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# Use case for Latin Modern Math (LM Math)

Since July 2011, there has been a beta release of LM Math. Some people seem to be quite excited about this. The release announcement states that this release

...completes the modernization of the Computer Modern family of typefaces .... the modernization was incomplete without the math fonts of the Comupter Modern family

I understand many reasons to use the lmodern text fonts:

As far as I can tell, none of these advantages apply to the LM Math fonts. So my question is, why did the GUST team feel the need to produce new math fonts? As far as I can tell the only advantage that they quote on their release announcement is that they are "modern", but I fail to find "new" a compelling argument. Why would I use a newer, less-tested font (that, if it is anything like the other LM fonts, is likely less well-hinted than bluesky/ams/y&y CM)?

I understand that this in an early beta release, so it is not suprising that there is not much (any?) documentation or rationale accompanying the fonts, but I hope someone here knows the reason for them.

EDIT: I looked a bit harder, and I saw that my installation of the lm package, version 2.004, 30.10.2009 already includes math fonts. In LaTeX I can use them by default \usepackage{lmodern}, or not by \usepackage[nomath]{lmodern}. The README says that these 20 math fonts are "at the moment a duplicate of PL math fonts", which presumably refers to these fonts. So, presumably, this new LM Math release is to replace these PL math fonts with something else. So, in addition to my question about what was wrong with the bluesky/ams/y&y CM math I'd also be interested to know: what was wrong with the PL math fonts?

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Well, you can use LM Math in Microsoft Office or other softwares. That's an advantage for many people. –  Leo Liu Aug 11 '11 at 10:16
@Leo, true, but the excited posting from Marcin Borkowski I linked at the beginning describes that as a "downside", so I'm presuming there is also some good advantages for TeX users. –  Lev Bishop Aug 11 '11 at 21:16
Hah, just found this post. Well, it was kind of a joke from my side, please don't take it too seriously! –  mbork Mar 7 '12 at 15:15

The LM Math fonts are available in OTF Math format. As such they are usable with unicode-math and ConTeXt Mk IV. So you can finally use LM when using Xe/LuaTeX + unicode-math or ConTeXt (as LM + STIX Math just looks horrible). Presumably they also added additional symbols as Unicode defines lots of math symbols that aren't available in the CM math fonts.

As I read the announcement, you can't use the fonts without an OTF capable TeX engine. So with pdflatex, you have to continue to use the old fonts anyway (see the comments for a qualification of this statement).

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Actually, pdftex can use OTF (although it will not subset the glyphs). For example, the LM-Math can be accessed in plain tex with \pdfmapline{+test LMMath-Regular <<LMMATH-v0903.otf}\font\tst=test\tst testing\bye. (I constructed test.tfm with LCDF typetools and the commandline otftotfm -n test --no-encoding --no-map LMMATH-v0903.otf). Presumably that's not enough to use the mathematical features of the font, but it does show that you can use otf with pdftex. –  Lev Bishop Aug 10 '11 at 20:08
The Xe/LuaTeX + unicode-math part, I think I understand. The ConTeXt part, I do not. As far as I can tell ConTeXt always supported LM and math: for example \starttext $x^2$\stoptext produces a PDF that uses the font LMMathItalic12-Regular (I used context live to check this). –  Lev Bishop Aug 10 '11 at 20:55
@Lev: though pdflatex can use OTF fonts in very basic way, it doesn't make use of any layout tables including MATH table in LM Math, and I don't think it can access more than 256 glyphs from the font. Regarding ConTeXt, context live is "MK II" which is pdflatex based, however "MK IV" can indeed use LM type1 math fonts, but it is kind of a hack until the proper OTF font is available (and the texlive 2011 version is now using it). –  Khaled Hosny Aug 10 '11 at 21:49
@Lev: I have to admit that I don't know much about TeX's fonts handling. For me the reason I'm happy to see the LM Math OTF fonts is that it gives me more choice when using unicode-math (and I use that quite a lot). Also having the TeX math font available might increase the interest in Unicode-enabled math typesetting, which in turn will hopefully cause the maintainers of related packages and engines to fix bugs faster. –  Caramdir Aug 11 '11 at 0:03

Let me provide a different take on this. This has more to do with the fact why unicode enabled math typesetting is useful, but does illustrate why opentype math fonts help.

ConTeXt allows you to export a TeX document to XML. In such an export math is exported to MathML.

Using an opentype math font allows you to look at the final glyph that is inserted into the pdf, and use that in the exported MathML. For example, $\Gamma$, which is normally typeset upright, is exported to $<mi>0x0393</mi>$. However, if you set an appropriate option so that uppercase Greek letters are typeset in italic, then $\Gamma$ is exported to $<mi>0x1D6E4</mi>$. The export functions do not care how the mapping from $\Gamma$ to the appropriate font glyph is done. It simply uses the opentype slot of the final glyph in the exported MathML.

Of course, it is still possible to implement such an export with Type1 fonts. However, in that case, the export functions will need to understand the mappings from macros (or input characters) to unicode slots. Thus, the macro package needs to implement two mappings: (i) input character to font glyph; (ii) input character to opentype location. With an opentype math font, these two mappings collapse to one. For this reason, even type1 math fonts are converted into a fake opentype math font on the fly in ConTeXt MkIV.

It is also possible (although not currently implemented) to embed the converted MathML into the pdf file so that a screen reader will read the mathematical expression correctly. This will improve the accessibility of TeX documents. Another advantage is that if you copy a math expression from a pdf and pastes it in a text file, the resulting MathML will be pasted. Wrap it around an appropriate environment, and then you can compile the expression with TeX again and recover the original math expression. (ConTeXt has a module for typesetting MathML; I haven't checked whether the LaTeX XML packages allow typesetting of MathML). This is similar to using an appropriate cmap file so that you can copy and paste text.

Using an opentype math fonts makes it easier to implement such features. And, of course, if you want a good looking pdf, you need a good (and preferably free) math fonts. Currently they are in short supply; so a new opentype math font is a welcome addition.

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Have you ever wished if you could just convert a logo you got in a PDF file into an SVG, so you can put up on your website and it would look good on Retina screens too?

Have you ever marvelled the beautifully formatted research papers in PDFs, but you didn’t want to invest into learning TeX or get tangled in installing TeX?

Nix to the rescue!

It’s worth emphasising that you DO NOT need NixOS to do the following. A macOS or most Linux kernel based systems will work.

You don’t have to be a guru to use Nix AND on top of that your knowledge is OS agnostic.

If you happen to hit some obstacles with Nix in your specific use-case, you can always go back to solving the problem in an OS / distro specific way, because having Nix on your system will not interfere with your underlying OS setup (beyond eating up extra disk space).

# PDF to SVG conversion

Here is a real-life example from the other day:

I had to convert a logo sent to me in a PDF file into SVG. I googled for “open source command line pdf to svg” Turns out there is indeed a pdf2svg tool, so I just did

$nix-shell -p pdf2svg --run 'pdf2svg --help' these paths will be fetched (5.74 MiB download, 31.79 MiB unpacked): /nix/store/2li666k7669n3i8n0wbgda41znxzif6s-libjpeg-turbo-1.5.3 ... Usage: pdf2svg <in file.pdf> <out file.svg> [<page no>]  then problem solved: $ nix-shell -p pdf2svg  --run 'pdf2svg logo.pdf logo.svg'

Everyone should know at least this much about Nix…

I forgot to mention in my previous article one more practicality. Install a faster, nicer package search utility with nix-env -i nox, so you can find package names easily:

$nox pdf2 1 pdf2djvu-0.9.9 (nixpkgs.pdf2djvu) Creates djvu files from PDF files 2 pdf2odt-20170207 (nixpkgs.pdf2odt) PDF to ODT format converter 3 pdf2svg-0.2.3 (nixpkgs.pdf2svg) PDF converter to SVG format 4 pdf2xml (nixpkgs.pdf2xml) 5 python2.7-pdf2image-0.1.14 (nixpkgs.python27Packages.pdf2image) A python module that wraps the pdftoppm utility to convert PDF to PIL Image object 6 python2.7-PyPDF2-1.26.0 (nixpkgs.python27Packages.pypdf2) A Pure-Python library built as a PDF toolkit 7 python3.6-pdf2image-0.1.14 (nixpkgs.python36Packages.pdf2image) A python module that wraps the pdftoppm utility to convert PDF to PIL Image object 8 python3.6-PyPDF2-1.26.0 (nixpkgs.python36Packages.pypdf2) A Pure-Python library built as a PDF toolkit Packages to install:  # Turn a Markdown document into PDF Here comes another, maybe even more powerful example. We wrote a grant application in Notion. Exported the document into markdown and turned it into a professional looking PDF: nix-shell -p texlive.combined.scheme-small pandoc \ --run 'pandoc grant.md -V geometry:margin=1in -o grant.pdf'  My colleague did it the 1st time on his Arch Linux, then after I amended the Notion page, did the export again and I ran the exact same command on my macOS and got an updated PDF. Now it’s your turn; give it a try! Let’s pick the ClojureScript README.md from github for example and see it for yourself how would it look if you would have printed it! $ curl -s https://raw.githubusercontent.com/clojure/clojurescript/master/README.md | \
nix-shell -p texlive.combined.scheme-small pandoc \

format: plain


text:
“Quotation is a serviceable substitute for wit.”

Tags: , ,
-Oscar Wilde

“I am so clever that sometimes I don’t understand a single word of what I am saying.”

Tags: ,
-Oscar Wilde

“I did then what I knew how to do. Now that I know better, I do better.”

Tags: , , , , ,
-Maya Angelou

“Quotation is a serviceable substitute for wit.”

Tags: , ,
-Oscar Wilde

“I am so clever that sometimes I don’t understand a single word of what I am saying.”

Tags: ,
-Oscar Wilde

“I did then what I knew how to do. Now that I know better, I do better.”

Tags: , , , , ,
-Maya Angelou

“The measure of intelligence is the ability to change.”

