114

In general, computing the extrema of a continuous function and rounding them to integers does not yield the extrema of the restriction of that function to the integers. It is not hard to construct examples. However, your particular function is convex on the domain $k>0$. In this case the extremum is at one or both of the two integers nearest to the ...


97

It depends on what you consider a “gap” in the rational numbers. As long as this is not a formally defined concept, we’re just talking about our everyday, geometrically informed conceptions of gaps. The mere fact that a certain equation doesn’t have a rational solution doesn’t seem like a basis for identifying a “gap&...


66

Yes. There is a geometric explanation. For simplicity, let me take $x=0$ and $h=1$. By the Fundamental Theorem of Calculus (FTC), $$ f(1)=f(0)+\int_{0}^{1}dt_1\ f'(t_1)\ . $$ Now use the FTC for the $f'(t_1)$ inside the integral, which gives $$ f'(t_1)=f'(0)+\int_{0}^{t_1}dt_2\ f''(t_2)\ , $$ and insert this in the previous equation. We then get $$ f(1)=f(0)+...


61

You have not translated the pages from Apostol's book into mathematical logic. What you have done is to transcribe them into your own idiosyncratic shorthand, which may be useful to you but is less than meaningless to anyone else. Let's start with the use of the symbol $\stackrel{\mathrm{def}}=.$ In normal mathematics, this tells us that the notation on the ...


57

With respect to the first part of your question: No, a function with two global minima does not necessarily have an additional critical point. A counterexample is $$ f(x, y) = (x^2-1)^2 + (e^y - x^2)^2 \, . $$ $f$ is non-negative, with global minima at $(1, 0)$ and $(-1, 0)$. If the gradient $$ \nabla f(x, y) = \bigl( 4x(x^2-1) - 4x(e^y - x^2) \, , \, 2e^y(...


50

Although the possibility of different axioms is a concern, I think the major objection the author is speaking of is largely about constructivism (i.e. intuitionistic logic). There really is a big gap between rational numbers and real ones: with enough memory and time, a computer can represent any rational number and can do arithmetic on these numbers and ...


47

There is a difference between a thing not existing in some set, and the existence of "gap" corresponding to that thing. For example, there is no rational number $p$ such that $p > q$ for all rational numbers $q$. Does this mean that there is a "gap" in the rationals corresponding to some "largest" rational number? I think that most people would argue ...


46

This is somewhat surprising if you're not used to this. But of course you're free to reject whatever mathematical statements you dislike. The real question is what else you are forced to reject with it, and what would remain of the mathematics that you know and love otherwise. The onus is on you, as someone who decided that "everyone else is wrong",...


41

Let $I$ be our integral. Substitute $t=-x \Rightarrow dt = -dx$. Then: $$I=\int_{-a\pi}^{a\pi} \frac{\cos^5 t+1}{e^{-t}+1}\,dt=\int_{-a\pi}^{a\pi} \frac{e^t(\cos^5 t+1)}{e^{t}+1}\,dt=\int_{-a\pi}^{a\pi} \frac{e^x(\cos^5 x+1)}{e^{x}+1}\,dx$$ Therefore: $$2I=\int_{-a\pi}^{a\pi} \frac{e^x(\cos^5 x+1)}{e^{x}+1}\,dx+\int_{-a\pi}^{a\pi} \frac{\cos^5 x+1}{e^{x}+...


41

Basically, there's a type error: "$\int f(x)\,dx$" is a perfectly meaningful thing, but that thing is not a single function - rather, it's a set of functions. The point is that a function doesn't have a unique antiderivative. For example, ${x^2\over 2}$ is an antiderivative of $x$ (with respect to $x$ of course), but so is ${x^2\over 2}-4217$. It's ...


39

The main question here seems to be "why can we differentiate a function only defined on integers?". The proper answer, as divined by the OP, is that we can't--there is no unique way to define such a derivative, because we can interpolate the function in many different ways. However, in the cases that you are seeing, what we are really interested ...


35

Here is a heuristic argument which I believe naturally explains why we expect the factor $\frac{1}{k!}$. Assume that $f$ is a "nice" function. Then by linear approximation, $$ f(x+h) \approx f(x) + f'(x)h. \tag{1} $$ Formally, if we write $D = \frac{\mathrm{d}}{\mathrm{d}x}$, then the above may be recast as $f(x+h) \approx (1 + hD)f(x)$. Now ...


35

The other answers have made good points about constants of integration but this is not actually what I meant, although it is related. What I meant is what lulu says in the comments: writing antiderivatives this way misleads you about the relationship between the $x$ on the LHS (which is a dummy variable) and the $x$ on the RHS (which is not). The "real&...


34

Nice question. Here is my solution. It would be nice to know if there is a simpler example. This is basically a "spline" example. First, if $A,B,C,D,r,s$ are all rational numbers, and $r < s$, then there exist rational numbers $a,b,c,d$ so that the cubic polynomial $f(x) = ax^3+bx^2+cx+d$ satisfies $f(r)=A,f'(r)=B, f(s)=C, f'(s)=D$. The ...


32

Weak continuous Sudoku: A weak continuous Sudoku can be constructed based on the ideas that you already provided. First, we construct a weak continuous Sudoku for the set $U=(0,1]$ instead of $U=[0,1]$. Here, a weak continuous Sudoku can be constructed by using the function $f$ from your attempt but as a function $f:(0,1]^2\to (0,1]$ (since one boundary is ...


