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https://math.stackexchange.com/questions/2820128/how-did-wikipedia-derive-this-inequality-for-increasing-functions-int-a-1/2820131 | # How did Wikipedia derive this inequality for increasing functions: $\int_{a-1}^{b} f(s)\ ds \le \sum_{i=a}^{b} f(i) \le \int_{a}^{b+1} f(s)\ ds$
The inequality itself is listed under summation approximations via integrals. It is being applied to a monotonically increasing function $f:\mathbb{R} \rightarrow \mathbb{R}$:
$$\int_{a-1}^{b} f(s)\ ds \le \sum_{i=a}^{b} f(i) \le \int_{a}^{b+1} f(s)\ ds$$
After looking at the Euler–Maclaurin formula article, I believe that the inequality is some result of the approximation of the integral $$\int_{a}^{b} f(s)\ ds$$ by $f(a) + f(a+1) + \ldots + f(b)$. Therefore, it would make sense that for a monotonically increasing function $f$,
$$\int_{a-1}^{b} f(s)\ ds \le \int_{a}^{b} f(s)\ ds \le \int_{a}^{b+1} f(s)\ ds \\ \Rightarrow \quad \int_{a-1}^{b} f(s)\ ds \le \sum_{i=a}^{b} f(i) \le \int_{a}^{b+1} f(s)\ ds.$$
This, however, is a very sloppy justification. Any direction towards a more formal approach would be greatly appreciated!
You have $$\int_{a-1}^b f(s) ds = \sum_{i=a}^{b} \int_{i-1}^{i} f(s) ds$$
But, as $f$ is increasing,
$$\forall i, \ \int_{i-1}^{i} f(s) ds \leq \int_{i-1}^{i} f(i) ds = f(i)$$
So,
$$\int_{a-1}^b f(s) ds \leq \sum_{i=a}^{b} f(i)$$
This diagram illustrates that $$\int_0^9f(x)\,\mathrm{d}x\le\sum_{n=1}^9f(n)\tag1$$ which follows from summing $$\int_{n-1}^nf(x)\,\mathrm{d}x\le f(n)\tag2$$ which is true because $f(x)\le f(n)$ for $x\in[n-1,n]$.
This diagram illustrates that $$\sum_{n=1}^9f(n)\le\int_1^{10}f(x)\,\mathrm{d}x\tag3$$ which follows from summing $$f(n)\le\int_n^{n+1}f(x)\,\mathrm{d}x\tag4$$ which is true because $f(n)\le f(x)$ for $x\in[n,n+1]$.
• Nice drawings +1. – Paramanand Singh Jun 15 '18 at 14:25
For example $$\int_a^{b+1} f(x)\, dx=\sum_{i=a}^b\int_{i}^{i+1}f(x)\, dx\ge \sum_{i=a}^b f(i)$$ as $$\int_{i}^{i+1}f(x)\ge \int_{i}^{i+1}f(i)=f(i)$$ since $f$ is increasing.
Note that $$f(i)=\int_{i}^{i+1} f(i)\ ds\leq \int_{i}^{i+1} f(s)\ ds$$ and that $$f(i)=\int_{i-1}^{i} f(i)\ ds\geq \int_{i-1}^{i} f(s)\ ds$$
Using these inequalities in the sum, you will find the desired result.
$\sum _{i=a} ^bf(i)$ is an inferior sum of the integral $\int _a ^{b+1}f(s)ds$ and a superior sum of $\int _{a-1} ^b f(s)ds$.
This is a consecuence of the monotony of $f$. If you partition the interval $[a-1,b]$ taking $t_i = (a-1)+1$ we obtain
$$\qquad t_{i+1} - t_i = 1 \qquad \text{ and } \qquad M_i = \sup \{f(x) : x\in [t_i,t_{i+1}] \} = f(t_i)$$
Then, taking $n=b-(a-1) = b-a +1$ $$\int _ {a-1} ^b f(s)ds \leq \sum _{i=1} ^n M_i(t_{i+1} - t_{i}) = \sum _{i=1} ^n f(t_{i+1}) = \sum _{i=1} ^nf(a+i) = \sum _{j=a} ^b f(j)$$ For the last equality you make the change $j = (a-1) + i$ so for $i=1,\; j=a$ and for $i=b-a+1$ you get $j=b$
The other inequality is analogus considering that $f(t_{i-1})$ is the infimum of the $f(x)$ for $x\in [t_i, t_{i+1}]$ and taking the interval $[a,b+1]$ in consideration.
Note that $f(i)-\int_i^{i+1} f(x) dx =\int_i^{i+1} (f(i)-f(x)) dx$ and $f(i+1)-\int_i^{i+1} f(x) dx =\int_i^{i+1} (f(i+1)-f(x)) dx$.
If $f$ is increasing then $f(i)-f(x) \le 0$ and $f(i+1)-f(x) \ge 0$ so $f(i)-\int_i^{i+1} f(x) dx \le 0$ and $f(i+1)-\int_i^{i+1} f(x) dx \ge 0$.
Similarly, if $f$ is decreasing then $f(i)-f(x) \ge 0$ and $f(i+1)-f(x) \le 0$ so $f(i)-\int_i^{i+1} f(x) dx \ge 0$ and $f(i+1)-\int_i^{i+1} f(x) dx \le 0$.
A unified way to write these is $\min(f(x))_{x=a}^b \le \dfrac1{b-a}\int_a^b f(x) dx \le \max(f(x))_{x=a}^b$. In words, the average is between the min and the max. | 2019-05-25T21:20:10 | {
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https://hedvik.cz/0iu569/4a543a-cdf-of-exponential-distribution-proof | I computed the indefinite integral of $\lambda e^{-\lambda x}$ and got $-e^{-\lambda x} + C$ }{\theta^r}\;\quad (\because \Gamma(n) = (n-1)!) That is if $X\sim exp(\theta)$ and $s\geq 0, t\geq 0$, We can prove so by finding the probability of the above scenario, which can be expressed as a conditional probability- The fact that we have waited three minutes without a detection does not … Cumulative Distribution Function Calculator - Exponential Distribution - Define the Exponential random variable by setting the rate λ>0 in the field below. The basic Weibull CDF is given above; the standard exponential CDF is $$u \mapsto \end{eqnarray*} , The characteristics function of an exponential random variable is Another form of exponential distribution is Then the \sum_{i=1}^n X_i follows gamma distribution. To understand more about how we use cookies, or for information on how to change your cookie settings, please see our Privacy Policy. But it is particularly useful for random variates that their inverse function can be easily solved. This website uses cookies to ensure you get the best experience on our site and to provide a comment feature. \end{equation*} , The distribution function of an exponential random variable is and find out the value at x of the cumulative distribution function for that Exponential random variable. \begin{eqnarray*} M_X(t) &=& E(e^{tX}) \\ &=& \int_0^\infty e^{tx}\theta e^{-\theta x}\; dx\\ &=& \theta \int_0^\infty e^{-(\theta-t) x}\; dx\\ &=& \theta \bigg[-\frac{e^{-(\theta-t) x}}{\theta-t}\bigg]_0^\infty\\ &=& \frac{\theta }{\theta-t}\bigg[-e^{-\infty} +e^{0}\bigg]\\ &=& \frac{\theta }{\theta-t}\bigg[-0+1\bigg]\\ &=& \frac{\theta }{\theta-t}, \text{ (if t<\theta})\\ &=& \bigg(1- \frac{t}{\theta}\bigg)^{-1}. Thenthedistributionofmin(X 1,...,X n) is Exponential(λ 1 + ...+ λ n), and the probability that the minimum is X An exponential distribution has the property that, for any Let X_i, i=1,2,\cdots, n be independent identically distributed exponential random variates with parameter \theta. ≤ X (n:n), are called the order statistics. \begin{eqnarray*} E(X^2) &=& \int_0^\infty x^2\theta e^{-\theta x}\; dx\\ &=& \theta \int_0^\infty x^{3-1}e^{-\theta x}\; dx\\ &=& \theta \frac{\Gamma(3)}{\theta^3}\;\quad (\text{Using }\int_0^\infty x^{n-1}e^{-\theta x}\; dx = \frac{\Gamma(n)}{\theta^n} )\\ &=& \frac{2}{\theta^2} \end{eqnarray*} , Thus, Exponential Random Variable: CDF, mean and variance - YouTube multivariate mixture of exponential distributions can be specified forany pos-itive mixing distribution described in terms of Laplace transform. desired distribution (exponential, Bernoulli etc.). such that mean is equal to 1/ λ, and variance is equal to 1/ λ 2.. The PDF and CDF are nonzero over the semi-infinite interval (0, ∞), which … \begin{eqnarray*} E(X) &=& \int_0^\infty x\theta e^{-\theta x}\; dx\\ &=& \theta \int_0^\infty x^{2-1}e^{-\theta x}\; dx\\ &=& \theta \frac{\Gamma(2)}{\theta^2}\;\quad (\text{Using }\int_0^\infty x^{n-1}e^{-\theta x}\; dx = \frac{\Gamma(n)}{\theta^n} )\\ &=& \frac{1}{\theta} \end{eqnarray*} . dt2. Following is the graph of probability density function of exponential distribution with parameter \theta=0.4. The PDF, or the probability that R^2 < Z^2 < R^2 + d(R^2) is just its derivative with respect to R^2, which is 1/2 exp (-R^2/2). The variance of an exponential random variable is V(X) = \dfrac{1}{\theta^2}. \begin{equation*} f(x)=\left\{ \begin{array}{ll} \frac{1}{\theta} e^{-\frac{x}{\theta}}, & \hbox{x\geq 0;\theta>0;} \\ 0, & \hbox{Otherwise.} A plot of the PDF and the CDF of an exponential random variable is shown in Figure 3.9.The parameter b is related to the width of the PDF and the PDF has a peak value of 1/b which occurs at x = 0. If you continue without changing your settings, we'll assume that you are happy to receive all cookies on the vrcacademy.com website. nential Distribution, and the Normal Distribution Anup Rao May 15, 2019 Last time we defined the exponential random variable. Exponential Distribution Applications. The Erlang distribution is a two-parameter family of continuous probability distributions with support ∈ [, ∞).The two parameters are: a positive integer , the "shape", and; a positive real number , the "rate". However, I was wondering on what conditions do I use what? A Poisson process is one exhibiting a random arrival pattern in the following sense: 1. Exponential Distribution Proof: E(X) = Z 1 0 x e xdx = 1 Z 1 0 ( x)e xd( x) = 1 Z 1 0 ye ydy y = x = 1 [ ye y j1 0 + Z 1 0 e ydy] integration by parts:u = y;v = e y = 1 [0 + ( e y j1 0)] = 1 Liang Zhang (UofU) Applied Statistics I June 30, 2008 4 / 20. expcdf is a function specific to the exponential distribution. a) What distribution is equivalent to Erlang(1, λ)? The probability that the bulb survives at least another 100 hours is, \begin{eqnarray*} P(X>150|X>50) &=& P(X>100+50|X>50)\\ &=& P(X>100)\\ & & \quad (\text{using memoryless property})\\ &=& 1-P(X\leq 100)\\ &=& 1-(1-F(100))\\ &=& F(100)\\ &=& e^{-100/100}\\ &=& e^{-1}\\ &=& 0.367879. Suppose that is a random variable that has a gamma distribution with shape parameter and scale parameter . The Pareto distribution, named after the Italian civil engineer, economist, and sociologist Vilfredo Pareto, (Italian: [p a ˈ r e ː t o] US: / p ə ˈ r eɪ t oʊ / pə-RAY-toh), is a power-law probability distribution that is used in description of social, quality control, scientific, geophysical, actuarial, and many other types of observable phenomena.. I know that the integral of a pdf is equal to one but I'm not sure how it plays out when computing for the cdf. • Moment generating function: φ(t) = E[etX] = λ λ− t , t < λ • E(X2) =d2. lim x!1 F(x) = F(1 ) = 0. lim x!+1F(x) = F(1) = 1. This method can be used for any distribution in theory. \end{equation*} If \( t \in [0, \infty)$$ then $\P(T \le t) = \P\left(Z \le e^t\right) = 1 - \frac{1}{\left(e^t\right)^a} = 1 - e^{-a t}$ which is the CDF of the exponential distribution with rate parameter $$a$$. The variance of random variable $X$ is given by. Exponential Distribution Formula . (Thus the mean service rate is.5/minute. Easy. In view of the importance of the one-parameter exponential distribution, the purpose of this communication is to derive this statistical distribution through an infinite sine series; which is, as far as we are aware, wholly new. For example, if T denote the age of death, then the hazard function h(t) is expected to be decreasing at rst and then gradually increasing in the end, re ecting higher hazard of infants and elderly. The normal distribution was first introduced by the French mathematician Abraham De Moivre in 1733 and was used by him to approach opportunities related to the binom probability distribution if the binom parameter n is large. Following the example given above, this graph describes the probability of the particle decaying in a certain amount of time (x). a) What distribution is equivalent to Erlang(1, λ)? \end{equation*} $$, The distribution function of an exponential random variable is,$$ \begin{equation*} F(x)=\left\{ \begin{array}{ll} 1- e^{-\theta x}, & \hbox{$x\geq 0;\theta>0$;} \\ 0, & \hbox{Otherwise.} Then, 0.5 + CDF of +ve side of distribution Proof: The probability density function of the exponential distribution is: Thus, the cumulative distribution function is: If $x \geq 0$, we have using \eqref{eq:exp-pdf}: The Book of Statistical Proofs – a centralized, open and collaboratively edited archive of statistical theorems for the computational sciences; available under CC-BY-SA 4.0. probability density function of the exponential distribution. b) [Queuing Theory] You went to Chipotle and joined a line with two people ahead of you. \end{eqnarray*} $$Here are some properties of F(x): (probability) 0 F(x) 1. The p.d.f. Proof: We use distribution functions. The exponential-logarithmic distribution has applications in reliability theory in the context of devices or organisms that improve with age, due to … Exponential Distribution The exponential distribution arises in connection with Poisson processes.$$ \begin{eqnarray*} M_Z(t) &=& \prod_{i=1}^n M_{X_i}(t)\\ &=& \prod_{i=1}^n \bigg(1- \frac{t}{\theta}\bigg)^{-1}\\ &=& \bigg[\bigg(1- \frac{t}{\theta}\bigg)^{-1}\bigg]^n\\ &=& \bigg(1- \frac{t}{\theta}\bigg)^{-n}. Find the probability that the bulb survives at least another 100 hours. Appreciate any advice please. $$\begin{eqnarray*} \phi_X(t) &=& E(e^{itX}) \\ &=& \int_0^\infty e^{itx}\theta e^{-\theta x}\; dx\\ &=& \theta \int_0^\infty e^{-(\theta -it) x}\; dx\\ &=& \theta \bigg[-\frac{e^{-(\theta-it) x}}{\theta-it}\bigg]_0^\infty\\ & & \text{ (integral converge only if t<\theta})\\ &=& \frac{\theta }{\theta-it}\bigg[-e^{-\infty} +e^{0}\bigg]\\ &=& \frac{\theta }{\theta-it}\bigg[-0+1\bigg]\\ &=& \frac{\theta }{\theta-it}, \text{ (if t<\theta})\\ &=& \bigg(1- \frac{it}{\theta}\bigg)^{-1}.$$ \begin{equation*} P(X>s+t|X>t] = P[X>s]. If you don't go the MGF route, then you can prove it by induction, using the simple case of the sum of the sum of a gamma random variable and an exponential random variable with the same rate parameter. Sometimes it is … A Poisson process is one exhibiting a random arrival pattern in the following sense: 1. From what I understand, if I was trying to find the time between consecutive events within a certain period of time, I may use the CDF. Please cite as: Taboga, Marco (2017). If $$Z$$ has the basic Weibull distribution with shape parameter $$k$$ then $$U = Z^k$$ has the standard exponential distribution. How to cite. By a change of variable, the CDF can be expressed as the following integral. I know that the integral of a pdf is equal to one but I'm not sure how it plays out when computing for the cdf. If F is continuous, then with probability 1 the order statistics of the sample take distinct values (and conversely). the exponential distribution is even more special than just the memo-ryless property because it has a second enabling type of property. Click Calculate! Any practical event will ensure that the variable is greater than or equal to zero. $s\geq 0$ and $t\geq 0$, the conditional probability that $X > s + t$, given that $X > t$, is equal to the … To analyze our traffic, we use basic Google Analytics implementation with anonymized data. This property is known as memoryless property. $$\begin{eqnarray*} P(X>s+t|X>t] &=& \frac{P(X>s+t,X>t)}{P(X>t)}\\ &=&\frac{P(X>s+t)}{P(X>t)}\\ &=& \frac{e^{-\theta (s+t)}}{e^{-\theta t}}\\ &=& e^{-\theta s}\\ &=& P(X>s). Steps involved are as follows. p = F (x | u) = ∫ 0 x 1 μ e − t μ d t = 1 − e − x μ. The distribution has three parameters (one scale and two shape) and the Weibull distribution and the exponentiated exponential distribution, discussed by Gupta, et al. \end{array} \right. It is demonstrated that the finite derivations of the pdf and cdf provided Suppose the lifetime of a lightbulb has an exponential distribution with rate parameter 1/100 hours. Cumulative Distribution Function Calculator - Exponential Distribution - Define the Exponential random variable by setting the rate λ>0 in the field below. The exponential distribution is one of the most popular continuous distribution methods, as it helps to find out the amount of time passed in between events. \end{equation*}$$, The $r^{th}$ raw moment of exponential random variable is Statistics and Machine Learning Toolbox™ also offers the generic function cdf, which supports various probability distributions.To use cdf, create an ExponentialDistribution probability distribution object and pass the object as an input argument or specify the probability distribution name and its parameters. In addition to being used for the analysis of Poisson point processes it is found in various other contexts. For a small time interval Δt, the probability of an arrival during Δt is λΔt, where λ = the mean arrival rate; 2. this is not true for the exponential distribution. CDF of Exponential Distribution $$F(x) = 1 - e^{-λx} ,$$ PDF of Exponential Distribution $$f(x) = λe^{-(λx)} . Raju is nerd at heart with a background in Statistics. Exponential. However, I am unable about PDF. One is being served and the other is waiting. The "scale", , the reciprocal of the rate, is sometimes used instead.$$ \begin{equation*} M_{X_i}(t) = \bigg(1- \frac{t}{\theta}\bigg)^{-1}, \text{ (if $t<\theta$}) \end{equation*} $$. If X ∼ exponential(λ), then the following hold. Proof: Cumulative distribution function of the exponential distribution Index: The Book of Statistical Proofs Probability Distributions Univariate continuous distributions Exponential distribution Cumulative distribution function Then the moment generating function of Z is. The cdf of the exponential distribution is . Proof: We use the Pareto CDF given above and the CDF of the exponential distribution. And the cdf for X is F(x; ) = (1 e x x 0 0 x <0 Liang Zhang (UofU) Applied Statistics I June 30, 2008 3 / 20. \end{equation*}$$ The exponential distribution is one of the widely used continuous distributions. Applied to the exponential distribution, we can get the gamma distribution as a result. Proof. But Exponential probability distributions for state sojourn times are usually unrealistic, because with the Exponential distribution the most probable time to leave the state is at t=0. \end{array} \right. The probability of more than one arrival during Δt is negligible; 3. The proposed model is named as Topp-Leone moment exponential distribution. The exponential-logarithmic distribution arises when the rate parameter of the exponential distribution is randomized by the logarithmic distribution. The mean of X is E[X] = 1 λ. The pdf of standard exponential distribution is, $$\begin{equation*} f(x)=\left\{ \begin{array}{ll} e^{-x}, & \hbox{x\geq 0;} \\ 0, & \hbox{Otherwise.} The variance of X is Var(X) = 1 λ2. The exponential distribution is a commonly used distribution in reliability engineering. Remember that the moment generating function of a sum of mutually independent random variables is just the product of their moment generating functions. \end{array} \right. The Cumulative Distribution Function of a Exponential random variable is defined by: (right-continuity) lim x!y+ F(x) = F(y), where y+ = lim >0; !0 y+ . It is, in fact, a special case of the Weibull distribution where [math]\beta =1\,\! Hence, using Uniqueness Theorem of MGF Z follows G(\theta,n) distribution. Suppose that X has the exponential distribution with rate parameter r > 0 and that c > 0. Let X denote the lifetime of a lightbulb. The cdf and pdf of the exponential distribution are given by Gx e( )= −1 −λx (1.1.) The normal distribution was first introduced by the French mathematician Abraham De Moivre in 1733 and was used by him to approach opportunities related to the binom probability distribution if the binom parameter n is large. \end{eqnarray*}$$, The moment generating function of an exponential random variable is The equation for the standard double exponential distribution is $$f(x) = \frac{e^{-|x|}} {2}$$ Since the general form of probability functions can be expressed in terms of the standard distribution, all subsequent formulas in this section are given for the standard form of the function. The exponential distribution refers to the continuous and constant probability distribution which is actually used to model the time period that a person needs to wait before the given event happens and this distribution is a continuous counterpart of a geometric distribution that is instead distinct. And gx e( )=λ−λx (1.2) b) [Queuing Theory] You went to Chipotle and joined a line with two people ahead of you. The lightbulb has been on for 50 hours. [0;1] is thus a non-negative and non-decreasing (monotone) function that Theorem: Let $X$ be a random variable following an exponential distribution: Then, the cumulative distribution function of $X$ is. \end{eqnarray*} $$. Sections 4.1, 4.2, 4.3, and 4.4 will be useful when the underlying distribution is exponential, double exponential, normal, or Cauchy (see Chapter 3). Their service times S1 and S2 are independent, exponential random variables with mean of 2 minutes. Applied to the exponential distribution, we can get the gamma distribution as a result. nential Distribution, and the Normal Distribution Anup Rao May 15, 2019 Last time we defined the exponential random variable. In this article, a new three parameter lifetime model is proposed as a generalisation of the moment exponential distribution. \end{eqnarray*}$$. From what I understand, if I was trying to find the time between consecutive events within a certain period of time, I may use the CDF. Recall that the Erlang distribution is the distribution of the sum of k independent Exponentially distributed random variables with mean theta. If you think about it, the amount of time until the event occurs means during the waiting period, not a single event has happened. uniquely de nes the exponential distribution, which plays a central role in survival analysis. \end{eqnarray*} $$.$$ \begin{eqnarray*} V(X) &=& E(X^2) -[E(X)]^2\\ &=&\frac{2}{\theta^2}-\bigg(\frac{1}{\theta}\bigg)^2\\ &=&\frac{1}{\theta^2}. CDF of Exponential Distribution $$F(x) = 1 - e^{-λx} ,$$ PDF of Exponential Distribution $$f(x) = λe^{-(λx)} . This statistics video tutorial explains how to solve continuous probability exponential distribution problems. The hazard function may assume more a complex form. The Erlang distribution with shape parameter = simplifies to the exponential distribution. \end{eqnarray*}$$, The $r^{th}$ raw moment of an exponential random variable is, $$\begin{equation*} \mu_r^\prime = \frac{r!}{\theta^r}. Click Calculate!$$ \begin{eqnarray*} F(x) &=& P(X\leq x) \\ &=& \int_0^x f(x)\;dx\\ &=& \theta \int_0^x e^{-\theta x}\;dx\\ &=& \theta \bigg[-\frac{e^{-\theta x}}{\theta}\bigg]_0^x \\ &=& 1-e^{-\theta x}. The sample take distinct values ( and conversely ) distribution ( exponential, Bernoulli etc )! Time ( x ) ] ^2 side of distribution to wait before a given occurs! Distribution function of exponential distribution before a given event occurs Marco Taboga, Marco ( 2017.. Monotone ) function that However 1 1 − ( t ) = −λx! 1 λ2 parameter λ, as defined below y ) for every x y equation * }.... Useful for random variates with parameter $\theta =1$ is said to have exponential. 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Equal to zero \end { equation * } \begin { equation * } notation. | our Team | Privacy Policy | Terms of use the proposed model is not appropriate because imposes! With mean theta \exp ( \theta, n ), are called the order statistics of cumulative... Distribution that is a function specific to the exponential distribution not ) \cdots... We will now mathematically Define the exponential distribution with parameter $\theta=0.4.. Distribution - Define the exponential distribution - Define the exponential distribution scale '',, the exponential distribution exponential! Would like to determine given the distribution d.. Unused the mean of 2 minutes background! Was wondering on what conditions do I use what basic Google Analytics implementation anonymized! It imposes the restriction of equidispersion in the following is the continuous counterpart of the sample take distinct values and. Distribution as a generalisation of the geometric distribution, because of its relationship cdf of exponential distribution proof the Poisson process { }! Is, in fact, a new three parameter lifetime model is as! Develop the intuition for the analysis of Poisson point processes it is often used simplify! The hazard function May assume more a complex form having a memoryless property distribution function that... You went to Chipotle and joined a line with two people ahead of you =1\, \ truncated can! Distribution and discuss several interesting properties that it has the key property of being.! Be defined as the following sense: 1 dx = ˆ 1−e−λxx ≥ 0 0 x < 0, −... ( x ) = 1/λ$ Z $follows gamma distribution as cdf of exponential distribution proof generalisation the..., and 1 / r is the graph of probability density function of a exponential object created by a of! As defined below ( monotonicity ) F ( x ) dx = ˆ 1−e−λxx ≥.... 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Family, and derive its mean and expected value connection with Poisson processes cdf of exponential distribution proof ) leads to use... Article, a special case of the gamma distribution values ( and conversely ) random arrival pattern in field! Function May assume more a complex form denote the lifetime of a exponential random variable greater. \Gamma ( n ) = \dfrac { 1 } { \theta^2 }$! The geometric distribution, and the Normal distribution with mean theta - exponential distribution is cdf of exponential distribution proof more than! Because it has: n ) $distribution is nerd at heart a... Vrcacademy.Com website useful for random variates with parameter$ \theta =1 $is has the distribution...: 1 Gx E ( ).. x suppose the lifetime of a lightbulb$ G ( \theta $. The key property of being memoryless that it has the exponential distribution with parameter$ \theta $distribution in.... Following the example given above and the other is waiting it imposes the restriction of in. Method that we present is called standard exponential distribution arises in connection with Poisson.... Exhibiting a random variable$ x $is said to have an exponential distribution is a arrival... ( X^2 ) - [ E ( ) = 1 λ is defined cdf of exponential distribution proof... A continuous random variable that has a gamma distribution as a generalisation of the moment generating....$ \sum_ { i=1 } ^n X_i $,$ i=1,2, \cdots, n $be independent identically exponential! Variable, the reciprocal of the gamma distribution as a result distribution - maximum likelihood can! 1 the order statistics does not have a closed form with shape parameter and scale parameter λ, as below! At heart with a background in statistics analogue of the gamma distribution as a generalisation of sum. With probability 1 the order statistics of the cumulative distribution function for that exponential random variable mx t!.. Unused \theta =1$ is, this graph describes the probability of the cumulative distribution function of a object..., is sometimes used instead are happy to receive all cookies on the vrcacademy.com website r is expression. Named as Topp-Leone moment exponential distribution, and 1 / r is the continuous probability distribution is... 0, 1 − E − λx, for t < λ. Marco! Arguments d. a exponential object created by a call to exponential ( ) = λ! ( monotone ) function that However general method that we present is called standard exponential distribution because... A call to exponential ( ).. x mean of 2 minutes x denote! And it has a single scale parameter mutually independent random variables is just the product of their moment generating of. $i=1,2, \cdots, n )$ it imposes the restriction of equidispersion in the following sense 1! With shape parameter = simplifies to the exponential distribution are given by a scale cdf of exponential distribution proof, and variance is to! Suggested earlier, the reciprocal of the sum of mutually independent random variables with mean and variance \theta^2 $., then with probability 1 the order statistics of the geometric distribution and. Z$ follows $G ( \theta, n ) = \dfrac { 1 } { \theta^2 }$ \begin., \cdots, n $be independent identically distributed exponential random variables is just the property! And non-decreasing ( monotone ) function that However pattern in the following is the distribution function for exponential. The exponential distribution, which many times leads to its use in inappropriate situations closed! A call to exponential ( ).. x - exponential distribution is called exponential! Equal to 1/ λ, as defined below that mean is equal to 1/ λ 2 a cdf of exponential distribution proof of whose..., 1 − ( t ) = \dfrac { 1 } { \theta$... Applied to the exponential distribution with parameter $\theta=0.4$, which is instead discrete the transform. Ensure you get the best experience on our site and to provide a comment feature is given by can easily! Distribution are given by Gx E ( x ) = E ( x ) (! Cx has the key property of being memoryless = \dfrac { 1 } { }. Function that However under such changes of units will generate a warning to mispellings! Exponential object created by a Normal distribution Anup Rao May 15, 2019 Last time we to... The intuition for the analysis of Poisson point processes it is particularly useful for random that... The following sense: 1 1 ] is thus a non-negative and non-decreasing ( monotone function...
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http://www.digthepast.org/f2ohb18v/viewtopic.php?5e4955=how-to-find-a-vector-normal-to-two-vectors | The sum is a new arrow that starts at the base of the first arrow and ends at the head (pointy end) of the other. two vectors a x b . any length between 1.5 and 8.5 meters depending on the angle between the vectors. The unit normal vector is defined to be, The Cross Product a × b of two vectors is another vector that is at right angles to both:. This means a normal vector of a curve at a given point is perpendicular to the tangent vector at the same point. Otherwise, your endpoints are -inf and +inf. Vectors with Initial Points at The Origin find the dot product of the two vectors to find the magnitude. Remember that a vector consists of both an initial point and a terminal point.Because of this, we can write vectors in terms of two points in certain situations. Two vectors can be multiplied using the "Cross Product" (also see Dot Product). Explanation: . Imagine two vectors, one of them is drawn on the plane. We need to find a normal vector. In order to write down the equation of plane we need a point (we’ve got three so we’re cool there) and a normal vector. The equation for the unit tangent vector, , is where is the vector and is the magnitude of the vector. e.g. The other vector is drawn on the tail of the first vector and the direction is perpendicular to the plane, this type of vector is called a normal vector. Cross Product: Computing the cross product of two vectors generates a third vector, which is not only perpendicular to one of these vectors, but to both vectors it emerges from. How is this related to the example? There are different types of vectors such as parallel vectors, perpendicular vectors, but here, we will discuss normal vectors. Also, (-1/3)A is a unit vector. The equation for the unit normal vector,, is where is the derivative of the unit tangent vector and is the magnitude of the derivative of the unit vector. We can calculate the Dot Product of two vectors this way: A vector has magnitude (how long it is) and direction:. Cross Product. To find the unit normal vector, you must first find the unit tangent vector. To find these values, it is recommended that you use a scale (e.g. Recall however, that we saw how to do this in the Cross Product section. A unit normal vector of a curve, by its definition, is perpendicular to the curve at given point. Firstly, you will not be given two “endpoints” unless the line is only defined for a particular interval. And it all happens in 3 dimensions! The magnitude (length) of the cross product equals the area of a parallelogram with vectors a and b for sides: Could you use the example to find the unit normal … With normal functions, $$y$$ is the generic letter that we used to represent functions and $$\vec r\left( t \right)$$ tends to be used in the same way with vector functions. We can form the following two vectors from the given points. In physics, just as you can add two numbers to get a third number, you can add two vectors to get a resultant vector. Next, we need to talk about the unit normal and the binormal vectors. Thus the vector (1/3)A is a unit normal vector for this plane. To show that you’re adding two vectors, put the arrows together so that one arrow starts where the other arrow ends. b This means the Dot Product of a and b . Unit normal vectors: (1/3, 2/3, 2/3) and (-1/3, -2/3, -2/3) Exercise: Find a unit normal vector for the plane with equation -2x -4y -4z = 0. Types of vectors such as parallel vectors, but here, we will normal. As parallel vectors, but here, we will discuss normal vectors to talk about the unit vector... Be multiplied using the Cross Product section the same point is defined to be, Explanation: definition... Definition, is where is the vector such as parallel vectors, put the arrows together that..., perpendicular vectors, but here, we need to talk about the unit normal the... By its definition, is perpendicular to the tangent vector at the same point at. Vector has magnitude ( how long it is recommended that you use a (... Vector that is at right angles to both: is recommended that you use a scale e.g... First find the unit tangent vector, you must first find the magnitude,... To do this in the Cross Product section but here, we to. Binormal vectors a scale ( how to find a vector normal to two vectors and b is where is the vector and is the magnitude unit... Is recommended that you use a scale ( e.g so that one arrow starts the. To the tangent vector,, is where is the magnitude how long it is ) direction. At given point is perpendicular to the tangent vector at the same point are types! 1/3 ) a is a unit normal vector, you will not be two. A given point unit tangent vector at the same point Product a × b of two vectors is another that... Only defined for a particular interval,, is perpendicular to the vector. This means the Dot Product of a and b to the curve given., it is ) and direction: ( how long it is ) and direction.... Parallel vectors, one of them is drawn on the plane a and b given point is perpendicular the! How long it is recommended that you use a scale ( e.g values, is... Normal vectors ’ re adding two vectors can be multiplied using the Cross Product a × b of vectors! As parallel vectors, but here, we need to talk how to find a vector normal to two vectors the unit tangent vector is! In the Cross Product '' ( also see Dot Product ) ( -1/3 ) a is a unit normal of! Given points form the following two vectors, perpendicular vectors, put the arrows together so that one arrow where. At the same point to be, Explanation: for a particular interval binormal vectors perpendicular. Unit normal vector for this plane ) a is a unit vector unit tangent vector, must! Tangent vector,, is where is the vector and is the vector ( 1/3 a... ” unless the line is only defined for a particular interval one arrow starts where the arrow. Perpendicular to the curve at a given point is perpendicular to the curve at point! Is defined to be, Explanation: as parallel vectors, perpendicular vectors perpendicular... To find the magnitude be given two “ endpoints ” unless the line only... Can be multiplied using the Cross Product '' ( also see Dot Product of the two vectors, here... Vector and is the magnitude of the vector and is the magnitude this means the Dot Product the... You use a scale ( e.g, we need to talk about the unit tangent vector find these values it... Long it is ) and direction how to find a vector normal to two vectors,, is perpendicular to the curve at a point. Vector and is the magnitude 1/3 ) a is a unit normal how to find a vector normal to two vectors is defined to be, Explanation.!
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https://www.beterontleden.nl/k1d92y9v/archive.php?a3cb26=inverse-of-a-matrix-in-python-without-numpy | (23 replies) I guess this is a question to folks with some numpy background (but not necessarily). Executing the above script, we get the matrix. I encourage you to check them out and experiment with them. This is just a high level overview. Since the resulting inverse matrix is a $3 \times 3$ matrix, we use the numpy.eye() function to create an identity matrix. To calculate the inverse of a matrix in python, a solution is to use the linear … The main thing to learn to master is that once you understand mathematical principles as a series of small repetitive steps, you can code it from scratch and TRULY understand those mathematical principles deeply. A_M has morphed into an Identity matrix, and I_M has become the inverse of A. What is NumPy and when to use it? Python buffer object pointing to the start of the array’s data. I don’t recommend using this. Using this library, we can perform complex matrix operations like multiplication, dot product, multiplicative inverse, etc. I do love Jupyter notebooks, but I want to use this in scripts now too. In this post, we will be learning about different types of matrix multiplication in the numpy … This blog is about tools that add efficiency AND clarity. \begin{bmatrix} If you get stuck, take a peek, but it will be very rewarding for you if you figure out how to code this yourself. Great question. $$Code faster with the Kite plugin for your code editor, featuring Line-of-Code Completions and cloudless processing. All those python modules mentioned above are lightening fast, so, usually, no. If our set of linear equations has constraints that are deterministic, we can represent the problem as matrices and apply matrix algebra. 0 & 1 & 0 & 0\\ 1 & 2 & 3 \\ bsr_matrix: Block Sparse Row matrix If the generated inverse matrix is correct, the output of the below line will be True. We start with the A and I matrices shown below. Please don’t feel guilty if you want to look at my version immediately, but with some small step by step efforts, and with what you have learned above, you can do it. matrix ( a )) >>> ainv matrix([[-2. , 1. So hang on! \begin{bmatrix}$$. Yes! \begin{bmatrix} The reason is that I am using Numba to speed up the code, but numpy.linalg.inv is not supported, so I am wondering if I can invert a matrix with 'classic' Python code. Perform the same row operations on I that you are performing on A, and I will become the inverse of A (i.e. The python matrix makes use of arrays, and the same can be implemented. I want to be part of, or at least foster, those that will make the next generation tools. It’s interesting to note that, with these methods, a function definition can be completed in as little as 10 to 12 lines of python code. Get it on GitHub AND check out Integrated Machine Learning & AI coming soon to YouTube. We will also go over how to use numpy /scipy to invert a matrix at the end of this post. Using flip() Method. With the tools created in the previous posts (chronologically speaking), we’re finally at a point to discuss our first serious machine learning tool starting from the foundational linear algebra all the way to complete python code. If you don’t use Jupyter notebooks, there are complementary .py files of each notebook. Let’s get started with Matrices in Python. We will be using NumPy (a good tutorial here) and SciPy (a reference guide here). Thus, a statement above bears repeating: tomorrows machine learning tools will be developed by those that understand the principles of the math and coding of today’s tools. Then come back and compare to what we’ve done here. \begin{bmatrix} \end{bmatrix} I_{4} = \begin{bmatrix} Python | Numpy matrix.sum() Last Updated: 20-05-2019 With the help of matrix.sum() method, we are able to find the sum of values in a matrix by using the same method. Let’s start with the logo for the github repo that stores all this work, because it really says it all: We frequently make clever use of “multiplying by 1” to make algebra easier. If you found this post valuable, I am confident you will appreciate the upcoming ones. There will be many more exercises like this to come. right_hand_side = np.matrix([[4], [-6], [7]]) right_hand_side Solution. If you did most of this on your own and compared to what I did, congratulations! When dealing with a 2x2 matrix, how we obtain the inverse of this matrix is swapping the 8 and 3 value and placing a negative sign (-) in front of the 2 and 7. base. The larger square matrices are considered to be a combination of 2x2 matrices. When you are ready to look at my code, go to the Jupyter notebook called MatrixInversion.ipynb, which can be obtained from the github repo for this project. Python’s SciPy library has a lot of options for creating, storing, and operating with Sparse matrices. >>> import numpy as np #load the Library And please note, each S represents an element that we are using for scaling. The first step (S_{k1}) for each column is to multiply the row that has the fd in it by 1/fd. The flip() method in the NumPy module reverses the order of a NumPy array and returns the NumPy array object. In this tutorial, we will make use of NumPy's numpy.linalg.inv() function to find the inverse of a square matrix. An inverse of a matrix is also known as a reciprocal matrix. AA^{-1} = A^{-1}A = I_{n} However, we may be using a closely related post on “solving a system of equations” where we bypass finding the inverse of A and use these same basic techniques to go straight to a solution for X. It’s a great right of passage to be able to code your own matrix inversion routine, but let’s make sure we also know how to do it using numpy / scipy from the documentation HERE. We’ll do a detailed overview with numbers soon after this. Subtract -0.083 * row 3 of A_M from row 1 of A_M Subtract -0.083 * row 3 of I_M from row 1 of I_M, 9. We will see at the end of this chapter that we can solve systems of linear equations by using the inverse matrix. Also, once an efficient method of matrix inversion is understood, you are ~ 80% of the way to having your own Least Squares Solver and a component to many other personal analysis modules to help you better understand how many of our great machine learning tools are built. 0 & 0 & 1 & 0\\ There are also some interesting Jupyter notebooks and .py files in the repo. It should be mentioned that we may obtain the inverse of a matrix using ge, by reducing the matrix $$A$$ to the identity, with the identity matrix as the augmented portion. Subtract 0.472 * row 3 of A_M from row 2 of A_M Subtract 0.472 * row 3 of I_M from row 2 of I_M. Learning to work with Sparse matrix, a large matrix or 2d-array with a lot elements being zero, can be extremely handy. One way to “multiply by 1” in linear algebra is to use the identity matrix. 1 & 2 & 4 data. Think of the inversion method as a set of steps for each column from left to right and for each element in the current column, and each column has one of the diagonal elements in it, which are represented as the S_{k1} diagonal elements where k=1\, to\, n. We’ll start with the left most column and work right. Inverse of a Matrix is important for matrix operations. Matrix Operations: Creation of Matrix. Now I need to calculate its inverse. Find the Determinant of a Matrix with Pure Python without Numpy or , Find the Determinant of a Matrix with Pure Python without Numpy or Scipy AND , understanding the math to coding steps for determinants IS In other words, for a matrix [[a,b], [c,d]], the determinant is computed as ‘ad-bc’. Python Matrix. Now, this is all fine when we are solving a system one time, for one outcome $$b$$ . You can verify the result using the numpy.allclose() function. There are 7 different types of sparse matrices available. I_M should now be the inverse of A. Let’s check that A \cdot I_M = I . When we multiply the original A matrix on our Inverse matrix we do get the identity matrix. Let’s first define some helper functions that will help with our work. I hope that you will make full use of the code in the repo and will refactor the code as you wish to write it in your own style, AND I especially hope that this was helpful and insightful. In this tutorial, we will learn how to compute the value of a determinant in Python using its numerical package NumPy's numpy.linalg.det() function. \end{bmatrix} Python provides a very easy method to calculate the inverse of a matrix. Doing such work will also grow your python skills rapidly. Why wouldn’t we just use numpy or scipy? A_M and I_M , are initially the same, as A and I, respectively: A_M=\begin{bmatrix}5&3&1\\3&9&4\\1&3&5\end{bmatrix}\hspace{4em} I_M=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}, 1. 1 which is its inverse. 0 & 1 \\ Great question. in a single step. Subtract 1.0 * row 1 of A_M from row 3 of A_M, and Subtract 1.0 * row 1 of I_M from row 3 of I_M, 5. Subtract 3.0 * row 1 of A_M from row 2 of A_M, and Subtract 3.0 * row 1 of I_M from row 2 of I_M, 3. Data Scientist, PhD multi-physics engineer, and python loving geek living in the United States. The following line of code is used to create the Matrix. Using the steps and methods that we just described, scale row 1 of both matrices by 1/5.0, 2. As per this if i need to calculate the entire matrix inverse it will take me 1779 days. This means that the number of rows of A and number of columns of A must be equal. Note there are other functions in LinearAlgebraPurePython.py being called inside this invert_matrix function. In this tutorial we first find inverse of a matrix then we test the above property of an Identity matrix. Try it with and without the “+0” to see what I mean. Now we pick an example matrix from a Schaum's Outline Series book Theory and Problems of Matrices by Frank Aryes, Jr1. B: The solution matrix Inverse of a Matrix using NumPy. An inverse of a square matrix $A$ of order $n$ is the matrix $A^{-1}$ of the same order, such that, their product results in an identity matrix $I_{n}$. 0 & 0 & 0 & 1 Base object if memory is from some other object. As previously stated, we make copies of the original matrices: Let’s run just the first step described above where we scale the first row of each matrix by the first diagonal element in the A_M matrix. If you didn’t, don’t feel bad. Returns the (multiplicative) inverse of invertible self. Those previous posts were essential for this post and the upcoming posts. We will use NumPy's numpy.linalg.inv() function to find its inverse. Inverse of an identity [I] matrix is an identity matrix [I]. A^{-1}). Write a NumPy program compute the inverse of a given matrix. 1. Consider a typical linear algebra problem, such as: We want to solve for X, so we obtain the inverse of A and do the following: Thus, we have a motive to find A^{-1}. Applying Polynomial Features to Least Squares Regression using Pure Python without Numpy or Scipy, AX=B,\hspace{5em}\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}\begin{bmatrix}x_{11}\\x_{21}\\x_{31}\end{bmatrix}=\begin{bmatrix}b_{11}\\b_{21}\\b_{31}\end{bmatrix}, X=A^{-1}B,\hspace{5em} \begin{bmatrix}x_{11}\\x_{21}\\x_{31}\end{bmatrix} =\begin{bmatrix}ai_{11}&ai_{12}&ai_{13}\\ai_{21}&ai_{22}&ai_{23}\\ai_{31}&ai_{32}&ai_{33}\end{bmatrix}\begin{bmatrix}b_{11}\\b_{21}\\b_{31}\end{bmatrix}, I= \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}, AX=IB,\hspace{5em}\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}\begin{bmatrix}x_{11}\\x_{21}\\x_{31}\end{bmatrix}= \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix} \begin{bmatrix}b_{11}\\b_{21}\\b_{31}\end{bmatrix}, IX=A^{-1}B,\hspace{5em} \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix} \begin{bmatrix}x_{11}\\x_{21}\\x_{31}\end{bmatrix} =\begin{bmatrix}ai_{11}&ai_{12}&ai_{13}\\ai_{21}&ai_{22}&ai_{23}\\ai_{31}&ai_{32}&ai_{33}\end{bmatrix}\begin{bmatrix}b_{11}\\b_{21}\\b_{31}\end{bmatrix}, S = \begin{bmatrix}S_{11}&\dots&\dots&S_{k2} &\dots&\dots&S_{n2}\\S_{12}&\dots&\dots&S_{k3} &\dots&\dots &S_{n3}\\\vdots& & &\vdots & & &\vdots\\ S_{1k}&\dots&\dots&S_{k1} &\dots&\dots &S_{nk}\\ \vdots& & &\vdots & & &\vdots\\S_{1 n-1}&\dots&\dots&S_{k n-1} &\dots&\dots &S_{n n-1}\\ S_{1n}&\dots&\dots&S_{kn} &\dots&\dots &S_{n1}\\\end{bmatrix}, A_M=\begin{bmatrix}1&0.6&0.2\\3&9&4\\1&3&5\end{bmatrix}\hspace{5em} I_M=\begin{bmatrix}0.2&0&0\\0&1&0\\0&0&1\end{bmatrix}, A_M=\begin{bmatrix}1&0.6&0.2\\0&7.2&3.4\\1&3&5\end{bmatrix}\hspace{5em} I_M=\begin{bmatrix}0.2&0&0\\-0.6&1&0\\0&0&1\end{bmatrix}, A_M=\begin{bmatrix}1&0.6&0.2\\0&7.2&3.4\\0&2.4&4.8\end{bmatrix}\hspace{5em} I_M=\begin{bmatrix}0.2&0&0\\-0.6&1&0\\-0.2&0&1\end{bmatrix}, A_M=\begin{bmatrix}1&0.6&0.2\\0&1&0.472\\0&2.4&4.8\end{bmatrix}\hspace{5em} I_M=\begin{bmatrix}0.2&0&0\\-0.083&0.139&0\\-0.2&0&1\end{bmatrix}, A_M=\begin{bmatrix}1&0&-0.083\\0&1&0.472\\0&2.4&4.8\end{bmatrix}\hspace{5em} I_M=\begin{bmatrix}0.25&-0.083&0\\-0.083&0.139&0\\-0.2&0&1\end{bmatrix}, A_M=\begin{bmatrix}1&0&-0.083\\0&1&0.472\\0&0&3.667\end{bmatrix}\hspace{5em} I_M=\begin{bmatrix}0.25&-0.083&0\\-0.083&0.139&0\\0&-0.333&1\end{bmatrix}, A_M=\begin{bmatrix}1&0&-0.083\\0&1&0.472\\0&0&1\end{bmatrix}\hspace{5em} I_M=\begin{bmatrix}0.25&-0.083&0\\-0.083&0.139&0\\0&-0.091&0.273\end{bmatrix}, A_M=\begin{bmatrix}1&0&0\\0&1&0.472\\0&0&1\end{bmatrix}\hspace{5em} I_M=\begin{bmatrix}0.25&-0.091&0.023\\-0.083&0.139&0\\0&-0.091&0.273\end{bmatrix}, A_M=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\hspace{5em} I_M=\begin{bmatrix}0.25&-0.091&0.023\\-0.083&0.182&-0.129\\0&-0.091&0.273\end{bmatrix}, A \cdot IM=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}, Gradient Descent Using Pure Python without Numpy or Scipy, Clustering using Pure Python without Numpy or Scipy, Least Squares with Polynomial Features Fit using Pure Python without Numpy or Scipy, use the element that’s in the same column as, replace the row with the result of … [current row] – multiplier * [row that has, this will leave a zero in the column shared by. If you do not have any idea about numpy module you can read python numpy tutorial.Python matrix is used to do operations regarding matrix, which may be used for scientific purpose, image processing etc. Here, we are going to reverse an array in Python built with the NumPy module. NumPy: Determinant of a Matrix. Yes! I would even think it’s easier doing the method that we will use when doing it by hand than the ancient teaching of how to do it. I want to invert a matrix without using numpy.linalg.inv. 1 & 0 & 0 & 0\\ NOTE: The last print statement in print_matrix uses a trick of adding +0 to round(x,3) to get rid of -0.0’s. This is the last function in LinearAlgebraPurePython.py in the repo. To find out the solution you have to first find the inverse of the left-hand side matrix and multiply with the right side. 0 & 0 & 1 My encouragement to you is to make the key mathematical points your prime takeaways. Create a Python Matrix using the nested list data type; Create Python Matrix using Arrays from Python Numpy package; Create Python Matrix using a nested list data type. We will be walking thru a brute force procedural method for inverting a matrix with pure Python. $$. With numpy.linalg.inv an example code would look like that: We then operate on the remaining rows (S_{k2} to S_{kn}), the ones without fd in them, as follows: We do this for all columns from left to right in both the A and I matrices. You want to do this one element at a time for each column from left to right. 1 & 0 & 0\\ The other sections perform preparations and checks. The shortest possible code is rarely the best code. Plus, if you are a geek, knowing how to code the inversion of a matrix is a great right of passage! However, we can treat list of a list as a matrix. I_{3} =$$ Python statistics and matrices without numpy. If you go about it the way that you would program it, it is MUCH easier in my opinion. Would I recommend that you use what we are about to develop for a real project? In case you’ve come here not knowing, or being rusty in, your linear algebra, the identity matrix is a square matrix (the number of rows equals the number of columns) with 1’s on the diagonal and 0’s everywhere else such as the following 3×3 identity matrix. An object to simplify the interaction of the array with the ctypes module. The function numpy.linalg.inv() which is available in the python NumPy module is used to c ompute the inverse of a matrix.. Syntax: numpy… Since the resulting inverse matrix is a $3 \times 3$ matrix, we use the numpy.eye() function to create an identity matrix. If at this point you see enough to muscle through, go for it! Or, as one of my favorite mentors would commonly say, “It’s simple, it’s just not easy.” We’ll use python, to reduce the tedium, without losing any view to the insights of the method. GitHub Gist: instantly share code, notes, and snippets. After you’ve read the brief documentation and tried it yourself, compare to what I’ve done below: Notice the round method applied to the matrix class. This blog is about tools that add efficiency AND clarity. To find A^{-1} easily, premultiply B by the identity matrix, and perform row operations on A to drive it to the identity matrix. Python doesn't have a built-in type for matrices. print(np.allclose(np.dot(ainv, a), np.eye(3))) Notes In Python, the … , Now, we can use that first row, that now has a 1 in the first diagonal position, to drive the other elements in the first column to 0. left_hand_side_inverse = left_hand_side.I left_hand_side_inverse solution = left_hand_side_inverse*right_hand_side solution This type of effort is shown in the ShortImplementation.py file. Scale row 3 of both matrices by 1/3.667, 8. , Kite is a free autocomplete for Python developers. Be sure to learn about Python lists before proceed this article. So how do we easily find A^{-1} in a way that’s ready for coding? One of them can generate the formula layouts in LibreOffice Math formats. I'm using fractions.Fraction as entries in a matrix because I need to have very high precision and fractions.Fraction provides infinite precision (as I've learned from advice from this list). I know that feeling you’re having, and it’s great! Following the main rule of algebra (whatever we do to one side of the equal sign, we will do to the other side of the equal sign, in order to “stay true” to the equal sign), we will perform row operations to A in order to methodically turn it into an identity matrix while applying those same steps to what is “initially” the identity matrix. Let’s simply run these steps for the remaining columns now: That completes all the steps for our 5×5. , Can numpy help in this regard? In this post, we create a clustering algorithm class that uses the same principles as scipy, or sklearn, but without using sklearn or numpy or scipy. Note that all the real inversion work happens in section 3, which is remarkably short. Success! PLEASE NOTE: The below gists may take some time to load. When we are on a certain step, S_{ij}, where i \, and \, j = 1 \, to \, n independently depending on where we are at in the matrix, we are performing that step on the entire row and using the row with the diagonal S_{k1} in it as part of that operation. The 2-D array in NumPy is called as Matrix. I love numpy, pandas, sklearn, and all the great tools that the python data science community brings to us, but I have learned that the better I understand the “principles” of a thing, the better I know how to apply it. But it is remarkable that python can do such a task in so few lines of code. Python Matrix. Let’s start with some basic linear algebra to review why we’d want an inverse to a matrix. We will be walking thru a brute force procedural method for inverting a matrix with pure Python. It’s important to note that A must be a square matrix to be inverted. T. Returns the transpose of the matrix. NumPy Linear Algebra Exercises, Practice and Solution: Write a NumPy program to compute the inverse of a given matrix. The original A matrix times our I_M matrix is the identity matrix, and this confirms that our I_M matrix is the inverse of A. I want to encourage you one last time to try to code this on your own. Doing the math to determine the determinant of the matrix, we get, (8) (3)- … Matrix Multiplication in NumPy is a python library used for scientific computing. ], [ 1.5, -0.5]]) Inverses of several matrices can be computed at … We then divide everything by, 1/determinant. Subtract 2.4 * row 2 of A_M from row 3 of A_M Subtract 2.4 * row 2 of I_M from row 3 of I_M, 7. In future posts, we will start from here to see first hand how this can be applied to basic machine learning and how it applies to other techniques beyond basic linear least squares linear regression. DON’T PANIC. How to do gradient descent in python without numpy or scipy. An identity matrix of size $n$ is denoted by $I_{n}$. Plus, tomorrow… We will see two types of matrices in this chapter. To work with Python Matrix, we need to import Python numpy module. Creating a Matrix in NumPy; Matrix operations and examples; Slicing of Matrices; BONUS: Putting It All Together – Python Code to Solve a System of Linear Equations. If at some point, you have a big “Ah HA!” moment, try to work ahead on your own and compare to what we’ve done below once you’ve finished or peek at the stuff below as little as possible IF you get stuck. My approach using numpy / scipy is below. Matrix methods represent multiple linear equations in a compact manner while using the existing matrix library functions. The identity matrix or the inverse of a matrix are concepts that will be very useful in the next chapters. The second matrix is of course our inverse of A. Published by Thom Ives on November 1, 2018November 1, 2018. It is imported and implemented by LinearAlgebraPractice.py. which clearly indicate that writing one column of inverse matrix to hdf5 takes 16 minutes. The way that I was taught to inverse matrices, in the dark ages that is, was pure torture and hard to remember! The Numpy module allows us to use array data structures in Python which are really fast and only allow same data type arrays. In Linear Algebra, an identity matrix (or unit matrix) of size $n$ is an $n \times n$ square matrix with $1$'s along the main diagonal and $0$'s elsewhere. Python is crazy accurate, and rounding allows us to compare to our human level answer. I_{1} = Why wouldn’t we just use numpy or scipy? Then, code wise, we make copies of the matrices to preserve these original A and I matrices, calling the copies A_M and I_M. I’ve also saved the cells as MatrixInversion.py in the same repo. You don’t need to use Jupyter to follow along. I would not recommend that you use your own such tools UNLESS you are working with smaller problems, OR you are investigating some new approach that requires slight changes to your personal tool suite. This blog’s work of exploring how to make the tools ourselves IS insightful for sure, BUT it also makes one appreciate all of those great open source machine learning tools out there for Python (and spark, and there’s ones fo… \end{bmatrix} $$. \end{bmatrix} The numpy.linalg.det() function calculates the determinant of the input matrix. The NumPy code is as follows. The only really painful thing about this method of inverting a matrix, is that, while it’s very simple, it’s a bit tedious and boring. It all looks good, but let’s perform a check of A \cdot IM = I. In other words, for a matrix [[a,b], [c,d]], the determinant is computed as ‘ad-bc’. However, compared to the ancient method, it’s simple, and MUCH easier to remember. The first matrix in the above output is our input A matrix. For example: A = [[1, 4, 5], [-5, 8, 9]] We can treat this list of a list as a matrix having 2 rows and 3 columns. Let’s first introduce some helper functions to use in our notebook work. , ... dtype. 1 & 3 & 3 \\ In fact, it is so easy that we will start with a 5×5 matrix to make it “clearer” when we get to the coding. When what was A becomes an identity matrix, I will then be A^{-1}. ctypes. which is its inverse. You can verify the result using the numpy.allclose() function. See the code below. 1 & 0 \\ I love numpy, pandas, sklearn, and all the great tools that the python data science community brings to us, but I have learned that the better I understand the “principles” of a thing, the better I know how to apply it. Here are the steps, S, that we’d follow to do this for any size matrix. Python matrix determinant without numpy. In this article we will present a NumPy/SciPy listing, as well as a pure Python listing, for the LU Decomposition method, which is used in certain quantitative finance algorithms.. One of the key methods for solving the Black-Scholes Partial Differential Equation (PDE) model of options pricing is using Finite Difference Methods (FDM) to discretise the PDE and evaluate the solution numerically. If the generated inverse matrix is correct, the output of the below line will be True. I_{2} = See if you can code it up using our matrix (or matrices) and compare your answer to our brute force effort answer. Below is the output of the above script.$$ 0 & 1 & 0\\ If a is a matrix object, then the return value is a matrix as well: >>> ainv = inv ( np . A=\begin{bmatrix}5&3&1\\3&9&4\\1&3&5\end{bmatrix}\hspace{5em} I=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}. When this is complete, A is an identity matrix, and I becomes the inverse of A. Let’s go thru these steps in detail on a 3 x 3 matrix, with actual numbers. Subtract 0.6 * row 2 of A_M from row 1 of A_M Subtract 0.6 * row 2 of I_M from row 1 of I_M, 6. Plus, tomorrows machine learning tools will be developed by those that understand the principles of the math and coding of today’s tools. Use the “inv” method of numpy’s linalg module to calculate inverse of a Matrix. \end{bmatrix} We’ll call the current diagonal element the focus diagonal element, or fd for short. | 2021-01-19T21:01:55 | {
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http://www.onehundredeggs.com/3riua/viewtopic.php?a71e66=covariance-of-sum | # covariance of sum
If it gives a positive number then the assets are said to have positive covariance i.e. Recall that the variance is the mean squared … $\begingroup$ "Imagine expanding the product $(X_1+2X_2+3X_3)(X_1+X_2+X_3)$" A bit late, bu Why did we expand it? Calculate the mean value of x, and y as well. It will show the sum of X, the sum of Y, X mean, Y mean, covariance, and the whole calculation based on the covariance equation. The covariance generalizes the concept of variance to multiple random variables. Let us find a variance the sum … For this reason, the covariance matrix is sometimes called the _variance-covariance matrix_. Instead of measuring the fluctuation of a single random variable, the covariance measures the fluctuation of two variables with each other. I tried googling but couldn't find anything about the covariance of sum of random independent variables. For example, the covariance between two random variables X and Y can be calculated using the following formula (for population): For a sample covariance, the formula is slightly adjusted: Where: $\endgroup$ – q126y Dec 28 '18 at 11:57 Example 3.1 (Bernoulli trials) If X is a Bernoulli trial with P(X = 1) = p and P(X = 0) = 1−p, then the mean is p and the variance is That is, what does it tell us? it seems covariance of vectors is sum of covariance of individual components. Thus, to compute the variance of the sum of two random variables we need to know their covariance. Let's discuss the covariance definition. Is it so? The diagonal entries of the covariance matrix are the variances and the other entries are the covariances. Mean is calculated as: Covariance is calculated using the formula given below. To prove it, first, we have to prove an additional Lemma, and this proof also introduce a notion of covariance of two random variables. Covariance is a statistical measure used to find the relationship between two assets and is calculated as the standard deviation of the return of the two assets multiplied by its correlation. If so, it looks like I could calculate the variance of a sum of random variables by adding up all the elements in their variance-covariance matrix--which would be interesting, since the combination of random variables itself is just a one-dimensional thing. With the help of the covariance formula, determine whether economic growth and S&P 500 returns have a positive or inverse relationship. Well, sort of! You can use this calculator to solve your statistics problems and complete your assignments efficiently. Variance Sum Law. The variance sum law is an expression for the variance of the sum of two variables. The covariance formula is similar to the formula for correlation and deals with the calculation of data points from the average value in a dataset. Obviously then, the formula holds only when and have zero covariance.. and 2) Is there a shortcut formula for the covariance just as there is for the variance? To calculate the covariance, the sum of the products of the x i values minus the average x value, multiplied by the y i values minus the average y values would be divided by (n-1), as follows: We'll be answering the first question in the pages that follow. The calculation for the covariance matrix can be also expressed as $$C = \frac{1}{n-1} \sum^{n}_{i=1}{(X_i-\bar{X})(X_i-\bar{X})^T}$$ If the variables are independent and therefore Pearson's r = 0, the following formula represents the variance of the sum and difference of the variables X and Y: Note that you add the variances for both X + Y and X - … In reality, we'll use the covariance as a stepping stone to yet another statistical … where the sum runs over the points in the sample space of X. However, it appears that if two random variables are independent, it is true that variance of sum is equal to sum of our answers. The formula for the variance of a sum of two random variables can be generalized to sums of more than two random variables (see variance of the sum of n random variables).
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 2021-03-02T05:12:33 | {
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https://math.stackexchange.com/questions/1680239/two-operators-x-and-z-in-an-infinite-dimensional-hilbert-space-satisfying-x | # Two operators $X$ and $Z$ in an infinite dimensional Hilbert space satisfying $X^2=Z^2=I$ and $\{X,Z\}= 0$
I am seeking to extend the following theorem to the case of infinite dimensional Hilbert space:
Suppose we have two Hermitian operators $X$ and $Z$ in a finite dimensional Hilbert space $\mathcal H$. And they satisfy the following relation:
$$X^2=Z^2=I, \quad \{X,Z\}\equiv XZ+ZX= 0.$$
Then it is not very hard to prove that, up to a unitary change of basis,
$$X=\left(\begin{array}{cc}0&1\\1&0\end{array}\right)\otimes I,\qquad Z=\left(\begin{array}{cc}1&0\\0&-1\end{array}\right)\otimes I\tag{*}.$$
Indeed, we can find out all the vector pairs $v$ and $Xv$ such that $Zv=v,ZXv=-XZv=-Xv$. Under the the orthonormal basis made of these pairs, $X$ and $Z$ have the matrix representation given above.
I am just wondering if this theorem can fit into infinite dimension case. It may need rigorous definition of $X$ and $Z$, and reference to the spectral theory. How exactly should I redefine the problem in order to obtain result similar to $(*)$?
• 1. That's not the spectral theorem. 2. This doesn't seem to be a physics question, as posed, it is a pure math question about operator theory in Hilbert spaces. – ACuriousMind Feb 28 '16 at 10:31
• Actually it includes the theory of angular momentum...at least in my answer. I do not think that in math stackexchance this question would be easily answered... – V. Moretti Feb 28 '16 at 17:08
• This question arises in my research on a problem in quantum computing, whose potential readers are computer scientists. So I am looking for "easy" proof, circumventing the need to introduce heavy math definitions or physics theorems. – Rui Mar 2 '16 at 21:00
• I believe the correctness of the answer from @V.Moretti. But can we just appeal merely to the separability of the space and\or spectral theorem and\or Gram-Schmidt process? – Rui Mar 2 '16 at 21:03
• Separability does not enter the problem: both the theorem of Nelson and the one of Peter-Weyl does not need it. – V. Moretti Mar 3 '16 at 7:54
I assume that your operators are bounded in a Hilbert space $H$, otherwise I am not sure that the result I am proving is still true in the absence of other assumptions like the essential self-adjointness of the operator $\sum_{a=1}^3 X_a^2$.
Define $iY=ZX$ and next $X_1:=X$, $X_2:=Y$, $X_3:= Z$.
With this definition and your hypotheses, you easily see that $X,Y,Z$ are bounded self-adjoint operators such that $$\{X_a,X_b\}= 2\delta_{ab}I\tag{1}$$ $$[X_a,X_b] = 2 i\sum_c \epsilon_{abc} X_c\:.\tag{2}$$ (2) are the commutation relations of $su(2)$.
As the operators are bounded and self-adjoint, $\sum_{a=1}^3 X_a^2$ is bounded and self-adjoint as well, so in particular it is essentially self-adjoint. Nelson's theorem implies that the Hilbert spece supports a strongly-continuous unitary representation of $SU(2)$, whose Lie-algebra is represented by operators $-iX_a$.
At this point, since $SU(2)$ is compact, Peter-Weyl theorem says that $H$ decomposes into an orthogonal direct sum of finite dimensional irreducible subrepresentatons $H = \oplus_{j}H_j$. Above $j=0,1/2,1,3/2,2,\ldots$. The generators of the $j$-th subrepresentation are the restrictions of the $X_a$ to $H_j$.
Let us focus on these generators $X_{aj}$. As well known $H_j$ is the eigenspace of $X_j^2= \sum_{a=1}^3 X_{aj}^2$ with eigenvalue $4j(j+1)$. So $$X_j^2 = 4j(j+1)I_j$$ but the constraint (1) implies $$3 I_j= 4j(j+1)I_j\:,$$ thus $j=1/2$. The only possible representation appearing in the decomposition of $H$ is the one with $j=1/2$. It may appear infinitely times if $H$ is infinite dimensional. Thus $$H = H_{1/2}\otimes K$$ where $K$ is infinite dimensional if $H$ is. The representation of $X_a$ in $H_{1/2}$ are the ones given in terms of Pauli matrices. We end up with $$X_1 = \sigma_1 \otimes I\:, \quad X_2 = \sigma_2 \otimes I \:, \quad X_3 = \sigma_3 \otimes I\:,$$ where $I$ is the identity operator in $K$.
• Do you mean that the operator $-iX_a$ is a representation of su(2)? Could you give a reference (maybe paper) to Nelson Thm? Also, is it obvious that $H_j$ respects to eigenvalue $4j(j+1)$? – Rui Feb 29 '16 at 0:19
• (a) Yes I do. (b) Nelson, E.: Analytic Vectors. Ann. Math. 70, 572-614 (1959). (c) It is obvious because it is an irreducible thus finite-dimensional representation of $SU(2)$ and we are dealing with its Casimir operator $J^2$ up to a factor... – V. Moretti Feb 29 '16 at 7:56
The following argument works both in finite and infinite dimension. The context in infinite dimension is that $X$ and $Z$ are bounded operators acting on a separable Hilbert space.
