Mathematics

12:37 AM

@TedShifrin standard metric, in $\mathbb{R}$.

4:03 AM

@TedShifrin All right! Means my main doubt arises from using Fundamental Theorem of Calculus when it should not be used blindly, okay!

Semiclassical

4:43 AM

@adeshmishra since you reference Feynman: the most obvious instance of a line integral in physics is the work to move a particle along a path through a force field

so $W=\int_a^b \vec{F}\cdot d\vec{r}$

In the particular case of $\vec{F}$ being a conservative force, one can find a (scalar) potential energy such that $\vec{F}=-\nabla U$

and therefore the work done is just $W=-\Delta U$

combine that with the work-energy theorem $W=\Delta K$, and you get energy conservation: $\Delta E = \Delta K+\Delta U = 0$

(also, in that case the work to move a particle along a closed path is zero: if the force is conservative, then when the particle returns to its initial state then the initial potential energy is the same as the final, i.e., no change in potential energy)

Alex D

5:15 AM

Hello all. I have a quick yes or no question I was hoping someone could answer for me

Keyenke

@Alex lets hear it

Alex D

I was asked the following: Let W be the set of all real valued polynomials of any degree and U be the set of all exponential functions. I'm asked whether U+W is a direct sum. I said yes because their intersection is empty. Is this correct reasoning?

It was an exam question but I thought it was really simple, so I'm having second thoughts maybe there's more to it

It's from linear algebra

Keyenke

@Alex im way out of my depth here ,I'm a high school student :p

Krull Dimension

5:37 AM

Alex aren't constants both exponential $ce^{0x}$ and polynomial $cx^0$

Leaky Nun

U is not a vector space

Krull Dimension

Is there a term for, like, linear fractional transformations except quadratic. Like z->2z/(1+z^2)

Ted Shifrin

The question is: What is the set of all exponential functions?

In order to have a vector space, you need all constant multiples, for starters, and that includes the zero function. You also need the zero function in $W$ (and it's not a polynomial of any (finite) degree). So the question is very sloppy. Did your exam question state it more carefully?

Jack Ohara

Hi chat

Im trying to solve this question

how many roots does x^y = 1 (mod p)

we know that there are p-1 elements in F_p so that is an upperbound

also we have that order of x has to divide y

the answer is supposed to be gcd (p-1, y-1)

But I do not see how

the original question is this x^y -x = 0 modulo n

but by the CRT I can split the solutions mod primes

Alex D

5:57 AM

Well yes, sorry. The question is: Consider the vector space $V$ the set of all real-valued continuous functions, and subspaces $U$, the set of all exponential functions, their sums and scalar multiples, and $W$, the set of all real-valued polynomials of any degree. Prove whether $U+W$ is a direct sum.

Krull Dimension

6:08 AM

Jack the multiplicative group mod p has p-1 elements, and any solution to that equation generates a subgroup of order y-1

Jack Ohara

@KrullDimension can you explain further?

the way i thought about it is that , those elements that have order y-1 or order dividing y-1 would be a solution

we have p-1 elements in total

I do not understand the gcd part

Krull Dimension

By Lagrange's theorem, the order of any subgroup must divide the order of the group, so if y-1 does not have a common divisor with p-1, we only get one solution

If there is a common divisor, then there must be as many solutions as that common divisor, because the multiplicative group mod p is cyclic

Jack Ohara

I see

@KrullDimension can you tell me how any solution generates a subgroup?

Krull Dimension

So for example, if you have p=13, y=21, you could just take all the powers of 2 (mod 13), pick every third one, and get four solutions to your equation

Jack Ohara

it can be the case that x^(y-1) = 1

but order of x is just a divisor of y-1

why would that be a subgroup of order y-1 ?

