# Tag Info

27

There is a case where the result is obvious: Suppose that $f,g$ are continuously differentiable at $a$, that $g'(a)\ne0$, and $f(a)=g(a)=0$. Under these assumptions, $$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f(x)-f(a)}{g(x)-g(a)}=\frac{\lim_{x\to a}\frac{f(x)-f(a)}{x-a}}{\lim_{x\to a}\frac{g(x)-g(a)}{x-a}}=\frac{f'(a)}{g'(a)}.$$ Of course, under ...

23

In my opinion, one of the best ways of developing intuition in topology is to study other branches of mathematics in which there are topological spaces. Many of these definitions, properties and theorems were imagined by people who were working in related branches of math, mostly analysis and geometry. These people stumbled upon spaces which had remarkable ...

20

Around $x=a$ each of these functions can be approximated by their tangent line: $$f(x)\approx f'(a)(x-a)+f(a)$$ $$g(x)\approx g'(a)(x-a)+g(a).$$ Thus $$\frac{f(x)}{g(x)}\approx \frac{f'(a)(x-a)+f(a)}{g'(a)(x-a)+g(a)}.$$ Taking the limit of the right hand side gives $\frac{f'(a)}{g'(a)}$.

13

Suppose $f(x),g(x)\to0$ as $x\to a$. Look at the point $(a,0)$ under a microscope. The graphs of $f$ and $g$ in that view look like straight lines. Recall that $dx$ is an infinitely small increment of $x$, that looks like $0$ unless you use this microscope. The slopes of those lines you see under the microscope are $$\frac{f(a+dx)-f(a)}{dx} = ... 7 If T is a monad on a category \mathcal{E}, then one can define T-algebras (I prefer the terminology T-modules) and this generalizes all the usual notions of algebraic structures. They are objects A equipped with a morphism T(A) \to A such that the two obvious compatibility conditions are satisfied. For example, if \mathcal{E}=\mathsf{Set} and ... 5 To simplify @Michael's answer even further, briefly put for infinitesimal \epsilon we have that f(\epsilon) is indistinguishable (in a suitable sense) from \epsilon f'(\epsilon), and similarly for g, and therefore \frac{f(\epsilon)}{g(\epsilon)} is indistinguishable from \frac{\epsilon f'(\epsilon)}{\epsilon ... 4 Prove it's true for a = 0, a=1. Then use induction on a to prove it holds for a\geq 10, in which you'll first need to show it holds for a = 10. This will help you develop intuition about the conjecture, as well as provide a mathematical proof as to it's truth. It is easy to see it holds for a = 0, since 2^0 = 1 > 0^3 = 0, and 2^1 = 2\gt 1^3 ... 4 Generally, powers always eventually get larger than polynomials. When a is large, increasing a doubles 2^a, but doesn't increase a^3 nearly that much. a^3 roughly increases by 3a^2, which is less than a^3. As others have suggested, you can turn this into an inductive proof: 2^{10}=1024\gt1000=10^3. Now assume 2^k \gt k^3 and k \ge ... 4 By the chain rule we shall have$$f'(tx)=\sum_{k=1}^n\frac{\partial f}{\partial x_k}(tx)x_k,$$put g_k(x):=\int_0^1 \frac{\partial f}{\partial x_k}(tx)dt. This is a form of Hadamard's lemma. I actually don't understand what you mean by your second question, which is very imprecise, but this lemma is used to prove, say, that every derivation of ... 3 In the theorem you have x^n, but in the series you have x^{2n} and x^{2n+1}. In the theorem you drop the n=0 term because that's the constant term (whose derivative is zero); in the case of cosine, the n=0 term is again the constant term, so it works out; but in the case of sine it's the x^1 term. To apply the theorem very carefully for cosine: ... 3 This is both a difficult and interesting question (for instance because it is difficult to say what intuition really is, among other things), and the only thing that I can offer is merely my opinion. I should like to point out from the beginning that I do not think that there is a very systematic way of obtaining intuition in general, only that there are ... 3 Imagine you're on an infinitely large, flat piece of paper with a transparent beach ball of diameter 1m. Assume the ball is perfectly round. You have a lazer pen. The paper is the complex plane. Place the beach ball down so that it rests on 0, its centre is 0.5m off the ground and the point directly above 0 is 1m off the ground. Take your lazer ... 2 Change the variable: set t=1/x, so you want to compute$$ \lim_{t\to0^+}\sqrt{\frac{1}{t^2}-\frac{4}{t}}-\frac{1}{t} = \lim_{t\to0^+}\sqrt{\frac{1-4t}{t^2}}-\frac{1}{t} = \lim_{t\to0^+}\frac{\sqrt{1-4t}-1}{t} $$Now it should be clearer why the limit can't be 0. The square root can be written$$ \sqrt{1-4t}=1+\frac{1}{2}(-4t)+o(t^2) $$so the limit ... 2 Expand f into its Taylor series,$$f(z) = \sum_{k=0}^\infty c_k(z-z_0)^k.$$Then you have$$\frac{f(z)}{(z-z_0)^{n+1}} = \sum_{k=0}^\infty c_k (z-z_0)^{k-n-1}.$$When you integrate termwise, the only term with a nonzero integral is the one with the exponent k-n-1 = -1, that is, k = n, so the integral is 2\pi i c_n. But$$c_k = ...