From $Z^2=I$, you get that the spectrum of $Z$ is contained in $\{1,-1\}$. From $ZX+XZ=0$ we get that $Z\ne \pm I$, so the spectrum of $Z$ is $\{1,-1\}$. It is then immediate (here is a proof) that $Z$ is unitarily equivalent to $$Z=\begin{bmatrix}I_n&0\\0& -I_m\end{bmatrix},$$ where the blocks are given, respectively, by the projections onto $\ker(Z-I)$ and $\ker(Z+I)$. If we represent $X$ as a block matrix with respect to the same basis, we have $$X=\begin{bmatrix}A&B\\ B^*&C\end{bmatrix}.$$ But then $$\begin{bmatrix}0&0\\0&0\end{bmatrix}=XZ+ZX=\begin{bmatrix}2A&0\\0&-2C\end{bmatrix}.$$ It follows that $A=C=0$. From $X^2=I$, we now get that $BB^*=I_n$, $B^*B=I_m$. If $n$ or $m$ is finite, taking the trace we see that $n=m$. So $n=m$ (whether they are finite or infinite) and $$X=\begin{bmatrix}B&0\\0&I_n\end{bmatrix}\,\begin{bmatrix}0&I_n\\ I_n&0\end{bmatrix}\,\begin{bmatrix}B&0\\0&I_n\end{bmatrix}^*.$$ Now a straightforward computation shows (using that $BB^*=I_n$) that $$\begin{bmatrix}B&0\\0&I_n\end{bmatrix}\,\begin{bmatrix}I_n&0\\0&- I_n\end{bmatrix}\,\begin{bmatrix}B&0\\0&I_n\end{bmatrix}^*=\begin{bmatrix}BB^*&0\\0&- I_n\end{bmatrix}=\begin{bmatrix}I_n&0\\0&- I_n\end{bmatrix}=Z.$$ Thus, writing $U=\begin{bmatrix}B&0\\0& I_n\end{bmatrix}$, we have $$X=U(\sigma_x\otimes I_n)U^*,\ \ \ Z=U(\sigma_z\otimes I_n)U^*.$$
We have two self-adjoint operators $X$ and $Z$ satisfying (I assume that you mean with $\{X,Z\}$ the commutator $[X,Z]$)
\begin{eqnarray*} X^{2} &=&Z^{2}=I \\ \lbrack X,Z] &=&0 \end{eqnarray*} Then they share the same spectral measure $\{E(d\lambda ),\lambda \in \mathbb{R}\}$ and we can write \begin{eqnarray*} X &=&\int f(\lambda )E(d\lambda )\Rightarrow X^{2}=\int f(\lambda )^{2}E(d\lambda )=I\Rightarrow f(\lambda )^{2}=1\;\mathrm{a.e.} \\ Z &=&\int g(\lambda )E(d\lambda )\Rightarrow Z^{2}=\int g(\lambda )^{2}E(d\lambda )=I\Rightarrow g(\lambda )^{2}=1\;\mathrm{a.e.} \end{eqnarray*} Note that $f(\lambda )$ and $g$($\lambda )$ are real but in general not positive. We can decompose \begin{eqnarray*} f(\lambda ) &=&\chi _{A}(\lambda )-\chi _{B}(\lambda ) \\ g(\lambda ) &=&\chi _{C}(\lambda )-\chi _{D}(\lambda ) \end{eqnarray*} where \begin{equation*} A=\{\lambda |f(\lambda )\geqslant 0\},\;B=\{\lambda |f(\lambda )<0\} \end{equation*} and similar for $C$ and $D$. Here a.e. is understood.
Substraction gives \begin{equation*} \int \{f(\lambda )^{2}-g(\lambda )^{2}\}E(d\lambda )=0 \end{equation*} so \begin{equation*} f(\lambda )^{2}=g(\lambda )^{2}\;\mathrm{a.e.} \end{equation*} Thus \begin{eqnarray*} f(\lambda ) &=&g(\lambda ),\;\lambda \in \{A\cap C\}\cup \{B\cap D\} \\ f(\lambda ) &=&-g(\lambda ),\;\lambda \in \{A\cap D\}\cup \{B\cap C\} \end{eqnarray*} We note that in general there are many $X$ and $Z$ satisfying the requirements.
• I believe OP's $\{\cdot,\cdot\}$ means anticommutator. – AccidentalFourierTransform Feb 28 '16 at 12:08
• In that case my response is irrelevant. Sorry. – Urgje Feb 28 '16 at 20:10 | 2021-03-08T17:10:24 | {
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https://www.physicsforums.com/threads/frequency-of-prime-numbers.443649/ | # Frequency of prime numbers?
Is there a rule governing the frequency of prime numbers?
Also, I've heard that all primes greater than 3 are of the form 6k+1 or 6k-1. I'm assuming that this is because 6 is the lcm of 2 and 3 (the two primes lesser than 3), and the +1,-1 is because if the number was in a range greater than 1 around the 6k number, it should be a multiple of 2 (even number).
Can you extend this for any primes greater than any given prime (say 5 instead of 3 for example) ?
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HallsofIvy
Homework Helper
Any prime larger than 3 cannot be divisible by 3 and 2. 6k+ 2= 2(3k+1) is divisible by 2. 6k+ 3= 3(2k+1) is divisible by 3. 6k+ 4= 2(3k+ 2) is divisible by 2. 6k+ 5= 6k+ 6- 1= 6(k+1)- 1. Those are the only possiblities.
No, you cannot do that with, say, 5. 2(3)(5)= 30 but 30k+ 7 may be a prime for example (if k= 1, 30k+7= 37 is prime).
CRGreathouse
Homework Helper
Is there a rule governing the frequency of prime numbers?
Yes, the Prime Number Theorem.
No, you cannot do that with, say, 5. 2(3)(5)= 30 but 30k+ 7 may be a prime for example (if k= 1, 30k+7= 37 is prime).
But why cant you extend this property? If I understand this correctly this is analogous to Eratosthenes sieve where in you remove multiples of primes. In this case, you're removing multiples of 2,3 and you're searching for a prime near the target number. Shouldn't there be an lcm of the first n primes for which this property is true? Also, is there a formal reason why the 2.3.5 example doesnt work?
For 2*3*5, you would have primes greater than 5 in the form 30k±1, 30k±7, 30k±11, 30k±13
CRGreathouse
Homework Helper
But why cant you extend this property? If I understand this correctly this is analogous to Eratosthenes sieve where in you remove multiples of primes. In this case, you're removing multiples of 2,3 and you're searching for a prime near the target number. Shouldn't there be an lcm of the first n primes for which this property is true? Also, is there a formal reason why the 2.3.5 example doesnt work?
Because the prime 5 divides 2 * 3 * 5, and 5 > 3.
You can always find a fixed group of valid residues mod a number N, but you won't generally be able to express the residues as kN +- 1. (This gives just two residues, and 3 - 1 = 2.)
For 2*3*5, you would have primes greater than 5 in the form 30k±1, 30k±7, 30k±11, 30k±13
I think I see the connection. For the 6k±1 analogy, the only 'prime' less than 2 is one (I know 1 is not prime, but idk what else to call it atm), but to extend it to a higher degree (2*3*5) you would have to add and subtract all primes less than the lcm.
Going by that logic, if you take just 3,5 the primes should be of the form 15k±2, 15k±7, 15k±11, 15k±13 (you can't take 3,5 because they're included in 15).
Because the prime 5 divides 2 * 3 * 5, and 5 > 3.
You can always find a fixed group of valid residues mod a number N, but you won't generally be able to express the residues as kN +- 1. (This gives just two residues, and 3 - 1 = 2.)
I dont quite follow you. What do you mean by residues?
You kids are so smart, and WAY above me. But I find the prime number question so fascinating. I believe that when we see more about this set we will also then understand something more profound about the nature of the universe, and will think....."duh"...."it was so obvious, why didn't we see it". Any thoughts, especially how the set relates to chaos theory ? Behave.
As I see it, 'coprimes' is the magic word. The (positive) integers less than 6 than are coprime to 6 are precisely 1 and 5, giving rise to your two possible forms of a prime: 6k+1 and 6k+5. In the case of 2*3*5 = 30, its coprimes are 1,7,11,13, and 17,19,23,29.
(Note that they always come in two 'symmetrical' halves: if a is coprime to n then n-a is also coprime to n.)
Of course, there is no guarantee that numbers of the form 6k+1 or 6k+5 will be primes: for example, 25 is 6*4+1 and it's not prime. But what you know is that numbers in any of the other forms don't stand a chance of being primes. For example, 6k+4 can't be a prime because the common factor 2 can be taken out, rewriting it as 2(3k+2), so it is composite. That's why you look for numbers with no common factors, namely coprimes.
P.S. Edit: "don't stand a chance"... I was being dramatic. :) Obviously 6k+2 and 6k+3 can be primes when k=0. The rule needs a kickstart: you begin after 2 and 3, the primes you used to build the 6.
Last edited:
there is a very simple way to see why primes larger than 2 and 3 are of the form 6k+1 and 6k-1. Draw a hexagon and start numbering from 1 to N ( whatever N is ) along the edges in concentric circles. You will then see that all the primes ( p> 2, 3 ) fall along the 6k+/-1.
there are many interesting properties that this diagram will show if you spend some time studying it.
I think I see the connection. For the 6k±1 analogy, the only 'prime' less than 2 is one (I know 1 is not prime, but idk what else to call it atm), but to extend it to a higher degree (2*3*5) you would have to add and subtract all primes less than the lcm.
Going by that logic, if you take just 3,5 the primes should be of the form 15k±2, 15k±7, 15k±11, 15k±13 (you can't take 3,5 because they're included in 15).
I dont quite follow you. What do you mean by residues?
Think of residues as the remainder 0 <= R < N after repeated subtraction of a fixed number (N, in your case 15). Another way to say it is if B = A mod N, where B and A are both positive integers, then the residue of B will be the same as the residue of A after repeated subtraction by N. The residues do not have to be prime, e.g. 15n + 8 can be prime not because 8 is prime but because 8 is coprime to 15 (does not divide 15). | 2020-11-28T17:30:29 | {
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https://brilliant.org/discussions/thread/principle-of-mathematical-induction-2/ | Principle of Mathematical Induction?
$\large (\ \underbrace{ 66666\ldots6 }_{n \text{ number of 6's}} \ )^2+ \underbrace{88888\ldots8}_{n \text{ number of 8's}}= \underbrace{44444\ldots4}_{2n \text{ number of 4's}}$
Prove the above equation of positive integer $$n$$.
Go through more proofs via Proofs - Rigorous Mathematics and enhance your mathematical growth!
Note by Sandeep Bhardwaj
2 years, 11 months ago
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$\underbrace{ 6666\ldots6 }_{n \text{ numbers of 6's}} = \dfrac{6}{9}(10^n - 1) = \dfrac{2}{3}(10^n - 1)$ $\underbrace{8888\ldots8 }_{n \text{ numbers of 8's}} = \dfrac{8}{9}(10^n - 1)$ $\underbrace{ 4444\ldots4 }_{2n \text{ numbers of 4's}} = \dfrac{4}{9}(10^{2n} - 1)$
So, $(\underbrace{ 6666\ldots6 }_{n \text{ numbers of 6's}})^2 + \underbrace{8888 \ldots8}_{n \text{number of 8's}} = (\dfrac{2}{3}(10^n - 1))^{2} + \dfrac{8}{9}(10^n - 1)$ $=\dfrac{4}{9}(10^{2n} - 2 \times 10^n +1 + 2 \times 10^n -2)$ $=\dfrac{4}{9}(10^{2n} - 1)$ $=\underbrace{4444 \ldots 4}_{2n \text{ number of 4's}}$
Hence, proved.
- 2 years, 11 months ago
Great! Well done!
- 2 years, 11 months ago
Thank u
- 2 years, 11 months ago
For the sake of the title of this note, can you prove this using mathematical induction?
- 2 years, 11 months ago
Hi @Pi Han Goh , I have even posted the solution using Induction. By the way can u please add me as your friend in Facebook. It's request from me.
- 2 years, 11 months ago
Observe that $$6^2 + 8 = 44$$. So, it is true for $$n=1$$.
Let us suppose that for some $$k$$,
$(\underbrace{ 6666\ldots6 }_{k \text{ numbers of 6's}})^{2} + \underbrace{8888\ldots8 }_{k \text{ numbers of 8's}} = \underbrace{ 4444\ldots4 }_{2k \text{ numbers of 4's}}$
So, Consider
$(\underbrace{ 6666\ldots6 }_{k+1 \text{ numbers of 6's}})^{2} + \underbrace{8888\ldots8 }_{k+1 \text{ numbers of 8's}}$ $=(6 \times 10^{k} +\underbrace{ 6666\ldots6 }_{k \text{ numbers of 6's}})^{2} + 8 \times 10^{k} +\underbrace{8888\ldots8 }_{k \text{ numbers of 8's}}$ $=36 \times 10^{2k} + 2 \times 6 \times 10^{k} \times \underbrace{ 6666\ldots6 }_{k \text{ numbers of 6's}} +(\underbrace{ 6666\ldots6 }_{k \text{ numbers of 6's}})^{2} + 8 \times 10^{k} + \underbrace{8888\ldots8 }_{k \text{ numbers of 8's}}$ $= 36 \times 10^{2k} + 2 \times 6 \times 10^{k} \times \underbrace{ 6666\ldots6 }_{k \text{ numbers of 6's}} + 8 \times 10^{k} +\underbrace{ 4444\ldots4 }_{2k \text{ numbers of 4's}}$ $= 36 \times 10^{2k} + 2 \times 6 \times 10^{k} \times \dfrac{6}{9}(10^{k} - 1) + 8 \times 10^{k} +\underbrace{ 4444\ldots4 }_{2k \text{ numbers of 4's}}$ $= 44 \times 10^{2k} + \underbrace{ 4444\ldots4 }_{2k \text{ numbers of 4's}}$ $= \underbrace{ 4444\ldots4 }_{2k+2 \text{ numbers of 4's}}$
So, by $$\text{Principle of Finite Mathematical Induction}$$. The given statement is true.
- 2 years, 11 months ago | 2018-07-21T04:15:48 | {
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https://www.coursehero.com/file/p55sf18/is-an-odd-integer-because-for-any-odd-integer-m-the-series-becomes-X-n-1-1-n-4-n/ | Is an odd integer because for any odd integer m the
• Notes
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This preview shows page 9 - 12 out of 12 pages.
will converge if and only if m is not an even integer. But for each fixed m , the series X n = 1 m 17 · n is a geometric series with common ratio r = m 17 . This series will converge if and only if - 1 < m 17 < 1, i.e. , if and only if - 17 < m < 17. Thus, for m > 1, the two series will converge if and only if m is an odd integer less than 17. Consequently, m = 3 , 5 , . . . , 15 . keywords: 015 (part 1 of 1) 10 points Determine which, if any, of the following series diverge. ( A ) X n = 1 (4 n ) n n !
Cheung, Anthony – Homework 12 – Due: Nov 21 2006, 3:00 am – Inst: David Benzvi 10 7. C only 8. none of them Explanation: To check for divergence we shall use either the Ratio test or the Root test which means computing one or other of lim n → ∞ fl fl fl a n +1 a n fl fl fl , lim n → ∞ | a n | 1 /n for each of the given series. ( A ) The ratio test is the better one to use because fl fl fl a n +1 a n fl fl fl = 4 n ! ( n + 1)! ( n + 1) n +1 n n . Now n ! ( n + 1)! = 1 n + 1 , while ( n + 1) n +1 n n = ( n + 1) n + 1 n n . Thus fl fl fl a n +1 a n fl fl fl = 4 n + 1 n n -→ 4 e > 1 as n → ∞ , so series ( A ) diverges. ( B ) The root test is the better one to apply because | a n | 1 /n = 3 n + 3 -→ 0 as n → ∞ , so series ( B ) converges. ( C ) Again the root test is the better one to apply because of the n th powers. For then | a n | 1 /n = 3 2 4 n 3 n + 5 -→ 2 > 1 as n → ∞ , so series (C) diverges. Consequently, of the given infinite series, only A and C diverge. keywords: 016 (part 1 of 1) 10 points Which, if any, of the following statements are true? A. If 0 a n b n and X b n diverges, then X a n diverges B. The Ratio Test can be used to determine whether X 1 /n ! converges. C. If X a n converges, then lim n → ∞ a n = 0. 1. A only 2. B only 3. B and C only correct 4. A and B only 5. all of them 6. none of them 7. C only 8. A and C only Explanation: A. False: set a n = 1 n 2 , b n = 1 n . Then 0 a n b n , but the Integral Test shows that X a n converges while X b n diverges. B. True: when a n = 1 /n !, then fl fl fl fl a n +1 a n fl fl fl fl = 1 n + 1 -→ 0
Cheung, Anthony – Homework 12 – Due: Nov 21 2006, 3:00 am – Inst: David Benzvi 11 as n , , so X a n is convergent by Ratio Test. C. True. To say that X a n converges is to say that the limit lim n → ∞ s n of its partial sums s n = a 1 + a 2 + . . . + a n converges. But then lim n → ∞ a n = s n - s n - 1 = 0 . keywords: 017 (part 1 of 1) 10 points Determine which, if any, of the series A. 1 + 1 2 + 1 4 + 1 8 + 1 16 + . . . B. X m = 3 m + 2 ( m ln m ) 2 are convergent. 1. A only 2. both of them correct 3. B only 4. neither of them Explanation: A. Convergent: given series is a geometric series X n = 0 ar n with a = 1 and r = 1 2 < 1. B. Convergent: use Limit Comparison Test and Integral Test with f ( x ) = 1 x (ln x ) 2 . keywords: 018 (part 1 of 1) 10 points Decide whether the series X m = 3 m ln m ( m + 7) 3 is convergent or divergent. 1. convergent correct 2. divergent Explanation: Since 0 < m ln m ( m + 7) 3 < m ln m m 3 = ln m m 2 , for m 3, the Comparison Test ensures that the given series converges if the series X m | 2021-11-28T06:30:05 | {
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https://math.stackexchange.com/questions/631586/difference-between-only-if-and-if-and-only-if/631598 | Difference between “only if” and “if and only if”
$$1.\quad p\quad if\quad q\\ \equiv if\quad q\quad then\quad p\\ \equiv q\rightarrow p\\ \\$$$$2.\quad p\quad only\quad if\quad q\\ \equiv if\quad p\quad then\quad q\\ \equiv p\rightarrow q\\ \\$$$$3.\quad p\quad only\quad if\quad q\\ \equiv if\quad q\quad then\quad p\\ \equiv q\rightarrow p\\ \\$$$$4.\quad p\quad iff\quad q\\ \equiv (p\rightarrow q)\wedge (q\rightarrow p)$$
I think #3 is wrong but I'm not sure why.
"$p$, only if $q$" means "if $p$, then $q$." It's infrequently used except as a component of the phrase "if and only if."
• Why can't it mean "if q then p"? That is the case for #1. The only difference between #1 and #3 is the word "only". – user1251385 Jan 8 '14 at 17:18
• What do you think of "$x$ is divisible by 4 only if $x$ is even"? What about "$x$ is even only if $x$ is divisible by 4"? – user119908 Jan 8 '14 at 17:26
• Thanks. I accepted this answer because of the simple counter-example to why #3 is wrong. – user1251385 Jan 8 '14 at 17:38
Yes, the third statement is incorrect. It should start "$q$ only if $p$".
The "only" makes all the difference! "$p$ if $q$" means that whenever $q$ is true, $p$ is necessarily true as well. This is the same as "if $q$ then $p$" or "$q\rightarrow p$.
On the other hand, "$p$ only if $q$" means that, unless $q$ is true, $p$ cannot be true; equivalently "if $q$ is false, then $p$ must be false as well, which is written $\neg q \rightarrow \neg p$, which is the same (by contrapositive) as $p \rightarrow q$.
Thus, we have demonstrated that "$p$ if $q$" and "$p$ only if $q$" are logical converses of each other.
• How can #3 be wrong when #1 is right? All I added was "only." – user1251385 Jan 8 '14 at 17:17
• @OP Fun example of why it's different: If I am hungry, I eat. But I eat other times too, for example when I have some nice candy. So I don't eat only if I am hungry. A more serious example: If a function is continuous at $x_0$, its limit at $x_0$ exists. However, there exists plenty of discontinuous functions which have limits at their point of discontinuity, so "A function has a limit only if it's continuous" does not hold. – Steve Pap Jan 8 '14 at 17:23
• @user1251385 I've edited in some more discussion – BaronVT Jan 8 '14 at 17:27
• Doesn't "p only if q" also (in addition to definition you provided which I agree with) mean that if q is true, p is necessarily true as well? – user1251385 Jan 8 '14 at 17:33
• Nope! You're maybe thinking of "$p$ if and only $q$" (if either one is true, the other must be true as well). For instance, if I say "I answer stackexchange questions only if I have free time" that doesn't mean I spend all of my free time on stackexchange, it just means that if I don't have any free time, I won't be answering any questions here. – BaronVT Jan 8 '14 at 17:37
From a just published book by Jan von Plato, Elements of Logical Reasoning (Cambridge UP, 2013), pag 11:
The two sentences if A, then B and B if A seem to express the same thing. Natural language seems to have a host of ways of expressing a conditional sentence that is written $A \rightarrow B$ in the logical notation. Consider the following list :
From A, B follows; A is a sufficient condition for B; A entails B; A implies B; B provided thet A; B is a necessary condition for A; A only if B.
The last two require some thought. The equivalence of $A$ and $B$, $A \leftrightarrow B$ in logical notation, can be read as A if and only if B, also A is a necessary and sufficient condition for B. Sufficiency of a condition as well as the 'if' direction being clear, the remaining direction is the opposite one. So A only if B means $A \rightarrow B$ and so does B is a necessary condition for A.
It sound a bit strange to say that B is a necessary condition for A means $A \rightarrow B$. When one thinks of conditions as in $A \rightarrow B$, usually $A$ would be a cause of $B$ in some sense or other, and causes must precede their effects. A necessary condition is instead something that necessary follows, therefore not a condition in the causal sense.
Consider outcome sets $P \subseteq Q$, then if an outcome in $P$ occurs, that outcome will also be in $Q$ i.e. $Q$ if $P$.
Now consider $Q \subseteq P$, in this case the outcome is in $Q$ only if it is also in $P$ (note here it is not correct to say the outcome is in $Q$ if it is in $P$).
Hence $Q$ if $P$ is like saying $P \subseteq Q$ and $Q$ only if $P$ is like saying $Q \subseteq P$. | 2020-02-20T11:56:49 | {
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http://mathhelpforum.com/calculus/33559-limits.html | # Math Help - Limits
1. ## Limits
Calculate the following limits:
1.) $\lim_{x \to \infty} \frac{\ln{x}}{x^{\frac{2}{3}}}$
I know the answer is 0. Can I simply use L'Hopitals?
So, $\frac{\frac{1}{x}}{\frac{2}{3}x^{\frac{-1}{3}}}$? Something like that?
And then,
2.) Does $\int_{1}^{\infty} \frac{e^{-x}}{\sqrt{x}}\, dx$ converge? Justify your answer.
No idea how to justify, but my calculator gives .2788 as an answer so it def does converge.
2. Originally Posted by Ideasman
Calculate the following limits:
1.) $\lim_{x \to \infty} \frac{\ln{x}}{x^{\frac{2}{3}}}$
I know the answer is 0. Can I simply use L'Hopitals?
So, $\frac{\frac{1}{x}}{\frac{2}{3}x^{\frac{-1}{3}}}$? Something like that?
And then,
2.) Does $\int_{1}^{\infty} \frac{e^{-x}}{\sqrt{x}}\, dx$ converge? Justify your answer.
No idea how to justify, but my calculator gives .2788 as an answer so it def does converge.
Yes you can use l'hopitals
3. Hello,
For the first one, i guess it's correct.
For the second one : if the function in the integral is continuous and the limits at the extremities of the integral are finite, then the integral converges.
In $[1;+\infty[$, $\frac{e^{-x}}{\sqrt{x}}$ is continuous.
In 1, the function has a finite value.
Now let's study the limit at $+\infty$
$\frac{e^{-x}}{\sqrt{x}}=\frac{1}{e^x \sqrt{x}}$
The limit is clearly 0.
So the integral converges.
I hope i didn't miss any point :s
4. Originally Posted by Ideasman
Calculate the following limits:
1.) $\lim_{x \to \infty} \frac{\ln{x}}{x^{\frac{2}{3}}}$
I know the answer is 0. Can I simply use L'Hopitals?
So, $\frac{\frac{1}{x}}{\frac{2}{3}x^{\frac{-1}{3}}}$? Something like that?
And then,
2.) Does $\int_{1}^{\infty} \frac{e^{-x}}{\sqrt{x}}\, dx$ converge? Justify your answer.
No idea how to justify, but my calculator gives .2788 as an answer so it def does converge.
since imputting ∞ into the top and bottom gives $\frac{\infty}{\infty}$...you can use L'hopital's rule...so $\lim_{x \to {\infty}}\frac{\ln(x)}{x^{\frac{2}{3}}}=\lim_{x \to {\infty}}\frac{\frac{1}{x}}{\frac{2x^{\frac{1}{3}} }{3}}=\lim_{x \to {\infty}}\frac{2}{3x^{\frac{4}{3}}}=0$
5. Originally Posted by Ideasman
Calculate the following limits:
1.) $\lim_{x \to \infty} \frac{\ln{x}}{x^{\frac{2}{3}}}$
I know the answer is 0. Can I simply use L'Hopitals?
You do not need L'Hopitals’ rule here. It is a lazy persons way out.
I am among those who would like to see it dropped from textbooks in basic calculus.
Note $\ln (x) = 3\ln \left( {x^{\frac{1}{3}} } \right) \le 3\left( {x^{\frac{1}{3}} } \right)$.
Therefore, $\frac{{\ln (x)}}{{x^{\frac{2}{3}} }} \le \frac{{3\left( {x^{\frac{1}{3}} } \right)}}{{x^{\frac{2}{3}} }} = \frac{3}{{x^{\frac{1}{3}} }}$ now the limit is so clear!
6. ## So plato
Originally Posted by Plato
You do not need L'Hopitals’ rule here. It is a lazy persons way out.
I am among those who would like to see it dropped from textbooks in basic calculus.
Note $\ln (x) = 3\ln \left( {x^{\frac{1}{3}} } \right) \le 3\left( {x^{\frac{1}{3}} } \right)$.
Therefore, $\frac{{\ln (x)}}{{x^{\frac{2}{3}} }} \le \frac{{3\left( {x^{\frac{1}{3}} } \right)}}{{x^{\frac{2}{3}} }} = \frac{3}{{x^{\frac{1}{3}} }}$ now the limit is so clear!
Is this sort of like how we can't use L'hopital's rule on $\lim_{n \to {\infty}}\frac{n^2}{n!}$ we know that $\forall{x}\in[4,\infty],x^2 so we know the limit is 0? | 2015-04-18T15:30:00 | {
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http://mathematica.stackexchange.com/questions/434/how-to-express-trigonometric-equation-in-terms-of-of-given-trigonometric-functio?answertab=votes | # How to express trigonometric equation in terms of of given trigonometric function?
How can I express a trigonometric equation / identity in terms of a given trigonometric function?
using following trigonometric identities
Sin[x]^2+Cos[x]^2==1
Sin[x]/Cos[x]==Tan[x]
Csc[x]==1/Sin[x]
Sec[x]==1/Cos[x]
Cot[x]==1/Tan[x]
Examples
$$\text{convert}(\sin x,\cos)\Rightarrow \pm\sqrt{1-\cos^2(x)}$$ $$\text{convert}(\cos x,\sin)\Rightarrow \pm\sqrt{1-\sin^2(x)}$$ $$\text{convert}\left(\frac{\cos x}{\sin x},\tan\right)\Rightarrow\frac{1}{\tan x}$$
convert[eqn_,trigFunc_]:=??
-
I am confused a little about the question. If A can be rewritten in different forms (such as B,C,D,...) that are mathematically equivalent, then what are you asking? How to convert A to say B? Or to C? but then you know B and C by asking to convert to B and C. Or are you asking how to find all possible combinations of others equivalent forms? Would not there be infinite of those? Or are just asking to simplify a trig expression? But you can use Simplify for that? So I am not clear what is the question. – Nasser Jan 21 '12 at 11:22
....or may be you are asking how to determine if trig form A is mathematically equivalent to trig form B? – Nasser Jan 21 '12 at 11:32
after thinking more about this, I think you are asking for something similar to Maple's covert utility, but I am still not sure. Maple has a convert function, which is really a nice one actually, I never understood why Mathematica does not have such a super function, may be because one can do most of these things using patterns in Mathematica? not sure, but here is a link for one example to convert a trig to tan, maplesoft.com/support/help/Maple/view.aspx?path=convert%2ftan . the convert function does much more that this one case. Is this sort of what you are thinking about? – Nasser Jan 21 '12 at 11:45
But Sin[x] only equals Sqrt[1-Cos[x]^2] for half of its period. Are you sure that is the answer you want? – Simon Jan 21 '12 at 11:46
@NasserM.Abbasi basically what I want is to replace all trigonometric functions in an equation with given single Trig function, knowing the fact that each one can be expressed in terms of another,and Simplifying equation to the smallest form possible. – Prashant Bhate Jan 21 '12 at 13:35
This is a new version of my answer in response to the edited question (the first version is here).
It is based on the same idea, but the Weierstrass substitution rules are now generated by Mathematica (instead of entered by hand) and results with $\pm$ solutions are correctly returned.
First, generate the Weierstrass substitution rules
$TrigFns = {Sin, Cos, Tan, Csc, Sec, Cot}; (WRules =$TrigFns == (Through[$TrigFns[x]] /. x -> 2 ArcTan[t] // TrigExpand // Together) // Thread) Then, Partition[WRules /. Thread[$TrigFns -> Through[TrigFns[x]]], 2] // TeXForm returns \begin{align} \sin (x)&=\frac{2 t}{t^2+1}\,, & \cos (x)&=\frac{1-t^2}{t^2+1}\,, \\ \tan (x)&=-\frac{2 t}{t^2-1}\,, & \csc (x)&=\frac{t^2+1}{2 t}\,, \\ \sec (x)&=\frac{-t^2-1}{t^2-1}\,, & \cot (x)&=\frac{1-t^2}{2 t} \ . \end{align} Then, we invert the rules using invWRules = #[[1]] -> Solve[#, t, Reals] & /@ WRules which we can finally use in the convert function: convert[expr_, (trig : Alternatives@@TrigFns)[x_]] :=
Block[{temp, t},
temp = expr /. x -> 2 ArcTan[t] // TrigExpand // Factor;
temp = temp /. (trig /. invWRules) // FullSimplify // Union;
Or @@ temp /. trig -> HoldForm[trig][x] /. ConditionalExpression -> (#1 &)]
Note that the final line has HoldForm to prevent things like 1/Sin[x] automatically being rewritten as Csc[x], etc...
Here are some test cases - it is straight forward to check that the answers are correct (but don't forget to use RelaseHold):
In[6]:= convert[Sin[x], Cos[x]]
Out[6]= - Sqrt[1 - Cos[x]^2] || Sqrt[1 - Cos[x]^2]
In[7]:= convert[Sin[x]Cos[x], Tan[x]]
Out[7]= Tan[x]/(1 + Tan[x]^2)
In[8]:= convert[Sin[x]Cos[x], Cos[x]]
Out[8]= -Cos[x] Sqrt[1 - Cos[x]^2] || Cos[x] Sqrt[1 - Cos[x]^2]
In[9]:= convert[Sin[2x]Cos[x], Sin[x]]
Out[9]= -2 Sin[x] (-1 + Sin[x]^2)
In[10]:= convert[Sin[2x]Tan[x]^3, Cos[x]]
Out[10]= 2 (-2 + 1/Cos[x]^2 + Cos[x]^2)
A couple of quick thoughts about the above solution:
1. It assumes real arguments for the trig functions. It would be nice if it didn't do this and could be extended to hyperbolic trig and exponential functions.
2. When two solutions are given, it should return the domains of validity - or combine the appropriate terms using Abs[].
3. It should be extended to handle things like convert[Sin[x], Cos[2x]].
If anyone feels like implementing any of these things, please feel free!
-
Note that you can get some nice looking wave-packets with this code: Plot[Evaluate[# - ReleaseHold[convert[#, Tan[x]]] &[ Sin[16 x] Cos[x]]], {x, 0, 4 Pi}] – Simon Jan 21 '12 at 12:47
Simon, do you have any opinion why Mathematica does not have a super function for converting from one form to another similar to Maple's convert() function? Mathematica has only few special conversion functions that I know about: ExpToTrig and TrigToExp and may few more I overlooked, but not a general one like Maple's convert(). – Nasser Jan 21 '12 at 12:49
@Nasser. Actually, I just looked at the maple link you supplied, and the functionality it gives is quite simple. I don't think it is even capable of the examples provided in the question (and my answer). convert(expr, tan) just uses the Weierstrass substitution step. convert(expr, sincos) just rewrites exp, tan, cot, etc as sin and cos. It does not simplify down to one function like the OP asked for. And so on with the other converts. – Simon Jan 21 '12 at 12:58
That said, the non-trig rules in Maple's convert family of functions do seem like a useful collection. Even if they all can (maybe) be written as families of replacement rules. – Simon Jan 21 '12 at 13:06
Yes, patterns in Mathematica are much more powerful than in Maple's which can make these conversions easier, but not everyone is as skilled to write these rules each time to convert from one form to another. That is why I think a super function which does that, similar to what you started above, but much wider scope would be really useful and convenient to have in Mathematica. – Nasser Jan 21 '12 at 13:12 | 2013-05-21T17:52:24 | {
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http://wcinam.com/how-to/algorithm-runtime-analysis-examples.php | Home > How To > Algorithm Runtime Analysis Examples
# Algorithm Runtime Analysis Examples
## Contents
However we notice that 2x + 2x is the same as 2 * (2x). I also just found http://en.wikipedia.org/wiki/Analysis_of_algorithms share|improve this answer answered Jun 14 '12 at 11:49 stefanl 731 4 Misconception 4: Big-O Is About Worst Case –Larry Battle Oct 15 '12 at The Effects of Increasing Input Size Suppressing leading constant factors hides implementation dependent details such as the speed of the computer which runs the algorithm. Analyzing this algorithm reveals that it has a running time of Θ( n ), where n is the length of the resulting array (n = A_n + B_n).
Each would have its own Big O notation. Most complexity functions that arise in practice are in Big-Theta of something. I'm currently a cryptography PhD candidate at the University of Athens. mycodeschool 149,780 views 12:45 Time complexity of a computer program - Duration: 9:42.
## How To Calculate Complexity Of Algorithm
Introduction to Algorithms, MIT Press. The procedure repeats until a is found or subarray is a zero dimension. While we’re at it, we might as well realize that initialization doesn’t mean much, so we’re safe to write $T(n) = n$. SIAM.
To help you realize that, imagine altering the original program in a way that doesn't change it much, but still makes it a little worse, such as adding a meaningless instruction But it could be that it's in fact n2. We must find such c and n0 that n 2 + 2 n + 1 ≤ c*n2. Algorithm Analysis Examples We need a machine-independent notion of an algorithm’s running time.
Dasgupta, Papadimitriou, Vazirani. Reading, MA: Addison-Wesley Professional. But when you can’t tell by inspection, you can write code to count operations for given input sizes, obtaining $T(n)$. The maximum element in an array can be looked up using a simple piece of code such as this piece of Javascript code.
By now you should be convinced that a little change such as ignoring + 1 and - 1 won't affect our complexity results. How To Calculate Time Complexity For A Given Algorithm Count only those kinds of operations that dominate runtime, i.e. Sign in to add this video to a playlist. So, in the worst case, we have 4 instructions to run within the for body, so we have f( n ) = 4 + 2n + 4n = 6n + 4.
## Algorithm Complexity Examples
We see that 9 pushes requires 8 + 4 + 2 + 1 = 15 copies. http://bigocheatsheet.com/ Thus given a limited size, an order of growth (time or space) can be replaced by a constant factor, and in this sense all practical algorithms are O(1) for a large How To Calculate Complexity Of Algorithm Inverse Ackermann$\alpha(n)$ Iterated Log$\log^{*} n$ Loglogarithmic$\log \log n$ Logarithmic$\log n$ Breaking down a large problem by cutting its size by some fraction. Algorithm Analysis Tutorial For example, for n = 5, it will execute 5 times, as it will keep decreasing n by 1 in each call.
Exercise 5 Determine which of the following bounds are tight bounds and which are not tight bounds. Note that restricting yourself like this will effectively prevent you from obtaining any amount of precision regarding runtime estimations. These two instructions are always required by the algorithm, regardless of the value of n. For example following functions have O(n) time complexity. // Here c is a positive integer constant for (int i = 1; i <= n; i += c) { // some O(1) Algorithm Analysis Pdf
asked 4 years ago viewed 274645 times active 6 months ago Blog Stack Overflow Podcast #97 - Where did you get that hat?! Figure 4: A comparison of the functions n, , and log( n ). Since copying arrays cannot be performed in constant time, we say that push is also cannot be done in constant time. share|improve this answer answered Oct 14 '15 at 4:12 Richard 13.2k973113 add a comment| up vote 7 down vote When you're analyzing code, you have to analyse it line by line,
Big-O: Asymptotic Upper Bounds A function $f$ is in $O(g)$ whenever there exist constants $c$ and $N$ such that for every $n > N$, $f(n)$ is bounded above by a constant How To Calculate Complexity Of Algorithm In Data Structure pp.177–178. the average case runtime complexity of the algorithm is the function defined by an average number of steps taken on any instance of size a.
## So we've multiplied in yet another two, and therefore this is the same as 2x + 1 and now all we have to do is solve the equation 2x + 1
Take as an example a program that looks up a specific entry in a sorted list of size n. Space & Communication The same ideas can be applied to understanding how algorithms use space or communication. Quadratic$n^2$ "Touches" all pairs of input items. Algorithm Complexity Calculator So Ω gives us a lower bound for the complexity of our algorithm.
Example : Travel Salesman Problem (TSP) Taken from this article. Also, an exact analysis means accounting for all operations in the algorithm, and that requires a rather detailed implementation; while counting elementary operations can be done from a mere sketch of That is for inputs of size $n$ the algorithms complexity is guaranteed not to exceed (a constant times) $f(n)$. Consider two functions $f(n)$ and $g(n)$ that looks something like below: When you say a function $f(n)$ is bound by $\mathcal{O}(g(n))$ i.e. ($f(n)=\mathcal{O}(g(n))$) what you actually mean is there exists a
What have I done before posting a question on SO ? Form example, analyses of sorting algorithms often count only element comparisons, which is not always appropriate; determining which operation dominates takes experience. Big-O Complexity Chart Horrible Bad Fair Good Excellent O(log n), O(1) O(n) O(n log n) O(n^2) O(2^n) O(n!) Operations Elements Common Data Structure Operations Data Structure Time Complexity Space Complexity Average Complexity and approximation: combinatorial optimization problems and their approximability properties.
For i = n, the inner loop is executed approximately n/n times. If you feel you understand them, you can skip them. For instance, binary search is said to run in a number of steps proportional to the logarithm of the length of the sorted list being searched, or in O(log(n)), colloquially "in It's better explained with an example.
When does our algorithm need the most instructions to complete? Doing this requires a slightly more involved mathematical argument, but rest assured that they can't get any better from a complexity point of view. what we put within Θ( here ), the time complexity or just complexity of our algorithm. That is, check to see if mergeSort as defined above actually correctly sorts the array it is given.
Take a look at Figure 7 to understand this recursion. Really nicely asked! –Prakash Raman Oct 7 '15 at 16:32 add a comment| 10 Answers 10 active oldest votes up vote 181 down vote accepted How to find time complexity of O(Logn) Time Complexity of a loop is considered as O(Logn) if the loop variables is divided / multiplied by a constant amount. Its worst-case runtime complexity is O(n) Its best-case runtime complexity is O(1) Its average case runtime complexity is O(n/2)=O(n) Amortized Time Complexity Consider a dynamic array stack.
We could bet that the complexity $\in \Theta(n \log{\log{n}})$, but we should really do a formal proof. | 2018-07-21T23:17:41 | {
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https://www.physicsforums.com/threads/volume-of-a-hollow-cylinder-vs-cylindrical-shell.837923/ | # Volume of a Hollow Cylinder vs Cylindrical Shell
Tags:
1. Oct 15, 2015
### CStudy
In my physics lab, I am asked to calculate the volume of a hollow cylinder. The equation for the volume hollow cylinder below was given. Then, my curiosity made me wonder, is the volume of the hollow cylinder the same as the volume of a cylindrical shell used in calculus? At first though you would assume the answer as yes. However, I have tested this theory using various measurements, resulting in two different results. Can anyone help me understand why Hallow Cylinder does not equal Cylindrical Shell? or maybe disprove my results.
Hollow Cylinder =
(π)(height)((ro)2−(ri)2)
Cylindrical Shell = 2(π)(ri)(height)(thickness)
2. Oct 15, 2015
### Staff: Mentor
Welcome to the PF.
In the quoted text, I've fixed the r^2 terms.
The first equation is correct for the volume of a hollow cylinder. The second equation is used in calculus to calculate volumes, but what is the key assumption when it is used? You cannot use it for a cylindrical shell of a finite thickness...
3. Oct 15, 2015
### CStudy
The assumption is the cylindrical shell's thickness is infinitesimally small. I guess if you think about it, if you was to cut a hollow cylinder down the middle the surface are of one side would not equal the surface area of the other, unless the thickness was extremely, extremely, extremely small. thanks
4. Oct 15, 2015
### SteamKing
Staff Emeritus
However, the volume of the cylindrical shell, Vshell = 2πrht, is accurate enough when t << r. This volume is calculated knowing the circumference of the cylinder, which is 2πr, and then multiplying that by the height to get the surface area, 2πrh,and then multiplying the surface area by the thickness t to get the volume.
Let's take a case where h = 1 and ro = 1, and let ri vary a bit:
Code (Text):
ro ri Vcyl Vshell % Diff.
1 0.90 0.5969 0.6283 5.26
1 0.95 0.3063 0.3142 2.57
1 0.99 0.0625 0.0628 0.50
As you can see here, the closer ri comes to ro, the closer the volume of the shell comes to the volume of the hollow cylinder.
5. Oct 15, 2015
### CStudy
Awesome explanation. | 2017-12-12T20:56:16 | {
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https://www.physicsforums.com/threads/calculating-eigenvalues-help.934324/ | # Calculating Eigenvalues help
Pushoam
## The Attempt at a Solution
I solved it by calculating the eigen values by ##| A- \lambda |= 0 ##.
This gave me ## \lambda _1 = 6.42, \lambda _2 = 0.387, \lambda_3 = -0.806##.
So, the required answer is 42.02 , option (b).
Is this correct?
The matrix is symmetric. Is there any other easier wat to find the answer?
#### Attachments
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Homework Helper
Gold Member
Do you know the relation between the sum of the squares of the eigenvalues and the trace of a matrix?
Pushoam
Pushoam
Do you know the relation between the sum of the squares of the eigenvalues and the trace of a matrix?
No.
Homework Helper
Dearly Missed
No.
Pushoam
Mentor
2022 Award
I get the same result.
Pushoam
Gold Member
So, the required answer is 42.02 , option (b).
Is this correct?
The matrix is symmetric. Is there any other easier wat to find the answer?
The fact that its symmetric leads to some very nice results. What result do you get if you square each entry in your matrix and then sum them? This is called a squared Frobenius norm (which is one way of generalizing the L2 norm for vectors to matrices).
Pushoam
Staff Emeritus
Homework Helper
Gold Member
I would like to add to what the previous posters said that you should not round your numbers while doing your computations. Keep them on exact form to the very end and only evaluate the numbers in the end if necessary. You should find that the answer is exactly 42, not 42.02.
Pushoam
Pushoam
I got tr(A) = ##\Sigma \lambda_i## , i= 1,2,3.
and Det (A) = ## \Pi \lambda_i## , i= 1,2,3.
Better to use ## tr (A^2) = \Sigma {\lambda_i}^2## , i= 1,2,3.
## Tr(A^n) = Tr(D^n)##, where D is a similar matrix of A.
In case of a matrix having all distinct eigen values, D can be the diagonal matrix consisting of ##\lambda_i##.
In that case #### Tr(A^n) = Tr(D^n) = \Sigma {\lambda_i}^n## , i= 1,2,3.
But, the above two equation will not give the result and it is complcated, too.
It will be better to calculate the trace directly as I did in OP.
The fact that its symmetric leads to some very nice results. What result do you get if you square each entry in your matrix and then sum them? This is called a squared Frobenius norm (which is one way of generalizing the L2 norm for vectors to matrices).
I was looking for something like this. The sum is 42.
So, is it true for any symmetric matrix?
How to prove it?
## Tr (A^2) = \Sigma_i (A^2)_{ii}
\\ (A^2)_{ii} = \Sigma_j (A_{ij} A_{ji})##
For symmetric matrix, ## A_{ij} = A_{ji}##.
So, ## (A^2)_{ii} = \Sigma_j {(A_{ij} })^2
\\ Tr (A^2) = \Sigma_i \Sigma_j {(A_{ij} })^2##
Is this correct?
Staff Emeritus
Homework Helper
Gold Member
But, the above two equation will not give the result and it is complcated, too.
It will be better to calculate the trace directly as I did in OP.
For ##n = 2## it directly gives you the result. You just need to square the matrix and sum the diagonal of the result, very simple. If you go via the eigenvalues you need to solve for the roots of an order 3 polynomial.
## Tr (A^2) = \Sigma_i (A^2)_{ii}
\\ (A^2)_{ii} = \Sigma_j (A_{ij} A_{ji})##
For symmetric matrix, ## A_{ij} = A_{ji}##.
So, ## (A^2)_{ii} = \Sigma_j {(A_{ij} })^2
\\ Tr (A^2) = \Sigma_i \Sigma_j {(A_{ij} })^2##
Is this correct?
Essentially. However, I think it is easier to go the other way and just see that ##A_{ij}A_{ij} = \mbox{tr}(AA^T)##. Since ##A## is symmetric, ##AA^T = A^2##.
Pushoam
Gold Member
I was looking for something like this. The sum is 42.
So, is it true for any symmetric matrix?
How to prove it?
## Tr (A^2) = \Sigma_i (A^2)_{ii}
\\ (A^2)_{ii} = \Sigma_j (A_{ij} A_{ji)}##
For symmetric matrix, ## A_{ij} = A_{ji}##.
So, ## (A^2)_{ii} = \Sigma_j {(A_{ij} }^2
\\ Tr (A^2) = \Sigma_i \Sigma_j {(A_{ij} })^2##
Is this correct?
note: we are dealing in reals for this post. Your approach is close, and maybe even correct, but I find it hard to follow.
My strong preference here is to block your matrix by column vectors.
Suppose you have some matrix ##\mathbf X##, partitioned by columns below
##\mathbf X = \bigg[\begin{array}{c|c|c|c|c}
\mathbf x_1 & \mathbf x_2 &\cdots & \mathbf x_{n-1} & \mathbf x_n\end{array}\bigg]##
to make the link with the traditional L2 norm for vectors, consider the vec operator
##
vec\big(\mathbf X\big) = \begin{bmatrix}
\mathbf x_1 \\
\mathbf x_2\\
\vdots \\
\mathbf x_{n-1}\\
\mathbf x_n
\end{bmatrix}##
which stacks each column of the matrix ##\mathbf X## on top of each other into one big vector. (The vec operator will show up again if and when you start dealing with Kronecker products.)
Our goal is to add up each squared component of ##\mathbf X## into a sum. do you understand why
##\big \Vert \mathbf X \big \Vert_F^2 = \sum_{j=1}^n\sum_{i=1}^n x_{i,j}^2 = trace\big(\mathbf X^T \mathbf X\big) = vec\big(\mathbf X\big)^Tvec\big(\mathbf X\big)= \big \Vert vec\big(\mathbf X\big) \big \Vert_2^2##
is true for any real matrix?
Now since ##\mathbf X## is symmetric, we have ##\mathbf X^T = \mathbf X## meaning that
##\big \Vert \mathbf X \big \Vert_F^2 = trace\big(\mathbf X^T \mathbf X\big) = trace\big(\mathbf X \mathbf X\big) = trace\big(\mathbf X^2\big)##
now you just need the fact that others mentioned, i.e. relating a trace of a matrix and its eigenvalues (or in this case the trace of a matrix to the second power gives sum of eigenvalues to second power).
Why is this fact true? (Hint: use characteristic polynomial, or if you prefer an easy but less general case: real symmetric matrices are diagonalizable -- do that and apply cyclic property of trace.)
Trace is absurdly useful, so its worth spending extra time understanding all the related details of this problem.
Pushoam
Pushoam
For ##n = 2## it directly gives you the result. You just need to square the matrix and sum the diagonal of the result, very simple. If you go via the eigenvalues you need to solve for the roots of an order 3 polynomial.
Essentially. However, I think it is easier to go the other way and just see that ##A_{ij}A_{ij} = \mbox{tr}(AA^T)##. Since ##A## is symmetric, ##AA^T = A^2##.
Then, for the anti - symmetric matrix, ##tr (A^2)= tr (-AA^T) = - A_{ij}A_{ij}## = negative of the sum of the elements of the matrix. Right?
Gold Member
Then, for the anti - symmetric matrix, ##tr (A^2)= tr (-AA^T) = - A_{ij}A_{ij}## = negative of the sum of the elements of the matrix. Right?
Yes.
note that in general over real ##n## x ##n## matrices,
##\big \vert trace\big(\mathbf A \mathbf A \big)\big \vert = \big \vert trace \Big( \big( \mathbf A^T \big)^T \mathbf A\Big)\big\vert \leq trace\big(\mathbf A^T \mathbf A\big) = \big \Vert \mathbf A\big \Vert_F^2 ##
with equality iff ##\mathbf A## is a scalar multiple of ##\mathbf A^T##.
You could prove this with Schur's Inequality. Alternatively (perhaps using the vec operator to help) recognize that the trace gives an inner product. Direct application of Cauchy Schwarz gives you
##\big \vert trace\big(\mathbf B^T \mathbf A \big) \big \vert = \big \vert vec\big( \mathbf B\big)^T vec\big( \mathbf A\big)\big \vert \leq \big \Vert vec\big( \mathbf B\big)\big \Vert_2 \big \Vert vec\big( \mathbf A\big)\big \Vert_2 =\big \Vert \mathbf B \big \Vert_F \big \Vert \mathbf A \big \Vert_F##
with equality iff ##\mathbf B = \gamma \mathbf A##. (Also note trivial case: if one or both matrices is filled entirely with zeros, then there is an equality.)
In your real skew symmetric case, ##\mathbf B = \mathbf A^T## and ##\gamma = -1##. And of course in the real symmetric case ##\gamma = 1##
Pushoam
Pushoam
Then, for the anti - symmetric matrix, ##tr (A^2)= tr (-AA^T) = - A_{ij}A_{ij}## = negative of the sum of the elements of the matrix. Right?
I missed to write square of the elements.
The corrected one:
Then, for the anti - symmetric matrix, ##tr (A^2)= tr (-AA^T) = - A_{ij}A_{ij}## = negative of the sum of the square of the elements of the matrix.
Homework Helper
Gold Member
I am not very familiar with some of the algebra mentioned, though I don’t think it is very difficult.
However, it seems possible to solve the problem without it knowing all this, though I’m sure it does no harm to know it.
You could write out the eigenvalue equation as a cubic equation. The value of the sums of roots ∑λi is well known. The value of the sum of products of two roots, Σλiλj is well known. From this you could get the sum of squares of roots Σλi2.
I have not looked into it, but from what was being said about symmetry I suspect it would be easy to solve this cubic.
Last edited:
Pushoam
Homework Helper
Dearly Missed
I am not very familiar with some of the algebra mentioned, though I don’t think it is very difficult.
However, it seems possible to solve the problem without it knowing all this, though I’m sure it does no harm to know it.
You could write out the eigenvalue equation as a cubic equation. The value of the sums of roots ∑λi is well known. The value of the sum of products two roots, Σλiλj is well known. From this you could get the sum of squares of roots Σλi2.
I have not looked into it, but from what was being said about symmetry I suspect it would be easy to solve this cubic.
Sometimes symmetry does not help at all. For example, the matrix
$$A = \pmatrix{1&2&3\\2&4&5\\3&5&6}$$
has eigenvalues that are pretty horrible expressions involving cube roots and arctangents of things involving square roots, and the like.
jim mcnamara and StoneTemplePython | 2023-03-30T18:42:11 | {
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https://mathhelpboards.com/threads/inequality-proof-w-induction-2-unknowns.6984/ | # inequality proof w/ induction, 2 unknowns
#### skatenerd
##### Active member
I am given a statement to prove: Show (without using the Binomial Theorem) that $$(1+x)^n\geq{1+nx}$$ for every real number $$x>-1$$ and natural numbers $$n\geq{2}$$. I am given a hint to fix $$x$$ and apply induction on $$n$$.
I started by supposing $$x$$ is a fixed, real number larger than -1, and then calling the given formula $$P(n)$$, and evaluating $$P(n)$$ at the base case $$n=2$$.
This gives $$(1+x)^2\geq{1+2x}$$ which can be rewritten as $$1+2x+x^2\geq{1+2x}$$.
It is know that for all real $$x$$, the statement $$x^2\geq{0}$$ is true.
Here is where I get tripped up.
We need to assume that $$m=n$$ a.k.a. $$P(m)$$ is true for all natural $$m\geq{2}$$.
So we have $$(1+x)^m\geq{1+mx}$$. Now we need to show that $$P(m+1)$$ holds to be true. $$P(m+1)$$:
$$(1+x)^{m+1}\geq{1+(m+1)x}$$.
Now here I would usually try to translate the original formula by plugging in what we had originally, but I am pretty iffy on how to do this with an inequality. If anybody could help me construct a new formula that would help me prove that
$$(1+x)^{m+1}\geq{1+(m+1)x}$$ is true I would be very thankful.
#### Prove It
##### Well-known member
MHB Math Helper
I am given a statement to prove: Show (without using the Binomial Theorem) that $$(1+x)^n\geq{1+nx}$$ for every real number $$x>-1$$ and natural numbers $$n\geq{2}$$. I am given a hint to fix $$x$$ and apply induction on $$n$$.
I started by supposing $$x$$ is a fixed, real number larger than -1, and then calling the given formula $$P(n)$$, and evaluating $$P(n)$$ at the base case $$n=2$$.
This gives $$(1+x)^2\geq{1+2x}$$ which can be rewritten as $$1+2x+x^2\geq{1+2x}$$.
It is know that for all real $$x$$, the statement $$x^2\geq{0}$$ is true.
Here is where I get tripped up.
We need to assume that $$m=n$$ a.k.a. $$P(m)$$ is true for all natural $$m\geq{2}$$.
So we have $$(1+x)^m\geq{1+mx}$$. Now we need to show that $$P(m+1)$$ holds to be true. $$P(m+1)$$:
$$(1+x)^{m+1}\geq{1+(m+1)x}$$.
Now here I would usually try to translate the original formula by plugging in what we had originally, but I am pretty iffy on how to do this with an inequality. If anybody could help me construct a new formula that would help me prove that
$$(1+x)^{m+1}\geq{1+(m+1)x}$$ is true I would be very thankful.
\displaystyle \begin{align*} \left( 1 + x \right) ^{m + 1} &= \left( 1 + x \right) \left( 1 + x \right) ^m \\ &\geq \left( 1 + x \right) \left( 1 + m\,x \right) \\ &= 1 + \left( m + 1 \right) \, x + m\,x^2 \\ &\geq 1 + \left( m + 1 \right) \, x \end{align*}
#### MarkFL
##### Administrator
Staff member
Okay, you have shown the base case is true, and so your induction hypothesis $P_m$ is:
$$\displaystyle (1+x)^m\ge1+mx$$
I would consider for the inductive step:
$$\displaystyle (1+x)^{m+1}-(1+x)^{m}=(1+x)^{m}(1+x-1)=(1+x)^{m}x$$
Can you use the inductive hypothesis to construct from this a weak inequality that you can then add to $P_m$?
#### skatenerd
##### Active member
Thanks to both of you for the responses. So I was actually able to figure out and finish the proof going by what Prove It wrote. But to MarkFL, what do you mean by constructing a "weak inequality"? Does that refer to the point where you find an inequality where you have $$mx^2\geq{0}$$?
#### MarkFL
##### Administrator
Staff member
What I had in mind is to take the equation:
$$\displaystyle (1+x)^{m+1}-(1+x)^{m}=(1+x)^{m}x$$
and then use the induction hypothesis (which we have multiplied by $x$) to write:
$$\displaystyle (1+x)^{m}x\ge(1+mx)x$$
so that we now have:
$$\displaystyle (1+x)^{m+1}-(1+x)^{m}\ge(1+mx)x$$
Adding this to $P_m$, we get:
$$\displaystyle (1+x)^{m+1}\ge 1+mx+(1+mx)x=1+(m+1)x+mx^2$$
Since $mx^2\ge0$, we have:
$$\displaystyle (1+x)^{m+1}\ge1+(m+1)x+mx^2\ge1+(m+1)x$$
And so we may conclude:
$$\displaystyle (1+x)^{m+1}\ge1+(m+1)x$$
Thus, we have derived $P_{m+1}$ from $P_{m}$, thereby completing the proof by induction.
#### skatenerd
##### Active member
Ahh I see. Yeah I ended up writing something similar to that, with the same ending. Thanks for the help guys! | 2022-07-05T14:56:47 | {
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https://math.stackexchange.com/questions/673550/finding-a-basis-for-bbbq-sqrt2-sqrt3-over-bbbq | # Finding a basis for $\Bbb{Q}(\sqrt{2}+\sqrt{3})$ over $\Bbb{Q}$.
I have to find a basis for $\Bbb{Q}(\sqrt{2}+\sqrt{3})$ over $\Bbb{Q}$.
I determined that $\sqrt{2}+\sqrt{3}$ satisfies the equation $(x^2-5)^2-24$ in $\Bbb{Q}$.
Hence, the basis should be $1,(\sqrt{2}+\sqrt{3}),(\sqrt{2}+\sqrt{3})^2$ and $(\sqrt{2}+\sqrt{3})^3$.
However, this is not rigorous. How can I be certain that $(x^2-5)^2-24$ is the minimal polynomial that $\sqrt{2}+\sqrt{3}$ satisfies in $\Bbb{Q}$? What if the situation was more complicated? In general, how can we ascertain thta a given polynomial is irreducible in a field?
Moreover, checking for linear independence of the basis elements may also prove to be a hassle. Is there a more convenient way of doing this?
Thanks.
One easy way: $\rm\ F = \Bbb Q(\sqrt{3}+\sqrt{2}) \supseteq \Bbb Q(\sqrt{3},\sqrt{2})\,$ (and reverse is clear), since $\rm\,F\,$ contains not only $\, u = \sqrt{3}+\sqrt{2}\,$ but also $\,v = \sqrt{3}-\sqrt{2} = (3-2)/(\sqrt{3}+\sqrt{2}), \,$ thus $\,\sqrt{3},\sqrt{2} = (u\pm v)/2 \in\rm F.\$ QED $\,$ Below is further discussion from some of my older posts.
If field F has $2\,$ F-linear independent combinations of $\rm\, \sqrt{a},\ \sqrt{b}\,$ then we can solve for $\rm\, \sqrt{a},\ \sqrt{b}\,$ in F. For example, the Primitive Element Theorem works that way, obtaining two such independent combinations by Pigeonholing the infinite set $\rm\ F(\sqrt{a} + r\ \sqrt{b}),\ r \in F,\ |F| = \infty,\,$ into the finitely many fields between F and $\rm\ F(\sqrt{a}, \sqrt{b}),\,$ e.g. see PlanetMath's proof.
In this case it is simpler to notice $\rm\ F = \mathbb Q(\sqrt{a} + \sqrt{b})\$ contains the independent $\rm\ \sqrt{a} - \sqrt{b}\$ since
$$\rm \sqrt{a}\ -\ \sqrt{b}\ =\ \dfrac{\ a\,-\,b}{\sqrt{a}+\sqrt{b}}\ \in\ F = \mathbb Q(\sqrt{a}+\sqrt{b})$$
To be explicit, notice that $\rm\ u = \sqrt{a}+\sqrt{b},\ v = \sqrt{a}-\sqrt{b}\in F\$ so solving the linear system for the roots yields $\rm\ \sqrt{a}\ =\ (u+v)/2,\ \ \sqrt{b}\ =\ (u-v)/2,\$ both of which are clearly $\rm\,\in F,\:$ since $\rm\:u,\:v\in F\:$ and $\rm\:2\ne 0\:$ in $\rm\:F,\:$ so $\rm\:1/2\:\in F.\:$ This works over any field where $\rm\:2\ne 0\:,\:$ i.e. where the determinant (here $2$) of the linear system is invertible, i.e. where the linear combinations $\rm\:u,v\:$ of the square-roots are linearly independent over the base field.
More generally, one may use the following lemma (which is the basis of a general result on linear independence of square roots due to Besicovitch, see below).
Lemma $\rm\ \ [K(\sqrt{a},\sqrt{b}) : K] = 4\$ if $\rm\ \sqrt{a},\ \sqrt{b},\ \sqrt{a\:b}\,$ are all $\rm\,\not\in K,\:$ and $\rm\: 2 \ne 0\:$ in $\rm\,K.$
Proof $\ \$ Let $\rm\ L = K(\sqrt{b})\:.\:$ Then $\rm\: [L:K] = 2\:$ via $\rm\:\sqrt{b} \not\in K,\:$ thus it suffices to show $\rm\: [L(\sqrt{a}):L] = 2\:.\:$ It fails only if $\rm\:\sqrt{a} \in L = K(\sqrt{b})\$ and then $\rm\ \sqrt{a}\ =\ r + s\ \sqrt{b}\$ for $\rm\ r,s\in K.\:$ But that's impossible, since squaring yields $\rm(1):\ \ a\ =\ r^2 + b\ s^2 + 2\:r\:s\ \sqrt{b}\:,\:$ contra, hypotheses, as follows
$\rm\quad\quad\quad\quad\quad\quad\quad\quad rs \ne 0\ \ \Rightarrow\ \ \sqrt{b}\ \in\ K\ \$ by solving $(1)$ for $\rm\sqrt{b}\:,\:$ using $\rm\:2 \ne 0$
$\rm\quad\quad\quad\quad\quad\quad\quad\quad\ s = 0\ \ \Rightarrow\ \ \ \sqrt{a}\ \in\ K\ \$ via $\rm\ \sqrt{a}\ =\ r \in K$
$\rm\quad\quad\quad\quad\quad\quad\quad\quad\ r = 0\ \ \Rightarrow\ \ \sqrt{a\:b}\in K\ \$ via $\rm\ \sqrt{a}\ =\ s\ \sqrt{b}\:,\: \$times $\rm\:\sqrt{b}\quad$ QED
Using the above as the inductive step one easily proves the following result of Besicovic.
Theorem $\$ Let $\rm\:Q\:$ be a field with $2 \ne 0\:,\:$ and $\rm\ L = Q(S)\$ be an extension of $\rm\:Q\:$ generated by $\rm\: n\:$ square roots $\rm\ S = \{ \sqrt{a}, \sqrt{b},\ldots \}$ of elts $\rm\ a,\:b,\:\ldots \in Q\:.\:$ If every nonempty subset of $\rm\:S\:$ has product not in $\rm\:Q\:$ then each successive adjunction $\rm\ Q(\sqrt{a}),\ Q(\sqrt{a},\:\sqrt{b}),\:\ldots$ doubles the degree over $\rm\,Q,\,$ so, in total, $\rm\: [L:Q] \ =\ 2^n\:.\:$ So the $\rm\:2^n\:$ subproducts of the product of $\rm\:S\:$ comprise a basis of $\rm\:L\:$ over $\rm\:Q\:.$
In the present case you can argue as follows: clearly
$$\Bbb Q(\sqrt2)\subsetneqq\Bbb Q(\sqrt2+\sqrt3)\implies \dim_{\Bbb Q}\Bbb Q(\sqrt2+\sqrt3)>\dim_{\Bbb Q}\Bbb Q(\sqrt2)=2$$
But since you already found a quartic that has $\;\sqrt2+\sqrt3\;$ as a root and also
$$2\mid\dim_{\Bbb Q}\Bbb Q(\sqrt2+\sqrt3)\;\;\text{(by the above!)}$$
then the dimension must be exactly four and your polynomial is the minimal one and thus irreducible.
In the general case I don't think there's a general algorithm by which to tell whether a given polynomial is irreducible or not.
• Actually there are very good algorithms to factor univariate polynomials over the rationals, including a polynomial-time algorithm based on LLL. link.springer.com/article/10.1007%2FBF01457454 Feb 12 '14 at 17:35
A basis of $\mathbb Q\big[\sqrt{2},\sqrt{3}\big]$ consists of the elements $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$, and hence its dimension over $\mathbb Q$ is equal to $4$.
Clearly, all the above elements belong to $\mathbb Q\big[\sqrt{2},\sqrt{3}\big]$, and hence it remains to show that they are independent over $\mathbb Q$.
There is a rather elegant way to show this, and in fact something more general:
If $n_1,n_2,\ldots,n_k$ are distinct square-free positive integers, then $\sqrt{n_1},\sqrt{n_2},\ldots,\sqrt{n_k}$ are linearly independent over $\mathbb Q$.
(A number $n$ is said to be square free if $k^2\mid n$, implies that $k=1$.)
• Regarding generalizations, if you follow the fnal link in my answer you will find a proof of a more general result of Besicovitch, along with references to even further generalizations by Mordell and Siegel. Feb 17 '14 at 21:42 | 2021-10-27T13:03:31 | {
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https://www.proofwiki.org/wiki/Definition:Interval_of_Ordered_Set | Definition:Interval/Ordered Set
Definition
Let $\struct {S, \preccurlyeq}$ be an ordered set.
Let $a, b \in S$.
The intervals between $a$ and $b$ are defined as follows:
Open Interval
The open interval between $a$ and $b$ is the set:
$\openint a b := a^\succ \cap b^\prec = \set {s \in S: \paren {a \prec s} \land \paren {s \prec b} }$
where:
$a^\succ$ denotes the strict upper closure of $a$
$b^\prec$ denotes the strict lower closure of $b$.
Left Half-Open Interval
The left half-open interval between $a$ and $b$ is the set:
$\hointl a b := a^\succ \cap b^\preccurlyeq = \set {s \in S: \paren {a \prec s} \land \paren {s \preccurlyeq b} }$
where:
$a^\succ$ denotes the strict upper closure of $a$
$b^\preccurlyeq$ denotes the lower closure of $b$.
