41 votes

Are there other kinds of bump functions than $e^\frac1{x^2-1}$?

There is even simpler example from this book of Loring Tu. You simply start with $$f(t)=\left\{\begin{array}{lr} 0&t\leqslant 0\\ e^{-1/t}&t>0 \end{array}\right..$$ Then you define $$g(t)=\...
Fallen Apart's user avatar
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29 votes
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Is there any simple set of properties which uniquely characterizes differentiation?

Claim Let $D : C^\infty(\mathbb{R}) \to C^\infty(\mathbb{R})$ be a nonzero linear operator such that $D(fg) = D(f)g + fD(g)$ and $D(f \circ g) = (D(f) \circ g)D(g)$ for all $f,g \in C^\infty(\mathbb{R}...
diracdeltafunk's user avatar
19 votes

Why does a function has to be differentiable so many times to be considered smooth?

It seems like your question is about how the term "smooth" in math corresponds to the term "smooth" as it's used in English. A once-differentiable function might "look smooth&...
Jair Taylor's user avatar
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15 votes

Are there other kinds of bump functions than $e^\frac1{x^2-1}$?

Here's how you can generate as many different kinds of bump functions as you want, for whatever definition of "kind" you may have: Start with any function $f(x)$ that grows faster than all ...
Abhimanyu Pallavi Sudhir's user avatar
15 votes
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If $f(x)$ is smooth and odd, must $f(x)/x$ be smooth?

$g$ will always be well defined at $0$ because: $$g(0) = \lim_{x\to 0} \frac{f(x)}{x} = \lim_{x\to 0} \frac{f(x)-0}{x-0} \equiv f'(0)$$ by the definition of the derivative. One can even prove via ...
Ninad Munshi's user avatar
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13 votes

Why does a function has to be differentiable so many times to be considered smooth?

I don't know whether this corresponds with the original coinage of the term, but I think it's fair to say that in this usage mathematicians have not sat down and thought, "OK, we want a ...
Steve Jessop's user avatar
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13 votes
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Particularly nice bump function

Similarly as in Bump function you can define $$ F(x)={\begin{cases}e^{-{\frac {1}{x}}}&{\text{if }}x>0,\\0&{\text{if }}x\leq 0,\end{cases}} $$ and $$ G(x)={\frac {F(1-x)}{F(x)+F(1-x)}},\...
Martin R's user avatar
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12 votes
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Completion of local frames for the tangent bundle of a smooth manifold

$\textbf{Hint: }$ After you choose vectors $v_{k+1},\dots,v_n$, you have linearly independent vectors on $T_pM$. Extend $\{v_{k+1},\dots,v_n\}$ around a neighbourhood of $p$, say to constant local ...
Kelvin Lois's user avatar
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12 votes
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Is there a simple way to characterize the smooth functions without using the derivative?

Consider data $(V,D)$ with the following properties: $$\tag1 V\text{ is a subalgebra of }\Bbb R^{\Bbb R}$$ $$\tag2 (\mathbf 1\colon x\mapsto1)\in V$$ $$\tag3 (\operatorname{id}\colon x\mapsto x)\in V$$...
Hagen von Eitzen's user avatar
11 votes
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Is there a smooth, preferably analytic function that grows faster than any function in the sequence $e^x, e^{e^x}, e^{e^{e^x}}...$

The link-only comment of metamorphy is actually a full answer, and gives an analytic function, instead of the smooth function of Michael's answer. Instead of hiding behind a link to Wikipedia, I give ...
Calvin Khor's user avatar
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11 votes
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Can a smooth map be extended to an open set?

Yes, and the construction is by means of a partition of unity. That is, consider a cover of $X$ by open sets $U_\alpha$, and for each $\alpha$, find a function $g_\alpha:U_\alpha\to\mathbb R^m$ such ...
Alex Ortiz's user avatar
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10 votes
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Does there exist smooth functions $f_i,g_i \in C^{\infty} (\mathbb R)$ such that $\sin (xy) = \sum\limits_{i = 1}^{n}f_i (x) g_i (y)$ for all $x,y\ $?

Taking $y = 1,..., n+1$ implies that $\sin x, \sin 2x, ... , \sin((n+1)x)$ are a subset an $n$-dimensional space of functions. This will contradict that these functions are linearly independent.
Zarrax's user avatar
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9 votes
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Can a non-zero smooth $f: [a,b] \rightarrow \mathbb{R}_{\geq 0}$ have infinitely many zeroes?

Let's start with $f_1(x) = \sin(1/x)$, which has infinitely many zeroes, but it's clearly not smooth (not even continuous). You can square it to get a nonnegative function $f_2(x) = \sin^2(1/x)$. As ...
Michał Miśkiewicz's user avatar
9 votes
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Why are coordinate maps diffeomorphisms? (A seeming counterexample?)

You have two different copies of $\Bbb{R}$ and they’re playing different roles, but you’re not distinguishing them. The first, lets call it $M$ to emphasize it is the ‘abstract manifold’ has a global ...
peek-a-boo's user avatar
  • 54.2k
8 votes

Can a function be smooth at a single point?

