I just found an interesting philosophical quote: For the things we have to learn before we can do them, we learn by doing them.” ― Aristotle, The Nicomachean Ethics
yeah, ofc we need some work to understand and remember. Sometimes working on something you can understand it, can assign a meaning from some perspective, so the things becomes "less abstract"
by example: to me sometimes the history of mathematics is key to understand, deeply, many topics
with history you can take the context, the limits of mathematicians of this time, why they create some things, etc... you can put yourself a bit into their eyes
How does this sound @BalarkaSen for the two room thing. Stretch the the two openings till they they are as large their respective rooms, each room has a "floor", which you then contract into middle wall. Now you have a "parallelepiped" with "two openings" for their corresponding rooms, now fold the rooms walls into the middle wall by "pushing" down on the top opening and up on the bottom, into the middle wall. Now you just have the middle wall which you squish to a point...
yeah, motivation is key to understand @skillpatrol, specially when you are trying to learn something very abstract...I learned some thing here, on mathexhange... the motivation concept is an efficient way to refer to context in a very specific way that is oriented to understand why something exist
I am looking at the following exercise:
Let the (linear) differential equation $y'+ay=b(x)$ where $a>0, b$ continuous on $[0,+\infty)$ and $\lim_{x \to +\infty} b(x)=l \in \mathbb{R}$.
Show that each solution of the differential equation goes to $\frac{l}{a}$ while $x \to +\infty$,
i.e. if $...
@SohamChowdhury Ok, so you have a map from each copy of $\mathbb{Z}$ to some group, and you want to show that this gives a unique map from the free group on two objects to that group
@SohamChowdhury Well, that is pretty much the entire idea here, to realize that having a homomorphism from each of two copies of $\mathbb{Z}$ corresponds to having a map from $A$
@SohamChowdhury Ok, let's try a simpler example. If you have a homomorphism from $\mathbb{Z}$ to a group $G$, can you find a map from a one-point set to $G$ (in some "natural" way)?
@SohamChowdhury Well, could you map it to something such that if you only knew the map from the one point to $G$ (i.e. that one point in $G$), you could figure out what the homomorphism from $\mathbb{Z}$ was?
Well. I'd like to say I know how to, but you'll give me some kind of Weierstrass function nonsense and then say "fact : you suck at calculus, get the hell out of here" :P
It's amazing how helpful the people in this chatroom are. I've practically been given short classes by a bunch of people from time to time.
All I know about varieties is that someone once asked Serre or someone what one was and he retorted "integral scheme of finite type over a field", if memory serves.
I did some commutative algebra from Atiyah-MacDonald months back, and sat on a class (which was really on algebraic number theory rather than commutative algebra).
I found it hard, so I gave up and learnt some topology instead. I guess I'd have to learn loads of that stuff at some point of time.
@BalarkaSen that's interesting. but i dont really know enough about chain maps/homotopies yet to say something meaningful right now (plus my mind is clogged w/ complex analysis)
@DanielFischer probably a dumb question, but if $f$ is an entire function such that $f^{(n)}(z)=0$ for all $z\in \Bbb{C}$, why does it follow that $f$ is a polynomial?
@SohamChowdhury a homomorphism $f:\Bbb{Z} \to G$ is determined by what it maps $1$ to
@iwriteonbananas If $f(z) = \sum_{k=0}^\infty a_k z^k$, then $f^{(n)}(z) = \sum_{k=n}^\infty \frac{k!}{(k-n)!}a_kz^{k-n}$. Thus $f^{(n)} \equiv 0$ means all coefficients of the series expansion of $f$ for powers $\geqslant n$ vanish, i.e. $f$ is a polynomial of degree $\leqslant n-1$.
It's the most important way to state it. Try to rephrase it as follows: "Maps out of the coproduct are just pairs of maps out of each of its components."
I can see it pretty clearly for $\sf Set$, but I think the coproduct is not as simple as the disjoint union in other categories, which is what gives me pause.
oh, wait.
yes, it makes a little sense now.
there are inclusions from $X$ and $Y$ to $X\sqcup Y$ "for free", so the only thing determining a map from $X\sqcup Y$ to $A$ is a pair of morphisms, from $X$ and $Y$ to $A$ resp.
so each morphism $X\sqcup Y \to A$ is equivalent to a pair of those from $X$ and $Y$.
When I said as sets I didn't mean in the category of sets. I meant show the isomorphism $\text{Hom}(X \sqcup Y,Z) \cong \text{Hom}(X,Z) \sqcup \text{Hom}(Y,Z)$, and the as sets was to remind that these are sets.
Find all the solutions of the form $y(x)= x^m \sum_{n=0}^{\infty} a_nx^n, \ x>0 (m \in \mathbb{R})$ of the differential equation $3x^2y''+5xy'+3xy=0$.
