353331
To answer OP's question it seems the author just conceived of the correct expression for the purposes of his proof himself. I don't like ad hoc mathematics like that much myself, either. Study algebraic geometry, learn the resl secrets of the universe, be a pythagorean m8

>>
Anonymous 2016-09-18 03:20:26 No.8353376
>>8353076
This is the right answer. After

int a,b=0;

the value of b is zero, but the value of a is undefined. After

static int c;

the value of c is zero.

>>
Anonymous 2016-09-18 03:49:34 No.8353434
I want to study physics at university. I have the grades to get into a very prestigious university. Can I expect good job prospects?

>>
Anonymous 2016-09-18 03:51:11 No.8353436
When an object reaches escape velocity, what happens to it's GPE?

>>
Anonymous 2016-09-18 03:55:25 No.8353448
>>8352026

>>
Anonymous 2016-09-18 04:20:06 No.8353481
>>8346571
>sacrificed your humanity for mathematics, there are still people who understand it better than you
*happy to creepy*hahahahAHAHAHAHAHAHAHAHAUAHAHAHAHA
>me, senior yr, pure math mjr.

>>
Anonymous 2016-09-18 04:22:53 No.8353485
>>8346571
>autism except worse
autism you're aware of and can't escape

>>
Anonymous 2016-09-18 04:26:07 No.8353491
>>8352171
good job
>fuck up that 80% of the world couldn't solve
seriously, gj m8

>>
Anonymous 2016-09-18 04:50:13 No.8353525
Rudin's first few pages are complete bullshit. I remember reading this and feeling dumb as a freshman. Just read a standard modern proof of the irrationality of sqrt2

>>
Anonymous 2016-09-18 04:50:29 No.8353526
>>8347434
What shit doesnt make sense anon? Im going through the MIT courseware calc 3 and he explains the stuff thats going on pretty well. I havent had a problem visualizing it yet. but im only on the 16th lecture. So I figure itll get harder.

>>
Anonymous 2016-09-18 04:57:25 No.8353533
>>8347434
hahahah nigga you're a brainlet haha like nigga it's just calc 1 with more independent variables hahaha like nigga it ain't that hard hahahahaha

>>
Anonymous 2016-09-18 07:30:43 No.8353729
i understand how d/dx == the slope at point f(x) for
what I don't understand is how one can do stuff like

y = f(x)
dy/dx = f'(x)
dy = f'(x)dx

how come you can do that??
what does the last statement exactly mean?

>>
Anonymous 2016-09-18 07:40:00 No.8353739
>>8353729
The last statement is wrong, but close to an abuse of notation
should be
$\int \, \mathrm{d} y = \int f'(x) \, \mathrm{d} x$
It says that the integral is the inverse of the derivative. That there is a function whose derivative is f'(x).
This is the fundamental theorem of calculus, and only started making sense to me after learning analysis

[  3  /  a  /  adv  /  an  /  c  /  cgl  /  ck  /  cm  /  co  /  diy  /  fa  /  fit  /  g  /  i  /  ic  /  jp  /  k  /  lit  /  m  /  mlp  /  mu  /  n  /  o  /  p  /  po  /  q  /  sci  /  sp  /  tg  /  toy  /  trv  /  tv  /  v  /  vg  /  vp  /  w  /  wg  /  wsg  /  x  ]

Contact me | All the content on this website come from 4chan.org. All trademarks and copyrights on this page are owned by their respective parties. Images uploaded are the responsibility of the Poster. Comments are owned by the Poster.

Dofus quêtes