What is the least number of square roots needed to express $\sqrt{1}+\sqrt{2}+\cdots+\sqrt{100}$?

What is the least number of square roots needed to express $\sqrt{1}+\sqrt{2}+\cdots+\sqrt{100}$ if it must be expressed in the form $a+b\sqrt{c}+d\sqrt{e}+\cdots$ where $a,b,c,d,e,\ldots$ are all integers?

Solution

In order to find the least number of square roots, we must express this number in simplest form. Thus, the numbers in the radicand that are in simplest form that will get counted are numbers with at most one of each prime factor. Then the question simplifies to "How many numbers are there between $1-100$ with at most one of each prime factor?

There are a few ways to proceed from here. I think the best way would be to complementary count. Then we are looking for numbers with at least $2$ factors of each prime, thus multiples of perfect squares. Our perfect squares are $1,4,9,16,25,36,49,64,81$. Thus we have $1+25+11-2+3+2 = 40$ such numbers by the principle of inclusion-exclusion. Thus, the answer is $100-40 = 60$.

Question

How is it that the number being expressed in simplest form will give the least number of radicals? The solution seems to imply that but why is it true?

• You mean expressing it as $a + b \sqrt{c} + d\sqrt{e} + \cdots$ where $a,b,c,\dots \in \mathbb N$ right? Because otherwise the question would be very problematic. – ThePortakal Jan 8 '16 at 3:45
• Yes, of course. Otherwise I wouldn't technically have to use any square roots. – user19405892 Jan 8 '16 at 3:45
• I added the extra condition in the question so that doesn't work as $3+\sqrt{3}+\sqrt{5}$ is not an integer. By least number of square roots, I mean the number of square root symbols. So $\sqrt{3}+\sqrt{3}$ would be two square roots while $2\sqrt{3}$ would be one. – user19405892 Jan 8 '16 at 4:07
• Why are you copying the exact solution here? – GohP.iHan Jan 9 '16 at 11:04
• @GohP.iHan It is actually my solution. – user19405892 Jan 9 '16 at 13:44

The square root of each square-free number less than $100$ contributes a term to the sum with a surd (a square root that cannot be simplified out), but all the square multiples of the number under the square root ($4$ times, $9$ times, $16$ times, etc.) can be combined with the term for the original square-free number, so they do not contribute any additional terms. For example,

$$\sqrt{11} + \sqrt{44} + \sqrt{99} = 6\sqrt{11}.$$

One way to count the integers whose square roots do not contribute new surds to the sum is as follows:

• $10$ perfect squares in the range $1$ to $100$, inclusive;
• $6$ cases of twice a perfect square in the range $8$ to $98$, inclusive;
• $4$ cases of $3$ times a perfect square in the range $12$ to $75$, inclusive;
• $3$ cases of $5$ times a perfect square in the range $20$ to $80$, inclusive;
• $3$ cases of $6$ times a perfect square in the range $24$ to $96$, inclusive;
• $6$ cases including $2$ cases each of $7$ times a perfect square, $10$ times a perfect square, and $11$ times a perfect square in the range $28$ to $99$, inclusive;
• $8$ cases including $13 \times 4$, $14 \times 4$, $15 \times 4$, $17 \times 4$, $19 \times 4$, $21 \times 4$, $22 \times 4$, and $23 \times 4$.

These add up to $40$ terms of the original sum that either are integers or can be combined with other terms, leaving $60$ unique surds. This is the same as your result, of course.

Here is the sum completely worked out, confirming that $60$ square roots are needed:

\begin{align} \sqrt1 + & \sqrt2 + \cdots + \sqrt{100} \\ =& \quad 1 + \sqrt2+\sqrt3 + 2 + \sqrt5+\sqrt6+\sqrt7 + 2\sqrt2 + 3 + \sqrt{10} \\ & + \sqrt{11} + 2\sqrt3 + \sqrt{13} + \sqrt{14} + \sqrt{15} + 4 + \sqrt{17} + 3\sqrt2 + \sqrt{19} + 2\sqrt{5} \\ & + \sqrt{21} + \sqrt{22} + \sqrt{23} + 2\sqrt6 + 5 + \sqrt{26} + 3\sqrt3 + 2\sqrt7 + \sqrt{29} + \sqrt{30} \\ & + \sqrt{31} + 4\sqrt2 + \sqrt{33} + \sqrt{34} + \sqrt{35} + 6 + \sqrt{37} + \sqrt{38} + \sqrt{39} + 2\sqrt{10} \\ & + \sqrt{41} + \sqrt{42} + \sqrt{43} + 2\sqrt{11} + 3\sqrt5 + \sqrt{46} + \sqrt{47} + 4\sqrt3 + 7 + 5\sqrt2 \\ & + \sqrt{51} + 2\sqrt{13} + \sqrt{53} + 3\sqrt6 + \sqrt{55} + 2\sqrt{14} + \sqrt{57} + \sqrt{58} + \sqrt{59} + 2\sqrt{15} \\ & + \sqrt{61} + \sqrt{62} + 3\sqrt7 + 8 + \sqrt{65} + \sqrt{66} + \sqrt{67} + 2\sqrt{17} + \sqrt{69} + \sqrt{70} \\ & + \sqrt{71} + 6\sqrt2 + \sqrt{73} + \sqrt{74} + 5\sqrt3 + 2\sqrt{19} + \sqrt{77} + \sqrt{78} + \sqrt{79} + 4\sqrt5 \\ & + 9 + \sqrt{82} + \sqrt{83} + 2\sqrt{21} + \sqrt{85} + \sqrt{86} + \sqrt{87} + 2\sqrt{22} + \sqrt{89} + 3\sqrt{10} \\ & + \sqrt{91} + 2\sqrt{23} + \sqrt{93} + \sqrt{94} + \sqrt{95} + 4\sqrt6 + \sqrt{97} + 7\sqrt2 + 3\sqrt{11} + 10 \\ =& \quad 55 + 28\sqrt2 + 15\sqrt3 + 10\sqrt5 + 10\sqrt6 + 6\sqrt7 + 6\sqrt{10} \\ & + 6\sqrt{11} + 3\sqrt{13} + 3\sqrt{14} + 3\sqrt{15} + 3\sqrt{17} + 3\sqrt{19} \\ & + 3\sqrt{21} + 3\sqrt{22} + 3\sqrt{23} + \sqrt{26} + \sqrt{29} + \sqrt{30} \\ & + \sqrt{31} + \sqrt{33} + \sqrt{34} + \sqrt{35} + \sqrt{37} + \sqrt{38} \\ & + \sqrt{39} + \sqrt{41} + \sqrt{42} + \sqrt{43} + \sqrt{46} + \sqrt{47} \\ & + \sqrt{51} + \sqrt{53} + \sqrt{55} + \sqrt{57} + \sqrt{58} + \sqrt{59} \\ & + \sqrt{61} + \sqrt{62} + \sqrt{65} + \sqrt{66} + \sqrt{67} + \sqrt{69} \\ & + \sqrt{70} + \sqrt{71} + \sqrt{73} + \sqrt{74} + \sqrt{77} + \sqrt{78} \\ & + \sqrt{79} + \sqrt{82} + \sqrt{83} + \sqrt{85} + \sqrt{86} + \sqrt{87} \\ & + \sqrt{89} + \sqrt{91} + \sqrt{93} + \sqrt{94} + \sqrt{95} + \sqrt{97} \\ \end{align}

• How do you know that we can't use nested radicals to reduce the number of radicals used? – GohP.iHan Jan 9 '16 at 11:03
• @GohP.iHan The problem statement stipulates that the answer must be in the form $a+b\sqrt{c}+d\sqrt{e}+\cdots$ where $a,b,c,d,e,\ldots$ are all integers. – David K Jan 9 '16 at 13:37
• Ohhh my bad... the author had a different question when I first viewed it. Thank you!! – GohP.iHan Jan 9 '16 at 13:51


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## AP® Calculus AB (2017 edition)

### Unit 1: Lesson 7

Determining limits using direct substitution

# Limits by direct substitution

AP.CALC:
LIM‑1 (EU)
,
LIM‑1.D (LO)
,
LIM‑1.D.1 (EK)
Sal explains how you can easily find limits of functions at points where the functions are continuous: simply plug in the x-value into the function! Later we will learn how to find limits even when the function isn't continuous.

## Want to join the conversation?

• I think Sal may have left something out. Correct me if I'm wrong, but on certain occasions, the entire function does not have to be continuous. If it is continuous at the value, a, then it should be fine to use direct substitution even though the function might be undefined elsewhere. This is because it isn't undefined at a so your output will be defined. Also, since your function is continuous at a, the limit there will exist meaning that you won't have to worry about jump discontinuities. Please tell me if this all seems reasonable.
• Because it is a continuous graph, every interval would be continuous provided it is a real number
• how would we know if the problem is to be solved by direct substitution method? or if the function given is continuous? without drawing its graph
• lim h(x) as x ->6 for h(x) =square root of (5x+6) should be 6 or -6 or both?
• y=√x is a function that returns only positive values. So in this context, we are only looking at the principal root, and the limit is positive 6.

Limits are unique: they cannot have multiple values. So "both" would never be an option.
• At , what's the difference between f(x) as x->a, and f(a)?
• One is a limit, the other is an evaluation of the function. If the function is continuous and defined at (in your example), a, then they're equivalent. But you can get some very interesting results if the function is not continuous or not defined.

The limit is basically saying what the function seems to be going to as x gets closer to closer to a, but the function may not be defined at that point.
• Is the limit zero or none since it is continuous?
• The limit is 0 as Sal has demonstrated in the video.
• Can a limit be a fraction?
• yes a limit can be a fraction. For example if I asked you to find the limit as X approaches 2 for the functions (1/X), when you use substitution to find the limit you find that the number it approaches is .5 or 1/2.
• I thought it like this : Lim x-> a f(x) = f(a) if f(x) is continuous at x=a, not vice versa? Can someone tell me why?
• That statement holds in both directions. A function is continuous at a point a if and only if the limit value equals the function value at a. Since the two statements are equivalent, this can be used as a definition of continuity.
• Ok so basically if a function, say f(x), is continuous at x=c, then the lim x-->c = f(x)? Is this why you can find limits of continuous functions by direct substitution?
• Yes, the limit as x->c of f(x) is f(c). This property is equivalent to the epsilon-delta definition of continuity, and it's why we can use direct substitution for most familiar functions.
• By Sal and almost all discussion about relation between continuity and limit: it always says that as long as the function is continuous, there is a limit.
i am wondering if there is a case that the function is continuous but the limit does not exist.

https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Active_Calculus_(Boelkins_et_al)/1%3A_Understanding_the_Derivative/1.7%3A_Limits%2C_Continuity%2C_and_Differentiability

in figure 1.7.2, the right graph, at point x=1. the left hand limit is 2. but the as x approaches 1 from the right, the function keeps oscillating, so the limit does not exist.

so the function is continuous by the graph, but limit does not exist.

am I wrong? or am I give some wrong example?
• Interesting question!

