# Sequence of Rationals Converging to a Limit

I’m trying to show that for every real number $r$, there exists a sequence of rational numbers $\{q_{n}\}$ such that $q_{n} \rightarrow r$.

Could I get some comments on my proof?

I know that between 2 reals $r, b$ there exists a rational number $m$ such that $r < m < b$.
So I can write

$r < q_{1} < b$ ; Now check if $q_{1} = r$ or not. If it does I’m done, and if not, I consider the interval $(a, q_{1})$.

$r < q_{2} < q_{1}$ check if $q_{2} = r$ or not. If it does I’m done, and if not, I consider the interval $(a, q_{3})$

If I continue in this manner, I see that $|r – q_{n+1}| < |r – q_{n}|$. So whether $r$ is rational or irrational, I’m making my the size of the interval $(r, q_{n})$ closer to 0 and as $n \rightarrow \infty$. And so given any $\epsilon > 0$, I know that $|q_{n} – r| < \epsilon$.

Revision

Let $\{q_{n} \}$ be a sequence of rational numbers and $q_{n} \rightarrow r$ where $r \in \mathbb{R}$.

If $r$ is rational, then let every element of $q_{n} = r$.

But if $r$ is irrational, then consider the interval $(r, b)$ where $b \in \mathbb{R}$. Since we can always find a rational number between two reals, consider

$r < q_{1} < b$. Now pick $q_{2}$ such that $q_{2}$ is the midpoint of $r$ and $q_{1}$. So we get that $r < q_{2} < q_{1}$. Then repeat the process so that $r < q_{n} < q_{n-1}$. Note that $|r – q_{n}| = \frac{1}{2}|r – q_{n-1}|$. As we take more values for $q_{n}$, it is clear that $|q_{n} – r| \rightarrow 0$. So given any $\epsilon > 0$, $|q_{n} – r| < \epsilon$.

#### Solutions Collecting From Web of "Sequence of Rationals Converging to a Limit"

You have the right idea but you are handwaving. You are not telling exactly how you will pick the rationals, and how their limit is $r$.

Also $|r – q_{n+1}| \lt |r – q_n|$ does mean $q_n \to r$. For instance, $\left|-1 – \frac{1}{n+1}\right| \lt \left|-1 – \frac{1}{n}\right|$, but $\frac{1}{n} \to 0$, and not $-1$.

Also, you don’t have to check if $q_i = r$, as you pick $r \lt q_i$.

(Also, the above writeup contains a possible hint, try to find it :-))

Since you are almost there (based on your Revision) (but note: you still are handwaving and don’t have a “proper” proof yet..)

Here is the hint I was referring to:

Since there is at least one rational in $(r, r+\frac{1}{n})$. Pick one and call it $q_n$. What is the limit of $q_n$?

A different proof:

Consider the set $S = \{q: q \ge r, q \in \mathbb{Q}\}$. Show that $\inf S = r$ and show that that implies there is a sequence $q_n \in S$ whose limit is $r$.

Here are some critiques to your proof.

1. When you write “So I can write $r<q_1<b$ “, what is $b$?

2. If $q_1 = r$, how are you done? (I agree that you are, but what would $q_2, q_3,…$ be in this case?)

3. “So whether $r$ is rational or irrational, I’m making my the (sic) size of the interval $(r,q_n)$ closer to $0$ and as $n\rightarrow\infty$.” As $n\rightarrow \infty$, what happens?

4. As noted in the comments, the condition $|r-q_{n+1}|<|r-q_n|$ does not guarantee that $|q_n-r|\rightarrow 0$.

Your idea for the proof is great, you just need a slight adjustment to force $|q_n-r|\rightarrow 0$. You could, for example, pick $q_n$ to be between $r$ and the midpoint of $(r,q_{n+1})$. This would force $|q_n-r|$ to be less than half of $|q_{n-1}-r|$, which will force $|q_n-r|\rightarrow 0$. Fix this, and fix up the general presentation a bit and you’ll have a great proof.

Your question set me thinking, and I came up with an argument inspired by it, but a bit different from your solution attempt. I do not imagine that the idea is new, but I do not recall seeing it done in this way (this may only mean that my education is incomplete- no pun intended). The idea is to take an irrational real number $r,$ and to recursively construct two sequences $(x_n)$ and $(y_n)$ of rational numbers which approach $r$ respectively from below and above (and are respectively non-decreasing and non-increasing). We start with a rational number $x_1 < r$ and a rational number $y_1 >r.$ Having found rational $x_n <r$ and rational $y_n >r,$ we set $z_n = \frac{x_n +y_n}{2}.$ Then $x_n < z_n < y_n.$ If $z_n <r,$ we set $x_{n+1} = z_n$ and $y_{n+1} = y_n,$ while if $z_n > r,$ we set $x_{n+1} = x_n$ and $y_{n+1} = z_n.$
(Note that $z_n \neq r$ as $z_n$ is rational, but $r$ is irrational). In the first case, we have $y_{n+1}-x_{n+1} = \frac{y_n – x_n}{2}$ and in the second case, we also have $y_{n+1} -x_{n+1} = \frac{y_n – x_n}{2}.$ Hence for each $n > 1,$ we have $y_n -x_n = \frac{y_1 – x_1}{2^{n-1}}.$ For large enough $n,$ we can make this less than any chosen real positive quantity $\varepsilon$. Then for large enough $n,$ we have both $0 < y_n -r < \varepsilon$ and $0 < r – x_n < \varepsilon.$ Hence the sequences $(x_n)$ and $(y_n)$ both have limit $r.$


# Is there an equation for the sum of alternating cubes?

The following sequences for sum of alternating cubes:

Odd cubes: [1, 28, 153, 496, 1225, 2556, 4753, 8128, 13041, 19900, 29161, 41328, 56953, 76636, 101025, 130816, 166753, 209628, 260281, 319600]

Even cubes: [8, 72, 288, 800, 1800, 3528, 6272, 10368, 16200, 24200, 34848, 48672, 66248, 88200, 115200, 147968, 187272, 233928, 288800]

Nothing on OEIS. Trying to find equations for generating such values.

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You may also want to look at: en.wikipedia.org/wiki/Cube_(algebra) in addition to the nice answer by Ross M. –  Amzoti Nov 26 '12 at 5:02
Not sure why there are three answers and yet none of them seem to answer the question directly -- figured it out: Odd sum: $a(n) = n^2(2n^2-1)^2$ Even sum: $a(n) = 2n^2(n+1)^2$ –  KaliMa Nov 26 '12 at 5:14
Are you sure the odd sum is correct? –  Amzoti Nov 26 '12 at 5:30
@Amzoti Typo'd it, shouldn't be squared at the end –  KaliMa Nov 26 '12 at 5:33
Still looks like there is a problem, shouldn't it be: type "Sum[(2*i+1)^3,{i,0,n}]" at woframalpha and see if it matches. You can remove the +1 for the even result. –  Amzoti Nov 26 '12 at 5:39

$$\sum\limits_{i=1}^n(2i)^3=8\sum\limits_{i=1}^ni^3$$ $$\sum\limits_{i=1}^n(2i-1)^3=\sum\limits_{i=1}^{2n}i^3-\sum\limits_{i=1}^n(2i)^3$$

Of course, the sum of cubes is (for some strange reason...) the square of the sum of consecutive integers.

$$\sum\limits_{i=1}^ni^3=\left(\sum\limits_{i=1}^ni\right)^2=\frac{n^2(n+1)^2}4$$

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Here is a question regarding the strange reason. –  robjohn Feb 3 '13 at 13:39

There is a very general way for producing formulae for functions $F:\mathbb{N} \rightarrow \mathbb{Z}.$ Let me explain.

Consider the set of functions $F:\mathbb{N} \rightarrow \mathbb{Z}.$ On such functions there is a discrete analog of differentiation called the shift operator, denoted $\Delta$, which given a function $F$ produces a function $\Delta F$ via the formula $\Delta F(x) := F(x+1) - F(x).$ Now classically, if one is given an infinitely differentiable function $f:\mathbb{R} \rightarrow \mathbb{R}$ one can consider the Taylor series $\displaystyle\sum_{i=0}^{\infty} \frac{f^{(k)}(0)}{k!}x^k$ and ask if this series converges and is equal to $f$ near $0.$ In the discrete case, note that iterated powers of the shift operator are always well defined. So it is natural to ask if there is a discrete analog of Taylor series. These analogs are called Mahler series. Given a function $F:\mathbb{N} \rightarrow \mathbb{Z},$ we define it's Mahler series via the formula

$$\sum_{i=0}^{\infty} \Delta F(0){{x}\choose{k}}.$$

The next question to ask is when is $F$ equal to it's Mahler expansion. And this is where the discrete case is nicer than the classical, for the answer is always. That is

$$F(x) = \sum_{i=0}^{\infty} \Delta F(0){{x}\choose{k}}$$

for all $x \in \mathbb{N}$

(Note if $k>x,$ we have ${x\choose k} = 0$ so there is no issue of convergence in the above series).

So if I have such a function $F:\mathbb{N} \rightarrow \mathbb{Z}$ which I want to express via a nice formula, a good place to begin is to try to write down it's Mahler expansion.

$$F(x) = \sum_{i=0}^{x} (2i)^3.$$

Then

$$\Delta F(x) = (2(x + 1))^3$$

$$\Delta^2 F(x) = (2(x + 2))^3 - (2(x + 1))^3 = 24x^2 +72x + 56$$

$$\Delta^3 F(x) = \Delta^2 F(x+1) - \Delta^2 F(x) = 192x + 208$$

$$\Delta^4 F(x) = \Delta^3 F(x+1) - \Delta^3 F(x) = 192$$

and

$$\Delta^k F(x) = 0 \text{ if } k\ge5.$$

It follows

$$F(x) = {{x}\choose{1}} + 56{{x}\choose{2}} + 208{{x}\choose{3}} + 192{{x}\choose{4}}.$$

Following the same mechanical procedure we can also produce a formula for the some of the cubes of the odd natural numbers and many other functions of this sort.