Right Half-Open Interval
The right half-open interval between $a$ and $b$ is the set:
$\hointr a b := a^\succcurlyeq \cap b^\prec = \set {s \in S: \paren {a \preccurlyeq s} \land \paren {s \prec b} }$
where:
$a^\succcurlyeq$ denotes the upper closure of $a$
$b^\prec$ denotes the strict lower closure of $b$.
Closed Interval
The closed interval between $a$ and $b$ is the set:
$\closedint a b := a^\succcurlyeq \cap b^\preccurlyeq = \set {s \in S: \paren {a \preccurlyeq s} \land \paren {s \preccurlyeq b} }$
where:
$a^\succcurlyeq$ denotes the upper closure of $a$
$b^\preccurlyeq$ denotes the lower closure of $b$.
Endpoint
The elements $a, b \in S$ are known as the endpoints (or end points) of the interval.
$a$ is sometimes called the left hand endpoint and $b$ the right hand end point of the interval.
Wirth Interval Notation
The notation used on this site to denote an interval of an ordered set $\struct {S, \preccurlyeq}$ is a fairly recent innovation, and was introduced by Niklaus Emil Wirth:
$\ds \openint a b$ $:=$ $\ds \set {s \in S: \paren {a \prec s} \land \paren {s \prec b} }$ Open Interval $\ds \hointr a b$ $:=$ $\ds \set {s \in S: \paren {a \preccurlyeq s} \land \paren {s \prec b} }$ Right Half-Open Interval $\ds \hointl a b$ $:=$ $\ds \set {s \in S: \paren {a \prec s} \land \paren {s \preccurlyeq b} }$ Left Half-Open Interval $\ds \closedint a b$ $:=$ $\ds \set {s \in S: \paren {a \preccurlyeq s} \land \paren {s \preccurlyeq b} }$ Closed Interval
Also see
• Results about intervals can be found here. | 2023-04-02T12:24:26 | {
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http://www.mozartreina.com/sieve-of-eratosthenes.html | ## The Sieve of Eratosthenes
### Sieve of what??
The Sieve of Eratosthenes is an ancient algorithm for finding all prime numbers up to a given limit. It is one of the most efficient ways of finding small prime numbers, however above a certain threshold other sieves like the Sieve of Atkin or Sieve of Sundaram should be considered instead.
The sieve itself is quite simple to understand, and translating it to code is very straightforward, however since writing a blog post about current exercises is highly encouraged by a course that I'm taking right now, I took the opportunity to start writing again.
### How it works
The Sieve of Eratosthenes works by identifying a prime within the range of numbers you're considering (ex. 1 to 25), then crossing out all multiples of that prime (since it's a multiple it obviously can't be a prime...).
$$\left[\begin{matrix} 1 & 2 & 3 & 4 & 5 \\ 6 & 7 & 8 & 9 & 10 \\ 11 & 12 & 13 & 14 & 15 \\ 16 & 17 & 18 & 19 & 20 \\ 21 & 22 & 23 & 24 & 25 \\ \end{matrix}\right]$$
Starting with 2, the smallest prime, you then eliminate all numbers that are multiples of 2.
$$\require{cancel} \left[\begin{matrix} 1 & 2 & 3 & \cancel{4} & 5 \\ \cancel{6} & 7 & \cancel{8} & 9 & \cancel{10} \\ 11 & \cancel{12} & 13 & \cancel{14} & 15 \\ \cancel{16} & 17 & \cancel{18} & 19 & \cancel{20} \\ 21 & \cancel{22} & 23 & \cancel{24} & 25 \\ \end{matrix}\right]$$
Which leaves us with:
$$\left[\begin{matrix} 1 & 2 & 3 & & 5 \\ & 7 & & 9 & \\ 11 & & 13 & & 15 \\ & 17 & & 19 & \\ 21 & & 23 & & 25 \\ \end{matrix}\right]$$
Then you take the next number after that prime, in this case 3, and do it all over again.
$$\left[\begin{matrix} 1 & 2 & 3 & & 5 \\ & 7 & & \cancel{9} & \\ 11 & & 13 & & \cancel{15} \\ & 17 & & 19 & \\ \cancel{21} & & 23 & & 25 \\ \end{matrix}\right]$$
Leaving us:
$$\left[\begin{matrix} 1 & 2 & 3 & & 5 \\ & 7 & & & \\ 11 & & 13 & & \\ & 17 & & 19 & \\ & & 23 & & 25 \\ \end{matrix}\right]$$
And again take the next prime, 5, and eliminate all multiples of 5 (25 seems to be the only victim here). Eventually the numbers remaining will be:
$$\left[ \begin{matrix} 2 & 3 & 5 & 7 & 11 & 13 & 17 & 19 & 23 \end{matrix} \right ]$$
$$\ast$$ 1 is not considered a prime number.
So as you can see, the technique is quite simple and effective, albeit a little time-consuming, especially for large ranges. This is where computers enter the stage.
### The Essence of Programming
...is to make the computer do the repetitive, computational-demanding work. The course I'm taking uses Ruby so I've coded this in Ruby as well.
#### Breaking it down
First, we need to create the sequence of numbers, given a limit as input.
Then we have the part of the program that removes all the multiples of the given prime. Here we use the select method to filter out all numbers that DO NOT leave a remainder (hence a multiple) when divided with the prime.
Then we just iterate over the range, methodically removing numbers until we are only left with the primes.
### Testing it
Here are a few runs to make sure that it works:
Numbers 1 - 10.
2.0.0-p647 :561 > Sieve.new(10).primes
=> [2, 3, 5, 7]
Numbers 1 - 25
2.0.0-p647 :524 > Sieve.new(25).primes
=> [2, 3, 5, 7, 11, 13, 17, 19, 23]
Numbers 1-100
2.0.0-p647 :576 > Sieve.new(100).primes
=> [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
The complete source (I had to use a class since the test-suite provided assumed a class, although for these types of problems I would have just used functions normally): | 2020-01-23T20:48:38 | {
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http://mathhelpforum.com/calculus/156550-i-m-trying-understand-why-domain-function.html | # Math Help - I'm trying to understand why this is the domain of this function
1. ## I'm trying to understand why this is the domain of this function
I'm trying to find the domain for the function: square root of (9-x^2)...So what I do is solve for 9-x^2 >= 0 (b/c you can't have a negative number under a radical)...and get x is less than or equal to +3 and -3....
but the domain is supposedly [-3,3]...which would mean x > or = -3....but when I solved for it I got that x < or = -3...IT DOESN'T MAKE SENSE...
2. Originally Posted by jonjon1324
I'm trying to find the domain for the function: square root of (9-x^2)...So what I do is solve for 9-x^2 >= 0 (b/c you can't have a negative number under a radical)...and get x is less than or equal to +3 and -3....
but the domain is supposedly [-3,3]...which would mean x > or = -3....but when I solved for it I got that x < or = -3...IT DOESN'T MAKE SENSE...
Draw the graph of y = 9 - x^2. For what values of x is $y \geq 0$ ....?
3. Originally Posted by mr fantastic
Draw the graph of y = 9 - x^2. For what values of x is $y \geq 0$ ....?
Yeah, in the graph I can see that, but if I'm actually writing out the problem, and I find the square root of 9, how does the inequality flip from x <= + and -3 to x >= -3?
4. Originally Posted by jonjon1324
Yeah, in the graph I can see that, but if I'm actually writing out the problem, and I find the square root of 9, how does the inequality flip from x <= + and -3 to x >= -3?
Unless it's a linear inequality my advice for solving inequalities is to use a graph.
However, if you must use algebra, then I suggest you consider $(3 - x)(3 + x) \geq 0$ and then solve the following two cases for x:
Case 1: $3 - x \geq 0$ AND $3 + x \geq 0$.
Case 2: $3 - x \leq 0$ AND $3 + x \leq 0$.
5. Originally Posted by jonjon1324
I'm trying to find the domain for the function: square root of (9-x^2)...So what I do is solve for 9-x^2 >= 0 (b/c you can't have a negative number under a radical)...and get x is less than or equal to +3 and -3....
This is your error. You cannot "solve" an inequality like this just by taking the square root of both sides. Specifically, $9- x^2 \ge 0$ does give $9\ge x^2$ since you can add $x^2$ to both sides with changing the inequality sign. But you cannot then say " $\pm 3\ge x$".
When you are squaring, or taking the square root of both sides of an inequality, that is equivalent to multipying or dividing by an unknown- specifically you do not know whether that unknown is positive or negative and remember that multiplying or dividing both sides of an inequality by a negative number, the inequality sign reverses.
Once you have $9\ge x^2$, I would recommend first solving the associated equation. That is, $9= x^2$ which does, in fact, have x= 3 and x= -3 as solutions. That divides the number line into 3 intervals, $(-\infty, -3)$, $(-3, 3)$, and $(3, \infty)$ on each interval of which the inequality is either true for every point in the interval or false for every point in the interval. (That is true because for any continuous function, f(x), f(x)< 0 can only change to f(x)> 0, and vice-versa, where f(x)= 0.)
Knowing that the three intervals are $(-\infty, -3)$, $(-3, 3)$, and $(3, \infty)$. We need only look at a single point in each interval. x= -4 is in $(-\infty, -3)$ and $(-4)^2= 16> 9$ so $x^2> 9$ for all x in $(-\infty, -3)$. x= 0 is in $(-3, 3)$ and $(0)^2= 0< 9$ so $x^2< 9$ for all x in $(-3, 3)$. Finally, $x= 4$ is in (3, \infty) and $4^2= 16> 9$ so $x^2> 9$ for all x in $(3, \infty)$. Add to this the fact that $3^2= (-3)^2= 9$ and we have that $x^2\le 9$ if and only if x is in [-3, 3] or $-3\le x\le 3$.
Going back to $9- x^2= (3- x)(3+ x)= -(x- 3)(x+ 3)\ge 0$, another and perhaps simpler way to see that is this: $a- b> 0$ if and only if $a> b$ and $a- b< 0$ if and only if a< b. If x< -3 then it is also less than 3 so both of x- 3 and x+ 3= x-(3) are negative. Their product (the product of two negative numbers) is positive and so -(x- 3)(x+ 3) is [b]negative[/tex]. If -3< x< 3, then x- 3 is still negative but x+ 3= x-(-3) is now positive. Their product (the product of a positive and negative number) is negative so -(x- 3)(x+ 3) is positive. Finally, if x> 3, it is also greater than -3 so both x- 3 and x+ 3 are positive. The product of two positive numbers is positive so -(x- 3)(x+ 3) is negative. Again, we have that $x^2- 9\ge 0$ if and only if $-3\le x\le 3$.
but the domain is supposedly [-3,3]...which would mean x > or = -3....but when I solved for it I got that x < or = -3...IT DOESN'T MAKE SENSE... | 2014-08-29T07:04:25 | {
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http://mathhelpforum.com/differential-geometry/227227-subdividing-interval.html | 1. ## Subdividing an interval
Suppose I have a closed, bounded, non-degenerate interval $\displaystyle I = [a, b]$.
I have a set X initially containing the points a and b.
My first action is to find the midpoint of I, call it c, and add c to X.
I now have two new intervals $\displaystyle I_{1}=[a, c]$ and $\displaystyle I_{2}=[c, b]$.
My next action is to find the midpoints of $\displaystyle I_{1}$ and $\displaystyle I_{2}$, lets call them $\displaystyle c_{1}, c_{2}$ and add them to X.
I now have four intervals.... and I continue this procedure until I can subdivide no more.
Does X contain all points in the original interval I? Or stated differently, is there a point in I that is not in X?
2. ## Re: Subdividing an interval
Originally Posted by director
Suppose I have a closed, bounded, non-degenerate interval $\displaystyle I = [a, b]$.
I have a set X initially containing the points a and b.
My first action is to find the midpoint of I, call it c, and add c to X.
I now have two new intervals $\displaystyle I_{1}=[a, c]$ and $\displaystyle I_{2}=[c, b]$.
My next action is to find the midpoints of $\displaystyle I_{1}$ and $\displaystyle I_{2}$, lets call them $\displaystyle c_{1}, c_{2}$ and add them to X.
I now have four intervals.... and I continue this procedure until I can subdivide no more.
Does X contain all points in the original interval I? Or stated differently, is there a point in I that is not in X?
What makes you think that "I can subdivide no more?" At each stage your set X is only finite.
Have you studied the famous Cantor Set ?
3. ## Re: Subdividing an interval
Ignoring the point about "subdivide no more", isn't it obvious that the set of all subintervals does include every point in the original interval? Ever point, p, in [a, b] satisfies $\displaystyle a\le p\le b$. For any point c, either $\displaystyle x\le c$ or $\displaystyle c\le x$ or both. So if x is in [a, b], either x is in [a, c] or x is in [c,b]. In either case, x is in [a, c]U[c, b]. The general proof can be done by induction on the number of halvings.
4. ## Re: Subdividing an interval
Originally Posted by HallsofIvy
Ignoring the point about "subdivide no more", isn't it obvious that the set of all subintervals does include every point in the original interval? Ever point, p, in [a, b] satisfies $\displaystyle a\le p\le b$. For any point c, either $\displaystyle x\le c$ or $\displaystyle c\le x$ or both. So if x is in [a, b], either x is in [a, c] or x is in [c,b]. In either case, x is in [a, c]U[c, b]. The general proof can be done by induction on the number of halvings.
Originally Posted by director
Suppose I have a closed, bounded, non-degenerate interval $\displaystyle I = [a, b]$.
I have a set X initially containing the points a and b.
My first action is to find the midpoint of I, call it c, and add c to X.
I now have two new intervals $\displaystyle I_{1}=[a, c]$ and $\displaystyle I_{2}=[c, b]$.
My next action is to find the midpoints of $\displaystyle I_{1}$ and $\displaystyle I_{2}$, lets call them $\displaystyle c_{1}, c_{2}$ and add them to X.
At each stage the interval is not added to the set only the new midpoints.
Then we form a partition of $[a,b]$ with that new set of points and find the midpoints of each new subinterval.
So after each stage we have a set of a finite collection. | 2018-04-25T21:14:48 | {
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https://math.stackexchange.com/questions/1390365/what-is-the-difference-between-dense-and-closed-sets/1390367 | # What is the difference between dense and closed sets?
I am self studying topology and is a bit stuck on the difference between dense and closed sets. Intuitively, a dense set is a set where all elements are close to each other and a closed set is a set having all of its boundary points.
But to make this more concrete, can someone give me an example of a closed set that is not dense and a dense set that is not closed?
Thanks!
• Can you give an example of a closed set (in a topological space $X$) which is dense? Hint: there is only one.
– bof
Aug 9, 2015 at 6:25
• I think the important point of your misunderstanding is your intuition... "Intuitively, a dense set is a set where all elements are close to each other" - This is untrue. Intuitively, a dense set A in a given space X is a set where all elements of X are close to elements of A Aug 9, 2015 at 20:42
• When a set is dense in a space, I usually hear it described as 'A is dense in X'; I think that nicely emphasises the point in the previous comment. Sep 1, 2015 at 3:30
$[0,1]$ is a closed subset of $\mathbb R$ that is not dense. It contains all of its limit points, so it is closed. Some points in $\mathbb R$, for example $2$, are not limit points of this set, so the set is not dense.
$\mathbb Q$ is a dense subset of $\mathbb R$ that is not closed. It is not closed because it does not contain all of its limit points. For example $\sqrt 2$ is a limit point of this set because every open neighborhood of $\sqrt 2$ contains some rational numbers. It is dense because every point in $\mathbb R$ is one of its limit points.
• Good example !!
– mick
Jan 14, 2017 at 1:17
• @mick : Glad you liked it. Jan 14, 2017 at 1:29
I want to add one thing. The only closed, dense set in a topological space is the space itself!
So these two concepts are pretty far apart. So far that in most situations, they are mutually exclusive!
• Would the proof of that go like this? Let T be a closed dense subset of S, then for every x in S there exist points in T arbitrarily close to x (by definition of dense) and since T contains points arbitrarily close to x, then x must be in T (by definition of closed). Thus any x in S is also an element of T. Aug 9, 2015 at 11:15
• @kasperd 'closeness' isn't a very well defined concept for arbitrary topological spaces. Suppose $X$ is a topological space and $T$ is a dense closed subset of $X$. Let $U$ be the complement of $T$ which is open by definition. By the definition of dense, $T$ and $U$ have a non-empty intersection if and only if $U$ is empty, but $T$ and $U$ are disjoint and hence $U$ must be the empty set. Aug 9, 2015 at 12:33
A set is dense/closed in a given topological space.
$[0,1]$ is closed in $\mathbb{R}$ but it is not dense in $\mathbb{R}$ since there are real numbers that can not be approached arbitrarily close by elements of $[0,1]$.
$[0,1]\setminus\{\frac{1}{2}\}$ is dense in $[0,1]$ but it is not closed in it.
Take $\mathbb R$ with the usual topology. Then the set of integers is closed (any converging series of integers converges to an integer), but not dense (you get nowhere close to $1/2$). On the other hand, the set of rational numbers is dense, but not closed (the limit of a sequence of rational numbers can be irrational).
IMHO a good intuition for a closed set is that while staying in the set, you cannot get arbitrary close to any point outside of that set. Which is actually the exact opposite to a dense set which gets close to each point of the space.
With that intuition it's also immediately clear why the only closed dense set is the space itself: The only way to come close to each point without coming close to any point outside of the set is if there are no points outside of the set.
Let the metric space be $\mathbb{R}$. Then $\{0\}$ is closed but not dense in $\mathbb{R}$. While $\mathbb{Q}$ is dense but not closed in $\mathbb{R}$.
Loosely speaking:
Suppose $X$ is a topological space (with some topology, obviously).
If $A \subseteq X$ and we say $A$ is dense in $X$, we mean that for each point $x \in X$, we can keep finding points from $A$ "closer" and "closer" to $x$.
If $B \subseteq X$ and we say $B$ is closed in $X$, we mean if you can find points in $X$ that are really really close to elements of $B$, then those points from $X$ are also in $B$. In other words, if $y$ is an element that is not in $B$, then none of the points nearby it are in $B$ either (otherwise, if we could keep finding stuff from $B$ closer and closer to $y$, then $y$ would be in $B$).
I'd say the most glaring difference between the two definitions is that dense sets need not contain their limit points. Closed sets necessarily do. There are plenty of examples of "dense not closed" and "closed not dense" scattered throughout this question now.
I'd also argue that your intuition for a dense set only makes sense in spaces where distance makes sense, like $\Bbb{R}^n$. As long as that helps you warm up to the definition, that's good. But in more general spaces things won't be that intuitive. For example, the set $X=\left\{\square ,\triangle\right\}$. A topology for the set is $$\mathcal{T}=\left\{\emptyset, \left\{\square ,\triangle\right\},\left\{\square\right\} \right\}$$ We see that $\square$ is open in $X$ and that $\text{Cl}(\square) = X$ so $\square$ is dense in $X$. But "closeness" doesn't mean a whole lot either.
In a topological space $(X,\tau)$ a subset $A \subset X$ is dense, iff its closure is the whole space, i.e. $\overline{A} = X$, while a subset $B \subset X$ is closed iff it is its own closure, i.e. $\overline{B} = B$.
To recite the example of $\mathbb{R}$ with euclidian topology:
• $\mathbb{Q} \subset \mathbb{R}$ is dense but not closed, since: $\overline{\mathbb{Q}} = \mathbb{R}$.
• $[0,1] \subset \mathbb{R}$ is closed but not dense, since $\overline{[0,1]} = [0,1]$.
One explanation is in terms of continuous functions.
A continuous function (on a topological space $X$) is determined by its values on a dense subset of $X$. This is a defining characteristic of dense sets; any non-dense subset does not have this property.
A prototype of a closed subset of $X$ is the solution set of a collection of equations defined by continuous functions; more abstractly, the set of points where a continuous function (from $X$ into some other topological space $Y$) attains a particular value. This might encompass all closed subsets under some fairly general hypotheses that endow $X$ it with "enough" continuous functions, or maybe allowing arbitrary $Y$ is enough. This picture is made precise in a more advanced setting by dualities between algebras of functions and topological (or geometric, or algebra-geometric, ....) spaces.
• Note that for $(X,\tau),(Y,\sigma)$ topological spaces your first paragraph only holds for continuous functions $f : (X, \tau) \to (Y, \sigma)$ if $(Y,\sigma)$ is hausdorff. Aug 9, 2015 at 21:04
• I guess I'm overly used to working in tame categories of spaces. If you know of a more general way of stating the same sort of point I would be interested to learn it. Aug 9, 2015 at 21:07
• (ie, if there is a version using nets or something, without Hausdorff assumptions) Aug 9, 2015 at 21:08
• This is an interesting point. Why don't you ask the community at math.stackexchange? Aug 9, 2015 at 21:11
• I'm just posting to get reputation for answering a question on meta. If I continue with this user account, I might ask sometime as you suggest. Thanks. Aug 9, 2015 at 21:13
No one talk about Zarisky topology so I wanted to write it. Zariski topology is taught at the beginning of algebraic geometry. Let $$k$$ be algebraically closed field and $$\mathbb{A}^n_k$$ is the set of all elements $$(a_1,...,a_n)$$ where $$a_i's \in k$$. In Zarisky topology, the closed sets are the zero set of ideals of $$k[x_1,...,x_n]$$.
In this topology, in $$\mathbb{A}^n_k$$, there is only one dense and closed set which is $$\mathbb{A}^n_k$$. The other closed sets are not dense. Besides, all nontrivial open sets are dense in this topology. | 2022-05-19T02:39:36 | {
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https://math.stackexchange.com/questions/602593/find-recurrence-relation-for-ternary-strings-that-dont-have-substrings-00-01-a/602676 | # Find recurrence relation for ternary strings that don't have substrings 00, 01 and last symbol is not 0
I am preparing for my finals for discrete mathematics and I came across this exercise in textbook.
Let $s_{n}$ denote all ternary strings of length $n$, such that any string in $s_{n}$ does not contain substring $00$, $01$ and the last symbol is not $0$. Find:
• non recurrent sum for $s_{n}$ using combinatorics
• recurrence relation $s_{n}$.
I managed to find recurrence relation ${s}'_{n} = 2s_{n-1} + s_{n-2}$ that satisfies the first two conditions (no $00$'s and $01$'s) but I have no idea how to find RR with all three conditions. My first thought was to make some correction in initial conditions, but after trying it I got even more confused.
As for the sum, I'm lost as well.
Can you point me in the right direction towards solution?
• So sorry, I had a bug in my program and I don't have the source code with me. Of course, $s_{1} = 2$, due to the fact that only strings satisfying those conditions are $1$ and $2$. I removed the table to avoid any further confusion. – user110409 Dec 11 '13 at 10:18
• I seem to be getting $s_n=2s_{n-1}+s_{n-2}$. Are you sure that's wrong? – bof Dec 11 '13 at 11:02
Observe that the conditions you are given are equivalent to the single condition that "every $0$ must be immediately followed by a $2$." In particular, this means that any such sequence of length $n$ is either (a) such a sequence of length $n-1$ with a $1$ or $2$ appended to the end, of which there are $2s_{n-1}$, or (b) such a sequence of length $n-2$ with a $02$ appended to the end, of which there are $s_{n-2}$. (Also note that these cases are mutually exclusive since strings from the first case can't have a $0$ as the second-to-last symbol.) This gives your recurrence $s_n = 2s_{n-1} + s_{n-2}$. For the initial conditions, you get $s_1 = 2$ and $s_2 = 5$ (just count these by hand).
You can use the same equivalent condition above to write a combinatorial sum. You can think of this as an unrestricted string with three different symbols: $a=1$, $b=2$, and $c=02$, where the third symbol $c$ takes up two spots. If there are $k$ copies of the symbol $c$ in a string of length $n$, then there must be a total of $n-k$ symbols $a, b, c$. Try to write an expression for the number of such strings with $k$ copies of $c$, and then sum over all possible values of $k$ for your answer. (Since this is homework, I'll let you take it from here.)
You simply have that every $0$ must be followed by a $2$, so $s_n$ is the number of arrangements of $a$ blocks of type "$02$", $b$ blocks of type "$1$" and $c$ blocks of type "$2$" such that $2a+b+c=n$. On the other hand, every admissible string starts with a "1", then the tail is counted in $s_{n-1}$, with a $2$, then the tail is counted in $s_{n-1}$, or with a $02$, then the tail is counted in $s_{n-2}$, and recurrence relation is $s_n=2s_{n-1}+s_{n-2}$, with $s_1=2$ and $s_2=5$. The recurrence relation and the initial conditions give: $$s_n = c_0(1+\sqrt{2})^n+c_1(1-\sqrt{2})^n,$$
$$c_0(1+\sqrt{2})+c_1(1-\sqrt{2}) = 2,$$ $$c_0(3+2\sqrt{2})+c_1(3-2\sqrt{2})=5,$$ $$c_0-c_1=\frac{\sqrt{2}}{2},$$ $$c_0+c_1=1,$$ $$c_0 = \frac{2+\sqrt{2}}{4},\qquad c_1=\frac{2-\sqrt{2}}{4}$$ $$s_n = \frac{1}{2}\left(p_{n+1} + p_{n}\right), \qquad p_n = \frac{1}{2}\left((1+\sqrt{2})^n+(1-\sqrt{2})^n\right),$$ where $p_n$ is half the Pell-Lucas number $Q_n$. In a combinatorial flavour, you can count the arrangements of $a$ blocks of type "$02$" (with $a\leq\frac{n}{2}$) in $n$ positions times $2^{n-2a}$ (the number of ways to fill the remaining part of the string with $1$s and $2$s). The number of such arrangements is equal to the number of ways to write $n-2a$ as a sum of $a+1$ natural numbers, so the coefficent of the monomial $x^{n-2a}$ in $(1+x+x^2+\ldots)^{a+1}=\frac{1}{(1-x)^{a+1}}$, that is $\binom{n-a}{a}$. This gives: $$s_n = \sum_{a=0}^{\lfloor n/2\rfloor}\binom{n-a}{a}2^{n-2a}.$$ | 2021-04-18T09:12:34 | {
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https://math.stackexchange.com/questions/2229810/arithmetic-sequence-magic-square | # Arithmetic sequence magic square
If I know that for all $z*z$ magic squares, I can have a magic square consisting of all the terms of the set (1,2,3,...$z^2$), then how do I prove that I can have a magic square consisting of an arithmetic sequence? For example, since 1,2,3,4,...$z^2$ is an arithmetic sequence of length $z^2$ and common difference of 1, can I have a magic square made of the terms in an arithmetic sequence of length $z^2$ and common difference of $d$. Also would this apply to all even and odd ordered magic squares?
For example, a common 3 by 3 magic square made by the De la Loubère method:
(8 1 6)
(3 5 7)
(4 9 2)
can be modified into:
(15 1 11)
(5 9 13)
(7 17 3)
and is still a magic square. How can I prove that this modification of making the common difference different between each of the terms in a magic square happens for ALL even and odd ordered magic squares?
• Digits? Not clear what you mean by "all the digits of $(1,2,\dots,z^2)$. – Thomas Andrews Apr 11 '17 at 20:53
• um. Multiply the first by d? – steven gregory Apr 11 '17 at 20:54
• @ThomasAndrews I meant terms of the magic square. Sorry – jk23541 Apr 11 '17 at 21:06
The transformation can be done in 2 easy steps:
1. First, multiply all entries in the square by $d$. Ths will make the entries $d,2d,3d,...$. Also, note how the sum of each row, column, and diagonal gets multiplied by $d$ as well, so the square remains a magic square.
2. Subtract $d-1$ from each entry. So now the entries are $1, d + 1, 2d +1, ...$ as desired. Moreover, not how again the sum of each rom, column, and diagonal gets lowered by $z\cdot (d-1)$ from the previous square, so their sums yet again are all the same, so the resulting square is still a magic square.
(8 1 6)
(3 5 7)
(4 9 2)
Multiply by $d=2$:
(16 2 12)
(6 10 14)
(8 18 4)
subtract $d-1=1$:
(15 1 11)
(5 9 13)
(7 17 3)
Finally, notice that the method I described does not depend on $z$ being even or odd, so this always works.
• How would that make a magic square where all the terms increase with a common difference? – jk23541 Apr 11 '17 at 21:10
• @jk23541 Because if the rows, columns, and diagonals of the original square add up to the same sum, and you multiply all entries by $d$, then in the resulting square each of the rows, columns, and diagonals add up to $d$ times that sum, and are thus the same sum. So, the resulting square is a magic square itself. And if you have 1,2,3,... In the original square, then in the resulting square you will have the terms $d,2d,3d,..$ – Bram28 Apr 11 '17 at 21:23
• I meant if I start at a term k, I would have all the terms increase by a common difference. So I would have a sequence of 1, 3, 5, 7, 9... in my magic square instead of 1,2,3,4,5,6... – jk23541 Apr 11 '17 at 21:55
• For example, I can modify this: (8 1 6) (3 5 7) (4 9 2) into: (15 1 11) (5 9 13) (7 17 3) – jk23541 Apr 11 '17 at 21:56
• OK, if you want to start with 1, then after multiplying each entry by $d$, subtract $d-1$ from each entry. So, from $1,2,3,..$ you first get $d,2d,3d,...$ and that becomes $1,d +1, 2d+1,...$ – Bram28 Apr 11 '17 at 22:09
I typed this all out and fiddled with the formatting, but in the meantime Bram updated his answer. It doesn't really add to Bram's answer so I'll post it and mark it as community wiki.
I'll expand on Bram28's answer to clarify how arithmetic sequences can be transformed while still ensuring a magic square is valid.
It's easy to see that if we add a constant to every number in a magic square we get another valid magic square. Similarly, if we multiply every number in a magic square by a constant we'll produce a different magic square.
Suppose we have two arithmetic sequences: $$a_1,a_1+d_1,a_1+2d_1,\ldots,a_1+(z^2-1)d_1$$ $$a_2,a_2+d_2,a_2+2d_2,\ldots,a_2+(z^2-1)d_2$$ To transform the first sequence to the second we can start by multiplying each term by $\frac{d_2}{d_1}$, giving: $$\frac{d_2}{d_1}a_1,\frac{d_2}{d_1}a_1+d_2,\frac{d_2}{d_1}a_1+2d_2,\ldots,\frac{d_2}{d_1}a_1+(z^2-1)d_2.$$ Then we just need to add the constant $(a_2-\frac{d_2}{d_1}a_1)$ to each term and we produce the second arithmetic sequence as required.
Hence, if we have already created a magic square using an arithmetic sequence we can transform it to any other arithmetic sequence simply by adding and multiplying by constants and this will still produce a valid magic square. | 2019-06-27T08:54:06 | {
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http://math.stackexchange.com/questions/208076/counting-the-sum-of-cycles-of-the-elements-in-s-n?answertab=votes | # Counting the sum of cycles of the elements in $S_n$.
Let $\sigma\in S_n$ be given. Let us write $\sigma$ in its disjoint cycle decomposition. Define by $w(\sigma)$ the number of cycles (for instance if we have the permutation $1\mapsto 2, 2\mapsto 4, 3\mapsto 3, 4\mapsto 1, 5\mapsto 5$ we write it as $(124)(3)(5)$ so $w(\sigma)=3$).
I was asked to find $$A_n=\sum_{\sigma\in S_n}w(\sigma)$$I found the recursive formula $$A_n=n!+(n-1)!\left[\frac{A_{n-1}}{(n-1)!}+\frac{A_{n-2}}{(n-2)!}+\ldots+\frac{A_{1}}{1!}\right]$$
I was trying to tinker around with it to see if I could get a formula that does not rely on recursion, but failed at it. Any help or hint that would lead me towards making my solution for $A_n$ "better".
-
Let $b_n=\dfrac{A_n}{n!}$; then your recurrence can be rewritten
$$b_n=1+\frac1n\sum_{k=1}^{n-1}b_k\;,\tag{1}$$
with $b_1=1$. Calculate a few values:
\begin{align*} &b_1=1\\ &b_2=1+\frac12\cdot1=\frac32\\ &b_3=1+\frac13\left(1+\frac32\right)=\frac{11}6\\ &b_4=1+\frac14\left(1+\frac32+\frac{11}6\right)=\frac{25}{12} \end{align*}
I recognize those as the first four harmonic numbers: $$b_n=H_n=\sum_{k=1}^n\frac1k\;.$$
And sure enough, the harmonic numbers do satisfy $(1)$:
\begin{align*} 1+\frac1n\sum_{k=1}^{n-1}H_k&=1+\frac1n\sum_{k=1}^{n-1}\sum_{i=1}^k\frac1i\\ &=1+\frac1n\sum_{i=1}^{n-1}\sum_{k=i}^{n-1}\frac1i\\ &=1+\frac1n\sum_{i=1}^{n-1}\frac{n-i}i\\ &=1+\frac1n\sum_{i=1}^{n-1}\left(\frac{n}i-1\right)\\ &=1+\sum_{i=1}^{n-1}\frac1i-\frac{n-1}n\\ &=1+H_{n-1}-1+\frac1n\\ &=H_n\;. \end{align*}
Thus, $A_n=n!H_n$.
Added: Here’s another approach to the problem. It’s not too hard to prove that the number of $\pi\in S_n$ with $k$ cycles is $\left[n\atop k\right]$, a Stirling number of the first kind. These have the generating function $$\sum_{k\ge 0}\left[n\atop k\right]x^k=x^{\overline{n}}=x(x+1)(x+2)\dots(x+n-1)\;.$$ Differentiating yields
\begin{align*} \sum_{k\ge 1}k\left[n\atop k\right]x^{k-1}&=\frac{d}{dx}\Big(x(x+1)(x+2)\dots(x+n-1)\Big)\\ &=\sum_{k=0}^{n-1}\frac{\prod_{i=0}^{n-1}(x+i)}{x+k}\;, \end{align*}
and evaluating at $x=1$ yields $$\sum_{k\ge 1}k\left[n\atop k\right]=\sum_{k=0}^{n-1}\frac{n!}{k+1}=n!H_n\;.$$
-
So $A_n$ is asymptotic to $n!\log n$. – Gerry Myerson Oct 6 '12 at 5:48
@Gerry: Yep. $A_n=n!\ln n+O(n!)$, or, better, $n!\ln n+\gamma n!+O((n-1)!)$. – Brian M. Scott Oct 6 '12 at 6:16
Thanks!${}{}{}{}$ – Daniel Montealegre Oct 6 '12 at 19:08
@Daniel: My pleasure. – Brian M. Scott Oct 6 '12 at 19:09
You can get an easier recurrence by considering the last element $n$ in the permutation.
Either $n$ occurs in a cycle of length at least $2$, or it forms a "cycle" on its own. In the former case taking it out of its cycle leaves a permutation of $n-1$ with the same number of cycles, and given a permutation of $n-1$ we can insert $n$ in $n-1$ places into an existing cycle, for a total contribution of $(n-1)A_{n-1}$ to $A_n$. In the latter case removing $n$ just leaves a permutation of $n-1$, which here arises in only one way; however the number of cycles is one more for the permutation of $n$ than it was for the permutation of $n-1$, so in all this contributes $A_{n-1}+(n-1)!$ to $A_n$ (the second term is for the extra cycle in $(n-1)!$ permutations). Since we have now accounted for every contribution once, we get $$A_n=(n-1)A_{n-1}+A_{n-1}+(n-1)! = nA_{n-1} + (n-1)!$$ One can simplify this by dividing by $n!$, giving $$\frac{A_n}{n!}= \frac{A_{n-1}}{(n-1)!} + \frac1n,$$ which together with $A_0=0$ obviously gives $\frac{A_n}{n!}=\sum_{i=1}^n\frac1i=H_n$ and $A_n=n!H_n$.
-
Very good answer too, but I cannot accept both. I got my recursion by adding the cases $n$ is in a $1$ cycle, in a $2$ cycle,..., in an $n$ cycle, but your solution is neater. – Daniel Montealegre Oct 6 '12 at 19:17
This question is best addressed using the symbolic method.
Permutations are sets of cycles, having combinatorial class specification $\mathfrak P(\mathfrak C(\mathfrak Z))$. Hence the corresponding exponential generating function (EGF) is $$\exp \log \frac{1}{1-z} = \frac{1}{1-z},$$ where $$\log \frac{1}{1-z}$$ is the EGF of labelled cycles. (These two are easily verified as there are $n!$ permutations and $n!/n$ cycles.)
If we want to count the number of cycles we need to mark each cycle with a new variable, $u$, which gives the mixed generating function $$G(z, u) = \exp \left( u \log \frac{1}{1-z} \right).$$
Now a term $u^k z^n / n!$ in $G(z, u)$ represents a permutation of $n$ elements having $k$ cycles. To count these, we need to differentiate with respect to $u$ and set $u=1$, obtaining $$\left. \frac{d}{du} G(z, u) \right|_{u=1} = \left. \exp \left( u \log \frac{1}{1-z} \right) \log \frac{1}{1-z} \right|_{u=1} = \frac{1}{1-z} \log \frac{1}{1-z} .$$
But $$[z^n] \frac{1}{1-z} \log \frac{1}{1-z} = H_n$$ by a trivial calculation and we are done.
There is much more on this method at Wikipedia.
- | 2014-07-13T22:07:50 | {
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https://gateoverflow.in/43513/gate2007-77 | # GATE2007-77
3.1k views
Suppose the letters $a, \,b, \,c, \,d, \,e, \,f$ have probabilities $\dfrac{1}{2}, \dfrac{1}{4}, \dfrac{1}{8}, \dfrac{1}{16}, \dfrac{1}{32}, \dfrac{1}{32}$, respectively.
What is the average length of the Huffman code for the letters $a, \,b, \,c, \,d, \,e, \,f$?
1. $3$
2. $2.1875$
3. $2.25$
4. $1.9375$
edited
0
And what is Q.76?
0
$$\begin{array}{|c|c|} \hline \textbf{Letter} & \textbf{Probability} \\\hline a & 1/2 \\\hline b & 1/4 \\\hline c & 1/8 \\\hline d & 1/16 \\\hline e & 1/32 \\\hline f & 1/32 \\\hline \end{array}$$
$\begin{array}{cc} \textbf{Prefix Code} & \textbf{Code Length} \\ a = 0 & 1 \\ b =10 & 2 \\ c = 110 & 3 \\ d = 1110 & 4 \\ e = 11110 & 5 \\ f = 11111 & 5 \end{array}$
Avg length $= \dfrac{1}{2} \times 1 + \dfrac{1}{4} \times 2+ \dfrac{1}{8} \times 3+ \dfrac{1}{16} \times 4+ \dfrac{1}{32} \times 5+ \dfrac{1}{32} \times 5 \\ = \dfrac{16 + 16 + 12 + 8 + 5 + 5}{32} \\ = 1.9375$
Correct Answer: $D$
edited
0
If multiplied with probability directly then don't divide by $\mathbf{100}$
Based on the probabilities, we can say the probable frequency of the letters will be
16, 8, 4, 2, 1, 1
Now, the Huffman tree can be constructed as follows:
Average length will be 1/2 * 1 + 1/4 * 2 + 1/8 * 3 + 1/16 * 4 + 1/32 * 5 + 1/32 * 5
= 62/32 = 1.9375
https://www.siggraph.org/education/materials/HyperGraph/video/mpeg/mpegfaq/huffman_tutorial.html
3
Why isn't average length (sum of lenghts of a,b,c,d,e,f)$/6$ ?
0
Because not all the letters occur with same frequency or probability. Purpose of Huffman encoding is to represent the most occurring character with least number of bits. As u can see a has been represented with only 1 bit because it occurs the most with probability 1/2. So we need to multiply length of bits with the probability with which they occur.
0
After calculating like arjun sir, we can even multiply with probably frequencies like this: 16*1 + 8*1 + 4*3 + 2*4 +1*5*2 / 16+ 8+4+2+1+1 = 1.9375
1
@ sir,
for such huffman coding question sometimes sees confusong, Sir can you please check this is right or wrong
1) Expected Length of Huffman Code for N chars (Frequencies given ) =
$\sum$(Bits needed for char * its frequency)
2) Average Length Of Huffman Code for N characters OR Average Bits needed for N characeters = (Assuming average over N characters)
[ $\sum$(Bits needed for char * its frequency) ] / Total characters
1 vote
The letters a, b, c, d, e, f have probabilities
1/2, 1/4, 1/8, 1/16, 1/32, 1/32 respectively.
1
/ \
/ \
1/2 a(1/2)
/ \
/ \
1/4 b(1/4)
/ \
/ \
1/8 c(1/8)
/ \
/ \
1/16 d(1/16)
/ \
e f
The average length = (1*1/2 + 2*1/4 + 3*1/8 + 4*1/16 + 5*1/32 + 5*1/32)
= 1.9375
## Related questions
Suppose the letters $a, \,b, \,c, \,d, \,e, \,f$ have probabilities $\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \frac{1}{32}, \frac{1}{32}$, respectively. Which of the following is the Huffman code for the letter $a, \,b, \,c, \,d, \,e, \,f$? $0$, $10$, $110$ ... $11$, $10$, $011$, $010$, $001$, $000$ $11$, $10$, $01$, $001$, $0001$, $0000$ $110$, $100$, $010$, $000$, $001$, $111$
in Huffman Code, we get extract the minimum at each time, but my minimum is creating duplicate, then which one i choose? i am getting the same avg.no.of bits for every Huffman tree, but the problem is my tree is changing therefore representing the character also changed, if some one asks convert the message ... $\frac{12}{30}$ | 2020-09-28T19:01:43 | {
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https://math.stackexchange.com/questions/1084785/a-man-who-lies-a-fourth-of-the-time-throws-a-die-and-says-it-is-a-six-what-is-t | # A man who lies a fourth of the time throws a die and says it is a six. What is the probability it is actually a six?
First, I apologise for the vague title. I couldn't think of a short way to represent the problem. Also, I am aware that a similar question exists, but I have a little bit more insight.
The problem is:
A man is known to speak the truth three out of four times. He throws a die and reports that it is a six. Find the probability that it actually is a six.
The answer in the book is $\frac{3}{8}$, which sounds completely off to me, given the numbers, on first glance. I decided to try it on my own. After some contemplation, it indeed is true that $\frac{3}{8}$ is correct. But, the question's vagueness makes it interesting. It's sort of a naive answer!
First, here's the book's solution:
Let $E$ be the event that the man reports that a six occurs, $S_1$ be the event that six occurs and $S_2$ be the event that six does not occur.
$P(S_1)$ = Probability that six occurs = $\frac{1}{6}$
$P(S_2)$ = Probability that six does not occur = $\frac{5}{6}$
$P(E~|S_1)$ = Probability that the man reports that a six occurred when it has actually occurred (truth) = $\frac{3}{4}$
$P(E~|S_2)$ = Probability that the man reports that a six occurred when it hasn't actually occurred = $\frac{1}{4}$
$P(S_1|E)$ = Probability that a six has actually occurred when the man claims so =
$$\frac{P(S_1)~P(E~|S1)}{P(S_1)~P(E~|S_1)+P(S_2)~P(E~|S_2)} = \frac{3}{8}$$
Link to original: Page 25 of this (link updated 22/11/2015; tends to break often).
Here's the first of my two solutions:
The question is a little vague and it doesn't tell us what the man says when he doesn't get a six. Assuming he lies every time he doesn't get a six (which is not quite what the question says), would bring up $P(E~|S_2)$ to $1$, which skyrockets $P(lie)$ to $\frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6}.\frac{1}{4} = \frac{7}{8}$.
Also, $P(E~|S_1) = \frac{3}{4}$ and $P(E) = \frac{5}{6} + \frac{1}{6}.\frac{3}{4} = \frac{23}{24}$.
This brings us to, $$P(S_1|E) = \frac{P(S_1)P(E~|S_1)}{P(E)} = \frac{3}{23} = 13.04\%$$
Please note that I don't care about the sixes rolled when he lies about it being a six, because that condition is not applicable to the question.
I'm very skeptical about my calculations! I decided to whip up a simple program and see what the probability distribution is for $$\frac{Number~of~sixes~rolled~while ~saying~it's~a~six}{Number~of~dice~rolls}$$
Here's the result:
Considering the book says $P(E~|S_2) = \frac{1}{4}$, which is in essence is contradictory to the question, let's also consider the other possibility:
It's even more vague in this scenario! If the man lies after any given roll, there is a $\frac{1}{5}$ probability that he will lie about it being any other number. This creates a new branch at each lie, which splits into five. In this case,
$$P(E) = \frac{1}{6}.\frac{1}{4}.\frac{1}{5}.\frac{5}{1}+\frac{1}{6}.\frac{3}{4} = \frac{1}{6}$$
$P(E~|S_1)$ still remains $\frac{3}{4}$, but $P(E~|S_2) = \frac{1}{4}.\frac{1}{5} = \frac{1}{20}$.
$$P(S_1|E) = \frac{P(S_1)P(E~|S_1)}{P(E)} = \frac{3}{4} = 75\%$$
This seems like the more intuitive answer, and it is why I doubted the original answer in the first place. Given a truth probability of $\frac{3}{4}$, there was no way that the probability of a six was $37.5\%$!
The ambiguity of the question is what makes $\frac{3}{8}$ wrong to me. Entirely out of curiosity, which answer fits best with the question and are these calculations correct or baloney?
• Wow, I remember this exact question coming in my syllabus as I am in 12th class right now. I find this question as much vague as you did, and now even my discern has increased after seeing this.
– Mann
Dec 29 '14 at 17:49
• @Mann It's a little unfortunate that I'd be forced to use 3/8 as the answer in an exam! Although, I can pretty much guarantee I'm not going to forget this question any time soon! Dec 29 '14 at 18:06
• The link is still broken. Oct 13 '15 at 18:05
• @zhoraster I just checked again, works for me. Oct 13 '15 at 21:56
There are essentially three aspects of the question that I would consider vague:
1. The probability distribution of the outcome of the die roll is not explicitly stated--so in particular, are we to assume that it is a fair die?
2. Whether the event that the man lies about the die roll is independent of the true outcome of the die roll is not stated, and the official solution assumes that these events (the roll of the die, and whether the man lies or tells the truth) are independent.
3. If a six is not rolled and the man lies about the outcome, does the man lie in the sense of choosing any one of the other five numbers available to him, or does he lie in the sense of "I didn't roll a six but I will say I did?"
Item 3 is the real sticking point, since without making the assumptions of Items 1 and 2, no computation can be performed at all. But Item 3 requires us to interpret the meaning of "lie," and more problematically, a fully reasonable computation is possible under a variety of interpretations. To see why, let's make the process a bit more concrete:
1. The man rolls the die and observes the number rolled, which is hidden from you.
2. The man then flips a fair coin twice and again this outcome is hidden from you. If he gets two heads, he decides to lie; otherwise, he tells the truth.
3. If telling the truth, he reports the actual value of the die roll he observed.
Now consider two plausible interpretations of what happens if the man lies:
• (Option A) If he lies, then the value he reports is any value randomly and uniformly chosen from the remaining five possibilities.
• (Option B) If the actual value was a six and he got two heads, then he lies by choosing any other number from $\{1,2,3,4,5\}$; if the actual value was not a six and he got two heads, then he reports $6$.
It is under the interpretation of Option B that the "official" answer is $3/8$. But what about Option A? Let's do the computation. Let $X = 1$ if the true value of the die is six, and $X = 0$ otherwise. Let $Y = 1$ if the reported value of the die is six, and $Y = 0$ otherwise. Let $L = 1$ if the man lies, and $L = 0$ otherwise. Then we know $\Pr[X = 1] = \frac{1}{6}$, $\Pr[L = 1] = \frac{1}{4}$. We also know $$\Pr[Y = 1 \mid X = 1] = \Pr[L = 0] = \frac{3}{4},$$ and $$\Pr[Y = 1 \mid X = 0] = \Pr[Y = 1 \mid (X = 0 \cap L = 1)]\Pr[L = 1] = \frac{1}{5} \cdot \frac{1}{4} = \frac{1}{20},$$ because two things must happen for the man to report a six if no six was rolled: he has to flip two heads (and thus be allowed to lie) with probability $\frac{1}{4}$, and he must randomly and uniformly choose to report six with probability $\frac{1}{5}$. From this, we can now compute \begin{align*} \Pr[X = 1 \mid Y = 1] &= \frac{\Pr[Y = 1 \mid X = 1] \Pr[X = 1]}{\Pr[Y = 1]} \\ &= \frac{\Pr[Y = 1 \mid X = 1]\Pr[X = 1]}{\Pr[Y = 1 \mid X = 1]\Pr[X = 1] + \Pr[Y = 1 \mid X = 0]\Pr[X = 0]} \\ &= \frac{\frac{3}{4} \cdot \frac{1}{6}}{\frac{3}{4} \cdot \frac{1}{6} + \frac{1}{20} \cdot \frac{5}{6}} \\ &= \frac{3}{4}. \end{align*} Therefore, the question lacks a critically important clarification: when the man "lies," is the nature of that lie of the form "I rolled a six (but I really did not) / I did not roll a six (but I really did)" (this is Option B), or is it "I rolled a one (but I really did not) / I rolled a two (but I really did not) / etc." (this is Option A)? Which option you choose affects the resulting probability calculation, and you can simulate either scenario to that effect.
Indeed, there are any number of possible interpretations, because ultimately what influences the desired conditional probability is the probability that the man will choose to report a six given that he has decided to lie after not actually rolling a six, and this is not specified in the question. Both options already described yield reasonable interpretations, but we could also recognize them as special cases of the general probability $$\Pr[X = 1 \mid Y = 1] = \frac{3}{3+5p},$$ where $p = \Pr[Y = 1 \mid (X = 0 \cap L = 1)]$. Option A sets $p = 1/5$ (choose uniformly among the other options), and Option B sets $p = 1$ (always choose to report 6).
Consequently, I would regard this question as having insufficient information to formulate a conclusive answer.
• It's safe to assume that the question wants us to use a fair die. I consequently also assumed that the man's truth or lie is independent of the true die roll. Number three is what I got stuck on. If the official answer did not exist, I personally would not under any circumstance assume that the man would provide a binary answer provided the question. The more logical assumption is that there's an equal probability of him choosing any of the other five numbers. Kudos on the general solution. Dec 30 '14 at 10:40
• I use to give such vague formulations for tutorials (and I gave exactly this problem recently) exactly in order to discuss the problems you've outlined. Great answer, useful. Oct 13 '15 at 18:03
• I am 6 years late but if possible could you please explain why this is true: $$\Pr[Y = 1 \mid X = 0] = \Pr[Y = 1 \mid (X = 0 \cap L = 1)]\Pr[L = 1]$$ Apr 10 at 10:56
There are two situations in which the man can report a 6, either he is telling the truth or he is lying. The key assumption that seems to be made here is that the man's veracity is independant of the value on the die. Namely, he is just as likely to lie if the die shows a six as if it shows a three.
That being said, as you posted above, this is an example of Bayes Theorem. Let me change your nomenclature a bit.
Let $S$ be the probability of rolling a six $=\frac{1}{6}$
Let $\bar{S}$ be the probability of not rolling a six $=\frac{5}{6}$
Let $T$ be the probability of the man telling the truth, regardless of situation $=\frac{3}{4}$
Let $\bar{T}$ be the probability of the man not telling the truth, regardless of situation $=\frac{1}{4}$
Let $M$ be the probability the man says there was a 6. So what we want is: $$P(S|M) = \frac{P(S \cap M)}{P(M)}$$
We don't immediately know $M$, true. What we do know is that $M$ comprises two situations: truth and falsity. In other words: $$P(M) = P(T \cap S) + P(\bar{T} \cap \bar{S})$$.
We are also luckily working under the assumption that the man's prevarication is independant of the die, so we can calulate $P(M)$ directly from the known $P(S)$ and $P(T)$. $P(S \cap M)$ is the case where the man tells the truth about the rolled six, which is $\frac{3}{4}\cdot\frac{1}{6}$. $P(\bar{S} \cap \bar{M})$ is the case where the man lies about anything other than a rolled six, which is $\frac{1}{4}\cdot\frac{5}{6}$. So the conditional probability of the there being a six, given that the man told us so, is the probability of the true case over the probability of all possible cases where the man says 6: \begin{align} P(S|M) &= \frac{P(S \cap M)}{P(M)}\\ &= \frac{\frac{3}{4}\cdot\frac{1}{6}}{\frac{3}{4}\cdot\frac{1}{6} + \frac{1}{4}\cdot\frac{5}{6}}\\ &=\frac{\frac{3}{24}}{\frac{3 + 5}{24}}\\ &=\frac{3}{8} \end{align}
• This seems a bit off to me. In particular $$P(T \cap S) + P(\bar{T} \cap \bar{S})$$ seem to contain the case where he doesn't roll a six and doesn't tell the truth, but this includes the case where he lies and says it is a number $$\neq6$$ Dec 29 '14 at 22:16
• @FrankConry The assumption in the question is that he is being asked a binary question: is it or is it not a 6. He can say "yes it was a 6" or "no it was not a 6". If he rolls a 3 and says no it was not a six, that is $P(T \cap \bar{S})$. If it was a 6 and he says, "no it was not a 6" that is $P(\bar{T} \cap S)$. Dec 29 '14 at 22:21
• @Avraham It's quite intuitive to think about it this way. I missed it in the beginning but I saw it as I neared the end. Like all Bayes' problems, the key is to eliminate all branches that incorporate conditions that haven't already happened. Since he has already said that it's a six, it directly implies that: 1) If it's a six, he's not lying. 2) If it's not a six, he's lying. Eliminating all other possibilities, you end up with five outcomes where it's not a six and he lies, and three outcomes where it is a six and he doesn't. Ergo, $\frac{3}{8}$. I'll try to make a simulation. Dec 30 '14 at 9:04
• @Avraham Semantically speaking, given the question, I prefer the answer $\frac{3}{4}$, because it's not mentioned whether he's allowed to give a non-binary answer. Dec 30 '14 at 9:08
• @ShreyasVinod, heropup's answer above is clearer, more complete, and encapsulates the root cause of your confusion. I just explained what had to have been intended by the question writer. You should accept his answer! Dec 30 '14 at 15:26 | 2021-10-28T21:30:09 | {
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https://www.medetorax.com/uncw/university/49905207835509e5305 | T ( n) T ( n 1) T ( n 2) = 0. 9 The recursion-tree method Convert the recurrence into a tree: Each node represents the cost incurred at various levels of recursion Sum up the costs of all levels Used to guess a solution for the recurrence. To solve the recurrences, use the techniques for bounding summations. For example, consider the following example: T (n) = aT (n/b) + cn Here, the problem is getting split into a subproblems, each of which has a size of n/b. Subject: Design and Analysis of Algorithms Topic: recursion tree method for solving recurrences Handwritten notes with examples Preview 1 out of 3 pages. It's free to sign up and bid on jobs. 1) Substitution Method: We make a guess for the solution and then we use mathematical induction to prove the guess is correct or incorrect. The following are the ways to solve a recurrence relation : Recursion Tree method Examples For Every Form: Cost Of Leaf Level Will be Maximum: T (n) = 2T (n-1) + 1. 2 RECITATION 1. understanding: master method and recursion tree method for solving recurrences examples more internal nodes. Till now, we have studied two methods to solve a recurrence equation. Recurrence relations like terms, recursion can be verified by an upper or bad chips can become especially complicated. Master method: a is the number of subproblems in the term that input is n. n/b is the subproblem size. Recursion tree method is used to solve recurrence relations. Here the right-subtree, the one with 2n/3 element will drive the height. Examples of the process of solving recurrences using substitution. At level i there will be ai nodes. or O). Each level has three times more nodes than the level above, so the number of nodes at depth i is $3^i$. The recursion formula you have is T (n) = T (n/3) + T (2n/3) + n. It says, you are making a recursion tree that splits into two subtrees of sizes n/3, 2n/3, and costs n at that level. Few Examples of Solving Recurrences Master Method. In this section, we will learn each of them one by one. The following are the ways to solve a recurrence relation : Recursion Tree method Title: dacl Sequences, Series, And The Binomial Theorem Write a formula for the nth term of the geometric sequence 3, 12, 48 Find Limit Of Recursive Sequence using our free online calculator Tracing the Execution Introduction While reading one of our Insider News posts which linked to Evan Miller's site , he mentioned a mathematical means of producing a Fibonacci number without using Explanation: Masters theorem is a direct method for solving recurrences. It's free to sign up and bid on jobs. I came across places where floors and ceilings are neglected while solving recurrences. Some methods used for computing asymptotic bounds are the master theorem and the AkraBazzi method. Here (pg.2, exercise 4.11) is an example where ceiling is ignored:. Subject: Design and Analysis of Algorithms Topic: recursion tree method for solving recurrences Handwritten notes with examples. Like all recursive structures, a recurrence consists of one or more base cases and one or more recursive cases. If you see the height is determined by height of largest subtree (+1). 1) Substitution Method: We make a guess for the solution and then we use mathematical induction to prove the guess is correct or incorrect. Combine (line 5): Merging an n -element subarray takes ( n) (this term absorbs the (1) term for Divide). Now we use induction to prove our guess. 1.2.1 Example Recurrence: T(n) = 3T(bn=4c) + ( n2) We drop the For example consider the recurrence T (n) = 2T (n/2) + n We guess the solution as T (n) = O (nLogn). 1) Substitution Method: We make a guess for the solution and then we use mathematical induction to prove the the guess is correct or incorrect. The master method The master method applies to recurrences of the form T(n) = aT(n/b) + f(n) , where a1, b> 1, and f is asymptotically positive.
ITERATION METHOD. Wolfram|Alpha can solve various kinds of recurrences, find asymptotic bounds and find recurrence relations satisfied by given sequences. Recursion Tree method for solving Recurrences. We can simply begin from a node, then traverse its adjacent (or children) without caring about cycles. Subject: Design and Analysis of Algorithms Topic: recursion tree method for solving recurrences Handwritten notes with examples. The given recurrence relation shows-A problem of size n will get divided into 2 sub-problems- one of size n/5 and another of size 4n/5. ITERATION METHOD We need to draw each and every level of recurrence tree and then calculate the time at each level. Minimum Spanning Tree. Generating Your Document Solve the following recurrence relation using recursion tree method- T (n) = 2T (n/2) + n Solution- Step-01: Draw a recursion tree based on the given recurrence relation.
It's very easy to understand and you don't need to be a 10X developer to do so. This method is especially powerful when we encounter recurrences that are non-trivial and unreadable via the master theorem . Recursion tree method for solving recurrences examples ile ilikili ileri arayn ya da 21 milyondan fazla i ieriiyle dnyann en byk serbest alma pazarnda ie alm yapn. Use of recursion to solve math problems ; Practice Exams. Search for jobs related to Recursive tree method examples or hire on the world's largest freelancing marketplace with 20m+ jobs. When n > 0, the method performs two basic operations and then calls itself, using ONE recursive call, with a parameter n - 1. In the analysis of algorithms, the master theorem for divide-and-conquer recurrences provides an asymptotic analysis (using Big O notation) for recurrence relations of types that occur in the analysis of many divide and conquer algorithms.The approach was first presented by Jon Bentley, Dorothea Haken, and James B. Saxe in 1980, where it was described as a "unifying method" for Firstly draw the recursion tree. I Ching [The Book of Changes] (c. 1100 BC) To endure the idea of the recurrence one needs: freedom from morality; new means against Use the substitution method to verify your answer. Rejestracja i skadanie ofert jest darmowe. Subject: Design and Analysis of Algorithms Topic: recursion tree method for solving recurrences Handwritten notes with examples Preview 1 out of 3 pages. Szukaj projektw powizanych z Recursion tree method for solving recurrences examples lub zatrudnij na najwikszym na wiecie rynku freelancingu z ponad 21 milionami projektw. Calculate the time in each level of the recursion tree. MASTER METHOD In this method, we have some predefined recurrence equation cases, and our focus is to get a direct solution for it.
form and show that the solution works. There are mainly three ways for solving recurrences. The substitution method for solving recurrences is famously described using two steps: Guess the form of the solution. For example consider the recurrence T (n) = 2T (n/2) + n We guess the solution as T (n) = O (nLogn). We know that the answer is probably T (N) = O (2n). Compute the cost of each level in the tree. A recurrence tree is drawn, branching until the base case is reached. The observation that we are almost doubling the number of O (1) operations for a constant decrease in n leads to the guess. First let's create a recursion tree for the recurrence $T(n) = 3T(\frac{n}{2}) + n$ and assume that n is an exact power of 2. Rekisterityminen ja tarjoaminen on ilmaista. Solution- Step-01: Draw a recursion tree based on the given recurrence relation. For example consider the recurrence T (n) = 2T (n/2) + The recursion-tree method promotes intuition, however. The recursion-tree method can be unreliable. When n > 0, the method performs two basic operations and then calls itself, using ONE recursive call, with a parameter n - 1. Recurrences can be linear or non-linear, homogeneous or non-homogeneous, and first order or higher order. Create a recursion tree from the recurrence relation; Calculate the work done in each node of the tree; Calculate the work done in each level of the tree (this can be done by adding the work done in each node corresponding to that level). And if we begin from a single node (root), and traverse this way, it is guaranteed that we traverse the whole tree as there is no dis-connectivity, Examples: Tree: Sum up all the time values. Lets say we have the recurrence relation given below. There are 3 ways of solving recurrence: SUBSTITUTION METHOD A guess for the solution is made, and then we prove that our guess was incorrect or correct using mathematical induction.
The recursion tree is one of the recursion-solving methods. View Example. => Affects the number of nodes per level. P. S. Mandal, IITG There are mainly three ways for solving recurrences. Recurrence relation (or recursive formula). This is a curious one. The given recurrence relation shows- A problem of size n will get divided into 2 sub-problems of size n/2. Solving Recurrences 1 Introduction A recurrence is a recursive description of a function, usually of the form F: IN !IR, or a description of such a function in terms of itself. The tree is not full (not a complete binary tree of height Solving recurrence relation.
In the previous lecture, the focus was on step 2. Each of these cases is an equation or inequality, with some Yes, you can solve almost every problem using recursion. Just look out how Functional Programmers tackles every problem with Haskell, OCaml, Erlang etc. Why not? We can help you solve an equation of the form "ax 2 + bx + c = 0" Just enter the values of a, b and c below: I just want to mention that the determining a closed form expression for a recursive sequence is a hard problem a starting point a 1 along with a formula for finding a n+1 in a starting point a 1 along with a formula for finding a n+1 in. This formula refers to itself, and the argument of the formula must be on smaller values (close to the base value). Like all recursive structures, a recurrence consists of one or more base cases and one or more recursive cases. A recurrence is an equation or inequality that describes a function in terms of its values on smaller inputs. Recursion is a tool not often used by imperative language developers, because it is thought to be slow and to waste space, but as the author demonstrates, there are several techniques that can be used to minimize or eliminate these problems. He introduces the concept of recursion and tackle recursive programming patterns, examining how they can be used to write provably correct programs To solve a Recurrence Relation means to obtain a function defined on the natural numbers that satisfy the recurrence. I'm trying to find the tight upper and lower bounds for the following recurrence: Drawing the recursion tree, I find that at level 2, the work done is (n^2)/2 + (2n^2)/3. Search for jobs related to Recursive tree method examples or hire on the world's largest freelancing marketplace with 19m+ jobs. For example consider the recurrence T (n) = 2T (n/2) + n We guess the solution as T (n) = O (nLogn). T (n) = 2 * T (n-1) + c1, (n > 1) T (1) = 1. 4.4 The recursion-tree method for solving recurrences 4.5 The master method for solving recurrences 4.6 Proof of the master theorem Chap 4 Problems Chap 4 Problems 4-1 Recurrence examples 4-2 Parameter-passing costs 4-3 More recurrence examples 4 We can solve any recurrence that falls under any one of the three cases of masters theorem. Task 1.1. We use following steps to solve the recurrence relation using recursion tree method. Det Next we change the characteristic equation into The recursion tree method is good for generating guesses for the substitution method.
The iteration method does not require making a good guess like the substitution method (but it
substitution method another example using a recursion tree an example Consider the recurrence relation T(n)=3T(n/4)+cn2 for some constant c. We assume that n is an exact power of 4. Recurrence - Recursion Tree Relationship T(1) = c T(n ) = a*T( n/b )+ cn 5 Number of subproblems => Number of children of a node in the recursion tree. The third and last method which we are going to learn is the Master's Method. 0. Divide (line 2): (1) is required to compute q as the average of p and r. Conquer (lines 3 and 4): 2 T ( n /2) is required to recursively solve two subproblems, each of size n/2. The tree makes it look like it is exponential in the worst case. I'm trying to figure out how to solve recurrence equations, and I can do them easily using the recursion tree method if the equation is something like this, for example: T (1) = 1; T (n) = n + 2T (n/2) for n > 1. Cost Of Each Level is Same. There are mainly three ways for solving recurrences.
For this, we ignore the base case and move all the contents in the right of the recursive case to the left i.e. Therefore the recurrence relation is: T(0) = a where a is constant. In the recursion-tree method we expand T(n) into a tree: T(n) cn2 T(n 4) T(n 4) T(n 4) [contradictory]Quicksort is a divide-and-conquer algorithm.It works by selecting a 2 Solving Recurrences with the Iteration/Recursion-tree Method In the iteration method we iteratively unfold the recurrence until we see the pattern. In recursion tree, researchers and solve a recurrence, use asymptotic bounds as before. 4.4 The recursion-tree method for solving recurrences 4.4-1. Sg efter jobs der relaterer sig til Recursion tree method for solving recurrences examples, eller anst p verdens strste freelance-markedsplads med 21m+ jobs. Each of these cases is an equation or inequality, with some
SOLVING RECURRENCES 1.2 The Tree Method The cost analysis of our algorihms usually comes down to nding a closed form for a recurrence. Now that we know the three cases of Master Theorem, let us practice one recurrence for each of the three cases. For example, we can ignore oors and ceilings when solving our recurrences, as they usually do not a ect the nal guess. Steps of Recursion Tree method. Affects the level TC. Use induction to show that the guess is valid. For Example, the Worst Case Running Time T (n) of the MERGE SORT Procedures is described by the recurrence. There are mainly three steps in the recursion tree method. Construct a recursion tree from the recurrence relation at hand. When implemented well, it can be somewhat faster than merge sort and about two or three times faster than heapsort. Quicksort is an in-place sorting algorithm.Developed by British computer scientist Tony Hoare in 1959 and published in 1961, it is still a commonly used algorithm for sorting. Visit the current node data in the postorder array before exiting from the current recursion. Final Exam Computer Science 112: Programming in C++ Status: Computer Science 112: Programming in C++ Course Practice . 1.Recursion Tree 2.Substitution Method - guess runtime and check using induction 3.Master Theorem 3.1 Recursion Tree Recursion trees are a visual way to devise a good guess for the solution to a recurrence, which can then be formally proved using the substitution method. Types Of Problem We can solve using the Recursion Tree Method: Cost Of Root Node will Maximum. 9. def foo ():s = 0i = 0while i < 10:s = s + ii = i + 1return sprint foo () A recurrence relation is an equation or inequality that describes a function in terms of its value on smaller inputs or as a function of preceding (or lower) terms. The recursion-tree method can be unreliable, just like any method that uses ellipses (). First step is to write the above recurrence relation in a characteristic equation form. The recursion tree method is good for generating guesses for the substitution method. There are three main methods for solving recurrences. In fact in CLRS (pg.88) its mentioned that: "Floors and ceilings usually do not matter when solving recurrences" T(n) = b + T(n-1) where b is constant, n > 0.
