just get rid of the unnecessarily large fractions and the large integral symbol, and you'll be fine:\documentclass{article} \usepackage{mathtools} \begin{document} \begin{align} \theta_f &= \pi - 2 \int_{r_\mathrm{min}}^\infty \frac{b}{r^2 \bigl[1 - (\frac{b}{r})^2 - \frac{2 v(r)} {m v_0^2}\bigr]^{1/2}}\, \mathrm{d}r \nonumber \end{align} \end{document} note especially the usage of 1/2 in the exponent, removal of the spacing-killing group and meaningless \displaystyle, removal of all illogical \cfrac, and addition of the thin space before \mathrm{d}r. and finally, the change of \text to \mathrm in r_{min} as it should not follow the text font, but rather be always upright rm.also note that i personally would not use [] at all and ...

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"is it generally just bad to leave fractions in either the numerator or denominator?" yes, exactly. intuitively, the simplified form of an expression is supposed to be the, well, simplest form. exactly what constitutes "fully simplified" is subjective and can vary from person to person (is $\sqrt{2}\over 2$ simpler than $1\over\sqrt{2}$?), but some things are pretty much constant across the board, one being that "fractions inside fractions" are to be avoided unless getting rid of them would add significant complexity to the expression.

i am working with qgis and in particular with river basins. if you see my picture below, each label shows a particular basin but actually only the biggest is shown (because i think its area covers the areas of the smaller ones).e.g. 4103600 is the largest basin and its boundaries are shown correctly, but within it there are 7 more nested basins (i.e. 4103631, 4103650, 4103615, 4103610, 4103750, 4103751 and 4103570).how can i show all the basins' boundaries (i.e. nested and non-nested) with qgis?i believe there should be a trick for doing so but i don't know which one!

this definition struct a { struct b; }; defines a struct a with a declaration of a nested struct b1. the fully qualified name of b is a::b, you could say b is inside the "namespace' of a. then this:struct a::b : a { // note i added spaces }; is the definition of a::b, and the single : specifies that it is derived from a.now, the interesting part is a::b::a::b. let's dissect it:a::b names the nested structure.a::b::a accesses the injected class name a inside b. the injection is due to the inheritance.a::b::a::b names the nested structure b in a again.and you can continue ad-infinitum, or at least until your compiler meets its translation limit2.a fun intellectual exercise, but avoid like the plague in actual code. [class.qual]/1 explains how the lookup worksif the nested-...

you will be given a nested array. your program has to visualize the array.but.. how?for example, let's assume we have a nested array, like [["1","2"],[["1","2"],"3"],"4",[[[["5"]]]],"6"].this nested array can be visualised as:->1 ->2 -->1 -->2 ->3 >4 ---->5 >6 examplesinput 1: ["atom",["proton",["up quark", "up quark", "down quark"], "neutron", ["up quark", "down quark", "down quark"], "electron"]] output 1: >atom ->proton -->up quark -->up quark -->down quark ->neutron -->up quark -->down quark -->down quark ->electron input 2: [["1","2"],["3","4"]] output 2: ->1 ->2 ->3 ->4 rulesyou may use string (or other types which work like a nested array) as input.the...

a commutative ring can be embedded in a field iff it is an integral domain.indeed, if a ring can be embedded in a field then it cannot have zero divisors because fields cannot have zero divisors.conversely, every integral domain can be embedded in a field, namely, its field of fractions.

ever since i learned about fractions in elementary school, i've known how to work with them. the problem is, although i remember the mathematical rules, i don't remember how i assimilated the equality between fractions and divisions when i was younger. this is really troubling to my mind because it feels like i partly understand this concept. i know that the denominator represents the amount of parts a whole was divided into and that the numerator represents the amount of parts i'm working with. let's say, i want to divide 3 by 4. the result of this division is 3/4. but, in this case, i'm dividing 3 wholes by 4. doesn't this come into conflict with the definition of the denominator, which is the amount of parts in which 1 whole is divided into? this confusion is really bugging me...

why is it so? isn't auto-property initializer actually translated into constructor code in il?the rules for automatically implemented property initializers are the same as those for field initializers, for the same reason. note that property initializers are executed before base class bodies, just like field initializers - so you're still in the context of a "somewhat uninitialized" object; more so than during a constructor body.so you should imagine that the property is being converted into this:private nested bar = new nested(this); // invalid public nested bar { get { return bar; } } in short, this restriction is to stop you from getting yourself into trouble. if you need to refer to this when initializing a property, just do it manually in a constructor, as per ...