Tags: , , , ,
-Albert Einstein

“It is not that I’m so smart. But I stay with the questions much longer.”

Tags: , ,
-Albert Einstein

“The measure of intelligence is the ability to change.”

Tags: , , , ,
-Albert Einstein

“It is not that I’m so smart. But I stay with the questions much longer.”

Tags: , ,
-Albert Einstein

“You are not entitled to your opinion. You are entitled to your informed opinion. No one is entitled to be ignorant.”

Tags: , , ,
-Harlan Ellison

“It is not that I’m so smart. But I stay with the questions much longer.”

Tags: , ,
-Albert Einstein

“The measure of intelligence is the ability to change.”

Tags: , , , ,
-Albert Einstein

“The man of knowledge must be able not only to love his enemies but also to hate his friends.”

Tags: , , , , , , ,
-Friedrich Nietzsche

“You might be poor, your shoes might be broken, but your mind is a palace.”

Tags: , ,
-Frank McCourt

“The measure of intelligence is the ability to change.”

Tags: , , , ,
-Albert Einstein

“It is not that I’m so smart. But I stay with the questions much longer.”

Tags: , ,
-Albert Einstein
|


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# (Fictive) story of a time where people reasoned only up to isomorphism

I seem to remember reading once a story that some mathematician had written to justify the use of categories, or isomorphisms or equivalences, or something like that. The story goes something like this:

Once upon a time, people did not know what equality was. Instead, they only thought about things up to isomorphism. For example, they did not say that two sets had the same number of elements, but that they were in bijection. Today with category theory go back to these roots.

Does anyone have an idea of who told this story and what the full story is?

• When we teach kids to count we create a bijection between the fingers on a hand and the objects that we are counting. Oct 6 '17 at 14:49
• That's the idea yes. I've actually seen a public lecture for non-mathematicians where the speaker started from there to the notion of bijection between infinite sets, to the proof of the fact that N is not in bijection with R. People seemed to really enjoy it. Oct 6 '17 at 14:54

This sounds an awful lot like TWF week 121:

To understand this, the following parable may be useful. Long ago, when shepherds wanted to see if two herds of sheep were isomorphic, they would look for an explicit isomorphism. In other words, they would line up both herds and try to match each sheep in one herd with a sheep in the other. But one day, along came a shepherd who invented decategorification. She realized one could take each herd and "count" it, setting up an isomorphism between it and some set of "numbers", which were nonsense words like "one, two, three,..." specially designed for this purpose. By comparing the resulting numbers, she could show that two herds were isomorphic without explicitly establishing an isomorphism! In short, by decategorifying the category of finite sets, the set of natural numbers was invented.

• See also the paper Categorification by Baez and Dolan. And don't overlook the same parable being told in Paul Taylor's Practical Foundations of Mathematics. Oct 6 '17 at 17:27
• @Todd have you a more precise reference as to where in Paul Taylor's book? It's a beast! Oct 7 '17 at 5:02
• @DavidRoberts "Bo Peep" makes her first appearance in Exercise 1.1, page 60, and later the assertion that any injective endofunction on a finite set is a bijection is called "Bo Peep's theorem". I've also heard Paul relate the parable at conference dinners. Oct 7 '17 at 8:02
• I think the story is Georges Ifrah's, available in English as From 1 to 0. "Let us imagine a shepherd, unable to count, who has a flock of sheep that he keeps in a cave. There are 55 of them, but he has no understanding of what "the number 55" means. He would like to be sure that all of them come back every evening. One day he has an idea. He sits down at the entrance of his cave and has the sheep go in one by one. Each time one of them passes, he makes a notch in a bone. When all the sheep have passed, he has made exactly 55 notches, without knowing the arithmetic meaning of that number." Oct 15 '17 at 22:00
• and a little later from the same book, now describing an archaeologist's discovery: "...without knowing it, the uneducated servant had repeated a procedure used by equally uneducated shepherds who lived in that region 3500 years earlier. The clay envelope had belonged to an accountant (who, unlike the shepherds, knew how to write). The shepherds had gone to see him before taking their master's flock to pasture. He made as many clay balls as there were animals in the flock and put them into the envelope, which was then closed." Oct 15 '17 at 22:08

As Todd has already written an answer for me, maybe I can claim it as an Answer:

Exercise 1.1 in my book Practical Foundations of Mathematics (CUP 1999) reads,

When Bo Peep got too many sheep to see where each one was throughout the day, she found a stick or a pebble for each individual sheep and moved them from a pile outside the pen to another inside, or vice versa, as the corresponding sheep went in or out. Then one evening there was a storm, and the sheep came home too quickly for her to find the proper objects, so for each sheep coming in she just moved ANY one object. She moved all of the objects, but she was still worried about the wolf. By the next morning she had satisfied herself that the less careful method of reckoning was sufficient. Explain her reasoning without the aid of numbers.

My reason for putting it in the book was to try to get some anthropologist to say when and what the original "proof" was, ie the cognitive basis of the long-universal belief that this is valid, which provides the justification of counting with numbers.

What I am trying to imagine is how one of our distant ancestors with the cognitive abilities but not the education of a mathematician might approach this. Of course they would not have formulated Peano Induction or Euclidean Infinite Descent. They would have an argument (that we would more or less accept as rigorous) that the result is true for three sheep, then four and five. After that they would use Induction in the naive epistemological sense to convince themselves that it is true for arbitrarily large sets.

It seems plausible that anyone who is challenged to come up with a proof would give the following (albeit non-constructive) proof.

Suppose that some sheep $s_0$ is missing in the evening. Then the pebble $p_0$ that served as its "name" in the morning was used for some other sheep $s_1$ in the evening. But then $s_1$ must have been named by a different pebble $p_1$ in the morning, which named yet another sheep $s_2$ in the evening. And so on. All of the sheep $s_0$, $s_1$, $s_2$, ... are different individuals, Likewise all of the pebbles $p_0$, $p_1$, $p_2$, ... But, essentially as Euclid says in Book VII, Proposition 31, this is impossible for a set of sheep.

By chance, this issue came up recently following an internal seminar by Martin Escardo in Birmingham (where I am now an Honorary Research Fellow). He was developing the foundations of arithmetic (in the setting of Homotopy Type Theory, though this was not essential) in such a way that $3\times 5=5\times 3$ could be seen in a primary-school fashion as transposing a rectangle.

He based this on a function $F:{\mathbb{N}}\to{\mathsf{Set}}$ with $F0=\emptyset$ and $F(\mathsf{succ} n)=F(n)\coprod{\mathbf{1}}$. In his treatment the most difficult Proposition is $$F(n)\cong F(m) \Longrightarrow n=m,$$ which he deduced from the Lemma $$X\coprod{\mathbf{1}}\cong Y\coprod{\mathbf{1}} \Longrightarrow X \cong Y.$$

This Lemma holds in any lextensive category, i.e. one with finite limits and stable disjoint coproducts. The Proposition follows using Peano induction, since then $$F(n+1)\cong F(m+1) \Longrightarrow F(n)\cong F(m) \Longrightarrow n=m \Longrightarrow n+1=m+1.$$

I think it is reasonable to suppose that Bo Peep could formulate this Lemma, but I feel it is more plausible that she would use the "infinite descent" argument than the Proposition.

I am not sure whether this answers the original question about justifying "the use of categories, or isomorphisms or equivalences", although maybe Martin's treatment of arithmetic does so.

• It seems that one could develop much of a theory of permutation groups before inventing the concept of number to ease the development. Does Martin Escardo's development resemble such a development of permutation groups? Gerhard "Who Needs These 'Numbers' Anyway?" Paseman, 2017.10.15. Oct 15 '17 at 16:21
• I see the temptation of permutation groups. Also, HoTT (Martin's setting) was partly motivated by (higher dimensional) groupoids. I think, however, Martin wanted to eliminate (as much as possible) the repeated use of induction in Giuseppe Peano's 1889 paper to prove things like commutativity, associativity and distributivity of $+$ and $\times$ in favour of (the more appropriate) primary-school set theory. (At least, Peano presumably used induction repeatedly: his treatment is extremely terse.) Oct 15 '17 at 17:53
• This suggests to me using Boolean circuit (flowchart) analysis to restructure proofs to minimize certain component usage (for example, building a three inverter circuit using ands, ors, and just two inverters). Do you know of anyone who has diagrammed the proofs and done a similar kind of analysis on them? Gerhard "Computer-Aided Proof Optimization Through Clones!" Paseman, 2017.10.15. Oct 15 '17 at 18:58
• Another simple argument (though I cannot speak as to its historical plausibility) which avoids explicit consideration of "infinite descent" might be like so: The sheep go out in some order, one pebble marked with corresponding sheep's name tossed onto the end of a row for each one, building up a row yea long. [cont] Oct 15 '17 at 21:39
• If the sheep were to return in precisely the opposite of the order in which they went out, then we could easily knock their corresponding pebble off the end of the row with each return; this would all play out in precise reverse of the build-up, and the row would return to its initial emptiness when, and precisely when, all the sheep had returned. BUT! [cont] Oct 15 '17 at 21:39


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## College Algebra (11th Edition)

$-3\sqrt{7}+2i$
Distribute signs to parentheses. $3\sqrt{7}-4\sqrt{7}+i-4i-2\sqrt{7}+5i$ Collect like terms, and add or subtract. $(3-4-2)\sqrt{7}+(1-4+5)i$ $-3\sqrt{7}+2i$


text:
# Is continuity a necessary condition to the next equivalence?

I had solved an exercise that assume continuity of a function. I solved without continuity, I think so. The problem is as follow: let $f:M\to N$ a continuous function, where $M$ and $N$ are metric spaces. Prove that the following two propositions are equivalent

I) If $(x_n)_n$ has no a convergent subsequence, then the same happens with the sequence $(f(x_n))_n$.

II) For every compact set $K\subseteq N$, $f^{-1}(K)$ is compact.

