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1

Becareful that you don't just remember STEPS ON HOW to do it, but WHY THE METHOD WORKS. Once things become common sense and then you practise-practise-practise it's hard to forget. Don't be scared to get it wrong if it helps you understand why it works. Not sure what level maths you are doing but try googling virtualb15 and wootube. NSW syllabus but the ...


3

From your question I surmise that we are given a certain sequence $(b_n)_{n\geq0}$ which is Cauchy, and an $\epsilon>0$. This means that there is an $N$ such that $$|b_n-b_m|\leq{\epsilon\over2}\qquad\forall \>m, \>n>N\ .$$ In addition we are told that there are arbitrarily large $m$ with $$|b_m|\geq\epsilon\ .\tag{1}$$ The claim then is that ...


2

First, as noted, it's not too hard to visualize this particular problem. When dealing with a few real numbers and their absolute values, a picture usually works. To be even more explicit, you might "shade" the region of the number line where $b_{n_0}$ could be (knowing that $|b_{n_0}|>\epsilon$) and then shade the region where $b_n$ could be (knowing ...


0

In learning there are two goals, one is obtaining knowledge and the other is obtaining skill. Having knowledge without skill means you can talk about it, but when it comes time to act, you lack the tools to produce results. Having skill without knowledge means you can generally produce results, but your limited understanding of the world has you relying on ...


1

Let me guess what is confusing you. In one variable calculus, if $u=u(x)$, then (on some interval) $$ \frac{d}{dx}u(x)=0\quad\iff\quad u(x)=C. $$ In several variable calculus (I use two variables below), if $u=u(x,y)$, then (in some domain) $$ \frac{\partial}{\partial x}u(x,y)=0\quad\iff\quad u(x,y)=\phi(y) $$ where $\phi$ only has $y$ as a variable. In ...


0

A variable is a quantity in a mathematical expression that can be assigned different values. For the purpose of a particular analysis, it is perfectly possible to consider that one variable, for the present, is fixed, while another is allowed to change freely. This is what is being done when taking partial derivatives. An arbitrary constant is a quantity ...


0

First, the only way for $D(a,b)=D=f_{xx}f_{yy}-f_{xy}^2$ to be positive is if both $f_{xx}$ and $f_{yy}$ have the same sign. So as you observe, a necessary condition to obtain a local max/min is that the concavity in both the $x$-direction and $y$-direction are the same. Now, the term $f_{xy}$ is giving the rate of change of $f_x$ in the $y$-direction (i.e. ...


0

I'm not quite sure if this counts as intuitive, but this is the way i make it clear to myself: Suppose our critial point is $0$, then the Taylor-expansion of $f$ at $0$ yields $$f(x) = f(0) + x^t H x + O(||x||^3)$$ where $H$ is the hessian at $0$. Now if you are close to $0$, then the term $x^tHx$ dominates over the term $O(||x||^3)$ and therefore ...


1

I don't teach that shortcut. Instead I teach the students to complete the square in the quadratic form $$ h^2 f''_{xx} + 2hk f''_{xy} + k^2 f''_{yy} $$ coming from the Taylor expansion of $f$ around the critical point. (Of course, the first order terms vanish.) If the quadratic form is positive definite, $f$ has a local minimum (since $f(x,y) > f(a,b)$ ...


0

I actually study analysis for my college using baby Rudin as a first reference book, but I've chosen another book as supplement e.g. now is watching S. Berberian' s book (the pdf-version) as a good support [http://www.amazon.com/Fundamentals-Analysis-Universitext-Sterling-Berberian/dp/0387984801]. I just can not find solutions...


0

Topological spaces are made out of open sets. Sometimes, you have the occasion to write your topological space as a union of open sets: e.g. because whatever you are trying to study is easy to understand when restricted to just one of the sets. If you can motivate that it is useful to do this sort of thing, then the usefulness of the usual definition of ...


0

A compact set is the "next best thing to finite"- every compact set is both closed and bounded (in a metric space), just like a finite set. In a metric space, given any point, p, there are both minimum and maximum distances from p to points in the compact set.


0

Well I like to think of compactness like this because it gives me the idea of a compact set in terms of closed sets. A collection $\mathcal{C}$ of subsets of $X$ is said to have the finite intersection property if for every finite subcollection $$\{C_1,\cdots C_n\}$$ of $\mathcal{C}$ , $\bigcap_{i=1}^{n} C_i$ should be nonempty. Now if in a topological ...


1

For boundedness: You can give an exercise so that the students need to show that a bounded metric can induce the same topology as an unbounded one (at least you can easily show that for metric spaces with $d(x,y)$ and $\frac{d(x,y)}{1+d(x,y)}$). So, boundedness is not really a topological property. If each open covering has a finite subcovering (and using ...


1

An equivalent def'n is that if $ F$ is a no-empty family of closed sets with the F.I.P. (Finite Intersection Property) then $\cap F \not = \phi $ . This generalizes the idea of limits , and you can show that many results, e.g. on bounded closed subsets of $ R^n$ , using this property, so it is seen to be a useful tool that a space is compact. Once you ...