somewhat as others have said: historically (pre-deimalization), the alternative was fractions such as 3/5, 2/7, 1/2, etc. comparisons are not entirely trivial; many people would not be able to reliably answer which of those are the largest. so by regulating partial calculations (as for taxes, interest, etc.) to parts of 100 made those comparisons much easier for common people.decimalization solves the issue of fractions with much the same idiom, but it is a surprisingly recent development (e.g., prices on the u.s. stock market didn't switch from fractions to decimals until the year 2001). but even with comparisons can be problematic. which is larger: 0.2 or 0.08? many people will answer that incorrectly because of the hidden place value. percentages make the like place values explicit and.

this definition struct a { struct b; }; defines a struct a with a declaration of a nested struct b1. the fully qualified name of b is a::b, you could say b is inside the "namespace' of a. then this:struct a::b : a { // note i added spaces }; is the definition of a::b, and the single : specifies that it is derived from a.now, the interesting part is a::b::a::b. let's dissect it:a::b names the nested structure.a::b::a accesses the injected class name a inside b. the injection is due to the inheritance.a::b::a::b names the nested structure b in a again.and you can continue ad-infinitum, or at least until your compiler meets its translation limit.a fun intellectual exercise, but avoid like the plague in actual code. 1 - note it's a completely other b. not at all related to ...

somewhat as others have said: historically (pre-decimalization), the alternative was fractions such as 3/5, 2/7, 1/2, etc. comparisons are not entirely trivial; many people would not be able to reliably answer which of those are the largest. so by regulating partial calculations (as for taxes, interest, etc.) to parts of 100 made those comparisons much easier for common people.decimalization solves the issue of fractions with much the same idiom, but it is a surprisingly recent development (e.g., prices on the u.s. stock market didn't switch from fractions to decimals until the year 2001). but even with comparisons can be problematic. which is larger: 0.2 or 0.08? many people will answer that incorrectly because of the hidden place value. percentages make the like place values explicit an.

somewhat as others have said: historically (pre-decimalization), the alternative was fractions such as 3/5, 2/7, 1/2, etc. comparisons are not entirely trivial; many people would not be able to reliably answer which of those are the largest. so by regulating partial calculations (as for taxes, interest, etc.) to parts of 100 made those comparisons much easier for common people.decimalization solves the issue of fractions with much the same idiom, but it is a surprisingly recent development (e.g., prices on the u.s. stock market didn't switch from fractions to decimals until the year 2001). but even with decimals, comparisons can be tricky. which is larger: 0.2 or 0.08? many people will answer that incorrectly because of the hidden place value. percentages make the like place values explic.

direct proof. suppose the function $f:[a,b]\rightarrow \mathbb{r}$ is increasing, with $f([a,b])=[c,d]$. pick $x$ and let $s$ be any sequence which converges to $x$. it suffices to show that the image of $s$ converges to $f(x)$.let $n_1 \supseteq n_2 \supseteq n_3 \supseteq \cdots $ be a strictly nested sequence of open intervals around $f(x)$ and moreover suppose the length of the intervals converges to 0.note that the inverse images are nested, too: $f^{-1}(n_i) \supseteq f^{-1}(n_{i+1})$. (think about the how the endpoints of the nested intervals $n_i$ are $ f(p_2)$. (if we had assumed the other case where the function value was less than the limit value, we would instead have gotten a contradiction from $p_1 < p$ but $f(p_1) > f(p)$.)so the limit doesn't exist...

json often returns sort of 'nested' associations like so:assoc = 1, "b" -> 2|>, "key2" -> 3, "b" -> 4|>|> it is often needed to apply f as follows.assoc = f[1], "b" -> 2|>, "key2" -> f[3], "b" -> 4|>|> i see some traditional ways, i.e. mapat, but i'm searching more specific way for the task, like keyvaluemap or something similar. which one would you recommend? (parsing apis results is a good example of usecase.)

this code works fine without aligned. i suggest to use the cellspace package to give cells some padding (otherwise, the fractions touch the horizontal lines), and the medium-sized fractions from nccmath – in my opinion text fractions look petty in this context. last remark: since version 3.9 of babel, it is recommended the language option be loaded with the document class, so language-dependent packages be aware of the main language in the document.\documentclass[english]{article} \usepackage{babel} \usepackage{array, cellspace} \setlength{\cellspacetoplimit}{4pt} \setlength{\cellspacebottomlimit}{4pt} \usepackage{amsmath, nccmath} \begin{document} \begin{table}[htb!] \centering \begin{tabular}{*{4}{sc|}} \cline{2-3} & \multicol...