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For the odd cubes, you are asking about $\sum_{i=0}^n (2i+1)^3=\sum_{i=0}^n 8i^3+12i^2+6i+1$. Now feed each term to Faulhaber's formula

For the even cubes, you want $\sum_{i=1}^n (2i)^3=8\sum_{i=1}^n i^3$

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I think I figured out the odd cubes via n^2*(2n^2 - 1). Having trouble with evens. –  KaliMa Nov 26 '12 at 5:02
@KaliMa: For the evens, the sum is just $\frac 16 n(n+1)(2n+1)$ –  Ross Millikan Nov 26 '12 at 5:03
I thought that equation is for the sum of first n general cubes (even and odd together) –  KaliMa Nov 26 '12 at 5:05
@KaliMa: True, but the upper index is half the last number that is cubed. The idea is if you want to do something with all the evens, just do it with all the numbers and double them. The factor 8 accounts for the doubling. –  Ross Millikan Nov 26 '12 at 5:07
@KaliMa: No, $40=8(1+4)=8(1+2^2), 8+64=8(1+2^3)=8(1+8)=72$ –  Ross Millikan Nov 26 '12 at 5:11


# Why does applying a small angle approximation to equivalent forms of $\frac{\cos^2 \theta}{\sin \theta\tan \theta}$ yield different results?

In a question about small angle approximations I had answered, I simplified the trigonometric expression using identities and then applied the small and approximation.

I hadn't made any arithmetic mistakes, but the answer I got was different to the mark scheme, where they applied the small angle approximation straight away with no simplification.

My question is: Why do these different approaches yield different answers?

Edit: The question is:

$$\frac{\cos^2 \theta}{\sin \theta\tan \theta}$$

$$\theta^{-2}-{\frac 12}$$

The mark scheme says:

$${\frac 14}{\theta^2}+\theta^{-2}-1$$

• It might help if you could post both approaches, including the simplifications you made. Also, why are you sure that you have no errors? – Dirk Apr 24 at 8:48
• @Dirk. My teacher and I went through it many times and we found nothing arithmetically wrong. I will add the question, although it may not look pretty – Adam Cummings Apr 24 at 8:50
• Your LaTeX looked reasonably pretty (congratulations!), but I think I made it a little prettier. :) Be that as it may ... You should also include your simplification, so that everyone can be as certain as you and the teacher that you made no mistake there. – Blue Apr 24 at 9:20
• We cannot judge your work if you don't explain it. – Yves Daoust Apr 24 at 9:22

The correct approximation using Taylor expansion is $$\theta^{-2}-\dfrac76+O(\theta^2)$$. However, since your approximation is not taking higher order terms into account, it undoubtedly depends on the expression you start with, even if they are equivalent.

For instance, you could write

$$\frac{\cos^2 \theta}{\sin\theta\tan\theta}\simeq\frac{(1-\frac12\theta^2)^2}{\theta\cdot\theta}=\theta^{-2}-1+\frac{\theta^2}4$$

Or

$$\frac{\cos^2 \theta}{\sin\theta\tan\theta}=\frac{\cos^3\theta}{\sin^2\theta}\simeq\frac{(1-\frac12\theta^2)^3}{\theta^2}=\theta^{-2}-\frac{3}{2}+\frac{3\theta^2}4-\frac{\theta^4}{8}$$

Note that only the leading term $$\theta^{-2}$$ is correct.

Here is a plot to compare the variants. The difference with $$\frac{\cos^2\theta}{\sin \theta\tan \theta}$$ is plotted.

From the top:

• orange: $$\theta^{-2}-\frac12$$
• green: $$\theta^{-2}-1+\frac{\theta^2}4$$
• blue: $$\theta^{-2}-\dfrac76$$
• purple: $$\dfrac{(1-\frac12\theta^2)^3}{1-(1-\frac12\theta^2)^2}$$, obtained from $$\dfrac{\cos^3\theta}{1-\cos^2\theta}$$, yet another equivalent expression
• red: $$\theta^{-2}-\frac{3}{2}+\frac{3\theta^2}4-\frac{\theta^4}{8}$$

Note that all these approximations are local: they are valid near $$0$$, and the farther from $$0$$, the worst the approximation. With Taylor expansion, you can add more terms to get a better approximation on a larger (but still finite) interval. Notice the behavior of the purple solution on a larger interval is better: it's because it's not a polynomial approximation but a rational function.

Note also that the difference, which is plotted, is (locally) much lower than the leading term $$\theta^{-2}$$.

If the goal is to get a better approximation on a bounded interval, there are other methods, such as uniform approximation by a polynomial (on a bounded interval) using Chebyshev's equioscillation theorem, Padé approximants, splines...

• I don't want to add a new answer that would be almost identical to this one, but I would try to emphasize the following point: $$\frac{\cos\theta}{\tan^2\theta}=\frac{\cos^2\theta}{\sin\theta\tan\theta}=\frac{\cos^3\theta}{\sin^2\theta}.$$ While the denominator is always approximated by $\theta^2$ when ignoring higher order terms, the numerator is obviously different in each case. – Ennar Apr 24 at 9:30
• @Ennar And the first is certainly the simplifiation the OP did. – Jean-Claude Arbaut Apr 24 at 9:32
• Yes. Hopefully looking at these expressions side by side will help them. – Ennar Apr 24 at 9:33
• Thanks you, this is very helpful. Apologies for the barrage of questions, but which of these result is "more correct"? (If that makes any sense). If there are multiple ways of obtaining the approximation, why has the mark scheme only specified one as correct? – Adam Cummings Apr 24 at 10:28
• @AdamCummings None of these approximations is very good (you can expect one correct term, and even that may fail), but if you want the "best", it's the one with the constant term closest to $-7/6$ (you could also try $\frac{\cos^3 x}{1-\cos^2 x}$, slightly better). If you want several terms, the good way is the Taylor formula. Now, regarding the mark scheme: as dumb as it seems, you had only to apply the approximation without simplification. – Jean-Claude Arbaut Apr 24 at 10:36


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# The ideals $\langle y-x-1\rangle$ and $\langle x-2,y-3\rangle$ in $\mathbb C[x,y]$ are prime

The following is a quote from Wolfram MathWorld article about prime ideals.

A maximal ideal is always a prime ideal, but some prime ideals are not maximal. In the integers, $\{0\}$ is a prime ideal, as it is in any integral domain. Note that this is the exception to the statement that all prime ideals in the integers are generated by prime numbers. While this might seem silly to allow this case, in some rings the structure of the prime ideals, the Zariski topology, is more interesting. For instance, in polynomials in two variables with complex coefficients $\mathbb C[x,y]$, the ideals $$\langle 0\rangle \subset \langle y-x-1\rangle \subset \langle x-2,y-3\rangle$$ are all prime.

How can we prove that $\langle y-x-1\rangle$ and $\langle x-2,y-3\rangle$ are prime ideals in $\mathbb C[x,y]$?

Perhaps showing that quotient rings are $\mathbb C[x,y]/\langle y-x-1\rangle\cong \mathbb C[x]$ and $\mathbb C[x,y]/\langle x-2,y-3\rangle\cong\mathbb C$ would be way to go? (This would mean the quotient ring is an integral domain and therefore the ideal is prime.)

• Yes to your questions at the very end. They are kernels of evaluation maps. – anon Apr 24 '16 at 5:22
• @user26857 Thanks for the edit. As you can see, I quoted the text from the link, so I written in in the same way. (As it was an exact quote.) What somewhat surprise me was the removal of (commutative-algebra) tag. According to the tag-excerpt this tag is for: "Questions about commutative rings, their ideals, and their modules." (But you are certainly more active in questions from this area as I am. So you probably know the norms of this math.SE subcommunity better.) – Martin Sleziak Apr 24 '16 at 6:42
• To see that $<y - x- 1>$ is a prime ideal, you can also note that, like every other degree $1$ polynomial, it is irreducible, and that irreducible elements are prime in, for example, UFDs (polynomial rings over $\mathbb{C}$ are certainly unique factorization domains). – Badam Baplan Aug 11 '18 at 15:50
• @BadamBaplan Maybe you could expand your comment a bit and post it as an answer. – Martin Sleziak Aug 11 '18 at 17:09

Let me try to expand anon's comment to an answer.

Let $R$ be a commutative ring. If $f(x)\in R[x]$ and $a\in x$ then there exist uniquely determined $q(x)\in R[x]$ and $b\in R$ such that $$f(x)=q(x)(x-a)+b.$$ Moreover, we have $b=f(a)$.

This follows from the fact that we can divide by monic polynomials in $R[x]$ for any commutative ring $R$. Using this observation we can also show that the map $f(x)\mapsto f(a)$ is a homomorphism from $R[x]$ to $R$ and the kernel is precisely the ideal $\langle x-a\rangle$.

Since we have $\mathbb C[x,y]=\mathbb C[x][y]$ we can use the above to get a homomorphism $$\varphi \colon \mathbb C[x,y] \to \mathbb C[x] \qquad \varphi \colon f(x,y) \mapsto f(x,x+1).$$ Kernel of this homomorphism is $\langle y-x-1 \rangle$. This shows that $$\mathbb C[x,y]/\langle y-x-1 \rangle \cong C[y].$$ So the quotient ring is an integrity domain and the ideal is a prime ideal.

Similarly, we can get, using the above observation, the homomorphisms $$\psi_1 \colon \mathbb C[x][y] \to \mathbb C[x] \qquad \psi_1 \colon f(x,y) \mapsto f(x,3)$$ and $$\psi_2 \colon \mathbb C[x] \to \mathbb C \qquad \psi_2 \colon g(x) \mapsto g(2).$$ By composing these two homomorphisms we get $\psi=\psi_2\circ\psi_1$ from $\mathbb C[x,y]$ to $\mathbb C$ given by $f(x,y)=f(2,3)$.

Again we have that $\operatorname{Ker} \psi_2 = \langle x-2 \rangle$.

