calculus - Textbooks that use notation with explicit argument variable in the upper bound $int^x$ for "indefinite integrals."

I dare to ask a question similar to a closed one but more precise.

Are there any established textbooks or other serious published work that use $\int^x$ notation instead of $\int$ for the so-called "indefinite integrals"?

(I believe I've seen it already somewhere, probably in the Internet, but I cannot find it now.)

So, I am looking for texts where the indefinite integral of $\cos$ would be written something like:
$$
\int^x\cos(t)dt =\sin(x) - C

$$
or
$$
\int^x\cos(x)dx =\sin(x) + C.
$$

(This notation looks more sensible and consistent with the one for definite integrals than the common one with bare $\int$.)

Some context.

IMO, the indefinite integral of $f$ on a given interval $I$ of definition of $f$ should not be defined as the set of antiderivatives of $f$ on $I$ but as the set of all functions $F$ of the form
$$
F(x) =\int_a^x f(t)dt + C,\qquad x\in I,
$$
with $a\in I$ and $C$ a constant (or as a certain indefinite particular function of such form).
In other words, I think that indefinite integrals should be defined in terms of definite integrals and not in terms of antiderivatives.
(After all, the integral sign historically stood for a sum.)

In this case, the fact that the indefinite integral of a continuous function $f$ on an interval $I$ coincides with the set of antiderivatives of $f$ on $I$ is the contents of the first and the second fundamental theorems of calculus:

1. the first fundamental theorem of calculus says that every representative of the indefinite integral of $f$ on $I$ is an antiderivative of $f$ on $I$, and




2. the second fundamental theorem of calculus says that every antiderivative of $f$ on $I$ is a representative of the indefinite integral of $f$ on $I$ (it is an easy corollary of the first one together with the mean value theorem).

real analysis - Is it true that $0.999999999dots=1$?

I'm told by smart people that
$$0.999999999\dots=1$$
and I believe them, but is there a proof that explains why this is?

Answer

What does it mean when you refer to $.99999\ldots$? Symbols don't mean anything in particular until you've defined what you mean by them.

In this case the definition is that you are taking the limit of $.9$, $.99$, $.999$, $.9999$, etc. What does it mean to say that limit is $1$? Well, it means that no matter how small a number $x$ you pick, I can show you a point in that sequence such that all further numbers in the sequence are within distance $x$ of $1$. But certainly whatever number you choose your number is bigger than $10^{-k}$ for some $k$. So I can just pick my point to be the $k$th spot in the sequence.

A more intuitive way of explaining the above argument is that the reason $.99999\ldots = 1$ is that their difference is zero. So let's subtract $1.0000\ldots -.99999\ldots = .00000\ldots = 0$. That is,

$1.0 -.9 = .1$

$1.00-.99 = .01$

$1.000-.999=.001$,

$\ldots$

$1.000\ldots -.99999\ldots = .000\ldots = 0$

calculus - derivative of $xcdot|sin x|$

I have the function $f(x)=x|\sin x|$, and I need to see in which points the function has derivatives.
I tried to solve it by using the definition of limit but it's complicated. It's pretty obvious that the function has no derivatives where $|\sin x|=0$, but I don't know how to show it.
I thought maybe calculate the derivative of $f(x)$ but I didn't learn how to calculate the derivative of $|\sin x|$.
How can I solve it withut knowing the derivative of $|\sin x|$? or better question how to calculate the derivative of $|\sin x|$?

edited:I didn't learn the derivative of $|f(x)|$.

Answer

I'm going to make this as simple as I can. So, ofcourse, I'll be assuming $x\in \mathbb R$

You're first question is for what values of $x$ is the function differentiable.

There are nice algebraic ways to find it but why waste time in an explicit proof if all one needs is to convince one's peers that one's logic is right.

Let's just observe the differentiability of $f(x) = x\cdot |\sin x|$ through it's graph.
But oh, wait, you must not know how to plot it's graph. So, let's take baby steps to get find it.

