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# Number of bit strings of length four do not have two consecutive 1s
I came across following problem:
How many bit strings of length four do not have two consecutive 1s?
I solved it as follows:
Total number of bit strings of length: $$2^4$$
Total number of length 4 bit strings with 4 consecutive 1s: 1
Total positions for three consecutive 1s in length 4 bit string: 2 (111X, X111)
Number of bit strings for each of above positions: 2 (X can be 0 or 1)
Total positions for two consecutive 1s in length 4 bit string: 3 (11XX, X11X, XX11)
Number of bit strings for each of above positions: 4
By inclusion exlcusion principle, the desired count $$=2^4-3\times 4+2\times 2-1=16-12+4-1=7$$
However the correct solution turns out to be 8. It seems that I incorrectly applied inclusion exclusion principle. Where did I go wrong?
• You should be doing inclusion-exclusion on the number of pairs of consecutive ones, not on the length of a string of consecutive ones. – Gerry Myerson Sep 16 '19 at 6:27
If I were doing this by inclusion-exclusion, I'd go: $$16$$ strings of length four; $$12$$ with at least one pair of consecutive ones ($$11xx,x11x,xx11$$ with $$x$$s arbitrary); five with at least two pair of consecutive ones ($$111x,1111,x111$$); one with three pair of consecutive ones; so $$16-12+5-1=8$$.
To count the number of bit strings with $$2$$ consecutive one bits (bad strings), I would let \begin{align} S_1&=11xx&4\\ S_2&=x11x&4\\ S_3&=xx11&4\\ N_1&=&12 \end{align} Then \begin{align} S_1\cap S_2&=111x&2\\ S_1\cap S_3&=1111&1\\ S_2\cap S_3&=x111&2\\ N_2&=&5 \end{align} and \begin{aligned} S_1\cap S_2\cap S_3&=1111&1\\ N_3&=&1 \end{aligned} The count of bad strings is $$N_1-N_2+N_3=8$$.
The count of good strings is $$16-8=8$$.
Generating Functions | {
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Generating Functions
Let $$x$$ represent the atom '$$0$$' and $$x^2$$ represent the atom '$$10$$' and build all possible strings by concatenating one or more atoms and removing the rightmost '$$0$$'. \begin{align} \overbrace{\vphantom{\frac1x}\ \ \ \left[x^4\right]\ \ \ }^{\substack{\text{strings of}\\\text{length 4}}}\overbrace{\ \quad\frac1x\ \quad}^{\substack{\text{remove the}\\\text{rightmost '0'}}}\sum_{k=1}^\infty\overbrace{\vphantom{\frac1x}\left(x+x^2\right)^k}^\text{k atoms} &=\left[x^4\right]\frac{1+x}{1-x-x^2}\\ &=\left[x^4\right]\left(1+2x+3x^2+5x^3+8x^4+13x^5+\dots\right)\\[9pt] &=8 \end{align} Note that the denominator of $$1-x-x^2$$ induces the recurrence $$a_n=a_{n-1}+a_{n-2}$$ on the coefficients.
Recurrence
Good strings of length $$n$$ can be of two kinds: a good string of length $$n-1$$ followed by '$$0$$' or a good string of length $$n-2$$ followed by '$$01$$'. That is, $$a_n=a_{n-1}+a_{n-2}$$ Starting with
$$a_0=$$ the number of good strings of length $$0=1$$.
$$a_1=$$ the number of good strings of length $$1=2$$.
we get $$a_4=8$$. | {
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# When is a function bounded using differentiation
1. Jan 16, 2008
### sara_87
1. The problem statement, all variables and given/known data
how do i determine whether a function is bounded using differentiation
eg: f(x)=x/(2^x)
2. Relevant equations
3. The attempt at a solution
i know it has something to do with maximums and minimums but i cant figure out how to do it.
any help would be appreciated. thank you
2. Jan 16, 2008
### Dick
You could look at the limits of the function as it approaches plus and minus infinity. If both exist and are finite, and if the function is defined and continuous for all x, then it is bounded.
3. Jan 16, 2008
### EnumaElish
You can use differentiation to investigate the behavior of f. Say, the function is f(x) = x/2^x on x > 0. Then f'(x) = 2^-x (1 - x Log[2]), which has roots 1/Log[2] and +infinity. At x = 1/Log[2], f''(x) = 2^-x Log[2] (-2 + x Log[2]) is < 0, so you have the maximum. Note that f(x) > 0 for x > 0 and f(0) = 0. As x --> +infinity, f(x) --> 0 from above; but f(0) = 0 so x = 0 is the minimum. Since you can "account for" both the maximum and the minimum, f is bounded on x > 0.
Last edited: Jan 16, 2008
4. Jan 16, 2008
### sara_87
thank you very much.
what if we have:
f(x)=(-2x^2)/(4x^2-1)
i know that it's not bounded but i dont know why
5. Jan 16, 2008
### Tom Mattson
Staff Emeritus
The graph of that function has two vertical asymptotes. Functions don't get much more "unbounded" than that!
6. Jan 16, 2008
### sara_87
actually i think it is bounded because there's no value of x that would make that function greater than 1
or is there?
7. Jan 16, 2008
### Tom Mattson
Staff Emeritus
Sure there is. As I said, the graph of that function has 2 vertical asymptotes. You can find values of x for which the function blows up to infinity, and down to negative infinity.
Do you know what I mean when I say "vertical asymptote"?
8. Jan 16, 2008
### sara_87 | {
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Do you know what I mean when I say "vertical asymptote"?
8. Jan 16, 2008
### sara_87
yes i do know what vertical assymptotes are.
umm but i still didnt understand what u meant. you can find values of x for which the function blows down to -ve infinity but not up.
?
9. Jan 16, 2008
### Tom Mattson
Staff Emeritus
The function certainly does blow up to positive infinity, as you approach -1/2 from the right and as you approach +1/2 from the left.
10. Jan 16, 2008
### sara_87
oh thank u very much
that helps.
just one last question:
same question as before but with function:
sqrt(x)/1000
is it not bounded since n continues to increase to infinity?
11. Jan 17, 2008
### Tom Mattson
Staff Emeritus
You're right that it's not bounded (on $[0,\infty)$ that is--we really should be specifying an interval when making these statements).
But what's "n"?
12. Jan 17, 2008
### HallsofIvy
Staff Emeritus
Now, I'm confused as to what function you are talking about. The original function was f(x)= x/(2^x) which is definitely bounded on $[0, \infty)$. It is bounded "above" but not bounded "below" so is not bounded. I don't see any asymptotes when I graph it.
13. Jan 17, 2008
### Tom Mattson
Staff Emeritus
Posts 1 through 3 pertain to f(x)=x/(2^x).
Posts 4 through 9 pertain to f(x)=(-2x^2)/(4x^2-1).
Posts 10 and 11 pertain to f(x)=sqrt(x)/1000.
14. Jan 17, 2008
### EnumaElish
Post 3 pertains to f(x)=x/(2^x) for x > 0. | {
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# Math Help - Integrate cos(x)cos(4x) dx
1. ## Integrate cos(x)cos(4x) dx
I'm trying to integrate:
$\int\cos(x)\cos(4x) dx$
Could some one tell me what I should be seeing about this to help me solve it. I thought I needed to change the $4x$ into $x$ and tried doing that using the double angle formula, but I have to use it twice and end up with a integrand that looks too complicated. I'd like to know what a good first step would be.
Thank you.
2. ## Re: Integrate cos(x)cos(4x) dx
this is easy...but you have to use a formula that transforms the product cosxcos4x into a sum.
HERE cos(4x)cos(x)=(1/2)[cos(5x)-cos(3x)}
Minoas
3. ## Re: Integrate cos(x)cos(4x) dx
Originally Posted by Furyan
I'm trying to integrate:
$\int\cos(x)\cos(4x) dx$
Could some one tell me what I should be seeing about this to help me solve it. I thought I needed to change the $4x$ into $x$ and tried doing that using the double angle formula, but I have to use it twice and end up with a integrand that looks too complicated. I'd like to know what a good first step would be.
Thank you.
\displaystyle \begin{align*} I &= \int{\cos{(x)}\cos{(4x)}\,dx} \\ I &= \frac{1}{4}\cos{(x)}\sin{(4x)} - \int{ -\frac{1}{4}\sin{(x)}\sin{(4x)}\,dx } \\ I &= \frac{1}{4}\cos{(x)}\sin{(4x)} + \frac{1}{4}\int{\sin{(x)}\sin{(4x)}\,dx} \\ I &= \frac{1}{4}\cos{(x)}\sin{(4x)} + \frac{1}{4} \left[ -\frac{1}{4}\sin{(x)}\cos{(4x)} - \int{ -\frac{1}{4}\cos{(x)}\cos{(4x)} \,dx} \right] \\ I &= \frac{1}{4}\cos{(x)}\sin{(4x)} - \frac{1}{16}\sin{(x)}\cos{(4x)} + \frac{1}{16}I \\ \frac{15}{16}I &= \frac{1}{4}\cos{(x)}\sin{(4x)} - \frac{1}{16}\sin{(x)}\cos{(4x)} \\ I &= \frac{4}{15}\cos{(x)}\sin{(4x)} - \frac{1}{15}\sin{(x)}\cos{(4x)} + C \end{align*}
4. ## Re: Integrate cos(x)cos(4x) dx
Thank you Prove It, | {
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4. ## Re: Integrate cos(x)cos(4x) dx
Thank you Prove It,
I haven't seen that before, integrating by parts twice and then letting the second integrand be I and solving that way. That's a very useful way of solving difficult integration problems. Thank you very much for showing me that. I tried this method with a similar problem that I had solved another way and found that I had to be very careful with signs and factors. I kept getting the wrong answer because I was getting a sign and factor wrong during the second integration. So I'm going to have to work on it, but it's great to know that if I can't see what else to do I can try this method.
Thanks again for all your efforts
5. ## Re: Integrate cos(x)cos(4x) dx
Originally Posted by Furyan
I'm trying to integrate:
$\int\cos(x)\cos(4x) dx$
Could some one tell me what I should be seeing about this to help me solve it. I thought I needed to change the $4x$ into $x$ and tried doing that using the double angle formula, but I have to use it twice and end up with a integrand that looks too complicated. I'd like to know what a good first step would be.
Thank you.
Hi Furyan!
As you can see here, we have the identity:
$\cos \theta \cos \varphi = {{\cos(\theta - \varphi) + \cos(\theta + \varphi)} \over 2}$
In other words:
$\int\cos(4x)\cos(x) dx = \int{{\cos(4x - x) + \cos(4x + x)} \over 2} = \int{{\cos(3x) + \cos(5x)} \over 2}$
This is what Minoas already suggested (except for the typo ).
6. ## Re: Integrate cos(x)cos(4x) dx
Hi ILikeSerena
Thank you very much for that. I did, in fact, also solve this problem using the factor formula, as Minoanman suggested, and it was very much simpler. However, although there are only four factor formulae in my book I'm worried that I might not remember them in an exam. Although I found Prove It's method more difficult I feel I'm more likely to remember it and at least get some method marks, even if I make some mistakes. | {
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I'm amazed how different the solutions look, depending on which method you use and yet they're equivalent. I solved one question using Prove It's method, the factor formula and a simple identity. All the solutions were equivalent, but they looked very different. I wouldn't have been able to rewrite any of them in terms of any of the others.
7. ## Re: Integrate cos(x)cos(4x) dx
Here's a trick that I like to use when I can't properly remember the formulas.
The only formula you need to remember is Euler's formula:
$e^{ix} = \cos x + i \sin x$
together with its rewritten forms (that you can both deduce from Euler's formula if you forget):
$\cos x = \dfrac 1 2 (e^{ix} + e^{-ix})$
$\sin x = \dfrac 1 {2i} (e^{ix} - e^{-ix})$
$\cos(4x)\cos(x) = \frac 1 2 (e^{i4x} + e^{-i4x}) \cdot \frac 1 2 (e^{ix} + e^{-ix})$
$= \frac 1 4 ((e^{i5x} + e^{-i5x}) + (e^{i3x} + e^{-i3x}))$
$= \frac 1 2 (\cos(5x) + \cos(3x)) \qquad \blacksquare$
8. ## Re: Integrate cos(x)cos(4x) dx
whoa! $\blacksquare$
That's a whole new level, but I'm liking it. I'm much better at deducing formualae than I am at remember them. Euler's formula, eh! I'm sure I came across that earlier today, but I glazed over when I saw $e^{ix}$. Is that i ,the i, $\sqrt{-1}?$. I'm going to look that up and try and find out why it's so fundamental. It looks very interesting.
Thank you very much for that
9. ## Re: Integrate cos(x)cos(4x) dx
Originally Posted by Furyan
whoa!
That's a whole new level, but I'm liking it. I'm much better at deducing formualae than I am at remember them. Euler's formula, eh! I'm sure I came across that earlier today, but I glazed over when I saw $e^{ix}$. Is that i ,the i, $\sqrt{-1}?$. I'm going to look that up and try and find out why it's so fundamental. It looks very interesting.
Thank you very much for that
Yep. It is that $i$. | {
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Thank you very much for that
Yep. It is that $i$.
According to wiki:
Euler's formula is ubiquitous in mathematics, physics, and engineering. The physicist Richard Feynman called the equation "our jewel" and "one of the most remarkable, almost astounding, formulas in all of mathematics."[2]
10. ## Re: Integrate cos(x)cos(4x) dx
Originally Posted by ILikeSerena
Yep. It is that $i$.
According to wiki:
Euler's formula is ubiquitous in mathematics, physics, and engineering. The physicist Richard Feynman called the equation "our jewel" and "one of the most remarkable, almost astounding, formulas in all of mathematics."[2]
Wow, it looks like a jewel. How exciting.
Thank you very much | {
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# Given a decreasing function s.t. $\int_0^\infty f(x)\,dx<\infty,$ prove $\sum_{n=1}^\infty f(na)$ converges
Let $f\in C([0,\infty))$ be a decreasing function such that $\int_0^\infty f(x)\,dx$ converges.
Prove $\sum_{n=1}^\infty f(na)$ converges, $\forall a>0$
My attempt:
By the Cauchy criterion, there exists $M>0,$ such that for $t-1>M:$
$$f(t)=\int_{t-1}^t f(t) \, dx \leq \int_{t-1}^t f(x)\,dx\xrightarrow{t \to \infty} 0$$
Hence, $f$ is non-negative.
$f$ is decreasing $\implies f(na)\leq f(a), \forall a>0, n\in \mathbb{N}.$
By integral monotonicity and non-negativity of $f$:
$$\int_1^\infty f(nx)\,dx \leq \int_1^\infty f(x)\,dx \leq \int_0^\infty f(x)\,dx$$
Hence $\int_1^\infty f(nx)\,dx$ converges and therefore $\sum_{n=1}^\infty f(na)$ converges.
Is that correct? If so, why is continuity necessary ? Is there a simpler way to prove it?
Any help appreciated.
• $$\int_{(n-1)a}^{na}f(x)dx\geq af(na)$$ – MAN-MADE Aug 22 '17 at 18:33
Hint: The following thing you wrote is the key:
$$f(t)=\int_{t-1}^{t}f(t)dx \leq \int_{t-1}^{t}f(x)dx$$
After that, just pick $t=an$ and sum over all $n$.
PS: and no, the assumption on continuity is not needed. Just integrability for the well-definedness.
As I commented before:
Since $f$ is decreasing and continuous, $$\int_{(n-1)a}^{na}f(x)dx\geq [na-(n-1)a]\inf_{(n-1)a< x\leq na}\{f(x)\}=af(na)$$
Then $$\int_{0}^{\infty}f(x)dx=\sum_{n=1}^{\infty}\int_{(n-1)a}^{na}f(x)dx\geq a\sum_{n=1}^{\infty}f(na)$$
Hence, since $a\in\mathbb{R}^+,$ $\sum_{n=1}^{\infty}f(na)<\infty$
\begin{align} & af(a) + af(2a) + af(3a) +\cdots \\[10pt] \le {} & \int_0^a f + \int_a^{2a} f + \int_{2a}^{3a} f + \cdots <\infty. \end{align}
This works if $f \ge0$ everywhere. If $f<0$ somewhere, then an easy argument shows the integral does not converge.
This is the integral test for convergence of series. Do a change of variable x --- to --- ax. The integral converges so does series g(n)=f(na). | {
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# Integral of periodic function over the length of the period is the same everywhere
I am stuck on a question that involves the intergral of a periodic function. The question is phrased as follows:
Definition. A function is periodic with period $a$ if $f(x)=f(x+a)$ for all $x$.
Question. If $f$ is continuous and periodic with period $a$, then show that $$\int_{0}^{a}f(t)dt=\int_{b}^{b+a}f(t)dt$$ for all $b\in \mathbb{R}$.
I understand the equality, but I am having trouble showing that it is true for all $b$. I've tried writing it in different forms such as $F(a)=F(b+a)-F(b)$. This led me to the following, though I am not sure how this shows the equality is true for all $b$,
$$\int_{0}^{a}f(t)dt-\int_{b}^{b+a}f(t)dt=0$$ $$=F(a)-F(0)-F(b+a)-F(b)$$ $$=(F(b+a)-F(a))-F(b)$$ $$=\int_{a}^{b+a}f(t)dt-\int_{0}^{b+a}f(t)dt=0$$
So, this leaves me with
$$\int_{a}^{b+a}f(t)dt-\int_{0}^{b+a}f(t)dt=\int_{0}^{a}f(t)dt-\int_{b}^{b+a}f(t)dt$$
I feel I am close, and I've made myself a diagram of a sine function to visualize what each of the above integrals might describe, but the power to explain the above equality evades me.
• See here for some proofs. – t.b. Jan 12 '12 at 8:09
• (Voted to close as duplicate) Even though this says continuous and the other says integrable, the proofs are the same, i.e. every proof here would apply over there. – 6005 Jul 17 '16 at 16:25
Let $H(x)=\int_x^{x+a}f(t)\,dt$. Then $$\frac{dH}{dx}=f(x+a)-f(x)=0.$$ It follows that $H(x)$ is constant. In particular, $H(b)=H(0)$.
We have $$\int_{0}^{a}f(t)\ dt+\int_{a}^{a+b}f(x)\ dx=\int_{0}^{b}f(y)\ dy+\int_{b}^{a+b}f(t)\ dt,$$ and setting $x=y-a$ turns the second integral into the third one.
No differentiation is needed: | {
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No differentiation is needed:
Pick the unique integer $n$ such that $b\leqslant na\lt b+a$, decompose the integral of $f(t)$ over $t$ from $b$ to $b+a$ into the sum of the integrals from $b$ to $na$ and from $na$ to $b+a$, apply the changes of variable $t=x+(n-1)a$ in the former and $t=x+na$ in the latter, then the periodicity of $f$ implies that $f(x)=f(t)$, hence the result is the sum of the integrals of $f(x)$ over $x$ from $b-(n-1)a$ to $a$ and from $0$ to $b-(n-1)a$...
...Et voilà !
You have made various false steps in your four line block and should have ended up with $$\int_{a}^{b+a}f(t)dt-\int_{0}^{b}f(t)dt=0$$ but this does not take you much further forward.
Instead note that somewhere in the interval $[b, b+a]$ is an integer multiple of $a$, say $na$. Then using $f(t)=f(t+a)=f(t+na)$: $$\int_{b}^{b+a}f(t)dt = \int_{b}^{na}f(t)dt+\int_{na}^{b+a}f(t)dt = \int_{b+a}^{(n+1)a}f(t)dt+\int_{an}^{b+a}f(t)dt = \int_{na}^{(n+1)a}f(t)dt = \int_{0}^{a}f(t)dt.$$
\begin{align} \int_{b}^{a+b}f(x)\ dx&= \int_{a}^{a+b}f(x)\ dx +\int_{b}^{a}f(x)\ dx\\&\overset{y=x-a}{=} \color{red}{\int_{0}^{a}f(y+a)\ dx} +\int_{b}^{a}f(x)\ dx\\&\overset{periodic}{=} \color{red}{\int_{0}^{b}f(y)\ dx} +\int_{b}^{a}f(x)\ dx\\&=\int_0^af(x)\ dx. \end{align}
• why a down vote here – Guy Fsone Jan 18 '18 at 8:15
• The integral is obtained in the second line itself. You have done a calculation mistake. – Robin Feb 17 '18 at 8:56
• Furthermore, this plagiarizes simultaneously three other answers, posted six years before. – Did Feb 18 '18 at 15:23 | {
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### Home > CCG > Chapter 9 > Lesson 9.1.3 > Problem9-35
9-35.
Are $ΔEHF$ and $ΔFGE$ congruent? If so, explain how you know. If not, explain why not.
What do the markings on the figures mean?
What other parts do these triangles have in common?
Yes, they are congruent. What is the reason? | {
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# Create a particular {x,y} list [duplicate]
How is it possible to create a list of {x,y} points as the following one:
list={{1,1},{1,2},...,{1,100},{2,1},{2,2},...,{2,100},...,{100,1},{100,2},...,{100,100}}
Thanks
• Table[{i,j}, {i, 100}, {j, 100}]? What have you tried? Feb 27 '17 at 16:23
• Hi, I was using Range, but it doesn't work for 2D lists. However, you answer gives my some additional brackets that are unwanted, like this: {{{1,1},{1,2},...,{1,100},{2,1},{2,2},...,{2,100},...,{100,1},{100,2},...,{100,100}}}. I've found the way with Flatten[]. Thanks for your help! Feb 27 '17 at 16:45
• Not an exact duplicate but a more general topic. Let me know if you disagree with closing.