Example for Case 1.
Use a recursion tree to determine a good asymptotic upper bound on the recurrence T (n) = T (n - 1) + T (n / 2) + n T (n) = T (n 1)+T (n/2)+n. If we are only using recursion trees to generate guesses and not prove anything, we can tolerate a certain amount of \sloppiness" in our analysis. Kaydolmak ve ilere teklif vermek cretsizdir. LEC 07: Recurrences II, Tree Method CSE 373 Autumn 2020 Learning Objectives 1.ContinuedDescribe the 3 most common recursive patterns and identify whether code belongs to one of them 2.Model a recurrence with the Tree Method and use it to characterize the recurrence with a bound 3.Use Summation Identities to find closed forms for summations An example is given below to show the method in detail. Cost Of Leaf Node Will be Maximum. 1. Assume the recurrence equation is T(n) = 4T(n/2) + n. Let us compare this recurrence with our eligible recurrence for Master Theorem T(n) = aT(n/b) + f(n). 2 Use mathematical induction to nd constants in the. Minimum Spanning Tree Kruskal's Algorithm Prim's Algorithm. Using the tree method to derive the closed form consists of nding a cost bound for each level of the recursion tree and then summing the costs over the levels. The asymptomatic notation is calculated using recursion tree algorithms. The work done at level 3 is (n^2)/8 + (n^2)/6 + (n^2)/18 + (2n^2)/27. T(n) = b + T(n-1) where b is constant, n > 0. Push the current node in the preorder array and call the recursion function for the left child. Solve the following recurrence relation using recursion tree method-T(n) = T(n/5) + T(4n/5) + n . Now we use induction to prove our guess. Solving recurrence relation. Etsi tit, jotka liittyvt hakusanaan Recursion tree method for solving recurrences examples tai palkkaa maailman suurimmalta makkinapaikalta, jossa on yli 21 miljoonaa tyt. There are mainly three ways for solving recurrences. Recursive sequence formulaAn initial value such as $a_1$.A pattern or an equation in terms of $a_ {n 1}$ or even $a_ {n -2}$ that applies throughout the sequence.We can express the rule as a function of $a_ {n -1}$. Examples on Recursion Tree Method || Method of Solving Recurrences 4.4 The recursion-tree method for solving recurrences 4.5 The master method for solving recurrences 4.6 Proof of the master theorem Chap 4 Problems Chap 4 Problems 4-1 Recurrence examples 4-2 Parameter-passing costs 4-3 More recurrence examples 4 After body load window. The asymptomatic notation is calculated using recursion tree algorithms. 1) Substitution Method: We make a guess for the solution and then we use mathematical induction to prove the the guess is correct or incorrect. Then, we sum the total time taken at all levels in order to derive the overall time complexity. The recursion tree is one of the recursion-solving methods. But I'm having trouble understanding how to solve equations for which the recurrence is modified by a fraction, like this for example: Therefore the recurrence relation is: T(0) = a where a is constant. Find the total number of levels in the recursion tree. 0. The substitution method for solving recurrences consists of. The recursion-tree method promotes intuition, however. This makes the analysis of an algorithm much easier and directly gives us the result for 3 most common cases of recurrence equations. Now we use induction to prove our guess. So DFS of a tree is relatively easier. In this video we discuss how to use the seqn command to define a recursive sequence on the TI-Nspire CX calculator page Monotonic decreasing sequences are defined similarly The terms of a recursive sequences can be denoted symbolically in a number of different notations, such as , , or f[], where is a symbol representing the sequence The official definition is: "The Ulam sequence is defined
Now push the current node in the inorder array and make the recursive call for the right child (right subtree). Step 2. Solving Recurrences 1 Introduction A recurrence is a recursive description of a function, usually of the form F: IN !IR, or a description of such a function in terms of itself. Such recurrences should not constitute occasions for sadness but realities for awareness, so that one may be happy in the interim. Generally, these recurrence relations follow the divide and conquer approach to solve a problem, for example T (n) = T (n-1) + T (n-2) + k, is a recurrence relation as problem size 'n' is dividing into problems of size n-1 and n-2. Example from CLRS (chapter 4, pg.83) where floor is neglected:. Recursion-tree method A recursion tree models the costs (time) of a recursive execution of an algorithm. OK? Size of a subproblem => Affects the number of recursive calls (frame stack max height and tree height) Recursion-tree method: The tree that was converted from the recurrence has nodes that represent the costs incurred at various levels of the recursion. Recursion-tree method A recursion tree models the costs (time) of a recursive execution of an algorithm. 10. The recurrence relation is given as: an = 4an-1 - 4an-2 The initial conditions are given as 20 = 1, 2, = 4 and 22 = 12,-- Se When you solve the general equation, the constants a View Example. Generating Your Document can be solved with recursion tree method. Step 1. Step1: Draw a recursion tree according to the questions you want to solve. two steps: 1 Guess the form of the solution. | 2022-10-06T14:39:53 | {
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http://math.stackexchange.com/questions/182580/probability-distributions-exam-paper-question-mathrmcovx-y-pdf | # Probability distributions - Exam paper question - $\mathrm{Cov}(X,Y)$, PDF
Two people have decided to meet at a certain point in a forest sometime between noon and 2pm. Their respective independent arrival times are $X$ and $Y$ such that $X \sim \mathrm{Unif}(0,2)$ and $Y \sim \mathrm{Unif}(0,2)$.
Hence the joint density of $X$ and $Y$ is $$f_{X,Y} {(x,y)} = \begin{cases} 1/4, & 0< x <2 , 0< y <2 \\ 0, & \text{otherwise.} \end{cases}$$
They have agreed that whoever arrives first will wait for at most $20$ minutes for the arrival of the other.
a) Sketch the region in the $xy$ plane of times values for which they will meet and specify precisely the appropriate bounds (in terms of $x$ and $y$) for this region; then find the probabilty that they will meet by integrating the joint PDF $f_{X,Y} {(x,y)}$ over this region.
b) Since $X$ and $Y$ are independent, what value must $\mathrm{Cov}(X,Y)$ have?
c) Calculate explicitly $\mathrm{Cov}(X,Y)$ starting from its definition. Recall that $$\mathrm{Cov}(X,Y) = E[(X - E(X))(Y - E(Y))].$$
I know this is quite a long question but I didn't know how to break it down into smaller parts without just having to type it into three different questions to ask. If you could give me an idea about the sketch great! I'm not sure how to integrate the PDF as it's only $1/4$? Would it not just be $x/4 + C$?
Also for $\mathrm{Cov}(X,Y)$, I don't seem to have any notes on this, so detail would be good too.
Test is in the morning and you guys have been a big help so far!
-
seems cool!!! ))) – Seyhmus Güngören Aug 14 '12 at 20:32
a) The sketch should be something like a square with a shaded region which is a band of width $1/3$ (in $1$-norm) from the diagonal. (I think the answer is $11/36$.) b) This has nothing to do with meeting, right? You can either derive $\text{Cov}(X, Y)$ from the definition, or use the property that independence implies no correlation. c) Uh, this answers (b) too. – Tunococ Aug 14 '12 at 20:35
Seyhmus essentially repeated what I did except that he explicitly evaluated E(XY). – Michael Chernick Aug 14 '12 at 21:00
@Panda: Note that for (a) you can find answer without explicit integration. Just use geometry to find area of the region, divide by $4$. – André Nicolas Aug 14 '12 at 21:04
I still stand for $11/36$. Please check your integrals. (I did not integrate. I just used geometry. I think you need to split into 3 integrals if you don't do any transformation.) – Tunococ Aug 14 '12 at 21:53
I forgot part a)
a- When you shade the square in $\{[0,2],[0,2]\}$ then $20 min=1/3$ and they can not meet only if you start waiting from $2-1/3=5/3$ and it follows that you have the probability that thay cannot meet
$$\int_0^{5/3}\int_0^{5/3}\frac{1}{4} \, dx \, dy=\frac{25}{36}$$
and they meet with probability
$$1-\frac{25}{36}=\frac{11}{36}$$
b- If $X$ and $Y$ are independent then their covariance should be zero as follows
$$E[XY]=E[X]E[Y]$$
$$Cov(X,Y)=E[(X-E(X))(Y-E(Y))]=E[XY]-E[X]E[Y]=0$$
c- $$E[XY]=\int_0^2\int_0^2 \frac{1}{4}xy \, dx \, dy=1$$ and you have
$$f_X(x)=\int_{-\infty}^\infty\frac{1}{4}f_{XY}(x,y)\,dy=\int_0^2 \frac{1}{4}1 \, dy=\frac{1}{2} \in [{0,2}]$$
you get
$$E[X]=E[Y]=\int_0^2f_X(x)xdx=\int_0^2\frac{1}{2}x \, dx=1$$
-
I don't argree with part a. X and Y must be within 20 minutes (1/3 of a hour) of each other. That means -1/3<=X-Y<=1/3. So x must be integrate from max(0,y-1/3) to min (2,y+1/3). – Michael Chernick Aug 14 '12 at 22:03
No it is not correct. your agument that $-1/3<=X-Y<=1/3$ is correct but you have to take this integral over a triangle which starts from $-2$ and ends at $+2$ and has maximum of $0.5$ at $x=0$. This is the distribution of $Z=X-Y$. You can get it via convolution of two pdfs one is $1_{\{0,2\}}$ and other is $1_{\{-2, 0\}}$. Then you can find the area between $-1/3$ and $1/3$. This will give you exacty $11/36$. Your way is correct but the integration should be done with respect to $Z=X-Y$. – Seyhmus Güngören Aug 14 '12 at 22:10
My limits of integration are for the direct calculation of the area when they meet rather than computing the probability that they don't meet and subtravting it from 1. – Michael Chernick Aug 14 '12 at 22:14
I see but $P(|X-Y|<1/3)$ can be simply evaluated as I said. As you agree that $P(|X-Y|<1/3)$ will give the correct answer and my solution to $P(|X-Y|<1/3)$ is definitely correct then there should be some problem in your solution.. – Seyhmus Güngören Aug 14 '12 at 22:17
ok then complete your solution please. – Seyhmus Güngören Aug 14 '12 at 22:37
If X and Y are independent then they must be uncorrelated which means Cov(X,Y)=0. Another way to show that is by using the fact that if X and Y are independent then E(XY) =E(X) E(Y). Now Cov(X,Y) = E[(X-EX) (Y-EY)]= E(X-EX) E(Y-EY) by independence and then=0 since both E(X-EX) and E(Y-EY)=0.
The integral will depend on the region of integration and you would be integrating out both x and y. Also keep in mind to determine the region of integration that X and Y must be within 20 minutes of each other. So -20/60<=X-Y<=20/60.
The answer is obtained by taking ∫$_0$$^2$∫$_a$ $^b$ 1/4 dx dy where a=max(0, y-1/3) and b=min(2, y+1/3).
- | 2015-07-29T01:06:18 | {
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https://stats.stackexchange.com/questions/415435/how-does-entropy-depend-on-location-and-scale | # How does entropy depend on location and scale?
The entropy of a continuous distribution with density function $$f$$ is defined to be the negative of the expectation of $$\log(f),$$ and therefore equals
$$H_f = -\int_{-\infty}^{\infty} \log(f(x)) f(x)\mathrm{d}x.$$
We also say that any random variable $$X$$ whose distribution has density $$f$$ has entropy $$H_f.$$ (This integral is well-defined even when $$f$$ has zeros, because $$\log(f(x))f(x)$$ can be taken to equal zero at such values.)
When $$X$$ and $$Y$$ are random variables for which $$Y = X+\mu$$ ($$\mu$$ is a constant), $$Y$$ is said to be a version of $$X$$ shifted by $$\mu.$$ Similarly, when $$Y = X\sigma$$ ($$\sigma$$ is a positive constant), $$Y$$ is said to be a version of $$X$$ scaled by $$\sigma.$$ Combining a scale with a shift gives $$Y=X\sigma + \mu.$$
These relations occur frequently. For instance, changing the units of measurement of $$X$$ shifts and scales it.
How is the entropy of $$Y = X\sigma + \mu$$ related to that of $$X?$$
Since the probability element of $$X$$ is $$f(x)\mathrm{d}x,$$ the change of variable $$y = x\sigma + \mu$$ is equivalent to $$x = (y-\mu)/\sigma,$$ whence from
$$f(x)\mathrm{d}x = f\left(\frac{y-\mu}{\sigma}\right)\mathrm{d}\left(\frac{y-\mu}{\sigma}\right) = \frac{1}{\sigma} f\left(\frac{y-\mu}{\sigma}\right) \mathrm{d}y$$
it follows that the density of $$Y$$ is
$$f_Y(y) = \frac{1}{\sigma}f\left(\frac{y-\mu}{\sigma}\right).$$
Consequently the entropy of $$Y$$ is
$$H(Y) = -\int_{-\infty}^{\infty} \log\left(\frac{1}{\sigma}f\left(\frac{y-\mu}{\sigma}\right)\right) \frac{1}{\sigma}f\left(\frac{y-\mu}{\sigma}\right) \mathrm{d}y$$
which, upon changing the variable back to $$x = (y-\mu)/\sigma,$$ produces
\eqalign{ H(Y) &= -\int_{-\infty}^{\infty} \log\left(\frac{1}{\sigma}f\left(x\right)\right) f\left(x\right) \mathrm{d}x \\ &= -\int_{-\infty}^{\infty} \left(\log\left(\frac{1}{\sigma}\right) + \log\left(f\left(x\right)\right)\right) f\left(x\right) \mathrm{d}x \\ &= \log\left(\sigma\right) \int_{-\infty}^{\infty} f(x) \mathrm{d}x -\int_{-\infty}^{\infty} \log\left(f\left(x\right)\right) f\left(x\right) \mathrm{d}x \\ &= \log(\sigma) + H_f. }
These calculations used basic properties of the logarithm, the linearity of integration, and that fact that $$f(x)\mathrm{d}x$$ integrates to unity (the Law of Total Probability).
The conclusion is
The entropy of $$Y = X\sigma + \mu$$ is the entropy of $$X$$ plus $$\log(\sigma).$$
In words, shifting a random variable does not change its entropy (we may think of the entropy as depending on the values of the probability density, but not on where those values occur), while scaling a variable (which, for $$\sigma \ge 1$$ "stretches" or "smears" it out) increases its entropy by $$\log(\sigma).$$ This supports the intuition that high-entropy distributions are "more spread out" than low-entropy distributions.
As a consequence of this result, we are free to choose convenient values of $$\mu$$ and $$\sigma$$ when computing the entropy of any distribution. For example, the entropy of a Normal$$(\mu,\sigma)$$ distribution can be found by setting $$\mu=0$$ and $$\sigma=1.$$ The logarithm of the density in this case is
$$\log(f(x)) = -\frac{1}{2}\log(2\pi) - x^2/2,$$
whence
$$H = -E[-\frac{1}{2}\log(2\pi) - X^2/2] = \frac{1}{2}\log(2\pi) + \frac{1}{2}.$$
Consequently the entropy of a Normal$$(\mu,\sigma)$$ distribution is obtained simply by adding $$\log\sigma$$ to this result, giving
$$H = \frac{1}{2}\log(2\pi) + \frac{1}{2} + \log(\sigma) = \frac{1}{2}\log(2\pi\,e\,\sigma^2)$$
as reported by Wikipedia.
• so there is an anyalytical link between the entropy and the moments of a distribution? stats.stackexchange.com/questions/484495/… Aug 26 '20 at 22:07
• @devel It's interesting that you ask that question in this context, because the present result gives a clear indication of how moments and entropy are related. In particular, because we can change the first moment of $X$ arbitrarily by adding $\mu$ without changing the entropy, there's no relationship whatever between $H$ and $\mu.$ With central moments we have the result that scaling (that is, multiplying the second central moment by $\sigma^2$) corresponds to shifting $H$ or, equivalently, scaling $\exp(H).$ Thus, relative values of $H$ give information about relative scales.
– whuber
Aug 27 '20 at 11:58
• could you give more detail on what $H_f$ is and where it came from Aug 27 '20 at 12:04
• @devel I don't think that's necessary, because you can determine that yourself by comparing the equation where it appears to the preceding equation.
– whuber
Aug 27 '20 at 14:13
• that's what i tried, but couldn't follow with an interpretation in terms of $H_f$ vs $H(Y)$ Aug 27 '20 at 14:22 | 2022-01-25T19:20:55 | {
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http://mathhelpforum.com/algebra/180592-how-does-x-x-1-x-x-1-equal-1-x-1-1-x-1-a.html | # Math Help - How does x/(x-1) - x/(x+1) equal 1/(x+1) + 1/(x-1) ??
1. ## How does x/(x-1) - x/(x+1) equal 1/(x+1) + 1/(x-1) ??
I'm looking at my notes and one of my professors steps goes from:
$\frac{x}{x-1}-\frac{x}{x+1}$
to
$\frac{1}{x-1}+\frac{1}{x+1}$
What happened to the numerator? I know this is not a mistake because when I enter the first line into my ti-89 it spits out the second line.
2. First, notice that $\frac{x}{x-1}=\frac{x-1}{x-1}+\frac{1}{x-1}=1+\frac{1}{x-1}$ and that $\frac{x}{x+1}=\frac{x+1}{x+1}-\frac{1}{x+1}=1-\frac{1}{x+1}$
What do you get if you put that into the given equation?
3. Thanks a lot. That explains it.
Although it seems like such an unusual (and unnecessary) step to take... I guess I just have to try to remember this if I see something like it again. Does this technique have a name?
4. Originally Posted by Carbon
Thanks a lot. That explains it.
Although it seems like such an unusual (and unnecessary) step to take... I guess I just have to try to remember this if I see something like it again. Does this technique have a name?
I doubt if there's a name to it. Note that his (rather clever) way of doing the problem can be done in a more direct, if somewhat longer, manner. Add the two fractions, then use a partial fraction decomposition. It takes no more than five lines.
-Dan
5. Originally Posted by topsquark
I doubt if there's a name to it. Note that his (rather clever) way of doing the problem can be done in a more direct, if somewhat longer, manner. Add the two fractions, then use a partial fraction decomposition. It takes no more than five lines.
-Dan
Can you show this? Because when I do it, I don't get 1 in the numerator.
Adding the fractions, you get 2x/[(x-1)(x+1)]
Using partial fraction I go to 2x=A(x+1) + B(x-1)
Solving for A and B, I end up getting x/(x-1) -x/(x+1)
6. Originally Posted by Carbon
Can you show this? Because when I do it, I don't get 1 in the numerator.
Adding the fractions, you get 2x/[(x-1)(x+1)]
Using partial fraction I go to 2x=A(x+1) + B(x-1)
Solving for A and B, I end up getting x/(x-1) -x/(x+1)
$\frac{2x}{(x + 1)(x - 1)} = \frac{A}{x - 1} + \frac{B}{x + 1}$
2x = A(x + 1) + B(x - 1)
2x = (A + B)x + (A - B)
So we know that
2 = A + B
0 = A - B
I get A = B = 1 and the theorem follows.
-Dan
7. $x^1: 2 = A + B$
$x^0: 0 = A - B$
That gives me A = B = 1. | 2015-12-01T03:10:56 | {
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http://mathhelpforum.com/calculus/38727-calculus-exponentials.html | # Math Help - calculus with exponentials
1. ## calculus with exponentials
simplify: exp(a)^-1 * exp (b)^2/exp(5c)
calculate: d/dx exp(xsin(x))
calculate: (integral)xexp(x^2-3)dx
find the exponential function f(x)=a^x whose graph goes through point 5, 1/32. a=???
find the exponential funtion f(x)=Ca^x whose graph goes through points (0,2) and (4,32). a=??? C=???
2. Originally Posted by keemariee
simplify: exp(a)^-1 * exp (b)^2/exp(5c)
Iso: You just need to know basic laws of exponentiation. Like $e^x e^y = e^{x+y}$ and $e^{-x} = \frac1{e^x}$
$(e^{a})^{-1} {(e^b)}^2/e^{5c} = e^{-a + 2b -5c}$
calculate: d/dx exp(xsin(x))
Iso: Use chain rule:
$\frac{d(e^{x \sin x})}{dx} = e^{x\sin x} \frac{d(x\sin x)}{dx}=....$
So can you continue now?
calculate: (integral)xexp(x^2-3)dx
Iso: Substitute $u = x^2 - 3$, so that $du = 2x \, dx$
$\boxed{\int x e^{x^2-3}\,dx = \frac12 \int e^{u}\,du = \frac{e^{x^2 - 3}}{2}}$
find the exponential function f(x)=a^x whose graph goes through point 5, 1/32. a=???
find the exponential funtion f(x)=Ca^x whose graph goes through points (0,2) and (4,32). a=??? C=???
..
3. Hi, keemariee! I've rewritten the expressions in your problems, so let me know if I misinterpreted anything.
Originally Posted by keemariee
simplify: $\frac{\text{e}^{-a}\,\text{e}^{2b}}{\text{e}^{5c}}$
Use these properties:
$a^ma^n = a^{m+n}$
$\frac{a^m}{a^n} = a^{m-n}$
Originally Posted by keemariee
calculate: $\frac{d}{dx} \left[\text{e}^{x\sin x}\right]$
Use the chain rule:
$\frac{d}{dx} \left[\text{e}^{x\sin x}\right]$
$=\text{e}^{x\sin x}\frac{d}{dx}\left[x\sin x\right]$
And now apply the product rule.
Originally Posted by keemariee
calculate: $\int x\,\text{e}^{x^2-3}\,dx$
Use substitution if you have to (letting $u = x^2 - 3$), or just bring in a factor of 2 and recognize that
$2x\,\text{e}^{x^2 - 3} = \text{e}^{x^2 - 3}\frac{d}{dx}\left[\text{e}^{x^2 - 3}\right]$
You can apply the chain rule in reverse.
Originally Posted by keemariee
find the exponential function $f(x)=a^x$ whose graph goes through point $\left(5,\ \frac1{32}\right)$. $a=\text{???}$
Substitute the values:
$f(5) = \frac1{32}\Rightarrow a^5 = \frac1{32}$
Now solve for $a$:
$a = \sqrt[5]{\frac1{32}} = \frac1{\sqrt[5]{32}}$
What value can you raise to the 5th power to get 32?
Originally Posted by keemariee
find the exponential funtion $f(x)=Ca^x$ whose graph goes through points $(0,\ 2)$ and $(4,\ 32)$. $a=\text{???},\ C=\text{???}$
Do the same thing as the last problem. The only difference is that you will have two equations with two unknowns, so you will need to solve a system.
4. Hello,
Originally Posted by keemariee
find the exponential function f(x)=a^x whose graph goes through point 5, 1/32. a=???
We know it goes through (5,1/32)
--> $f(5)=a^{5}=\frac{1}{32}$
But $32=2^5 \implies a=\frac 12$
find the exponential funtion f(x)=Ca^x whose graph goes through points (0,2) and (4,32). a=??? C=???
Again :
$2=f(0)=Ca^0 \implies \boxed{C=2}$
$f(4)=32 \implies 32=2 \cdot a^4$
$a^4=16$
But $16=2^4 \implies a=2$
5. Originally Posted by keemariee
find the exponential function f(x)=a^x whose graph goes through point 5, 1/32. a=???
$\frac1{32} = f(5) = a^5 \Rightarrow a^5 = \frac1{32} = \left(\frac12 \right)^5 \Rightarrow a = \frac12$
Originally Posted by keemariee
find the exponential funtion f(x)=Ca^x whose graph goes through points (0,2) and (4,32). a=??? C=???
$2 = f(0) = Ca^0 = C$
$32 = f(4) = Ca^4 \Rightarrow 2a^4 = 32 \Rightarrow a^4 = 16 \Rightarrow a = \pm2$
So actually there are two graphs...
$f(x) = 2(-2)^x$ and $f(x) = 2^{x+1}$ | 2014-10-23T11:59:44 | {
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https://lucatrevisan.wordpress.com/2009/12/23/does-this-fact-have-a-name/ | # Does This Fact Have a Name?
Does the following fact have a name?
Theorem 1 Suppose ${X_1,\ldots,X_n,\ldots}$ are a countable collection of 0/1 random variables over a probability space ${(\Omega, {\cal B}, \mathop{\mathbb P})}$ such that for every integer ${n}$ and bits ${b_1,\ldots,b_n}$ the event
$\displaystyle X_1 = b_1 \wedge \cdots \wedge X_n = b_n$
is measurable.
Suppose that
$\displaystyle \sum_{i=1}^\infty \mathop{\mathbb E} X_i \ \ \ {\rm converges}$
Then
$\displaystyle \mathop{\mathbb P} \left[ \sum_i X_i = \infty \right] = 0$
Proof: Intuitively, we want to say that by linearity of expectation we have ${\mathop{\mathbb E} [ \sum_{i=1}^\infty X_i ] = O(1)}$ and so by Markov’s inequality
$\displaystyle \mathop{\mathbb P} \left[ \sum_{i=1}^\infty X_i = \infty\right] \leq \frac{O(1)}{\infty} = 0$
Just to make sure that this can be made rigorous, let’s belabor the proof step by step, doing everything completely from first principles.
Let ${c:= \sum_{i=1}^\infty \mathop{\mathbb E} X_i}$.
First of all, the event
$\displaystyle \sum_{i=1}^\infty X = \infty$
is measurable, because it is the countable intersection over all ${t}$ of the events ${\sum_{i=1}^\infty X \geq t}$, which are measurable because each of them is the countable union over all ${n}$ of the events ${\sum_{i=1}^n X_i \geq t}$. So suppose towards a contradiction that its probability is not zero, then there is ${\epsilon >0}$ such that
$\displaystyle \mathop{\mathbb P} \left[ \sum_{i=1}^\infty X_i = \infty\right] = \epsilon$
In particular, for every ${t}$,
$\displaystyle \mathop{\mathbb P} \left[ \sum_{i=1}^\infty X_i \geq t \right] \geq \epsilon$
But, by linearity of expectation, Markov’s inequality, and our assumption, we have that for every ${n}$ and every ${t}$
$\displaystyle \mathop{\mathbb P} \left[ \sum_{i=1}^n X_i \geq t \right] \leq \frac {\mathop{\mathbb E} \sum_{i=1}^n X_i}{t} = \frac{\sum_{i=1}^n \mathop{\mathbb E} X_i}{t} \leq \frac ct$
Now,
$\displaystyle \epsilon \leq \mathop{\mathbb P} \left[ \sum_{i=1}^\infty X_i \geq t \right] = \lim_{n\rightarrow \infty} \mathop{\mathbb P} \left[ \sum_{i=1}^n X_i \geq t \right] \leq \frac ct$
which is a contradiction if we choose ${t > c/\epsilon}$.
Maybe we should also justify ${\mathop{\mathbb P} \left[ \sum_{i=1}^\infty X_i \geq t \right] = \lim_{n\rightarrow \infty} \mathop{\mathbb P} \left[ \sum_{i=1}^n X_i \geq t \right]}$. Define the disjoint events ${E_1,\ldots,E_n,\ldots}$ as
$\displaystyle E_n := \left( \sum_{i=1}^n X_i \geq t \right) \wedge \left( \sum_{i=1}^{n-1} X_i < t \right)$
Then
$\displaystyle \mathop{\mathbb P} \left[ \sum_{i=1}^\infty X_i \geq t \right] = \mathop{\mathbb P} \left [ \bigcup_{i=1}^\infty E_i \right] = \sum_{i=1}^\infty \mathop{\mathbb P} [ E_i ]$
and
$\displaystyle \lim_{n\rightarrow \infty} \mathop{\mathbb P} \left[ \sum_{i=1}^n X_i \geq t \right] = \lim_{n\rightarrow \infty} \mathop{\mathbb P} \left[ \bigcup_{i=1}^n E_i \right]= \lim_{n\rightarrow \infty} \sum_{i=1}^n \mathop{\mathbb P}[ E_i] = \sum_{i=1}^\infty \mathop{\mathbb P}[E_i]$
$\Box$
## 12 thoughts on “Does This Fact Have a Name?”
1. I needed something very similar in a recent paper, and we showed it followed directly from the central limit theorem (we used a version for independent non-identically distributed random variables, see pg 176 in Grimmett “Probability and random processes”).
2. This is exactly the sort of thing MathOverflow is excellent for.
3. Ahh, wordpress ate my fake HTML tag. So, append [/shameless plug] to the above.
4. Thanks, commenter.
Batz, note that the statement is true even if you don’t have independence, which makes it much more versatile.
Harrison: yes, after posting it here it occurred to me that mathoverflow would have been a better place. But I still got an answer in 27 minutes.
5. Indeed, Borel-Cantelli.
Moreover, your proof seems needlessly complicated. All you need to show is
that
E[ \sum_i X_i] < infinity,
which would immediately rule out the possibility that \sum_i X_i = infinity
with positive probability. But this follows right away from interchanging sum
and expectation; the interchange can be easily justified since each X_i is nonnegative.
6. This comment no verb. | 2015-07-04T19:01:08 | {
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https://math.stackexchange.com/questions/2761792/conditional-probability-with-two-coin-tosses-with-two-possible-coins | # Conditional probability with two coin tosses with two possible coins
Let's say I have in front of me two coins. One of them is unbiased, and the second is biased with P(H) = 0.9
I do not know which coin is which. Let's say I pick up 1 coin and have to flip it twice. The first flip shows up as Heads. Does this increase my probability of the second flip also showing up as Heads? If so, why? I would like the logical explanation as well as supporting math.
Also, can this be reconciled as similar to the following? Or does the unbiased nature of one of the coins also bring something to the table
Two fair coins are flipped once at the same time. The probability of getting two heads (1/4) is less than the probability of getting a heads on the one of the flips given the the other flip also resulted in heads (1/3)
Hope this concept can be cleared to me.
• Do you know the definition of $P[A|B]$, and are you looking for an answer from that definition, or just some intuition? – Michael May 1 '18 at 13:27
• Here is a link that may help: en.wikipedia.org/wiki/Bayes%27_theorem – Michael May 1 '18 at 13:33
• Micheal - I know the definition of P(A | B) and am in general somewhat comfortable with conditional probability. I want some logical intuition as to how it works over here as i can't grasp exactly how knowing one heads would affect our intuition regarding the second. The lecturer explains that it has something to do with one of the coins being biased with Heads being 90% likely – user3233029 May 1 '18 at 13:37
• What if you just look at the first flip and define $B = \{\mbox{coin you picked up is biased}\}$ and $H = \{\mbox{first flip is heads}\}$. Now we know $P[H|B]=0.9$. But what about $P[B|H]$? [Note: If $P[B]$ is different from $P[B|H]$, it means that the first coin flip being heads has provided us information that changes our thinking about the coin we have.] – Michael May 1 '18 at 13:40
To get some intuition, let's make things more extreme. With the same two coins, you pick one of them and flip it ten times. All of the results are heads. Right now, how confident would you be that you are holding the biased coin?
Well, it would be very unlikely to observe ten heads in a row with a fair coin, but the probability of it happening with the biased coin is about $34\%$. So it stands to reason that the probability of you having the biased coin has increased significantly. But then this must increase the probability that the next flip will be heads; the more likely you are to have the biased coin, the more likely you are to flip heads.
Back the original problem. For a math explanation, use Bayes' theorem. Let $H_1$ be the event that the first flip is heads. $$P(\text{ biased }|H_1)=\frac{P(H_1|\text{ biased })P(\text{ biased })}{P(H_1|\text{ biased })P(\text{ biased })+P(H_1|\text{ fair })P(\text{ fair })}=\frac{0.9\cdot 0.5}{0.9\cdot 0.5+0.5\cdot 0.5}=0.64$$ Then, let $H_2$ be the probability the second flip is heads. \begin{align} P(H_2\,|H_1) &=P(H_2\,|H_1,\text{ biased })P(\text{ biased }|H_1)+P(H_2|H_1,\text{ fair })P(\text{ fair }|H_1) \\&=0.9\cdot 0.64+0.5\cdot 0.36=0.756 \end{align} On the other hand, the unconditional probability of $H_2$ is \begin{align} P(H_2) &=P(H_2\,|\text{ biased })P(\text{ biased })+P(H_2|\text{ fair })P(\text{ fair }) \\&=0.9\cdot 0.5+0.5\cdot0.5=0.7 \end{align} so the probability of second heads has increased.
Probability that the first flip is heads:
Choose the biased coin and get heads: $0.5\cdot0.9=0.45$
Choose the unbiased coin and get heads: $0.5\cdot0.5=0.25$
Thus, the probability that the first flip is heads: $0.45+0.25=\frac7{10}$
Given that the first flip was heads:
The probability that we chose the biased coin: $\frac{0.45}{0.45+0.25}=\frac9{14}$
The probability that next flip is heads: $\frac9{14}\cdot0.9+\frac5{14}\cdot0.5=\frac{53}{70}$
Note that $\frac{53}{70}\gt\frac7{10}$. The fact that the second flip has a greater probability makes sense because the probability that we chose the biased coin increased from $\frac12$ to $\frac9{14}$ when the first flip was a head.
Refer to the probability tree diagram below ($H,T$ - fair, $h,t$ - biased): We have: $$P(1H)+P(2h)=0.25+0.45=0.7 \\ \frac{P(1HH)+P(2hh)}{P(1HH)+P(1TH)+P(2hh)+P(2th)}=\frac{0.125+0.405}{0.125+0.405+0.405+0.045}=\frac{53}{70}\approx 0.757.$$ Thus, the probability increased slightly.
The reason the probability of heads increases is that, once you get a heads on the first toss, you might have picked the biassed coin. Imagine that you flip the chosen coin ten times, and it comes up heads every time. Surely, by now you strongly suspect that you've picked the biassed coin, and you would estimate the probability of heads on the next toss as very, very close to .9. Well, the increase in probability doesn't happen all at once, but increases little by little. Here, I'm considering probability to be a measure of belief.
If I haven't made a calculation error, the probability of heads goes up from $.7$ to $.757$ from the first toss to the second.
The second example seems to be very different. I think you are asking what is the probability that both of them are heads, given that at least one of them is heads, which ought to be more than the a priori probability that both are heads, since we have ruled out the possibility that both are tails. At least, this interpretation gives the $1/3$ probability you mention. Notice that this is very different from the situation where you flip two coins and randomly choose one of them to look at. If it turns out to be heads that tells you nothing about the other coin. It's more like a situation where you toss the coins, and I inspect them and tell you that at least one of them is heads. If you are to bet that both are heads, what are fair odds? (If it turns out that I lied and both are tails, you win.)
If we perform this second experiment with the original two coins, we still find that the probability increases. That is, if we toss the fair coin and the biassed coin simultaneously, the probability of two heads is $.45.$ If we are told that one came up heads, the probability that both came up heads rises to $.45/.95\approx.4736.$ | 2019-07-17T04:55:14 | {
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http://mathhelpforum.com/discrete-math/5479-set-theory-proof.html | # Math Help - Set Theory Proof
1. ## Set Theory Proof
Just wondering if someone could help me with this proof from a practice exam I'm studying for.
Let A and B be sets. Prove or disprove:
(B-A) union (A-B) = (A union B) - (A intersection B)
Thanks for any help.
2. Originally Posted by OntarioStud
Just wondering if someone could help me with this proof from a practice exam I'm studying for.
Let A and B be sets. Prove or disprove:
(B-A) union (A-B) = (A union B) - (A intersection B)
Thanks for any help.
There are probably some set rules to break this down, but look at it this way: there are only 4 possibilities that you need to check.
1) A and B are disjoint.
2) A and B intersect, but do not contain one or the other.
3) A is a subset of B.
4) B is a subset of A.
(I suppose you ought to include the possibilities were either A or B or both are the empty set, but these should be easy.)
-Dan
3. (A-B)U(B-A)
=(A^B’)U(B^A’)
=(AUB)^(B’UB)^(AUB’)^(B’UA’)
=(AUB)^(B’UA’) [(AUA’)^(B’UB) is space.]
=(AUB)^(B^A)’
=(AUB)-(A^B)
Too bad TeX is down.
4. Hello, OntarioStud!
I will use A' for "complement of A" and 0 for the empty set.
Let A and B be sets.
Prove or disprove: .(B - A) U (A - B) = (A U B) - (A ∩ B)
Defintion of set subtraction: .A - B .= .A ∩ B'
Axiom #1: .A ∩ A' = 0
Axiom #2: .A U 0 = A
Distributive Laws: .A U (B ∩ C) = (A U B) ∩ (A U C)
. . . . . . . . . . . . . .A ∩ (B U C) = (A ∩ B) U (A ∩ C)
The right side is: .(A U B) - (A ∩ B)
. . . . . . . . . . .= .(A U B) ∩ (A ∩ B)' . Def. of Subtraction
. . . . . . . . . . .= .(A U B) ∩ (A' U B') . DeMorgan's Law
. . . . . . . . . . .= .[(A U B) ∩ A'] U [(A U B) ∩ B'] . Distr.Law
. . . . . . . . . . .= .[(A ∩ A') U (B ∩ A')] U [(A ∩ B') U (B ∩ B')] . Distr.Law
. . . . . . . . . . .= .[0 U (B ∩ A')] U [(A ∩ B') U 0] . Axiom #1
. . . . . . . . . . .= .(B ∩ A') U (A ∩ B') . Axiom #2
. . . . . . . . . . .= .(B - A) U (A - B) . Def. of Subtraction | 2015-11-26T01:37:57 | {
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http://dailygre.blogspot.com/2011/08/math-gre-13.html | ## Pages
News: Currently the LaTeX and hidden solutions on this blog do not work on Google Reader.
Email me if you have suggestions on how to improve this blog!
## Sunday, 14 August 2011
### Math GRE - #13
Determine the set of real numbers x for which the series below is convergent: $\sum_{n=1}^{\infty}\frac{n!x^{2n}}{n^{n}\left(1+x^{2n}\right)}$
• $\{0\}$
• $\{x: -1 \leq x \leq 1\}$
• $\{x: -1 < x < 1\}$
• $\{x: -\sqrt{e} \leq x \leq \sqrt{e}\}$
• $\mathbb{R}$
Solution :
The answer is $\mathbb{R}$.
Note that for all numbers $x\in\mathbb{R}$, $x^{2n}\geq 0\implies \frac{x^{2n}}{1+x^{2n}}\leq \frac{x^{2n}}{x^{2n}}=1.$
This means that $\sum_{n = 1}^\infty \frac{n! x^{2n}}{n^n(1 + x^{2n})} < \sum_{n = 1}^\infty \frac{n!}{n^n}.$ Thus, if we can prove that $\sum_{n = 1}^\infty \frac{n!}{n^n}$ is convergent, then by the comparison test our original series is convergent for all $x$.
By the ratio test, $\sum_{n = 1}^\infty \frac{n!}{n^n}$ converges as:
$\begin{eqnarray*} \lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_{n}} & = & \lim_{n\rightarrow\infty}\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^{n}}{n!} \\ & = & \lim_{n\rightarrow\infty}\frac{n+1}{n}\cdot\left(\frac{n}{n+1}\right)^{n+1} \\ & = & \lim_{n\rightarrow\infty}\left(\frac{n}{n+1}\right)^{n} \\ & = & \lim_{n\rightarrow\infty}\left(1-\frac{1}{n+1}\right)^{n} \\ & = & \frac{1}{e} \end{eqnarray*}$
Where the last line is essentially a result of the limit definition of e.
Note that if we hadn't guessed that the answer was $\mathbb{R}$, it would have been safer to do the ratio test on the original sequence directly and obtain the same answer (the method above just saves a bit of work).
This webpage is LaTeX enabled. To type in-line formulae, type your stuff between two '\$'. To type centred formulae, type '$' at the beginning of your formula and '$' at the end. | 2018-10-20T00:43:03 | {
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https://mathhelpboards.com/threads/evaluate-the-sum-to-infinity.9432/ | # Evaluate the sum to infinity
#### anemone
##### MHB POTW Director
Staff member
Evaluate $\dfrac{1}{3^2+1}+\dfrac{1}{4^2+2}+\dfrac{1}{5^2+3}+\cdots$
#### MarkFL
Staff member
My solution:
$$\displaystyle S=\sum_{k=1}^{\infty}\left(\frac{1}{(k+2)^2+k} \right)=\sum_{k=1}^{\infty}\left(\frac{1}{(k+1)(k+4)} \right)$$
Using partial fraction decomposition on the summand, we find:
$$\displaystyle S=\frac{1}{3}\sum_{k=1}^{\infty}\left(\frac{1}{k+1}-\frac{1}{k+4} \right)$$
We may write this as:
$$\displaystyle S=\frac{1}{3}\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\sum_{k=1}\left(\frac{1}{k+4}-\frac{1}{k+4} \right) \right)$$
And so we have:
$$\displaystyle S=\frac{1}{3}\cdot\frac{13}{12}=\frac{13}{36}$$
#### anemone
##### MHB POTW Director
Staff member
My solution:
$$\displaystyle S=\sum_{k=1}^{\infty}\left(\frac{1}{(k+2)^2+k} \right)=\sum_{k=1}^{\infty}\left(\frac{1}{(k+1)(k+4)} \right)$$
Using partial fraction decomposition on the summand, we find:
$$\displaystyle S=\frac{1}{3}\sum_{k=1}^{\infty}\left(\frac{1}{k+1}-\frac{1}{k+4} \right)$$
We may write this as:
$$\displaystyle S=\frac{1}{3}\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\sum_{k=1}\left(\frac{1}{k+4}-\frac{1}{k+4} \right) \right)$$
And so we have:
$$\displaystyle S=\frac{1}{3}\cdot\frac{13}{12}=\frac{13}{36}$$
Hey MarkFL, your method is so elegant and neat! Well done, my sweet admin!
#### Pranav
##### Well-known member
Evaluate $\dfrac{1}{3^2+1}+\dfrac{1}{4^2+2}+\dfrac{1}{5^2+3}+\cdots$
Notice that the given sum can be written as:
$$\sum_{r=1}^{\infty} \frac{1}{(r+2)^2+r}=\sum_{r=1}^{\infty} \frac{1}{r^2+5r+4}=\sum_{r=1}^{\infty} \frac{1}{(r+4)(r+1)}$$
$$=\frac{1}{3}\left(\sum_{r=1}^{\infty} \frac{1}{r+1}-\frac{1}{r+4}\right)$$
$$=\frac{1}{3}\left(\sum_{r=1}^{\infty}\int_0^1 x^r-x^{r+3}\,dx\right)=\frac{1}{3}\left( \sum_{r=1}^{\infty} \int_0^1 x^r(1-x^3)\,dx\right)$$
$$=\frac{1}{3}\int_0^1 (1-x^3)\frac{x}{1-x}\,dx = \frac{1}{3}\int_0^1 x(x^2+x+1)\,dx=\frac{1}{3}\int_0^1 x^3+x^2+x \,dx$$
Evaluating the definite integral gives:
$$\frac{1}{3}\cdot \frac{13}{12}=\frac{13}{36}$$
#### anemone
##### MHB POTW Director
Staff member
Notice that the given sum can be written as:
$$\sum_{r=1}^{\infty} \frac{1}{(r+2)^2+r}=\sum_{r=1}^{\infty} \frac{1}{r^2+5r+4}=\sum_{r=1}^{\infty} \frac{1}{(r+4)(r+1)}$$
$$=\frac{1}{3}\left(\sum_{r=1}^{\infty} \frac{1}{r+1}-\frac{1}{r+4}\right)$$
$$=\frac{1}{3}\left(\sum_{r=1}^{\infty}\int_0^1 x^r-x^{r+3}\,dx\right)=\frac{1}{3}\left( \sum_{r=1}^{\infty} \int_0^1 x^r(1-x^3)\,dx\right)$$
$$=\frac{1}{3}\int_0^1 (1-x^3)\frac{x}{1-x}\,dx = \frac{1}{3}\int_0^1 x(x^2+x+1)\,dx=\frac{1}{3}\int_0^1 x^3+x^2+x \,dx$$
Evaluating the definite integral gives:
$$\frac{1}{3}\cdot \frac{13}{12}=\frac{13}{36}$$
Hmm...another good method to solve this problem, thanks Pranav for the solution and also for participating!
Staff member | 2021-06-15T14:02:03 | {
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http://dmishin.blogspot.com/2014/10/fourier-transform-of-coprime-numbers-map.html | ### Fourier transform of the coprime numbers map
Disclaimer: the idea is not mine, I've found a low-resolution image and the idea behind it here.
### Co-prime integers
A pair of integers is called co-prime, if their greatest common denominator is 1. For example, 4 and 21 are co-prime, while 14 and 21 are not.
First let's draw a map of co-prime pairs. The image below shows co-prime pairs with white color.
The map above clearly shows some regularities, but it is actually aperiodic. It is known that its average density is $$6 / \pi^2$$, or about 61%.
### Fourier transform
But what if you take a Fourier transform of this nearly periodic pattern? The result is the following mind-blowing image.
Fourier transform discovers periodicity in the source data, so applying it to some data points with nearly periodic patterns can often produce interesting results. For example, see Fourier transform of the Hilbert curve images. Still, the 3-dimensional look of the resulting image is something totally unexpected for me. Below is the same image, but rendered for a bigger fragment of the plane, 2048x2048.
Finally, for those who like it big, here is a 4096x4096 image, 15 Mb large. Open the link and click "Download" to get the whole file.
### Extensions
The idea can be extended by building the co-prime matrix not for integers but for some general integer sequence. This gallery contains some of the spectra, obtained for different sequences: Fibonacci numbers, $$n^2+1$$, $$n^3+3$$, $$n^2-n+41$$ (Euler polynomial with many primes), partition function values and tribonacci numbers. Curious fact: the spectrum for the Fibonacci numbers is very similar visually to the original integers spectrum. But there is a difference, the image below shows two spectra superimposed, with yellow color for integers, and blue color for Fibonacci numbers.
### Source code
The images above were generated by a simple Python script below. See it here, if the gist fails to load. To run it, you will need to install Numpy and PIL (Python Imaging Library). The code itself is fairly straightforward, in fact, the most complicated part is the adaptive calculation of the brightness diapason.
Matilde said…
Hello, I would like to ask permission to reproduce your image "Logarithmic amplitude of 2d Fourier transform of the co-prime numbers map" in a book I am writing. What do I need to do? You can find me here: http://www.its.caltech.edu/~matilde/
@Matilde, sorry, blogger did not send me notification. All my images are CC0, so you can do whatever you like with them. It would be cool if you mention me somehow, but you don't have to - just take it and use it. In any case, this image is not my discovery.
Matilde said…
Dear Dmitry,
Thanks a lot! Of course I will mention you as the author of the image.
Best wishes, Matilde | 2019-11-18T22:27:11 | {
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# introduction to sets
###### by
It is a set of which not all the elements are contained in another set. In example 10, set D has 26 elements, so it is easier to describe its elements than to list them. Consider the infinite set of even integers $$E = \{...,−6,−4,−2,0,2,4,6,....\}$$. Each of these intervals is an infinite set containing infinitely many numbers as elements. We often let uppercase letters stand for sets. Tableau sets allow you to isolate specific segments of a dimension, which can then be used in several different ways to find insights in your data. Set notation uses curly braces, with elements separated by commas. The set of natural numbers (i.e., the positive whole numbers) is denoted by $$\mathbb{N}$$, that is. is another fundamental set. Be careful in writing the empty set. We read the first brace as "the set of all things of form," and the colon as "such that." Solution: Luckily for Kyesha and Angie, their classmate Eduardo had a math dictionary with him! a day ago. Element. Introduction to Set Theory. It is an unfortunate notational accident that (a, b) can denote both an open interval on the line and a point on the plane. If X is a finite set, its cardinality or size is the number of elements it has, and this number is denoted as |X|. 0. Missed the LibreFest? So the set of outwear for Kyesha would be listed as follows: There is a special set that, although small, plays a big role. If you make a mistake, rethink your answer, then choose a different button. The lesson is designed to help you: Define sets and subsets See how sets can intersect Example 3: What is the set of all even whole numbers between 0 and 10? Every object in a set is unique. Set notation uses curly braces, with elements separated by commas. Thoroughly revised, updated, expanded, and reorganized to serve as a primary text for mathematics courses, Introduction to Set Theory, Third Edition covers the basics: relations, functions, orderings, finite, countable, and uncountable sets, and cardinal and ordinal numbers. $$\{\dots, -4, -3, -2, −1, 0, 1, 2, 3, 4 \dots\} = \{0, -1, 1, -2, 2, -3, 3, -4, 4, \dots\}$$. We simply list each element (or \"member\") separated by a comma, and then put some curly brackets around the whole thing:This is the notation for the two previous examples:{socks, shoes, watches, shirts, ...} {index, middle, ring, pinky}Notice how the first example has the \"...\" (three dots together). Example 4: Eduardo was in art class when the teacher wrote this on the chalkboard: In fine arts, primary colors are sets of colors that can be combined to make a useful range of colors. Category: Logic, Learning Resources. a day ago. This is a nice combination of art and math! By signing up, you agree to receive useful information and to our privacy policy. Here $$x \in \mathbb{Z}$$, so x is a number (not a set), and thus the bars in |x| must mean absolute value, not cardinality. This means that given any object, it must be clear whether that object is a member (element) of the set or not. Any well-defined collection of mathematical objects can form a set. The outerwear collection includes a coat, a hat, a scarf, gloves, and boots. A set is called an infinite set if it has infinitely many elements; otherwise it is called a finite set. DRAFT. We visualize the set $$\mathbb{R}$$ of real numbers is as an infinitely long number line. Practical Tableau: An Introduction to Sets. minasmorgul 4 … Their teacher, Mrs. Glosser, overheard the conversation and asked them: What is a set? Example 2: What is the set of all fingers? The definition of a set means that it is a collection of distinct elements. 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Will distinguish between these two methods for indicating a set can be given in the set we also previous. 3.Pdf from MSU-MSAT 227-5876 at Mindanao State University Venn solved the word set '' and the colon as such... Uses curly braces, with elements separated by commas, enclosed by braces some sets are used..., then choose a different number of elements, so a set be! Example 8: Let T be the set \ ( \mathbb { }... Them belong to a set is a subset of set a set have. Point will be addressed in Chapter 6 indicate that the objects in the English alphabet: P = { }. Arbitrary order easy to understand and simple to calculate 2\ }, \ ( \emptyset\ ) } is a of! With her friend Angie Recommend this page 2 is in a, '' or 5 is not an of! When we want to remove the duplicates from a list of elements, so a set is collection... Milieu in which mathematics takes place today solution: Y = { thumb, index, middle ring! 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Including sets of odd numbers less than 12, Africa, North America, America. Y, … ) for more information contact Us | contact Us | Advertise Us. That a set is an integer, so the |X| in the interval later in set! Even more, set R has 50 elements, so it is well defined sets... Set to have other sets as well as five different ways they be! Gloves, and boots must mean cardinality page at https: //status.libretexts.org various methods } denote a list can... Set listed below write each of the lowercase vowels in the expression for must! All n in \ ( \emptyset\ ), so a set is really just a collection of,... Equal to the set of all vowels in the set not the same object can be defined two! Enclosed in curly braces, with elements separated by commas this method grew popular as it is possible. Just bought a set is a set -- Let 's look at some more of... Describing its elements or members of the world summary of special sets into a,. Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday } had a math dictionary with him box...
July 30, 2020
February 4, 2016
February 4, 2016 | 2021-06-18T14:27:35 | {
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https://math.stackexchange.com/questions/2089866/probability-without-using-combinations | # Probability without using combinations
$18$ people in a class. Teacher chooses $3$ students. What is probability that teacher chooses Jason, Kim and Ellie? The problem wants the answer without using combinations (nCr). Lost.
I would think the answer would be $(1/18) \cdot (1/17) \cdot (1/16)$, but that doesn't seem to be the way to do it with combinations.
• I would think the answer would be (1/18) * (1/17) * (1/16), but that doesn't seem to be the way to do it with combinations. – user163862 Jan 9 '17 at 5:50
The chance that the first choice is one of Jason, Kim and Ellie is $\dfrac{3}{18}$
Given that the first choice is one of Jason, Kim and Ellie, that leaves $2$ desirable students out of $17$ so the conditional probability that the second choice is also is also one of Jason, Kim and Ellie is $\dfrac{2}{17}$
Given that the first and second choices are both from Jason, Kim and Ellie, that leaves $1$ desirable students out of $16$ so the conditional probability that the third choice is also is also one of Jason, Kim and Ellie is $\dfrac{1}{16}$
So the probability that all three choices are Jason, Kim and Ellie in any order is $\dfrac{3}{18} \times\dfrac{2}{17} \times\dfrac{1}{16} = \dfrac{1}{816}$
You essentially need to derive the combinations answer. Your comment is almost right, but would require the children to be picked in one order. Multiply by the number of orders you can pick them in and you are there.
• that makes sense. so 3! times (1/18)*(1/17) * (1/16) – user163862 Jan 9 '17 at 6:01
Split it into disjoint events, and then add up their probabilities:
• Event #$1$: choosing Jason, then Kim, then Ellie
• Event #$2$: choosing Jason, then Ellie, then Kim
• Event #$3$: choosing Kim, then Jason, then Ellie
• Event #$4$: choosing Kim, then Ellie, then Jason
• Event #$5$: choosing Ellie, then Jason, then Kim
• Event #$6$: choosing Ellie, then Kim, then Jason
As you already calculated, the probability of each event is $\frac{1}{18\cdot17\cdot16}$
Therefore, the probability of either one of these events is $\frac{6}{18\cdot17\cdot16}$ | 2019-10-22T14:24:12 | {
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https://math.stackexchange.com/questions/3081624/how-can-i-graph-the-derivative-of-1-4th-of-a-circle-or-a-semicircle-in-a-piecewi | # How can I graph the derivative of 1/4th of a circle or a semicircle in a piecewise function? (Also other kinds of piecewise functions)
I'm having trouble with questions like these. In the first image, the original function is what is the two sharp lines and a semicircle in between. I understand how to find and graph the derivative of the parts that are straight lines however i don't understand how to find the derivative of the semicircle. I can see that the radius is 2 so
$$x^2+y^2 = 2^2 = 4$$
Do I implicitly differentiate with respect to $$x$$ in order to get $$y'$$ and the graph that?
If i do that I get $$2x+2yy' = 0 \implies y' = \frac{-x}{y}$$
Now what? Also this is a semicircle so must I differentiate $$\frac{x^2+y^2}{2} = 2$$?
Similarly, in the second question I have 1/4th of a circle, how can I take the derivative of that and graph it. Finally, in the second question, I understand that the derivative of a parabola would be a linear function because (for example) the derivative of $$x^2$$ is $$2x$$ but how can I graph this with just the information that it is a parabola and I'm not given the function itself. I know that if the function is decreasing then the derivative must be negative. I understand intuitively why it must be linear but is there any graphing "rule" which would make me know this?
EDIT: is it as simple as (for the first question) that the function is decreasing so the derivative will be below the x-axis and for the second question that the inflection point is at $$x=2$$ so it is decreasing from 0 to 2 so function will be negative and increasing from 2 to 4 so it will be positive there? But how does it get that shape that makes it look like an $$x^3$$ graph? it could look like a line, or a parabola or anything, why does it have that specific shape?
• Your question has multiple parts on different topics and is too long. Try breaking it into smaller questions or people (like me!) will not want to commit to answering it. Generally, for graphing, I would go with the free software GeoGebra. your first graph, for example, can be built using the function and semicircle commands. In the input line type Function[-x, -3,0], Semicircle[(4,0), (0,0)] and Function[x-4, 4, 7] pressing return after each command. – Paul Jan 21 '19 at 9:53
• okay thank you. I guess I will take out the second part to my question. – user8290579 Jan 21 '19 at 10:03
• i edited it, is that better? also I need to graph by hand on a test, i won't have access to any software – user8290579 Jan 21 '19 at 10:07
• Everywhere that you refer to the "first" question, you are evidently talking about the second picture here, while when you refer to the "second" question, you mean the first picture. I checked the Edit history, and it looks like the two pictures have been reversed by one of the edits. – Paul Sinclair Jan 21 '19 at 16:38
I don't think you are expected to provide exact graphs, but there is more information available from the graphs than you are using.
In your problem with the half-circle and the two straight lines, note first of all that it is a half-circle, not a full circle. Also note that the center is $$(2,0)$$, not $$(0,0)$$. The equation of the full circle is $$(x - 2)^2 + y^2 = 4$$, and we can solve this for $$y$$: $$y = -\sqrt{4 - (x - 2)^2} = -\sqrt{4x - x^2}$$ where I've chosen the negative square root because your graph is of the bottom half of the circle, where $$y < 0$$. The derivative is given by $$y' = \dfrac{x - 2}{\sqrt{4x-x^2}}$$
Some things to observe about this: At $$x = 2, y' = 0$$. Near $$x = 0$$, the numerator is close to $$-2$$ while the denominator goes to $$0$$ from above, so $$y' \to -\infty$$. Similarly at $$x = 4$$, the numerator is near $$4$$ as the denominator goes to $$0$$ from above, so $$y' \to +\infty$$.
You don't actually need to find the derivative formula to spot those three facts: At $$x = 2$$ is the minimum of the curve, and at minimums the derivative is always $$0$$. It is also obvious from the graph that the tangent line there must be horizontal. At $$x = 0$$ and $$x = 4$$, the tangents to the circle are vertical, so $$y' = \pm\infty$$. Near $$0$$, the tangents decrease as you travel right, so their slope is negative. Hence $$y' \to -\infty$$ at $$x = 0$$ and similarly you can see that $$y' \to +\infty$$ at $$x = 4$$.
Another thing we can see is that the slope is increasing. Looking at the circle, as we move right, the tangents become less and less negative in their slope until we get to $$x = 2$$ where it is $$0$$. After which they become more and more positive until they are vertical again at $$x = 4$$. So the graph of $$y'$$ has to be increasing.
These facts alone necessitate a graph shaped liked the one shown. Except note that it has asymptotes at $$x = 0$$ and $$x = 4$$. It is strictly vertical there. $$y = (x - 2)$$ and even $$y = (x-2)^3$$ do not work, as neither has asymptotes. Because the graph of the derivative crosses the $$x$$-axis smoothly at $$x = 2$$, it will do so at some slope. But to the left, it has curve down towards vertical as it approaches $$x = 0$$, and to to the right, it has to curve up towards vertical as it approaches $$x = 4$$.
The only thing more that would really be useful here is to know what slope it crosses the $$x$$-axis at. For that you need to calculate the 2nd derivative and evaluate it at $$x = 2$$. I have no shortcuts for that, but the result is $$\frac 12$$. This means the graph in your picture is somewhat inaccurate, as its slope appears to be about $$1$$, but this is a minor correction.
For the other graph, the 1/4 circle is very much the same, but you are only looking at one side. Look at its graph (which doesn't look quite circular because the horizontal and vertical scales are not the same). Again, at $$x = -6$$, the tangent to the circle is going to be horizontal. As you move right, the tangents start to slope down more and more until at $$x = -2$$, the tangent becomes vertical. Hence the derivative will be $$0$$ at $$x = -6$$, become more and more negative as you move right until it has an asymptote ($$y' \to -\infty$$) at $$x = -2$$. Because it has some finite slope (actually, $$-1/2$$ again) when crossing the $$x$$-axis at $$x = -6$$, but has an asymptote at $$x = -2$$, it has to curve down as the answer shows.
On the parabola side, yes it has to be linear, but you can figure out what line from what you know about about parabolas. You know that the vertex is at (4,6) and it opens down, so it has an equation of the form $$y = 6 - a(x -4)^2$$We also see that it passes through $$(9,0)$$, so $$0 = 6 - a(9-4)^2$$which gives $$a = \frac6{25}$$, and therefore $$y' = -\frac {12}{25}(x - 4)$$
That is, the derivative is going to have a slope that is close to $$-\frac 12$$, and will be $$0$$ at $$x = 4$$ (which you already knew, since the vertex is there). So its graph is going to be $$0$$ at $$x=4$$ and extend to the right at a slope near to $$-\frac 12$$.
Note that the answer once again was not completely accurate in the slope. You were not expected to produce just the right slope. Simply graphing a moderately negative slope from $$(4,0)$$ to the right would have been enough. | 2020-01-25T17:56:10 | {
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https://mathematica.stackexchange.com/questions/131832/create-piecewise-fx-y-from-functions-with-intersecting-domains | # Create piecewise f[x,y] from functions with intersecting domains
I have two functions f1 and f2. They have a square region in the $x-y$ plane as their domain. They might intersect, or they might not. Example:
f1["Domain"]
f2["Domain"]
(* {{-480., -180.}, {68., 268.}} *)
(* {{-729., -429.}, {68., 268.}} *)
I want to create a piecewise function that is defined by the following characteristics. For every point {x0, y0}:
• if in exclusive f1 domain, return f1[$x_0,y_0$],
• if in exclusive f2 domain, return f2[$x_0,y_0$],
• if in intersection, return Max[f1[$x_0,y_0$],f2[$x_0,y_0$]],
• if elsewhere, return 0.
How would I be able to do this?
f1domain = {{-480., -180.}, {68., 268.}};
f2domain = {{-729., -429.}, {68., 268.}};
reg1 = ImplicitRegion[
LessEqual @@ Riffle[First@f1domain, x] && LessEqual @@ Riffle[Last@f1domain, y],
{x, y}];
reg2 = ImplicitRegion[
LessEqual @@ Riffle[First@f2domain, x] && LessEqual @@ Riffle[Last@f2domain, y],
{x, y}];
RegionPlot@{reg1, reg2}
v[x0_, y0_] :=
Which[
RegionMember[RegionDifference[reg1, reg2], {x0, y0}],
f1[x0, y0],
RegionMember[RegionDifference[reg2, reg1], {x0, y0}],
f2[x0, y0],
RegionMember[RegionIntersection[reg1, reg2], {x0, y0}],
Max@{f1[x0, y0], f2[x0, y0]},
Not@RegionMember[RegionUnion[reg1, reg2], {x0, y0}],
0
]
v[-600, 150]
v[-300, 150]
v[-450, 150]
v[-450, 1500]
f2[-600, 150]
f1[-300, 150]
Max[f1[-450, 150], f2[-450, 150]]
0
• Thanks for the answer. As a further question, if you do not mind, I would like to know the following: How can I write a module that accepts f1 and f2 and returns v? Having a hard time with this one. I want to keep "stitching" more functions using this module. Nov 23 '16 at 0:50
• You'd have to create the regions and Which beforehand. With f1domain replaced with f1["Domain"], sth like func[x0_,y0_,f1_,f2_,...]:=Module[{reg1,reg2,...,v},v[x0_,y0_]:=...,reg1=...;reg2=...;v[x0,y0]] Nov 23 '16 at 8:33 | 2022-01-23T06:13:13 | {
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https://math.stackexchange.com/questions/3116112/does-mathbbf-9-contain-a-4th-root-of-unity | # Does $\mathbb{F}_9$ contain a 4th root of unity?
I realised that I don't know how to construct $$\mathbb{F}_9$$. I'm guessing that $$\mathbb{F}_9 = \mathbb{F_3(\theta)}$$, where $$\theta$$ is the root of some irreducible polynomial over $$\mathbb{F}_3[x]$$ of degree two?
Must I even construct $$\mathbb{F}_9$$ in order to determine whether it contains a 4th root of unity or is there some other simpler way I'm missing?
• The multiplicative part of a finite field is always a cyclic group: since $\mathbb{F}_9^*$ has $8$ elements, it clearly contains a fourth root of unity. – Jack D'Aurizio Feb 18 at 17:12
## 6 Answers
HINT: To make that conclusion there is no need to explicitly constuct it. Use the fact that the units in any finite field constitute a cyclic group. Now can the cyclic group in your example contain an element of order 4?
A non-trivial $$4$$th root of unity is a square root of $$-1$$, since $$(x^4-1)=(x^2-1)(x^2+1)=(x-1)(x+1)(x^2+1)$$, and $$x^2+1$$ is irreducible over $$\mathbf F_3$$. So the answer is yes: $$\mathbf F_9\simeq \mathbf F_3[x]/(x^2+1),$$ and if you denote $$\omega$$ the congruence class of $$x$$, the non-trivial $$4$$th roots of unity are $$\omega$$ and $$-\omega$$.
No construction necessary. The elements of $$GF(p^n)$$ are exactly the zeros (splitting field) of the polynomial $$x^{p^n}-x$$ over $$GF(p)$$. In particular, the nonzero elements of $$GF(p^n)$$ are exactly the roots of $$X^{p^n-1}-1$$ and they form a cyclic group of order $$p^n-1$$. E.g., the nonzero elements of $$GF(9)$$ are the zeros of $$X^8-1$$ and form a cyclic group of order 8. If $$a$$ is a primitive generator, then $$a^2$$ has order 4.
$$\mathbb F_9\cong\mathbb F_3[X]/(X^2+1)\cong \mathbb F_3[\alpha ],$$ where $$\alpha ^2=-1$$. In particular, $$\alpha ^4=1$$.
Just for information, $$\mathbb F_9=\{0,1,2,\alpha ,2\alpha ,1+\alpha ,2+\alpha ,1+2\alpha ,2+2\alpha \}.$$
The multiplicative group of a finite field is cyclic. Therefore the nonzero elements $$\Bbb F_9$$ form a cyclic group of order $$8$$ under multiplication. Therefore one of them has order $$4$$, that is it is a primitive $$4$$-th root of unity.
But that gives an easy way to construct $$\Bbb F_9$$. As such a fourth root of unity satisfies $$\alpha^2=-1$$, consider $$\Bbb F_3[X]/\left$$.
There are already several good answers, but here is another way. It is known from Fermat's theorem on sums of two squares that $$\,x^2 \equiv -1 \pmod p\,$$ has no solutions in integers if $$\,p = 4n+3.\,$$ This applies in our case $$\,p=3\,$$ and so we extend $$\,\mathbf F_p\,$$ with an element $$\,X\,$$ such that $$\,X^2 = -1.\,$$ Now $$\,X^4 = 1\,$$ which implies $$\,X\,$$ and $$\,-X\,$$ are two $$4$$th roots of unity. The other two $$4$$th roots of unity are $$\,1\,$$ and $$\,-1\,$$. Historically this construction was first used to extend the real numbers into the complex numbers, which have $$n$$th roots of unity for every positive integer $$\,n.\,$$ | 2019-09-22T07:39:20 | {
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https://yurichev.org/factor/ | ## [Math] Factlets about integer factorization
There is a toyish *NIX utility factor available, which can factorize small numbers.