By definition, a function f(x) is continuous at x=x_0 if and only if for every epsilon>0, there exists delta>0 such that whenever x is within delta of x_0, f(x) is always within epsilon of f(x_0).

Though the function might look continuous at x=1, it is really discontinuous at x=1 due to the oscillatory behavior. For example, if we choose epsilon = 1/2, there does not exist delta>0 such that whenever x is within delta of 1, f(x) is always within epsilon of f(1). This is because there are values of x arbitrarily close to 1 (from the right) such that f(x) differs from f(1) by as much as 1 unit.
(1 vote)
• I'm not entirely sure if this statement is always true. But my argument bases on the assumption that the function f(x) = |x| is not continuous at x = 0, which I think is true but correct me if I'm wrong.

So, the limit lim_{x -> 0}f(x) for f(x) = |x| is definitely defined since it approaches 0 from both sides. The function itself is usually defined so that |x| equals 0 at x = 0. Which means that lim_{x -> a}f(x) equals f(a) at a = 0, but as said AFAIK the absolute value function |x| is NOT continuous for x = 0.

And that would disprove the statement that if lim_{x -> a}f(x) equaling f(a) will mean that f is continuous at a (and vice versa).
(1 vote)
• Hey Raphael! I think you're getting continuity and differentiation confused. For a function, f(x), to be continuous, f(c) = lim_{x->c}f(x). This means that f(c) = lim_{x->c-}f(x) = lim_{x->c+}f(x) = lim_{x->c}f(x). As long as this holds true the function will be continuous at the given x-value, c.

Differentiation on the other hand is different. Differentiation requires that a function is continuous at the given x-value along with being absent of cusps, sharp turns (like x=0 for f(x) = |x|), and vertical tangents.

So to answer your question, f(x) = |x| is continuous at x = 0, however, it is not differentiable at x = 0.


# Order of magnitudes comparasions

I have a list of order of magnitudes I want to compare.

My only idea is using calculus methods (limits , integral, etc...) to assert the functions relation.

I need your help with the following.

I need to determine (again by order of magnitude) how to order the following (by asceneding order):

$log(n) / log(log(n))$

$log(log(n))$

$log^3(n)$

$n/log(n)$

$n^{log(n)}$

Is it even possible to use limits on such functions?

I mean if I take $f(n)$ to be $log(n) / log(log(n))$ and $g(n)$ to be $n^{log(n)}$, It seems very hard to find out the limit.

-

There are quite a number of comparisons to be made. It is in most cases relatively straightforward to decide about the relative long-term size.

Let's start with your pair. We have $f(n)=\log(n)/\log(\log (n))$ and $g(n)=n^{\log(n)}$.

For large $n$ (and it doesn't have to be very large), we have $f(n)\lt \log(n)$.

Also, for large $n$, we have $\log(n)\gt 1$, and therefore $g(n)=n^{\log(n)}\gt n$.

So for large $n$, we have $$\frac{f(n)}{g(n)} \lt \frac{\log(n)}{n}.$$ But we know that $\lim_{n\to\infty}\dfrac{\log(n)}{n}=0$. This can be shown in various ways. For instance, we can consider $\dfrac{\log(x)}{x}$ and use L'Hospital's Rule to show this has limit $0$ as $x\to\infty$.

It takes less time to deal with the pair $n/\log(n)$ and $n^{\log(n)}$. If $n$ is even modestly large, we have $n/\log(n)\lt n$. But after a while, $\log(n)\gt 2$, so $n^{\log(n)}\gt n^2$. It follows that in the long run, $n^{\log(n)}$ grows much faster than $n/\log(n)$.

As a last example, let us compare $\log^3(n)$ and $n/\log(n)$. Informally, $\log$ grows glacially slowly. More formally, look at the ratio $\dfrac{\log^3(n)}{n/\log(n)}$. This simplifies to $$\frac{\log^4 (n)}{n}.$$ We can use L'Hospital's Rule on $\log^4(x)/x$. Unfortunately we then need to use it several times. It is easier to examine $\dfrac{\log(x)}{x^{1/4}}$. Then a single application of L'Hospital's Rule does it. Or else we can let $x=e^y$. Then we are looking at $y^4/e^y$, and we can quote the standard result that the exponential function, in the long run, grows faster than any polynomial.

Remark: The second person in your list is the slowest one. Apart from that, they are in order. So you only need to prove four facts to get them lined up. A fair part of the work has been done above.

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Your answer is very very good and understood. I will work on these tomorrow, make sure I understand anything. Thank you very much! – SyndicatorBBB Nov 3 '12 at 20:49
@Guy: I don't mind your accepting my answer. However, it is useful in general to wait somewhat longer, in order to elicit a variety of approaches. – André Nicolas Nov 3 '12 at 20:54
I commented below. Can you please advise ?Thank you. – SyndicatorBBB Nov 4 '12 at 10:01

For instance: $$\log(n^{\log n})=\log^2(n)$$ And since asymptotically: $$\frac{\log(n)}{\log(\log(n))} < \log^2(n)$$ We know that: $$n^{\log n}$$ Is at least exponentially larger than: $$\frac{\log(n)}{\log(\log(n))}$$

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How did you calculate this inequalties? – SyndicatorBBB Nov 3 '12 at 20:46
Just observe: a $\log(n)$ cancels out, and $1/\infty$ is obviously smaller than $\infty$. – nbubis Nov 3 '12 at 20:47
Undestood! Thank you very much! – SyndicatorBBB Nov 3 '12 at 20:50

@André Nicolas I really liked the comprasion of the functions with simplier functions.

I was trying to use this technique on some of these functions.

Eventually I found a couple which I was not able to solve using this technique and I don't understand why.

Assume $f(n) = log(log(n))$ and $g(n) = n/log(n)$

Now one might say, $f(n) < log(n)$ and $g(n) > 1/n$

If I look at the ratio/limit of $f(n)/g(n)$, I find out that $f(n)$ is much faster.

I completly understand that I chose unwisely the lower bound of g(n). Does it mean that on such functions I cannot use this technique?

If you "give away" stuff to make things simpler, have to be sensible about it. Replacing $g(n)$ by something that approaches $0$ is unreasonable. For example, could use $f(n)\lt \log(n)$, and keep $g(n)$. How one "weakens" an inequality has to be guided by basic intuition/experience about the sizes of things. The one thing that will be constantly useful is (i) $\log(n)$ is long-run slower than any positive power of $n$ and its twin (ii) For constant $a\gt 1$, $a^n$ is long-run faster than any polynomial in $n$. – André Nicolas Nov 4 '12 at 14:26


# If all points of a real function with positive values would be local minimum, can one say it is constant function?

During my studies I faced a function $$f:\mathbb{R} \to \mathbb{R}^+$$ with the property: for all $$x \in \mathbb{R}$$ and all $$y$$ in open interval $$(x-\frac{1}{f(x)} ,x+\frac{1}{f(x)})$$ we have $$f(x) \leq f(y)$$. At first I guessed maybe it is necessarily constant function but I could not show this. I tried to generate a counter example for my guess and unfortunately I could not again. would anyone please help me?

• The answer to the question in the title is (of course) no. [Define $f(x) = 2$ for $x \ne 0$ and $f(0)=1$.] But the question in the text is more complicated. – Gerald Edgar Aug 6 '20 at 9:31
• What is $\mathbb{R}^+$? Is $f(x)=0$ allowed? – Wilberd van der Kallen Aug 6 '20 at 9:52
• $\mathbb{R}^+$ is denoted for the open interval $(0,+\infty)$ – M. Reza. K Aug 6 '20 at 10:12

My answer is: Such a function must be constant. Suppose (for puoposes of contradiction) $$f$$ is a nonconstant function $$f : \mathbb R \to (0,+\infty)$$ such that $$\forall x\in\left(a-\frac{1}{f(a)},a+\frac{1}{f(a)}\right),\quad f(x) \ge f(a) .$$

For $$m > 0$$, define $$U_m := \{x : f(x)>m\}$$.

Lemma 1: For all $$m \in \mathbb R$$, the set $$U_m$$ is open.
Proof. Let $$x \in U_m$$. Then $$(x-1/f(x),x+1/f(x)) \subseteq U_m$$. So $$U_m$$ is open.

Define $$\mathcal G = \{(a,b,m) : a < b, f(a)\le m, f(b)\le m, \text{ and }\forall x\in(a,b),f(x)> m\}$$.

Lemma 2: Let $$a,b \in \mathbb R$$ with $$f(a) \ne f(b)$$, and $$m>0$$. Then there exists $$a_1, b_1$$ with $$a < a_1 < b_1 < b$$ and $$m_1 \ge m$$ such that $$(a_1,b_1,m_1) \in \mathcal G$$.
Proof. One of $$f(a), f(b)$$ is smaller, assume WLOG that $$f(a) < f(b)$$. Now $$U_{f(a)}$$ is open, $$a \notin U_{f(a)}$$, and $$b \in U_{f(a)}$$. The maximal open interval $$(c,d) \subseteq U_{f(a)}$$ with $$c is nonempty, in fact $$(c,b) = (c,d) \cap (a,b) \ne \varnothing$$. By maximality, $$f(c) \le f(a)$$. Choose $$e$$ with $$c < e < b$$. For $$x \in (c,e)$$ we have $$f(c) < f(x)$$ so $$c \le x-1/f(x)$$ and thus $$f(x) \ge 1/(x-c)$$. Thus $$f(x)$$ is unbounded on $$(c,e)$$. Let $$m_1$$ be such that $$m_1 \ge m, m_1 > f(c), m_1 > f(e)$$. The open set $$U_{m_1}$$ has $$U_{m_1} \cap (c,e) \ne \varnothing$$ and $$c,e \notin U_{m_1}$$. So let $$(a_1,b_1)$$ be a maximal open subinterval of the open set $$U_{m_1} \cap (c,e)$$.

Lemma 3: Let $$(a_1, b_1, m_1) \in \mathcal G$$, and let $$m > m_1$$. Then there exist $$a_2, b_2$$ with $$a_1 < a_2 < b_2 < b_1$$ and $$m_2 \ge m$$ such that $$(a_2, b_2, m_2) \in \mathcal G$$.
Proof: If $$f(a_1) \ne f(b_1)$$, apply Lemma 2 directly. Otherwise, pick any $$c \in (a_1,b_1)$$ and apply Lemma 2 to $$a_1, c, m$$.