The Weierstrass function is continuous everywhere and nowhere differentiable (see this proof). Given parameters $0<a<1$ and $b$ an odd integer such that $ab>1+\frac{3}{2}\pi$, this is $$w(x)=\...
obscurans's user avatar
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8 votes
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Dirac's $\delta$ distribution smooth approximation

After reading the comments in hyperplane's answer, here's a very standard theorem: Let $\zeta\in L^1(\Bbb{R}^n)$, and define $c:=\int_{\Bbb{R}^n}\zeta(x)\,dx$, and for each $t>0$, let $\zeta_t(x)=\...
peek-a-boo's user avatar
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8 votes
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How to compute $\displaystyle\lim_{n\to\infty} \frac1{n+1}\sum_{k=1}^n \left|X+\frac kn\right|-\left|X-\frac kn\right|$?

$\frac1{n}\sum_{k=1}^n \left|x \pm\frac kn\right|$ are Riemann sums for $\int_0^1 |x \pm t|\, dt$, respectively, so that $$ \lim_{n \to \infty} f_n(x) = f(x) = \int_0^1 (|x+t|-|x-t|) \, dt \, . $$ ...
Martin R's user avatar
  • 112k
7 votes

Are there other kinds of bump functions than $e^\frac1{x^2-1}$?

Even if that was not specifically asked for by the OP's question, two features that some people are looking for in a bump function are : a roughly flat top "plateau", as in @FallenApart answer a ...
Noé AC's user avatar
  • 1,588
7 votes

Constant Rank Theorem for Manifolds with Boundary

Lee's Introduction to Smooth Manifolds deals with the case of local immersions for manifolds with boundary in Theorem 4.15. So let us suppose that $F:\mathbb{H}^m \rightarrow \mathbb{R}^n$ has $\...
user886204's user avatar
7 votes
Accepted

Can $\frac{\mathrm d}{\mathrm dx}$ increase the support of a function?

It is the contrary. The derivative "decreases" the support. If $x\notin{\rm supp}f$, then for some neighborhood $N_x$ of $x$, $f\equiv0$ in $N_x\implies f'\equiv0$ in $N_x\implies N_x\subset({\rm supp}...
trisct's user avatar
  • 5,211
7 votes

Is there a smooth, preferably analytic function that grows faster than any function in the sequence $e^x, e^{e^x}, e^{e^{e^x}}...$

This gives details to my comment: Let $\{f_k\}_{k=1}^{\infty}$ be a sequence of functions $f_k:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy the following for all $k \in \{1, 2, 3, ...\}$: $f_k(x)&...
Michael's user avatar
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7 votes
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Existence of Taylor series and Analyticity

No. A function $f(x)$ is analytic at a point $x_0$ if its Taylor series exists at $x_0$ and converges to $f(x)$ in a neighborhood of $x_0$. The Taylor series may exist but not converge, or it may ...
Qiaochu Yuan's user avatar
7 votes
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Discontinuous function in Smooth infinitesimal analysis

tl;dr It is not the case that all $x \in R$ satisfy either $x = \varepsilon$ or $x \neq \varepsilon$, so your function is not well-defined. Let $R$ denote the real line of Smooth Infinitesimal ...
Z. A. K.'s user avatar
  • 10.9k
7 votes

Particularly nice bump function

A function that fulfill your requirements is the Rvachëv function $R(x)$, which is a displaced version of the first lobe of the Fabius function $F(x)$: $$R(x)=\begin{cases} 0,\quad |x|\geq 1,\\ F(x+1),...
Joako's user avatar
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6 votes
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Connecting smooth functions in a smooth way

There is a well known result of Borel that will be useful here. Borel's theorem on power series: Let $x_0\in \mathbb R.$ Given $a_0,a_1, \dots \in \mathbb R,$ there exists $f\in C^\infty(\mathbb R)$ ...
zhw.'s user avatar
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6 votes
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Can a function be smooth at a single point?

As I suggested yesterday, I will try to fix the example Paul Sinclair made in this post. Let $f_0$ be a continuous and nowhere differentiable function. We define recursively for $n\geq 0$ as follows: ...
Severin Schraven's user avatar
6 votes

Can a function be smooth at a single point?

I think I have a simpler solution. It doesn't depend on nowhere differenitable functions and seems less computational to me. For $f\in C^n=C^n(\mathbb R),$ define $$\|f\|_n = \max_{|k|\le n} \sup_{\...
zhw.'s user avatar
  • 105k
6 votes
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Why the multiplication map of a Lie group has constant rank?

It has constant rank because it is a submersion. It is a submersion because if you fix a coordinate, it is a diffeomorphism. More precisely, fix $(x,y)\in G\times G$, we wish to show that $d_{(x,y)}...
Maxime Ramzi's user avatar
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6 votes
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Is the space of maps which satisfy this vanishing condition finite-dimensional?

Let us write the condition $(df_x)^T(h(x))=0$ more explicitly. We can write $$ (df_x)^T=\bigg(\nabla f^1(x)\,\bigg|\,...\,\bigg|\,\nabla f^k(x)\bigg), $$ where $\nabla f^i(x)$ is the column vector ...
PozzPlot's user avatar
  • 1,283
6 votes
Accepted

Continuous function for day/night with night being $c$ times longer than day

In the notation of my "answer" to my still-unanswered question Almost simple Hermite interpolation, we can compute a quintic polynomial $l_a(x)$ such that $l_a(0) = l_a(1) = 0,$ $l_a(a) = 1,$...
Calum Gilhooley's user avatar

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