Post your questions that don't deserve their own thread in here.
Tips for good questions:
>provide context
>check stackexchange first
>if stuck half-way into a question, show your work so far
Previous thread:
Which vitamin/nutrient deficiencies can be self-diagnosed relatively easily/cheaply?
fat jon snow
I need solid problems on electromagnetism, electric fields and waves.
pls help.
will post hot mongolian figure skater as a rewars
...
Does anyone know how I would solve these questions, I can't figure out if i'm supposed to set it up using an integral or how I would find the total amount of water from the tank.
How do I go about integrating [eqn] -\frac { 1 } { 4 } \int d^4 x \left ( F_ { \mu \nu } \right ) ^2 [/eqn] P&S says that it's [eqn] \frac { 1 } { 2 } \int d^4 x A_{ \mu } ( x ) \left ( \partial ^2 g^{ \mu \nu } + k^{ \mu } k^ { \nu } \right ) A_{ \nu } ( x ) [/eqn]Unless I've missed something the only way I can see to do it is to write it out and integrate term by term.
First time reading this.
I know this is going to sound bad but, what regime do you guys follow when doing exercises?
All of them before next chapter? Odd numbered only? How many exactly?
Its not that I wouldn't like to do them but... there's a shitload of them and I would like to finish the book someday so can I get a recommendation?
youtube.com
Somewhat funny.
Jackson is a good start, as user above suggested. You could also just google any applied EM book for inspiration. It's hard to recommend something unless you make your request a tad more precise.
Integral, yes. In a) you just integrate it (ans = -360). In b) you should divide your time interval into the periods where V' is monotonic, and then just add up absolute values (ans = 920). In c) you just find how much water has flowed out until it started to fill (at t = 2 sec, dV = -640, V = 60). For d) it is t=0 // V=700, and e) is just the indefinite integral of V'.
I would just go to momentum space where it is (almost) trivial. Or use a better book, I fucking hate Peskin/Schroeder with a passion. Srednicki, for instance, comes with a solution manual.
Define a binary operation [math]\star[/math] on [math]\mathbb{Q}[/math] by [math]a\star b = (a-2)(b-2)[/math]. Find [math]\star[/math]'s identity, if it exists, and state which elements are invertible.
As far as I can tell, there is no inverse, as the formula for an inverse to [math]a[/math] is [math]\frac{3a-4}{a-2}[/math], and an inverse has to be constant. But this would mean no elements are invertible, which makes the second part of the question pointless. Am I missing something?
I usually do a couple and stop when I feel like I've grasped the subject and move on to the next chapter. The ones I don't do I mark or write down so I can solve them when I'm revising.
>and an inverse has to be constant
Sure. But it is constant, for fixed a.
You wouldn't say addition isn't invertible because -a isn't "constant".
At a minimum, the identity is always invertible.
>an inverse has to be constant.
But the inverse of x is allowed to depend on x. For example the inverse of 5 in (Z,+) is -5, and more generally the inverse of x is -x.
If you didn't allow the inverse of x to depend on x, then x^{-1} would have the same value for any x' in the group, and by the uniqueness of inverse that means...
Finding an invertible element requires having found the identity for [math]\star[/math]. The identity element, if it exists, must satisfy [math]x = (x-2)(x-2)[/math], ie. [math]x^2 - 5x + 4 = 0[/math], ie. x = 1 or 4. But neither 1 nor 4 are identity elements for [math]\star[/math], since [math]x \star 1 = 2-x[/math], which is not x in general and [math] x \star 4 = 2(x-2)[/math], which is also not x in general. Hence, there is no identity and therefore no invertible elements (because the notion does not make sense without an identity)
"Let G be a finite group. Explain why each row and column of the multiplication table is a permutation of the elements in G."
My thinking so far is as follows:
>a permutation is a reordering of the elements in a set S, or more formally a bijection from S to S
>most rows and columns of multiplication tables contain elements which are not in S, e.g. if S = {1, 2, 3, 4, 5} then row 2 would be {2, 4, 6, 8, 10} and clearly 6, 8, 10 are not in S
>therefore the row is not a permutation, as it does not map back to S
>therefore the question makes no sense, I can't explain something that's wrong
>...