Main Proof: We assume $$f$$ is not constant. So there exist $$a_1 < b_1$$ with $$f(a_1) \ne f(b_1)$$. By Lemma 2 there exist $$a_2, b_2$$ with $$a_1 < a_2 < b_2 < b_1$$ and $$m_2 > 2$$ so that $$(a_2,b_2,m_2) \in \mathcal G$$. Then by Lemma 3, there exist $$a_3, b_3$$ with $$a_2 < a_3 < b_3 < b_2$$ and $$m_3 > 3$$ so that $$(a_3,b_3,m_3) \in \mathcal G$$. Continuing recursively, we get sequences $$(a_k), (b_k), (m_k)$$ so that $$\forall k:\quad a_k < a_{k+1} < b_{k+1} < b_k,\quad m_k > k,\quad\text{and } (a_k,b_k,m_k)\in\mathcal G .$$ The sequence $$(a_k)$$ is increasing and bounded above, so it converges. Let $$a = \lim_{k\to\infty} a_k$$. Then $$a_k < a_{k+1} \le a \le b_{k+1} < b_k$$. From $$(a_k,b_k,m_k) \in \mathcal G$$ we conclude $$f(a) > m_k > k$$. This is true for all $$k$$, so $$f(a)$$ is not a real number, a contradiction.

• Your answer was enjoyable and very nice to me. It really engaged me and I am thinking if instead of $\mathbb{R}$ in domain of the $f$, replace a complete metrically convex metric space, does your answer still work? – M. Reza. K Aug 7 '20 at 12:11
• @M.Reza.K for every $x,y$ in the complete metrically convex space there is an isometry $g:[a,b]\to X$ such that $g(a)=x, g(b)=y$. Apply the result to $h=f\circ g$. It follows that $f(x)=h(a)=h(b)=f(y)$. I am interested if we can drop the completeness: the proof seems to work for rational numbers... Perhaps it also works for intrinsic metric spaces? – erz Aug 7 '20 at 12:40
• @erz My proof certainly fails for rational numbers. A decreasing sequence of closed intervals may have empty intersection in $\mathbb Q$. – Gerald Edgar Aug 7 '20 at 13:32

This is not an answer but it greatly narrows down the class of functions $$f$$ with the described property.

1. If $$c>0$$, then the sets $$f^{-1}([c,+\infty))$$ and $$f^{-1}((c,+\infty))$$ are both open.

Indeed, every $$x$$ in any of these sets comes with an open ball of radius $$\frac{1}{f(x)}$$.

Consequently, the sets $$A_c=f^{-1}((-\infty, c))$$ and $$B_c=f^{-1}((-\infty, c])$$ are both closed. In particular, $$f$$ is lower semi-continuous.

1. Let $$x\in B_c$$. We know that if $$d(y,x)<\frac{1}{c}\le \frac{1}{f(x)}$$, then $$f(x)\le f(y)$$. On the other hand, if $$y\in B_c$$ and $$d(y,x)<\frac{1}{c}$$, then $$d(y,x)<\frac{1}{f(y)}$$, from where $$f(y)\le f(x)$$. Hence, if $$y\in B_c$$ and $$d(y,x)<\frac{1}{c}$$, we get $$f(y)= f(x)$$.

Not only this means that $$f$$ is locally a constant on $$B_c$$, but also that $$f^{-1}(c)$$ is closed.

1. Let us prove the following strengthening of the Baire's theorem: let $$X$$ be a complete metric space, and let $$B_n$$ be a sequence of closed sets such that $$X=\bigcup B_n$$. Then $$X=\overline{\bigcup int B_n}$$.

Indeed, if $$U\subset X\backslash \bigcup int B_n$$ is open and nonempty, it is metrizable with a complete metric, and $$B_n\cap U$$ is closed in $$U$$. Since $$U=\bigcup (B_n\cap U)$$, from the usual Baire's theorem, there is $$m$$ such that $$V=int (B_m\cap U) \ne \varnothing$$. But then $$V=int (B_m\cap U)=int B_m\cap U\subset int B_m \backslash \bigcup int B_n=\varnothing$$. contradiction.

1. So, combining these observation we get the following picture: there is a dense open set $$U$$ such that $$f$$ is locally constant on $$U$$; every level set of $$f$$ is closed, and the distance between elements of $$f^{-1}(c)$$ and $$f^{-1}(d)$$ is at least $$\frac{1}{\max \{c,d\}}$$.

The only way it's not a constant function is when the components of $$U$$ are like the gaps in the Cantor set. So the possible counterexample may be constructed like the Cantor's staircase, but the smaller the component, the larger is the value on it. I was unable to perform this fine tuning though, maybe it's impossible.


# Problems about Countability related to Function Spaces

Suppose we have the following sets, and determine whether they are countably infinite or uncountable .

1. The set of all functions from $\mathbb{N}$ to $\mathbb{N}$.

2. The set of all non-increasing functions from $\mathbb{N}$ to $\mathbb{N}$.

3. The set of all non-decreasing functions from $\mathbb{N}$ to $\mathbb{N}$.

4. The set of all injective functions from $\mathbb{N}$ to $\mathbb{N}$.

5. The set of all surjective functions from from $\mathbb{N}$ to $\mathbb{N}$.

6. The set of all bijection functions from from $\mathbb{N}$ to $\mathbb{N}$.

My thoughts on this:

1. The first set is uncountable by using diagonalization argument.

2. I have read something about it saying this is countable since we can see this as a set of finite sequences of natural numbers. But I have hard time following the argument.

3. The article I read says this is uncountable but I couldn't follow the argument either.

4. For (4),(5) and (6), I am not even sure how to approach problems like these.

Could someone please explain how to approach this kind of problems in general?

And does it matter when I change all the $\mathbb{N}$ to $\mathbb{R}$?

-
Recall the definition of a function as a set of ordered pairs. That should help. –  Potato Dec 3 '11 at 2:27
Maybe you can argue that (2) is countable because any given sequence that is non-increasing eventually stops decreasing at some finite natural number or hits zero and then is forced to be constant at 0. Then come up with a clever way to list all the functions, maybe by listing them first by starting position then by "length before it is constant" or something like that. –  tomcuchta Dec 3 '11 at 2:28
@tomcuchta Your idea for (2) is alright, but a small nitpick. The function need not go to $0$ but might stop at $1$ (say). What is true (and important) is that the function is eventually constant. –  Srivatsan Dec 3 '11 at 2:39
I removed the (functional-analysis), the (set-theory) and the (proof-theory) tag. Each of these tags is intended for subjects usually taught at an advanced undergraduate or graduate level. –  t.b. Dec 3 '11 at 2:42
@t.b.Thanks, I guess I am confused about what function spaces really study. –  geraldgreen Dec 3 '11 at 3:12

Another approach to the 6th part.

Let us denote the set of all bijections from $\mathbb N$ to $\mathbb N$ by $\operatorname{Bij}(\mathbb N,\mathbb N)$.

Clearly $$|\operatorname{Bij}(\mathbb N,\mathbb N)| \le \aleph_0^{\aleph_0} = 2^{\aleph_0} = \mathfrak c.$$

Let us try to show the opposite inequality.

For arbitrary function $f\in\operatorname{Bij}(\mathbb N,\mathbb N)$ we define $$\operatorname{Fix}(f)=\{n\in\mathbb N; f(n)=n\}.$$ (That is, $\operatorname{Fix}(f)$ is the set of all fixed points of the map $f$.) Let us try to answer the question, whether for any $A\subseteq\mathbb N$ it is possible to find a bijection ${f_A}:{\mathbb N}\to {\mathbb N}$ such $\operatorname{Fix}(f_A)=A$. Let us consider two cases.

If the complement of the set $A$ is infinite, i.e. $\mathbb N\setminus A=\{b_n; n\in\mathbb N\}$ (and we can assume that the elements $\mathbb N\setminus A$ are written in the increasing order, i.e. $b_1<b_2<\dots<b_n<b_{n+1}<\dots$ ), then we can define a function $f_A$ as follows: $$f_A(x)= \begin{cases} x, & \text{if }x\in A, \\ b_{2k+1}, &\text{ak x=b_{2k} for some k\in\mathbb N},\\ b_{2k}, &\text{ak x=b_{2k+1} for some k\in\mathbb N}. \end{cases}$$ In the other words, we did not move the elements of $A$ and the elements from the complement were paired and we interchanged it pair. Such map is a bijection from $\mathbb N$ to $\mathbb N$ such that $\operatorname{Fix}(f_A)=A$.

Next we consider the case that $\mathbb N\setminus A$ is finite.

If $\mathbb N\setminus A=\emptyset$, then $f=id_{\mathbb N}$ is a function fulfilling $\operatorname{Fix}(f_A)=\mathbb N=A$.

If the set $\mathbb N\setminus A$ is a singleton $\{a\}$, then is is impossible to find a bijection $f$ such that $\operatorname{Fix}(f)=A=\mathbb N\setminus\{a\}$. No element of $\mathbb N\setminus A$ can be mapped on $a$ (since all such elements are fixed). But $a$ cannot be mapped on $a$ since this would mean that $a$ is a fixed point.

But if the set $\mathbb N\setminus A$ has at least two elements, then it is possible to construct such a map. We again assume that the elements $b_k$ are written in the increasing order, i.e. $\mathbb N\setminus A=\{b_0,\dots, b_n\}$ and $b_0<b_1<\dots<b_n$. We put $f_A(x)=x$ for $x\in A$, $f_A(b_k)=b_{k+1}$ for $0\le k<n$ and $f_A(b_n)=b_0$. (I.e. we made a cycle consisting of elements of $\mathbb N\setminus A$.)

This defines a bijection ${f_A}:{\mathbb N}\to{\mathbb N}$ such that $\operatorname{Fix}(f_A)=A$.

The assignment $A\mapsto f_A$ that we described above is a map from the set $\mathcal P({\mathbb N})\setminus\{\mathbb N\setminus\{a\}\in\mathbb N\}$ consisting of all subsets of $\mathbb N$, whose complement is not a singleton, to the set $\operatorname{Bij}(\mathbb N,\mathbb N)$. This map is injective; since from $f_A=f_B$ we get $A=\operatorname{Fix}(f_A)=\operatorname{Fix}(f_B)=B$.

From the set $\mathcal P({\mathbb N})$ with the cardinality $\mathfrak c$ we have omitted a countable set $\{\mathbb N\setminus\{a\}\in\mathbb N\}$. Therefore the cardinality of this difference $\mathcal P({\mathbb N})\setminus\{\mathbb N\setminus\{a\}\in\mathbb N\}$ is again $\mathfrak c$.

So we found an injection from a set of cardinality $\mathfrak c$ to the set $\operatorname{Bij}(\mathbb N,\mathbb N)$. This yields the opposite inequality $$\mathfrak c\le|{\operatorname{Bij}(\mathbb N,\mathbb N)}|$$ and by Cantor-Bernstein theorem we get $|{\operatorname{Bij}(\mathbb N,\mathbb N)}|=\mathfrak c$.

-
Thanks for doing the details. It pointed out a flaw in my answer. –  David Mitra Dec 3 '11 at 14:07
I did not notice that we are using basically the same argument until you pointed it out. I like your answer, but I am out of votes for today. –  Martin Sleziak Dec 3 '11 at 14:14

This is not a complete answer. [I am taking $\mathbb N$ to include $0$.] I will give you a proof that there are uncountably many bijections from $\mathbb N$ to itself. This obviously subsumes the questions (4)-(6).