I solved as follows:

I) implies II). Suppose that $K$ is a compact set such that $f^{-1}(K)$ is not compact. Then there is a sequence $(x_n)_n\subset f^{-1}(K)$ that has no convergent subsequence, and by hypothesis $(f(x_n))_n\subset K$ has no convergent subsequence but this contradics the fact that $K$ is compact.

II) implies I). Let $(x_n)_n$ be a sequence that have no a convergent subsequence. Now suppose that $(f(x_n))_n$ have a convergent subsequence, $(f(x_{n_k}))_k$, and let $y$ be its limit. Is a result that the set $K=\{f(x_{n_1}), f(x_{n_2}),\dots,y\}$ is compact. Then $(x_{n_k})_k\subset f^{-1}(K)$, where $f^{-1}(K)$ is compact by hypothesis. Then $(x_{n_k})_k$ must have a convergent subsequence that is also a subsequence of $(x_n)_n$, contradiction.

I would like to know if there is something wrong on this proof.

Take $f:\mathbb{R}\to\mathbb{R}$, defined as $$f(x) = \begin{cases} x& x\neq 0\\ 1& x = 0 \end{cases}\text{.}$$ I) holds:

Take $(x_n)_n$ that has no accumulating point ($\Leftrightarrow$ has no convergent subsequence). Suppose that $(f(x_n))_n$ has an accumulating point $x$.

Case $x\neq 1$: is impossible because that would mean that there are infinitely many $n$s such that $f(x_n) = x_n\in (x-\varepsilon, x + \varepsilon)$ for all $\varepsilon > 0$ small enough so that $1\notin (x-\varepsilon, x + \varepsilon)$.

Case of $x= 1$: if there are finitely many $n\in\mathbb{N}$, such that $x_n = 0$, $f(x_n) = x_n$ for $n\geq n_0$ for some $n_0$, then we can repeat the upper argument. Otherwise, $0$ is an accumulating point of $(x_n)_n$. Hence, this case is also impossible.

II) does not hold: $f^{-1}([-1/2,1/2]) = [-1/2,1/2]\setminus\{0\}$.

That means that (at least) I does not imply II. You went wrong when you claim that Then there is a sequence $(x_n)_n\subset f^{-1}(K)$ that has no convergent subsequence: take $K = [-1/2,1/2]$. Since $f^{-1}(K) = K \setminus\{0\}$, we also have $(x_n)_n\subset K$.

If $f$ is continuous, II automatically holds, since $K$ must be a closed set.


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# Q.8 Half of a herd of deer are grazing in the field and three fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.

Let us assume the number of deer in the herd to be x.

Half of a herd of deer are grazing in the field i.e.,   $\frac{x}{2}$.

Now half are remaining.

Three fourths of the remaining are playing nearby, implies :

$\frac{3}{4}\times \frac{x}{2} = \frac{3x}{8}$

Remaining number of deer will be :

$\frac{x}{2} - \frac{3x}{8} = \frac{4x - 3x}{8} = \frac{x}{8}$

Given that rest are 9, so

$\frac{x}{8} = 9$

Hence number  of deer in the herd = 72

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text:
# The DTFT of a sequence $$x\left[ n \right]$$ is given by $$X\left( {{e^{iω }}} \right)$$. Since $$X\left( {{e^{iω }}} \right)$$ is period function of ω, it can be expressed classical Fourier series as,$$X\left( {{e^{iω }}} \right) = \sum\limits_{n = - \infty }^\infty {{C_n}{e^{in{ω _0}ω }}}$$Where ω0 is a fundamental frequency. Which of the following statement is correct?

1. $${\omega _0} = \pi$$$${C_n} = - x\left[ n \right]$$
2. $${\omega _0} = \pi$$$${C_n} = x\left[ -n \right]$$
3. ω0 = 1, $${C_n} = x\left[ -n \right]$$
4. ω0 = 1, $${C_n} = - x\left[ -n \right]$$

Option 3 : ω0 = 1, $${C_n} = x\left[ -n \right]$$

## Definition of Fourier Series MCQ Question 1 Detailed Solution

Concept:

If f(n) is a signal then its DTFT will be defined As.

$$F\left( {{e^{j\omega }}} \right) \leftrightarrow \;\mathop \sum \nolimits_{n = - \infty }^\infty f\left( n \right){e^{ - j\omega n}}$$

Analysis:

x(n) ↔ X(e)

$$X\left( {{e^{j\omega }}} \right) = \mathop \sum \nolimits_{n = - \infty }^\infty x\left( n \right){e^{ - j\omega n}} \ ..(i)$$

It is given that:

$$X\left( {{e^{j\omega }}} \right) = \mathop \sum \nolimits_{n = - \infty }^\infty {C_n}{e^{jn{\omega _0}\omega }} \ ..(ii)$$

On Replacing (n) by (-n) in equation (i)

$$X\left( {{e^{j\omega }}} \right) = \mathop \sum \nolimits_{n = - \infty }^\infty x\left( { - n} \right){e^{j\omega n}} \ ..(iii)$$

On comparing equation (ii) and (iii)

Cn = x(-n)

ω0 = 1

# A periodic function of half wave symmetry is necessarily

1. an even function
2. an odd function
3. neither odd nor even
4. either odd or even

Option 4 : either odd or even

## Definition of Fourier Series MCQ Question 2 Detailed Solution

Half-wave even symmetry:

$$f\left( t \right) = f\left( {t \pm \frac{T}{2}} \right)$$

where T is the period of the periodic waveform

Half-wave odd symmetry:

$$f\left( t \right) = \; - f\left( {t \pm \frac{T}{2}} \right)$$

Even function symmetry:

f(t) = f(-t)

Graphically, the waveform is symmetrical about the vertical axis (dependent axis) as shown:

Odd function symmetry:

f(t) = -f(-t)

or

f(-t) = -f(t)

Graphically, the waveform is symmetric about the origin.

Important Points

• Signals with even symmetry, contain cosine terms in their Fourier series representation. It may or may not have a D.C term, but sine terms will always the zero.
• Odd function symmetry contains only sine terms
• Half-wave even symmetry contains even harmonics only.
• Half-wave odd symmetry contains odd harmonics only

# The Fourier series expansion of sgn(cos(t)) has _________. Where sgn represents the signum function

1. only sine terms with even harmonies.
2. only sine terms with odd harmonies.
3. only cosine terms with odd harmonies.
4. only cosine terms with even harmonies.

Option 3 : only cosine terms with odd harmonies.

## Definition of Fourier Series MCQ Question 3 Detailed Solution

Concept:

Even Symmetricities:

⇒ FS expansion of an even signal does not contents sin terms

ODD symmetricities:

⇒ FS expansion of an odd signal contains only sin term.

Half wave symmetricities:

$$\begin{array}{*{20}{c}} {x\left( t \right) = - x\left( {t + \frac{{{T_o}}}{2}} \right)}\\ {\begin{array}{*{20}{l}} \downarrow \\ {{c_n} = - {c_n}.{e^{\frac{{Jn{\omega _0}{T_0}}}{2}}}} \end{array}} \end{array}$$

Time shifting

$$\begin{array}{*{20}{c}} \;\\ {x\left( {t - {t_0}} \right)\rightarrow{c_n}e} \end{array} - jn{\omega _0}{t_0}$$

$$1 = - {e^{\frac{{Jn{\omega _0}{T_o}}}{2}}},\frac{{{\omega _o}{T_o}}}{2} = \pi$$

1 = -eJnπ

1 + (e)n = 0

∴ eπj = -1

1 + (-1)n = 0      -----(1)

Equation (1) will be satisfied for n = odd integer

0  = ODD harmonic

∴ F.S expansion of an H.W.S contains only ODD harmonic

Even + H.W.S:

⇒ F.S expansion of even H.W.S signal contains cos terms with ODD harmonic.

Cos terms with ODD harmonic

Odd + H.W.S

⇒ F.S expansion of odd H.W.S. signal contains sin terms with ODD harmonics.

Analysis:

$$\begin{array}{l} \;sgn\left( {cos{\rm{\;}}t} \right){\rm{\;}} = {\rm{\;}}1;{\rm{\;}}cost{\rm{\;}} > {\rm{\;}}0\\ = {\rm{\;}} - 1;{\rm{\;}}cost{\rm{\;}} < {\rm{\;}}0 \end{array}$$

It represents square wave, which is an even and half-wave symmetry function, it contains cosine terms for all odd harmonics.

# The waveform is given by v(t) = 10 sin (2π100t). What will be the magnitude of the second harmonic in its Fourier series representation?

1. 0 V
2. 20 V
3. 100 V
4. 200 V

Option 1 : 0 V

## Definition of Fourier Series MCQ Question 4 Detailed Solution

Concept:

The exponential representation of the Fourier series is given by:

$$x\left( t \right) = \;\mathop \sum \limits_{ - \infty }^\infty {C_k}\;{e^{jk{ω _0}t}}$$

Where, Ck in the Fourier coefficient given by:

$${C_k} = \frac{1}{{{T_0}}}\smallint x\left( t \right)\;{e^{ - jk{ω _0}t}}dt$$

Given:

v(t) = 10 sin (2π100t)

ω0 = 200π

v(t) = (10 ej200πt - 10 e-j200πt )/2j

∴ The second harmonic = 0

# A 4 kHz square wave of duty cycle 50% and p-p (0.2V to +0.2V) is applied as input to 20 kHz narrow bandpass filter which is followed by an amplifier of voltage gain 20. Find out the frequency components present in the output. Given that cut off frequencies of a narrow bandpass filter is 20 kHz and 40 kHz.