The polynomials in $\operatorname{Ker} \psi$ are precisely the polynomials such that $\psi_1(f)\in \operatorname{Ker} \psi_2$. So $f\in \operatorname{Ker} \psi$ if and only if $$f(x,y) = q(x,y)(y-3) + h(x)(x-2).$$ Every such polynomial belongs to $\langle x-2,y-3 \rangle$. And since both $x-2$ and $y-3$ belong to the kernel of $\psi$, we get $$\operatorname{Ker} \psi = \langle x-2,y-3 \rangle.$$

From this we get $\mathbb C[x,y]/\langle x-2,y-3 \rangle\cong\mathbb C$. Again, since the quotient ring is an integrity domain, we get that the ideal is prime.


# Math nomenclature

I kinda feel ashamed about asking this, but could someone explain me what this means? $$\binom{k}{i}$$

That should indicate the number of nodes at a certain depth in a binomial heap..but I can't remember what that actually means and I wasn't successful with Google!

Thanks

==================================

so if i had $$\binom{0}{0}$$ would that be $$\dfrac{0!}{0!(0-0)!}$$ ...what would that be...zero?

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Search for binomial coefficient and use \binom{k}{i} to get MathJax to format it properly for display as $\binom{k}{i}$. –  Dilip Sarwate Feb 9 '13 at 15:47
No need to feel ashamed, Seb! That's what we're here for...ask away! Your question was good, and showed context, was clear, and it's hard to google for something when you don't know what to call it! –  amWhy Feb 9 '13 at 16:07
thanks for the edits and for the patient :)! –  TriRook Feb 9 '13 at 16:07
@Seb Once you know that $0! = 1$, you can answer your own question about $\binom00$. In general $\binom{k}{i}$ is only zero when $i < 0$ or when $i > k$. –  Erick Wong Feb 9 '13 at 19:26

I assume you mean $$\dbinom{n}k$$ $\dbinom{n}k$ is a short hand for the number of ways in which you can choose $k$ objects from $n$ distinguishable objects. These number are called the binomial coefficients. Some textbooks and articles also denote this as $C(n,k)$.

As an example, if we have three colored balls, $\color{red}{\text{red}}$, $\color{blue}{\text{blue}}$ and $\color{brown}{\text{brown}}$, then there are three ways of choosing two balls. \begin{matrix} \color{red}{\text{red}} & \color{blue}{\text{blue}}\\ \color{blue}{\text{blue}} & \color{brown}{\text{brown}}\\ \color{brown}{\text{brown}} & \color{red}{\text{red}} \end{matrix} Note that when we say we choose, we are not interested in the order in which these are picked i.e. $\color{red}{\text{red}} \,\, \color{blue}{\text{blue}}$ and $\color{blue}{\text{blue}} \,\, \color{red}{\text{red}}$ refer to the same choice of two balls. Hence, we have $\dbinom{3}2 = 3$.

Similarly, if we have $5$ colored balls, say $\color{red}{\text{red}}$, $\color{blue}{\text{blue}}$, $\color{brown}{\text{brown}}$, $\color{orange}{\text{orange}}$, and $\color{lightgreen}{\text{green}}$, there are $10$ ways of choosing $3$ balls. \begin{matrix} \color{red}{\text{red}} & \color{blue}{\text{blue}} & \color{brown}{\text{brown}}\\ \color{red}{\text{red}} & \color{blue}{\text{blue}} & \color{orange}{\text{orange}}\\ \color{red}{\text{red}} & \color{blue}{\text{blue}} & \color{lightgreen}{\text{green}}\\ \color{red}{\text{red}} & \color{orange}{\text{orange}} & \color{brown}{\text{brown}}\\ \color{red}{\text{red}} & \color{orange}{\text{orange}} & \color{lightgreen}{\text{green}}\\ \color{red}{\text{red}} & \color{brown}{\text{brown}} & \color{lightgreen}{\text{green}}\\ \color{blue}{\text{blue}} & \color{brown}{\text{brown}} & \color{orange}{\text{orange}}\\ \color{blue}{\text{blue}} & \color{brown}{\text{brown}} & \color{lightgreen}{\text{green}}\\ \color{blue}{\text{blue}} & \color{orange}{\text{orange}} & \color{lightgreen}{\text{green}}\\ \color{orange}{\text{orange}} & \color{brown}{\text{brown}} & \color{lightgreen}{\text{green}} \end{matrix}

In general, $$\dbinom{n}r = \dfrac{n!}{r!(n-r)!}$$ where $k! = k \times (k-1) \times (k-2) \times \cdots \times 2 \times 1$.

The name binomial coefficient arises from binomial theorem. When we expand $(x+y)^n$, the coefficient of $x^k y^{n-k}$ is given by $\dbinom{n}k$ i.e. $$(x+y)^n = \sum_{k=0}^n \dbinom{n}k x^k y^{n-k}$$ There are a lot of wonderful properties these binomial coefficients satisfy and I highly recommend you to go through the wiki-page for these properties.

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Nice answer! Your examples would probably look better as matrixes than as aligns. –  Rahul Feb 9 '13 at 16:36
@ℝⁿ. Thanks. Yes. These indeed look better. I have updated them. –  user17762 Feb 9 '13 at 16:39
awesome answer thanks for taking time to go through it so well!! I have a question but i'll post it below cause i'm trying to put a formula in.. –  TriRook Feb 9 '13 at 17:02

$$\binom{k}{i}\; \text{ is called a \color{blue}{\bf{binomial\;coefficient}}, which is read as "k choose i"}$$

"k choose i" comes from the fact that if gives you the number of ways to choose $i$ elements from a set of $k$ elements. The term "binomial coefficient" makes explicit its relationship to the binomial theorem. When we expand $(x+y)^n$, the coefficient of $x^k y^{n-k}$ is given by $\large\binom{n}k$ i.e. $$(x+y)^n = \sum_{k=0}^n \dbinom{n}k x^k y^{n-k}$$

You can compute your binomial coeffient by noting that $$\displaystyle \;\binom{k}{i} = \frac{k!}{i!(k-i)!} = \frac{k\cdot (k-1)\cdots (k - i + 1)}{i\cdot (i-1) \cdots 2 \cdot 1}$$

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I've not in recent times ever heard $\binom{k}{i}$ read as anything other than "$k$ choose $i$". (But maybe I don't get out enough ...) –  Peter Smith Feb 9 '13 at 16:12
@PeterSmith Good point; "often" should have been "usually", which should probably be omitted altogether! Edited accordingly. –  amWhy Feb 9 '13 at 16:16

$k \choose i$ is a binomial coefficient. It is the coefficient of $x^i$ in expanding $(1+x)^k$. It is also the number of $i$-element subsets of a $k$-element set. It can be computed via $\frac{k!}{i!(k-i)!}$ or with several recursive formulas. Values are often listed with help of Pascal's triangle.

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For the question in the edit: We have $0! = 1$, so $\binom{0}{0} = 1$ which is the first entry in Pascal's triangle.

A previous question on this site asked about the definition of zero factorial; there are several excellent answers there. As Zhen Lin points out in a comment, the best explanation of why $0! = 1$ depends on how you define the factorial.

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# Logical functions with no arguments

I am trying to understand how logical functions are evaluated if no arguments are passed to them.

{Or[], Nor[], And[], Nand[], Xor[], Xnor[]}

{False, True, True, False, False, True}

I tried Trace but it didn't offer many clues. Mathematica (12.2.0 on Win7-x64) evaluates these expressions without any messages.

• If you have a look in the Properties and relations section you will see that it zero-argument evaluates to False. reference.wolfram.com/language/ref/Or.html
– kcr
Feb 28, 2022 at 8:39
• Similar situation in Java developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…
– kcr
Feb 28, 2022 at 8:39
• Perhaps this is the way they are supposed to be constructed based on theoretical coding principles? Sorry for the triple comment...I am horrible
– kcr
Feb 28, 2022 at 8:40
• These are simply the trivial cases for these functions. And, for example, is True unless one of it's arguments is False. The zero-argument case is just an extension of that. Feb 28, 2022 at 11:05
• One sees similar behavior with Plus and Times (where it might be more obviously plausible). Symbols with the OneIdentity attribute will, when given an empty argument list, evaluate to their respective identity elements. Feb 28, 2022 at 15:57

I am pretty sure that Syed figured out the question by now, but for future visitors I thought it a good idea to write something.

1. Why does Or[] return False?

From the docs we read

It evaluates its arguments in order, giving True immediately if any of them are True, and False if they are all False.

which is another way of saying that it returns True if any or all of its arguments is true and False otherwise.

1. Why does Nor[] return True?

The easiest way to realize this -I think- is again to have a look at the documentation where we find that it is equivalent to Not[Or[]]. So, Not[False] is True.

1. Why does And[] return True?

Well, again this can be seen directly from the docs and also from the comment by @Sjoerd Smit.

It evaluates to False if any of its arguments are false, and to True otherwise.

1. Why does Nand[] return False?

Similar to the explanation of Nor in this case. Nand[] is equivalent to Not[And[]] and hence Not[True] which is False.

1. Why does Xor[] return False?

Well, it has to evaluate all of its argument and yields True if an odd number of them is True and False otherwise. Since, we do not have an odd number of True statements, it returns False

1. Why does Xnor[] return True?

Xnor[] is equivalent to Not[Xor[]]

• Thanks for the write up.
– Syed
Jan 14 at 12:56
• @Syed if you want to add more to it, let me know and I will turn it into a wiki :-)
– bmf
Jan 14 at 12:58
• It seems that in your description of Nor True and False should be reversed. Jan 14 at 19:26
• @yarchik yes, of course you are right. thanks for spotting this :-)
– bmf
Jan 15 at 1:38

The answer of @bmf is perfect, but I would like to bring some arguments that allow us to better remember the results. It also shows a connection to other MA functions.

The Xor function can be seen as Sum modulo 2. In fact

Sum[1, {i, 0}]
(* 0 *)

which explains why Xor[] == False.