Consider functions $f : \mathbb N \to \mathbb N$ of the following special form: For every $k$, the pair of elements $\{ 2k, 2k+1 \}$ is mapped to itself. However, we have two ways of doing this:

• either: map $2k$ to itself, and similarly for $2k+1$.

• or: map $2k$ and $2k+1$ to each other.

The key point is that for each $k$, you could pick one of the two possibilities independently. Since there are $2$ options for each $k$, you have $2^{\mathbb N}$ options; i.e., there are uncountably many such functions satisfying this restriction. And each such function is a bijection from $\mathbb N$ to itself.

The above needs a little bit of work to make it completely rigorous. In particular, try to formalise what I mean by "$2^\mathbb N$ possibilities".

For (3), the idea is similar. For each $k$, try mapping $k$ to either $2k$ or $2k+1$. Details are left as an exercise.

For (2), does the following hint help? Every non-increasing function $f : \mathbb N \to \mathbb N$ is eventually constant. So the function is uniquely specified if we describe the finite initial "portion" of the function.

Another approach (due to tb) for (2): If you plot the graph of a non-increasing function, it looks like a staircase that goes down as we move from left to right -- a staircase with finitely many steps because you the function can go down only finitely many times. One can also reconstruct the function if we record the upper right corners of each of those steps. Since the latter corresponds to a finite subset of $\mathbb N \times \mathbb N$ (which is countable), it follows that our set of non-increasing functions is countable as well.

-

I'll use this opportunity to mention here a very nice (at least in my opinion) proof that the cardinality of all bijections $\mathbb N\to\mathbb N$ is $\mathfrak c=2^{\aleph_0}$, which is basically the part 6. I making it a community wiki, since the idea is not mine. I believe I've seen this at sci. math but I am not sure about it and I cannot find the link now. (Feel free to add some pointers and links, if you know some.)

HINT: If someone wants to try to develop the idea by himself, the basic idea is Riemann's rearrangement theorem.

Let us denote the set of all bijections from $\mathbb N$ to $\mathbb N$ by $\operatorname{Bij}(\mathbb N,\mathbb N)$.

Clearly $$|\operatorname{Bij}(\mathbb N,\mathbb N)| \le \aleph_0^{\aleph_0} = 2^{\aleph_0} = \mathfrak c.$$

Now we will show the opposite inequality. Let us choose a series $\sum_{n=0}^\infty x_n$, which is convergent but not absolutely convergent (e.g. $\sum_{n=0}^\infty x_n = \sum_{n=0}^\infty (-1)^n\frac1{n+1}$). By Riemann's theorem, for any real number $r$ there exists a permutation $\pi\in \operatorname{Bij}(\mathbb N,\mathbb N)$ such that $\sum_{n=0}^\infty x_{\pi(n)}=r$. If we choose for each $r$ one such permutation, we get an injective map from $\mathbb R$ to $\operatorname{Bij}(\mathbb N,\mathbb N)$. Thus $|\operatorname{Bij}(\mathbb N,\mathbb N)| \ge |\mathbb R| = \mathfrak c = 2^{\aleph_0}$.

-
+1, nice approach. I don't think you have to make an answer CW just because it isn't your idea. Writing an answer is contribution enough and one deserves some credit for that. =) –  Srivatsan Dec 4 '11 at 0:24

Let \eqalign{ A_1&=\text{the set of all functions from }\Bbb N\text{ to }\Bbb N. \cr A_2&=\text{the set of all non-increasing from }\Bbb N\text{ to }\Bbb N. \cr A_3&=\text{the set of all non-decreasing }\Bbb N\text{ to }\Bbb N. \cr A_4&=\text{the set of all 1-1 functions from}\Bbb N\text{ to }\Bbb N. \cr A_5&=\text{the set of all onto functions from }\Bbb N\text{ to }\Bbb N. \cr A_6&=\text{the set of all bijections from }\Bbb N\text{ to }\Bbb N. \cr }

Consider $A_6$:

$A_6$ may be thought of as the set $\bigl\{\,\pi :\pi \text{ is a permutation of } (1,2,3,\ldots)\,\bigr\}$.

For any permutation $\pi$, we may associated a binary sequence $(x_n)$ by setting $$x_n=\cases{ 0,& \text{if } \pi_n=n\cr 1,& \text{otherwise} }.$$ This induces an onto map from $A_6$ to the set of infinite binary sequences (Edit: not quite, as Martin Sleziak points out in his answer, the binary sequences that have "only one 1" have no element in $A_6$ mapped to them. But, as the set of such sequences is countable, the rest of the argument will go through. Martin also describes nicely how to obtain this map.). Since the latter set is uncountable, so is the former.

So, $A_6$ is uncountable.

From this, it follows that $A_5$, $A_4$, and $A_1$ are uncountable.

Now, consider $A_2$:

If $f$ is a non-increasing function, then $f$ must be eventually constant. Thus, the cardinality $A_2$ would be at most the cardinality of the set $$\bigcup_{j,m} \bigl\{\,(a_i)_{i=1}^\infty : a_i=m, i\ge j\,\bigr \}.$$ But this latter set is a countable union of countable sets and is, thus, countable.

So, $A_2$ is countable.

Now, consider $A_3$:

The cardinality of the set of non-decreasing functions is at least the cardinality of the set $$\{\, A\subset\Bbb N : A \text{ is an infinite subset of } \Bbb N\, \}.$$

Since this latter set uncountable, $A_3$ is uncountable.

-
"If f is a non-increasing function, then f must be eventually constant" Why ? it can also be increasing. –  GinKin Mar 11 at 12:45
@GinKin I took "non-increasing" to mean $f(b)\le f(a)$" whenever $a\le b$ (that is, (perhaps non-strictly) decreasing). This is standard terminology. What do you mean by "it can also be increasing"? –  David Mitra Mar 11 at 13:11
Whoops mixed the terms, I meant that if it doesn't increase then it doesn't have to be eventually constant, it can be decreasing. –  GinKin Mar 11 at 13:24
@GinKin But the range is $\Bbb N$. If it were strictly decreasing, it would eventually be constantly $0$. –  David Mitra Mar 11 at 13:25


# The number of solutions of a polynomial system using a Gröbner basis

The following extract from this Wikipedia page explains that we can easily deduce the number of solutions of a polynomial system using Gröbner basis:

Given the Gröbner basis G of an ideal I, it has only a finite number of zeros, if and only if, for each variable x, G contains a polynomial with a leading monomial that is a power of x (without any other variable appearing in the leading term). If it is the case the number of zeros, counted with multiplicity, is equal to the number of monomials that are not multiple of any leading monomial of G. This number is called the degree of the ideal.

Question: Is there a function in Sage computing this number of zeros from a given Gröbner basis?

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Yes, these

monomials that are not a multiple of any leading monomial of G

form a basis of the quotient ring $R/I$ (by the reduction algorithm), also called a normal basis for $I$.

sage: R.<x,y,z> = PolynomialRing(QQ,3)
sage: I = Ideal([x^2+y+z-1,x+y^2+z-1,x+y+z^2-1])
sage: B = I.groebner_basis(); B
[x^2 + y + z - 1, y^2 + x + z - 1, z^2 + x + y - 1]
sage: I.normal_basis()
[x*y*z, y*z, x*z, z, x*y, y, x, 1]

The number of elements in this basis, or the vector space dimension of $R/I$, can also be found as follows:

sage: I.vector_space_dimension()
8

The zero set, or variety, of the ideal is the following:

sage: I.variety(QQbar)
[{z: 0, y: 0, x: 1},
{z: 0, y: 1, x: 0},
{z: 1, y: 0, x: 0},
{z: -2.414213562373095?, y: -2.414213562373095?, x: -2.414213562373095?},
{z: 0.4142135623730951?, y: 0.4142135623730951?, x: 0.4142135623730951?}]

The number of elements is not 8, because we are missing the "multiplicities". Indeed,

False

So perhaps the more interesting quantity is

5

which agrees with the number of zeros without multiplicity.

(There is an easy algorithm to compute the radical of a zero-dimensional ideal.)

more

Is it required to compute the Gröbner basis? Does the function normal_basis() recompute the Gröbner basis?

( 2020-01-26 20:05:55 -0500 )edit

The method normal_basis() internally requires a Gröbner basis. If one has not yet been computed, then it computes one. If one has been computed already, then it uses that one; it does not recompute it. Note also that the result / speed depends on the chosen monomial ordering (degrevlex by default). Of course the dimension does not depend on this choice.

( 2020-01-27 03:57:43 -0500 )edit


# Solving a Eigenvalue Problem

$$\textbf{Eigenvalue problem on a unit square \Omega = [0,1]^2 :}$$

Consider the eigenvalue problem with the Dirichlet boundary condition that is, $$-Lu = \lambda u$$ where $$L = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}$$.

The boundary condition is that $$u=0$$ on $$\partial \Omega$$.

I am computing the $$\textbf{eigenvalues}$$ and $$\textbf{eigenfunctions}$$ of Laplacian numerically in Mathematica. Specially, I am interested about the $$\textbf{multiplicity}$$ of a specific eigenvalue. I already know that Tally or Count can certainly count the occurrances of eigenvalues in the list called vals.

{ℒ, ℬ} = {-Laplacian[u[x, y], {x, y}],
DirichletCondition[u[x, y] == 0, True]};

{vals, funs} =
DEigensystem[{ℒ, ℬ},
u[x, y], {x, 0, 1}, {y, 0, 1}, _];

Tally[vals]

In the eigenvalue problem there are infinite numbers of eigenvalues. So, We can not list all of them. I want to find out $$\textbf{multiplicity}$$ of a specific eigenvalue that counts the total number of occurrances. For example, the multiplicity of the eigenvalue $$5 \pi^2$$ is $$2$$. Similarly, I want to find out the multiplicity of the eigenvalue $$50 \pi^2$$ is $$3$$.