1. 12 kHz, 20 kHz, 28 kHz
2. 20 kHz, 28 kHz, 36 kHz
3. 20 kHz, 28 kHz, 32 kHz
4. 20 kHz, 24 kHz, 28 kHz

Option 2 : 20 kHz, 28 kHz, 36 kHz

## Definition of Fourier Series MCQ Question 5 Detailed Solution

$$f = \frac{1}{T}$$

$$T = \frac{1}{f} = \frac{1}{4}msec$$

The given

Narrow bandpass filter

T = 0.25 msec

Now applying Fourier series formula on x(t), we get:

$${a_k} = \frac{1}{T}\mathop \smallint \limits_T^\; x\left( t \right)\;{e^{ - jk{\omega _0}t}}dt$$

$${a_k} = \frac{1}{T}\mathop \smallint \limits_0^{T/2} 0.2\;\;\;{e^{ - jk{\omega _0}t}}dt - \frac{1}{T}\mathop \smallint \limits_{T/2}^T 0.2\;{e^{ - jk{\omega _0}t}}dt$$

$${a_k} = \frac{{0.2}}{T}\left[ {\frac{{{e^{ - jk{\omega _0}t}}}}{{ - jk{\omega _0}}}} \right]_0^{T/2} - \frac{{0.2}}{T}\left[ {\frac{{{e^{ - jk{\omega _0}t}}}}{{ - jk{\omega _0}}}} \right]_{\frac{T}{2}}^T$$

$${a_k} = \frac{{0.2}}{{jk{\omega _0}T}}\left[ {1 - {e^{ - \frac{{jk{\omega _0}T}}{2}}}} \right] - \frac{{0.2}}{{jk{\omega _0}T}}\left[ { - {e^{ - jk{\omega _0}T}} + {e^{ - jk{\omega _0}T/2}}} \right]$$

ω0T = 2π

$${e^{ - jk{\omega _0}T}} = {e^{ - jk2\pi }} = {\left( 1 \right)^k} = 1$$

$${e^{ - \frac{{jk{\omega _0}T}}{2}}} = {e^{ - jk\pi }} = {\left( { - 1} \right)^k}$$

$${a_k} = \frac{{0.2}}{{j2\pi k}}\left[ {1 - {{\left( { - 1} \right)}^k}} \right] - \frac{{0.2}}{{j2\pi k}}\left[ {{{\left( { - 1} \right)}^k} - 1} \right]$$

$${a_k} = \frac{{2 \times 0.2}}{{j2\pi k}}\left[ {1 - {{\left( { - 1} \right)}^k}} \right]$$

$${a_k} = \frac{{0.2}}{{j\pi k}}\left[ {1 - {{\left( { - 1} \right)}^k}} \right]$$

$${a_k} = \left\{ {\begin{array}{*{20}{c}} {\frac{{0.4}}{{j\pi k}}\;\;\;\;\;;\;\;\;k\;\;odd}\\ {0\;\;\;\;;\;\;\;\;k\;even} \end{array}} \right.$$

So, odd components are a1, a3, a5, a7, a9…., and so on.

Since the filter passes frequencies only between 20 kHz and 40 kHz, the components lying between this band of frequency will be passed by the filter.

The fundamental frequency given is 4 kHz.

a1 = 4 kHz

a3 = 12 kHz

$$\left. {\begin{array}{*{20}{c}} {{a_5} = 20\;kHz}\\ {{a_7} = 28\;kHz}\\ {{a_9} = 36\;kHz} \end{array}} \right\}$$

These 3 components of 20 kHz, 28 kHz, and 36 kHz will be passed and the rest of the components will be rejected by the filter.

a11 → 44 kHz

a13 → 52 kHz

And so on

Option 2 is correct.

# The exponential Fourier series representation of a continuous-time periodic signal x(t) is defined as:$$x(t)=\displaystyle\sum_{k=-\infty}^\infty a_k e^{jkω_0 t}$$Where ω0 is the fundamental angular frequency of x(t) and the coefficients of the series are ak. The following information is given about x(t) and ak.I. x(t) is real and even having a fundamental period of 6II. The average value of x(t) is 2III. $$a_k = \left\lbrace \begin{matrix} k, \ \ 1 \le k \le 3 \\\ 0, \ \ \ k > 3 \end{matrix} \right.$$The average power of the signal x(t) (rounded off to one decimal place) is _______.

## Definition of Fourier Series MCQ Question 6 Detailed Solution

Concept:

The Exponential Fourier Series (EFS) of the function g(t) is represented as:

$$g\left( t \right) = \mathop \sum \limits_{n = - \infty }^\infty {C_n}{e^{jn{\omega _0}t}}$$

Where Cn is the EFS coefficient and it is defined by:

$${C_n} = \frac{1}{T}\mathop \smallint \limits_0^T g\left( t \right){e^{ - jn{\omega _0}t}}dt$$

Parseval’s theorem

This tells that the total power in a periodic signal is equal to the sum of the squared amplitudes of each harmonic.

$$\frac{1}{T}\mathop \smallint \limits_0^T {\left| {x\left( t \right)} \right|^2}dt = \mathop \sum \limits_{n = - \infty }^\infty {\left| {{C_n}} \right|^2}$$

Calculation:

The given signal is defined as:

$$x\left( t \right) = \mathop \sum \limits_{n = - \infty }^\infty {a_k}{e^{jk{\omega _0}t}}$$

Where ak is the Fourier Series coefficient.

Given that the x(t) is real and even. So, this implies that the spectrum also real and even.

$$a_k = \left\lbrace \begin{matrix} k, \ \ 1 \le k \le 3 \\\ 0, \ \ \ k > 3 \end{matrix} \right.$$

Average power is:

$$\frac{1}{T}\mathop \smallint \limits_0^T {\left| {x\left( t \right)} \right|^2}dt = \mathop \sum \limits_{n = - 3}^3 {\left| {{a_k}} \right|^2}$$

$${P_{avg}} = {\left| {{a_{ - 3}}} \right|^2} + {\left| {{a_{ - 2}}} \right|^2} + {\left| {{a_{ - 1}}} \right|^2} + {\left| {{a_0}} \right|^2} + {\left| {{a_1}} \right|^2} + {\left| {{a_2}} \right|^2} + {\left| {{a_3}} \right|^2}$$

Pavg = 9 + 4 + 1 + 4 + 1 + 4 + 9

Pavg = 18 + 8 + 2 + 4

Pavg = 32 W

# Fourier series of an odd periodic function contains only

1. odd harmonics
2. even harmonics
3. cosine harmonics
4. sine harmonics

Option 4 : sine harmonics

## Definition of Fourier Series MCQ Question 7 Detailed Solution

Fourier series:

The Fourier series for the function f(x) in the interval α < x < α + 2π is given by

$$f\left( x \right) = \frac{{{a_o}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\cos nx + \mathop \sum \limits_{n = 1}^\infty {b_n}\sin nx$$

where

$${a_o} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)dx;\;{a_n} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)\cos nxdx;\;{b_n} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)\sin nxdx$$

An even function is any function f such that f(-x) = f(x)

Example: cos x, sec x, x2, x4, x6 …….., x-2, x-4 ……..

An odd function is any function f such that f(-x) = -f(x)

Example: sin x, tan x, cosec x, cot x, n, x3 ……., x-1, x-3 ……..

$$\mathop \smallint \limits_{ - L}^L f\left( x \right)dx = \left\{ {\begin{array}{*{20}{c}} {2\mathop \smallint \limits_0^L f\left( x \right)dx,\;\;when\;f\left( x \right)\;is\;an\;even\;function}\\ {0,\;\;when\;f\left( x \right)\;is\;an\;odd\;function} \end{array}} \right.$$

When f is an even periodic function of period 2L, then its Fourier series contains only cosine (include possibly, the constant term) terms.

$$f\left( x \right) = \frac{{{a_o}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\frac{{\cos n\pi x}}{L}$$

$${a_o} = \frac{1}{L}\mathop \smallint \limits_{ - L}^L f\left( x \right)dx = \frac{2}{L}\mathop \smallint \limits_0^L f\left( x \right)dx$$

$${a_n} = \frac{1}{L}\mathop \smallint \limits_{ - L}^L f\left( x \right)\cos \frac{{n\pi x}}{L}dx = \frac{2}{L}\mathop \smallint \limits_0^L f\left( x \right)\cos \frac{{n\pi x}}{L}dx$$

When f is an odd periodic function of period 2L, then its Fourier series contains only sine terms.

$$f\left( x \right) = \mathop \sum \limits_{n = 1}^\infty {b_n}\sin \frac{{n\pi x}}{L}$$

$${b_n} = \frac{1}{L}\mathop \smallint \limits_{ - L}^L f\left( x \right)\sin \frac{{n\pi x}}{L}dx = \frac{2}{L}\mathop \smallint \limits_0^L f\left( x \right)\sin \frac{{n\pi x}}{L}dx$$

• Based on the symmetry of the periodic signal given we can conclude the Fourier series coefficients and the type of harmonics that are present in the signal given.
• With the Fourier series, the non-sinusoidal periodic waveform can be converted into the sinusoidal wave.

The below table shows the type of Fourier coefficient terms corresponding to the symmetry.

 Symmetry Condition a0 an bn Terms Even x(t) = x(-t) Non zero Non zero Zero DC and Cosine Odd x(t) = - x(-t) Zero Zero Non zero Only Sine Half wave $$x\left( t \right) = - x\left( {t \pm \frac{T}{2}} \right)$$ Zero Zero ; n even Non zero; n odd Zero; n even Non zero; n odd Only odd harmonics

# If a function f(t) is odd then its Fourier coefficients satisfy

1. a0 ≠ 0, an ≠ 0 bn = 0
2. a0 ≠ 0 an ≠ 0 bn ≠ 0
3. a0 = 0 an ≠ 0 bn ≠ 0
4. a0 = 0 an = 0 bn ≠ 0

Option 4 : a0 = 0 an = 0 bn ≠ 0

## Definition of Fourier Series MCQ Question 8 Detailed Solution

Fourier series:

The Fourier series for the function f(x) in the interval α < x < α + 2π is given by

$$f\left( x \right) = \frac{{{a_o}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\cos nx + \mathop \sum \limits_{n = 1}^\infty {b_n}\sin nx$$

where

$${a_o} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)dx;\;{a_n} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)\cos nxdx;\;{b_n} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)\sin nxdx$$

An even function is any function f such that f(-x) = f(x)

Example: cos x, sec x, x2, x4, x6 …….., x-2, x-4 ……..