The And function can be regarded as the Product modulo 2. In fact

Product[1, {i, 0}]
(* 1 *)

which explains why And[] == True.

See also a comment of @DanielLichtblau above, which argues in terms of Plus and Times.

• (+1) that's a very neat description!!!
– bmf
Jan 15 at 1:39


# Is the following generalization of Strong Law of Large Numbers valid?

According to SLLN, if $$X_1, X_2, \ldots$$ is an infinite sequence of i.i.d. random variables with expected value $$\mu$$ and $$S_n := \sum_{i=1}^n X_i/n$$ then $$S_n \to \mu$$ almost surely.

If the sequence is instead $$X_{1,1}, X_{2,1}, X_{2,2}, \ldots, X_{n,1}, \ldots,X_{n,n}, X_{n+1,1}, \ldots$$ whose terms are i.i.d. random variables with expected value $$\mu$$ and $$S_n := \sum_{i=1}^n X_{n,i}/n$$, can we still say $$S_n \to \mu$$ almost surely? If not are there any extra conditions that make this true?

I will give an example for why this is not trivial. Consider distribution of variables is Bernoulli with $$p=1/2$$ and the sample $$0, 1, 1, 0, 0, 0, 1, \ldots$$. That is $$(2n-1) \times 0$$ are followed by $$2n \times 1$$ and that is repeated for every $$n > 0$$. For this sequence $$S_n$$ converges to $$1/2$$ in the first case and the limit does not exist in the second. If the probability of all such examples is $$0$$ then convergence will be almost sure, but this is something that can probably be proven on a case by case with union bound inequality. I just wonder whether there is some general result that makes such analysis easier.

First observe that in this setting, the sequence $$\left(S_n\right)_{n\geqslant 1}$$ is independent. For an independent sequence, in view of the Borel-Cantelli lemma, almost sure convergence and complete convergence are equivalent. Hence $$S_n\to \mu$$ almost surely if and only if for all positive $$\varepsilon$$, $$\tag{*} \sum_{n\geqslant 1}\mathbb P\left(\lvert S_n-\mu\rvert \gt\varepsilon\right)<+\infty.$$ Let $$(Y_i)_{i\geqslant 1}$$ be an i.i.d. sequence such that $$Y_1$$ has the same law as $$X_{1,1}$$. Then $$(*)$$ is equivalent to $$\forall \varepsilon>0, \sum_{n\geqslant 1}\mathbb P\left(\left\lvert \sum_{j=1}^n(Y_j-\mu)\right\rvert \gt n\varepsilon\right)<+\infty.$$ By Theorem 3 in this paper by Baum and Katz, this is equivalent to $$\mathbb E[Y_1^2]<\infty$$, hence we do need extra conditions.
Yes, the result holds equally well in both cases. More generally, Let $$(X_i:i=1,2,...)$$ be any infinite sequence of i.i.d. random variables with expected value $$μ$$, and let $$F: \Bbb Z_+\to\mathrm P(\Bbb Z_+)$$ such that $$|F(k)|\in\Bbb Z_+$$ with $$|F(k)|\to\infty$$ as $$k\to\infty$$. Then$$\frac1{|F(n)|}\sum_{i\in F(n)}X_i\;\to_{\text{a.s.}}\;\mu$$as $$n\to\infty$$. The proof is essentially the same as that for the simple average of an initial segment of the original sequence, because the i.i.d. property of any (finite or infinite) subsequence of the original sequence $$(X_i:i=1,2,...)$$ follows from that of the original sequence.


subset of binary space countable?

I know that the set of all binary sequences is uncountable. Now consider the subset of this set, that whenever a digit is $1$, its next digit must be $0$. Is this set countable?

I think it is not because it is like "half" of the set of binary sequences. By that I mean if I just delete the sequences with a $1$ followed by a $1$, then I'm left with the required sequence. But I'm not sure how to show this clearly. Can someone think of a bijection that would prove this? Or am I wrong? Please help.

Thanks

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If you insist to have a bijection, here is one. Your set $S$ is the set of all binary infinite words containing no factor $11$. Any such word can be uniquely written as an infinite word on the alphabet $\{0, 10\}$, that is, $S = \{0, 10\}^\omega$. For instance, $$0100100010010(100)^\omega = \color{blue}{0}\color{red}{10}\color{blue}{0}\color{red}{10}\color{blue}{0}\color{blue}{0}\color{red}{10}0\color{red}{10}(\color{red}{10}\color{blue}{0})^\omega$$

It follows that the map from $\{0,1\}^\omega \to S$ consisting to replace every $1$ by $\color{red}{10}$ is a bijection. No need of the Cantor-Bernstein-Schroeder theorem.

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You can explicitly write an injective function from the set of all binary sequences to the set you are considering: just map every sequence to one with $0$ inserted in every other position.

So no, this is definitely not countable.

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But how does an injective function suffice here? Don't you need a bijective function to show the same cardinality? – verticese Feb 25 at 8:02
@verticese: it shows that it has at least the same cardinality as the bijective sequences (that is, the continuum), which is all you need to show that it is uncountable. But it is not hard to get a bijection if you want: of course you also have an injection going the other way: the identity. This gives you a bijection by Cantor-Bernstein-Schroeder. In this case, you can probably define a bijection explictly without much difficulty, although there would be no nice and concise formula, I think, but rather a procedure. – tomasz Feb 25 at 8:05
As @tomasz notes: If $A\subseteq B$ then $|A|\le|B|$. If there's an injection $B\to A$, then $|B|\le|A|$, so by Cantor-Schroeder-Bernstein there's a bijection $A\to B$. – BrianO Feb 25 at 8:12

Your set has an uncountable subset: consider the infinite sequences of $a$ and $b$ where $a=10$ and $b=00$, the set of these sequences has the same cardinality of the set of sequences of $0$ and $1$ (just map $a \mapsto 0$ and $b \mapsto 1$), and it is is clearly a subset of your set.

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This is the same as what I suggested in my answer. – tomasz Feb 25 at 9:34
@tomasz I only see that now after reading your comment. I would say this answer is another viewpoint on the same thing you suggested, and sometimes an alternate viewpoint helps (as in how this one helped me understand your answer better). – Todd Wilcox Feb 25 at 16:53

Here's another (possibly) interesting way of proving uncountability, since you tagged this as a topology question. You can inject your set into $[0,1]$ under the evaluation map

$$(x_i) \mapsto \sum_{i = 1}^\infty \frac{x_i}{2^i}$$

This is injective on your set (but not the set of all binary sequences, since $0.0111\dots = 0.100\dots$). We shall show the image of this map is a perfect set, and hence uncountable (see Rudin, or some other source on metric spaces). Surely no point in the set is isolated, because if we have an expansion $x = 0.a_0 a_1 \dots a_2$, and if this expansion does not eventually end in all zeroes, we may truncate the expansion to find another number in the image which is as close to this number as possible. If the number does end in all zeroes, you may add a one to an arbitrary position to get another number as closed as possible. Now suppose $\lim x_i = x$, where each $x_i$ is in the image of the evaluation. We must have $0 < x < 1$, for if $x = 1$, the $x_i$ must eventually be of the form $0.11\dots$, which is impossible. Write $x = 0.a_0 a_1 \dots$. Suppose $a_k = 1$. If we force $|x_i - x| < 1/2^{k+2}$, then the first $k+2$ digits of $x_i$ must be the same as the first $k+1$ digits of $x$, hence $a_{k+1} = 0$. Thus the image is perfect, and the set is uncountable.

Intuition tells me this set is probably fractal, similar to the Cantor set, but I can't visualize it very well.

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You can inject the set of all sequences of integers into this your set (call it $A$): $$(n_i)_{i<\omega} \mapsto (1 + 0^{n_0+1} + 1 + 0^{n_1+1} + \dotsc)\colon \omega^\omega\to A.$$ where the value is meant as an infinite regular expression. That is, the $i$-th $1$ is followed by $n_i+1$ many $0$s. (The "$+ 1$" guarantees that the value under this function is a member of A.)

Thus, $A$ has cardinality $\lvert A\rvert \ge |\omega^\omega| = \lvert\Bbb R\rvert$. But $A\subseteq 2^\omega$, so $\lvert A\rvert \le \lvert 2^\omega\rvert = \lvert\Bbb R\rvert$. By the Cantor-Schroeder-Bernstein Theorem, $\lvert A\rvert = \lvert\Bbb R\rvert$.

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Notation: $\chi_A (x)=1$ if $x\in A$, and $\chi_A (x)=0$ if $x\not \in A.$

Let $E=\{2 n:n\in N\}.$ For $S\subset E \;$ let $S^*=\{n :2 n\in S\}$ and let $f(S)=(\chi_{S^*} (n))_{n\in N}.$

Then $f$ is a bijection from the power-set $P(E)$ to the set $B$ of all binary sequences.

For $S\subset E$ let $g(S)=(\chi_S (n))_{n\in N}.$ Then $g$ is an injection from $P(E)$ into the set $G$ of binary sequences $(b_n)_{n\in N}$ that satisfy $\forall n\in N\;(b_n=1\implies b_{n+1}=0)\}.$

Thus $g f^{-1}$ is an injection from $B$ into $G.$ So $G$ is uncountable because it has the uncountable subset $\{g f^{-1}(x): x\in B\}.$

The Cantor-Schroeder-Bernstein Theorem implies that for any $B,G$ with $G\subset B,$ there is a bijection $h:B\to G$ if there is an injection $i:B\to G.$

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# Variance of number of heads in three flips of a coin selected from a set of fair and unfair coins

We have $10$ coins, $2$ are two-tailed, $2$ are two-headed, the other $6$ are fair ones. We (randomly) pick a coin and we flip it $3$ times. Find the variance of the number of gotten heads.