– Kuba
Feb 27 '17 at 16:46
• @Michele You need to Flatten the result; consider Flatten[Table[{i, j}, {i, 1, 100}, {j, 100}], 1]. More in general, also take a look at the proposed duplicate. Feb 27 '17 at 16:49
• Join @@ Array[List , {100, 100} ] Feb 27 '17 at 20:24
Tuples[Range[100], 2]
or using Table as mentioned in the comment
It is worth noting that the method relying on Tuples is 10-12 times faster than Table | {
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# Notes on Dijkstra's Algorithm
## Setup
We are given a directed graph $G=(V,E)$ with a designated start node $s$. Each edge $(u, v) \in E$ has a weight $w(u,v) \geq 0$ indicating its cost. For each vertex $v \in V$, it has a predecessor $v.\pi$ that is another vertex or NIL.
## Goal
Find the shortest path from $s$ to every other vertex in the graph.
## Pseudocode
$Q$ is a min-priority queue of vertices, keyed by their $d$ values.
#### 1.DIJKSTRA$(G, s)$
$S = \varnothing$
$Q = G.V$
while $Q \neq \varnothing$
$u =$ EXTRACT-MIN(Q)
$S = S \cup \{u\}$
for each vertex $v \in G.Adj[u]$
RELAX(u,v,w)
#### 2.INITIALIZE-SINGLE-SOURCE$(G,s)$
for each vertex $v \in V[G]$
$v.d = \infty$
$v.\pi$ = NIL
$d[s] = 0$
#### 3. RELAX$(u,v,w)$
if $v.d > u.d + w(u,v)$
$v.d = u.d + w(u,v)$
$v.\pi = u$
## Proof of correctness
We want to show that the algorithm terminates with $u.d$ = $\delta(s,u)$ for all $u \in V$, where $\delta(s,u)$ is the shortest path weight from $s$ to $u$, and is defined as
$\delta(s, u) = \begin{cases} min\{w(p):s\xrightarrow{\text{p}}u\} , & \quad \text{if there is a path from s to u}\\ \infty , & \quad \text{otherwise} \end{cases}$
$w(p)$ is the sum of weights of the path $p$
## We first show
At the start of each iteration of the while loop in Dijkstra’s algorithm, $v.d = \delta(s, v)$ for each vertex $v \in S$ (*)
1) Base Case:
|S| = 0, i.e when $S = \varnothing$, the statement is vacuously true.
2) Inductive Step:
Suppose the claim holds when $|S| = k$ for some $k \geq 1$. Now, let $S$’s size grow to $k+1$ by adding vertex $v$. We want to show $v.d = \delta(s,v)$.
If $v=s$ then $s.d = \delta(s,s) =0$. If $v \neq s$, it must be $S \neq \varnothing$. If there’s no path form $s$ to $v$, $v.d = \delta(s,v)= \infty$. If there are some path from $s$ to $v$, there sholud be one shortest path $p$ from $s$ to $v$. | {
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Let us consider the first vertex $y$ on $p$ that is not in $S$, and let $x$ on $p$ be $y’s$ predecessor. By induction hypothesis, $x.d = \delta(s,x)$. Because edge $(x, y)$ is relaxed in the previous iteration, by convergence property (if $s\rightarrow u\rightarrow v$ is a shortest path in G for some $u, v \in V$, and if $u.d = \delta(s,u)$ at any time prior to relaxing edge $(u,v)$, then $v.d = \delta(s,v)$ at all times afterwards), we have $y.d = \delta(s,y)$. Since $y$ is before $v$ on $p$ and all edge weights are nonnegative, we have $\delta(s,y) \leq \delta(s,v)$, and thus $y.d = \delta(s,y) \leq \delta(s,v)\leq v.d$
Because both vertices $v$ and $y$ are in $V-S$ when $v$ is selected by the algorithm, $v.d \leq y.d$. Thus, $y.d = \delta(s,y) = \delta(s,v) = v.d$, and so $v.d = \delta(s,v)$, and so complete the proof of (*).
## Now
When the algorithm terminates, it means $Q = \varnothing$. Since the algorithm maintains the invariant that $Q = V - S$ (when initialization, $S = \varnothing$ and $Q$ contains all vertices of graph; for every time through while loop, a vertex is extracted from $Q$ and added to $S$, so the invariant is maintained), it means $S = V$, so $u.d = \delta(s,u)$ for all vertices $u \in V$
## Time Complexity
Using priority queue implemented in min heap, the time to initialize the queue with setting all vertices’ distances to $\infty$ is $O(V)$. Inside the while loop, we will do $V$ times EXTRACT_MIN operation, each of them will take $O(logV)$. Besides, we will do at most $E$ times of updating associated distance key for vertices in the queue, which will cost $O(ElogV)$. In total, the time will be $O((E+V)logV)$
## Java implementation | {
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## Java implementation
In the code below, vertices is the hashmap containing the mapping between GeographicPoint and MapNode. MapNode is made up of GeographicPoint and list of MapEdge. MapNodes in vertices are converted to DiNode in hashmap diPoints by using the method convertMap to facilitate implementation of the algorithm. The constructPath method will return the path found by the algorithm from start to goal.
See the full framework of this project here (incomplete). Note that the code may be ugly as I mainly pay attention to the algorithm itself.
public List<GeographicPoint> dijkstra(GeographicPoint start,
GeographicPoint goal, Consumer<GeographicPoint> nodeSearched){
if(!vertices.containsKey(start) || !vertices.containsKey(goal)){
System.out.println("input location does not exist in graph");
}
HashMap<GeographicPoint, DiNode> diPoints = convertMap(vertices);
HashMap<GeographicPoint, GeographicPoint> parentMap = new HashMap<>();
Set<GeographicPoint> visitSet = new HashSet<>();
PriorityQueue<DiNode> queue = new PriorityQueue<DiNode>();
DiNode cur = diPoints.get(start);
cur.sumDist = 0.0;
queue.offer(cur);
while(!queue.isEmpty()){
cur = queue.poll();
nodeSearched.accept(cur.mapNode.location);
if(!visitSet.contains(cur.mapNode.location)){
}
if(cur.mapNode.location.equals(goal)) {
return constructPath(start, goal, parentMap);
}
for (MapEdge e : cur.mapNode.edges) {
if(!visitSet.contains(e.end)){
double nbSumDist = diPoints.get(e.end).sumDist;
if(cur.sumDist + e.distance < nbSumDist){
// update cur as neightbor's parent in parent map
parentMap.put(e.end, cur.mapNode.location);
DiNode next = diPoints.get(e.end);
// update next neighbor's sum of distance
next.sumDist = cur.sumDist + e.distance;
// put neighbor with updated distance into priority queue
queue.offer(new DiNode(next.mapNode, next.sumDist));
}
}
}
}
System.out.println("There is no such path");
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}
System.out.println("There is no such path");
}
Above is one part of the back-end code of a project in UCSD’s Advanced Data Structure course. It utilizes the Google Map’s API, and can demonstrate route planning in a real world’s map as shown below. Note that the algorithm is a bit different from the one described above (instead of initializing the priority queue with inserting all vertices and later updating their sum of weighted distance, it enqueues the vertex after its distance is updated), but the general idea is the same.
## References
[1] T. Cormen, C. Stein, R. Rivest, and C. Leiserson. Introduction to Algorithms (3rd ed.). MIT press,2009.
[2] J. Kleinberg and E. Tardos. Algorithm Design. Addison-Wesley, 2005 | {
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# Which is the “fastest” way to compute $\sum \limits_{i=1}^{10} \frac{10i-5}{2^{i+2}}$?
I am looking for the "fastest" paper-pencil approach to compute $$\sum \limits_{i=1}^{10} \frac{10i-5}{2^{i+2}}$$
This is a quantitative aptitude problem and the correct/required answer is $3.75$
In addition, I am also interested to know how to derive a closed form for an arbitrary $n$ using mathematica I got $$\sum \limits_{i=1}^{n} \frac{10i-5}{2^{i+2}} = \frac{5 \times \left(3 \times 2^n-2 n-3\right)}{2^{n+2}}$$
Thanks,
-
Does someone feel like writing an "abstract duplicate" for this kind of question? I think I've seen several of them, but I don't know whether we have one that deals with this more generally and would serve as a good duplicate for all of them. – joriki Nov 12 '11 at 15:55
The actual correct answer to this question is $\approx 3.7219$. I do not know what "quantitative aptitude test" means, but would not 3.7 be a better answer than 3.75? – Aleksey Pichugin Nov 12 '11 at 16:36
We'll split the sum you're looking for up into
$$10 \sum_{i=1}^{10} {i \over 2^{i+2}} - 5 \sum_{i=1}^{10} {1 \over 2^{i+2}}.$$
Call this $10S_1 - 5S_2$.
We can write $$S_1 = \sum_{i=1}^{10} {i \over 2^{i+2}} = {1 \over 4} \sum_{i=1}^{10} {i \over 2^i}$$ and the result $\sum_{i=1}^\infty i/2^i = 2$ is well-known; thus $S_1 \approx 1/2$.
Similarly, $S_2 \approx \sum_{i=1}^\infty 1/2^{i+2} = 1/4$ by the usual sum of a geometric series.
So your sum is approximately $15/4$. In fact the infinite sum
$$\sum_{i=1}^\infty {10i-5 \over 2^{i+2}}$$
is exactly 15/4. We've left off some small positive terms so your sum is a bit less than $15/4$.
An exact form for the sum
As for getting an exact form for the sum: call it $f(n)$. Then we have
$$f(n) = {15 \over 4} - \left( 10 \sum_{i=n+1}^\infty {i \over 2^{i+2}} - 5 \sum_{i=n+1}^\infty {1 \over 2^{i+2}} \right).$$
Write this as $f(n) = 15/4 - g(n) + h(n)$. | {
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Write this as $f(n) = 15/4 - g(n) + h(n)$.
$h(n)$ is easy -- it's the sum of a geometric series with its first term $5/2^{n+3}$ and ratio $1/2$, so $h(n) = 5/2^{n+2}$.
$g(n)$ is a bit harder. So that we don't have so many constants floating around, consider
$$G(n) = \sum_{i={n+1}}^\infty {i \over 2^i}$$
and you can see $g(n) = (5/2) G(n)$. Now you can write
$$G(n) = {(n+1) \over 2^{n+1}} + {{n+2} \over 2^{n+2}} + {{n+3} \over 2^{n+3}} + \cdots$$
and this is just
$$G(n) = \left( {n \over 2^{n+1}} + {n \over 2^{n+2}} + {n \over 2^{n+3}} + \cdots \right) + {1 \over 2^n} \left( {1 \over 2^1} + {2 \over 2^2} + {3 \over 2^3} + \cdots \right).$$
The first sum is a geometric series, summing to $n/2^n$; the second sum is $2$. Thus $G(n) = (n+1)/2^n$. Therefore you get
$$f(n) = {15 \over 4} - {5 \over 2} {(n+2) \over 2^n} + {5 \over 2^{n+2}}.$$
A sum that everybody should know, but lots of people don't
Finally, I used the result $\sum_{i=1}^\infty i/2^i$ twice here, both in getting the approximation and in getting the exact form. How can we prove that? One way is to write $${1 \over 2} + {2 \over 4} + {3 \over 8} + {4 \over 16} + \cdots$$ as a sum of one $1/2$, two $1/4$s, three $1/8$s, and so on; then regroup those terms as $$\left( {1 \over 2} + {1 \over 4} + {1 \over 8} + \cdots \right) + \left( {1 \over 4} + {1 \over 8} + {1 \over 16} + \cdots \right) + \left( {1 \over 8} + {1 \over 16} + {1 \over 32} + \cdots \right) + \cdots$$ Each pair of parentheses contains a geometric series; summing those gives $1 + 1/2 + 1/4 + 1/8 + \cdots$, another geometric series, which has sum $2$.
Alternatively, note that $${1 \over 1-z} = 1 + z + z^2 + z^3 + \cdots$$ and differentiating both sides gives $${1 \over (1-z)^2} = 1 + 2z + 3z^2 + \cdots$$. Multiply both sides by $z$ to get $${z \over (1-z)^2} = z + 2z^2 + 3z^3 + \cdots$$ and plug in $z = 1/2$ to get $${1/2 \over (1-1/2)^2} = {1 \over 2} + {2 \over 2^2} + {3 \over 2^3} + \cdots.$$
- | {
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-
I would do it like this. Using $x \frac{\mathrm{d}}{\mathrm{d} x}\left( x^k \right) = k x^k$, and $\sum_{k=1}^n x^k = x \frac{x^n-1}{x-1}$. Then $$\begin{eqnarray} \sum_{k=1}^n \left( a k +b\right) x^k &=& \left( a x \frac{\mathrm{d}}{\mathrm{d} x} + b\right) \circ \sum_{k=1}^n x^k = \left( a x \frac{\mathrm{d}}{\mathrm{d} x} + b\right) \circ \left( x \frac{x^n-1}{x-1} \right) \\ &=& x \left( a x \frac{\mathrm{d}}{\mathrm{d} x} + a + b\right) \circ \left(\frac{x^n-1}{x-1} \right) \\ &=& x \left( (a+b) \frac{x^n-1}{x-1} + a x \frac{ n x^{n-1}(x-1) - (x^n-1) }{(x-1)^2} \right) \\ &=& x \left( (a+b) \frac{x^n-1}{x-1} + a x \frac{ (n-1) x^{n} - n x^{n-1} + 1 }{(x-1)^2} \right) \end{eqnarray}$$
Now applying this: $$\begin{eqnarray} \sum_{k=1}^n \frac{10 k -5}{2^{k+2}} &=& \frac{5}{4} \sum_{k=1}^n (2k-1)\left(\frac{1}{2}\right)^k \\ &=& \frac{5}{4} \frac{1}{2} \left( -(2^{1-n} - 2) + (n-1) 2^{2-n}- n 2^{1-n} + 4 \right) \\ &=& \frac{5}{4} \left( 3 + (n-3) 2^{-n} \right) \end{eqnarray}$$
-
Thanks for doing this in general form. If you don't mind I'll use this question for duplicates in the future and point to this answer :-) – joriki Nov 13 '11 at 8:04
@joriki:Question is tagged as algebra-precalculus. – Quixotic Nov 18 '11 at 7:31
@MaX: I see. Thanks for pointing that out. Fortunately there are so many answers here, including N.S.' that shows an elementary way to sum $i/q^i$, that it will still be a good duplicate in many future circumstances. I wonder which answer you'll be accepting... – joriki Nov 18 '11 at 7:39
@joriki:You are welcome, I am accepting Michael Lugo's answer as it provides the fastest method to reach the solution. – Quixotic Nov 19 '11 at 14:17
You presented this beautifully. Thank you. – 000 Jul 14 '12 at 9:48
First note that $\displaystyle\frac{10i-5}{2^{i+2}} = \frac{5(2i-1)}{2^{i+2}}$
Secondly, using the sum of a geometric series you can show that $2^0 + 2^1 + 2^2 + ... + 2^k = 2^{k+1}-1$. | {
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\displaystyle \begin{align*} \text{So } \sum \limits_{i=1}^{n} \frac{10i-5}{2^{i+2}} &= 5(\frac{2\times1-1}{2^{1+2}}+\frac{2\times2-1}{2^{2+2}}+\frac{2\times3-1}{2^{3+2}}+...+\frac{2\times n-1}{2^{n+2}})\\ &= 5(\frac{2^{n-1}(2\times1-1)+2^{n-2}(2\times2-1)+2^{n-3}(2\times3-1)+...+2^{0}(2\times n-1)}{2^{n+2}}) \\ &= 5(\frac{2(2^{n-1}\times1+2^{n-2}\times2+2^{n-3}\times3+...+2^{0}\times n)-(2^{n-1}+2^{n-2}+2^{n-3}+...+1)}{2^{n+2}})\\ &= 5(\frac{2(2^{n-1}\times1+2^{n-2}\times2+2^{n-3}\times3+...+2^{0}\times n)-(2^n-1)}{2^{n+2}})\\ &= 5(\frac{2((2^{n-1}+2^{n-2}+2^{n-3}+...+1)+(2^{n-2}+2^{n-3}+2^{n-4}+...+1)+...+(2+1)+(1))-(2^n-1)}{2^{n+2}})\\ &= 5(\frac{2((2^{n}-1)+(2^{n-1}-1)+(2^{n-2}-1)+...+(2^{2}-1)+(2-1))-(2^n-1)}{2^{n+2}})\\ &= 5(\frac{2((2^{n}+2^{n-1}+2^{n-2}+...+2) + (-1)\times n)-(2^n-1)}{2^{n+2}})\\ &= 5(\frac{2(2\times (2^n - 1))- n)-(2^n-1)}{2^{n+2}})\\ &= 5(\frac{4 \times 2^n - 4 - 2n -2^n + 1}{2^{n+2}})\\ &= 5(\frac{3 \times 2^n - 2n - 3}{2^{n+2}})\text{ which is the closed form you got from Mathematica}\end{align*} | {
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-
The equations are a lot more readable in display style, which you get either by switching it on individually with \displaystyle or by making the entire formula a displayed one by enclosing it in double dollar signs. Also note that you can use the eqnarray and align environments to align consecutive equations with each other. – joriki Nov 13 '11 at 8:03
@joriki Thanks for the tips and I think my answer has been formatted correctly but I believe my 5th line is a tad too long? I had to trim is down a bit and it is still quite long... – Sp3000 Nov 13 '11 at 9:50
Yes, that's a problem actually, which I didn't foresee when I wrote my comment :-) You can alleviate it somewhat by not putting the "So" into the equation but on a line of its own. Also, though it looks nicer with a full equation in the first line, in this case you could break the line before the first equals sign; it will still look better with eqnarray since you can align the expression in the first line with the other expressions instead of with the equals signs. – joriki Nov 13 '11 at 10:03
The sum $S= \sum \limits_{i=1}^n \frac{1}{2^i}=1-\frac{1}{2^n}$ is geometric, thus easy to calculate.
Here is a simple elementary way of calculating
$$T=\sum_{i=1}^n \frac{i}{2^i} \,.$$:
$$T=\sum_{i=1}^n \frac{i}{2^i} =\frac{1}{2}+ \sum_{i=2}^n \frac{i}{2^i} =\frac{1}{2}-\frac{n+1}{2^{n+1}}+ \sum_{i=2}^{n+1} \frac{i}{2^i} \,.$$
Changing the index in the last sum yields:
$$T= \frac{2^n-n-1}{2^{n+1}}+\sum_{i=1}^{n} \frac{i+1}{2^{i+1}}=\frac{2^n-n-1}{2^{n+1}}+\sum_{i=1}^{n} \frac{i}{2^{i+1}}+\sum_{i=1}^{n} \frac{1}{2^{i+1}} \,.$$
Thus
$$T= \frac{2^n-n-1}{2^{n+1}}+\frac{1}{2}T+ \frac{1}{2}-\frac{1}{2^{n+1}}$$
Thus
$$\frac{1}{2}T=\frac{2^{n+1}-n-2}{2^{n+1}}\,.$$
Hence
$$\sum_{i=1}^n \frac{i}{2^i} =\frac{2^{n+1}-n-2}{2^{n}} \,.$$
-
$\sum \limits_{i=1}^{10} \frac{10i-5}{2^{i+2}} =\frac{10}{4} \sum_{i=1}^{10} \frac{i}{2^i}-\frac{5}{4}\sum_{i=1}^{10} \frac{1}{2^i}$
Second $\sum$ is obviously $1-\frac{1}{1024}$. | {
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Second $\sum$ is obviously $1-\frac{1}{1024}$.
First $\sum$:
$\frac{1}{2} +$
$\frac{1}{4} + \frac{1}{4} +$
$\frac{1}{8} + \frac{1}{8} + \frac{1}{8} +$
$\vdots$
$\frac{1}{1024} + \frac{1}{1024} + \dots + \frac{1}{1024}$
Summing by columns, $(1-\frac{1}{1024})+(\frac{1}{2}-\frac{1}{1024})+\dots+(\frac{1}{512}-\frac{1}{1024})=(1+\frac{1}{2}+\dots+\frac{1}{512})-\frac{10}{1024}=\frac{1023}{512}-\frac{10}{1024}$
Rest is boring and easy.
- | {
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# Ring of polynomials as a module over symmetric polynomials
Consider the ring of polynomials $\mathbb{k} [x_1, x_2, \ldots , x_n]$ as a module over the ring of symmetric polynomials $\Lambda_{\mathbb{k}}$. Is $\mathbb{k} [x_1, x_2, \ldots , x_n]$ a free $\Lambda_{\mathbb{k}}$-module? Can you write down "good" generators explicitly? (I think that it has to be something very classical in representation theory).
Comment: My initial question was whether this module flat. But since all flat Noetherian modules over polynomial ring are free (correct me if it is wrong), it is the same question.
There is much more general question, which seems unlikely to have good answer. Let $G$ be a finite group and $V$ finite dimensional representation of G. Consider projection $p: V \rightarrow V/G$. When is $p$ flat?