I just checked my notes, I'm a fucking idiot please ignore this post. Except the image, which in retrospect was a great choice.
The multiplication of a group maps into the group (it's the definition of a composition law). The example you chose was not a group law
"le redpill xD" me on biomedical engineering.
Earlier today, there was a discussion about race and intelligence, and I made the point that the races must be similar in intelligence because natural selection does not seem to favor reasoning ability.
Is this correct? I dont really care that much, since I'm not going to change my beliefs about this, but I'm curious what Veeky Forums has to say about it.
I just completed all of Khan Academy math, looking for advice of what to learn next.
Yeh, I should have started with a book but I am but a poor factory hand brainlet and didn't know.
Help me out anons. I'm going to college this fall. Should I choose EE or engineering physics?
EE:
Pros:
+ Taught in English (as I'm from a non-English speaking country.)
+ Smaller class size.
+ Better job prospect after graduating in case I can't get into grad school immediately afterwards.
Cons:
+ Higher tuitions than ENPH (160%)
+ I don't know it EE is what I really want to do with my life.
ENPH:
Pros:
+ A breed of physics and engineering. Jack of all trade. Get to study QM and Electromagnetic Field Theory.
=> Does this really make me a more well-rounded candidate for just any science/engineering grad programs?
Cons:
+ Nobody knows that ENPH exists.
=> Hard to find a job afterwards.
+ Taught in my native language.
+ Class size can be as three times as large.
Can anyone explain what happened in the red box, logically?
I see that they both have the same base, but I don't understand how or why it's okay to drop the base entirely.
Since the base is the same, would it matter if you replaced them with other numbers with the same value?
Took a base 3 logarithm of both sides
The exponential function is injective. This means the two (equivalent) things:
1) x != y implies that a^x != a^y
2) a^x = a^y implies x = y
The reason it is invective is because it is strictly increasing. You should have studied that in high school.
Was it for academic prep or just fun? Props either way.
I would keep practicing the fundamentals periodically, since they undergird so much other math. A good book for fundamentals is the classic Princeton book of mathematics.
>The reason (...) is *that*
Fuck, I'm writing like a illiterate person.
On my Calc 2 quiz yesterday I had to expand (is that the right word?) a maclaurin series for the function 2/(1+x^2). I was running out of time and didn't want to take the derivatives so I just used 1/1-x as the summation of x^n and substituted (-x^2) for x. I checked later that I got the right summation but will my professor take points off because I did it a different way?
>
Just for fun, I have an interest in machine learning and a lot of computer science so last year I decided to teach myself some mathematics.
Maybe one day I wont be working in a factory for just above min wage :)
I think this would depend on the professor. It is not a really good idea to use results not proved in class, especially if they making solving the question considerably easier, but he should give you at least some partial credit.
>with other numbers with the same value
I don't know what you mean by this. It sounds like you ask if I can replace 3 with any other number besides 3, which my answer to this is no, only 3 equals 3.
My current level of mathematical understanding is that I'd try to remove one of the 3's from the other side, or one of the exponents.
I don't know if a negative cube root is possible, because when I attempted the equation by myself the first time I went:
3^4x = 3^-3
cuberoot(3^4x)=3
Which obviously took me nowhere.. I think I'm missing a fundamental.
I'm aware it's something to to with log, but I'm trying to learn to solve it by hand without calculator.
This makes some sense. So I should be solving 4x=3, but I don't know why it was okay to remove the 3's entirely. This is what I think the first reply was telling me about.
Like everyone, I probably did learn it in high school but it was more than 8 years ago/ Being a teenager you know, you don't hold every fact in your brain, especially ones you didn't think mattered.
You don't need a calculator to calculate the log of expressions like that
Since magnetic fields are products of moving charges, does this mean the existence of magnetic fields depend on the frame of reference?
Wait, I figured what you mean.
This makes sense - thank you!
