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In later centuries mathematicians found more formal algebraic ways to enlarge the system of rational numbers to include what we now call all the "real numbers". The new ones are the irrationals. There are several ways to do this. One of the most common is to make precise the notion of an infinite decimal, along with rules for arithmetic with them. Others come with the names "Cauchy sequences" or "Dedekind cuts".
• To add a bit to your answer, note that the very notion of compass and straightedge constructions assumes with no justification that one can construct ideal circles/lines and somehow they always intersect when certain conditions are met (such as closer than the sum of radii). When one thinks about it, that is not at all obvious. Perhaps it is not even physically meaningful, for all we know! If so, it may very well be that $\sqrt{2}$ can't be constructed in any suitable physical sense! Furthermore, the computable reals are elementarily equivalent to the standard reals. Interesting at least. – user21820 May 29 '17 at 15:26
Using definite descriptor notation, we can define:
$$\sqrt{x} = (\iota y \in \mathbb{R})(y \geq 0 \wedge y^2=x)$$
In words: $\sqrt{x}$ is the unique $y \geq 0$ such that $y^2=x$.
From this, you can show that for all $y$ and all $x \geq 0$, we have: $$y^2 = x \iff y \in \pm\sqrt{x}.$$
But in some sense, this is a purely logical construct, at least insofar as we haven't explained how to approximate $\sqrt{x}$ to arbitrary precision. There's ways of doing this, but we haven't given one.
I think the author is simply wrong. There is another meaningful way of defining $\sqrt{2}$. Consider the following sequence: $$x_{n+1} := \frac{1}{2}\left(x_n + \frac{2}{x_n}\right)$$ with $x_0=1$.
Then $\lim_{n\to\infty} x_n=\sqrt{2}$, even though all $x_n\in\mathbb{Q}$ for arbitrary $n\in\mathbb{N}$. Now instead of defining $\sqrt{2}$ as the solution of $x^2-2=0$ we could have also defined it as the limit on $x_n$. | {
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Proof for the sequence can be found here (German only).
• This is still circular, in a deeper way. How do you define the notion of a limit without some underlying structure of the real numbers? In the rationals that sequence has no limit. What Cauchy showed is that It is possible to define the limit as the sequence itself. It's the necessity for some definition that's the crux of Gelfand's cryptic comment. – Ethan Bolker May 27 '17 at 17:10
• There is no need to use $\mathbb{R}$ with the above sequence. For example no transcendental numbers (like $\pi$) are needed. And in some sense $37$ is also just the result of multiple (albeit finite) application of Peano's axioms. The "only" difference with the above definition of $\sqrt{2}$ is that you need to apply the appropriate axioms an infinite number of times. – Hannebambel May 27 '17 at 17:18
• No, the author is right. As @EthanBolker says, your sequence has no limit within $\mathbb{Q}$. The key difference vs. the construction of $37$ lies exactly in the finite number of steps: no matter how large the $n$, no number $x_n$ in your sequence will ever equal $\sqrt{2}$. And the fact that $\sqrt{2}$ is not in $\mathbb{Q}$ is equivalent to the notion that $x^2 - 2 = 0$ is unsolvable with the standard algebraic operations. – Roland May 28 '17 at 7:55
• I somehow still don't get how the author is correct. The author says "$\sqrt{2}$ means nothing except the positive solution of the equation $x^2−2=0$", whereas I think that $\sqrt{2}$ also means the limit of the aforementioned sequence. – Hannebambel May 28 '17 at 18:23
• This answer shows that, in a certain sense, it is possible to "solve" for what $\sqrt{2}$ is. But Gelfand never claimed it wasn't possible to do so--that was not his point at all. His point is that merely naming the answer as $\sqrt{2}$ doesn't solve anything, because all you've done is give a name to what the answer is. – Eric Wofsey May 29 '17 at 21:48 | {
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This is a great question (+1) and judging by the quote, Gelfand must be a great writer at least for algebra textbooks (I confess that I have not read any of his books and hence judging only from the quote in question). In fact he has pinned down the essence of solution of polynomial equations via algebra.
If we restrict ourselves to algebra then the solution of polynomial equations via radicals is equivalent to replacing one polynomial equation with a set of binomial equations of type $x^{n} - a=0$. And for binomial equations we don't do anything apart from inventing symbols like $a^{1/n}$.
Next there are some equations which can't be solved via radicals. How does an algebraist deal with the situation now? He smartly handles the situation using field extensions and for any polynomial $f(x)$ over a field $F$ one can create an extension field $E$ which contains all the roots of $f(x)$. The technique of construction of this extension field is so simple that one is tempted to call such mechanism of solution of equations as "cheating". Thus if $f$ is irreducible then one can get a root in quotient $F[x] /(f(x))$ and guess what the root of $f(x)$ is the coset $x+(f(x))$.
In effect algebraic methods never try to find the value of root of an equation, but they are rather concerned with the structure of the field extensions which contains the roots. In some very special cases such extensions are radical and these are the familiar ones where we use the quadratic formula, or Cardano's formula (or more complicated stuff of similar type). | {
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Let's now come back to the usual scenario where the equations have coefficients in specific fields $\mathbb{R}$ and $\mathbb{C}$ and lets first handle the case when coefficients are real. Real numbers are the product of a non-algebraic process and their existence is almost on same footing as those of rational numbers. Thus they can and are used to measure magnitude and it is agreed by convention that there are sufficient real numbers to locate every point on a geometric line so that all the geometric measurement is possible via real numbers.
But then the real numbers are a very powerful system and offer us the following justification for the symbol $a^{1/n}$:
Theorem 1: If $a$ is a positive real number and $n$ is a positive integer then there is a unique positive real number $b$ such that $b^{n} =a$ and it is denoted by symbol $a^{1/n}$.
Thus at least some binomial equations of type $x^{n} - a=0$ have a root which has existence in the real number system. By existence we mean that it is possible to give some concrete idea about this root in comparison to the integers and rationals. In other words we can provide some approximation to the value of the root using the rational numbers. And moreover the approximation can be as accurate as we want. It is in this sense of approximation via rationals that every real number exists. And it applies to all sorts of irrational numbers including the famous ones $e, \pi, \sqrt{2}$. Thus irrational numbers are not meaningless, they are as meaningful as the rationals and perhaps mathematically more significant, but unfortunately algebraic techniques cannot help in discovering their true nature. | {
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Let's also note that there are some real numbers which are the roots of certain polynomial equations with rational coefficients and these we call algebraic real numbers. The reason for extending our number system from $\mathbb{Q}$ to $\mathbb{R}$ is not to give some sort of existence to these algebraic numbers (like $\sqrt{2}$). As far roots of polynomials are concerned the algebraic mechanism of field extensions is a sufficient and beautiful approach. Even if we tried to add such numbers to $\mathbb{Q}$ it does not give us $\mathbb{R}$, but it rather gives $\overline{\mathbb{Q}}$ which we call the algebraic closure of $\mathbb{Q}$. It includes all the algebraic numbers both real and complex, but it does not capture all the real numbers or all the complex numbers. In fact one can prove that like $\mathbb{Q}$, the set $\overline{\mathbb{Q}}$ is also countable whereas both $\mathbb{R}, \mathbb{C}$ are uncountable.
Real numbers were invented to deal with the essential/pressing need to arithmetize geometry. The idea was to develop of a system of numbers which could correspond to the points on a line and it was known from the time of Pythagoras (or perhaps even earlier) that there were some points on the line which did not correspond to a rational number. Another need for real numbers was coming from the very powerful techniques of calculus which were based on geometrical intuition, but there was no rigorous justification of these methods using arithmetical means. | {
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To answer your question, we have two ways to look at the symbol $\sqrt{2}$. One is via algebra which says that it is a solution to the equation $x^{2}-2=0$ and in this approach we can't distinguish between $\sqrt{2}$ and $-\sqrt{2}$. Another way is to think of it as a real number represented by its rational approximations. Thus we can say that $\sqrt{2}$ is a positive real number which is greater than all the positive rationals whose square is less than $2$ and it is less than all the positive rationals whose square is greater than $2$. Further in this particular case we are lucky to have a mechanism of locating a point on the number line corresponding to it via geometrical constructions with ruler and compass.
Let's also discuss a bit about the complex number system. But before we do that we need to note that there is another set of equations which the real number system can handle very well:
Theorem 2: If $f(x)$ is polynomial of odd degree with real coefficients then there is at least one real number $c$ such that $f(c) =0$.
But there are many equations with real coefficients for which there is no root in the real number system. The simplest and most famous such equation is $x^{2}+1=0$. Once again an algebraist comes to the rescue and creates a field extension $\mathbb{C}$ as the quotient $\mathbb{R} [x] /(x^{2}+1)$ and surprisingly in one shot the algebraist succeeds in solving all polynomial equations whatsoever :
Fundamental Theorem of Algebra: If $f(x)$ is a polynomial of positive degree with complex coefficients then there is a complex number $c$ such that $f(c) =0$.
But this is all cheating. The power of the above theorem is not due to the algebraic technique of creating field extensions via quotients, but rather due to the properties of real numbers mentioned in theorem 1 and 2 above. | {
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• Technically, this kind of justification can only reach the computable reals and also the computable complex numbers, which includes the algebraic closure of the rationals. Notice that the computable complex numbers is algebraically closed, so we can't justify going beyond it because we can't actually demonstrate the existence of anything beyond it! Also, you are right that the fact that the complex numbers are algebraically closed is at its crux not really a matter of algebra but of analysis. Even the field-theoretic proof at one point requires IVT for odd-degree polynomials. – user21820 May 29 '17 at 16:05
• In other words the only reason we have for believing the existence of all standard real numbers (whether by Cauchy sequences or Dedekind cuts) is that we must believe the power-set axiom or equivalent, which is purely set-theoretic and has no ontological justification. Actually, even function types are not enough to grant the classical power-set, because if we use type theory we can have function types that are compatible with the universe being computable. What breaks is that type membership may not be always true or false. – user21820 May 29 '17 at 16:35
• Of course, ZFC set theory is very elegant and convenient for modern mathematics and so, justified or not, it has found acceptance. =) – user21820 May 29 '17 at 16:36 | {
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# Summary of my understanding of sequences and series' convergence and divergence?
I'm trying to summarise my understanding of infinite sequences, series, and their relationships with respect to convergence at the fundamental level. Here is what I know. How much of this is correct?
First off, here's a table of the notations that I use, and their corresponding meaning.
My understanding is that:
• The sequence $\lbrace a_n \rbrace _{n=0}^{\infty}$ converges if $$\lim\limits_{n\to\infty}a_n=L_{a}.$$
• The infinite series $\sum\limits_{n=0}^{\infty}a_n$ converges if its sequence of partial sums, $\lbrace s_n \rbrace _{n=0}^{\infty}$, has a limit, i.e.$$\lim\limits_{n\to\infty}s_n=L_{s}.$$
• If the infinite series $\sum\limits_{n=0}^{\infty}a_n$ converges, then the limit of the sequence $\lbrace a_n \rbrace _{n=0}^{\infty}$ is $0$, i.e. $$\sum\limits_{n=0}^{\infty}a_n \: converges \rightarrow \lim\limits_{n\to\infty}a_n=0.$$
• The divergence test:
If the the limit of the sequence $\lbrace a_n \rbrace _{n=0}^{\infty}$ is NOT $0$ or does not exist, then the infinite series diverges, i.e. $$\lim\limits_{n\to\infty}a_n\neq0 \rightarrow \sum\limits_{n=0}^{\infty}a_n \: diverges$$
Would seriously appreciate it if anyone could verify whether the above is accurate or incorrect in any way.
EDIT: I've modified the two limits notation that were mentioned in the comments and answers below, as well as adding the additional condition (limit does not exist or does not equal zero) for the divergence test. I appreciate all the answers/comments.
• I would say that the notation $L_{a_n}, L_{s_n}$ looks pretty bad, these terms should not depend on $n$. – user99914 Dec 4 '17 at 5:08
• BTW one useful equivalent of $\lim_{n\to \infty}a_n=L,$ that students often don't think of , is : For every $r>0$ the set $\{n:a_n\not \in [-r+l,r+L]\}$ is finite. – DanielWainfleet Dec 4 '17 at 5:49 | {
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Everything is correct, though the notation $\lim\limits_{n\to\infty}a_n=L_{a_n}$ is nonstandard. Perhaps an improvement would be to write it as $\lim\limits_{n\to\infty}a_n=L_{a}.$ Normally just an $L$ will suffice but if you're working with multiple sequences (such as $a_n, b_n, c_n$) then $L_a$ is a good notation for the limit of the sequence $a$.
The reason why $L_{a_n}$ is not so good is because $n$ is just a free variable that has no importance whatsoever; if you write $a_n$ or $a_k$ it's the same thing. What matters is the sequence whose name is "$a$".
Also note that your last two statements are trivially equivalent; they are contrapositives of each other. In general "If $p$ then $q$" is stating the same thing as "If not $q$ then not $p$".
Therefore, just from knowing
$$\sum\limits_{n=0}^{\infty}a_n \: converges \rightarrow \lim\limits_{n\to\infty}a_n=0.$$
you can deduce
$not (\lim\limits_{n\to\infty}a_n=0 ) \rightarrow not (\sum\limits_{n=0}^{\infty}a_n \: converges)$
or in other words
$\lim\limits_{n\to\infty}a_n \not =0 \text{ or the limit doesn't exist} \rightarrow \sum\limits_{n=0}^{\infty}a_n \: diverges$
• Regarding the redundancy in my understanding, although $\lim\limits_{n\to\infty}a_n \not =0 \text{ or the limit doesn't exist} \rightarrow \sum\limits_{n=0}^{\infty}a_n \: diverges,$ it does not mean that $\lim\limits_{n\to\infty}a_n=0 \rightarrow \sum\limits_{n=0}^{\infty}a_n \: converges,$ am I right? – user98937 Dec 4 '17 at 13:11
• @user98937 Yes that's perfectly right. What you wrote is the converse of $\sum\limits_{n=0}^{\infty}a_n \: converges \rightarrow \lim\limits_{n\to\infty}a_n=0.$. Converses are sometimes true, but not in this case as you can see from $\sum_{n=1}^{\infty} \dfrac 1n$ which diverges, in spite of the fact that $\lim_{n \to \infty} \dfrac 1n=0$ – Ovi Dec 4 '17 at 15:33
Everything you wrote looks good to me. | {
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# Confusion regarding pdf of circularly symmetric complex gaussian rv
Considering a random variable $x$ that takes in values from a complex domain. Its real and imaginary components are totally uncorrelated. I am following this link and also studying this document.
In the wikipedia link, the pdf for a single observation is given and $k =2$ for bivariate gaussian as my assumption is that the real and imaginary are totally uncorrelated and are gaussian respectively. There is a 2 in the denominator in the pdf but the log likelihood for the complex case does not have any 2 in the denominator. The log-likelihood for $k=2$ would be $$-N \ln| \Sigma| - (x_n) {|\Sigma|}^{-1}(x_n)^H - k N \ln \pi$$ where $\mu=0$ as I am assuming zero mean r.v $x$
• Confusion 1:
I was thinking that the variance of $x$ was $$\Sigma = \begin{bmatrix}\sigma_1^2 & 0\\ 0& \sigma_2^2\end{bmatrix}\,\text.$$ So for $N$ samples (observation), the joint pdf would turn out to be $$P_x(x_1,x_2,...,x_N) = \prod_{n=1}^N\frac{1}{\pi \sigma^2_x} \exp \bigg(\frac{-{({x_n})}^H ({x_n})}{\sigma^2_x} \bigg)$$ where $\sigma^2_x = [\sigma_1^2,\sigma_2^2]$. Would there be a 2 in the denominator of the pdf and what is the correct pdf expression?
• Confusion 2:
In the document, the expression for the pdf for complex case looks different from the wikipedia link. Are they the same but maybe I am missing some link between them? Which pdf should I use?
• Confusion 3:
When simulating the r.v in Matlab with zero mean and variance 1, I am halving the variance because the real and imaginary part's variance should add up to 1. Then, would the pdf also contain half of the variance as $\sigma^2_x/2$?
Let me try to establish the relation between the univariate PDF for a real Gaussian and the univariate PDF for a complex proper (i.e. circular symmetric) Gaussian. | {
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You know that
$p_x(x)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp(\frac{-x^2}{2\sigma^2})$ is the PDF of a real-valued Gaussian with variance $\sigma^2$. We write $x\sim\mathcal{N}(0,\sigma^2)$ to denote that $x$ is a random variable that follows a real-valued Gaussian with zero mean and variance $\sigma^2$.
Now, let $z=a+jb$ be a circular symmetric random variable with real part $a$ and imaginary part $b$. In a circularly symmetric Gaussian random variable, the real and imaginary part are i.i.d., i.e. $a\sim\mathcal{N}(0,\sigma^2)$ and $b\sim\mathcal{N}(0,\sigma^2)$. Since $a,b$ are independent, their joint PDF is the product of their PDFs.
$$p_z(z=a+jb)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(\frac{-a^2}{2\sigma^2}\right)\cdot\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(\frac{-b^2}{2\sigma^2}\right)$$
which gives the PDF of the complex variable $z$ at the value $a+jb$. We can reformulate this to \begin{align}p_z(z=a+jb)&=\frac{1}{2\pi\sigma^2}\exp\left(\frac{-(a^2+b^2)}{2\sigma^2}\right)\\&=\frac{1}{2\pi\sigma^2}\exp\left(\frac{-z^*z}{2\sigma^2}\right)\\&=\frac{1}{2\pi\sigma^2}\exp\left(\frac{-\|z\|^2}{2\sigma^2}\right)\end{align}
We also write for this case that $z$ follows a circular Gaussian distribution and denote this by $z\sim\mathcal{CN}(0,2\sigma^2)$. Why is it suddenly $2\sigma^2$ (i.e. the double variance compared to the uni-variate, real case? Well, because $E[\|z\|^2]=2\sigma^2$ since it consists of a real and an imaginary part (which are independent and hence their variance adds up together).
So, to sum up:
• if $x\sim\mathcal{N}(0,\sigma^2)$, then $p_x(x)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{x^2}{2\sigma^2}\right)$
• if $z\sim\mathcal{CN}(0,\sigma^2)$, then $p_z(z)=\frac{1}{\pi\sigma^2}\exp\left(-\frac{z^*z}{\sigma^2}\right)$ | {
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• Thank you so much, cannot express my gratitude in words here.....just to clarify 2 things from you - (1) if I consider design where the r.v $z \sim CN(0,2\sigma^2)$ then the pdf has the 2 in the denominator, otherwise if $z \sim CN(0,\sigma^2)$ then there is no 2 in the pdf's denominator. So, either design is acceptable. Is my understanding right? (2) notations - in my question I wrote the $[.]^H$ complex conjugate transpose but you have written * which is only the conjugate. Why the $[.]^H$ is wrong? – Ria George Apr 18 '17 at 15:33
• sure, if $z\sim CN(0,2\sigma^2)$, then there is the two in the denominator, if its $z\sim CN(0,\sigma^2)$, then it's not there. Here, the two comes from the description of the variance (e.g. if we had $z\sim CN(0,1.23456\sigma^2)$, then in the denominator there would be $1.23456$). For a scalar variable $z$, $[]^H$ and $[]^*$ are equal. When it comes to vector-variables, you'd need $[]^H$. However, note that in this case, also the denominator changes to $\pi^k\sigma^{2k}$, since the PDF is the product of the $k$ PDFs of each component (k is the vector size); each component has var. $\sigma^2$. – Maximilian Matthé Apr 18 '17 at 17:58
• This answer has the correct ideas but is marred badly by a confusion of the concepts of probability and probability density. The probabiity that a continuous random variable equals a given number is $0$, regardless of the choice of number. Indeed, the probability that $z$ equals $a+jb$ (where we choose $a=b=0$) is claimed to be $$P(z=0+j0)=\frac{1}{2\pi\sigma^2}$$ which has value greater than $1$ if $\sigma^2 < \frac{1}{2\pi}$. -1 pending clean-up of the answer to be correct. – Dilip Sarwate Apr 18 '17 at 22:26 | {
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# Integral of $\text{Log}(z)$
I'd like to check my work. I'm trying to integrate $f(z)=\text{Log}(z)$ on the right half of the unit circle from $-i$ to $i$. $\text{Log} = \ln(r)+i\theta$ according to my book. So, \begin{align} \int_C f(z)\,dz&=\int_{z_1}^{z_2}f(z)\,dz\\\\& = \int_{-i}^i \text{Log}\,dz \\\\ &= \bigg[z\text{Log}-z\bigg]_{-i}^i \\\\ &= (i\text{Log}(i)-i)-\left(-i\text{Log}(-i)+i\right) \\\\ &= i\left(\ln(1)+\frac{\pi}{2}i\right)-i \ + i\left(\ln(1)-\frac{\pi}{2}i \right) \ -i \\\\&= -2i \\\\ \therefore \int_{-i}^i \text{Log}\,dz=-2i \end{align}
This may be incorrect, I'm not really sure how to incorporate the branch cut here.
Thanks!