Take what we know. The standard graph of $y = \sin x$

Note that the roots (ie, $\sin x = 0$) are $x = n\pi,\quad n\in\mathbb Z$

Now, let's graph y = $|\sin x|$ . How?
There's a method to get $|f(x)|$ from $f(x)$ and it goes something like this:

Step 1: Make $y = 0$ a mirror which reflect all $y<0$ into the plane $y>0$
Step 2: Eliminate the portion of the graph which lies in $y < 0$
Step 3: Be amazed that by executing the above two steps precisely, you got the right graph.

^{Learn why this works}

Now we have to multiply this with $x$. There's no real method for this, it only takes a slight bit of thinking and understanding of what multiplication does to a graph.

Usually when we multiply a function with a constant, say $c$.

• The graph diminishes for $c\in(0,1)$


• Enlarges for $c>1$


• Turns upside down for $c<0$ and follows the above two observations once again.

Since we're multiplying by a variable and not a pure scalar, the graph is distorted such that all the above can be seen at once and in increasing degree with increase in the magnitude of $x$.

Now, it is obvious that the roots of this graph are the same as that of $\sin x$

and you know we can't differentiate a function at sharp points. (Why?)

Notice that the sharp point at $x = 0$ has been smoothed over by the inversion in the graph for $x<0$. But is it differentiable here at $x=0$?

To prove that it's differentiable at this point,

$$f'(0) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} , \quad \text{ where } x =0\\
= \lim_{h \to 0} \frac{\sin (0+h) - \sin 0}{h}\\
= \lim_{h \to 0} \frac{\sin h}{h} = 1
$$

(Why?)

$\therefore $ Derivative exists @ $x = 0$ $\implies$ Differentiable @ $x = 0$

So, we can now safely say that we can differentiate $f(x) = x\cdot|\sin x|\quad \forall \space x\in\mathbb R - \{n\pi\}, \quad n\in\mathbb Z-\{0\}$

Or more easily in words,
$f(x) = x|\sin x|$ is differentiable at $x \neq n\pi ,\quad n \neq 0$

---

The following is how I would differentiate the function:

$$
\frac{d}{dx} x\cdot|\sin x|\\
= \frac{d}{dx} x\cdot\sqrt{\sin^2 x}
\quad ,\quad \{\because\space|x| = \sqrt{x^2}\} \\
= x\frac{d}{dx}\sqrt{\sin^2 x} + |\sin x|\frac{dx}{dx} \quad ,\quad \{\because\space (uv)' = uv' + u'v\}\\
= x\cdot\frac{1}{2\sqrt{\sin^2x}}\cdot (2\sin x)\cdot (\cos x) + |\sin x|
\quad , \quad \{\because\text{Chain Rule }\}\\
=\frac{x\cdot\sin 2x}{2|\sin x|} + |\sin x|$$

This isn't totally simplified but hopefully this is helpful.

Now, to further clarify on the derivitive of $|x|$:
$$\frac{d}{dx} |x|
= \frac{d}{dx} \sqrt{x^2}
= \frac{1}{2}(x^2)^{\frac{1}{2} - 1} \cdot 2x
= \frac{2x\cdot (x^2)^{-\frac{1}{2}}}{2}
=\frac{1}{2\sqrt{x^2}}
= \frac{x}{|x|}
\equiv \frac{|x|}{x}
= \text{sgn}(x)$$

Here is more information on $\text{sgn}(x)$ and a better more explicit way of finding the derivative of the absolute value

Exercise: Can you try to get the derivative of $x|\sin x|$ with the sign-function included?

polynomials - Quadratics: Intuitive relation between discriminant and derivative at roots

While working with quadratics that have real roots, I realized an interesting fact:

The slope of a quadratic at its roots is equal to $\pm \sqrt{D}$ where $D=b^2-4ac$

Proof:

$$f(x) = ax^2 + bx +c$$

$$f'(x) = 2ax+b$$

Roots:

$$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

So, if we try to find the slope at any root ($r$):

$$f’(r) = \pm \sqrt D$$

where the sign ($\pm$) can be determined by whether the root is on the right of the vertex or the left.