• Edit your question to include your Mathematica code rather than pictures of the code. Feb 15 '21 at 17:51
• Convert your cells to InputForm and look at the Markdown help Feb 15 '21 at 18:24
• Feb 15 '21 at 19:30

With the exact solutions

u[{nx_, ny_}, {x_, y_}] = Sin[nx π x] Sin[ny π y];

and $$\{n_x,n_y\}\in\mathbb{N}^2$$, the eigenvalues are

λ[nx_, ny_] = (nx^2 + ny^2) π^2;

Test:

Assuming[Element[nx | ny, PositiveIntegers],
{u[{nx, ny}, {x, 0}], u[{nx, ny}, {x, 1}],
u[{nx, ny}, {0, y}], u[{nx, ny}, {1, y}]} // FullSimplify]
(*    {0, 0, 0, 0}    *)

-D[u[{nx, ny}, {x, y}], {x, 2}] - D[u[{nx, ny}, {x, y}], {y, 2}] ==
λ[nx, ny] * u[{nx, ny}, {x, y}] // FullSimplify
(*    True    *)

The number of pairs $$\{n_x,n_y\}$$ equal to $$\lambda/\pi^2$$ is therefore given by OEIS A063725. You can calculate the multiplicity of a given value of $$i=\lambda/\pi^2$$ directly from the generating function,

g[q_] = (EllipticTheta[3, q] - 1)^2/4;
m[i_Integer?NonNegative] := SeriesCoefficient[g[q], {q, 0, i}]

For example, the multiplicity of $$\lambda=50\pi^2$$ is 3, as you know:

m[50]
(*    3    *)

There is no eigenvalue at $$326\pi^2$$, on the other hand:

m[326]
(*    0    *)

More generally, the $$d$$-dimensional generating function

g[d_, q_] = ((EllipticTheta[3, q] - 1)/2)^d;

allows us to find the multiplicity of the eigenvalue $$\lambda=i\pi^d$$ directly:

m[d_Integer?NonNegative, i_Integer?NonNegative] :=
SeriesCoefficient[g[d, q], {q, 0, i}]

For example, the $$d=17$$-dimensional hypercube has $$1\,416\,786\,753\,216$$ eigenvalues $$\lambda=250\pi^{17}$$:

m[17, 250]
(*    1416786753216    *)

If you want the actual solutions, not just the number of solutions, PowersRepresentations is a good starting point (but it gives solutions containing zeros, which we need to filter out):

s[d_Integer?NonNegative, i_Integer?NonNegative] :=
Select[PowersRepresentations[i, d, 2], Min[#] > 0 &]

For example, the eigenvalue $$\lambda=50\pi^2$$ in $$d=2$$ dimensions can be composed in two ordered ways, as you know:

s[2, 50]
(*    {{1, 7}, {5, 5}}    *)

Enumerating all permutations of these ordered combinations gives the full multiplicity of 3:

t[d_Integer?NonNegative, i_Integer?NonNegative] :=
Join @@ Permutations /@ s[d, i]

t[2, 50]
(*    {{1, 7}, {7, 1}, {5, 5}}    *)

In $$d=17$$ dimensions the eigenvalue $$\lambda=250\pi^{17}$$ has 999 ordered solutions,

s[17, 250]
(*    {{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 6, 14},
{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 10, 10},
{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 15},
...
{3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5},
{3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 6}}    *)

which then become the aforementioned $$1\,416\,786\,753\,216$$ unordered solutions from t[17, 250].

• We should work with the generating function explicitly because it gives a very efficient recipe, especially in the high-dimensional case. Just look at the given example for the 17-dimensional hypercube: manually summing and counting is going to be very tedious and difficult, whereas the call to m[17, 1000] takes less than half a second. The same is true, to some extent, for your case $d=2$ when you work with large values of $\lambda$. Feb 16 '21 at 8:32
• Sorry, this is not the right place for an introduction to generating functions or Jacobi theta functions. Feb 16 '21 at 9:31


# Show why the value converges to $$\pi$$

$$a_0=1$$

$$a_{n+1}=a_n+\sin{(a_n)}$$

Explain why the following occurs:

$$a_0=1$$

$$a_1=1+\sin{(1)}\approx 1.841470985$$

$$a_2=1+\sin{(1)}+\sin{(1+\sin{(1)})}\approx 2.805061709$$

$$a_3=1+\sin{(1)}+\sin{(1+\sin{(1)})}+\sin{(1+\sin{(1)}+\sin{(1+\sin{(1)})})}\approx 3.135276333$$

$$a_4\approx 3.141592612$$

$$a_5\approx 3.141592654\approx\pi$$

Note by Jack Han
3 years, 7 months ago

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## Comments

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Let's see here...

The interesting thing I was talking about is the fact that the given series always converges into a value $$a_n$$ such that $$sin(a_n)=0$$ for different values of $$a_0$$. To put it in more exact terms, it always converges to a value of $$a_n$$ such that $$cos(a_n)=-1\Rightarrow a_n=m\pi$$, where $$m$$ is an odd integer. ($$m$$ can also be even, but that is a degenerate case where all terms are same)

I'm going to use something here that I actually learned from gradient descent. If you don't know what it is, you can google it. But, the mathematics used below is an extremely tame form and is easy to understand with little knowledge of calculus.

Consider the function $$f\left( x \right) =\cos { (x) } \\ \Rightarrow \frac { df\left( x \right) }{ dx } =-\sin { (x) }$$

Now see what happens when we take some arbitrary value of $$x$$ (say $$x=1$$)and then do the following repeatedly:

$$x:=x-\frac { df\left( x \right) }{ dx }=x+\sin { (x) }$$ (":=" is the assignment operator )

In the above figure, we can see two points marked. One is red, which represents the first value of $$x$$($$=1$$). The other is brown, and is after one iteration of above step.

We can see that when we do $$x:=x+sin(x)$$, what is actually happening is that $$x$$ is surfing along the slope of the curve $$cos(x)$$. We move the value of $$x$$ down the tangent. Change $$x$$ little by little, so that finally, after many iterations it moves closer and closer to the minima, i.e $$x=\pi$$.

I know this is not a definitive proof of what happens... I'm sure you will realize the importance of this once you understand what is happening.

In general, series defined as $$a_n=a_{n-1}-\alpha\frac { df\left(a_{n-1} \right) }{ da_{n-1} }$$

Will converge to the nearest value of $$a$$ (nearest to $$a_0$$) such that $$f(a)$$ is minimum, provided the value of $$\alpha$$ is not too large.

- 3 years, 7 months ago

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Good work.

- 3 years, 7 months ago

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Good work! Newton's method for estimating roots.

Pretty much seals the deal. Great solution +1.

- 3 years, 7 months ago

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Yes, that was awesome, that is basically the newton Rhapsody method of estimation of roots, doing the following iteration for any curve will eventually lead us to the nearest root, that is great, actually i think this is pretty much the solution +1

- 3 years, 7 months ago

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- 3 years, 7 months ago

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As usual, since the series converges.. this means that when $$n\to\infty$$ ,

$$a_{n+1}=a_{n}$$. But $$a_{n+1}=a_{n}+sin(a_{n})$$

$$\Rightarrow sin(a_{n})=0$$

Now how do we know that $$a_{n}=\pi$$? We know this since $$a_0=1$$ and the series is constantly increasing. Therefore, it converges onto the first value of $$x>1$$ such that $$sin(x)=0$$.

- 3 years, 7 months ago

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Bro , But It is not always true $$\lim _{ n\rightarrow \infty }{ ({ a }_{ n }) } =\lim _{ n\rightarrow \infty }{ { (a }_{ n+1 }) }$$ .

- 3 years, 7 months ago

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Why not? Do you have a counter-example?

- 3 years, 7 months ago

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Comment deleted Mar 13, 2015

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But can't the value of zero be discarded? Because here, $$a_{n}=1$$ for all n.

There is no question of convergence at all as all the terms have a fixed value.

- 3 years, 7 months ago

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My Bad , I wrongly co-relate two diffrent situations , Actually this example is given when I got two answwer to an single recursion , which is something Like that $${ a }_{ n+1 }={ \left( \sqrt { 2 } \right) }^{ { a }_{ n } }$$ , I misinterpreate this current situation, So I deleted that irrelevant comment , But Yes this is good recurance , and I'am trying to prove convergence of this recurance. If I got correct than I report you.

- 3 years, 7 months ago

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Thanks for joining the party :D

- 3 years, 7 months ago

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No, the sequence only increases, only that the rate of increase decreases till it reaches 0 at $$a_n= \pi$$

you can easily see it from the graph if x+sin(x), it is a non decreasing function and reaches inflection point at x= pi

but yes convergence remains unproven, i am trying

- 3 years, 7 months ago

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My bad. Sorry I jumped to a wrong conclusion. Thanks for replying.

But the rate at which the value of $$a_{n}$$ increases isn't the same as the rate at which $$x+sinx$$ is increasing. So how do you directly conclude from the graph?

- 3 years, 7 months ago

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it is not the increasing nature that confirms, it is the fact that the graph lies above x=y as well

- 3 years, 7 months ago

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I sort of understand what you're saying...But isn't it easier to say this: The increment in the variable in every iteration is $$sina_{n}$$ and as $$a_{n}$$ tends to $$\pi$$, $$\quad sina_{n}$$ becomes smaller and smaller. (I think that is what you were saying initially about rate of change slowing down).

I have deleted my previous comment as it might only muddle things up for someone else.

Also, please check out my response to your post on this problem of mine.

- 3 years, 7 months ago

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Indeed it is ,simply that sin(an) is positive, in the range and hence increment is positive or sequence rises :)

- 3 years, 7 months ago

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Can someone prove the convergence of this recursion?

@Ronak Agarwal @Mvs Saketh @Azhaghu Roopesh M @Pratik Shastri @Raghav Vaidyanathan @Deepanshu Gupta or anyone else who's interested.

- 3 years, 7 months ago

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This is turning out to be very interesting...

I want to know if we can find a general form for a function $$f(x)$$ such that the series $$a_1,a_2,...$$ defined by:

$$a_{n+1}=a_{n-1}+f(a_{n-1})$$

Converges for a given value of $$a_0$$.

Further, is it true that all of the values of such $$a_n$$ as $$n\to \infty$$ satisfy $$f(a_n)=0$$?

- 3 years, 7 months ago

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I think the answer to this is going to be extremely interesting.... I have a feeling... Is anyone else thinking what I'm thinking?

- 3 years, 7 months ago

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What are you thinking?

- 3 years, 7 months ago

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are you sure you have differentiated it correctly ? there should be atleast one $$x$$ in your $$f'(x)$$ i think

- 3 years, 7 months ago

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If you call L the value of the limit you obtain sin(L)=0. Now L can be pi or zero but zero is impossible because of the initial condition. More precisely you can say that the value of the sequence is LOW bounded

- 3 years, 7 months ago

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Which value converges to pi ??

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# Integration of piecewise function

I was doing a question that involved integration of a piecewise function.

The question, as can be seen above, asks to evaluate the integral from 0 to pi. I worked it out by breaking the function into 2 integrals where the first had lower and upper limits of 0 and pi/2 respectively and adding it to the integral with lower and upper limits of pi/2 and pi respectively.

Now, my question is, how can we use FTC 2 on the first integral if it is not continuous over the domain [0, pi/2] as the pi/2 is not included in the sin (x).

I have tried my best to articulate my question but if there is any confusion, please do let me know.

Is there something I am missing? Can someone kindly clarify.

You can still evaluate the integrals individually and then add them together. There is a jump discontinuity at $$\frac{\pi}{2}$$ due to the interval given in the piecewise, but this point will not impact the sum. That is because the thickness of a single point is intuitively $$0$$.

Consider the point $$c$$ (Note: c is finite):

$$\int_c^{c} f(x) dx= F(c)-F(c)=0$$ by the FTC.