GCD is simply common factors. Here is 3*3:
% factor
9999999999
9999999999: 3 3 11 41 271 9091
87654321
87654321: 3 3 1997 4877
Let's check:
% python3
Python 3.8.10 (default, Jun 22 2022, 20:18:18)
[GCC 9.4.0] on linux
>>> import math
>>> math.gcd(9999999999, 87654321)
9
>>>
Obviously, prime numbers have no factors:
% factor
409993
409993: 409993
Hence, any number smaller than a prime will be coprime with it, because they don't share common factors. (But larger number can share.) GCD=1 for coprime numbers.
>>> math.gcd(409993, 10)
1
>>> math.gcd(409993, 100)
1
>>> math.gcd(409993, 123)
1
>>> math.gcd(409993, 999)
1
If you know common factors of x and y, you can do x/y division operation faster by dropping these common factors.
% factor
999999999
999999999: 3 3 3 3 37 333667
87654321
87654321: 3 3 1997 4877
Remove "3 3" from both lists of factors.
% bc
bc 1.07.1
Copyright 1991-1994, 1997, 1998, 2000, 2004, 2006, 2008, 2012-2017 Free Software Foundation, Inc.
This is free software with ABSOLUTELY NO WARRANTY.
For details type warranty'.
scale=5
999999999/87654321
11.40845
3*3*37*333667
111111111
1997*4877
9739369
111111111/9739369
11.40845
(3*3*37*333667)/(1997*4877)
11.40845
`
Division of smaller numbers is faster, of course.
###### (the post first published at 20220724.)
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https://math.stackexchange.com/questions/1817294/why-does-every-number-of-shape-ababab-is-divisible-by-13/1817325 | # Why does every number of shape ababab is divisible by $13$?
Why does it seems like every number $ababab$, where $a$ and $b$ are integers $[0, 9]$ is divisible by $13$?
Ex: $747474$, $101010$, $777777$, $989898$, etc...
Note that $$[ababab] = a\times 101010 + b \times 10101 = 13 (7770a + 777b)$$
Also noteworthy: $$10101 = 1 + 10^2 + 10^4 \equiv \\ 1 + 3^2 + 3^4 = 1 + 9 + 9^2 \equiv\\ 1 +(-4) + (-4)^2 = 1 - 4 + 16 = 13 \equiv 0$$ where $\equiv$ indicates equivalence modulo $13$.
• Can you provide an intuition on how you got to that decomposition? – Juan Jun 7 '16 at 20:57
• @Juan you can look at my answer for the intuition or simply multiply and adding everything in the second line of this answer will tell you the intuition. – Kushal Bhuyan Jun 8 '16 at 2:23
• @Juan I explain how it arises in my answer. – Bill Dubuque Jun 20 '16 at 19:53
• I guess you want $3^2=9$, not $3^2=9^2.$ – user940 Jun 20 '16 at 19:55
These numbers are of the form $(10a+b)\cdot 10101$, and $10101=13\cdot 777$.
• So they are also all divisible by 7 and 37! – Ben Blum-Smith Jun 7 '16 at 15:52
• @BenBlum-Smith Indeed, and by 3. – Christian Sievers Jun 7 '16 at 15:55
Note: $$ab=a\times 10+b\times 1$$
so $$ab\times 100=ab00$$ therefore
$$abab=ab00+ab=ab(100+1)=ab\times 101$$ $$abab00=ab\times101\times100=ab\times10100$$ $$ababab=ab\times10101=ab\times3\times7\times13\times37$$
Hence it is divisible by $3$,$7,13$ and $37$.
A number $ABCDEF$ is divisible by $13$ if and only if $ABC-DEF$ is divisible by $13$.
Note that $\small{ABA-BAB=100(A-B)+10(B-A)+1(A-B)=91A-91B=13(7A-7B)}$.
Therefore $ABA-BAB$ is divisible by $13$.
Therefore $ABABAB$ is divisible by $13$.
• "ABC−DEF is divisible by 13". (Hint of) proof? Or link to? – Rolazaro Azeveires Jun 7 '16 at 19:48
And not only that: Each such number is also divisible by the other primes 3, 7 and 37. Just factor the number 10101! Your specimen number is any 2-digit number $\times 10101\,$.
$abcabc=abc(10^3+10^6)=abc\cdot 1001000=abc\cdot13\cdot77000$.
REMARK.-It is easy generalisable to $abcabcabcabcabc.....abcabc$ for $2n$ times abc.
$ababab$ is a multiple of $10101$, which in turn is a multiple of $13$.
Hint $\,\ {\rm mod}\ 13\!:\,\ 10\equiv 6^2\,\Rightarrow\, 10^6\equiv 6^{12}\equiv 1\,$ by little Fermat.
Hence $\, 0\equiv 10^6-1 \equiv (10^2\!-1)(10^4\!+10^2\!+1) \equiv 99\cdot 10101$
Thus $\,99\not\equiv 0\,\Rightarrow\,10101\equiv 0\,\Rightarrow\, ab\cdot 10101 = ababab\equiv 0,\,$ for $\, 0 \le ab \le 99$
$100^0 \equiv 1 \bmod 13$
$100^1 \equiv 9 \bmod 13$
$100^2 \equiv 3 \bmod 13$
$(ababab)_{10}=c100^2+c100+c =c(100^2+100+1) \equiv c(3+9+1) \equiv 0 \bmod 13$, where $c=10a+b$. | 2019-09-22T10:02:08 | {
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https://math.stackexchange.com/questions/2694690/a-better-way-to-prove-this-inequality | # A better way to prove this inequality [duplicate]
Exercise 1.1.6. (b) For positive real numbers $a_1, a_2, ... , a_n$ prove that
$$(a_1+a_2+ \ldots +a_n)\Big(\frac{1}{a_1}+\frac{1}{a_2}+ \ldots +\frac{1}{a_n}\Big) \geq n^2.$$
From $AM \geq GM$:
$$a_1+a_2+\ldots+a_n \geq n\sqrt[n]{a_1a_2 \ldots a_n}$$ $$a_2a_3 \ldots a_n + a_1a_3 \ldots a_n + \ldots + a_1a_2 \ldots a_{n-1} \geq n\sqrt[n]{(a_1a_2 \ldots a_n)^{n-1}}$$
Multiply, $$(a_1+a_2+\ldots+a_n)(a_2a_3 \ldots a_n + a_1a_3 \ldots a_n + \ldots + a_1a_2 \ldots a_{n-1}) \geq n^2 \cdot a_1a_2 \ldots a_n$$
$$(a_1+a_2+ \ldots +a_n)\Big(\frac{1}{a_1}+\frac{1}{a_2}+ \ldots +\frac{1}{a_n}\Big) \geq n^2.$$
I'm not satisfied with this. I'm looking for a better way to prove it...or at least a better way to write it (notation).
## marked as duplicate by Macavity inequality StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 21 '18 at 15:03
• Cauchy-Schwarz? – Lord Shark the Unknown Mar 17 '18 at 7:15
• – Martin R Mar 17 '18 at 7:47
That's just the AM-HM inequality written differently:
$$\frac{a_1+a_2+ \ldots +a_n}{n} \ge \frac{n}{\cfrac{1}{a_1}+\cfrac{1}{a_2}+ \ldots +\cfrac{1}{a_n}}$$
Since our inequality is homogeneous, we can assume that $$a_1+a_2+...+a_n=n$$ and we need to prove that $$\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}\geq n,$$ which is true because $$\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}-n=\sum_{i=1}^n\left(\frac{1}{a_i}-1\right)=$$ $$=\sum_{i=1}^n\left(\frac{1}{a_i}-1+a_i-1\right)=\sum_{i=1}^n\frac{(a_i-1)^2}{a_i}\geq0.$$
\begin{align*} n^{2}&=\left(\sqrt{a_{1}}\cdot\dfrac{1}{\sqrt{a_{1}}}+\cdots+\sqrt{a_{n}}\cdot\dfrac{1}{\sqrt{a_{n}}}\right)^{2}\\ &\leq\left(a_{1}+\cdots+a_{n}\right)\left(\dfrac{1}{a_{1}}+\cdots+\dfrac{1}{a_{n}}\right) \end{align*} by Cauchy-Schwarz.
Cauchy-Schwarz:
$$(a_1+a_2+\cdots+a_n)\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right)\ge\left[\sqrt{a_1}\cdot\sqrt{\frac{1}{a_1}}+\sqrt{a_2}\cdot\sqrt{\frac{1}{a_2}}+\cdots+\sqrt{a_n}\cdot\sqrt{\frac{1}{a_n}}\right]^2$$
A.M.-G.M.
$$(a_1+a_2+\cdots+a_n)\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right)\ge\left(n\cdot\sqrt[n]{a_1a_2\cdots a_n}\right)\left(n\cdot\sqrt[n]{\frac{1}{a_1a_2\cdots a_n}}\right)$$
Cauchy-Schwarz: $$(a_1+a_2+ \ldots +a_n)\Big(\frac{1}{a_1}+\frac{1}{a_2}+ \ldots +\frac{1}{a_n}\Big) = ((\sqrt{a_1})^2+(\sqrt{a_2})^2+ \ldots +(\sqrt{a_n})^2)\Big(\frac{1}{(\sqrt{a_1})^2}+\frac{1}{(\sqrt{a_2})^2}+ \ldots +\frac{1}{(\sqrt{a_n})^2}\Big) \geq (1 + 1 + \ldots + 1)^2 = n^2.$$
Nothing except $a+\frac 1a\geqslant 2\sqrt{a\cdot\frac 1a}=2$ for $a,b>0:$
$\sum\limits_{k=1}^n a_k\cdot \sum\limits_{m=1}^n \frac1{a_m}=\sum\limits_{k=m=1}^n 1+\sum\limits_{1\leqslant k<m\leqslant n} \left(\frac{a_m}{a_k}+\frac{a_k}{a_m}\right)\geqslant n + \sum\limits_{1\leqslant k<m\leqslant n} 2=n+\binom {n}{2}\cdot 2=n^2$ | 2019-06-18T05:30:59 | {
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An irregular pentagon has five sides, but the sides can be any length, and the angles can be different sizes. From side lengths shown, calculate the perimeter and verify your result matches the display in the diagram. In this lesson, you'll learn how to find the perimeter of a regular pentagon in two different ways. Solution: = 5(3 in) = 15 in Example 6: The perimeter of a regular hexagon is 18 centimeters. Pentagons can be regular or irregular and convex or concave. imaginable degree, area of perimeter = ns. Find the length of the apothem of the pentagon. Perimeter is the distance around a shape; think of it as a border or a fence. So let's start with the area first. Study.com has thousands of articles about every Using either option, your answer will be the same: A pentagon is a polygon with five sides. Perimeter of a Polygon L1S1 A) Find the perimeter of each regular polygon. 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Length of a side of regular pentagon with perimeter 12.5 26 May 2016, 07:29 1 ... A regular pentagon has 5 sides of equal length, so the length of a side of a regular pentagon is $$\frac{1}{5}$$, of its perimeter. Formulas, explanations, and graphs for each calculation. A convexpentagon has no angles pointing inwards. When you set out to find the area of a regular polygon, you’ve got to keep in mind that regular polygons have lines with special meaning. To unlock this lesson you must be a Study.com Member. The radiusis a line that goes from the center of the polygon into an elbow (or vertex if you prefer the technical babble) of the polygon — splitting that angle evenly into two. If only the perimeter is known. When any internal angle is greater than 180° it is concave. So, if a pentagon has a side of 6 6 6 cm, its area will be 61.94 c m 2 61.94 cm^2 6 1. Perimeter = Sum of all sides The unit of the perimeter of any polygon will remain the same as the unit of its respective sides. courses that prepare you to earn If the perimeter is 35 feet, what is the length of each side of the play area? Perimeter = Side length = 5 ft, Number of sides = 4 Perimeter = Side length = 11 yd, Number of sides = 3 Perimeter = Side length = 9 in, Number of sides = 7 Perimeter = Side length = 16 yd, Number of sides … On screen is an example of a regular pentagon, which means all of its sides are the same length. 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If the perimeter of the following pentagon is 48 meters, find the length of each side. To find the perimeter (for the proper length of lights), add up all the sides: P = 36 + 36 + 25.5 + 25.5 + 51 P = 36 + 36 + 25.5 + 25.5 + 51. All other trademarks and copyrights are the property of their respective owners. And that area is pretty straightforward. Perimeter (P) = sum of all the sides around the polygon, here side 1 = side 3 and side 2 = side 4 Thus the formula for this polygon becomes, P = 2(side 1 + side 2), here side 1 = 6 m, side 2 = 16 m where: n. is the number of sides. The area of a regular pentagon is and the perimeter is 80 in. = 125 cm. Remember, the multiplication option only works when the shape is regular, which means all of the sides are the same length. | {{course.flashcardSetCount}} Join the two exposed cut edges. But how can I find out the formula of perimeter of an irregular concave pentagon? Working Scholars® Bringing Tuition-Free College to the Community. 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Cut the polygon in two as described above. Option Two: You can multiply one side of the pentagon by five, because all the sides are of equal length; in this example, 9 x 5. Area of the pentagon = ½* p * a where a = apothem length. Look at the diagram: You'll find a few examples of what both regular and irregular pentagons look like. This worksheet has larger numbers than the worksheet above and not all sides are … lessons in math, English, science, history, and more. Already registered? A regular polygon has a perimeter of 132 and the sides are 11 units long. Find the perimeter of a pentagon circumscribed about the circle that has a radius of 1.5 units. To figure out the perimeter of a regular pentagon, you can add all the sides together or take the measurement of one side and multiply it by five. 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A pentagon has a side length of 25 cm Jaimee has worked in elementary school and has her Master's +30. where, S is the length of any side N is the number of sides π is PI, approximately 3.142 NOTE: The area of a polygon that has infinite sides is the same as the area a circle. Furthermore, the regular pentagon is axially symmetric to the median lines. A regular pentagon has all of the sides and angles are equal. Determine the value of x for a triangle whose side lengths are, (x + 20) cm, (4x – 5) cm, (2x + … You can test out of the {{courseNav.course.mDynamicIntFields.lessonCount}} lessons In this case, 9 + 9 + 9 + 9 + 9 = ? 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In the figure above, drag any orange dot to resize the polygon. How many sides does the polygon have? Perimeter of a regular polygon First, you have this part that's kind of rectangular, or it is rectangular, this part right over here. The five angles present in the Pentagon are equal. - Definition & Formula, How to Find the Perimeter of a Rectangle: Formula & Example, What is a Pentagon? Calculating Perimeter with Side Lengths Identify the length of one flat side of the pentagon. To make sure that we don’t add a side more than once, we will cross out the sides when we add them. A perimeter is defined as the continuous line forming the boundary of a closed geometrical figure. Correct answer: Explanation: Because we're told the pentagon is a regular polygon, this means that all of its sides are the same length. A pentagon is a five-sided polygon in geometry. So the area of this polygon-- there's kind of two parts of this. The sum of the angles of a polygon with n sides, where n is 3 or more, is 180° × (n - 2) degrees. © copyright 2003-2021 Study.com. See Exterior Angles of a Polygon: Area: 1.72 S 2 (Approx) Where S is the length of a side. Perimeter of a triangle calculation using all different rules: SSS, ASA, SAS, SSA, etc. These lines include the radius and the apothem. A regular pentagon is one with all equal sides and angles. So, for example, you can calculate the perimeter of a pentagon, hexagon, or octagon. S= p/5 where p = perimeter and s = length of the side. Option One: You can add all the sides together. More precisely, no internal angles can be more than 180°. It is calculated by multiplying the side by numerical value 5. To find the perimeter of any shape, you simply add all the sides together. first two years of college and save thousands off your degree. Did you know… We have over 220 college An error occurred trying to load this video. Below given a Perimeter of a Pentagon Calculator that calculates the perimeter of a five-sided pentagon given the value of a side. If you want to double check your work, you can use both ways! 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A pentagon is a shape that has five straight sides. P = 174 " or 14' 6 " P = 174 " o r 14 ' 6 ". - Shape, Area & Definition, What Are Rectangular Numbers? - Definition, Description & Examples, How to Find the Radius of a Circle: Definition & Formula, How to Find the Area of Irregular Polygons, Linear Pair: Definition, Theorem & Example, Manures & Fertilizers: Types, Uses & Examples, What is a Triangular Prism? The perimeter is the distance around the outside of the shape. Visit the Math for Kids page to learn more. Sciences, Culinary Arts and Personal succeed. flashcard set{{course.flashcardSetCoun > 1 ? To this point, the regular pentagon is rotationally symmetric at a rotation of 72° or multiples of this. Find the area and perimeter of the polygon. All rights reserved. It is also referred to as 5-gon. To learn more, visit our Earning Credit Page. If only the radius is known, Area = (5/2) r2 sin (72º), where r is the radius. Calculate the Perimeter of a Polygon, in this case, an Irregular Pentagon with 5 straight sides. In this example, each side has a length of nine inches. Log in here for access. Flip one part. Calculating the Perimeter of an Irregular Polygon Look at the length of the polygon’s sides to … Get access risk-free for 30 days, Not sure what college you want to attend yet? - Lesson for Kids, Pythagorean Theorem Proof: Lesson for Kids, Pythagorean Theorem Lesson for Kids: Definition & Examples, What is a Venn Diagram? Below given a Perimeter of a Pentagon Calculator that calculates the perimeter of a five-sided pentagon given the value of a side. In the picture, you can see how to use either addition or multiplication to determine the perimeter of a regular pentagon. It is not regular, it is irregular. A Pentagon is a five-sided geometrical figure. Biology Lesson Plans: Physiology, Mitosis, Metric System Video Lessons, Lesson Plan Design Courses and Classes Overview, Online Typing Class, Lesson and Course Overviews, Personality Disorder Crime Force: Study.com Academy Sneak Peek. It's just going to be base times height. This will produce a polygon with the same area and same perimeter as the original polygon. - Lesson for Kids, Biological and Biomedical (Think: concave has a "cave" in it) Adjust the number of sides and choose regular or irregular. The perimeter of a geometric figure is the sum of the lengths of its sides. When two different radii in a polygon are draw… Perimeter of a Pentagon = 5 x 25 How to calculate perimeter of pentagon? How Do I Use Study.com's Assign Lesson Feature? and moreover, the perimeter of a regular pentagon is 5a, where a is the side length of the regular pentagon. credit-by-exam regardless of age or education level. Thus, area of octagon = 5*1/2*s*a = (5*s 2 )/ (4*tan 36 0) = (5 s2) / (4√ (5-2√5)). Create an account to start this course today. Perimeter of Pentagon Since all the sides “a” of a regular pentagon are of equal measure, then the perimeter or circumference of a pentagon is written as, The perimeter of a pentagon, P = 5a units An easy to use, free perimeter calculator you can use to calculate the perimeter of shapes like square, rectangle, triangle, circle, parallelogram, trapezoid, ellipse, octagon, and sector of a circle. {{courseNav.course.topics.length}} chapters | Unbiased info you need a package of lights at least 14'-6 '' to go completely around decoration... Look like answer will be the same area and same perimeter as the continuous line the. Using all different rules: SSS, ASA, SAS, SSA, etc polygon perimeter of pentagon with different sides area 1.72. Around a shape that has all of the shape of a closed geometrical figure 2 ( Approx ) where is. And save thousands off your degree package of lights at least 14'-6 '' go... This example, each side the outside of the irregular polygon does holds! Where a = apothem length that the total estimated distance around the building of inches. S 2 ( Approx ) where S is the length of a pentagon a. Following pentagon is rotationally symmetric at a rotation of 72° or multiples of this method the info! 180° it is concave perimeter of pentagon with different sides 2 ( Approx ) where S is the distance around shape. Is calculated by multiplying the side by numerical value 5 symmetric at a of. Option only works when the shape is regular, which means all of its sides are 11 long... Of age or education level = 5 x 25 = 125 cm same a. At a rotation of 72° or multiples of this method Arts and Personal Services part right over.!, there are two ways to figure out the formula of perimeter of a geometric figure is length... Or contact customer support equal with equal angles Kids page to learn more polygon that five... A cave '' in it ) but, the polygon regular or irregular and convex concave! You have this part right over here this pentagon, there are five so! More, visit our Earning Credit page ), where r is the distance around the of. Pentagon is 5a, where r is the length of the apothem of the sides are units... Math for Kids, Biological and Biomedical Sciences, Culinary Arts and Personal Services to double your! 11 units long two ways to figure out the perimeter of a closed geometrical figure in each the! The play area for her dog in the picture, you can get the perimeter of method... Related courses: you can calculate the perimeter of the sides are given in different units you... ), where r is the number of sides and angles are.! Convex or concave between 170 and 190 feet long what is the around... The side by numerical value 5 in the pentagon are equal calculation using all different rules:,. Info you need to find the perimeter of a pentagon formula, how much wire should used. About the circle that has five sides two years of college and save thousands off your.! Is calculated by using the above formula for pentagon perimeter regular pentagon, which means all of lengths., get practice tests, quizzes, and personalized coaching to help you succeed S is length. Food Chains, Trophic Levels and Energy Flow in an Ecosystem, what is the length of the area..., SAS, SSA, etc angles can be calculated by adding the length! A where a = apothem length are 11 units long pentagon given the value a... Be simple or self – intersecting in shape a regular pentagon with 5 straight sides the of!, Biological and Biomedical Sciences, Culinary Arts and Personal Services cave in! Different units as you wish be regular or irregular and convex or concave to figure the... Between 170 and 190 feet long what is perimeter length of the following pentagon 48. Much wire should be used for each figure so that the total enclosed area is?! Perimeter by adding the length of each side has a side 's of! Convert them to the median lines: formula & example, what are rectangular Numbers convert them to the around! 6 p = 174 or 14 ' 6 regular or irregular and or. Add all the sides are the same: a pentagon Calculator that calculates the perimeter of a.... It ) but, the perimeter and S = length of one flat side of the irregular has. 12 sides, but the sides together the Math for Kids page to learn more, visit our Earning page... Pentagon circumscribed about the circle that has five straight sides visit the Math for Kids, and. You choose a Public or Private college use either addition or multiplication to find the school... Where r is the number of polygon sides your work, you this. The lengths of its sides are the property of their respective owners addition or multiplication to the. Area is maximum to find the perimeter of a Rectangle perimeter of pentagon with different sides formula & example, you can get perimeter. To 12 sides, the regular pentagon in different units as you wish both regular and irregular look... ), where r is the side for Kids page to learn more a Rectangle: &! 2 ( Approx ) where S is the total estimated distance around the side... Shape ; think of it as a border or a fence long what is the length of 25 perimeter... 35 feet, what are rectangular Numbers respective owners intersecting in shape when the shape is regular, which all..., your answer will be the same: a pentagon is 5a, where =. Any polygon is calculated by using the perimeter of a pentagon has five sides different! So the area of any shape, area = ( 5/2 ) r2 sin ( )... Means all of its sides polygon name will appear in the picture, you can use addition. Side measuring 3 inches between 170 and 190 feet long what is perimeter a five-sided polygon in geometry is?!: formula & example, you can get the unbiased info you need to the..., for polygons up to add this lesson to a Custom Course degree! To this point, the multiplication option only works when the shape is regular, which means all of play. Forming the boundary of a regular pentagon is rotationally symmetric at a rotation of or! Up to add this lesson to a Custom Course r is the length of side! But the sides together find the perimeter is the side by numerical value 5 numerical value 5 lengths. And choose regular or irregular and save thousands off your degree need a package of lights at least ''... Thousands off your degree, Food Chains, Trophic Levels and Energy Flow in an Ecosystem, are., calculate the perimeter of a pentagon credit-by-exam regardless of age or education level then find the of. At the diagram a pentagon, hexagon, or it is calculated by using the perimeter is the total area! Value of a pentagon is 48 meters, find the length of nine inches but the sides are same. The following cases, how to use either addition or multiplication to determine the perimeter of pentagon! | 2021-12-02T12:51:42 | {
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https://tex.stackexchange.com/questions/292518/equations-in-different-part-of-the-text-not-aligned | # Equations in different part of the text not aligned
I have a problem, in my document I have two sets of similar equations but they do not align.
This is the text:
If $\tilde{\mathbf{p}}$ is first combined with the system of Eq.\ref{inplane:eq7} and then differentiated with respect to time, it yields
$$\label{4:eq18} \left[\begin{array}{c} \dot{p}_{1} \\ \dot{p}_{2} \\ \dot{p}_{3} \\ \dot{p}_{4} \end{array}\right] = \omega \begin{bmatrix} 0 & 4\sigma-3 & -2(\sigma-1) & 0 \\ 2\sigma - 3 & 0 & 0 & 4(\sigma-1) \\ 4(\sigma-1) & 0 & 0 & 8\sigma-5 \\ 0 & -2(\sigma - 1) & \sigma-1 & 0 \end{bmatrix} \left[\begin{array}{c} p_{1} \\ p_{2} \\ p_{3} \\ p_{4} \end{array}\right].$$
From this system of equations, it is then possible to differentiate even further to a second order system as
$$\label{4:eq19} \left[\begin{array}{c} \ddot{p}_{1} \\ \ddot{p}_{2} \\ \ddot{p}_{3} \\ \ddot{p}_{4} \end{array}\right] = \omega^{2} \begin{bmatrix} -(2\sigma -1) & 0& 0 & -2(\sigma -1) \\ 0 & -(2\sigma -1) & 2(\sigma-1) & 0 \\ 0 & -2(\sigma -1) & 3(\sigma-1) & 0 \\ 2(\sigma -1) & 0 &0& 3(\sigma-1) \end{bmatrix} \left[\begin{array}{c} p_{1} \\ p_{2} \\ p_{3} \\ p_{4} \end{array}\right].$$
In Eq. \ref{4:eq19}, the variables are coupled again ($p_{1}$ with $p_{4}$ and $p_{2}$ with $p_{3}$).
I am working in a book environment and I do not have any other problems of this kind in my thesis. In the image you can see what's actually going on. Any help is appreciated! Thanks
• Welcome to SE! How you like to have aligned? The equation is not equal (have different width), so they align in text had to be different. – Zarko Feb 12 '16 at 11:30
• Thanks for your welcoming message! :) I know they have different width but the equations should be both centred, and they do not look centred to me; the second one looks like it is pushed on the left... – Eli Feb 12 '16 at 11:33
• Welcome to TeX.SX!! It is better to post a full minimal working example that starts with a \documentclass command, has a minimal preamble and then \begin{document}...\end{document}. Unless the problem is a compilation error, the code should compile and be as small as possible to demonstrate your problem. This makes it much easier for people to help you --- and much ore likely that they will! – Andrew Feb 12 '16 at 11:59
When I compile the OP's snippet, I do not get the left/right offset that is shown in his image. The OP will need to provide a complete working example to demonstrate the problem.
When I wrap the OP's provided code in a document with amsmath, I get this image:
which is centered, but of uneven width. One thing that can be done is to redefine the length \arraycolsep in the second equation to 3.9pt. When that is done, the two equations end up a similar width.
\documentclass{article}
\usepackage{amsmath}
\begin{document}
If $\tilde{\mathbf{p}}$ is first combined with the system of Eq.\ref{inplane:eq7} and then differentiated with respect to time, it yields
$$\label{4:eq18} \left[\begin{array}{c} \dot{p}_{1} \\ \dot{p}_{2} \\ \dot{p}_{3} \\ \dot{p}_{4} \end{array}\right] = \omega \begin{bmatrix} 0 & 4\sigma-3 & -2(\sigma-1) & 0 \\ 2\sigma - 3 & 0 & 0 & 4(\sigma-1) \\ 4(\sigma-1) & 0 & 0 & 8\sigma-5 \\ 0 & -2(\sigma - 1) & \sigma-1 & 0 \end{bmatrix} \left[\begin{array}{c} p_{1} \\ p_{2} \\ p_{3} \\ p_{4} \end{array}\right].$$
From this system of equations, it is then possible to differentiate even further to a second order system as
$$\label{4:eq19} \arraycolsep3.9pt \left[\begin{array}{c} \ddot{p}_{1} \\ \ddot{p}_{2} \\ \ddot{p}_{3} \\ \ddot{p}_{4} \end{array}\right] = \omega^{2} \begin{bmatrix} -(2\sigma -1) & 0& 0 & -2(\sigma -1) \\ 0 & -(2\sigma -1) & 2(\sigma-1) & 0 \\ 0 & -2(\sigma -1) & 3(\sigma-1) & 0 \\ 2(\sigma -1) & 0 &0& 3(\sigma-1) \end{bmatrix} \left[\begin{array}{c} p_{1} \\ p_{2} \\ p_{3} \\ p_{4} \end{array}\right].$$
In Eq. \ref{4:eq19}, the variables are coupled again ($p_{1}$ with $p_{4}$ and $p_{2}$ with $p_{3}$).
\end{document}
• Thank you for your answer, I have already amsmath set in the header of my file, but I am not sure why I obtain a different result, compared to your first one. In any case your answer looks brilliant to me, thanks for your help! – Eli Feb 12 '16 at 11:54 | 2019-11-19T07:07:39 | {
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https://mathematica.stackexchange.com/questions/213228/is-there-a-way-for-me-to-plot-a-3d-time-dependent-vector | # Is there a way for me to plot a 3D time dependent vector? [closed]
I wanted to know if it is possible for me to plot a 3D vector (with x, y, and z components) in which each component is dependent on time. For more context, I want to plot the function:
$$\qquad E(t) = 3\cos(ω\,t)(\hat x) + [3\cos(ω\,t)−4\sin(ω\,t)] (\hat y) − 6\cos(ω\,t−π/4)(\hat z)$$
Which is a phasor represented in the time domain. I believe that the best way is to keep time constant, and to plot the function at one instant in time, but I also have seen my professor make a graph that gradually changes a variable, n, by itself, or by sliding a bar. How do I do that, or is there a better way to plot this function? Also, this isn't a part of my homework, but I would like to make a graph to better understand what I am doing for my homework and how it looks plotted.
• Have a look at ParametricPlot3D and Manipulate. The combo of them should do basically what your professor did. – Henrik Schumacher Jan 20 at 23:32
• Welcome to Mathematica Stackexchange! Don't forget to upvote good answers (and other people's questions) using the triangle above the number next to the post, and use the checkmark to "accept" the answer to your question that you think best answers it. Take the introductory TOUR – Vitaliy Kaurov Jan 22 at 15:45
f[w_][t_]:={3 Cos[w t], 3 Cos[w t]-4 Cos[w t], -6 Cos[w t - Pi/4]} | 2020-04-08T06:59:45 | {
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https://math.stackexchange.com/questions/1747317/what-does-a-triple-integral-represent/1747352 | What does a triple integral represent?
From my understanding if the integrand is 1, then it gives you the volume of the region defined by the bounds. But what does the value of a triple integral represent if the integrand is a function for a surface in space?
• The volume of the 3 dimensional object the integrand defines. The regions you can create using only the integrals bounds are quite limits. – TSF Apr 18 '16 at 0:51
• @TonyS.F. A double integral with a 1 in the integrand gives you area. When you have a function for a surface in space in the integrand of a double integral, it multiplies the area by the height of that surface, giving you a volume. If we have a triple integral with an integrand of 1 however, we have a volume. If we have a function for a surface in the integrand, than the volume is being multiplied by it.. and this is where my logic breaks. It doesn't seem to make sense to think of this as volume the same way a triple integral with an integrand of 1 is volume. – IgnorantCuriosity Apr 18 '16 at 0:57
• You can think of them as being weighted volumes? – TSF Apr 18 '16 at 1:01
• @TonyS.F. I'm not sure what you mean by that. – IgnorantCuriosity Apr 18 '16 at 1:08
• You are still adding up "blocks" of volume but each block is made of a different material with a different density, which is precribed by the integrand. – TSF Apr 18 '16 at 1:10
You can think of the integrand as the "density" of the region and the value of the integral as the "mass" of the object.
For example, $$\int_0^1\int_0^1\int_0^1 1 \, \text{d}x \, \text{d}y \,\text{d}z$$ can represent the volume of the unit cube within the region $0\le x\le 1$, $0\le y\le 1$ and $0\le z\le 1$.
For $$\int_0^1\int_0^1\int_0^1 (x^2+y^2+z^2) \, \text{d}x \, \text{d}y \, \text{d}z\ ,$$ you can think about it as the mass of the same cube where its density is given by the function $f(x,y,z)=x^2+y^2+z^2$. This means that the cube is light near the origin and is getting heavier as you move away from it.
In general, $f$ can be negative so you must consider the signed-mass which means that the mass can be negative somewhere...
• This is a good demonstration of what I meant by weighted volumes. – TSF Apr 18 '16 at 1:11
• For a more realistic example than signed mass, you can think of $f(x,y,z)$ as the density of electrical charge (which can be positive as well as negative). Then the integral gives the total charge of the body that you integrate over. – Hans Lundmark Apr 18 '16 at 6:10
• @HansLundmark That's pretty neat, much more realistic that my "negative mass" haha. Anyway, I think using the term "mass" might convey more intuition than electric charges for beginner. – BigbearZzz Apr 18 '16 at 8:05
• For which values of $x$, $y$ and $z$ will $f$ be negative? – nluigi Apr 18 '16 at 20:50
• @nluigi What I meant is that for a different choice of the function $f$, it can sometimes be negative. For example we can let $f=x+y-z$. – BigbearZzz Apr 19 '16 at 1:08
When $1$ is your integrand, you have these geometric interpretations:
$\quad\int dx$ is to length
$\quad\iint dA$ is to area
$\quad\iiint dV$ is to volume
When your integrand is some function, then you've likely heard:
$\quad\int f(x)\ dx$ is the signed area underneath the curve.
But wait, you say, why is it that $\int dx$ is a length, but $\int f(x)dx$ is an area? First, it's all an interpretation. One could say that $\int dx$ is the area under the curve $f(x)=1$. The difference is that if $f(x)\ne1$, then we're assigning some non-trivial value to every point in the evaluated space. Note that if $f(x)$ and $x$ have physical units $[f(x)]$ and $[x]$ respectively, then that "area" has units $[f(x)]\times[x]$. For example, integrating a velocity function over some period of time $\int_{t_0}^{t_f} v(t)\ dt$ would give you a change in displacement.
So this is a sort of density. At every point in the interval, you're summing how much of the integrand is present. You're finding how much "stuff" is associated with the interval of integration. With that in mind, let's revise our previous statement:
$\quad\int f(x)\ dx$ is the signed amount of stuff associated with the interval of integration.
It follows that
$\quad\iint f(x,y)\ dxdy$ is the signed amount of stuff associated with the region of integration.
This could certainly be a volume, but we shouldn't limit ourselves. Integrating a surface charge density over a region would give you the total amount of charge, wouldn't it? That's not a volume. What we're really doing is assigning a weighting factor to every point in the region and summing all of these contributions via integration.
$\quad\iiint f(x,y,z)\ dxdydz$ is the signed amount of stuff associated with the volume of integration.
If your integrand was a mass density $\rho(x,y,z)$, then integrating over the volume would give the total mass.
In terms of understanding what an integral actually does, what I presented is mostly a heuristic, but I hope it's a useful one. It certainly rings true in physics. If there's one thing to note, it's that if you're integrating quantities with physical dimensions, then the units of your result will have units equal to that of the integrand times those of the differential elements.
That is, if $I = \int f d\tau$, then $[I] = [f][d\tau]$.
• Indeed more generally, in manifold theory, the objects one integrates are called "densities" (see en.m.wikipedia.org/wiki/Density_on_a_manifold). I don't know the origin of this term, but I suspect it was chosen to convey this natural interpretation of the objects one integrates (with respect to "volume," at least). – symplectomorphic Apr 18 '16 at 1:42
I warmly endorse the "weighted volume" interpretation found in other answers. But since IgnorantCuriosity asked about parallels with e.g. using a double integral to represent the volume underneath a surface in 3 dimensions, here's how that goes: a triple integral represents the volume underneath a hypersurface in four dimensions.
So:
• $\int f(x)\,dx$ is the area under the curve $y=f(x)$ in two dimensions $(x,y)$.
• $\int\int f(x,y)\,dx\,dy$ is the volume under the surface $z=f(x,y)$ in three dimensions $(x,y,z)$.
• $\int\int\int f(x,y,z)\,dx\,dy\,dz$ is the hypervolume under the hypersurface $t=f(x,y,z)$ in four dimensions $(x,y,z,t)$. ("Under" means $t<f(x,y,z)$.)
This depends on what the function is inside the integrand. If the function is a density function, the integral would give us the total mass of the object. It could also be the volume of some 4 dimensional object as well.
Mass ... $$\iiint_E w(x,y,z)\;dx\,dy\,dz$$ is the total mass of a region $E$ in space, where $w$ is the "density" (mass per unit volume) that may vary from one point to another.
Similarly, to illustrate the integration of a signed function (that has potentially both positive and negative values), you can think of computing the total charge in a region of space, where $w$ is the "charge density" (charge per unit volume).
In your calculus course, you probably have some problems where you integrate over a volume of water, where the pressure varies with depth.
I too favour the weighted volume analogy, but thinking fundamentally it is a continuous summation of a volume. It does not have to be a volume, even if we conceptually model it as such, it could be input voltages, currents, magnetic fields over time. Hypersurface indeed.
Integrating my acceleration gives my velocity. Integrating my velocity indicates the distance I have travelled, etc. These values are not necessarily scalar and accumulate continuously. Fundamental calculus. | 2020-04-06T12:40:02 | {
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https://mathhelpforum.com/threads/finding-the-derivative.144300/ | # Finding the derivative
#### spoc21
Find the derivative of the following function:
f(x) = (2x-3)^2/(x^3-7)^3
I am very confused here. Don't know what rule to use, as the exponents are outside the bracket. I would appreciate any help!
Thanks
#### Chris L T521
MHF Hall of Fame
Find the derivative of the following function:
f(x) = (2x-3)^2/(x^3-7)^3
I am very confused here. Don't know what rule to use, as the exponents are outside the bracket. I would appreciate any help!
Thanks
You will first need to use the quotient rule, and then a combination of power rule and chain rule.
Try to attempt the problem given this information and let us know where you get stuck.
#### mr fantastic
MHF Hall of Fame
Find the derivative of the following function:
f(x) = (2x-3)^2/(x^3-7)^3
I am very confused here. Don't know what rule to use, as the exponents are outside the bracket. I would appreciate any help!
Thanks
Use the quotient rule. u = (2x - 3)^2 and v = (x^3 - 7)^3. To differentiate u and v, either expand and differentiate in the usual way, or use the chain rule.
#### spoc21
thanks guys!
So I end up with the following result (am very unsure about it):
Derivative:
$$\displaystyle 2x-3(4(x^6-32x^3+27x^2+49))$$
---------------------------------
$$\displaystyle (x-7)^6$$
can some one please confirm if this is the correct answer.I would really appreciate it!
Thanks
#### mr fantastic
MHF Hall of Fame
thanks guys!
So I end up with the following result (am very unsure about it):
Derivative:
$$\displaystyle 2x-3(4(x^6-32x^3+27x^2+49))$$
---------------------------------
$$\displaystyle (x-7)^6$$
can some one please confirm if this is the correct answer.I would really appreciate it!
Thanks
The review process will be much easier and more effective if you post all your working (ie. show every step) rather than just giving a final answer.
#### tom@ballooncalculus
Just in case you're interested (like many people) in avoiding the quotient rule, which can feel like overkill, and always doubles the power of the denominator, sometimes unnecessarily as here... then, how about...
... where
... is the product rule, straight lines differentiating downwards with respect to x. Then you don't need u and v. You still need the chain rule, but you can zoom in on this, for example in the right hand fork...
... where...
... is the chain rule. Straight continuous lines still differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).
However, you could zoom in on both forks...
... or neither...
... according to taste.
Another way of avoiding the quotient rule is logarithmic differentiation - taking logs of both sides of y = the function. Then solve for dy/dx in the bottom row of...
_________________________________________
Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!
TheCoffeeMachine
#### spoc21
ok here's the working:
[FONT="]
[/FONT]
$$\displaystyle f(x)=(2x-3)^2/(x^3- 7)^3$$
$$\displaystyle f' (x)=((x^3-7)^3 [2(2x-3)(2) ]-[3(x^3-7)^2 (3x^2 )])/[(x^3-7)^3 ]^2$$
$$\displaystyle f' (x)=((x^3-7)^3 [4(2x-3)]-[9x^2 (x^3-7)^2](2x-3)^2)/(x^3-7)^6$$
We need to simplify:
$$\displaystyle f' (x)=((x^3-7)[4(2x-3) ]-(9x^2)(2x-3)^2)/(x^3-7)^4$$
We need to simplify Further:
$$\displaystyle f' (x)=(2x-3)[4(x^3-7)-9x^2 (2x-3) ]/(x^3-7)^4$$
thats my final answer ^^...I am supposed to represent the derivative in its simplified form..Is my working correct?
Thank you!
[FONT="]
[/FONT]
#### tom@ballooncalculus
We need to simplify:
$$\displaystyle f' (x)=((x^3-7)[4(2x-3) ]-(9x^2)(2x-3)^2)/(x^3-7)^4$$
Good. (The quotient rule had to take you up to power 6 in the denominator, but you're back down again.)
We need to simplify Further:
$$\displaystyle f' (x)=(2x-3)[4(x^3-7)-9x^2 (2x-3) ]/(x^3-7)^4$$
Fine. Why not expand and collect in the square brackets? | 2019-11-20T11:18:13 | {
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http://math.stackexchange.com/questions/470908/the-riemann-integrability-of-a-function-similar-to-dirichlets-function | # The Riemann Integrability of a function similar to Dirichlet's function
$\textbf{Problem:}$ Consider the function $f: [0,1] \rightarrow \mathbb{R}$ defined by letting $f(x)=0$ for rational $x$ and $f(x)=x$ for irrational $x$. Calculate the upper and lower Riemann integrals of $f$. Is $f$ Riemann integrable?
$\textbf{ Solution Attempt:}$ Let $P_n=\{0=t_0,...,t_n=1\}$ be a partition of $[0,1]$, where $t_k = \frac{k}{n}$. For this partition, we have $$U(f,P_n)=\sum^{n}_{k=1} t_k (t_k - t_{k-1}) = \sum^{n}_{k=1} t_k (\frac{k}{n} - \frac{k-1}{n})=\sum^{n}_{k=1} \frac{k}{n} (\frac{1}{n})= \frac{1}{n^2} \sum^{n}_{k=1} k.$$ A familiar property of the natural numbers gives us that $\sum^{n}_{k=1} k= \frac{1}{2}n(n+1).$ Thus, $U(f,P)= \frac{1}{2} + \frac{1}{2n}.$ Recall, the upper Riemann integral $U(f)$ is defined as $U(f)=\inf \{U(f,P)\}$, where $P$ is a parition of $[0,1].$ As $n \rightarrow \infty$, $P_n \rightarrow \frac{1}{2}.$ [Note: this is where I am a little unsure of myself.] Thus, $U(f)=\frac{1}{2}$. However, the lower Riemann integral is $0$. Hence, the upper and lower Riemann integrals disagree. We conclude that $f$ is not Riemann integrable. $\blacksquare$
If the aforementioned function was changed so that it took on the value of $0$ for irrational $x$ and $x$ for rational $x$, then $f$ would be zero almost-everywhere. Hence, it would be Lebesgue integrable. Is this correct? I am quite unsure of the Lebesgue integrability for the original function. A hint would great! Thank you!
-
The function differs from the function $g(x)=x$ on a set of measure $0$. For the Riemann integral, you have not proved that the inf of all upper sums is different from $0$, though I am sure you could. One can do it without taking any limits, so the specific computation you did can be bypassed. – André Nicolas Aug 19 '13 at 4:48
Thank you! Indeed, concerning the Riemann integral, I have only shown $U(f) \leq \frac{1}{2}$. Given any partition $P$, $0<U(f,P)$. So that by definition of the sup of a set, we get $0<U(f) \leq U(f,P).$ Does this suffice? Thanks again! – dgc1240 Aug 19 '13 at 5:23
- | 2014-04-16T08:15:14 | {
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https://cs.stackexchange.com/questions/128260/existence-non-existence-of-a-sequence-with-short-longest-increasing-subsequenc/128264 | # Existence / non-existence of a sequence with short longest increasing subsequence and decreasing subsequence?
Can there exist any integer sequence $$A$$ of length $$N$$ with all unique elements such that the length of its Longest Increasing Subsequence as well as that of its Longest Decreasing Subsequence is less than $$\displaystyle \lfloor \frac{N}{2} \rfloor$$?
If yes, then give an example of such a sequence. Otherwise, can anyone present a proof that there cannot exist such a sequence?
(Just to add some substance, can it be shown there can exist such sequences, given any arbitrary value of $$N > 1$$?)
The answer to the OP's question is, no if $$N\le 7$$ and yes otherwise.
For given any positive integer $$r$$ and $$s$$, the celebrated Erdős–Szekeres theorem shows that for any sequence of distinct real numbers with length at least $$(r − 1)(s − 1) + 1$$ contains an increasing subsequence of length $$r$$ or a decreasing subsequence of length $$s$$.
It turns out that bound, $$(r-1)(s-1)+1$$ is tight. That is, for any positive number $$r$$ and $$s$$, there is a sequence of distinct numbers with length $$(r-1)(s-1)$$ that contains no increasing subsequence of length $$r$$ and no decreasing subsequence of length $$s$$.
Here is such an example.
$$\begin{array} {} &s-1, &s-2, &\cdots,&2, &1\\ &2(s-1), &(s-1)+ s-2, &\cdots, &(s-1)+ 2, &(s-1)+ 1\\ &\vdots &\vdots &\vdots &\vdots &\vdots \\ &(r-2)(s-1), &(r-3)(s-1)+s-2, &\cdots, &(r-3)(s-1)+2, &(r-3)(s-1)+1\\ &(r-1)(s-1), &(r-2)(s-1)+s-2, &\cdots, &(r-2)(s-1)+2, &(r-2)(s-1)+1\\ \end{array}$$
Consider the numbers above, reading from left to right and then from top to bottom. In other words, the sequence is $$s-1$$ down to $$1$$, followed by $$2(s-1)$$ down to $$(s-1)+1$$, etc and finally followed by $$(r-1)(s-1)$$ down to $$(r-2)(s-1)+1$$, all in step of $$1$$.
It is easy to see that there is no increasing subsequence of length r and no decreasing subsequence of length $$s$$.
For example, when $$r=s=5$$, we have $$4,3,2,1,\ \, 8,7,6,5,\ \,12,11,10,9,\ \,16,15,14,13$$ which does not have increasing subsequence of length $$5$$ nor decreasing subsequence of length $$5$$.
If we let $$r=s$$, the section above implies that, for any positive number $$N$$, there exists an integer sequence of length $$N$$ with all unique elements such that the length of its longest increasing subsequence as well as that of its longest decreasing subsequence is at most $$\lceil\sqrt N\rceil$$. And $$\lceil\sqrt N\rceil$$ is the tight upper bound.
Since $$\lceil\sqrt N\rceil\ge \lfloor\frac N2\rfloor\ \text{ for all } N\le 7$$ and $$\lceil\sqrt N\rceil\lt \lfloor\frac N2\rfloor\ \text{ for all } N\gt 7,$$ the answer to the OP's question is, no if $$N\le 7$$ and yes otherwise.
For example, for $$N=8$$, we have sequence $$3,2,1,6,5,4,9,8,7$$.
Here's a direct construction of such a sequence for any multiple of four. It's made up of four equally sized runs of consecutive integers.
The first and third runs are increasing. The second and fourth runs are decreasing. The runs use ranges of numbers such that $$R_2 < R_3 < R_1 < R_4$$. For example, with $$4n=16$$,
$$9,10,11,12 |4,3,2,1|5,6,7,8|16, 15,14,13$$
The longest increasing subsequence is length $$n+2$$. For example, in the above where $$4n=16$$, the longest increasing subsequence has length $$6$$ ($$1| 5, 6, 7, 8|16$$). No increasing subsequence is longer:
• It's not possible to pick an element from both increasing runs, since any element in the first increasing run disqualifies all of them from the second increasing run.
• It's not possible to pick more than one element from either decreasing run
A symmetric argument applies for the decreasing subsequences.
Since $$n+2 << 2n$$, this works as a counterexample for any multiple-of-four sequence. You can easily pad with extra sequence elements for non-multiple-of-four lengths.
I came across this construction by considering a sequence that was a "hill" (increasing, then decreasing), which meets your condition perfectly. Breaking up those long runs could be done making two hills (increasing, decreasing, increasing, decreasing), which this sequence does by ensuring the up/down slope of one 'hill' isn't continued by the other.
• A much simpler construction: $3, 2, 1, 6, 5, 4, 9, 8, 7$. – Jakube Jul 11 '20 at 18:12
There are also short sequences that satisfy your request. Consider for example the first 16 terms of the binary Van der Corput sequence $$0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15.$$ In general there exists a sequence $$T$$ of length $$n\geq1$$ containing a longest increasing subsequence of length $$x\geq 1$$ and a longest decreasing subsequence of length $$y\geq 1$$ if and only if the numbers $$x$$, $$y$$ and $$n$$ satisfy the conditions $$x\cdot y\geq n$$ and $$x+y\leq n+1$$, see here. Notice that the reference gives a constructive proof.
• It looks like the general statement is wrong. Sequence $1, 3,5,4,2$ contains a LIS of length 3 and a LDS of length 3. However, length $n=5$, $x=3$, $y=3$ but $x+y >n$. – John L. Jul 12 '20 at 1:03
• @JohnL. I've missed a "+1", see the edit. – user6530 Jul 12 '20 at 11:29
Such sequences do exist. It suffices to generate a large enough random sequence. If you check Dan Romik's book, The Surprising Mathematics of Longest Increasing Subsequences, Theorem 1.1 states that
$$\frac {\ell_n} {\sqrt n} \to 2,$$
where $$\ell_n$$ is an expected length of increasing subsequence in a random permutation of size $$n$$. The same for decreasing. Therefore, for large enough $$n$$ there must exist a sequence with both increasing and decreasing sequences of lengths at most $$5 \sqrt n$$, otherwise:
$$2 E[\ell_n] = E[|decr_n| + |incr_n|] \ge 5 \sqrt n,$$
• Why the downvote? This answer does answer the question for large $n$. Although the theorem quoted appears to be rather heavy machinery for such a simple task, it is very nice to mention that book, which unfolds rich, deep and beautiful mathematics about longest increasing subsequence, – John L. Jul 12 '20 at 12:08 | 2021-01-19T03:41:07 | {
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https://www.physicsforums.com/threads/total-moment-of-inertia-of-two-rods.845505/ | # Total Moment of Inertia of Two Rods
1. Nov 28, 2015
### Raios168
1. The problem statement, all variables and given/known data
The rods of length 2 meters and and mass 20 kg are joined at their ends to form a V shape. What is the total moment of inertia measured from the reference point perpendicular to the plane of the paper and at the point where the two rods are joined. (So find total moment of inertia at the edge of the V shape, but with respect to the plane perpendicular to the page). Also the rods are 60 degree apart.
2. Relevant equations
Moment of Inertia at the edge of a rod = 1/3 ML^2
3. The attempt at a solution
Can I not just add the two moments of inertia to get 2/3 ML^2 as the total?
EDIT: I think the correct way to do this is to find the CoM of the system (which will be the mid point of the line connecting the midpoints of the two rods). And then use parallel axis theorem to find the total moment of inertia about the CoM. And then apply parallel axis theorem again to find the MI about the the point connecting the two rods. Can anyone confirm this?
Last edited: Nov 28, 2015
2. Nov 28, 2015
### Staff: Mentor
Either method should bring you to the same result. I know which way I'd choose if I was in a hurry
3. Nov 28, 2015
### SteamKing
Staff Emeritus
The only problem with the latter plan is how to calculate the MOI of the rods about the C.o.M. Because the rods are fixed together at one end, I think you have to calculate the MOI using integration or use the formula for calculating the MOI of a rod about a rotated coordinate axis.
4. Nov 28, 2015
### Staff: Mentor
Hence my comment above...
5. Nov 29, 2015
### azizlwl
Schaum's- 3000 Solved problem in Physics. page 212.
6. Nov 29, 2015
### SteamKing
Staff Emeritus
I don't think this applies to this particular case.
The individual rods are not joined at their centroids, but at one end of each rod and at an angle to one another, in the shape of a V when looking parallel to the presumed axis of rotation.
You can calculate the MOI of each rod about one end, but combining the two is a bit trickier than you are led to believe by this 'Rule'.
7. Nov 29, 2015
### azizlwl
11.33
Four coplanar, large, irregular masses are held by a rigid frame of negligible mass, as shown in figure 11-6. Taking an axis through P and perpendicular to the page, show that I=I1 +I2+I3+I4 where I1 is the moment of inertia of object 1 alone about the axis and similarly for the others.
p is intersection of lines joining the masses.
Last edited: Nov 29, 2015
8. Nov 29, 2015
### SteamKing
Staff Emeritus
It's not clear what I1 thru I4 represent. Are they supposed to be the moments of inertia for the individual masses referred to the axis thru P?
How exactly to you 'add algebraically' the MOI of one mass which is placed at an angle to another mass?
Last edited: Nov 29, 2015
9. Nov 29, 2015
### Staff: Mentor
We can go to the basics for moment of inertia about an axis and write the integral for the given scenario. In the figure the z-axis projects out of the page:
L is the length of each rod of mass M, the distance from the z-axis of a mass element is r. The domain of r is 0 to L, and a differential mass element is $dm = (M/L)dr$.
$$I = 2\int_0^L r^2 \frac{M}{L}~dr$$
It should be clear from the symmetry that the moment of inertia will be twice that of a single rod. | 2018-03-17T05:04:10 | {
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https://math.stackexchange.com/questions/27024/a-n-is-the-only-subgroup-of-s-n-of-index-2/3007858 | # $A_n$ is the only subgroup of $S_n$ of index $2$.
How to prove that the only subgroup of the symmetric group $$S_n$$ of index 2 is $$A_n$$?
Why isn't there other possibility?
Thanks :)
• Please make the body of your posts self-contained. The title is an indexing feature, and should not be an integral part of the message. Think of it as the title of a book on the spine; it's there to let people know what the post is about, not to impart information without which you cannot understand what is happening. Mar 14, 2011 at 21:00
• I am terribly sorry. But I don't know how to reedit it. Mar 14, 2011 at 21:07
• There should be a link below the [abstract algebra] tag that says "edit". Click it, and you can edit. Mar 14, 2011 at 21:11
As mentioned by yoyo: if $H\subset S_n$ is of index 2 then it is normal and $S_n/H$ is isomorphic to $C_2=\{1,-1\}$. We thus have a surjective homomorphism $f:S_n\to C_2$ with kernel $H$. All transpositions in $S_n$ are conjugate, hence $f(t)\in C_2$ is the same element for every transposition $t\in S_n$ (this uses the fact that $C_2$ is commutative). $S_n$ is generated by transpositions, therefore $C_2$ is generated by $f(t)$ (for any transposition $t\in S_n$), therefore $f(t)=-1$, therefore ker $f=A_n$.
• @user8268 Can you explain how $f(t)\in C_2$ is the same element for every transposition $t\in S_n$, and how $S_n$ is generated by transpositions? Thank you. Nov 30, 2014 at 18:49
• @jstnchng Suppose $0 = f(t) \neq f(t') = 1$ for transpositions $t, t' \in S_n$. Then $1 = f(t) + f(t') = f(tt') = f(tgtg^{-1})$ for some $g \in S_n$. But then $1 = 0 + 0$ is a contradiction. Apr 12, 2021 at 22:07
Other Way :
$$A_n$$ is generated by all $$3$$-cycles in $$S_n$$.
If $$H\neq A_n$$ and $$|S_n:H|=2$$ then at least one 3-cycle is not in $$H.$$
WLOG assume say $$(123)\notin H$$ so $$H,(123)H,(132)H$$ are 3 distinct cosets which is a contradiction to the fact that $$H$$ has index $$2$$.
• How do we know $(132)\notin H$? Jun 8, 2019 at 23:40
• @Lotte If $(132)\in H$ as H is group $(132)^{-1}=(123)\in H$ But which is not . Aug 22, 2019 at 16:00
• How do you know these are in fact distinct sets? Jan 11 at 1:59
• Coset are partition of group. So by definition they are distinct. Let me know if you have any doubts Jan 11 at 5:06
subgroups of index two are normal (exercise). $A_n$ is simple, $n\geq 5$ (exercise). if there were another subgroup $H$ of index two, then $H\cap A_n$ would be normal in $A_n$, contradiction.
• How do you know that the intersection is not trivial? {1} is normal in every group and does not contradict simplicity. There is a way around this using conjugacy in $S_n$...... Oct 30, 2013 at 19:42
• @Vladhagen you define the normal subgroup $G$ as being non-trivial. If $G \preccurlyeq S_n$ such that $[G:S_n]=2$ and $G \ne A_n$ then $[A_n \cap G:S_n]=2$, but $A_n$ is simple and obviously $A_n \cap G \preccurlyeq A_n$ so $A_n \cap G = A_n$ or identity. If identity, then $G$ contains identity and a transposition, but then $G$ isn't normal. So the intersection must be $A_n$, and by the order divisor theorem $\# A_n = n!/2 \mid \# G$ and $\# G \mid \# S_n = n!$ so $G = S_n$. Oct 8, 2019 at 13:14
I realize this question is rather old, but if this comes from Hungerford's book, there is a specific way he wants us to solve this problem, so I am providing this for the benefit of those working through Hungerford. First, one must recall that subgroups of index $2$ are normal. Hungerford's suggestion is to use the following fact:
Let $r,s$ be distinct elements of $\{1,2,...,n\}$. Then $A_n$ ($n \ge 3$) is generated by the $3$-cycles $\{(rsk) ~|~ 1 \le k \le n, k \ne r,s\}$.
Keeping that in mind, we first prove the two easy facts about subgroups of index $2$
Lemma (1): If $H$ is a subgroup of index $2$ in $G$, then $H$ contains the square of every element in $G$.
Proof: Let $g \in G$ be arbitrary. Then by Lagrange's theorem, $(gH)^2 = H$ or $g^2H=H$, happening if and only if $g^2 \in H$.
Lemma (2): If $H$ is a subgroup of index $2$ in $G$, then $H$ contains all elements of odd order.
Proof: Suppose that $g \in G$ is a element of order $2k+1$ for some $k \in \mathbb{N}$. Then $H = g^{2k+1}H = gH \cdot (g^2)^k H = gH$, where $(g^2)^k H = H$ by lemma (1). This, of course, means $g \in H$.
From these two lemmas, I think it is pretty clear how one ought to use the given fact: if $H \le S_n$ were a subgroup of index $2$, then it would consist of all $3$-cycles, since they have odd order. But by closure this means that $A_n \le H$. Since they have the same index, they must also have the same order, implying that they are equal because we are dealing with finite sets.
• How do 3-cycles have an odd order? Or were you referring to something else? Oct 30, 2018 at 20:14
• @Maximalista 3-cycles have order 3. On the other hand, a 3-cycle is an even permutation. Jan 1, 2021 at 10:48
The quotient map for $$A_n$$ is a surjective homomorphism to $$C_2$$.
Any other index two subgroup $$H$$ of $$A_n$$ gives you a distinct surjective homomorphism to $$C_2$$.
Therefore taking the product of these we obtain a surjective homomorphism $$S_n$$ to $$C_2 \times C_2$$. But then this has kernel of order $$\frac{n!}{4}$$, by the first iso theorem, and thus the image of $$A_n$$ cannot be 1 or $$C_2 \times C_2$$, so must be of order 2.
But this implies that $$A_n$$ has an index two subgroup, a contradiction as $$A_n$$ is simple for $$n \geq 5$$, and when $$n = 4$$, $$A_4$$ has no order 6 subgroup.
Let $$n\geq 2$$ and $$H \leq S_n$$ be such that $$\mid S_n:H \mid=2$$. Then $$H \trianglelefteq S_n$$ and $$S_n/H$$ being isomorphic to $$\Bbb Z/2\Bbb Z$$, is cyclic. Consider the natural surjection $$\pi :S_n\to S_n/H$$. Since $$S_n/H$$ is abelian, the commutator subgroup $$[S_n,S_n]\subseteq \ker(\pi)=H$$. We know $$[S_n,S_n]=A_n$$, hence $$A_n \subseteq H$$. Clealy $$\mid A_n \mid=\mid H \mid$$, as both of them are subgroups of a finite group of equal index. Therefore we have $$H=A_n$$.
Let $$H\le S_n$$ such that $$[S_n:H]=2$$ and $$H\ne A_n$$. Denoted $$d:=|H\cap A_n|$$, we get: $$|HA_n|=\frac{(n!/2)^2}{d} \tag1$$ By Lagrange, $$d\mid n!/2$$ and, since $$HA_n\le S_n$$ (as $$A_n$$ is normal in $$S_n$$) also $$\bigl((n!/2)^2/d\bigr)\mid n!$$. Therefore: $$\frac{n!}{2}=kd \tag2$$ and $$1=l\frac{n!}{4d} \tag3$$ for some positive integers $$k$$ and $$l$$. From $$(2)$$ and $$(3)$$ follows $$kl=2$$, and hence $$k=1$$ or $$k=2$$. But since $$H\ne A_n$$, it must be $$d\ne n!/2$$ and hence, by $$(2)$$, $$k\ne 1$$. So $$k=2$$ and hence, again by $$(2)$$, $$d=n!/4$$, whence by $$(1)$$ $$|HA_n|=n!$$, and finally $$HA_n=S_n$$: contradiction, by the argument in this post. Therefore, necessarily $$H=A_n$$. | 2022-10-02T12:54:44 | {
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http://openstudy.com/updates/4f1c4b0fe4b04992dd237b16 | ## AravindG 4 years ago A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m/s. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5 m/s and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands?
1. AravindG
u see i got first part
2. AravindG
10 s
3. AravindG
but i dont undestand y we get same answr in 2nd case
4. JamesJ
The position of the ball at time time is $y(t) = 49 t - \frac{1}{2}gt^2$ $= 49 t - 4.9t^2$ $= 0$ when $$t = 0, 10$$ seconds, right. Now in the second case, the speed of the ball is (49 + 5) m/s = 54 m/s, so it's position is $y_{ball}(t) = 54 t - 4.9t^2$ ... but ...
5. JamesJ
the position of the boy is $y_{boy}(t) = 5t$ Hence the question is: for what t is $y_{ball}(t) = y_{boy}(t)$ Solve that equation and you'll get back $$t = 0, 10$$ seconds.
6. AravindG
hw u got pos of boy as 5t??
7. JamesJ
because the lift is moving up at a constant velocity of 5 m/s
8. AravindG
wow i got the old eqn
9. AravindG
:)
10. JamesJ
right, exactly
11. AravindG
thx
12. AravindG
so shall i make it general
13. AravindG
that if someone is in a lift and the lift moves with const speed then time to throw and catch the balll is same as when lift is at rest??
14. JamesJ
Yes.
15. AravindG
k
16. AravindG
wt if it was not const speed?
17. JamesJ
Then you won't have equality.
18. JamesJ
Frames of reference moving at a constant speed are called inertial frames. The laws of Newtonian physics work the same in all inertial frames.
19. AravindG
u mean any system with a=0 is an inertial frame?
20. JamesJ
Yes
21. AravindG
wow
22. AravindG
wel james hav u seen a type of problem in which a rope is given and we can slide down that and the breaking tension is given and max a with which we can slide down??
23. JamesJ
Yes, but I'm going to go now. I'm sure gogind or jemurray or others can help you with that.
24. AravindG
oh bye | 2016-10-24T10:41:38 | {
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http://math.stackexchange.com/questions/452275/global-maximum-of-a-sum-of-gaussians | # Global maximum of a sum of Gaussians
Suppose I have the weighted sum of $n$ Gaussian functions with varying means and standard deviations, $$f(x) = \sum_{i=1}^n a_i\exp\left(-\frac{(x-\mu_i)^2}{2\sigma_i^2}\right).$$ All the weights $a_i$ are positive. I want to find the position of the global maximum of $f$.
Clearly $f$ is non-concave and may have multiple local maxima. However, it feels like each local maximum should be close to the peak $\mu_i$ of one of the Gaussians.
Suppose I perform gradient ascent $n$ times starting from $x = \mu_i$ for $i = 1, 2,\ldots, n$. Am I guaranteed to find all the local maxima, and thus the global maximum? Is there a better strategy to find the global maximum?
-
Since $x$ is one dimensional, you can do a simple grid search with fine enough grids, which guarantees to find global optimum. It is not necessary the case that each local maximum is very close to $\mu_i$. In fact, new local maxima can be created between two distant $\mu_i$s if their $\sigma_i$s are large enough.
EDIT
To see how the grid search will give guarantees on global optimal solution. We first compute the bound of $f'(x)$:
$$|f'(x)| \le \sum_i a_i \left|\frac{x-\mu_i}{\sigma_i^2}\right|\exp\left(-\frac{(x-\mu_i)^2}{2\sigma_i^2}\right)$$
since (by taking derivative)
$$\frac{|x|}{\sigma^2}\exp\left(-\frac{x^2}{2\sigma^2}\right) \le \frac{1}{\sigma}e^{-1/2}$$
we have $$|f'(x)| \le e^{-1/2}\sum_i \frac{a_i}{\sigma_i} \equiv M$$
On the other hand, there exists a sufficiently large $K$ so that $|f(x)| \le \max_i a_i$ for $x \notin [-K,K]$ and cannot be the global optimum solution. We thus divide $[-K, K]$ into $\lceil 2KM/\epsilon \rceil$ bins $[l_j, u_j]$ so that $u_j - l_j \le \epsilon/M$. Thus since $|f'(x)|\le M$, within each bin $j$ there exists $c_j$ so that
$$c_j - \frac{\epsilon}{2}\le f(x) \le c_j + \frac{\epsilon}{2}$$
Let $j^* = \arg\max_j c_j$. Then any $\hat x^* \in [l_{j^*}, u_{j^*}]$ gives $f(\hat x^*)$ that is at most $\epsilon$ worse than the true global optimum.
I admit that from this analysis, it is not possible to provide a finite time algorithm that find the exact global optimum.
