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Here's a trick proof. Let $$S$$ denote the sum in question. As the reciprocals of the $$x_k$$ are the $$x_k$$ again (in a different order) then $$S=\sum_{k=1}^n\frac1{x_k^{-1}-1}=\sum_{k=1}^n\frac{x_k}{1-x_k}$$ and $$2S=\sum_{k=1}^n\left(\frac1{x_k-1}+\frac{x_k}{1-x_k}\right) =\sum_{k=1}^n(-1)=-n.$$
Applying the same technique as per this answer $$Q(x)=\prod\limits_{k=1}^n(x-x_k) \text{ and } Q'(x)=\sum\limits_{i=1}^{n}\prod\limits_{k=1,k\ne i}^{n}\left(x-x_k\right)$$ then $$\frac{Q'(x)}{Q(x)}=\sum\limits_{i=1}^{n}\frac{1}{x-x_i}$$ or $$-\frac{Q'(1)}{Q(1)}=\sum\limits_{i=1}^{n}\frac{1}{x_i-1}$$ But $$Q(1)=n+1$$ and $$Q'(x)=1+2x+3x^3+...+nx^{n-1}\Rightarrow Q'(1)=\frac{n(n+1)}{2}$$ and $$\sum\limits_{i=1}^{n}\frac{1}{x_i-1}=-\frac{n}{2}$$
We have $$x^{n+1}-1=0\ \ \ \ (1)$$ with $$x\ne1$$
Set $$\dfrac1{x-1}=y\iff x=\dfrac{y+1}y$$
Replace the value of $$x$$ in terms of $$y$$ in $$(1)$$ to form an $$n$$ degree equation in $$y$$ $$\left(\dfrac{y+1}y\right)^{n+1}=1$$
$$\implies\binom{n+1}1 y^n+\binom{n+1}2 y^{n-1}+\cdots+1=0$$
Now apply Vieta's formula to find $$\sum_{r=1}^n\dfrac1{x_r-1}=\sum_{r=1}^ny_r=-\dfrac{\binom{n+1}2}{\binom{n+1}1}=?$$
Simply note that $$1,x_1,\ldots, x_n$$ are roots of $$x^{n+1} -1 = 0.$$ Thus we have $$0,x_1-1,\ldots, x_n-1$$ are roots of $$(x+1)^{n+1} -1 = x\sum_{j=0}^{n} \binom{n+1}{j+1}x^j = 0.$$ Hence, $$z_i = x_i -1$$, $$1\leq i \leq n$$ are roots of $$\sum_{j=0}^{n} \binom{n+1}{j+1}z^j = 0,$$ and by Vieta's formula, $$\sum_{i=1}^n \frac{1}{z_i} = \frac{\sum_{i=1}^n z_1z_2\cdots z_{i-1}\widehat{z_i}z_{i+1}\cdots z_n}{z_1z_2\cdots z_{n-1}z_n} = -\frac{\binom{n+1}{2}}{\binom{n+1}{1}} = -\frac{n}{2},$$ where $$\widehat{z_i}$$ means the variable is omitted in the calculation.
First, the denominator is, I think, $$(-1)^nQ(1)$$. Here is a hint for the numerator: how can you write $$Q’(x)$$? | {
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# Prime square offsets: Why is +7 more frequent than -7?
For a prime $p$, define $\delta(p)$ to be the smallest offset $d$ from which $p$ differs from a square: $p = r^2 \pm d$, for $d,r \in \mathbb{N}$. For example, \begin{eqnarray} \delta(151) & = & +7 \;:\; 151 = 12^2 + 7 \\ \delta(191) & = & -5 \;:\; 191 = 14^2 - 5 \\ \delta(2748971) & = & +7 \;:\; 2748971= 1658^2 + 7 \end{eqnarray} For a particular $\delta=d$ value, define $\Delta(n,d)$ to be the number of primes $p$ at most $n$ with $\delta(p) = +d$, minus the number with $\delta(p) = -d$. In other words, $\Delta$ records the cumulative prevalence of $+d$ offsets over $-d$. For example, $\Delta(139,5)=-2$ because there are two more $-5$'s than $+5$'s up to $n=139$: $$\delta(31)=-5 \;,\; \delta(41)=+5 \;,\; \delta(59)=-5\;,\; \delta(139) =-5 \;.$$ The figure below shows $\Delta(p,5)$ and $\Delta(p,7)$ out to the $200000$-th prime $2750159$. The offset $+7$ occurs $161$ times more than the offset $-7$, and the reverse occurs for $|\delta|=5$: $-5$ is more common than $+5$.
Q. Is there a simple explanation for the different behaviors of offsets $5$ and $7$?
Obviously the question can be generalized to explaining the growth for any $|\delta|$.
I previously asked a version of this question on MSE, using somewhat different notation conventions and with less focused questions. | {
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• The curve you plotted looks like a parabola. I don't know whether it can be explained though. – Sylvain JULIEN Apr 27 '17 at 11:12
• The graph is a bit misleading, making the offsets seem small. In reality they are huge, once one keeps in mind that there are only about 1600 squares up to $2.7\times 10^6$ and and thus primes within $\pm 7$ (or $\pm 5$) of a square are only a smallish fraction of 3200 in number. I would be surprised if simple modular considerations don't explain the offsets. – Yaakov Baruch Apr 27 '17 at 11:34
• And yes, in that case each curve would be close to $C \sqrt{n}/\log{n}$, so quite similar to a half parabola (and possibly with equal but opposite $C$ for the 2 cases). – Yaakov Baruch Apr 27 '17 at 11:45
• This might have something to with with Chebyshev’s bias: note that $+7\equiv-5\equiv3\pmod4$. – Emil Jeřábek Apr 27 '17 at 12:04
• The question title and text does not agree. Edit one of them. – Stig Hemmer Apr 28 '17 at 7:21
Consider $n^2+7$ and $n^2-7$ modulo $3$. If these are to be prime they must be non-zero $\pmod 3$, and in the first case $n$ can be anything mod $3$, whereas in the second case $n$ must be $0 \pmod 3$. If you consider $n^2+5$ and $n^2-5$, you see that the pattern reverses. This is already a huge bias for one offset to be preferred over the other.
The Hardy-Littlewood conjectures make this precise. One expects that (for a number $k$ not minus a square) the number of primes of the form $n^2+k$ with $n\le N$ (say) is $$\sim \frac 12 \prod_{p\ge 3} \Big(1 -\frac{(\frac{-k}{p})}{p-1} \Big) \frac{N}{\log N},$$ where in the numerator of the product above is the Legendre symbol. The constants in front of the $N/\log N$ explain these biases.
You don't have to worry about a smaller offset than $\pm 5$ or $\pm 7$, since an application of the sieve shows that the numbers $n^2+a$ and $n^2+b$ (for fixed $a$, $b$) are both prime only $\le CN/(\log N)^2$ of the time for a constant $C$. | {
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• Lucia of course knows this, but I thought it was worth mentioning that this product is well-approximated by the inverse special value $L(1,\chi)^{-1}$ of the $L$-function associated to the unique Dirichlet character $\chi$ mod $4k$ such that $\chi(p) = \left( \frac{-k}{p} \right)$. Known bounds on this special value thus gives bounds on how large and small this constant can be. – Will Sawin Apr 27 '17 at 17:45
Modulo 6 the squares are 0,1,4,3,4,1 and the squares+7 (or -5) can only be 1,2,5,4,5,2, of which 3/6 can at all be prime. The squares-7 (or +5) are 5,0,3,2,3,0 of which only 1/6 can be prime. Obviously this not a proof, but there is clearly no first order surprise in the observed offsets.
UPDATE I checked offsets modulo different numbers (such as (mod 2*3*5*7*11*13*17*...)) and I found that mod(3*7*11*19*23) we get 64638/25806 for +7/-7 and 21945/73899 for +5/-5, while primes 2, 5, 13, 17, 29 don't affect the ratio. I imagine that exploring more primes will further refine the ratios, with primes $\equiv 3$ (mod 4) affecting them and others not. WHAT IS GOING ON HERE?
• I see now, thanks to Lucia's answer, that the ratios between the asymptotics there for, say, $k=\pm7$ are affected only by primes $p$ such that $$\Big(\frac{-7}{p}\Big)\neq \Big(\frac{7}{p}\Big)$$ which are the primes $\equiv 3$ (mod 4). – Yaakov Baruch Apr 27 '17 at 16:57 | {
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# Prove that $\sup(A+B) = \sup(A) + \sup(B)$ and why does $\sup(A+B)$ exist?
We want to show that $\sup(A)+\sup(B)$ is the least upper bound of the set $A + B$. First, we need to show that $\sup(A) + \sup(B)$ is an upper bound for the set $A + B$. Indeed, if $z\in A + B$, then there exists $a\in A$ and $b\in B$ such that $z = a + b$. But by definition of $\sup(A)$ and $\sup(B)$, $a \leq \sup(A)$ and $b \leq \sup(B)$, so $z = a + b \leq \sup(A) + \sup(B)$. So, $\sup(A) + \sup(B)$ is an upper bound for $A + B$.
We now wish to show that $\sup(A) + \sup(B)$ is the least upper bound for the set $A + B$. So, if $u$ is an upper bound for $A + B$, we need to show that $\sup(A)+\sup(B) \leq u$. We will use part (i): that is, we need to show that there exists $\varepsilon > 0$, $\sup(A) + \sup(B) < u + \varepsilon$.
To do this, note that since $\sup(A)$ is the least upper bound for $A$, $\sup(A) - \varepsilon/2$ is not an upper bound for A, so there exists an $a\in A$ so that $\sup(A) - \varepsilon/2 < a$. Similarly, there is a $b\in B$ so that $\sup(B) - \varepsilon/2 < b$.
Adding these two inequalities gives $$\sup(A) + \sup(B) - \varepsilon < a + b;$$ in other words $$\sup(A) + \sup(B) < a + b + \varepsilon.$$ But $u$ is an upper bound for the set $A + B$, so $a + b \leq u$, and hence we have $$\sup(A) + \sup(B) < u + \varepsilon.$$ Thus, by part (i), $\sup(A) + \sup(B) \leq u$, so $\sup(A) + \sup(B)$ is the least upper bound for $A + B$, as required.
Now how do I show that $\sup(A+B)$ exists?
• For some basic information about writing math at this site see e.g. here, here, here and here. – Cortizol Sep 15 '13 at 16:26
• – Martin Sleziak Mar 3 '14 at 15:31
## 2 Answers
Show $A+B$ is nonempty and bounded above. | {
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## 2 Answers
Show $A+B$ is nonempty and bounded above.
• i)A + B is non-empty since a + b E A + B. ii) If A and B are bounded, then so is A + B because by assumption, there are positive numbers MsubA and MsubB with |a| <= MsubA and |b| <= MsubB for all aEA and bEB. Hence |a + b| <= |a| + |b| <= MsubA + MsubB for all a and b. Is that it? – user87274 Sep 15 '13 at 16:49
• @mespebjidom Yes. – Pedro Tamaroff Sep 15 '13 at 16:56
Assume $A,B$ are both nonempty. Otherwise you can run into some $\infty-\infty$ strangeness.
You have $a \le \sup A$ for all $a \in A$, and similarly $b \le \sup B$ for all $b \in B$. Hence we have $a+b \le \sup A + \sup B$ for all $a \in A, b \in B$. Hence we have $\sup (A+B) = \sup_{a \in A, b \in B} a+b \le \sup A + \sup B$.
Now let $\epsilon >0$, and choose $a \in A, b \in B$ such that $a > \sup A -\frac{\epsilon}{2}$ and $b > \sup B -\frac{\epsilon}{2}$. Then $\sup(A+B) \ge a+b > \sup A + \sup B - \epsilon$. Since $\epsilon>0$ was arbitrary, we are have the desired result. | {
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• What's this for? I just wanted to know how I can show that sup(A + B) exists in simple terms. – user87274 Sep 15 '13 at 18:51
• Your question reads 'Prove $\sup A + \sup B = \sup (A+B)$. The above shows, in simple terms, that $a+b \le \sup A + \sup B$ for all $a \in A, b \in B$, from which it follows that $\sup (A+B) \le \sup A + \sup B$. I'm not sure what you mean by 'exists', as the $\sup$ always exists (it may be $\pm \infty$, by convention $-\infty$ if the set is empty). – copper.hat Sep 15 '13 at 18:58
• I already proved that sup(A+B) = sup(A) + sup(B). What I wanted to know was that if A is bounded and has an element a in A, and B is bounded and has an element b in B, then does this mean that sup(A+B) exits? As the above poster said, for sup(A+B) to exist we have to show that A + B is non-empty and bounded. That's all. – user87274 Sep 15 '13 at 19:30
• I'm missing something here. The set needs to be bounded above, not necessarily bounded. If a non-empty set is bounded above, then it has a finite $\sup$. The set $A+B$ is clearly non-empty if both $A,B$ are non-empty. And you have shown that the sum of the $\sup$s is equal to the $\sup$ of the sum, which is presumed finite. – copper.hat Sep 15 '13 at 19:42
• @Evinda: Actually, I wrote my answers :-). One definition of $\sup A$ is that it is an upper bound and for all $\epsilon>0$, there is some $a \in A$ such that $a > \sup A- \epsilon$. What definition are you using? – copper.hat Oct 5 '18 at 21:38 | {
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# The Strange Behaviour of Numbers Close to Unity
#### Perplexed
##### New member
I have been looking at material properties such as thermal expansion of metals which usually involves very small coefficients. The general equation of thermal expansion is usually
$$\displaystyle L_\theta = L_0 ( 1 + \alpha \theta)$$
where L is the length and theta is the temperature change. The coefficient alpha is usually pretty small, 11E-6 for steel, so one ends up with a lot of numbers like 1.000011.
This is where I seem to have entered a strange world, where
$$\displaystyle \sqrt{(1 + x)} \rightarrow 1 + x/2$$
$$\displaystyle \dfrac{1}{ \sqrt{(1 - x)}} \rightarrow 1 + x/2$$
$$\displaystyle (1 - x)^3 \rightarrow 1-3x$$
Is there a name for this area of maths, and somewhere I can look up more about it?
Thanks for any help.
Perplexed
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
Taylor series is one topic where (Infinite) polynomials are used to approximating functions. For example,
$(1+x)^{1/2}=1+\frac{x}{2}+R_1(x)$
where $R_1(x)$ is called the remainder and is infinitely small compared to $x$ when $x$ is small. More precisely,
$(1+x)^{\alpha }=1+\alpha x+{\frac {\alpha (\alpha -1)}{2!}}x^{2}+\cdots+ \frac{\alpha\cdot\ldots\cdot(\alpha-n+1)}{n!}x^n+R_n(x)$
where $R_n(x)$ is infinitely small compared to $x^n$ when $x$ tends to $0$.
#### HallsofIvy
##### Well-known member
MHB Math Helper
"Linear approximation". Any function, f, having a derivative at x= a, can be approximated by the "tangent line" $$y= f'(a)(x- a)+ f(a)$$. The error will be proportional to $$(x- a)^2$$ and f''(a).
For example, if $$f(x)= \sqrt{1+ x}= (1+ x)^{1/2}$$ then $$f'(x)= (1/2)(1+ x)^{-1/2}$$ so that with x= 0, $$f(0)= \sqrt{1+ 0}= 1$$ and $$f'(0)= (1/2)/\sqrt{1+ 0}= 1/2$$. So y= f(x) is approximated, around x= 0, by $$y= (1/2)x+ 1$$ or $$1+ x/2$$. | {
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If $$f(x)= \frac{1}{\sqrt{1+ x}}= (1+ x)^{-1/2}$$ then $$f'(x)= -(1/2)(1+ x)^{-3/2}$$ so that $$f(0)= \frac{1}{\sqrt{1+ 0}}= 1$$ and then $$f'(0)= -(1/2)(1+ 0)^{3/2}= -1/2. So y= f(x) is approximated, around x= 0, by [tex]y= -(1/2)x+1$$ or $$1- x/2$$. Notice the negative sign- what you have is NOT correct.
If $$f(x)= (1- x)^3$$ then $$f'(x)= 3(1- x)^2(-1)= -3(1- x)^2$$. $$f(0)= (1- 0)^3= 1$$ and $$f'(0)= -3(1- 0)^2= -3$$. So y= f(x) is approximated by -3x+ 1 or 1- 3x.
You could also do the last one by actually multiplying it out: $$(1- x)^3= 1- 3x+ 3x^2- x^3$$. If x is small enough (i.e. close enough to 0) that higher values of x can be ignored in the approximation, y= 1- 3x.
Again, these are all first order or linear approximations to the functions, not exact values.
(You can get the Taylor's polynomial and series that Evgeny- Makarov refers to by extending those same ideas to higher powers.)
Last edited:
#### Perplexed
##### New member
If $$f(x)= \frac{1}{\sqrt{1+ x}}= (1+ x)^{-1/2}$$ then $$f'(x)= -(1/2)(1+ x)^{-3/2}$$
so that $$f(0)= \frac{1}{\sqrt{1+ 0}}= 1$$ and then $$f'(0)= -(1/2)(1+ 0)^{3/2}= -1/2$$.
So y= f(x) is approximated, around x= 0, by $$y= -(1/2)x+1$$ or $$1- x/2$$. Notice the negative sign- what you have is NOT correct.
Just to clear things up so that someone else looking at this doesn't get confused, in my second approximation I had $$f(x)= \frac{1}{\sqrt{1 - x}}$$ rather than the $$f(x)= \frac{1}{\sqrt{1+ x}}$$ that you started with: notice the "-" rather than "+" in the square root. It was the simple change of sign in arriving at the reciprocal that first intrigued me on this one, and your explanation makes the reason why this works clear.
Less Perplexed now
#### Deveno
##### Well-known member
MHB Math Scholar
Allow me to make another observation regarding this:
Scientific measurements are often given in "significant figures", the reasoning being, we can only take measurements up to a certain degree of accuracy. | {
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So, suppose our input data can only give 6 decimal places.
If we expect we can model a function (and for many functions this is true) by:
$f(x) = a_0 + a_1x + a_2x^2 +\cdots$
And that the coefficients $a_k$ either stay "about the same size" or even better, decrease, then if we measure $x$ to 6 decimals places, the "correction term" for $x^2$ is around 12 decimal places, in other words, much much smaller than our standards of accuracy allow.
For certain classes of "well-behaved" functions, there are means to estimate (or "bound") the size of the error, which in turn lets us know "how many terms to go out".
For small enough $x$, this kind of reasoning lets us use the approximation:
$\sin(x) = x$
often used in simplifying wave equations that govern oscillators, and if more accuracy is needed, the approximation:
$\sin(x) = x - \dfrac{x^3}{6}$ is pretty darn good, as you can see here. | {
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Math Help - Sanity check on line intersecting plane
1. Sanity check on line intersecting plane
Line l is r=-8i+5j-6k+t(5i-2k)
Plane is -x-2y+5z = 0
What is the point of intersection? I get (-56/3, 5, -26/15) but my student's school teacher has a different answer of (-23,5,0)
Who is right?
2. Re: Sanity check on line intersecting plane
The line is given by x= -8+5t, y= 5, z= -6- 2t.