Additionally, note that the function is integrable from that interval.

$$\int_0^{\pi/2} \sin(x) dx = [-\cos(x)]_0^{\pi/2} = 0-(-1) = 1$$

• Hi, yes, I can see why. However, does this discontinuity affect our ability to use FTC 2 as it requires continuity over [a, b]? Aug 14 at 12:30
• It does not because the sum of one point is, intuitively, $0$. In other words, a point has no thickness/width to sum. Aug 14 at 12:34
• Oh ok. Thank you. So I would be correct in saying that in a case like this, there is some flexibility to the theorem? Aug 14 at 12:37
• @Oofy2000 No because $\sin(x)$ is integrable/well defined/continuous from $0 \le x \lt \frac{\pi}{2}$ Aug 14 at 12:41
• @Oofy2000 Added one. Here is a rigorous explanation, if interested: math.stackexchange.com/questions/511197/area-under-a-point Aug 14 at 13:13

First prove this, from the definition of the integral:

Prop. Suppose $$p\in[a,b]$$, and $$f:[a,b]\to\Bbb R$$ satisfies $$f(x)=0$$ for all $$x\in[a,b]\setminus\{p\}$$. Then $$f$$ is integrable on $$[a,b]$$ and $$\int_a^bf=0$$.

Hence, even though $$f(x)-\sin(x)$$ does not quite vanish at every point, it follows that $$\int_0^{\pi/2}(f(x)-\sin(x))\,dx=0,$$hence $$\int_0^{\pi/2}f(x)\,dx=\int_0^{\pi/2}\sin(x)\,dx.$$

(Technicality: Since $$\sin(x)$$ and $$f(x)-\sin(x)$$ are both integrable on $$[0,\pi/2]$$, so is $$f$$.


# Continuous process vs Discrete process!

I encountered two statements in Tibshirani's article Regression Shrinkage and Selection via the Lasso, p.58 (pdf).

1. Subset selection is a "discrete process" which leads to unstable coefficients.
2. ridge regression is a "continuous process" leading to stable coefficients.

My question is, why ridge is called continuous process while subset selection is a discrete process. Also, why ridge leads to stability and the Subset selection to instability.

• 1. Do you have a particular book or something that says this? It might be useful to answerers to have context. 2. Are you asking "what makes it discrete or continuous?" or are you asking "why would subset selection be unstable while ridge regression is stable?" or are you asking "why would discreteness/continuity be described as stable or unstable?" – Glen_b Jul 25 '16 at 1:54
• @ Glen_b, my question is why ridge is called continuous process, and subset selection is discrete process, and why the first one leads to stability and the later to instability. – jeza Jul 25 '16 at 2:03
• You still need to provide some context / a quote. – gung Jul 25 '16 at 17:33
• @ gung, now, it is ok – jeza Jul 25 '16 at 18:42
• Selecting a subset of variables is a binary decision and in that sense discrete. Ridge regression, on the other hand, performs regularization by putting stronger emphasis on certain variables (without merely making the decision to include or remove them) as a result of a modified residual term. A more precise explanation can be found in Wikipedia. – Igor Jul 25 '16 at 21:18

For subset selection, in its most general form, you pick some metric to optimize and try all possible subsets of variables to include in your linear regression model. Suppose you have $p$ covariates, then there are $2^p$ models, i.e. there are $2^p$ vectors $\beta$ to choose from. Hence, jumping from model to model is a discrete process. You cannot "interpolate" between models.
For ridge regression, you add an $l_2$ penalty to the coefficients, so that the model you choose is $$\min_\beta \| X\beta - y\|^2 + \frac{\lambda}{2}\|\beta\|^2$$ In this case, for any $\lambda \ge 0$ you get a different model (that is solution of $\beta(\lambda)$). In fact, $\beta$ will be a smooth function of $\lambda$: change $\lambda$ a little bit, and your fitted $\beta$ will change a little bit. Hence, there are an infinite number of possible models. Thus, this is a continuous process and you can smoothly go from one model to the other by changing $\lambda$.
This is also the reason why subset selection is unstable, and ridge regression is stable. Suppose you change the metric you use to fit the subset selection process a little bit, and you can end up with a very different model (here with different model I mean what elements of $\beta$ are non-zero). As I noted above, this is not true for ridge regression; a small change in $\lambda$ leads to a small change in $\beta$.
Also worth noting that in general all coefficients in Ridge regression are non-zero for all values of $\lambda$, but they converge monotonically to $0$ as $\lambda \to \infty$.


# Calculate $\int_0^\infty\frac{\sin(x)\log(x)}{x}\mathrm dx$.

Calculate $\displaystyle\int_0^\infty\dfrac{\sin(x)\log(x)}{x}\mathrm dx$.

I tried to expand $\sin(x)$ at zero, or use SI(SinIntegral) function, but it did not work. Besides, I searched the question on math.stackexchange, nothing found.

Mathematica tells me the answer is $-\dfrac{\gamma\pi}{2}$, I have no idea how to get it.

Notice that

\begin{align*} \int_{0}^{\infty} \frac{\sin x}{x^s} \, dx &= \frac{1}{\Gamma(s)} \int_{0}^{\infty} \left( \int_{0}^{\infty} t^{s-1}e^{-xt} \, dt \right) \sin x \, dx \\ &= \frac{1}{\Gamma(s)} \int_{0}^{\infty} \left( \int_{0}^{\infty} e^{-tx} \sin x \, dx \right) t^{s-1} \, dt \\ &= \frac{1}{\Gamma(s)} \int_{0}^{\infty} \frac{t^{s-1}}{1+t^2} \, dt \\ &= \frac{1}{2\Gamma(s)} \beta\left(\frac{s}{2}, 1-\frac{s}{2}\right) \\ &= \frac{\pi}{2\Gamma(s)\sin\left(\frac{\pi s}{2}\right)}. \end{align*}

(This heuristic computation is valid line-by-line on the strip $1 < \Re(s) <2$ in view of Fubini's theorem, and then extends to the larger strip $0 < \Re(s) < 2$ by analytic continuation.) Differentiating both sides, we get

$$\int_{0}^{\infty} \frac{\sin x \log x}{x^s} \, dx \stackrel{(*)}{=} -\frac{d}{ds} \frac{\pi}{2\Gamma(s)\sin\left(\frac{\pi s}{2}\right)} = \frac{\pi}{2\Gamma(s)\sin\left(\frac{\pi s}{2}\right)} \left( \psi(s) + \frac{\pi}{2}\cot\left(\frac{\pi s}{2}\right) \right).$$

Now the answer follows by plugging $s = 1$:

$$\int_{0}^{\infty} \frac{\sin x \log x}{x} \, dx = -\frac{\gamma \pi}{2}.$$

Justification of $\text{(*)}$. Let us prove that

$$F(s) := \int_{0}^{\infty} \frac{\sin x}{x^s} \, dx \tag{1}$$

is analytic on the open strip $S = \{ s \in \Bbb{C} : 0 < \Re(s) < 2 \}$ and its derivative can be computed by the Leibniz's integral rule. A major issue of $\text{(1)}$ is that the defining integral converges only conditionally for $0 < \Re(s) \leq 1$ and thus raises some technical difficulties. In order to circumvent this, we improve the speed of convergence using integration by parts:

$$F(s) = \underbrace{\left[ \frac{1-\cos x}{x^s} \right]_{0}^{\infty}}_{=0} + s \int_{0}^{\infty} \frac{1-\cos x}{x^{s+1}} \, dx.$$

Notice that the resulting integral is absolutely convergent on $S$.

Now we claim that $g(s) := F(s)/s$ is differentiable and its derivative can be computed by the Leibniz's integral rule. Let $\epsilon$ be such that $\bar{B}(s, \epsilon) \subset S$. Then whenever $0 < |h| < \epsilon$,

\begin{align*} &\frac{g(s+h) - g(s)}{h} - \int_{0}^{\infty} \frac{(1-\cos x)(-\log x)}{x^{1+s}} \, dx \\ &\hspace{9em} = \int_{0}^{\infty} \frac{1-\cos x}{x^{1+s}} \left( \frac{x^{-h} - 1}{h} + \log x \right) \, dx. \end{align*}

Notice that the integrand is dominated by

$$\left| \frac{1-\cos x}{x^{1+s}} \left( \frac{x^{-h} - 1}{h} + \log x \right) \right| \leq \frac{1-\cos x}{x^{1+\Re(s)}} (\max\{ x^{\epsilon}, x^{-\epsilon} \} + 1) |\log x|$$

This dominating function is integrable. Hence by the dominated convergence theorem, as $h \to 0$, we have

$$\lim_{h \to 0} \frac{g(s+h) - g(s)}{h} = \int_{0}^{\infty} \frac{(1-\cos x)(-\log x)}{x^{1+s}} \, dx.$$

Plugging this back, we know that $F(s)$ is differentiable with

$$F'(s) = \int_{0}^{\infty} \frac{1-\cos x}{x^{s+1}} \, dx + s \int_{0}^{\infty} \frac{(1-\cos x)(-\log x)}{x^{1+s}} \, dx.$$

Finally, performing integration by part to the latter integral yields the desired conclusion:

\begin{align*} &s \int_{0}^{\infty} \frac{(1-\cos x)(-\log x)}{x^{1+s}} \, dx \\ &\hspace{5em} = \underbrace{\left[ \frac{(1-\cos x)\log x}{x^s} \right]_{0}^{\infty}}_{=0} - \int_{0}^{\infty} \left( \frac{\sin x \log x}{x^s} + \frac{1-\cos x}{x^{1+s}} \right) \, dx \end{align*}

• @JackD'Aurizio, Thank you! – Sangchul Lee Sep 26 '16 at 13:35
• Nothing to thank me for, it is a really well-written answer, credits to you. – Jack D'Aurizio Sep 26 '16 at 13:38
• +1. Not too many people care to write a justification $^{*}$ as you did it ( included me !!! ). – Felix Marin Oct 2 '16 at 8:37

Hereafter the $\ds{\ln}$-branch-cut runs along $\ds{\left(-\infty,0\right]}$ with $\ds{\ln\pars{z}\ \,\mrm{arg}}$ given by $\ds{-\pi < \mrm{arg}\pars{z} < \pi}$.

\begin{align} \mc{J} & \equiv \Im\int_{0}^{\infty}\ln\pars{x}\expo{-\pars{t - \ic}x}\,\dd x = \Im\bracks{{1 \over t - \ic}\int_{0}^{\pars{t - \ic}\infty} \ln\pars{x \over t - \ic}\expo{-x}\,\dd x} \\[5mm] & = -\,\Im\braces{{t + \ic \over t^{2} + 1}\int_{\infty}^{0}\bracks{\ln\pars{x \over \root{t^{2} + 1}} + \arctan\pars{1 \over t}\ic}\expo{-x}\,\dd x} \\[5mm] & = \Im\braces{{t + \ic \over t^{2} + 1}\bracks{% \int_{0}^{\infty}\ln\pars{x}\expo{-x}\,\dd x - {1 \over 2}\ln\pars{t^{2} + 1}+ \arctan\pars{1 \over t}\ic}} \end{align}

Note that
$\ds{\int_{0}^{\infty}\ln\pars{x}\expo{-x}\,\dd x = \left.\partiald{}{\mu}\int_{0}^{\infty}x^{\mu}\expo{-x}\,\dd x\, \right\vert_{\ \mu\ =\ 0} = \Gamma\,'\pars{1}\ =\ \overbrace{\Gamma\pars{1}}^{\ds{=\ 1}}\ \overbrace{\Psi\pars{1}}^{\ds{-\gamma}}\ =\ -\gamma\quad}$ where $\ds{\Gamma}$

and $\ds{\Psi}$ are the Gamma and Digamma Functions, respectively. Then,

\begin{align} \mc{J} & \equiv \bbox[8px,#ffe,border:0.1em groove navy]{% \Im\int_{0}^{\infty}\ln\pars{x}\expo{-\pars{t - \ic}x}\,\dd x} = -\,{\gamma \over t^{2} + 1} - {1 \over 2}\,{\ln\pars{t^{2} + 1} \over t^{2} + 1} + {t\arctan\pars{1/t} \over t^{2} + 1} \\[5mm] & = \bbox[8px,#ffe,border:0.1em groove navy]{-\,{\gamma \over t^{2} + 1} + \totald{}{t}\bracks{{1 \over 2}\,\arctan\pars{1 \over t}\ln\pars{t^{2} + 1}}} \label{2}\tag{2} \end{align}

Note that$\ds{{1 \over 2}\,\arctan\pars{1 \over t}\ln\pars{t^{2} + 1}}$ vanishes out when $\ds{t \to \infty}$ and $\ds{t \to 0^{+}}$.