• the circuit is a circle not the segment [-i,i]. – zwim Jun 15 '17 at 5:13
• So, I'm required to use the formula $\int_C f(z)dz=\int_a^b f[z(\theta)]z'(\theta)d\theta$? Where $a$ and $b$ are $\frac{\pi}{2}$ and $\frac{-\pi}{2}$? What would my choice for $z'(\theta)$ be in this case? – Kosta Jun 15 '17 at 5:14
• Have a look at these answers : math.stackexchange.com/questions/1602614/… – zwim Jun 15 '17 at 5:34
I thought it might be instructive to present two alternative approaches to evaluate the integral of interest. To that end we proceed.
Let $\text{Log}(z)=\log(|z|)+i\text{Arg}(z)$ where $-\pi < \text{Arg}(z)\le \pi$. In addition, let $C$ denote the semicircular contour in the right-half plane of radisu $1$ and that begins at $z=-i$ and ends at $z=i$.
METHODOLOGY $1$:
On $C$, $z=e^{i\theta}$, $-\pi/2\le \theta \le \pi$. Hence, the integral $\int_C \text{Log}(z)\,dz$ is given by
\begin{align} \int_C \text{Log}(z)\,dz&=\int_{-\pi/2}^{\pi/2} \text{Log}(e^{i\theta})\,ie^{i\theta}\,d\theta\\\\ &=-\int_{-\pi/2}^{\pi/2} \theta e^{i\theta}\,d\theta\\\\ &=-2i\int_0^{\pi/2} \theta\sin(\theta)\,d\theta\\\\ &=-2i \end{align}
METHODOLOGY $2$: | {
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METHODOLOGY $2$:
Another approach is to exploit Cauchy's Integral Theorem. Inasmuch as $\text{Log}(z)$ is analytic for $z\in \mathbb{C}\setminus (-\infty,0]$. Then, the integral over $C$ can be deformed to connect $-i$ to $i$ along any rectifiable path that does not intersect the branch cut. Therefore, we write
\begin{align} \int_C \text{Log}(z)\,dz&= \lim_{\epsilon\to 0^+}\left(\int_{-1}^{-\epsilon }\text{Log}(iy)\,i\,dy+\int_{-\pi/2}^{\pi/2}\text{Log}(\epsilon e^{i\theta})\,ie^{i\theta}\,d\theta+\int_{\epsilon }^1\text{Log}(iy)\,i\,dy\right)\\\\ &=i \lim_{\epsilon\to 0^+}\int_\epsilon^1 \left(\text{Log}(-iy)+\text{Log}(iy)\right)\,dy\\\\ &=2i\int_0^1 \text{Log}(y)\,dy\\\\ &=-2i \end{align}
as expected!
• This is very helpful, thank you Mark! – Kosta Jun 16 '17 at 16:21
• You're welcome. Pleased to hear that this is helpful! – Mark Viola Jun 16 '17 at 16:25
An alternate way to check, without using contours, is to use $$\int \ln(x) \, dx = x \, \ln(x) - x$$ which provides \begin{align} \int_{-i}^{i} \ln(x) \, dx &= \left( i \, \ln(e^{\pi \, i/2}) - i \right) - \left( -i \, \ln(e^{-\pi \, i/2}) + i \right) \\ &= \frac{\pi \, i^2}{2} - \frac{\pi \, i^2}{2} - 2 i \\ &= -2i. \end{align} With this it can be stated that the real line integral result is the same as the contour integral result.
looks good to me .Fundamental theorem of calculus wins again | {
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# Thread: Trigonometric Identity Problem
1. ## Trigonometric Identity Problem
How do you show that:
2sinxcosx=
(4tan{x/2})/(1+tan2{x/2}) . (1-tan2{x/2})/(1+tan2{x/2})
Thanks a lot for helping!!
2. ## Re: Trigonometric Identity Problem
Originally Posted by JiaGeng
How do you show that:
2sinxcosx=
(4tan{x/2})/(1+tan2{x/2}) . (1-tan2{x/2})/(1+tan2{x/2})
$1+\tan^2(\frac x 2) = \sec^2(\frac x 2)$
$1-\tan^2(\frac x 2) = \dfrac{2 \tan(\frac x 2)}{\tan(x)}$
$\sin(2x)=2\sin(x)\cos(x)$
-------------------------
$\dfrac{4 \tan(\frac x 2)}{1+\tan^2(\frac x 2)}\dfrac{1-\tan^2(\frac x 2)}{1+\tan^2(\frac x 2)}=$
$\dfrac{4 \tan(\frac x 2)}{\sec^2(\frac x 2)}\dfrac{ \dfrac{2 \tan(\frac x 2)}{\tan(x)}}{\sec^2(\frac x 2)}=$
$4\cos(\frac x 2)\sin(\frac x 2)\dfrac{2\sin(\frac x 2)\cos(\frac x 2)}{\tan(x)}=$
$2\sin(x)\dfrac{\sin(x)}{\tan(x)}=$
$2\sin(x)\cos(x)$
3. ## Re: Trigonometric Identity Problem
Hello, JiaGeng!
$\text{Prove: }\:2\sin x\cos x\:=\:\frac{4\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}} \cdot \frac{1-\tan^2\frac{x}{2}}{1 + \tan^2\frac{x}{2}}$
The first fraction is:
. . $\frac{4\tan\frac{x}{2}}{1 + \tan^2\frac{x}{2}} \;=\;\frac{4\frac{\sin\frac{x}{2}}{\cos\frac{x}{2} }}{\sec^2\frac{x}{2}} \;=\;\frac{4\frac{\sin\frac{x}{2}}{\cos\frac{x}{2} }}{\frac{1}{\cos^2\frac{2}{2}}} \;=\;4\sin\tfrac{x}{2}\cos\tfrac{x}{2} \;=\;2\sin x$
The second fraction is:
. . $\frac{1-\tan^2\frac{x}{2}}{1 + \tan^2\frac{x}{2}} \;=\;\frac{1-\frac{\sin^2\frac{x}{2}}{\cos^2\frac{x}{2}}}{\sec^ 2\frac{x}{2}} \;=\; \frac{1-\frac{\sin^2\frac{x}{2}}{\cos^2\frac{x}{2}}}{\frac {1}{\cos^2\frac{x}{2}}} \;=\;\cos^2\tfrac{x}{2} - \sin^2\tfrac{x}{2} \;=\;\cos x$
Therefore, the RHS becomes: . $2\sin x\cos x$.
4. ## Re: Trigonometric Identity Problem
Thanks romsek and soroban!
But I was wondering if there is a way to do it from LHS to get RHS because that's what the question requires.... Would that be even harder?
5. ## Re: Trigonometric Identity Problem | {
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5. ## Re: Trigonometric Identity Problem
Originally Posted by JiaGeng
Thanks romsek and soroban!
But I was wondering if there is a way to do it from LHS to get RHS because that's what the question requires.... Would that be even harder?
Equality works in both directions. If you want, write the equations in reverse order and equality still holds at every step.
6. ## Re: Trigonometric Identity Problem
Hello, JiaGeng!
I was wondering if there is a way to do it from LHS to get RHS.
Would that be even harder?
If you would rather not run the steps in reverse,
it can be done ... with a little imagination.
$2\sin x\cos x \;=\;\frac {2(2\sin \frac{x}{2}\cos\frac{x}{2})}{1}\cdot \frac{\cos^2\!\frac{x}{2} - \sin^2\!\frac{x}{2}}{1}$
. . . . . . . . . $=\;\frac{4\sin\frac{x}{2}\cos\frac{x}{2}}{\cos^2\! \frac{x}{2} + \sin^2\!\frac{x}{2}} \cdot \frac{\cos^2\!\frac{x}{2} - \sin^2\!\frac{x}{2}}{\cos^2\!\frac{x}{2} + \sin^2\!\frac{x}{2}}$
In both fractions, divide numerator and denominator by $\cos^2\!\tfrac{x}{2}\!:$
$\dfrac{\frac{4\sin\frac{x}{2} \cos\frac{x}{2}}{\cos^2\!\frac{x}{2}}} {\frac{\cos^2\!\frac{x}{2}}{\cos^2\!\frac{x}{2}} + \frac{\sin^2\!\frac{x}{2}}{\cos^2\!\frac{x}{2}}} \cdot \dfrac{\frac{\cos^2\!\frac{x}{2}}{\cos^2\!\frac{x} {2}} - \frac{\sin^2\!\frac{x}{2}}{\cos^2\!\frac{x}{2}}} {\frac{\cos^2\!\frac{x}{2}}{\cos^2\!\frac{x}{2}} + \frac{\sin^2\!\frac{x}{2}}{\cos^2\!\frac{x}{2}}} \;=\; \dfrac{4\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}} {1 + \left(\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} \right)^2} \cdot \dfrac{1 - \left(\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} \right)^2} {1 + \left(\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} \right)^2}$
. . . . $=\;\frac{4\tan\frac{x}{2}}{1 + \tan^2\!\frac{x}{2}} \cdot \frac{1-\tan^2\!\frac{x}{2}}{1 + \tan^2\!\frac{x}{2}}$ | {
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# A Sample of Ask Dr. Math, Part 2: Questions Outside of School
In the first post, I gave a small sampling of questions we’ve had from students, parents, and teachers, all related to school, and discussed how we like to deal with these. But we also get many questions with no direct relation to school. These may come from people who actually use math in their work (computer programming, plumbing, and anything in between), or from students who are trying to deeply understand what they are studying, or who are just curious. Often, of course, these too are from students or teachers, but they are going beyond that role. These are perhaps the most interesting of all. They are really why we are here.
## A real-life question
This question arrived in 2014 from a flooring inspector, Tim, asking about how high a floor would buckle due to a given increase in length if the ends are fixed:
Modeling How Much Wood Expands
Say I place a six foot long yardstick wide side down. One end is butted against a wall (fixed object). When the other end of the yardstick is pushed towards the wall, by say 1/4", the middle area rises -- and it rises much, much higher than 1/4".
I would like to know if there is a fixed formula for this exponential gain of "flat" vs. "up."
I come across this daily, seeing floors buckle up and away from substrates because they are a tight fit.
(He misused the word exponential; I suppose he was just thinking that the rise is surprisingly high, as in exponential growth.)
Dr. Rick and I both recognized this problem as related to previous questions we had answered, and referred to those in our replies. His old answer assumed a circular arc for the new shape, and did a sanity check by comparing the surprising answer to one obtained by supposing that the board just bends sharply in the middle. The circular arc version does not lend itself to easy calculation, but the sharp bend could be turned into a formula. | {
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There is a way to calculate the height, but it isn't a formula, as such.
...
For more on this topic, including a way to do a *rough* approximation with a simple formula, see:
http://mathforum.org/library/drmath/view/62613.html
I had been thinking while Dr. Rick wrote, and found a different old answer (by Dr. Jeremiah) that didn’t actually come up with a method, just a starting point. But I “played” with numbers using a spreadsheet much as Dr. Rick had for his earlier answer, and saw a pattern that suggested an approximate formula he might try. When I saw what Dr. Rick had written, I wrote to propose this formula.
I see Dr. Rick beat me to most of what I was going to say, except that my link to our archives is not quite as helpful:
Rail Bend in Hot Weather
http://mathforum.org/library/drmath/view/61465.html
But I made a spreadsheet carrying out the calculations for a circular arc; and have found experimentally that the rise seems to be quite accurately approximated by a simple formula. I have to study more to try to justify it, but here goes.
For the actual length of the stick, s, and the (small) decrease in the space available, x,
h = 0.605 sqrt(sx)
But people who enjoy math can’t be fully satisfied with a guess; I continued pursuing my approximation, hoping to justify it. The patient, Tim, didn’t need a proof, but I did. By the end of the day, I could write up a derivation of the formula and send it along, and then, the next day, apply it to some examples.
I've verified my simple approximation, which turns out to be
h = sqrt(6sx)/4
This changes my 0.605 to 0.612, which fits better for very small x but not quite as well for larger ones.
I'm sure you don't need to see this, but for my records, here is the derivation of my approximation: ... | {
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A missing formula for a counter-intuitive observation (a half-inch shift can result in almost a 7-inch buckle) can be a great motivation for some fun with math. And we also have a reminder that in the real world, an approximation can be both good enough, and as good as you’ll get. (Be sure, as always, to read the whole story in the original page!)
## A curious question
(Well, that last one arose from curiosity, too, but this is pure-math curiosity.)
Within months of my starting with Ask Dr. Math, I wrote this answer, to a question wondering about place value in decimal numbers:
The Oneths Place
Why is there not a oneths place? My classmates and I were wondering. It is weird that decimals begin with tenths, hundredths, and so on. We appreciate your help.
I love it when children are wondering, especially when a teacher encourages it. So I was happy to think about this, even though I had probably never considered the question before. (Since then, we have had some form of this question at least a dozen more times! Often, I have answered by saying, “Believe it or not, you’re not the first to ask this; if you search our site for the word ‘oneth’, you’ll find this answer …”)
I’m sure my answer is not all that could be said about it, but I managed not only to give the quick answer (oneths are just another name for ones, since 1/1 = 1, and we don’t want two places with the same value), but to relate it to some other perceived asymmetries in the way we write numbers, and bring in signed exponents (which may be beyond the students’ knowledge, or may be something they were just about to learn). I closed with a general comment about curiosity:
I hope that helps. It's interesting how we tend to expect things to be symmetrical. Mathematicians and scientists often expect it too, and when things don't seem balanced, they try to find out why. So keep wondering about things like this.
## A philosophical question | {
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## A philosophical question
Another question we’ve received a number of times involves how a line segment can have a finite length when it is composed of infinitely many points. As a result, we’ve referred to the following question a dozen times or so, in addition to giving fresh answers from different perspectives:
How Can a Line Have Length?
We have a line, OR a line segment, OR a ray (it really doesn't matter). If this particular linear entity (let us say that it is a line segment, for the sake of hypothetical simplicity) consists only of infinite points, all with zero volume, mass, diameter, cross-sectional area, etc., how can it have length? Why are we able to take a particular line segment and give it a length of 5? If a substance can only be composed of volumeless points, how can it have volume in and of itself?
Dr. Daniel’s answer (from 1998) touches on several key ideas in philosophical thinking about this topic, making a concise summary without digging in too deeply. (I try to avoid digging too deeply, because of the possibility of cave-ins … .) When I answer related questions, I often refer to that answer, and also to this one:
Point and Line
I'm in high school, but this problem has really nothing to do with school, it's just been bothering me for a while.
A point has no dimension (I'm assuming), and a line, which has dimension, is a bunch of points strung together. How does something without dimension create something with dimension?
Going even farther, a point essentially is nothing, because it has no dimension. My question is, How does a bunch of nothings (a point) create a something (a line)?
Here Dr. Jordi got into a long discussion about the problem, perhaps not very effectively in the end. After referring to these two answers, I generally add something like the following as my own brief perspective: | {
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Neither will be a fully satisfying answer, largely because we humans are finite beings, not able to directly observe infinity. (It's amazing enough that we can THINK about infinity, and get some of the answers right!) But the basic idea is that we don't define a line as something made by putting points together; rather, we start by defining a line, and then note that it consists of points. So we can find points on a line, without having to _make_ the line _out of_ points (or make a cube out of stacked squares).
Note also that we do not define length in terms of points; since 0, the width of a point, times infinity, the number of points, is an indeterminate quantity (we can't assign any one value to it), that would be useless to attempt. So length is an independent concept, defined only in relation to the location of the endpoints of a segment, not to the points that make it up.
In choosing these examples of “curious questions”, I have chosen to avoid questions with really interesting answers, which deserve fuller treatment on their own! Keep watching, and I’ll get to some of them.
### 2 thoughts on “A Sample of Ask Dr. Math, Part 2: Questions Outside of School”
1. Hi – I would like to know the probability of this occurring – this past January I celebrated my 74th birthday which means I was born in 1947 (1947/74 numbers reversed). My daughter was born in April, 1974. She will be turning 47 this year (1974/47 numbers reversed) What is the probability of this occurring in a parent/son-daughter relationship?
1. Hi, Gwen.
I suppose you asked this in connection with this particular post just because it’s an example of curiosity! In fact, there is another post that deals with this very kind of question: Should Rare Events Surprise Us? You’ll find there that an important question we need to ask is, “What do you mean by ‘this’?” | {
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I don’t think you are asking what percentage of all people now alive were born in 1947, and had a daughter born in 1974; and if you did, I’d have to pass, as it would require digging deeper into birth statistics than I feel like going. (It would not be an interesting question mathematically.)
I think your main interest is in the coincidence of the numbers. I may not get to a probability in the end, but there may be some interesting observations along the way!
My first observation is that the two reversals are closely related. The reason you are 74 this year is that 2021-1947 = 74; that is, 2021 = 1947 + 74. But that implies that 2021 also equals 1974 + 47, because both sums can be written as 1900 + 47 + 74, just rearranging the numbers. So that fact that this year your age is the reversal of your birth year implies that the same will be true this year for anyone born in 1974. To put it another way, her being born in 1974, the reversal of your birth year, predestined everything you observe this year. It all had to happen sooner or later!
Let’s test that out. I was born in 1952. I couldn’t have a daughter born in 1925 or 2025, but in fact it turns out that my father was born in 1925! My claim is that this by itself implies that something like your situation would occur at some point. So if we add 52 to 1925, we find that in 1977, my father was 52, and I was 25. Of course that happened once in a lifetime, but that it would happen to us at all depended only on our two birthdates and our living long enough.
Will everyone have some year in his life when his age is the reverse of his birth year? That just depends the reverse of his birth year being a likely age. For someone born in 1900, it happens immediately (and again if he reaches 100); born in 1901, it happens when he reaches 10. If you were born in 1909, you’ve have to live to 90 to have it happen. So basically, lifespan is the only restriction. | {
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How likely is it that one will have a child in his “reverse year”? That restricts it more, since the resulting age has to be within the childbearing years. For example, in your case your age at her birth was 1974 – 1947 = 27. If we suppose that, say, a child can be born to a parent aged 16 to 45, then your scenario is feasible for people born in about 30% of all years, and considerably more likely for, say, 10-15%. (Actually, this is a little more subtle than what I just said, and I checked by making a spreadsheet to actually count years that would work.)
With appropriate statistical information, one might come up with a probability of actually having a child in a given year. But that’s beyond my level of interest.
But I recalled my twin brother pointing out something similar based on our father’s age, and was able to find it: Reversal of Age Digits Every Eleven Years. What he describes there will also apply to you and your daughter: Because you were a multiple of 9 years old when she was born, every 11 years your ages have reversed digits, as they are this year. So not only is your situation this year not unique, it is not unique to you: It also happened when she was 3, and 14, and 25, and 36! But it will stop happening when you pass 100 …
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# Integral of the entrywise square of the exponential of a matrix
Note: I posted my question on math.stackexchange but got no answer. That is why I am asking it here.
Let $A$ be a $n\times n$ square matrix such that the real part of all eigenvalues are negative. For each $i,j$, let $\exp(At)_{ij}$ be the element $(i,j)$ of the matrix. It is well known that: $$\int_0^\infty \exp(At)_{ij}dt = -(A^{-1})_{ij}$$ Is it possible to simplify a similar expression where each element is squared: $$\int_0^\infty (\exp(At)_{ij})^2 dt = ??$$
I am wondering if it is possible to simplify the above expression. If it helps, I can assume that $A$ is diagonalizable. Note that unless for one-dimensional matrices, $(\exp(At)_{ij})^2\ne\exp(2At)_{ij}$.