My attempt:

$X$ - number of heads that we got

$\mathbb{P}\left(X=0\right)=\frac{2}{10}\cdot1\cdot1\cdot1 + \frac{6}{10}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}$ - we picked two-tailed coin or fair one

$\mathbb{P}\left(X=1\right)=\frac{6}{10}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot 3$ - we picked fair one and we have 3 possibilites: HHT, HTH, THH

$\mathbb{P}\left(X=2\right)=\frac{6}{10}\cdot{3\choose2}\cdot\left(\frac{1}{2}\right)^2\cdot\left(\frac{1}{2}\right)^1$ - again we picked fair one and we have to have 2 successes in 3 tries

$\mathbb{P}\left(X=3\right)=\frac{2}{10}\cdot1\cdot1\cdot1+\frac{6}{10}\cdot\left(\frac{1}{2}\right)^3$ - we can pick two-headed or fair one coin

And the rest is simple,

$\text{Var}X=\sum_{i=0}^3i^2\cdot\mathbb{P}\left(X=i\right)-\left(\sum_{i=0}^3i\cdot\mathbb{P}\left(X=i\right)\right)^2$

Is my solution correct?

• The information is symmetric, so the expressions for the probability of 1 head or 2 heads should be the same. – Paul Feb 2 '17 at 18:42
• Yes, your approach is correct. $\checkmark$ – callculus Feb 2 '17 at 18:47
• @Paul well, yea, I should have decided whether I use probability mass function or just standard combinatorics. But the result is the same – SantaXL Feb 2 '17 at 18:50

That's one approach.   It is good.   Here's another,

We know $X\mid C \sim \mathcal{Bin}(3, C)$ and $\mathsf P(C=c)=\tfrac 15\mathbf 1_{c=0}+\tfrac 35\mathbf 1_{c=1/2}+\tfrac 15\mathbf 1_{c=1}$

So $\mathsf E(C)= \tfrac 1{2}$ and $\mathsf {Var}(C)=\tfrac 1{10}$ and $\mathsf E(C^2)=\frac 7{20}$

By the Law of Total Probability: $\mathsf E(X) \\ = \mathsf E(\mathsf E(X\mid C)) \\ = \mathsf E(3C) \\ = \frac 32$

Then by the Law of Total Variance: $\quad\mathsf {Var}(X) \\ = \mathsf{Var}(\mathsf E(X\mid C))+\mathsf E(\mathsf{Var}(X\mid C)) \\ = \mathsf {Var}(3C)+\mathsf E(3C(1-C)) \\ = \frac 9{10}+\frac 32-\frac {21}{20} \\ = \frac {27}{20}$

• Hmm, your result is different from mine. In my approach: $\mathbb{P}(X=0)=\mathbb{P}(X=3)=\frac{9}{40}, \ \mathbb{P}(X=1)=\mathbb{P}(X=2)=\frac{11}{40} \\ EX=1\cdot\frac{9}{40}+2\cdot\frac{9}{40}+3\cdot\frac{11}{40}=\frac{9+18+33}{40}=\frac{60}{40}=\frac{3}{2} \\EX^2=1^2\cdot\frac{9}{40}+2^2\cdot\frac{9}{40}+3^2\cdot\frac{11}{40}=\frac{9+36+99}{40}=\frac{144}{40}=\frac{18}{5} \\ VarX=EX^2-E^2X=\frac{18}{5}-\frac{9}{4}=\frac{27}{20}$ – SantaXL Feb 3 '17 at 13:52
• Where is the mistake? – SantaXL Feb 3 '17 at 13:53
• In mine. @SantaXL corrected. – Graham Kemp Feb 3 '17 at 14:49
• ok, thanks for showing me this approach, I wasn't aware such Law exists. – SantaXL Feb 3 '17 at 15:22


# Proving $24\mid5^{2n}-1$ using modular arithmetic

So, in general I am aware of how to use modular arithmetic to prove a divisibility. But I have the following problem:

Prove that $24\mid5^{2n}-1$ for all $n\in\mathbb Z$.

I know that theoretically, I could show 23 different cases that the expression is congruent to $0\bmod24$, but that seems like it might be excessive to me.

Is there a faster way to show this?

• $5^{2n}=(5^2)^n=25^n$ Now, note that $25^n=(24+1)^n$ and use something that should be familiar to you. Dec 11, 2017 at 3:01

You want to prove what $5^{2n}$ is congruent to in modulo 24

Notice that $5^{2n} = (5^2)^n$

$5^2=25 \equiv 1 \pmod {24}$

So $5^{2n} \equiv 1^n \equiv 1 \pmod {24}$

• Edited, and thanks Dec 11, 2017 at 3:05
• Makes sense, thank you! Dec 11, 2017 at 3:19

You could use $5^{2n}-1=(5^2-1)(5^{2n-2}+5^{2n-4}+\ldots +1)$ where you are summing the geometric series.

Use induction. It is true for $n=1$. Assuming it is true for $n$, show it is true for $n+1$: $$5^{2(n+1)}-1=25(5^{2n}-1)+24 = 0 (mod \ 24).$$

1)

Prove $8|5^{2n} - 1$ and $3|5^{2n} - 1$

a) $5^{2n} - 1 = (5^n -1)(5^n + 1)$. $5^n \pm 1 \equiv 1^n\pm 1 \equiv 0\mod 2$ so $2|5^2 - 1$ and $2|5^2 + 1$ and so $4|5^{2n}-1$. If $5^n - 1 = 2k$ then $5^{2n} + 1= 2k + 2 = 2(k+1)$. Either $k$ is even or $k+1$ is even so either $2k$ is even and $2(k+1)$ is divisible by $4$, or $2k$ is divisible by $4$ and $2(k+1)$ is even. Either way. $2k*2(k+1)= 5^{2n} -1$ is divisible by $8$.

b)$5^{2n}-1\equiv (-1)^{2n}-1 \equiv 1^n - 1 \equiv 0 \mod 3$.

So $3|5^{2n}-1$.

2)

Or notice $5^{2n} -1 = 25^n -1 \equiv 1^n -1 \equiv 0 \mod 24$.

(I will admit, I did not see the obvious right away.)


# Does $i^4$ equal $1?$

I can't seem to find a solution to this for the life of me. My mathematics teacher didn't know either.

Edit: I asked the teacher that usually teaches my course today, and she said it was incredible that the other teacher didn't know.

My logic goes as follows:

any real number: $x$ to the fourth power is equal to $(x^2)^2$. Using this logic, $i^4$ would be equal to $(i^2)^2$. This would result in $(-1)^2$, and $(-1)^2 = 1$.

Obviously, this logic can be applied to any real numbers, but does it also apply to complex numbers?

• Yes, $i^4=1$. But what else could it equal ??
– user65203
Jan 7 '17 at 15:00
• To check if $i^4$ is real or not, why not try conjugating it and see what happens? Jan 7 '17 at 15:11
• Yes, @J.M.isn'tamathematician, but before you proceed to prove something, I'm going to get some paper and work it out for myself Jan 7 '17 at 15:21
• "I'm going to get some paper and work it out for myself" - that was my intention behind asking you those questions. Jan 7 '17 at 15:24
• If you're working within the naive approach to complex numbers often used when they're first introduced the students, "Suppose $i^2=-1$ but that other than that all the usual rules of arithmetic apply", then basically the bottom line is that this is one of the "usual rules" which is simply assumed to apply. Specifically, we assume that complex numbers are associative, that is that for any $a, b, c$, $a(bc)=(ab)c$. You might try showing in detail why this assumption implies $i^4=1$. Jan 12 '17 at 21:03

Yes. The powers of $i$ are cyclic, repeating themselves ever time the exponent increases by 4: $$i^0 = 1$$ $$i^1=i$$ $$i^2 = -1$$ $$i^3 = -i$$ $$i^4 = 1$$ $$i^5 = i$$ $$i^6 = -1$$ $$i^7 = -i$$ $$i^8 = 1$$ etc.

Your reasoning is excellent, and you should feel good about the fact that you figured this out on your own. The fact that your math teacher didn't know this is, in my professional opinion as a mathematics educator, a disgrace.

Edited to add: As Kamil Maciorowski notes in the comments, the pattern persists for negative exponents, as well. Specifically, $$i^{-1}= \frac{1}{i} = -i$$ If $\frac{1}{i}=-i$ seems odd, notice that $i(-i) = -i^2 = -(-1) = 1$, so $i$ and $-i$ are multiplicative inverses; therefore $i^{-1} = -i$. Once you know that, you can extend the pattern: $$i^{-1} = -i$$ $$i^{-2} = -1$$ $$i^{-3} = i$$ $$i^{-4} = 1$$ and so on.

Second update: The OP asks for some additional discussion of the property $\left( x^a \right)^b = x^{ab}$, so here is some background on that:

First, if $a$ and $b$ are natural numbers, then exponentiation is most naturally understood in terms of repeated multiplication. In this context, $x^a$ means $(x\cdot x\cdot \cdots \cdot x)$ (with $a$ factors of $x$ appearing), and $\left( x^a \right)^b$ means $(x\cdot x\cdot \cdots \cdot x)\cdot(x\cdot x\cdot \cdots \cdot x)\cdot \cdots \cdot (x\cdot x\cdot \cdots \cdot x)$, with $b$ sets of parentheses, each containing $a$ factors of $x$. Since multiplication is associative, we can drop the parentheses and recognize this as a product of $ab$ factors of $x$, i.e. $x^{ab}$.

Note that this reasoning works for any $x$, whether it is positive, negative, or complex. It even applies in settings were multiplication is noncommutative, like matrix multiplication or quaternions. All we need is that multiplication is associative, and that $a$ and $b$ be natural numbers.

Once we have established that $\left( x^a \right)^b = x^{ab}$ for natural numbers $a,b$ we can extend the logic to integer exponents. If $a$ is a positive number, and if $x$ has a multiplicative inverse, then we define $x^{-a}$ to mean the same thing as $\left(\frac1x\right)^a$, or (equivalently) as $\frac1{x^a}$. With this convention in place, it is straightforward to verify that for any combination of signs for $a,b$, the formula $\left(x^a\right)^b = x^{ab}$ holds.

