most or at least a majority of the lectures that are part of the standard syllabus get recorded here, but they're just for internal use and not uploaded to YT
In $R^2,$ let $L$ be the line $y = mx,$ where $m\neq 0.$ Find an expression for $T(x, y),$ where $T$ is the projection on $L$ along the line perpendicular to $L.$ (See the definition of projection in the exercises of Section $2.1.$)
This was a problem given in the book, "Linear Algebra" by Steph...
thomas: in your question, you write: "Complying with this definition, I considered the subspace of R^2 consisting of all the points in line L and the subspace consisting of all the points lying on the line perpendicular to L" (emphasis added), which makes sense, as the book only defines the projection on W_1 along W_2 for complementary subspaces W_1, W_2 of a vector space.
you then begin talking about the set of all lines perpendicular to L. this is where you get distracted. there are indeed infinitely many lines perpendicular to L, but only one of them is a subspace of R^2 - the one going through the origin.
the exercise is asking you to consider that line only, and not any of the other ones.
when A has more than one element, the "diagonal" {(a,a): a in A} is a good example of a subset of A x A that is not a cartesian product of subsets of A
Hmm.. I see. Well, if you have some construction that allows you to equip the set of continuous automorphisms of a Lie group with a structure of a Lie group, I see no issue
i recently came across a neat result, if a function $f : U \subset \mathbb{C} \rightarrow \mathbb{C}$ is an open map ($U$ is open), and has partial derivatives a.e. (in both $x$ and $y$ directions), then its also differentiable a.e.
@porridgemathematics There is no people that know everything. I'm ignorant at many things too, so I often try to not formulate an opinion about subjects and things I know nothing about. Most people don't even notice when they're doing this.
@Jakobian yeah, I knew I was ignorant to the minutiae of what professional general topologists do, Id like to think I had some understanding of what it might look like though
I agree that as you move along the constraint $g=c$ either you cross a contour of $f$ or you are tangent to a contour of $f$. However I dont see why $f$ is always at a maximum on the constraint when they are tangent. e.g. this example on desmos (the blue curve is the constraint) desmos.com/3d/80883fd44d
I am looking for trig identities of the form
$\sin(a+b)^{2k} = \sum a_i \sin^{2 n_i}(b_i a) \cos^{2 m_i}(c_i a) \sin^{2 l_i}(d_i b) \cos^{2 j_i}(e_i b)$
where $k,n_i,m_i,l_i,j_i$ are integers $> -1$ and $a_i,b_i,c_i,d_i,e_i$ are rational numbers$>0$.
Im not sure they exist for all $k$.
For insta...
@SineoftheTime I've noticed that you were browsing questions about sequentially closed sets. Maybe I should add that "sequence has too few points" is only one possible reason for why sequences are not enough. If you only consider nets indexed by ordinals, this might still not be enough to describe the convergence in a topological space.
Even though, you could say, such nets have as many points as you would want
@Jakobian I was looking at the proof that $A^{\perp}$ is closed using a sequence in $A^{\perp}$ and proving that its limit point is in $A^{\perp}$. then I've seen that sequentially closed does not imply closed
You can check by hand that a function defined to agree with the derivative where the latter is defined (and to be whatever on the point of nonsmoothness) satisfies the definition of weak derivative
@onepotatotwopotato If $c\in [a, b]$ is the point where it lacks smoothness, then you can consider some small neighbourhood around $c$, outisde of it everything will work fine, inside of it, the integral can be made arbitrarily small
finitely many points doesn't affect the integral, so I can just perform integration by parts if the derivative already exists so the weak and the classic are the same?
basically yes, you can move the $\partial$ operator to $f$ as long as you are integrating over a region where $\partial f$ makes sense and you can apply Stoke's theorem on
As you wrote, you have already proved $\lim_{n \rightarrow \infty} \mu^*(A_n) \leq \mu^*(A)$. It remains to show that
$\mu^*(A) \leq \lim_{n \rightarrow \infty} \mu^*(A_n)$.
Your idea to complete the proof is essentially correct, all it needs is a small adjustment.
Given any $\epsilon > 0$. We...
For any real numbers $a$ and $b$ and $1 \leq p < \infty$, prove that
$$|a+b|^p \leq 2^p \{ |a|^p +|b|^p \}$$
This inequality is given in the the book Real Analysis by Royden, Chapter $7$, page $136$. I don't understand how the author comes to this inequality. Can anyone provide some hints?
$f(z)=1/\sin(z)$, I have to find the Laurent decomposition of $f$ on the discs $\{n\pi<|z|<(n+1)\pi \}$ for $n\ge 0$. By Laurent decomposition I mean $f=h+g$ where $h$ is analytic for $|z|>n\pi$ and $g$ is analytic for $|z|<(n+1)\pi$
@Jakobian Correction, $f(x) = x^p+1-(x+1)^p$ is increasing. For this you simply state that $f$ is continuous and $f'(x) = p(x^{p-1}-(x+1)^{p-1})\geq 0$ for $x > 0$
Prove that if the derivative $f'(x)$ of a function exists on the measurable set $E$, then $f'(x)$ is measurable on $E$.
We are told to only consider 1 dimensional spaces,that f is a measurable function in one variable.that is, f is a measurable function in one variable.
Following is my s...
Let $x,y,z$ be real.
Consider a scalar field
$$z = f(x,y)$$
More specific; $f(x,y)$ is a (given) real polynomial in $x,y$ of degree at most $5$ such that
For all $x,y$
$$f(x,y)> 0$$
For a given pair of points $(a,b),(c,d)$ ($a,c$ are the x-part and $b,d$ are the y-part) consider the line integral...
If $f:[a, b]\to\mathbb{R}$ arbitrary function, then the $$\text{upper derivative }\overline{D}f(x) := \limsup_{y\to x} \frac{f(x)-f(y)}{x-y}$$ and $$\text{lower derivative }\underline{D}f(x) := \liminf_{y\to x} \frac{f(x)-f(y)}{x-y}$$ are always measurable. Thus the set $E = \{x\in [a, b] : \over...
here's my answer that the set of points where a function is differentiable is measurable, and the derivative on that set is measurable too
It uses Vitali's covering theorem
The tricky part is $\bigcup_n \bigcap_k E_n^k\subseteq E$. For this write $\frac{F(d)-F(c)}{d-c}$ as a convex combination of $\frac{F(d)-F(x)}{d-x}$ and $\frac{F(c)-F(x)}{c-x}$
it sounds like you are looking for a path with minimal potential between two fixed points which indeed is calculus of variation as ted indicated. Maybe you can check gelfand's calculus of variations.
Why you have the limit to the maximum degree of the polynomial being five?