• – anon Nov 3 '14 at 20:52
• Sorry. I really tried to find something before asking and did not manage. – quinque Nov 3 '14 at 21:56
Let $s_i= \sum x_1 \ldots x_i$ be the fundamental symmetric polynomials. We have a sequence of free extensions $$k[s_1, \ldots, s_n] \subset k[s_1, \ldots, s_n][x_1]\subset k[s_1, \ldots, s_n][x_1][x_2] \subset \cdots \\ \subset k[s_1,\ldots ,s_n] [x_1] \ldots [x_n] = k[x_1, \ldots ,x_n]$$ of degrees $n$, $n-1$, $\ldots$ ,$1$. At step $i$ the generators are $1, x_i, \ldots, x_i^{n-i}$. Therefore $$k[s_1, \ldots, s_n] \subset k[x_1, \ldots, x_n]$$ is free of degree $n!$ with generators $x_1^{a_1} x_2^{a_2} \cdots x_n^{a_n}$ with $0 \le a_i \le n-i$.
More generally for a finite reflection group of transformations $G$ acting on a vector space $V$ over a field $k$ of characteristic $0$ (to be safe) the algebra of invariants $k[V]^G$ is a polynomial algebra and $k[V]$ is a free $k[V]^G$ module of rank $|G|$ --see the answer of @stephen: . | {
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• I think some more details are in order: why $1,x_1,\dots,x_1^{n-1}$ is a basis of the first ring extension? – user26857 Nov 3 '14 at 19:41
• And why don't satisfy any equation of degree $<n$? – user26857 Nov 3 '14 at 19:50
• @user26857: Another useful observation using just basic Galois theory is : $k(s_1, \ldots, s_n) \subset k(x_1, \ldots, x_n)$ is Galois with group $S_n$. – orangeskid Nov 3 '14 at 19:57
• Great answer! Just a remark: The basis orangeskid describes is different from the Schubert basis - geometrically, the images of the $x_i$ in the integral cohomology of the flag variety are the Chern classes of the $n$ tautological line bundles (and the iterated computation above has a geometric counteprart, by viewing the flat variety as an iterated projective bundle and using Leray-Hirsch at each step), while as I said above the Schubert polynomials correspond to Bruhat cells. – Hanno Nov 3 '14 at 20:02
• Regarding your last paragraph, it can certainly happen that in characteristic $p$ the ring of invariants is not polynomial even when the group is generated by reflection; on the other hand polynomials invariants is enough to imply that the group is generated by reflections in any characteristic. – Stephen Nov 3 '14 at 20:34
Yes, that's indeed a classical result of representation theory: ${\mathbb Z}[x_1,...,x_n]$ is graded free over ${\Lambda}_n$ of rank $n!$ (the graded rank is the quantum factorial $[n]_q!$), and a basis is given by Schubert polynomials defined in terms of divided difference operators.
See for example the original article of Demazure, in particular Theorem 6.2. | {
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See for example the original article of Demazure, in particular Theorem 6.2.
Passing to the quotient, one obtains the graded ring ${\mathbb Z}[x_1,...,x_n]/\langle\Lambda_n^+\rangle$ which is isomorphic to the integral cohomology ring of the flag variety of ${\mathbb C}^n$, and the ${\mathbb Z}$-basis of Schubert polynomials coincides with the basis of fundamental classes of Bruhat cells in the flag variety. This is explained in Fulton's book 'Young Tableaux', Section 10.4.
The answer by @orangeskid (+1) is the most classical and direct, and the answer by @Hanno connects your question to Schubert calculus and the geometry of flag varieties (+1). Hoping this isn't too self-promoting, you might also have a look at my paper Jack polynomials and the coinvariant ring of $G(r,p,n)$ (I worked in somewhat more generality)
http://tinyurl.com/l8u9wh5
where I showed that certain non-symmetric Jack polynomials give a basis as well (I was working with more general reflection groups and over the complex numbers, but a version should work over any field of characteristic $0$; I have not thought much about this in characteristic $p$). The point of my paper was really to connect the descent bases (yet another basis!) studied much earlier by Adriano Garsia and Dennis Stanton in the paper Group actions of Stanley-Reisner rings and invariants of permutation groups
http://tinyurl.com/lhhtney
to the representation theoretic structure of the coinvariant algebra as an irreducible module for the rational Cherednik algebra. Of course this structure becomes much more complicated in characteristic smaller than $n$; in particular the coinvariant algebra will in general not be irreducible as a module for the Cherednik algebra. | {
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Towards your second question: by definition $p$ is flat if and only if $k[V]$ is a flat $k[V]^G$-module. This certainly holds if $k[V]$ is free over $k[V]^G$; a sufficient condition for this is given in Bourbaki, Theorem 1 of section 2 of Chapter 5 of Lie Groups and Lie algebras (page 110): in case the characteristic of $k$ does not divide the order of $G$, it suffices that $G$ be generated by reflections. This is false (in general) in characteristic dividing the order of the group. But see the paper Extending the coinvariant theorems of Chevalley, Shephard-Todd, Mitchell, and Springer by Broer, Reiner, Smith, and Webb available for instance on Peter Webb's homepage here
http://www.math.umn.edu/~webb/Publications/
for references and what can be said in this generality (this is an active area of research so you shouldn't expect to find a clean answer to your question).
Conversely, assuming $p$ is flat and examining the proof of the above theorem in Bourbaki, it follows that $k[V]$ is a free $k[V]^G$-module. Now Remark 2 to Theorem 4 (page 120) shows that $G$ is generated by reflections. So to sum up: if $p$ is flat then $G$ is generated by reflections; in characteristic not dividing the order of $G$ the converse holds. | {
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• MO level, neat. +1! – orangeskid Nov 3 '14 at 20:00
• I have a problem about your links. Article by Webb and others is just about connected topics. They do not answer my question (or I just do not see it, then please explain). About Bourbaki book I have a deeper trouble. It sounds silly but I even can not find what is called theorem 1 in this section. Could you post here what it is about (proof is not necessary). – quinque Nov 3 '14 at 23:32
• @user167762, The statement is Theorem 1 on page 110 of my English edition---are you able to locate it now? I referenced the paper of of BRSW not because it answers your question, but because it is the most recent thing I can think of that deals with the subject in detail and I thought the references it contained might prove useful for you or other future readers. I will edit my answer a bit to give an indication of what I believe to be the state of the art. As far as I know, in positive characteristic there is no clean characterization of groups such that the projection is flat. – Stephen Nov 4 '14 at 12:26
Another perspective:
The ring extension $S=k[s_1,\dots,s_n]\subset k[x_1,\dots,x_n]=R$ is finite, and since $R$ is Cohen-Macaulay then it is free of rank say $m$. We thus have a Hironaka decomposition $R=\oplus_{i=1}^m S\eta_i$ for some homogeneous $\eta_i\in R$. Using this the Hilbert series of $R$ is $$H_R(t)=\sum_{i=1}^mH_S(t)t^{\deg\eta_i}=\sum_{i=1}^mt^{\deg\eta_i}/\prod_{i=1}^n(1-t^i).$$ On the other side, $H_R(t)=1/(1-t)^n$ and therefore $\sum_{i=1}^mt^{\deg\eta_i}=\prod_{i=1}^{n-1}(1+t+\cdots+t^i)$ which for $t=1$ gives $m=n!$.
• You used a theorem about Cohen-Macaulay. I do not know it. Could you please write it? – quinque Nov 3 '14 at 22:26 | {
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# Equation of one branch of a hyperbola in general position
Given a generic expression of a conic:
$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F=0$$
is there a way to write an expression for one of the branches as a function of the coefficients? I tried using the quadratic formula to get an expression for $y$: $$y=\frac{-(Bx+E)\pm \sqrt{(Bx+E)^2 - 4(C)(Ax^2 + Dx + F)}}{2C}$$
but this doesn't always work. Consider:
$$xy=1$$
Here, $A=0, B=1, C=0, D=0, E=0, F=0$, so $y=\frac{\cdot}{0}$, which isn't particularly helpful. In other cases it is not as bad, but still not what I'm looking for. E.g.
$$x^2 - y^2 - 1=0$$
Using the formula above, we get $$y=\pm \sqrt{x^2-1}$$ which seems nice, but $y=\sqrt{x^2-1}$ it is actually one half of each branch rather than one entire branch, as can be seen here.
http://www.wolframalpha.com/input/?i=plot%28x^2+-+y^2+-+1%3D0%29
http://www.wolframalpha.com/input/?i=plot%28y%3Dsqrt%28x^2-1%29%29
I am trying to draw one of these branches, so I need an ordered set of points along a predefined "grid" of either of the variables. Is it possible to do this?
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Your equation doesn't work for all hyperbolas since you found it as a solution of quadratic equation. For $xy = 1$ it won't work since it's not quadratic equation. You can't solve $0x^2+2x+1=0$ as quadratic equation, since it's not. – Kaster Mar 7 '13 at 23:56
First of all, if you want to draw a portion of the hyperbola, you typically just need parametric equations of the form $x = x(t)$, $y = y(t)$; you don't necessarily need to get $y$ as a function of $x$.
But, ignoring that quibble, I'll try to answer the question you asked.
The key is to first eliminate the $xy$ term in the implicit equation. In effect, you do this by rotating the coordinate system. This is really the same idea as Will Jagy used in his answer, but it might be easier to understand if I don't mention eigenvectors. | {
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Suppose we introduce a $uv$ cooordinate system that is rotated by an angle $\theta$ (counterclockwise) from the $xy$ one. Then we have
$$x = u\cos\theta - v\sin\theta \quad ; \quad y = u\sin\theta + v\cos\theta$$
You can plug these $x$ and $y$ expressions into your original equation, and you'll get
$$\bar A u^2 + \bar B uv + \bar C v^2 + \bar Du + \bar Ev + \bar F = 0$$
where
$$\bar A = A \cos^2\theta + 2B \sin \theta\cos\theta + C \sin^2\theta$$
$$\bar B = 2 B \cos 2\theta - (A - C) \sin 2\theta$$
$$\bar C = A \sin^2\theta - 2B \sin\theta\cos\theta + C \cos^2\theta$$
and so on. Now we just have to cleverly choose $\theta$ so that the $uv$ term disappears. Clearly, we'll get what we want if we choose $\theta$ so that $\bar B = 0$, and this means
$$\tan2\theta = \frac{2B}{A - C}$$
After using this technique, we can now assume that the equation has the form
$$a u^2 + c v^2 - 2adu - 2cev + f = 0$$
Note that I'm using $-2ad$ in place of $\bar D$ and $-2ce$ in place of $\bar E$, just to make the next step more convenient. And the next step is just some "complete the square" tricks. The equation can be written:
$$a (u^2 - 2du) + c (v^2 - 2ev) + f = 0$$
In other words:
$$a (u^2 - 2du + d^2) + c (v^2 - 2ev + e^2) + f - ad^2 - ce^2 = 0$$
which is
$$a (u - d)^2 + c (v - e)^2 = ad^2 + ce^2 - f$$
Now it should be clear that we have either an ellipse (if the signs of $a$ and $c$ are the same) or a hyperbola (if the signs are different). I'm ignoring the parabola case and several degenerate cases. The ellipse/hyperbola has its center at the point $(d,e)$ in the $uv$ coordinate system, and its axes of symmetry are parallel to the $u$ and $v$ axes.
At this point, if our curve turns out to be a hyperbola, you can use standard parametric equations (like the ones Wil Jagy gave) to trace out one branch or the other.
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I think you are going to have a more satisfactory experience doing this: find the center $\vec{P} = (x_0, y_0)$ of your hyperbola. Find the eigenvectors of the matrix $$\left( \begin{array}{cc} A & B/2 \\ B/2 & C \end{array} \right)$$ and normalize and choose order and $\pm$ so that with basis $\vec{u},\vec{v}$ you can then write your branch as $$g(t) = \vec{P} + \vec{u} \cosh t + \vec{v} \sinh t.$$ Note that while $\vec{u},\vec{v}$ are perpendicular to each other they are not particularly of any length, in the way I write it above. If you prefer an orthonormal basis you then just put one constant scalar multiplication in front of the $\cosh$ term and one in front of the $\sinh$ term. It will all work out if you actually do have a hyperbola, which happens when $B^2 > 4 A C.$
I did teach this about 20 years ago, a whole section on translations and rotations of conic sections for a engineering calculus course. The memories are dim, but it does appear that it is best to find an orthonormal basis for the above matrix first, then find the center expressed in that basis, and end up with constant scalar coefficients of the hyperbolic trig terms.
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I posted an answer following your suggestion with an example. The only part I didn't follow was "and normalize and choose order and $\pm$ so that with the new basis you can then write your branch as...". What do you mean "choose order" and "choose $\pm$"? Also, how do you determine the values of the parameter that constitute each branch? – David Doria Mar 8 '13 at 13:31
Following Will Jagy's suggestion, here are some examples:
Example #1
Consider
$$x^2 - y^2 -1 = 0$$
($A=1$, $B=0$, $C=-1$, $D=0$, $E=0$, $F=-1$). From
$$p_c = \begin{pmatrix}x_c\\y_c\end{pmatrix} = \begin{pmatrix}\frac{BE-2CD}{4AC-B^2}\\ \frac{DB-2AE}{4AC-B^2}\end{pmatrix}$$
we have
$$p_c = \begin{pmatrix}0\\0\end{pmatrix}$$ | {
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we have
$$p_c = \begin{pmatrix}0\\0\end{pmatrix}$$
Now, the eigenvalues/vectors of the matrix $$\begin{pmatrix}A & \frac{B}{2}\\\frac{B}{2} & C\end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix}$$ are $\lambda_1 = -1$, $\lambda_2 = 1$, $v_1 = \begin{pmatrix}0\\1\end{pmatrix}$, $v_2 = \begin{pmatrix}1\\0\end{pmatrix}$
From the parametric form
$$g(t)=p_c + v_1 cosh(t) + v_2 sinh(t)$$
we have
$$g(t) = \begin{pmatrix}0\\1\end{pmatrix} cosh(t) + \begin{pmatrix}1\\0\end{pmatrix} sinh(t)$$
which looks like
Now how would we draw the other branch? That is, how do you determine the values of the parameter that constitute each branch?
Example #2
Consider
$$7x^2 - 3y^2 - 25 = 0$$
($A=7$, $B=0$, $C=-3$, $D=0$, $E=0$, $F=-25$). From
$$p_c = \begin{pmatrix}x_c\\y_c\end{pmatrix} = \begin{pmatrix}\frac{BE-2CD}{4AC-B^2}\\ \frac{DB-2AE}{4AC-B^2}\end{pmatrix}$$
we have
$$p_c = \begin{pmatrix}0\\\frac{-25}{6}\end{pmatrix}$$
Now, the eigenvalues/vectors of the matrix $$\begin{pmatrix}A & \frac{B}{2}\\\frac{B}{2} & C\end{pmatrix} = \begin{pmatrix}7 & 0 \\ 0 & -3 \end{pmatrix}$$ are $\lambda_1 = -3$, $\lambda_2 = 7$, $v_1 = \begin{pmatrix}0\\1\end{pmatrix}$, $v_2 = \begin{pmatrix}1\\0\end{pmatrix}$
From the parametric form
$$g(t)=p_c + v_1 cosh(t) + v_2 sinh(t)$$
we have
$$g(t) = \begin{pmatrix}0\\ \frac{-25}{6}\end{pmatrix} + \begin{pmatrix}0\\1\end{pmatrix} cosh(t) + \begin{pmatrix}1\\0\end{pmatrix} sinh(t)$$ | {
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-
David, the other branch is $p_c -v_1 \cosh t + v_2 \sinh t$ – Will Jagy Mar 8 '13 at 20:51
David, please do another example, something along the lines of $x^2 - x y - y^2 =1,$ where the eigenvectors will not naturally come out length one, and if you choose an orthonormal basis you will need scalar multiples, $p_c + \alpha v_1 \cosh t + \beta v_2 \sinh t.$ Note that the bit about the scalar multipliers would apply even for, say, $7 x^2 - 3 y^2 = 25,$ where one branch is $$x = \frac{5}{\sqrt 7} \cosh t, y = \frac{5}{\sqrt 3} \sinh t.$$ – Will Jagy Mar 8 '13 at 21:04
@WillJagy How do you choose the eigenvector that goes with cosh vs sinh? Even in my Example #1, the hyperbola opens up/down (instead of left/right) if I use x=sinh(t) and y=cosh(t) as the eigenvectors they way I have written them imply (I guess I thought I had made a typo so I switched them (x=cosh(t) and y=sinh(t) when I produced that graph): wolframalpha.com/input/… – David Doria Mar 11 '13 at 12:04
@WillJagy I worked out the $7x^2 - 3y^2 - 25 = 0$ example in my answer. How did you get from where I am to the $\frac{5}{\sqrt{7}}$ and $\frac{5}{\sqrt{3}}$ coefficients? And did you just accidentally omit the $center + ...$ term? My comment above holds on this problem as well - how do you choose the eigenvector to associate with the cosh vs sinh term? – David Doria Mar 11 '13 at 12:11
The center is actually at $(0,0)$ because $D=0,E=0.$ If you choose unit length eigenvectors, then you expect scalar coefficients to be necessary. In particular, try your recipe at $t=0$ (after replacing the -25/6 by 0) and you will find the point does not lie on the actual hyperbola. How about if you try doing $xy = 1?$ You already know exactly how the graph looks. One test for correctness is, once you write $x = x_c + x_1 \cosh t + x_2 \sinh t,$ then $y = y_c + y_1 \cosh t + y_2 \sinh t,$ you must be able to plug those into the original equation and get truth. – Will Jagy Mar 11 '13 at 20:38 | {
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# Proof $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$
Apparently $$1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2$$.
How? What's the proof? Or maybe it is self apparent just looking at the above?
PS: This problem is known as "The sum of the first $$n$$ positive integers".
• This an important example of a finite integral, a good tutorial here: stanford.edu/~dgleich/publications/finite-calculus.pdf – anon Aug 12 '10 at 19:46
• Note that we do not define this sum to be (x+1)x/2, we prove it to be so. It is a very useful formula. You could check out en.wikipedia.org/wiki/Triangle_number, which unfortunately, does not prove the formula you cite. – Ross Millikan Apr 21 '11 at 22:14
• Although we already have three proofs of this here, I think it would be really interesting if we amassed more (perhaps excluding the typical intro-to-induction styled proof). – davidlowryduda Apr 22 '11 at 1:15
• It could be proved by mathematical induction – J. W. Tanner Sep 5 '19 at 1:10
• Not appropriate for an answer, but you've asked either a very easy or a very difficult question. If by "why" you mean, "Can I see a proof of this fact?" the question is fairly easy to answer. If by "why" you mean, "Why should this be true?" you've asked a very deep kind of question that mathematicians make entire careers out of. Sure it can be shown to be true, but how does it connect to other true results? Why is it a polynomial of degree $2$? What about the sum of $k^2$? And on and on... – Charles Hudgins Sep 5 '19 at 3:05
Let $$S = 1 + 2 + \ldots + (n-1) + n.$$ Write it backwards: $$S = n + (n-1) + \ldots + 2 + 1.$$ Add the two equations, term by term; each term is $$n+1,$$ so $$2S = (n+1) + (n+1) + \ldots + (n+1) = n(n+1).$$ Divide by 2: $$S = \frac{n(n+1)}{2}.$$ | {
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• Well, you lost me at "each term is n+1...". As far as I can see, if you add the two equations term by term it will be: n+n + (n-1)+(n-1) + ... + 2+2 + 1+1. How did you get (n+1) + (n+1) + ... + (n+1)? – b1_ Aug 12 '10 at 18:27
• I got it! It's 2S = (1+2+...+(n-1)+n) + (n+(n-1)+...+2+1) - so you write one backwards, then match up each term. 2S = (1+n) + (2+n-1)+...+(n-1+2)+n+1, and so 2S=(n+1)+(n+1)+...+(n+1)+(n+1) etc – b1_ Aug 12 '10 at 19:04
• This trick is usually attributed to Gauss (when he was a schoolboy... though it's unclear if the story is true or not). – Fixee Feb 27 '11 at 15:17
• This helped make it clear for me youtu.be/1wnIsgUivEQ – vexe Nov 27 '15 at 8:17
• @john this can be made rigorous using sigma notation. It's simply a change of index. – Brevan Ellefsen Jul 29 '18 at 1:34
My favourite proof is the one given here on MathOverflow. I'm copying the picture here for easy reference, but full credit goes to Mariano Suárez-Alvarez for this answer.
Takes a little bit of looking at it to see what's going on, but it's nice once you get it. Observe that if there are n rows of yellow discs, then: | {
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1. there are a total of 1 + 2 + ... + n yellow discs;
2. every yellow disc corresponds to a unique pair of blue discs, and vice versa;
3. there are ${n+1 \choose 2} = \frac 12 n(n+1)$ such pairs.
• great proof!!!! – anon Aug 12 '10 at 21:22
• Maybe I lack imagination, but to me it's clearer to just make a square of yellow discs ($n^2$ of them), duplicate the diagonal ($n^2+n$), then cut in half to get the answer ($(n^2+n)/2$). This is how Knuth does this (and much more intricate) summations in "Concrete Mathematics." – Fixee Feb 27 '11 at 15:27
• @Fixee: I don't know why you're comparing them; this is an entirely different proof. Unlike the other proof (also good), this doesn't require computing areas, cutting, or duplicating -- in fact this doesn't even involve the number $n(n+1)/2$ directly; what this gives is a proof that $1 + 2 + \dots + n = \binom{n+1}{2}$, and it so happens that the latter is $n(n+1)/2$. It's a bijection proof, rather than an area proof (vaguely speaking). – ShreevatsaR Nov 25 '11 at 6:13
• @Vaughn Climenhaga - is there an analog of this in three dimensions and higher? – Vincent Tjeng Apr 1 '13 at 9:43
• Wow. That is simply amazing. I didn't get it at first glance and then it hit me. Ingenious. – Karl Feb 3 '16 at 20:30
What a big sum! This is one of those questions that have dozens of proofs because of their utility and instructional use. I present my two favorite proofs: one because of its simplicity, and one because I came up with it on my own (that is, before seeing others do it - it's known).