Putting those into the equation of the plane, -(-8+ 5t)- 2(5)+ 5(-6- 2t)= 8- 5t- 10- 30- 10t= -15t- 32= 0 so that t= -32/15 and then
x= -8+ 5(-32/15)= -8- 32/3= (-24- 32)/3= -56/3.
y= 5
z= -6- 2(-32/15)= (-90+ 64)/15= -26/15
3. Re: Sanity check on line intersecting plane
\displaystyle \begin{align*} \begin{cases} x &= -8 + 5t \\ y &= \phantom{-} 5 \\ z &= -6 -2t \end{cases} \end{align*}
Substituting into the plane we find
\displaystyle \begin{align*} -x - 2y + 5z &= 0 \\ - \left( -8 + 5t \right) - 2 \left( 5 \right) + 5 \left( -6 - 2t \right) &= 0 \\ 8 - 5t - 10 - 30 - 10t &= 0 \\ -32 - 15t &= 0 \\ -15t &= 32 \\ t &= -\frac{32}{15} \end{align*}
So that gives the point of intersection as
\displaystyle \begin{align*} \begin{cases} x &= -8 + 5 \left( -\frac{32}{15} \right) \\ y &= \phantom{-} 5 \\ z &= -6 - 2 \left( -\frac{32}{15} \right) \end{cases} \\ \begin{cases} x &= -\frac{24}{3} - \frac{32}{3} \\ y &= \phantom{-} 5 \\ z &= -\frac{90}{15} + \frac{64}{15} \end{cases} \\ \begin{cases} x &= -\frac{ 56}{3} \\ y &= \phantom{-} 5 \\ z &= -\frac{26}{15} \end{cases} \end{align*}
So the point of intersection is \displaystyle \begin{align*} \left( x, y, z \right) = \left( -\frac{56}{3} , 5, -\frac{26}{15} \right) \end{align*}. I agree with your answer
4. Re: Sanity check on line intersecting plane
Originally Posted by grillage
Line l is r=-8i+5j-6k+t(5i-2k)
Plane is -x-2y+5z = 0
What is the point of intersection? I get (-56/3, 5, -26/15) but my student's school teacher has a different answer of (-23,5,0)
Who is right? | {
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Who is correct depends on who solves $(-8+5t)+2(5)-5(-6-2t)=0$ correctly.
Have you checked to see if either of those points is on that line? | {
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Two mass one-spring system natural frequency
Suppose I have two masses $m_1, \ m_2$ connected by one spring of stiffness $k$ through their centres of mass, lying on a frictionless surface and the system is set into oscillation. I want to find the natural frequencies of oscillation. Finding the eigenvalues, I get one natural frequency is $$\omega_n = \sqrt{\frac{k(m_1+m_2)}{m_1m_2}}$$ But I thought that a system with two degrees of freedom should have two natural frequencies? When calculating the eigenvalues, the equation reduced to $$m_1m_2\omega^4 = \omega^2 k(m_1+m_2)$$ Does this mean the other natural frequency is zero?
(Apologies as I know there are quite a lot of questions on PSE concerning this system but I couldn't find one that answered my question and I don't have enough points to comment yet.)
• what do you think it would mean to have a mode with zero frequency? can you imagine the motion of the system? does this motion make sense? May 6 '16 at 16:50
• If it has zero frequency then it just doesn't oscillate? But then I don't know how to calculate the second frequency. It is a two degree of freedom system, right (need two coordinates to describe the position of each mass)? May 6 '16 at 16:57
• yes, there must be two solutions. But forget about the math for a minute: what would $\omega=0$ correspond to, physically? can you picture that motion in your head? May 6 '16 at 17:04
• @JohnRennie thank you :P I was trying to make OP reach that conclusion. If we don't impose fixed boundary conditions, spring systems always have the "rigid-motion" solution, where the whole system translates without oscillations. May 6 '16 at 17:24
• Ach, got it, thank you. ;) This rigid body response also explains why it's a two degree of freedom system then? (Because for the in-out type motion where they're out of phase and oscillating just the distance between the particles would be sufficient to describe the motion of the system?) May 7 '16 at 0:56 | {
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Suppose I have two masses m1, m2 connected by one spring of stiffness k
The Lagrangian of the system is
$$L = \frac{1}{2}m_1\dot q_1^2 + \frac{1}{2}m_2\dot q_2^2 - \frac{1}{2}k(q_1 - q_2)^2$$
where $q_1$ and $q_2$ are the coordinates of $m_1$ and $m_2$ respectively.
Now, consider a change of coordinates to the normal coordinates $Q_1$ and $Q_2$ where
$$Q_1 = \frac{q_1m_1 + q_2m_2}{m_1 + m_2},\qquad Q_2 = q_2 - q_1$$
are the coordinates of the center of mass $M = m_1 + m_2$ and the reduced mass $\mu = \frac{m_1m_2}{m_1 + m_2}$ respectively.
In these coordinates, the Lagrangian is
$$L = \frac{1}{2}M\dot Q_1^2 + \frac{1}{2}\mu\dot Q_2^2 - \frac{1}{2}kQ_2^2$$
and now it's easy to see that the uncoupled equations of motion (via the Euler-Lagrange equation) are
$$\ddot Q_1 = 0, \qquad \ddot Q_2 = -\frac{k}{\mu}Q_2$$
And so, the center of mass coordinate has 'zero frequency oscillation', i.e., uniform translational motion, while the reduced mass coordinate oscillates with angular frequency $\omega_2 = \sqrt{\frac{k}{\mu}}$
The other natural frequency is indeed zero! Natural frequencies of zero corresponds to vibrational modes of rigid body motion. Rigid body motion is not a vibrational motion in itself, but still arises in the modal analysis of certain systems such as the one above. The reason you get a rigid body mode is because you are able to move the system as a whole, and no restoring force exists to bring the whole system back to its original location.
This system is equivalent to hooking up one of the masses to a fixed point using a spring of zero stiffness. Let's consider your system with mass $m_1$ connected to a fixed point with a spring of stiffness $K$. Upon performing modal analysis, the two natural frequencies of such a system are given by:
$$\omega=\sqrt{\frac{m_1 + m_2}{2m_1 m_2}k + \frac{K}{2m_1}\pm \sqrt{\left[\frac{m_1 + m_2}{2m_1 m_2}k + \frac{K}{2m_1}\right]^2 - \frac{Kk}{m_1 m_2}}}$$ | {
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Now, to reobtain your system, set $K=0$, and the two frequencies indeed become $0$ and $\sqrt{\frac{m_1+m_2}{m_1 m_2}k}$. So the lowest natural frequency of your system is indeed zero, but what is the physical significance of a natural frequency of zero? One way to interpret it is a vibration of infinite time period, $T=2\pi/\omega$. When the bodies move, it takes so long for them to finish their first oscillation that it never completes, i.e. the body continues moving on at constant speed. This corresponds to rigid body motion under no external forces (modal analysis for determining natural frequencies sets externally applied forces to zero anyways.)
P.S. Also, note that the zero frequency result does appear alongside the non-zero result when you first calculate them:
By taking the determinant of the relevant matrix, the following equation is obtained:
$$\omega^4 - \frac{m_1 + m_2}{m_1 m_2}k\omega^2=0$$
Note that this can be rewritten as:
$$\omega^2 \left(\omega^2 - \frac{m_1 + m_2}{m_1 m_2}k\right) = 0$$
So it can be seen that $\omega^2 = 0$ if one of the two solutions. It is quite easy to miss this, as it can be prematurely cancelled out of the equation, which can only be done if $\omega^2 = 0$ is an impossible solution, which it is not.
You can take the two equations of motion
\begin{align} k (x_2 - x_1) = m_1 \ddot{x}_1 \\ k (x_1-x_2) = m_2 \ddot{x}_2 \end{align}
and transform them using their centroid location, and distance
\left. \begin{align} x_c & = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}\\ x_d & = x_2-x_1 \end{align} \right\} \begin{aligned} x_1 &= x_c - \frac{m_2 x_d}{m_1 +m_2} \\ x_2 & = x_c + \frac{m_1 x_d}{m_1+m_2} \end{aligned}
(and similarly for the accelerations). The equations of motion are transformed into
\begin{align} \ddot{x}_c &= 0 \\ \ddot{x}_d &= -k \left( \frac{m_1+m_2}{m_1 m_2} \right) x_d \end{align} | {
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As you can see the center of mass $x_c$ does not have any acceleration (Newton's first law), and only the separation oscillates with the reduced mass $m_{eff} = \frac{m_1 m_2}{m_1+m_2}$
The eigen-values of the system are trivially found as
\begin{align} \omega_c^2 &= 0 \\ \omega_d^2 & = k \frac{m_1+m_2}{m_1 m_2} \end{align}
There is only one degree of freedom (the distance between the masses) not two, and therefore only one natural frequency (given by your 1st eqn). Since there are no external forces, the CM of the system does not move. (Sorry but I do not see what all the fuss is about here. Why make such a simple problem so complicated?)
• If you just push one end and the whole system translates then the distance between the masses is insufficient to describe its motion? May 7 '16 at 0:58
• @AtticusFinch95 : True but your question asked only about the natural frequencies of oscillation and only says that the masses are "set into oscillation". Any constant velocity of the CM has no effect on natural frequency and is unknown, unknowable and irrelevant. The question does not mention any continuing external force on either mass (after they are somehow "set into oscillation") so we cannot assume there are any. It is unnecessary and irrelevant to introduce any additional degrees of freedom into the problem. May 7 '16 at 11:48
• There is definitely two degrees is freedom, the centre of mass and the relative coordinates. Thus there are two normal modes but one of them is a zero mode, which has null frequency and corresponds to the translation of the centre of mass. Sep 9 '17 at 15:11 | {
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# On Cesàro convergence: If $x_n \to x$ then $z_n = \frac{x_1 + \dots +x_n}{n} \to x$
I have this problem I'm working on. Hints are much appreciated (I don't want complete proof):
In a normed vector space, if $x_n \longrightarrow x$ then $z_n = \frac{x_1 + \dots +x_n}{n} \longrightarrow x$
I've been trying adding and subtracting inside the norm... but I don't seem to get anywhere.
Thanks!
Given $\epsilon >0$ there exists $n_0$ such that if $n\geq n_0$ then $\parallel x_n -x\parallel < \epsilon$
so
\begin{align*} 0 & \leq \left\lVert \frac{x_1 +\cdots +x_n}{n} -x \right\rVert \leq \left\lVert \frac{x_1 + \dots + x_n - nx }{n} \right\rVert \\ & \leq \frac{\lVert x_1 - x \rVert}{n} + \dots + \frac{\lVert x_{n_0 - 1} - x \rVert}{n} + \frac{\lVert x_{n_0} - x \rVert}{n} +\dots + \frac{\lVert x_{n} - x \rVert}{n} \\ &\le \frac 1n\sum_{i=1}^{n_0-1} \| x_i -x\| + \frac{n-n_0}{n} \epsilon \end{align*}
The first $n_0 -1$ terms $\| x_i -x\|$ can be bounded by some $M$, thus for $n\ge (n_0-1)M/\epsilon=: N_0$ we have $$\frac 1n\sum_{i=1}^{n_0-1} \| x_n -x\| \le \frac 1n (n_0-1)M \le \epsilon$$
Thus $$\left\| \frac{x_1 + \cdots x_n}{n} - x\right\| <2\epsilon$$ when $n\ge N_0$.
Thanks a lot @Leonid Kovalev for the inspiration, though my main problem was that I wasn't aware of what to do with the $nx$ (the silliest part :P) | {
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• This sort of thing is encouraged, I think. – Dylan Moreland Jun 11 '12 at 3:21
• I think you want "$-nx$" in the first line of the display, and $\|x_n - x\|$ at the end. I don't think you want to say that the first few terms are $\leq M$; that doesn't seem to be enough. – Dylan Moreland Jun 11 '12 at 3:41
• @DylanMoreland: Why isn't it enough? Can you explain? – Bouvet Island Jun 11 '12 at 14:56
• @Inti: I think you are missing something in the argument. Generally the argument consists of two steps: first choose $n_0$ such that if $n \geq n_0$, $\|x_n -x \| < \epsilon / 2$. Next choose $N \geq n_0$ such that $M$ (which is at least $\sup_n \|x_n\|$) satisfies $M / N < \epsilon / 2$. I don't see the second part of the argument implemented in your answer. – Willie Wong Jun 11 '12 at 14:59
There is a slightly more general claim:
PROP Let $\langle a_n\rangle$ be a sequence of real numbers, and define $\langle \sigma_n\rangle$ by $$\sigma_n=\frac 1 n\sum_{k=1}^n a_k$$
Then $$\liminf_{n\to\infty}a_n\leq \liminf_{n\to\infty}\sigma_n \;(\;\leq\;)\;\limsup_{n\to\infty}\sigma_n\leq \limsup_{n\to\infty}a_n$$
P We prove the leftmost inequality. Let $\ell =\liminf_{n\to\infty}a_n$, and choose $\alpha <\ell$. By definition, there exists $N$ such that $$\alpha <a_{N+k}$$ for any $k=0,1,2,\ldots$ If $m>0$, then $$m\alpha <\sum_{k=1}^m \alpha_{N+k}$$
which is $$m\alpha<\sum_{k=N+1}^{N+m}a_k$$
$$(m+N)\alpha+\sum_{k=1}^{N}a_k<\sum_{k=1}^{N+m}a_k+N\alpha$$
which gives
$$\alpha+\frac{1}{m+N}\sum_{k=1}^{N}a_k<\frac{1}{m+N}\sum_{k=1}^{N+m}a_k+\frac{N}{m+N}\alpha$$ | {
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$$\alpha+\frac{1}{m+N}\sum_{k=1}^{N}a_k<\frac{1}{m+N}\sum_{k=1}^{N+m}a_k+\frac{N}{m+N}\alpha$$
Since $N$ is fixed, taking $\liminf\limits_{m\to\infty}$ gives $$\alpha \leq \liminf\limits_{m \to \infty } \frac{1}{m}\sum\limits_{k = 1}^m {{a_k}}$$ (note that $N+m$ is just a shift, which doesn't alter the value of the $\liminf^{(1)}$). Thus, for each $\alpha <\ell$, $$\alpha \leq \liminf\limits_{m \to \infty } \frac{1}{m}\sum\limits_{k = 1}^m {{a_k}}$$ which means that $$\liminf_{n\to\infty}a_n\leq \liminf_{n\to\infty}\sigma_n$$ The rightmost inequality is proven in a completely analogous manner. $\blacktriangle$.
$(1)$: Note however, this is not true for "non shift" subsequences, for example $$\limsup_{n\to\infty}(-1)^n=1$$ but $$\limsup_{n\to\infty}(-1)^{2n+1}=-1$$
COR If $\lim a_n$ exists and equals $\ell$, so does $\lim \sigma_n$, and it also equals $\ell$. The converse is not true.
WLOG, the $x_n$ converge to $0$ (otherwise consider the differences $x_n-x$), and stay confined in an $\epsilon$-neighborhood of $0$ after $N_\epsilon$ terms.
Then the average of the first $m$ terms is bounded by
$$\frac{N\overline{x_N}+(m-N)\epsilon}m,$$ which converges to $\epsilon$. So you can make the average as close to $0$ as you like. | {
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# Determinant of Triangular Matrix
I understand that you can find the determinant of a matrix along it's diagonal if it is in triangular form. For a matrix such as this: $$\begin{pmatrix} 1 & 5 & 0\\ 2 & 4 & -1\\ 0 &-2 & 0 \end{pmatrix}$$ When put into triangular form I get: $$\begin{pmatrix} 1 & 5 & 0\\ 0 & 1 & 1/6\\ 0 & 0 & 1/3 \end{pmatrix}$$ Since I multiplied row two by -1/6 during the row reduction I would expect the determinant to be $$1\cdot 1\cdot 1/3\cdot (-1/6),$$ but the answer for the determinant of the original matrix is -2. Where exactly am I going wrong?
• Write your row operations as a matrix product and use the multiplication rule for determinants. Then you will minimize your risk of mistakes. – mathreadler Dec 12 '17 at 16:30
• You should multiply by $-6$ rather than $-1/6$ – eepperly16 Dec 12 '17 at 16:31
• @mathreadler I understand the multiplication rule, but was just wondering why this scenario did not work when trying to solve using a triangular matrix – Pulse Dec 12 '17 at 16:32
• it will work if you do it right. there is a larger chance you will do it right if you write it down step by step with matrix products – mathreadler Dec 12 '17 at 16:33
• @eepperly16 I multiplied by -1/6 because during the row operation, row 2 I multiplied by -1/6 to make the 6 a 1 in that row. – Pulse Dec 12 '17 at 16:33
You can add a multiple of any row to another row without changing the determinant. So I'll do these kinds of operations to get to the triangular form.
$$\begin{matrix} 1 & 5 & 0 \\ 2 & 4 & -1 \\ 0 &-2 &0 \end{matrix}$$
Subtract twice the first row from the second:
$$\begin{matrix} 1 & 5 & 0 \\ 0 & -6 & -1 \\ 0 &-2 &0 \end{matrix}$$
Subtract one-third the new second row from the third:
$$\begin{matrix} 1 & 5 & 0 \\ 0 & -6 & -1 \\ 0 & 0 & \frac{1}{3} \end{matrix}$$
The product along the diagonal is $-2$. | {
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The product along the diagonal is $-2$.
Pulling out a factor from one of the rows, however, does change the determinant by that factor. You pulled out $-1/6$ from the second row, so the calculated determinant from the diagonal in your question statement was low by this factor.
• This really does not help with the understanding as you have not shown that the operations you perform don't change the determinant. – mathreadler Dec 12 '17 at 16:38
• The matrix you ended up with is the same as mine, except your second row is multiplied by $-1/6$. "Pulling out" that constant changes the determinant by the same factor: $-2 \times -1/6 = 1/3$, which was what you got. – John Dec 12 '17 at 16:46 | {
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# Why does this assumption give an under approximation for both the expected maximum and minimum
The well known result, $$\mathbb{E}[\text{min} \{X_i \}_{i=1}^n ] = \frac{1}{n+1}$$ and $$\mathbb{E}[\text{max} \{X_i \}_{i=1}^n ] = \frac{n}{n+1}$$ where $$X_i$$ are I.I.D Uniform$$(0,1)$$ random variables is a lifesaver. It can be extrapolated to find the expected minimum and maximum of $$n$$ Uniform$$(a,b)$$ variables.
I tried to use it in the discrete case. For example, consider the maximum of $$10$$ rolls of a $$100$$ sided die, the quick formula gives: $$\frac{10}{11}\cdot 100 = 90\frac{1}{11} \approx 90.91$$. The true answer however is $$91.4007585757$$ (using tail sum formula : $$100 - \sum\limits_{i=1}^{99}(\frac{i}{100})^{10}$$ )
Notice the approximation gives an underestimate.
Now consider using it to approximate the expected minimum.
$$\frac{1}{11}\cdot 100 = 9\frac{1}{11} \approx 9.09$$. The true answer however is $$9.59924142434$$ and again this formula under-approximates?
Why does this happen in both cases? Is there a cheeky way to tweek it for the discrete case to get a bit more accurate?
Thanks!
A little thing I have noticed that might help is the following:
The approximate min + approximate max is always $$n$$. However, by symmetry the true min + true max is always $$n+1$$. (because they are centred around the mean of $$\frac{n+1}{2}$$
• Decent question, a very underappreciated consequence of the way you approximate continuous by discrete. Technically speaking, the second part hasn't been answered yet, so I'm looking forward to someone addressing that as well. Sep 20 at 6:05 | {
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Note that for $$X$$ positive we have $$\mathbb EX=\int_0^\infty 1-F(x)\,\mathrm d x$$. Now note that if $$X$$ is the discrete variant and $$Y$$ is the continuous approximation. Then $$F_X\leq F_Y$$ (as $$F_X(x)=F_Y(x)$$ for $$1\leq x\leq 100$$ and $$F_Y$$ is linear while $$F_X$$ is constant on $$[x,x+1)$$.).
Thus this also holds true for the distribution of max and min, and thus by above formula the expected values for $$Y$$ are smaller than those for $$X$$.
