220

Here is an approach. We give some preliminary results. The poly-Hurwitz zeta function The poly-Hurwitz zeta function may initially be defined by the series $$ \begin{align} \displaystyle \zeta(s\mid a,b) := \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s}(n+b)}, \quad \Re a >-1, \, \Re b >-1, \, \Re s>0. \tag1 \end{align} $$ This special ...


209

Haskell's answer does a great job of outlining conditions that a derivative $f'$ must satisfy, which then limits us in our search for an example. From there we see the key question: can we provide a concrete example of an everywhere differentiable function whose derivative is discontinuous on a dense, full-measure set of $\mathbb R$? Here's a closer look ...


163

EDIT: Pardon me, but it has been shown in the comments by robjohn and Michael that these are not linearly independent. Indeed:$$91a_1-10a_2=10$$     — Akiva Weinberger Think of a series of real numbers with decimal expansions like 0.1100110000110000001100000000110000000000110000000... 0....


163

$\phantom{}$ Dear MSE users, this is the new episode of Mister Feynman and Monsieur Laplace versus contour integration. Tonight we have a scary integral, but we may immediately notice that $$ \mathcal{L}(x^{-ia})(s) = s^{ia-1}\Gamma(1-ia),\qquad \mathcal{L}^{-1}\left(\frac{1}{x^2+bx+1}\right)(s) =\frac{e^{Bs}-e^{\overline{B}s}}{\sqrt{b^2-4}}$$ where $B$ is ...


160

The probabilistic way: This is $P[N_n\leqslant n]$ where $N_n$ is a random variable with Poisson distribution of parameter $n$. Hence each $N_n$ is distributed like $X_1+\cdots+X_n$ where the random variables $(X_k)$ are independent and identically distributed with Poisson distribution of parameter $1$. By the central limit theorem, $Y_n=\frac1{\sqrt{n}}(...


156

Consider this example: $$ 3-\frac12,\quad 5+\frac13,\quad 3-\frac14,\quad 5+\frac15,\quad 3-\frac16,\quad 5+\frac17,\quad 3-\frac18,\quad 5+\frac19,\quad\ldots\ldots $$ It alternates between something approaching $3$ from below and something approaching $5$ from above. The lim inf is $3$ and the lim sup is $5$. The inf of the whole sequence is $3-...


150

Here’s one to get things started. Let $U$ be a non-empty open subset of $\Bbb R$. For $x,y\in U$ define $x\sim y$ iff $\big[\min\{x,y\},\max\{x,y\}\big]\subseteq U$. It’s easily checked that $\sim$ is an equivalence relation on $U$ whose equivalence classes are pairwise disjoint open intervals in $\Bbb R$. (The term interval here includes unbounded ...


145

Definitions tend to be minimalistic, in the sense that they don't include unnecessary/redundant information that can be derived as a consequence. Same reason why, for example, an equilateral triangle is defined as having all sides equal, rather than having all sides and all angles equal.


139

The pictorial version. (But it is the same as your inequality version, actually.) Suppose you want to prove continuity at $a$. Choose points $b,c$ on either side. (This fails at an endpoint, in fact the result itself fails at an endpoint.) By convexity, the $c$ point is above the $a,b$ line, as shown: Again, the $b$ point is above the $a,c$ line, as ...


136

I suggest $$\sum_{k=1}^n \frac{1}{k^2} \leqslant 1 + \sum_{k=2}^n \frac{1}{k^2 - \frac14} = 1 + \sum_{k=2}^n \left(\frac{1}{k-\frac12} - \frac{1}{k+\frac12}\right) = 1 + \frac23 - \frac{1}{n+\frac12}.$$


135

Because that suggests that there might be functions which are discontinuous at $a$ for which it is still true that the limit$$\lim_{t\to0}\frac{f(a+t)-f(a)}t$$exists. Besides, why add a condition that it always holds?


128

$$ \begin{align} \sum_{n=1}^\infty\frac{\zeta(2n)}{2^{2n-1}} &=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac2{k^{2n}2^{2n}}\tag{1}\\ &=2\sum_{k=1}^\infty\sum_{n=1}^\infty\frac1{(4k^2)^n}\tag{2}\\ &=2\sum_{k=1}^\infty\frac1{4k^2-1}\tag{3}\\ &=\sum_{k=1}^\infty\left(\frac1{2k-1}-\frac1{2k+1}\right)\tag{4}\\[6pt] &=1\tag{5} \end{align} $$ ...