That's what I have tried:
Since $x>0$ the differential equation can be written as follows.
$$y''+ \frac{5}{3x}y'+ \frac{1}{x}y=0$$
$$p(x)=\fra...
@evinda Only if $c_1 = c_2 = 0$. In a power series, you have only integer powers of $x$, and due to the $x^{\pm\frac{1}{2}}$ factors here, you have non-integer powers. But it is of the form $x^{-\frac{1}{2}}\cdot P(x)$ where $P$ is a power series. That is for many things almost as good.
@evinda Yes, then all coefficients must be $0$. It may be best to say that if $c_1 y_1(x) + c_2y_2(x) \equiv 0$, then $c_1 \sqrt{x} y_1(x) + c_2 \sqrt{x} y_2(x) \equiv 0$, and then you have power series.
@DanielFischer Why can we write it like that:$c_1 \sqrt{x} y_1(x) + c_2 \sqrt{x} y_2(x) \equiv 0$ although $y_1(x)$ contains $x^{\frac{1}{2}}$ and $y_2(x)= x^{-\frac{1}{2}}$ ?
@evinda We have $c_1\sqrt{x}y_1(x) + c_2\sqrt{x}y_2(x) = \sqrt{x}(c_1y_1(x) + c_2y_2(x)) = \sqrt{x}\cdot 0 = 0$. I multiplied with $\sqrt{x}$ to get rid of the $x^{\pm\frac{1}{2}}$ in $y_1$ resp. $y_2$, so that from the multiplication we obtain a nice power series for each.
Since $c_1 \sqrt{x} y_1(x)+ c_2 \sqrt{x} y_2(x)$ is a power series, there are $d_n \in \mathbb{R}$ such that $c_1 \sqrt{x} y_1(x)+ c_2 \sqrt{x} y_2(x)= \sum_{n=0}^{\infty} d_n x^n$ and since $c_1 \sqrt{x} y_1(x)+ c_2 \sqrt{x} y_2(x)=0 \forallx \in (0,R)$, $d_n=0 \foral…
It's well known that the sum of measurable functions is measurable, if they are real or complex valued. However, the proofs I've seen heavily rely on the usage of the countable set of rational numbers. Which made me wonder, what happens if we don't have luxury of having a nice, countable and dens...
@DanielFischer Do you maybe know why the vector space $\mathbb{C}$ is considered a one-dimensional vector space but $\mahtbb{R}^{2}$ is a two-dimensional vector space? If we take $(a,b) \in \mathbb{C}$ then can we not state $(a,b) = a(1,0) + b(0,1) = a(1,0) + ib$ where we can take $(1,0)$ and $(0,1)$ as basis vectors?
@JohnDoe It depends on the scalar field. $\mathbb{C}$ is one-dimensional when the scalar field is $\mathbb{C}$, and two-dimensional when the scalar field is $\mathbb{R}$. If we take $\mathbb{Q}$ as the scalar field, $\mathbb{C}$ is infinite-dimensional.
@BalarkaSen I guess I meant to say that you can retract the diagonal walls into the edge they are connected to on the middle walls, so squish it flat. You could also thing of unfolding it completely by pushing out till the diagonal walls are parallel with the middle wall, and you will have some triangle flaps, and then contract all that.
I had a question about normal distribution curves or distribution curves in general. I understand the basic formula, but I was wondering if there was a way to give a min, max, and mean to find the covariance you need to use to create that certain curve. Is this possible?
The differential equation Laguerre $xy''+(1-x)y'+ay=0, a \in \mathbb{R}$ is given.
Show that the equation has $0$ as its singular regular point .
Find a solution of the differential equation of the form $x^m \sum_{n=0}^{\infty} a_n x^n (x>0) (m \in \mathbb{R})$
Show that if $a=n$, where $n \in ...
Hi @Danielfischer I just wanted to know if the following is correct: To show that a normed space is a topological vector space (as was done [math.stackexchange.com/questions/167890/… this post)) we first have to show that vector addition $V \times V \rightarrow V$ is continuous.
Is this equivalent to showing that $\forall \epsilon > 0$, $\exists \delta > 0$ such that if $\Vert x-x_{0} \Vert \leq \delta$ and $\Vert y-y_{0} \Vert \leq \delta$ then $\Vert (x+x_{0}) - (y-y_{0})\Vert \leq \epsilon$?
I am looking at the following exercise:
Let the (linear) differential equation $y'+ay=b(x)$ where $a>0, b$ continuous on $[0,+\infty)$ and $\lim_{x \to +\infty} b(x)=l \in \mathbb{R}$.
Show that each solution of the differential equation goes to $\frac{l}{a}$ while $x \to +\infty$,
i.e. if $...