Krull Dimension

6:20 AM

Yeah, that's true

I misspoke, you generate a subgroup of order dividing y-1

Jack Ohara

Still not seeing the full picture of the proof

Krull Dimension

6:50 AM

The group of solutions to x^(y-1)=1 (mod p) is isomorphic to the group of solutions to x*(y-1)=0 (mod p-1). These solutions are all multiples of (p-1)/gcd(p-1,y-1), of which there are precisely gcd(p-1,y-1) less than (p-1)

@JackOhara Sorry, took me a moment to collect my thoughts

Jack Ohara

@KrullDimension thanks and no worries

let me see if that makes sense to me

Jack Ohara

7:15 AM

@KrullDimension works thanks!

Krull Dimension

Glad I could help!

Jack Ohara

:)

northerner

9:46 AM

Does the calculator that comes with Microsoft Windows have a bug? It evaluates 2^3^4 to 4,096 but WolframAlpha evaluates it to something much greater wolframalpha.com/input/?i=2%5E3%5E4

Isn't it normally accepted that $^$ is right associative?

Peter

In fact, a^b^c is usually considered to be a^(b^c)

northerner

@Peter so put simply MS calculator is wrong

adesh mishra

10:08 AM

@Semiclassical Thank you

Royi

12:02 PM

Hello,
I encountered a nice thing about matrices but I can't prove it.
For block matrices of the form [I, A; A' I] (I use MATLAB's notation) there is some special property to the $ V $ matrix. So if C = [I, A; A', I] then the $ V $ matrix of the SVD of $ C $ has the property that $ V(1:n / 2, :)' V(1:n / 2, :) = 0.5 I $.
Anyone have seen this?

Astyx

12:13 PM

What's Matlab notation and what is the V matrix ?

Royi

OK. $ C = U S {V}^{T} $.

So $ {V}_{1: \frac{n}{2}, :}^{T} {V}_{1: \frac{n}{2}, :}^{T} $ has $ 0.5 $ in its diagonal.

Astyx

What is U ? and S ?

Royi

I use 1:r, 1:q to mean a sub set of the 1 to r rows and 1 to q columns.

$ U $ ans $ S $ are from the SVD decomposition of C.

Astyx

So U and V are unitary and S is diagonal ?

Royi

It is an interesting property I haven't seen anywhere.

Yep. We are on real matrices so $ U $ and $ V $ are orthonormal.

Astyx

12:23 PM

Is I the identity matrix ?

Royi

Now, let's say $ \boldsymbol{v}_{i} $ is the vector composed by the $ i $ column of $ V $. What I see is that the norm of its first $ n $ elements is $ 0.5 $ (Where the vector length is $ 2 n $). Always.

Actually the squared norm.

It comes form the special structure of $ C $ but I can't prove why.

Astyx

C is symmetrical, which means it U = V

Royi

No necessarily.

Astyx

What do you mean ?

Royi

I think $ U $ and $ V $ are equal for matrices which are not only symmetric but are result of some quadratic form.

Have a look at - pastebin.com/FbcRfg23.

Astyx

12:43 PM

Spectral theorem

In mathematics, particularly linear algebra and functional analysis, a spectral theorem is a result about when a linear operator or matrix can be diagonalized (that is, represented as a diagonal matrix in some basis). This is extremely useful because computations involving a diagonalizable matrix can often be reduced to much simpler computations involving the corresponding diagonal matrix. The concept of diagonalization is relatively straightforward for operators on finite-dimensional vector spaces but requires some modification for operators on infinite-dimensional spaces. In general, the spectral...

Royi

I know the spectral theorem. do you want to pin point me to what you meant?

Astyx

The spectral theorem tells you that if your matrix is symmetrical it decomposes as $USU^T$ with S diagonal and U unitary

ie U=V in your decomposition

Thorgott

That's not the SVD

Hamzalihi

henlo people can i ask a question here to recieve a tip to solve a problem

Royi

The SVD has a lot to do with the Eigen Decomposition. They collide for the case the matrix is in the form $ {A}^{T} A $ or $ A {A}^{T} $.