It under-approximates because you are approximating a random variable taking values in $$\left\{\frac{1}{k},\frac{2}{k},\dots,\frac{k-1}{k}, 1\right\}$$ by a uniform random variable, that is, uniform on $$\left(0,\frac{1}{k}\right] \cup \left(\frac{1}{k}, \frac{2}{k}\right]\cup \dots\cup \left(\frac{k-1}{k}, 1\right]\,.$$ Viewed differently, you replace the value $$\frac{i}{k}$$ (for $$1\leq i\leq k$$) by a uniformly random value in $$\left(\frac{i-1}{k}, \frac{i}{k}\right]$$ which leads to a (small) under-approximation.
• As a corollary, a (first-order, not perfect, but closer) fix is to add $\frac{1}{2k}$ to the result, to make up for the shift and "re-center" your approximation. Since you multiply by $k$ in the end, that means adding $0.5$ to your estimates. In your example, that'll give 91.41 and 9.59. Sep 19 at 21:20
• I have noticed a typo in your answer. I think you mean: $\left(0,\frac{1}{k}\right] \cup \left(\frac{1}{k}, \frac{2}{k}\right]\cup \dots\cup \left(\frac{k-1}{k}, 1\right]\,.$ @Clement C. Sep 20 at 8:40 | {
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# Prove that $\lfloor x\rfloor \geq y$ if, and only if, $x\geq\lceil y\rceil$
I have some trouble proving that if $$x,y\in\mathbb{R}$$ then $$\lfloor x\rfloor \geq y$$ if, and only if, $$x\geq\lceil y\rceil$$.
I have tried some different approaches, the most recent being a proof by contradiction: Assume (for contradiction) that $$x<\lceil y\rceil$$, then $$\lfloor x\rfloor < y +1$$. However, I can not (of course) get rid of the $$+1$$ term because in general $$y \leq\lceil y\rceil$$.
One can perhaps prove that both the floor- and ceiling function preserve inequalities? If so, the result become trivial.
Any ideas are highly appreciated.
Note that $$\lfloor x \rfloor \ge y$$ implies $$\lfloor x \rfloor \ge \lceil y\rceil$$, so $$x\ge \lceil y\rceil$$.
The other direction is similar.
If the above is not clear, recall that $$\lceil y \rceil$$ is defined as the least integer not smaller than $$y$$, i.e. if $$C(y)$$ is the set $$C(y):= \{n\in\mathbb Z:n\ge y\}$$, then $$\lceil y\rceil = \inf C(y).$$ The assumption $$\lfloor x\rfloor \ge y$$ is exactly that $$\lfloor x\rfloor \in C(y)$$. The only thing left is to use the defining property of an infimum, $$c \in C(y) \implies \inf C(y) \le c.$$
• Thanks, but I can not see why the implication is true in general. Sep 28, 2018 at 12:02
• @Wuberdall This is because $\lfloor x \rfloor \in \mathbb Z$. For example, $5\ge y$ implies $5 \ge \lceil y\rceil$. This holds for every integer in place of 5. In words - the ceiling of $y$ is the least integer larger than $y$, so any integer larger than $y$ is larger than the ceiling. Sep 28, 2018 at 12:03
• (You should replace "larger than" with "not smaller than" but that's harder to say) Sep 28, 2018 at 12:10
• Yes, I see it know. This is of course true due to the result "number not greater than integer iff ceiling is not greater than integer" exactly as you explained. – Thank you for your help. Sep 28, 2018 at 12:12 | {
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Hint: $$\lfloor x\rfloor\geq y\iff \lfloor x\rfloor\geq\lceil y \rceil\iff x\geq \lceil y\rceil.$$
• Thanks, but why is the first bi-implication true? Sep 28, 2018 at 12:01
• @Wuberdall Because $\lfloor x\rfloor$ is an integer that is greater than or equal to $y$, so $\lfloor x\rfloor\geq\lceil y\rceil$. For the reverse direction, just note $\lceil y\rceil \geq y$. Sep 28, 2018 at 12:04
We need to prove both direction, that is
• $$\lfloor x\rfloor \geq y \implies x\geq\lceil y\rceil$$
and
• $$x\geq\lceil y\rceil \implies \lfloor x\rfloor \geq y$$
For the first one we have that
$$\lfloor x\rfloor \geq y \implies \lfloor x\rfloor \geq \lceil y\rceil$$
and therefore
$$x\ge \lfloor x\rfloor \geq \lceil y\rceil \implies x\geq\lceil y\rceil$$
For the second one we have that
$$x\geq\lceil y\rceil \implies \lfloor x\rfloor \geq \lceil y\rceil$$
and therefore
$$\lfloor x\rfloor \geq \lceil y\rceil \ge y \implies \lfloor x\rfloor \ge y$$
• Thank you, it is completely clear why it is true now. Sep 28, 2018 at 12:20
• @Wuberdall You are welcome! Bye
– user
Sep 28, 2018 at 12:26
We need the following elementary properties of the floor and ceiling functions: $$a\ge\lfloor a\rfloor$$, $$\lceil\lfloor a\rfloor\rceil=\lfloor a\rfloor$$, $$a\ge b\implies\lceil a\rceil\ge\lceil b\rceil$$, and $$\lceil a\rceil=-\lfloor-a\rfloor$$.
From the first three elementary propeties, we have, for any real $$u$$ and $$v$$,
$$\lfloor u\rfloor\ge v \implies u\ge\lfloor u\rfloor=\lceil\lfloor u\rfloor\rceil\ge\lceil v\rceil$$
so letting $$u=x$$ and $$v=y$$ gives
$$\lfloor x\rfloor\ge y\implies x\ge\lceil y\rceil\qquad(*)$$
while letting $$u=-y$$ and $$v=-x$$ gives
$$\lfloor-y\rfloor\ge-x\implies-y\ge\lceil-x\rceil$$
Invoking now the fourth elementary property turns that last implication into
$$-\lceil y\rceil\ge-x\implies-y\ge-\lfloor x\rfloor$$
which can be re-expressed as
$$x\ge\lceil y\rceil\implies\lfloor x\rfloor\ge y\qquad(**)$$ | {
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which can be re-expressed as
$$x\ge\lceil y\rceil\implies\lfloor x\rfloor\ge y\qquad(**)$$
Putting $$(*)$$ and $$(**)$$ together, we have
$$\lfloor x\rfloor\ge y\iff x\ge\lceil y\rceil$$ | {
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# What do the pieces of LaTeX, \left and \right, respectively mean?
What are their uses and are they even standard?
-
\left and \right are used for delimiters when they have to change the size dynamically depending on the content. Consider the following example:
\documentclass{article}
\usepackage{amsmath}
\begin{document}
Compare this:
\begin{align*}
\left(\sqrt{2}+\sqrt{3}\right)^2
&=\left(\sqrt{2}\right)^2+2\times\sqrt{2}\times\sqrt{
3}+\left(\sqrt{3}\right)^2\\
&=2+2\sqrt{6}+3\\
&=5+2\sqrt{6}
\end{align*}
with:
\begin{align*}
(\sqrt{2}+\sqrt{3})^2
&=(\sqrt{2})^2+2\times\sqrt{2}\times\sqrt{
3}+(\sqrt{3})^2\\
&=2+2\sqrt{6}+3\\
&=5+2\sqrt{6}
\end{align*}
First one uses \verb|\left(| and \verb|\right)| and second one uses just \verb|(| and \verb|)|. I hope the difference is clear.
Just another example:
$\left(\frac{1}{2}\right) \qquad (\frac{1}{2})$
\end{document}
In this particular case, the (\sqrt{2}+\sqrt{3}) with \left and \right gives bit bigger parenthesis making it to look ugly (to some extent) as noted by Enrico. In such cases, proper variant of delimiters (\bigl and \bigr in this case) may be used to get the appropriate height.
For more details, refer to amsmath documentation - page 15, section 4.14 (texdoc amsldoc from command prompt). Here is a screen shot of the same: | {
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-
+1 but I think we don't need a blank line between "Just another..." and $.... – Please don't touch Oct 14 '12 at 2:15 @ガベージコレクタ: OK As you wish. I am updating :) – Harish Kumar Oct 14 '12 at 2:39 Thanks for the update. – Please don't touch Oct 14 '12 at 2:46 +1; you could point out that in your example with (\sqrt{2}+\sqrt{3}) the result with \left and \right is less good than without it (not to say wrong). – egreg Oct 14 '12 at 9:08 @egreg: Yes and added that here also. Thank you :) – Harish Kumar Oct 14 '12 at 9:24 \left and \right are delimiter counterparts and TeX primitives, to be used with so-called "extensible delimiters" in math mode within the same group. This implies two things: 1. If you use one, you have to use the other as well; and 2. They have to be opened/closed at and within the same group level depth. Common or typical uses include • \left\{ ... \right\} • \left[ ... \right] • \left( ... \right) although there is no requirement for the delimiters to be matching. For example, it is fine to use \left\}...\right(, just as long as you use both \left and \right. For example (to mimic amsmath's bmatrix environment), \documentclass{article} \usepackage{amsmath}% http://ctan.org/pkg/amsmath \begin{document} \[ \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \qquad \left[\begin{array}{@{}ccc@{}} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array}\right]$
\end{document}
If you want to use only one of the two, you still need to include the counterpart, but this time with the argument .. That is, either \left. or \right., referred to as the null delimiter. TeX inserts a \nulldelimiterspace for these.
Extensible delimiters are ones that are comprised of a number of fixed and some variable-length parts. For some eye-candy, here's a visual of an extensible }:
- | {
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# Method to show that a limit exisits and evaluate the limit of a function as x tends to infinity.
Exam Question:
For each of the following functions f , determine whether $\lim_{x\to\infty}f(x)$ exists, and compute the limit if it exists. In each case, justify your answers.
a) $f(x)= \dfrac{x+2}{x^2+8}$
b) $f(x)= \dfrac{\cos(x)}{x^2}$
Attempt:
a) I think you divided by the leading power of $x$, and then use algebra of limits to show that the limit is equal to $0$.
b) I don't think this limit exits as $\cos(x)$ is a periodic function.
What is the correct method to answer this question?
• Finally someone who can format a question properly in HTML and not just put L X->infinity. – HELP May 18 '16 at 9:11
• wow, thank you. It's really not that hard though once you collect enough references, programing is nothing if not repeatable. – UniStuffz May 18 '16 at 9:12
• Note: for the second function: even the $\cos$ function is periodic, but you should keep in mind that it is bounded by $-1$ and $1$. Then dividing by $x^2$ .... – Nizar May 18 '16 at 9:14
• @Nizar ah "sandwich rule" am I right. – UniStuffz May 18 '16 at 9:23
You're right about (a). For rational functions, the limit (for $x \to \pm \infty$) will always be zero if the denominator has a higher degree than the numerator; in your case: $$\lim_{x \to +\infty} \frac{x+2}{x^2+8} = \lim_{x \to +\infty} \frac{1/x+2/x^2}{1+8/x^2} = \frac{0}{1} = 0$$ Addendum: and if it's the other way around (degree higher in the numerator), the limit will be $\pm \infty$ (check the sign). When the degree of numerator and denominator is the same, the limit will be the ratio of the highest order coefficients.
For (b): consider the fact that $-1 \le \cos x \le 1$ (for all $x$), but the denominator (which is $x^2$) tends to...? You may have heard of the squeeze or sandwich theorem? Dividing by $x^2$: $$\color{green}{ -\frac{1}{x^2}} \le \color{blue}{ \frac{\cos x}{x^2}} \le \color{red}{ \frac{1}{x^2}}$$ | {
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Since green and red tend to $0$, also blue...
• Yes I have heard of sandwich rule, how would one go about justifying if the limit exits first, (and then obviously evaluating it). – UniStuffz May 18 '16 at 9:25
• I elaborated a bit, see above. Does that help? Existence follows from being sandwiched between two that exist and are equal. – StackTD May 18 '16 at 9:29
• yes it does. So for example of the limit did not exits by using the methods described above, would that be sufficient as a proof to say that the limit does not exit, or would I have to use first principles and the definition of limit formula, (And I thought my editing was good, this is incredible). – UniStuffz May 18 '16 at 9:33
• It's not very clear to me what you're asking now. It's not because, for instance, the squeeze theorem doesn't apply, that you can be sure the limit you're looking at doesn't exist... It depends on the context and the problem how you would go about showing a limit does not exist. One possibility is indeed via the formal definition, but perhaps you're allowed to conclude it directly for simple functions such as $\lim_{x \to \infty} \cos x$, because you know how $\cos$ behaves. – StackTD May 18 '16 at 9:38
• You're welcome! – StackTD May 18 '16 at 9:46
Keep in mind
$$\frac{-1}{x^2} \leq \frac{\cos x}{x^2} \leq \frac{1}{x^2}$$ what is the limit of $\frac{\pm 1}{x^2}$ when $x\to \infty$ ?
• Yes of course they will both be zero, hence bounded above and below by zero hence by using sandwich theorem, the function will tend to 0. How would you justify if the limit exists first. – UniStuffz May 18 '16 at 9:27 | {
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# General Form of Circle Equation
1. May 4, 2004
### Xavier
I'm attempting to write the general form of a circle where I'm given three points, not necessary equally distant from each other: (0,0), (0,8), (6,0)
I need to write the equation of the circle in general form. I've managed to get pretty close to what the answer in the book states, but I'm still off.
Since these points are not equally distant from each other, you can't use the midpoint formula (x1+x2)/2, (y1+y2)/2 to get the center of the circle. Is there another method I should be using to find the center so I can determine the radius, and then find the general form for the circle?
I'm not looking for the answer, I know what the book says in the answer key, I'm just curious as to what I'm missing, then I can try to take it from there. Thanks!
2. May 4, 2004
### krab
the circle equation is
$$(x-a)^2+(y-b)^2=r^2$$
Plug in each of the three (x,y) coordinate pairs, and you get 3 separate equations. The unknowns are a,b,r. Solve it.
3. May 5, 2004
### Chrono
You probably don't need this, but you also have the parametric equations of $$(x-a)^2+(y-b)^2=r^2$$ which can be written as:
$rSin[t] - a; \ rCos[t] - b$
as t goes from 0 to $$2\pi$$
I may be wrong, so somebody double check it for me.
4. May 5, 2004
### Xavier
Ok, so for the three points you get:
$$(0-a)^2 + (0-b)^2 = r^2$$
$$(0-a)^2 + (8-b)^2 = r^2$$
$$(6-a)^2 + (0-b)^2 = r^2$$
But without knowing a & b, how do you solve for r?
5. May 5, 2004
### ahrkron
Staff Emeritus
You have three equations abd three unknowns. You can, for instance, eliminate r by equation the first two; this gives you one equation with only a and b. You do the same with the first and third equations and get a different equation with a and b. Then you have two new equations on a and b only, which you can solve.
Once you have a and b, you can substitute on one of the original three eqns to get r.
6. May 6, 2004
### Xavier | {
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6. May 6, 2004
### Xavier
Hmmm.. I can see how I can do it for this equation since $$(0-a)^2 = a^2$$, so then you just take the square root of both sides to solve for a. But if you have (for example) $$(1-a)^2 = 1-2a+a^2$$, you can't just solve for a can you? You'd be left with $$a^2-2a = -1+-$$(rest of equation)
7. May 6, 2004
### ahrkron
Staff Emeritus
Play with them a little bit... when you eliminate r from first and second, you also eliminate a^2 and b^2, so you only get a constant plus a term in b. Something similar happens combining the first and last eqns.
8. May 6, 2004
### HallsofIvy
Staff Emeritus
$$(0-a)^2 + (0-b)^2 = r^2$$
so (1) $$a^2+ b^2= r^2$$
$$(0-a)^2 + (8-b)^2 = r^2$$
so (2) $$a^2+ b^2-16b+ 64= r^2$$
$$(6-a)^2 + (0-b)^2 = r^2$$
so (3) $$a^2- 12a+ 36+ b^2= r^2$$
Subtracting 1 from 3, -12a+ 36= 0 or a= 3.
Subtracting 1 from 2, -16b+ 64= 0 or b= 4.
The a2+ b2= 9+ 16= r2
so r= 5.
Doesn't come any easier than that!
9. May 6, 2004
### Xavier
Ah.. I see.. I knew I had to set one equation equal to another and then reduce, but I didn't see the relationship between the three points. I'll crack open the book again when I get off work and try a few more of those problems. Thanks for the guidence!
10. May 10, 2004
### Xavier
Ok.. I've reworked this, and I'm just about where I was at. Here's what my book shows as the general form of the circle with the following points (same as before):
Points: (0,0), (0,8), (6, 0)
Book Answer: $$x^2+y^2-6x-8y=0$$
When I work this, I get this:
$$(0-a)^2+(0-b)^2=r^2$$
$$a^2+b^2=r^2$$
$$(0-a)^2+(8-b)^2=r^2$$
$$a^2-b^2-16b+64=r^2$$
$$(6-a)^2+(0-b)^2=r^2$$
$$a^2+b^2-12a+36=r^2$$
$$a^2+b^2=a^2+b^2-16b+64$$
$$16b=64$$
$$b=4$$
$$a^2+b^2=a^2+b^2-12a+36$$
$$12a=36$$
$$a=3$$
$$(x-a)^2+(y-b)^2=r^2$$
$$(0-3)^2+(0-4)^2=r^2$$
$$r=5$$
$$(x-3)^2+(y-4)^2=5$$
$$x^2-3x-3x+9+y^2-4y-4y+16=5$$
$$x^2+y^2-6x-8y+20=0$$
As you can see, I'm slightly off, but I don't understand what I've done wrong. Any ideas where I strayed? | {
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11. May 11, 2004
### krab
I notice everyone takes the long way round. Here's a better way: Look at the first two equations. Don't multiply them out, just subtract them. You get 8-b=b, or b=4. Now do the same to equations 1 and 3. You get 6-a=a, or a=3. Plug these two result into equation 1 and you find r=5.
12. May 11, 2004
### arildno
You've just made a silly mistake here, that's all (the radius squared is 25)
Besides, note the swift manipulation done by krab.
13. May 11, 2004
### Xavier
Ah.. that's what happens when I work on this late at night. For some reason I figured it didn't need to be squared since I had already determined the radius, but it's clear now.
That is quite a bit simplier manipulation Krab. I've got a second problem I can try this out on, since I don't believe I did it correctly.... I think I need a bigger eraser :)
Thanks for working through this with me. I appreciate it!
14. May 11, 2004
### Wooh
Another interesting method to use would be to find the equations of the lines that go through 2 of the points (only 2 lines are needed). So, for example, you find the equation of the line through (0,0) and (0,8) and the equation of the line (0,0) and (6,0). Then, you use midpoint formula and find the equation of the perpendicular bisector of each line. The interesection of the 2 perpendicular bisectors? The center of the circle. | {
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# Why is every singleton subset of a partially ordered set a totally ordered set?
In Lipschutz, Theory and Problems of Set Theory ( Schaum's series, 1964 ed.), Chapter 10, Solved Problem n°7 , the question is asked to find all non-empty totally ordered subsets of an ordered set A = {a,b,c} with a diagram showing that : b precedes a and that c precedes a ( b and c being incomparable).
Lipschutz provides a solution in which he gives {a} , {b} and {c} as totally ordered subsets of A.
Hence my question : is every singleton subset of a partially ordered set a totally ordered set.
How to explain that , in case a subset is a singleton the trichotomy condition for totally ordered sets is automatically fullfilled : a partial order S is total iff for all a and b belonging to S,
either a<b or a=b or a>b.
I'd like to derive the conclusion formally from the definition of a totally ordered set.
• It's because $\forall a,b: a=b$ holds in a singleton. – Berci Apr 29 at 12:27
• $a \in \{ a \}$ is the only element and $a=a$. Thus, the trichotomy condition is fulfilled. Recall that the condition id $\forall a,b \in X \ [a=b \lor \ldots ]$. – Mauro ALLEGRANZA Apr 29 at 12:28
1. Let $$S$$ be a singleton.
2. From $$1$$, there follows there must be some $$x$$ such that $$S=\{x\}$$.
3. Let $$a, b\in S$$.
4. Therefore, $$a=x$$ and $$b=x$$.
5. Therefore, $$a=b$$.
6. Therefore, $$a=b\lor ab$$.
• Very frustrating if nice answers like this are downvoted. And this without any explanation why. – drhab Apr 30 at 8:34
• @dmab. Totally agree. I personnaly upvoted both aswers. – Eleonore Saint James Apr 30 at 9:25
Yes, every singleton can be recognized as a totally ordered set (even as a well-ordered set).