7

One of my favorite textbooks is Klaus Janich's Topology, and he has a nice motivation for compactness I feel, namely why we should care about. This is in addition to my comment about compact subsets of a Hausdorff space being essentially like finite point sets. But he writes: In compact spaces, the following generalization from "local" to "global" ...


0

I think the better way to motivate compactness is in real analysis, a set $F$ is said be a compact set if it bounded and closed. Is very easy to imagine something compact like this one. The general definition, in general topological spaces, is motivate by Bolzano-Weierstrass theorem.


0

According to Munkres, the original definition of compactness is a space which satisfies the Bolzano-Weierstrauss property holds. That is, if every infinite subset has a limit-point. Unfortunately, it turns out, this conception of compactness,sometimes called limit point compactness, doesn't have all the useful properties that compactness has. For ...


0

The motivation for the definition of compactness is that that condition is extremely useful. Essentially every proof of every fact about the Riemann integral on the line, for example, depends on it. Definitions capture useful properties which allow us to prove useful things — the good ones, at least. If you want to motivate the definition, give the ...


0

Yes that is the question! It´s the historical dataset from a german weather station. I also think it´s hard to describe, because the data are in the past. So in your opinion I can define my own intervals for the three threshold? That was my next idea if find a definition of the temperature I would compare it to other values for example windy or humidity to ...


0

Is the problem "how to define the thresholds for 'hot', 'mild', and 'cold'?" If so, you can define them any way you like! These are vague terms and can be used in any number of ways. I would argue that, for weather, the answer is probably more biological and individual preference. In Singapore, the lowest temperature on record is somewhere around 19.5C, ...


0

There are only two greatly fruitful extensions of real numbers: the complex numbers and affine reals. And the both are heavily used in analysis. Take for instance the formula for inverse Mellin's transform: $$\left\{\mathcal{M}^{-1}\varphi\right\}(x) = f(x)=\frac{1}{2 \pi i} \int_{c-i \infty}^{c+i \infty} x^{-s} \varphi(s)\, ds$$ But one thing that I ...


4

It's an amazing book, but it's not the sort of book most people would want to read from cover to cover, but rather to dip into occasionally. Why don't you read the parts most relevant to your course? Or read about something you have never looked at before? It's not exactly a textbook, more of a resource.


0

I would recommend Khan Academy: https://www.khanacademy.org/math


1

If you want to learn about algorithms, it is a very good idea to learn programming. It always helped me to understand the algorithms, if I also programmed them. And if you are good with understanding algorithms, programming will not be very hard. As programming languages i recommend python and C/C++


0

Mathematical maturity is a departure from the way we typically would percieve lower level mathematics. In multivariable calculus or real analysis for example you would typically study Euclidean space which typically is not argued nor is there any trouble making familiar connections with this space to things we feel we understand. Higher levels of ...


1

Let's try a concrete example: $f(x)$ is a $3$rd-degree polynomial in $x$ and has a maximum at $x=5$ and a minimum at $x=8$. Could there be another extremum? First, let us see that the two extrema described above imply $f(x)\to+\infty$ as $x\to+\infty$. To see this, suppose the other alternative holds: $f(x)\to-\infty$ as $x\to+\infty$. Then the curve ...


0

Comparison test: Note that $$-1 \leq \sin(x) \leq 1 ~~ \forall ~x$$ Therefore, $$\dfrac{\sin(x)}{f(n)} \leq \dfrac{1}{f(n)} ~~ \forall ~x$$ So for your case: $f(n)=n^p$. This converges by $p$-series when $p>1$. Therefore, your original series converges for the same. You cannot use alternating series test, since: $$\lim_{n\rightarrow \infty} ...


3

Limit comparison test. $$\lim \frac{\sin(1/n)/n^p}{1/n^{p+1}}=1,$$ $\sum\frac{1}{n^{p+1}}$ converges when $p>0$, diverges when $p\leq 0$.


2

Since $$ \sin \frac1n \approx \frac1n, $$ your series is equivalent to $$ \sum_n \frac{1}{n^{p+1}}. $$


1

$$ \frac{\sin 1/n}{n^p} \sim \frac{1/n}{n^{p}} \,\,\, \text{if }n\to \infty $$ so I guess every $p>0$ :)


1

Term-by-term integration and differentiation of series is very useful when it works, but nailing down precisely when it does and doesn't work needs more mathematical sophistication than starting to learn calculus. From the point of view of applications (i.e. science and engineering as compared with mathematics), learning how to use "dangerous" mathematical ...


8

Even if one starts studying via series (And some courses do), one usually doesn't go to term-by-term integration and differentiation for a few reasons: It doesn't always work, and you have to deal with convergence issues, and uniform convergence, and this just makes it more difficult at times, especially for a student who is just starting. It is not always ...


2

At the very, veery heart functional analysis is about topological vector spaces (TVS). So basically the answer is almost 'no'. Browse through your notes and look for theorems that concern more than continuity. There are however concepts, i.e. techniques, that require seminorms resp. norms like integration resp. differention. Now, there are TVS which are not ...