If the quadratic has only $1$ root (or $2$ roots that are the same) then it means the quadratic is at a stationary point so the slope must be $0$. This is backed by the fact that quadratics have only 1 distinct root when $b^2 - 4ac = 0$.

What geometric/intuitive approach can be applied to explain this interesting phenomenon?

linear algebra - Intuitive perspective of eigenvalues and rank of a matrix

Assuming a matrix $A$, $n\times n$, with $n$ non-repeated and non-zero eigenvalues;

1. If we calculate the matrix $A-\lambda I$ for one of its $n$ eigenvalues, we see that its rank has been decreased by one. If the eigenvalue has repetitiveness of $k$, then the rank decreases again by $k$. What would be an intuitive explanation for it?




2. By $Ax=\lambda x$ one could argue that we try to find the values of $\lambda$ for which an $n\times n$ matrix with $\text{rank}(A)=n$ to have the same impact on $x$ as a:

   
   
   

   2a. a scalar $\lambda$?

   
   
   
   

   or

   
   
   

   2b. a $n\times n$ diagonal matrix of rank $n$?




3. In the relationship $(A-\lambda I)x=0$, given that we want a nontrivial solution for the vector $x$, could we declare the matrix $A-\lambda I$ as zero, without the determinant, following directly the above relationship?

Answer

For 1, it depends on what you mean by “intuitive”, but here’s a shot at it. A matrix either sends a vector to 0 or it doesn’t. The amount of vectors that it sends to 0 is related to the amount of vectors that it doesn’t by the rank it’s theorem. The more vectors it sends to 0, the fewer it sends to non-zero values. If A has an eigenvector then $A-\lambda I$ sends a vector to 0. Therefore, it can’t send as many to non-zero values (meaning it’s rank has been reduced).

EDIT: This is not always true as Widawensen points out in another answer.

For 2, the answer is a. But we normally think about it the other way around. We are trying to find a vector such that A operating on that vector simply scales it.

For 3, if $A-\lambda I$ is 0 then $A=\lambda I$. This is the special case where A has a single eigenvalue of multiplicity equal to its rank. But there are cases where A can have eigenvector a but A is not diagonal. Those cases can be found with the determinant because we know that the determinant of a singular matrix is 0 and we want $A-\lambda I$ to be singular.

combinatorics - Why are permutations (nPr) called variations in non-English languages?

First of all, you should be at least a little familiar with combinatorics to understand that question.
Some often used calculator keys in stochastic are the nCr and nPr ones.

Edit: I've first asked this question for the German and English Stackexchange version (both places where "mathematics" tags exist), but as it turned out this is likely also the case for all non-English languages and the community here may be better suited to answer the question, I've also posted it here. Despite that, below the German language is referred to as an example, where it is called differently.
Edit2: Also posted at History of Science and Mathematics and linguistics.

nCr is quite obvious. The "C" stands for "combinations" (actually those without repetition) and this is how they are called in German and English. That is just the binomial coefficient:

$$\binom{n}{k}=\frac{n!}{k!(n-k)!}=\text{n nCrk}$$

Keeping that knowledge in mind, as a German, you would assume nPr is for calculating the permutations (without repetition, again), i.e. just:

$$n!$$

However, that's not the case, actually it calculates the "variation", as it's called in German:

$$\frac{n!}{(n-k)!} =\text{n nPr k}$$

And it is true: Actually the "P" does stand for "permutation" in English. So the last formula is what they call "permutation".

Just different names?

So we could say, these are just different names, but no, it gets more complicated, because – using the German terms here again – permutations are just a special kind of variations. Essentially, it's the last formula, where k=n, i.e. you choose all items and do not select a subset when arranging them.

Obviously the English mathematics do not use the term "permutations" for the specific version we name it in German, but for the general version.
Essentially this leads to another problem, however, when we look at nPr with repetition. All examples before where without repetition, but you have formulas for the ones with repetition, too.

So the "permutation with repetition"/"Variation mit Wiederholung" and is easy to calculate, you just:

$$n^k$$

Wikipedia does not seem to want to acknowledge the English term for that saying they have "sometimes been referred to" in this way… (Or is this actually something different as the formula is k^n?)