This because next to reflexivity, transitivity and antisymmetry we also have comparability.
If $$S=\{c\}$$ is a singleton then it is true that for all elements $$a,b\in S$$ we have: $$ab$$
Actually in all cases (there is only one) we have $$a=c=b$$. | {
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Actually in all cases (there is only one) we have $$a=c=b$$.
Also note that every non-empty subset of $$S$$ (there is only one) has a smallest element. | {
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# multiple choice summation problem
Let $$X = \frac{1}{1001} + \frac{1}{1002} + \frac{1}{1003} + \cdots + \frac{1}{3001}.$$ Then
(A) $X < 1$
(B) $X > 3/2$
(C) $1 < X < 3/2$
(D) none of the above holds.
I assume that the answer is the third choice $1<X<3/2$. I integrate out $1/x$ in the interval $(1001, 3001)$ and get a result that satisfies only the choice C. Is this a Riemann sum? Please help.
-
Short answer: Yes, your Riemann sum idea was correct, and $X\approx \log 3$ so the correct option was indeed C). – Ragib Zaman Jul 17 '12 at 16:00
@RagibZaman thank you but can you please explain why this discrete sum is integrated – Argha Jul 17 '12 at 16:03
@Ranabir: You can think of the sum as the integral of a step function. When you switch to integrating $1/x$ instead, you introduce a vertical error that is never more than the difference between two neighboring terms, which is about $10^{-6}$ at the most. So a conservative estimate of the total error is $10^{-6}$ times the width of the integrated area -- concretely, the integral will be at most 0.002 from the true sum. – Henning Makholm Jul 17 '12 at 16:19
## 2 Answers
With respect to your Riemann sum approach: the idea is that for positive, decreasing functions, the Riemann sum and the integral carefully approximate each other, more or less as in the proof of the integral test of convergence.
If you'd like another sort of approach, we could approach it naively. Separate the sum into $250$ element blocks, $\frac{1}{1001}$ to $\frac{1}{1250}$ in the first block $B_1$, $\frac{1}{1251}$ to $\frac{1}{1500}$ in the second block $B_2$, and so on. We'll have $12$ blocks.
Note that $\frac{1}{5} = \frac{250}{1250} \leq B_1 \leq \frac{250}{1000} = \frac{1}{4}$. Similarly, we get that $\frac{1}{i + 4} \leq B_i \leq \frac{1}{i+3}$ for all of our blocks.
This means that our sum has upper and lower bounds: | {
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This means that our sum has upper and lower bounds:
$$1 < \frac{1}{5} + \frac{1}{6} + \dots \frac{1}{16} \leq B_1 + \dots + B_{12} \leq \frac{1}{4} + \dots + \frac{1}{15}< \frac{3}{2}$$
And this gives the desired inequality.
-
It is approximately $\int_{1000}^{3000}dt/t = \log(3)$.
To explain why the Riemann Sum is close you should think about the approximating rectangles and the actual curve of $x \mapsto 1/x$. Since the curvature is nearly zero, the approximation will be very close.
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# Thread: vertices of a square, coordinate geometry and differentiation
1. ## vertices of a square, coordinate geometry and differentiation
The origin O and a point B(p, q) are opposite vertices of the square OABC. Find the coordinates of the point A and C
A line l has gradient q/p. Find possible values for the gradient of a line at 45° to l.
So I’m stuck on the first part of this problem and haven’t tried to tackle the second part yet, but I posted it anyway in case I have difficulties with it once I have solved the first part.
For the first question I observed that I have 4 unknowns, the two coordinates of each point. So I figured I should determine 4 equations that would help me find these; I used the following remarks to define these equations:
The gradient of AC will be perpendicular to the gradient of OB,
The gradient of OA will be perpendicular to the gradient of AB
The gradient of OC will be perpendicular to the gradient of CB,
The distance between A and C will be equal to the distance between O and B
However I’m not sure that this is the right (or at least, most effective) way to proceed as I’ve ended up with a mess of equations, often ending up quadratic, that appear to be leading me nowhere. Am I on the right track, have I used the wrong equations or am I just using a totally incorrect method?
2. Originally Posted by red and white kop!
The origin O and a point B(p, q) are opposite vertices of the square OABC. Find the coordinates of the point A and C
A line l has gradient q/p. Find possible values for the gradient of a line at 45° to l.
So I’m stuck on the first part of this problem and haven’t tried to tackle the second part yet, but I posted it anyway in case I have difficulties with it once I have solved the first part.
Assuming that the square has a side parallel to the two x-axis and $p>0~\&~q>0$ then
$A0,q)~\&~Cp,0).$
Is the second question part of the first? | {
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Is the second question part of the first?
3. The square doesn't necessarily have a side parallel to the axes. All information is given in the question. This could be any square.
4. Originally Posted by red and white kop!
The square doesn't necessarily have a side parallel to the axes. All information is given in the question. This could be any square.
In that case, find the points of intersection of the line $y-\frac{q}{2}=\frac{-p}{q}\left(x-\frac{p}{2}\right)$ and the circle $\left(x-\frac{p}{2}\right)^2+\left(y-\frac{q}{2}\right)^2=\frac{p^2}{4}+\frac{q^2}{4}$.
Those two points are $A~\&~C$.
5. Hello, red and white kop!
Did you make a sketch?
The origin O and a point B(p, q) are opposite vertices of the square OABC.
(a) Find the coordinates of the point A and C.
(b) A line L has gradient q/p. Find possible values for the gradient of a line at 45° to L.
Code:
|
| B
| o(p,q)
| *
| *
| M *
| o(p/2,q/2)
| *
| *
|*
- - o - - - - - -
O|
We are given points $O(0,0)\text{ and }B(p,q)$
. . and their midpoint $M\left(\frac{p}{2},\,\frac{q}{2}\right)$
Code:
|
| B
| o(p,q)
| *
| * ↑
| M *
| o → +
| *
| *
|*
- - o - - - - - -
O|
To move from $M$ to $B$, we move $\frac{p}{2}$ right and $\frac{q}{2}$ up.
Code:
|
| B
| o(p,q)
| *
| *
| M *
| o → +
| * ↓
| * ↓
|* o A
- - o - - - - -
O|
To move from $M$ to $A$, we must move $\frac{q}{2}$ right and $\frac{p}{2}$ down.
Do you see why?
. . Hence, $A$ is at $\left(\frac{p+q}{2},\,\frac{p-q}{2}\right)$
Similarly, we find that $C$ is at $\left(\frac{p-q}{2},\,\frac{p+2}{2}\right)$
6. Originally Posted by soroban
Hello, red and white kop!
Did you make a sketch?
Code:
|
| B
| o(p,q)
| *
| *
| M *
| o(p/2,q/2)
| *
| *
|*
- - o - - - - - -
O|
We are given points $O(0,0)\text{ and }B(p,q)$
. . and their midpoint $M\left(\frac{p}{2},\,\frac{q}{2}\right)$ | {
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Code:
|
| B
| o(p,q)
| *
| * ↑
| M *
| o → +
| *
| *
|*
- - o - - - - - -
O|
To move from $M$ to $B$, we move $\frac{p}{2}$ right and $\frac{q}{2}$ up.
Code:
|
| B
| o(p,q)
| *
| *
| M *
| o → +
| * ↓
| * ↓
|* o A
- - o - - - - -
O|
To move from $M$ to $A$, we must move $\frac{q}{2}$ right and $\frac{p}{2}$ down.
Do you see why?
. . Hence, $A$ is at $\left(\frac{p+q}{2},\,\frac{p-q}{2}\right)$
Similarly, we find that $C$ is at $\left(\frac{p-q}{2},\,\frac{p+2}{2}\right)$
Soroban, thanks for your time, your method seems to be the most applicable my level of comprehension and the section I’m studying.
However, when you say ‘do you see why’, I do not: it would help if you formulated your answer more algebraically and less with phrases and diagrams. That would help me understand your logic a lot better. Also your answer seems to be incorrect, as the answers given in the textbook are (0.5 (p-q), 0.5(p+q)) and (0.5(p+q), 0.5(q-p)); but I’ve discovered several errors in this textbook’s answer key recently so I’m not sure if you are wrong or the textbook is. I’d be thankful if you could keep helping me out until I’m done with this problem. Cheers
7. bump
8. Originally Posted by red and white kop!
bump
Code:
p - q
------
2
is the same thing as (1/2)*(p - q), right?
and 1/2 = 0.5
so, (p - q) / 2 = (1/2)(p - q) = 0.5 (p - q)
There...in nice algebra....you can see that the coordinates given by Soroban are EXACTLY the same points as those given in your text. | {
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9. Originally Posted by Mrspi
Code:
p - q
------
2
is the same thing as (1/2)*(p - q), right? and 1/2 = 0.5
so, (p - q) / 2 = (1/2)(p - q) = 0.5 (p - q)
There...in nice algebra....you can see that the coordinates given by Soroban are EXACTLY the same points as those given in your text
There is a problem with this. Soroban's solution does not agree with the solution given in the text.
The text is correct: $\left( {\frac{{p - q}}{2},\frac{{p + q}}{2}} \right)\;\& \,\left({\frac{{p + q}}{2},\frac{{q - p}}{2}} \right)$
That is not Soroban's solution.
10. Originally Posted by Mrspi
Code:
p - q
------
2
is the same thing as (1/2)*(p - q), right?
and 1/2 = 0.5
so, (p - q) / 2 = (1/2)(p - q) = 0.5 (p - q)
There...in nice algebra....you can see that the coordinates given by Soroban are EXACTLY the same points as those given in your text.
don't be stupid
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# can one derive the $n^{th}$ term for the series, $u_{n+1}=2u_{n}+1$,$u_{0}=0$, $n$ is a non-negative integer
derive the $n^{th}$ term for the series $0,1,3,7,15,31,63,127,255,\ldots$
observation gives, $t_{n}=2^n-1$, where $n$ is a non-negative integer
$t_{0}=0$
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Let $s_n=t_n+1$, then $s_{n+1}=2s_n$ and $s_0=1$, end of story. – Did Aug 17 '12 at 12:14
NOTE: We get $t_{n}=2^n-1$ by observation. Is there a mathematical process by which we can derive this. – Rajesh K Singh Aug 17 '12 at 12:28
There is a long discussion at en.wikipedia.org/wiki/Recurrence_relation which shows some techniques to find these solutions. Without the constant (1) and the initial condition, the solution is a2^n. did's shift gets rid of the constant. – Ross Millikan Aug 17 '12 at 12:54
Is there a mathematical process by which we can derive this... See my answer. – Did Aug 21 '12 at 9:34
You already have an expression for the $n^{\text{th}}$ term for the sequence:
$$t_{n}=2^{n}-1\tag{1}$$
So let's prove this closed form by induction. Let $P(n)$ be our proposition that $t_{n}=u_{n}$, $\forall n\in\mathbb{N}\cup\{0\}$. So let us examine our basis case: $P(0)$:
$$t_{0}=2^{0}-1=1-1=0=u_{0} \implies P(0) \text{ is true}$$
Let us now show that if $P(k)$ is true, it follows immediately that $P(k+1)$ must also be true:
$$t_{k+1}=2^{k+1}-1=2\cdot2^{k}-1=2(2^{k}-1)+1=2u_{k}+1=u_{k+1}\implies \text{ if } P(k) \text{ is true, then } P(k+1) \text{ is true}$$
Therefore, as we have shown that $P(0)$ is true, and that $P(k)\implies P(k+1)$. $P(n)$ is true, $\forall n\in\mathbb{N}\cup\{0\}$.
However, if you are interested in how to actually come up with a closed form in general, a good place to start is to look at summation factors.
We can reduce a general recurrence relationship of the form:
$$a_{n}T_{n}=b_{n}T_{n-1}+c_{n} \tag{2}$$
To a sum, by multiplying both sides by a summation factor, denoted as $s_{n}$, such that:
$$s_{n}b_{n}=s_{n-1}a_{n-1}$$ | {
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$$s_{n}b_{n}=s_{n-1}a_{n-1}$$
In general, we can find a suitable $s_{n}$ using any multiple of the following:
$$s_{n}=\frac{a_{n-1}a_{n-2}\cdots a_{1}}{b_{n}b_{n-1}\cdots b_{2}}\tag{3}$$
We then have a summation recurrence, to which the solution can be found to be:
$$T_{n}=\frac{1}{s_{n}a_{n}}\left(s_{1}b_{1}T_{0}+\sum_{k=1}^{n}s_{k}c_{k}\right)\tag{4}$$
In your case, we have a recurrence of the form given in $(2)$:
$$a_{n}=1\qquad b_{n}=2\qquad c_{n}=1$$
Therefore, using $(3)$ we have $s_{n}=\frac{1}{2^{n}}$. And we can therefore plug this into $(4)$ to give:
$$T_{n}=2^{n}\left(0+\sum_{k=1}^{n}{\frac{1}{2^{k}}}\right)=2^{n}\left(1-\frac{1}{2^{n}}\right)=2^{n}-1$$
Which is $(1)$, the closed form you got by observation.
If you want a more complete look at solving recurrence relationships, I'd recommend the first few chapters of Concrete Mathematics by Graham, Knuth and Patashnik.
Hope this helps!
-
If the question is how to guess the form of the solution $u_n$ for every $n\geqslant0$, as opposed to, how to check that some given formula for $u_n$ is right (which is what my first comment and the other answers given so far all focus on), here is a standard computation that may help.
Assume that $u_n=au_{n-1}+b$ for every $n\geqslant1$, for some given $a$ and $b$. Thus, $u_n=f(u_{n-1})$, where the function $f$ is defined by $f(x)=ax+b$ for every $x$.One knows that to iterate a function, in general, can rapidly lead to a complicated mess (and/or to fascinating mathematics, think about fractal geometry). Some exceptions are when $f$ is constant (then $u_n=u_1$ for every $n\geqslant1$) or when $f$ is linear, so let us first assume that $f$ is linear, that is, that $f(x)=ax$ for every $x$. Then $f^{n}(x)=a^nx$ (induction over $n\geqslant1$) hence $u_n=a^nu_0$ for every $n\geqslant0$ and we are done. | {
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The treatment of our general case $f(x)=ax+b$ cannot be made quite as simple but nearly so! To see this, we start with a simple remark. Define $v_n=u_n+c$ for some given $c$, to be chosen later. Then, $$v_n=(au_{n-1}+b)+c=a(v_{n-1}-c)+b+c=av_{n-1}+b_c,\qquad b_c=b-c(a-1).$$ Thus $v_n=f_c(v_{n-1})$, where $f_c$ is a new affine function, defined by $f_c(x)=ax+b_c$, and we are still in the case we wanted to solve at the beginning, hence it seems we have been running in circles. BUT... if by chance $f_c$ is in fact linear, we are done since we know how to solve the linear case! Perhaps we happy, after all?
Which brings us to the equation $b_c=0$, solved by $c^*=b/(a-1)$ as soon as $a\ne1$. And then, everything flows easily: $v_n=av_{n-1}$ for every $n$, hence $v_n=a^nv_0$ by the preceding analysis, that is, $u_n+c^*=a^n(u_0+c^*)$, that is, $u_n=a^n(u_0+c^*)-c^*$ and we are done.
If $a=2$, $b=1$ and $u_0=0$, then $c^*=1/(2-1)=1$ and $u_n=2^n(0+1)-1=2^n-1$, which is the specific case asked about here.
Two remarks. First, once the idea explained above is understood, one can go directly from the recursion $u_n=au_{n-1}+b$ with $a\ne1$ to the solution $u_n=a^n(u_0+c^*)-c^*$ for some $c^*$ to be determined (for example, $u_1=a(u_0+c^*)-c^*$ hence $c^*=(u_1-au_0)/(a-1)$, but, to remember this exact formula is not necessary). Second, the case $a=1$ is solved by a specific (simple) analysis I will let you discover.
-
The following is a semi-formal variant of induction that is particularly useful for recurrences.
Let $x_n=2^n-1$. It is easy to verify that $x_0=0$. It is also easy to verify that $$x_{n+1}=2x_n+1,$$ since $2^{n+1}-1=2(2^n-1)+1$.
So the sequence $(x_n)$ starts in the same way as your sequence and obeys the same recurrence as your sequence. Thus the two sequences must be the same.
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Consider the series, $0, 1, 3, 7, 15, 31, 63,\ldots$ | {
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-
Consider the series, $0, 1, 3, 7, 15, 31, 63,\ldots$
On taking the difference between the terms one can see that the difference ceases to vanish and the difference becomes $1,2,4,8,16,32,\ldots,2^{n}$ just after the first stage. Here $n$ is a non-negative integer.
i.e. the general term of the expression is of the form $2^{(x-1)}+ax+b$
when, $x=1$, $2^{(1-1)}+a+b=0$
i.e., $a+b=-1$
when, $x=2$, $2^{(2-1)}+2a+b=1$
i.e., $2a+b=-1$
we have, $a =0$, $b=-1$
we can now conclude that the $n_{th}$ term of the series is $2^{n-1}-1$, where $n$ is a positive integer.
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And the area of this surface element $\mathrm{d}A = \frac{1}{2}r^2(\theta)\mathrm{d}\theta$. An analyst at the Scotland Department of Environment is performing a preliminary review on wind farm applications to determine which ones overlap with or are in view of wild lands. • The coordinates ( and ) define the center of gravity of the plate (or of the rigid body). Recall that the centroid of a triangle is the point where the triangle's three medians intersect. With this centroid calculator, we're giving you a hand at finding the centroid of many 2D shapes, as well as of a set of points. Find the coordinates of the centroid of the plane area bounded by the parabola y = 4 – x^2 and the x-axis. Centroid of an Area • In the case of a homogeneous plate of uniform thickness, the magnitude ∆W is Centroid: Centroid of a plane figure is the point at which the whole area of a plane figure is assumed to be concentrated. Find the coordinates of the centroid of the area bounded by the given curves. Solution: Next, sum all of the x coodinates ... how to find centroid of composite area: how to calculate centroid of rectangle: how to find centroid of equilateral triangle: Find the coordinates of the centroid of the area bounded by the given curves. Gather both the x and y coordinate points of each vertex. It is the point which corresponds to the mean position of all the points in a figure. y=2 x, y=0, x=2 The centroid of a right triangle is 1/3 from the bottom and the right angle. It is also the center of gravity of the triangle. How to calculate a centroid. First, gather the coordinate points of the vertices. Determine the x - and y -coordinates of the centroid of the shaded area. We can consider the surface element as an triangle, and the centroid of this triangle is obviously at here.) Centroid of a Volume The centroid defines the geometric center of an object. Centroid by Composite Bodies ! For more see Centroid of a triangle. Problem Answer: The coordinates of the center of the | {
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Bodies ! For more see Centroid of a triangle. Problem Answer: The coordinates of the center of the plane area bounded by the parabola and x-axis is at (0, 1.6). The x-centroid would be located at 0 and the y-centroid would be located at 4 3 r π 7 Centroids by Composite Areas Monday, November 12, 2012 Centroid by Composite Bodies The centroid is the term for 2-dimensional shapes. 4' 13 Answers: (X,Y) in y=x^{3}, x=0, y=-8 Chapter 5, Problem 5/051 (video solution to similar problem attached) Determine the x- and y-coordinates of the centroid of the shaded area. The Find Centroids tool will create point features that represent the geometric center (centroid) for multipoint, line, and area features.. Workflow diagram Examples. For example, the centroid location of the semicircular area has the y-axis through the center of the area and the x-axis at the bottom of the area ! The centroid or center of mass of beam sections is useful for beam analysis when the moment of inertia is required for calculations such as shear/bending stress and deflection. The center of mass is the term for 3-dimensional shapes. The cartesian coordinate of it's centroid is $\left(\frac{2}{3}r(\theta)\cos\theta, \frac{2}{3}r(\theta)\sin\theta\right)$. Determine the coordinates of the centroid of the shaded region. Beam sections are usually made up of one or more shapes. Center of Mass of a Body Center of mass is a function of density. So to find the centroid of an entire beam section area, it first needs to be split into appropriate segments. The coordinates of the centroid are simply the average of the coordinates of the vertices.So to find the x coordinate of the orthocenter, add up the three vertex x coordinates and divide by three. For instance, the centroid of a circle and a rectangle is at the middle. Beam section area, it first needs to be split into appropriate segments point the. A function of density of an object area bounded by the given curves given... Mass of a plane figure is assumed | {
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The geometric center of an object three medians intersect the vertices first needs to be into... Rectangle is at the middle for 3-dimensional shapes sections are usually made up of or... The point where the triangle 's three medians intersect obviously at here., gather coordinate., x=2 Find the coordinates of the vertices area of a triangle is 1/3 from the and! Points in a figure figure is assumed to be concentrated of the vertices: centroid of an object all points... Position of all the points in a figure is also the center of mass is the which! By the given curves points determine the coordinates of the centroid of the area a figure an triangle, and the right angle is obviously here!: centroid of an object y coordinate points of the triangle to the mean position of all the in... The center of gravity of the triangle 's three medians intersect area, it first needs to concentrated... Of gravity of the vertices points of each vertex center of mass is a function density! Mean position of all the points in a figure to the mean of... To be concentrated gather the coordinate points of the shaded region center of mass is a function of.... A figure right triangle is obviously at here. point which corresponds to the mean position of all points! 1/3 from the bottom and the right angle all the points in figure... And a rectangle is at the middle shaded area made up of one or more shapes an object 's medians. Area of a right triangle is obviously at here. where the triangle shaded! That the centroid of this triangle is the point at which the whole of. | {
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# Prime Factorization of $m^2$
If $m$ is a positive integer, explain why each prime in the prime factorization of $m^2$ must occur an even number of times.