• Note that OP wants an elementwise square, not the usual matrix-multiplication square. – Federico Poloni Nov 24 '16 at 8:19
Inspired strongly by Anthony's answer, here is a formula that works for arbitrary $A$. Let $M$ be the $n^2 \times n^2$ square matrix given by $$M= A \otimes I_n + I_n \otimes A_n$$ i.e. in terms of indices
$$M_{ij,kl}=A_{i,k} \delta_{j,l} + \delta_{i,k}A_{j,l}$$
Then because $A \otimes I_n$ and $I_n \otimes A$ commute, $$e^{Mt} = (e^{At} \otimes I_n) (I_n \otimes e^{At}) = e^{At} \otimes e^{At}$$ i.e. in indices $$(e^{M t})_{ij,kl} = (e^{A t})_{i,j} (e^{At})_{k,l}$$ so in particular $$(e^{At})_{i,j}^2 = (e^{Mt})_{ii,jj}$$ and $$\int_t (e^{At})_{i,j}^2 = \int_t (e^{Mt})_{ii,jj} = - (M^{-1})_{ii,jj}$$
(Here I am using commas to separate the two indices of a matrix entry) | {
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(Here I am using commas to separate the two indices of a matrix entry)
• Thanks for this answer. I am not sure to understand why the two summands would commute and why $e^{Mt}_{ij,kl}$ can be expressed in terms of $e^{At}$ (with these indices, it seems false). – N. Gast Nov 24 '16 at 16:18
• I think this is problematic because each of the summands is of rank $n$, so the sum is of rank at most $2n$. For $n>2$, $M$ will not be invertible. – Anthony Quas Nov 24 '16 at 16:39
• @AnthonyQuas That was a stupid mistake, fixed now. – Will Sawin Nov 24 '16 at 18:09
• @N.Gast Sorry, that was wrong, is this clearer? – Will Sawin Nov 24 '16 at 18:09
• @WillSawin: does M^{-1} exist? – Anthony Quas Nov 24 '16 at 18:42
You can do something. Here's a computation for diagonalizable $A$. Let $A=BDB^{-1}$ and let the elements of $D$ be $-\lambda_1,\ldots,-\lambda_d$. Then \begin{align*} \int_0^\infty (e^{At})_{ij}^2\,dt&= \int_0^\infty (Be^{Dt}B^{-1})_{ij}^2\,dt\\ &=\int_0^\infty \sum_{k,k'} B_{ik}e^{-\lambda_k t}B^{-1}_{kj} B_{ik'}e^{-\lambda_k' t}B^{-1}_{k'j}\,dt\\ &=\int_0^\infty \sum_{k,k'} B_{ik}B^{-1}_{kj} B_{ik'}B^{-1}_{k'j}e^{-(\lambda_k+\lambda_k') t} \,dt\\ &=\sum_{k,k'} \frac{1}{\lambda_k+\lambda_k'}B_{ik}B^{-1}_{kj} B_{ik'}B^{-1}_{k'j}. \end{align*}
Given a Hurwitz matrix $\mathrm A \in \mathbb R^{n \times n}$, let
$$\Phi (t) := \exp(\mathrm A t)$$
be the state transition matrix, and let its $(i,j)$-th entry be denoted by
$$\varphi_{ij} (t) := \mathrm e_i^{\top} \Phi (t) \, \mathrm e_j$$
Hence, | {
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$$\varphi_{ij} (t) := \mathrm e_i^{\top} \Phi (t) \, \mathrm e_j$$
Hence,
$$\begin{array}{rl } \displaystyle\int_0^{\infty} \left( \varphi_{ij} (t) \right)^2 \, \mathrm d t &= \displaystyle\int_0^{\infty} \left( \mathrm e_i^{\top} \Phi (t) \, \mathrm e_j \right)^2 \, \mathrm d t\\\\ &= \displaystyle\int_0^{\infty} \mathrm e_i^{\top} \Phi (t) \, \mathrm e_j \mathrm e_j^{\top} \Phi^{\top} (t) \, \mathrm e_i \, \mathrm d t\\\\ &= \mathrm e_i^{\top} \underbrace{\left( \displaystyle\int_0^{\infty} \Phi (t) \, \mathrm e_j \mathrm e_j^{\top} \Phi^{\top} (t) \, \mathrm d t \right)}_{=: \mathrm W_c} \mathrm e_i = \mathrm e_i^{\top} \mathrm W_c \mathrm e_i\end{array}$$
where $\mathrm W_c$ is the controllability Gramian of the pair $(\mathrm A, \mathrm e_j)$ and is the solution to the following controllability Lyapunov equation
$$\boxed{\mathrm A \mathrm W_c + \mathrm W_c \mathrm A^{\top} + \mathrm e_j \mathrm e_j^{\top} = \mathrm O_n}$$
Thus, the $n$ columns of the integral of the entrywise product
$$\int_0^{\infty} \left( \Phi (t) \circ \Phi (t) \right) \mathrm d t$$
are the diagonals of $\mathrm W_c^{(1)}, \mathrm W_c^{(2)}, \dots, \mathrm W_c^{(n)}$, where $\mathrm W_c^{(1)}, \mathrm W_c^{(2)}, \dots, \mathrm W_c^{(n)}$ are the solutions to the following $n$ controllability Lyapunov equations
$$\begin{array}{cl} \mathrm A \mathrm W_c^{(1)} + \mathrm W_c^{(1)} \mathrm A^{\top} + \mathrm e_1 \mathrm e_1^{\top} &= \mathrm O_n\\ \mathrm A \mathrm W_c^{(2)} + \mathrm W_c^{(2)} \mathrm A^{\top} + \mathrm e_2 \mathrm e_2^{\top} &= \mathrm O_n\\ \vdots & \\ \mathrm A \mathrm W_c^{(n)} + \mathrm W_c^{(n)} \mathrm A^{\top} + \mathrm e_n \mathrm e_n^{\top} &= \mathrm O_n\end{array}$$
• In MATLAB, the Gramian can be computed using function gram. – Rodrigo de Azevedo Nov 25 '16 at 22:18 | {
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# Is $n^2 + n + 1$ prime for all n?
I recently stumbled across this question in a test.
Paul says that "$n^2+n+1$ is prime $\forall\:n\in \mathbb{N}$".
• Paul is correct, because...
• Paul is wrong, because...
The answers sheet says that "all answers providing a valid counterexample, e.g. $n=4$, are valid; however, I'm interested in understanding why - is there some sort of theorem which can predict whether a polynomial is prime or not?
What I tried was:
$n^2+n+1$ can be factorized as $(n+\frac{1+\sqrt{3}i}{2})(n+\frac{1-\sqrt{3}i}{2})$. Neither of these factors is a suitable factor (i.e. is natural, different from $1$, and different from $n^2+n+1$) for any $n\in\mathbb{N}$, therefore $n^2+n+1$ must be prime $\forall\:n\in\mathbb{N}$.
Instead, $n^2-1$ can be factorized as $(n+1)(n-1)$. The factor $(n+1)$ is a suitable factor (i.e. natural, different from $1$, and different from $n^2-1$) for $n>2$; therefore, $n^2-1$ is not prime for $n>2$.
However, the above reasoning is clearly wrong, because "$n^2+n+1$ is prime $\forall\:n\in\mathbb{N}$" doesn't hold for the case $n=4$.
What is wrong in the reasoning? Can one tell whether a polynomial $p(n)$ is prime for all $n$ a priori?
• Counterexample clearly is a formal proof. Furthermore, for quite a lot of small $n$ (like $n \leq 40$, IIRC - could be wrong), such numbers are prime. – Marcin Łoś May 15 '14 at 14:32
• @MarcinŁoś thanks for pointing that out. I edited the question to better suit what I'm asking for. – Giulio Muscarello May 15 '14 at 14:36
• It was established by Goldbach that there is no polynomial with integer coefficients that can give a prime for all integer inputs. Ref: mathworld.wolfram.com/Prime-GeneratingPolynomial.html – Joel May 15 '14 at 14:44
Hint $\ \ f(1)=3\,\Rightarrow\, 3\mid f(1\!+\!3n),\$ e.g. $\,3\mid f(4)=21$
Remark $\$ Above we used the Polynomial Congruence Theorem | {
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Remark $\$ Above we used the Polynomial Congruence Theorem
$\ {\rm mod}\ 3\!:\ \color{}{1\!+\!3n\equiv 1}\,\Rightarrow\, f(\color{}{1\!+\!3n})\equiv f(\color{}1)\equiv 0,\$ for all $\,f\in\Bbb Z[x]$
Alternatively $\,\ 3n\mid f(1\!+\!3n)-f(1)\$ by the Factor Theorem $\,x-y\mid f(x)-f(y)$
The idea generalizes to a proof that any nonconstant polynomial $\in\Bbb Z[x]\,$ has a non-prime value.
Just because a polynomial $f$ doesn't factor doesn't mean that the number $f(m)$ you get when evaluating $f$ at a number $m$ doesn't factor. The simplest example is something like $f(x) = x+1$. This polynomial is irreducible, hence doesn't factor, but $f(3) = 4$ is clearly composite.
Here is a link showing that no polynomial with integer coefficients can evaluate to primes for all integers.
You want to disprove a statement of the form $\forall n\colon f(n)\text{ is prime}$, that is you want to prove $\neg\forall n\colon f(n)\text{ is prime}$, which is equivalent to $\exists n\colon \neg(f(n)\text{ is prime})$. Therefore, a counterexample is often the easiest and most straightforward way to disprove a general statement. In fact, any proof that merely shows that a $\forall$ statement is wrong without constructing a counterexample (or essentially doing so, but leaving out the details) is frowned upon by some mathematicians as being non-constructive. That being said, maybe you'd like a proof that there is no polynomial $f$ such that $f(n)$ is prime for all $n\in \mathbb N$ (except constant polynomials): | {
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Let $$f(n)=a_dx^d+a_{d-1}x^{d-1}+\ldots+a_1x+a_0$$ be a polynomial of degree $d\ge 1$ with coefficients $a_i\in\mathbb C$, $0\le i\le d$, and $a_d\ne 0$. First we observe tat in fact $a_i\in \mathbb Q$ for all $i$: By plugging in $x=1, 2, \ldots, d+1$, we obtain $d+1$ rational equations in the unknowns $a_i$. Since the powers of $x$ are linearly independant (or look up Vandermonde matrix), there exists a unique solution, which must be rational. Then there exists $M\in\mathbb N$ sich that $a_iM\in\mathbb Z$ for all $i$. By assumption, $Mf(n)$ is $M$ times a prime for all $n\in\mathbb N$. For each prime $p$, there are at most $d$ values of $n$ such that $f(n)=p$. Since $M$ has only finitely many prime factors, $f(n)\mid M$ for only finitely many $n$. Therefore there exists $n$ such that $p:=f(n)\not\mid M$. By reducing $Mf(X)$ modulo $p$, we see that $Mf(n+kp)\equiv 0\pmod p$ for all $k\in\mathbb Z$. This implies that the prime $f(n+kp)$ is $p$ for all $k$. Especially, there are more than $d$ values of $n$ for which $f(n)=p$, contradiction. $_\square$
Trivia remark: On the other hand, there exists a polynomial in several variables with the property that its values are either negative or prime, and all primes occur!
• Your trivia remark is really cool! Do you have a link/reference? – André 3000 May 15 '14 at 15:22
• Here is a link to an early paper by Jones et al, not long after Matiyasevich proved existence. The $25$ variables has gone down a lot since then, I think to about $9$. – André Nicolas May 15 '14 at 15:38
• @AndréNicolas That's cool - so instead of letters we can now use digits to index the variables :) – Hagen von Eitzen Dec 7 '16 at 17:09
No. For $n=43$, $n^2+n+1=1893$ which is divisible by $3$. | {
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No. For $n=43$, $n^2+n+1=1893$ which is divisible by $3$.
• "What is wrong in the reasoning? What is a correct proof that a given polynomial $p(n)$ is (or isn't) prime for all natural numbers n?" Please read the entire question before answering. What I'm looking for is not a counterexample, but rather a proof. – Giulio Muscarello May 15 '14 at 14:32
• @GiulioMuscarello I understand you don't find this answer satisfying, but it's in fact perfectly valid proof. – Marcin Łoś May 15 '14 at 14:34
• @MarcinŁoś Sure, but the OP already had a much smaller counterexample with $n=4$. I don't see what this answer really adds. – André 3000 May 15 '14 at 14:38
• Jika is probably thinking of $n^2-n+41$, which is prime for $0 \le n \le 40$ – Robert Israel May 15 '14 at 15:06
• @RobertIsrael Yes you are right. Even though, I was thinking of Euler's formula $n^2-n+41$, I found that $43$ is a counter example. I hesitate to post my answer but since it is a valid mathematical proof I did post it. – Jika May 15 '14 at 15:15
Modulo 3, that polynomial is equivalent to some well known polynomial $(n-1)^2$ $$(n-1)^2=n^2-2n+1\equiv n^2+n+1 [3]$$
Hence if $n=1+3k$ for any $k$ (so each time $n-1\equiv 0[3]$) then $n^2+n+1$ will have 3 as a divisor. | {
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# Is my inductive proof correct?
Trying this again.
Given $f(n) = 2f(n-1) + 1$ with $f(0) = 0$, I guess that $f(n) = 2^n-1$.
Base case: $f(0) = 2^0 - 1 = 1 - 1 = 0$, true.
Inductive step: Suppose $f(n) = 2^n-1$ for some $n \geq 0$. I will show that $f(n+1) = 2^{n+1}-1$.
$f(n+1) = 2f(n) + 1$
$f(n+1) = 2(2^n-1) + 1$
$f(n+1) = 2^{n+1} - 1$
This completes the proof.
My questions:
1. Is this proof correct? Awkward? Backwards?
2. It would help me to get the terminology right. Which piece is the inductive hypothesis? Or the "ansatz"?
• This one is good, clean, "forward." Jan 8 '16 at 19:53
As far as terminology, the inductive hypothesis is the thing you assume is true at some $n$, and is used to prove the statement for $n + 1$. So your inductive hypothesis is that $f(n) = 2^n - 1$.
The ansatz is the educated guess that the solution is $2^n - 1$, likely based on some experimentation. The ansatz then became your induction hypothesis (as it is wont to do).
Looks good to me. The inductive hypothesis is the step where you assume $f(n)=2^n-1$. We assume the inductive hypothesis because we have shown the existence of such an $n$ in our base case. | {
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# Does Liebniz-Criteria become a necessary condition for convergence if $a_n$ is monotonically decreasing?
The Leibniz Criterion says that if the sequence $a_n$ is monotonically decreasing then the following statements are equivalent:
\begin{align} 1) & & & \sum_{n=0}^\infty (-1)^na_n \text{ converges} \\[6pt] 2) & & & \lim_{n\rightarrow\infty}(a_n) = 0 \end{align}
so this has to mean that if $a_n$ is monotonically decreasing, the following is true
$$\sum_{n=0}^\infty (-1)^na_n \text{ converges} \Leftrightarrow \lim_{n\rightarrow\infty}(a_n) = 0$$
I know that $\lim_{n\rightarrow\infty}(a_n) = 0$ is not a necessary condition for convergence if $a_n$ is not monotonically decreasing, but the way I am reading the rule stated in my script, it becomes a necessary condition if $a_n$ is monotonically decreasing.
For example to prove the non-convergence of
$$\sum_{n=0}^\infty (-1)^n(1)_n$$
I can just point out to the fact that
$$\lim_{n\rightarrow\infty}(1)_n = 1$$
because $(1)_n$ is a monotonically decreasing sequence. Is this interpretation correct? | {
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because $(1)_n$ is a monotonically decreasing sequence. Is this interpretation correct?
• The condition $b_n\to 0$ is always necessary to the convergence of $\sum b_n$. When $b_n=(-1)^n a_n$, one has the equivalence $b_n\to 0\ \iff\ a_n\to 0$. So $a_n\to 0$ is a necessary condition for the convergence of $\sum (-1)^n a_n$. – Giuseppe Negro May 31 '15 at 11:49
• Yes the statement is true. Proof: $$b_n\to 0\ \iff\ \lvert b_n\rvert \to 0 \iff \lvert (-1)^n a_n\rvert\to 0 \iff a_n\to 0.$$ If you are not convinced please post a counterexample. – Giuseppe Negro May 31 '15 at 11:59
• if it were so why would leibniz go through the trouble of putting the additional monotonic decreasing condition there – Gravity May 31 '15 at 11:59
• I guess you are mixing up "necessary" and "sufficient" here. $a_n\to 0$ is necessary to the convergence of $\sum (-1)^na_n$. $a_n\to 0$ AND $a_n$ monotonically decreasing is sufficient to the convergence of $\sum (-1)^n a_n$. – Giuseppe Negro May 31 '15 at 12:01
• I know realize that you were right – Gravity May 31 '15 at 12:33
As Giuseppe Negro has noticed, for the convergence of $\displaystyle \sum a_n$,$$\lim_{n\rightarrow\infty}(a_n) = 0$$ is a necessary condition anyway.
The convergence test of Leibniz is only applicable to alternating series, i.e. series where every positive term is followed by a negative term, and every negative term is followed by a positve term ( + - + - + - + - ...).
For a general series, the condition $\displaystyle\lim_{n \to \infty} a_n= 0$ is a necessary condition for convergence, but not sufficient. This is the main point of Leibniz test. He could prove that for alternating series this condition is both necessary and sufficient. Thus if you've got an alternating series $\sum (-1)^n a_n$ with $\displaystyle\lim_{n \to \infty} a_n= 0$, than you are certain it converges. | {
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I don't know why you are referring to the condition of "monotonically decreasing". This doesn't matter for Leibniz test. It's only an issue in the integral test.
• I understand your first point, but on your second point$\sum (-1)^n a_n$ with $\displaystyle\lim_{n \to \infty} a_n= 0$ you are certain that is converges, only when a is monotonically decreasing. The rule is defined for monotonically decreasing a – Gravity May 31 '15 at 12:36
• @Steven Van Geluwe: Alternation of signs and terms going to $0$ is not sufficient for convergence. Example, let $a_n=\frac{1}{n}$ if $n$ is odd, and let $a_n=\frac{1}{2^n}$ when $n$ is even. Then $\sum_1^\infty (-1)^n a_n$ diverges. – André Nicolas May 31 '15 at 12:52
• @Gravity: There are many series $\sum (-1)^n a_n$ that converge even though the $a_n$ are not monotonically decreasing. For example, let $a_n=\frac{1}{n^2}$ when $n$ is odd, and let $a_n=\frac{1}{n^3}$ when $n$ is even. – André Nicolas May 31 '15 at 12:59
• @Andre: that is correct and that is why Leibniz says the condition is sufficient but not necessary, but still we can only apply the Leibniz rule if $a_n$ is monotonically decreasing. we cannot in all Circumstances say $\sum (-1)^n a_n$ converges when $\displaystyle\lim_{n \to \infty} a_n= 0$ is true, and that is the point – Gravity May 31 '15 at 13:05
• I down voted, because I am afraid that the main point of this answer is wrong. Leibniz's test requires monotonicity of the $a_n$ term. – Giuseppe Negro May 31 '15 at 21:25 | {
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# Is $K_1$ bipartite?
I'm thinking that it could be trivially bipartite since it only has one vertex and no edges but I am still a little bit unsure about it being trivially bipartite.
• Ummm... what is "k sub 1"? Do you instead mean $K_1$? – David G. Stork Jun 10 at 19:01
• not only is it bipartite, it is also unipartite – Yorch Jun 10 at 19:02
• @DavidG.Stork Yes, did not know how to make it like that in my title. Seemed to fix itself. – ThreeRingBinder Jun 10 at 20:23
Depends on your definitions. (As pointed out in Matthew Daly's answer, $$K_1$$ should be bipartite, because it has no odd cycles, so it would otherwise be an awkward exception. But not all definitions will play nicely with this desired property.)
West's Introduction to Graph Theory says
1.1.10. Definition. A graph $$G$$ is bipartite if $$V(G)$$ is the union of two disjoint (possibly empty) independent sets called partite sets of $$G$$.
So under this definition, if $$V(K_1) = \{v\}$$, then we let $$\{v\}$$ be one partite set, and $$\varnothing$$ be the other; $$K_1$$ is bipartite.
Bondy and Murty write
A graph is bipartite if its vertex set can be partitioned into two subsets $$X$$ and $$Y$$ so that every edge has one end in $$X$$ and one end in $$Y$$
which still works just fine, setting $$X = \{v\}$$ and $$Y = \varnothing$$. They do not specifically point out that $$X$$ or $$Y$$ could be empty, but they do not rule it out either. (Elsewhere, Bondy and Murty talk about "nontrivial partitions" or "partitions into nonempty parts", which make it clear that a partition without this qualifier is allowed to have an empty part.)
They are in better shape than Diestel, who has
A graph $$G = (V, E)$$ is called $$r$$-partite if $$V$$ admits a partition into $$r$$ classes such that every edge has its ends in different classes | {
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with an earlier qualification that the classes of a partition may not be empty. Since $$K_1$$ should be bipartite by any reasonable definition, Diestel is in the wrong here. (Diestel later claims that all graphs with no odd cycles are bipartite, with no mention of $$K_1$$ as a special exception.)
If we treat "bipartite" as synonymous with "$$2$$-colorable", then $$K_1$$ is happily bipartite, since any function on its vertex set is a $$2$$-coloring (and also a $$1$$-coloring).
• Quick question. Lets say we have 𝑋={𝑣} and 𝑌=∅, in reference to Bondy and Murty's definition. Would the edge connecting both subsets have vertices that are the empty set? Ie one empty set from X and the other from Y – ThreeRingBinder Jun 11 at 4:36
• The definition says every edge has a certain property; it doesn't say there exist any such edges. Since $K_1$ has no edges, the definition holds. – Misha Lavrov Jun 11 at 5:24
• Ah, makes sense now. Thanks! – ThreeRingBinder Jun 11 at 6:51
My graph theory professor spent a while teaching on this, because it's a mess either way.
It probably should be trivially considered bipartite. Otherwise you have to put disclaimers in all of your theorems (like the fact that $$K_1$$ doesn't contain any odd cycles). But $$\{\{v\},\emptyset\}$$ is not a valid partition of the vertex set, since it contains the empty set as a member. | {
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• A reasonable definition of "bipartite" does not run into trouble if one partite set is empty. – Misha Lavrov Jun 10 at 19:24
• @MishaLavrov On the other hand, requiring the parts of a partition to be non-empty isn't entirely unheard of either. So defining bipartite as a partition of the set of vertices into two can run into issues with $K_1$. – Arthur Jun 10 at 19:33
• Yeah, I think that Dr. Schäffer's solution was to have us write in the margin of Bondy and Murty "... or $K_1$". – Matthew Daly Jun 10 at 19:44
• But Bondy and Murty are totally fine with empty parts in a partition, as far as I can tell... – Misha Lavrov Jun 10 at 20:22
• @MishaLavrov As I say, my professor assumed that "partitioned into two subsets $X$ and $Y$" means that $\{X,Y\}$ was a partition of the vertex set, which could not be legally done if one of the sets was empty. Bondy and Murty clearly either thought that it was okay or they hadn't considered the edge case, but pedantry like that got under my professor's skin. ^_^ – Matthew Daly Jun 10 at 20:37 | {
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# Do we have to expand the brackets before applying De Morgan's Laws?