By replacing \eqref{2} in \eqref{1}: \begin{align} &\color{#f00}{\int_{0}^{\infty}{\sin\pars{x}\ln\pars{x} \over x}\,\dd x} = -\gamma\int_{0}^{\infty}{\dd t \over t^{2} + 1} = \color{#f00}{-\,{1 \over 2}\,\gamma\pi} \end{align}

• (+1) That is how I just did this. Then I looked on MSE to see if it's already been discusses and found this page. Well done my friend! – Mark Viola Apr 23 at 3:36


# All real numbers in $[0,2]$ can be represented as $\sqrt{2 \pm \sqrt{2 \pm \sqrt{2 \pm \dots}}}$

I would like some reference about this infinitely nested radical expansion for all real numbers between $0$ and $2$.

I'll use a shorthand for this expansion, as a string of signs, $+$ or $-$, with infinite periods denoted by brackets.

$$2=\sqrt{2 + \sqrt{2 + \sqrt{2 + \dots}}}=(+)$$

$$1=\sqrt{2 - \sqrt{2 - \sqrt{2 - \dots}}}=(-)$$

$$0=\sqrt{2 - \sqrt{2 + \sqrt{2 + \dots}}}=-(+)$$

$$\phi=\sqrt{2 + \sqrt{2 - \sqrt{2 + \sqrt{2 - \dots}}}}=(+-)$$

$$\frac{1}{\phi}=\sqrt{2 - \sqrt{2 + \sqrt{2 - \sqrt{2 + \dots}}}}=(-+)$$

In general, the expansion can be found by a very easy algorithm:

• take any number in $(0,2)$, square it
• if the result $>2$ write $+$, if the result $<2$ write $-$
• subtract $2$ from the result, square, repeat

If on some step we get $2$ exactly, we just write $(+)$ and the expansion is finished.

Examples:

$$\pi-2=--+-++-+-+++++++-+-+---------+-+--+--+--+++---++++ \dots=1.141592653589793 \dots$$

Basically, $50$ terms of our expansion gave only $15$ correct decimal digits for $\pi$. But considering the expansion can be coded as binary, it's not so bad.

The convergence plot, and two binary plots for this $50$ terms can be seen below:

$$e-1=+-----+++-++-+---++-++++-+---++-+++-++++-++++---++ \dots=1.71828182845905 \dots$$

Do you know any reference about this expansion? Can every real number between $0$ and $2$ be expanded this way?

Is number $2$ special in this case, or can we make a similar expansion using some other number (and other power for the root)?

Edit

Now that I think about it, we can use the general expansion for $x \in [0,a]$:

$$x=\left(a \pm \left(a \pm \left(a \pm \dots \right)^p \right)^p \right)^p$$

$$a=2^{\frac{p}{1-p}}$$

For example:

$$\frac{1}{4}=\left(\frac{1}{4} + \left(\frac{1}{4} + \left(\frac{1}{4} + \dots \right)^2 \right)^2 \right)^2$$

$$\frac{3}{4}-\frac{\sqrt{2}}{2}=\left(\frac{1}{4} - \left(\frac{1}{4} - \left(\frac{1}{4} - \dots \right)^2 \right)^2 \right)^2$$

etc.

However, this case $p=2,~~~a=\frac{1}{4}$ is not just a random example, it's the only rational expansion of this kind. So I would say it's more important than the titular root expansion.

Edit

An interesting article that connects the nested roots of this kind to Chebyshev polynomials: http://www.sciencedirect.com/science/article/pii/S0022247X12003344

• Wow. Great question, +1 – user223391 Apr 6 '16 at 21:33
• Beautiful idea. One could make a binary-like number system based on it. Let a "bit" code 0 for minus or 1 for plus. I wonder if it could have unique representation contrary to the usual way to write numbers, for example $0.1b = 0.011\dots b = 1/2$. – mathreadler Apr 7 '16 at 8:01
• I wonder if there is some class of numbers which have periodic "number expansions" in that number system. Like the rationals have for ordinary number systems using the division algorithm. – mathreadler Apr 7 '16 at 8:27
• @mathreadler, that would be a subset of algebraic numbers – Yuriy S Apr 7 '16 at 8:30
• the $50$ bits gives $50/\log_{10}(2) \approx 15$ decimals which is just one or two bits short of mantissa of a ordinary double precision floating point number system. – mathreadler Apr 7 '16 at 15:36

Here is a possible explanation. Let $\alpha \in [0, \pi/2]$ and define $\epsilon_1, \epsilon_2, \cdots$ by $\epsilon_i = \operatorname{sgn}( \cos ( 2^i \alpha )) \in \{-1, 1\}$. Here, we take the convention that $\operatorname{sgn}(0) =1$. Then applying the identity $2\cos\theta = \operatorname{sgn}(\cos\theta) \sqrt{2 + 2\cos(2\theta)}$ repeatedly, we have

$$2\cos \alpha = \sqrt{2 + \epsilon_1 \sqrt{2 + \epsilon_2 \sqrt{ \cdots + \epsilon_n \sqrt{2 + \smash[b]{2\cos(2^{n+1} \alpha)} }}}}.$$

This can be used to show that, with an appropriate definition of infinite nested radical, the following identity

$$2\cos \alpha = \sqrt{2 + \epsilon_1 \sqrt{2 + \epsilon_2 \sqrt{ 2 + \cdots }}}$$

is true. This shows that any real number between $[0, 2]$ can be written as an infinite nested radical of the desired form. Moreover, if we denote $x = 2\cos\alpha$, then

• $\epsilon_1 = \operatorname{sgn}(2\cos (2\alpha)) = \operatorname{sgn}(x^2 - 2)$,
• $\epsilon_2 = \operatorname{sgn}(2\cos (4\alpha)) = \operatorname{sgn}((x^2 - 2)^2 - 2)$,

and likewise. This explains why signs are determined by OP's algorithm.

• Wow, this is a great explanation, which also shows where $2$ comes from – Yuriy S Apr 6 '16 at 22:47
• @YuriyS, Thank you! And yes, this explains why the range is $[0, 2]$. Unfortunately this seems not work for other cases with $a > 2$ instead of $2$, and I even suspect that the analogous claim may be false. (For an an analogous situation, we can in fact check this. Numbers of the form $\pm \frac{1}{2} \pm \frac{1}{4} \pm \frac{1}{8} \pm \cdots$ represents any real number in $[-1, 1]$, but the range of $\pm \frac{1}{3} \pm \frac{1}{9} \pm \frac{1}{27} \pm \cdots$ is the (shifted) Cantor set.) – Sangchul Lee Apr 6 '16 at 22:59
• Nice answer (+1). This is theme of problems 183 - 185 in the classic Problems and Theorems in Analysis I by G. Polya and G. Szegö – Markus Scheuer Apr 9 '16 at 16:02
• This is incredible. – 6005 Jul 26 '16 at 21:08

Peculiar observation

If we define a binary number $b = b_1b_2\cdots b_n$ with digits mapped to the symbols like this: $$b_k = \begin{cases}0 \text{ if } (-) \text{ at position } k\\1 \text{ if } (+) \text{ at position } k\end{cases}$$ Then if we run the algorithm proposed in the question, looping

x(k) = x(k-1)^2-2;
b(k) = (x(k)>0);

the vector b will get logical values corresponding to bits 1 and 0 of the binary number above and we can calculate it for the linear space of $x\in[0,1]$. If we do this we can then calculate each number as the scalar product $$[1/2,1/4,\cdots,1/2^k]b$$ and if we then plot it, it will look like

Which is kind of a peculiar plot having a bit of a discontinuous and fractal structure. I think the largest discontinuity is around $x = \sqrt{1/2}$ but I have no theoretical explanation why..

edit as pointed out by Sangchul Lee this seems similar to Tent Map

• Good observation. I am not expert of symbolic dynamics, but this is related to the tent map orbits. Basically this is because the string of signs are produced by the doubling transform on the circle via the correspondance I explained in my answer. – Sangchul Lee Apr 8 '16 at 2:58

Your algorithm pretty much shows that there exists an expansion for every number in $[0,2]$.

If we replace $2$ by $a>2$ (and keep the square root), we will fail because we need that squaring a number from the interval $[u,v]$ produces a number that is either in $a+[u,v]$ or in $a-[u,v]$. So we must have $u=0$ and $v\ge a$ and $v^2\le a+v$. The last two imply $a\le v\le 2$.

If we additionally switch to $k$th roots, the condition becomes that $v\ge a$ and $v^k\le a+v$, hence $a\le v\le\sqrt[k-1]2$.


# An example of an algebra but not a sigma algebra?

Can someone give me an example of an algebra but not a sigma algebra, with a $\sigma$-finite measure $\mu$ on it?

• As far as I'm aware, $\sigma$-finiteness is only defined on sigma algebras - how are you defining it here? – B. Mehta Apr 14 '18 at 22:43
• @B.Mehta I understood this to mean a measure on the sigma-algebra generated by the algebra. – Severin Schraven Apr 14 '18 at 22:44

Let $X=[0, \infty)$ and consider $\mathcal{A}=\{ \text{finite union of }[a, b)\}$.

It's clear that $[a, b)\cap [c, d) = [\max\{a, c\}, \min\{b, d\})$ and $[a, b)^C = [0, a)\cup [b, \infty)$.

So $\mathcal{A}$ is an algebra but clearly not a $\sigma$-algebra since it doesn't contain any open interval.

Take the obvious measure $m([a, b)) = b-a$.