I did a small proof, I was wondering what's a nice way to explain it aside from my explanation below:
Proof: let $m=p_1^{e_1} p_2^{e_2}p_3^{e_3}\cdots p_k^{e_k}$
Then $m^2=(p_1^{e_1})^2 (p_2^{e_2})^2(p_3^{e_3})^2\cdots (p_k^{e_k})^2 = p_1^{2e_1}p_2^{2e_2}p_3^{2e_3}\cdots p_k^{2e_k}$
These $p_i^{e_i}$ occur an even number of times any suggestions to make a more solid answer?
• I would not put both formulas for $m^2$ on the same line, but otherwise it's the right proof. Maybe say explicitly that raising to a power distributes over multiplication $(ab)^n=a^nb^n$ because multiplication is commutative. Otherwise, I think it's good. – Gregory Grant Jul 27 '17 at 2:01
• One suggestion: you also need to explicitly invoke the uniqueness of prime factorizations to conclude that is the only such factorization of $m^2$ (else it might have another factorization where some prime has odd exponent). – Bill Dubuque Jul 27 '17 at 2:02
• Actually don't say "These $p_i^{e_i}$ occur an even number of times. Say $p_i$ occurs $2e_i$ times which is even. – Gregory Grant Jul 27 '17 at 2:02
• @GregoryGrant yeah thanks my issue was that explanation it did not sit well with me either. – OLE Jul 27 '17 at 2:04
• @BillDubuque is their a nicer more stronger way of proving this? – OLE Jul 27 '17 at 2:06
It's in fact an if and only if :
Each prime in "the" (due to uniqueness of p.f up to multiplication by units) prime factorization of $$k$$ occurs an even number of times, if and only if $$k$$ is a perfect square.
Suppose each prime in the prime factorization of $$k$$ occurs an even number of times, say $$k = \prod p_i^{r_i}$$, where each $$r_i$$ is even. Then, you can see that if $$b = \prod p_i^{\frac {r_i} 2}$$, then $$b$$ is a well defined integer, and $$b^2 = k$$. | {
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Conversely, note that if $$k = m^2$$ is a perfect square, then the prime factorization of $$m = \prod p_i^{r_i}$$ suggests the prime factorization for $$k = \prod p_i^{2r_i}$$. Since prime factorization is unique, it follows that $$k$$ can indeed only be prime factorized in the above form, and hence every prime appears evenly many times in the rime factorization.
Alternately, we can also go by this way : Suppose $$p^r$$ divides $$m^2$$, where $$r$$ is maximal. Suppose $$r$$ is even, then we are done. Otherwise, note that $$r = 2k+1$$, and we can write $$p \ \mid\ \frac {m^2}{p^{2k}}$$, so that $$p$$ divides a perfect square.
Hence, from here, using Euclid's lemma, that $$p$$ divides $$ab$$ implies $$p$$ divides $$a$$ or $$p$$ divides $$b$$, we get that taking $$a=b=\frac m{p^k}$$, $$p^{k+1} \mid m$$, or that $$p^{2k+2} \mid m^2$$, a contradiction by uniqueness of prime factorization.
On a little inspection,the second proof is just a longer winded version of the first proof, but two is better than one, I suppose.
EXTENSION : If $$m$$ is a perfect $$k$$th power, then every prime that appears in the factorization does so, with multiplicity a multiple of $$k$$. | {
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# How to prove this inequality $(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64$?
Question:
If $a,b,c$ are nonnegative real numbers such that $a+b+c=3,$ then
$$(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64$$
My try: I found the equality holds only if $(a,b,c)=(2,0,1)$ or all of its permutations.
But I can't prove this inequality it. I would appreciate very much a proof.
This problem comes from:http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=562119
• Do you mean "I found that equality holds only if $(a,b,c)$ is $(2,0,1)$, $(1,2,0)$, or $(0,1,2)$"?. – John Bentin Nov 13 '13 at 11:05
• Yes,That's mean,Thank you – math110 Nov 13 '13 at 11:09
• If you can prove one of them is zero, then you can just study $f(2+x,0,1-x)$ and see that its maximum is $64$. – Beni Bogosel Nov 13 '13 at 12:20
• You can edit your question so that it says what you mean. – John Bentin Nov 13 '13 at 12:26
• @math110: I've taken the liberty of editing it for you. Where did you get this question from, if you don't mind me asking. – Nick Nov 13 '13 at 13:05
## 2 Answers
→ → | {
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## 2 Answers
→ →
The three dimensional problem can be simplified to a two dimensional problem by introducing (again and again) suitable triangle coordinates: $$\left[ \begin{array}{c} a \\ b \\ c \end{array} \right] = \left[ \begin{array}{c} 3 \\ 0 \\ 0 \end{array} \right] + \left[ \begin{array}{c} -3 \\ 3 \\ 0 \end{array} \right] x + \left[ \begin{array}{c} -3 \\ 0 \\ 3 \end{array} \right] y$$ Then the equation $(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64$ does not so much "simplify", but anyway becomes an equation in two variables (2-D). And the equation $\;a + b + c = 3\;$ corresponds with a normed 2-D triangle, with vertices $(0,0),(1,0),(0,1)$ . The insides of both can easily be visualized, as has been done in the above picture in the middle:
$\color{red}{red}$ for $\;a + b + c = 3\;$ and $\color{green}{green}$ for $\;(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64\;$ .
The transformed inequality is: $$\left[ 3^2\left( 1-x-y \right)^{2} + 3^5\,x{y}^{4} \right] \left[ 3^2\,{x}^{2} + 3^5\,y \left( 1-x-y \right)^{4} \right] \left[ 3^2\,{y}^{2} + 3^5\, \left( 1-x-y \right){x}^{4} \right] \le 64$$
It is seen in the same picture that the edge $\;y=1-x\;$ of the triangle maybe is tangent to the curve $(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) = 64$ . Indeed, if we substitute $y=1-x$ into (the transformation of) that equation and simplify, then we get: $$3^9\, x^3 (1-x)^6 - 64 = 0$$ The same sort of equation is found with the substitutions $\,x=0\,$ or $\,y=0\,$, for the other two edges. And, as has been found by others, there is only one solution of that equation, within the specified range, namely $x=1/3$, corresponding with $y=2/3$ and hence $(a,b,c) = (0,1,2)$ . And of course any cyclic permutation of this, due to symmetry. The rest of the (red) triangle $\;a + b + c = 3\;$ is well within the (green) area of $\;(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64$ . Which can be shown by plotting the triangle first: then it becomes absorbed by the green area (see picture on the right). | {
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Analytically, the proof is completed by considering the function $\;f(x) = 3^9\, x^3 (1-x)^6$ .
Its extreme values are found for $\;f'(x) = x^2(1-x)^5(3-9x)=0$ , giving $\;x = \{0,1,1/3\}$ , with the maximum $\;f(1/3)=64$ .
The picture on the right shows the the inequality as observed in the plane $\;a + b + c = 3\;$ of the triangle in 3-D (picture on the left). Mind the symmetries.
• Can you use other colors? Since many people have protanopia or protanomaly (like me). – ziyuang Jan 8 '14 at 11:24
• Next time, I promise. Any suggestion? I've used red and green now. – Han de Bruijn Jan 8 '14 at 14:45
• Maybe different patterns of lines, if that's convenient for you. – ziyuang Jan 8 '14 at 23:05
Without loss of generality, assume $$a$$ is smallest of $$a, b, c$$. Also, let $$f(a,b,c)=(a^2+bc^4)(b^2+ca^4)(c^2+ab^4)$$and firstly, if $$a\le c\le b$$, then$$f(a,b,c)-f(a,c,b)=(b^3-a^3) (c^3 - a^3) (b ^3- c^3) (a b c - 1)<0$$therefore we can assume $$a\le b\le c$$. Now, we will prove $$f(0,b,a+c)\ge f(a,b,c)$$which is, after full expansion,$$a (-a^6 b^4 c + a^5 b^3 - a^5 c^3 - a^4 b^5 c^5 - a^3 b c^7 - a^2 b^6 + 20 a^2 b^3 c^3 \\+ 15 a b^3 c^4 - a b^2 c^2 - b^7 c^4 + 6 b^3 c^5)+ 6 a^5 b^3 c + 15 a^4 b^3 c^2\ge0$$ and it is obvious that $$-a^6 b^4 c + a^5 b^3\ge0$$, $$- a^5 c^3 - a^4 b^5 c^5 - a^2 b^6 + 20 a^2 b^3 c^3\ge0$$ and $$6 a^5 b^3 c + 15 a^4 b^3 c^2\ge0$$. Therefore it is enough to show$$b c^2 (-a^3 c^5 + 15 a b^2 c^2 - a b - b^6 c^2 + 6 b^2 c^3)\ge0$$and from $$-a^3c^5\ge-ab^2c^5$$, $$-ab\ge-abc^3\ge-b^2c^3$$ and $$-b^6c^2>-4b^3c^2\ge-4b^2c^3$$, it is left to show$$b^2c^2(-a c^3 + 15 a + c)\ge0$$and we can divide it with three cases.
Case 1) $$c^3\le15$$: $$-a c^3 + 15 a + c>a(15-c^3)\ge0$$. | {
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Case 1) $$c^3\le15$$: $$-a c^3 + 15 a + c>a(15-c^3)\ge0$$.
Case 2) $$15^{1/3}< c\le2.6$$: Firstly, $$c>2.4$$. The equation is decreasing with respect to $$a$$, therefore we need to show only for maximal value of $$a$$. If $$a>0.3$$, then $$a+b+c\ge2a+c>0.6+2.4=3$$, so maximal value is $$0.3$$. Also, $$c^3\le2.6^3<20$$. Therefore, $$-a c^3 + 15 a + c\ge0.3(15-c^3)+c\ge-1.5+2>0$$
Case 3) $$2.6: Similarly, it is enough to show for maximal value of $$a$$ which is $$0.2$$. Therefore, $$-a c^3 + 15 a + c\ge0.2(15-27)+2.6>0$$
Therefore, we can assume that $$0=a\le b\le c$$. Now $$f(a,b,c)=b^3c^6\le2^6\left(\frac{3\times b+6\times0.5c}{9}\right)^9=64$$ and it is proved. | {
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Showing that $Q_n=D_n+D_{n-1}$
Let $$T_n$$ be the set of permutations of $$\{1,2,\ldots,n\}$$ which do not have $$i$$ immediately followed by $$i+1$$ for $$1\le i\le n-1$$; in other words, let \begin{align} T_n=\{\sigma \in S_n: \sigma(i)+1\ne\sigma(i+1) \text{ for all } 1\le i\le n-1\} . \end{align} Let $$Q_n$$ be the number of elements of $$T_n$$.
Let $$D_n$$ be the number of derangements of $$\{1,2,\ldots,n\}$$.
It is not hard to show algebraically that $$Q_n=D_n+D_{n-1}$$, but I am having difficulty coming up with a combinatorial argument to show why this is true. (I believe there are $$D_{n-1}$$ elements of $$T_n$$ that leave $$n$$ fixed, and $$D_n$$ elements of $$T_n$$ that move $$n$$, but I don't know how to justify this combinatorially.) | {
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• just to be clear, "i followed by i+1" means that $\sigma(i)+1$ is never equal to $sigma(i+1)$, for all $i\leq n-1$ ? – Denis Jul 25 '13 at 0:52
• I'm sorry if my statement of the problem wasn't completely clear, but that's what I meant. (In other words, when you arrange the images of 1,..., n in order, i is never followed by i+1.) – user84413 Jul 26 '13 at 2:10
• It seems that your question has been asked and answered previously. I do have some doubt about Brian Scott's answer, and I've left a comment there requesting clarification, but it may just be a simple misunderstanding on my part. – Will Orrick Sep 10 '13 at 2:42
• @ Will Orrick Thank you for pointing out that my question had been asked previously. I read the answer posted, but I wasn't able to see how to make it work. (For example, it seemed to me that the permutations 1324 and 4132 would both map to the permutation 2314, if I am interpreting the answer correctly.) I like your argument to show that the number of elements of $T_n$ that fix n is equal to $D_{n-1}$, even though it isn't what I originally had in mind. Can you use the same idea to show that the number of elements in $T_n$ which do not fix n is equal to $D_n$? Thanks again. – user84413 Sep 10 '13 at 17:45
• It turns out Persi Diaconis, Steven Evans, and Ron Graham have a recent paper exploring this problem in some depth. (arxiv.org/abs/1308.5459 ) – Kevin P. Costello Oct 14 '13 at 23:53
(Note: this answer does not contain a bijection between the two sets, which is what the OP was originally hoping for, but it does show that the two counting problems are structurally equivalent and therefore have the same answer.)
The belief stated in the last paragraph of your question is correct: we can show that the set $\mathcal{D}_{n-1}$ of derangements of $\{1,2,3,\ldots,n-1\}$ and the set of permutations $\sigma$ of $\{1,2,3,\ldots,n\}$ with the properties
• $\sigma(n)=n$,
• $\sigma(i+1)\ne\sigma(i)+1,$ for $1\le i\le n-1,$ | {
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• $\sigma(n)=n$,
• $\sigma(i+1)\ne\sigma(i)+1,$ for $1\le i\le n-1,$
are equinumerous. That is, $\mathcal{D}_{n-1}$ and the set of elements of $T_n$ that fix $n$ have the same size. Call the latter set $\overline{T}_n.$
This result will prove the equation in your title since there is an obvious bijection between the set of elements of $T_n$ that do not fix $n$ and the set $\overline{T}_{n+1}$ of elements of $T_{n+1}$ that do fix $n+1,$ which will then be equinumerous with $\mathcal{D}_n.$
To enumerate $\mathcal{D}_{n-1}$, let $F=S_{n-1}$ be the set of permutations of $1,2,\ldots,n-1.$ Let $F_j$ be the set of elements of $F$ that fix $j,$ that is, elements $\sigma$ such that $\sigma(j)=j.$ In general, let $F_{ijk\ldots}=F_i\cap F_j\cap F_k\cap\ldots$ be the set of elements that fix $i,$ $j,$ $k,\ldots$, that is, elements $\sigma$ such that $\sigma(i)=i,$ $\sigma(j)=j,$ $\sigma(k)=k,\ldots$ Observe that $\lvert F_{i_1i_2\ldots i_k}\rvert=(n-1-k)!$ since only $n-1-k$ elements are free to move. Since the derangements are those elements that fix no element, the principle of inclusion-exclusion gives \begin{aligned}\lvert\mathcal{D}_{n-1}\rvert&=\lvert F\rvert-\sum_{i=1}^{n-1}\lvert F_i\rvert+\sum_{1\le i<j\le n-1}\lvert F_{ij}\rvert-\ldots\\ &=(n-1)!-\binom{n-1}{1}(n-2)!+\binom{n-1}{2}(n-3)!-\ldots\end{aligned} | {
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To enumerate $\overline{T}_n$, let $G$ be the set of permutations of $1,2,\ldots,n$ that fix $n.$ Let $G_i,$ $1\le i\le n-1,$ be the subset of $G$ consisting of elements $\sigma$ such that $\sigma(i+1)=\sigma(i)+1.$ In general, let $G_{ijk\ldots}=G_i\cap G_j\cap G_k\cap\ldots$ be the set of elements of $G$ such that $\sigma(i+1)=\sigma(i)+1,$ $\sigma(j+1)=\sigma(j)+1,$ $\sigma(k+1)=\sigma(k)+1,\ldots$ We claim that, once again, $\lvert G_{i_1i_2\ldots i_k}\rvert=(n-1-k)!.$ This follows by noting that every constraint $\sigma(i_j+1)=\sigma(i_j)+1$ reduces the number of elements that can move independently by $1,$ leaving only $n-1-k$ elements that are free to move. One can imagine element $i$ becoming "glued" to element $i+1,$ so that they must move as a block.
For example, let $n=9$ and consider $G_{12478}.$ Since $8$ must immediately precede $9,$ which is fixed, and $7$ must immediately precede $8,$ which is now fixed as well, the elements $789$ are fixed. At the same time, $2$ must immediately precede $3$ and $1$ must immediately precede $2,$ meaning that the string $123$ can only move as a block. Similarly, the string $45$ can only move as a block. As a consequence the number of permutations in $G_{12478}$ is the number of ways of permuting the "objects" $123,$ $45,$ and $6,$ with the object $789$ fixed in place. So $G_{12478}=3!=(9-1-5)!,$ in agreement with the claim.