The first involves the above picture: | {
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The first involves the above picture:
In short, note that we want to know how many boxes are in the outlined region, as the first column has 1 box, the second 2, and so on (1 + 2 + ... + n). One way to count this quickly is to take another copy of this section and attach it below, making a $n*(n+1)$ box that has exactly twice as many squares as we actually want. But there are $n*(n+1)$ little squares in this area, so our sum is half that: $$1 + 2 + ... + n = \dfrac{n(n+1)}{2}.$$
Second proof, same as the first but a little bit harder and a little bit worse:
Let us take for granted the finite geometric sum $1 + x + x^2 + ... + x^n = \dfrac{x^{n+1} - 1}{x-1}$ (If you are unfamiliar with this, comment and I'll direct you to a proof). This is a polynomial - so let's differentiate it. We get $$1 + 2x + 3x^2 + ... + nx^{n-1} = \dfrac{ (n+1)x^n (x-1) - x^{n+1} + 1}{ (x-1)^2 }$$ Taking the limit as x approaches 1, we get
$$\lim_{x \to 1} \dfrac{ (n+1)x^n (x-1) - x^{n+1} + 1}{ (x-1)^2} = \dfrac{ (n+1) [ (n+1)x^n - nx^{n-1} ] - (n+1)x^n }{2(x-1)} =$$ $$\lim_{x \to 1} \dfrac{ (n+1)[(n+1)(n)x^{n-1} - n(n-1)x^{n-2}] - (n+1)(n)x^{n-1} } {2}$$
where we used two applications of l'Hopital above. This limit exists, and plugging in x = 1 we see that we get $$\dfrac{1}{2} * (n+1)(n) [ (n+1) - (n-1) - 1] = \dfrac{ (n)(n+1)}{2}.$$
And that concludes the second proof.
• I came up with the first proof by myself a long time ago :). It was an awesome moment where the concept of "area" was redefined. – Jacob Apr 22 '11 at 1:50
• @mixedmath A bit overkill, but nice anyways. (+1) – Franklin Pezzuti Dyer Jan 20 '18 at 15:40
How many ways are there to choose a $2$-element subset out of an $n$-element set?
On the one hand, you can choose the first element of the set in $n$ ways, then the second element of the set in $n-1$ ways, then divide by $2$ because it doesn't matter which you choose first and which you choose second. This gives $\frac{n(n-1)}{2}$ ways. | {
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On the other hand, suppose the $n$ elements are $1, 2, 3, ... n$, and suppose the larger of the two elements you choose is $j$. Then for every $j$ between $2$ and $n$ there are $j-1$ possible choices of the smaller of the two elements, which can be any of $1, 2, ... j-1$. This gives $1 + 2 + ... + (n-1)$ ways.
Since the two expressions above count the same thing, they must be equal. This is known as the principle of double counting, and it is one of a combinatorialist's favorite weapons. A generalization of this argument allows one to deduce the sum of the first $n$ squares, cubes, fourth powers...
Gathering as many proofs as we can? Write the series recursively:
$$S(n) = S(n - 1) + n \tag{1}$$
Substitute $n \to n+1$ :
$$S(n + 1) = S(n) + n + 1\tag{2}$$
Equation (2) subtract Equation (1):
$$S(n+1) - S(n) = S(n) + 1 - S(n - 1) \tag{3}$$
And write it up:
$$\begin{cases} S(n+1) &= 2S(n) -S(n-1) + 1 \\ S(n) &= S(n) \end{cases} \tag{4}$$
Which can now be written in matrix form:
$$\begin{bmatrix} S(n+1) \\ S(n) \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} S(n) \\ S(n-1) \end{bmatrix} + \begin{bmatrix} 1 \\ 0 \end{bmatrix} \tag{5}$$
And then converting the affine equation (5) to the linear equation (6):
$$\begin{bmatrix} S(n+1) \\ S(n+0) \\ 1\end{bmatrix} = \begin{bmatrix} 2 & -1 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix} \begin{bmatrix} S(n) \\ S(n-1) \\ 1\end{bmatrix} \tag{6}$$
And closing the equation:
$$\begin{bmatrix} S(n+1) \\ S(n) \\ 1\end{bmatrix} = \begin{bmatrix} 2 & -1 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}^n \begin{bmatrix} S(1) \\ S(0) \\ 1\end{bmatrix} \tag{6}$$
Then finding the Jordan form of the 3x3 matrix: | {
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Then finding the Jordan form of the 3x3 matrix:
\begin{align} \begin{bmatrix} S(n+1) \\ S(n) \\ 1\end{bmatrix} &= \left(\begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}^{-1}\right)^n \begin{bmatrix} S(1) \\ S(0) \\ 1\end{bmatrix} \\ &= \begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}^n \begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}^{-1} \begin{bmatrix} S(1) \\ S(0) \\ 1\end{bmatrix} \\ &= \begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix} \begin{bmatrix} 1 & \binom{n}{1} & \binom{n-1}{2} \\ 0 & 1 & \binom{n}{1} \\ 0 & 0 & 1\end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \\ 1 & -1 & 0 \\ 0 & 0 & 1\end{bmatrix} \begin{bmatrix} S(1) \\ S(0) \\ 1\end{bmatrix} \end{align} \tag{7}
And multiplying the matrices out:
$$\begin{bmatrix} S(n+1) \\ S(n) \\ 1\end{bmatrix} = \begin{bmatrix} \frac{ (2n+2)S(1) - 2nS(0) + {n}^{2} + n}{2} \\ \frac{ 2nS(1) + (2 - 2n)S(0)+{n}^{2}-n}{2} \\ 1 \end{bmatrix} \tag{8}$$
And given that $S(0) = 0$ and $S(1) = 1$, we get that:
$$S(n) = \frac{n^2 + n}{2} \tag{9}$$
For example, $$X = 1+2+3+4+5+6$$ Then twice $X$ is $$2X = (1+2+3+4+5+6) + (1+2+3+4+5+6)$$ which we can rearrange as $$2X = (1+2+3+4+5+6) + (6+5+4+3+2+1)$$ and add term by term to get $$2X = (1+6)+(2+5)+(3+4)+(4+3)+(5+2)+(6+1)$$ to get $$2X = 7+7+7+7+7+7 = 6*7 = 42$$ | {
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• Legend has it that Gauss used this method to find the sum of all numbers from 1 to 100 without actually summing them. – lhf Apr 21 '11 at 22:43
• In class, I think. A teacher set him in a corner because he was misbehaving (apparently having already finished that day's work), and told him to add these numbers. Almost immediately, Gauss went to play because he was done - and when the teacher got angry it took her nearly 10 minutes to verify that his answer was correct! Or at least, that's how the legend I heard went. – davidlowryduda Apr 21 '11 at 22:46
• @mixedmath, Gauss's Day of Reckoning, which was cited in wikipedia, discusses the legend. – lhf Apr 21 '11 at 23:01
• @mixedmath: I heard a different version - teacher asked all the kids to add the numbers from 1 to 1,000, because he (late 18th century Germany, therefore "he") needed some time to do some work. And CFG had the answer within 30 seconds... – gnasher729 Dec 22 '16 at 16:56
• Nobody should take nearly ten minutes. When you add up the single digits, you should notice that you are ten times adding 1+2+3+4+5+6+7+8+9 = 45. And the tens, you have ten 10s, ten 20s, ten 30s, ten 90s. Plus a hundred. All in all, should be done in a minute. – gnasher729 Dec 22 '16 at 16:59
My favourite proof of this fact involves counting the edges of the complete graph $K_n$ in two different ways.
On the one hand, any vertex, $v_1$ say, is connected to $n-1$ other vertices, thus contributing $n-1$ edges. Moving clockwise, the next vertex $v_2$ contributes $n-2$ edges (not counting the edge connecting $v_1$ and $v_2$), $v_3$ adds $n-3$ edges, ... , $v_{n-1}$ contributes 1 edge and $v_n$ adds no new edges.
Thus the total number of edges in the complete graph $K_n$ is:
$$E = \sum\limits_{i=1}^{n-1} i$$
But clearly, any edge connects two vertices, so the number of edges is the number of ways to choose two distinct elements from the set $\{1,...,n\}$ and hence: | {
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$$E = \sum\limits_{i=1}^{n-1} i = \binom{n}{2} = \frac{n(n-1)}{2}$$
• You can easely explain this proof without graph theory... $n+1$ people went to a party, and everyone shakes hands with everyone else... How many handshakes were there? :) – N. S. Sep 29 '12 at 14:03
Once you have a formula like this, you can prove it by induction. But that begs the question as to how you get such a formula. In this case you might ask: (a) what's the "average" term? and (b) how many terms are there?
Draw a triangular pyramid of base $n+1$. We get a unique coordinate for any of the $\sum_{i=1}^ni$ elements of the pyramid not in the bottom row by choosing two of the elements in the bottom row of $n+1$. This gives a bijection from the $\binom{n+1}{2}$ coordinate pairs to the $\sum_{i=1}^ni$ elements of the pyramid not in the bottom row.
Image from Mariano Suárez-Alvarez's answer on Math Overflow:
• @Mixedmath: No he means something quite different. A picture can be found in this MO-topic mathoverflow.net/questions/8846/proofs-without-words/8847#8847 – Myself Apr 22 '11 at 0:39
• This is a beautiful proof and doesn't seem to be getting enough appreciation, so I added the image from the MO answer to make it clearer. Feel free to remove the image if you don't want it. – ShreevatsaR Nov 25 '11 at 5:55
• @ShreevatsaR: That image is to powerful! – String Sep 25 '14 at 10:22
Basically same proof as Yoyo's, just purely combinatorial (no picture needed):
How many ways can we chose two distinct numbers between $1$ and $n+1$?
We pick first the largest, which is of the form $i+1$ for some $1 \leq i \leq n$, and then we have exactly $i$ distinct choices for the smallest one.
Thus we have $\sum_{i=1}^n i$ choices.
Here is another idea:
Using $(i+1)^2-i^2=2i+1$ we get a telescopic sum:
$$\sum_{i=1}^n 2i+1 = \sum_{i=1}^n (i+1)^2-i^2 = (n+1)^2-1=n^2+2n \,.$$
Then
$$n^2+2n= 2\left[\sum_{i=1}^n i\right] +n \,.$$ | {
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Then
$$n^2+2n= 2\left[\sum_{i=1}^n i\right] +n \,.$$
• Note that the second idea is useful for generalization in that we can find the sums of any powers is we know the sums of the powers less than. It can take a lot of work but using this method, you would be able to find 1^10 + 2^2 + ... + n^10 without too much work... – quasicompactscheme Sep 18 '11 at 15:07
If you knew of the geometric series, you would know that
$$\frac{1-r^{n+1}}{1-r}=1+r+r^2+r^3+\dots+r^n$$
If we differentiate both sides, we have
$$\frac{nr^{n+2}-(n+1)r^{n+1}+r}{(1-r)^2}=1+2r+3r^2+\dots+nr^{n-1}$$
Letting $r\to1$ and applying L'Hospital's rule on the fraction, we end up with
$$\frac{n(n+1)}2=1+2+3+\dots+n$$
HINT Pair each summand $$k$$ with its "reflection" $$n+1-k$$. This is simply a discrete analog of the method of computing the area of the triangle under the diagonal of a square by reflecting a subtriangle through the midpoint of the diagonal to form an $$n$$ by $$n/2$$ rectangle.
Like the analogous proof of Wilson's theorem, the method exploits the existence of a nontrivial symmetry. In Wilson's theorem we exploit the symmetry $$n \mapsto n^{-1}$$ which exists due to the fact that $${\mathbb F}_p^*$$ forms a group. Here we exploit a reflection through a line - a symmetry that exists due to the linear nature of the problem (which doesn't work for nonlinear sums, e.g. $$\sum n^2$$). Symmetries often lead to elegant proofs. One should always look for innate symmetries when first pondering problems.
Generally there are (Galois) theories and algorithms for summation in closed form, in analogy to the differential case (Ritt, Kolchin, Risch et al.). A very nice motivated introduction can be found in the introductory chapter of Carsten Schneider's thesis Symbolic Summation in Difference Fields.
Here is an easy way to visualize it:
Draw a rectangular grid with a height of $n$ squares and width of $n+1$ squares. Obviously it has $n(n+1)$ squares in it. | {
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In the first row, color the leftmost square red and the other $n$ squares blue; in the second row, color the leftmost $2$ squares red and the other $n-1$ squares blue; and so forth (in the last row, there will be $n$ red squares and one blue square). Clearly, there are $\sum_{i=1}^n i$ red squares and the same number of blue squares.
Adding the red and blue squares together, we get $2 \sum_{i=1}^n i = n(n+1)$, or $\sum_{i=1}^n i = n(n+1)/2$.
Let us denote the sum as $S_1(n)$. This function must be a second degree polynomial in $n$ because the first order difference $S_1(n)-S_1(n-1)=n$ is a linear polynomial in $n$. So it suffices to construct the Lagrangian polynomial by three known points, let $(0,0), (1,1), (2,3)$.
$$S_1(n)=0\frac{(n-1)(n-2)}{(0-1)(0-2)}+1\frac{(n-0)(n-2)}{(1-0)(1-2)}+3\frac{(n-0)(n-1)}{(2-0)(2-1)}=\frac{n(n+1)}2.$$
The most general case of this is called an arithmetic progression or (finite) arithmetic series. There are many, many, many proofs. An easy one: write all the summands in a row; write them again just below, but from right to left now (so $1$ is under $n$, $2$ is under $n-1$, etc). Add them up, and figure out how it relates to the quantity you are looking for.
• Yep, that's it, write one backwards. I read this post but it didn't click - I guess I needed to write it out to see it. Thx – b1_ Aug 12 '10 at 19:43
Here are two ways to calculate this sum. First is by symmetry of another sum:
\begin{aligned} \displaystyle & \sum_{0 \le k \le n}k^2 = \sum_{0 \le k \le n}(n-k)^2 = n^2\sum_{0 \le k \le n}-2n\sum_{0 \le k \le n}k+\sum_{0 \le k \le n}k^2 \\& \implies 2n\sum_{0 \le k \le n}k = n^2(n+1) \implies \sum_{0 \le k \le n}k = \frac{1}{2}n(n+1).\end{aligned}
The second is writing it as double sum and switching the order of summation: | {
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The second is writing it as double sum and switching the order of summation:
\begin{aligned}\displaystyle & \begin{aligned}\sum_{1 \le k \le n}k & = \sum_{1 \le k \le n}~\sum_{1 \le r \le k} = \sum_{1 \le r \le n} ~\sum_{r \le k \le n} = \sum_{1 \le r \le n}\bigg(\sum_{1 \le k \le n}-\sum_{1 \le k \le r-1}\bigg) \\& =\sum_{1 \le r \le n}\bigg(n-r+1\bigg) = n\sum_{1 \le k \le n}-\sum_{1 \le k \le n}k+\sum_{1 \le k \le n}\end{aligned} \\& \implies 2\sum_{1 \le k \le n}k = n^2+n \implies \sum_{1 \le k \le n}k = \frac{1}{2}n(n+1), ~ \mathbb{Q. E. D.} \end{aligned}
Note I started using k back on the third line for convenience because r is just a dummy vairable at this point, and our sum no longer depends on k. Note that the first trick can easily be generalised:
\begin{aligned} & \hspace{0.5in}\begin{aligned}\displaystyle \sum_{0 \le k \le n}k^{2p} &= \sum_{0 \le k \le n}(n-k)^{2p} \\& = \sum_{0 \le k \le n}~\sum_{0 \le r \le 2p}\binom{2p}{r}n^r(-1)^{2p-r}k^{2p-r}\\& = \sum_{0 \le k \le n}k^{2p}-2pn\sum_{0 \le k \le n}k^{2p-1}+\sum_{0 \le k \le n}~\sum_{2 \le r \le 2p}\binom{2p}{r}n^r(-1)^{2p-r}k^{2p-r} \end{aligned} \\& \implies \sum_{0 \le k \le n}k^{2p-1} = \frac{1}{2pn}\sum_{0 \le k \le n}~\sum_{2 \le r \le 2p}\binom{2p}{r}n^r(-1)^{2p-r}k^{2p-r}. \end{aligned}
You can take the power series
$$f(x)=\sum_{n=0}^\infty\left(\sum_{j=0}^{n}j\right)x^n$$
and you can check that it has a postive convergence ratio, and changing the order of the series you can deduce that
$$f(x)=\frac{x}{(1-x)^3}.$$
On the other hand the taylor series for $\frac{x}{(1-x)^3}$ is precisely
$$\frac{x}{(1-x)^3}=\sum_{n=0}^\infty \frac{n(n+1)}{2}x^n$$
so $$\sum_{j=0}^nj=\frac{n(n+1)}{2}.$$
$\dfrac{k+1}2-\dfrac{k-1}2=1 \implies\dfrac{k(k+1)}2-\dfrac{(k-1)k}2=k$
$\implies \sum_{k=1}^n\Bigg(\dfrac{k(k+1)}2-\dfrac{(k-1)k}2\Bigg)=\sum_{k=1}^nk\implies\dfrac{n(n+1)}2-\dfrac{1(1-1)}2=\sum_{k=1}^nk$
$\implies\sum_{k=1}^nk=\dfrac{n(n+1)}2$ | {
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$\implies\sum_{k=1}^nk=\dfrac{n(n+1)}2$
$$\sum_{i=0}^ni-\sum_{i=0}^{n-1}i=S_1(n)-S_1(n-1)=n,$$ so that $S_1(n)$ must be a polynomial of the second degree in $n$. By the method of undeterminate coefficients, noting that there is no constant term as $S_1(0)=0$: $$S_1(n)-S_1(n-1)=n=(an^2+bn)-(a(n-1)^2+b(n-1))=\\=2an+b-a,$$ and by identification $$a=b=\frac12.$$ $$S_1(n)=\frac{n^2+n}2.$$ Let us generalize to the sum of squares, $$S_2(n)-S_2(n-1)=n^2=(an^3+bn^2+cn)-(a(n-1)^3+b(n-1)^2+c(n-1))=\\=3an^2+(-3a+2b)n+a-b+c,$$ giving $$a=\frac13,b=\frac12,c=\frac16.$$ $$S_2(n)=\frac{2n^3+3n^2+n}6.$$ For any power, you get a triangular system of equations where you recognize a part of Pascal's triangle, with alternating signs, as in the sum of cubes: $$S_3(n)-S_3(n-1)=n^3=4an^3+(-6a+3b)n^2+(4a-3b+2c)n+(-a+b-c+d),$$ $$\color{blue}{4}a=1\\-\color{blue}{6}a+\color{blue}{3}b=0\\\color{blue}{4}a-\color{blue}{3}b+\color{blue}{2}c=0\\-\color{blue}{1}a\ +\color{blue}{1}b-\color{blue}{1}c+\color{blue}{1}d=0,$$ giving $$a=\frac14,b=\frac12,c=\frac14,d=0.$$ $$S_3(n)=\frac{n^4+2n^3+n^2}4.$$
• You can also retrieve the polynomial by Lagrange interpolation with $(0,0), (1,1),(2,3)$: $$S_1(n)=0\frac{(n-1)(n-2)}{(1-0)(2-0)}+1\frac{(n-0)(n-2)}{(1-0)(1-2)}+3\frac{(n-0)(n-1)}{(2-0)(2-1)},$$ but this leads to much longer calculation. – Yves Daoust Jul 26 '14 at 10:22
Above is an image representing Pascals Triangle. What I want to draw attention to is the hockey stick formation, particularly, the blue hockey stick. Notice how the entries in the stick of the blue hockey stick are in arithmetic progression, and that the entry in the blade represents the sum of the entries in the stick.
To prove this inductively we have as a bootstrap condition
$$1=\frac{1(1+1)}{2}=\binom{1+1}{2} = \sum\limits_{i=1}^1\binom{i}{1}=1$$
and for the general case | {
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$$1=\frac{1(1+1)}{2}=\binom{1+1}{2} = \sum\limits_{i=1}^1\binom{i}{1}=1$$
and for the general case
$$\begin{array}{lll} \sum\limits_{i=1}^{n+1}&=&(n+1)+\sum\limits_{i=1}^{n}i\\ &=&\binom{n+1}{1}+\binom{n+1}{2}\\ &=&\binom{n+2}{2}\\ &=&\binom{(n+1)+1}{2}\\ &=&\frac{(n+1)((n+1)+1)}{2} \end{array}$$
Of course, we assumed that $\binom{n}{k}+\binom{n}{k+1} = \binom{n+1}{k+1}$ holds. $$\begin{array}{lll} \binom{n}{k}+\binom{n}{k+1}&=&\frac{n!}{k!(n-k)!} + \frac{n!}{(k+1)!(n-(k+1))!}\\ &=&\frac{n!(k+1)}{k!(k+1)(n-k)!} + \frac{n!(n-k)}{(k+1)!(n-(k+1))!(n-k)}\\ &=&\frac{n!k+n!+n!n-n!k}{(k+1)!(n-k)!}\\ &=&\frac{n!+n!n}{(k+1)!((n+1)-(k+1))!}\\ &=&\frac{n!(n+1)}{(k+1)!((n+1)-(k+1))!}\\ &=&\frac{(n+1)!}{(k+1)!((n+1)-(k+1))!}\\ &=&\binom{n+1}{k+1}\\ \end{array}$$
• I feel like this is a very intuitive answer, generalizes to the sum of $n^k$ well, and is easy to remember. – Simply Beautiful Art Nov 29 '16 at 0:50
Let us denote the sum as $S_1(n)$. This function must be a second degree polynomial in $n$ because the first order difference $S_1(n)-S_1(n-1)=n$ is a linear polynomial in $n$. So it suffices to verify the formula for three different values of the argument.
$$S_1(0)=0=\frac{0(0+1)}2,$$ $$S_1(1)=1=\frac{1(1+1)}2,$$ $$S_1(2)=3=\frac{2(2+1)}2.$$ QED.
