Dumb questions thread

you can still have unit vectors with rectangular coordinates

vectors can be described in any coordinate system you like.
the coordinates don't even have to be linearly independent.. it's just most useful if they are.

Using Bernoulli I get x = (1/2)^[1/(n+1)]

which seems retarded but it does solve it.. help

Assuming that's correct, you found a constant solution.
Follow the hint and consider what the chain rule tells you about the left hand side after the suggested multiplication. (Compare to what you'll have on the right hand side to pick a new variable)

But wouldnt "a" have unit while "b" would be unitless? If i make a dot product of that vector with itself, wouldnt it result in a sum of two things with different units? thanks in advance

how much training in logical training would I need to understand Gödel's Incompleteness Theorem?

I know BASIC set theory, we had some in calculus
but the symbols in the proofs look like hieroglyphics to me
any free resource?

You gonna call the police?

What you wrote in that picture is a tangent vector. Which means it is actually takes the form of a map [math] \nu :C_{M,p}^\infty \to \mathbb{R} [/math].

can someone describe Gauss's Law to me in layman's terms, and an example of its use?

>But wouldnt "a" have unit while "b" would be unitless?
a is units, b is units/radian

Not entirely sure this is what you're looking for but read up on Dirac Bra-Ket Notation.

en.wikipedia.org/wiki/Bra–ket_notation

I fucking love bra-ket, don't get me wrong, but,
>elementary question about vectors
>here, go read about matrix mechanics
????

I'm lost.
That's a [math] v [/math] not a [math] \nu [/math]
And you end up with something [math] \to \mathbb{R}^n [/math] (prob n=3)?
Does [math] C^\infty [/math] the family of smooth functions?

Start by reading "How to Prove It"

The way an electric field moves away from a given charge density, is proportional to that charge density.

It's not a map to R^3 it's a map to R and yes it is the space of smooth functions.

>And you end up with something...

No you don't understand. An element of [math]{\mathbb{R}^n}[/math] should be interpreted as a map of this type because [math] {T_p}{\mathbb{R}^n} \cong {\mathbb{R}^n}[/math].

> the family of smooth functions?

[math] C_{M,p}^\infty = \coprod\limits_{U \ni p} {C_M^\infty \left( U \right)} / \sim [/math]. Where [math] {C_M^\infty \left( U \right)} [/math] is the set of smooth functions [math]f:U \to \mathbb{R}[/math] and [math] \sim [/math] is two functions being equal when restricted to some neighborhood of the point p.

Spherical and cylindrical coordinates are very basic and common.

Yeah, you can have a vector as written above. The most intuitive way to think about a vector is in ractangular coordinates where each element encodes a magnitude in a dimension (thereby giving rise to slopes in each direction). You can also use vectors to encode polar coordinates, spherical coordinates, any of the previous examples including an element in the time dimension, etc. Basically, vectors are just matrices with either height or width of 1 and are used to concisely hold information.

For an interesting look at the beginning of this look at the Kahn Academy lessons in linear algebra (specifically when he starts going over changes in basis, where your vectors can be composed of a linear combination of vectors (where the standard basis is [1, 0, 0, ... , 0], [0, 1, 0, ... , 0], [0, 0, 1, ... , 0], ... , [0, 0, 0, ... , 1])). This doesn't touch on different coordinate systems, but it begins to lay the groundwork.

Note also that this is categorical direct limit.
i.e. [math]C_{M,p}^\infty = \mathop {\underrightarrow {\lim }}\limits_{U \ni p} C_M^\infty \left( U \right)[/math]

Here OP, improve the mind.
en.wikipedia.org/wiki/Curvilinear_coordinates

Why do you need all this to explain to them that a vector can be written in terms of any basis of the vector space? Isn't this all, a bit of a show-off?

>Isn't this all, a bit of a show-off?
Yes. You get used to ignoring the pompous math majors responding to pre-college questions.

Not at all. OP's question is surprisingly deep. I think it's impossible to really understand what's going on without a solid understanding of sheaves of germs of smooth functions. Given a smooth manifold (the smoothness not being essential; we could just as well consider the C^k case), we may consider a local splitting of the tangent bundle into a direct sum of line bundles. Choosing a non-vanishing section of each constituent line bundle, we may express each section of the tangent bundle as a unique linear combination of these sections of the constituent bundles. Specializing to the manifold R^2-0 yields the situation in the OP, modulo some trivial and self-explanatory notational differences.

Coordinate systems and vectors are studied extensively in Vector Calculus and Mulivariate calculus. This isn't really a dumb question at all.

If OP is unaware of anything but rectangular coordinates, OP has no fucking clue what any of those words mean, you fucking oblivious tool.

OP here, I'm still in doubt, if I were to write a position vector in polar coordinates, would the first component be the radius and the second an angle? How can them have different units?

wow please help poor OP here, thats my only question, I know about the existence of curvilinear coordinates, I've done vecotr calculus still I have a huge gap..

I just want to know how to write a position vector in polar coordinates for example... would a componente be the radius and the other an angle?? or radius times angle?? idk

cant be angle, units wont match, pls 4chen

Please OP asked the wrong question, just point me the right one or a new one if both are wrong

I answered you here:

so if i have a particle with the following position: (r,theta)= (1, pi) my vector would be
[math]\vec{r}=1m \hat{r}+\pi rad \frac{m}{rad} \hat{\theta}[/math]

is that what you're saying? if so why? is it defined this way?

The technical answer is that we treat angles as unitless because problems arise from, for example, attempting to carry radian "units" through trigonometric functions. However, in your simpler case, I find it fairly logical to write it similar to how you did and find that radians cancel leaving a sum of meters, and hope that maybe that helps your intuition a bit.

So if i calculate the norm of it, I'll end up with
[math]\left| \vec{r} \right|=\sqrt {r^2 +\theta^2}[/math]
which is intuitively wrong? i dont even know if this word exists, but shouldnt the norm be the radius?

kek'd audibly

The norm of that 1m.

but how do i get this result with some formalism?

Since you are using polar coordinates the norm is defined as the modulus of the term in [math]r[/math]

So is that thing about taking the square root of the dot product of the vector with itself is only valid in rectangular coordinates? How does dot product work in polar coordinates? Any source on that? thanks in advance

It's 6 a.m. here I'm drunk and I can barely read, I only dropped by to tell you the norm of a an euclidean vector in polar form is by definition what I said in my previous post.

ittc.ku.edu/~jstiles/220/handouts/The Position Vector.pdf

thank you so much thats all i needed, love you user /thread

polar coordinates are fine, theyre just not vectors.