If we have the example:
$$\overline{(A + B + C)D}$$
Then can we apply De Morgan's Law as is, or do we first need to expand out the brackets?
If I expand the terms first, I get:
$$original: \overline{(A + B + C)D}$$ $$expanded: \overline{AD + BD + CD}$$ $$applying De Morgan's: \overline{AD}.\overline{BD}.\overline{CD}$$ $$simplifying: \overline{A}\overline{B}\overline{C}\overline{D}$$
In contrast, if I don't expand the brackets, I get something like:
$$original: \overline{(A + B + C)D}$$ $$applying De Morgan's: \overline{(A + B + C)}+\overline{D}$$ $$applying De Morgan's: (\overline{A} . \overline{B + C})+\overline{D}$$ $$applying De Morgan's: (\overline{A} . \overline{B} . \overline{C})+\overline{D}$$ $$simplifying: \overline{A} . \overline{B} . \overline{C}+\overline{D}$$
So you can see, I'm getting different answers. I'm curious which method is correct - I think the first seems correct, but perhaps the 2nd is correct or both are wrong.
If anyone is able to explain why one particular method is wrong, I'd appreciate it too.
• MathJax tip: when writing text in mathematics code, use the \text{} command to get proper spacing and formatting. For example, $$\text{applying De Morgan's} : \overline{(A + B + C)} + \overline{D}$$ produces$$\text{applying De Morgan's} : \overline{(A + B + C)} + \overline{D}$$ Aug 22, 2021 at 0:54
• Thanks. Now I will do that in future :) Aug 22, 2021 at 5:58
• @RandomUser123 You can also do that now. Questions and answers can be edited, and indeed should be edited if they can be improved. Aug 22, 2021 at 8:45 | {
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Both lead to the same result; your error is in the first one when you simplify from $$\overline{AD}.\overline{BD}.\overline{CD}$$ to $$\overline{A}\overline{B}\overline{C}\overline{D}$$.
If you apply De Morgan's law to each pair, you get: $$(\overline{A}+\overline{D})(\overline{B}+\overline{D})(\overline{C}+\overline{D}).$$ From here, you can apply the distributive law to get what you got by the second method in the next to last line. The simplifcation there is also incorrect because you cannot simply drop the parenthesis; you must apply the distributive property.
• Makes sense. I was viewing it as $$\overline{ADBDCD}$$ then just cancelling removing the unneccessary duplicates. Because the NOTs were over multiple variables, I shouldn't be able to do that. I also tried myself, expanding out those 3 sets of brackets and surpisingly managed to simplify down that mess to the correct answer...seeing that all of the brackets contain D' and using your suggestion of using the distributive law is much easier though...I guess it's a good way to check...since I worry if I use these shortcuts, I will apply it incorrectly/in a situation that's invalid or something Aug 22, 2021 at 6:17 | {
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# What's the probability that a given permutation has exactly $k$ fixed points. [duplicate]
Given a random permutation $$\sigma \in S_n$$ from $$[n] \to [n]$$ in a uniform probability space, what is the probability that $$\sigma$$ has exactly $$k$$ fixed points for a given $$k$$ between $$1$$ and $$n$$?
In other words: what is the probability that $$\exists x_1 ,...,x_k \in [n] : \sigma (x_i) = x_i$$ for $$\ i\in \{1,...,k\}$$ and for every $$y \notin \{x_1 , ... , x_k\}$$ we get $$\sigma(y) \neq y$$.
I saw that $$\lim_{n \to \infty } prob(A_0) = e^{-1}$$ using Inclusion–exclusion principle and i belive that for a given k : $$\lim_{n \to \infty} prob(A_k) = \frac{e^{-1}}{k!}$$ but I am not sure how to show it.
*$$A_k$$ stands for the event "k".
• The number $D_{n,k}$ of permutations with exactly $k$ fixpoints of an $n$-set is called a rencontres number. – Jeppe Stig Nielsen Oct 10 '17 at 11:36
There are $\binom{n}{k}$ possibilities for the $k$ fixed points.
If they are established then there are $!(n-k)$ derangements for the other points.
That gives $\binom{n}{k}\left[!(n-k)\right]$ permutations with exactly $k$ fixed points on a total of $n!$ permutations.
Also we have the formula: $$!(n-k)=(n-k)!\sum_{i=0}^{n-k}\frac{(-1)^i}{i!}$$ and we end up with probability: $$\frac1{k!}\sum_{i=0}^{n-k}\frac{(-1)^i}{i!}$$
• Very cool! As $n$ grows (for fixed $k$), this approaches $\frac1{k! e}$ – MichaelChirico Nov 6 '19 at 11:31
• And also by corollary, the probability of at least one fixed point approaches $1-\frac1{e}$ – MichaelChirico Nov 6 '19 at 11:38
1. Number of dearrangements of $k$-elements set is $$!k = k!\sum\limits_{m=0}^{k}\frac{(-1)^m}{m!}$$
2. In n-elements set we can select $(n-k)$ elements to dearrange them (all remaining $k$ points are fixed) in $\binom{n}{k}$ ways | {
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Thus $$A_k^n = \binom{n}{k}(n-k)!\sum\limits_{m=0}^{n-k}\frac{(-1)^m}{m!}$$ And probability is $$P(A_k^n) = \frac{\binom{n}{k}(n-k)!\sum\limits_{m=0}^{n-k}\frac{(-1)^m}{m!}}{n!}=\frac{\frac{n!}{k!}\sum\limits_{m=0}^{n-k}\frac{(-1)^m}{m!}}{n!}=\frac{1}{k!}\sum\limits_{m=0}^{n-k}\frac{(-1)^m}{m!}$$
By way of enrichment here is an alternate formulation using combinatorial classes. The class of permutations with fixed points marked is
$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \textsc{SET}(\mathcal{U} \times \textsc{CYC}_{=1}(\mathcal{Z}) + \textsc{CYC}_{=2}(\mathcal{Z}) + \textsc{CYC}_{=3}(\mathcal{Z}) + \cdots).$$
This gives the generating function $$G(z, u) = \exp\left(uz + \frac{z^2}{2} + \frac{z^3}{3} + \frac{z^4}{4} + \frac{z^5}{5} + \cdots\right)$$
which is
$$G(z, u) = \exp\left(uz-z+\log\frac{1}{1-z}\right) = \frac{\exp(uz-z)}{1-z}.$$
Now for $$k$$ fixed points we get
$$[u^k] \frac{\exp(uz-z)}{1-z} = [u^k] \frac{\exp(uz)\exp(-z)}{1-z} = \frac{z^k}{k!} \frac{\exp(-z)}{1-z}.$$
This is the EGF of permutations having $$k$$ fixed points. We extract the count by computing (the factor $$n!$$ is canceled because we require the average)
$$[z^n] \frac{z^k}{k!} \frac{\exp(-z)}{1-z} = \frac{1}{k!} [z^{n-k}] \frac{\exp(-z)}{1-z}.$$
We find
$$\bbox[5px,border:2px solid #00A000]{ \frac{1}{k!} \sum_{q=0}^{n-k} \frac{(-1)^q}{q!}.}$$
We can identify this as choosing the $$k$$ fixed points and combining them with a derangement of the rest:
$$\frac{1}{n!} {n\choose k} (n-k)! \sum_{q=0}^{n-k} \frac{(-1)^q}{q!}$$
which is the combinatorial class
$$\textsc{SET}_{=k}(\mathcal{Z}) \times \textsc{SET}(\textsc{CYC}_{\ge 2}(\mathcal{Z})).$$ | {
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# Show that $\sqrt{1+x}<1+\frac{x}{2}$ for all $x>0$
I am a little stuck on this question and would appreciate some help. The question asks me to prove that $$\sqrt{1+x}<1+\frac{x}{2}$$ for all $$x>0$$.
I squared both sides of the question to get $$1+x<\frac{x^2}{4}+x+1$$ for all $$x>0$$. Then, I multiplied both sides by $$4$$ to get $$4+4x for all $$x>0$$.
I am a little stuck and was wondering what to do after this step and how to actually provide sufficient proof to say that this statement is true.
You did well. Let's finish it: we have $$x^2>0$$ thus $$\dfrac{x^2}{4}>0$$. Adding $$x+1$$ to both sides, we have $$\dfrac{x^2}{4}+x+1> x+1$$ or $$(\frac{x}{2}+1)^2>x+1$$ or $$\frac{x}{2}+1>\sqrt{x+1}$$
Squaring the equation was sufficient. After you cancel the $$1+x$$ on each side, you have
$$\frac{x^2}{4} > 0$$
which is true for all real $$x \ne 0$$ since $$x^2 \ge 0$$ (with equality only when $$x=0$$). Thus, the inequality is proved.
You can also prove the inequality using the Mean Value Theorem. Define $$f(t)=\sqrt{t+1}$$ and apply the theorem on the interval $$[0,x]$$. Then, there exists $$c\in(0,x)$$ s.t. $$\frac{\sqrt{x+1}-\sqrt{1}}{x-0}=\frac{1}{2\sqrt{c+1}}>\frac{1}{2}.$$
I am answering less to provide a list of steps to follow for proving the inequality, and more to discuss how I believe that the argument should be presented. This seems appropriate, as the question has been tagged with and . If I asked a student to prove some inequality and their first step was to assume the inequality and square both sides, I would almost certainly deduct points unless the student were careful about justifying such a step. The goal here is to start with known true statements and to then deduce the desired result.
Claim: For any $$x > 0$$, $$\sqrt{1+x} < 1 + \frac{x}{2}.$$ | {
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Claim: For any $$x > 0$$, $$\sqrt{1+x} < 1 + \frac{x}{2}.$$
Proof: Assume that $$x > 0$$. The square of any real number is positive, hence $$0 < \left( \frac{x}{2} \right)^2.$$ By additive cancelation (i.e. by "adding the same number to both sides"), this implies that $$x + 1 < \left( \frac{x}{2} \right)^2 + x + 1 = \left( \frac{x}{2} + 1 \right)^2.$$ The square root function is increasing on $$[0,\infty)$$ (that is, if $$0 \le a < b$$, then $$0 \le \sqrt{a} \le \sqrt{b}$$), and the assumption that $$x > 0$$ ensures that $$1+x > 0$$, thus $$1 + x$$ is in the domain of the square root function (i.e. it has a well-defined real square root). Hence $$\sqrt{x+1} < \sqrt{ \left( 1 + \frac{x}{2} \right)^2 } = \left|1 + \frac{x}{2}\right| = 1 + \frac{x}{2},$$ where the final equality follows from the assumption that $$x > 0$$.
### Discussion
Remember that if you start with false assumptions, you can prove anything—false implies true. Thus it is poor practice to begin a proof by asserting the statement which you are trying to prove and then applying algebraic manipulation. You can do this if you are careful—be very observant of two-sided implications and check the hypotheses at each step. Indeed, this is a very reasonable way of deducing a correct proof in the first place.
However, when presenting an argument, such a proof is typically poor style. Personally, I think that it is better (stylistically, which is a matter of taste) for every statement in a proof to follow from previous statement, and not depend on future statements via "if and only ifs". | {
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I will also note that the proof presented above is very elementary—it doesn't rely on any deep theorems. It is, however, somewhat tedious. More powerful theorems give quicker, more elegant proofs. For example, El31's proof, via the mean value theorem, is quite slick. However, when first learning a topic, I think that one should focus on finding an elementary proof first: find such a proof first. Then, if need be, look for something more elegant.
The important principle here is that if $$a,b$$ are nonnegative real numbers, then $$a iff $$a^2 This property is used all the time. If we grant this, then all we need to show is that
$$(\sqrt{1+x})^2 < (1+x/2)^2,$$
which is the same as saying $$1+x < 1+x + x^2/4.$$ That is obviously true, and we're done.
Observe that : $$\left(\forall x>0\right),\ \sqrt{1+x}-1-\frac{x}{2}=-\frac{x^{2}}{4}\int_{0}^{1}{\frac{1-t}{\left(1+tx\right)\sqrt{1+tx}}\,\mathrm{d}t} < 0$$
• No offense meant, but this is perhaps one of the most obtuse correct answers I've seen on this site. – jawheele Mar 30 at 5:51 | {
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# Math Help - Solve Equations 0< or = x < or = 2pi
1. ## Solve Equations 0< or = x < or = 2pi
How would you solve these
Solve Equation cos(squared)x - sin(squared) x = - cos x, 0 < or = to x < or equal 2 pi.
Answear is in form of this
pi divided by 2, 2pi divided by 3, 4 pi divided by 3, 3 pi divided by 2 this is probably not the right answer however I don't know how to solve this.. any hints?
2. Originally Posted by j0nath0n3
How would you solve these
Solve Equation cos(squared)x - sin(squared) x = - cos x, 0 < or = to x < or equal 2 pi.
Answear is in form of this
pi divided by 2, 2pi divided by 3, 4 pi divided by 3, 3 pi divided by 2 this is probably not the right answer however I don't know how to solve this.. any hints?
Note that $\sin^2 x=1-\cos^2 x$
So we can rewrite the equation as $\cos^2 x -\left[1-\cos^2 x\right]=-\cos x$
This simplifies to $2\cos^2 x-1=-\cos x\implies 2\cos^2x+\cos x-1=0$
Now, if we make the substitution $z=\cos x$, the equation now becomes the quadratic equation $2z^2+z-1=0\implies (2z-1)(z+1)=0$
This tells us that $z=-1$ or $z=\tfrac{1}{2}$
But $z=\cos x$, so now that means either $\cos x=\tfrac{1}{2}$ or $\cos x=-1$
Now can you take it from here and find the values of x that satisfy these two equations, keeping in mind that x has restricted values?
3. Originally Posted by Chris L T521
Note that $\sin^2 x=1-\cos^2 x$
So we can rewrite the equation as $\cos^2 x -\left[1-\cos^2 x\right]=-\cos x$
This simplifies to $2\cos^2 x-1=-\cos x\implies 2\cos^2x+\cos x-1=0$
Now, if we make the substitution $z=\cos x$, the equation now becomes the quadratic equation $2z^2+z-1=0\implies (2z-1)(z+1)=0$
This tells us that $z=-1$ or $z=\tfrac{1}{2}$
But $z=\cos x$, so now that means either $\cos x=\tfrac{1}{2}$ or $\cos x=-1$ | {
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But $z=\cos x$, so now that means either $\cos x=\tfrac{1}{2}$ or $\cos x=-1$
Now can you take it from here and find the values of x that satisfy these two equations, keeping in mind that x has restricted values?
Thats really good and I understood however my answear choices are in radian... and I dont understand why..
4. Originally Posted by j0nath0n3
Thats really good and I understood however my answear choices are in radian... and I dont understand why..
In general, we prefer to use radians in math. Degrees can be used as well. Keep in mind that the relationship between degrees and radians is $360\text{ degrees}=2\pi\text{ radians}$.
5. ## Re :
Originally Posted by Chris L T521
In general, we prefer to use radians in math. Degrees can be used as well. Keep in mind that the relationship between degrees and radians is $360\text{ degrees}=\pi\text{ radians}$.
Chris , is it a typo , i thought $360\text{ degrees}=2\pi\text{ radians}$
Chris , is it a typo , i thought $360\text{ degrees}=2\pi\text{ radians}$ | {
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In fact, permutation is another term used to describe bijective functions from a finite set to itself. a permutation is an arrangement of $n$ elements without repetition. A big part of discrete mathematics is about counting things. Permutations with Repetition | Discrete Mathematics. Combinations - Permutions. P R (4, 2) = 4 2 = 16. A phone number is an example of a ten number permutation; it is drawn from the set of the integers 0-9, and the order in which they are arranged in matters. I Pascal's triangle is perfectly symmetric I Numbers on left are mirror image of numbers on right I Why is this the case? IApermutationof a set of distinct objects is anordered arrangement of these objects. Likewise, [triangle, melon, airplane] is a permutation of three objects as well. Discrete Mathematics - Quick Guide, Mathematics can be broadly classified into two categories A permutation is an arrangement of some elements in which order matters. Permutations are used when we are counting without replacing objects and order does matter. Therefore, the number of different permutations of a You only need the decomposition in disjoint cycles. Case1: Let G={ 1 } element then permutation are S n or P n = Case 2: Let G= { 1, 2 } elements then permutations are . About the journal. It is used to create a pairwise relationship between objects. Combinatorics is the study of arrangements of objects, it is an important part of discrete mathematics. Cyclic Notation In the former article, we saw various ideas behind multiple formulas and theorems in discrete math concerning permutations. functions in discrete mathematics ppthank aaron rookie cards. So total ways are. We must count objects to solve many different types of problems, like the Incycle notation, we represent a cyclic permutation by this cycle. including groups, rings, and fields. The Mathematics Department of the Rutgers School of Arts and Sciences is one of the oldest mathematics departments in the United States, graduating its | {
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Arts and Sciences is one of the oldest mathematics departments in the United States, graduating its first major in 1776 Cornette, Discrete Mathematics - Counting Theory The Rules of Sum and Product. The combination is a way of selecting items from a collection, such that (unlike permutations) the order of selection does not matter. North East Kingdoms Best Variety best order to read the old testament; sandman hotel victoria bed bugs; functions in discrete mathematics ppt }\) There are $$3! One byte consists of 8 bits. A permutation refers to a selection of objects from a set of objects in which order matters. }$$ Suppose that $$A = \{1, 2, 3\}\text{. Permutation and combination are the ways to represent a group of objects by selecting them in a 4! A permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement. = 5!/3! Open-source computer software for working with permutations includes the GAP suite of Wednesday, December 28, 2011. In computer science and discrete mathematics, an inversion in a sequence is a pair of elements that are out of their natural order . For example, suppose we want to multiply (1;5)(2;3;6)(1;6;4)(3;5). Discrete Mathematics provides a common forum for significant research in many areas of discrete mathematics and combinatorics . MATH 3336 Discrete Mathematics Combinations and Permutations (6.3) Permutations Definition: A permutation of a set of distinct objects is an ordered arrangement of these objects. Permutation: In mathematics, one of several ways of arranging or picking a set of items. Discrete Mathematics MCQ Questions with Answers is a PDF booklet containing 50 MCQ questions and answers on topics such as counting, sets, sequences and permutations, Permutations. A permutation of \(n$$ distinct objects is just a listing of the objects in some order. Input: N = 7668. DISCRETE MATHEMATICS Permutations and combinations Book arrangement problems Combinations and Permutations Worksheet 9 | {
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Permutations and combinations Book arrangement problems Combinations and Permutations Worksheet 9 Permutation Word Problems Explained the Easy B9. In general P(n, k) means the number of permutations of n objects from which we take k objects. MATH 25400. IExample: S = fa;b;cg. Permutations Permutations Example = (4 5)(2 3) = (4 5)(2 1) 1 A classic example asks how many different words can be obtained by re-ordering the letters in = n (n - 1) (n - 2) (n - 3) . 1. We are going to pick (select) r objects from the urn in sequence. In other words a Permutation is an ordered Combination of elements. Malte Helmert, Gabriele R oger (University of Basel)Discrete Mathematics in Computer Science October 21, 2020 10 / 20 B9. For example: Math 3336 Section 6. Calculate the permutations for P R (n,r) = n r. For n >= 0, and r >= 0. It's free to sign up and bid on jobs. 1 Discrete Math Basic Permutations and Combinations Slide 2 Ordering Distinguishable Objects When we have a group of N objects that are Use Wolfram|Alpha to apply and understand these and related concepts. INo object can be selected more than once. Before we discuss permutations we are going to have a look at what the words combination means and permutation. You should practice these MCQs for 1 Recommended: Please try your approach on {IDE} first, before moving on to the solution. Honors Discrete Mathematics. Discrete mathematics deals with areas of mathematics that are discrete, as opposed to continuous, in nature. For instance, in how many ways can a panel of jud A permutation is an arrangement of some elements in which order matters. Combinatorics . Case 3: Let G={ 1, 2, 3 } elements then permutation are 3!=6. In several IOrder of arrangement matters. Find the factorial n! Alternatively, the permutations formula is expressed as follows: We study generating functions for the number of even (odd) permutations on n letters avoiding 132 and an arbitrary permutation on k letters, or containing exactly once. Example: | {
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letters avoiding 132 and an arbitrary permutation on k letters, or containing exactly once. Example: How many different ways can 3 students line up to purchase a new textbook reader? Permutations. - These MCQs cover theoretical concepts, true-false(T/F) statements, fill $. Summary of permutations. Such kind of finite studies are involved in discrete mathematics. One well-known zero-inflated model is Diane Lambert's zero-inflated Poisson model, which concerns a random event containing excess zero-count data in unit time. To permute a list is to rearrange its elements. You must consider also all the ways to arrange the people into the positions. The formula for Permutations Replacement or Repetition is P R (n,r)=n r. Substituting the values of n, r in the formula and we get the equation as follows. The Rule of Sum and Rule of Product are used to decompose difficult counting problems into Permutations. Solution: n-factorial gives the number of permutations of n items. Combinations. An injective function (mapping) of a finite set into itself is called a permutation.There follows from the definition of finite sets that shows that such a function is necessarily also surjective and consequently a permutation is always bijective. Permutations. Below is the few Discrete mathematics MCQ test that checks your basic knowledge of Discrete mathematics Discrete Math Textbook Solutions and Answers Discrete MathChapter 14 Combination: A combination of a set of distinct objects is just a count of the number of ways a specific number of elements can be selected from a set of a certain size. Permutations Permutations Cycle Notation { Algorithm Let be a permutation of nite set S. 1: function ComputeCycleRepresentation(, S) 2: remaining = S 3: cycles = ; 4: while remaining is not empty do 5: Remove any element e from remaining. Let A be a ith n elements. To multiply permutations, trace through the images of points, and build a new permutation from the images, as when translating into | {
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through the images of points, and build a new permutation from the images, as when translating into cycle structure. Therefore, we are choosing a sequence of 60 dice rolls from a set size of 6 possible numbers for Suppose that a permutation is . Types of Sets in Discrete Structure or Discrete Mathematics. V.N. Independent events Consider a quiz with four true/false and three multiple choice questions, (a){(e). Discrete Mathematics Discrete Mathematics, Study Discrete Mathematics Topics. Modus Ponens and Modus Tollens Understand your high school math homework by watching free math videos online from your own free math help tutor may be used as a test or review for Unit Instructor: Is l Dillig. Permutation A rearrangement of the elements in an ordered list S into a one-to-one correspondence with S itself. Combinations and Permutations. Set Theory & Algebra; Combinatorics; Graph Theory; Linear Algebra; Probability; Discrete Mathematics | Types of Recurrence Relations - Set 2. (Multiplication Principle) But what if we only want the number of permutations of r distinct objects from a collection of n? Problems Discrete Mathematics Book I Used for Self Study Discrete Math 6.3.2 Counting, Permutation and Combination Practice DM-16- Propositional Logic -Problems related to In document Discrete mathematics (Pldal 67-70) It is well-known that 1 bit can represent one of two possible distinct states (e.g. Our 1000+ Discrete Mathematics MCQs (Multiple Choice Questions and Answers) focuses on all chapters of Discrete Mathematics covering 100+ topics. of a number, including 0, up to 4 digits long. As such, it is a remarkably broad subject Since combinatorics is widely accessible, this Exercise :: Permutation and Combination - General Questions. MCQ (Multiple Choice Questions with answers about Discrete Mathematics Circular Permutations Determine Abstract Universal cycle for k-permutations is a cyclic arrangement in which each k-permutation appears exactly once as k consecutive | {
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is a cyclic arrangement in which each k-permutation appears exactly once as k consecutive elements. Introduction Permutations Discrete Mathematics Andrei Bulatov Discrete Mathematics - Permutations What is Combinatorics Combinatorics, the Study Resources Main Menu A permutation is an arrangement in a definite order of a number of objects taken, some or all at a time. Wolfram|Alpha is useful for counting, generating and doing algebra with permutations. Malte Helmert, Gabriele R oger (University of Basel)Discrete Mathematics in Computer Science October 21, 2020 17 / 20 B9. In mathematics, a permutation of a set is, loosely speaking, an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements.The word "permutation" also refers to the act or process of changing the linear order of an ordered set. A graph is determined as a mathematical structure that represents a particular function by connecting a set of points. A finite sequence of length$ n \$ in which all the elements are different, i.e. Permutations Permutations Cyclic Permutation De nition (Cyclic Permutation) A permutation iscyclicif it has a single k-cycle with k >1. permutations and combinations is the another topic included in discrete mathematics which also refers to the finite calculations. 100 Units. Permutations of Objects Ppt - Free download as Powerpoint Presentation (.ppt), PDF File (.pdf), Text File (.txt) or view presentation slides online. permutation arrangement An r-permutation of n objects is an ordered arrangement of r objects from the n objects A permutation is a (possible) rearrangement of objects. Rolling Dice. Question and Answers related to Discrete Mathematics Circular Permutations. For example, the number of insurance claims within a population for a certain type of risk would be zero-inflated by those people who have not taken out insurance against the risk and thus are unable to claim. | {
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In Section VO:Vector Operations we defined vector addition and scalar multiplication. These two operations combine nicely to give us a construction known as a linear combination, a construct that we will work with throughout this course.