I didn't see an answer using the method that I used, so I'm posting an answer here to 'spread the knowledge!'
This can be applied to higher powers such $1^2+2^2+\cdots n^2$ or $1^3+2^3+\cdots n^3$.
Solving by the use of Indeterminate Coefficients:
Assume the series$$1+2+3+4+5\ldots+n\tag1$$ Is equal to the infinite series$$1+2+3+4+5+\ldots+n=A+Bn+Cn^2+Dn^3+En^4+\ldots\&c\tag2$$ If we 'replace' $n$ with $n+1$, we get$$1+2+3+4+\ldots+(n+1)=A+B(n+1)+C(n+1)^2+\ldots\&c\tag3$$ And subtracting $(3)-(2)$, gives\begin{align*} & n+1=B+C(2n+1)\tag4\\n & +1=2Cn+(B+C)\tag5\end{align*} Therefore, $C=\dfrac 12,B=\dfrac 12,A=0$ and $(1)$ becomes$$1+2+3+4+5\ldots+n=\dfrac n2+\dfrac {n^2}2=\dfrac {n(n+1)}{2}$$ | {
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With the Euler-Maclaurin summation formula, one can easily derive that
$$\sum_{k=1}^nk=\int_0^nx\ dx+\frac12n=\frac12n^2+\frac12n$$
More generally, one may derive that
$$\sum_{k=1}^n k^p = {1 \over p+1} \sum_{j=0}^p (-1)^j{p+1 \choose j} B_j n^{p+1-j},\qquad \mbox{where}~B_1 = -\frac{1}{2}$$
where we use the Bernoulli numbers. This more general formula is more commonly known as Faulhaber's formula.
Another "picture proof" I just thought of... but without a picture, since I can't draw:
Suppose you want to add up all the integers from 1 to n. Then draw n rows on a board, put 1 unit in the first row, 2 in the second, and so on. If you draw a right triangle of height and length n to try and contain this shape, you will cut off half of each unit on the diagonal. So let's add all of these up; you have $$n^2/2$$ units inside the triangle, and you have $$n/2$$ units cut off on the diagonal (there are n squares on the diagonal, and half of each one is cut off). Adding these gives $$n^2/2 + n/2 = n(n+1)/2$$.
If you draw it out, it makes more sense, and I think it's geometrically a bit more straightforward than the other picture proof.
(Side note: I have no idea how to write math on this site... I'll go consult the meta, I'm sure there's something there about it.)
All these proofs seem very complicated. The way I remember it is:
The sequence is: 1, 2, 3, ..... (n-2), (n-1), n.
Taking the last and first term, 2nd and (n-2)th term and so on, we form n/2 pairs of (n+1). So the sum of n/2 pairs of (n+1) is n/2 * (n+1)
Example: 1, 2, 3, 4, 5, 6 = (1+6) + (2+5) + (3+4) = 3x7 =21
This still holds for an odd number of terms | {
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This still holds for an odd number of terms
• But this is not a proof. This is an explanation. – Asaf Karagila Sep 29 '12 at 12:00
• @user929404: As Asaf points out, this is not a proof. You say that these proofs seem very complicated, but Joe's accepted answer for example uses the exact same idea as your explanation, but does so in a rigorous way which makes it a proof. Sometimes it is difficult to see the idea behind a proof, but it is also not enough to just give the idea; we need proof to be sure that our ideas actually work. – Michael Albanese Sep 30 '12 at 4:21
• This seems like a valid proof to me, at least when $n$ is even. (When $n$ is odd, you do need to say that you have $(n-1)/2$ pairs that sum to $(n+1)$, together with an additional term of $(n+1)/2$.) – Jesse Madnick Dec 17 '12 at 19:41
• As almost all things that call themselves proofs are not actually proofs, but rather outlines of proofs, this is certainly an acceptable outline of a proof. If you want to argue over what actually is a genuine proof, then you have to be ready to accept an almost inhumane level of pedantry and precision that only computers can realistically handle. – DanielV Dec 2 '14 at 11:01
• This is how a young Gauß did it, the story goes... – Chris Custer Sep 22 '18 at 13:01
You can also prove it by induction, which is nice and easy, although it doesn't give much intuition as to why it works.
It works for $1$ since $\frac{1\times 2}{2} = 1$.
Let it work for $n$.
$$1 + 2 + 3 + \dots + n + (n + 1) = \frac{n(n+1)}{2} + (n + 1) = \frac{n(n+1) + 2(n+1)}{2} = \frac{(n+1)(n+2)}{2}.$$
Therefore, if it works for $n$, it works for $n + 1$. Hence, it works for all natural numbers.
For the record, you can see this by applying the formula for the sum of an arithmetic progression (a sequence formed by constantly adding a rate to each term, in this case $1$). The formula is reached pretty much using the method outlined by Carl earlier in this post. Here it is, in all its glory: | {
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$$S_n = \frac{(a_1 + a_n) * n}{2}$$
($a_1$ = first term, $a_n$ = last term, $n$ = number of terms being added).
This is the shortest proof (without words):
is my proof showing that $$\sum_{j=1}^n{j}=\frac{n(n+1)}{2}.$$
I would like to resubmit the first 'backwards and forwards' proof above. My issue with the proof is the use of "$\dots$" which translates to and so on. Here is a more mathematically rigorous proof using properties of summation $\Sigma$. $$2\sum_{k=1}^{n} k =\sum_{k=1}^{n} 2k = \sum_{k=1}^{n} (k+k) = \sum_{k=1}^{n}k + \sum_{k=1}^{n}k = \sum_{k=1}^{n}k+ \sum_{k=1}^{n}(n-(k-1))= n(n+1)$$
• If you want a rigorous proof using sigma, you should go full boar and use ${\displaystyle \sum _{n\in B}f(n)=\sum _{m\in A}f(\sigma (m))}$ from en.wikipedia.org/wiki/Summation – CopyPasteIt Jul 2 '17 at 1:56
Some time ago I saw someone explain it as follows:
The average value of $$1,2,3,\dots,n$$ is simply $$\frac{n+1}2$$. Thus $$1+2+3+\dots+n=\frac{n(n+1)}2$$.
Of course the proof behind this leads to Gauss's proof quite directly, but nonetheless I really like this restatement of it as it is easy to understand even if one does not know much math. And it quickly gives the sum of terms in arithmetic progression as well. Such a sum is simply the number of terms times the average of the first and last terms.
I stumbled across this identity
$$\quad ab - 1 = (a - 1) (b - 1) + (a - 1) + (b - 1)$$
while working with modulo arithmetic a couple of days ago. I was upset that I couldn't use it in any (really) interesting ways, but now I've discovered that I can 'pile onto' this question!
If $$n \ge 1$$ is an integer we can write
$$\tag 1 n^2 = (n - 1)^2 + 2 (n - 1) + 1$$
Observe that the first term of the rhs of $$\text{(1)}$$ is the square of an integer, just like the lhs of $$\text{(1)}$$. So you can use 'downward finite induction formula recursion' (not sure what to call this) and conclude that
$$\quad n^2 = [\displaystyle 2 \sum_{k=1}^{n-1}\, k] + n$$ | {
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$$\quad n^2 = [\displaystyle 2 \sum_{k=1}^{n-1}\, k] + n$$
and this even holds for $$n = 1$$.
Rewriting the formula we obtain
$$\quad \displaystyle \frac{n(n-1)}{2} = \sum_{k=1}^{n-1}\, k$$ | {
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# corbett maths probability trees | {
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The Corbettmaths Practice Questions on Venn Diagrams answers – Corbettmaths. Some questions to delve a little deeper into the understanding of probability and tree diagrams. Probability Trees. Download the medium term plan by clicking on the button above. "The teacher selects two students at random to go on a trip. I want to help you achieve the grades you (and I) know you are capable of; these grades are the stepping stone to your future. One ball is picked out, and not replaced, and then another ball is picked out. Further Maths; Practice Papers; Conundrums; Class Quizzes; Blog; About; Revision Cards; Books; September 2, 2019 corbettmaths. Use the fact that probabilities add up to 1 to work out the probabilities of the missing branches. is written alongside the line. 1) Complete the probability tree diagram. A card game in which students determine an optimal strategy for playing the game. Click here for Answers . A running club has 160 members. Instructions Use black ink or ball-point pen. There are 6 girls and 5 boys to choose from. To be able to assign a probability to each number, an experiment would need to be conducted. Fill in the boxes at the top of this page with your name, centre number and candidate number. Tree Diagrams Practice Questions Click here for Questions . Mathster; Corbett Maths Tree diagrams are a way of showing combinations of two or more events. corbettmaths exam style questions linear graphs / corbettmaths exam style questions direct and inverse proportion / corbettmaths exam style questions answers / corbettmaths exam papers / corbettmaths exam style questions simultaneous equations / corbettmaths exam style questions probability / corbettmaths exam style questions equation of a line answers / corbettmaths exam style questions … The first two are fairly standard (I)GCSE fare but subsequent questions become more complex with questions 5 and 6 requiring the solution of quadratic equations if an algebraic approach is used. PROBABILITY & TREE | {
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6 requiring the solution of quadratic equations if an algebraic approach is used. PROBABILITY & TREE DIAGRAMS Materials required for examination Items included with question papers Ruler graduated in centimetres and Nil millimetres, protractor, compasses, pen, HB pencil, eraser. For example, the probability of rolling a 6 on a die will not affect the probability of rolling a 6 the next time. Aim for at least 8%. Maths Genie ¦ Corbett Maths ¦ Mr Barton Maths Takeaway ¦ Mr Barton Maths Topic Search ¦ Just Maths 13.1 Calculating Probability - Calculate simple probabilities from equally likely events - Understand mutually exclusive and exhaustive outcomes Just Maths Probability – F – Probability 1 v3 ¦ Probability – F – Probability 1 v3 – SOLUTIONS Probability – F – Probability 2 v3 ¦ Probability – F – Probability 2 … They have kindly allowed me to create 3 editable versions of each worksheet, complete with answers. PROBABILITY & TREE DIAGRAMS Materials required for examination Items included with question papers Ruler graduated in centimetres and Nil millimetres, protractor, compasses, pen, HB pencil, eraser. Each branch is labelled at the end with its outcome and the probability is written alongside the line. 29 Tree Diagrams 252 30 Venn Diagrams 380. These are the Corbettmaths Textbook Exercise answers to Fractions to Percentages Exam Style Questions Corbett Maths Answers. Tree diagrams. The probability of rain in the village is 0.3. Answer all questions. Guestbook. Tree Diagrams - conditional / without replacement: Worksheets with Answers. We aim … Each branch is a possible outcome and is labelled with a probability. Print the worksheet by clicking the blue button above the video. With assessment results and gap analyses to download required resources from below first happening... Heads and Tails ) selects both a Boy and a girl, 5-a-day, and more! most )! Basic facts About equally likely outcomes can help to solve more complicated problems to a. Two biscuits were | {
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1. (10%) Use generating functions to answer the following questions.
(A) Find the solution of the recurrence relation $a_n = 4a_{n-1} – 3a_{n-2} + 2^n + n + 3 \text{ with } a_0 = 1 \text{ and } a_1 = 4.$
(B) Find the coefficient of $x^{10}$ in the power series of $x^4 / (1 – 3x)^3.$
(A) $a_n = -4 \cdot 2^n + \dfrac{39}{8} \cdot 3^n + \dfrac{19}{8} – \dfrac{7}{4}(n+1) – \dfrac{1}{4}(n+2)(n+1), \, n \geq 0.$
(B) $\binom{3+6-1}{6} \cdot 3^6 = 28 \cdot 3^6 = 20412$
2. (10%) How many relations are there on a set with n elements that are
(A) both reflexive and symmetric?
(B) neither reflexive nor irreflexive?
(A) $2^{\frac{n(n-1)}{2}}$
(B) $2^{n^2} – 2^{n(n-1)+1}$
3. (5%) How many nonisomorphic unrooted trees are there with five vertices?
4. (5%) Multiple answer question. (It is possible that more than one of the choices are correct. Find out all correct choices.)
A hash table of length 10 uses the hash function $h(k) = k \, mod \, 10$ and the linear probing for handling overflow. After inserting 6 values into an initially empty hash table, the table is as shown below. Which one(s) of the following choices gives a possible order in which the key values could have been inserted in the table?
(A) 46, 42, 34, 52, 23, 33
(B) 34, 42, 23, 52, 33, 46
(C) 46, 34, 42, 23, 52, 33
(D) 42, 46, 33, 23, 34, 52
(E) 42, 23, 34, 46, 52, 33
5. (5%) Fill in the six black (I, II, …, and VI) in the following program that implements a queue by using 2 stacks.
class MyQueue<T> {
private:
stack<T> stack1;
stack<T> stack2;
public:
MyQueue()
{
stack1 = new stack<T>();
stack2 = new stack<T>();
}
// enqueue(): Add an element at the rear side of MyQueue
void enqueue(T, e)
{
stack1.push(e);
}
// dequeue(): Remove the front element from MyQueue
T dequeue(T, e)
{
if((__I__).isEmpty())
while(!(__II__).isEmpty())
(__III__).push((__IV__).pop());
T temp = null;
if(!(__V__).isEmpty())
temp = (__VI__).pop();
return temp;
}
}
6. (5%) AVL Tree. | {
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6. (5%) AVL Tree.
(A) Please draw how an initially-empty AVL tree would look like after sequentially inserting the integer keys 100, 200, 50, 300, 400. There is no need to show it in a step-by-step fashion; you only need to draw the final result.
(B) Continue the previous sub-problem. Suppose that the integer keys 25, 250, 225, 500, 240, 260 are sequentially inserted into the AVL tree of the previous sub-problem. Draw the AVL tree after all of these integer keys are inserted.
(A)
(B)
7. (5%) Reconstruct and draw the maximum binary heap whose in-order traversal is 2, 16, 7, 62, 5, 9, 188, 14, 78, 10. There is no need to show it in a step-by-step fashion; you only need to draw the final result.
8. (5%) The following algorithm takes an array as input and returns the array with all the duplicate elements removed. For example, if the input array is {1, 3, 3, 2, 4, 2}, the algorithm returns {1, 3, 2, 4}.
S = new empty set
A = new empty dynamic array
for every element x in input array
if not S.member(x) then
S.insert(x)
A.append(x)
return A
Suppose that the input array has n elements. What is the Big-O complexity of this algorithm, if the set S is implemented as:
(A) a hash table (with the assumption that overflow does not occur)?
(B) a binary search tree?
(C) an AVL tree?
9. (10%) The recurrence $T(n) = 7T(\dfrac{n}{2}) + n^2$ describes the running time of an algorithm A. A completing algorithm A’ has a running time of $T'(n) = aT'(\dfrac{4}{n}) + n^2.$ What is the largest integer value for $a$ such that A’ is asymptotically faster than A?
10. (15%) Consider the following undirected graph G = (V,E). | {
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10. (15%) Consider the following undirected graph G = (V,E).
(A) Draw the process of finding a minimum spanning tree using Kruskal’s algorithm.
(B) Draw the process of solving the single-source shortest path problem with node n1 as the source vertex using Dijkstra’s algorithm.
(C) Starting from n1, find the Depth-First Search (DFS) traversal sequence of G (the priority of node is inversely proportional to the weight of incident edge).
(A)
(B)
(C) $n1 \to n4 \to n6 \to n5 \to n2 \to n3$
11. (18%) Given an ordered file with keys 1, 2, …, 16, determine the number of key comparisons made by a search algorithm A while searching for a specific key K.
(A) A is the binary search algorithm and K is 2.
(B) A is the binary search algorithm and K is 10.
(C) A is the binary search algorithm and K is 15.
(D) A is the Fibonacci search algorithm and K is 2.
(E) A is the Fibonacci search algorithm and K is 10.
(F) A is the Fibonacci search algorithm and K is 15.
12. (7%) Given a store of n items, what’s is the least upper bound (in Big-O notation) of the running time of the solutions to the following problems:
(A) Fractional knapsack problem;
(B) General 0/1 knapsack problem.
### 8 留言
1. #### 流動性
您好，我想知道第11題的DEF小題您在fibonacci search的求解過程是如何呢?因為我在網路上爬文有找到一篇討論的答案和這裡不一樣，謝謝!
以下是我找到的討論:
• 文章作者的留言
#### mt
Fibonacci search的步驟網路上真的有很多版本，解出來的答案也都不太一樣，我自己是用這部影片的方法。所以答案不同有可能是因為方法不一樣，可以去看看那所學校的大學部上課的時候的用書，然後看那本書是用什麼方法，這樣比較保險，加油喔！
2. #### w
第七題max heap圖是不是錯了
3. #### TNF
請問第八題的A選項
題目是假設沒有overflow的發生
但如果幾乎全部都collision且採linear probing
判斷元素是否在hash table的比較次數變成1+2+…+(n-1)
是否最差時間複雜度就是O(n^2)了呢
另外O(n)是平均狀況嗎??
• 文章作者的留言
#### mt
我這題是假設說每個bucket只有一個slot然後題目又說沒有overflow（等於沒有collision，因為只有一個slot時，collision等於overflow），所以我才寫O(n)。然後就算collision的話也不會overflow所以不會有採用哪種probing的問題，不過這題也可以想成是用chaining，這樣的話是O(n^2)沒錯。希望有回答到你的問題～
4. #### aaa
嗨，第10題(b)的第2個iteration的node n3應該是75吧？因為第1個iteration是選node n4，所以更新n3從無限大->75 | {
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# Cardinality of all sequences of non-negative integers with finite number of non-zero terms. (NBHM 2012)
Consider the set $S$ of all sequences of non-negative integers with finite number of non-zero terms.
1. Is the set $S$ countable or not?
2. What is the cardinality of the set $S$ if it is not countable?
My intuition is the set is countable. The sequence has only finitely many non-zero terms. For any fixed $N$ consider the set $A_N$ which contains all sequences $\{a_n\}$ s.t. $a_k = 0$ $\forall$ $k >N$. The set $A_N$ is countable as the first $N$ terms of a sequences can be filled up by non-negative integers in a countable number of ways. So $A_N$ is countable and $S$ is a countable union of countable sets. So $S$ is countable.
I do not know if it is true or false. If it is false please identify the mistake. Thank you for your help.
Please suggest me a book where I shall get sufficient number of such type of problem to clear basic ideas on cardinal number.
• Would you add "integer" before "numbers" in the title and in the first sentence? Your intuition is correct. – egreg Dec 29 '13 at 17:09
• You can try Hrbacek, Jech - Introduction to set theory – Giulio Bresciani Dec 29 '13 at 17:27
You can think a sequence as a finite subset of $\mathbb{N}^2$: for all $a_n\neq 0$, take the point $(n,a_n)$. This way, you can inject $S$ in the set $\mathcal{P'}(\mathbb{N}^2)$ (with $\mathcal{P'}$ I mean the set of finite subsets) and this is in bijection with $\mathcal{P'}(\mathbb{N})$. But $\mathcal{P'}(\mathbb{N})$ is countable, because it is a countable union of countable sets (subsets with $0$ elements, subsets with $1$ element, subsets with $2$ elements...).
• Thank you for the answer supporting my approach. – Dutta Dec 30 '13 at 1:46 | {
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• Thank you for the answer supporting my approach. – Dutta Dec 30 '13 at 1:46
Let $\mathbb{N}$ be the set of non-negative integers. If $s=s_0,s_1,s_2,\dots,s_n,\dots$ is a sequence in $S$, define $\psi(s)$ by $$\psi(s)=\left(\prod_{i=0}^\infty p_i^{s_i}\right)-1,$$ where $p_i$ is the $i$-h prime. By the Unique Factorization Theorem, $\psi$ is a bijection from $S$ to $\mathbb{N}$.
• Nicolas: Thank you for introducing a new concept. Let me know why you are subtracting 1 from the product in the definition of $\phi(s)$. – Dutta Dec 30 '13 at 1:46
• That is because I am using as $\mathbb{N}$ the non-negative integers. If we are making a bijection to the positive integers, there is no $-1$ term. Of course it makes no real difference, since there is an obvious bijection between the non-negatives and the positives. – André Nicolas Dec 30 '13 at 1:57
• It is clear now. – Dutta Dec 30 '13 at 2:02
• I think the answer by Giulio Bresciani is more "versatile," and therefore more worthy of accepting. Mine is a little too cute, too tailored to the specific situation. – André Nicolas Dec 30 '13 at 2:06
Put a decimal point in front of any of these sequences and you have a rational number between 0 and 1. The set is countable. There is a bit more to it than just that. You will need to consider repetitions. You end up with a countable number of equivalence classes that are each at most countable. 1,1,0,0,0,0... becomes .110000... and 11,0,0,0,0... also becomes .110000...
• Happy new year. This is also a nice answer. – Dutta Jan 1 '14 at 16:15 | {
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# Checking whether a number is prime or composite
This is a question that came up while I was doing an exercise. I ended up with the number
$$200! + 1$$
and I want it to be composite but I don't know of any methods to check whether a number is prime or not.