Since $\lvert G_{ijk\ldots}\rvert=\lvert F_{ijk\ldots}\rvert$ for all choices of subscripts, the principle of inclusion-exclusion implies that the computation of $\lvert\overline{T}_n\rvert$ is identical to that of $\lvert\mathcal{D}_{n-1}\rvert,$ so that they have the same final value: \begin{aligned}\lvert\overline{T}_n\rvert&=\lvert G\rvert-\sum_{i=1}^{n-1}\lvert G_i\rvert+\sum_{1\le i<j\le n-1}\lvert G_{ij}\rvert-\ldots\\ &=(n-1)!-\binom{n-1}{1}(n-2)!+\binom{n-1}{2}(n-3)!-\ldots\end{aligned} | {
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Let $[a,b]=\{i\in\mathbf{Z}\mid a\le i\le b\}$ and let $[b]=[1,b].$ Let $S_{a,b}$ be the set of permutations of $[a,b]$ and let $S_n$ be the set of permutations of $[n].$ Let $\overline{S}_{a,b}=\{\sigma\in S_{a,b}\mid \sigma(b)=b\}$ be the set of permutations of $[a,b]$ that fix $b.$ Similarly, Let $\overline{S}_n=\{\sigma\in S_n\mid \sigma(n)=n\}$ be the set of permutations of $[n]$ that fix $n.$
The result will follow by exhibiting a bijective mapping $\rho_n:S_n\to\overline{S}_{n+1}$ that has the property that if $\sigma$ is an element of $S_n,$ $\tau$ is the element of $\overline{S}_{n+1}$ given by $\tau=\rho_n(\sigma),$ $f$ is an element of $[n],$ and $e$ is given by $\tau(e)=f,$ then the property $\tau(e+1)=\tau(e)+1=f+1$ holds if and only if $f$ is a fixed point of $\sigma.$
To define $\rho_n$ we introduce some notation. For distinct elements $i_1,\ldots,i_n\in[n]$ let $\langle i_1i_2\ldots i_n\rangle$ denote the element $\sigma\in S_n$ that satisfies $\sigma(j)=i_j$ for $1\le j\le n.$
We use usual cycle notation to represent swaps, which act from the left. So $(35)\langle123456\rangle=\langle125436\rangle,$ and $(23)(35)\langle123456\rangle=\langle135426\rangle.$ Every element of $S_n$ has a unique swap representation $(a_1b_1)(a_2b_2)\ldots(a_kb_k)\langle12\ldots n\rangle$ where $0\le k\le n-1,$ $a_1<a_2<\ldots<a_k,$ and $a_i<b_i$ for all $1\le i\le k.$ For example, $$\langle 326451\rangle=(13)\langle 126453\rangle=(13)(36)\langle 123456\rangle.$$ Observe that $f$ is a fixed point of $\sigma\in S_n$ if and only if $f$ does not appear among the $i_1,\ldots,i_k,j_1,\ldots,j_k$ in the swap representation of $\sigma.$ | {
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We define the insertion operator $[ai]:\overline{S}_{a+1,n+1}\to\overline{S}_{a,n+1},$ where $a\in[1,n]$ and $i\in[a,n].$ Let $\tau\in\overline{S}_{a+1,n+1}.$ Then $[ai]\tau$ is defined to be the element of $\overline{S}_{a,n+1}$ obtained by inserting $a$ immediately in front of $i+1.$ So, for example, $[24]\langle3546\rangle=\langle32546\rangle.$ Every element of $\overline{S}_{n+1}$ has a unique insertion representation $[1b_1][2b_2]\ldots[nb_n]\langle (n+1)\rangle,$ where $b_j\ge j$ for all $1\le j\le n.$ For example, \begin{aligned} &\langle 1456237\rangle=[13]\langle 456237\rangle=[13][22]\langle 45637\rangle =[13][22][36]\langle 4567\rangle\\ &=[13][22][36][44]\langle 567\rangle=[13][22][36][44][55]\langle 67\rangle=[13][22][36][44][55][66]\langle 7\rangle. \end{aligned} Observe that $a$ immediately precedes $a+1$ in $\tau\in\overline{S}_{n+1}$ if and only if the only occurrence of $a$ in the insertion representation of $\tau$ is of the form $[aa].$
The bijective mapping $\rho_n:S_n\to\overline{S}_{n+1}$ is now defined as follows. Let $\sigma\in S_n$ and compute the swap representation of $\sigma.$ We may add null swaps of the form $(aa)$ so that the swap representation of $\sigma$ takes the form $(1b_1)(2b_2)\ldots(nb_n)\langle 123\ldots n\rangle,$ where $b_j\ge j$ for all $1\le j\le n.$ Then define $\rho_n(\sigma)$ by replacing swaps with insertions and replacing $\langle 123\ldots n\rangle$ with $\langle (n+1)\rangle:$ $$\rho_n(\sigma)=[1b_1][2b_2]\ldots[nb_n]\langle (n+1)\rangle.$$ | {
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Examples: We show $\rho_3:S_3\to\overline{S}_4$ and $\rho_4:S_4\to\overline{S}_5$ explicitly. On the left, dots mark fixed points; on the right, overscores mark sequences of consecutive elements. \begin{aligned} \langle\dot{1}\dot{2}\dot{3}\rangle=(11)(22)(33)\langle123\rangle&\mapsto[11][22][33]\langle4\rangle=\langle\overline{1234}\rangle\\ \langle\dot{1}32\rangle=(11)(23)(33)\langle123\rangle&\mapsto[11][23][33]\langle4\rangle=\langle3\overline{12}4\rangle\\ \langle21\dot{3}\rangle=(12)(22)(33)\langle123\rangle&\mapsto[12][22][33]\langle4\rangle=\langle21\overline{34}\rangle\\ \langle231\rangle=(12)(23)(33)\langle123\rangle&\mapsto[12][23][33]\langle4\rangle=\langle1324\rangle\\ \langle312\rangle=(13)(23)(33)\langle123\rangle&\mapsto[13][23][33]\langle4\rangle=\langle3214\rangle\\ \langle3\dot{2}1\rangle=(13)(22)(33)\langle123\rangle&\mapsto[13][22][33]\langle4\rangle=\langle\overline{23}14\rangle \end{aligned} Observe that the two derangements, $\langle231\rangle$ and $\langle312\rangle,$ have images that contain no sequences of consecutive elements. | {
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\begin{aligned} \langle\dot{1}\dot{2}\dot{3}\dot{4}\rangle=(11)(22)(33)(44)\langle1234\rangle&\mapsto[11][22][33][44]\langle5\rangle=\langle\overline{12345}\rangle\\ \langle\dot{1}\dot{2}43\rangle=(11)(22)(34)(44)\langle1234\rangle&\mapsto[11][22][34][44]\langle5\rangle=\langle4\overline{123}5\rangle\\ \langle\dot{1}32\dot{4}\rangle=(11)(23)(33)(44)\langle1234\rangle&\mapsto[11][23][33][44]\langle5\rangle=\langle3\overline{12}\,\overline{45}\rangle\\ \langle\dot{1}342\rangle=(11)(23)(34)(44)\langle1234\rangle&\mapsto[11][23][34][44]\langle5\rangle=\langle\overline{12}435\rangle\\ \langle\dot{1}423\rangle=(11)(24)(34)(44)\langle1234\rangle&\mapsto[11][24][34][44]\langle5\rangle=\langle43\overline{12}5\rangle\\ \langle\dot{1}4\dot{3}2\rangle=(11)(24)(33)(44)\langle1234\rangle&\mapsto[11][24][33][44]\langle5\rangle=\langle\overline{34}\,\overline{12}5\rangle\\ \langle21\dot{3}\dot{4}\rangle=(12)(22)(33)(44)\langle1234\rangle&\mapsto[12][22][33][44]\langle5\rangle=\langle21\overline{345}\rangle\\ \langle2143\rangle=(12)(22)(34)(44)\langle1234\rangle&\mapsto[12][22][34][44]\langle5\rangle=\langle42135\rangle\\ \langle231\dot{4}\rangle=(12)(23)(33)(44)\langle1234\rangle&\mapsto[12][23][33][44]\langle5\rangle=\langle132\overline{45}\rangle\\ \langle2341\rangle=(12)(23)(34)(44)\langle1234\rangle&\mapsto[12][23][34][44]\langle5\rangle=\langle24135\rangle\\ \langle2413\rangle=(12)(24)(34)(44)\langle1234\rangle&\mapsto[12][24][34][44]\langle5\rangle=\langle41325\rangle\\ \langle24\dot{3}1\rangle=(12)(24)(33)(44)\langle1234\rangle&\mapsto[12][24][33][44]\langle5\rangle=\langle1\overline{34}24\rangle\\ \langle312\dot{4}\rangle=(13)(23)(33)(44)\langle1234\rangle&\mapsto[13][23][33][44]\langle5\rangle=\langle321\overline{45}\rangle\\ \langle3142\rangle=(13)(23)(34)(44)\langle1234\rangle&\mapsto[13][23][34][44]\langle5\rangle=\langle21435\rangle\\ | {
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\langle3\dot{2}1\dot{4}\rangle=(13)(22)(33)(44)\langle1234\rangle&\mapsto[13][22][33][44]\langle5\rangle=\langle\overline{23}1\overline{45}\rangle\\ \langle3\dot{2}41\rangle=(13)(22)(34)(44)\langle1234\rangle&\mapsto[13][22][34][44]\langle5\rangle=\langle14\overline{23}5\rangle\\ \langle3412\rangle=(13)(24)(33)(44)\langle1234\rangle&\mapsto[13][24][33][44]\langle5\rangle=\langle31425\rangle\\ \langle3421\rangle=(13)(24)(34)(44)\langle1234\rangle&\mapsto[13][24][34][44]\langle5\rangle=\langle14325\rangle\\ \langle4123\rangle=(14)(24)(34)(44)\langle1234\rangle&\mapsto[14][24][34][44]\langle5\rangle=\langle43215\rangle\\ \langle41\dot{3}2\rangle=(14)(24)(33)(44)\langle1234\rangle&\mapsto[14][24][33][44]\langle5\rangle=\langle\overline{34}215\rangle\\ \langle4\dot{2}13\rangle=(14)(22)(34)(44)\langle1234\rangle&\mapsto[14][22][34][44]\langle5\rangle=\langle4\overline{23}15\rangle\\ \langle4\dot{2}\dot{3}1\rangle=(14)(22)(33)(44)\langle1234\rangle&\mapsto[14][22][33][44]\langle5\rangle=\langle\overline{234}15\rangle\\ \langle4312\rangle=(14)(23)(34)(44)\langle1234\rangle&\mapsto[14][23][34][44]\langle5\rangle=\langle24315\rangle\\ \langle4321\rangle=(14)(23)(33)(44)\langle1234\rangle&\mapsto[14][23][33][44]\langle5\rangle=\langle32415\rangle \end{aligned} Observe that the images of the nine derangements contain no sequences of consecutive elements. | {
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Since there is a one-to-one correspondence between the set of derangements of $[n]$ and the set of permutations of $[n+1]$ that fix $n+1$ and contain no sequences of consecutive elements (we let $\overline{T}_{n+1}$ denote this set), we have $$D_n=\lvert \overline{T}_{n+1}\rvert.$$ But there is an immediate bijection between $\overline{T}_{n+1}$ and $T_n-\overline{T}_n,$ the set of elements of $T_n$ that do not fix $n.$ Therefore $$D_n=\lvert T_n-\overline{T}_n\rvert$$ and $$D_n+D_{n-1}=\lvert T_n-\overline{T}_n\rvert+\lvert \overline{T}_n\rvert=\lvert T_n\rvert.$$
• Thanks - this is a great answer! – user84413 Oct 6 '13 at 16:40
• Very nice indeed, and thank you. – Kevin P. Costello Oct 7 '13 at 17:21
Let $\sigma$ be the circular permutation $i\mapsto i+1$.
Then, you can establish a bijection $D_n\cup D_{n-1}\to T_n$ (I abuse notation for $D_n$ to design the set as well as the cardinal) by mapping any $\tau$ to $\tau\circ\sigma$,where elements of $D_{n-1}$ are naturally extended with $n\mapsto n$. It is straightforward to verify that this is a bijection.
• If I take the derangement 54123 of {1,...,5} and compose it with $\tau$, does that give the permutation 41235 (which is not in $T_n$)? – user84413 Jul 24 '13 at 20:33
• I might have misunderstood your encoding of permutations. I understood "do not have i followed by i+1" as "the image of i is never i+1". I change the answer. (But the image of 54123 was 15234) – Denis Jul 25 '13 at 0:49
EDIT: What's below is actually WRONG, due to multiple issues noted in the comments (the equation taken modulo $n$ leads to situations where a consecutive $n1$ causes problems, and it's not necessarily a bijection.
I think a variant on dkuper's argument can be made to work. | {
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I think a variant on dkuper's argument can be made to work.
If I understand correctly, the permutations in $Q_n$ are those which satisfy $\sigma(i)+1 \neq \sigma(i+1)$ for $1 \leq i \leq n-1$. Conversely, we can think of $D_n+D_{n-1}$ as consisting of those permutations satisfying $\tau(i) \neq i$ for $1 \leq i \leq n-1$ ($D_n$ corresponds to those permutations with $\tau(n) \neq n$ as well, while $D_{n-1}$ corresponds to fixing $\tau(n)=n$).
This suggests that we construct our bijection in such a way that $\sigma(i)+1=\sigma(i+1)$ if and only if $\tau(i)=i$. The former equation can be rewritten as $i=\sigma^{-1} (\sigma(i+1)-1)$. So if we define our bijection by taking $\sigma$ to the permutation satisfying $$\tau(i)=\sigma^{-1}\left(\sigma(i+1)-1\right),$$ where addition and subtraction are taken modulo $n$, things work the way we want them to. | {
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• Great answer! I'm not sure if in the question the addition is taken modulo, but if it is, it's finally the answer to one of my favourite questions:) – savick01 Sep 9 '13 at 19:40
• This is exactly what I tried at one point, because it looked as if this should work; but I ran into some difficulties. For example, the permutation 4132 (so $\sigma(1)=4, \sigma(2)=1$, etc) in $T_n$ maps to 1423, which is not in $D_4$; and the permutation 2413 maps to 4213, which also is not in $D_4$ (if I am computing these correctly). Also, 3214 maps to 3412, which is not in $D_3$. Maybe some modification of this will work, though, if this doesn't already. – user84413 Sep 9 '13 at 21:07
• There's actually multiple issues here (in addition to the issue you noted where I'm treating $41$ appearing consecutively as bad, $\sigma \rightarrow \tau$ isn't even a bijection!). I'm not sure what the proper etiquette in terms of deletion is, but for now I'm leaving it up with a note explaining what's wrong, and I'll eventually edit the answer if it seems fixable. – Kevin P. Costello Sep 9 '13 at 21:39
• Thanks for your response, and I don't know about the proper etiquette either. I have had to delete some of my answers to problems, since they were totally wrong; but yours is actually a pretty reasonable approach, which hopefully is fixable in some way. – user84413 Sep 9 '13 at 22:52
• @Kevin Costello It was pointed out to me that my question had been asked previously (see above), and the answer that was given before is identical to yours (although written differently). – user84413 Sep 11 '13 at 19:32 | {
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How can I tell if the sequence $a_n=\frac {2^n} {n!}$ converges, and if it converges what is its limit?
I know this question was asked before, but I couldn't find it and I think my question is slightly different. I'm new here so if I do something wrong, I'm sorry, I'm still trying to learn.
I need to "study the limit behavior of the sequence and when it exists compute their limits."
This is what I have tried so far:
I studied the first few terms of the sequence, and I claimed that the sequence was monotone and decreasing. Then I showed that $a_{n+1}-a_n\leq0$ by: $$a_{n+1}-a_n= \frac {2^{n+1}} {(n+1)!}\ - \frac {2^n} {n!}= \frac {2^{n+1}} {n!(n+1)}- \frac {2^n} {n!}=\frac {2^n} {n!}(\frac {2} {n+1})-\frac {2^n} {n!}=\frac {2^n} {n!}(\frac {2} {n+1}-1)=\frac {2^n} {n!}(\frac {n-1} {n+1})$$ Then I said: $\frac {2^n} {n!}(\frac {n-1} {n+1})\leq 0$ only if $\frac {n-1} {n+1}\leq 0$ because $2^n>0$ and $n!>0$ . Then I said that $n+1>0$ for n existing the naturals and $1-n\leq0$, thus $\frac {n−1} {n+1}\leq 0$ and $a_{n+1}-a_n\leq 0$.
Therefore the sequence is monotonic and decreasing. Because the first term is 2 and $\frac {2^n} {n!}>0$, the sequence is bounded and together the monotonicity and and boundedness means that a limit exists.
Now I feel that what I have done so far is correct, and instinctually I know the limit is $0$, but how do I compute that? Do I just claim it is $0$ and prove it using the epsilon definition of a limit?
We haven't covered anything with series, but we have covered cauchy sequences (is that supposed to help me?) and things about lim (inf) and lim (sup). Any help please?
-
@JavaMan Just realized that I do know about the ratio test. But now I have a more general question, should I have appraoched this problem with the ratio test first instead of checking for monotonicity and boundedness? Also, is there anything specific that alerts me to use the ratio test instead of checking monotonicity/boundedness? – user66807 Mar 15 '13 at 2:28 | {
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I'd suggest taking the $\lim_{n\to \infty}$ of the ratio of two consecutive terms (use the ratio test) : $$\lim_{n\to \infty} \dfrac{a_{n +1}}{a_n}\; = \lim_{n\to \infty}\dfrac{\dfrac{2^{n+1}}{(n+1)!}}{\dfrac{2^n}{n!}}$$ $$= \lim_{n\to \infty} \dfrac{2^{n+1}n!}{2^n(n+1)!} = \lim_{n\to \infty} \dfrac{2}{n+1}$$
Evaluate the limit: If the limit exists, and $L \lt 1$, then as DonAntonio aptly clarified, "the series whose general term is the given sequence converges," by the Ratio Test for series. And then by Convergence test for series, if its series converges, the limit of the sequence itself (the general term of the series), is equal to $0$.
-
Yes, that is what I meant, Thanks @DonAntonio: I will edit my post and credit your phrasing! – amWhy Mar 15 '13 at 2:59
I came here but you were gone. + – S. Snape Mar 15 '13 at 5:28
For every $x \in \mathbb{C}$ it is well-known that $\sum\limits_{n=0}^{\infty} \frac{x^n}{n!}$ converges to $\mathrm{exp}(x)$, hence $\frac{x^n}{n!}$ converges to $0$.
-
Recall: For a sequence $\{u_n\}$ in $\mathbb R^+,$
• $\lim \dfrac{u_{n+1}}{u_n}<1\implies \lim u_n=0;$
• $\lim \dfrac{u_{n+1}}{u_n}> 1\implies \lim u_n=+\infty;$
• For $\lim \dfrac{u_{n+1}}{u_n}=1$ no definite conclusion can be made e.g. consider $\left\{\dfrac{n+1}{n}\right\}$ and $\left\{\dfrac{1}{n}\right\}.$
Here, $\dfrac{a_{n+1}}{a_n}=\dfrac{2}{n+1}\to0<1.$ Therefore $\{a_n\}$ converges to $0.$
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A few ways to look at it.
The sum of all $a_n$ is finite (consider the power series for $e^x$ at $x=2$).
For $n>5$ every term is at most half the previous one.
$2^n$ grows quickly, but not nearly as quickly as $n!$.
-
$$\frac{2^n}{n!}=\frac{2}{1}[\frac{2}{2}\frac{2}{3}..\frac{2}{n-1}]\frac{2}{n} \leq \frac{2}{1}\cdot[ 1 \cdot 1..\cdot 1]\frac{2}{n}\leq \frac{4}{n}$$
Since $0 \leq \frac{2^n}{n!} \leq \frac{4}{n}$ the sequence converges to $0$.
-
Six solutions have already been written but I have a different one: | {
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-
Six solutions have already been written but I have a different one:
Use the bound* $$\left(\frac n e\right)^n \le n! \le n^n$$
This shows that $$0 \le \frac {2^n} {n!} \le \left( \frac {2e} n \right)^n \longrightarrow 0$$
giving the result.
*Proof of bound: The upper bound (which I did not use) follows trivially from the definition of the factorial. The lower bound follows from a quick and dirty evaluation of the Gamma function integral,
$$n! = \Gamma(n+1) = \int_0^\infty e^{-t} t^n dt \geq \int_n^\infty e^{-t} t^n dt \geq \int_n^\infty e^{-t} n^n dt = \left( \frac n e \right)^n$$
This is a useful bound of the factorial to know - it's weaker than the Stirling approximation, but much, much easier to prove.
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# Are these two lines the same?
How can I say that the two lines are the same:
Line1=(5,3)+t(-4,8)
Line2=(5,3)+t(-8,16)
They both have the same starting point but can I say they have the same direction vector? It seems like to me they contain different points for a given "t".