>
>I'm aware it's something to to with log, but I'm trying to learn to solve it by hand without calculator.
3^(4x)=3^(-3)
Log[3^(4x)]=Log[3^(-3)]
Property of the logarithms (independent of the base of the log): Log[a^b]=b*Log[a], thus
4x*Log[3]=-3*Log[3]
So you can remove the two logs from both sides of the equation without computing them.
How come the mods allow all those hurr durr what's ur iq threads?
>Since magnetic fields are products of moving charges, does this mean the existence of magnetic fields depend on the frame of reference?
Yes, that's true. You have a set of "Lorentz transformations" for E and B.
[math] \vec{B}'=\gamma (\vec{B} -\frac{1}{c^2} \vec{v}\times \vec{E}) [/math]
where [math]\gamma[/math] is the usual relativistic factor.
The next logical step is to ask "then, can I travel so fast that all magnetic field transform into an electric field and vice-versa?"
The answer is no to both. This is because E^2-B^2 remains constant under this transformation.
>This makes some sense. So I should be solving 4x=3, but I don't know why it was okay to remove the 3's entirely. This is what I think the first reply was telling me about.
Look, when x > y, then 2^x > 2^y or, in other words, when x != y, then 2^x != 2^y. That means the only time when 2^{expression 1} is equal to 2^{expression 2} is when {expression 1} equals {expression 2}. That's why you can drop the base.
Thinking about logs here, although not incorrect, is not necessary, and may create bad habits for you, since there are conditions for when you are allowed to apply things like logs to equations in order to solve them.
> (You)
>>Property of the logarithms (independent of the base of the log): Log[a^b]=b*Log[a], thus
>No, that only works if the logarithm is base a
>loga(ab)=b∗loga(a)=b∗1=b
No, [math]log_c(a^b)=b*log_c[a][/math], the first equality is true for any base. If you want the rest, of course you need c=a, but there is no need for that (as you can cancel the logarithms from both sides of the equation OP asked for).
If I parametrize a sphere into x=a*cosu*sinv, y=a*sinu*sinv and z=a*cosv, but only want the frontal hemisphere, should the intervals for both u and v be π?
Why are a function satisfying the Cauchy-Riemann equations a sufficient condition for complex differentiability? For a function to be complex differentiable, shouldn't the derivative have to be equal in every direction, not just in the real and imaginary directions?
A 160-lb man carries a 25-lb can of paint up a helical staircase that encircles a silo with a radius of 20 ft. Suppose also there is a hole in the can of paint so that 9-lb of paint leaks steadily out of the can during the man's ascent. If the silo is 90 ft. high and the man makes exactly three complete revolutions, how much work is done by the man against gravity in climbing to the top?
Depends what you mean by the front.
Your intervals give you the hemisphere protruding along the positive y-axis (assuming a>0).
Asorbic acid deficiency can be diagnosed by watching their gums bleed.
Those directions give you a basis for the whole tangent space.
Positive in the x-axis
"Recall that when there are N active nodes, the efficiency of slotted ALOHA is Np(1 – p)^N–1. Find the value of p that maximizes this expression."
honestly I'm drawing a blank here. I'm not asking to be told how to do this, but anybody give me a hint? or point me in the right direction for a refresher on how to do this? What would I look for - maximizing expressions? maximizing probabilities?
Then the interval for u is [-pi/2,pi/2]
Without making a thread for it, I want to take this opportunity to point out that science fiction writer Arthur C. Clarke held brainlets in low regard:
"Bowman had never found it possible to focus his interest exclusively on any subject; despite the dark warnings of his instructors, he had insisted on taking his Master's degree in General Astronautics-a course with a vague and woolly syllabus, designed for those whose IQs were in the low 130s and who would never reach the top ranks of their profession." -Arthur C. Clarke, /2001: A Space Odyssey/
Brainlets BTFO
Anyone know how to integrate this?
[math]\frac{1}{2a}\int _{-a}^a\sum _{n=0}^a\cos \left(nx\right)dx[/math]
how relevant is the air for the Archimedes principle? How would affect the objects if this was removed? (not considering a change in the properties of the fluid.)