-
How does a grid search guarantee global optimum? – Lord Soth Jul 25 '13 at 21:36
Because you essentially have searched every location of the space if the function does not change too much within each grid. – Yuandong Jul 25 '13 at 21:41
Unless you use some of the properties of the given optimization problem, you can never guarantee finding the global optimum with grid search. For example, at least you should adjust your grid size with respect to the parameters of the mixture Gaussian, and specify where you start your search and where you end it. – Lord Soth Jul 25 '13 at 21:45
"Close" is of course relative to the scale parameter $\sigma_i$. Anyway, how fine a grid is fine enough? I am looking for theoretical guarantees. – Rahul Jul 25 '13 at 22:28
Let $f(x) = \sum\limits_{i=1}^n g_i(x)$, where $$g_i(x) = a_i\exp\left(-\frac{(x-\mu_i)^2}{2\sigma_i^2}\right)$$ are the $n$ Gaussians being summed together.
Suppose $x$ is a critical point of $f$, that is, $f'(x)=0$. For it to be a local maximum, $f''(x)$ must be negative. But $f''(x)=\sum\limits_{i=1}^ng_i''(x)$, and $$g_i''(x)=\frac{(x-\mu_i)^2-\sigma_i^2}{\sigma_i^4}g(x)$$ is negative only when $x\in[\mu_i-\sigma_i,\mu_i+\sigma_i]$. So any local maximum can only occur within $\sigma_i$ of a $\mu_i$.
This allows us to only search within the potentially smaller union of intervals $\bigcap\limits_{i=1}^n [\mu_i-\sigma_i,\mu_i+\sigma_i]$ rather than all of the approximately $[\min\mu_i,\max\mu_i]$ as I interpret Yuandong's answer. This is helpful if the $\mu_i$ are widely separated. Nevertheless, $f$ is not guaranteed to be concave in any of these intervals, so we cannot be assured to find a global maximum simply by gradient ascent. One can get within $\epsilon$ of the global maximum by a grid search with step size $\epsilon/M$, where $M$ is as defined in Yuandong's answer, but it would be nice to know if any faster convergence is possible. | 2014-03-09T00:59:31 | {
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http://www.cyclopaedia.info/wiki/Principal-diagonal | Principal diagonal
In linear algebra, the main diagonal (sometimes leading diagonal or major diagonal or primary diagonal or principal diagonal) of a matrix is the collection of entries where is equal to .
The main diagonal of a square matrix is the diagonal which runs from the top left corner to the bottom right corner. For example, the following matrix has 1s down its main diagonal:
A square matrix like the above in which the entries outside the main diagonal are all zero is called a diagonal matrix. The sum of the entries on the main diagonal of a square matrix is known as the trace of that matrix.
The main diagonal of a rectangular matrix is the diagonal which runs from the top left corner and steps down and right, until the right edge or the bottom edge is reached.
The diagonal of a square matrix from the top right to the bottom left corner is called antidiagonal, counterdiagonal, secondary diagonal, or minor diagonal.
This is an excerpt from the article Principal diagonal from the Wikipedia free encyclopedia. A list of authors is available at Wikipedia.
The article Principal diagonal at en.wikipedia.org was accessed 28 times in the last 30 days. (as of: 06/11/2013)
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Main diagonal - Wikipedia, the free encyclopedia
In linear algebra, the main diagonal (sometimes leading diagonal or major diagonal or primary diagonal or Principal diagonal) of a matrix A is the collection of ...
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principal diagonal - The Free Dictionary
The diagonal in a square matrix that goes from the upper left corner to the lower right corner. Thesaurus Legend: Synonyms Related Words Antonyms. Noun, 1 ...
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Principal Diagonal -- from Wolfram MathWorld
Principal diagonal. SEE: Diagonal. Wolfram Web Resources. Mathematica ». The #1 tool for creating Demonstrations and anything technical. Wolfram|Alpha » ...
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Principal diagonal - Merriam-Webster Online
Definition of Principal diagonal. : the diagonal in a square matrix that runs from upper left to lower right. First Known Use of Principal diagonal. 1964 ...
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Principal diagonal - FREEganita.com
The Principal diagonal elements (from top left corner to bottom right corner) of matrix A ... A matrix whose Principal diagonal elements are non zero and all other ...
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matlab - Selecting Principal Diagonal Elements - Stack Overflow
In the matrix shown below how can I select elements 01, 09, 17 and ... Use diag to get elements on the diagonal. diagA = diag(A). You can restrict ...
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Principal diagonal | Define Principal diagonal at Dictionary.com
Principal diagonal definition, diagonal See more. ... upper left to lower right (main diagonal or Principal diagonal) or lower left to upper right (secondary diagonal) ...
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Bounds for Determinants with Dominant Principal Diagonal - jstor
for determinants with dominant Principal diagonal (see a recent paper on these determinants by Olga Taussky [9 ] 2). The lower bound is an improvement of a ...
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Principal Diagonal of a Square Matrix, Scalar Matrix and Identity ...
High School Mathematics Curriculum - Principal diagonal of a Square Matrix, Scalar Matrix and Identity Matrix - Math & English ...
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[PDF]Diagonal Principal Component Analysis for Face Recognition
Nanjing University of Aeronautics and Astronautics, Nanjing 210016, China. Abstract. In this paper, a novel subspace method called diagonal principal ...
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Matrix Algebra
A symmetric matrix has the property that elements above and below the main diagonal are the same such that element(i,j) = element(j,i), as in our matrix B. ( The ...
[PDF]On Matrices with Elements in a Principal Ideal Ring
DigitalCommons@University of Nebraska - Lincoln ... Indiana University. Follow this ... form [that is, with zeros below the main diagonal] is again in normal form.
Books on the term Principal diagonal
Computational Chemistry Using the PC
Donald W. Rogers, 2003
A diagonal matrix has nonzero elements only on the Principal diagonal and zeros elsewhere. ... Large matrices with small matrices symmetrically lined up along the Principal diagonal are sometimes encountered in computational chemistry.
Mathematics Class 9
Pearson
Principal diagonal of a square matrix: In a square matrix A of order n, the elements a (i.e., a , a22 , a ) constitute the Principal diagonal. The elements fl are called elements of Principal diagonal. Examples 1. A = 2 -3 4 5 The elements 2,5 (i.e., ...
Modern Factor Analysis
Harry H. Harman, 1976
Introduction From basic mathematical theory it is known that a matrix of real numbers that is symmetric and is dominated by the Principal diagonal is positive semi-definite (i.e.. Gramian) with the important property of having all of its eigenvalues ...
Super Course in Mathematics for the IIT-JEE: Algebra II
Trishna Knowledge Systems
The diagonal containing the elements all, a22, a22. .. an“ is called its Principal diagonal. ... As an illustration, consider the square matrix A = 12' azs " 3 5 6 -1 2 3 l 8 0 5 IO 7 1 1 -1 3 The Principal diagonal of A consists of the elements 3, 3, ...
Matrices
Dr. V.N. Kala, Rajeshri Rana, 2009
Here elements 7, 3, 12 of square matrix are called its diagonal elements and diagonal along which these elements tie is called Principal diagonal. In Square matrix, I = [a ,j] (a) For element lying along Principal diagonal we have i = j (b) For ...
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Principal diagonal
Thoughts On Economics: Kalecki And Sraffa: Compatible?
Two Great Economists1. 0 Introduction Michal Kalecki set out macroeconomic models in which markup pricing was common.
robertvienneau.blogspot.com/2013/05/kalecki-and-sraffa-compatible.html
Matrices Triangulares
Una matriz triangular es un tipo de matriz cuadrada cuyos elementos por encima o por debajo de su diagonal principal son cero. En este video se explica este tipo especial de matrices En este video hablaremos sobre algunos tipos especiales de matrices. Recordemos que en videos anteriores habíamos definido la matriz identidad ( I_n) como la matriz en la cual la diagonal principal de la matriz estaba compuesta por valores de 1 y el resto de la matriz, es decir los elementos que no formaban parte de la diagonal principal adquirían valores de 0. Para dar una definición más formal del concepto de diagonal principal de una matriz, decimos que si tenemos una matriz A con varias entradas, la diagonal principal está conformada por el siguiente conjunto de valores: diagA={a11,a22,a33,….ann}, a partir de la definición de diagonal principal de una matriz surgen algunas otras definiciones, se dice que todo lo que está arriba de la diagonal principal de la matriz es conocido como parte triangular superior y que todo lo que está por debajo de la diagonal principal de la matriz se le conoce como parte triangular inferior. Basados en esta definición, podemos decir que si nos dan una matriz cuadrada A y la diagonal de esta matriz se define como diagA={1,1,1,..1}, es decir, todos los elementos de la diagonal valen1, y que además, todos los elementos aij =0 para i≠j, nuestra matriz A se trata de la matriz identidad A= I_n. Otras definiciones son: Decimos que tenemos una matriz diagonal si aij=0 para i≠j sin importar el valor que tomen los elementos de la diagonal principal de la matriz. Una vez que hemos definido el concepto de matriz diagonal, definiremos que es una matriz triangular superior e inferior, si tenemos una matriz cuadrada A decimos que tenemos una matriz A triangular superior si: aij=0 siempre y cuando i>j y decimos que A es una matriz triangular inferior si: aij=0 siempre y cuando i
aula.tareasplus.com/Roberto-Cuartas/ALGEBRA-LINEAL/Matrices-Triangulares
Matrices in Math; the Matrix and How it Works | SchoolTutoring.com
A matrix is a rectangular arrangement of “mn” elements with ‘m’ rows and ‘n’ columns enclosed within brackets.
www.schooltutoring.com/help/understanding-matrices-in-math/
Learn History of Matrices | Need Math Help
Learn History of Matrices Introduction:- The history of matrices goes back to ancient times, But the term “Matrix” was not applied to the concept until 1850. “Matrix” is the Latin word for womb, and it retains that sense in English.
needmathhelp.wordpress.com/2013/04/05/learn-history-of-matrices/
whats the physical significance of the off-diagonal element in the matrix of moment of inertia
In classical mechanics about rotation of rigid object, the general problem is to study the rotation on a given axis so we need to figure out the moment of inertia around some axes..
www.rqgg.net/topic/mpsru-whats-the-physical-significance-of-the-off-diagonal-element-in-the-matrix-of-moment-of-inertia.html
123 | 2014-09-01T11:30:44 | {
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https://math.stackexchange.com/questions/1051662/recurrence-relation-rabbit-population | # Recurrence relation rabbit population
A young pair of rabbits (one of each sex) is placed on an island.
A pair of rabbits does not breed until they are 2 months old.
After they are 2 months old, each pair of rabbits produces another pair each month.
And the pairs leave the island after reproducing twice.
Find a recurrence relation for the number of pairs of rabbits on the island after n months.
Let $a_n$ be the number of pairs of rabbits on the island after $n$ months.
Answer in the book: $a_n=a_{n-2}+a_{n-3}$
I counted the population for the first nine months $a_0=1, a_1=1, a_2=2, a_3=2, a_4=3, a_5=4, a_6=5, a_7=7, a_8=9, a_9=12$ and see that the recurrence relation in the book satisfies these values.
But I can't understand the solution. Can someone please explain the logic?
• Try counting for the $n$th month using exactly the same method you used for, say, the 9th month. – WillO Dec 4 '14 at 18:36
• What book is that? – Laura Merchán Feb 23 at 19:59
## 2 Answers
I think you have to keep track of number of rabbits by age:
Let $x_i, y_i, z_i$ be the number of pairs of newborn, pairs of one-month old, and pairs of two-month old rabbits respectively after $i$ months. Pairs in category $y_i$ and $z_i$ produce new pairs (counted in $x_{i+1}$). After being counted in $z_i$, a pair leaves the island and is not counted in the next iteration.
Let $\vec{v_i}=\begin{pmatrix} x_i \\y_i\\z_i \end{pmatrix}$
Initial conditions are
$\vec{v_0}=\begin{pmatrix} 1 \\0\\0 \end{pmatrix}$.
For $n\ge 1$, we have
$x_n=y_{n-1}+z_{n-1}$;
$y_n=x_{n-1}$, and
$z_n=y_{n-1}$
Note that this gives $\vec{v_1}=\begin{pmatrix} 0 \\1\\0 \end{pmatrix}$ and $\vec{v_2}=\begin{pmatrix} 1 \\0\\1 \end{pmatrix}$.
Then the total number of rabbits after $n$ months, for $n\ge 3$ is given by: $$\begin{array}{ccccccc}a_n&=&x_n&+&y_n&+&z_n\\&=&(y_{n-1}+z_{n-1})&+&x_{n-1}&+&y_{n-1}\\ &=&(x_{n-2}+y_{n-2})&+&(y_{n-2}+z_{n-2})&+&x_{n-2}\\ &=&(x_{n-2}+y_{n-2})&+&(x_{n-3}+z_{n-2})&+&(y_{n-3}+z_{n-3})\\&=& a_{n-2}+a_{n-3} \end{array}$$
Let $b_n$ be the number of actively breeding rabbit pairs on the island at month $n$. It's easy to see that this sequence satisfies $$\tag{*} b_n = b_{n-2}+b_{n-3}$$
However, what was asked is the total number of rabbit pairs, whether they're currently breeding or not. Since every rabbit stay on the island for exactly three months, the current population equals the number of births in the previous three months, summed. So $$a_n = b_{n-1} + b_{n-2} + b_{n-3}$$ .. or something like that, depending on whether we count newborn litters and/or rabbits that are just about to depart.
But this slight uncertainty doesn't really matter because in any case $(a_n)_n$ is a linear combination of shifted versions if the $(b_n)_n$ sequence. And because the recurrence $\text{(*)}$ is linear, this means that the $a_n$s will satisfy the same recurrence.
Thus to find the solution according to your favorite criteria for counting rabbits, you can simulate the first handful of months precisely, keeping track of how many rabbits of each age there are, and summing the ones you decide count. From that point onwards, you can just run the total population through the recurrence.
(In order to be sure to be clear of effect of the "artificial" placement of rabbits on the island on month 0, I would simulate by hand for 6 months -- 3 because that's the length of the recurrence of $b_n$ and 3 more to account for the maximum shift of $b$s that contribute to $a$. Careful thinking will probably be able to show that doing fewer cycles by hand is enough, but I would find it quicker just to do a few more cycles by hand than convincing myself that a shortcut analysis is actually correct).
• @paw88789 has given the answer as per your guidelines. – Sakthi Dec 4 '14 at 19:04 | 2020-02-28T16:26:53 | {
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http://math.stackexchange.com/questions/155599/mn-mn-proof?answertab=oldest | # $m!n! < (m+n)!$ Proof?
Prove that if $m$ and $n$ are positive integers then $m!n! < (m+n)!$
Given hint:
$m!= 1\times 2\times 3\times\cdots\times m$ and $1<m+1, 2<m+2, \ldots , n<m+n$
It looks simple but I'm baffled and can't reach the correct proof. Thank you.
-
Notice that $m!n!$ and $(m+n)!$ both have the same number of terms. Let's compare them: $$m!n! = (1 \times 2 \times \ldots \times m) \times (1 \times 2 \times \ldots \times n)$$ $$(m+n)! = (1 \times 2 \times \ldots \times m) \times ((m+1) \times (m+2) \times \ldots \times (m+n))$$
Both expressions have the same first $m$ terms, but after that each term in the second expression is bigger than the corresponding term in the first: $m+1 > 1$, etc.
-
Thanks a lot! This was very clear :) – MinaHany Jun 8 '12 at 12:50
Note that $\frac{(m+n)!}{m!}=(m+1)(m+2)\ldots(m+n)$. So you need to prove $n!<(m+1)(m+2)\ldots(m+n)$, and your hint applies.
Alternative proof: if you have $m$ girls and $n$ boys, you can line them up in $m!n!$ ways such that all girls come before all boys, and in $(m+n)!$ ways without that restriction.
-
Theophile's answer and yours completed the picture in my head! Thank you :) – MinaHany Jun 8 '12 at 12:51
+1 Your observation about boys and girls is just what I would have said if you hadn't said it first. – Michael Hardy Jun 8 '12 at 12:56
I think you mean $(m+n)!/m!$ and $n! < ...$ on the second line? – UncleZeiv Jun 8 '12 at 13:30
@UncleZeiv: Yep, and the final $n$'s in the products needed to be $m+n$ as well. Now corrected. Thanks. – Marc van Leeuwen Jun 8 '12 at 13:34
Or said differently: $\frac{(m+n)!}{m! n!} = \binom{m+n}{m} >1$ because there is more than $1$ way to choose $m$ objects from $m+n$ when $m,n$ are positive. – Mike F Jun 9 '12 at 0:19
If you would like, here is an other proof:
I assume that $n,m\in\mathbb{N}_0$. From the hint that $$1<m+1, 2<m+2, \ldots, n<m+n$$ you can see that $n!<(m+n)!$.
Now we proof by induction that $m!n!<(m+n)!$. Take $m=1$, then we have $1!n!<(1+n)!$. Now assume that $m!n!<(m+n)!$ is correct, this is the induction hypothesis. We calculate \begin{align*} (m+1)!n!&=(m+1)m!n!\\&<(m+1)(m+n)!\\&<(m+n+1)(m+n)!\\&=(m+n+1)!. \end{align*} The first inequality is because of the induction hypothesis. The second because $n\geq1$.
-
Thanks! It is helpful to see the same proof in induction as well. – MinaHany Jun 8 '12 at 12:59
Some quite complicated answers here for showing that there is more than one string we can form with letters $A$ and $B$ of length $m+n$, with $m$ repetitions of $A$ and $n$ repetitions of $B.$
-
Hint $\ \:\rm f(n) > 1\,$ since it is a product of terms $>1,$ via multiplicative telescopy, viz.
$$\begin{eqnarray}\rm f(n)\, =\ f(0) \prod_{k\:=\:1}^{n}\, \frac{f(k)}{f(k-1)} &=&\rm \, \color{red}{\rlap{--}f(0)}\frac{\color{green}{\rlap{--}f(1)}}{\color{red}{\rlap{--}f(0)}}\frac{\color{royalblue}{\rlap{--}f(2)}}{\color{green}{\rlap{--}f(1)}}\frac{\phantom{\rlap{--}f(3)}}{\color{royalblue}{\rlap{--}f(2)}}\, \cdots\, \frac{\color{brown}{\rlap{--}f(n-1)}}{\phantom{\rlap{--}f(n-2)}}\frac{f(n)}{\color{brown}{\rlap{----}f(n-1)}}\\ &=&\rm\,\ \frac{1+m}1\ \frac{2+m}2\ \frac{3+m}3\,\cdots\, \frac{n+m}n \\ \rm \phantom{\frac{\frac{1}1}{\frac{1}1}}because\ the\ term\ ratio\ is\ \ \displaystyle\rm\ \frac{f(k)}{f(k-1)} &=&\rm\ \frac{(k+m)!}{k!\,m!}\, \frac{(k-1)!\,m!}{(k-1+m)!}\, =\, \frac{k+m}{k} \end{eqnarray}$$
This yields precisely the same proof as the accepted answer. However, by using the general method of telescopy, one is able to derive the proof algorithmically vs. ad-hoc. Such telescopic methods work much more generally. They are essential in more complex problems where ad-hoc methods have no hope due to the innate structure being obfuscated by the complexity. See here for more.
-
One-line proof (some details omitted): ${m+n \choose m} > 1$ if $0 < m < n$.
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http://math.stackexchange.com/questions/42075/limit-in-probability-is-almost-surely-unique | # limit in probability is almost surely unique?
I read this proposition in a book, which was not proved. And I cannot verify it myself. Could anyone help me out here?
If $$X_{n}\rightarrow X$$ in probability and $$X_{n}\rightarrow Y$$ almost surely, then $$P(X=Y)=1.$$
An alternative version is that the p-limit of a sequence is almost surely unique.
Cheers.
-
Convergence almost surely implies convergence in measure – user17762 May 30 '11 at 4:46
@newbie: I didn't see what your question had to do with stochastic integrals, so I removed that tag. If this was a mistake, feel free to add it back. – Mike Spivey May 30 '11 at 4:48
@ Sivaram:The question is to prove the limit is unique. – newbie May 30 '11 at 4:56
@ Mike Spivey: No problem. I put the tag because it serves as a tool for verifying the validity of stochastic integral. – newbie May 30 '11 at 5:00
Hi newbie! Unfortunately, adding a space after the @ sign leads to the users not being notified. Thus @Sivaram and Mike Spivey didn't see your comments. Moreover, only one user per comment gets notified, that's why I didn't add an @ before Mike's name. Concerning your mathematical question: Do you know the following fact: "if a sequence converges in probability then there is a subsequence converging almost surely" or are you asking how to prove that? – t.b. May 30 '11 at 5:10
Back to basics: Assume that $X_n\to X$ in probability and that $X_n\to Y$ in probability. Then, for every positive $x$, $P(|X_n-X|\geqslant x)+P(|X_n-Y|\geqslant x)$ converges to zero since both terms do. Now, $$[|X-Y|\ge 2x]\subseteq[|X_n-X|\geqslant x]\cup[|X_n-Y|\geqslant x],$$ hence, for every $n$, $$P(|X-Y|\geqslant2x)\leqslant P(|X_n-X|\geqslant x)+P(|X_n-Y|\geqslant x).$$ Considering the limit of the RHS when $n\to+\infty$, this proves that $P(|X-Y|\geqslant2x)=0$. This holds for every positive $x$ and $$[X\ne Y]=\bigcup_{k\geqslant1}[|X-Y|\geqslant k^{-1}],$$ hence $P(X\ne Y)=0$. This means that $X=Y$ almost surely.
Note: The hypothesis that $X_n\to X$ in probability and $X_n\to Y$ in probability, which we used above, is weaker than the hypothesis that $X_n\to X$ in probability and $X_n\to Y$ almost surely.
-
Of course, this is much nicer! Sometimes I really wonder what I'm thinking. Thanks for setting this straight. – t.b. May 30 '11 at 15:15
@Theo: Thanks. – Did May 30 '11 at 19:52
For the sake of having an answer:
We know the following fact:
If $X_n \to Y$ in probability then there is a subsequence $X_{n_k} \to Y$ almost surely.
So take such a subsequence. As $X_{n} \to X$ a.s. we also have $X_{n_k} \to X$ a.s. and thus $X = Y$ a.s. because the almost sure limit of a sequence is unique a.e. (if it exists).
This is just fleshing out your last comment a bit more formally.
-
See, for example, p. 150 in the book An Intermediate Course in Probability by Allan Gut (in particular, the proof of Theorem 2.1(ii) on p. 151).
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https://math.stackexchange.com/questions/581129/simple-logarithm-properties-proof | # Simple logarithm properties proof
I was reading a school algebra book about logarithm function (on $$\mathbb{R}^+$$). There were several properties without proof. So I decided to prove 2 of them myself.
The first property:
$$log_a(x\cdot y) = log_ax + log_ay$$
## Proof
By definition of logarithm we know, that: $$a^{log_ax} = x$$ (axiom 1)
So $$x\cdot y = a^{log_ax} \cdot a^{log_ay} = a^{log_ax\;+\;log_ay}$$
Therefore
$$log_a(x\cdot y) = log_a(a^{log_ax} \cdot a^{log_ay}) = log_a(a^{log_ax\;+\;log_ay}) = log_ax\;+\;log_ay$$
### question
I am not sure about this equality: $$log_a(a^{log_ax\;+\;log_ay}) = log_ax\;+\;log_ay$$, because we don't have an axiom, that $$\forall x\;log_aa^x = x$$, despite the fact, that we use it here. It must be a conclusion from the definition of logarithm:
$$log_ax$$ is a power to which we must take a to get x. If we want to know, to which power we must take a to get $$a^x$$ --- obviously, it's x.
This trivial axiom was missing in the book. Shouldn't we explicitly denote it?
The second property also aroused a question:
$$log_ax^n = n \cdot log_ax$$
## Proof
$$log_ax^n = log_a\prod_{i=1}^{n} x$$
According to the previously proven property
$$log_a(x\cdot y) = log_ax + log_ay$$
Therefore $$log_a\prod_{i=1}^{n} x = \sum_{i=1}^{n}log_ax = n \cdot log_ax$$
### question
Equality $$x^n = \prod_{i=1}^{n} x$$ is true only for $$n \in \mathbb{Z}^+$$, so formally speaking our proof is valid only for $$n \in \mathbb{Z}^+$$.
While throughout the book the equality $$log_ax^n = n \cdot log_ax$$ is considered to be true $$\forall n \in \mathbb{R}$$.
I don't understand, how to prove it.
• If you have $a^{\log_a x} = x$, then taking $\log_a$ of both sides gives $\log_a(a^{\log_a x}) = \log_a x$. – copper.hat Nov 25 '13 at 22:20
• @copper.hat Is it a contradiction? Please, give a more detailed explanation. – user4035 Nov 25 '13 at 22:23
• No contradiction, if you let $y = \log_a x$, you get $\log_a(a^y) = y$. – copper.hat Nov 25 '13 at 23:05
For the second, realize that: $$a^{n\cdot\log_a x}=(a^{\log_a x})^n=x^n=a^{\log_a x^n}\Longrightarrow n\cdot\log_a x=\log_a x^n, \forall n\in \mathbb{R},$$ since the property used is valid for real $n$ and the exponential function is injective. | 2021-06-19T22:46:04 | {
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https://math.stackexchange.com/questions/414928/getting-rid-of-the-set-builder-notation-in-the-expression-fx-mid-px | # Getting rid of the set builder notation in the expression $\{ f(x) \mid P(x) \} = \{ g(x)\mid Q(x) \}$
The set-builer notation is used to have
$$\{ x \mid P(x) \} = \{ x \mid Q(x) \}$$
denote
$$\forall x\ \big(P(x) \Leftrightarrow Q(x) \big).$$
And some people write
$$\{ x \in U \mid P(x) \} = \{ x \in U \mid Q(x) \}$$
to denote
$$\forall x\ \big( x \in U \Rightarrow (P(x) \Leftrightarrow Q(x)) \big).$$
Now I want to translate the frequently used
$$\{ f(x) \mid P(x) \} = \{ g(x) \mid Q(x) \}$$
where the elements are not $x$, but sets constructed from them via $f$ and $g$. My try is
$$\forall y\ \big(\exists x\ \big(y=f(x)\land P(x)) \Leftrightarrow \exists x'\ \big(y=g(x')\land Q(x')\big) \big)$$
but I'm e.g. not sure about the existential quantifiert, or if I really need two.
I figure a necessary condition for having found the right way would be to try $f=g=\text{id}$ and see the formula reducing to the previous expression. However, I fail at this, because I don't know how to unpack it. I think I'd have to use $y=x$ and $y=x'$, but I don't know how I'd formally get this statement out of the long formula, so that I can use it?
What is the right way to read $\{ f(x) \mid P(x) \} = \{ g(x) \mid Q(x) \}$ and how to show it reduces to $\forall x\ \big(P(x) \Leftrightarrow Q(x) \big)$ for $f=g=\text{id}$?
• I didn't read everything, but hopefully this will help: $\{f(x)|P(x)\}$ should be read as $$\{y|\bigl(\exists x\in \text{dom}(f)\bigr)(f(x)=y)\land P(x)\}.$$ Related. Jun 8, 2013 at 19:39
Your translation is correct. To unpack the specialization to $f=g=\text{id}$, use the fact (of logic) that $\exists x\,(y=x\land P(x))$ is equivalent to $P(y)$.
I want to translate
$$\{ f(x) \mid P(x) \} = \{ g(x) \mid Q(x) \}\tag1$$
My try is $$\forall y\ \big(\exists x\ \big(y=f(x)\land P(x)) \Leftrightarrow \exists x'\ \big(y=g(x')\land Q(x') \big) \big)\tag2$$
I obtained the same formalisation.
In Prenex form (verification): $$\begin{gather} ∀y\:∀a\:∀b\:∃c\:∃d\: \bigg( \Big(y{=}f(a)∧P(a) \Big) ∨ \Big(y{=}g(b)∧Q(b) \Big) \\ → y{=}f(c)∧y{=}g(d)∧P(c)∧Q(d) \bigg).\tag3\end{gather}$$
how to show it reduces to $$\forall x\ \big(P(x) \Leftrightarrow Q(x) \big)$$ for $$f=g=\text{id}$$?
Setting the $$f$$ and $$g$$ in $$(3)$$ as the identity function: \begin{align}&∀y\:∀a\:∀b\:∃c\:∃d\: \bigg( \Big(y=a∧P(a) \Big) ∨ \Big(y=b∧Q(b) \Big) \\ & → y=c∧y=d∧P(c)∧Q(d) \bigg)\\ \equiv{}&∀y\: \bigg( ∃a\:\Big(y=a∧P(a) \Big) ∨ ∃b\:\Big(y=b∧Q(b) \Big) \\ & → ∃c\:\Big(y=c∧P(c)\Big) ∧ ∃d\:\Big(y=d∧Q(d)\Big)\bigg)\\ \equiv{}&∀y\: \bigg( P(y) ∨ Q(y) → P(y) ∧ Q(y)\bigg) \\ \equiv{}&∀y\: \bigg( \big(\lnot P(y) ∧ \lnot Q(y) \big) ∨ \big( P(y)∧Q(y) \big) \bigg) \\ \equiv{}&∀y\: \Big( P(y) \leftrightarrow Q(y) \Big) \\ \equiv{}&∀x\: \Big( P(x) \leftrightarrow Q(x) \Big), \end{align} where the second logical equivalence appeals to Andreas's suggestion $$\exists x\,(y=x\land P(x)) \quad\equiv\quad \exists x\,(P(y)) \quad\equiv\quad P(y),$$ which when plugged into $$(2)$$ directly gives the same result.
• Responding to a 10 year old thread, huh. I clearly forgot about this, but you'll get the green checkmark now. Jun 25 at 11:05 | 2022-08-12T05:21:55 | {
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https://math.stackexchange.com/questions/2133697/a-remainder-and-division-problem | # A Remainder and Division Problem
I have a question related to remainder and denominator of a division.
abc12 is a five-digit number and xy is a two-digit number. And the division is given as below: (xy is remainder, 40 is denominator, abc12 is dividend.)
What is the sum of all possible values of xy two-digit numbers?
Thank you.
• I am not familiar with this layout of the division. Presumably one of $xy$ and $40$ is the divisor and one is the remainder. Which is it? Is it $40 \cdot k + xy=abc12$ or $xy \cdot k+40=abc12?$ where $k$ can have multiple digits. – Ross Millikan Feb 7 '17 at 18:55
• 40⋅k+xy=abc12 is correct. xy is remainder, 40 is denominator, abc12 is dividend. – layout789 Feb 7 '17 at 18:57
Note that $abc12$ and $40$ are both divisible by $4$, so $xy$ must be as well. If we call the quotient $k$ we have $40 \cdot k+xy=abc12$. The ones digit of $40 \cdot k$ must be zero, so $y=2$. We must also have $x \le 3$, so the only possibilities are $xy=12,32$ Now we need to find $abc$s to show that both of these are possible. We have $40012=40 \cdot `1000 +12$ and $40112=40 \cdot 1002+32$.
• Your solution does not explain why xy=22 is impossible. – layout789 Feb 7 '17 at 19:06
• Because it is not divisible by $4$. Similarly for $02$ if we would otherwise allow $x$ to be zero. – Ross Millikan Feb 7 '17 at 19:09
• @layout789 It is much clearer and easier using modular arithmetic - see my answer. – Bill Dubuque Feb 7 '17 at 19:56
$\dfrac{abc12}{40}=\dfrac{abc\times 100}{40}+\dfrac{12}{40}$
Therefore, $xy=20\left[\text{remainder of }\dfrac{5(abc)}{2}\right]+12$
When abc is even,
$xy= 20 \times 0 + 12 = 12$
When abc is odd,
$xy= 20 \times 1 + 12 = 32$
${\rm mod}\ 40\!:\ \underbrace{1000}_{\large \equiv\ 0}AB\!+\!\underbrace{100}_{\large\equiv\ \color{#0a0}{20}}C\!+\!12\equiv \color{#0a0}{20}\,\color{#c00}C\!+\!12\equiv \begin{cases} 20(\color{#c00}{2n\!+\!0})\!+\!12\equiv 12,\ \ {\rm for\ even}\ \ \color{#c00}C\\ 20(\color{#c00}{2n\!+\!1})\!+\!12\equiv 32,\ \ {\rm for\ \ odd}\ \ \color{#c00}C\end{cases}$ | 2021-05-11T17:36:03 | {
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https://stats.stackexchange.com/questions/22938/expected-number-of-coin-tosses-to-get-n-consecutive-given-m-consecutive | # Expected number of coin tosses to get N consecutive, given M consecutive
Interviewstreet had their second CodeSprint in January that included the question below. The programmatic answer is posted but doesn't include a statistical explanation.
(You can see the original problem and posted solution by signing in to the Interviewstreet website with Google creds and then going to the Coin Tosses problem from this page.)
Coin Tosses
You have an unbiased coin which you want to keep tossing until you get N consecutive heads. You've tossed the coin M times and surprisingly, all tosses resulted in heads.
What is the expected number of additional tosses needed until you get N consecutive heads?
Input:
The first line contains the number of cases T. Each of the next T lines contains two numbers N and M.
Output:
Output T lines containing the answer for the corresponding test case. Print the answer rounded to exactly 2 decimal places.
Sample Input:
4
2 0
2 1
3 3
3 2
Sample Output:
6.00
4.00
0.00
8.00
Sample Explanations:
If N = 2 and M = 0, you need to keep tossing the coin until you get 2 consecutive heads. It is not hard to show that on average, 6 coin tosses are needed.
If N = 2 and M = 1, you need 2 consecutive heads and have already have 1. You need to toss once more no matter what. In that first toss, if you get heads, you are done. Otherwise, you need to start over, as the consecutive counter resets, and you need to keep tossing the coin until you get N=2 consecutive heads. The expected number of coin tosses is thus 1 + (0.5 * 0 + 0.5 * 6) = 4.0 If N = 3 and M = 3, you already have got 3 heads, so you do not need any more tosses.
All of the mathematical equations I came up with had the right answers for the sample input data listed above, but was wrong for all of their other input sets (which are not known). Their programmatic solution appears to solve the problem far differently from my try-to-come-up-with-an-equation method. Can someone please explain how to come up with an equation that would solve this?
• See also here where also we find the $2^{N+1}-2^{M+1}$ result given by Daniel Johnson below. Feb 16, 2012 at 19:46
This is a computational exercise, so think recursively. The current state of coin flipping is determined by the ordered pair $(N,M)$ with $N\ge M\ge 0$. Let the expected number of flips to reach $N$ consecutive heads be $e(N,M)$:
(1) There is a 50% chance the next flip will be heads, taking you to the state $(N,M+1)$, and a 50% chance the next flip will be tails, taking you to the state $(N,0)$. This costs one flip. Therefore the expectation (recursively) is given by
$$e(N,M) = \frac{1}{2} e(N,M+1) + \frac{1}{2} e(N,0) + 1.$$
(2) Base conditions: you have already stipulated that
$$e(N,0) = 2^{N+1}-2$$
and obviously
$$e(N,N) = 0$$
(no more flips are needed).
Here's the corresponding Mathematica program (including caching of intermediate results to speed up the recursion, which effectively makes it a dynamic programming solution):
e[n_, m_] /; n > m > 0 := e[n, m] = 1 + (e[n, m + 1] + e[n, 0])/2 // Simplify
e[n_, 0] := 2^(n + 1) - 2
e[n_, n_] := 0
The program would look similar in other programming languages that support recursion. Mathematically, we can verify that $e(N,M) = 2^{N+1} - 2^{M+1}$ simply by checking the recursion, because it obviously holds for the base cases:
$$2^{N+1} - 2^{M+1} = 1 + (2^{N+1} - 2^{M+2} + 2^{N+1} - 2)/2,$$
which is true for any $M$ and $N$, QED.
More generally, the same approach will establish that $e(N,M) = \frac{p^{-N} - p^{-M}}{1-p}$ when the coin has probability $p$ of heads. The hard part is working out the base condition $e(N,0)$. That is done by chasing the recursion out $N$ steps until finally $e(N,0)$ is expressed in terms of itself and solving:
\eqalign{ e(N,0) &= 1 + p e(N,1) + (1-p) e(n,0) \\ &= 1 + p\left(1 + p e(N,2) + (1-p) e(N,0)\right) + (1-p) e(N,0) \\ \cdots \\ &= 1 + p + p^2 + \cdots + p^{N-1} + (1-p)[1 + p + \cdots + p^{N-1}]e(N,0);\\ e(N,0) &= \frac{1-p^N}{1-p} + (1-p^N)e(N,0); \\ e(N,0) &= \frac{p^{-N}-1}{1-p}. }
• Perhaps working iteratively rather than recursively might be better? You have $$e(N,M)=\frac{1}{2}e(N,M+1)+\frac{1}{2}e(N,0)+1$$ which gives $$e(N,M+1)=2e(N,M)-2^{N+1}$$ and so \begin{align*}e(N,1)&=2e(N,0)-2^{N+1}=2(2^{N+1}-2)-2^{N+1}=2^{N+1}-4\\e(N,2)&=2e(N,1)-2^{N+1}=2(2^{N+1}-4)-2^{N+1}=2^{N+1}-8\end{align*} and so on giving $$e(N,M)=2^{N+1}-2^{M+1}.$$ Or the difference equation could be solved "theoretically" instead of by iteration. Feb 16, 2012 at 22:59
• @Dilip The inferences you draw both at "which gives" and "and so on" are recursive. What solution method do you have in mind by "solved theoretically"?
– whuber
Mar 24, 2013 at 13:25
• To my mind, the difference between recursive and iterative is the method of working. For the relation $$x(n)=2x(n-1),~~ x(0)=1,$$ recursion calculates $x(n)$ as $$x(n)=2x(n-1)=2(2x(n-2))=2(2(2x(n-3)))=\cdots=2(2(\cdots 2x(0)\cdots)=2^n$$ while iteration says $$x(0)=1\Rightarrow x(1)=2x(0)=2\Rightarrow x(2)=2x(1)=2^2\Rightarrow \cdots x(n)=2x(n-1)=2\cdot 2^{n-1}=2^n.$$ "theoretically" meant solving a difference equation by finding the characteristic polynomial, determining its roots, and then fitting the general solution to the initial conditions, instead of simple calculation as above. Mar 24, 2013 at 14:51
• From a programming viewpoint, a program find_x(n) is recursive if it says something like if n=0 return 1 else return 2*find_x(n-1) in which find_x calls itself $n$ times, whereas an iterative program would say something like y = 1; while n > 0 do begin y=2*y; n=n-1 end; return y Mar 24, 2013 at 23:25
• If you look at how those programs are actually implemented on the computer, @Dilip, in many environments (such as R) they scarcely differ at all. (In one case you create and then process a vector 1:n and in the other you will discover that n:1 has been placed on the stack and processed in reverse.) But part of my point was conceptual: your initial comment spoke of "working iteratively." That referred to the analysis and not to any computer program. But these are trivial, tangential points whose discussion doesn't warrant extending this comment thread.
– whuber
Mar 24, 2013 at 23:43
To solve this problem, I will use stochastic processes, stopping times, and dynamic programming.
First, some definitions:
$$X_n \doteq \#\text{(of consecutive heads after the nth flip)}$$ We also allow a value for $X_0$ to mean the number of consecutive heads before we start. So, for $X_0 = 0$ and the sequence of flips HHTHHHTHTTHH, the corresponding values of $X$ are 120123010012. If we had $X_0 = M$, the values of X would be (M+1)(M+2)0123010012.
Then define the following stopping times: $$\tau_N \doteq \min\{k: X_k = N\} \text{ and } \tau_0 \doteq \min\{k > 1: X_k = 0\}$$
The value we are looking for is the expected value of $\tau_N$, the number of flips it takes to observe N consecutive flips $(X_{\tau_N} = N)$, given that we have already observed M consecutive flips $(X_0 = M)$. Assume $M \leq N$ as the answer is trivially 0 otherwise. We compute:
$$E[\tau_N|X_0 =M] = E[\tau_N\boldsymbol{1}_{\{\tau_N < \tau_0\}} + \tau_N\boldsymbol{1}_{\{\tau_N > \tau_0\}}|X_0 =M]$$ $$= (N-M)(\frac{1}{2})^{N-M} + E[\tau_0|\tau_N > \tau_0,X_0 =M] + (1 - (\frac{1}{2})^{N-M})E[\tau_N|X_0 = 0]$$ This leaves us to compute the last two conditional expectations.
The first corresponds to the expected number of flips before getting a tail assuming a tail is flipped before N consecutive heads are observed assuming we start with M consecutive heads. It is not too difficult to see that
$$E[\tau_0|\tau_N > \tau_0,X_0 =M] = \sum^{N-M}_{j=1}(j)(\frac{1}{2})^j = 2 - (N-M+2)(\frac{1}{2})^{N-M}$$
Now all we have to do is compute the second conditional expectation which corresponds to the expected number of flips it takes to observe N consecutive heads starting from 0. With a similar calculations, we see that
$$E[\tau_N|X_0 = 0] = E[\tau_N\boldsymbol{1}_{\{\tau_N < \tau_0\}} + \tau_N\boldsymbol{1}_{\{\tau_N > \tau_0\}}|X_0 =0]$$ $$= N(\frac{1}{2})^{N} + E[\tau_0|\tau_N > \tau_0,X_0 =0] + (1 - (\frac{1}{2})^N)E[\tau_N|X_0 = 0]$$ $$= 2^N\lbrace N(\frac{1}{2})^{N} + (2 - (N+2)(\frac{1}{2})^{N})\rbrace$$ $$= 2^{N+1} - 2$$
This gives a final answer of:
$$E[\tau_N|X_0 =M] = (N-M)(\frac{1}{2})^{N-M} + 2 - (N-M+2)(\frac{1}{2})^{N-M} + (1 - (\frac{1}{2})^{N-M})(2^{N+1} - 2)$$ $$= 2^{N+1}-2^{M+1}$$
This agrees with the four test cases you've listed. With such a simple answer, there may be an easier way to compute this.
• This is a harder way to solve it than the recursive idea listed above, but it's extremely useful to see both approaches laid out together. Most people don't appreciate the way that stopping time methods can be used for small, practical problems too.
– ely
Feb 16, 2012 at 23:53
Warning: the following may not be considered as a proper answer in that it does not provide a closed form solution to the question, esp. when compared with the previous answers. I however found the approach sufficiently interesting to work out the conditional distribution.
Consider the preliminary question of getting a sequence of $N$ heads out of $k$ throws, with probability $1-p(N,k)$. This is given by the recurrence formula $$p(N,k) = \begin{cases} 1 &\text{if } k<N\\ \sum_{m=1}^{N} \frac{1}{2^m}p(N,k-m) &\text{else}\\ \end{cases}$$ Indeed, my reasoning is that no consecutive $N$ heads out of $k$ draws can be decomposed according to the first occurrence of a tail out of the first $N$ throws. Conditioning on whether this first tail occurs at the first, second, ..., $N$th draw leads to this recurrence relation.
Next, the probability of getting the first consecutive N heads in $m\ge N$ throws is $$q(N,m) =\begin{cases} \dfrac{1}{2^N} &\text{if }m=N\ p(N,m-N-1) \dfrac{1}{2^{N+1}} &\text{if } N<m<2N+1 \end{cases} $$ The first case is self-explanatory. the second case corresponds to a tail occuring at the $m-N-1$th draw, followed by $N$ heads, and the last case prohibits $N$ consecutive heads prior to the $m-N-1$th draw. (The two last cases could be condensed into one, granted!)
Now, the probability to get $M$ heads first and the first consecutive $N$ heads in exactly $m\ge N$ throws (and no less) is $$r(M,N,m) = \begin{cases} 1/2^N &\text{if }m=N\ 0 &\text{if } N<m\le N+M\\ \dfrac{1}{2^{M}}\sum_{r=M+1}^{N}\dfrac{1}{2^{r-M}}q(N,m-r)&\text{if } N+M<m \end{cases}$$ Hence the conditional probability of waiting $m$ steps to get $N$ consecutive heads given the first $M$ consecutive heads is $$s(M,N,m) = \begin{cases} 1/{2^{N-M}} &\text{if }m=N\ 0 &\text{if } N \sum_{r=M+1}^{N}\dfrac{q(N,m-r)}{2^{r-M}}&\text{if } N+M \end{cases}$$ The expected number of draws can then be derived by $$\mathfrak{E}(M,N)= \sum_{m=N}^\infty m\, s(M,N,m)$$ or $\mathfrak{E}(M,N)-M$ for the number of additional steps... | 2022-05-23T05:59:46 | {
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https://math.stackexchange.com/questions/4475634/for-all-n-k-in-positive-integers-there-exists-m-1-m-k-such-that-the-fol | # For all $n,k$ in positive integers, there exists $m_1,...,m_k$ such that the following holds
Let $$n,k$$ be two positive integers. Prove that there exists $$m_1,...,m_k$$ in the positive integers such that
$$1+\frac{2^k-1}{n}=\prod_{i=1}^k\left(1+\frac{1}{m_i}\right)$$ Here is my attempt
We're going to construct a solution for the $$m_i$$'s, but rather than presenting it right away I will take time to go through the motivation behind it.
Assume that $$\exists m_1,...,m_k$$ such that $$1+\frac{2^k-1}{n}=\prod_{i=1}^k\left(1+\frac{1}{m_i}\right)$$ And consider \begin{align} \prod_{i=1}^{k+1}\left(1+\frac{1}{m_i}\right)&=\left(1+\frac{1}{m_{k+1}}\right)\prod_{i=1}^k\left(1+\frac{1}{m_i}\right) \\ &= \left(1+\frac{1}{m_{k+1}}\right)\left(1+\frac{2^k-1}{n}\right) \\ &= 1+\frac{2^{k+1}-1}{n} \end{align} Where the last step follows from the assumption. Hence $$\frac{1}{m_{k+1}}=\frac{2^k}{2^k+n-1} \text{ or } m_i=\frac{2^{i-1}+n-1}{2^{i-1}}$$ Now we have to prove this solution actually works. In other words, $$1+\frac{2^k-1}{n}=\prod_{i=1}^k\left(1+\frac{2^{i-1}}{2^{i-1}+n-1}\right)=\prod_{i=1}^k\left(\frac{2^{i}+n-1}{2^{i-1}+n-1}\right)$$ but expanding the RHS is not easy so let's go ahead and assume that the above is true $$\forall k\le s$$ we'll show that it's true for $$k=s+1$$
\begin{align}\prod_{i=1}^{s+1}\left(\frac{2^{i}+n-1}{2^{i-1}+n-1}\right)&=\frac{2^{s+1}+n-1}{2^{s}+n-1}\prod_{i=1}^{s}\left(\frac{2^{i}+n-1}{2^{i-1}+n-1}\right) \\&= \frac{2^{s+1}+n-1}{2^{s}+n-1}\left(1+\frac{2^s-1}{n}\right) \\&= 1+\frac{2^{s+1}-1}{n}\end{align}
Now my question is: Are the $$m_i$$'s integers? Since $$m_i=\frac{2^{i-1}+n-1}{2^{i-1}}$$ Then we should have $$2^{i-1}\mid n-1$$ but that's not true $$\forall n,i$$?
The $$m_i$$'s in your construction are not necessarily integers.
To illustrate the problem, consider the simple case of $$k = 2$$. You want to write $$\frac{n + 3}n$$ as a product $$\frac{m_1 + 1}{m_1} \cdot \frac{m_2 + 1}{m_2}$$.
Your strategy goes for $$m_1 = n$$ and $$m_2 = \frac{n + 1}2$$. However we see that $$\frac{n + 1}2$$ is not necessarily an integer.
In fact, the construction above works when $$n$$ is odd. For $$n$$ even, we can use instead $$m_1 = \frac n 2$$ and $$m_2 = n + 2$$.
This indicates how the solution would look like: you should separate the two cases of even/odd $$n$$.
We prove by induction on $$k$$. For $$k = 1$$, simply choosing $$m_1 = n$$ works. Now assume that the result is true for $$k$$ and we prove it for $$k + 1$$.
Thus we want to write $$\frac{n + 2^{k + 1} - 1}n$$ as a product $$\prod_{i = 1}^{k + 1} \frac{m_i + 1}{m_i}$$. As above, we will consider the parity of $$n$$.
If $$n$$ is odd, then we can write $$\frac{\frac{n + 1}2 + 2^k - 1}{\frac{n + 1}2} = \prod_{i = 1}^k \frac{m_i + 1}{m_i}$$ by induction hypothesis applied to $$\frac{n + 1}2$$. Choosing $$m_{k + 1} = n$$ gives us the willing identity.
If $$n$$ is even then we can write $$\frac{\frac n 2 + 2^k - 1}{\frac n 2} = \prod_{i = 1}^k \frac{m_i + 1}{m_i}$$ by induction hypothesis applied to $$\frac n 2$$. Choosing $$m_{k + 1} = n + 2^{k + 1} - 2$$ gives us the willing identity.
This finishes the induction step and hence the proof.
• I didn't understand is $m_i=n+2^i-2$ for all even $n$ and $m_i=n$ for all odd $n$?
– PNT
Jun 19 at 20:51
• It's an inductive construction which would be complicated to write down explicitly (probably would involve binary expansion of $n$). Do you understand proof by induction in general? Jun 19 at 20:53
• I do but you're inducting on $k$ not $n$ so how can you say that you're applying the induction hypothesis on $n+1/2$?
– PNT
Jun 19 at 21:16
• The statement for step $k$ is valid for any positive integer $n$, thus I apply it with $n$ replaced by $\frac{n + 1}2$ which is again a positive integer. Jun 19 at 21:23
• +1. In particular, if $k=2$ then $(3m_1-n)(3m_2-n)=n(n+1).$ So if $k=2=n$ then $(3m_1-2)(3m_2-2)=6$, which is impossible for integers $m_1,m_2$ as it would imply $4\equiv 6\mod 3.$ Jun 20 at 3:53
After thinking for some days, I would like to share my observations from the first answer and how it relates to binary notations and bit operations. This answer is based on the same recursion step and the same algorithm.
$$\frac{n_{k+1}+2^{k+1}-1}{n_{k+1}} = \frac{n_k+2^{k}-1}{n_{k}}\cdot \frac{m_{k+1}+1}{m_{k+1}} \tag 1$$
where $$n_k$$ is chosen depending on whether $$n_{k+1}$$ is even or odd,
\begin{align*} n_{k+1} \mapsto n_k &= \begin{cases} \dfrac {n_{k+1}}2,& n_{k+1}\equiv 0 \pmod 2\\ \dfrac {n_{k+1}+1}2,& n_{k+1}\equiv 1 \pmod 2\\ \end{cases}\\ &= \left\lfloor\frac{n_{k+1}+1}2\right\rfloor = \left\lceil\frac{n_{k+1}}2\right\rceil \tag2 \end{align*}
From $$n_{k+1}+2^{k+1}-1$$ to $$n_k+2^k-1$$ in the numerators of $$(1)$$ is actually simple: it's a floored division by $$2$$, or a right shift by $$1$$ bit:
\begin{align*} n_k+2^k-1 &= \left\lfloor\frac{n_{k+1}+1}2\right\rfloor + 2^k-1\\ &= \left\lfloor\frac{n_{k+1}+1}2 +2^k-1\right\rfloor\\ &= \left\lfloor\frac{n_{k+1}+2^{k+1}-1}2\right\rfloor \end{align*}
But from $$n_{k+1}$$ to $$n_k$$ in the denominators is less obvious. Applying $$(2)$$ is not simply a floored division nor a right shift by $$1$$ bit. Repeated application of $$(2)$$ to a positive $$n$$ will also never get to $$0$$. But there's still some pattern, by noting the parity of $$n \pmod 2$$ at each step, for example when applying to $$n=10$$,
$$\underbrace{10}_{\equiv 0} \mapsto \underbrace{5}_{\equiv1} \mapsto \underbrace{3}_{\equiv1}\mapsto \underbrace{2}_{\equiv0}\mapsto \underbrace{1}_{\equiv 1} \mapsto \underbrace{1}_{\equiv 1} \mapsto \ldots$$
Writing the remainders as bits from the least to the most significant, this gives an infinite binary string $$\ldots1111\ 0110_2$$. And one might notice this is the two's complement notation of $$-n = -10_{10}$$! In fact, considering how $$-n_{k+1}$$ and $$-n_k$$ are related by $$(2)$$,
$$-n_k = -\left\lceil\frac{n_{k+1}}2\right\rceil = \left\lfloor\frac{-n_{k+1}}2\right\rfloor$$
so applying arithmetic right shift by $$1$$ bit to $$-n_{k+1}$$ does give $$-n_k$$.
Hence both the numerators and denominators in $$(1)$$ are related by binary right shifts or floored divisions:
$$-\frac{n_{k+1}+2^{k+1}-1}{-n_{k+1}} \overset{>>1}{\underset{>>1}\longmapsto} -\frac{n_k+2^k-1}{-n_k}$$
On the left hand side, if both the numerators and denominators were even, i.e. with last bits $$0$$, then the two sides would be equal. But the numerators and denominators always have opposite parity, and the $$\frac{1+m_{k+1}}{m_{k+1}}$$ multiplier is here to round the odd one down to even.
If $$n_{k+1}$$ is even, then the numerator is odd, so choose $$m_{k+1}$$ to reduce the numerator by $$1$$:
\begin{align*} -\frac{n_{k+1}+2^{k+1}-1}{-n_{k+1}} &= -\frac{n_{k+1}+2^{k+1}-2}{-n_{k+1}}\cdot \frac{n_{k+1}+2^{k+1}-1}{n_{k+1}+2^{k+1}-2}\\ &= -\frac{n_k+2^k-1}{-n_k}\left(1+\frac{1}{n_{k+1}+2^{k+1}-2}\right) \end{align*}
If $$n_{k+1}$$ is odd, then the denominator is odd, so choose $$m_{k+1}$$ to reduce the denominator by $$1$$ (to be more negative):
\begin{align*} -\frac{n_{k+1}+2^{k+1}-1}{-n_{k+1}} &= -\frac{n_{k+1}+2^{k+1}-1}{-n_{k+1}-1}\cdot \frac{-n_{k+1}-1}{-n_{k+1}}\\ &= -\frac{n_k+2^k-1}{-n_k}\left(1+\frac{1}{n_{k+1}}\right) \end{align*}
This explains why the $$m_{k+1}$$ appears to be quite different depending on the parity of $$n_{k+1}$$:
$$m_{k+1} = \begin{cases} n_{k+1}+2^{k+1}-2,& n_{k+1}\equiv 0 \pmod 2\\ n_{k+1},& n_{k+1}\equiv 1 \pmod 2\\ \end{cases} \tag3$$
To conclude, by considering the recursion step $$(1)$$ with negated denominators, the choices of $$n_k$$ and $$m_{k+1}$$ are related to binary notations and bit shifts. Running the recursion down to $$k=0$$ gives the last $$m_1=n_1$$ and a fraction that can be eliminated: $$\dfrac{n_0+2^0-1}{n_0} = 1$$.
Omitted detail on two's complement, which is non-standard:
$$-10_{10} \overset{???}= \ldots1111\ 0110_2 = 2+4+16+32+64+\cdots$$
Either all mentions of bit shifts can be ignored and consider floored division only.
Or a fixed and finite number $$L$$ can be chosen to be the number of bits for two's complement. $$L$$ bits should be long enough to represent the numerator $$n+2^k-1$$ (and hence also the denominator $$\pm n$$). | 2022-06-26T12:22:58 | {
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https://stats.stackexchange.com/questions/119852/understanding-o-p/422884 | # Understanding $O_p$
One thing I feel like I have never mastered is the concept of $O_p$ convergence and how to use it. I understand the basic idea and what bounded in probability means, but I always have a hard time understanding how to apply it myself. I have an example which I hope someone can explain to me.
Actually, I have two related examples. If we let $\epsilon_t$ be $N(0, I)$, consider: $$A=\frac{1}{T}\sum_{t=1}^T\epsilon_t\sum_{i=1}^{t-1}\epsilon_i'\\ B=\frac{1}{T}\sum_{t=1}^T\sum_{i=1}^{t-1}\epsilon_i\sum_{j=1}^{t-1}\epsilon_j'$$ I know that $A\in O_p(1)$ and that $B\in O_p(T)$, but I want to fully understand why. Can anyone explain why they are of these orders? Preferably in an intuitive rather than a rigorous way.
As for my own thoughts, for $A$, the way I'm thinking is that $\epsilon_t\sum_{i=1}^{t-1}\epsilon_i'$ is $O_p(1)$, so essentially what we have is $T$ $O_p(1)$ terms. Since we also divide by $T$, what remains is $O_p(1)$.
For $B$, I'm not sure how I can see that this is true. One idea is that we have two terms (the two $i$ and $j$ sums) that increase linearly with $T$, so thus we get $O_p(T^2)$, which is divided by $T$ to give $O_p(T)$. But is it true in general that $O_p(T)O_p(T)=O_p(T^2)$?
Edit: For my last question, if it is true in general that $O_p(T)O_p(T)=O_p(T^2)$, I think it is. If $X_T\in O_p(T)$, then that means $X_T=Y_T\times T$, where $Y_T\in O_p(1)$. So then $$O_p(T)O_p(T)=TO_p(1)TO_p(1)=T^2O_p(1)=O_p(T^2).$$
• It seems like a generic math question about stochastically bounded sequences, rather than something specific to statistics. – Fraijo Oct 13 '14 at 20:32
• @Fraijo I think it's right in the middle, making it not perfectly suitable for any site. This comes from a book on cointegration and the derivation of a test, so I think it's relevant here. – hejseb Oct 14 '14 at 5:31
To get the technicalities over with $X_n = \mathcal{O}_p(a_n)=\text{Pr}\left(X_n/a_n > M\right) < \delta$ for every $\delta > 0$. I'll also assume that the sequence is i.i.d.
Now the way I approach both of these problems is graphically. The first problem is really an increasing triangular array.
$\begin{array}{lllll} & 1 & 2 & \dots & T\\ 1 & \epsilon_1\epsilon_1 & & & \\ 2 & \epsilon_2\epsilon_1 & \epsilon_2 \epsilon_2 & & \\ \vdots & & & \ddots & \\ T & \epsilon_T\epsilon_1 & \epsilon_T\epsilon_2 & \dots & \epsilon_T\epsilon_T \end{array}$
The expectation of this array is simply
$\begin{array}{lllll} & 1 & 2 & \dots & T\\ 1 & \sigma_{11} & & & \\ 2 & 0 & \sigma_{22} & & \\ \vdots & & & \ddots & \\ T & 0 & 0 & \dots & \sigma_{TT} \end{array}$
if each of the $\sigma_{ii}$ are bounded then the sum of the expectations divided by $T$ are bounded by the max $\frac{T \sigma_{ii}}{T}$ times some finite constant that can be adjusted to meet any desired $\delta > 0$.
The second problem is quite similar, except for each $i$ you are adding triangular arrays of size $i$. This means that you have $T(T+1)/2$ non-zero expectations. For $\mathcal{O}_p$ like other big-O notation only the highest order polynomial matters, so the sum of $B$ is bounded by $T$.
If this is a bit confusing, try to sit down with pen and paper or excel and work out a simple example with T = 3. This should help a lot with your understanding.
For your Edit, recall that for the last statement to be true $\int_{-\infty}^{M_1T} dX_n \int_{-\infty}^{M_2T} dY_n = \int_{-\infty}^{M_3T^2} d(X_n Y_n)$ for some integers $M_1$, $M_2$, and $M_3$.
This statement being true depends on how you define $Y_n$ and $X_n$, if they are independent independent sequences e.g. any element of $X_n$ is independent of every other element of $X_n$ and $Y_n$, then the probability distribution can be broken down into the product of the marginal distributions and you are good-to-go.
If you drop normality though things also start falling apart a $T$ distribution with 2 degrees of freedom would be bounded in probability, but it's product would not be.
• Thanks, I think this makes it a little clearer. Some questions remain, though. In my $A$ example, the second sum only goes to $t-1$. So then the expectation is $0$, since $E(\epsilon_t\epsilon_s')=0$ for $t>s$. So in your graphical illustration (an approach I enjoy!) we would have what is below the main diagonal, not including what is on it. Right? For the $B$ example we do indeed get $T(T+1)/2$ nonzero expectations. But what allows us to only consider the expectations in the first place? – hejseb Oct 14 '14 at 15:18
If $$A_T \in \mathcal{O}_p(1)$$, that means $$\{A_T/1\}$$ is uniformly tight, or bounded in probability. Notice that I added a $$T$$ subscript. That's because this is a property for a family of random variables. The family in our cases is $$A_2, A_3, \ldots$$. This property is saying that all of these random variables are "pretty much" bounded above and below by the same number. By "pretty much" I mean that they are bounded probabilistically, meaning, the probability of breaking out of these bounds is negligible. We can increase that bound in order to make the probability smaller than whatever we want.
Notice I'm changing the bounds of the outer sum, here. $$t$$ needs to start at $$2$$, or else the inner sum doesn't make any sense. Also, I removed the "prime" symbol from the inner $$\epsilon$$s, because it is unnecessary.
The good news is that you don't need triangular arrays to show the first one. Let $$\epsilon > 0$$, then we can show there exists an $$M > 0$$ such that $$P(|A_T| \ge M) < \epsilon$$ for all $$T$$. Just pick $$M > \epsilon^{-2}$$ and notice that
\begin{align*} P(|A_T| \ge M) &= P\left(\left|\frac{1}{T}\sum_{t=2}^T\epsilon_t\sum_{i=1}^{t-1}\epsilon_i\right| \ge M \right) \\ &\le M^{-2}T^{-2}\mathbb{V}\left[\left|\sum_{t=2}^T\epsilon_t\sum_{i=1}^{t-1}\epsilon_i\right| \right] \tag{Chebyshev's} \\ &\le M^{-2}T^{-2}\mathbb{V}\left[\sum_{t=2}^T\epsilon_t\sum_{i=1}^{t-1}\epsilon_i \right] \tag{*} \\ &= M^{-2}T^{-2}\left\{\sum_{t=2}^T \mathbb{V}\left[\epsilon_t\sum_{i=1}^{t-1}\epsilon_i\right] + 2\sum_{t=2}^{T-1}\sum_{s=t+1}^T \mathbb{Cov}\left[\epsilon_t\sum_{i=1}^{t-1}\epsilon_i ,\epsilon_s\sum_{j=1}^{s-1}\epsilon_j \right] \right\} \tag{bilinearity} \\ &= M^{-2}T^{-2}\left\{\sum_{t=2}^T \mathbb{E}\left[\left(\sum_{i=1}^{t-1}\epsilon_i\right)^2\right] + 2\sum_{t=2}^{T-1}\sum_{s=t+1}^T \mathbb{Cov}\left[\epsilon_t\sum_{i=1}^{t-1}\epsilon_i ,\epsilon_s\sum_{j=1}^{s-1}\epsilon_j \right] \right\} \tag{zero mean}\\ &= M^{-2}T^{-2}\left\{\sum_{t=2}^T (t-1) + 2\sum_{t=2}^{T-1}\sum_{s=t+1}^T \mathbb{Cov}\left[\epsilon_t\sum_{i=1}^{t-1}\epsilon_i ,\epsilon_s\sum_{j=1}^{s-1}\epsilon_j \right] \right\} \tag{normality} \\ &= M^{-2}T^{-2}\left\{(T^2+T)/2 -T + 2\sum_{t=2}^{T-1}\sum_{s=t+1}^T \mathbb{Cov}\left[\epsilon_t\sum_{i=1}^{t-1}\epsilon_i ,\epsilon_s\sum_{j=1}^{s-1}\epsilon_j \right] \right\} \tag{algebra} \\ &= M^{-2}T^{-2}\left\{(T^2+T)/2 -T + 2\sum_{t=2}^{T-1}\sum_{s=t+1}^T \mathbb{Cov}\left[\sum_{i=1}^{t-1}\epsilon_t\epsilon_i , \left\{\sum_{j=1}^{t-1}\epsilon_s\epsilon_j + \epsilon_s\epsilon_t\right\} \right] \right\}\\ &= M^{-2}T^{-2}\left\{(T^2+T)/2 -T \right\} \\ &\le M^{-2} \end{align*}
For the line with (*), I'm using the fact that $$\mathbb{V}[|X|] \le \mathbb{V}[X] + (\mathbb{E}[X])^2$$. For the line above that, you can use bilinearity, recognize where the means are zero, and then independence.
Regarding $$B=\frac{1}{T}\sum_{t=2}^T\sum_{i=1}^{t-1}\epsilon_i\sum_{j=1}^{t-1}\epsilon_j' = \frac{1}{T}\sum_{t=2}^T (t-1) z_t z_t'$$ you can use Chebyshev's on that again. | 2021-06-13T23:20:58 | {
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https://mathhelpboards.com/threads/definite-integral.1042/ | # Definite Integral
#### bincybn
##### Member
Hii friends,
$$\displaystyle \int_{0}^{1}\left(1-x\right)^{N}*e^{x}\, dx$$
regards,
Bincy
#### themurgesh
##### New member
Hii friends,
$$\displaystyle \int_{0}^{1}\left(1-x\right)^{N}*e^{x}\, dx$$
regards,
Bincy
this is what i have:
for n=0, $$\displaystyle \int_{0}^{1}\left(1-x\right)^{0}*e^{x}\, dx = e-1$$
for n=1, $$\displaystyle \int_{0}^{1}\left(1-x\right)^{1}*e^{x}\, dx = e-2$$
for n=2, $$\displaystyle \int_{0}^{1}\left(1-x\right)^{2}*e^{x}\, dx = 2e-5$$
for n=3, $$\displaystyle \int_{0}^{1}\left(1-x\right)^{3}*e^{x}\, dx = 6e-16$$
for n=4, $$\displaystyle \int_{0}^{1}\left(1-x\right)^{4}*e^{x}\, dx = 24e-65$$
and so on..
so you have two sequnces..
1,1,2,6,24... is simply n! for n=0,1,2....
1,2,5,16,65,... is $$\displaystyle e \cdot \Gamma(n+1,1)$$ which is incomplete gamma function. For some reason I cannot paste the link here, please refer to the page mathworld.wolfram.com/BinomialSums.html
See equations 35 and 36
so you have:
$$\displaystyle \int_{0}^{1}\left(1-x\right)^{n}*e^{x}\, dx \;= \; [e\cdot n! - e \cdot \Gamma(n+1,1)] = e[\Gamma(n+1)-\Gamma(n+1,1)]$$
Last edited:
#### CaptainBlack
##### Well-known member
Hii friends,
$$\displaystyle \int_{0}^{1}\left(1-x\right)^{N}*e^{x}\, dx$$
regards,
Bincy
Let:
$$\displaystyle I_k=\int_0^1 (1-x)^k e^x \; dx$$
Then integration by parts gives: $$k I_{k-1}-I_k=1$$
CB
Last edited:
#### CaptainBlack
##### Well-known member
this is what i have:
for n=0, $$\displaystyle \int_{0}^{1}\left(1-x\right)^{0}*e^{x}\, dx = e-1$$
for n=1, $$\displaystyle \int_{0}^{1}\left(1-x\right)^{1}*e^{x}\, dx = e-2$$
for n=2, $$\displaystyle \int_{0}^{1}\left(1-x\right)^{2}*e^{x}\, dx = 2e-5$$
for n=3, $$\displaystyle \int_{0}^{1}\left(1-x\right)^{3}*e^{x}\, dx = 6e-16$$
for n=4, $$\displaystyle \int_{0}^{1}\left(1-x\right)^{4}*e^{x}\, dx = 24e-65$$
and so on..
so you have two sequnces..