Is there any general rule about $n+1$ or $n! + 1$ to determine if or when these are prime or composite?
The exercise I was doing when I ended up stuck at this question was this:
Show that there exists $n \in \mathbb N$ such that $n, n+1 \dots, n+200$ are all composite.
I am hoping for a solution not using calculators, software or the internet. I expect there to be a short and (computationally) simple answer. At least that's what I hope.
• $n+1$: definitely not, since every integer is of this form. – vadim123 Oct 20 '15 at 1:47
• Wilson's theorem gives a useful result: if $n+1$ is prime then it divides $n!+1$. Unfortunately $201$ is not a prime so it does not apply here. – Winther Oct 20 '15 at 1:50
• Wolfram was able to answer this, and gives $200!+1 = 1553\times 826069\times 353297821\times k$ where $k$ is a very large prime number that won't fit on this page. – JMoravitz Oct 20 '15 at 1:52
• There is an ongoing project here that searches for primes on that form (known as factorial primes). – Winther Oct 20 '15 at 1:56
• If you use $210!$ instead of $200!$ you can apply Wilson's theorem since $211$ is prime. – Paul Hankin Oct 20 '15 at 4:04
As to the original question, primes of the form $n!\pm 1$ are known as factorial primes and not all are known. It is in general a complicated question to determine if a number is prime or not, and only partial results are known. For example, if $n+1$ is prime then $n!+1$ is not.
As for the exercise which prompted this question, proving that there exists some $n$ such that $n,n+1,n+2,\dots,n+200$ are all composite consider the following: | {
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Suppose we want to force each $n+i$ to be composite. If we want to force $2\mid n$ and $3\mid (n+1)$ and $5\mid (n+2)$, etc... that would correspond to the system of congruencies:
$\begin{cases} n\equiv 0\pmod{2}\\ n+1\equiv 0\pmod{3}\\ n+2\equiv 0\pmod{5}\\ \vdots\\ n+200\equiv 0\pmod{p_{201}}\end{cases}$
Consider then the Chinese Remainder Theorem.
The Chinese remainder theorem states that we can find such an $n$ that satisfies all of those congruencies since each of what we are modding out by are relatively prime to one another in every case.
Note: there is nothing intrinsically special about ordering these as being modulo $2$ followed by $3$ followed by $5$, etc... So long as we pick a list of length 200 where each of the entries on the list are coprime to one another, this will work.
Edit: Minor missing detail. It is possible that $n+i=p_i$ in one of those cases. To account for this possibility, technically chinese remainder will give us a solution to $n\equiv k\pmod{\prod p_i}$, so we can avoid this by instead of taking the smallest positive integer $n$ that works, by instead taking $n+\prod p_i$.
• So the question isn't asking you for a specific $n$? The Chinese Remainder Theorem solves this, as we're only looking for the existence of such an $n$, just put the numbers you're adding by, on the other side, because there are infinitely many primes, which are definitely coprime to one another. – Almentoe Oct 20 '15 at 2:32
• That is the way it is worded. In fact, the same proof can be modified to show the existence of arbitrarily long prime gaps. If you wanted to find an exact value for n, it could be done but would be incredibly tedious to do. – JMoravitz Oct 20 '15 at 2:41 | {
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As others have mentioned, it is not easy in general to check whether numbers of the form $n! + 1$ are prime. But as you may have noticed, the numbers $200! + 2$, $200! + 3, \dots, 200! + 200$ are all composite, as they are divisible by the numbers $2, 3, \dots, 200$ respectively. This gives only 199 consecutive composite numbers rather than the 201 required by your problem, but that's nothing that can't be fixed by increasing $n$ a little.
• Or noting that $200! + 201$ and $200! + 202$ are composite since they're divisible by 3 and 2 respectively. – Paul Hankin Oct 20 '15 at 4:56 | {
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# [SOLVED]How to use Taylor series to represent any well-behaved f(x)
#### DeusAbscondus
##### Active member
Does one assess $x$ at $x=0$ for the entire series? (If so, wouldn't that have the effect of "zeroing" all the co-efficients when one computes?)
only raising the value of $k$ by $1$ at each iteration?
and thereby raising the order of derivative at each iteration?
$$\sum_{k=0}^{\infty}\frac{f^{k}(0)}{k!} x^k= f(0)+\frac{df}{dx}|_0 \ x + \frac{1}{2!}\frac{d^2f}{dx^2}|_0 \ x^2+ \frac{1}{3!}\frac{d^3f}{dx^3}|_0 \ x^3+ ....$$
I have no experience with series or sequences, so, I know I have to remedy this gap in my knowledge.
In the interim, however, I am currently enrolled in a Math course that looks at Calculus by beginning with Taylor series.
I am an adult beginner at Math, having done an introductory crash course in Calculus last year; I wanted to flesh this out: hence my current enrolment.
But I am at a loss to know how to manipulate the notation above and would appreciate a worked solution for some simple $f(x)$ (I won't nominate one, so as to preclude the possibility of my cheating on some set work)
I just need to see this baby in action with a "well-behaved function" of someone else's choosing, with some notes attached if that someone would be so kind.
Thanks,
Deo Abscondo
Last edited:
#### Bacterius
##### Well-known member
MHB Math Helper
Re: How to use Taylor series to represent any well-behaved $f(x)$
Yes, the idea is that as you add more and more terms to the Taylor series, the series approximation becomes better and better and fits the function more closely for values farther away from $x = 0$.
For instance, let's take the venerable $y = \sin(x)$ function. Let's plot it:
The first term of the Taylor series (i.e. the Taylor series approximation at $k = 0$) is just $f(0) = 0$, hence: | {
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Of course, this approximation sucks. Let's try $k = 1$. Then the Taylor series approximation for $k = 1$ is:
$$\sum_{k = 0}^1 \frac{f^k (0)}{k!} x^k = \sin(0) + x \cos(0) = x$$
And our "first-order approximation" for $\sin(x)$ is the curve $y = x$, as illustrated below:
Still not an awesome approximation, but it works pretty well for $x$ close to zero. In fact this is known as the small-angle approximation which says that $\sin(x) \approx x$ for small $x$.
What about the second order approximation $k = 2$, which is given by:
$$\sum_{k = 0}^2 \frac{f^k (0)}{k!} x^k = \sin(0) + x \cos(0) - \frac{sin(0)}{2!} x^2 = x$$
It turns out that the new term becomes zero. Ok.. fine.. that happens, so what about $k = 3$? Now we see that:
$$\sum_{k = 0}^3 \frac{f^k (0)}{k!} x^k = x - \frac{\cos(0)}{3!} x^3 = x - \frac{x^3}{3!}$$
And let's plot this:
Wow! That's a really good approximation for all $|x| < 1$. And we see a pattern: repeatedly differentiating $\sin(x)$ will end up giving you $\cos(x)$, $- \sin(x)$, $- \cos(x)$, $\sin(x)$ endlessly. Every second term will become zero because of the $\sin(0)$ term, so only odd-numbered terms actually matter. We conclude that:
$$\sin(x) = \sum_{k = 0}^\infty \frac{f^k (0)}{k!} x^k = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$$
Here's the Taylor series approximation for $k = 7$. Check it out:
And here's the one for $k = 9$:
As you add more and more terms, the approximation tends towards the function for larger and larger values of $x$. In the limit, the Taylor series is equal to the original function.
So in other words, a Taylor series is another representation of a function, as an infinite series (a sum of infinitely many terms). A Taylor series approximation is the Taylor series truncated at a finite number of terms, which has the nice property of approximating the function around $x = 0$, and is often easier to calculate and work with, especially in physics where approximations are often used. | {
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There is in fact a theorem that gives the maximal error of the Taylor series approximation at any point $x$ of the function in terms of $k$. Of course, as $k$ tends to infinity, the error tends to zero. This is Taylor's Theorem.
EDIT: uploaded images to imgur for perennity. W|A does strange things to externally linked images.
Last edited:
#### DeusAbscondus
##### Active member
Re: How to use Taylor series to represent any well-behaved $f(x)$
Thanks kindly Bacterius.
But how would this work for a polynomial?
I mean, wouldn't the evaluation $x=0$ lead to "zeroing" all the co-efficients and resulting in non-sense?
If I take the case of $y=x^2+2x$ is this amenable to being described by the same series?
$$\sum_{k=0}^{\infty}\ \frac{f^k (0)}{k!}=0+ 2(0) + \frac{1}{2!}(2\cdot 0) .....$$
This doesn't seem to work! All I get is an infinite string of 0s!
So in other words, a Taylor series is another representation of a function, as an infinite series (a sum of infinitely many terms). A Taylor series approximation is the Taylor series truncated at a finite number of terms, which has the nice property of approximating the function around $x = 0$, and is often easier to calculate and work with, especially in physics where approximations are often used.
There is in fact a theorem that gives the maximal error of the Taylor series approximation at any point $x$ of the function in terms of $k$. Of course, as $k$ tends to infinity, the error tends to zero. This is Taylor's Theorem.
Last edited:
#### Bacterius
##### Well-known member
MHB Math Helper
Re: How to use Taylor series to represent any well-behaved $f(x)$
Taylor series of polynomials have the interesting property that they are in fact equal to the polynomial itself, and all extra terms are zero. In other words, the Taylor series for $x^2 + 2x$ is $x^2 + 2x + 0 + 0 + \cdots$. To see why:
$$f(x) = x^2 + 2x$$
$$f'(x) = 2x + 2$$
$$f''(x) = 2$$
$$f'''(x) = 0$$
.. and any further differentiation still gives zero | {
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So:
$$\sum_{k = 0}^\infty \frac{f^{(k)} (0)}{k!} x^k = \frac{0^2 + 2 \cdot 0}{0!} x^0 + \frac{2 \cdot 0 + 2}{1!} x^1 + \frac{2}{2!} x^2 + 0 + \cdots = 0 + 2x + x^2 + 0 + \cdots = x^2 + 2x$$
So, yes, it still works, although it would seem to be less useful for polynomials than for other functions (but then, polynomials are easy to compute and are already pretty simple. I don't think "reducing" them to infinite series generally helps).
To calculate Taylor series I recommend you write down the iterated derivative of your function $f(x)$ and then plug in the numbers. Remember, $f^{(k)} (0)$ means "the $k$th derivative of $f$ evaluated at $x = 0$". The $x$ in your Taylor series is a different $x$ and is *not* equal to zero.
Also, there is a generalization of Taylor's series which is centered on arbitrarily values of $x$ instead of $x = 0$. I'll let you work out the general expression, though, it's an interesting exercise.
#### DeusAbscondus
##### Active member
Re: How to use Taylor series to represent any well-behaved $f(x)$
That has nailed it for me Bacterius!
Mightily obliged to you.
As per usual, the problem has vanished under the gaze of fresh eyes.
(Going now to chew on this strong meat with a cup of medicinal wine "to aid the digestion")
Cheers,
D'abs
Taylor series of polynomials have the interesting property that they are in fact equal to the polynomial itself, and all extra terms are zero. In other words, the Taylor series for $x^2 + 2x$ is $x^2 + 2x + 0 + 0 + \cdots$. To see why:
$$f(x) = x^2 + 2x$$
$$f'(x) = 2x + 2$$
$$f''(x) = 2$$
$$f'''(x) = 0$$
.. a.
#### DeusAbscondus
##### Active member
Re: How to use Taylor series to represent any well-behaved $f(x)$
The x in your Taylor series is a different x and is *not* equal to zero.
Okay, basically, I'm in the clear .....
but just what is this "different x"?
If it is distinct from the $x$ of my polynomial, by what principle/rule do I distinguish the two when I come to compute the polynomial?
D'abs | {
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D'abs
#### Bacterius
##### Well-known member
MHB Math Helper
Re: How to use Taylor series to represent any well-behaved $f(x)$
Okay, basically, I'm in the clear .....
but just what is this "different x"?
If it is distinct from the $x$ of my polynomial, by what principle/rule do I distinguish the two when I come to compute the polynomial?
D'abs
The $x$ in the Taylor series is the $x$ at which you are evaluating your Taylor series (or your approximation of it). The derivative is always evaluated at 0 for this version of the Taylor series (sorry I agree it was a bit confusing, there is only one $x$, it's just the derivative is evaluated at a constant).
Essentially you are numerically evaluating derivatives of your original function, and using the resulting values as coefficients for your Taylor series (this is not a particularly useful way to think of it but it may be more intuitive to you to understand what is what)
#### DeusAbscondus
##### Active member
Re: How to use Taylor series to represent any well-behaved $f(x)$
That makes sense.
Meanwhile, I've been plugging and chugging a few simple polynomial functions through the T. series: hehe! what fun! it works!!! (*broadly grins)
You made my day: not understanding was getting me down (as usual);
and, again as usual, once some fresh light comes, and the aha! moment arrives, and one feels exuberant rather than dispirited.
Have a great weekend Bacteriu!
[QUOTEEssentially you are numerically evaluating derivatives of your original function, and using the resulting values as coefficients for your Taylor series (this is not a particularly useful way to think of it but it may be more intuitive to you to understand what is what)[/QUOTE] | {
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# Minimum number of states in dfa accepting binary number with decimal equivalent divisible by $n$
I was aware of the fact that, if DFA needs to accept binary string with its decimal equivalent divisible by $n$, then it can have minimum $n$ states.
However recently came across following text:
• If $n$ is power of $2$
Let $n=2^m$, so number of minimum states $=m+1$. For $n=8=2^3$, we need $3+1=4$ states.
• Else If $n$ is odd
Number of states $=n$. For $n=5$, we need $5$ states.
• Else if $n$ is even
Keep dividing it with $2$ until you get odd quotient. The result is final odd quotient plus number of divisions done. For $n=20,20/2=10,10/2=5→5+2=7$ states
I was guessing:
Q1. From where all these facts came. (I know this must have come from general pattern that such DFAs follow, but then I have below doubts)
Q2. Are these points indeed correct?
Q3. If yes (for Q2), then what can be the reason for each point?
Q4. Do these three bullet points cover all cases? (I guess yes, since there is even and odd, but I am new to this, and I was guessing if any such more point is left unmentioned in my textbook, especially since I did not find any reference book (by Peter Linz and Hopcroft-Ullman) talking about this topic) | {
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• There are a few things missing here. First, is the number input LSB to MSB or MSB to LSB? (Looks like LSB to MSB.) Second, are you allowing the empty string to represent zero? Dec 23 '17 at 22:41
• The way to prove such results is to twofold: first you prove an upper bound by constructing a DFA accepting the given language, then you prove a lower bound using Myhill–Nerode theory. Dec 23 '17 at 22:42
• For an example, see this: cs.stackexchange.com/questions/85785/…. Dec 23 '17 at 22:53
• Regarding Q4, yes, they cover all cases. I'm certain you can prove this in your own, even if you're new to "this" – this has nothing to do with automata theory. Dec 23 '17 at 22:55
• Can you provide any link discussing these two points: (1) LSB to MSB and MSB to LSB number input (2) Representing zero by empty string and possibly these two points in the context of "divisible by $n$ DFA" . Also I guess first point wont make number of states in min DFA to change. Second point might change it.
– anir
Dec 24 '17 at 7:02
## Case 1: $n$ is a power of 2
If you wanted to check whether a decimal number is divisible by some power of 10, you can just look at the number of trailing zeros. For example, all numbers that are divisible by $100 = 10^2$ end with 2 zeros (this is of course including numbers ending with more than 2 zeros). The same idea can be applied here for binary numbers and powers of 2. Specifically to check for multiples of $n = 2^m$, you can simply check if the string ends in $m$ zeros. For example, if $n = 16 = 2^4$, all multiples of $n$ will end in 0000 (4 zeros). Thus we can create the following DFA for $n = 16$:
This construction can be extended to all multiples of 2, where you have $m+1$ states to ensure that your input ends in at least $m$ zeros (reading a 0 takes you one state forward, reading a 1 takes you back to the start). As suggested in the comments on your question, you can show that you can't do any better than this using a Myhill–Nerode argument. | {
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## Case 2: $n$ is odd
The general idea here is that when a number is divided by $n$, the possible remainders are $0, 1, \ldots, n-2$ or $n-1$. We can give our DFA one state for each of these possible remainders so as we process a string, we're keeping track of the remainder of what we've read so far and transitioning to the appropriate remainder based on what character we read. Then, we accept if we finish in the remainder 0 state.
Design DFA accepting binary strings divisible by a number 'n' walks through a detailed method for constructing such a DFA, and you can find out more about this by searching for "DFA based division". This gives a total of $n$ states, which is optimal by Myhill-Nerode because these remainders are exactly the equivalence classes given by the relation $\equiv_L$ for the language $L = \{\text{binary strings divisible by }n\}$.
## Case 3: $n$ is even (and not a power of 2)
The technique in case 2 also works here but it isn't optimal. Every $n$ in this case can be expressed as $2^km$ where $m$ is odd (this is the division procedure described in your question). Thus, to check whether a binary string is divisible by $n$, we check whether it is divisible by both $2^k$ and $m$.
We know that it takes $k$ states to check divisibility by $2^k$ (case 1) and $m+1$ states to check divisibility by $m$ (case 2). We can take a DFA for divisibility by $m$, unmark the accepting state and make it the start state for a DFA for divisibility by $2^k$, giving our final DFA $k+m$ states.
As an example, here's the DFA for $n = 6$:
The top three states ensure that the number is a multiple of 3, and the final accept state ensures that it is a multiple of 2. | {
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• Revisiting this problem, I dont get how to combine two DFAs in case 3. You said "unmark the accepting state and make it the start state for a DFA for divisibility by $2^k$". But you didnt explain how you came up with transition outgoing (labeled 1) from final state of last DFA. If we are to combine "divisible by $2^k$ DFA" with "divisible by $m$ ($m$ being any odd number)" DFA, "all" outgoing transitions on reading "1" from "all" states of divisible by $2^k$ DFA will go to same second state of divisible by $m$ DFA, (Q1.) right? [to next comment...]
– anir
Dec 3 '19 at 13:03
• [..from earlier comment] For divisible by $2^k$, we need string to end with $k$ zeroes. But, your case 3, divisible by 6 DFA will also let $k=1$ zeroes appear anywhere. (Q2.) How this fits in divisible by 6 string? For example in your divisible by 6 DFA, for 110110, first 11 moves through divisible by 3 DFA, next 01 moves through divisible by 2 DFA, next 1 moves through divisible by 3 DFA and final 0 moves through divisible by 2 DFA. (Q3.) How running through different component DFAs one after other still ensures that the whole string will still be divisible by $2^km$
– anir
Dec 3 '19 at 13:09 | {
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# Math Help - Circle questions -- Secants, chords and tangents
1. ## Circle questions -- Secants, chords and tangents
Hello again, just got a few questions pertaining to circles. I'll just post one to start with;
"The radius of the earth is approximately 6371 km. If the international space station (ISS) is orbiting 353 km above the earth, find the distance from the ISS to the horizon (x).
So solving this.. according to the segments of secants and tangents theorem...
x2 = (2r + 353)(353)
x2 = (12742 + 353)(353)
x2 = 4622535
x = $\sqrt{4622535}$ = 2150km (approx.)
Did I do that right?
2. ## Re: Circle questions -- Secants, chords and tangents
Originally Posted by StonerPenguin
Hello again, just got a few questions pertaining to circles. I'll just post one to start with;
"The radius of the earth is approximately 6371 km. If the international space station (ISS) is orbiting 353 km above the earth, find the distance from the ISS to the horizon (x).
So solving this.. according to the segments of secants and tangents theorem...
x2 = (2r + 353)(353)
x2 = (12742 + 353)(353)
x2 = 4622535
x = $\sqrt{4622535}$ = 2150km (approx.)
Did I do that right?
Computation looks right to me. But simply memorizing theorems does not promote understanding.
Let's see how we get that theorem. The radius through a point of tangency and the tangent at the same point are perpendicular.
$Let\ r = length\ of\ radius\ of\ given\ circle,$
$u + r = length\ from\ the\ given\ circle's\ center\ to\ a\ given\ point\ outside\ the\ circle,$
$x = length\ of\ the\ line\ that\ is\ tangent\ to\ the\ given\ circle\ and\ runs\ through\ the\ given\ point.$
By the Pythagorean Theorem, we have:
$(u + r)^2 = x^2 + r^2 \implies u^2 + 2ru + r^2 = x^2 + r^2 \implies x^2 = u^2 + 2ru = u(u + 2r) \implies x = \sqrt{u(u + 2r)}.$
3. ## Re: Circle questions -- Secants, chords and tangents | {
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3. ## Re: Circle questions -- Secants, chords and tangents
ISS is6724 km above center of earth.At that point the angle between straight down and the visible horizon has a sin of 6371/6724 or 71.35 deg
cos 71.35 =d/6724
d km to horizon =2150 km
4. ## Re: Circle questions -- Secants, chords and tangents
Hello, StonerPenguin!
The radius of the earth is approximately 6371 km.
If the international space station (ISS) is orbiting 353 km above the earth,
find the distance from the ISS to the horizon (x).
Code:
o
|\
| \
352 | \
| \ x
| \
* * * \
* | *\
* | o
* 6371| * *
| *6371
* | * *
* o *
* *
* *
* *
* *
* * *
Note the right triangle.
The equation is: . $x^2 + 6371^2 \:=\:6723^2$
5. ## Re: Circle questions -- Secants, chords and tangents
Thank you JeffM, bjhopper and Soroban! It's nice to see different perspectives and the image drawn in code is really cool.