Note however that in extending the formula to cover a larger set of exponents, we have also made it necessary to restrict the domain of values $x$ over which this property holds. If $a$ and $b$ are just natural numbers then $x$ can be almost any object in any set over which an associative multiplication is defined. But if we want to allow $a$ and $b$ to be integers then we have to restrict the formula to the case where $x$ is an invertible element. In particular, the formula $x^{a}$ is not really well-defined if $x=0$ and $a$ is negative.

Now let's consider the case where the exponents are not just integers but arbitrary rational numbers. We begin by defining $x^{1/a}$ to mean $\sqrt[a]{x}$. ( See Why does $x^{\frac{1}{a}} = \sqrt[a]{x}$? for a short explanation of why this convention makes sense.)

In this definition, we are assuming that $a$ is a natural number, and that $x$ is positive. Why do we need $x$ to be positive? Well, consider an expression like $x^{1/2}$. If $x$ is positive, this is (by convention) defined to be the positive square root of $x$. But if $x$ is negative, then $x^{1/2}$ is not a real number, and even if we extend our number system to include complex numbers, it is not completely clear which of the two complex square roots of $x$ this should be identified with. More or less the same problem arises when you try to extend the property to complex $x$: while nonzero complex numbers do have square roots (and $n$th roots in general), there is no way to choose a "principal" $n$th root.

Things get really crazy when you try to extend the property $\left(x^a\right)^b=x^{ab}$ to irrational exponents. If $x$ is a positive real number and $a$ is a real number, we can re-define the expression $x^a$ to mean $e^{a\ln x}$, and it can be proved that this re-definition produces the same results as all of the conventions above, but it only works because $\ln x$ is well-defined for positive $x$. As soon as you try to allow negative $x$, you run into trouble, since $\ln x$ isn't well-defined in that case. One can define logarithms of negative and complex numbers, but they are not single-valued, and there are all kinds of technicalities about choosing a "branch" of the logarithm function.

In particular -- and this is very important for the question at hand -- the identity $\left(x^a\right)^b=x^{ab}$ does not hold in general if $x$ is not a positive real number or if $a,b$ are not both integers. A lot of people misunderstand this, and indeed there are many, many, many, many questions on this site that are rooted in this misunderstanding.

But with respect to the question in the OP: It is perfectly reasonable to argue that $i^4 = \left(i^2 \right)^2$, because even though $i$ is a complex number, the exponents are integers, so the basic notion of exponentiation as repeated multiplication is reliable.

• Can you add please the values for $i^9, i^{10}, i^{11}, i^{12}$ ? Jan 5 '17 at 20:54
• @Hexacoordinate-C I can't tell if you're joking or not. Jan 5 '17 at 20:57
• Note that this post uses the field properties of the complex numbers (namely that the nonzero complex numbers form a group under multiplication). See math.stackexchange.com/a/2085631 for the answer to the last part of the question about trying to apply logic that works for real numbers to complex numbers. Jan 6 '17 at 3:48
• @user21820 Nice comment, +1
– user284001
Jan 6 '17 at 12:50
• You might also want to explain why the powers are cyclic because it wasn't instantly obvious why to me. Jan 6 '17 at 13:02

I'm surprised that none of the other answers pointed out the most important point in your question:

Obviously, this logic can be applied to any real numbers, but does it also apply to complex numbers?

This attitude is the right way to go. The logic you speak of is more precisely:

$x^{ab} = (x^a)^b$ for any real number $x$ and natural numbers $a,b$.

If you want the more general fact for integer exponents:

$x^{ab} = (x^a)^b$ for any real number $x \ne 0$ and integers $a,b$.

In fact it turns out that 'miraculously' we have an even more general fact for real exponents:

$x^{ab} = (x^a)^b$ for any real number $x > 0$ and reals $a,b$.

Notice that all these precise statements about real exponentiation show you clearly that you must know exactly what the objects are before you can apply any operations to them, not to say claim any properties about the resulting values.

For this reason it is actually an important question to ask whether there are corresponding rules for complex numbers.

Yes, but not as nice.

$x^{ab} = (x^a)^b$ for any complex number $x \ne 0$ and integers $a,b$. (*)

Here exponentiation is simply the result of starting from $1$ and repeatedly multiplying/dividing by $x$ where the number of times is specified by the exponent (multiplying for positive; dividing for negative). This fact holds in any structure that has invertible multiplication, including the field of rationals, the field of reals, and the field of complex numbers.

$x^{ab},x^a$ are well-defined since $x \ne 0$.

However, in general "$x^{ab} = (x^a)^b$" does not hold for complex $x$ even if $a,b$ are both rational. For instance (according to standard conventions):

$i = (-1)^{1/2} = (-1)^{(2 \times 1/4)} \ne ((-1)^2)^{1/4} = 1^{1/4} = 1$.

So it's excellent that you ask whether some new structure (complex numbers) have the same properties as some other structure (real numbers) instead of just blindly assuming it does.

The question was recently edited to ask for including an explanation of (*). Actually, there is nothing much to explain intuitively, since it boils down to the fact that an $ab$-fold repetition of an operation is the same as a $b$-fold repetition of an $a$-fold repetition of that operation. One can either stop there, but if one wants to ask why then one would need to fix a foundational system first, and in particular the rules concerning integers and induction/recursion. The below proof will use associativity of integer addition and multiplication, and distributivity of multiplication over addition for integers, which correspond to basic facts about repetition.

Suppose we have a field $S$ (such as the complex numbers) and an exponentiation operation that satisfies the following: $\def\lfrac#1#2{{\large\frac{#1}{#2}}}$

$x^0 = 1$ for every $x \in S$.

$x^{k+1} = x^k x$ for every $x \in S$ and integer $k$.

Note that any reasonable foundational system is capable of defining such an operation recursively (you need one direction for positive $k$ and another for negative $k$), and can easily prove by induction the following two theorems.

$x^{a+b} = x^a x^b$ for every nonzero $x \in S$ and integers $a,b$.

Take any nonzero $x \in S$ and integer $a$.

Then $x^{a+0} = x^a = x^a x^0$.

Given any integer $b$ such that $x^{a+b} = x^a x^b$:

$x^{a+(b+1)} = x^{(a+b)+1} = x^{a+b} x = ( x^a x^b ) x = x^a ( x^b x ) = x^a x^{b+1}$.

$x^{a+(b-1)} = x^{(a+b)-1} = x^{a+b} \div x = ( x^a x^b ) \div x = x^a ( x^b \div x ) = x^a x^{b-1}$.

Therefore by induction $x^{a+b} = x^a x^b$ for every integer $b$.

$x^{ab} = (x^a)^b$ for every nonzero $x \in S$ and integers $a,b$.

Take any nonzero $x \in S$ and integer $a$.

Then $x^{a \times 0} = x^0 = 1 = (x^a)^0$.

Given any integer $b$ such that $x^{ab} = (x^a)^b$:

$x^{a(b+1)} = x^{ab+a} = x^{ab} x^a = (x^a)^b (x^a) = (x^a)^{b+1}$.

$x^{a(b-1)} = x^{ab-a} = x^{ab} \div x^a = (x^a)^b \div (x^a) = (x^a)^{b-1}$.

Therefore by induction $x^{ab} = (x^a)^b$ for every integer $b$.

Notice that we did not use commutativity here, which in fact shows that the argument holds in any division ring. If you restrict the exponents to natural numbers, then it clearly holds in any group when "nonzero" is deleted.

Finally, there are some nice properties that arise from the above properties such as:

$i^{k+4} = i^k i^4 = i^k (i^2)^2 = i^k (-1)^2 = i^k$ for any integer $k$.

In short, powers of $i$ (a square-root of $-1$ in the complex field) are cyclic.

• You give a very important message. All too often, a “rule” is taught as something universally true, without any mention of the domain of applicability. Jan 6 '17 at 4:06
• @Lubin: Thank you! This is one of the things that makes teaching type checking (borrowed from CS) an extremely attractive notion to me, since it forces students to know what they are doing rather than mimicking their teacher and hoping what they write on paper gets the marks. =) Jan 6 '17 at 4:12
• The answer is misleading: roots are the inverse of exponents so $i^4 = 1 = 1^{\frac{1}{4}}$. The problem is that the root operation is not one-to-one and this answer only took the principal root of $i^{\frac{1}{4}}$. It has 4 roots: $1, -1, i, -i$ Jan 7 '17 at 20:39
• @brent.payne: It is not misleading. That is the usual definition of the complex exponentiation operation. The point in my post is very clear; some properties of real exponentiation do not carry over to complex exponentiation. If you want a relation instead of a function, you cannot use the exponentiation notation. Also , even if you attempt to define exponentiation to be on sets of complex numbers, this same property will still fail; $\{1\} = \{1\}^{2 \times \frac12} \ne (\{1\}^2)^{\frac12} = \{-1,1\}$. You cannot anyhow choose a root you like. Jan 8 '17 at 3:34
• The answer is very good except for the $i = (−1) ^ {1/2}$ and $1 ^ {1/4} = 1$ parts. This is not a standard convention. You own comment above says "You cannot anyhow choose a root you like.", yet you have chosen roots twice. Jan 13 '17 at 9:05

Geometrically, multiplication by $i$ does the following to a complex number:

• Scales length by a factor of $1$
• Rotates 90 degrees

If you rotate by 90 degrees four times in the same direction, where do you end up?

• You should clarify this with a complex number plotted on the complex plane. Jan 5 '17 at 23:47
• Obviously depending on where you start the rotation, so no one knows what $i^4$ is actually Jan 13 '17 at 4:56
• @BradPitt Neal is thinking of multiplication by a number as a function. Multiplication by $i^4$ is the identity function- scale by 1 and rotate by 0. Sep 12 '17 at 7:15
• @BradPitt The answer is, "The same way you were pointing when you started." In other words, nothing changed, so you multiplied by $1$. :)
– Neal
Sep 12 '17 at 10:26

Seeing as the current best answer that I understand doesn't contain proof at my level (and likely the level of the people this question would help), I'm going to answer.