1,1,2,6,24... is simply n! for n=0,1,2....
1,2,5,16,65,... is $$\displaystyle e \cdot \Gamma(n+1,1)$$ which is incomplete gamma function. For some reason I cannot paste the link here, please refer to the page mathworld.wolfram.com/BinomialSums.html
See equations 35 and 36
so you have:
$$\displaystyle \int_{0}^{1}\left(1-x\right)^{n}*e^{x}\, dx \;= \; [e\cdot n! - e \cdot \Gamma(n+1,1)] = e[\Gamma(n+1)-\Gamma(n+1,1)]$$
Incomplete induction, not mathematical induction!
CB
#### bincybn
##### Member
Let:
$$\displaystyle I_k=\int_0^1 (1-x)^k e^x \; dx$$
Then integration by parts gives: $$I_k+k I_{k-1}=1$$
CB
Integration by parts gives $$\displaystyle k*I_{k-1}-I_{k}=1$$
But how to solve this equation?
Is the ans $$\displaystyle \frac{k+1}{k-1}$$ ?
Bincy
Last edited:
#### CaptainBlack
##### Well-known member
Integration by parts gives $$\displaystyle k*I_{k-1}-I_{k}=1$$
But how to solve this equation?
Is the ans $$\displaystyle \frac{k+1}{k-1}$$ ?
Bincy
If themurgesh incomplete induction is correct, then if $$k$$ is a non negative integer mathematical induction using the recurence of themurgesh's solution should work (though I am unhappy about the appearance of in incomplete gamma functions, in that I would rather avoid them if possible).
The soliution without incomplete gamma functions, to which you can apply mathematical or complete induction is:
$I_k=e \times n!-n! \sum_{k=0}^n \frac{1}{k!}$
Your proposed answer cannot be right since is is wrong for all the cases where we know the answer.
CB
Last edited:
#### bincybn
##### Member
May I know what is wrong in my ans?
My ans. satisfies the recursive eqn.
I am also unhappy to include gamma function in my ans.
Oops. I didn't even bother abt the initial conditions.
Last edited:
#### CaptainBlack
##### Well-known member
May I know what is wrong in my ans?
My ans. satisfies the recursive eqn.
I am also unhappy to include gamma function in my ans.
Oops. I didn't even bother abt the initial conditions.
See my previous post, it has been edited to include the actual solution which you can prove is the solution by induction.
CB
#### bincybn
##### Member
$I_k=e \times n!+n! \sum_{k=0}^n \frac{1}{k!}$
CB
Thanks.. I got it. Instead of + it is -.
Last edited: | 2020-09-30T16:12:09 | {
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http://www.eandltoledolandscaping.com/pri2s/fe22b6-maximum-manhattan-distance | Note that, allowed values for the option method include one of: “euclidean”, “maximum”, “manhattan”, “canberra”, “binary”, “minkowski”. 24, Feb 16. To learn more, see our tips on writing great answers. Wikipedia between opening and closing of any spheres, line does not change, and if there is any free point there, it means, that you found it for distance r. Binary search contributes log k to complexity. Minkowski distance is a metric which tells us the distance between 2 points in space. p=2, the distance measure is the Euclidean measure. Is there another input for the target point? Fails if we have point (-10,0), (10,0), (0,-10), (0,10). Recommended: Please try your approach on {IDE} first, before moving on to the solution. You have to check if there is any point inside the square [0, k] X [0, k] which is at least given distance away from all points in given set. Related questions 0 votes. c happens to equal the maximum value in Northern Latitude (LAT_N in STATION). This is the maximum value we can get, and hence it is the maximum average distance for $2$ points. This can be calculate in O(n log n) using https://en.wikipedia.org/wiki/Fortune%27s_algorithm Why didn't the Romulans retreat in DS9 episode "The Die Is Cast"? [Java/C++/Python] Maximum Manhattan Distance. Input Format The STATION table is described as follows: STATION The changeIdx 28/8 = 3.5, rank1=[7,5.5,3.5,3.5,8,1,2] You should draw "Manhattan spheres of radius r" around all given points. In the above picture, imagine each cell to be a building, and the grid lines to be roads. To learn more, see our tips on writing great answers. HAMMING DISTANCE: We use hamming distance if we need to deal with categorical attributes. Here, the maximum # of picks an item can get is 7, as it sees every other item once. Intersection of two Jordan curves lying in the rectangle. Manhattan Distance is a very simple distance between two points in a Cartesian plane. This distance is defined as the Euclidian distance. Change coordinate to a u-v system with basis U = (1,1), V = (1,-1). euclidean:. The difference depends on your data. Why did postal voting favour Joe Biden so much? Maximum Manhattan distance between a distinct pair from N coordinates. The set of vectors whose 1-norm is a given constant forms the surface of a cross polytope of dimension equivalent to that of the norm minus 1. Mismatch between my puzzle rating and game rating on chess.com, How to mount Macintosh Performa's HFS (not HFS+) Filesystem. Manhattan distance (L1 norm) is a distance metric between two points in a N dimensional vector space. Left borders will add segment mark to sweeping line, Left borders will erase it. Euclidean distance: Manhattan distance: Where, x and y are two vectors of length n. 15 Km as calculated by the MYSQL st_distance_sphere formula. Change coordinate to a u-v system with basis U = (1,1), V = (1,-1). La geometría del taxista, considerada por Hermann Minkowski en el siglo XIX, es una forma de geometría en la que la métrica usual de la geometría euclidiana es reemplazada por una nueva métrica en la que la distancia entre dos puntos es la suma de las diferencias (absolutas) de sus coordenadas. This is an M.D. Maximum distance between two components of x and y (supremum norm). I think this would work quite well in practice. Count paths with distance equal to Manhattan distance. How this helps. Brüngger used the branch and bound algorithm with the Manhattan distance heuristic and a pre-generated table of move sequences up to length 14. How do I express the notion of "drama" in Chinese? Is it unusual for a DNS response to contain both A records and cname records? Manhattan distance. How to select or copy all text in InDesign document What does “God said” mean in Gen. 1:3,6,9,11,14,20,24,26? Manhattan distance on Wikipedia. It is the sum of the lengths of the projections of the line segment between the points onto the coordinate axes. For high dimensional vectors you might find that Manhattan works better than the Euclidean distance. The distance between two points measured along axes at right angles.The Manhattan distance between two vectors (or points) a and b is defined as ∑i|ai−bi| over the dimensions of the vectors. What do you mean by "closest manhattan distance"? Intuition. Manhattan distance is a well-known distance metric inspired by the perfectly-perpendicular street layout of Manhattan. Use MathJax to format equations. It seems that for the first few rounds, we simply divide the circumference by the number of points (i.e. And the manhatten distance is the largest of abs (u1-u2), abs (v1-v2). Author: PEB. The set of vectors whose 1-norm is a given constant forms the surface of a cross polytope of dimension equivalent to that of the norm minus 1. Maximum average Manhattan distance to nearest neighbor. While moving line you should store number of opened spheres at each point at the line in the segment tree. 21, Sep 20. maxl1=manhattan_distance([1,2,3,4,5,6,7,8],[8,7,6,5,4,3,2,1]), which is 32. What happened in Rome, when the western empire "fell"? the maximum average distance is $\frac{60}{n}$). Coords of the two points in this basis are u1 = (x1-y1)/sqrt (2), v1= (x1+y1), u2= (x1-y1), v2 = (x1+y1). Join Stack Overflow to learn, share knowledge, and build your career. [33,34], decreasing Manhattan distance (MD) between tasks of application edges is an effective way to minimize the communication energy consumption of the applications. The result may be wrong in any directions. And you have to check if there is any non marked point on the line. Usual distance between the two vectors (2 norm aka L_2), sqrt(sum((x_i - y_i)^2)).. maximum:. fly wheels)? Stack Overflow for Teams is a private, secure spot for you and a linear transformation of) the planar Manhattan distance. using Manhattan distance. If the variable “closest” is the min distance to the corner, which is the max one? After some searching, my problem is similar to. Maximum average Manhattan distance to nearest neighbor. Is it possible to make a video that is provably non-manipulated? Generally, Stocks move the index. The maximum Manhattan distance is found between (-4, 6) and (3, -4) i.e., |-4 – 3| + |6 – (-4)| = 17. 2.2 Manhattan Distance Manhattan distance computes the absolute differences between coordinates of pair of objects 2.3 Chebychev Distance Chebychev Distance is also known as maximum value distance and is computed as the absolute magnitude of the This is because ranking effectively 'stretches' out the frequency counts to map them to a range. The distance between two array values is the number of indices between them. @D3r0X4 Computing an L1 Voronoi diagram absolutely would work, but it would require more implementation effort than the other answer and not be worth it unless the points are sufficiently sparse. For example, a list of frequency of wins could be, freq1=[6,5,2,2,5,7,0,1] In the simple case, you can set D to be 1. Intuition. Is there a better way to measure an incremental change in frequency count lists? As the name itself suggests, Clustering algorithms group a set of data points into subsets or clusters. The reason for this is quite simple to explain. I am using the normalized manhattan distance (L1-norm) between two lists as a metric to measure how much change has happened. Why does the U.S. have much higher litigation cost than other countries? In the end, when no more moves can be done, you scan the array dist to find the cell with maximum value. In the example below the points are (1, 1), (6,1), (6,6), (3,4) and the smallest maximal Manhattan distance (equal to 5) is achieved from points (4,3), (5,2) (marked with E). It has real world applications in Chess, Warehouse logistics and many other fields. For example, one change in a list of 8 items equals 0.25, and if every rank changes, then changeIdx = 4. Find P(x,y) such that min{dist(P,P1), dist(P,P2), Can you please include an example? Realistic task for teaching bit operations. rank2=[2,3.5,5.5,7,5.5,3.5,8,1]. all paths from the bottom left to top right of this idealized city have the same distance. No, we need to find target point. Euclidean, Manhattan & Maximum(Chebychev) distance. (Manhattan Distance) of 1. Let's call this changeIdx .This measure is basically changeIdx=L1Norm/n Making statements based on opinion; back them up with references or personal experience. We know that the maximum manhattan distance of a ranked list can be obtained by reversing the list. For k = 3, assuming 1 <= x,y <= k, P1 = (1,1), P2 = (1,3), P3 = (2,2). Manhattan distance # The standard heuristic for a square grid is the Manhattan distance [4]. I don't understand your output requirement. There are two matching pairs of values: and .The indices of the 's are and , so their distance is .The indices of the 's are and , so their distance is . Realistic task for teaching bit operations, What's the meaning of the French verb "rider". Coords of the two points in this basis are u1 = (x1-y1)/sqrt(2), v1= (x1+y1), u2= (x1-y1), v2 = (x1+y1). How to pull back an email that has already been sent? Lets try a. Construct a Voronoi diagram 101. lee215 79746. We can turn a 2D problem into a 1D problem by projecting onto the lines y=x and y=-x. The program can be used to calculate the distance easily when multiple calculations using the same formula are required. Usual distance between the two vectors (2 norm aka L_2), sqrt(sum((x_i - y_i)^2)).. maximum:. These are set of points at most r units away from given point. Manhattan distance is often used in integrated circuits where wires only run parallel to the X or Y axis. It has complexity of O(n log n log k). I can divide this by n to get a changeIdx from 0-max distance. The reason for this is quite simple to explain. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. The result may be wrong in any directions. Do rockets leave launch pad at full thrust? Noun . Thus, the heuristic never overestimates and is admissible. Let’s say point $P_1$ is $(x_1, y_1)$ and point $P_2$ is $(x_2, y_2)$. The use of Manhattan distance depends a lot on the kind of co-ordinate system that your dataset is using. I'm not sure if my solution is optimal, but it's better than yours. Sort by u-value, loop through points and find the largest difference between pains of points. 1 answer. The following paths all have the same taxicab distance: The rest of the states for a pair of blocks is sub-optimal, meaning it will take more moves than the M.D. your coworkers to find and share information. rev 2021.1.11.38289, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, maximum Manhattan Distance of ranked list, Taxicab distance , city block distance, Manhattan distance, Voronoi diagram boundaries with Manhattan distance, Average Manhattan Distance on a Hypercube, Average ratio of Manhattan distance to Euclidean distance, Maximum average Manhattan distance to nearest neighbor. What game features this yellow-themed living room with a spiral staircase? PKD. Minimum Manhattan Distance Approach to Multiple Criteria Decision Making in Multiobjective Optimization Problems Wei-Yu Chiu, Member, IEEE, Gary G. Yen, Fellow, IEEE, and Teng-Kuei Juan Abstract—A minimum Manhattan distance (MMD) approach to multiple criteria decision making in multiobjective optimiza-tion problems (MOPs) is proposed. manhattan: Faster solution, for large K, and probably the only one which can find a point with float coordinates, is as following. Let's call this changeIdx .This measure is basically changeIdx=L1Norm/n therefore, it gives an idea of the # of rank changes. Why is there no spring based energy storage? Thanks for contributing an answer to Mathematics Stack Exchange! Then, you have to check if there is any non marked point on the line inside the initial square [0,k]X[0,k]. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. then you will never process a cell (that has already been processed that you can get to quicker so you never process any already processed cells. And the manhatten distance is the largest of abs(u1-u2), abs(v1-v2). To begin with, a lot of this material in this section has been referred from the now offline page of divingintodatascience, the site had been of great help. Manhattan Distance is the sum of absolute differences between points across all the dimensions. So, again, overall solution will be binary search for r. Inside of it you will have to check if there is any point at least r units away from all given points. The social distance requirement in Manhattan is 2 metres. Voronoi diagram would be another fast solution and could also find non integer answer. It is obvious, that if there is such point for some distance R, there always will be some point for all smaller distances r < R. For example, the same point would go. Maximum distance between two components of x and y (supremum norm). Example. How this helps. to get the block in the right place. In mathematics, Chebyshev distance (or Tchebychev distance), maximum metric, or L∞ metric[1] is a metric defined on a vector space where the distance between two vectors is the greatest of their differences along any coordinate dimension. This is your point. The main question that students struggled with was whether the e-value represents Manhattan or Euclidean distance. Manhattan distance metric can be understood with the help of a simple example. We need to find the greatest of these distances, so the solution would be to minimize ( x 1 , y 1 ) and maximize ( x 2 , y 2 ) . Last Edit: August 7, 2020 6:50 AM. You have to sort all vertical edges of squares, and then process them one by one from left to right. Alternatively, the Manhattan Distance can be used, which is defined for a plane with a data point p 1 at coordinates (x 1, y 1) and its nearest neighbor p 2 at coordinates (x 2, y 2) as You can implement it using segment tree. Related questions 0 votes. euclidean:. Machine Learning Technical Interview: Manhattan and Euclidean Distance, l1 l2 norm. lestari, suci kurnia (2018) algoritma k-means manhattan distance dan chebysyev (maximum value distance) pada sertifikasi hospitality pt.xyz. In other words, entities within a cluster should be as similar as possible and entities in one cluster should be as dissimilar as possible from entities in another. Now turn the picture by 45 degrees, and all squares will be parallel to the axis. It was introduced by Hermann Minkowski. The Taxicab norm is also called the 1 norm.The distance derived from this norm is called the Manhattan distance or 1 distance. Thanks for contributing an answer to Stack Overflow! It only takes a minute to sign up. Active 7 years, 6 months ago. Make yourself known to an official member of staff and/or call the national coronavirus helpline number on 800-232-4636 . But it is much much harder to implement even for Manhattan measure. Euclidean distance computes the root of square difference between co-ordinates of pair of objects. It was introduced by Hermann Minkowski. What does the phrase "or euer" mean in Middle English from the 1500s? ... and its weight in the union is the maximum of the two weights. Look at your cost function and find the minimum cost D for moving from one space to an adjacent space. How do airplanes maintain separation over large bodies of water? site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. for processing them all. This is a total of 28 comparisons (unique combinations). Or the final results is … The further you are from the start point the bigger integer you put in the array dist. A point P(x, y) (in or not in the given set) whose manhattan distance to closest is maximal and max(x, y) <= k. But I feel it run very slow for a large grid, please help me to design a better algorithm (or the code / peseudo code) to solve this problem. Is it unusual for a DNS response to contain both A records and cname records? The l1 of these is 25. and changeIdx is 25/8= 3.125. There is psudo-code for the algorithm on the wikipedia page. Kth Manhattan Distance Neighbourhood: Given a matrix M of size nxm and an integer K, find the maximum element in the K manhattan distance neighbourhood for all elements in nxm matrix. Manhattan distance is often used in integrated circuits where wires only run parallel to the X or Y axis. Chebyshev distance is a distance metric which is the maximum absolute distance in one dimension of two N dimensional points. Manhattan distance [edit | edit source] More formally, we can define the Manhattan distance, also known as the L 1-distance, between two points in an Euclidean space with fixed Cartesian coordinate system is defined as the sum of the lengths of the projections of the … Based on the gridlike street geography of the New York borough of Manhattan. One would expect this to perform at least as well as A*, and likely better. And as you can see, the maximum difference in the short paths to each of the corners is max{1, 4, 1, 4} which is 4. How do you run a test suite from VS Code? The two dimensional Manhattan distance has "circles" i.e. Will 700 more planes a day fly because of the Heathrow expansion? The difference depends on your data. [33,34], decreasing Manhattan distance (MD) between tasks of application edges is an effective way to minimize the communication energy consumption of the applications. p = ∞, the distance measure is the Chebyshev measure. How to increase spacing of uppercase subscript. Minimize maximum manhattan distance of a point to a set of points. Details. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Minkowski distance is a metric which tells us the distance between 2 points in space. While Euclidean distance gives the shortest or minimum distance between two points, Manhattan has specific implementations. Now, how to fast check for existence (and also find) a point which is at least r units away from all given points. My mean is that the closest point (the point which have min manhattan dist) to target point. Last Edit: August 7, 2020 6:50 AM. lestari, suci kurnia (2018) algoritma k-means manhattan distance dan chebysyev (maximum value distance) pada sertifikasi hospitality pt.xyz. What are the earliest inventions to store and release energy (e.g. As shown in Refs. Asking for help, clarification, or responding to other answers. An algorithm of my own design. site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. We can represent Manhattan Distance as: Since the above representation is 2 dimensional, to calculate Manhattan Distance, we will take the sum of absolute distances in both the x and y directions. Now you can check for existence of any point outside such squares using sweeping line algorithm. Query the Manhattan Distance between points P 1 and P 2 and round it to a scale of 4 decimal places. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. We can just work with the 1D u-values of each points. Count of obtuse angles in a circle with … The Manhattan distance (aka taxicab distance) is a measure of the distance between two points on a 2D plan when the path between these two points has to follow the grid layout. level sets in the form of squares, with sides of length √ 2 r, oriented at an angle of π/4 (45°) to the coordinate axes, so the planar Chebyshev distance can be viewed as equivalent by rotation and scaling to (i.e. https://en.wikipedia.org/wiki/Fortune%27s_algorithm, Podcast 302: Programming in PowerPoint can teach you a few things, Minimize maximum manhattan distance of a point to a set of points, Calculate distance between two latitude-longitude points? (Haversine formula), All points with minimum Manhattan distance from all other given points [Optimized], Find the set S which maximize the minimum distance between points in S union A, Find two farthest points in 2D space (Manhattan distance), maximum of minimum of difference in subsequence of k size, Minimal sum of distances in Manhattan metric in c, Book, possibly titled: "Of Tea Cups and Wizards, Dragons"....can’t remember. Alas does not work well. Manhattan distance metric can be understood with the help of a simple example. [Java/C++/Python] Maximum Manhattan Distance. Making statements based on opinion; back them up with references or personal experience. 15, Feb 19. Details. Do that by constructing "manhattans spheres of radius r" and then scanning them with a diagonal line from left-top corner to right-bottom. Broadly speaking there are two ways of clustering data points based on the algorithmic structure and operation, namely agglomerative and di… Book about young girl meeting Odin, the Oracle, Loki and many more. You might need to adapt this for Manhattan distance. Manhattan distance is a metric in which the distance between two points is calculated as the sum of the absolute differences of their Cartesian coordinates. Thus you can search for maximum distance using binary search procedure. (Manhattan Distance) of 1. Contribute to schneems/max_manhattan_distance development by creating an account on GitHub. Given , find the minimum distance between any pair of equal elements in the array.If no such value exists, return .. Euclidean, Manhattan & Maximum(Chebychev) distance. 27.The experiments have been run for different algorithms in the injection rate of 0.5 λ full. Station ) agree to our terms of service, privacy policy and cookie policy item in inventory. Do we use approximate in the simple case, you can set D to ordered! As an index of change need to deal with categorical attributes DNS response to contain both a and... With a spiral staircase Romulans retreat in DS9 episode the Die is Cast '' the difference. } $) bodies of water complexity of O ( n log n log maximum manhattan distance... *, and all squares will be parallel to the corner, which is the Chebyshev measure and... Planar Manhattan distance dan chebysyev ( maximum value in Western Longitude ( LONG_W in STATION.... Licensed under cc by-sa each point at the line segment between the are! Frequency l1 as an index of change throws Stack with the help of a with. Stack with the Bane spell contributing an answer to mathematics Stack Exchange and an anthropologist be ordered copy paste... Stack maximum manhattan distance the 1D u-values of each points CEO of Stack OverflowMinimizing the maximum average distance is a distance which! Than the Euclidean distance drama '' in Chinese be understood with the help of a tree stump such... A private, secure spot for you and your coworkers to find and share information, privacy policy cookie. Solution and could also find non integer answer checking procedure is n log for. Hamming distance: we use hamming distance if we need to deal with categorical.... ( i.e fly because of the lengths of the projections of the states for a square grid is min! }$ ) the point which have min Manhattan dist ) to target point 2020 AM... And likely better 25. and changeIdx is 25/8= 3.125 metric inspired by the perfectly-perpendicular street layout of Manhattan the... Express the notion of drama '' in Chinese professionals in related fields yourself. Array.If no such value exists, return select or copy all text in InDesign what. And is admissible also called the 1 norm.The distance derived from this norm is also called the 1 distance... Each other externally or euer '' mean in Gen. 1:3,6,9,11,14,20,24,26 45 degrees, all. Are in the array dist meeting Odin, the distance measure is the Ogre 's damage...... and its weight in the above picture, imagine each cell to be 1 because... See our tips on writing great answers points across all the input points at most r units from. The center French verb rider '' function and find the cell with maximum.! Problem by projecting onto the lines y=x and y=-x segment mark maximum manhattan distance sweeping line, left borders will add mark... Other item once from having a specific item in their inventory rings to be,. Of drama '' in Chinese sweeping line, left borders will erase it puzzle rating game! This by n to get a changeIdx from 0-max distance 0.5 λ full ) a! Given point or responding to other answers query the Manhattan distance query the Manhattan distance [ 4 ] solution! Machine Learning Technical Interview: Manhattan and Euclidean distance happened in Rome, when the empire. Deep can they be built components of x and y ): problem is similar to of! Does the phrase or euer '' mean in Gen. 1:3,6,9,11,14,20,24,26 more.. Edit: August 7, 2020 6:50 AM works better than the.... Much much harder to implement even for Manhattan distance between bit vectors root of square difference between of. Therefore, it is important to know the range non marked point on the line understood the. Onto the lines y=x and y=-x and if every rank changes in Middle English from the 1500s clusters that delivered! And round it to a scale of 4 decimal places to cut a out... P=1, the heuristic never overestimates and is admissible it valid to use the frequency l1 as index! With a spiral staircase data points into subsets or clusters students struggled with was whether e-value. Have been run for different algorithms in the rectangle with basis U = ( )! Between my puzzle rating and game rating on chess.com, how to back., see our tips on writing great answers -10,0 ), ( 0,10.! Edit: August 7, as it sees every other item once in Chinese between puzzle... Been run for different algorithms in the simple case, you agree to our terms of service privacy... Game rating on chess.com, how to select or copy all text in InDesign document what does the ! Non integer answer be another fast solution and could also find non integer answer many fields. '' around all given points is there a better way to measure an change... About young girl meeting Odin, the other items do not have to be roads 's... 7 years, 6 months ago be another fast solution and could also find non integer answer adjacent.
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https://www.jiskha.com/questions/156164/is-y-x-3-a-solution-to-the-differential-equation-xy-3y-0-how-do-i-go-about-solving | calculus
is y = x^3 a solution to the differential equation xy'-3y=0??
how do i go about solving this??? also, is there a trick to understanding differential equations? i'm really struggling with this idea, but i'm too embarassed to ask my professor for help.
1. 👍
2. 👎
3. 👁
1. To find out if your answer is correct, differentiate it and see if you get the original back.
y = x^3
dy/dx = 3 x^2
now put that in d dy/dx - 3 y
x(3 x^2)- 3 x^3 = ?
3 x^3 - 3 x^3 = 0
sure enough it works
1. 👍
2. 👎
2. Well, here try getting all the y stuff on one side and all the x stuff on the other side (separating variables)
x dy/dx = 3 y
dy/y = 3 dx/x
integrate both sides
ln y = 3 ln x
but 3 ln x = ln x^3
so
ln y = ln x^3
so
y = x^3 will work.
1. 👍
2. 👎
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https://www.physicsforums.com/threads/sum-of-an-infinite-series.596924/ | # Sum of an infinite series?
1. Apr 15, 2012
### Sean1218
1. The problem statement, all variables and given/known data
Ʃ 4/(n(n+2)) from n=1 to n=infinity
2. Relevant equations
3. The attempt at a solution
I tried using partial fractions to get A/n + B/(n+2), and I solved for A and B to get A=2 and B=-2
I tried summing them up, so everything would cancel except the first & last term, but nothing cancels.
I'm not sure of any other methods for finding sums or if I'm not just using this one wrong.
Any help?
2. Apr 15, 2012
### HallsofIvy
Staff Emeritus
Nothing cancels? Yes, partial fractions gives the addend as 2/n- 2/(n+2).
When n= 1, that is 2- 2/3.
When n= 2, that is 1- 2/4.
When n= 3, that is 2/3- 2/5.
When n= 4, that is 2/4- 2/6.
When n= 5, that is 2/5- 2/7.
When n= 6, that is 2/6- 2/8.
When n= 7, that is 2/7- 2/9
etc.
I see a lot cancelling!
3. Apr 15, 2012
### LCKurtz
You have the right idea. Try writing out the first 6 or so terms of$$\sum_1^\infty(\frac 2 n - \frac 2 {n+2})$$leaving in the parentheses and not simplifying as you go. You will see some terms that cancel in the middle. Once you do that write the first $n$ terms.
I see Halls types faster than I do. | 2017-12-13T00:26:39 | {
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https://startups.mathsgee.com/35842/when-are-two-nonzero-vectors-orthogonal-to-each-other | # arrow_back When are two nonzero vectors orthogonal to each other?
25 views
When are two nonzero vectors orthogonal to each other?
Remember $$\theta=\cos ^{-1}\left(\frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\|\|\mathbf{v}\|}\right)$$ It follows from this that $\theta=\pi / 2$ if and only if $\mathbf{u} \cdot \mathbf{v}=0$. Thus, Two nonzero vectors $\mathbf{u}$ and $\mathbf{v}$ in $R^{n}$ are said to be orthogonal (or perpendicular) if $\mathbf{u} \cdot \mathbf{v}=0 .$ We will also agree that the zero vector in $R^{n}$ is orthogonal to every vector in $R^{n}$.
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## Related questions
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Show that two nonzero vectors $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ in $R^{3}$ are orthogonal if and only if their direction cosines satisfy
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Let $U, V, W$ be orthogonal vectors and let $Z=a U+b V+c W$, where $a, b, c$ are scalars.
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https://www.e-pandu.com/2019/06/differentiation-of-exponential-and.html | # Differentiation of Exponential and Logarithmic Functions
The exponential function derivative is not easily solved directly with the definition of the derived function. However, it can be solved using the inverse function derivative, namely the logarithmic function. So, in this article the derivative of the logarithmic function will take precedence before obtaining the derivative function derivative.
### Derivatives of Logarithmic Functions
For example, a function f is stated by the following formula,
f(x) = alog x, x > 0
then the derivative of the function can be obtained by the following steps,.
Note that for h → 0, then u → 0.
if it is known that then the form above can be simplified to become,
So, the function f(x) = alog x, x > 0 is differentiable and the derivative is expressed by a formula.
Especially for the logarithmic function with a base number e expressed by f(x) = ln x, x > 0, the derivative formula is as follows.
In general, the derivative of function f(x) = alog h(x) is
The function derivative f (x) = ln g (x) is
### Derivative Function f(x) = ex
To obtain a function formula in the form of f(x) = ex, we can use the derivative formula of the logarithmic function. However, we can also use the inverse function f(x) = ln x first. Suppose y = ex then applies x = ln y.
By using Leibniz notation in derivatives, the following form is obtained,
So, the derivative of the exponential function f(x) = ex is expressed by the formula:
f’(x) = ex
Whereas, for exponential functions with numbers and past bases, namely f(x) = ax in the same way as above we can find the formula as follows.
f’(x) = ax nlog a or f’(x) = ax ln a
In general, the derivative of the function f(x) = eh(x) is
f’(x)’=h’(x) . eh(x)
the derivative of function f(x) = ag(x) is
f’(x) = ag(x) . g’(x) ln a
Example 1.
Find the function derivative f(x) = 3log (2x – 4).
Answer:
Example 2.
Find the function derivative f(x) = ln (2x3 + 7).
Answer:
Example 3.
Find the function derivative f(x) = log 3x.
Answer:
Example 4.
Find the function derivative f(x) = e2x +3
Answer:
### 1 Response to "Differentiation of Exponential and Logarithmic Functions"
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https://martinacoogan.com/p1b7e/zig9g.php?page=determinant-of-transpose-6f78f5 | Select Page
# determinant of transpose
In particular, if all entries of a square matrix are zero except those along the diagonal, it is a diagonal matrix. So we can then say that the determinant of A transpose is equal to this term A sub 11 times this, but this is equal to this for the n-by-n case. But the columns of AT are the rows of A, so the entry corresponds to the inner product of two rows of A. Introduction to matrices. ', then the element B(2,3) is also 1+2i. Here you can calculate a determinant of a matrix with complex numbers online for free with a very detailed solution. B = A.' and enables operator overloading for classes. Determinant of a Identity matrix is 1. Therefore, det(A) = det(), here is transpose of matrix A. Recall that matrices can be placed into a one-to-one correspondence with linear operators. tB(y, x) = tu(Ψ(y))(x), we find that B(x, y) = tB(y, x). Let be an square matrix: where is the jth column vector and is the ith row vector (). The entry pj i is also obtained from these rows, thus pi j = pj i, and the product matrix (pi j) is symmetric. Example: Python code to find the determinant of a transpose matrix However, there remain a number of circumstances in which it is necessary or desirable to physically reorder a matrix in memory to its transposed ordering. Reduce this matrix to row echelon form using elementary row operations so that all the elements below diagonal are zero. In other words, for a matrix [[a,b], [c,d]], the determinant is computed as ‘ad-bc’. Correspondence Chess Grandmaster and Purdue Alumni. En effet, si A est inversible, det(A) ≠ 0, donc det( t A) ≠ 0 puisque det( t A) = det(A). If A is an m × n matrix and AT is its transpose, then the result of matrix multiplication with these two matrices gives two square matrices: A AT is m × m and AT A is n × n. Furthermore, these products are symmetric matrices. Let A and B be matrices and c be a scalar. EduRev, the Education Revolution! B = A.' If the determinant of a matrix is zero, it is called a singular determinant and if it is one, then it is known as unimodular. Moreover, if the diagonal entries of a diagonal matrix are all one, it is the identity matrix: Rank. does not affect the sign of the imaginary parts. For example, with a matrix stored in row-major order, the rows of the matrix are contiguous in memory and the columns are discontiguous. 2. involving many infinite dimensional vector spaces). Here, it refers to the determinant of the matrix A. Linear Algebra: Determinant of Transpose Proof by induction that transposing a matrix does not change its determinant Linear Algebra: Transposes of sums and inverses. We can also say that the determinant of the matrix and its transpose are equal. For example, if A(3,2) is 1+2i and B = A. So, by calculating the determinant, we get det(A)=ad-cb, Simple enough, now lets take A T (the transpose). By that logic, because I have shown it to be true for the nxn case, it will then be true for the 3x3 case, 4x4 case, 5x5 case, etc...you get the idea. If, we have any given matrix A then determinant of matrix A is equal to determinant of its transpose. A determinant is a real number or a scalar value associated with every square matrix. To begin with let’s look into the role of Adjoint in finding the Inverse of a matrix and some of its theorems. Determinant is a very useful value in linear algebra. The adjoint allows us to consider whether g : Y → X is equal to u −1 : Y → X. Up Next. A series of linear algebra lectures given in videos to help students learn about determinant of transpose. Our mission is to provide a free, world-class education to anyone, anywhere. To find the transpose of a matrix, we change the rows into columns and columns into rows. Matrix definitions involving transposition, Implementation of matrix transposition on computers, Transposes of linear maps and bilinear forms, https://en.wikipedia.org/w/index.php?title=Transpose&oldid=991607775, Creative Commons Attribution-ShareAlike License, This page was last edited on 30 November 2020, at 23:05. This proof is largely one of induction. One of the easiest and more convenient ways to compute the determinant of a square matrix is based on the LU decomposition where , and are a permutation matrix, a lower triangular and an upper triangular matrix respectively.We can write and the determinants of , and are easy to compute: returns the nonconjugate transpose of A, that is, interchanges the row and column index for each element. Determinants and transposes. Determinant of a Identity matrix is 1. So far, every-thing we’ve said about determinants of matrices was related to the rows of the matrix, so it’s some-what surprising that a matrix and its transpose have the same determinant. Of course, probably not, but that is the reason behind those joke proofs such as 0=1 or -1=1, etc. This is one of the key properties in Linear Algebra and is being used in major parts of Matrix and Determinants. B = transpose(A) is an alternate way to execute A.' By defining the transpose of this bilinear form as the bilinear form tB defined by the transpose tu : X## → X# i.e. I mean, one could assume that 2=3, and then construct a proof that 3=4. For example, software libraries for linear algebra, such as BLAS, typically provide options to specify that certain matrices are to be interpreted in transposed order to avoid the necessity of data movement. We can prove this by taking variable elements within a matrix. Created by the Best Teachers and used by over 51,00,000 students. Proof. In particular, if all entries of a square matrix are zero except those along the diagonal, it is a diagonal matrix. If , is a square matrix. defined by ⟨z, h⟩ := h(z)). We’ll prove that, and from that theorem we’ll automatically get corre-sponding statements for columns of matrices that we have for rows of matrices. If A contains complex elements, then A.' det uses the LU decomposition to calculate the determinant, which is susceptible to floating-point round-off errors. 3. If rows and columns are interchanged then value of determinant remains same (value does not change). Back to Course. Autrement dit, le déterminant d’une matrice ou celui de sa transposée est le même. If , is a square matrix. Every linear map to the dual space u : X → X# defines a bilinear form B : X × X → F, with the relation B(x, y) = u(x)(y). Comme dans le cas des matrices et , on a les résultats fondamentaux . First in the case where the rank of Ais less than n, then the case where the rank of A is n, and for the sec- $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $\begin{bmatrix} a & c \\ b & d \end{bmatrix}$, $\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{1m} \\ a_{21} & a_{22} & a_{23} & a_{2m} \\ a_{31} & a_{32} & a_{33} & a_{3m} \\ .... & .... & .... & .... \\ .... & .... & .... & .... \\ .... & .... & .... & .... \\ a_{m1} & a_{m2} & a_{m3} & a_{mm} \\ \end{bmatrix}$, $\begin{bmatrix} a_{22} & a_{23} & a_{2m} \\ a_{32} & a_{33} & a_{3m} \\ .... & .... & .... \\ .... & .... & .... \\ .... & .... & .... \\ a_{m2} & a_{m3} & a_{mm} \\ \end{bmatrix}$, $\begin{bmatrix} a_{11} & a_{21} & a_{31} & a_{m1} \\ a_{12} & a_{22} & a_{32} & a_{m2} \\ a_{13} & a_{23} & a_{33} & a_{m3} \\ .... & .... & .... & .... \\ .... & .... & .... & .... \\ .... & .... & .... & .... \\ a_{1m} & a_{2m} & a_{3m} & a_{mm} \\ \end{bmatrix}$, $\begin{bmatrix} a_{22} & a_{32} & a_{m2} \\ a_{23} & a_{33} & a_{m3} \\ .... & .... & .... \\ .... & .... & .... \\ .... & .... & .... \\ a_{2m} & a_{3m} & a_{mm} \\ \end{bmatrix}$, In the calculation of det(A), we are going to use co-factor expansion along the, Additionally, in the calculation of det(A, However, lets keep pressing on with a more 'concrete' approach (if the above logic was too abstract). The definition of the transpose may be seen to be independent of any bilinear form on the modules, unlike the adjoint (below). Well, for this basic example of a 2x2 matrix, it shows that det(A)=det(A T). Determinant of any square matrix is equal to determinant of its transpose.Lets take an example of any square matrix and find value of its determinant.Then transpose this matrix and again find value of determinant of transpose of matrix.We will note that determinant of matrix is equal to determinant of its transpose.. In particular, this allows the orthogonal group over a vector space X with a quadratic form to be defined without reference to matrices (nor the components thereof) as the set of all linear maps X → X for which the adjoint equals the inverse. Learn more about definition, determinant and inverse matrix at BYJU’S. Prepared at the University of Colorado Boulder … Ideally, one might hope to transpose a matrix with minimal additional storage. If every element in a row or column is zero, then the determinant of the matrix is zero. In linear algebra, the transpose of a matrix is an operator which flips a matrix over its diagonal; The determinant calculation is sometimes numerically unstable. Cela permet de montrer que si une matrice est inversible, sa transposée l’est aussi. that is, it switches the row and column indices of the matrix A by producing another matrix, often denoted by AT (among other notations). The transpose of a matrix A, denoted by AT,[1][4] A′,[5] Atr, tA or At, may be constructed by any one of the following methods: Formally, the i-th row, j-th column element of AT is the j-th row, i-th column element of A: If A is an m × n matrix, then AT is an n × m matrix. The determinant is extremely small. Demonstrates how to transpose matrices and calculate determinants. As I had proved in the beginning 2x2 case, we could just as easily said that it would hold for any (n+1)x(n+1) matrix. The determinant of the transpose can thus be written as: $$\det(A^{T}) =\sum_{\pi}\operatorname{sign}(\pi)\prod_{i=1}^{n}A_{\pi(i),i}$$ So, to prove that the determinant of the transpose is the same, we have move the permutation from the second index to the first in (1). https://www.projectrhea.org/rhea/index.php?title=Determinant_Transpose_Proof&oldid=51894. Khan Academy is a 501(c)(3) nonprofit organization. Determinant of a Matrix; Transpose Matrix; Here, we will learn that the determinant of the transpose is equal to the matrix itself. If repeated operations need to be performed on the columns, for example in a fast Fourier transform algorithm, transposing the matrix in memory (to make the columns contiguous) may improve performance by increasing memory locality. This article is about the transpose of matrices and. Have questions? [6.2.5, page 265. In this context, many authors use the term transpose to refer to the adjoint as defined here. We can do this as follows. If all the elements of a row (or column) are zeros, then the value of the determinant is zero. So if we assume for the n-by-n case that the determinant of a matrix is equal to the determinant of a transpose-- this is the determinant of the matrix, this is the determinant of its transpose-- these two things have to be equal. If pi j is the entry of the product, it is obtained from rows i and j in A. This is one of the key properties in Linear Algebra and is being used in major parts of Matrix and Determinants. This page was last modified on 3 July 2012, at 06:19. Although the determinant of the matrix is close to zero, A is actually not ill conditioned. Determinants and matrices, in linear algebra, are used to solve linear equations by applying Cramer’s rule to a set of non-homogeneous equations which are in linear form.Determinants are calculated for square matrices only. After some linear transformations specified by the matrix, the determinant of the symmetric matrix is determined. The determinant of a matrix is equal to the determinant of its transpose. Here, we will learn that the determinant of the transpose is equal to the matrix itself. To avoid confusing the reader between the transpose operation and a matrix raised to the tth power, the AT symbol denotes the transpose operation. The map tu is called the transpose[10] of u. In addition, as a disclaimer, and food for thought, it is wise in general to explain why a preliminary inductive assumption should be convincing. Next. The determinant of a matrix can be arbitrarily close to zero without conveying information about singularity. Determinant evaluated across any row or column is same. ', then the element B(2,3) is also 1+2i. We can verify from example that both comes out to be equal. The resulting functional u#(f) is called the pullback of f by u. But what was that? The determinant of a square matrix is the same as the determinant of its transpose. In the first step we determine the A T with the help of the definition of the transposed matrix, that says A T = ( a... What happens next? Let X# denote the algebraic dual space of an R-module X. It calculated from the diagonal elements of a square matrix. Here, Ψ is the natural homomorphism X → X## into the double dual. For example, the determinant of the complex conjugate of a complex matrix (which is also the determinant of its conjugate transpose) is the complex conjugate of its determinant, and for integer matrices: the reduction modulo m of the determinant of such a matrix is equal to the determinant of the matrix reduced modulo m (the latter determinant being computed using modular arithmetic). The determinant of the product of two square matrices is equal to the product of the determinants of the given matrices. The determinant and the LU decomposition. These bilinear forms define an isomorphism between X and X#, and between Y and Y#, resulting in an isomorphism between the transpose and adjoint of u. If any two row (or two column) of a determinant are interchanged the value of the determinant … We first calculate determinant of matrix A and then we calculate determinant of transpose of matrix A. La transposée du produit de deux matrices est égale au produit des transposées de ces deux matrices, mais dans l'ordre inverse : =. If the matrix A describes a linear map with respect to bases of V and W, then the matrix AT describes the transpose of that linear map with respect to the dual bases. returns the nonconjugate transpose of A, that is, interchanges the row and column index for each element. The continuous dual space of a topological vector space (TVS) X is denoted by X'. If rows and columns are interchanged then value of determinant remains same (value does not change). Matrix Transpose The transpose of a matrix is used to produce a matrix whose row and column indices have been swapped, i.e., the element of the matrix is swapped with the element of the matrix. [1][2], The transpose of a matrix was introduced in 1858 by the British mathematician Arthur Cayley.[3]. A square matrix whose transpose is equal to itself is called a symmetric matrix; that is, A is symmetric if, A square matrix whose transpose is equal to its negative is called a skew-symmetric matrix; that is, A is skew-symmetric if, A square complex matrix whose transpose is equal to the matrix with every entry replaced by its complex conjugate (denoted here with an overline) is called a Hermitian matrix (equivalent to the matrix being equal to its conjugate transpose); that is, A is Hermitian if, A square complex matrix whose transpose is equal to the negation of its complex conjugate is called a skew-Hermitian matrix; that is, A is skew-Hermitian if, A square matrix whose transpose is equal to its inverse is called an orthogonal matrix; that is, A is orthogonal if, A square complex matrix whose transpose is equal to its conjugate inverse is called a unitary matrix; that is, A is unitary if. These results may not hold in the non-commutative case. transpose and the multiplicative property of the determinant we have detAt = det((E 1 Ek) t) = det(Et k Et 1) = det(Et k) det(Et 1) = detEk detE1 = detE1 detEk = det(E1 Ek) = detA. does not affect the sign of the imaginary parts. I have taken an example and have proved that determinant of matrix is equal to determinant of its transpose. A T = $\begin{bmatrix} a & c \\ b & d \end{bmatrix}$ So, det(A T)=ad-cb. In this lesson we will learn about some matrix transformation techniques such as the matrix transpose, determinants and the inverse. About. A quick proof of the symmetry of A AT results from the fact that it is its own transpose: On a computer, one can often avoid explicitly transposing a matrix in memory by simply accessing the same data in a different order. Rank, trace, determinant, transpose, and inverse of matrices. The Hermitian adjoint of a map between such spaces is defined similarly, and the matrix of the Hermitian adjoint is given by the conjugate transpose matrix if the bases are orthonormal. Suppose 3 x 3 matrix . Rank, trace, determinant, transpose, and inverse of matrices. For example, if A(3,2) is 1+2i and B = A. Proportionality or Repetition Property . The determinant of a matrix can be arbitrarily large or small without changing the condition number. In this lesson we will learn about some matrix transformation techniques such as the matrix transpose, determinants and the inverse. Note that this article assumes that matrices are taken over a commutative ring. Over a complex vector space, one often works with sesquilinear forms (conjugate-linear in one argument) instead of bilinear forms. For a 2x2 matrix, it is simply the subtraction of the product of the top left and bottom right element from the product of other two. To go through example, have a look at the file present below. Therefore, A is not close to being singular. The transpose of a linear operator can be defined without any need to consider a matrix representation of it. This leads to the problem of transposing an n × m matrix in-place, with O(1) additional storage or at most storage much less than mn. This leads to a much more general definition of the transpose that can be applied to linear operators that cannot be represented by matrices (e.g. Remember, we're doing the n plus 1 by n … This definition also applies unchanged to left modules and to vector spaces.[9]. Let be an square matrix: where is the jth column vector and is the ith row vector (). This page has been accessed 32,375 times. B = transpose(A) is an alternate way to execute A.' Site Navigation. Therefore, efficient in-place matrix transposition has been the subject of numerous research publications in computer science, starting in the late 1950s, and several algorithms have been developed. If X and Y are TVSs then a linear map u : X → Y is weakly continuous if and only if u#(Y') ⊆ X', in which case we let tu : Y' → X' denote the restriction of u# to Y'. The following relation characterizes the algebraic adjoint of u[8], where ⟨•, •⟩ is the natural pairing (i.e. A tolerance test of the form abs(det(A)) < tol is likely to flag this matrix as singular. Let A be the symmetric matrix, and the determinant is denoted as “ det A” or |A|. Theorem 6. Linear Algebra: Determinant of Transpose . Having said that I would also like to bring your attention to the fact that the Inverse of a Matrix exists if and only if the value of its determinant is equal to zero. If u : X → Y is a linear map, then its algebraic adjoint or dual,[7] is the map #u : Y# → X# defined by f ↦ f ∘ u. A symmetric matrix is a square matrix when it is equal to its transpose, defined as A=A^T. We first calculate determinant of matrix A and then we calculate determinant of transpose of matrix A. Set the matrix (must be square). Read the instructions. Every square matrix can be represented as the product of an orthogonal matrix (representing an isometry) and an upper triangular matrix (QR decomposition)- where the determinant of an upper (or lower) triangular matrix is just the product of the elements along the diagonal (that stay in their place under transposition), so, by the Binet formula, $A=QR$ gives: \det(A^T)=\det(R^T … Multiply the main diagonal elements of the matrix - determinant is calculated. Part 5 of the matrix math series. Determinant of a transposed matrix Ok. and enables operator overloading for classes. Let X and Y be R-modules. The determinant of the transpose of a square matrix is equal to the determinant of the matrix, that is, jAtj= jAj. Did you know that the Inverse of a Matrix can be easily calculated using the Adjoint of a Matrix? To calculate a determinant you need to do the following steps. Determinant is calculated by reducing a matrix to row echelon form and multiplying its main diagonal elements. Matrix Transpose; Matrix Multiplication; Matrix Addition/Subtraction; Determinant Calculator. Similarly, the product AT A is a symmetric matrix. Determinant of transpose. Use with caution, and enjoy. Best Videos, Notes & Tests for your Most Important Exams. Determinant of transpose. Donate or volunteer today! The matrix of the adjoint of a map is the transposed matrix only if the bases are orthonormal with respect to their bilinear forms. For n ≠ m, this involves a complicated permutation of the data elements that is non-trivial to implement in-place. All-zero Property. We’ll prove this like the last theorem. Therefore, det(A) = det(), here is transpose of matrix A. In other words, the determinant of a linear transformation from R n to itself remains the same if we use different coordinates for R n.] Finally, The determinant of the transpose of any square matrix is the same as the determinant of the original matrix: det(A T) = det(A) [6.2.7, page 266]. Indeed, the matrix product A AT has entries that are the inner product of a row of A with a column of AT. Matrix Transpose The transpose of a matrix is used to produce a matrix whose row and column indices have been swapped, i.e., the element of the matrix is swapped with the element of the matrix. Minor of a Matrix. The dot product of two column vectors a and b can be computed as the single entry of the matrix product: [ a ⋅ b ] = a T b , {\displaystyle \left [\mathbf {a} \cdot \mathbf {b} \right]=\mathbf {a} ^ {\operatorname {T} }\mathbf {b} ,} We say that σ ( i) = j, i = σ − 1 ( j) and change i → j in the product sign. The determinant of a square matrix is the same as the determinant of its transpose. Les propriétés essentielles des déterminants sont résumées dans le théorème fondamental suivant. Determinants of Products & Transposes Determinants of products & transposes of matrices can easily be found once the determinants of the matrices themselves are known: Theorem (Determinants of Products & Transposes) Let A;B be n n square matrices and 6= 0. If A contains complex elements, then A.' If the vector spaces X and Y have respectively nondegenerate bilinear forms BX and BY, a concept known as the adjoint, which is closely related to the transpose, may be defined: If u : X → Y is a linear map between vector spaces X and Y, we define g as the adjoint of u if g : Y → X satisfies. Transpose, and inverse matrix AT BYJU ’ S 3,2 ) is called the of., interchanges the row and column index for each element if every in. 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http://mathhelpforum.com/calculus/172340-solids-revolution-question.html | # Thread: Solids of Revolution Question
1. ## Solids of Revolution Question
The Problem
The finite region bounded by the curve $y\ =\ e^x\ +\ 1$, the y-axis, and the line $x\ =\ ln\ 2$, is rotated through 360 degrees about the x-axis. Show that the volume of the solid formed is $\frac{\Pi}{2}(7\ +\ ln\ 4)$.
Attempt
Volume = $Integral(\Pi\ y^2\ dx)$, with limits $0\ and\ ln\ 2.$
This becomes: Volume = $\Pi\ Integral((e^x\ +\ 1)^2)$, with limits $0\ and\ ln\ 2.$
(We can take pi outside the integral and multiply the result by it later).
Integrating term by term:
$Integral((e^x\ +\ 1)^2)\ =\ \frac{e^{2x}}{2}\ +\ 2e^x + 1$.
Substituting $ln\ 2$:
$\frac{e^{2\ ln\ 2}}{2}\ +\ 2e^{ln\ 2}\ +\ 1\ =\ 7.$
Substituting $0$:
$\frac{e^{2(0)}}{2}\ +\ 2e^{0}\ +\ 1\ =\ 3\ \frac{1}{2}.$
Subtracting:
$7\ -\ 3\ \frac{1}{2}\ =\ 3\ \frac{1}{2}$.
So my answer is:
$3\ \frac{1}{2}\ \Pi$.
So I don't know how they've reached the answer given.
Any hints?
2. I don't think that you have enough bounds on your region...
3. Originally Posted by Prove It
I don't think that you have enough bounds on your region...
I'm not exactly sure what you mean by this. This is from a textbook, so do you mean that the information given does not give enough bounds to determine the volume of the region, or do you mean that the bounds of 0 and ln 2 that I have used are wrong?
Also, I found that if you multiply out the answer given, you get $\frac{7\Pi}{2}\ +\ \frac{\Pi\ ln\ 4}{2}$. My answer is equivalent to $\frac{7\Pi}{2}$, so I just can't see how they got the second part.
Anyone else have any ideas?
4. Sorry I misread what you posted, I see now that the $\displaystyle y$ axis is also a bound.
Your integral $\displaystyle \int_0^{\ln{2}}{\pi y^2\,dx} = \pi \int_0^{\ln{2}}{(e^x + 1)^2\,dx}$ is correct, however you evaluated this integral incorrectly.
$\displaystyle \int{(e^x + 1)^2\,dx} = \int{e^{2x} + 2e^x + 1\,dx}$
$\displaystyle = \frac{e^{2x}}{2} + 2e^x + x + C$.
Substituting ln 2:
$\frac{e^{2\ ln\ 2}}{2}\ +\ 2e^{ln\ 2}\ +\ ln\ 2\ =\ 6\ +\ ln\ 2$
Substituting 0:
$\frac{e^{2(0)}}{2}\ +\ 2e^0\ +\ 0\ =\ 2\ \frac{1}{2}$
Subtracting:
$6\ +\ ln\ 2\ -\ 2\ \frac{1}{2}\ =\ 3\ \frac{1}{2}\ +\ ln\ 2\ =\ \frac{7}{2}\ +\ ln\ 2$
Multiplying by $\Pi$:
$\frac{7\Pi}{2}\ +\ \Pi\ ln\ 2$
Factorising:
$\frac{\Pi}{2}(7\ +\ 2\ ln\ 2)$.
Getting closer, but I still don't understand how they got:
$\frac{\Pi}{2}(7\ +\ ln\ 4)$
6. Basic logarithm law: $\displaystyle n\log{x} = \log{(x^n)}$...
7. Originally Posted by Prove It
Basic logarithm law: $\displaystyle n\log{x} = \log{(x^n)}$... | 2016-12-06T21:17:30 | {
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https://mathematica.stackexchange.com/questions/87263/find-the-area-of-a-contour-surface-plotted-by-contourplot3d | # Find the area of a contour surface plotted by ContourPlot3D
I am looking for a way to find the surface area of a contour from ContourPlot3D. Here is an example where I might like to find the area of the contour surface to G that equals -5:
G[x_, y_, z_] :=
-2 Log[(2 + 2 Sqrt[((y)^2 + (z)^2)^2 + (-1 - Abs[x])^2] + 2 Abs[x])/
(-2 + 2 Sqrt[((y)^2 + (z)^2)^2 + (1 - Abs[x])^2] + 2 Abs[x])] -
2 Log[(1. + 2 Sqrt[((x)^2 + (y)^2)^2 + (-0.5 - Abs[-0.5 + z])^2] +
2 Abs[-0.5 + z])/
(-1. + 2 Sqrt[((x)^2 + (y)^2)^2 + (0.5 - Abs[-0.5 + z])^2] +
2 Abs[-0.5 + z])]
ContourPlot3D[G[x, y, z], {x, -2, 2}, {y, -2, 2}, {z, -2, 2},
Contours -> {-5}]
I would prefer a general method that would work even with more complicated functions than this. I don't know if there is a way to derive the area from the contour plot itself, or some other way accomplish this. Any ideas?
• Welcome to Mathematica.SE! I hope you will become a regular contributor. To get started, 1) take the introductory Tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise. – bbgodfrey Jun 30 '15 at 20:40
This problem can be solved out of the box in V10 using the ImplicitRegion function. Lets use your function G and define the bounded 3D region. We can use RegionPlot3D to visualize the object.
region = ImplicitRegion[G[x, y, z] <= -5, {x, y, z}];
RegionPlot3D[region, PlotRange -> 1.5, PlotPoints -> 60]
Now we will use DiscretizeRegion to form a MeshRegion. The RegionBoundary will give the 3D surface mesh from which we can easily extract the surface area using the Areaproperty.
dis = DiscretizeRegion[region,{{-1.5, 1.5},{-1.5,1.5},{-1.5, 1.5}},
AccuracyGoal -> 9,MaxCellMeasure -> {"Area" -> 0.01}];
HighlightMesh[dis, Style[1, Thin, Red]]
Area@RegionBoundary[dis]
18.7526
However a considerably slower Method -> "ContourPlot3D" will generate a more accurate mesh resulting in a more precise area calculation.
disFine = DiscretizeRegion[region,{{-1.5, 1.5},{-1.5, 1.5},{-1.5,1.5}},
Method -> "ContourPlot3D",AccuracyGoal -> 9,MaxCellMeasure -> {"Area" -> 0.01}]
Area@RegionBoundary[disFine]
18.7894
You can also use DiscretizeGraphics with some clever options in your ContourPlot3D. Just add the Option Lighting -> "Neutral":
gr = ContourPlot3D[G[x, y, z], {x, -2, 2}, {y, -2, 2}, {z, -2, 2},
Contours -> {-5}, Lighting -> "Neutral"] // Quiet;
Then you can compute the area with:
Area @ DiscretizeGraphics @ gr
18.8018
Here is the image from DiscretizeGraphics`: | 2019-09-22T06:52:01 | {
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https://math.stackexchange.com/questions/1627613/if-x0-19-x0-20-x0-21-cdots-x0-91-546-find-the-value-of-100x | # If $[x+0.19] +[x+0.20] +[x+0.21] +\cdots [x+0.91] =546$ find the value of $[100x]$..
Problem :
If $[x+0.19] +[x+0.20] +[x+0.21] +\cdots [x+0.91] =546$ find the value of $[100x]$ where [.] represents the greatest integer function less than equal to x.
My approach :
$x +1.19 = x + \frac{19}{100} = \frac{100x+19}{100}$
Similarly other terms
Not getting any clue further please suggest will be of great help.
• $x\in\mathbb{Z}$? – barak manos Jan 26 '16 at 12:48
First note that the number of terms is $73$. Also, if we look at $[x+a]$ where $a$ ranges from $0.19$ to $0.91$, then we see it can only reach two values; those are $n=[x+0.19]$ and possibly, but not necessarily, $n+1$. We know though that $$73n\leq [x+0.19]+[x+0.20]+\cdots+[x+0.91]<73(n+1)$$ and so we deduce $n=7$. We can also find exactly where the value of $[x+a]$ changes from $n$ to $n+1$ - since, $546-73\cdot 7=35$, so there are $35$ terms $[x+a]=n+1$, so the last $a$ such that $[x+a]=n$ is $a=0.56$. So, $[x+0.56]=7$ but $[x+0.57]=8$. This means that $7.43\leq x<7.44$. So, $743\leq 100x<744$, so $[100x]=743$.
Hope this helped!
For a clue note that the original equation is the sum of 73 terms. The first and last term differ by at most $1$, so the sum is the total of a number of terms at the lower value and a number of items at the higher value. You should be able to work out how many of each, and this will tell you where the value steps up by $1$. This will in turn give you information to bound $x$ sufficiently to answer the question.
There are 91-18=73 terms. And 73*7=511 < 546 < 73*8. So maybe x should be between 7 and 8.
How do you manage the number of 7s and 8s?
There's a way to solve this ( for positive x ) by first observing that your sum is the difference of two sums of the same form :
$$S_n=\sum_{k=0}^{n}\lfloor x+ak\rfloor$$
Let's start by defining $x_0=\lfloor x\rfloor$ and $f!=x-x_0$ so we have the integer and fractional part of $x$ handy.
Now we can see that :
$$|x+ak|=x_0+\lfloor f+ak \rfloor$$
Let's assume $an < 1$ for convenience ( as it suffices in this problem ).
There is some $k_0$ where $f+ak >= 1$ that contributes $+1$ to the sum and that other values contribute nothing. This $k_0$ is given by :
$$k_0=\left\lceil \frac{1-f}{a} \right\rceil$$
And we can write the sum $S_n$ easily as :
$$S_n = nx_0 + ( n+1-k_0)$$
when $n >= k_0$ and
$$S_n = nx_0$$
when $n < k_0$
And now we have a way to solve the problem at hand.
We can see that, for $a=0.01$ either $S_{91}-S_{18}=(91-18)(x_0+1)=546$ which is not the case as $73$ does is not a factor of $546$ or :
$$S_{91}-S_{18}=x_0(91-18)+(91-18)-k_0+1=546$$
We get :
$$545=73(x_0+1)-k_0$$
Taking the modulus 73 we get :
$$k_0\,mod\,73 =73-( 545\, mod\, 73 )=39$$
And as this is the only possible value in the range $18$ to $91$ we can say $k_0=39$.
Using this we get for $f=1-ak_0=1-(0.01)(39)=0.61$ and using this in our formula for the sum we can get :
$$546=73(x_0+1)-k_0+1=73(x_0+1)-38$$
So finally :
$$x_0=\frac{546+38}{73}=8$$
An so $x=(8-1)+0.61=7.61$ and the final result for $\lfloor 100x\rfloor=761$
• Unfortunately, there must be some mistake somewhere in your calculations, because $[x + 0.19] + \dots + [x + 0.91] = 564$ when $x = 7.61$, while the sum should have been $546$. The Mathematica code does the sum for you: f[x_] := Sum[Floor[x + k], {k, 0.19, 0.91, 0.01}]; f[7.61]. For the correct value $x = 7.43$ see vrugtehagel's answer. – Alex M. Jan 26 '16 at 18:11 | 2020-01-20T03:32:54 | {
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https://math.stackexchange.com/questions/2920199/fx-fracx2-1x-1-is-different-from-fx-x-1-in-spite-of-the | # $f(x) = \frac{x^2 - 1}{x-1}$ is different from $f(x) = x + 1$, in spite of the fact that rules of algebra is followed, why?
Suppose there be a function, $$f(x) = \frac {x^2-1}{x-1}$$
For $x=1$, the function becomes un defined. But, in Algebra we know it is allowed to Cancel Denominator and numerator by the common factor and this would result in the same expression which is equivalent to the first one. But,
$$\require{cancel} f(x) = \frac{(x+1)\cancel{(x-1)}}{\cancel{(x-1)}} =y$$ $$\Rightarrow y = x + 1$$
But, here when I see that Graph of both the equation they look same however the first one is undefined for x = 1 which is not with the case of the second one. But, just canceling a common term in numerator and denominator or multiplying them, changes the whole function why?
So, rules of algebra don't work? or there is a problem with my understanding?
• Could you add pictures of the graphs? You might be making some mistake, like actually plotting $x^2 - \frac1x - 1$ instead of $\frac{x^2-1}{x-1}$ (which is what happens if you write x^2-1/x-1 into any reasonable graphing tool), or any of several other possible mistakes, and we can't tell if you don't give us more to go by. – Arthur Sep 17 '18 at 13:43
• @Arthur Okay let me add them, nice idea, btw :) – Abhas Kumar Sinha Sep 17 '18 at 13:44
• @AbhasKumarSinha What do you mean? Both are straight lines: wolframalpha.com/input/?i=(x%5E2-1)%2F(x-1) wolframalpha.com/input/?i=x%2B1 – Piotr Wasilewicz Sep 17 '18 at 13:45
• @AbhasKumarSinha they are not equivalent because of domain. There are two different functions with the same graph in range $\mathbb{R}-\{1\}$ – Piotr Wasilewicz Sep 17 '18 at 13:53
• @AbhasKumarSinha No you don't (followed the rules of algebra). You can't short fraction if there is possible to have $0$ in denominator. You can do it if and only if you are sure that there is no $0$ in denominator. For example: $\frac{(x+1)(sinx + 2)}{sinx + 2} = x + 1$ But not in this case. – Piotr Wasilewicz Sep 17 '18 at 14:00
The functions are different because they have different domains.
A function is defined by its domain, codomain, and graph or, alternatively, by its domain, codomain, and a rule that specifies how elements of the domain are mapped to elements in the codomain.
The implied domain of the function defined by $$f(x) = \frac{x^2 - 1}{x - 1}$$ is the largest set of real numbers that do not make the denominator equal to zero, which is $$\{x \in \mathbb{R} \mid x \neq 1\} = \mathbb{R} - \{1\}$$
Thus, we should write that $f$ is the function $f: \mathbb{R} - \{1\} \to \mathbb{R}$ defined by $$f(x) = \frac{x^2 - 1}{x - 1}$$ Its graph is the line $y = x + 1$ with a hole at the point $(1, 2)$ since $$\lim_{x \to 1} f(x) = \lim_{x \to 1} = \frac{x^2 - 1}{x + 1} = \lim_{x \to 1} (x + 1) = 1 + 1 = 2$$
Its graph is the punctured line shown below.
Notice that at every point in the domain of $f$, the denominator does not equal to zero. Thus, we may divide by $x - 1$ to obtain $$f(x) = \frac{x^2 - 1}{x - 1} = \frac{(x + 1)(x - 1)}{x - 1} = x + 1$$ for each $x \in \text{Dom}_f = \mathbb{R} - \{1\}$.
The function $g: \mathbb{R} \to \mathbb{R}$ defined by $g(x) = x + 1$ is defined for every real number. Its graph is just the line $y = x + 1$.
While the two functions agree on the intersection of their domains, they have different domains. Therefore, they are different functions.
We can prove that both of these functions have the same graph for domain $D=\mathbb{R}-\{1\}$:
let $a \in D$ be any number from domain. Then we have:
$f(a) = a + 1 = a + 1 \cdot \frac{a-1}{a-1} = \frac{a^{2}-1}{a-1} = f(a)$
So the graph is the same for $D=\mathbb{R}-\{1\}$: but in general these two functions are different because of Domain.
• There are no two functions given. To speak about a function we have to know the image and domain. – Cornman Sep 17 '18 at 13:55
• @Cornman But you can compute the widest domain for some formula and what I meant was: these two the widest domain are different for those formulas. Now right? – Piotr Wasilewicz Sep 17 '18 at 13:58
• In my opinion the widest domain which fits $f(x)=\frac{x^2-1}{x-1}$ is indeed $\mathbb{R}$, since the pole can be "fixed". It is $\lim_{x\to 1} f(x)=2$. – Cornman Sep 17 '18 at 14:03
• @Cornman No it isn't. Becouse there are two different limits for $1^{-}$ and $1^{+}$ – Piotr Wasilewicz Sep 17 '18 at 14:05
• But you can put $f(x) = \frac{x^{2}-1}{x-1}$ for $x \neq 1$ and $f(x) = 2$ for $x = 1$. Of course you can do it but the formula will be different. – Piotr Wasilewicz Sep 17 '18 at 14:06
The problem is, that $f(x)=\frac{x^2-1}{x-1}$ is not a function, for the simple reason, that a function has to be defined with a domain and image like:
$f:\mathbb{R}\setminus\{1\}\to\mathbb{R}$.
One might consider $\mathbb{R}\setminus\{1\}$ as the domain of $f$, since for $x=1$ we get $1-1=0$ in the denominator, which leads to an illegal operation (dividing by zero). So this is a pole.
Or is it?
It is completly fine to define $f:\mathbb{R}\to\mathbb{R}$, since the pole is "liftable". A fake pole, if you please.
The only difference is that $\frac{(x^2-1)}{(x-1)}$ is not defined at x=1, although the limit as x tends to 1 exists and is equal to 2. Rest everything is same. | 2019-11-20T20:17:14 | {
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http://mathhelpforum.com/statistics/5248-combinations.html | # Math Help - Combinations
Here is the problem:
A sequence of letters of the form abcba is an example of a palindrome of five letters.
a. If a letter may appear more than twice, how many palindromes of five letters are there? of six letters?
b. Repeat part a under the condition that no letter appears more than twice.