2

For your first example, note that $T(0)\neq 0$. This violates a necessary property satisfied by all linear transformations. For your second example, because it is a linear combination of the inputs, we see that it is linear. Your third example involves multiplying two of the input values. Anything that does this will probably not be linear. Here is a ...

2

With respect to the transformation $T(v) = Av + b$, it is an unfortunate coincidence that this is the equation of a line. This is not what it means to be a linear equation - instead, linearity means that changes in the input result in proportional changes in the output. In this context, we call equations of the sort $T(v) = Av + b$ to be an affine ...

2

Basically, the idea is in the previous answers. Nevertheless, I think the following proof is easier to understand. Let's prove the following: Fact: Suppose $\lambda_0$ is an eigenvalue of $A$ and with geometric multiplicity $k$, then its algebraic multiplicity is at least $k$. Proof: By the assumption, we can find an orthonormal basis for the subspace ...

2

This is something I'm still trying to figure out & just wish I knew fully so as to be able to answer this question properly - as far as I understand it at this moment any first order linear or non-linear PDE may be (equivalently) re-expressed in the form of a Hamilton-Jacobi equation by a change of variables, (No. 37), which in turn can be interpreted as ...

2

The Minkoswki content is a rather simplistic way to define an $m$-dimensional measure of an object immersed in the euclidean space (however mathematically speaking Minkowski content is not a measure). If $\mathcal A$ is a regular $m$-dimensional surface in $\mathbb R^n$ it is easy to understand that the Minkowski content (both lower and upper) gives the ...

2

It seems like what you're getting into is topological graph theory, which mainly concerns embedding graphs into surfaces. The study of planar graphs is a special case of this where the graphs are being embedded into the plane. So in topological graph theory, for example, you might also study graphs which could be linklessly embedded on a torus. I'm not by ...

2

You misunderstood why you go from $\sum\limits_{n=0}^\infty$ to $\sum\limits_{n=1}^\infty$. It's not that you're getting rid of the lowest power term, as you have done. But rather that you're getting rid of the $x^0$ term, which is constant. So when you differentiate $\sin(x)$, you don't get rid of the $n=0$ term, because that's the $x^1$ term.

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The action of $f$ does indeed define a partial function from $S$ to itself. But that's intentional. Category actions are equivalent to diagrams. The "picture" you should have in mind is that $S$ is the disjoint union of all of the objects in the diagram, so that you really do want the action of $f$ to be a partial operation, defined only on those objects of ...

1

In some approaches, such as Fraleigh's, it is natural to define isomorphisms before homomorphisms so that one can identify structures that look different but are really the same. This then leads to homomorphisms by relaxing the requirements of injectiveness and surjectiveness, leaving just the important property of preserving structure, in the sense of ...

1

In general, a homomorphism is a function between sets which respects some sort of algebraic structure on the sets (the appropriate language here is category theory, but we'll just stick to the basics). In the example you mentioned, a linear transformation respects the vector space structure of a vector space, i.e., you can perform the vector space operations ...

1

This is wrong: $$\sin'(x)= \left(\sum_{n=0}^{+ \infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} \right)'=\sum_{n=1}^{+ \infty} \frac{(-1)^n(2n+1)x^{2n}}{(2n+1)!}$$ The point of discarding the derivative of the first term $a_0x^0$ is simply that the derivative of a constant term is zero, so it can be left out. Otherwise the first term would be $0 \times a_0x^{-1}$, ...

1

I am not sure if I speak from the same perspective as you do as I look at these objects from the analytic side and not the geometric side but there is a ton of number theoretic information. Here are two examples I can think of off the top of my head. Fourier Coefficients Many Modular forms of half weight or integral weight tend to have Fourier coefficients ...

1

If you really have a good idea of what a homeomorphism is, you can understand most of these definitions pretty easily. A homeomorphism is the equivalent of an isomorphism for topological spaces: if two spaces $X$ and $Y$ are homeomorphic, then they essentially "look the same." For example, any closed interval $[a,b]$ is homeomorphic to the unit interval ...

1

Your argument is very interesting. I'm not sure how to best adapt it yet. The standard arguments would say that eigenfunctions in $L^{2}$ with different eigenvalues are orthogonal. Because $L^{2}[0,1]$ is separable (has a countable orthonormal basis), then there can be at, most, a countably-infinite number of eigenvalues. For non-singular Sturm-Liouville ...

1

Consider a triangle ABC. Let D be the midpoint of $\overline{AB}$, E be the midpoint of $\overline{BC}$, F be the midpoint of $\overline{AC}$, and O be the . By definition, $AD=DB$, $AF=FC$, $BE=EC$ \,. Thus $[ADO]=[BDO]$, $[AFO]=[CFO]$, $[BEO]=[CEO]$, and $[ABE]=[ACE]$ \, where $[ABC]$ represents the area of triangle \triangle $ABC$ ; these hold because in ...

1

Spent the day learning PCA, hope my cartoon translates the intuition over to you! I have also tried to briefly explain the utility of PCA and related it to an analogy (no maths) to help give that feeling of "learning closure". Visual Intuition (zoom in) Intuition via utility I think the main usage for PCA is to be able to categorise different distinct ...

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