## Linear Combinations
Definition LCCV (Linear Combination of Column Vectors) Given $n$ vectors $\vectorlist{u}{n}$ from $\complex{m}$ and $n$ scalars $\alpha_1,\,\alpha_2,\,\alpha_3,\,\ldots,\,\alpha_n$, their linear combination is the vector \begin{equation*} \lincombo{\alpha}{u}{n} \end{equation*}
So this definition takes an equal number of scalars and vectors, combines them using our two new operations (scalar multiplication and vector addition) and creates a single brand-new vector, of the same size as the original vectors. When a definition or theorem employs a linear combination, think about the nature of the objects that go into its creation (lists of scalars and vectors), and the type of object that results (a single vector). Computationally, a linear combination is pretty easy.
Example TLC: Two linear combinations in $\complex{6}$.
Example ABLC: Archetype B as a linear combination.
With any discussion of Archetype A or Archetype B we should be sure to contrast with the other.
Example AALC: Archetype A as a linear combination.
There's a lot going on in the last two examples. Come back to them in a while and make some connections with the intervening material. For now, we will summarize and explain some of this behavior with a theorem. | {
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Theorem SLSLC (Solutions to Linear Systems are Linear Combinations) Denote the columns of the $m\times n$ matrix $A$ as the vectors $\vectorlist{A}{n}$. Then $\vect{x}$ is a solution to the linear system of equations $\linearsystem{A}{\vect{b}}$ if and only if $\vect{b}$ equals the linear combination of the columns of $A$ formed with the entries of $\vect{x}$, \begin{equation*} \vectorentry{\vect{x}}{1}\vect{A}_1+ \vectorentry{\vect{x}}{2}\vect{A}_2+ \vectorentry{\vect{x}}{3}\vect{A}_3+ \cdots+ \vectorentry{\vect{x}}{n}\vect{A}_n = \vect{b} \end{equation*}
In other words, this theorem tells us that solutions to systems of equations are linear combinations of the $n$ column vectors of the coefficient matrix ($\vect{A}_j$) which yield the constant vector $\vect{b}$. Or said another way, a solution to a system of equations $\linearsystem{A}{\vect{b}}$ is an answer to the question "How can I form the vector $\vect{b}$ as a linear combination of the columns of $A$?" Look through the archetypes that are systems of equations and examine a few of the advertised solutions. In each case use the solution to form a linear combination of the columns of the coefficient matrix and verify that the result equals the constant vector (see exercise LC.C21).
## Vector Form of Solution Sets
We have written solutions to systems of equations as column vectors. For example Archetype B has the solution $x_1 = -3,\,x_2 = 5,\,x_3 = 2$ which we now write as \begin{equation*} \vect{x}=\colvector{x_1\\x_2\\x_3}=\colvector{-3\\5\\2} \end{equation*} Now, we will use column vectors and linear combinations to express all of the solutions to a linear system of equations in a compact and understandable way. First, here's two examples that will motivate our next theorem. This is a valuable technique, almost the equal of row-reducing a matrix, so be sure you get comfortable with it over the course of this section.
Example VFSAD: Vector form of solutions for Archetype D. | {
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Example VFSAD: Vector form of solutions for Archetype D.
This is such an important and fundamental technique, we'll do another example.
Example VFS: Vector form of solutions.
Did you think a few weeks ago that you could so quickly and easily list all the solutions to a linear system of 5 equations in 7 variables?
We'll now formalize the last two (important) examples as a theorem.
Theorem VFSLS (Vector Form of Solutions to Linear Systems) Suppose that $\augmented{A}{\vect{b}}$ is the augmented matrix for a consistent linear system $\linearsystem{A}{\vect{b}}$ of $m$ equations in $n$ variables. Let $B$ be a row-equivalent $m\times (n+1)$ matrix in reduced row-echelon form. Suppose that $B$ has $r$ nonzero rows, columns without leading 1's with indices $F=\set{f_1,\,f_2,\,f_3,\,\ldots,\,f_{n-r},\,n+1}$, and columns with leading 1's (pivot columns) having indices $D=\set{d_1,\,d_2,\,d_3,\,\ldots,\,d_r}$. Define vectors $\vect{c}$, $\vect{u}_j$, $1\leq j\leq n-r$ of size $n$ by
\begin{align*} \vectorentry{\vect{c}}{i}&= \begin{cases} 0&\text{if $i\in F$}\\ \matrixentry{B}{k,n+1}&\text{if $i\in D$, $i=d_k$} \end{cases}\\ \vectorentry{\vect{u}_j}{i}&= \begin{cases} 1&\text{if $i\in F$, $i=f_j$}\\ 0&\text{if $i\in F$, $i\neq f_j$}\\ -\matrixentry{B}{k,f_j}&\text{if $i\in D$, $i=d_k$} \end{cases}. \end{align*}
Then the set of solutions to the system of equations $\linearsystem{A}{\vect{b}}$ is \begin{equation*} S=\setparts{ \vect{c}+\alpha_1\vect{u}_1+\alpha_2\vect{u}_2+\alpha_3\vect{u}_3+\cdots+\alpha_{n-r}\vect{u}_{n-r} }{ \alpha_1,\,\alpha_2,\,\alpha_3,\,\ldots,\,\alpha_{n-r}\in\complex{\null} } \end{equation*} | {
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Note that both halves of the proof of Theorem VFSLS indicate that $\alpha_i=\vectorentry{\vect{x}}{f_i}$. In other words, the arbitrary scalars, $\alpha_i$, in the description of the set $S$ actually have more meaning --- they are the values of the free variables $\vectorentry{\vect{x}}{f_i}$, $1\leq i\leq n-r$. So we will often exploit this observation in our descriptions of solution sets.
Theorem VFSLS formalizes what happened in the three steps of Example VFSAD. The theorem will be useful in proving other theorems, and it it is useful since it tells us an exact procedure for simply describing an infinite solution set. We could program a computer to implement it, once we have the augmented matrix row-reduced and have checked that the system is consistent. By Knuth's definition, this completes our conversion of linear equation solving from art into science. Notice that it even applies (but is overkill) in the case of a unique solution. However, as a practical matter, I prefer the three-step process of Example VFSAD when I need to describe an infinite solution set. So let's practice some more, but with a bigger example.
Example VFSAI: Vector form of solutions for Archetype I.
This technique is so important, that we'll do one more example. However, an important distinction will be that this system is homogeneous.
Example VFSAL: Vector form of solutions for Archetype L.
## Particular Solutions, Homogeneous Solutions
The next theorem tells us that in order to find all of the solutions to a linear system of equations, it is sufficient to find just one solution, and then find all of the solutions to the corresponding homogeneous system. This explains part of our interest in the null space, the set of all solutions to a homogeneous system. | {
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Theorem PSPHS (Particular Solution Plus Homogeneous Solutions) Suppose that $\vect{w}$ is one solution to the linear system of equations $\linearsystem{A}{b}$. Then $\vect{y}$ is a solution to $\linearsystem{A}{b}$ if and only if $\vect{y}=\vect{w}+\vect{z}$ for some vector $\vect{z}\in\nsp{A}$.
After proving Theorem NMUS we commented (insufficiently) on the negation of one half of the theorem. Nonsingular coefficient matrices lead to unique solutions for every choice of the vector of constants. What does this say about singular matrices? A singular matrix $A$ has a nontrivial null space (Theorem NMTNS). For a given vector of constants, $\vect{b}$, the system $\linearsystem{A}{b}$ could be inconsistent, meaning there are no solutions. But if there is at least one solution ($\vect{w}$), then Theorem PSPHS tells us there will be infinitely many solutions because of the role of the infinite null space for a singular matrix. So a system of equations with a singular coefficient matrix never has a unique solution. Either there are no solutions, or infinitely many solutions, depending on the choice of the vector of constants ($\vect{b}$).
Example PSHS: Particular solutions, homogeneous solutions, Archetype D.
The ideas of this subsection will be appear again in Chapter LT:Linear Transformations when we discuss pre-images of linear transformations (Definition PI). | {
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Question
# Obtain all other zeroes of $$3x^4+6x^3-2x^2-10x-5$$, if two of its zeros are $$\sqrt{\dfrac{5}{3}}$$ and $$-\sqrt{\dfrac{5}{3}}$$.
Solution
## Since the two zeroes of the given polynomial $$p(x)=3x^{ 4 }+6x^{ 3 }-2x^{ 2 }-10x-5$$ are $$\sqrt { \dfrac { 5 }{ 3 } } ,-\sqrt { \dfrac { 5 }{ 3 } }$$, therefore, $$\left( x-\sqrt { \dfrac { 5 }{ 3 } } \right) \left( x+\sqrt { \dfrac { 5 }{ 3 } } \right) =x^{ 2 }-\dfrac { 5 }{ 3 }$$ is a factor of the given polynomial.now, we divide the polynomial $$p(x)=3x^{ 4 }+6x^{ 3 }-2x^{ 2 }-10x-5$$ by $$x^{ 2 }-\dfrac { 5 }{ 3 }$$ as shown below:Therefore, $$3x^{ 4 }+6x^{ 3 }-2x^{ 2 }-10x-5=\left( x^{ 2 }-\dfrac { 5 }{ 3 } \right) (3x^{ 2 }+6x+3)$$Now, we factorize $$3x^{ 2 }+6x+3$$ as follows:$$3x^{ 2 }+6x+3=0\\ \Rightarrow 3(x^{ 2 }+2x+1)=0\\ \Rightarrow x^{ 2 }+2x+1=0\\ \Rightarrow (x+1)^{ 2 }=0\quad \quad \quad \quad (\because \quad (a+b)^{ 2 }=a^{ 2 }+b^{ 2 }+2ab)\\ \Rightarrow (x+1)(x+1)=0\\ \Rightarrow x+1=0,\quad x+1=0\\ \Rightarrow x=-1,\quad x=-1$$Hence, the zeroes of the polynomial $$p(x)=3x^{ 4 }+6x^{ 3 }-2x^{ 2 }-10x-5$$ are $$\sqrt { \dfrac { 5 }{ 3 } } ,-\sqrt { \dfrac { 5 }{ 3 } } ,-1,-1$$.Maths
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### TheScrasse's blog
By TheScrasse, history, 7 weeks ago,
Hello everyone,
here is a very simple idea that can be useful for (cp) number theory problems, especially those concerning multiples, divisors, $\text{GCD}$ and $\text{LCM}$.
Prerequisites: basic knowledge of number theory (divisibility, $\text{GCD}$ and $\text{LCM}$ properties, prime sieve).
## Idea
Let's start from a simple problem.
You are given $n$ pairs of positive integers $(a_i, b_i)$. Let $m$ be the maximum $a_i$. For each $k$, let $f(k)$ be the sum of the $b_i$ such that $k | a_i$. Output all pairs $(k, f(k))$ such that $f(k) > 0$.
An obvious preprocessing is to calculate, for each $k$, the sum of the $b_i$ such that $a_i = k$ (let's denote it as $g(k)$). Then, there are at least $3$ solutions to the problem.
#### Solution 1: $O(m\log m)$
For each $k$, $f(k) = \sum_{i=1}^{\lfloor m/k \rfloor} g(ik)$. The complexity is $O\left(m\left(\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{m}\right)\right) = O(m\log m)$.
#### Solution 2: $O(n\sqrt m)$
There are at most $n$ nonzero values of $g(k)$. For each one of them, find the divisors of $k$ in $O(\sqrt k)$ and, for each divisor $i$, let $f(i) := f(i) + g(k)$.
If $m$ is large, you may need to use a map to store the values of $f(k)$ but, as there are $O(n\sqrt[3] m)$ nonzero values of $f(k)$, the updates have a complexity of $O(n\sqrt[3] m \log(nm)) < O(n\sqrt m)$.
#### Solution 3: $O(m + n\sqrt[3] m)$
Build a linear prime sieve in $[1, m]$. For each nonzero value of $g(k)$, find the prime factors of $k$ using the sieve, then generate the divisors using a recursive function that finds the Cartesian product of the prime factors. Then, calculate the values of $f(k)$ like in solution 2.
Depending on the values of $n$ and $m$, one of these solutions can be more efficient than the others.
Even if the provided problem seems very specific, the ideas required to solve that task can be generalized to solve a lot of other problems. | {
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Hint 1
Hint 2
Hint 3
Solution
## agc038_c - LCMs
Hint 1
Hint 2
Hint 3
Solution
Implementation (C++)
## abc191_f - GCD or MIN
Hint 1
Hint 2
Hint 3
Hint 4
Solution
Implementation (C++)
## Conclusions
We've seen that this technique is very flexible. You can choose the complexity on the basis of the constraints, and $f(k)$ can be anything that can be updated fast.
Of course, suggestions/corrections are welcome. In particular, please share in the comments other problems that can be solved with this technique.
I hope you enjoyed the blog!
• +105
» 7 weeks ago, # | +15 Nice blog! I like solving problems which use such ideas, so here are a few more problems with a similar flavour:1436F - Sum Over SubsetsChefsums1493D - GCD of an Array
» 10 days ago, # | +8 The Solution spoiler for 1154G - Minimum Possible LCM seems to be broken, I see this:
• » » 10 days ago, # ^ | +6 Fixed. I don't know the reason of the failed parsing, though. The original text was Spoiler$\text{GCD}(a_i, a_j) = h$.and I had to change it to Spoiler$\text{GCD}(a_i, a_j) = h$ .
» 10 days ago, # | +13 Solution 1 can be easily optimized to $O(m \log{\log{m}})$ using the same idea as calculating sums-over-submasks. The code looks something like this: vector f = g; for (int p : primes) for (int i = m / p; i > 0; --i) f[i] += f[i*p]; I suspect there are very few (if any) problems in practice for which solution 3 is directly applicable, but this optimized solution 1 is not. | {
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# Testing for Linearity of Vectors in a Subspace - Examples with Solutions
Given a set of vectors in a subspace, how do we test whether the vectors are linearly independent ?
## What are Linearly Dependent or Independent Vectors?
We first use vectors in two and three dimensional spaces to visualize the concept of dependence and independence of vectors.
In figure 1 below we have have two vectors that are parallel such that $v_2 = 2 v_1$. We say that these two vectors are dependent because we can express one vector in terms of the other as follows:
$v_2 = 2 v_1$ or $v_1 = \dfrac{1}{2} v_1$.
In figure 2 below we have have two vectors that are parallel such that $v_2 = - v_1$. These two vectors are dependent because we can express one vector in terms of the other.
In figure 3, all vectors are parallel to each other. All These vectors are dependent because we can express one vector in terms of the other(s).
In figure 4, we can never express one vector in terms of the other because they are not parallel. These vectors are independent because we cannot express one vector in terms of the other.
In figure 5, using the geometrical sums of vectors, we can write
$3 v_4 = 2 v_1 + 3 v_2 + v_3$
and therefore these vectors are linearly dependent because we can express one vector in terms of the others.
In figure 6, the 3 vectors $v_1$ , $v_2 \}$ and $v_3$ are in the same plane $P$ and are therefore dependent because we can express any of these vectors in term of the other two vectors using linear combinations.
In figure 7, the pairs of vectors $\{v_1 , v_2\}$ , $\{v_2 , v_3\}$ and $\{v_1 , v_3\}$ are in different planes and are therefore independent because we cannot express any of these vectors in term of the other two vectors using linear combinations.
## Formal Definiton of Linearity of Vectors | {
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## Formal Definiton of Linearity of Vectors
Vectors $v_1, v_2 .... v_n$ are linearly dependent when at least one of the vectors may be expressed as a linear combinations of the remaining vectors as follows
$v_1 = r_2 v_2 + r_3 v_3 + .... + r_n v_n$
Writing the above with the zero on the right side, we obtain
$v_1 - r_2 v_2 - r_3 v_3 - .... - r_n v_n = 0$
Hence the following definition
Given a set of vectors $S = \{\textbf{v}_1 , \textbf{v}_2, ... , \textbf{v}_n \}$ ,
If the equation
$r_1 \textbf{v}_1 + r_2 \textbf{v}_2 + ... + r_n\textbf{v}_n = \textbf{0}$ (I)
has only one trivial solution $r_1 = 0 , r_2 = 0 , ... , r_n = 0$, we say that $S$ is a set of linearly independent vectors.
If the above equation has other solutions where not all $r_i$ equal to zero, then $S$ is a set of linearly dependent vectors.
In the examples below, matrices are row reduced in order to test for linearity. This may done using the row reduce augmented matrices calculator included.