The solutions of $i^n$ are repeating in a simple pattern. The pattern goes as follows

$$i^0 = 1$$ $$i^1 = i$$ $$i^2 = i\cdot i = -1$$ $$i^3 = (i\cdot i)\cdot i = -i$$ $$i^4 = i\cdot i\cdot i\cdot i = (i\cdot i)(i\cdot i) = (-1)^2 = 1$$

This pattern is repeated infinitely.

edit: As tomazs pointed out, this only works because multiplication by pure and simplified complex numbers is associative.

$$i\cdot i\cdot i\cdot i = (i\cdot i)(i\cdot i) = (i\cdot i\cdot i)\cdot i = ((i\cdot i)\cdot i)\cdot i$$ etc.

• Good answer. Just be exact with parenthesis. In the last step it is $(-1)^2=1$. Since otherwise one reads it as $-(1)^2=-1$ and this can lead to confusion. Jan 5 '17 at 21:39
• @MarvinF, thanks for pointing that out, didn't even think about it. Jan 5 '17 at 21:42
• I think you were setting up a nice pattern with $i^2$ and $i^3$: To get to the next power of $i$, multiply the previous one by $i$. But you didn't continue that pattern for $i^4$, and compute it as $i^3 \cdot i$. There's nothing wrong with what you have, but "thematically," you might consider the "multiply by another copy of $i$" approach. Jan 6 '17 at 0:41
• @pjs36 I had that originally, but it looked even messier than my current proof of $i^4$ Jan 6 '17 at 0:56
• A fair point, and a rather sophisticated one :) I just wanted to mention it Jan 6 '17 at 0:58

Alternatively:

$$i^4=i^{2+2}=i^2i^2=(-1)(-1)=1$$

Raising to positive integer power is the same as repeated multiplication, thus you don't even have to think whether $i^4=(i^2)^2$ is true, just expand the power:

$$i^4=i\cdot i\cdot i\cdot i=(i\cdot i)(i\cdot i)=(-1)(-1)=1.$$

The second equality works due to associativity of multiplication for complex numbers.

• I prefer @Tomasz answer, that explicitly invokes associativity. Your parenthesing isn't justified.
– user65203
Jan 7 '17 at 15:02
• @YvesDaoust you're right. I was thinking of mentioning it, but for some reason decided to not do when I was writing the answer. Anyway, tomasz's answer is way more complete than mine, and it was written later, so I couldn't have had it as an example :). Jan 7 '17 at 15:18
1. $$i^4=(i^2)^2=(-1)^2=1$$
2. $$i^4=\left(|i|e^{\arg(i)i}\right)^4=\left(e^{\frac{\pi i}{2}}\right)^4=e^{\frac{4\pi i}{2}}=e^{2\pi i}=1$$
• Pretty sure that if OP is at the level of wondering whether $i^4 = 1$, they won't be familiar with the complex exponential. I'm quite often wrong though, so who knows... Jan 5 '17 at 20:54
• Proofing this with exponentiation looks like a cyclic proof to me Jan 5 '17 at 21:43

The property of multiplication of real numbers which you used and referred to, but did not precisely quote, is associativity (i.e. $z_1\cdot (z_2\cdot z_3)=(z_1\cdot z_2)\cdot z_3)$). Multiplication of complex numbers is associative, so you have: $$i^4=i\cdot(i\cdot(i\cdot i))=(i\cdot i)\cdot(i\cdot i)=(-1)\cdot (-1)=1.$$ (Note that if multiplication was not associative, an expression like $i^4$ would not immediately make sense: it is not at all obvious if it means $i\cdot(i\cdot(i\cdot i))$ or $((i\cdot i)\cdot i)\cdot i$. There are expansions of complex numbers with non-associative multiplication for which the rule you are using here would not apply.)

You can also verify it directly using the definition of multiplication of complex numbers: \begin{align} i\cdot(i\cdot(i\cdot i))&=(0+1i)\cdot ((0+1i)\cdot(-1+0i))\\ &=(0+1i)\cdot(0+(-1)i)\\ &=-(-1)+0i\\ &=1 \end{align}

Checking associativity in general is a bit more troublesome (in fact, it is almost always tiresome to check directly), but still workable. Recall that the definition of multiplication in complex numbers is $$(a_1+b_1i)\cdot(a_2+b_2i)=(a_1a_2-b_1b_2)+i(a_1b_2+b_1a_2).$$ Now you can just compute directly \begin{multline*} (a_1+b_1i)\cdot \left( (a_2+b_2i)\cdot (a_3+b_3i)\right)=(a_1+b_1i)\cdot((a_2a_3-b_2b_3)+i(a_2b_3+b_2a_3))=\\ =(a_1a_2a_3-a_1b_2b_3-(b_1a_2b_3+b_1b_2a_3))+i(a_1a_2b_3+a_1b_2a_3+b_1a_2a_3-b_1b_2b_3). \end{multline*} On the other hand, \begin{multline*} ((a_1+b_1i)\cdot (a_2+b_2i))\cdot (a_3+b_3i)=((a_1a_2-b_1b_2)+i(a_1b_2+b_1a_2))\cdot(a_3+b_3i)=\\ =(a_1a_2a_3-b_1b_2a_3-(a_1b_2b_3+b_1a_2b_3))+i(a_1a_2b_3-b_1b_2b_3+a_1b_2a_3+b_1a_2b_3).\end{multline*} Finally, $$a_1a_2a_3-a_1b_2b_3-(b_1a_2b_3+b_1b_2a_3)=a_1a_2a_3-b_1b_2a_3-(a_1b_2b_3+b_1a_2a_3),$$ and $$a_1a_2b_3+a_1b_2a_3+b_1a_2a_3-b_1b_2b_3=a_1a_2b_3-b_1b_2b_3+a_1b_2a_3+b_1a_2a_3,$$ so, since two complex numbers are equal when their real and complex parts are equal, the multiplication is associative. (This argument can be made to be a little bit shorter if you apply commutativity of multiplication of complex numbers.)

Yes you are right. Rotation by 90 degree in complex plane and multiplication by $i$ are very much same. If you raise $i$ to 4,8,12th ... powers you get 1. Raising to 1, 5, 9 .. you get back $i$; raising to 2,6,10 powers gives you $-1$ and so on.

The property $$(x^{m})^n=x^{mn}\tag{1}$$ for $m$ and $n$ nonnegative integers holds whenever we are dealing with an associative operation with a neutral element $1$, like for complex multiplication. Indeed it holds for $n=1$, because $(x^{m})^0=1=x^{m0}$ by definition.

Recall that $x^n$ is defined recursively: $x^0=1$, $x^{k+1}=x^k\cdot x$.

Suppose property $(1)$ holds for $n$; then \begin{align} (x^{m})^{n+1}&= (x^m)^n\cdot x^m &&\text{definition of powers}\\ &=x^{mn}\cdot x^m &&\text{induction hypothesis}\\ &=x^{mn+m} && \text{rule of powers $(2)$}\\ &=x^{m(n+1)} && \text{property of integers} \end{align} The mentioned rule of powers is where associativity is used: $$x^{h+k}=x^h\cdot x^k\tag{2}$$

Again, this is true by definition when $k=1$. Suppose it holds for $k$; then \begin{align} x^{h+(k+1)}&=x^{(h+k)+1} && \text{property of integers}\\ &=x^{h+k}\cdot x &&\text{definition of powers}\\ &=\bigl(x^h\cdot x^k\bigr)\cdot x &&\text{induction hypothesis}\\ &=x^h\cdot\bigl(x^k\cdot x\bigr) &&\text{associativity}\\ &=x^h\cdot x^{k+1} &&\text{definition of powers} \end{align}

Therefore you are certainly allowed to say that $$i^4=(i^2)^2=(-1)^2=1$$

Light comes from the definition of the complex multiplication, which only involves real arithmetic:

$$(a+ib)(c+id):=ac-bd+(ad+bc)i.$$

Then

$$(0+1i)^2=(0+1i)(0+1i)=(\bar1+0i),\\ (0+1i)^3=(\bar1+0i)(0+1i)=(0+\bar1i),\\ (0+1i)^4=(0+\bar1i)(0+1i)=(1+0i).$$

• I'm not the downvoter, so I've no idea about the reason for the downvote, but if you ask me, you definitely should not use "$\overline{1}$" for "$-1$" since in complex analysis "$\overline{z}$" denotes the conjugate of $z$. Also, I don't think your answer quite addresses the underlying question of what complex exponentiation satisfies. You can make it relevant by stating that in your answer the integer exponent is defined to mean repeated multiplication, and so you are simply evaluating the powers one by one. Makes sense? Jan 13 '17 at 15:56
• $\bar 1$ makes a much nicer alignment than $-1$. Purely cosmetic. I didn't think of the confusion with conjugation, which I use to denote $^*$. The accepted answer just says that the powers are cyclically $i,-1,-i,1$ without any justification. IMO, it is the one that deserves a downvote.
• I also use "$z^*$" to denote the conjugate of $z$, but the overline notation is actually very common. As I said, I didn't downvote your answer, but I do think it doesn't address the underlying question. And I don't know who downvoted my answer either..... Not you right? Jan 14 '17 at 8:52


# Eigenvectors of harmonic series matrix

I have a finite upper triangular $n$ by $n$ matrix $A$ which has the following form:

$$A = \begin{pmatrix} 1 & 1\over 2 & 1\over 3 & \cdots & 1\over n \\ 0 & 1\over 2 & 1\over 3 & \cdots & 1\over n \\ 0 & 0 & 1\over 3 & \cdots & 1\over n \\ \vdots &\vdots &\vdots &\ddots &\vdots \\ 0 & 0 & 0 & \cdots& 1\over n \end{pmatrix}$$

Or alternatively $A =TH$, where $T$ is the all-ones upper $n$ by $n$ triangular matrix and $H$ is the first $n$ terms of the harmonic series on the diagonal.

