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12

My experience is that if you spend enough time both actively and passively (subconsciously) thinking about the material, these mathematical concepts will become as natural as breathing. It is a slow process at first (it took me until my junior year of undergrad to really start to understand what math was really about) but once you have a sufficient ...


10

Here is some intuition motivated by applications. In many applications, we have a system that takes some input and produces an output. A special case of this situation is when the inputs and outputs are vectors (or signals) and the system effects a linear transformation (which can be represented by some matrix $A$). So, if the input vector (or input ...


6

I don't know how intuitive this will be, but here is how I think of it. It is similar to what Travis mentioned in the comments. We need the Euler characteristic to not change when we change the triangulation. Consider the following triangulation of the plane. This graph has 4 vertices, 5 edges, and 3 faces, which gives us the $4-5+3=2$ we expected. ...


5

There are several good answers here, one accepted. Nevertheless I'm surprised not to see the $L^2$ norm described as the infinite dimensional analogue of Euclidean distance. In the plane, the length of the vector $(x,y)$ - that is, the distance between $(x,y)$ and the origin - is $\sqrt{x^2 + y^2}$. In $n$-space it's the square root of the sum of the ...


5

Let me give a geometric explanation in 2D. The first fact that is almost always glossed over in class is that a "linear operator" or a matrix $T$ acting on a space $V \to V$ looks like a combination of rotations, flips, and dilations. To image what I mean, think of a checker-board patterned cloth. If I apply a transform to the space, it stretches (or ...


5

I'm a high school student, so I have no idea what a Jacobian or a manifold is, but as someone who's self-studied linear algebra and abstract algebra, I think it's pretty complex and takes a rather smart/dedicated person to pass these classes, so you're definitely smart enough to understand these concepts. In my opinion, these fields are kind of intuitive in ...


4

The answer, as you might guess, is b). But still, work hard. It is worth it! I have only very basic intuition and only in the area I am beginning to specialize in, but every time I gain a little insight and really feel it, the fun and the rush are worth it! Do lots of problems. Ask each and every question you have, multiple times, to multiple people. Compare ...


3

I wrote a blog post addressing this question here. In that post I show that this alternating behavior (a simpler but less general version of it coming from counting "cells") is forced by a few natural axioms, the most important of which is an inclusion-exclusion axiom $$\chi(A \cup B) = \chi(A) + \chi(B) - \chi(A \cap B)$$ where some restrictions on $A$ ...


3

I think that one point is missing, that "nearby" is invariant under any definition of "nearness" (within the context of the given topology). In the case of a metric space, this means that for any $\varepsilon$, we can find a rational number whose distance from $r$ is less than $\varepsilon$. So no matter what "near" means to you, because it's a fickle ...


3

You can visualize it by looking this identity as identity of polynomials: $$ (1-x)(1+x)=1-x^2 $$ $$ (1-x)(1+x+x^2)=1-x^3 $$ and in general $$ (1-x)(1+x+x^2+x^3+\ldots+x^{n-1})=1-x^n $$


3

Provided that the domain that $x$ varies over is $\mathbb{R}$, and that $<,>,5,7$ all have their normal meanings in this domain, the statement $$(x< 5) \land (x > 7)$$ is not true for any choice of $x$. I'm not sure you'd want to call this a logical impossibility though, as there is nothing about logic that restricts $x\in \mathbb{R}$, or fixes ...


3

The example given by Noah is rather paradigmatic. That is, algebras all of whose congruence relations permute have a lot of nice structure around. The proof he gave that groups are congruence-permutable cam be greatly generalized to the following characterization. TFAE for an equational class of algebras $\mathcal{V}$: 1) every algebra in $\mathcal{V}$ ...


3

For a general integer $x$, consider the following eight quantities: $$ 20!x + 2, 20!x + 3, 20!x + 5, 20!x + 7, 20!x + 11, 20!x + 13, 20!x + 17, 20!x + 19. $$ I claim that these are all pairwise relatively prime. Consider $20!x + p$ and $20!x + q$, for $p < q$: \begin{align*} (20!x + p, 20!x + q) &= (20!x + p, q-p) \\ &= (p,q-p) \text{ since } q-...


3

I'm not sure how intuitive is this, but the norm $||f||_{L^2([a,b])}$ of an integrable function defined on $[a,b]$ is the square root of the "area under the graph" of $|f|^2$ on the interval $[a,b]$. For example, if $f \equiv C$ is a constant function, then the area under the graph of $f^2 = C^2$ is the area of a rectangle given by $(b - a)C^2$ and so $||f||...


3

If you have some physics background, then $L^2$ norm can often be interpreted as the "energy" of the wave functions. Physical interpretation of L1 Norm and L2 Norm In quantum physics, the $L^2$ norm represents the probability of detecting a particular pure state amount many mixed states. In statistic, minimizing the $L^2$ norm of the difference between 2 ...