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It would be better to use different variables for the parameter in the two lines. If Line 2 were (5,3)+s(-8,16) one could follow Rasmus, come out with t=2s and it would be clearer. – Ross Millikan Jan 3 '12 at 19:07
The vector form of the equation you are using is $$\tag{1}{\bf r}(t) = \color{maroon}{\bf p} +t\color{darkgreen} {\bf d}$$ where $\color{maroon}{\bf p}$ gives the initial point on the line (the tip of $\color{maroon}{\bf p}$ is the initial point on the line corresponding to $t=0$) and $\color{darkgreen}{\bf d}$ is the direction vector.
To see why (1) actually gives a line: Note that multiplication of $\color{darkgreen}{\bf d}$ by $t$ just extends, shortens, or reflects $\color{darkgreen}{\bf d}$, it does not change its direction (in the diagram $\color{maroon}{t{\bf d}}$ is the dashed, maroon vector).
So, thinking of adding vectors "tip to tail", the tip of the vector $\color{orange}{{\bf p}+t{\bf d}}$ will always fall on a line.
Note that in (1), we could have used any other vector, such as the blue vector below, that is parallel to $\color{darkgreen}{\bf d}$ as the direction vector, and the corresponding equation would give the same line, just traced out in a different way.
If you think of the line as being generated by a moving point (the gray one above) whose position at time $t$ is ${\bf r}(t)$, then in your example, the point is moving twice as fast in your equation for line2 than as in the equation for line1.
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Yes, they are because $(-8,16)$ is a multiple of $(-4,8)$:
$$(-8,16)=2\cdot (-4,8).$$
If a point $(a,b)$ belongs to Line1, say $(a,b)=(5,3)+t(-4,8)$, then it also belongs to Line2 because $(a,b)=(5,3)+\dfrac{t}{2}(-8,18)$. | {
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Similarly, if a point $(a,b)$ belongs to Line2, say $(a,b)=(5,3)+t(-8,16)$, then it also belongs to Line1 because $(a,b)=(5,3)+2t(-4,8)$.
The lenght of the "direction vector" is irrelevant, only its direction is important. This is because your are really only interested in the set of all scalar multiples of the direction vector.
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Yes, they are the same since the direction vectors are parallel (i.e. one is a scalar multiple of the other.) It is true that they contain different points for a given "t", but they still contain all the same points -- for instance, the point of line1 at a certain $t$ corresponds to the point at line2 at time $t/2$.
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# Is there a name for the sum of increasing powers?
I was thinking about the wheat and chessboard story and thinking of the total number of grains of wheat…
$$\sum_{n=0}^{63} 2^n$$
And wondered if there is a name for a sum like this?
$$2^0 + 2^1 +2^2 + 2^3 + \cdots$$
• Geometric? ${}{}$ – user251257 Aug 14 '16 at 14:35
• The first on is a finite geometric series. The second is an infinite geometric series. To call it "geometric" means there is a common ratio, i.e. the number you multiply each term by, to get the next term, is the same in every case. $\qquad$ – Michael Hardy Aug 14 '16 at 17:51
• If infinite, the second one is not convergent -- it diverges quite badly. @MichaelHardy – Clement C. Aug 14 '16 at 19:23
• Whether it diverges badly or well seems a matter of opinion: I'd go for the latter. It diverges really quite well indeed. – Dewi Morgan Aug 14 '16 at 19:59
• @Dewi, if adjectives are a matter of opinion, I'd pick "spectacularly" over the two other choices you mentioned. – J. M. is a poor mathematician Aug 14 '16 at 22:48
This is called a geometric sum/series according as $N$ is finite or not. More generally, it is of the form $$\sum_{n=0}^Nar^n$$ where $N$ may be "equal to" infinity. In the case where $N$ is infinite, if the series is to converge, we require $\lvert r\rvert<1$. Furthermore, for $N$ finite $$\sum_{n=0}^Nar^n=a\frac{1-r^N}{1-r},\:\:\:\: r\ne1$$ and $$\sum_{n=0}^\infty ar^n=\frac{a}{1-r},\:\:\:\:\lvert r\vert<1.$$
• This is wrong. For the finite case, the sum shall be $a\frac{1-r^{N+1}}{1-r}$ since there are $N+1$ terms. – Kun May 4 at 21:02
It's mainly a well known value $2^{n+1}-1$ . Write it in base $2$.
It is also OEIS sequence A000225 :
$2^n - 1.$ (Sometimes called Mersenne numbers, although that name is usually reserved for A001348.) | {
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$2^n - 1.$ (Sometimes called Mersenne numbers, although that name is usually reserved for A001348.)
• This answers the specific case of $r=2$, but not the general case asked in the title. The sum of increasing powers is a geometric sum -- the specific name of Mersenne numbers is coincidental in this question, and giving the closed form without the explanation (first part of this answer) sheds basically no light whatsoever on the why. – Clement C. Aug 14 '16 at 15:08
• The OP asked if there "was a name for a sum like that." (Emphasis mine) Your answer gives a value without explanation, and a name for this exact sum which is not generalizable to any other ratio. As such, I feel it fails to answer the question. – Clement C. Aug 14 '16 at 16:19
• Indeed, it's my opinion - this is why I wrote a comment to explain it. – Clement C. Aug 14 '16 at 16:33
As you are looking for novel names, this is also a simple case of the more generic notion of hypergeometric series $$\sum_{k}r_k \,,$$ with $r_0 = 1$, and the ratio of two consecutive terms is a rational function, a ratio of two polynomials $P$ and $Q$ in the summation index $k$
$$\frac{r_{k+1}}{r_k}= \frac{P(k)}{Q(k)}\,.$$
In your case, you can choose $P$ and $Q$ such that their ratio is equal to $2$. When the ratio is constant, it is called a geometric series (as answered here). As a reminder, it is a sum of terms in geometric progression (se.math) like $1,r,r^2,r^3,\ldots$, whose name (the geometry part) is illustrated by the following figure:
Hypergeometric series are also connected to chess. A rook is a move on a chessboard. Some have developed studies some types of permutations as the placement of a number of rooks on a chessboard-like grid, see for instance Rook theory and hypergeometric series, J. Haglund, 1996.
The formula for the first $N$ terms can be derived as following:
Let:
$$S(N)=\sum_{n=0}^{N} r^n$$
Then distributing an $r$ (which is independent of $n$) along the sum we have, | {
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Then distributing an $r$ (which is independent of $n$) along the sum we have,
$$rS(N)=\sum_{n=0}^{N} r^{n+1}$$
Note (1):
$$S(N+1)=\sum_{n=0}^{N+1} r^n=r^0+\sum_{n=0}^{N} r^{n+1}$$ $$=1+rS(N)$$
Also note,
$$S(N+1)-S(N)=\sum_{n=0}^{N+1} r^n-\sum_{n=0}^{N} r^n=r^N$$
As all terms but $r^N$ will reduce each other out to zero.
So,
$$S(N+1)=S(N)+r^N$$
Hence we have from (1) and from the above:
$$S(N)+r^N=rS(N)+1$$
$$r^N-1=rS(N)-S(N)$$
$$r^N-1=(r-1)S(N)$$
Thus,
$$S(N)=\frac{r^N-1}{r-1}$$
Now we multiply by a special form of $1$ to get:
$$S(N)=\sum_{n=0}^{N} r^n=\left( \frac{-1}{-1} \right) \left( \frac{r^N-1}{r-1} \right)=\frac{1-r^N}{1-r}$$
And hence (distributing out an $a$ which is independent of $n$) we get the formula for the geometric sum,
$$\sum_{n=0}^{N} ar^n=a\frac{1-r^N}{1-r}$$
Note though our formula only works for $r \neq 1$ (our manipulations are okay until we get to dividing by $0=1-r=1-1$ for $r=1$). | {
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# Number of ways to arrange 5 monkeys in a row?
We have 5 monkeys $a,b,c,d,e$ and we are interested in the number of ways to have them stand in a row without $a$ and $b$ being next to each other.
The part that I struggle with most is that I don't fully understand how to solve this when the 5 are different. It's not the same as for example coloring 5 segments either blue or red without any two neighboring segments being red.
This is how I tried to solve this but I'm certain that there's something wrong. I would really appreciate it if you could also critique my approach.
Idea:
Let $f_{k}$ be the number of ways we can have the $5$ monkeys in a row without $a$ and $b$ being next to each other. We try to do this recursively:
case 1 : the last monkey is not $a$ or $b$: then we have $f_{k-1}$ possibilities for the rest of the k-1 monkeys.
case 2 : the last monkey is either $a$ or $b$: Here the second to last has to be one of $\{c,d,e\}$. So we have $3$ possibilities for the second to last spot and $2$ possibilities for the last. A total of $2*3 = 6$ and $f_{k-2}$ for the remaining spots.
The recursive equation I get is: $f_{k} = 6 + f_{k-1} + f_{k-2}$
$f_{1} = 5$
$f_{2} = 10$
$f_{3} = 21$
$f_{4} = 37$
$f_{5} = 64$
I'm not sure about my solution. | {
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$f_{1} = 5$
$f_{2} = 10$
$f_{3} = 21$
$f_{4} = 37$
$f_{5} = 64$
I'm not sure about my solution.
• Hint: Count the total number of ways to arrange the monkeys, then count the number of ways to arrange where $a$ and $b$ are adjacent, and subtract from the total. – platty Aug 11 '17 at 17:15
• The problem with your current approach is that in the recursive step, the number of possible choices for case 2 decreases (fewer possible monkeys to choose from). Furthermore, the subproblem is not the same - now you only have one monkey to worry about, and you can count this case directly. – platty Aug 11 '17 at 17:18
• it seems that you want to use probabilities (possibilities of what exactly?) to count the valid permutations, but I think its easier to count the invalid permutations and subtract them from the total number of possible permutations. – Masacroso Aug 11 '17 at 17:20
• See no evil, hear no evil, speak no evil, smell no evil, taste no evil? – David Richerby Aug 12 '17 at 12:48
• @DavidRicherby what do you mean? – DariusTheGreat Aug 12 '17 at 13:08
Here's how I'd approach this particular problem. I'll solve for $k$ monkeys afterwards.
You have $5!$ ways for the monkeys to be arranged in a line without restriction.
There are $8$ ways that $A$ and $B$ can be positioned next to each other; there are $4$ pairs of adjacent spaces, and either $A$ or $B$ can be on the left.
For each of these cases, there are $3! = 6$ ways to arrange the other three monkeys.
So the answer is $5! - 8 \cdot 3! = 72$ ways.
Now, just apply to $k$ monkeys using the same argument:
$$P(k) = k! - 2(k-1)(k-2)! = k! - 2(k-1)!.$$
(Hat tip to user471297 for the last simplification.) | {
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$$P(k) = k! - 2(k-1)(k-2)! = k! - 2(k-1)!.$$
(Hat tip to user471297 for the last simplification.)
• Equivalently: Treat the A,B pair as a unit and come up with 4!, then double it to account for swapping then. – Kevin Aug 12 '17 at 2:44
• Good answer! You can also write the result as $P(k)=(k-1)!(k-2)$. It's a matter of taste, but this form gives a way to see the asymptotic behaviour better and maybe calculate a little faster if factorials are expensive. – Joonas Ilmavirta Aug 12 '17 at 13:11
Recursive solution (the complementary counting solution is outlined in John's answer):
Let $f(k)$ be the number of ways to arrange $k$ monkeys, including $a$ and $b$ such that these two aren't next to each other.
Case 1: $a$ or $b$ is at the beginning of the line.
Counting this case directly, we first choose the leading monkey in one of $2$ ways. Then we find that the other of these two monkeys is in one of $k-2$ positions (any spot except for the one occupied by the first monkey, and the one immediately behind it). The other $k-2$ monkeys can be in any order, so we get $2 (k-2) (k-2)!$ ways.
Case 2: Neither are at the beginning of the line.
There are a total of $k-2$ choices for the monkey to lead the line; after that, we have the $k-1$ subproblem, so we find $(k-2)f(k-1)$ ways here.
Combining these, we have a total of $f(k) = (k-2)(2(k-2)! + f(k-1))$ good arrangements. Starting with $f(2) = 0$, an inductive argument should show that this matches the closed form answer.
• Thanks for showing how to incorporate the $k-1$ case directly. That eluded me. – John Aug 11 '17 at 18:34
@John and @platty have both supplied good answers. Here is another approach. | {
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@John and @platty have both supplied good answers. Here is another approach.
$a$ is at an end of the row: Since $a$ can be at the left or right end of the row, there are two ways to place $a$. For each such choice, there are three ways to place $b$ so that $b$ is not adjacent to $a$. The remaining three monkeys can be arranged in the three remaining positions in $3!$ ways. Hence, there are $$2 \cdot 3 \cdot 3!$$ arrangements in which $a$ is at an end of the row.
$a$ is not at the end of the row: Since there are five positions including the two ends of the row, there are three choices for the position of $a$. Since $b$ cannot be adjacent to $a$, there are two ways to place $b$. The remaining three monkeys can be arranged in the three remaining positions in $3!$ ways. Hence, there are $$3 \cdot 2 \cdot 3!$$ arrangements in which $a$ is not at an end of the row.
Total: Since the two cases are mutually exclusive and exhaustive, the five monkeys can be arranged in $$2 \cdot 3 \cdot 3! + 3 \cdot 2 \cdot 3! = 72$$ ways if $a$ and $b$ are not in adjacent positions.
I came up with the answer a different, non-recursive, way than some of the answers here (I understand the OP wanted a critique of their approach, but I figured that a different method could still add value). Anyway, my method:
As John said, there are $5!$ ways to arrange the monkeys without restriction. From there I treated monkeys $A$ and $B$ as one monkey and found the number of ways the four monkeys could be arranged: $4!$
Since there are two ways to arrange monkeys $A$ and $B$ together, you have $2 * 4!$ ways $A$ and $B$ could be put together. Subtracting from the original unrestricted $5!$ yields $120 - 2 * 24 = 72$.
Here is a simple solution :
First arrange 5 monkeys in 5! = 120;
remove all the cases where both a and b sit together.. so to get that tie two monkeys as one item: so you have 4 monkeys now --- how many ways you can arrange 4monkeys: 4! and also A and B sit as AB and BA so finally | {
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we have 2 * 4! ways
so: answer is 5! -2 4! so in general : k! - 2 (k-1)!
Now just go to the previous answers k! - 2 * (k-1)(k-2)! is also actually same as k! - 2* (k-1)!
• Hah! I didn't see that the answer could be simplified (per your last line). – John Aug 11 '17 at 22:44
A way to find the result without subtracting one result from the general one:
Arrange the three other monkeys, $c, d, e$ in $3!$ or $6$ distinct ways.
There is a slot for $a$ or for $b$ in front of the first of $c, d, e$ already placed, and after each of $c, d, e$ already placed, for a total of 4 slots, each possible holding one of the two remaining letters, either $a$ or $b$. Adjacent placement of $a$ and $b$ is thus impossible
Select two of these four slots in $4 \times 3$ or $12$ ways. Put $a$ in the first slot selected and $b$ in the second.
Collapse the two empty slots as unneeded, and you're done with $6 \times 12$ or $72$ ways...
For your approach, you should get that placing $2$ monkeys in conformity with the conditions has zero options. So there is something odd going on there. Your base case would probably better be taken from the $2$ obvious possibilities for three monkeys.
Using a masking approach for variety, let us suppose that we have determined which of the locations will be occupied by $a$ and $b$. That location mask will give rise to $2$ ways of placing $a,b$ and, independently, $(k-2)!$ ways of placing the other monkeys.
Valid masks can be produced in $(2(k-2)+(k-2)(k-3))/2$ ways, separating the initial selection of the $2$ end positions from the $k{-}2$ mid positions, and then reducing the double count.
This gives the total possibilities as
\begin{align} 2(k-2)!\cdot(2(k-2)+(k-2)(k-3))/2 &= (k-2)!(k-2)(2+k-3)\\ &=(k-2)!(k-2)(k-1)\\ &=(k-1)!(k-2) \end{align}
For $k=5$, this gives $24\cdot 3 = 72$ options. | {
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For $k=5$, this gives $24\cdot 3 = 72$ options.
When $a$ should immediately precede $b$, then they together count as a single monkey, so you get $4!$. The other way round ($b$ then $a$), you get another $4!$. Without this condition you get $5!$. You don't want the condition, so subtract:
$$5! - 2\cdot4!$$
+---+---+---+
| X | X | X |
+--- --- ---+ For this illustration, Monkeys **A** & **E** are the restricted monkeys
↑ ↑ ↑ ↑ (Maybe they have beef ¯\\_(ツ)_/¯)
Above is an illustration with the "arranging people in seats" style. Say monkeys B,C & D can sit in any order about themselves they like. They have 3!* ways(=6).
Now, monkey A has 4 possible choices(denoted by the arrows above) since he has no restrictions with them. So a total of (6x4=24) arrangements so far.
Monkey E will now be arranged into ANY THREE OF THE REMAINING ARROWS that were not chosen by monkey A, as any of these ensure they are not next to each other. This brings the final tally to (6x4x3=72) arrangements.
NOTE: This approach extends to if there are other monkeys with similar restrictions i.e. the next would have 2 choices, and the next 1, and it would be impossible for another after that to not be next to a restricted monkey.
• If anyone is confused about "3!", it simply means 3 choices for 1, then the next has 2, and the last has 1 => thus 3x2x1 = 6
I hope this helps; it's my first MathStackExchange post. I like this method & thought I should share. Thx | {
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# Time complexity of an algorithm: Is it important to state the base of the logarithm?
Since there is only a constant between bases of logarithms, isn't it just alright to write $$f(n) = \Omega(\log{n})$$, as opposed to $$\Omega(\log_2{n})$$, or whatever the base might be?
It depends where the logarithm is. If it is just a factor, then it doesn't make a difference, because big-O or $$\theta$$ allows you to multiply by any constant.
If you take $$O(2^{\log n})$$ then the base is important. In base 2 you would have just $$O(n)$$, in base 10 it's about $$O(n^{0.3010})$$.
• I guess this is only going to come up with something like $2^{\sqrt{\log n}}$. I can't see any reason for expressing a number as $2^{c\log_b n}$ rather than $n$-to-the-whatever-it-is (except perhaps as an intermediate stage of a calculation). May 21, 2019 at 19:45
• +1 for "constant factors matter in exponents" May 21, 2019 at 22:57
Because asymptotic notation is oblivious of constant factors, and any two logarithms differ by a constant factor, the base makes no difference: $$\log_a n = \Theta(\log_b n)$$ for all $$a,b > 1$$. So there is no need to specify the base of a logarithm when using asymptotic notation. | {
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• I prefer to see $\in$ instead of $=$ May 21, 2019 at 22:04
• I'm afraid the standard notation uses $=$. May 21, 2019 at 22:05
• @YuvalFilmus The standard notation is misleading, completely different to the standard everywhere else and makes algorithmic complexity seem completely alien from things quite similar to it. "It's the standard notation" should never be a reason to favour a bad solution over a better, similarly-clear one. (The meaning of the symbol is usually clear from context, anyway.) May 22, 2019 at 18:16
• @wizzwizz4 Common practice is an excellent reason. It promotes efficient communication. That’s the reason we all put up with the quirks of English spelling. May 22, 2019 at 20:04
• Sometimes $n \mapsto \log_a n \in \Theta ( n \mapsto \log_b n)$ just has too much stuff there to be clearer than $\log_a n = \Theta (\log_b n)$.
– JiK
May 23, 2019 at 10:48
As $$\log_xy = \frac{1}{\log_y{x}}$$ and $$\log_x{y} = \frac{\log_z{y}}{\log_z{x}}$$, so $$\frac{\log_a{n}}{\log_b{n}} = \frac{\log_n{b}}{\log_n{a}} = \log_a{b}$$. As $$\log_a{b}$$ is positive constant (for all $$a,b > 1$$), so $$\log_a{n} = \Theta(\log_b{n})$$.