## Examples with Solutions
Example 1
Are the vectors in the set $\left \{ \begin{bmatrix} -2 \\ 1 \end {bmatrix} , \begin{bmatrix} 6 \\ -3 \end {bmatrix} \right \}$ linearly independent?
Solution to Example 1
Let $v_1 = \begin{bmatrix} -2 \\ 1 \end {bmatrix}$ and $v_2 = \begin{bmatrix} 6 \\ -3 \end {bmatrix}$
The question could be answered by noticing that $v_2 = - 3 v_1$ and therefore the given vectors $v_1$ and $v_2$ are dependent (not independent)
The above was possible because we are dealing with vectors of small dimension. (2 components only)
We will now use a method that may be applied in any situation.
According to the definition given above, we need to find $r_1$ and $r_2$ such that
$r_1 v_1 + r_2 v_2 = 0$
The above may be written in matrix form as follows
$[ v_1 \;\; v_2] \begin{bmatrix} r_1 \\ r_2 \end {bmatrix} = 0$
where $[ v_1 \;\; v_2]$ is a matrix whose columns are $v_1$ and $v_2$ | {
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Write the system of equations in augmented matrix form
$\begin{bmatrix} -2 &6&|&0\\ 1 & -3&|&0 \end{bmatrix}$
Use the Gauss Jordan method to row reduce the above augmented matrix and obtain
$\begin{bmatrix} 1 &-3&|&0\\ 0 & 0&|&0 \end{bmatrix}$ (I)
The solution to the above reduced system (which is also a solution to the system before reduction) is found as follows
The second equation gives: $0 r_2 = 0$ therefore $r_2$ may take any real value
The first equation gives: $r_1 = 3 r_2$
The solution set may be written as: $\left \{ \begin{bmatrix} 3 r_2\\ r_2 \end{bmatrix} \right \} , r_2 \in R$
which therefore means that we have an infinite number of solutions and therefore the two vectors are linearly dependent.
Note that you do not have to solve the system in order to decide whether the given vectors are dependent or independent.
Construct the augmented matrix using the vectors as columns of the matrix and the constant column on the right is all zeros which in fact may be omitted. We then row reduce the augmented matrix. Then the following conclusions may easily be drawn:
1) The columns with a pivot are independent of each other
2) The columns with no pivots are dependent on the ones with the pivot.
3) The coefficients in the columns without pivot gives the coefficients of dependence on the independent columns.
Let us apply the above to the reduced matrix (I) above
1) Column 1 has a pivot and may be considered as independent
2) Column 2 has no pivot and may therefore be considered as dependent on column (1)
3) The coefficient $- 3$ in column 2 is telling us that $v_2 = - 3 v_1$
Example 2
Are the vectors in the set $\left \{ \begin{bmatrix} -2 \\ 1 \\ 4 \end {bmatrix} , \begin{bmatrix} 1\\ 0 \\ 5 \end {bmatrix} , \begin{bmatrix} 1\\ 2 \\ -1 \end {bmatrix} \right \}$ linearly independent? | {
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Solution to Example 2
step 1: Construct the augmented matrix
Construct the augmented matrix whose columns are the given vectors and zeros on the right column
$\begin{bmatrix} -2 &1&1&|&0\\ 1 & 0&2&|&0 \\ 4 & 5 & -1& |& 0 \end{bmatrix}$
step 2: Row reduce the above matrix (you may use the row reduce augmented matrices calculator included).
$\begin{bmatrix} 1 &0&0&|&0\\ 0 & 1&0&|&0 \\ 0 & 0 & 1& |& 0 \end{bmatrix}$
step 3: Draw conclusions from the row reduced matrix
All 3 columns have a pivot each and therefore all 3 given vectors are independent
Example 3
Are the vectors in the set $\left \{ \begin{bmatrix} 2 \\ -1 \\ 3\\ 1 \end {bmatrix} , \begin{bmatrix} 0 \\ 2 \\ -1\\ 1 \end {bmatrix} , \begin{bmatrix} 4 \\ -8 \\ 9\\ -1 \end {bmatrix} \right \}$ linearly independent?
Solution to Example 3
step 1: Construct the augmented matrix using the given vectors as columns and zeros on the right column
$\begin{bmatrix} 2&0&4&|&0\\ -1&2&-8&|&0\\ 3&-1&9&|&0\\ 1&1&-1&|&0 \end{bmatrix}$
step 2: Row reduce the above matrix
$\begin{bmatrix} 1&0&\color{red}{2}&|&0\\ 0&1&\color{blue}{-3}&|&0\\ 0&0&0&|&0\\ 0&0&0&|&0 \end{bmatrix}$
step 3: Draw conclusions
Only the first and second columns (from the left) have a pivot and therefore the given vectors are not independent.
The coefficients $\color{red}{2}$ and $\color{blue}{-3}$ in the third column give the dependence of the original third columns as a linear combination of the other first and the second columns as follows
$\begin{bmatrix} 4 \\ -8 \\ 9\\ -1 \end {bmatrix} = \color{red}{2} \begin{bmatrix} 2 \\ -1 \\ 3\\ 1 \end {bmatrix} - \color{blue}{3} \begin{bmatrix} 0 \\ 2 \\ -1\\ 1 \end {bmatrix}$ | {
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# Big List of Erdős' elementary proofs
Paul Erdős was one of the greatest mathematicians of all time and he was famous for his elegant proofs from The Book. I posted a question about one of his theorem and got a reference, and I have other questions I want to know the answer to too. But, instead of requesting a reference for each theorem he gave with an elementary proof, I've decided to make a thread for a big list of all his elementary proofs.
I'm excited. Let's make an index of the pages of the Book shown to us!
To get you guys started, I will make a wish list of his theorems who's references I want to see. I encourage you to add to my wish list if you so desire.
Wish list : | {
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Wish list :
• The product of two or more consecutive positive integers is never a square or any other higher power.
• A connected graph with a minimum degree $d$ and at least $2d+1$ vertices has a path of length at least $2d+1$.
• Let $d(n)$ be the number of divisors of $n$. Then the series $\sum_{n=1}^\infty d(n)/2^n$ converges to an irrational number
• Let $g(n)$ be the minimal number of points in the general position in the plane needed to ensure a subset exists that forms a convex $n$-gon. Then $$2^{n-2} + 1 \leq g(n) \leq \frac{(2n-4)!}{(n-2)!^2} + 1$$
• Erdős-Mordell inequality
• You need to define what you mean by "elementary". I think the list of elementary proofs will be very big, e.g. it will contain all his results in set theory, all his results on finite and infinite graphs. – bof Jul 6 '16 at 9:31
• @bof Hi bof. It will be big but I don't think all of them will be posted here. For starters, I'm looking for the questions in the wish list. Please post if you have any results – user230452 Jul 6 '16 at 9:35
• Is it fair that 4 people get to close a thread when 50 people have upvoted it ? Please open the thread for further posts – user230452 Jul 7 '16 at 10:59
• I voted to reopen. – Spenser Jul 7 '16 at 13:00
• @Spenser Thanks, Spencer. I appreciate it. I have collected another theorem of Erdos' to post. – user230452 Jul 7 '16 at 13:24
I think this is worth posting here, mostly because I really enjoy the simplicity of this proof but also because I have no idea how well it is known. The result is not deep or important, so the main interest is in the simplicity of the argument. Erdős proved a lower bound on the number of primes before an integer $n$.
Wacław Sierpiński, in his Elementary Theory of Numbers, attributes to Erdős the following elementary proof of the inequality $$\pi(n)\geq\frac{\log{n}}{2\log{2}}\quad\text{for }n=1,2,\ldots.$$ | {
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Please note that I have adapted the argument from the text of the book to make things, in my opinion, a bit clearer. Note also that the only tools used in the below proof are some basic combinatorial facts and some results about square-free numbers, which can, for example, be proved with the Fundamental Theorem of Arithmetic.
Let $n\in\mathbb{N}=\{1,2,3,\ldots\}.$ Consider the set $$S(n) = \{(k,l)\in\mathbb{N}^{2}:l\text{ is square-free and }k^{2}l\leq n\}.$$ It is a standard fact that every natural number has a unique representation in the form $k^{2}l,$ where $k$ and $l$ are natural numbers and $l$ is square-free. This gives $\lvert S(n)\rvert = n.$
Now if we have a pair $(k,l)$ with $k^{2}l\leq n,$ then we must have $k^{2}\leq n$ and $l\leq n$, since $k$ and $l$ are positive. Note that this gives $k\leq\sqrt{n}.$ Since $l$ is square-free, $l$ can be expressed as a product of distinct primes, each of which must be not-greater-than $n$ since $l\leq n$. That is, $l$ can be expressed as a product of the primes $p_{1},p_{2},\ldots,p_{\pi(n)}.$ There are $2^{\pi(n)}$ such products.
Therefore, if we know $(k,l)\in S(n)$ then there are at most $\sqrt{n}$ possibilities for what $k$ might be and at most $2^{\pi(n)}$ possibilities for what $l$ might be (independent of $k$, of course). It follows that $\lvert S(n)\rvert \leq 2^{\pi(n)}\sqrt{n},$ so $n\leq2^{\pi(n)}\sqrt{n}.$
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Taking $\log$s and rearranging gives the result.
• How can you say a lower bound on the Prime Number Theorem is not important ! – user230452 Jul 6 '16 at 6:05
• @user230452: Better bounds are known through elementary arguments (I think the most famous is by Chebyshev, and has the advantage of providing an upper bound of the same order of growth). So this result is not so important since a little more work can give much better results, with still only elementary ideas. – Will R Jul 6 '16 at 17:23
• @Therkel: I think I understand, but, in this situation, I felt it best to make clear a) how I knew this argument was from Erdos and b) that chasing up the source would lead readers to a superficially different-sounding argument. – Will R Jul 6 '16 at 17:27
• This is brilliant! – Vinícius Novelli Jul 20 '16 at 1:41
Here is an exposition of the proof that made Erdos famous by David Galvin. An elementary proof of Bertrand's postulate, which states that there is a prime number in between every $n$ and $2n$. The essence of this proof is in noticing that the lower bound of $$\binom{2n}{n} \geq \frac{4^n}{2n + 1}$$ The binomial expression is the middle term (and the largest) of $(1+x)^{2n}$. The lower bound is the average of the sum of all binomial coefficients. This is obtained by putting $x=1$, and then dividing by the number of terms. This gives us the average. Obviously, the largest term should be bigger than the average. If the postulate does not hold, there is an upper bound that is smaller than this lower bound for large $n$. The postulate can easily be verified for the smaller values of $n$.
https://www3.nd.edu/~dgalvin1/pdf/bertrand.pdf
One very simple, and yet one of my favorites is the Erdős-Anning theorem:
Let $A \subseteq \mathbb C$ be an infinite set of points, such that
$$\forall x, y\in A \quad |x-y| \in \mathbb N$$
then there exists some $c,k \in \mathbb C$, such that all $a \in A$ is of the form $a = cx + k$ for some $x \in \mathbb R$. | {
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It was proved in 1945 in the American Mathematical Bulletin. http://www.ams.org/journals/bull/1945-51-08/S0002-9904-1945-08407-9/S0002-9904-1945-08407-9.pdf
A more colorful formulation is that an infinite sets of points on the Cartesian plain with mutual integer distances must lie on a straight line.
To prove it, we need an upper bound on the number of non collinear points possible in a set with integral distances. More specifically, if there is a set of non collinear points have integer distances, all at most $d$, then at most $4(d + 1)^2$ points with integer distances can be added to the set.
• I have a doubt. Can you help me ? The theorem says that for any $n$, it is possible to have $n$ points on a plane, not all collinear such that they have integral distances but not infinitely many points. As I understand it, infinity is not a number, but a concept. To say that infinitely many points are possible it means that for any $n \geq l$, it is possible, where $l$ is an arbitrarily large natural number. It seems to me the first part of the theorem is stating it, and the second is contradicting it. – user230452 Jul 6 '16 at 3:45
• @user230452 No, to say that there is an infinite set with integral distances is not the same as saying for any $n\ge l$ there is a set with $n$ points and integral distances. Compare: For any $n\ge l$ there is a bounded subset of $\Bbb R$ with integral distances, but there is no bounded infinite subset of $\Bbb R$ with integral distances. $\Bbb N$ is an unbounded infinite subset of $\Bbb R$ with integral distances. – Mario Carneiro Jul 6 '16 at 8:05
• @bof There was a typo; the constraint $0 \in A$ was lacking. It is equivalent to adding some constant, s.t. $a = cx + k$. – user347499 Jul 6 '16 at 16:23
Erdos' proof of Infinite primes
The following proof is taken from the book - "Proofs from THE BOOK" by Martin Aigner and Gunter Ziegler. This proof is attributed to Erdos. | {
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This proves that there are infinitely many primes and that the series of the sum of prime reciprocal steps diverges. Let us assume that the infinite series $\sum\frac{1}{p}$, where $p$ denotes the prime numbers is a convergent one. Then, there must exist some natural number $k$ such that, $$\sum_{i > k} \frac{1}{p_i} < \frac{1}{2}$$
For an arbitrary natural number $N$, we get the inequality $$\sum_{i > k} \frac{N}{p_i} < \frac{N}{2}$$
Now, we call all the primes $p_1, p_2, \dots , p_k$ the small primes, and all the other primes the big primes.
Let $N_b$ denote the number of positive integers $n \leq N$ that have at least one divisor that is a big prime. And $N_s$ denote the number of positive integers less than $N$ that have only small prime divisors.
We will show that for a suitable $N$, $N_b + N_s < N$, which is a contradiction because their sum should equal $N$.
First, we estimate $N_b$ $$N_b \leq \sum_{i > k}\big\lfloor \frac{N}{p_i} \big\rfloor < \frac{N}{2}$$
Now, we look at $N_s$. We write every $n \leq N$ as $a_nb_n^2$ where $a_n$ is the square free part. $a_n$ is a product of different small primes. There are only $k$ small primes, and each prime may either be chosen or not chosen. So, there are only $2^k$ different values of $a_n$. Furthermore, \begin{align} b_n \leq \sqrt n \leq \sqrt N \\ \text{There are at most $\sqrt N$ different values of $b_n$} \\ \implies N_s \leq 2^k\sqrt N \end{align}
Since, $N_b$ is always less than $N/2$, we just have to find a value of $N$ such that $N_s \leq N/2$. This happens when $N = 2^{2k + 2}$. When $N = 2^{2k + 4}$, $N_s < N/2$. This proves our contradiction that $$N_b + N_s < N$$ which proves that the series diverges and that there are infinitely many primes.
Erdos' favourite questions. The following are not research papers but popular questions Erdos used to ask children.
• If $n+1$ integers are chosen from the first $2n$ integers, there will always be two that are co prime. | {
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There will be two numbers that are consecutive. These two numbers will be relatively prime. To see that this is not true when $n$ integers are chosen, just select all the even numbers.
• If $n+1$ integers are chosen from the first $2n$ integers, there will always be two such that one divides the other.
Every number is expressible as the product of a power of two and an odd number. There are only $n$ odd numbers in the first $2n$ integers. Two of the numbers are multiplied by the same odd number. One of them has a smaller power of two. This number divides the other. To show that it is not necessarily true for $n$ integers, choose $n+1,n+2, \dots 2n$. | {
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• Suppose we have $n$ natural numbers none of which is greater than $2n$ such that the least common multiple of any two is greater than $2n$. Then, all $n$ numbers are greater than $2n/3$.
• Regarding the first one, you can easily see that "all the even numbers" is actually the only solution -- we can't select 1, because it's coprime to everything; and we can't have any consecutive numbers. If we take $n-1$ numbers, a bit of thinking/excluding a few small cases shows that for $n>3$, the only solutions are "all even numbers but one". For $n=3$, we have $\{3,6\}$. In general, any set of $n-k$ must be all even unless $n-k \ge 2n/3$, because we have an odd number at least $3$, which immediately narrows us to $1/3$ of the numbers. I wonder how # of solutions grows with $n$ and $k$? – Alex Meiburg Jul 7 '16 at 23:21
• @Alex, your final conclusion is false. In fact, for all $\epsilon > 0$ there exists an $n$ and a set $S \subset \{1,2,\ldots, 2n\}$ with $|S| > (1 - \epsilon)n$ such that $S$ contains an odd integer and for all $x,y \in S$ we have $\gcd(x,y) > 1$. For example, choose $n$ to be the product of the first $m$ odd primes and define $S = \{2k | k < n \text{ and } \gcd(k, n) > 1 \} \cup \{n\}$. You will see that $|S| = n - \varphi(n) + 1 > n(1 - \frac{c}{\log \log(n)})$ for some absolute constant $c$. For $m \ge 8$ we have $|S| > \frac{2n}{3}$. – Woett Oct 11 '18 at 12:00
• @Woett, right, how silly of me. I figured that if $k$ is odd then the density of numbers sharing a factor with $k$ is at most $1/3$ -- totally neglecting the possibility of multiple factors. :) Taking $n=105=3\cdot 5\cdot 7$ gives a density of $0.54$, of course. – Alex Meiburg Oct 12 '18 at 2:46
Erdos answered the following question in the affirmative - Are there infinitely many odd numbers that are not expressible as the sum of a prime number and a power of $2$. The proof is explained in this paper : http://www.maa.org/sites/default/files/3004416309960.pdf.bannered.pdf | {
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The essence of the proof is in showing that for every integral value of $k$, there is a $2^k$ which has a certain residue with certain moduli. By the Chinese remainder theorem, there are infinitely many odd numbers that satisfy those same congruences. Whenever they do, $a - 2^k$ is composite for all integer values of $k$.
• Surely this is clear because the set of all numbers $p+2^k$ is density zero? Although perhaps that is not elementary. – Mario Carneiro Jul 7 '16 at 1:20
• The density should be positive, not zero. There are n/log(2) choices of (prime, power of 2) both less than n. If most, or a positive fraction, of those are unique representations then a positive density subset of integers less than 2n have such a representation. – zyx Jul 7 '16 at 2:08
• @zyx Oh, indeed if $|A\cap[n]|\sim f(n)$, and $B=\{a+2^k:a\in A,k\in\Bbb N\}$, then $|B\cap[n]|\le\sum_k f(n-2^k)\le f(n)\log_2n$, which is not strong enough to conclude density zero with $A=\Bbb P$, $f(n)=n/\log n$. A simulation of the quantity $\sum_{p\le n}\log_2(n-p)$ appears to converge to about $1.468n$, which is obviously useless as an upper bound on $|B\cap[n]|$ since it's larger than the trivial bound. – Mario Carneiro Jul 7 '16 at 4:59
• That is close to n/log(2) = 1.443 n . – zyx Jul 7 '16 at 5:08
• Does the simulation appear to converge to 1/log(2) when using integer logarithms, $\lfloor \log_2 ( \dots ) \rfloor$? @MarioCarneiro – zyx Jul 7 '16 at 19:10
The proof of the Littlewood-Offord lemma for sums of real numbers.
Erdos noticed that under the correspondence between sequences of $\pm$ signs and finite sets, Sperner's theorem applies and gives the optimal bound for the Littlewood-Offord lemma in dimension $1$ (on how many signed sums of $n$ given numbers of absolute value at least $1$, can have absolute value at most $1$). This observation is a few lines of very basic algebra and combinatorics. | {
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The rest of the paper is more difficult, and is an extension of Sperner's theorem to handle the case of larger bounds on the sums, once again reading optimal bounds from the combinatorics. This gave evidence that the higher dimensional Littlewood-Offord problem on sums of vectors might also be essentially combinatorial, which was later shown to be true.
Erdos - Mordell Inequality : For a point $O$ inside a given triangle $ABC$, the perpendiculars $OP$, $OQ$ and $OR$ are drawn to the side $$OA + OB + OC \geq 2(OP + OQ + OR)$$
The product of consecutive integers is never a power
This was proved by Erdos and Selfridge.
http://www.renyi.hu/~p_erdos/1975-46.pdf
Suppose $A$ is a set of $r$ subsets on the set $\{1,2,\dots,n\}$ such that any two sets in $A$ have a non empty subset and $n/2\geq r$, the maximal size of $A$ is $\binom{n-1}{r-1}$ i.e., $$|A| \leq \binom{n-1}{r-1}$$with equality holding if and only if all the sets share a common element.
A family of sets may also be called a hypergraph. When every set in $A$ has the same size $r$, it is called a $r$-uniform hypergraph.