3

All of these topics can be understood intuitively, they just don't tend to be taught intuitively. I remember when starting linear algebra, I already knew what a vector is so it made sense to me, but I pity someone who was trying to understand what a vector is from the axioms of a vector space, which were what was being written onto the blackboard in front of ...


3

Your proof above isn't a proof that $f(x,y)$ is differentiable at $(x,0)$. Your proof is a proof that $f(x,y)$ is partial differentiable at $(0, 0)$. This is the regular definition of derivative: $$\lim_{h \to 0} \ (f(x+h)-f(x))\frac 1 h$$ If I wanted to prove that $f(x)$ is differentiable at $x=0$, I would need to substitute $x=0$ into the limit and then ...


2

Well I will try to give you a simple example. An eigenvalue equation can be found in different areas. Since you are learning linear algebra, I will give you an example of this one. Suppose you are working in $\Bbb{R}^3$ and you are given a linear transformation $T:\Bbb{R}^3 \to \Bbb{R}^3$. Then, one says that $v \in \Bbb{R}^3$ is an eigenvector of $T$ with ...


2

OK let's see if this helps you. Suppose you have two functions $f,g:[a,b]\to \mathbb{R}$. If someone asks you what is distance between $f(x)$ and $g(x)$ it is easy you would say $|f(x)-g(x)|$. But if I ask what is the distance between $f$ and $g$, this question is kind of absurd. But I can ask what is the distance between $f$ and $g$ on average? Then it is $$...


2

It is believed, that $\pi$, $e$ and every irrational algebraic number are normal to every base. The truth is, that none of those numbers have been proven to be normal to a single base.


2

EDIT: I believe the terminology is as follows: two congruences $\sigma$ and $\rho$ on an algebra $A$ are permutable if $\sigma\rho=\rho\sigma$, that is, if we have $$\{(a, b): \exists c(a\rho c, c\sigma b)\}=\{(a, b): \exists c'(a\sigma c', c'\rho b)\}.$$ See e.g. https://en.wikipedia.org/wiki/Congruence-permutable_algebra. Let me give a somewhat degenerate ...


2

In general the differences are very subtle, and I don't know of any good way to visualize them. For instance, here is an example of a function that is $C^2$ but is not differentiable at $x=0$, the function $y=|x|^3$: (It does not have a third derivative at $x=0$ because the second derivative is $3|x| + 3x$.) Can you tell visually that this function does not ...


2

For any $x \in \mathbb{R}$, the statement $x < 5 \wedge x > 7$ is logically false. Equivalently, the statement There exists an $x \in \mathbb{R}$ such that $x < 5$ and $x > 7$ is logically false.


1

Yes, it happens, but you shouldn't take $1$ but rather $0.01$. Then really $6.01^2$ is close to $36+2\times 6\times 0.01$. This picture might help. The idea is that if $dS$ is small, then the "neglected" part is even much smaller.


1

Your intuition is correct. What is not correct is what you call the usual definition: Let $U$ be any non-empty open subset of $X$. A set $A$ is dense in $X$ iff $A\cap U\neq \emptyset $. Of course there is a problem of quantification. Instead $A$ is dense in $X$ iff for any $U$ be any non-empty open subset, $A\cap U\neq \emptyset $


1

Yes, you are right, this is impossible. Though, if you talk about sets : $$(x<5)\cap (x>7)=\varnothing.$$


1

The Euler characteristic can be computed from the number of cells of each dimension. The non-alternating sum (or other functions of the ranks) cannot. The Euler characteristic has nice formulas when $X=X_1\cup X_2$, or when $X=X_1\times X_2$, or when $X$ is a bundle etc. The non-alternating sum (or other functions of the ranks) have not.


1

At first note, that $i=\sqrt{-1}$ is the imaginary unit in (9) and we see a plain multiplication with $i$. We start with the representation (5) and use (7) to obtain \begin{align*} y&=x^{\frac{1}{2}}Z_{\frac{1}{3}}\left(\frac{2}{3}ix^{\frac{3}{2}}\right)\\ &=x^{\frac{1}{2}}\left[AJ_{\frac{1}{3}}\left(\frac{2}{3}ix^{\frac{3}{2}}\right) +B\left(\...


1

In 2, the sets $U$ and $A$ are both in $\mathfrak{T}$, i.e. they are both open sets of $X$. But the intersection of two open sets is open, so $C=U \cap A$ is open. A similar logic will work for 1, but you must use that a set is closed if and only if its complement is open, making the argument somewhat more complicated.


1

For 1: Let $C\subseteq A$ be closed in $(A,\mathfrak{T}|_A)$ and let $A$ be closed in $(X,\mathfrak{T})$ Then $A\setminus C$ is open in $(A,\mathfrak{T}|_A)$ by the definition of being closed. Then, $(A\setminus C)\cap A$ is open in $(X,\mathfrak{T})$ by definition of the induced topology. Furthermore, $X\setminus A$ is open in $(X,\mathfrak{T})$ by the ...



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