Chubby Chef
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Note that logical implication (aka ‘material conditional’) does not translate directly into our intuitive understanding of implication. We often think of implication as synonymous with causality, ...

Kunz's Einführung in die kommutative Algebra und algebraische Geometrie is fairly standard, and (as far as I can tell) corresponds quite neatly to the book by Eisenbud. If you're looking for something ...

At $x=0$ we have $\ln (1-0) = \ln (1) = 0$ in the denominator, which is undefined.

I believe this is called the imbalance of a vertex, see for example this paper: https://core.ac.uk/download/pdf/42932978.pdf Note that this is defined as indegree - outdegree, but clearly this can be ...

In this case you can't multiply both sides by $(x+2)$ as it might be negative for certain values of $x$ and that would require reversing the inequality sign. Try multiplying both sides by $(x+2)^2$ ...

Let $b_n$ converge to $l_1$, and suppose $\frac{a_n}{b_n}$ converges to $l_2$. Then by the Algebraic Limit Theorem we have$$\lim_{n\to \infty}b_n\cdot \frac{a_n}{b_n} = l_1l_2.$$ But $$b_n \cdot \frac{... View answer Accepted answer 3 votes$$\lim_{x\to 0} \lvert \ln (x) \rvert = \infty\lim_{x\to 0^+} \frac{1}{x} = \infty$$Whether this limit really exists is a matter of convention. Remember that \infty is just a convenient ... View answer Accepted answer 2 votes Suppose for contradiction that A\cap E \neq \varnothing. Then there exists some x\in A\cap E. By definition, x\in A and x\in E. But since B\cap E = \varnothing and A\cap C = \varnothing, ... View answer 2 votes I have never heard of this book's publisher and, searching for the authors, couldn't find a university page or anything (as a matter of fact, this and one other book seem to be their only meaningful ... View answer 2 votes It could be the case that you've only studied propositional (also called zeroth-order) logic so far. If so, then we may reword your statement slightly and write something like "if p is even and ... View answer 2 votes I'm not sure what you're trying to do with your example, but one way of expressing f(n) with a general formula is$$f(n) = \sum_{i=0}^{n} (n-i+1)\rho^i.$$View answer 2 votes Note that, by definition of a derivative, for \lvert x \rvert to be differentiable at 0 the limit$$\lim_{x\to 0} \frac{\lvert x \rvert - \lvert 0 \rvert}{x - 0} = \lim_{x\to 0} \frac{\lvert x \...

A discontinuous function is still defined on every point in its domain, discontinuity is defined in terms of limits. What you’re describing is called a ‘partial function’, I believe the term sees some ...

So, I think the misunderstanding you have is that $\varepsilon - \delta$-type proofs cannot be applied to one-sided limits. The first step, when given a function as simple is this, is to sketch it. ...

You've already essentially answered your own question, although the lower bound you give for $z$ seems somewhat arbitrary. You've already realised that $x \geq 9$, so you can easily substitute $9$ ...

I believe your concern regarding the supplied proof is valid. As a matter of fact, I think the chosen proof strategy isn't the optimal one - it is often said that a direct proof is better than a proof ...

As stated, the problem doesn’t make sense because $6(3)^6 = 4374 < 6561$ and $6(3)^7 = 13122 > 6561$. Taking the logarithm you’d expect $T_n$ to be about the $7.6$th term, but that’s of course ...

Hint: if you're at all familiar with logic, this proof is (perhaps superficially) very similar to a standard proof of the unique readability theorem in propositional calculus. So, the first step is ...

The reciprocal of a function $f(x)$ is simply $\frac{1}{f(x)}$. So let's denote it by $r(x)$ for convenience. Then, to find the point(s) of intersection, you want to find all $x$ that would satisfy $f(... View answer Accepted answer 1 votes I don't think your proof is quite right. Instead, try expanding the definition of a limit a little bit. The key will be to pick some$\varepsilon >0$that is strictly less than the limit$L$, say$\...

I think you're a little confused about the steps in this problem. Note that, after you've multiplied both sides by the denominator, you need to try and solve the resulting equation, in this case $$7 = ... View answer Accepted answer 1 votes Two ways to look at it: one is that given a (bijective) function f\colon X\to Y we define the inverse to be a function f^{-1}\colon Y \to X. But your f is a function of the form 1/k and this ... View answer 1 votes This is a late answer (3 years late!), but I thought I'd put my two cents in just in case somebody ever stumbles across this question. One book that matches your description perfectly (and, I'd ... View answer 1 votes Your first example is technically correct. I say technically because there seems to be a subtle difference between the actual definition of \implies and the way most mathematicians use it. Of course,... View answer 1 votes Using the standard limit properties:$$\begin{align} \lim_{x\to 0} (1 - \frac{1}{x}(\sqrt{x^2+1}-1)) &= \lim_{x\to 0}1 - \lim_{x\to 0} \frac{\sqrt{x^2+1} - 1}{x} \\ &= 1 - \lim_{x\to 0}\left(...

You’ve got the right idea, but the notation is a little confused. Using your counterexample, let $g(x) = f(x) = x^2$. Trivially, $f, g \in O(h(x))$ where $h(x) = x^2$. Naturally, $f \circ g (x) = x^4$,...

The simplest method would be to use the ceiling function: that is, $\lceil x \rceil$ is the smallest integer which is not smaller than $x$. So, supposing the squares are numbered $1$ to $10$, a pair ...
A function is two sets with a mapping between them. Its domain could indeed consist of a single element, much like how its output could also be a single number. Think of the identity function $f:\{1\} ... View answer 1 votes I’ll start with the second statement, as I think it makes more sense that way.$\exists x, \forall y, P(x,y)$means that for all$y$there exists such an$x$that the predicate$P(x,y)$is true. ... View answer Accepted answer 1 votes Let$x$stand for the amount of postcards that Ron had before trading, and let$y$stand for the amount of postcards that Matt had before trading. So, to find$x$and$y\$ we can form simple equations ...