so i tried doing the following integral:$$i=\int_0^{+\infty}\sin(2^x)\ dx$$which is quite similar to the famous fresnel integral. first, i rewrote $\sin$ using its complex exponential definition, then i let $u=2^x$:$$i=\int_0^{+\infty}\frac{e^{i2^x}-e^{-i2^x}}{2i}\ dx=\frac1{2i\ln(2)}\int_1^{+\infty}\frac{e^{iu}-e^{-iu}}u\ du$$(so close to letting me use frullani's integral $\ddot\frown$)but where do i go from here? it looks very close to a place where i could use the exponential integral or something like that, but not quite...

because $42$ is the fifth catalan number then using the known integral representation of the catalan's numbers we have that$$c_5=\frac1{2\pi}\int_0^4x^5\sqrt{\frac{4-x}x}\,\mathrm dx=42$$playing with some change of variable above you can get different integral expressions.

because $42$ is the fifth catalan number then using the known integral representation of the catalan's numbers we have that$$c_5=\frac1{2\pi}\int_0^4x^5\sqrt{\frac{4-x}x}\,\mathrm dx=42$$playing with some change of variable above you can get different integral expressions. by example with the change of variable $(4-x)/x=t$ we can build the following improper integral$$c_5=\frac{16}{\pi}\int_0^\infty\sqrt t\left(\frac12+\frac{t}2\right)^{-7}\mathrm dt=42$$maybe more interesting integral representations can be achieved considering first the integral representations for $21$, what is a number with more interesting properties.

all the numbers you list are algebraic.lemma.if $\alpha$ and $\beta$ are algebraic, then $\alpha+\beta$ and $\alpha\beta$ are algebraic.proof. since $\beta$ is algebraic (over $\mathbb{q}$), it is also algebraic over $\mathbb{q}(\alpha)$. hence $\mathbb{q}(\alpha,\beta)$ is a finite extension of $\mathbb{q}(\alpha)$, which in turn is finite over $\mathbb{q}$. therefore $\mathbb{q}(\alpha,\beta)$ is finite by the dimension formula: $$ [\mathbb{q}(\alpha,\beta):\mathbb{q}]= [\mathbb{q}(\alpha,\beta):\mathbb{q}(\alpha)] [\mathbb{q}(\alpha):\mathbb{q}] $$ any finite extension of $\mathbb{q}$, say $n$-dimensional, consists of algebraic elements, because if $\gamma$ is an element, $\{1,\gamma,\gamma^2,\dots,\gamma^n\}$ is linearly dependent, so we find a polynomial with coefficients in $\mathbb{

generally, what are the rules for simplifying answers?i had this question:given $ f(x) = \frac{1}{x}$, evaluate $\frac{f(x) - f(a)}{x-a}.$why is the following a bad answer:$\frac{\frac{1}{x} - \frac{1}{a}}{x-a}$the better answer is:$\frac{-1}{xa}$why is this? is it generally just bad to leave fractions in either the numerator or denominator? so we should try to eliminate the fraction in the numerator by solving the equation first on the top by using a common denominator?

given a fraction in the format m/n (where m and n are coprime integers), output the corresponding unicode fraction. you will not be expected to allow any input that does not correspond to a unicode character. arrays, e.g. [2, 3] as opposed to 2/3, are accepted. m / n as opposed to m/n is also fine. two separate inputs m and n are also valid.the unicode fractions that must be handled are as follows:½, ⅓, ⅔, ¼, ¾, ⅕, ⅖, ⅗, ⅘, ⅙, ⅚, ⅐, ⅛, ⅜, ⅝, ⅞, ⅑, ⅒ thus, the possible inputs are as follows:1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 2/5, 3/5, 4/5, 1/6, 5/6, 1/7, 1/8, 3/8, 5/8, 7/8, 1/9, 1/10 1/2 -> ½ 1/3 -> ⅓ 2/3 -> ⅔ 1/4 -> ¼ 3/4 -> ¾ 3/8 -> ⅜ 1/10 -> ⅒ make your code as short as possible; this is code-golf.