Protip: you can switch the order of summation and integration
HOW THE FUCK DO YOU SOLVE THIS FFS?!
fuckin a
thanks breh
Why binary? Why do we only use 1's and 0's to store information and not 1's and 2's and 3's, etc.?
It's already solved you mental wet fart
there are ternary computers but it didn't catch on in the same way en.wikipedia.org
much has been built on the foundation of binary, from classical logic to circuits. a detail of history, wish i had a more in depth answer for you
electronics are made of digital switches that can only have one of two states: on and off, 1 and 0
ternary computers have been designed to have 1, 0, -1; 3 states. but they're not being used for whatever reason
Notice that on the right you literally have a number.
It is a retarded form, but that is a number.
Thanks brehs
Find quantity of whole solutions.
HOW THE FUCK DO YOU SOLVE THIS FFS?!
use log you fucking brainlet
STFU I ALREADY DID YOU FUCKING RETARD. I SOLVED IT FASTER THAN IT TOOK YOU TO POST THIS RETARDED """""""""""""""""""""""""ADVICE"""""""""""""""""""""" SUCK MY DICK BRAINLET FAGGOT
FFS HOW THE FUCK DO YOU SOLVE THIS FUCKING BULLSHIT??!
Pls help with abstract algebra hw
Let F be a field with four elements. Show that char F = 2. (Hint: You know that the group structure of F is an abelian group, and you know how to construct all abelian groups of order 4. Show that one leads to a field and the other doesn't)
The hint refers to the Fundamental Theorem of Finite Abelian Groups, which says any group of order 4 is isomorphic to both Z mod 4 and Z mod 2 cross Z mod 2. But neither are fields, so he suggested using the distributive property to redefine multiplication for each such that one IS a field. SOS
Newtons method.
Nope. It's a fucking high school test. Must be a simpler way.
>The hint refers to the Fundamental Theorem of Finite Abelian Groups, which says any group of order 4 is isomorphic to both Z mod 4 and Z mod 2 cross Z mod 2.
that should be Z mod 4 *or* Z mod 2 cross Z mod 2.
really stupid question but i haven't been able to answer it. why do people talk about the convergence or divergence of a series [math]\sum^{\infty}_{n=1}a_n[/math]? isn't it just a limit and a fixed number? suppose [math]\lim a_n=\ell[/math] then it doesn't make sense to say [math]\ell[/math] converges. idk it probably is right but something about it is just bugging me
What is the term for a function broken down into its most inefficient and basic form?
What does this look like for the partition counting function? I tried to figure it out myself, but I ended up with something that I don't know how to properly notate.
You have two choices: [math]G=\mathbb{Z}_4[/math] or [math]H=\mathbb{Z}_2\times\mathbb{Z}_2[/math]. Now, [math]G[/math] can't be a field because [math][2]_4\neq [0]_4[/math], but [math][2]_4\cdot [2]_4=[4]_4=[0]_4[/math], making it a zero-divisor. It follows that the field must be [math]H[/math], and then the unit element is (you may need to prove this) [math]([1]_2, [1]_2)[/math], which proves the claim.
Shouldn't Newton's method be HS?
Well it fucking isn't and also I don't need APPROXIMATE FUCKING VALUES I need the exact stuff ffs.
I have your answer, but I don't like your attitude. So fuck you :^).
>Well it fucking isn't and also I don't need APPROXIMATE FUCKING VALUES I need the exact stuff ffs.
cringe
x=1 is an obvious solution. Then argue that the quadratic equation and the terms with logarithms can't intersect anywhere else.
You sure it can't be Z mod 4? I thought I just constructed a valid Cayley table for multiplication for it via distributive property of rings
* | 0 | 1 | 2 | 3
0 | 0 | 0 | 0 | 0
1 | 0 | 1 | 2 | 3
2 | 0 | 2 |?_1|?_2
3 | 0 | 3 |?_3|?_4
?_1 = 2 * 2 = 2(1 + 1) = 2 + 2 = 0
?_2 = 2 * 3 = (1 + 1)3 = 3 + 3 = 2
?_3 = 3 * 2 = 3(1 + 1) = 3 + 3 = 2
?_4 = 3 * 3 = 3(2 + 1) = 3*2 + 3 = 2 + 3 = 1
Nice try butthurt faggot, but insecure retards like you NEVER have the fucking answer, so you can chill and maybe cool your ass.