Here's another question I've had trouble with:
"Explain how you know $\overline{AB}$ $\overline{CD}$ given E is the center of the circle. (Include theorem numbers.)"
And here's some pertinent theorems;
Theorem 10.4
If one chord is a perpendicular bisector of another chord, then the first chord is a diameter.
Theorem 10.5
If a diameter of a circle is perpendicular to a chord, then the diameter bisects the chord and its arc.
Theorem 10.6
In the same circle, or in congruent circles, two chords are congruent if and only if they are equidistant from the center.
Obviously from the diagram $\overline{AB}$ $\overline{CD}$ by theorem 10.6, but I can't really word this well. Any help? Theorems and proofs are my weakest areas :/
6. ## Re: Circle questions -- Secants, chords and tangents
t is a triangle a is an angle
t AED congruent to BEC isosceles t's same legs and equal altitudes
a DEC = 180- 2* 1 / 2 *AED
a AEB 180-2*1/2 * BEC a AED = aBEC
a AEB = a DEC
AB =DC equal arcs = equal chords | {
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Is this inequality true for all k ? $\sum_{n=k}^{n=+\infty} \frac{1}{n^4} \leq (\sum_{n=k}^{n=+\infty} \frac{1}{n^2})^3$
Can it be generalized for other powers ? Wolfram seems to say it is true for k below 20000.
I stumbled upon it randomly when trying to approximate $\sum_{n=1}^{n=+\infty} \frac{1}{n^4}$.
My reasoning was :
$$\left(\sum_{n=k}^{n=+\infty} \frac{1}{n^2}\right)^2=\sum_{n=k}^{n=+\infty} \frac{1}{n^4} + (\text{double products}) \geq\sum_{n=k}^{n=+\infty} \frac{1}{n^4}$$
So
$$\sum_{n=1}^{n=+\infty} \frac{1}{n^4} \leq \sum_{n=1}^{n=k-1} \frac{1}{n^4}+\left(\sum_{n=k}^{n=+\infty} \frac{1}{n^2}\right)^2 \leq \left(\sum_{n=1}^{n=k-1} \frac{1}{n^4}\right)+\left(\frac{1}{k-\frac{1}{2}}\right)^2$$
where the last inequality comes from An inequality: $1+\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt\frac53$.
Then I noticed that, perhaps, I could raise the last term to the power of 3 instead of just 2, making the inequality stronger.
• The LHS behaves like $\int_{k}^{+\infty}\frac{dx}{x^4}=\frac{1}{3k^3}$ while the RHS behaves like $\left(\int_{k}^{+\infty}\frac{dx}{x^2}\right)^3=\frac{1}{k^3}$, so that is not surprising. – Jack D'Aurizio Apr 21 '16 at 15:36
For $k > 1$, $$\sum_{n=k}^\infty \dfrac{1}{n^4} < \int_{k-1}^\infty \dfrac{dx}{x^4} = \dfrac{1}{3(k-1)^3}$$
$$\left(\sum_{n=k}^\infty \dfrac{1}{n^2}\right)^3 > \left(\int_{k}^\infty \dfrac{dx}{x^2}\right)^3 = \dfrac{1}{k^3}$$
$\dfrac{1}{k^3} > \dfrac{1}{3(k-1)^3}$ when $3(k-1)^3 > k^3$, which is true for $k > 3.2612$.
Let's see what we can say about comparing $s_1(k) =\sum_{n=k}^{\infty} \frac1{n^a}$ with $s_2(k) =\left(\sum_{n=k}^{\infty} \frac1{n^b}\right)^c$ for large enough $k$, where $a > 1$ and $b > 1$ so the sums converge.
Using the integral approximation, $s_1(k) =\sum_{n=k}^{\infty} \frac1{n^a} \approx \int_k^{\infty} \frac{dx}{x^a} =-\frac1{(a-1)x^{a-1}}\big|_k^{\infty} =\frac1{(a-1)k^{a-1}}$. | {
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Therefore $s_2(k) =\left(\sum_{n=k}^{\infty} \frac1{n^b}\right)^c \approx \left(\frac1{(b-1)k^{b-1}}\right)^c =\frac1{(b-1)^c k^{c(b-1)}}$, so $\dfrac{s_1(k)}{s_2(k)} \approx \dfrac{\frac1{(a-1)k^{a-1}}}{\frac1{(b-1)^c k^{c(b-1)}}} = \dfrac{{(b-1)^c k^{c(b-1)}}}{{(a-1)k^{a-1}}} = \dfrac{{(b-1)^c }}{{(a-1)}}k^{c(b-1)-(a-1)}$.
Therefore if $c(b-1) > a-1$, then $s_1(k) > s_2(k)$ for all large enough $k$; if $c(b-1) < a-1$, then $s_1(k) < s_2(k)$ for all large enough $k$;
If $c(b-1) = a-1$, then $\dfrac{s_1(k)}{s_2(k)} \approx \dfrac{{(b-1)^c }}{{(a-1)}}$, so the result depends on this ratio.
For your case of $a=4$ and $b=2$, the key difference is $c(b-1)-(a-1) =c-3$.
If $c > 3$, then $s_1(k) > s_2(k)$ for large enough $k$; if $c < 3$, then $s_1(k) < s_2(k)$ for large enough $k$.
If $c=3$, which is your case, the ratio is $\dfrac{(b-1)^c }{(a-1)} =\dfrac{1}{3} < 1$, so $s_1(k) \approx \frac13 s_2(k) < s_2(k)$ for large enough $k$, which confirm's Robert Israel's result (good thing too, because any result of mine that differs from a result of his is probably wrong). | {
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# For which values of $\alpha$ is {$z_n$} a bounded sequence?
Where $$\alpha$$ is a real constant, consider the sequence {$$z_n$$} defined by $$z_n=\frac{1}{n^\alpha}$$. For which value of $$\alpha$$ is {$$z_n$$} a bounded sequence?
How do I start with this kind of question? I think that $$\forall\space \alpha\in\Bbb{R}_{\geq0}$$ the sequence is convergent and therefore bounded, but how do I write it out?
• In order for us to tell you how to write things out, it would be helpful if you explained why you believe that the answer is what it is. Why do you think that $\frac 1{n^{\alpha}}$ converges for $\alpha \geq 0$? Are you saying that $\frac 1{n^{\alpha}}$ is not bounded when $\alpha < 0$? If so, then you must say so explicitly in your answer. Also, why do you believe that this is the case? – Ben Grossmann Aug 16 '20 at 10:16
• Because it is clear that for $\alpha\geq0$ the sequence converges to 0. If $\alpha<0$ then the value of $\frac{1}{n^\alpha}$ will become very big unless $\alpha>-\frac{1}{n}$. I might be wrong, but this is what I think. I don't know how to approach this question. – Jess Aug 16 '20 at 10:20
• By giving the answer that you have given, you have not only "approached" the problem correctly, but also have given an almost complete answer. It seems that your only question, then, is how to write this up with sufficient "formality." – Ben Grossmann Aug 16 '20 at 10:25
## 2 Answers
If $$\alpha=0$$, $$(z_n)$$ is constant, hence bounded.
If $$\alpha>0$$, $$(z_n)$$ converges to 0 and is thus bounded.
If $$\alpha<0$$, $$(z_n)$$ diverges to $$+\infty$$ and is thus unbounded.
As I state in the comment, you have the correct answer. The only remaining task is to give a formal explanation of the answer. One way to write an answer is as follows: | {
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First, we note that the function $$f: \Bbb [1,\infty) \to \Bbb R$$ defined by $$f(x) = x^{\beta}$$ satisfies $$\lim_{x \to \infty}f(x) = \begin{cases} 0 & \beta < 0\\ 1 & \beta = 0\\ \infty & \beta > 0. \end{cases}$$ I suspect that you do not need to prove this statement formally: it is likely that there is a statement in the textbook that you can refer to.
With that established, address the problem in $$3$$ cases: in the case that $$\alpha < 0$$, conclude using the above fact that $$\lim_{n \to \infty} z_n = \infty$$, which means that the sequence is not bounded. In the case that $$\alpha = 0$$, conclude that $$z_n \to 0$$, which means that the sequence is convergent and is therefore bounded. Similarly, if $$\alpha > 0$$, conclude that $$z_n \to 0$$, which means that the sequence is convergent and therefore bounded.
Thus, we conclude that the sequence is bounded if and only if $$\alpha \geq 0$$.
• If the sequence is bounded, does that mean that the limit of the sequence exists? Because I just noticed that the next question asks me to find the values of $\alpha$ for which lim$_{n\to\infty}z_n$ exists. I assume the values of $\alpha$ are the same as which $z_n$ is a bounded sequence? – Jess Aug 16 '20 at 10:40
• It is not necessarily the case that a bounded sequence will have a limit. However, we can clearly see by exhausting the possibilities that for this problem, the sequence $z_n$ converges whenever it is bounded. – Ben Grossmann Aug 16 '20 at 10:42 | {
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# A fair coin is tossed untill head appears for the first time. What is probability that the number of tosses required is odd? [duplicate]
Q. A fair coin is tossed untill head appears for the first time. What is probability that the number of tosses required is odd?
My work:
suppose that head comes in first toss so probability of getting head in the first toss $$=\dfrac{1}{2}$$
suppose that first & second tosses show tails & third toss shows head so probability of getting head in the third toss $$=(1-\dfrac12)(1-\dfrac12)\dfrac{1}{2}$$ $$=\dfrac1{2^3}$$
suppose that first 4 tosses show tails & fifth toss shows head so probability of getting head in the fifth toss $$=(1-\dfrac12)^4\dfrac{1}{2}$$ $$=\dfrac1{2^5}$$
suppose that first 6 tosses show tails & seventh toss shows head so probability of getting head in the fifth toss $$=(1-\dfrac12)^6\dfrac{1}{2}$$ $$=\dfrac1{2^7}$$
…………….
and so on
But I am not able to find the final probability of getting head first time so that the number of tosses required is odd. what should do I next to it? please help me.
• You're almost there. Just recognize that you have a geometric series and take its sum. To allow you to check your work, the answer will be $\frac 23$. – Robert Shore Jul 3 at 4:06
• This is an exact duplicate of math.stackexchange.com/q/834344/117057 , which unfortunately doesn't have an accepted answer. – shoover Jul 3 at 22:27
You have a geometric series,
$$\frac12+\frac1{2^3}+\frac1{2^5}+\frac1{2^7}+\ldots=\sum_{n\ge 0}\frac12\cdot\left(\frac14\right)^n=\frac{\frac12}{1-\frac14}=\frac23\;.$$
Alternatively, if $$p$$ is the desired probability, then $$p=\frac12+\frac14p$$: with probability $$\frac12$$ you get a head on the first toss, and with probability $$\frac14$$ you start with two tails and are now in exactly the same position that you were in at the beginning. Solving this for $$p$$ again yields $$p=\frac23$$. | {
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• Your second method is very simple. I have never thought of it. Hats off to you sir – user801303 Jul 3 at 4:44
• @PaulAldrin: You’re welcome. – Brian M. Scott Jul 3 at 4:44
• @TShiong: When you can find a recurrence like that, it usually makes things a lot simpler! – Brian M. Scott Jul 3 at 4:45
Your approach is good and will get you the right answer. Just realize that you're building a geometric series and you want its sum.
I'm drafting an answer to give you an alternative approach. Let $$p$$ be the probability you're looking for. Then your first flip will be heads with probability $$0.5$$. If it's tails, then you will solve your original problem (the first heads occurs on an odd toss) exactly when, from your new starting point, your first heads occurs on an even toss, which happens with probability $$1-p$$.
That means $$p = 0.5 + 0.5(1-p) \Rightarrow 1.5 p = 1 \Rightarrow p = \frac 23$$.
• haven't you used sum of infinite GP? – user805532 Jul 3 at 4:38
• No, I think I've provided an alternative approach that avoids the need, just as does the alternative solution provided by Brian M Scott. – Robert Shore Jul 3 at 8:35
Here's another way to solve the problem by considering pairs of tosses at a time instead of single tosses.
Let $$\sigma$$ be an arbitrary infinite sequence of heads and tails. $$\sigma$$ uses 1-based indexing.
$$\text{e.g.}\;\;\; \sigma = HTHTHTHTHTTTTTHHHH\cdots$$
Imagine grouping the elements of $$\sigma$$ into pairs.
$$\sigma = HT,HT,HT,HT,HT,TT,TT,HH,HH\cdots$$
Let's imagine we have three states, $$S$$, $$E$$, and $$O$$.
• $$S$$ is the start state, we haven't seen a head yet.
• $$E$$ is the state marking that we saw a head at an even index first.
• $$O$$ is the state marking that we saw a head at an odd index first.
$$E$$ and $$O$$ are both absorbing states. Once we enter one of those states, we are never going to leave it. | {
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Our state at the beginning of our process is always $$S$$ because, initially, we haven't observed any tosses at all of our coin.
Next let's consider what happens when we read our first toss-pair from $$\sigma$$.
There are twice as many ways to transition from $$S$$ to $$O$$ than there are to transition from $$S$$ to $$E$$.
TT
S -----> S
TH
S -----> E
HT
S -----> O
HH
S -----> O
As the number of pairs processed approaches infinity, the probability that the current state is $$S$$ approaches zero.
However, the ratio of the probability that that the current state is $$O$$ is always twice the probability that the current state is $$E$$.
Therefore, the limiting probability that the state is $$O$$ is $$2/3$$
You are almost done. Add all the terms
$$\frac12+\frac1{2^3}+\frac1{2^5}+\frac1{2^7}+\ldots$$ above series is an infinite GP with first term $$a=\dfrac{1}{2}$$ and common ratio $$r=\dfrac{1}{4}$$ $$\dfrac{a}{1-r}$$
$$=\frac{\frac{1}{2}}{1-\frac{1}{4}}$$$$=\frac23$$
• does this really have infinite terms which can be added? – user805532 Jul 3 at 4:38
• Of course it has uncountable number of terms. I 've added them in my answer – user801303 Jul 3 at 4:40
• it's of course countable – RiaD Jul 3 at 12:58
• @PaulAldrin none of the terms are infinite in value, but they are infinitely many. Name a large odd number n and I can respond with a term $1/2^n$. – neptun Jul 3 at 13:09
• r should be 1/4, not 1/2 – Mark Pattison Jul 3 at 13:32
Another way to avoid summing a GP.
The probability of the first head happening on toss $$n$$ is $$(1/2)^{n}$$. Let $$A$$ be the event that the first head occurs on an odd toss, and $$E_k$$ be the event that it occurs either on toss $$2k+1$$ or $$2k+2$$. Now $$P(A\mid E_k)=\frac{(1/2)^{2k+1}}{(1/2)^{2k+1}+(1/2)^{2k+2}}=\frac{2}{3},$$ independently of $$k$$. Since exactly one of the $$E_k$$ will occur almost surely, we have $$P(A)=\sum_kP(A\mid E_k)P(E_k)=\sum_k\frac23P(E_k)=\frac23\sum_kP(E_k)=\frac23.$$ | {
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# How to average cyclic quantities?
Looking on Internet, I mostly found this definition:
Given quantities on a cyclic domain D, first rescale the domain to [0;2$\pi$[, then, find $z_n$ the point on the unit circle corresponding to the $n$th value, and compute the average by:
$$z_m = \sum_{n=1}^N z_n$$
The average angle is then $\theta_m = \arg z_m$ and to obtain the average value you scale back to your original domain D.
I must say, I have problems with this definition. For simplicity, I will use oriented angles in degree for my examples (i.e. D = [0;360[). With this formula, having angles -90, 90 and 40 will give 40 as mean angle, when I would expect 13.33 as an answer (i.e. (90-90+40)/3).
For my own problems, I usually use:
$$v_m = \mathop{\rm arg\,min}_{v\in D} \sum_{n=1}^{N} d(v_n,v)^2$$
Where $d(x,y)$ is the distance in the cyclic domain, and $\{v_1, v_2, \ldots, v_n\}$ is the set of cyclic data I want to average of.
It has the advantage to work the same way whatever the domain (replace D by a non-cyclic domain and $d$ with the usual euclidean distance, and you find the usual definition of an average). However, it is expensive to compute and I don't know any exact method to do it in general.
So my question is: what is the appropriate way to deal with average of cyclic data? And do you have good pointers that explain the problem and its solutions? | {
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-
en.wikipedia.org/wiki/Circular_mean – Rasmus Dec 16 '10 at 19:35
I know about this page. However, there is no justification for it. Also, I left a comment in the discussion of this page in the hope of understanding. But for now, I still disagree with this method to calculate the average. – PierreBdR Dec 17 '10 at 17:38
The choice of distance metric depends crucially on the application. Bearing data, for instance, might be derived from estimates of X and Y with normally distributed errors, and this leads naturally to the circular mean. For other cases this might not be a good choice. – wnoise Apr 4 '11 at 18:52
Like all averages, the answer depends upon the choice of metric. For a given metric $M$, the average of some angles $a_j \in [-\pi,\pi]$ for $j \in [1,N]$ is that angle $\bar{a}_M$ which minimizes the sum of squared distances $d^2_M(\bar{a}_M,a_j)$. For a weighted mean, one simply includes in the sum the weights $w_j$ (such that $\sum_j w_j = 1$). That is,
$$\bar{a}_M = \mathop{\rm arg\,min}_{x} \sum_{j=1}^{N}\, w_j\, d^2_M(x,a_j)$$
Two common choices of metric are the Frobenius and the Riemann metrics. For the Frobenius metric, a direct formula exists that corresponds to the usual notion of average bearing in circular statistics. See "Means and Averaging in the Group of Rotations", Maher Moakher, SIAM Journal on Matrix Analysis and Applications, Volume 24, Issue 1, 2002, for details. http://lcvmwww.epfl.ch/new/publications/data/articles/63/simaxpaper.pdf
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Thank you, it is exactly the kind of reference I was looking for! – PierreBdR May 9 '11 at 13:54 | {
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Thank you, it is exactly the kind of reference I was looking for! – PierreBdR May 9 '11 at 13:54
The problem with expecting the mean of 90°, −90°, and 40° to be (90°−90°+40°)/3 = 13.33° is that you would then expect the mean of 10° and 350° to be (10°+350°)/2 = 180°, and not 0° which is the more reasonable answer. It only gets worse when you have more than two angles (What is the mean of 340°, 350°, 360°, 10°, and 20°? What about 340°, 350°, 0°, 10°, and 20°?). Essentially, what you're doing there is equivalent to setting $z_n = e^{i\theta_n}$ and computing $$\bar z = (z_1 z_2 \cdots z_N)^{1/N},$$ and the problem is of course that it's not obvious a priori which of the $N$ possible roots of that equation is the right one, if any.
The "circular mean" definition is not so bad. In fact, it corresponds to the point which minimizes the sum of its squared distances to the points corresponding to the data, $$\bar z = \underset{\lvert z \rvert = 1}{\arg\min} \sum_{n=1}^N \lvert z - z_n \rvert^2.$$ So this is almost the same as the formula you like to use; you only have to define the "distance" between angles as the distance between the corresponding points on the unit circle. That is, $d(\theta, \phi) = \sqrt{2 - 2\cos(\theta-\phi)} = 2 \sin(\lvert\theta-\phi\rvert/2)$. This metric is close to $\lvert\theta-\phi\rvert$ when $\theta$ and $\phi$ are close, and has the advantage of being really easy to find the solution to.
-
Of the $N$ possible roots, one will be the global minimum of his distance functions, so checking all of them is sufficient. (See the reference Rob Johnson gave). – wnoise Apr 4 '11 at 19:02 | {
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The angle is supposed to be the independent variable, not the dependent variable. If your function is cyclic (using degrees), z(-90)=z(270) so it doesn't matter which you use. Then the average value of the function is $$z_m=\sum_{n=1}^Nz(\theta_n)$$ You are right that if you average the angles, it matters which lap of the circle you use (-90 vs 270) because the difference of 360 gets divided by N.
-
This is what I am reading on Internet, but I have issues with this definition (as described in my post) and there is never a good explanation on why we should use this also for cyclic data. Could you expand? – PierreBdR Dec 16 '10 at 17:54
The definition of a cyclic function is that f(x+T)=f(x) for some period T. The recommendation you saw was to rescale the units of x so that the period is 2pi radians or 360 degrees. I don't think that is important, but was trying to stay with it. As the function is cyclic, it doesn't matter which cycle you take the data from. So if you want the average value over a cycle it would be (if we take T to be 360) $$\frac{1}{360}\int_a^{a+360}f(x)dx$$ You can then approximate this by a sum just by taking equally spaced steps like $$\frac{1}{n}\sum_{i=0}^{n-1}f(\frac{360i}{n})$$ – Ross Millikan Dec 16 '10 at 22:06
It looks like I am working on a different problem than you. Did you check the Wikipedia page that Rasmus suggested? Is that what you are after? – Ross Millikan Dec 16 '10 at 22:27
For angles, one can adapt the iterative way of computing means to angles, that is:
given angles v[1] .. v[n]
m[1] = v[1]
m[i] = remainder( m[i-1] + remainder( v[i]-m[i-1], C)/i, C) (i=2..n)
where remainder(x,y) gives the signed remainder on dividing x by y and C is the measure of a circle.
I suspect this gives your vm, but haven't been able to prove it. | {
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I suspect this gives your vm, but haven't been able to prove it.