An odd function is any function f such that f(-x) = -f(x)

Example: sin x, tan x, cosec x, cot x, n, x3 ……., x-1, x-3 ……..

$$\mathop \smallint \limits_{ - L}^L f\left( x \right)dx = \left\{ {\begin{array}{*{20}{c}} {2\mathop \smallint \limits_0^L f\left( x \right)dx,\;\;when\;f\left( x \right)\;is\;an\;even\;function}\\ {0,\;\;when\;f\left( x \right)\;is\;an\;odd\;function} \end{array}} \right.$$

When f is an even periodic function of period 2L, then its Fourier series contains only cosine (include possibly, the constant term) terms.

$$f\left( x \right) = \frac{{{a_o}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\frac{{\cos n\pi x}}{L}$$

$${a_o} = \frac{1}{L}\mathop \smallint \limits_{ - L}^L f\left( x \right)dx = \frac{2}{L}\mathop \smallint \limits_0^L f\left( x \right)dx$$

$${a_n} = \frac{1}{L}\mathop \smallint \limits_{ - L}^L f\left( x \right)\cos \frac{{n\pi x}}{L}dx = \frac{2}{L}\mathop \smallint \limits_0^L f\left( x \right)\cos \frac{{n\pi x}}{L}dx$$

When f is an odd periodic function of period 2L, then its Fourier series contains only sine terms.

$$f\left( x \right) = \mathop \sum \limits_{n = 1}^\infty {b_n}\sin \frac{{n\pi x}}{L}$$

$${b_n} = \frac{1}{L}\mathop \smallint \limits_{ - L}^L f\left( x \right)\sin \frac{{n\pi x}}{L}dx = \frac{2}{L}\mathop \smallint \limits_0^L f\left( x \right)\sin \frac{{n\pi x}}{L}dx$$

• Based on the symmetry of the periodic signal given we can conclude the Fourier series coefficients and the type of harmonics that are present in the signal given.
• With the Fourier series, the non-sinusoidal periodic waveform can be converted into the sinusoidal wave.

The below table shows the type of Fourier coefficient terms corresponding to the symmetry.

 Symmetry Condition a0 an bn Terms Even x(t) = x(-t) Non zero Non zero Zero DC and Cosine Odd x(t) = - x(-t) Zero Zero Non zero Only Sine Half wave $$x\left( t \right) = - x\left( {t \pm \frac{T}{2}} \right)$$ Zero Zero ; n even Non zero; n odd Zero; n even Non zero; n odd Only odd harmonics

# The DTFT of a sequence $$x\left[ n \right]$$ is given by $$X\left( {{e^{iω }}} \right)$$. Since $$X\left( {{e^{iω }}} \right)$$ is period function of ω, it can be expressed classical Fourier series as,$$X\left( {{e^{iω }}} \right) = \sum\limits_{n = - \infty }^\infty {{C_n}{e^{in{ω _0}ω }}}$$Where ω0 is a fundamental frequency. Which of the following statement is correct?

1. $${\omega _0} = \pi$$$${C_n} = - x\left[ n \right]$$
2. $${\omega _0} = \pi$$$${C_n} = x\left[ -n \right]$$
3. ω0 = 1, $${C_n} = x\left[ -n \right]$$
4. ω0 = 1, $${C_n} = - x\left[ -n \right]$$

Option 3 : ω0 = 1, $${C_n} = x\left[ -n \right]$$

## Definition of Fourier Series MCQ Question 9 Detailed Solution

Concept:

If f(n) is a signal then its DTFT will be defined As.

$$F\left( {{e^{j\omega }}} \right) \leftrightarrow \;\mathop \sum \nolimits_{n = - \infty }^\infty f\left( n \right){e^{ - j\omega n}}$$

Analysis:

x(n) ↔ X(e)

$$X\left( {{e^{j\omega }}} \right) = \mathop \sum \nolimits_{n = - \infty }^\infty x\left( n \right){e^{ - j\omega n}} \ ..(i)$$

It is given that:

$$X\left( {{e^{j\omega }}} \right) = \mathop \sum \nolimits_{n = - \infty }^\infty {C_n}{e^{jn{\omega _0}\omega }} \ ..(ii)$$

On Replacing (n) by (-n) in equation (i)

$$X\left( {{e^{j\omega }}} \right) = \mathop \sum \nolimits_{n = - \infty }^\infty x\left( { - n} \right){e^{j\omega n}} \ ..(iii)$$

On comparing equation (ii) and (iii)

Cn = x(-n)

ω0 = 1

# A periodic function of half wave symmetry is necessarily

1. an even function
2. an odd function
3. neither odd nor even
4. either odd or even

Option 4 : either odd or even

## Definition of Fourier Series MCQ Question 10 Detailed Solution

Half-wave even symmetry:

$$f\left( t \right) = f\left( {t \pm \frac{T}{2}} \right)$$

where T is the period of the periodic waveform

Half-wave odd symmetry:

$$f\left( t \right) = \; - f\left( {t \pm \frac{T}{2}} \right)$$

Even function symmetry:

f(t) = f(-t)

Graphically, the waveform is symmetrical about the vertical axis (dependent axis) as shown:

Odd function symmetry:

f(t) = -f(-t)

or

f(-t) = -f(t)

Graphically, the waveform is symmetric about the origin.

Important Points

• Signals with even symmetry, contain cosine terms in their Fourier series representation. It may or may not have a D.C term, but sine terms will always the zero.
• Odd function symmetry contains only sine terms
• Half-wave even symmetry contains even harmonics only.
• Half-wave odd symmetry contains odd harmonics only


text:
IntMath Home » Forum home » Fourier Series » Discontinous functions - fourier series and taylor series

Discontinous functions - fourier series and taylor series [Solved!]

My question

can a discontinous function can be develop in fourier series and taylar series?

Relevant page

2. Full Range Fourier Series

What I've done so far

Just wondering because I tried to do a few

X

can a discontinous function can be develop in fourier series and taylar series?
Relevant page

2. Full Range Fourier Series

What I've done so far

Just wondering because I tried to do a few

Re: Discontinous functions - fourier series and taylor series

Hi Gaurav

A function has to be continuous before you can expand it using Fourier or Taylors, and for Fourier the function needs to be periodic.

With Taylors though, you can limit the domain to the continuous part and then expand for that. For example, on this page:

1. Taylor Series

The example on f(x)=log x shows that even though it is discontinuous at x = 0 and not defined for x < 0, we can still do a Taylor's series expansion.

Hope that makes sense

X

Hi Gaurav

A function has to be continuous before you can expand it using Fourier or Taylors, and for Fourier the function needs to be periodic.

With Taylors though, you can limit the domain to the continuous part and then expand for that. For example, on this page:

1. Taylor Series

The example on f(x)=log x shows that even though it is discontinuous at x = 0 and not defined for x < 0, we can still do a Taylor's series expansion.

Hope that makes sense

Re: Discontinous functions - fourier series and taylor series

Yes it does. thanks

X

Yes it does. thanks


text:
# Simultaneous detection of ozone and nitrogen dioxide by oxygen anion chemical ionization mass spectrometry: a fast-time-response sensor suitable for eddy covariance measurements

We report on the development, characterization, and field deployment of a fast-time-response sensor for measuring ozone (inline-formulaO3) and nitrogen dioxide (inline-formulaNO2) concentrations utilizing chemical ionization time-of-flight mass spectrometry (CI-ToFMS) with oxygen anion (inline-formula $M3inlinescrollmathmlchem{\mathrm{normal O}}_{normal 2}^{-}$ 17pt16ptsvg-formulamathimg26d368ad2bb5d49e9b215b03d31563a9 amt-13-1887-2020-ie00001.svg17pt16ptamt-13-1887-2020-ie00001.png ) reagent ion chemistry. We demonstrate that the oxygen anion chemical ionization mass spectrometer (Ox-CIMS) is highly sensitive to both inline-formulaO3 (180 counts sinline-formula−1 pptvinline-formula−1) and inline-formulaNO2 (97 counts sinline-formula−1 pptvinline-formula−1), corresponding to detection limits (inline-formula3σ, 1 s averages) of 13 and 9.9 pptv, respectively. In both cases, the detection threshold is limited by the magnitude and variability in the background determination. The short-term precision (1 s averages) is better than 0.3 % at 10 ppbv inline-formulaO3 and 4 % at 10 pptv inline-formulaNO2. We demonstrate that the sensitivity of the inline-formulaO3 measurement to fluctuations in ambient water vapor and carbon dioxide is negligible for typical conditions encountered in the troposphere. The application of the Ox-CIMS to the measurement of inline-formulaO3 vertical fluxes over the coastal ocean, via eddy covariance (EC), was tested during the summer of 2018 at Scripps Pier, La Jolla, CA. The observed mean ozone deposition velocity (inline-formulavd(inline-formulaO3)) was 0.013 cm sinline-formula−1 with a campaign ensemble limit of detection (LOD) of 0.0027 cm sinline-formula−1 at the 95 % confidence level, from each 27 min sampling period LOD. The campaign mean and 1 standard deviation range of inline-formulaO3 mixing ratios was inline-formula41.2±10.1 ppbv. Several fast ozone titration events from local NO emissions were sampled where unit conversion of inline-formulaO3 to inline-formulaNO2 was observed, highlighting instrument utility as a total odd-oxygen (inline-formula $M23inlinescrollmathmlchem{\mathrm{normal O}}_{x}\phantom{\rule{0ex}{0ex}}=\phantom{\rule{0ex}{0ex}}{\mathrm{normal O}}_{normal 3}\phantom{\rule{0ex}{0ex}}+\phantom{\rule{0ex}{0ex}}{\mathrm{normal NO}}_{normal 2}$ 75pt13ptsvg-formulamathimgd5b8caa296edd90f045d885f0525d6c6 amt-13-1887-2020-ie00002.svg75pt13ptamt-13-1887-2020-ie00002.png ) sensor. The demonstrated precision, sensitivity, and time resolution of this instrument highlight its potential for direct measurements of inline-formulaO3 ocean–atmosphere and biosphere–atmosphere exchange from both stationary and mobile sampling platforms.