given a fraction in the format m/n (where m and n are coprime integers), output the corresponding unicode fraction. your program/function will not be expected to take in any input that does not correspond to a unicode character. arrays, e.g. [2, 3] as opposed to 2/3, are accepted. m / n as opposed to m/n is also fine. two separate inputs m and n are also valid.the unicode fractions that must be handled are as follows:½, ⅓, ⅔, ¼, ¾, ⅕, ⅖, ⅗, ⅘, ⅙, ⅚, ⅐, ⅛, ⅜, ⅝, ⅞, ⅑, ⅒ thus, the possible inputs are as follows:1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 2/5, 3/5, 4/5, 1/6, 5/6, 1/7, 1/8, 3/8, 5/8, 7/8, 1/9, 1/10 1/2 -> ½ 1/3 -> ⅓ 2/3 -> ⅔ 1/4 -> ¼ 3/4 -> ¾ 3/8 -> ⅜ 1/10 -> ⅒ make your code as short as possible; this is code-golf.

given a fraction in the format m/n (where m and n are coprime integers), output the corresponding unicode fraction. your program/function will not be expected to take in any input that does not correspond to a unicode character. arrays, e.g. [2, 3] as opposed to 2/3, are accepted. m / n as opposed to m/n is also fine. two separate inputs m and n are also valid.the unicode fractions that must be handled are as follows:½, ⅓, ⅔, ¼, ¾, ⅕, ⅖, ⅗, ⅘, ⅙, ⅚, ⅐, ⅛, ⅜, ⅝, ⅞, ⅑, ⅒ whose unicode codepoints are as follows:188 ¼, 189 ½, 190 ¾, 8528 ⅐, 8529 ⅑, 8530 ⅒, 8531 ⅓, 8532 ⅔, 8533 ⅕, 8534 ⅖, 8535 ⅗, 8536 ⅘, 8537 ⅙, 8538 ⅚, 8539 ⅛, 8540 ⅜, 8541 ⅝, 8542 ⅞ thus, the possible inputs are as follows:1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 2/5...

i have a block of code that gives me a list that has some triple nested lists within it:my_list = [[['item1','item2']], [['item3', 'item4']]] and i would like to make it:my_list = [['item1','item2'], ['item3', 'item4']] any suggestions?

the long division algorithm is a way of taking the numerator, and split it into multiples of the denominator, separate those out into their own fractions, and simplify away the denominator. for instance, we have $x(x^2 - 2x+1) = x^3-2x^2+x$, so $$ \frac{x^3 - 12x^2 + 0x-42}{x^2-2x+1} = \frac{(x^3-2x+x) - 10x^2 - 1x-42}{x^2-2x+1}\\ = \frac{x(x^2-2x+1)}{x^2-2x+1} + \frac{-10x^2-1x-42}{x^2-2x+1}\\ = x + \frac{-10x^2-1x-42}{x^2-2x+1} $$ so now we've gotten an $x$, and we're left with a new fraction where the degree of the numerator has gone down by 1 compared to where we started. we can keep going, as long as the degree of the numerator is larger than or equal to the degree of the denominator, which in this case means one more step. this time, we see that $-10(x^2-...

the problem with using an anonymous pure function inside of inversefunction is that you get nested anonymous pure functions when evaluating the second derivative. compare the following 2 traceprint outputs:traceprint[inversefunction[function[x, log[x]+exp[x]]]'', derivative[1][_]]; traceprint[inversefunction[log[#]+exp[#]&]'', derivative[1][_]]; ((1/(function[x,log[x]+exp[x]]^[prime])[inversefunction[function[x,log[x]+exp[x]]][#1]]&)^[prime])((1/((log[#1]+exp[#1]&)^[prime])[inversefunction[log[#1]+exp[#1]&][#1]]&)^[prime])notice how the the first trace output has a mixture of x and #1, while the second trace output replaces x with #1. the latter trace is an example of a nested pure function using anonymous pure functions, which is why it doesn't work. the issue of nested anonymo...

now you can use multiple levels of nested teams to reflect your group or company's hierarchy within your github organization, making your organization's permissions structure clearer and easier to manage.child teams inherit their parent's access permissions, so repository permissions and @mentioning among nested teams work from top to bottom. if your team structure is employees > engineering > application engineering > identity, granting engineering write access to a repository means application engineering and identity also get that access. and if you @mention the identity team or any other team at the bottom of the organization hierarchy, they're they only ones who will receive a notification.membership inheritance from parent to child teams isn't automatic. if you're a member of e...