Note : The condition of $n/2 \geq r$ is imposed because if $r$ is more than half of $n$, then any two sets have a non empty intersection. The maximal size of the family is then $\binom nr$. | {
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Sum of sin waves with same frequency and different amplitudes and phase
For the equation:
$$A_1 \sin (\omega t + \theta_1) + A_2 \sin (\omega t + \theta_2) = A_3 \sin (\omega t + \theta_3)$$
I've been able to show that the amplitude of the sum is (I believe this is a standard problem):
$$A_3 = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\theta_1 - \theta_2)}$$
and the new phase is:
$$\theta_3 = \arctan \left(\frac{A_1 \sin \theta_1 + A_2 \sin \theta_2}{A_1 \cos \theta_1 + A_2 \cos \theta_2}\right)$$
My question is what happens when the phases $\theta_1$ and $\theta_2$ are zero (or just equal to each other). Most books and internet sites (eg http://mathworld.wolfram.com/HarmonicAdditionTheorem.html) state that under these conditions:
$$A_3 = \sqrt{A_1^2 + A_2^2}$$
However, if I set $\theta_1$ and $\theta_2$ to zero one gets instead (Since $cos(0) = 1$):
$$A_3 = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2}$$
I've been staring at this for a while and I don't understand why the published answer is different from mine. I am sure it is something simple but I've not been able to spot it (Addendum: Wikipedia site: https://en.wikipedia.org/wiki/List_of_trigonometric_identities has the same result under section Linear Combinations).
• I realized the problem: $A_3 = \sqrt{A^2_1 + A^2_2 + 2 A_1 A_2}$ applies to the relation $cos (wt + \theta_1) + sin(wt + \theta_2)$ not $sin (wt + \theta_1) + sin(wt + \theta_2)$. For the latter it works fine. This is an error on my part. ok to delete this question, as its not a valid one? – rhody Jul 13 '18 at 18:23
• Although your Question contained and was motivated by an error, it brings up some useful content, and I'd like to retain it. Even the error, articulated in your Comment, bears pointing out as it is the sort of thing anyone might overlook at first glance. – hardmath Jul 14 '18 at 13:02
The formula $\sqrt{A_1^2+A_2^2}$ applies to sinusoids in quadrature ($\frac\pi2$ phase difference), such as when summing a sine and a cosine. | {
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Signals in phase just add up their amplitudes, $A_1+A_2$, while signals in opposition subtract, $|A_1-A_2|$.
• I've been staring at this for hours yesterday, and then this morning after I posted the question and suddenly realized my error. – rhody Jul 13 '18 at 18:26
You wrote:
$$A_1 \sin (\omega t + \theta_1) + A_2 \sin (\omega t + \theta_2) = A_3 \sin (\omega t + \theta_3)$$
The Wolfram article you cite says:
It is always possible to write a sum of sinusoidal functions $$f(\theta)=acos(\theta)+bsin(\theta )$$
as a single sinusoid the form $$f(\theta)=ccos(\theta+\delta)$$ $$A_3 = \sqrt{A_1^2 + A_2^2}$$
So you have sine plus sine, while Wolfram has sine plus cosine. To convert Wolfram's cosine to a sine, you need to shift the phase by $\frac{\pi}2$, which then makes the cosine of the phase difference equal to zero.
Waves with no phase difference (or even pi's) directly add up their amplitudes to form a new wave.
$$A_1 \sin (\omega t) + A_2 \sin (\omega t) = (A_1+A_2) \sin (\omega t )$$
The $A_3$ you prescribed is for waves with phase difference $(\theta_1 - \theta_2)=\frac{\pi}{2}$.
The equation you got putting $\theta_1=\theta_2=0$ is correct and simplifies to $A_3=(A_1+A_2)$.
The result given in Equation (21) of Wolfram site
for adding in-phase waves of same frequency with zero relative phase angle gives
$$A_3= \sqrt{A_1^2+A_2^2+ 2 A_1 A _2}= \pm (A_1+A_2)$$
The other result came about when the phase difference is $\pi/2$ obtained directly by Pythagoras thm as diagonal length of a rectangle of vectors or phasors.
The parallelogram of amplitudes can be drawn and the situation becomes clear. There are three situations. In-phase,add; Out of phase, subtract, or take the difference; when relative phase difference is $\pi/2$ take root square sum. | {
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# How to find the elongation of a spring when it is tied to a brass bar over a tilted wall?
The problem is as follows:
An homogeneous brass bar which goes from point $$A$$ to point $$B$$ has a weight of $$300\,N$$. The bar is suported over two frictionless surfaces as it is shown in the picture from below. The system is in static equilibrium by the action of the force excerted by the spring tied to the end at point $$B$$ on the bar which in turn is fixed to the ceiling. Using this information find the elongation of the spring. Assume that the constant of the spring is $$k=500\,\frac{N}{m}$$.
The alternatives given in my book are as follows:
$$\begin{array}{ll} 1.&28\,cm\\ 2.&23\,cm\\ 3.&25\,cm\\ 4.&26\,cm\\ 5.&30\,cm\\ \end{array}$$
What I did to solve this problem was to use a "trick" which I seen in a similar problem about a ladder standing next to a wall.
The rationale was that when an horizontal object is leaning against two surfaces there will be a reaction from both. But if it is asked a force in between. You may use any point to establish the torque about that point and use Varignon's theorem? to find the unknown force. This is done to "cancel out" the reaction force from both surfaces.
Therefore:
Torque about the point specified in the figure from below results into:
$$\sum_{i=1}^{n}\tau=0$$
$$(l\cos 30^{\circ})\times(300)+500\times x \times 2l =0$$
simplifying terms:
$$-\frac{\sqrt {3}}{2}\times 3 + 10 x = 0$$
$$x= \frac{3 \sqrt {3}}{20}\approx .2598076212 \,m$$
which transformed into cm would be about $$26\,cm$$ and this appears in the alternatives as option four. But is my answer the right approach to this problem?.
My source of confusion is that in the problem is not indicated if the spring is paralell to the incline or if it does make another angle. | {
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It is not indicated in the problem but how can I find let's say the Reaction in both the floor and in the wall?. How can I find these?. Am I using Varignon's theorem in this problem?. Can someone help me with this matter please?.
• @Cesareo I'm sorry but I still don't get to the answer you reached. This is not homework. Can you post a solution or more hints?. How exactly you got to $30\,cm$.? – Chris Steinbeck Bell Jul 11 '20 at 3:27
First, if you are referring to Varignons' theorem about quadrilaterals and the parallelogram formed by the midpoints of the sides, I don't think you are usinng it.
Second, to determine both the reactions in the floor and the wall we can apply Newton's law and impose that the sum af all forces is zero, and then impose that the torque about any point we want is zero. You already did the second part, so you could just apply Newton's law: $$\vec{N_A}+\vec{N_B}+\vec{P}+k\vec{\Delta l}=0$$ If the spring is parallel to the wall we know the directions of all this vectors and by decomposing along two axes $$x$$ and $$y$$ (I choose the $$x$$-axis parallel to the ground, but it's arbitrary) we obtain two equations that allow us to find the moduluses of the reaction forces: $$N_A+\frac{N_B}{2}-P+\frac{\sqrt{3}}{2}k\Delta l=0$$ $$-\frac{\sqrt{3}}{2}N_B+\frac{1}{2}k\Delta l=0$$
Personally I solved the problem calculating the torque about the point B because there are 4 forces and two are applied in B, so the equations are particularly simple, but any other point is just as good and gives the same solutions. I too found that the answer is 26cm
Fig 2 has made things easy for taking force/moment equilibrium to evaluate forces.
Take moments about top left corner call it $$I$$
$$k x \cdot 2 l = l \cos 30^{\circ} 300$$
$$x = 0.2598$$ meters i.e., the fourth option:
$$x= 25.98 cm$$
Btw, I is the center about which the brass bar is instantaneously rotating in dynamic problems.
Calling $$\alpha = 30^{\circ},\ \beta = 60^{\circ}$$ and | {
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Calling $$\alpha = 30^{\circ},\ \beta = 60^{\circ}$$ and
$$\cases{ A=(0,0)\\ B=A+l(\cos\alpha,\sin\alpha)\\ G = \frac 12\left(B+A\right) }$$
with the forces
$$\cases{ f_A = (0,N_A)\\ f_B = k \Delta x (\cos\beta,\sin\beta)+N_B(-\cos\alpha,\sin\alpha)\\ f_G = (0,-W) }$$
we have
$$\cases{ f_A+f_B+f_G = 0\\ (A-B)\times f_A+(G-B)\times f_G = 0 }$$
from which we obtain
$$N_A = \frac W2,\ N_B = \frac W4,\ \Delta x = \frac{\sqrt{3}W}{4k}$$
so $$\Delta x\approx 0.26$$ cm
Another way using virtual displacements.
Calling
$$\cases{ f_{B_1} = k \Delta x (\cos\beta,\sin\beta)\\ f_{B_2} = N_B(-\cos\alpha,\sin\alpha) }$$
In equilibrium, we have
$$\delta A\cdot f_A + \delta G\cdot f_G + \delta B \cdot\left(f_{B_1}+f_{B_2}\right) = 0$$
and due to orthogonality
$$\delta A\cdot f_A = \delta A\cdot f_G = \delta B\cdot f_{B_2} = 0$$
and following
$$\frac 12\left(\delta A + \Delta x(\cos\beta,\sin\beta)\right)\cdot f_G + k (\Delta x)^2\left(\cos\beta,\sin\beta\right)\cdot \left(\cos\beta,\sin\beta\right)=0$$
or
$$-\frac 12 W\sin\beta + k\Delta x = 0\Rightarrow \Delta x = \frac{\sqrt{3}W}{4k}$$
• The bar should be $2l$ long, not $l$. – enzotib Jul 11 '20 at 8:06 | {
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# Why both data have same standard deviation?
But I am curious why the standard deviation for both data is same. I think because data 2 is "100 - data 1".
• $var(X+a)=var(X)$ and $var(aX)=a^2var(X)$ for a real number $a$ and a random variable $X$. With those, can you prove why $var(100-X)=var(X)$ where X is your data set? (Equal variances is the same as equal standard deviations.) – Dave Feb 15 '20 at 0:24
I still would like you to try to prove it, but I’ll give some intuition about those two identities I gave.
Remember that variance (and standard deviation) have something to do with how spread out your data are.
$$var(X+a)=var(X)$$ means that if you take data with some amount of spread and slide them up or down the number line, you do not change the spread.
$$var(X)=var(-X)$$ is not quite what I gave, but it’ll be part of your proof. The intuition here is that you’re just spinning it around, taking a mirror image. The data are as spread out as their reflection.
Together, $$var(100-X)=var(X)$$ means that you flip the data to the mirror image and then slide along the number line, but you do not change the spread.
Capisce?
• what is the meaning of Capisce? – aan Feb 16 '20 at 11:41
• that's Italian for "do you comprehend ?" – IrishStat Feb 16 '20 at 13:20
I believe the answer to this question can be expressed in more elemental terms. Beginning with the formula for the sample standard deviation
$$s_{x} = \sqrt{\frac{ \sum_{i=1}^{n}{(x_i - \overline{x})^2} }{n - 1}},$$ | {
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$$s_{x} = \sqrt{\frac{ \sum_{i=1}^{n}{(x_i - \overline{x})^2} }{n - 1}},$$
I want to draw your attention to the expression $$(x_i - \overline{x})$$ in the numerator. Ignore the summation notation for the purposes of this answer. Note, we decrement the sample mean (i.e., $$\overline{x}$$) from each realization of $$x_{i}$$. If we were to increase or decrease each $$x_{i}$$ by the same amount, the mean will change. However, the distance of each realization of $$x_{i}$$ from that central tendency remains the same. In other words, each deviation from the mean is the same.
Subtracting each $$x_{i}$$ from a constant (e.g., $$100 - x_{i}$$) flips the vector of values to its mirror image, then slides it along the number line by a constant amount. Suppose the first realization of $$x_{1}$$ = 20 and $$\overline{x}$$ = 4. In keeping with the foregoing expression,
$$(20 - 4) = 16,$$
which is the deviation from the sample mean. Now, assume we don't move along the number line just yet and we simply put a negative sign in front of each $$x_{i}$$, such that we have $$-(x_{i})$$; this flips the sign of the mean as well. The first realization, $$x_{1}$$, is now $$-20$$. Again, substituting $$-20$$ into the expression,
$$(-20-(-4)) =-16,$$
which is the same number of units away from the mean. Note, squared deviations result in a positive number, so the numerator remains the same. You can run this code in R which builds upon @knrumsey's insightful response, but breaks it down further:
x <- rnorm(10000, 30, 5) # Simulates 10,000 random deviates from the normal distribution
par(mfrow = c(2, 2))
hist(x, xlim = c(0, 100), col = "blue")
hist(-x, xlim = c(-100, 0), col = "red") # Note the -x, it simply flips the sign (mirror image)
hist(x, xlim = c(0, 100), col = "blue")
hist(100-x, xlim = c(0, 100), col = "red") # Subtracting x from 100 shifts values along the x-axis | {
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The first row of plots shows how the realizations translate when we negate each $$x_{i}$$. The second row shows what happens when each $$x_{i}$$ is subtracted from 100; the translation first flips then slides across the number line. The spread of the distribution is unaffected by this.
To supplement @Dave's answer, take a look at the following histograms which have the same x-axis. Data2 is just a shifted version of Data1 and therefore the standard deviation, which is a measure of spread, shouldn't change.
R code to generate histograms:
x <- 100*rgamma(1000, 6, 2) #Simulate some data
hist(x, xlim=c(0,100))
hist(100-x, xlim=c(0,100))
• Thanks. well-plotted histogram. May I know how do you plot the histogram? – aan Feb 15 '20 at 5:29
• @aan see the update. – knrumsey Feb 15 '20 at 17:49
• @knrumsey Good illustration. However, simulating from the gamma distribution would produce realizations that cluster around a mean that is much lower than what I see here given the parameters you specified. Right? – Thomas Bilach Feb 16 '20 at 20:24
• @Tom, that is correct. I didn't put too much thought into which distribution I simulated from. Just wanted draws between 0 and 100. – knrumsey Feb 16 '20 at 20:45
• @Thomas Bilach, realizing that I misunderstood your question. Thanks for pointing this out, I've updated the post to reflect the fact that I multiplied the data by 100. – knrumsey May 3 '20 at 0:48
changes period to period in DATA 1 identical to the negative changes in DATA 2
OBS DATA 1 FIRST DIF DATA 2 FIRST DIFF
1 30 NA 70 NA
2 40 10 60 -10
3 80 40 20 -40
4 30 -50 70 50
5 20 -10 80 10
6 33 13 67 -13
7 33 0 67 0
8 33 0 67 0
9 33 0 67 0
CONTINUING STEPS TO UNCOVER THE BASIC RELATIONSHIP OF DATA1 AND DATA2:
I took the series DATA1 and computed the ACF ... here it is ..
AND for DATA2 here
...note that not only are the standard deviations the same but the ACF's are the same. | {
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...note that not only are the standard deviations the same but the ACF's are the same.
• This is an interesting perspective, but I believe it would need more explanation to be understood by many readers. – whuber Feb 15 '20 at 22:53
• @IrishStat, thanks. do you mind to explain more? – aan Feb 16 '20 at 11:42
• i added some material to my response detailing how differenced data period to period ( val2-val1,val3-val2,.. val9-val8) for both original series DATA1 and DATA2 were complements of each other – IrishStat Feb 16 '20 at 13:19
• @IrishStat. Noted. what is the meaning of Q.E.D and OBS? – aan Feb 16 '20 at 15:58
• An explicit connection between those assertions and the result is needed. What is interesting is your use of successive differences ("changes period to period" in an otherwise unordered dataset) to characterize the standard deviation of the dataset. In other words, you appear to be claiming that the set of equalities $$x_{i+1}-x_i = -(y_{i+1}-y_i)$$ for $i=1, 2, \ldots, n-1$ implies $$\sum_{i=1}^n(x_i-\bar x)^2=\sum_{i=1}^n(y_i-\bar y)^2.$$ That's true, but it does need to be shown: you can't just exhibit a table of numbers and declare "QED"! – whuber Feb 16 '20 at 20:15 | {
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# Wanna practice Induction ?
The principle of mathematical induction is a very very useful tool in many proofs, and also in proving some useful formulas. $\color{#D61F06}{\text{The principle states that any given set of positive integers}}$ has ALL natural numbers if it follows the following conditions -
$(i)$ The first natural number, i.e. $1$ is in the set.
$(ii)$ Whenever the integer $k$ is in the set, then $k+1$ is also in the set.
This looks so obvious, see that if you have been given the two statements simultaneously, then by using $(ii)$ on $(i)$, you can say that $2$ is in the set. Then again applying the statement $(ii)$ for $2$, we say that $(3)$ will be in the set, so on.
$\color{#3D99F6}{\textbf{Thus if you prove a certain result for 1}}$ , $\color{#3D99F6}{\textbf{and then assume that the result is true}}$ for $k$, just prove that it is true for $k+1$ and you're done !
Same thing you can do for 0, then prove for whole numbers !!!
Problems for practice:-
Prove that the following results hold $\forall n \in \mathbb{N}$
$(a) \displaystyle \sum_{k=0} ^n 2^k = 2^{n+1}-1$
$(b) \displaystyle \sum_{k=1} ^n k = \dfrac{n(n+1)}{2}$
$(c) \displaystyle \sum_{k=1} ^n k^2 = \dfrac{n(n+1)(2n+1)}{6}$
$(d) \displaystyle \sum_{k=1}^n k^3 = \biggl( \sum_{k=0} ^n k \biggr) ^2$
$(e) \displaystyle 24 \mid 2\cdot 7^n + 3\cdot 5^n -5$
$(f) \displaystyle 1\cdot 2+2\cdot 3+3\cdot 4+ .... + n(n+1) = \dfrac{n(n+1)(n+2)}{3}$
I felt like sharing because though it's very well known to all, I am willing to find some good level problems for practicing this foundation builder concept once more... $\color{#20A900}{\text{Isn't this a good revision ?}}$
5 years, 4 months ago | {
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5 years, 4 months ago
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@Aditya Raut this is really a nice note for practice...... could you please add more divisibility kinda problems to give a more diverse experience to readers? ........ thanks :)
- 5 years, 4 months ago
- 5 years, 3 months ago
Thanks - they look like a good selection of induction problems
- 4 years, 10 months ago | {
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Thanks - they look like a good selection of induction problems
- 4 years, 10 months ago
I actually haven't tried $(d)$ and $(f)$ before since doing math XD
Here's the solution for $(e)$ if someone got stuck.
$2\times 7^{m+1} + 3 \times 5^{m+1} - 5$
$= 2\times 7^{m}\times 7 + 3\times 5^{m}\times 5 - 5$
$= 14\times 7^{m} + 15 \times 5^{m} - 5$
$= 12\times 7^{m} + 12\times 5^{m} + 2\times 7^{m} + 3\times 5^{m} - 5$
$= 12(7^{m} + 5^{m}) + 24k$ for some $k$
$= 24n + 24k$ for some $n$
- 5 years, 3 months ago
$(a)$ Base Case $n = 1$
$\displaystyle \sum_{k = 0}^{1}2^{k} = 3$
$2^{n+1} - 1 = 3$
True.
Propose it is true $\forall n \in N$
Then it must be true for $n + 1$
$\displaystyle\sum_{k = 0}^{n+1}2^{k} = \sum_{k = 0}^{n}2^{k} + 2^{n+1}$
Substituting the value from our proposal,
$\sum_{k = 0}^{n}2^{k} + 2^{n+1} = 2^{n+1} - 1 + 2^{n+1}= 2^{n+2} - 1$
Thus, $LHS = RHS$
Thus, $LHS = RHS \forall n \in N$
Sorry for ugly arrangement.
- 5 years, 3 months ago
@Aditya Raut Can you add these to the Induction Wiki page? Thanks!
Staff - 5 years, 1 month ago
Nice! I'm too into creating notes and problems set for inequality :) Why not take a look at it @Aditya Raut ? :D
- 5 years, 3 months ago
What is e) ?... in d) second summation is it k=0 or k=1 ? Nice collection.
- 5 years ago
Thanks, but Just think ! Will adding $0$ make any change? In that summation, starting from $k=0$ will yield same thing as starting with $k=1$, won't it ? @Niranjan Khanderia
- 5 years ago
Thanks.
- 5 years ago
And e) means prove that for all positive integers $n$, $(2\cdot 7^n + 3\cdot 5^n -5)$ is divisible by $24$.
$a\mid b$ is the symbol of "a divides b" or "b is divisible by a".
Latex code for the symbol $\mid$ is "\mid"
- 5 years ago
Thanks. I did not get it since I did not assume the parenthesis.
- 5 years ago
/(good/)
- 4 years, 10 months ago | {
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- 5 years ago
/(good/)
- 4 years, 10 months ago
A good level problem is here: Fro every positive integer $n$, prove that $\sqrt{(4n+1)}<\sqrt n +\sqrt{(n+1)}<\sqrt{(4n+2)}$. Hence or otherwise, prove that [$\sqrt{n}+\sqrt{n+1}$]=[$\sqrt{4n+1}$] where [.] denotes floor function. It is an IITJEE problem.