### Zitieren

Zitierform:

Novak, Gordon A. / Vermeuel, Michael P. / Bertram, Timothy H.: Simultaneous detection of ozone and nitrogen dioxide by oxygen anion chemical ionization mass spectrometry: a fast-time-response sensor suitable for eddy covariance measurements. 2020. Copernicus Publications.

### Zugriffsstatistik

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### Rechte

Rechteinhaber: Gordon A. Novak et al.

Nutzung und Vervielfältigung:


text:
Please note that internally all infill criteria are minimized. So for some of them, we internally compute their negated version, e.g., for EI or also for CB when the objective is to be maximized. In the latter case mlrMBO actually computes the negative upper confidence bound and minimizes that.

setMBOControlInfill(control, crit = NULL,
interleave.random.points = 0L, filter.proposed.points = NULL,
filter.proposed.points.tol = NULL, opt = "focussearch",
opt.restarts = NULL, opt.focussearch.maxit = NULL,
opt.focussearch.points = NULL, opt.cmaes.control = NULL,
opt.ea.maxit = NULL, opt.ea.mu = NULL, opt.ea.sbx.eta = NULL,
opt.ea.sbx.p = NULL, opt.ea.pm.eta = NULL, opt.ea.pm.p = NULL,
opt.ea.lambda = NULL, opt.nsga2.popsize = NULL,
opt.nsga2.generations = NULL, opt.nsga2.cprob = NULL,
opt.nsga2.cdist = NULL, opt.nsga2.mprob = NULL,
opt.nsga2.mdist = NULL)

## Arguments

control [MBOControl] Control object for mbo. [MBOInfillCrit] How should infill points be rated. See infillcrits for an overview of available infill criteria or implement a custom one via makeMBOInfillCrit.# Default is “(lower) confidence bound” (see makeMBOInfillCritCB). [integer(1)] Add interleave.random.points uniformly sampled points additionally to the regular proposed points in each step. If crit="random" this value will be neglected. Default is 0. [logical(1)] Design points located too close to each other can lead to numerical problems when using e.g. kriging as a surrogate model. This may solve the 'leading minor of order ...' error during model fit. This parameter activates or deactivates a heuristic to handle this issue. If TRUE, proposed points whose distance to design points or other current candidate points is smaller than filter.proposed.points.tol, are replaced by random points. If enabled, the column entry for prop.type is set to “random_filter” in the resulting opt.path, so you can see whether such a replacement happened. This does only work for numeric parameter sets without any discrete parameters. Default is FALSE. [numeric(1)] Tolerance value filtering of proposed points. We currently use a maximum metric to calculate the distance between points. Default is 0.0001. [character(1)] How should SINGLE points be proposed by using the surrogate model. Possible values are: “focussearch”: In several iteration steps the parameter space is focused on an especial promising region according to infill criterion. “cmaes”: Use CMA-ES (function cmaes from package cmaesr to optimize infill criterion. If all CMA-ES runs fail, a random point is generated instead and a warning informs about it. “ea”: Use an (mu+1) EA to optimize infill criterion. “nsga2”: NSGA2 for multi obj. optimizations. Needed for mspot. Default is “focussearch”. Alternatively, you may pass a function name as string. [integer(1)] Number of independent restarts for optimizer of infill criterion. If opt == "cmaes" the first start point for the optimizer is always the currently best point in the design of already visited points. Subsequent starting points are chosen according to the CMA-ES restart strategy introduced by Auger and Hansen. For details see the corresponding paper in the references and the help page of the underlying optimizer cmaes. Default is 3. [integer(1)] For opt = "focussearch": Number of iteration to shrink local focus. Default is 5. [integer(1)] For opt = "focussearch": Number of points in each iteration of the focus search optimizer. Default is 1000. [list] For opt = "cmaes": Control argument for cmaes optimizer. For the default see the help page of the underlying optimizer cmaes. [integer(1)] For opt = "ea": Number of iterations / generations of EA. Default is 500. [integer(1)] For opt = "ea": Population size of EA. The default is 10 times the number of parameters of the function to optimize. [numeric(1)] For opt = "ea": Distance parameter of crossover distribution , see sbx_operator. Default is 15. [numeric(1)] For opt = "ea": Probability of 1-point crossover, see sbx_operator. Default is 0.5. [numeric(1)] For opt = "ea": Distance parameter of mutation distribution, see pm_operator. Default is 15. [numeric(1)] For opt = "ea": Probability of 1-point mutation, see pm_operator. Default is 0.5. [numeric{1}] For opt.ea = "ea". Number of children generated in each generation. Default is 1. [numeric{1}] For opt.multiobj.method = "nsga2". Population size of nsga2. Default is 100. [numeric{1}] For opt.multiobj.method = "nsga2". Number of populations for of nsga2. Default is 50. [numeric{1}] For opt.multiobj.method = "nsga2". nsga2 param. Default is 0.7. [numeric{1}] For opt.multiobj.method = "nsga2". nsga2 param. Default is 5. [numeric{1}] For opt.multiobj.method = "nsga2". nsga2 param. Default is 0.2. [numeric{1}] For opt.multiobj.method = "nsga2". nsga2 param. Default is 10.

## Value

[MBOControl].

Other MBOControl: makeMBOControl, setMBOControlMultiObj, setMBOControlMultiPoint, setMBOControlTermination


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G = C2×S3×D5order 120 = 23·3·5

Direct product of C2, S3 and D5

Aliases: C2×S3×D5, C15⋊C23, C61D10, C101D6, C30⋊C22, D305C2, D15⋊C22, (C6×D5)⋊3C2, (C5×S3)⋊C22, C51(C22×S3), (C3×D5)⋊C22, (S3×C10)⋊3C2, C31(C22×D5), SmallGroup(120,42)

Series: Derived Chief Lower central Upper central

 Derived series C1 — C15 — C2×S3×D5
 Chief series C1 — C5 — C15 — C3×D5 — S3×D5 — C2×S3×D5
 Lower central C15 — C2×S3×D5
 Upper central C1 — C2

Generators and relations for C2×S3×D5
G = < a,b,c,d,e | a2=b3=c2=d5=e2=1, ab=ba, ac=ca, ad=da, ae=ea, cbc=b-1, bd=db, be=eb, cd=dc, ce=ec, ede=d-1 >

Subgroups: 260 in 64 conjugacy classes, 28 normal (14 characteristic)
C1, C2, C2, C3, C22, C5, S3, S3, C6, C6, C23, D5, D5, C10, C10, D6, D6, C2×C6, C15, D10, D10, C2×C10, C22×S3, C5×S3, C3×D5, D15, C30, C22×D5, S3×D5, C6×D5, S3×C10, D30, C2×S3×D5
Quotients: C1, C2, C22, S3, C23, D5, D6, D10, C22×S3, C22×D5, S3×D5, C2×S3×D5

Character table of C2×S3×D5

 class 1 2A 2B 2C 2D 2E 2F 2G 3 5A 5B 6A 6B 6C 10A 10B 10C 10D 10E 10F 15A 15B 30A 30B size 1 1 3 3 5 5 15 15 2 2 2 2 10 10 2 2 6 6 6 6 4 4 4 4 ρ1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 trivial ρ2 1 1 1 1 -1 -1 -1 -1 1 1 1 1 -1 -1 1 1 1 1 1 1 1 1 1 1 linear of order 2 ρ3 1 -1 -1 1 -1 1 1 -1 1 1 1 -1 1 -1 -1 -1 -1 -1 1 1 1 1 -1 -1 linear of order 2 ρ4 1 -1 -1 1 1 -1 -1 1 1 1 1 -1 -1 1 -1 -1 -1 -1 1 1 1 1 -1 -1 linear of order 2 ρ5 1 -1 1 -1 -1 1 -1 1 1 1 1 -1 1 -1 -1 -1 1 1 -1 -1 1 1 -1 -1 linear of order 2 ρ6 1 -1 1 -1 1 -1 1 -1 1 1 1 -1 -1 1 -1 -1 1 1 -1 -1 1 1 -1 -1 linear of order 2 ρ7 1 1 -1 -1 1 1 -1 -1 1 1 1 1 1 1 1 1 -1 -1 -1 -1 1 1 1 1 linear of order 2 ρ8 1 1 -1 -1 -1 -1 1 1 1 1 1 1 -1 -1 1 1 -1 -1 -1 -1 1 1 1 1 linear of order 2 ρ9 2 2 0 0 -2 -2 0 0 -1 2 2 -1 1 1 2 2 0 0 0 0 -1 -1 -1 -1 orthogonal lifted from D6 ρ10 2 -2 0 0 2 -2 0 0 -1 2 2 1 1 -1 -2 -2 0 0 0 0 -1 -1 1 1 orthogonal lifted from D6 ρ11 2 2 0 0 2 2 0 0 -1 2 2 -1 -1 -1 2 2 0 0 0 0 -1 -1 -1 -1 orthogonal lifted from S3 ρ12 2 -2 0 0 -2 2 0 0 -1 2 2 1 -1 1 -2 -2 0 0 0 0 -1 -1 1 1 orthogonal lifted from D6 ρ13 2 -2 2 -2 0 0 0 0 2 -1-√5/2 -1+√5/2 -2 0 0 1-√5/2 1+√5/2 -1-√5/2 -1+√5/2 1-√5/2 1+√5/2 -1-√5/2 -1+√5/2 1-√5/2 1+√5/2 orthogonal lifted from D10 ρ14 2 -2 2 -2 0 0 0 0 2 -1+√5/2 -1-√5/2 -2 0 0 1+√5/2 1-√5/2 -1+√5/2 -1-√5/2 1+√5/2 1-√5/2 -1+√5/2 -1-√5/2 1+√5/2 1-√5/2 orthogonal lifted from D10 ρ15 2 2 2 2 0 0 0 0 2 -1-√5/2 -1+√5/2 2 0 0 -1+√5/2 -1-√5/2 -1-√5/2 -1+√5/2 -1+√5/2 -1-√5/2 -1-√5/2 -1+√5/2 -1+√5/2 -1-√5/2 orthogonal lifted from D5 ρ16 2 2 2 2 0 0 0 0 2 -1+√5/2 -1-√5/2 2 0 0 -1-√5/2 -1+√5/2 -1+√5/2 -1-√5/2 -1-√5/2 -1+√5/2 -1+√5/2 -1-√5/2 -1-√5/2 -1+√5/2 orthogonal lifted from D5 ρ17 2 2 -2 -2 0 0 0 0 2 -1+√5/2 -1-√5/2 2 0 0 -1-√5/2 -1+√5/2 1-√5/2 1+√5/2 1+√5/2 1-√5/2 -1+√5/2 -1-√5/2 -1-√5/2 -1+√5/2 orthogonal lifted from D10 ρ18 2 -2 -2 2 0 0 0 0 2 -1-√5/2 -1+√5/2 -2 0 0 1-√5/2 1+√5/2 1+√5/2 1-√5/2 -1+√5/2 -1-√5/2 -1-√5/2 -1+√5/2 1-√5/2 1+√5/2 orthogonal lifted from D10 ρ19 2 -2 -2 2 0 0 0 0 2 -1+√5/2 -1-√5/2 -2 0 0 1+√5/2 1-√5/2 1-√5/2 1+√5/2 -1-√5/2 -1+√5/2 -1+√5/2 -1-√5/2 1+√5/2 1-√5/2 orthogonal lifted from D10 ρ20 2 2 -2 -2 0 0 0 0 2 -1-√5/2 -1+√5/2 2 0 0 -1+√5/2 -1-√5/2 1+√5/2 1-√5/2 1-√5/2 1+√5/2 -1-√5/2 -1+√5/2 -1+√5/2 -1-√5/2 orthogonal lifted from D10 ρ21 4 4 0 0 0 0 0 0 -2 -1+√5 -1-√5 -2 0 0 -1-√5 -1+√5 0 0 0 0 1-√5/2 1+√5/2 1+√5/2 1-√5/2 orthogonal lifted from S3×D5 ρ22 4 -4 0 0 0 0 0 0 -2 -1+√5 -1-√5 2 0 0 1+√5 1-√5 0 0 0 0 1-√5/2 1+√5/2 -1-√5/2 -1+√5/2 orthogonal faithful ρ23 4 4 0 0 0 0 0 0 -2 -1-√5 -1+√5 -2 0 0 -1+√5 -1-√5 0 0 0 0 1+√5/2 1-√5/2 1-√5/2 1+√5/2 orthogonal lifted from S3×D5 ρ24 4 -4 0 0 0 0 0 0 -2 -1-√5 -1+√5 2 0 0 1-√5 1+√5 0 0 0 0 1+√5/2 1-√5/2 -1+√5/2 -1-√5/2 orthogonal faithful