Astyx

12:58 PM

Oh right

I misread what the SVD was my bad

Thorgott

in an Eigendecomposition $USU^T$, the diagonal of $S$ are the eigenvalues whereas in an SVD $USV^T$, the diagonal of $S$ are the singular values

the singular values are the square roots of the Eigenvalues of $A^TA$ and the SVD is obtained from the Eigendecomposition of $A^TA$ (which is always symmetric), explaining the relationship

Royi

@Thorgott, do you have any idea about what I saw?

Thorgott

I don't really follow your notation

Royi

1:15 PM

Was the structure of $ C $ clear?

I will try again. $ C \in \mathbb{R}^{2 n \times 2 n} $ where $ C = \begin{bmatrix}
I & A \\
{A}^{T} & I
\end{bmatrix} $

The matrix $ I $ is the identity matrix (Size $ n \times n $) and $ A $ is any $ n \times n $ matrix.

The SVD of $ C $ is given by $ C = U S V $.
So far so good?

Thorgott

yeah, I just don't follow the ${V}_{1: \frac{n}{2}, :}^{T} {V}_{1: \frac{n}{2}, :}^{T}$, it looks like these don't even have compatible dimensions to me

Royi

Now, let's have $ \boldsymbol{v}_{i} $ as the $ i $ -th column of V.

Clearly $ {v}_{i} \in \mathbb{R}^{2n} $.

What's amazing in that case is: $ \sum_{j = 1}^{n} {\left( \boldsymbol{v}_{i}[j] \right)}^{2} = 0.5 $.

Huy

Theorem: Let x0, y0 and t0 be real and F(x,y,t) and G(x,y,t) be continuous with continuous partial derivatives. Then, there are unique functions x(t), y(t) defined on an interval around t0 satisfying:

x'(t) = F(x,y,t)
y'(t) = G(x,y,t)
x(t0) = x0
y(t0) = y0

Does this theorem have a name? Is there a name for the general, n-dimensional version of this (this would be the 2-dimensional version)? I can only recall Picard-Lindelöf from analysis, but I don't need it for Lipschitz-continuous, just continuously differentiable is sufficient.

Royi

Where $ \boldsymbol{v}_{i}[j] $ is the $ j $ -th element of the vector.

Needless to say $ \sum_{j = n + 1}^{2 n} {\left( \boldsymbol{v}_{i}[j] \right)}^{2} = 0.5 $ as well.

Is it clear now? The indices?

Thorgott

1:56 PM

But what if $A=0$? Then you can take $C=U=S=V=1$ and then the property isn't satisfied.

@Huy continuously differentiable implies locally Lipschitz

Huy

I never said anything else

Royi

You're right.

Let's say it is not the case.

As in my actual case $ C $ has no zero entries off the block diagonal.

Thorgott

@Huy well, it's a corollary of Picard-Lindelöf then and that has generalizations to $n$ dimensions

@Royi a column of $V$ will be an eigenvector of $C^TC$ to some Eigenvalue $\lambda$, if you write the column as $\begin{pmatrix}v\\w\end{pmatrix}$, then $\lVert v\rVert^2+\lVert w\rVert^2=1$ by normality and then your observation is equivalent to $\lVert v\rVert^2=\lVert w\rVert^2$.
What I am able to deduce from the properties is that we always have $\lambda\lVert v\rVert^2=\lVert v\rVert^2+\lVert A^Tv\rVert^2+2\langle Aw,c\rangle$ and $\lambda\lVert w\rVert^2=\lVert w\rVert^2+\lVert A^Tw\rVert^2+2\langle Aw,v\rangle$. Not sure how to proceed.

Royi

I'm not sure how you got the property you mention.

Thorgott

2:13 PM

I split the Eigenvalue equation into parts for $v,w$, then paired them up with $v,w$ and rearranged

Royi

2:23 PM

Could you please elaborate?