- 4 years, 7 months ago | {
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all. I ran into this problem and was wondering where my logic failed. I was solving an absolute value involving a quotient and went about it the following way:
$$\begin{eqnarray} \lvert \frac{x+1}{x-2} \rvert &<& 3\\ \Rightarrow \frac{x+1}{x-2} &>& -3 \hspace{1mm}and\hspace{1mm}\frac{x+1}{x-2} < 3\\ x+1&>&-3(x-2)\hspace{2cm}\text{(Solving for the left inequality)}\\ x+1&>&-3x+6\\ 4x&>&5\\ x&>&\frac{5}{4} \end{eqnarray}$$ which cannot be true, as $$x$$ can equal $$2$$ with these restrictions which is obviously not allowed. Doing the left side the correct way:
$$\begin{eqnarray} \frac{x+1}{x-2} &>& -3\\ \frac{x+1}{x-2} + 3 &>& 0\\ \frac{x+1}{x-2} + \frac{3x-6}{x-2} &>& 0\\ \frac{4x-5}{x-2} &>& 0\\ \Rightarrow x < \frac{5}{4}\lor x>2\\ \end{eqnarray}$$
leads to the solution set $$(-\infty,\frac{5}{4})\cup(2,\infty)$$. My question is, what happened with the first method where I failed to come up with a solution for $$(2,\infty)$$, (and why the signage for $$x>\frac{5}{4}$$ is backwards/incorrect in the first example, does the multiplication by -3 reverse the inequality even if I'm not introducing a new negative term to one side?). I assume since you cannot divide by zero, you're not allowed to multiply the denominator to the other side without restrictions, but I am not quite sure. Thank you, I greatly appreciate it.
• The use of the Union symbol $\cup$ is certainly not the correct way, as it is used to denote union of sets. You can use a $\textrm{or}$, or the or symbol $\vee$ (\vee). Jul 7, 2021 at 23:31
• @ultralegend5385: When writing in the context of mathematical logic, I think \lor, \land and \lnot are easier to remember for $\lor,\land,\lnot$ respectively. Then again, sometimes I prefer writing \neg over \lnot for the logical not $\neg$ Jul 7, 2021 at 23:44
• @ultralegend5385 fixed. Jul 7, 2021 at 23:54 | {
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In the first method, when you go from $$\dfrac{x+1}{x-2}\gt -3$$ to $$x+1\gt -3(x-2)$$, what you're doing is multiplying both sides of the inequality by $$x-2$$ and you assume that $$x-2\gt 0$$ when doing that because otherwise the inequality sign flips to $$\le$$
So, assuming $$x\gt 2$$, you get from that inequality that $$x\gt 5/4$$ and the second inequality $$\dfrac{x+1}{x-2}\lt 3$$ gives $$x\gt 7/2$$
So, assuming $$x\gt 2$$, you get $$x\gt 5/4$$ and $$x\gt 7/2$$; together they imply $$x\gt\max\{2,5/4,7/2\}=7/2$$
Similarly, assume $$x-2\lt 0$$, ie, $$x\lt 2$$ and solve $$x+1\lt -3(x-2)$$ and $$x+1\gt 3(x-2)$$ simultaneously to obtain the other solution set. You get $$x\lt\min\{2,5/4,7/2\}=5/4$$
The complete solution set is thus $$(-\infty,5/4)\cup (7/2,\infty)$$
The answer you arrived at is wrong. Note that $$x=3\in (2,\infty)$$ doesn't satisfy the original equation.
In the second method, you're only solving one of the two inequalities that are supposed to hold. If you solve the second inequality $$\dfrac{x+1}{x-2}\lt 3$$ in a similar way, you get $$\frac{x+1}{x-2}-3\lt 0\iff\frac{7-2x}{x-2}\lt 0\iff x\in (-\infty, 2)\cup (7/2,\infty)$$
and the solution set to your original problem is the intersection of this with $$(-\infty,5/4)\cup (2,\infty)$$ so that
$$[(-\infty,2)\cup (7/2,\infty)]\cap[(-\infty,5/4)\cup (2,\infty)]=(-\infty,5/4)\cup (7/2,\infty)$$
which is the solution set to the original problem $$\left|\dfrac{x+1}{x-2}\right|\lt 3$$
• I understand now. Thank you for explaining it so well. Jul 7, 2021 at 23:55
• @js2822: You're welcome. Also, nice work for asking a good question and being responsive to others! +1 ;) Jul 7, 2021 at 23:58
HINT | {
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HINT
Since both sides are nonnegative, you can square them in order to obtain an equivalent inequation: \begin{align*} \left|\frac{x+1}{x-2}\right| < 3 & \Longleftrightarrow \left(\frac{x+1}{x-2}\right)^{2} < 9\\\\ & \Longleftrightarrow \frac{(x^{2} + 2x + 1) - 9(x^{2} - 4x + 4)}{(x-2)^{2}} < 0\\\\ & \Longleftrightarrow \frac{-8x^{2} + 38x - 35}{(x-2)^{2}} < 0\\\\ & \Longleftrightarrow \begin{cases} 8x^{2} - 38x + 35 > 0\\\\ x\neq 2 \end{cases} \end{align*}
Can you take it from here?
• One question: Does squaring an absolute value add any extraneous solutions in certain circumstances? Or do they never appear since the absolute value is always nonnegative? Jul 8, 2021 at 0:00
• As long as both sides are nonnegative, you can proceed as suggested so that there won't be any extraneous solutions. Jul 8, 2021 at 0:46
We can only "flip the sign" if what is inside the absolute value is negative. And similarly, we "keep the sign", if what is inside the absolute value bracket is positive.
We could write it this way...
$$-3<\frac {x+1}{x-2}<0$$ or $$0\le \frac {x+1}{x-2}<3$$
Or say, the left-hand equation only applies when $$-1
Then we can solve the two sets of inequalities:
$$-1 or $$x > \frac {7}{2}$$ or $$x \le -1$$
Now we can combine the intervals
$$(-\infty, \frac 54)\cup (\frac 72,\infty)$$ | {
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# Why doesn't the derivative of integration by trig substitution match the original function?
For one of my assignments in Calc II, I had to solve ${\displaystyle\int} \frac{\sqrt{x^2-4}}{x} \text{ d}x$. By using Trig substitution where $x = 2 \text{ sec } \theta$, my solution was $\sqrt{x^2-4} - 2\text{ arcsec}\left(\frac{x}{2}\right) + C$. I tested my solution by graphing $\frac{\text{d}}{\text{d}x}\left[\sqrt{x^2-4} - 2\text{ arcsec}\left(\frac{x}{2}\right)\right]$, and it only matched the original function where $x \geq 2$. To match for $x \leq -2$, I had to use $\frac{\text{d}}{\text{d}x}\left[\sqrt{x^2-4} + 2\text{ arcsec}\left(\frac{x}{2}\right)\right]$.
Assuming I solved this correctly, what causes this discrepancy? How can we know if, when, and where this kind of thing will happen? Also, how would we notate the difference? Is it necessary to write the solution as a piecewise function, or does none of this matter anyway? I would think that, if it was because of the square root, the change in sign would be in front of it, but this seems to be because of the arc secant.
Edit Since a comment mentioned using WolframAlpha, I decided to double-check through there, and they do have my same answer under the Step-by-Step Solution, although they then simplified it to arctan somehow. Graphing the derivative of their solution with arctan matches the original function even less, with both a different C and shape. For explicitness, here's my work: $x = 2\sec \theta \Rightarrow \text{d}x = 2 \sec\theta \tan\theta$
$\int \frac{\sqrt{4 \sec^2\theta - 4}}{2 \sec\theta}(2 \sec\theta \tan\theta \text{ d}\theta) = \int 2 \sqrt{\sec^2 \theta-1} \tan\theta \text{ d}\theta = 2 \int \tan^2 \theta \text{ d}\theta$
With trig identity replacement: $2 \int \sec^2 \theta \text{ d}\theta - 2 \int 1 \text{ d}\theta = 2 \tan\theta - 2\theta$
With replacement of $\theta$ back to its original value: $\sqrt{x^2 - 4} - 2\text{ arcsec} \left(\frac{x}{2}\right) + C$ | {
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Edit 2 I understand that the square root causes sign issues, but even switching to absolute value doesn't seem to have any effect in the graphing. Here's a link to everything graphed on desmos.
• I beat you to it, deleting my comment! – Namaste Oct 21 '16 at 20:17
• According to Wolfram, your antiderivative is incorrect. – Ken Duna Oct 21 '16 at 20:17
• I hadn't checked with Wolfram before, but in their Step-By-Step, their answer is mine, but with different work, so I updated my answer in response and showed my work. – BrainFRZ Oct 21 '16 at 20:53
The trouble is when getting rid of the square root. In general we have $\sqrt{x^2} = |x|$, so $$\sqrt{\sec^2\theta-1} = \sqrt{\tan^2\theta} = \lvert\tan\theta\rvert.$$ So the integrand is really $$\lvert\tan\theta\rvert \tan \theta = \begin{cases} \sec^2\theta - 1 & 0\le\theta<\pi/2,\\ -(\sec^2\theta - 1) & \pi/2<\theta<\pi\\ \end{cases}$$ (where you take $\theta = \operatorname{arcsec} (x/2)$ with $0 \le \theta < \pi$). When reversing the substitution there is also the issue of $$\tan\theta = \begin{cases} \sqrt{\sec^2\theta - 1} & 0\le\theta<\pi/2,\\ -\sqrt{\sec^2\theta - 1} & \pi/2<\theta<\pi\\ \end{cases}$$ so the correct integral has two parts: $$\int \frac{\sqrt{x^2-4}}{x}\,dx = \begin{cases} \sqrt{x^2-4} - 2 \operatorname{arcsec}(x/2) + C_1 & x \ge 2, \\ \sqrt{x^2-4} + 2 \operatorname{arcsec}(x/2) + C_2 & x \le -2. \\ \end{cases}$$
• Sorry, I had an error in my post which I just fixed. And generally you will have to deal with things piecewise when sign changes occur (e.g. across $x=0$ in $\sqrt{x^2} = |x|$). – arkeet Oct 21 '16 at 21:25 | {
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# Given $A^2+cA+cI=0$, how to find inverse of $A+(c-1)I$?
Suppose a square matrix $A$ such that $A^2+cA+cI=0$ for all $c \in \mathbb{Z}$. How can I show that $A+(c-1)I$ is invertible and find its inverse?
I started off this way: $A+(c-1)I = A+cI-I$
Then $(A+cI-I)(d_1A+d_2I)=I$, where $d_1, d_2 \in \mathbb{Z}$.
Expand it and it becomes: $d_1A^2+d_1cA-d_1A+d_2A+(c-1)d_2I=I$
$\Rightarrow (c-1)d_2=1 \;\; and \; \; d_1A^2+d_1cA-d_1A+d_2A=0$
$\Rightarrow d_2=\frac{1}{(c-1)}$
Continue from $d_1A^2+d_1cA-d_1A+d_2A=0$, after some manipulation, I got $d_1(A^2+cA+cI)-d_1cI-d_1A+d_2A=0$.
Since given that $A^2+cA+cI=0$,
$d_1(A^2+cA+cI)-d_1cI-d_1A+d_2A=0$
$\Rightarrow -d_1cI-d_1A+d_2A=0$
$\Rightarrow -d_1cI-d_1A+\frac{1}{c-1}A=0$
$\Rightarrow -d_1cI-d_1A=-\frac{1}{c-1}A$
$\Rightarrow d_1(cI+A)=\frac{1}{c-1}A$
At this point, since I don't know if $A$ is invertible yet, I cannot do it as $d_1=\frac{1}{c-1}\frac{A}{(cI+A)}$.
Even if I did this, I still cannot find a value for $d_1$ and $d_2$ to find the inverse of $A+(c-1)I$. How should I continue from here?
-
Did you really mean for all $c \in \mathbb{Z}$? Wouldn't then $A^2 = A+I = 0$, hence $I^2 = 0$? – Niels Diepeveen Sep 25 '11 at 18:23
Instead of all this, rewrite your starting equation as $A^2+(c-1)A+A+(c-1)I=-I$. Then factor the left-hand-side...
But if you favor a more systematic approach you can also proceed as you do until $$d_1A^2+d_1cA-d_1A+d_2A+(c-1)d_2I=I$$ At this point you decide to set $(c-1)d_2=1$, but that is too early! First get rid of the $A^2$ using $A^2=-c(A+I)$, to get $$-d_1c(A+I)+d_1cA - d_1A + d_2A + (c-1)d_2I = I$$ which simplifies to $$(d_1c-d_1+d_2-d_1c)A + ((c-1)d_2-d_1c-1)I = 0$$ $$(d_2-d_1)A + ((c-1)d_2-cd_1-1)I$$ Now set both coefficients to 0. This gives immediately $d_1=d_2$, and we then need to solve $$(c-1)d-cd-1=0$$ in which the $cd$'s cancel and give us $d=-1$. So the sought inverse is $-A-I$. | {
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-
hmm...What happens if I couldn't see this and didn't start from this equation? I remember I had another problem to find inverses and I used a similar method and worked. I had the problem posted here too at math.stackexchange.com/questions/58841/… – xenon Sep 25 '11 at 15:23
I've added a more systematic approach. You have to start from the given relation, though -- otherwise you have no way to know there is even an inverse anywhere. – Henning Makholm Sep 25 '11 at 15:36
Thanks! But how's $A^2=-(A+I)$ and not $A^2=-(cA+cI)$? – xenon Sep 25 '11 at 15:57
At one part, you set both the coefficients to be $0$. Is there a reason for doing this? Because I was thinking why wouldn't it be $((c-1)d_2-cd_1-1)=1$ and $(d_2-d_1)A=-I$? Of course, what you have done by having the coefficients zero makes the whole thing simpler but I was just thinking what if I set it something else other than zero. – xenon Sep 25 '11 at 16:21
It follows from some experience with quotients of polynomial rings. Your original relation can be used to prove that any linear combination of powers of $A$ can be rewritten into the form $pA+qI$, and if that is all we know about $A$, then there is only one such form. So if we have $pA+qI=0$ then either it must be because $p=q=0$ or else $A$ happened to satisfy a nicer property than the one we can depend on. (In your example the nicer property is "is a multiple of $I$"). It's always worth it to try a general method first and try more complex tactices if that fails. – Henning Makholm Sep 25 '11 at 16:31
Perhaps a more transparent way: "change of variables". Let $B = A + (c-1) I$, or $A = B - (c-1) I$. Then $0 = A^2 + c A + c I = (B - (c-1) I)^2 + c (B - (c-1) I) + c I= B^2 + (2-c) B + I$. Multiply by $B^{-1}$ to get $B + (2-c) I + B^{-1} = 0$, i.e. $B^{-1} = -B + (c-2) I$ or $(A + (c-1) I)^{-1} = -A - (c-1) I + (c-2) I = -A - I$. | {
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You may find it a bit dodgy to multiply by $B^{-1}$ before you know that $B^{-1}$ exists, but once you have the result $-A-I$ it's easy to verify that this works by multiplying it by $A + (c-1) I$.
-
$$\begin{eqnarray*} x^2+cx+c = \left[x+(c-1)\right](x+1)+1\\ \therefore 0 = A^2+cA+cI = \left[A+(c-1)I\right](A+I)+I\\ \therefore \left[A+(c-1)I\right](A+I)=-I\\ \therefore \left[A+(c-1)I\right]^{-1} = -(A+I). \end{eqnarray*}$$
-
NEW ANSWER. In the "long" division $$X^2+cX+c=(X+c-1)(X+1)+1,$$ replace $X$ by $A$: $$0=A^2+cA+cI=\Big(A+(c-1)I\Big)\Big(A+I\Big)+I,$$ $$\Big(A+(c-1)I\Big)^{-1}=-A-I.$$
[EDIT. I'm realizing that this answer is the same as user1551's, who posted it before. Sorry...]
OLD ANSWER. Let $A$ be an $n$ by $n$ matrix with coefficients in a field $K$, let $f\in K[X]$ be a polynomial annihilating $A$, and let $g\in K[X]$ be any polynomial.
If $g$ is prime to $f$, then $g(A)$ is invertible, and the inverse of $g(A)$ is given by $h(A)$ where $h\in K[X]$ is an inverse of $g$ mod $f$.
Moreover, there is a closed formula for such an $h$ (see this answer).
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# Which payoff do you want to go for?
Payoff 1.
Toss 5 coins. You get $1 for each consecutive HT that you get. Payoff 2. Toss 5 coins. You get$1 for each consecutive HH that you get.
For example, if you tossed HTHHH, under payoff 1 you will get $1, but under payoff 2 you will get$2.
Note by Calvin Lin
3 years, 2 months ago
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This is a trick question, both payoffs have same expected value of profit, which is: $1. Consider this: We will count the number of ways HT can occur. We can have: HT_ _ _ : 8 ways _ HT _ _ _ : 8 ways _ _ HT _ : 8 ways _ _ _ HT : 8 ways Total number of ways: 32 Similarly for HH, the total number of ways in which it can occur is 32. What this means is that the sum of the payoffs of HT and HH over the 32 possible outcomes is$32 for both. The payoffs may be distributed in different ways for HT and HH, but their expected value is same. | {
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Mathematically, the expected value is same. But there is one more thing left to do: analyze the probability distribution. We can easily see that payoff 2 might be desirable because it tends to give large profits as compared to payoff 1 for certain events. eg: HHHHH gives $4 in p2 and 0 in p1. HHHHT gives$3 in p2 and $1 in p1. But the thing is that these events are not very common. Using method of reflection and analyzing the 16 possibilities , we see that there are many events for which p2 gives$0 payoff. There are 13 outcomes for which p2 gives $0 payoff, but only 6 where p1 gives$0 payoff.
So If I'm given this choice, I'd take payoff 1, as then I have more chance of leaving with at least a dollar in my pocket.
Since this is in quantitative finance, should I talk about risk and stuff? I do not know.
- 3 years, 2 months ago | {
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Yes, the expected payoff for both scenarios is $1. The easiest way to see this is to look at Indicator Variables, which is essentially what Raghav did (though not phrased in that language). Well, everyone has their own "payoff preference". For example, you stated that you would prefer to "be more likely to leave with at least a dollar in your pocket". As such, this tells me that you are very risk adverse, and that you would prefer the certainty of a positive payoff. If you want to talk about "risk and stuff", what do you need to consider, and why? Staff - 3 years, 2 months ago Log in to reply I am not formally educated in these topics, so I'm just saying all this from a logical standpoint. After a bit of thinking, I think I agree with you on the fact that everyone has their own "payoff preference". One can take a risk to win more, or take less risk to win less. In the scenario mentioned here, the aspects of risk are not conspicuous. The amount to be won is not significant, and there is no penalty for losing. Hence, both the payoffs are inherently attractive offers. When we come to real life situations, I think there are many more things that contribute to the risk: 1. The probability that you will get a payoff. 2. The probability that you stand to lose money. 3. The security of the winnings. Money once won shouldn't be taken away from you. 4. The effects of said decisions on long term/ on your ability to take other decisions. According to me, every decision of ours is based on weighing in these risk factors against the probable prize. The balance may tip either way, and so may our decisions. I don't even know if I'm thinking in the right direction... Help me out here @Calvin Lin - 3 years, 2 months ago Log in to reply Part of this series of posts is for you to figure out what kind of risk appetite you have. It is a personal preference, and there isn't necessarily a "right" answer. Your understanding will help guide you in decisions that you make in future. For the | {
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"right" answer. Your understanding will help guide you in decisions that you make in future. For the points that you raised, you should answer for yourself if those should matter, and why. Hopefully, you will come up with a consistent, logical risk-reward structure. For example, if you would choose payoff 1 over 2 and payoff 3 over 4, but would prefer a combined 2 and 4 over a combined 1 and 3, then it would be very easy for someone to sell you things in part, and make you overpay for them. Staff - 3 years, 2 months ago Log in to reply I think that there may be some double-counting here. For example, you've counted HTHTT and HTHTH twice each. There are only $$2^{5} = 32$$ possible sequences in total, and there are some, such as TTHHH, that have no HT payouts, so there cannot be $$32$$ HT outcomes. I'm getting an average payout of $$\dfrac{28}{32}$$ dollars in the first game type and $$\dfrac{31}{32}$$ dollars in the second, but I am too tired to double-check. I'll do that tomorrow. :) Edit: Oh. now I see what you've done; very clever. I must have missed out on my count somewhere. I'll still wait to confirm in the morning. And yes, there does seem to be an advantage to choosing p1, (even though the expected winnings are the same), in the sense that you are more likely to at least win some money. - 3 years, 2 months ago Log in to reply Is "more likely to at least win some money" the main consideration that you will use if the expected payoff is equal? For example,what would you choose between: Payoff 1: 50% of$0, 50% of $100 Payoff 2: 90% of$1.01, 1% of $4900 Staff - 3 years, 2 months ago Log in to reply If Payoff 2 was 99% of$40.40, 1% of $1000, (to give it the same expected payoff as Payoff 1), I would go with Payoff 2 since I would be assured of money and would at least have a slim chance of a large amount of money. If Payoff 2 was 99% of$0, 1% of $5000, then it gets interesting, (at least for me). Even a such a slight chance of winning$5000 would be hard to pass | {
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interesting, (at least for me). Even a such a slight chance of winning$5000 would be hard to pass up, so I would probably go with Payoff 2, unless I absolutely needed that $100 right away. If Payoff 2 was 99.9% of 0$, 0.1% of$50000 .... hmmmm .... That's a lot of dough, so I'd go with Payoff 2. However, if the options were these last two I've listed, then I might choose the$5000 option. | {
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