Permutation representations of C2×S3×D5
On 30 points - transitive group 30T21
Generators in S30
(1 19)(2 20)(3 16)(4 17)(5 18)(6 21)(7 22)(8 23)(9 24)(10 25)(11 26)(12 27)(13 28)(14 29)(15 30)
(1 9 14)(2 10 15)(3 6 11)(4 7 12)(5 8 13)(16 21 26)(17 22 27)(18 23 28)(19 24 29)(20 25 30)
(1 19)(2 20)(3 16)(4 17)(5 18)(6 26)(7 27)(8 28)(9 29)(10 30)(11 21)(12 22)(13 23)(14 24)(15 25)
(1 2 3 4 5)(6 7 8 9 10)(11 12 13 14 15)(16 17 18 19 20)(21 22 23 24 25)(26 27 28 29 30)
(1 18)(2 17)(3 16)(4 20)(5 19)(6 21)(7 25)(8 24)(9 23)(10 22)(11 26)(12 30)(13 29)(14 28)(15 27)

G:=sub<Sym(30)| (1,19)(2,20)(3,16)(4,17)(5,18)(6,21)(7,22)(8,23)(9,24)(10,25)(11,26)(12,27)(13,28)(14,29)(15,30), (1,9,14)(2,10,15)(3,6,11)(4,7,12)(5,8,13)(16,21,26)(17,22,27)(18,23,28)(19,24,29)(20,25,30), (1,19)(2,20)(3,16)(4,17)(5,18)(6,26)(7,27)(8,28)(9,29)(10,30)(11,21)(12,22)(13,23)(14,24)(15,25), (1,2,3,4,5)(6,7,8,9,10)(11,12,13,14,15)(16,17,18,19,20)(21,22,23,24,25)(26,27,28,29,30), (1,18)(2,17)(3,16)(4,20)(5,19)(6,21)(7,25)(8,24)(9,23)(10,22)(11,26)(12,30)(13,29)(14,28)(15,27)>;

G:=Group( (1,19)(2,20)(3,16)(4,17)(5,18)(6,21)(7,22)(8,23)(9,24)(10,25)(11,26)(12,27)(13,28)(14,29)(15,30), (1,9,14)(2,10,15)(3,6,11)(4,7,12)(5,8,13)(16,21,26)(17,22,27)(18,23,28)(19,24,29)(20,25,30), (1,19)(2,20)(3,16)(4,17)(5,18)(6,26)(7,27)(8,28)(9,29)(10,30)(11,21)(12,22)(13,23)(14,24)(15,25), (1,2,3,4,5)(6,7,8,9,10)(11,12,13,14,15)(16,17,18,19,20)(21,22,23,24,25)(26,27,28,29,30), (1,18)(2,17)(3,16)(4,20)(5,19)(6,21)(7,25)(8,24)(9,23)(10,22)(11,26)(12,30)(13,29)(14,28)(15,27) );

G=PermutationGroup([[(1,19),(2,20),(3,16),(4,17),(5,18),(6,21),(7,22),(8,23),(9,24),(10,25),(11,26),(12,27),(13,28),(14,29),(15,30)], [(1,9,14),(2,10,15),(3,6,11),(4,7,12),(5,8,13),(16,21,26),(17,22,27),(18,23,28),(19,24,29),(20,25,30)], [(1,19),(2,20),(3,16),(4,17),(5,18),(6,26),(7,27),(8,28),(9,29),(10,30),(11,21),(12,22),(13,23),(14,24),(15,25)], [(1,2,3,4,5),(6,7,8,9,10),(11,12,13,14,15),(16,17,18,19,20),(21,22,23,24,25),(26,27,28,29,30)], [(1,18),(2,17),(3,16),(4,20),(5,19),(6,21),(7,25),(8,24),(9,23),(10,22),(11,26),(12,30),(13,29),(14,28),(15,27)]])

G:=TransitiveGroup(30,21);

C2×S3×D5 is a maximal subgroup of
D6⋊F5  C20⋊D6  D10⋊D6
C2×S3×D5 is a maximal quotient of
D205S3  D20⋊S3  D12⋊D5  D60⋊C2  D15⋊Q8  D6.D10  D125D5  C12.28D10  C20⋊D6  Dic5.D6  C30.C23  Dic3.D10  D10⋊D6

Matrix representation of C2×S3×D5 in GL4(𝔽31) generated by

 30 0 0 0 0 30 0 0 0 0 30 0 0 0 0 30
,
 1 0 0 0 0 1 0 0 0 0 1 10 0 0 9 29
,
 1 0 0 0 0 1 0 0 0 0 30 21 0 0 0 1
,
 19 1 0 0 11 30 0 0 0 0 1 0 0 0 0 1
,
 30 30 0 0 0 1 0 0 0 0 1 0 0 0 0 1
G:=sub<GL(4,GF(31))| [30,0,0,0,0,30,0,0,0,0,30,0,0,0,0,30],[1,0,0,0,0,1,0,0,0,0,1,9,0,0,10,29],[1,0,0,0,0,1,0,0,0,0,30,0,0,0,21,1],[19,11,0,0,1,30,0,0,0,0,1,0,0,0,0,1],[30,0,0,0,30,1,0,0,0,0,1,0,0,0,0,1] >;

C2×S3×D5 in GAP, Magma, Sage, TeX

C_2\times S_3\times D_5
% in TeX

G:=Group("C2xS3xD5");
// GroupNames label

G:=SmallGroup(120,42);
// by ID

G=gap.SmallGroup(120,42);
# by ID

G:=PCGroup([5,-2,-2,-2,-3,-5,168,2404]);
// Polycyclic

G:=Group<a,b,c,d,e|a^2=b^3=c^2=d^5=e^2=1,a*b=b*a,a*c=c*a,a*d=d*a,a*e=e*a,c*b*c=b^-1,b*d=d*b,b*e=e*b,c*d=d*c,c*e=e*c,e*d*e=d^-1>;
// generators/relations

Export

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ISSN 1000-1239 CN 11-1777/TP

• •

### 使用最大熵模型进行中文文本分类

1. (复旦大学计算机与信息技术系 上海 200433) (lironglu@163.net)
• 出版日期: 2005-01-15

### Using Maximum Entropy Model for Chinese Text Categorization

Li Ronglu, Wang Jianhui, Chen Xiaoyun, Tao Xiaopeng, and Hu Yunfa

1. (Department of Computing and Information Technology, Fudan University, Shanghai 200433)
• Online: 2005-01-15

Abstract: With the rapid development of World Wide Web, text classification has become the key technology in organizing and processing large amount of document data. Maximum entropy model is a probability estimation technique widely used for a variety of natural language tasks. It offers a clean and accommodable frame to combine diverse pieces of contextual information to estimate the probability of a certain linguistics phenomena. This approach for many tasks of NLP perform near state-of-the-art level, or outperform other competing probability methods when trained and tested under similar conditions. However, relatively little work has been done on applying maximum entropy model to text categorization problems. In addition, no previous work has focused on using maximum entropy model in classifying Chinese documents. Maximum entropy model is used for text categorization. Its categorization performance is compared and analyzed using different approaches for text feature generation, different number of feature and smoothng technique. Moreover, in experiments it is compared to Bayes, KNN and SVM, and it is shown that its performance is higher than Bayes and comparable with KNN and SVM. It is a promising technique for text categorization.


text:
PCA plot, less points than samples
2
0
Entering edit mode
13 months ago
eridanus ▴ 30

Hello. I am working with RNA seq data, and I am trying to do differential expression analysis. I am trying to make a PCA plot with Deseq2. Howevere although I have 12 samples, the PCA plot appears to have 11 dots. When I am making MDS plot with limma voom I see the same (11 dots instead of 12). I have checked dds object (and rlog transformed respectively) and its dimensions is 12. What could be the cause? Thank you!

dds <- DESeqDataSetFromMatrix(countData=countData, colData=colData, design=~group)
keep <- rowSums(counts(dds)) >= 10
dds <- dds[keep,]
rld <- rlog(dds)
plot_pca<-plotPCA(rld, intgroup = c("group"))
plotPCA(rld, intgroup = c("group"))
plot_pca <- plot_pca + geom_text(aes_string(label = "name"), color = "black")
print(plot_pca)

RNA-Seq • 432 views
0
Entering edit mode

This is my PCA plot. I checked and finally all the samples are included. I wonder if I should remove some samples from the differential expression analysis. WHat do you think?