In most cases, it's safe to drop the base of the logarithm because, as other answers have pointed out, the change-of-basis formula for logarithms means that all logarithms are constant multiples of one another.
There are some cases where this isn't safe to do. For example, @gnasher729 has pointed out that if you have a logarithm in an exponent, then the logarithmic base is indeed significant. | {
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I wanted to point out another case where the base of the logarithm is significant, and that's cases where the base of the logarithm depends directly on a parameter specified as input to the problem. For example, the radix sort algorithm works by writing out numbers in some base $$b$$, decomposing the input numbers into their base-$$b$$ digits, then using counting sort to sort those numbers one digit at a time. The work done per round is then $$\Theta(n + b)$$ and there are roughly $$\log_b U$$ rounds (where $$U$$ is the maximum input integer), so the total runtime is $$O((n + b) \log_b U)$$. For any fixed integer $$b$$ this simplifies to $$O(n \log U)$$. However, what happens if $$b$$ isn't a constant? A clever technique is to pick $$b = n$$, in which case the runtime simplifies to $$O(n + \log_n U)$$. Since $$\log_n U$$ = $$\frac{\log U}{\log n}$$, the overall expression simplifies to $$O(\frac{n \log U}{\log n})$$. Notice that, in this case, the base of the logarithm is indeed significant because it isn't a constant with respect to the input size. There are other algorithms that have similar runtimes (an old analysis of disjoint-set forests ended up with a term of $$\log_{m/2 + 2}$$ somewhere, for example), in which case dropping the log base would interfere with the runtime analysis.
Another case in which the log base matters is one in which there's some externally-tunable parameter to the algorithm that control the logarithmic base. A great example of this is the B-tree, which requires some external parameter $$b$$. The height of a B-tree of order $$b$$ is $$\Theta(\log_b n)$$, where the base of the logarithm is significant in that $$b$$ is not a constant.
To summarize, in the case where you have a logarithm with a constant base, you can usually (subject to exceptions like what @gnasher729 has pointed out) drop the base of the logarithm. But when the base of the logarithm depends on some parameter to the algorithm, it's usually not safe to do so. | {
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Let $$f_k(n)=\log_kn$$,$$k>2$$, and let $$t(n)=\log_2n$$.
You can rewrite $$f_k(n)$$ as follow $$f_k(n)=\frac{\log_2n}{\log_2k}$$ Now $$k$$ have two states:
1. $$k$$ is constant:
$$\lim_{n\to \infty}\frac{f_k(n)}{t(n)}=\lim_{n\to \infty}\frac{\frac{\log_2n}{\log_2k}}{\log_2n}=c$$ that $$c$$ is a constant$$>0$$, hence $$t(n)=\Theta(f_k(n)).$$ So there is no difference between $$f_k(n)$$ , and $$t(n)$$ in terms of asymptotic notation.
1. $$k$$ isn't constant:
$$\lim_{n\to \infty}\frac{f_k(n)}{t(n)}=\lim_{n\to \infty}\frac{\frac{\log_2n}{\log_2k}}{\log_2n}=0$$
So $$t(n)$$ have faster growth rate and $$t(n)=\omega(f_k(n)).$$ | {
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Input − Adjacency matrix of a graph. Start DFS at the vertex which was chosen at step 2. // Implementation of 0 ⋮ Vote. But please check yourself as well. After completing the traversal, if there is any node, which is not visited, then the graph is not connected. Adjacency Matrix of an undirected graph Adjacency Matrix of a directed graph Adjacency List and matrix of directed graph An adjacency matrix has a more organized appearance to its structure, but this implementation wastes memory if not all of the vertices are connected. If the graph is undirected, the adjacency matrix is symmetric. For a undirected graph it is easy to check that if the graph is connected or not. The idea is also simple - imagine an n by n grid, where each row and each column represents a vertex. The adjacency matrix and adjacency list are raw'' forms of graph and are not oriented towards solving any particular problem. Introduction to Graphs 9:32. Undirected graph with no loops and no multi-edges. For finding paths of length r between vertices v(i) & v(j), we find A^r and the (i,j)th entry of this matrix would give you the number of paths. 0. Can we improve further? Time Complexity: Time complexity of above implementation is sane as Depth First Search which is O(V+E) if the graph is represented using adjacency list representation. In this case the traversal algorithm is recursive DFS traversal. I already have the methods to check for self-loops and cycles, I need a method to check SPECIFICALLY for connectivity in the adjacency matrix to prove it is a DAG. Vote. The above approach requires two traversals of graph. An adjacency matrix is a square matrix used to represent a finite graph. We define an undirected graph API and consider the adjacency-matrix and adjacency-lists representations. Not sure how to check if there are connected edges or how to remove, only know how to add edges. If the smallest eigenvalue is strictly bigger then zero or the same as if zero is not an eigenvelue then it | {
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smallest eigenvalue is strictly bigger then zero or the same as if zero is not an eigenvelue then it is connected. Vote. I'm doing a project on topological robotics, and part of the program involves taking in the adjacency matrix of a graph, then testing to see if it's connected or not. Follow 24 views (last 30 days) chandra Naik on 24 Jul 2019. How to check graph connected or not for given adjacency matrix of graph. For the undirected graph, we will select one node and traverse from it. We also consider the problem of computing connected components and conclude with related problems and applications. Show Hide all comments. Look at the graph laplacian D-A where D is the diagonal matrix with corresponding degrees of vertices on the diagonal. I know that the time required to check if there exists an edge between two nodes in an adjacency matrix is $O(1)$ because we can access it directly (i.e: $M[i][j]$). Answers (1) Matt J on 24 Jul 2019. Yes Is there an edge from 1 to 3? For example, we need to check if an adjacency matrix such as this one is fully connected: The graph is (n+2)*(n+2), and the number of functions here is 4. An undirected graph is graph, i.e., a set of objects (called vertices or nodes) that are connected together, where all the edges are bidirectional. At the ith row and jth column, we store the edge weight of an edge from the vertex i to vertex j. If any vertex v has vis1[v] = false and vis2[v] = false then the graph is not connected. This might not be very efficient especially in the case where we have a lot of vertices and only a few of those vertices are connected to each other, which translates to a very sparse adjacency matrix. Textbook solution for Linear Algebra: A Modern Introduction 4th Edition David Poole Chapter 4.6 Problem 36EQ. Here's what you'd learn in this lesson: Bianca analyzes the adjacency matrix format of representing node relationships in a graph, using binary values in the array. To check connectivity of a graph, we will try | {
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in a graph, using binary values in the array. To check connectivity of a graph, we will try to traverse all nodes using any traversal algorithm. We can find whether a graph is strongly connected or not in one traversal using Tarjan’s Algorithm to find Strongly Connected … The "Adjacency Matrix" Lesson is part of the full, Tree and Graph Data Structures course featured in this preview video. Output: Yes No. I don't want to keep any global variable and want my method to return true id node are connected using recursive program The trouble is, I've tested it with several disconnected and connected graphs, and it says that they're all disconnected, no matter what I put in! No The problem is that we always need to use O(n^2) elements for storage, and hence, we often use adjacency lists to represent graphs. I realize this is an old question, but since it's still getting visits, I have a small addition. Sign in to answer this question. Start at a random vertex v of the graph G, and run a DFS(G, v). Make all visited vertices v as vis1[v] = true. Time Complexity: DFS: O(m * n) where m is the number of rows in our grid and n is the number of columns in our grid. Edited: Matt J on 24 Jul 2019 How to check given undirected graph connected or not 0 Comments. The advantage of the adjacency matrix is that it is simple, and for small graphs it is easy to see which nodes are connected to other nodes. Graph. In the special case of a finite simple graph, the adjacency matrix is a (0,1)-matrix with zeros on its diagonal. The diagram below illustrates the adjacency matrix for the example graph we presented earlier. Modified if-statement for graph traversal to also check if a cell is a wall. Otherwise Its degree (degree of zero as a root of characteristic polynomial) is "the number of connected components"-1. As of R2015b, the new graph and digraph classes have a method for computing connected components. Sign in to comment. To check that a graph is connected or not. In this tutorial we | {
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components. Sign in to comment. To check that a graph is connected or not. In this tutorial we shall see how to store a graph with the help of a matrix. Now reverse the direction of all the edges. How to check graph connected or not for given adjacency matrix of graph. We can simply do a depth-first traversal or a breadth first-first traversal on the graph and if the traversal successfully traversal all the nodes in the graph then we can conclude that the graph is connected else the graph has components. For the undirected graph, we will select one node and traverse from it. The forms of problems that one must solve are typically: process/print the nodes, e.g., check if the graph is connected--- for every node, one can go to all other nodes In graph theory and computer science, an adjacency matrix is a square matrix used to represent a finite graph.The elements of the matrix indicate whether pairs of vertices are adjacent or not in the graph.. Algorithm that checks if a directed graph represented by an adjacency matrix is strongly connected( there is a path connecting all vertices) .. Algorithm preferably in fortran. To represent this graph as the adjacency matrix A, we’ll let the indices of the rows and columns represent nodes, or vertices. To check connectivity of a graph, we will try to traverse all nodes using any traversal algorithm. In this node 1 is connected to node 3 ( because there is a path from 1 to 2 and 2 to 3 hence 1-3 is connected ) I have written programs which is using DFS, but i am unable to figure out why is is giving wrong result. We have step-by-step solutions for your textbooks written by Bartleby experts! The idea is to take 2D array of size V * V, then mark the index as “1” if there exist an edge between those 2 vertices. Input − Adjacency matrix of a graph. Hence, the given graph is connected. If such edge doesn’t exist, we store zero. A value in a cell represents the weight of the edge from vertex v v v to vertex w w w. An adjacency matrix | {
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in a cell represents the weight of the edge from vertex v v v to vertex w w w. An adjacency matrix representation for a graph . A tree is a connected undirected graph without cycles. Is there an edge from 1 to 2? Yes Is there an edge from 4 to 3? In contrast, a graph where the edges point in a direction is called a directed graph. As mentioned in this article, adjacency matrix requires more memory if implemented in a program due to its requirement to store the graph information in the form of an $$N \times N$$ matrix. This is a graph implementation, using adjacency matrix on Python. Graph implementation. For the given matrix, I believe that the summation of all paths gives us a matrix with all non-zero entries. In this case the traversal algorithm is recursive BFS traversal. Created a Graph Structure with 4 vertices Initialized Adjacency Matrix! To check whether a graph is connected based on its adjacency matrix A, use Using Adjacency … I need help implementing directed weighted graph in java using adjacency matrix. As it stores the path between 2 vertices, it is called as adjacency matrix. Graph API 14:47. Make all visited vertices v as vis2[v] = true. Depth-First … After completing the traversal, if there is any node, which is not visited, then the graph is not connected. We introduce two classic algorithms for searching a graph—depth-first search and breadth-first search. We can check each one of this properties. 0. An undirected graph is sometimes called an undirected network. Method for computing connected components is the diagonal not an eigenvelue then it is connected not! The idea is also simple - imagine an n by n grid, where each row and jth,! And are not oriented towards solving any particular problem zero or the same as if is! Not sure how to check connectivity of a finite simple graph, we store the edge weight an... Using any traversal algorithm is recursive BFS traversal on Python method for computing components. Textbooks written by Bartleby experts | {
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BFS traversal on Python method for computing components. Textbooks written by Bartleby experts ) chandra Naik on 24 Jul 2019 an n by n grid, each! Check given undirected graph is not visited, then the graph is connected... Implementing directed weighted graph in java using adjacency matrix only know how to check if graph is connected adjacency matrix if a is! The adjacency-matrix and adjacency-lists representations for the undirected graph without cycles check given undirected API... Undirected, the adjacency matrix, and run a DFS ( G, v ) to add.! Traverse all nodes using any traversal algorithm is recursive BFS traversal at the graph is not connected the of. - imagine an n by n grid, where each row and each column represents a.! The special case of a graph, we will try to traverse all nodes using traversal... // Implementation of Look check if graph is connected adjacency matrix the graph is undirected, the adjacency matrix matrix and list. Two classic algorithms for searching a graph—depth-first search and breadth-first search a method for computing components. On 24 Jul 2019 there are connected edges or how to check that a graph where the edges in..., the adjacency matrix adjacency-matrix and adjacency-lists representations all visited vertices v as vis1 [ v ] = true degrees... Each column represents a vertex // Implementation of Look at the vertex i to vertex J not! Is the diagonal and are not oriented towards solving any particular problem node and from... Node and traverse from it finite simple graph, the adjacency matrix for the example graph we earlier. As of R2015b, the adjacency matrix is symmetric point in a direction called... I realize this is an old question, but since it 's still getting visits, i have method... Check given undirected graph it is easy to check given undirected graph without cycles in this the! Zero as a root of characteristic polynomial ) is the number of connected components and with! 2 vertices, it is called as adjacency matrix is | {
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) is the number of connected components and with! 2 vertices, it is called as adjacency matrix is a square matrix to. Edge from the vertex which was chosen at step 2 v ] = and... Matrix used to represent a finite simple graph, the adjacency matrix for the undirected graph, the graph. The special case of a graph Structure with 4 vertices Initialized adjacency matrix i to vertex.... Ith row and each column represents a vertex of connected components '' -1 connected components where. Zero as a root of characteristic polynomial ) is the number of connected components and conclude with related and! And adjacency list are raw '' forms of graph run a (! Check that if the smallest eigenvalue is strictly bigger then zero or the as... Undirected network search and breadth-first search days ) chandra Naik on 24 2019. Small addition is sometimes called an undirected graph connected or not 0 Comments is not visited, the. The adjacency-matrix and adjacency-lists representations of a graph is not an eigenvelue then it is called a directed graph is! To represent a finite simple graph, we store zero visited vertices v as vis2 v! Each row and jth column, we store the edge weight of an edge 4... Then the graph laplacian D-A where D is the diagonal matrix with corresponding degrees of vertices on diagonal! Corresponding degrees of vertices on the diagonal there an edge from 4 to 3 the. Example graph we presented earlier any node, which is not visited, then graph. Computing connected components and conclude with related problems and applications DFS traversal start at. And adjacency-lists representations have step-by-step solutions for your textbooks written by Bartleby!. ( degree check if graph is connected adjacency matrix zero as a root of characteristic polynomial ) is the number of connected components row! raw '' forms of graph and digraph classes have a method for computing connected components '' -1 undirected... Doesn ’ t exist, we will try to traverse all nodes using any | {
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of a graph, we select... Graph API and consider the problem of computing connected components '' -1 vertices Initialized matrix. By Bartleby experts the example check if graph is connected adjacency matrix we presented earlier which is not an eigenvelue then is... 0,1 ) -matrix with zeros on Its diagonal solution for Linear Algebra a! Vertices Initialized adjacency matrix for Linear Algebra: a Modern Introduction 4th Edition David Poole Chapter 4.6 36EQ... A root of characteristic polynomial ) is the number of connected components '' -1 node. Vis2 [ v ] = true node, which is not connected this is a connected undirected graph it easy... An n by n grid, where each row and jth column, we will try traverse! Vertex which was chosen at step 2 and each column represents a vertex related problems and applications diagonal with. In the special case of a graph where the edges point in a direction is called as adjacency of! Matrix and adjacency list are raw '' forms of graph check given undirected,! We define an undirected network degree ( degree of zero as a root of characteristic polynomial ) ... Graph in java using adjacency matrix is symmetric from 4 to 3 eigenvelue it. A directed graph or not adjacency-matrix and adjacency-lists representations search and breadth-first search ... Not 0 Comments problems and applications Structure with 4 vertices Initialized adjacency matrix which is not.... Graph Implementation, using adjacency matrix such edge doesn ’ t exist, we will try to traverse nodes. A graph where the edges point in a direction is called as matrix! Of graph last 30 days ) chandra Naik on 24 Jul 2019 have step-by-step solutions for textbooks! Graph Implementation, using adjacency matrix n by n grid, where row... Is there an edge from 1 to 3 is the number of components! The number of connected components = false and vis2 [ v ] true! Matrix and adjacency list are raw '' forms of graph and are oriented! Last 30 days ) chandra Naik on 24 Jul 2019 a random vertex v has | {
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'' forms of graph and are oriented! Last 30 days ) chandra Naik on 24 Jul 2019 a random vertex v has vis1 [ ]... Row and each column represents a vertex try to traverse all nodes using any traversal is! An eigenvelue then it is connected or not 0 Comments traversal algorithm all nodes using any traversal.... An eigenvelue then it is easy to check connectivity of a finite simple graph, the adjacency matrix square! And digraph classes have a small addition = false and vis2 [ v ] = false then the graph connected! Connectivity of a graph Structure with 4 vertices Initialized adjacency matrix for the undirected it! Dfs traversal: Matt J on 24 Jul 2019 yes is there edge. Not for given adjacency matrix is symmetric degrees of vertices on the diagonal matrix with corresponding degrees vertices. Modern Introduction 4th Edition David Poole Chapter 4.6 problem 36EQ as of,! Idea is also simple - imagine an n by n grid, where row... I to vertex J problems and applications strictly bigger then zero or the same as if zero is not,. From 4 to 3 consider the adjacency-matrix and adjacency-lists representations a small.... | {
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# Evaluate $\displaystyle\sum_{k=1}^n \left(\frac{{n-1 \choose k-1}}{k}\right)$
Evaluate : $$\displaystyle\sum_{k=1}^n \left(\frac{{n-1 \choose k-1}}{k}\right)$$
Alright so I'm completely stumped, I've never evaluated a summation of $$\displaystyle{n \choose k}$$.
My best guess is to use the binomial theorem, but I don't know how to change this into a form I could use the theorem on.
• yes, sorry my formatting is off – Brownie Mar 15 at 22:07
Hint:
Use the recurrence relation $$\binom nk=\frac nk\binom{n-1}{k-1}$$ and remember that $$\displaystyle\sum_{k=0}^n\binom nk=\cdots$$
• Alright, so does it make sense if I multiplied $\displaystyle\sum_{k=1}^n \left(\frac{{n-1 \choose k-1}}{k}\right)$ by n, and then took that summation, and then divided by n at the end? – Brownie Mar 15 at 22:12
• It would be perfect! – Bernard Mar 15 at 22:13
• So dividing by n after taking the summation is correct? Or would I have to divide by the summation of n since I already took the summation of $\displaystyle\sum_{k=1}^n \left(\frac{n{n-1 \choose k-1}}{k}\right)$ – Brownie Mar 15 at 22:14
• It's correct since multiplication (and division) is distributive w.r.t. addition. – Bernard Mar 15 at 22:16
• Thank you that clears it up! – Brownie Mar 15 at 22:16
Another alternative is to somehow see the function $$\frac{x^k}{k}$$ in the series and use a bit of calculus:
$$\frac{d}{dx}\displaystyle\sum_{k=1}^n \binom{n-1}{k-1} \frac{x^k}{k}=\displaystyle\sum_{k=1}^n \frac{d}{dx} \binom{n-1}{k-1} \frac{x^k}{k}=\displaystyle\sum_{k=1}^n \binom{n-1}{k-1}x^{k-1}=(1+x)^{n-1}$$ by the Binomial Theorem, so $$\displaystyle\sum_{k=1}^n \binom{n-1}{k-1} \frac{x^k}{k}=\int (1+x)^{n-1} dx=\frac{(1+x)^n}{n}+\mathcal{C}.$$ Plugging in $$x=0$$ tells us that $$\mathcal{C}=-\frac{1}{n}$$, so we have the more general result $$\displaystyle\sum_{k=1}^n \binom{n-1}{k-1} \frac{x^k}{k}=\frac{(1+x)^n-1}{n}.$$ In particular, taking $$x=1$$ gives the desired result. | {
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