0

Yes, insisting upon a normed vector space rather than a metric vector space introduces important additional properties. For instance, $\mathbb{R}^n$ is always complete when equipped with a norm, because every norm on a finite-dimensional real (complex) vector space is equivalent, and $\mathbb{R}^n$ has at least one norm under which it is complete, namely ...


4

A metric space doesn't come with a vector space structure, without which there can't be much analysis going on. It woulds just be a corner of topology, then. You can imagine equipping a vector space with a separate metric, but by the time you require that the metric be compatible with the linear structure -- $d(v+w,u+w)=d(v,u)$ and $d(λv,λu)=|λ|d(v,u)$ are ...


0

Your solution is correct, I had to do this for a programming contest, and the problem was extremely similar in it's mathematical nature.


0

Proof is not the first thing that comes to mind as being important for learning semi-advanced calculus. For those who already have experience with proof, it becomes another tool for expositing the theory, and many questions become clearer if precise definitions and proofs are at hand. But the historical route and the one that even today is used by most ...


1

Actually you are right except for one case. $\gcd(0,0) = 0$ The divisors of $0$ are $D_0 = \{0, \pm 1, \pm 2, \dots \}$ and there is no greatest element of this set. But clearly $d'|0$ and $d'|0 \implies d'|0$. So $\gcd(0,0) = 0$. What the second definition captures that the first does not is that $D_m \cap D_n = D_{\gcd(m,n)}$ where $D_x$ represents the ...


1

a. The answer given by Andre Nicolas is wrong, even for 3x3 matrices. The matrix 1 2 2 a 1 1 b 1 1 has a determinant of 0 no matter what values a and b have. ... b. I am working on writing a "cookbook" for random generation of matrices with certain properties. For the problem posted by Hooked, I generate: a lower triangular matrix L whose diagonal ...


5

I had completely forgotten about this question and came across it again recently, so I decided to email Dr. Strang. To my surprise, he responded almost immediately. The chalk he used is called Railroad Chalk, and you can find it at many popular online retailers.


0

We are arguing about numbers $x\in{\mathbb Z}$. When $x\leq 0$ then $2^x>0\geq 4x$, and a case analysis shows that for $1\leq x\leq3$ one has $2^x<4x$. When $x=4$ then obviously $2^x=4x$. It is therefore sufficient to prove that for $x\geq4$ one has $$2^x\geq 4x\quad\Longrightarrow\quad 2^{x+1}>4(x+1)\ .$$ But this is immediate: $$2^{x+1}=2\cdot ...


0

$$2^x=4x$$ $$\frac{2^x}{4}=x$$ $$2^{x-2}=x.$$ Assume that $x\in \mathbb{Z}$. Clearly, $x\geq 2$, otherwise RHS is non-integral. But the LHS is convex while RHS is linear, hence there can only be $2$ solutions, one for which $\frac{d}{dx}2^{x-2}<\frac{d}{dx} x$, which must occur first, and one for which $\frac{d}{dx} 2^{x-2} > \frac{d}{dx} x$, which ...


2

For positive $n$, we have two growing sequences $$1,2,4,8,\color{green}{16},32,64,128,256\cdots\\ 0,4,8,12,\color{green}{16},20,24,28,32\cdots$$ This shows them that the "curves" cross each other at $16$, and it seems that the first grows faster. Indeed, taking the ratios of successive terms $$\frac{2^{n+1}}{2^n}=2>\frac{4(n+1)}{4n}=1+\frac1n.$$


0

Solutions can occur only for $x>0$ when both sides have the same sign. Take the derivative of $f(x)=2^x-4x$ and see that $f(4)=0$ and $f'(x)>0$ for $x>4$, so the function $f(x)>0$ for $x>4$ since it is growing, hence, no more solutions for $x>4$. Testing other integers $x=1,2,3$ is easy.


1

First of all, the solution of $2^x=4x$ has to be positive. We have $$\frac{2^x}{x}=4\tag1$$ and let $f(x)=\frac{2^x}{x}$. Then, we have $$f'(x)=\frac{2^x(x\ln 2-1)}{x^2}$$ So, we know that $f(x)$ is increasing for $x\gt \frac{1}{\ln 2}$ where $1=\frac{1}{\ln e}\lt\frac{1}{\ln 2}\lt \frac{1}{\ln\sqrt e}=2$. Since $f(1)=2,f(2)=2,f(3)=8/3,f(4)=4$, there is ...


5

Hint: plot the graph of $y=2^x$ and $y=4x$ and shows that the only other solution is between $0$ and $1$.


3

If your intention is to 'convince' and not to prove, I'd draw a graph with the functions $y=2^x$ and $y=4x$. The growth rate of each function should make clear that they intersect only at two points, being the first between $0$ and $1$ (and hence, not being an integer).


0

That is a good idea. pay attention for postulates that you are accepted for doing proof. If students are familiar with postulates then it is very suitable to understand.


0

You have a function $y=f(x)$ and want to rotate the graph by angle $\theta$ use a rotation matrix. For your example, you want to rotate by $-\pi/4$. Start with a point on your graph $\mathbf{x}=(x,f(x))$ and use the rotation matrix: $$ R(\theta)=\left[\begin{array}{cc} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{array}\right]. $$ You ...



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