Anyway, if we assume the term is used like that, we've got another way to have German "Permutationen" "with repetition". This time, however, as in the German definition of permutations we do not select items, we just have multiple of the same items. So e.g. you have r, s, …, t same elements in n elements you get a formula like that:

$$\frac{k!}{r!\cdot s! \cdots t!}$$

And this is what we call "Permutation mit Wiederholung" in German. But what term is then used in the English for this kind of "repetition"?

So how did this inconsistent naming across languages happen? Is there any "correct" term or has one term been invented before another one, so someone adapted it wrong?
Do other languages possibly also name it differently, i.e. is the German naming the exception or the English one?
And what term is used for "Permutationen mit Wiederholung"/same elements in a set in English then?

If you need some more understanding:

---

Edit: I found something: The English Wikipedia describes the term "variations" as:

• Variations without repetition, an archaic term in combinatorics still commonly used by non-English authors for k-permutations of n


• Variations with repetition, an archaic term in combinatorics still commonly used by non-English authors for n-tuples

Despite that sounding a little pejorative to me as a German speaker, it raises the question of whether this is really (internationally?) deprecated/outdated? Or what term is supposed to be used?
Also the relation to tuples, which are – I thought – just a different concept of a list of numbers, is not clear to me. After all, I could not found any of the formulas I've just mentioned in the linked article.

Answer

$(\boldsymbol{1})\quad$ We call $\,P(n,k)\,$ k-permutations. Order matters, repetitions not allowed. The number of permutations of the n objects taken k at a time:

$$P(n,k)=n(n-1)...(n-k+1)=\frac{n!}{(n-k)!}$$
So, these are ordered arrangements/selections/choices. We also use $\;_n P_k\;$notation.

$\quad$

$(\boldsymbol{2})\quad$ When we permute all objects we simply call them permutations and write $\,n!$

$\quad$

$(\boldsymbol{3})\quad$ If repetitions are allowed and order matters, we refer to such arrangements as permutations with repetitions or distinguishable permutations:

$${{n}\choose{n_1, n_2, n_3…,n_p}} = \frac {n!}{n_1!\, n_2!\, n_3!…n_p!}$$

$$\quad$$

$(\boldsymbol{4})\quad$ When we have $n^k$ ordered arrangements, replacement allowed -- we call them permutations with replacements or k-tuples.

$$\quad$$

$(\boldsymbol{5})\quad$ When repetitions are not allowed and order doesn't matter, we call such arrangements combinations: ${{n}\choose{k}}\;$ or $\;_n C_k\;$ or $\;C(n,k).\;$ We choose n objects taken k at a time without regard to order. We read it "n choose k" and write:
$${{n}\choose{k}}=\frac{n(n-1)...(n-k+1)}{k!}=\frac{n!}{k!\,(n-k)!}$$

$(\boldsymbol{6})\quad$ And finally, when we deal with unordered arrangements, repetitions allowed -- we call them combinations with repetitions:
$${{n+k-1}\choose{k}}=\frac{(n+k-1)\cdot...\cdot n}{k!}=\frac{(n+k-1)!}{k!\,(n-1)!}$$

Please note it doesn't matter what these are called in German or French since each language has its own rules. An English manual of style is not applicable to other languages and vice versa; likewise you can't replace permutations or combinations with German variations in English. Yet we sometimes have different notations even within one language as authors may have their own preferences in terms of notation and terminology. There's nothing to worry about here.

Despite sounding a little pejorative to me, a German language speaker, it raises a question of whether such usage is really (internationally?) deprecated and considered outdated? -- No. That's more of a speculation on terminology used in other languages by some Wikipedia writers. See their editing history. Do not blindly trust something which is not a hard science in Wikipedia.