Clearly they fucking can since the second root is 3 wolframalpha.com
>Nice try butthurt faggot, but insecure retards like you NEVER have the fucking answer, so you can chill and maybe cool your ass.
cringe
HAHAHAHA what a dumb fucking faggot you could literally be substituted by a fucking program written by a 10 yo.
And I'm not even him.
Maybe what I'm asking is, how do we prove we cannot redefine multiplication in Z mod 4 such that it can be a field? Idk what I'm doing
>HAHAHAHA what a dumb fucking faggot you could literally be substituted by a fucking program written by a 10 yo.
cringe
Z mod 4 has zero divisors, fields don't
kek basement dweller enjoying his basic logarithm knowledge power trip.
Veeky Forums never change
Zmod2 cross Zmod2 also has no multiplicative inverses so it's also not a field though...?
>Zmod2 cross Zmod2 also has no multiplicative inverses
but that's wrong, every non-zero element is invertible
That is a ring, yes, but you should remember one thing: a (commutative) field is an integral domain. In integral domains there are no non-trivial zero-divisors.
And this, this one's a bit harder. But, you are lucky. You only have 4 elements. Assuming you were to redefine it that way, then you would have to assign an element to each pair of elements in your group in a way that it is associative with your group operation. You can go through all the various possibilities because there are not too many elements in the group. It sucks, but you can bruteforce this thing. On the other hand, you could just try to find (=prove) a result saying that the characteristic of a finite field is a prime number, but the characteristic in the other case would be 4.
Why is it so common for people to say that organic chemistry is the hardest class you'll ever take? I'm under the impression that it's some kind of impossible class where everyone fails.
Que? What can you multiply, say, (1,0) by to produce (1,1)? Does this require the methodology I used in ?
Hmm, so is it even necessary to prove this (that there doesn't exist a redefining of multiplication in Z mod 4 such that it is a field)? Or, once we prove that Zmod2 cross Zmod2 is indeed a field, we're done? ty based animu poster
The characteristic of a ring A can be thought of as a positive generator for the kernel of the map from the integers to A which takes 1 to 1. If A is a domain, then this kernel must be prime. Sort through the details and your result falls out from this.
HOW?!!
how what?
Suppose there was a field made from [math]\mathbb{Z}_4[/math]. Then (the group operation is still the addition of those remainder class thingies) [math]m([1]_4)=0[/math] iff this [math]m[/math] is divisible by 4. Therefore, the characteristic can not be 2, so your characteristic (all fields have either a prime or 0) must be 0, but this is impossible if your field is finite. Np buddy
It's easy, can you not see it? Factor the quadratic and the rest is trivial.
> Factor the quadratic
What do you mean?
.... are you serious? how old are you? do you not know how to factor a quadratic?
isn't this website 18+?
Part of it factors to [math] (x-1)(x-4) [/math]
Why the hell does Cantor's diagonalization proof work?
I'm talking about that example with "x will have a different nth digit than the nth element of an infinite sequence built with elements of R"
can't you also say that you can create an x that is +something bigger than any other element in a list made of natural numbers, therefore proving that you can't map a bijection between N and N, therefore leading to a contradiction?
>can't you also say that you can create an x that is +something bigger than any other element in a list made of natural numbers
there's no largest natural number, so no
>I'm talking about that example with "x will have a different nth digit than the nth element of an infinite sequence built with elements of R"
because the claim is you can list the real numbers, but you can just explicitly construct a number different than every number in such a list, so the list couldn't have had every real number
>therefore proving that you can't map a bijection between N and N
also the identity map is a bijection from N to N, so you just seem confused here, try to formalize what you're doing and you'll probably see why it doesn't work