-
A quick check indicates it might not work. Applying this formula to the same set of values, simply taking the elements in various order leads to different results. – PierreBdR Dec 17 '10 at 17:44 | {
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How to derive the identity $\sin x/x=\prod_{n=1}^\infty \cos(x/2^n)$ without using telescoping?
I am wondering how to derive the following equality $$\frac{\sin x}{x}=\prod_{n=1}^\infty \cos(x/2^n)\tag{1}$$ without using the method of telescoping. I know that there is already a question on deriving this infinite product representation of $$\sin x/x$$, but all the answers in the link telescope the product. Here is the method, for completeness.
First, we can use the trigonometric identity $$\sin x=2\cos (x/2)\sin(x/2)$$ to yield $$\cos(x/2)=\frac{\sin x}{2 \sin(x/2)}$$. More generally, this implies that $$\cos(x/2^{n})=\frac{\sin (x/2^{n-1})}{2 \sin(x/2^{n})}$$
Our infinite product is thus $$\prod_{n=1}^\infty\cos(x/2^n)=\frac{\sin (x)}{2 \sin(x/2)}\cdot \frac{\sin (x/2)}{2 \sin(x/4)} \cdot \frac{\sin (x/4)}{2 \sin(x/8)} \cdots$$ Treating the product as a limit of a finite product $$f_k(x)=\prod_{n=1}^k \cos(x/2^n)$$, we notice that $$f_k(x)=\frac{\sin(x)}{2^k\sin(x/2^k)},$$ with $$\lim_{k\to\infty} f_k(x)=\sin x/x$$. Thus, $$\frac{\sin x}{x}=\prod_{n=1}^\infty \cos(x/2^n).$$
Question:
How to show that $$(1)$$ is true without using telescoping?
• Why do you want to avoid telescoping? – Lucas Henrique Jan 2 at 23:33
• Note the Weierstrass factorization theorem:\begin{align*} \frac{\sin x}{x}&=\prod_{n=1}^\infty\left(1-\frac{x^2}{n^2\pi^2}\right)\\ \cos\frac{x}{2^n}&=\prod_{m=0}^\infty\left(1-\frac{x^2}{(2m+1)^2 2^{2(n-1)}\pi^2}\right) \end{align*} – Edward H Jan 2 at 23:44
• You could use the fact that $\sum_{n\ge1}\pm 2^{-n}$ (where the signs are chosen independently and uniformly) is uniformly distributed on $[-1,1]$; the desired equality relates two formulas for the characteristic function of such a random variable. – kimchi lover Jan 2 at 23:50
• @kimchilover very nice – Sandeep Silwal Jan 3 at 0:10
• @Lucas Henrique I am curious to know if there are other ways to evaluate this product. – Zachary Jan 3 at 2:13 | {
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You can do this by a trick that is essentially the same as looking at this product in frequency domain.
To avoid any analytical difficulty, let's examine finite sums; we have $$\prod_{n=1}^{k}\cos(x/2^n)=\prod_{n=1}^k\left(\frac{e^{ix/2^n}+e^{-ix/2^n}}{2}\right)$$ We can expand the sum on the right as $$\frac{1}{2^k}\sum_{\sigma\in\{-1,1\}^k}\exp\left(ix\cdot \left(\sigma_1\cdot \frac{1}2+\sigma_2\cdot \frac{1}{2^2}+\ldots+\sigma_k\cdot \frac{1}{2^k}\right)\right)$$ where $$\sigma$$ is a string of $$k$$ terms in $$\{-1,1\}$$ representing which side of the sum within the former product was followed.
One can see that for $$n=1$$, the angular frequencies encountered (i.e. the coefficient of $$ix$$) are $$1/2$$ and $$-1/2$$ . For $$n=2$$, the frequencies are $$-3/4,\,-1/4,\,1/4,\,3/4$$. We can prove via induction that the possible values of that coefficient are just the set of numbers of the form $$a/2^k$$ for odd integers $$a$$ between $$-2^k$$ and $$2^k$$. Thus, the partial sum works out to: $$\frac{1}{2^k}\cdot \sum_{\substack{a\text{ odd}\\ -2^k < a < 2^k}}\exp\left(ix \cdot \frac{-a}{2^k}\right)$$ We could bail out at this step and recognize that the sum is actually a geometric series (with ratio $$\exp\left(\frac{ix}{2^{k-1}}\right)$$), which would lead us back to the expression you derived for the partial sums. However, we could also recognize this an average of equally spaced evaluations of the function $$z\mapsto \exp(ix\cdot z)$$ over the interval $$[-1,1]$$ with more evaluations as $$k$$ increases; thus, in the limit, this product becomes an integral giving the average value of $$\exp(ixt)$$ over the interval $$[-1,1]$$: $$\lim_{k\rightarrow\infty}\prod_{n=1}^k\cos(x/2^n) = \frac{1}2\int_{-1}^1\exp(ixt)\,dt$$ Of course, this is just integrating an exponential function, which can be done easily, and works out to $$\frac{\sin(x)}x$$.
Here is a different packaging of the same basic argument presented in Milo Brandt's answer. | {
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Here is a different packaging of the same basic argument presented in Milo Brandt's answer.
Let $$X_k=\sum_{j=1}^k \sigma_j 2^{-j}$$, where the $$\sigma_i$$ are iid $$\pm1$$ random variables. This has uniform distribution on the $$2^k$$ points uniformly spaced $$2^{1-k}$$ apart in the range from $$-1+2^{-k}$$ to $$1-2^{-k}$$, as can be seen from the binary expansions of the integers from $$0$$ to $$2^k$$. One can verify directly that $$X_k$$ converges in distribution to the continuous uniform distribution on $$[-1,1]$$.
The characteristic function of $$X_k$$, namely the function $$\phi_k(t)=E[\exp(itX_k)]$$ is given by $$\prod_{j=1}^k E[\exp(i \sigma_j 2^{-j})] = \prod_{j=1}^k \cos(t2^{-j})$$.
By Lévy's continuity theorem, for each $$t$$, one has $$\lim_{k\to\infty}\varphi_k(t)=\varphi(t)$$, where $$\varphi(t)$$ is the characteristic function of the uniform distribution on $$[-1,1]$$, which is $$\varphi(t)=\frac 12\int_{-1}^1 \exp(itx)\,dx = \frac{\sin(t)}t.$$ | {
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# Number of times $2^k$ appears in factorial
For what $n$ does: $2^n | 19!18!...1!$?
I checked how many times $2^1$ appears:
It appears in, $2!, 3!, 4!... 19!$ meaning, $2^{18}$
I checked how many times $2^2 = 4$ appears:
It appears in, $4!, 5!, 6!, ..., 19!$ meaning, $4^{16} = 2^{32}$
I checked how many times $2^3 = 8$ appears:
It appears in, $8!, 9!, ..., 19!$ meaning, $8^{12} = 2^{36}$
I checked how many times $2^{4} = 16$ appears:
It appears in, $16!, 17!, 18!, 19!$ meaning, $16^{4} = 2^{16}$
In all,
$$2^{18} \cdot 2^{32} \cdot 2^{36} \cdot 2^{16} = 2^{102}$$
But that is the wrong answer, its supposed to be $2^{150}$?
• Note that, for example, $6!$ contributes 4 factors of 2 - one from 2, one from 6 and two from 4. You only count 3 of these. – Wojowu Apr 24 '15 at 16:37
A simple trick to compute $k$ such that $2^k|n!$ is to compute $\sum_{i=1}^\infty \left\lfloor\frac{n}{2^i}\right\rfloor$, this is because $n$ has $[n/2]$ numbers divided by $2$, if we pick out these numbers and find out that there're $[n/4]$ numbers divided by $4$.. If we continue this procedure, we see that $$k=1\cdot\left(\left\lfloor\frac{n}{2}\right\rfloor-\left\lfloor\frac{n}{4}\right\rfloor\right)+2\left(\left\lfloor\frac{n}{4}\right\rfloor-\left\lfloor\frac{n}{8}\right\rfloor\right)+\ldots=\sum_{i=1}^\infty \left\lfloor\frac{n}{2^i}\right\rfloor$$. In this case, we have to sum $$0+1+1+3+3+4+4+7+7+8+8+10+10+11+11+15+15+16+16=150.$$ Your fault is that your did not count the contribution of those which is not the power of $2$. For instance, there's $14$ in $14!$..
How many times $2$ divides the product $\prod_{i=1}^{19}i!$ ?
Let's call each term inside a factorial $i$. That way, $i = 1$ occurs in 19 factorials, $i = 2$ occurs in 18 factorials, and $i = 3$ occurs in 17 factorials etc. | {
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$i = 2$ occurs 18 times. $1 \times 18 = 18$
$i = 4$ occurs 16 times. $2 \times 16 = 32$
$i = 6$ occurs 14 times. $1 \times 14 = 14$
$i = 8$ occurs 12 times. $3 \times 12 = 36$
$i = 10$ occurs 10 times. $1 \times 10 = 10$
$i = 12$ occurs 8 times. $2 \times 8 = 16$
$i = 14$ occurs 6 times. $1 \times 6 = 6$
$i = 16$ occurs 4 times. $4 \times 4 = 16$
$i = 18$ occurs 2 times. $1 \times 2 = 2$
$$18 + 32 + 14 + 36 + 10 + 16 + 6 + 16 + 2 = 150$$
It might be more helpful to do this recursively.
Let $T(n) = \prod_{k=1}^n k!$.
We will use the notation: $2^{r} \| m$ to mean that $2^r$ is the largest power of $2$ that divides $m$.
Then we have $2 \| 2! = T(2)$. We also know that $2 \| 3!$, so $2^2 \| T(3) = 3! T(2)$. Continuing:
$$2^3 \| 4!$$ $$2^3 \| 5!$$ $$2^4 \| 6!$$ $$2^4 \| 7!$$ $$2^7 \| 8!$$ $$2^7 \| 9!$$ $$2^8 \| 10!$$ $$2^8 \| 11!$$ $$2^{10} \| 12!$$ $$2^{10} \| 13!$$ $$2^{11} \| 14!$$ $$2^{11} \| 15!$$ $$2^{15} \| 16!$$ $$2^{15} \| 17!$$ $$2^{16} \| 18!$$ $$2^{16} \| 19!$$
If we take the sum of all of those powers, $$2\cdot 1 + 2\cdot 3 + 2\cdot 4 + 2\cdot 7 + 2\cdot8 + 2 \cdot 10 + 2 \cdot 11 + 2 \cdot 15 + 2 \cdot 16$$
$$=2(1+3+4+7+8+10+11+15+16) = 2(75) = 150.$$ | {
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1. Aug 6, 2008
### laura_a
1. The problem statement, all variables and given/known data
I am working on a topic called differential equations, and Im stuck on some working out.
2. Relevant equations[/b]
I have to integrate it, and I know I can without factorising, but it is so messy, my professor said to factorise the fraction so it looks a bit like a1/(bu+ c)+ a2/(du+ e)
Well thats what the professor said, I have no idea what it means... I can factorise as far as
= u/(-u^2 +3u +2) - 2/(-u^2 +3u +2)
And I know that -u^2 +3u +2 = (-u + 1)(u-2) +4
But not sure how to put it all together
Here is the working out for the whole question just in case you're interested
\begin{align*} y' &= \frac{y+2x}{y-2x} \\ Let y &= ux \\ \text{Then we have} y' &= \frac{ux + 2x}{ux - 2x} \\ &= \frac{u+2}{u-2} \\ \text{Now } y' &= \frac{dy}{dx} = \frac{d(ux)}{dx} = x \frac{dy}{dx} + u \\ \text{So we can say that} x \frac{du}{dx} + u& = \frac{u+2}{u-2} \\ x \frac{du}{dx} &= \frac{u+2}{u-2} - u \\ x \frac{du}{dx} &= \frac{u+2}{u-2} - \frac{u^2-2u}{u-2} \\ x \frac{du}{dx} &= \frac{u+2- u^2+2u}{u-2} \\ x \frac{du}{dx} &= \frac{-u^2+3u+2}{u-2} \\ \end{align*} \bigskip
Then I have to integrate both sides of this equation...which is what I think I need to factorise in order to make it nice and neat
$$\frac{u-2}{-u^2 + 3u + 2}du &= \frac{1}{x} dx$$
Last edited: Aug 6, 2008
2. Aug 6, 2008
### HallsofIvy
Staff Emeritus
That can't be factored using integer coefficients but you know that if an expression factors as (x-a)(x-b) then a and b must be roots of (x-a)(x-b)= 0. Use the quadratic formula to find the roots of x2- 3x- 2= 0.
3. Aug 6, 2008
### Defennder
Try completing the square of $$-u^2+3u+2$$. Then use a simple algebraic identity to decompose it into partial fractions.
4. Aug 7, 2008
### laura_a
I think i've gotten a little closer, The question I'm trying to do is to solve a d.e. using change of variables, so once I get this part worked out I should be able to get it | {
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So this is how far I've gone
Since $$u^2- 3 u -2= 0$$
has two roots
$$u_1 = (3+ 17^{0.5})/2, u_2= (3- 17^{0.5})/2$$
so
$$u^2- 3 u -2= (u - (3+ 17^{0.5})/2 ) (u- (3- 17^{0.5})/2 )$$
$$\frac{u-2}{-u+3u+2}du = \frac{1}{2}dx$$ - I'll call this equation (2)
So now I've got the LHS as
$$S = -u /((u - (3+ 17^{0.5})/2 ) +2 /(u- (3- 17^{0.5})/2 )$$
I changed the u-2 to -u+2 since I made a similar change on the denominator in order to solve it...
Anyhow, I integrated that and ended up with something really messy invloving logs
$$-2(u + 17^{0.5} ln(u-17^{0.5}-3)+3ln(u-17^{0.5}-3)-17^{0.5}-3+4ln(u+17^{0.5}-3) = ln(x) + C$$
(I integrated both sides of equation (2) above)
Well the prob is now, I'm probably wrong anyway, but the next step is to solve for u, and then I have to sub back in the original change of varaible which was y=xu and end up with an expression for y in terms of x and C... before I go on, can anyone tell me if I should bother working through those logs, or is it wrong? Thanks :)
5. Aug 8, 2008
### arildno
Let us take this as a GENERAL case, shal we?
We are to decompose:
$$\frac{u-u_{0}}{(u-u_{1})(u-u_{2})}=\frac{A}{u-u_{1}}+\frac{B}{u-u_{2}}$$
where the unindexed u is our variable, the indexed u's known numbers, and A and B constants to be determined.
We therefore must have:
$$u-u_{0}=A(u-u_{2})+B(u-u_{1})$$, or by comparing coefficients, we get the system of equations:
$$A+B=1$$
and
$$u_{2}A+u_{1}B=u_{0}$$
whereby we arrive at the solutions:
$$A=\frac{u_{0}-u_{1}}{u_{2}-u_{1}},B=\frac{u_{2}-u_{0}}{u_{2}-u_{1}}$$
This is much simpler than using specific numbers! | {
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# Epsilon-delta definition of limit : set an upper bound to delta?
Am I allowed to set an upper bound to delta in the epsilon-delta definition of limit? Why would it still be equivalent to the original definition ? For example, if the function is defined for all $x$ in $\Bbb R$ :
$f : D(f) = \Bbb R \rightarrow \Bbb R$.
Let $\epsilon > 0, \exists \delta > 0 \text{ s.t. } \forall x : 0 < |x-x_o| < \delta < 3 \Rightarrow |f(x) - l| < \epsilon$
$\Leftarrow\text{?}\Rightarrow$
Let $\epsilon > 0, \exists \delta > 0 \text{ s.t. } \forall x : 0 < |x-x_o| < \delta \Rightarrow |f(x) - l| < \epsilon$.
• Lets suppose you found an $\; \delta \;$ then the definition still holds for any $\delta^* \le \delta$. Hence, you can choose your $\; \delta^* \le 3$ – XPenguen Dec 26 '15 at 20:16
• If you find an epsilon which beats a small delta, you've found an epsilon which beats a big one. – Mark Bennet Dec 26 '15 at 20:16
$$\forall\varepsilon > 0\ \exists \delta > 0\ \forall x \Big( 0 < |x-x_o| < \delta < 3 \Rightarrow |f(x) - \ell| < \varepsilon\Big)$$
I would phrase this differently:
$$\forall\varepsilon > 0\ \exists \delta \in(0,3)\ \forall x \Big( 0 < |x-x_o| < \delta \Rightarrow |f(x) - \ell| < \varepsilon\Big)$$
Now the question is whether that is equivalent to this:
$$\forall\varepsilon > 0\ \exists \delta>0\ \forall x \Big( 0 < |x-x_o| < \delta \Rightarrow |f(x) - \ell| < \varepsilon\Big)$$
If it is true that $\forall\varepsilon>0\ \exists\delta\in(0,3)\ \cdots\cdots$ then it is true that $\forall\varepsilon>0\ \exists\delta>0\ \cdots\cdots$, simply because every number in $(0,3)$ is $>0$. | {
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Now the question is whether the converse holds. If it is true that $\forall\varepsilon>0\ \exists\delta>0\ \cdots\cdots$, does it necessarily follow that $\forall\varepsilon>0\ \exists\delta\in(0,3)\ \cdots\cdots\,{}$? Here the answer in general is “no”. I.e. there are some things you could put in place of $\text{“}\cdots\cdots\text{''}$ for which the first statement would be true and the second false. However, in the definition of "limit" the thing in place of $\text{“}\cdots\cdots\text{''}$ is “if a certain thing is $<\delta$, then a certain thing follows.” Given a certain $\varepsilon>0$ suppose we know there exists $\delta>0$ such that if a certain thing is less than $\delta$ then a certain conclusion follows. If the given value of $\delta$ is small enough, then the minimum of that value of $\delta$ and $2.9$ is small enough, because everything less than $\min\{\delta,2.9\}$ is less than $\delta$. So when the statement in place of $\text{“}\cdots\cdots\text{''}$ has the form “if a certain thing is $<\delta$, then a certain thing follows.”, then it is true that if $\forall\varepsilon>0\ \exists\delta>0\ \cdots\cdots$ then $\forall\varepsilon>0\ \exists\delta\in(0,3)\ \cdots\cdots$.
Yes, of course. If the first condition is satisfied, then clearly the second is too. Conversely, if the second condition is satisfied, then one can replace $\delta$ by $\min(\delta, 3)$ and so the first condition is satisfied.
Yes, as you have to exhibit an $\delta$ that works for every $\epsilon$, then you can be as artificial as you want if that works for you. Always in existence proofs the same rule applies, construct all you need and show that it does work.
Yes you can. The idea is that some proposition $P(x,\epsilon)$ is true for all $x$ that are sufficiently close to $x_0$. So if it holds for all $x\in (-d+x_0,d+x_0)$ then it holds for all $x\in (-d'+x_0,d'+x_0)$ whenever $0<d'<d$. Sometimes it is useful to know how large $d$ can be, for a given $\epsilon.$ | {
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# Calculus : Study Guide for AP Calculus AB/BC or University Calculus
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1.25 MB | 300 pages
### PRODUCT DESCRIPTION
Calculus Study Guide suitable for AP Calculus or First Year College Calculus Students. Main topics covered : Functions Limits and Continuity, Derivatives, Application of the Derivative, Indefinite integration, Definite Integration, Techniques of Integration, Sequences and Series, L’Hopitals Rule, Improper Integrals, Taylor Series, Short Introduction to Differential Equations.
Calculus without Limits is a self-study guide that can be used as a textbook/lecture supplement by students, as a source of homework problems by the instructor or as a source of lecture material. Covers most topics discussed in a first year calculus course (or AP Calculus). The emphasis is on solved problems that show students how to attack the types of problems they will encounter in homework and on exams. Discussion of concepts is informal followed by example problems. Many practice problems are also included for students. | {
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1 - Function Review
What is a function - Graphing - Even and Odd Functions - Increasing and Decreasing Functions - One to One Functions - Powers and Polynomials - Absolute Value Function - Exponential Function - Logarithm - Trigonometric Functions
2- Limits
Definition - Tricks for Doing Limits - Properties of Limits and Rules - Limits at Infinity - One Sided Limits Continuity
3- Derivative
Notation and Definition in terms of limits - Slope of Tangent Line - Power Rule and Derivative of a constant - Chain Rule - Derivatives of Trig Functions - Product and Quotient Rules
4- Applications of the Derivative
Derivative of Inverse Function - Implicit Differentiation - Application: Rates of Change, Velocity and Acceleration - Finding Minima and Maxima First and Second Derivative Test - Inflection Points - The Mean Value Theorem - Related Rates
5- The Integral
Notation - Power Rule of Integration - Integration of Trig Functions - Integration of inverse function - Integrating Exponential Function - The substitution technique - Integrating Hyperbolic Functions - Definite Integration - Sums and Area Under a Curve - Area Between Two Curves -
6- U-Substitution
Solved Problems
7- Integration by Parts
Definition and When to Use - Examples
8- Integration of Rational Functions
Introduction - Examples
9- Integration Using the Method of Partial Fractions
Discussion - Examples
10- Integrating Powers of Trig Functions
Examples
11- Trig Substitution
Three Cases - Basic Technique -Examples
12 - Improper Integrals
Definition - Examples
13 - Sequences and Infinite Series
Sequences - Limits - Infinite Series
14 Ratio, Comparison, and Integral Tests
Examples
15 - Alternating Series
16 - Power Series and Radius of Convergence
17- Taylor Series Expansions
18- L’Hopital’s Rule
When to use and Definition - Examples
19- Introduction to Differential Equations
First Order Differential Equations - Second Order Differential Equations
Total Pages
300
N/A
Teaching Duration
1 Year | {
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