2
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13 months ago
ATpoint 55k

Use returnData=TRUE for plotPCA and see whether there are 12 or 11 samples in the returned data.frame. If 12 then two points are overlapping perfectly.

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Entering edit mode

thank you. I tried and it is 12. I thought about overlapping, but then I added the labels, and I thought that if there was overlapping it would be obvious in the labels (that was not the case). what can I do about overlapping? Thank you a lot!

1
Entering edit mode

You can try to increase the number of genes used for PCA, maybe use 1000 and hope they separate better. ?plotPCA for details.

2
Entering edit mode

Or, I'd doublecheck that your 12 samples really are 12 samples, and that you don't have a goof where the same fastq was analyzed under two different names.

0
Entering edit mode

Good point. For this I guess a simple cor() would do and you should get a Pearson correlation of 1 in that case.

0
Entering edit mode

This is my PCA plot. The clustering is not that good, but in differential expression analysis I can see differences in expression. Should I remove some samples? What do you think? Thank you!

0
Entering edit mode

You can't remove samples just because they don't cluster how you like. It might be nice if you could figure out what's driving that first principle component.

0
Entering edit mode

Yes you are right. Thank you for your response!


text:
# Should {tex-core} questions always have stand-alone answers?

Questions tagged seem to demand answers in plain TeX (or at least plain TeX macros + some extensions as required), even if the same solution is entirely applicable in LaTeX, ConTeXt or some other situation. In questions such as reversing the order of expansion, we (currently) have one answer using expl3 and a second which loads some pre-packaged material to do the job. Both are good answers, but I wonder how other people see this. Should we encoruage answers to which are

1. Plain TeX (i.e. they finish `\bye`)
2. Stand-alone (no pre-packaged code used)

I ask because if people really want to learn TeX internals then they'll need to have things demonstrated in primitives (or at least primitives plus a small set of basic macros such as `\@firstofone`).

• I understand the concern. In some cases, still, it is much quicker to just reuse some code than to recode it (for completeness, the non-expl3 answer is in fact just using code that can be found elsewhere on `{TeX}`, although it is not very explicitly stated). – Bruno Le Floch Feb 16 '11 at 19:00
• @Bruno: Yes, I can see that too (after all, one reason I do use `expl3` is it avoids re-coding the same things again and again). I guess it partly depends on whether the question is asking for a particular idea to be explained at a low level. – Joseph Wright Feb 16 '11 at 19:16
• Aargh, three good answers: which one to accept! – Joseph Wright Feb 17 '11 at 7:50

My two cents: In the same way as a we want ConTeXt answers to LaTeX specific questions, I see no harm in non-plain-TeX answers to questions.

• Additionally, the question-asker (or someone coming via a search engine) might not be aware of pre-packaged solutions. – Caramdir Feb 16 '11 at 23:31

As I see it, `tex-core` refers fairly specifically to low-level TeX functionality (such as the behaviour of primitives) and it's not necessarily appropriate to ask ‘how can I program this complicated thing’ as part of that — I don't think it's that helpful to spread around lots of code written only in TeX primitives since as far as I know no-one is writing packages for iniTeX only.

So when a programming question comes up in this scenario to ask, say, ‘is it possible to expandably do some things?’, I think it's perfectly acceptable to say ‘yes—here is how expl3/ConTeXt/etoolboox does it’. Maybe in some cases it's helpful to restrict an answer to plain TeX only, but it shouldn't be mandated in my opinion.

To agree with Hendrik, it is good to see alternative ways of doing things and having Context/expl3 answers increases the reference value of the Q&A thread. But such answers should advertise that they do not meet the exact requirements the question was after.

Use your judgement. If someone is asking how to do something fundamental in a fairly format-agnostic way, giving them an answer only in terms of expl3 is probably not the best way to go (assuming you want your answer accepted!), but it won't likely cause much harm.

Answering in terms of e.g. `manmac.tex` or `texinfo.tex` would probably be better, though; they have much less code (or at least a smaller number of files) to wade through in order to see how something works. And, being basically self-contained packages, there is a much smaller amount of patching than in LaTeX; what remains consists mostly of overriding/renaming stuff from `plain`, and is greatly simplified by not having to worry about stepping on other packages' toes.

However, if the question is really about something big and complicated, suggest whatever you think best! ConTeXt, LaTeX3, even using LuaTeX to invoke METAPOST in-process, if that's what it takes!

(Of course, if the question makes it clear that the asker really just wants to know how to do something in LaTeX, maybe the `tex-core` tag should just be removed?)


text:
## Error = Irreducible Error + Bias + Variance

A fundamentally important concept from machine learning and statistics is the decomposition of prediction error.

Very informally: the total “wrongness” of a predictor is the sum of (at least) three subtypes of “wrongness”:

1. Irreducible error. Some prediction tasks are intrinsically more difficult than others. If a system is complex, chaotic, or only partially observed, then we will have a hard time making accurate predictions about it. But if a system is simple, linear, or fully observed, then we will have much less difficulty making accurate predictions. This irreducible error is a property of the prediction task itself—any prediction technique we might choose will be subject to it.
2. Bias. When we choose a model for a prediction task, we impose a set of corresponding assumptions—informally, we can call these assumptions biases. For example: if we choose to use linear regression, then we are in effect assuming that our predicted value $$y$$ is generated by a linear combination of input variables $$X$$: that is, $$y = Xw + b$$. If this assumption is false (and it almost always is), then any linear regression that we choose will make errors when we use it for prediction. This bias represents the predictor’s systemic mistakes.
3. Variance. Our predictor may have an element of randomness to it; it may make different predictions for the same inputs if we run it multiple times. This isn’t usually a good design for a predictor (unless it operates in an adversarial setting where some unpredictability is useful), so we usually implement fully deterministic predictors. However, even if our predictor is totally deterministic, it will be subject to the randomness in the dataset we fit it to. That is, if we replaced our dataset with a new dataset and re-fit the predictor, then it might make radically different predictions. We use the term variance to refer to the predictor’s randomness. It represents the predictor’s unreliability.

Some interesting phenomena that data scientists observe in practice:

• There is usually a tradeoff between a predictor’s bias and variance. That is: predictors with high bias usually have low variance; and predictors with high variance usually have low bias.
• The parameter which controls this tradeoff is the complexity of the predictor. I.e., simple predictors tend to have high bias/low variance, while complicated predictors usually have low bias/high variance.
• The optimal amount of predictor complexity is usually proportional to the amount of data available for training that predictor. That is: we are better off using simple models when we have little data; and complicated models when we have a lot of data.

## Human Fallibility

Most of us go through life with some concept of human fallibility echoing in our heads:

I’m not perfectI’m only human.

or

Humans make mistakes.

This is with good reason. History is replete with examples of poor judgment. And when we rely on humans—whether it be ourselves or someone else—we are often met with disappointment.

What happens when we view human fallibility through the lens of bias, variance, and irreducible error?

### Irreducible Error

Many human mistakes result from fundamental difficulties: the world is complex and mostly unobserved. There are unknown unknowns. We are subjected to certain unavoidable challenges as we navigate this uncertain environment.

One could invoke religious language, and call this kind of error original sin.

A classic concept from Stoic philosophy: we can divide our circumstances into those within our control, and those outside of our control.

Irreducible error stems from the circumstances that are out of our control.

Stoicism makes the useful suggestion that we focus on the things within our control, and not attach emotional significance to the things outside our control.

My pessimism inclines me to think irreducible error is large.

My optimism inclines me to think that humans do pretty well in spite of it, all things considered.

### High Bias

A lack of subtlety or nuance. Black and white thinking.

This is—perhaps to a small degree—within our control.

Some worthwhile questions to ask ourselves, in order to understand or address our personal biases:

• Should I trust my reflexive reaction in this situation? Or is deeper deliberation warranted?
• What incorrect assumptions am I taking for granted?
• What nuance am I failing to see in the present situation?
• What bullshit stories am I telling myself?
• Who can I disagree with in a constructive manner, to help correct my own bias?

### High Variance

I wager that any human who thinks much about their own behavior will eventually reach the same, puzzling conclusion: I fail to do the things I know I should be doingeven when I claim that I want to do them.

An undisciplined mind. Impulsiveness.

What Aristotle called “incontinence”.

Freudian view: the human mind is composed of conflicting subconscious wills. They may be poorly integrated—behavior will depend on which of these wills happens to be dominant at a given time.

Some worthwhile questions to ask, in order to examine our own “variances”:

• What values or ideals do I hold, consciously or unconsciously?
• Is my behavior consistent with my own values?
• What impulses drive my behavior? Can I improve my control over them?
• What is the smallest, simplest, easiest action I can take to behave more consistently with my values?

$$\blacksquare$$