hake's theorem, due to heinrich hake of düsseldorf in 1921, says that an improper henstock–kurzweil integral (aka generalized riemann integral, gauge integral, perron integral, or denjoy integral) on a bounded interval is already proper. that is, if $f$ is defined on a half-open interval $[a,c)$, $f$ is hk-integrable on $[a,b]$ for each $b$ satisfying $a \leq b < c$, and the limit of $\int_a^b f$ as $b \to c^-$ exists, then we may define $f(c)$ however we like and find that $f$ is integrable on $[a,c]$ and that $\int_a^c f$ equals the aforementioned limit. (the converse is also true; if $\int_a^c f$ exists, then it may be calculated as a limit.)i can't find anything about this hake. most references just say ‘hake's theorem’, a few say ‘h. hake’, and bartle's book on the...

yes, this holds in general. recall that an extension $f/\bbb q$ is algebraic iff for each $\alpha\in f$ we have$$[\bbb q(\alpha):\bbb q]=\dim_{\bbb q} \bbb q(\alpha)< \infty$$that is, if every element in it generates a finite extension. similarly any finite extension is algebraic since if $\beta\in f$ then $\bbb q(\beta)\subseteq f\implies [\bbb q(\beta):\bbb q]\le [f:\bbb q]$. for a nested radical expression, this is easy then, take the most nested radical, adjoin the needed roots and proceed by induction.in your first example $\alpha=\sqrt{11+2\sqrt 7}$, then we see that the field $\bbb q(\sqrt 7)$ has a polynomial for this which is ${1\over 2}(x^2-11)-\sqrt 7 = 0$ so the splitting field of this polynomial, which has degree at most $2$ over $\bbb q(\sqrt 7)$, and therefore degree at m

so it's given this indefinite integral $$\int \frac{1}{\sin ^{\frac{1}{2}}x \cos^{\frac{7}{2}}x}dx$$is there anyone could solve this integral? thanks in advance.

i would suggest you to have a look at this answer where i have discussed the fundamental theorem of calculus for functions which are not necessarily continuous. next note that the integrand here has a discontinuity at $x=1$ and as far as the interval of integration is concerned the discontinuity is removable and hence we just remove it by changing value of integrand at $x=1$ to $0$. doing this has no impact on the value of the integral (because the value of a riemann integral does not depend on the value of the integrand at a finite number of points in the interval under consideration) and thus the desired integral is equal to $\int_{1/2}^{1}0\,dx=0$.in general if we have to evaluate an integral of the form $\int_{a} ^{b} f(x) \, dx$ where $f$ has a finite number of discontinuities at ...

you can obtain a correct vertical padding with the cellspace package, which lets you define minimal vertical paddings at top and bottom of cells in column with specifier prefixed with the letter s (or c if you load siunitx). for matrices, which have no specifier, this is done with the [math] option of the package.if you want a left or right alignment of the elements in a matrix, you can use the xmatrix* environments defined by mathtools. however, cellspace does not work with these environments . so, either you use a xmatrix environment (centred) and the classic trick of \phantom{...} elements where necessary, or you use the \setcellgapes{xxx}\makegapedcells construct defined by makecell.last point: i used the medium-sized fractions from nccmath, as i don't the size discrepancy between ...

i have this trigonometric equality to prove:$$\frac{\cos x}{1-\tan x}-\frac{\sin x}{1+\tan x}=\frac{\cos x}{2\cos^2x-1}$$ i started with the left hand side, reducing the fractions to common denominator and got this: $$\frac{\cos x+\cos x\tan x-\sin x+\sin x\tan x}{1-\tan^2x}\\=\frac{\cos x+\cos x(\frac{\sin x}{\cos x})-\sin x+\sin x(\frac{\sin x}{\cos x})}{1-\left(\frac{\sin^2x}{\cos^2x}\right)}\\=\frac{\cos x+\left(\frac{\sin^2x}{\cos x}\right)}{1-\frac{\sin^2}{\cos^2x}}$$and by finding common denominator top and bottom and then multiplying the fractions i got: $$\frac{\cos^2x}{\cos^3x-\cos x\sin^2x}$$ which is far from the right hand side and i don't know what am i doing wrong.what is the correct way to prove this equality?

{a[0, 0, 1], a[0, 0, 2], a[1, 1, 1], a[1, 1, 2], a[2, 2, 1], a[2, 2, 2]}

{#[[1]], #[[2]], #[[1]] + #[[2]]} & /@ {{a, b}, {c, d}, {e, f}}{{a, b, a + b}, {c, d, c + d}, {e, f, e + f}}