• +1 Neat example. – Severin Schraven Apr 14 '18 at 22:58
• Thanks Jacky that’s very understandable! :D – JINGYA HAN Apr 14 '18 at 23:35

Take all subsets of $\mathbb{N}$ which are either finite or have finite complement. It is an algebra, but not a sigma-algebra. For the sigma-finite measure you can pick the zero measure.

• Or for the measure pick counting measure. – GEdgar Apr 14 '18 at 23:01
• Thanks for your example. In this context I don’t think the measure could be replaced by a counting measure since that would not always be finite. But say if we replace the measure by a counting measure, does it mean this measure does not have a unique extension on the whole natural number set, since it would meet the requirement of using Carathéodory Extension Theorem? – JINGYA HAN Apr 14 '18 at 23:34
• @JINGYAHAN We could replace the measure with the counting measure, it would still be sigma-finite as we can write $$\mathbb{N}=\bigcup_{n\in \mathbb{N}} \{n\}$$ and the singeltons have counting measure equal to $1$. – Severin Schraven Apr 15 '18 at 0:26
• @SeverinSchraven oh yes you’re right. I confused the definition. The point is you cannot assign a singleton with an infinity measure. Thanks for your help! – JINGYA HAN Apr 15 '18 at 0:31
• @JINGYAHAN Exactly, we don't want concentration of mass on singeltons. By the way you should accept Jacky's answer (on the left you of the question you can find the respective button), unless you are not yet satisfied with the answers given and want to wait for another one. – Severin Schraven Apr 15 '18 at 0:36


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# Writing a series using Sigma notation

How do I write $$2+ \frac{3x}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}?$$

I have been struggling with these types of problems, so please, an explanation of how to get the result will be appreciated.

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$$2+ \frac{3x}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}$$

Since everything else has a denominator, let's try putting a denominator for $2$. Since anything divided by one itself, we get:

$$\frac{2}{1}+ \frac{3x}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}$$

Everything else has an $x$. Let's give $2$ an $x$ too! $x^0 = 1$, and anything multiplied by $1$ is itself, so:

$$\frac{2x^0}{1}+ \frac{3x}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}$$

All the other $x$s have an exponent. Let's give one to that $x$ next to the $3$. $x^1 = x$, so we get:

$$\frac{2x^0}{1}+ \frac{3x^1}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}$$

However, these denominators are just powers of two. Therefore, we have:

$$\frac{2x^0}{2^0}+ \frac{3x^1}{2^1} + \frac{4x^2}{2^2}+\frac{5x^3}{2^3}+\frac{6x^4}{2^4}$$

Now we see some patterns:

• $2, 3, 4, 5, 6$ (numerator of coefficient of $x$)
• $0, 1, 2, 3, 4$ (exponents of $x$)
• $1, 2, 4, 8, 16$ (denominator of coefficient of $x$. Powers of $2$, viz. $2^0, 2^1, 2^2, 2^3, 2^4$)

We'll make the sum range from $0$ to $4$, although in theory we could actually make it range over anything.

$$\sum_{k=0}^4$$

We need exponents of $x$, so we'll put those in there.

$$\sum_{k=0}^4 x^k$$

We need those powers of two in the denominator, so add that too:

$$\sum_{k=0}^4 \frac{x^k}{2^k}$$

However, there are still those numerators we haven't checked off our list yet. We can't have them ranging from $0$ to $4$, we need them from $2$ to $6$. The solution is to add two!

$$\sum_{k=0}^4 \frac{(k+2)x^k}{2^k}$$

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I'd have gone on to write $\dfrac{2x^0}{1}+ \dfrac{3x^1}{2} + \dfrac{4x^2}{4}+\dfrac{5x^3}{8}+\dfrac{6x^4}{16}$ in the form $\dfrac{2x^0}{2^0}+ \dfrac{3x^1}{2^2} + \dfrac{4x^2}{2^2}+\dfrac{5x^3}{2^3}+\dfrac{6x^4}{2^4}$. –  Michael Hardy Apr 17 '13 at 23:34
@MichaelHardy, good idea, I added it. –  George V. Williams Apr 18 '13 at 0:05

It's much like finding a pattern, and a way to express the pattern. Starting point: $2 = \dfrac 21$. Numerator coeffienct of the $x$-term, increases by one per term , starting at $2$ (add 1 to starting numerator); the exponent for $x$ increases by one (starting, say, at $x^0,...x^1 = x,...x^2...$ (Add one to the x-exponent starting at $0$. Also, the denominator represents powers of $2$, starting at $2^0 = 1$...Increase (add) by one the exponent of the term $2$ in the denominator $2^0, ... 2^1 = 2, ...2^{1+1} = 2^2 = 4, ...$

Try writing out the first few terms of the following and see if it matches, starting index $0$, and on...:

$$\sum_{k = 0}^\infty \frac{(k+2)x^k}{2^k}$$

If things don't work out (here they happen to work just fine), there is often a little trial-and-error involved. For example, we can "tweak" if we want to represent the series with a starting index $1$, by "subtracting $1$ from each of the expressions given in terms of $k$.

NOTE: The above will give you an infinite series. To get the first five terms, we start at $k = 0$, and need to sum through (inclusive) $k = 4$. But you can obtain any positive number $n$ of terms by summing from $k = 0$ to $k = n-1)$

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The best way to get a "knack" for "cracking" a sequence is through practice. Try to approach sequences of numbers like you'd approach a puzzle: what fits? what doesn't fit? What patterns do I see...start writing down what you do see, and try creating a "model" term (a "work in progress," so to speak)...Figure out if starting with $0$ or $1$ for one of the varying terms works best. If a term appears constant through the sequence, represent it as a constant, etc. –  amWhy Apr 17 '13 at 23:47
Such great advice! +1 –  Amzoti Apr 18 '13 at 0:24
Such a great answer +1 –  Babak S. Aug 21 '13 at 6:14

Note that the power of $x$ increase in steps of $1$ starting from $x^0$. Hence the $(n+1)^{th}$ term has $x$ raised to the power $n$. The denominators are powers of $2$ starting from $2^0$ and the numbers in the numerator also increase in steps of $1$ starting from $2$. $$\sum_{n=0}^4 \dfrac{(n+2)x^n}{2^n}$$

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Notice that $2=\frac {2x^0}1$, so you want $$2+ \frac{3x}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}=\frac {2x^0}1+ \frac{3x}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}=\sum_{i=1}^5\frac {(i+1)x^{i-1}}{2^{i-1}}$$

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Start off by looking for any patterns. I would start off by looking at how the powers on $x$ change. Starting at $0$ ($x^0=1$) it looks like the power goes up by one at each "step." So we can start off with the "skeleton" sum $$\sum_{i=0}^4 x^i.$$ Similarly observe the coefficient patters, it looks like the numerator starts at $2$ and goes up by one at each step and the denominator are powers of $2$ starting at the $2^0$. So putting that all together we get $$\sum_{i=0}^4 \frac{(i+2)x^i}{2^i}$$

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# Is there a family of processes centred on the Poisson process?

I am looking for a model, characterized continuously by a single parameter, to describe the arrival times of buses with unit expected interarrival time. At one extreme of the parameter (say $\theta=1$), the process is deterministic, with all interarrival times equal (to $1$). When $\theta=0$, the process is pure Poisson (with unit expected interarrival or waiting time). As $\theta$ approaches $-1$, the arrivals tend to cluster, with long intervals between the clusters. At the (unattainable) extreme of $\theta=-1$, all the buses arrive in a simultaneous convoy, after an infinite wait, and you have to wait forever for the next convoy.

The choice of $\{1,0,-1\}$ for the extreme and central parameter values isn't important: they could be $\{-\infty, 0, \infty\}$, $\{0,\frac12,1\}$, ..., whatever. At this stage, simplicity and naturalness, subject to the above conditions, is more important than realism.

In a pure Poisson process, the interarrival times (times between each bus) has an exponential distribution. So you want some family of distributions which includes the exponential as a special case. The gamma distribution (see: https://en.wikipedia.org/wiki/Gamma_distribution). When then shape parameter is 1, you have the exponential. When the shape parameter ($k$ or $\alpha$ in wiki) approaches zero, you are approaching constant interarrival times, and growing above 1 you get interarrival times more variable than for the poisson process.
• Yes, this could be the model if we set $k\theta=1$. But I think that you have it the wrong way round: We approach a constant (unit) interarrival time as $k\rightarrow\infty$, and long intervals between clusters as $k\rightarrow0$. – John Bentin Feb 12 '15 at 16:06


# Number of real roots of a quartic

#### CaptainBlack

##### Well-known member
rayman's question from another place:

Could someone help me with this problem, I have no idea how to start with it

How many real roots does this polynomial have p(x)=x^4-x^3-1?

Clearly state the argument that explains the number of real roots.

Thank you for any help

Descartes rule of signs tells you this has exactly 1 positive root, and exactly 1 negative root, so it has two real roots.

CB

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#### chisigma

##### Well-known member
The Descartes rule establishes the maximum number of positive/negative real roots that a polynomial can have but it gives no information about the effective number of the real roots of a polynomial. In our case is $\displaystyle p(x)= x^{4}-x^{3} -1$ and, in my opinion, the number of its real root can be found considering the polynomial $\displaystyle q(x)= x^{4}-x^{3}$. It is easy enough to see that $q(x)$ has a root of order 3 in x=0 and a root of order 1 in x=1. Furthermore q(x) has a minimum in $x=\frac{3}{4}$ and here is $q(x)=- \frac{27}{256}$. Now if we consider the quartic equation $\displaystyle q(x)+a=0$, on the basis of consideration we have just done, it is easy to find that the quartic equation has two real roots for $a<\frac{27}{256}$, one real root of order 2 for $a=\frac{27}{256}$ and no real roots for $a>\frac{27}{256}$...

Kind regards

$\chi$ $\sigma$

#### CaptainBlack

##### Well-known member
The Descartes rule establishes the maximum number of positive/negative real roots that a polynomial can have but it gives no information about the effective number of the real roots of a polynomial. In our case is $\displaystyle p(x)= x^{4}-x^{3} -1$ and, in my opinion, the number of its real root can be found considering the polynomial $\displaystyle q(x)= x^{4}-x^{3}$. It is easy enough to see that $q(x)$ has a root of order 3 in x=0 and a root of order 1 in x=1. Furthermore q(x) has a minimum in $x=\frac{3}{4}$ and here is $q(x)=- \frac{27}{256}$. Now if we consider the quartic equation $\displaystyle q(x)+a=0$, on the basis of consideration we have just done, it is easy to find that the quartic equation has two real roots for $a<\frac{27}{256}$, one real root of order 2 for $a=\frac{27}{256}$ and no real roots for $a>\frac{27}{256}$...

Kind regards

$\chi$ $\sigma$
In this case Descartes rule of signs does tell us exactly how many real roots we have.

The number of positive roots is equal to the number of changes of signs of the coefficients less a multiple of 2. In this case the number of sign changes is 1, and as there is no multiple of 2 other than 0 which leaves the number of roots non-negative there is exactly one positive real root. The same argument applies to the negative roots.

CB