I'm interested in diagonalizing this matrix, as I'm interested in understanding and computing $A^k$ (specifically the final column). Since all elements on the diagonal are distinct the eigenvalues are simply the first $n$ terms of the harmonic series.

Doing some numerical analysis using numpy gave fairly opaque numbers, however Mathematica produced very interesting eigenvectors. E.g. for $n = 5$ the matrix with the eigenvectors as columns is:

$$P = \begin{pmatrix} 1 & -1 & 1 & -1 & 1 \\ 0 & 1 & -2 & 3 & -4 \\ 0 & 0 & 1 & -3 & 6 \\ 0 & 0 & 0 & 1 & -4 \\ 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix}, P^{-1} = \begin{pmatrix} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 & 4 \\ 0 & 0 & 1 & 3 & 6 \\ 0 & 0 & 0 & 1 & 4 \\ 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix}$$

This seems to be some alternating sign Pascal triangle. In fact, $P^{-1}$ seems the be the actual Pascal triangle! And indeed $PHP^{-1} = A$ (no PHP pun intended), for the cases I've tried.

### I have two questions:

1. Is there a good explanation for this seeming connection between the eigenvectors of the truncated harmonic series matrix and Pascal's triangle, assuming my numerical exploration is not a coincidence?

2. Can we use this to say anything interesting about (the final column of) $A^k$? A closed form that does not require constructing the Pascal triangles perhaps?

• For curiosity, it is the probability matrix of a markov chain for standing on a stairs $n-1$ steps left to the top and with equal $1/n$ probability to stay on the same stage or go to any other step (but only upwards). Apr 19 '17 at 10:51
• Maybe there is something with the equidistribution of uniform variables and pascals triangle making successive better approximations to normal (law of large numbers). Apr 19 '17 at 10:54
• @mathreadler Yes! The origin is a study of the probability distribution of a $k$-throws dice game where the size of the dice you throw is determined by your previous throw (and initially $n$).
– orlp
Apr 19 '17 at 10:55
• This earlier question seems relevant: math.stackexchange.com/q/69925/137524. Apr 19 '17 at 16:00

Let's indicate your matrix as $${\bf A}_{\,h} = \left\| {\;a_{\,n,\,m} \quad \left| {\;0 \le n \le m \le h} \right.\;} \right\| = \left\| {\;{{\left[ {n \le m} \right]} \over {m + 1}}\;} \right\|_{\,h}$$ so that it is a $(h+1) \times (h+1)$ matrix, indexed from $0$, and where the square brackets indicate the Iverson's bracket.

Then let's define \eqalign{ & {\bf E}_{\,h} = \left\| {\;\left[ {n + 1 = m} \right]\;} \right\|_{\,h} \cr & {\bf S}_{\,h} = \left\| {\;\left[ {n \le m} \right]\;} \right\|_{\,h} = {{{\bf I}_{\,h} } \over {{\bf I}_{\,h} - {\bf E}_{\,h} }}\quad \cr & \left( {f(n) \circ {\bf I}_{\,h} } \right) = \left\| {\;f(n)\left[ {n = m} \right]\;} \right\|_{\,h} \cr} so that
${\bf E}_{\,h}$ is the Shifting matrix, with $1$ on the first upper sub-diagonal and null otherwise;
${\bf S}_{\,h}$ is the Summing matrix, upper triangular with all $1$ 's;
$\left( {f(n) \circ {\bf I}_{\,h} } \right)$ is the diagonal matrix, with diagonal elements $f(0),f(1), \cdots, f(h)$.

Then, as you correctly pointed out, we can write: $${\bf A}_{\,h} = {\bf S}_{\,h} \left( {{1 \over {n + 1}} \circ {\bf I}_{\,h} } \right)$$ which means $${\bf A}_{\,h} ^{\; - \,{\bf 1}} = \left( {\left( {n + 1} \right) \circ {\bf I}_{\,h} } \right)\left( {{\bf I}_{\,h} - {\bf E}_{\,h} } \right)$$

It is not difficult to demonstrate that \eqalign{ & \left( {f(n) \circ {\bf I}_{\,h} } \right)\;{\bf E}_{\,h} = {\bf E}_{\,h} \;\left( {f(n - 1) \circ {\bf I}_{\,h} } \right) = \cr & = \left( {\prod\limits_{0\, \le \,k\, \le \,n - 1} {f(k)} \circ {\bf I}_{\,h} } \right)^{\; - \,{\bf 1}} \;{\bf E}_{\,h} \left( {\prod\limits_{0\, \le \,k\, \le \,n - 1} {f(k)} \circ {\bf I}_{\,h} } \right) \cr} and that, given in general $${\bf A}_{\,h} (f) = {\bf S}_{\,h} \left( {f(n) \circ {\bf I}_{\,h} } \right)\quad \left| {\;f(m) \ne f(n)} \right.$$ the relevant matrix of eigenvectors $\mathbf W$ is : $${\bf W}_{\,h} (f) = \left\| {\;\left[ {n \le m} \right]{{f(m)^{\,m - n} } \over {\prod\limits_{n\, \le \,k\, \le \,m - 1} {\left( {f(m) - f(k)} \right)} }}\;} \right\|_{\,h} = \left\| {\;\left[ {n \le m} \right]\prod\limits_{n\, \le \,k\, \le \,m - 1} {\left( {{{f(m)} \over {f(m) - f(k)}}} \right)} \;} \right\|_{\,h}$$ while the inverse is: \eqalign{ & {\bf W}_{\,h} ^{\; - \,{\bf 1}} (f)\quad \left| {\;0 \ne f(0)} \right.\quad = \cr & = \left\| {\;\left[ {n \le m} \right]{{f(m)f(n)^{\,m - n - 1} } \over {\prod\limits_{n + 1\, \le \,k\, \le \,m} {\left( {f(n) - f(k)} \right)} }}\;} \right\|_{\,h} = \left\| {\;\left[ {n \le m} \right]{{f(m)} \over {f(n)}}\prod\limits_{n + 1\, \le \,k\, \le \,m} {{{f(n)} \over {\left( {f(n) - f(k)} \right)}}} \;} \right\|_{\,h} \cr}

Now, in the present case, we obtain \eqalign{ & {\bf W}_{\,h} = \left\| {\;\left[ {n \le m} \right]\prod\limits_{n\, \le \,k\, \le \,m - 1} {\left( {{{k + 1} \over {k - m}}} \right)} \;} \right\|_{\,h} = \cr & = \left\| {\;\left[ {n \le m} \right]{{m!} \over {n!}}\prod\limits_{1\, \le \,j\, \le \,m - n} {\left( { - {1 \over j}} \right)} \;} \right\|_{\,h} = \left\| {\;\left( { - 1} \right)^{\,m - n} \left( \matrix{ m \cr n \cr} \right)\;} \right\|_{\,h} = \overline {\bf B} _{\,h} ^{\; - \,{\bf 1}} \cr} where $\overline {\bf B} _{\,h}$ is the transpose of the Lower Triangular Pascal array.

Finally $${\bf A}_{\,h} = {\bf S}_{\,h} \left( {{1 \over {n + 1}} \circ {\bf I}_{\,h} } \right) = \overline {\bf B} _{\,h} ^{\; - \,{\bf 1}} \;\left( {{1 \over {n + 1}} \circ {\bf I}_{\,h} } \right)\;\overline {\bf B} _{\,h}$$ which demonstrates your assumption.

Concerning your last question instead, we have of course that $${\bf A}_{\,h} ^{\;{\bf q}} = \overline {\bf B} _{\,h} ^{\; - \,{\bf 1}} \;\left( {{1 \over {\left( {n + 1} \right)^{\;q} }} \circ {\bf I}_{\,h} } \right)\;\overline {\bf B} _{\,h}$$ whose single components are: \eqalign{ & a^{\left( q \right)} _{\,n,\,m} = \sum\limits_{0\, \le \,\left( {n\, \le } \right)\,k\,\left( { \le \,m} \right)\, \le \,h} {\left( { - 1} \right)^{\,k - n} \left( \matrix{ k \cr n \cr} \right){1 \over {\left( {k + 1} \right)^{\;q} }}\left( \matrix{ m \cr k \cr} \right)} = \cr & = \sum\limits_{0\, \le \,\left( {n\, \le } \right)\,k\,\left( { \le \,m} \right)\, \le \,h} {\left( { - 1} \right)^{\,k - n} \left( \matrix{ m \cr n \cr} \right)\left( \matrix{ m - n \cr k - n \cr} \right)\left( {k + 1} \right)^{\; - \,q} } = \cr & = \left( \matrix{ m \cr n \cr} \right)\sum\limits_{0\, \le \,\left( {n\, \le } \right)\,k\,\left( { \le \,m} \right)\, \le \,h} {\left( { - 1} \right)^{\,k - n} \left( \matrix{ m - n \cr k - n \cr} \right)\left( {k - n + n + 1} \right)^{\; - \,q} } = \cr & = {{m!} \over {n!}}\sum\limits_{\,0\, \le \,j\, \le \,m - n} {{{\left( { - 1} \right)^{\,j} } \over {j!\left( {m - n - j} \right)!}}\left( {j + n + 1} \right)^{\; - \,q} } \cr} Cannot find at the moment a "more compact" expression.

Since the given matrix is upper triangular, the eigenvalues are trivially $\frac{1}{1},\frac{1}{2},\frac{1}{3},\ldots$ and the shown decomposition is just an instance of the following identity: $$\frac{1}{x(x+1)(x+2)\cdots(x+n)} = \frac{1}{n!}\sum_{k=0}^{n}\frac{(-1)^k}{x+k}\binom{n}{k}$$ that follows from the residue theorem.

• Could you elaborate a bit on how the decomposition is an instance of the identity? I did manage to find a closed form for the last column of $A^k$ though, if $x$ goes from $1$ to $n$ denoting the row number, then $f_{k, n}(x) = \frac{x}{n} \left | \sum_{i=1}^n \frac{(-1)^i}{i^k}\binom{n}{i}\binom{i}{x} \right |$.
– orlp
Apr 19 '17 at 12:47