31

Remember that the book you are reading is on axiomatic set theory. Any time you do pure mathematics, you have to start with axioms. You can't prove them, you just specify them. And then you use them to prove other things. The famous example of this is the parallel postulate. People were surprised when it was realized that you could have a perfectly ...


31

There are several key ingredients I will briefly describe here. I won't go into too much detail as you've mentioned that you don't have a graduate real analysis background yet. But indeed a full description of the theory is a standard part of a graduate course in linear PDE. So I hope that answers your side question as well. We start with a strongly ...


31

Let $g(x)=P(x)e^{-x^2}$. Then $$g'(x)=P'(x)\cdot e^{-x^2}+P(x)\cdot (-2x)e^{-x^2}= (P'(x)-2xP(x))e^{-x^2}. $$ We may consider $g$ to be extended to the infinities per $g(\pm\infty)=0$ because the exponential function outweighs the polynomial factor. First assume that the roots of $P$ are distinct. Then $g$ has $n+2$ distinct zeroes, namely the $n$ zeroes of $...


30

This isn't a field property, it's a property of the underlying logical framework within which we're defining fields in the first place. Specifically, the main property is that if $a=b$ then any sentence involving $a$ is equivalent to the same sentence gotten by replacing some of the $a$s with $b$s; we also use the simpler property that "$=$" is reflexive. ...


28

The polynomials $$p_k(h):=\frac{h^k}{k!}$$ have two remarkable properties: they are derivatives of each other, $p_{k+1}'(h)=p_k(h)$, their $n^{th}$ derivative at $h=0$ is $\delta_{kn}$ (i.e. $1$ iff $n=k$, $0$ otherwise). For this reason, they form a natural basis to express a function in terms of the derivatives at a given point: if you form a linear ...


27

@IsaacBrown 's answer is concise and correct. Here's another way to understand what is going on. If you think about physically tracing the graph of $|x|$ between $(-1,1)$ and $(1,1)$ as you move at unit speed along the $x$-axis then the corner at the origin requires an instantaneous change of direction. If you slow down so that your speed approaches $0$ as ...


26

You're correct that it doesn't really make sense to write $\lim\limits_{h\to 0}\frac{f(x+h)-f(x)}{h}$ unless we already know the limit exists, but it's really just a grammar issue. To be precise, you could first say that the difference quotient can be re-written $\frac{f(x+h)-f(x)}{h}=2x+h$, and then use the fact that $\lim\limits_{h\to 0}x=x$ and $\lim\...


25

If we choose $$ (a_1, \ldots, a_{n-1}, a_n) = (a, \ldots, a, \frac{P}{a^{n-1}}) $$ for some $a > 0$ then $\prod a_i = P$ is satisfied, and we need that $$ f(a) = \sum a_i = (n-1)a + \frac{P}{a^{n-1}} = S \, . $$ This equation has a solution because $f$ is continuous on $(0, \infty)$ with $$ \min_{a > 0} f(a) = f(\sqrt[n]P) = n \sqrt[n]P \le S $$ ...


24

Most humans find it much easier to understand proofs written in a natural language (assuming, of course, it is a language that they are fluent in) with logic symbols kept to a minimum. You may find it easy to deal with a proof of four or five lines written in logic symbols, but I suspect it would be quite a different matter with a $100$-page proof. Natural ...


24

When first think about machine I can think of many useless machines which behave arbitrarily. For example, a random collection of levers, wheels, and pulleys connected by ropes doesn't do anything (i connect these pieces randomly). It seems that the number of useless machines is much more than that of useful machine. So, does the concept of machine naturally ...


23

Here's a weak solution. Using your favorite bijection, replace $[0,1]$ with the Cantor group $2^\mathbb N$ of infinite binary sequences. Then let $f(x,y)=x+y$. That is, just use the group operation: pointwise XOR.


21

The equality in $dy = f\, dx$ is very misleading, because strictly speaking it's not true. To see why, note that $dy$ is a differential $1$-form defined on $\Bbb{R}^2$, which means for each $p \in \Bbb{R}^2$, $dy_p : T_p \Bbb{R}^2 \to \Bbb{R}$ is linear. Similarly, $dx$ is also a differential $1$-form on $\Bbb{R}^2$. Let's for the sake of concreteness say $f:...


21

The best option here is to read Dedekind's original Continuity and irrational numbers or its exposition in Hardy's A Course of Pure Mathematics. Expansion of numbers systems can be seen driven by algebraic needs as one moves along the path $\mathbb {N}\to\mathbb{Z} \to\mathbb {Q} $. But the next step to $\mathbb {R} $ is totally non-algebraic and not based ...


21

This is false in general, take $x=\frac{\ln 3}{\ln 2}$, then $2^x=e^{x\ln 2}=e^{\ln 3}=3\in\mathbb{Q}$ but $x\notin\mathbb{Q}$. Otherwise, there would exist $p,q\geqslant 1$ coprimes such that $x=\frac{p}{q}$, that is to say $q\ln 3=p\ln 2$ and thus $3^q=2^p$ which is not because $2$ and $3$ are coprimes.


21

Your conjecture is wrong. $\Vert(x, y)\Vert = \max(|x|, |x+y|)$ is a norm on $\Bbb R^2$, but the ball $D = \{ (x, y) \mid \Vert(x, y)\Vert \le 1 \}$ is not symmetric with respect to the $x$-axis or $y$-axis: (Image created with Desmos.) One problem in your argument is that the set $\{t \in \mathbb{R}^n : \|t\| \neq \|t_x\|\}$ is not closed, and therefore ...


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