Do other languages possibly also name it differently? -- Yes. You can see it in the comments from people of other countries. There should be a lot of subtle examples. Let me make a rough guess based on "googling":

1. Disposizioni semplici=Variation ohne Wiederholung=l'arrangement=k-permutations




2. Arranjo com repetição=Variation mit Wiederholung=permutations with replacement.

So we should have been able to say these are just different technical terms -- but no -- it gets more complicated because in German terminology permutations are just a special kind of "variations". -- Yes, to some extent at least. Permutations is a broad term in English. What's more you can view these formulas from different "angles", e.g., combinations being just a special case of distinguishable permutations, permutations being just ordered combinations; and permutations with replacement can be called permutations with repetitions (it might create some confusion!) and so on.

The German flow chart with the German nomenclature? -- I checked it. It's really nice and logical. Quite commendable. I don't see how it may be inferior to any other nomenclature. It is rather on the contrary.

Nomenclatures, notations, and terminology differ from country to country. In biology, any species receives a binomial name (Latin name) and there's no ambiguity across the world about that species. In math we also have universal symbols and notations but they are not so rigid. You can find $cot^{-1}$, $arccot$, $arcctg $ used to denote the same and so forth. You can find in some countries analytic geometry is almost never part of calculus but always part of linear algebra. Sometimes you can find calculus being called mathematical analysis and being confused with analysis or real analysis. You may come across calques (or verbatim translations) of higher algebra, general algebra, etc. You may see how in English we coined words Calc I, II, III, IV as well as precalculus. Things are not clear cut, and there will be variations, just like the difference in meaning the word gift has in English and in German. While the word variation may have very similar meanings in English and German, there will also be differences, maybe subtle differences. And it is exactly the words with minor differences in meaning that cause most of the confusion. People expect them to be the same but they are not. One final example. We have books on vector calculus but it is a bit of a misnomer, as these books are just enhanced versions of Calculus III/IV. And it may have no bearing whatsoever on what might be the case in other languages.

CONCLUSION:
Now we can answer the "title" question: Why are permutations P(n,r) called variations in languages other than English? -- That is simply not the case! While European languages may use math variations in a similar way, the usage will diverge to some extent.

IMPORTANT:
Please note that not only technical terms are quite different from country to country but notations, too, may vary. Thus, in France, Russia, etc. permutations are often denoted $A^{k}_n$, and $C^{k}_n$ is used for combinations, which means the upper and bottom indexes are reversed. It may lead to mistakes in translation.

analysis - Existence of smooth function $f(x)$ satisfying partial summation

We know that $\sum_{i=1}^{n}i=\frac{n(n+1)}{2}$, $\sum_{i=1}^{n}\frac{1}{i}=\psi_{0} (n+1)-\psi_{0} (1)$, where $\psi_{0}(x)$ is the digamma function.

My problem is,

(1).Is there a transformation such that it maps
$x \to \frac{x(x+1)}{2}$ and $\frac{1}{x} \to \psi_{0}(x+1)$, and map a smooth $f(x)$ into another smooth function $g(x)$, such that $g(x)-g(x-1)=f(x)$ ? When I mention transformation, I mean an operator or algorithm for me to get $g(x)$ from $f(x)$.

(2). Surely $g(x)$ if exists, it is not unique because $g(x)+C$ also satisfy the condition. Let's take $g(x)+C$ and $g(x)$ as the same case. Is there another smooth $h(x)\not = g(x)+C$ satisfying this condition?

The problem came when I tried to evaluate $\sum_{i=1}^{n} \sqrt{i}$, I'd like to represent it by integral form. Thanks for attention!

Answer

You can use the zeta function and the Hurwitz zeta function to write your sum. As a more general case of your summation, we can have the following representation

$$\sum_{i=1}^{n} i^s = \sum_{i=1}^{\infty}i^s - \sum_{i=0}^{\infty}(i+n+1)^s = \zeta(-s) - \zeta(-s, n+1)\,.$$
Now, substituting $s= -\frac{1}{2}$ in the above identity yields

$$\sum_{1}^{n} i^{\frac{1}{2}} = \zeta\left(-\frac{1}{2}\right) - \zeta\left(-\frac{1}{2}, n+1\right) \,.$$

See convergence issues of the zeta function and the Hurwitz zeta function.