127

$$x^2+y^2-xy=\frac{x^2}{2}+\frac{y^2}{2}+\frac{(x-y)^2}{2}$$


125

Since $$\zeta(2n) = \frac{1}{(2n-1)!}\int_{0}^{\infty}\frac{x^{2n-1}}{e^x-1}\,dx $$ we have: $$\sum_{n=1}^{\infty}\frac{\zeta(2n)}{2^{2n-1}} = \int_{0}^{\infty}\frac{\sinh(x/2)}{e^x-1}\,dx =\frac12\int_{0}^{\infty}e^{-x/2}\,dx = \color{red}{1}.$$


124

A Parable (that makes no claims of historical accuracy): Suppose your mathematical universe contains only the rational numbers. You study rational-valued functions of rational numbers, and discover the $\varepsilon$-$\delta$ definition of continuity. Then you discover the real numbers, and the fact that the rationals are dense in the reals. It's natural to ...


123

The real "gist" of continuity, in its various forms, is that it's the "property that makes calculators and measurements useful". Calculators and measurements are fundamentally approximate devices which contain limited amounts of precision. Special functions, like those which are put on the buttons of a calculator, then, if they are to be useful, should have ...


120

Update 2/19/2018: It appears that this answer has received a lot of attention, which I'm very glad to know about. When you're reading through this answer and you're trying to learn about $\delta$-$\epsilon$ proofs for the first time, I would recommend skipping the sections labeled Addendum. on your first read. Please let me know of any other clarifications ...


120

An easy way to visualize why this can't be true is to try putting some points on a number line. Start with 1 point in [0, 1): 2 points in [1, 2): And so on: Now you have a sequence that grows to infinity but keeps getting closer together.


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 ...


113

if $a_k< a_{k+1}$ then $x := \frac{a_{k+1}+a_k}{2}$ is a rational number between those two, so no.


111

We can write: $$1+ x + x^2 = \frac{1-x^3}{1-x}$$ Therefore: $$f(x) = \frac{1-x}{1-x^3} $$ We can then expand this in powers of $x$: $$f(x) = (1-x)\sum_{k=0}^{\infty}x^{3 k}$$ which is valid for $\left|x\right|<1$. The coefficient of $x^{36}$ is thus equal to $1$, so the 36th derivative at $x = 0$ is $36!$ .


106

$\newcommand{\QQ}{\mathbb{Q}}$ Derivatives don't really go wrong, it's antiderivatives. (EDIT: Actually, the more I think about it, this is just a symptom. The underlying cause is that continuity on the rationals is a much weaker notion than continuity on the reals.) Consider the function $f : \QQ \to \QQ$ given by $$f(x) = \begin{cases} 0 & x < \pi \...


105

I'll try to give a soft answer to what I see as the spirit of the question, which is not why you get exactly log, but why the behaviour is different when integrating $x^{k}$ for $k=-1$. The way I see it is that the logarithm would actually be there for other powers of $x$ too, but it's "hidden" by the fact that in a geometric series one term "gobbles up" ...


101

Here's another intuitive justification. The obvious graphical intuition says that when $a \leq b \leq c$, then $\int_a^b f(x) dx + \int_b^c f(x) dx = \int_a^c f(x) dx$. If we want this formula to hold for arbitrary $a,b,c$, then we should be able to take $a=c$, so that $\int_a^b f(x) dx + \int_b^a f(x) dx = \int_a^a f(x) dx$. But $\int_a^a f(x) dx = 0$, so ...


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&...


95

Here is an approach that seems rather natural, based on applying the fundamental theorem of calculus successively to $f(x)$, $f'(t_1)$, $f''(t_2)$, etc.: \begin{eqnarray*} && f(x)=f(a)+\int_a^x f'(t_1)\,dt_1 \\&& = f(a)+ \int_a^x f'(a)\,dt_1 + \int_a^x \!\! \int_a^{t_1} f''(t_2)\,dt_2\,dt_1 \\&& = f(a)+ \int_a^x f'(a)\,dt_1 + \int_a^...


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