I do not know how to go about doing this problem. Any help would be appreciated. thanks.
2. Hello, ashkash!
I used Brute Force . . . and started making a list.
A sequence of letters of the form ABCBA is an example of a palindrome of five letters.
(a) If a letter may appear more than twice,
how many palindromes of five letters are there? of six letters?
With five letters, consider the first three letters.
3 different letters: . $ABC\;\;\Rightarrow\;\;ABCBA$
2 different letters: . $\begin{array}{ccc}ABA\;\;\Rightarrow\;\;ABABA \\ AAB\;\;\Rightarrow\;\;AABAA \\ ABB\;\;\Rightarrow\;\;ABBBA\end{array} \begin{array}{ccc}* \\ * \\ *\end{array}$
. . Only one letter: . $AAA\;\;\Rightarrow\;\;AAAAA\;\;\;\:*$
With six letters, again consider the first three letters.
3 different letters: . $ABC\;\;\Rightarrow\;\;ABCCBA$
2 different letters: . $\begin{array}{ccc}ABA\;\;\Rightarrow\;\;ABAABA \\ AAB\;\;\Rightarrow\;\;AABBAA \\ ABB\;\;\Rightarrow\;\;ABBBBA\end{array} \begin{array}{ccc}* \\ * \\ * \end{array}$
. . Only one letter: . $AAA\;\;\Rightarrow\;\;AAAAAA\;\;\;\:*$
(b) Repeat part (a) under the condition that no letter appears more than twice.
Disregard answers in (a) marked with an asterisk $(*).$
3. for the question you need to consider all 26 letters of the alphabet and not just a,b,and c.
4. Originally Posted by ashkash
for the question you need to consider all 26 letters of the alphabet and not just a,b,and c.
Would $AAAAA$ be a valid combination?
5. yes, it would.
6. As soroban pointed out all you have to consider is the first three letters, each is independent of the other...
Here's an example problem. Flip a coin. After one flip you have only two possible outcomes (2^1):
$\boxed{\begin{array}{cc}H\\T\end{array}}$
After two flips you get four possible outcomes (2^2):
$\boxed{\begin{array}{cccc}HH\\HT\\TH\\TT\end{array }}$
After three flips you get eight possible outcomes (2^3):
$\boxed{\begin{array}{cccccccc}HHH\\HHT\\HTH\\HTT\\ THH\\THT\\TTH\\TTT\end{array}}$
You can see that after each flip the outcome multiplies by two (because a flip can give 2 different results)
Now your problem can give 26 different results per letter, and only 3 letters matter. Therefore we get the answer of 26^3
you get the same answer with 6 places as well
7. Hello again, ashkash!
If we are considering all 26 letters of the alphabet,
. . we simply append a permutation factor.
With five letters, consider the first three letters.
3 different letters: . $ABC\;\;\Rightarrow\;\;ABCBA$ . . . one way
There are $26$ choices for the " $A$",
. . $25$ choices for the " $B$",
. . and $24$ choices for the " $C$".
Hence, there are: . $26\cdot25\cdot24 \times 1 \:=\:15,600$ ways.
2 different letters: . $\begin{array}{ccc}ABA\;\;\Rightarrow\;\;ABABA \\ AAB\;\;\Rightarrow\;\;AABAA \\ ABB\;\;\Rightarrow\;\;ABBBA\end{array}$ . . . three ways
There are $26$ choices for the " $A$"
. . and $25$ choices for the " $B$".
Hence, there are: . $26\cdot25 \times 3 \:=\:1950$ways.
Only one letter: . $AAA\;\;\Rightarrow\;\;AAAAA$ . . . one way
There are $26$ choices for the " $A$".
Hence, there are: . $26 \times 1 \:=\:26$ ways.
Therefore, there are: . $15,600 + 1950 + 26 \:=\:\boxed{17,576\text{ ways.}}$
8. just to clear confusion: $\underbrace{\overbrace{15600+1950+26}^{\text{Sorob }\!\!\text{an's Way}}=17576=\overbrace{26^3}^{\text{my way}}}_{\text{both ways give the same answer}}$ | 2014-09-01T11:41:40 | {
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https://math.stackexchange.com/questions/1541986/what-is-the-best-way-to-compute-the-pseudoinverse-of-a-matrix | # What is the best way to compute the pseudoinverse of a matrix?
Mathematica gives the pseudo-inverse of a matrix almost instantaneously, so I suspect it is calculating the pseudo-inverse of a matrix not by doing singular value decomposition. Since the pseudo-inverse of a matrix is unique, is there a good formula that we can use to simplify our calculation in obtaining the pseudo-inverse, in place of compact singular value decomposition?
I'm thinking about the property that $A^\dagger = (A^\ast A)^{-1} A$, and I think this should give the unique pseudo-inverse of $A$. But I'm not too sure. Thanks for any insight!
• Have you compared with the time it takes for Mathematica to compute the svd? I think for medium sized matrices this can seem almost instantaneous. – littleO Nov 23 '15 at 0:00
Without being able to see your results, the following comments may lack full context.
The implementation of the pseudoinverse is modern and efficient, and will bypass the SVD if possible. Computation with numerical matrices are quite fast and are based at BLAS-level optimizations for your hardware.
Examples where $\mathbf{A}^{+}$ is constructed without the SVD are presented by user1551 in Find the pseudo-inverse of the matrix A without computing singular values of A.
Sequence of Jordan normal forms
These example bring your points to life. The SVD takes much longer to compute that the SVD.
$$\begin{array}{cc} % \mathbf{A} = \left( \begin{array}{cccc} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) % & \mathbf{A}^{+} = \left( \begin{array}{crcc} 1 & -1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) \\ % \text{SVD time} = 0.006 \text{ s} & \mathbf{A}^{+}\text{ time} = 0.0003 \text{ s} % \end{array}$$
$$\begin{array}{cc} % \mathbf{A} = \left( \begin{array}{cccc} 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) % & \mathbf{A}^{+} = \left( \begin{array}{crrc} 1 & -1 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) \\ % \text{SVD time} = 0.008 \text{ s} & \mathbf{A}^{+}\text{ time} = 0.0003 \text{ s} % \end{array}$$
$$\begin{array}{cc} % \mathbf{A} = \left( \begin{array}{cccc} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) % & \mathbf{A}^{+} = \left( \begin{array}{crrr} 1 & -1 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) \\ % \text{SVD time} = 0.011 \text{ s} & \mathbf{A}^{+}\text{ time} = 0.0004 \text{ s} % \end{array}$$
The SVD is an invaluable theoretical tool as it resolved the four fundamental spaces, aligns them, and computes the scale factors. As seen in How does the SVD solve the least squares problem? it provides a natural expression for the solution of the least squares problem. It is quite useful for understand different forms of the Moore-Penrose pseudoinverse: What forms does the Moore-Penrose inverse take under systems with full rank, full column rank, and full row rank?, Pseudo-inverse of a matrix that is neither fat nor tall?. Yet as you point out, there are oftern faster ways to compute $\mathbf{A}^{+}$.
If you think the SVD is too slow
A common issue is when users create matrices with integers and special constants like $\pi$ and complain the SVD is too slow. You will find relief if you request instead of an arbitrary precision result with SingularValueDecomposition[A] you request a numeric precision result with SingularValueDecomposition[N[A]].
Let $$A\in M_{m,n}(K)$$ with $$rank(A)=r\leq \min(n,m)$$ and let $$u=\min(m^2n,n^2m)$$. Note that $$A^+\in M_{n,m}(K)$$.
Case 1. You want the exact value of $$A^+$$.
Then you calculate a rank factorisation of $$A$$ ($$A=BC$$ where $$B\in M_{m,r},C\in M_{r,n}$$ have full rank), using the reduced row echelon form of $$A$$ (complexity $$\approx u$$. Then $$A^+=C^+B^+=C^*(CC^*)^{-1}(B^*B)^{-1}B^*$$ (complexity $$\approx 7u$$).
Total complexity $$\approx 8u$$.
Beware, here the word "complexity" measures the number of operations; in an exact calculation, the number of digits can become very large, which substantially increases the true complexity. If you cannot work with a large number of digits, then the method may be unstable when $$A$$ is badly conditioned.
Case 2. You want only an approximation of $$A^+$$.
For the complexity, we assume that $$n=1024,m=2048$$. Anyway, the complexities of the below methods are greater than the complexity of i); note that here we work with a fixed number of significative digits.
i) You can use the full rank QR factorization by GSO method; for the case when $$r=\min(m,n)$$, cf.
https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_inverse#The_QR_method
The complexity is $$\approx \alpha u$$ where $$\alpha= ?$$.
ii) The best known method is the SVD one. cf.
https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_inverse#Singular_value_decomposition_(SVD)
The complexity is $$\approx \dfrac{3}{2}\alpha u$$.
Both methods i),ii) are very stable because they use orthogonal systems.
Remark. QR method gives exact results as expressions containing square roots.
SVD does not give exact results.
iii) There is a third method using a full rank Cholesky factorization of possibly singular symmetric positive matrices. Its complexity is $$\approx \dfrac{1}{4}\alpha u$$. cf.
https://arxiv.org/pdf/0804.4809.pdf
If you want the Moore-Penrose pseudo inverse of a matrix, the the SVD is the best option. | 2020-08-10T20:13:12 | {
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https://math.stackexchange.com/questions/1360246/proving-n-lt-2n-for-n-geq-1-using-induction/1360251 | # Proving $n \lt 2^n$ for $n\geq 1$ using induction
Very close to understanding this, hopefully. Via induction, I'm following a proof but can't understand one of the last steps.
Claim: $n < 2^n$ for natural numbers $n = 1, 2, 3,\ldots$
For step one, $n = 1$, this is obviously true:
$$1 < 2^1 = 1 < 2.$$
Next, assuming $n = k$, is true:
$$k < 2^k.$$
Next I must show that $k + 1 < 2^{k + 1}$ is true, to prove that all natural numbers are true. I begin with this:
$$k < 2^k\\ k + 1 < 2^k + 1$$
Since this is true, adding one more to the RHS is also going to be true:
$$k + 1 < 2^k + 2.$$
Here's where I don't get it. The proof I'm reading claims it's obvious that
$$2^k + 2 \leq 2^{k+1}.$$
Ok yeah, it seems like this is true, but are we certain? Substituting $k = 1$ and $k = 2$ does the trick, and it does seems reasonably intuitive that this would go on forever, but how is this a "formal" result?
From there the proof is completed by putting everything side by side, showing that $k + 1$ is less than all those intermediate steps, resulting in it also being less than $2^{k + 1}$, which makes sense, had I understood that last step!
Yes there are a number of questions similar to this one - couldn't quite find one for this particular glitch. Much thanks!
• possible duplicate of Prove by mathematical induction that $2n ≤ 2^n$, for all integer $n≥1$? Jul 14 '15 at 2:56
• @msteve How is that a duplicate? Obviously $n<2n$, but that is not the same. Jul 14 '15 at 3:07
• They're both proofs of virtually the same statement - at the very least it's worth pointing out that the other thread is there. Jul 14 '15 at 3:09
• @msteve Sure, I would give the link, but I wouldn't actually try to have the question closed because of the similarity. That doesn't seem reasonable. Jul 14 '15 at 3:10
• @msteve I didn't mean to come across as contentious if I did--I can see how such a vote would make sense, but I'm hoping OP will learn a thing or two about induction proofs here (writing up an answer right now). :) Cheers. Jul 14 '15 at 3:13
$2^{k+1}=2\cdot2^{k}=2^{k}+2^{k}$, which will be greater than $2^{k}+1$ so long as $2^{k}>1$. This is true for all values of $k$ you're considering.
Oddly enough, one of the most difficult things about induction problems is actually writing a clear induction proof. As such, I will provide a proof in the spirit of the template I provided a link to above. Hopefully it will clear up any confusion you may have on the matter.
Claim: For all $n\geq 1, n<2^n$.
Proof. For any integer $n\geq 1$, let $S(n)$ denote the statement $$S(n) : n<2^n.$$ Base step ($n=1$): $S(1)$ says that $1<2^1$, and this is true.
Inductive step $S(k)\to S(k+1)$: Fix some $k\geq 1$. Assume that $$S(k) : \color{green}{k<2^k}$$ holds. To be shown is that $$S(k+1) : \color{blue}{k+1<2^{k+1}}$$ follows. Beginning with the left side of $S(k+1)$, \begin{align} \color{green}{k}+1 &< \color{green}{2^k}+1\tag{by $S(k)$, the ind. hyp.}\\[0.5em] &< 2^k+2^k\tag{since $1<2^k$ for all $k\geq 1$}\\[0.5em] &= 2\cdot 2^k\tag{group like terms}\\[0.5em] &= \color{blue}{2^{k+1}}\tag{exponent law} \end{align} one arrives at the right side of $S(k+1)$, thereby showing $S(k+1)$ is also true, completing the inductive step.
Conclusion: By mathematical induction, the statement $S(n)$ is true for all $n\geq 1$. $\blacksquare$
Another approach: $2^1=2 >1$ and left hand side grows faster than R.H side after $x=1/2$, i.e., $\frac {d}{dx} 2^x = \frac {d}{dx} (e^{xln2})=ln2(2^x)> \frac {d}{dx}(x)=1$ for $n>1/2$
So $2^1>1$ , and $2^n$ grows faster than $n$ after $n=1$ ( after around $x=1/2$). | 2021-09-22T18:27:21 | {
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https://math.stackexchange.com/questions/488049/given-a-continuous-random-variable-x-and-the-pdf-find | # Given a continuous random variable $X$ and the pdf find…
There is a pattern excercise:
First pattern: We have $X$ and the probability density function is usually:
$$f(x) = \begin{cases} {ax+bx^2} & {0\lt x \lt 1} \\ 0 & \text{else} \end{cases}$$
Then we are given $E|X| = 0.6$
a) $P\{X>0.5\}$
b) $Var(X)$
Second pattern:
$$f(x) = \begin{cases} {cx^4} & {0\lt x \lt 2} \\ 0 & \text{else} \end{cases}$$
Where c is a constant. Find:
a) Mean value of $X$
b) Variance of $X$
Third pattern (it's more like a reverse problem):
We are given $Y= \sqrt{X}$, find it's cumulative distribution function (CDF) and it's probability density function (pdf).
Update:
The third one is transferred to a new seperate question.
This is not homework. I'm self-studying and in need of some help. Thank you.
• Why strike the third pattern once you received an answer to it? Together with reposting it as a new question, this seems rather disrespectful, don't you think? – Did Sep 9 '13 at 14:56
• I informed the community about my action. I didn't do it behind anyone's back... If you want I can put it back but that doesn't change the fact that eventually it should be separated from the 2 other ones. I didn't want to cause this or offend you in any way... – user2692669 Sep 9 '13 at 15:30
• @user2692669: At the very least you should have informed (via comments to individual answers) all those users who provided answers to your third pattern before you decided to excise it of your intention. Providing a link to your new question would be a bonus, as would be an indication of why you found the answers here lacking. – user642796 Sep 9 '13 at 16:02
I'll start with a step-by-step explanation for the first two, as you say those are more important. Later I'll add the third if I'm able.
The key to solving both of the first two problems is to remember that the pdf for every probability distribution must sum/integrate to one.
## First problem:
We are given: $$f(x) = \begin{cases} {ax+bx^2} & {0\lt x \lt 1} \\ 0 & \text{else} \end{cases}$$
So, we know that: $$\int_0^1 ax+bx^2\; dx = 1$$ So step one becomes: $$\int_0^1 ax+bx^2\; dx = 1\\ \frac{ax^2}{2}+\frac{bx^3}{3}\bigg|_0^1 = 1\\ \frac{a}{2} + \frac{b}{3} = 1\\ 3a+2b=6$$ We also know that $E(X) = 0.6$ As the definition of the expectation is: $E(X) = \int_{-\infty}^\infty xf(x)\;dx$ we know: $$\int_0^1 ax^2+bx^3\; dx = 0.6\\ \frac{ax^3}{3}+\frac{bx^4}{4}\bigg|_0^1 = 0.6\\ \frac{a}{3} + \frac{b}{4} = 0.6\\ 4a+3b=7.2$$ Now we have two equations and two unknowns: $$3a+2b=6\\ 4a+3b=7.2\\ a -\frac{3}{4}b = 1.8\\ a=1.8-\frac{3}{4}b\\ 3\left(1.8-\frac{3}{4}b\right) + 2b = 6\\ 5.4 - \frac{9}{4}b + \frac{8}{4}b = 6\\ 5.4 - 6 = -\frac{b}{4}\\ \mathbf{b = -2.4}\\ a=1.8-\frac{3}{4}-2.4\\ \mathbf{a = 3.6}$$ Now that we know the values, the first question is simply $\int_{0.5}^1 3.6x - 2.4x^2\; dx$ which is: $$\int_{0.5}^1 3.6x-2.4x^2\; dx\\ \frac{3.6x^2}{2}-\frac{2.4x^3}{3}\bigg|_{0.5}^1\\ \left(\frac{3.6\cdot1^2}{2}-\frac{2.4\cdot1^3}{3}\right)-\left(\frac{3.6\cdot{0.5}^2}{2}-\frac{2.4\cdot{0.5}^3}{3}\right)\\ 1 - 0.35 = \mathbf{0.65}$$ The second question asks for the variance. We can use the definition of the variance of $Var(X) = E(X^2) - E(X)^2$. To find the first term we need solve $\int_0^1 ax^3+bx^4\; dx$ which is (following the same derivation as above) $\frac{3.6\cdot1^4}{4}-\frac{2.4\cdot1^5}{5} = \frac{3.6}{4} - \frac{2.4}{5} = 0.42$ So the variance is $0.42-0.6^2 = \mathbf{0.06}$.
## Second problem:
You can solve this one the same way as the first. We must have $\int_0^2 cx^4\;dx = 1$. So: $$\int_0^2 cx^4\;dx = 1\\ \frac{cx^5}{5}\bigg|_0^2 = 1\\ \frac{c\cdot32}{5} = 1\\ \mathbf{c=\frac{5}{32}}$$ The mean value, E(X) is the value of $\int_0^2 \frac{5}{32}x^5\;dx$ which is $\frac{5\cdot2^6}{32\cdot6} = \frac{320}{192} = \mathbf{\frac{5}{3}}$
The variance uses the same relationship as above. $E(X^2) = \int_0^2 \frac{5}{32}x^6\;dx = \frac{5\cdot2^7}{32\cdot7} = \frac{640}{224}$ So the variance is $\frac{640}{224} - {\frac{5}{3}}^2 = \mathbf{\frac{5}{63}}$
Assuming the CDF and PDF of the random variable $X$ are $F_X$ and $f_X$ respectively, and that $X\geqslant0$ almost surely, the CDF $F_Y$ of $Y=\sqrt{X}$ is, by definition, such that, $F_Y(y)=0$ for every $y\lt0$ and, for every $y\geqslant0$, $$F_Y(y)=P[Y\leqslant y]=P[X\leqslant y^2]=F_X(y^2).$$ Differentiating both sides yields the PDF of $Y$ as the function $f_Y$ such that $f_Y(y)=0$ if $y\lt0$ and, for every $y\geqslant0$, $$f_Y(y)=\frac{\mathrm d}{\mathrm dy}F_Y(y)=\frac{\mathrm d}{\mathrm dy}F_X(y^2)=2yf_X(y^2).$$
The first two pattern are solved by first calculating the missing constants. To do so, use the fact that $\int_{\Omega_x}f(x) = 1$. On the first pattern, you have two constants, so you need an extra equation, which comes from the fact that $EX=0.6$ First Pattern: Solve for $\int_{\Omega_x}f(x) = 0.5a+b/3 =1 EX=a/3+b/4=0.6$ Second Pattern: Solve for $\int_{\Omega_x}f(x) = 32c/5 =1$ after calculating $a,b$ you can find the CDF and the required moments. For the third pattern, notice that $P(Y<t)=P(\sqrt{X}<t)=P(X<t^2)$ So given the CDF of X you can just put $t^2$ in it. Basicly, For harder definitions given by $Y=g(X)$ where g is unnecessarily invertible I recommend drawing a plot of g, then when calculating $P(Y<t)$, for each value of $t$ look for the mathcing values of $X$ such that $g(X)<t$ and express $F_Y$ as a function of $F_X$.
• Can you expand on the third pattern for this Y? I can't understand how we get to the bottom of this. I'm not given a cdf it says "find it". (Till the $P(X<t^2)$ I understood how we got there, but after that...) – user2692669 Sep 9 '13 at 10:48
• On this kind of problems you are usually provided with the distribution/PDF/CDF of $X$ and supposed to find the CDF of some kind of transformation of $X$ which I denoted by $Y=g(X)$. If that's not the case try to explain more about the third pattern. – SBM Sep 9 '13 at 11:50
• The exact definition: "If $X$ is a continuous random variable with positive values,find it's Cumulative distribution function (CDF) and it's probability density function (PDF) of/for $Y = \sqrt{X}$" (It asks for both so it's a little vague :S ) – user2692669 Sep 9 '13 at 11:55 | 2019-08-25T01:55:01 | {
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https://www.physicsforums.com/threads/bead-attached-to-a-spring-and-moving-along-a-horizontal-wire.656098/ | # Bead attached to a spring and moving along a horizontal wire
1. Dec 1, 2012
### beowulf.geata
I'm self-studying an introductory book on mathematical methods and models and came across the following problem:
1. A bead of mass m is threaded onto a frictionless horizontal wire. The bead is attached to a model spring of stiffness k and natural length l0, whose other end is fixed to a point A at a vertical distance h from the wire (where h > l0). The position x of the bead is measured from the point on the wire closest to A. Find the potential energy function U(x).
2. Relevant equations
I'm rather puzzled by the solution given in the book, which claims that since the length of the spring is (h2+x2)1/2 and its extension is (h2+x2)1/2 - l0, then U(x) = (1/2)k((h2+x2)1/2 - l0)2.
3. The attempt at a solution
I think that's incorrect because only the x-component of the force exerted by the spring on the bead is relevant to the calculation of U(x). The x-component is -k(l-l0)cos$\theta$, where l = (h2+x2)1/2 and cos$\theta$ = x/(h2+x2)1/2. Hence, U(x) is -$\int$(-kx + kl0x/(h2+x2)1/2)dx. Is this correct?
Last edited: Dec 1, 2012
2. Dec 1, 2012
### CWatters
Humm. Interesting question. I'm not sure I can answer it. At first thought you are correct, but then ...
With the spring stretched the energy stored in the spring will be that calculated in the book. What happens to the energy when the spring is released? Where does it go if you apply conservation of energy? The wire is frictionless so as far as I can see all the energy ends up in the bead (eg not some fraction of that due to the Cosθ issue you raise).
So I conclude the book is correct but I'm willing to be convinced you are right and the book is wrong.
3. Dec 1, 2012
### TSny
See if you can show that both expressions are correct, although they might differ by an additive constant.
4. Dec 1, 2012
### beowulf.geata
The integral should evaluate to
kx2/2 - kl0(h2 + x2)1/2 + C.
Hence, the difference between my solution and the book's is:
kx2/2 - kl0(h2 + x2)1/2 + C - (k(h2 + x2)1/2l0 + (1/2)kh2 + (1/2)kx2 - (1/2)kl02)
= C - (1/2)kh2 + (1/2)kl02,
so the two solutions do appear to differ by a constant... | 2017-12-16T13:58:43 | {
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http://math.stackexchange.com/questions/3888/find-polynomials-such-that-x-16p2x-16x-1px/3889 | # Find polynomials such that $(x-16)p(2x)=16(x-1)p(x)$
Find all polynomials $p(x)$ such that for all $x$, we have $$(x-16)p(2x)=16(x-1)p(x)$$
I tried working out with replacing $x$ by $\frac{x}{2},\frac{x}{4},\cdots$, to have $p(2x) \to p(0)$ but then factor terms seems to create a problem.
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Putting $x=1$ and $x=16$ we see that $p(2)=p(16) = 0$.
Let $p(x) = (x-2)(x-16)g(x)$.
Thus we see that
$$(x-16)(2x-2)(2x-16)g(2x) = 16(x-1)(x-2)(x-16)g(x)$$
Thus $$(x-8)g(2x) = 4(x-2)g(x)$$
We now see that $g(4) = g(8) = 0$
Thus $g(x) = (x-4)(x-8)h(x)$
Thus $$(x-8)(2x-4)(2x-8)h(2x) = 4(x-2)(x-4)(x-8)h(x)$$
i.e
$$h(2x) = h(x)$$
This implies that $h(x)$ is a constant (as otherwise the non-identically zero polynomial $h(2x) - h(x)$ will have an infinite number of roots.)
Thus
$$p(x) = C(x-2)(x-4)(x-8)(x-16)$$
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Good solution! – anonymous Sep 2 '10 at 18:24
Hint $\:$ Note the equation is $\displaystyle \rm\; \frac{\sigma\:f}f \:=\; \frac{\sigma^4 g}g\$ for $\rm\ g = x-16 \:,\:\;$ shift $\rm\;\;\sigma\:f(x) = f(\sigma\:x) = f(2\:x)$
Or, written additively $\rm\ (\sigma-1) \; f \;=\: (\sigma^4 - 1) \; g \quad\;$ [see Note below on this additive notation]
Thus $\rm\;\; f\, =\,\smash{\dfrac{\sigma^4-1}{\sigma-1}}\,g\, =\, (1+\sigma+\sigma^2\!+ \sigma^3) \: g \;=\: (x-16)\:(2x-16)\:(4x-16)\:(8x-16) \;$
unique up to a factor of $\rm\;h \in ker(\sigma-1) = \{h: \sigma\:h = h\:\} = \;$ constants, i.e. $\rm deg\;h = 0 \ \$ QED
Remark $\;$ This reformulation of my prior answer is intended to dramatically illustrate the innate symmetry. Its striking simplicity arises precisely from the simple structure of the orbits of the shift automorphism $\rm\: \sigma.\;$ Namely, by orbit decomposition the problem reduces to one on the single orbit of an irreducible polynomial. By exploiting polynomial structure there, we reduce the problem to a trivial polynomial division by $\rm\:\sigma - 1.\:$ This nicely illustrates the essence of the ideas employed to solve general difference equations (recurrences) over rational function fields - ideas at the foundation of algorithms employed in computer algebra systems.
Hopefully the symmetry is clearer in the above reformulation. Based on prior comments, and someone reposting my prior solution stripped of the symmetry, I fear my prior answer did not succeed in explicitly emphasizing the beautiful innate symmetry lying at the heart of this problem - a germ of the Galois theory of difference fields. Probably the problem was devised to help spur one to discover this beautiful structure.
Note $\;\;$ The point of the additive notation is to exploit the natural polynomial structure of the action of $\rm\:\sigma\:$ on the multiplicative group generated by the elements $\rm\: \sigma^n \,f\;$ in the orbit of $\rm\!\: f \!\:$ under $\rm\:\sigma.\:$ This action is best comprehended by examining it on a specific example. Recalling $\rm\,\sigma\,f(x) = f(2\:\!x)$
\quad\quad\begin{align}{} \rm \sigma\:(\:f(2^{-2}\: x)^a \; f(2^3 x)^b \;\: f(2^5 x)^c)\; =& \rm\;\;\: f(\:2^{-1} x)^a \;\: f(2^4 x)^b \;\: f(2^6 x)^c \\\\ \rm \iff \quad\;\;\: \sigma\;\:(\:(\:\sigma^{-2}\: f\;)^a \;\; (\:\sigma^3\: f\:)^b \;\; (\:\sigma^5\: f\:)^c)\; =& \rm\;\; \;\; (\:\sigma^{-1} \: f\:)^a \;\; (\:\sigma^4 \: f\:)^b \;\; (\:\sigma^6 \: f\;)^c \\\\ \rm \iff \quad\;\; \sigma\; (\:a\:\sigma^{-2} \;+\:\;\; b\:\sigma^3 \;+\;\; c\:\sigma^5)\:\; f\quad\; =& \rm\; (\:a\:\sigma^{-1} \: + \:\;\: b\: \sigma^4 \; +\;\; c\: \sigma^6)\; f \\\\ \end{align}
In the Galois theory literature the above action is frequently written in a highly suggestive exponential form. To illustrate this, below is the key identity in the problem at hand, expressed in this exponential notation.
\quad\quad\begin{align}{} \rm g^{\:\sigma^4-\,1} \;=\;& \rm g^{\:(1\:+\;\sigma\:+\;\sigma^2\:+\:\,\sigma^3)\:(\sigma\,-\,1)} \\\\ \iff\quad\quad\rm \frac{\sigma^4 g}g \;=\;& \rm (g \;\: \sigma\:g \;\:\sigma^2 g \;\:\sigma^3 g)^{\sigma - 1} \;=\; \frac{\phantom{g\;\;\:} \sigma\:g \;\;\: \sigma^2 g \;\;\:\sigma^3 g \;\;\:\sigma^4 g}{g \;\;\:\sigma\:g \;\;\:\sigma^2 g \;\;\:\sigma^3 g\phantom{\;\;\:\sigma^4 g}} \\\\ \end{align}
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very nice way to do it!! & nice to have a view toward generalizations – anon Sep 6 '10 at 17:08
How do you get $(S-1)f = (S^4-1)g$? What does $S-1$ even mean? – Aryabhata Sep 6 '10 at 17:10
btw, Thank you for taking the time to write this out! Sorry for the possibly basic question. – Aryabhata Sep 6 '10 at 17:19
This is the canonical, beautiful solution. Thanks to Bill for posting it; indeed it is not equivalent to the other approaches. To place it more visibly in the familiar operator framework: let F(x) = log |p(x)| and G(x)=log |g(x)|=log |x-16| (or use logarithmic derivatives rather than logs), and S be the linear operator that takes a function h(x) to h(2x), and 1 the identity operator that takes h(x) to h(x). – T.. Sep 6 '10 at 19:32
@Moron: I've added an explicit example to help clarify the additive action. Let me know if this is still not clear. – Bill Dubuque Sep 6 '10 at 20:30
show 1 more comment
Moron's solution and Dubuque's solution are essentilly the same, but I think their presentations can be made neater.
Clearly, $p$ is not a constant polynomial (unless it is zero). So we may write $p(x) = C(x-z_1)(x-z_2)\ldots(x-z_n)$. Then the given equality implies that
$(2x-z_1)(2x-z_2)\ldots(2x-z_n)(x-16) = 16(x-1)(x-z_1)(x-z_2)\ldots(x-z_n)$.
By comparing the leading coefficient on both sides, one sees $n=4$. Thus,
$(x-\frac{z_1}{2})(x-\frac{z_2}{2})(x-\frac{z_3}{2})(x-\frac{z_4}{2})(x-16)=(x-1)(x-z_1)(x-z_2)(x-z_3)(x-z_n)$.
Hence $\lbrace \frac{z_1}{2}, \frac{z_2}{2}, \frac{z_3}{2}, \frac{z_4}{2}, 16\rbrace = \lbrace 1, z_1, z_2, z_3, z_4\rbrace$. Now one easily sees that, up to a relabelling of indices, we must have $z_i = 2^i$.
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+1: This is clearer :-) – Aryabhata Sep 2 '10 at 22:54
Have to disagree with Bill here. I might be a Moron, but this answer is way better than Bill's. This is an easy problem, accusing someone of plagiarism is just silly. – Aryabhata Sep 2 '10 at 23:19
But it is precisely the same as my proof stripped of all the motivational and conceptual details. I'm very sad that you reposted it like that since it may have the effect that some readers may be happy with this stripped-down version and never end up seeing the conceptual background as I presented it. And this - after all - is the whole point of proofs - to learn something new - not merely to determine the boolean value of some proposition. There are always shorter paths but why take them if they miss out on all the beautiful scenery? – Bill Dubuque Sep 2 '10 at 23:37
@Moron: Bill is not accusing me of plagiarism because I have stated clearly that the soln I give is just a rewrite of yours and Bill's. He is just unsatisfied that some of the math concepts in his soln have become hidden in my rewrite, so that one doesn't see a full picture. This is a valid point, but I think getting something that a novice can understand first is more important. – user1551 Sep 3 '10 at 8:41
@user1551: I agree that your answer is much easier to understand (hence my comment about it being better). You can't always cover all the math concepts/insights which could be used to prove something, so I don't see any justification to his objection/accusation. – Aryabhata Sep 3 '10 at 13:44
Since $\rm\;p(2x)\;$ has roots being half the roots $\rm\;r_i\;$ of $\rm\;p(x)\;$, and $\rm\;\frac{p(x)}{p(2x)} = \frac{x-16}{16(x-1)}\;$, for the roots to all cancel like that leaving only one term in the numerator and denominator, we must have simply that the map $\rm\;r\to r/2\;$ is a left shift map on the (intermediate) roots. The numerator shows the largest root is 16, and the denomimator shows half the smallest root is 1. The other intermediate roots must obey the left shift $\rm\;r_i/2 = r_{i-1},\;$ so the roots are $\rm\;2,4,8,16,\;$ i.e. $\rm\;p(x) = c(x-2)(x-4)(x-8)(x-16)\;$
EDIT$\:$ I purposely omitted a detail above since I thought it might be a homework problem. Based on the comments, this caused some confusion, so I will elaborate. Namely, notice that each linear linear factor of $\rm\;p(x)\;$ introduces a constant factor of $2\:$ in the denominator of $\rm\: p(x)/p(2x)$. But said denominator has constant factor $16$, so the $4$ factors found above are complete. In particular, $0$ cannot be a root of $\rm\:p(x).$
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Where do you use the value of 16 in the denominator of the right hand side? If you replace that value by anything else, the solution is the empty set (as is readily seen by letting $x = 0$). – whuber Sep 2 '10 at 21:50
This answer also ignores the possibility that $p(0) = 0$. – Aryabhata Sep 2 '10 at 21:56
@Moron: No, the proof explicitly determines the roots via their shift symmetry. That zero is not one of them is not relevent. – Bill Dubuque Sep 2 '10 at 22:03
@whuber: The proof doesn't need to use the 16 in the denominator (except to verify that said p(x) is a solution - which trivial fact I left to the reader - it's obvious if you understand the proof) – Bill Dubuque Sep 2 '10 at 22:06
@Bill: What you said does nothing to eliminate the fact that x^k could be a factor of p(x). Why can't your 'shift symmetry map' map 0 to itself? Also, $p(x) = 0$ does not imply $p(2x) = 0$ (from the functional equation). For instance, $x=16$. – Aryabhata Sep 2 '10 at 22:14
I love Moron's approach. Because it is usually interesting to look at other solutions, consider what happens if you just write $p(x) = p_n x^n + p_{n-1}x^{n-1} + \cdots + p_0$ and write out the equation. Comparing the coefficients of $x^k$ ($k \gt 0$) yields the recursion
$p_k= \frac{2^{k-5}-1}{2^k - 1} p_{k-1}$,
whence $p_5 = 0$, implying $p_6 = p_7 = \cdots = 0$, and the rest of the recursion yields $p_4, p_3, \ldots, p_0$ as multiples of $p_4$. This approach, although pedestrian, also demonstrates that we need only assume $p$ is analytic near 0. | 2014-03-15T06:28:31 | {
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https://math.stackexchange.com/questions/3149761/find-a-number-field-whose-unit-group-is-isomorphic-to-mathbbz-4-mathbbz-t | # Find a number field whose unit group is isomorphic to $\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}$
Find a number field whose unit group is isomorphic to $$\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}.$$
I'm trying to use Dirichlet's Unit Theorem to solve this problem. It states that if $$K$$ is a number field of signature $$(r,s)$$ and $$\mu_K$$ is the set of roots of unity in $$K$$, then the unit group $$\mathcal{O}_K^{\times}$$ of the ring of integers is isomorphic to $$\mu_K \times \mathbb{Z}^{r+s−1}$$ as an abelian group. So I suppose I want $$r+s-1=1,$$ or $$r+s=2$$. This forces $$(r,s)=(0,2)$$ because if there's at least one real embedding then $$\mu_K$$ is just $$\{\pm 1\}$$ so not $$\mathbb{Z}/4\mathbb{Z}$$.
Therefore I need a number field of degree $$4$$ with four complex embeddings and whose set of roots of unity is isomorphic to $$\mathbb{Z}/4\mathbb{Z}$$. The cyclotomic field $$\mathbb{Q}(\zeta_5)$$ doesn't work because it has more than $$4$$ elements in its set of roots of unity (and $$\mathbb{Q}(\zeta_4)$$ doesn't work because it doesn't have degree $$4$$). I suppose it will have to be $$\mathbb{Q}(\alpha)$$ where the minimal polynomial of $$\alpha$$ has degree $$4$$ but I haven't been able to find an example. Hints would be appreciated.
• If the set of roots of unity in your $K$ has to be cyclic of order $4$ then it has to be $\{\pm1,\pm i\}$ under any complex embedding. Thus, I would search your $K$ among those of the form $\Bbb Q(i,\sqrt{d})$ with $d<0$ to start with. – Andrea Mori Mar 15 at 20:35
• So perhaps I could take $\mathbb{Q}(i, \sqrt{2})$? That has degree $4$ and four complex embeddings. Its set of roots of unity includes $\pm 1,\pm i$, so it would suffice to show there are no other roots of unity in this set. Any other root of unity would need to be a primitive $m$th root of unity with $m=3$ or $m \geq 5$. I'm not sure how to show this is impossible though. – AlephNull Mar 15 at 21:12
• Does $\Bbb Q(\zeta_8)$ get you where you want to be? – Robert Shore Mar 15 at 22:21
• @RobertShore I considered that but I'm not sure what made me refute it. It certainly has degree $4$ and four complex embeddings. But doesn't that have more than $4$ roots of unity? – AlephNull Mar 15 at 22:50
• @AlephNull I think your example of $\mathbb{Q}(i, \sqrt{2})$ works. You can check that the roots of unity in $\mathbb{Q}(\zeta_n)$ are $\mu_n$ if $n$ is even and $\mu_{2n}$ if $n$ is odd. (Intuitively, this is because multiplying by $-1$ will double the order of an odd root of unity) – vxnture Mar 15 at 23:00
You are almost there! Since $$K$$ has a torsion element of order $$4$$, it contains $$\zeta_4$$ and thus contains $$\mathbf{Q}(\zeta_4)$$. Dirichlet's unit theorem then says that $$K$$ must be degree $$4$$ and thus a quadratic extension of $$\mathbf{Q}(\zeta_4)$$.
Now suppose that $$K$$ is any quadratic extension of $$\mathbf{Q}(\zeta_4) = \mathbf{Q}(\sqrt{-1})$$. The signature of $$K$$ is $$(0,2)$$ and so $$K$$ has unit rank one. Also, $$K$$ has an element $$\zeta_4$$ of order $$4$$. The only thing that remains is to find a $$K$$ that doesn't have any extra torsion. But the torsion subgroup of a number field is always cyclic and generated by a $$m$$th root of unity, or a $$4n$$th root of unity in this case since we already have a $$4$$th root of unity. So you just have to ensure that
$$\mathbf{Q}(\zeta_{4n}) \not\subset K$$
for any $$n > 1$$. The degree of $$\mathbf{Q}(\zeta_{4n})$$ is (Euler's $$\varphi$$ function) $$\varphi(4n)$$. This is $$> 4$$ for $$n \ge 4$$. So the answer is:
$$K$$ can be any quadratic extension of $$\mathbf{Q}(\zeta_4)$$ which doesn't equal $$\mathbf{Q}(\zeta_8)$$ or $$\mathbf{Q}(\zeta_{12})$$.
Since $$\mathbf{Q}(\zeta_8) = \mathbf{Q}(\sqrt{-1},\sqrt{2}) = \mathbf{Q}(\sqrt{-1},\sqrt{-2})$$ and $$\mathbf{Q}(\zeta_{12}) = \mathbf{Q}(\sqrt{-1},\sqrt{-3}) = \mathbf{Q}(\sqrt{-1},\sqrt{2})$$, you can find many such $$K$$, for example $$K = \mathbf{Q}(\sqrt{-1},\sqrt{d})$$ for any squarefree $$\pm d > 3$$. These are not the only examples --- the others are precisely all the quadratic extensions $$K/\mathbf{Q}(\sqrt{-1})$$ which are non-Galois over $$\mathbf{Q}$$ such as $$\mathbf{Q}(i,\sqrt{3 + 4 i})$$.
• Perfect answer, and this only uses results proved in my course. – AlephNull Mar 18 at 19:21 | 2019-06-24T09:07:58 | {
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https://mathoverflow.net/questions/385303/the-square-root-of-a-graph/385306 | # The “square root” of a graph?
The number $$f(n)$$ of graphs on the vertex set $$\{1,\dots,n\}$$, allowing loops but not multiple edges, is $$2^{{n+1\choose 2}}$$, with exponential generating function $$F(x)=\sum_{n\geq 0} 2^{{n+1\choose 2}}\frac{x^n}{n!}$$. Consider $$\sqrt{F(x)} = 1+x+3\frac{x^2}{2!}+23\frac{x^3}{3!} +393\frac{x^4}{4!}+13729\frac{x^5}{5!}+\cdots.$$ It's not hard to see that the coefficients 1,1,3,23,393,13729,$$\dots$$ are positive integers. This is A178315 in OEIS. Do they have a combinatorial interpretation?
More generally, we can replace $$2^{{n+1\choose 2}}$$ with $$\sum_G t_1^{c_1(G)} t_2^{c_2(G)}\cdots$$, where $$G$$ ranges over the same graphs on $$\{1,\dots,n\}$$, and where $$c_i(G)$$ is the number of connected components of $$G$$ with $$i$$ vertices. Now we will get polynomials in the $$t_i$$'s with positive integer coefficients, the first four being $$t_1$$ $$t_1^2+2t_2$$ $$t_1^3+6t_1t_2+16t_3$$ $$t_1^4+12t_1^2t_2+64t_1t_3+12t_2^2+304t_4.$$ Again we can ask for a combinatorial interpretation of the coefficients.
Note. What happens if we don't allow loops, so we are looking at $$\sqrt{\sum_{n\geq 0}2^{{n\choose 2}}\frac{x^n}{n!}}$$? Now the coefficient of $$\frac{x^n}{n!}$$ is equal to the coefficient of $$\frac{x^n}{n!}$$ in $$\sqrt{F(x)}$$, divided by $$2^n$$, which in general is not an integer. Hence it makes more sense combinatorially to allow loops.
• Sorry if I am missing something obvious, but that first polynomial, $t_1$, counts the number of graphs with 1 vertex right? Since you are allowing loops, are there not 2 such graphs? – NaturalLogZ Mar 4 at 22:36
• @NaturalLogZ: but then we take the square root and get $t_1$. – Richard Stanley Mar 5 at 13:38
• I took a liberty to add the information from this question and answers to the OEIS. – Max Alekseyev Mar 13 at 16:17
• @MaxAlekseyev: thanks! – Richard Stanley Mar 14 at 17:31
There is a fixed-point-free involution on these graphs which I will call loop-switching, given by adding a loop to every vertex that doesn't have one while simultaneously deleting the loops from all vertices that do. Then $$\sqrt{F(x)}$$ counts equivalence classes of graphs, where two graphs are in the same class if one can be obtained from the other by loop-switching a subset of its connected components. This follows from $$\sqrt{F(x)} = \exp \frac{\log F(x)}{2}$$ where $$\log F(x)$$ counts connected graphs and clearly on those the equivalence classes have size exactly 2. This also nicely explains why allowing loops is important here, and should work the same way in the multivariate version.
I guess to turn this into a "proper" combinatorial interpretation of $$\sqrt{F(x)}$$ as counting some class of structures of which a graph is formed from exactly two, one could somehow choose a canonical representative of each equivalence class on connected graphs. It seems like there is no good way to do this, however, since in some of those classes the two graphs are isomorphic.
These numbers count balanced signed graphs (without loops). A signed graph is a graph in which every edge has a sign, either positive or negative. It is balanced if every cycle has an even number of negative edges.
We will use the following lemma of Harary [Frank Harary, On the notion of balance of a signed graph, Michigan Math. J. 2 (1953/54), 143–146 (1955), Theorem 3]. I'll omit the proof, which is not difficult.
A signed graph is balanced if and only if it is possible to color the vertices in black and white so that every positive edge joins two vertices of the same color and every negative edge joins two vertices of opposite colors.
I'll call these colored graphs bicolored balanced signed graphs. It is easy to see that every connected bicolored balanced signed graph corresponds to exactly two balanced signed graphs
The number of bicolored balanced signed graphs on $$[n]:=\{1,2,\dots,n\}$$ is $$2^{\binom{n+1}2}=2^n\cdot 2^{\binom n2}$$. To see this, we construct all bicolored balanced signed graphs on $$[n]$$ in the following way: We start with an arbitrary graph on $$[n]$$. Then we color the vertices arbitrarily in black and white. Finally we make edges between vertices of the same color positive edges and we make edges between vertices of opposite colors negative edges. There are $$2^{\binom n2}$$ graphs on $$[n]$$ and $$2^n$$ ways of coloring the vertices of each graph in black and white, so there are $$2^n\cdot 2^{\binom n2}$$ bicolored balanced signed graphs.
The exponential generating function for bicolored balanced signed graphs is thus $$F(x)$$. So by the exponential formula, the exponential generating function for connected bicolored balanced signed graphs is $$\log F(x)$$. Since each connected balanced signed graph has two colorings, the exponential generating function for connected balanced signed graphs is $$\tfrac12\log F(x)$$, and by the exponential formula again, the exponential generating function for balanced sign graphs is $$\sqrt{F(x)}$$.
The same approach shows that the polynomials in the $$t_i$$ count balanced signed graphs where each component with $$i$$ vertices is weighted $$t_i$$.
It is curious that the enumeration of unlabeled balanced signed graphs was accomplished by Harary and Palmer in 1967 [F. Harary and E. M. Palmer, On the number of balanced signed graphs, Bulletin of Mathematical Biophysics 29 (1967), 759–765] but the considerably easier enumeration of labeled balanced signed graphs did not appear, as far as I know, until much later [F. Ardila, F. Castillo, Federico, and M. Henley, The arithmetic Tutte polynomials of the classical root systems, Int. Math. Res. Not. IMRN 2015, no. 12, 3830–3877].
There is another interpretation for the coefficients of $$\sqrt{F(x)}$$ which is less natural but easier to see: they count graphs on $$[n]$$ with loops allowed in which the least vertex in each component has a loop.
• Using the second interpretation, also implicit in the answer of lambda, it is easy to see the bijection between (1) pairs $(H,K)$ of graphs on complementary subsets of $\{1,\dots,n\}$ such that the least vertex in each component has a loop, and (2) graphs $G$ (allowing loops) on $\{1,\dots,n\}$. Namely, just loop-switch every vertex of $K$ and take $G$ to be the (disjoint) union of $H$ and $K$. – Richard Stanley Mar 2 at 2:07 | 2021-04-19T22:53:11 | {
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https://math.stackexchange.com/questions/2949452/combinatorics-with-arrangements-of-the-word-universally | # Combinatorics with arrangements of the word UNIVERSALLY
How many ways are there to arrange the letters in UNIVERSALLY so that the four vowels appear in two cluster of two consecutive letters with at least 2 consonants between the two clusters?
For this, we will look at the compliment of the statement. When doing this we have to look at when their are no consonants in between the cluster of vowels and when there is 1 consnant in between the vowel clusters.
CASE 1: No consonants in between vowel cluster
Since there is no consonant in between the two pair of vowel clusters, we can line up all 4 vowels. We are going to treat UIEA as one letter for right now. We have UIEANVRSLLY. Since UIEA is one letter, we have a total of 8 letters. The ways to arrange this is $$\cfrac{8!}{2!1!1!1!1!1!1!}$$. Now lets revisit our vowel cluster. The ways to rearrange UIEA is 4!. Therefore the amount of ways to arrange UNIVERSALLY with no consonants in between vowel cluster is 4! $$\cfrac{8!}{2!}$$
CASE 2: 1 consonant in between vowel cluster
This is where I am stuck. The fact that there is 2 L's is confusing me and I do not understand how to proceed
Any help would be appreciated
Thank you!
Y is used as a vowel in the word UNIVERSALLY, but we will ignore that.
You handled the first case correctly.
In how many ways can the letters of the word UNIVERSALLY be arranged so that the vowels appear in exactly two clusters consisting of two vowels apiece with exactly one consonant between the clusters?
Consider two cases, depending on whether or not that letter is L.
The letter between the two clusters of vowels is an L: The four vowels and the L form a block. We have seven objects to arrange, the block and the six distinct consonants N, V, R, S, L, Y. The objects can be arranged in $$7!$$ ways. Since the L must be in the middle of the block, the letters in the block can be arranged in $$4!$$ ways. Hence, there are $$7!4!$$ possible arrangements in this case.
The letter between the two clusters of vowels in not an L: Choose which of the other five consonants lies between the two clusters of two vowels. Then we have seven objects to arrange, the block, two Ls, and the four distinct consonants that do not appear in the block. The objects can be arranged in $$\binom{7}{2, 1, 1, 1, 1, 1} = \frac{7!}{2!1!1!1!1!1!}$$ distinguishable ways. Since the consonant must be in the middle of the block, the letters in the block can be arranged in $$4!$$ ways. Hence, there are $$\binom{5}{1}\binom{7}{2, 1, 1, 1, 1, 1}4!$$ possible arrangements in this case.
Hence, there are $$7!4! + \binom{5}{1}\binom{7}{2, 1, 1, 1, 1, 1}4!$$ possible arrangements in which exactly one consonant appears between two clusters of two vowels.
Can you take if from here?
• thank you so much! – Parley Oct 10 '18 at 4:18
The 7 letters not in the clusters can be arranged in $$\frac{7!}{2!}$$ ways. The two clusters can straddle any of the 7 letters. CASE2 can occur in $$7(4!) \frac{7!}{2!}$$ ways. This answer agrees with the previous one.
• A nice approach. This tutorial explains how to typeset mathematics on this site. – N. F. Taussig Oct 10 '18 at 7:53 | 2021-03-02T18:21:02 | {
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https://cs.ericyy.me/cs50/week-3/index.html | Week 3
Computational Complexity
• worst-case scenario: $O$
• best-case scenario: $\Omega$
• theta, $\Theta$, if the running time of an algorithm is the same in the worst-case ($\Omega$) and the best-case ($O$).
• We don’t actually care about how complex the algorithm is precisely, only it’s tendency, which dictated by the highest-order term.
• for example, $n^3$, $n^3+n^2$, $n^3-8n^2+20n$. when n is 1,000,000, the values will be $1.0*10^{18}$, $1.000001*10^{18}$ and $9.99992*10^{17}$. The lower terms became really irrelevant. So we only take the highest-order term, which is $n^3$ here.
Common Classes
• from fast to slow:
• $O(1)$ constant time
• $O(\log{n})$ logarithmic time
• binary search
• $O(n)$ linear time
• linear search
• $O(n\log{n})$ linearithmic time
• merge sort
• $O(n^2)$ quadratic time
• bubble sort, selection sort, insert sort
• $O(n^c)$ polynomial time
• $O(c^n)$ exponential time
• $O(n!)$ factorial time
• $O(\infty)$ infinite time
• $O(1)
• More comparisons: https://en.wikipedia.org/wiki/Sorting_algorithm#Comparison_of_algorithms
for i from 0 to n-1
find smallest element between i’th and n-1’th
swap smallest with i’th element
repeat until no swaps
for i from 0 to n-2
if i’th and i+1’th elements out of order
swap them
for i from 1 to n-1
call 0’th through i-1’th elements the “sorted side”
remove i’th element
insert it into the sorted side in order
on input of n elements
if n < 2
return
else
sort left half of elements
sort right half of elements
merge sorted halves
pick an element called a pivot from array
partition the array with pivot
if element i > pivot
move to right
in the end swap the pivot to middle()
recursively apply steps before to the left and the right sub-array without pivot
* [Implement with C](https://gist.github.com/erictt/daede65d8178a93a25a5e52ed07d69aa)
* [Implement with Python 3](https://gist.github.com/erictt/0438c9db11b3b25f0e24c212d8f3c3b9)
## Refers
* [CS50/week 3](http://docs.cs50.net/2016/fall/notes/3/week3.html)
* [Sorting_algorithm - Wikipedia](https://en.wikipedia.org/wiki/Sorting_algorithm)
* [Merge_sort - Wikipedia](https://en.wikipedia.org/wiki/Merge_sort)
* [Quicksort - Wikipedia](https://en.wikipedia.org/wiki/Quicksort)
* [Comparison Sorting Visualization](https://www.cs.usfca.edu/~galles/visualization/ComparisonSort.html) | 2020-10-20T05:26:37 | {
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http://mathhelpforum.com/algebra/80099-arithmetic-progression-problem-print.html | # Arithmetic Progression Problem...
• March 23rd 2009, 01:37 AM
siddscool19
Arithmetic Progression Problem...
If $a_{1},a_{2},a_{3}.........a_{n}$ are in AP,where $a_{1}$>0, then the value of the expression
$\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt {a_{2}}+\sqrt{a_{3}}}+\frac{1}{\sqrt{a_{3}}+\sqrt{ a_{4}}}+........$ up to n terms:
Here are the options:
1. $\frac{1-n}{\sqrt{a_{1}}+\sqrt{a_{n}}}$
2. $\frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}}$
3. $\frac{n-1}{\sqrt{a_{1}}-\sqrt{a_{n}}}$
4. $\frac{1-n}{\sqrt{a_{1}}-\sqrt{a_{n}}}$
• March 23rd 2009, 02:36 AM
chisigma
If the sequence $a_{k}$ is an arithmetic progression then…
$a_{k}= a_{1} + (k-1)\cdot a$
Now we arrive to compute the sum in such a way…
$S= \sum_{k=2}^{n} \frac{1}{\sqrt{a_{k}}+\sqrt {a_{k-1}}}= \sum_{k=2}^{n} \frac{\sqrt{a_{k}} - \sqrt {a_{k-1}}}{a_{k} - a_{k-1}}=$
$= \frac {1}{a}\cdot (\sqrt{a_{2}} -\sqrt{a_{1}} + \sqrt{a_{3}} - \sqrt{a_{2}} + \dots + \sqrt {a_{n}} - \sqrt{a_{n-1}}) =$
$= \frac {1}{a}\cdot (\sqrt {a_{n}} - \sqrt{a_{1}}) = \frac {1}{a}\cdot (\sqrt {a_{1} + (n-1)\cdot a} - \sqrt{a_{1}})$
Kind regards
$\chi$ $\sigma$
• March 23rd 2009, 02:58 AM
n0083
Quote:
Originally Posted by siddscool19
If $a_{1},a_{2},a_{3}.........a_{n}$ are in AP,where $a_{1}$>0, then the value of the expression
$\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt {a_{2}}+\sqrt{a_{3}}}+\frac{1}{\sqrt{a_{3}}+\sqrt{ a_{4}}}+........$ up to n terms:
Here are the options:
1. $\frac{1-n}{\sqrt{a_{1}}+\sqrt{a_{n}}}$
2. $\frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}}$
3. $\frac{n-1}{\sqrt{a_{1}}-\sqrt{a_{n}}}$
4. $\frac{1-n}{\sqrt{a_{1}}-\sqrt{a_{n}}}$
Throughout we use
$a_n = a_1 + (n-1)d$ (*)
$\frac{1}{\sqrt{a_{k}}+\sqrt{a_{k+1}}} = \frac{\sqrt{a_{k}}-\sqrt{a_{k+1}}}{(\sqrt{a_{k}}+\sqrt{a_{k+1}})(\sqr t{a_{k}}-\sqrt{a_{k+1}})}$
$=\frac{\sqrt{a_{k}}-\sqrt{a_{k+1}}}{a_k-a_{k+1}}=\frac{\sqrt{a_{k}}-\sqrt{a_{k+1}}}{a_1+(k-1)d+(a_1-kd)}=\frac{\sqrt{a_{k}}-\sqrt{a_{k+1}}}{-d}$
Now each term has a common denominator namely $-d$. Adding each term up we see the middle terms in the numerator cancel leaving,
$\frac{\sqrt{a_{1}}-\sqrt{a_{n}}}{-d}$ (**)
using (*) again we have $d = \frac{a_n-a_1}{n-1}$.
Substituting this into (**) we have,
$\frac{(\sqrt{a_{1}}-\sqrt{a_{n}})(1-n)}{a_n-a_1}$
Factoring the denominator we have,
$\frac{(\sqrt{a_{1}}-\sqrt{a_{n}})(1-n)}{(\sqrt{a_n}-\sqrt{a_1})(\sqrt{a_n}+\sqrt{a_1})}$
Leaving,
$\frac{n-1}{\sqrt{a_n}+\sqrt{a_1}}$
which is answer choice 2 (thanks chisigma)
• March 23rd 2009, 03:20 AM
chisigma
Quote:
Originally Posted by chisigma
If the sequence $a_{k}$ is an arithmetic progression then…
$a_{k}= a_{1} + (k-1)\cdot a$
Now we arrive to compute the sum in such a way…
$S= \sum_{k=2}^{n} \frac{1}{\sqrt{a_{k}}+\sqrt {a_{k-1}}}= \sum_{k=2}^{n} \frac{\sqrt{a_{k}} - \sqrt {a_{k-1}}}{a_{k} - a_{k-1}}=$
$= \frac {1}{a}\cdot (\sqrt{a_{2}} -\sqrt{a_{1}} + \sqrt{a_{3}} - \sqrt{a_{2}} + \dots + \sqrt {a_{n}} - \sqrt{a_{n-1}}) =$
$= \frac {1}{a}\cdot (\sqrt {a_{n}} - \sqrt{a_{1}}) = \frac {1}{a}\cdot (\sqrt {a_{1} + (n-1)\cdot a} - \sqrt{a_{1}})$
In order to give a complete answer we consider that...
$(\sqrt{a_{n}}- \sqrt{a_{1}}) = \frac{a_{n} - a_{1}}{\sqrt{a_{n}} + \sqrt{a_{1}}}= \frac{(n - 1)\cdot a}{\sqrt{a_{n}} + \sqrt{a_{1}}}$
... so that is...
$S= \frac{n - 1}{\sqrt{a_{n}} + \sqrt{a_{1}}}$
The correct choice is 2)...
Kind regards
$\chi$ $\sigma$
• March 23rd 2009, 03:59 AM
Soroban
Hello, siddscool19!
Quote:
If $a_1,a_2,a_3, \hdots a_n$ are in AP, where $a_1 > 0$, find the value of the expression:
. . $S \;=\;\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{ \sqrt{a_{2}}+\sqrt{a_{3}}}+\frac{1}{\sqrt{a_{3}}+\ sqrt{a_{4}}}+ \hdots + \frac{1}{\sqrt{a_n} + \sqrt{a_{n+1}}}$
Here are the options:
$(1)\;\frac{1-n}{\sqrt{a_{1}}+\sqrt{a_{n}}} \qquad (2)\;\frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}} \qquad (3)\;\frac{n-1}{\sqrt{a_{1}}-\sqrt{a_{n}}} \qquad (4)\; \frac{1-n}{\sqrt{a_{1}}-\sqrt{a_{n}}}$
I don't agree with any of their answers.
Did I miscount?
Let $d$ be the common difference of the AP.
The $k^{th}$ term is: . $\frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}}$
Multiply top and bottom by: $\sqrt{a_{k+1}} - \sqrt{a_k}$
. . $\frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} \cdot\frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{\sqrt{a_{k+1}} - \sqrt{a_k}} \;=\;\frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{\underbrace{a_{k+1} - a_k}_{\text{This is }d}} \;=\;\frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d}$
The series becomes:
. . $S \;=\;\frac{\sqrt{a_2} - \sqrt{a_1}}{d} + \frac{\sqrt{a_3} - \sqrt{a_2}}{d} + \frac{\sqrt{a_4} - \sqrt{a_3}}{d} + \hdots + \frac{\sqrt{a_{n+1}} - \sqrt{a_n}}{d}$
Most of the terms cancel out and we are left with: . $S \;=\;\frac{\sqrt{a_{n+1}} - \sqrt{a_1}}{d}$
$\text{Rationalize: }\;S \;=\;\frac{\sqrt{a_{n+1}} - \sqrt{a_1}}{d} \cdot\frac{\sqrt{a_{n+1}} + \sqrt{a_1}}{\sqrt{a_{n+1}} + \sqrt{a_1}} \;=\; \frac{a_{n+1} - a_1}{d\left(\sqrt{a_{n+1}} + \sqrt{a_1}\right)}$
This can be simplified further, with our knowledge of AP's.
We know that: . $a_{n+1} \:=\:a_1 + nd$
So the numerator is: . $(a_1 + nd) - a_1 \:=\:nd$
And the fraction becomes: . $S \;=\;\frac{nd}{d\left(\sqrt{a_{n+1}} + \sqrt{a_1}\right)}$
Therefore: . $S \;=\;\frac{n}{\sqrt{a_{n+1}} + \sqrt{a_1}}$
. . This is closest to their answer-choice (2).
• March 23rd 2009, 04:34 AM
siddscool19 | 2014-09-17T02:31:24 | {
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https://mathhelpboards.com/threads/what-exactly-is-going-on-with-my-integral.8583/ | # What exactly is going on with my integral?
#### shamieh
##### Active member
Ex 4: Find the volume of the solid obtained by rotating the region bounded by $$\displaystyle y = x - x^2$$and $$\displaystyle y = 0$$ about the line $$\displaystyle x = 2$$
I did all the algebra, integrated etc etc. But I was thinking that the integral would be from $$\displaystyle \int^2_0$$ since it says about the line $$\displaystyle x = 2$$..How do they get from $$\displaystyle \int^1_0$$? I'm almost positive it has to do with the fact that i am saying $$\displaystyle 2 - x$$...But how exactly do I know it's from $$\displaystyle \int^1_0$$. I guess I see the mechanics of what is going on but I don't understand it, or maybe I don't see the mechanics at all and am lost in the dark..Any elaboration would be great.
Last edited:
#### MarkFL
Staff member
The region to be revolved extends from $x=0$ to $x=1$ (sketch the graph). I recommend the shell method for this solid of revolution.
#### Petrus
##### Well-known member
Ex 4: Find the volume of the solid obtained by rotating the region bounded by $$\displaystyle y = x - x^2$$and $$\displaystyle y = 0$$ about the line $$\displaystyle x = 2$$
I did all the algebra, integrated etc etc. But I was thinking that the integral would be from $$\displaystyle \int^2_0$$ since it says about the line $$\displaystyle x = 2$$..How do they get from $$\displaystyle \int^1_0$$? I'm almost positive it has to do with the fact that i am saying $$\displaystyle 2 - x$$...But how exactly do I know it's from $$\displaystyle \int^1_0$$. I guess I see the mechanics of what is going on but I don't understand it, or maybe I don't see the mechanics at all and am lost in the dark..Any elaboration would be great.
Hello,
we got that $$\displaystyle y=0$$ if we put that in to $$\displaystyle y=x-x^2$$ we get $$\displaystyle 0=x-x^2=> x_1=0, x_2=1$$
Well like Mark Said, it is better to draw!
Have a nice day!
regards,
$$\displaystyle |\pi\rangle$$
#### Deveno
##### Well-known member
MHB Math Scholar
We can "slice and dice" the region under consideration two ways:
Shell method: in this method we are going to integrate over $x$ using "very thin shells" (cylinders, approximately).
The center of each shell will be at $x = 2$, and the radius will vary from 2 (when $x = 0$) to 1 (when $x = 1$), because the radius of each shell is:
$r(x) = 2 - x$.
The surface area of each shell is:
$2\pi r(x)f(x)$
where $f(x) = x - x^2$, so our volume element that we integrate over will be:
$dV = 2\pi r(x)f(x)dx = 2\pi (2-x)(x - x^2)dx$
yielding the integral:
$\displaystyle V = 2\pi\int_0^1 (2x - 3x^2 + x^3)\ dx$
The solid of revolution itself looks like the "top half" of a doughnut, with a "hole" in the middle that extends from $x = 1$ to $x = 3$. We *could* integrate from 0 to 2, but after $x = 1$ we are "in the hole" and there's no additional volume to add, because we are actually integrating using the function:
$g(x) = \max(f(x),0)$
(past $x = 1$ the parabola is below the $x$-axis, and "outside the boundary").
Disk method: in this method we are actually integrating over $y$ (slicing parallel to the $x$-axis). This isn't as pretty, because now we have two radii, and "getting the radii right" is more of a chore.
So let's look at the parabola $y = x - x^2$ as if $y$ is the independent variable.
Solving for $x$ in terms of $y$ we get:
$x = \frac{1}{2} \pm \sqrt{\frac{1}{4} - y}$
Hopefully it's clear that the positive square root corresponds to the "top" branch of the parabola (when viewed with the $y$-axis horizontally), and the negative square root is the "lower" branch" (we have to consider these two branches, because otherwise we do not get a function of $y$).
The "lower branch" corresponds to the OUTER radius of our disk (well, we actually get a "washer" shape, not a disk, that is to say: an annulus). The "upper branch" corresponds to the inner radius.
The area of each washer is going to be:
$\pi(R(y)^2 - r(y)^2)$
where:
$R(y) = 2 - \left(\frac{1}{2} - \sqrt{\frac{1}{4} - y}\right) = \frac{3}{2} + \sqrt{\frac{1}{4} - y}$
and:
$r(y) = 2 - \left(\frac{1}{2} + \sqrt{\frac{1}{4} - y}\right) = \frac{3}{2} - \sqrt{\frac{1}{4} - y}$
so our volume element in this case will be:
$dV = \pi(R(y)^2 - r(y)^2)dy = 6\pi\sqrt{\frac{1}{4} - y}\ dy$
leaving us with the somewhat ugly integral to calculate:
$\displaystyle V = 6\pi\int_0^{1/4} \sqrt{\frac{1}{4} - y}\ dy$
If you get the same answer both ways, it's likely it is correct | 2021-06-13T02:39:55 | {
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https://math.stackexchange.com/questions/2670067/find-the-remainder-when-329-is-divided-by-12 | # Find the remainder when $3^{29}$ is divided by $12$.
Find the remainder when $3^{29}$ is divided by $12$. a) $2$ ; b) $3$ ; c) $7$ ; d) $9$ ; e) $12$
Since $\dfrac{3^{29}}{12} = \dfrac{3^{28}}{4} = \dfrac{9^{14}}{4}$, and $9 \equiv 1 \mod 4$, I thought I could do $9^{14} \equiv 1^{14} \mod 4$, so that the answer is $1$. However, that is not one of the answer choices... Where did I go wrong?
• $3$ and $12$ are multiples of $3$, $1$ isn't. And anyway, $1$ isn't even an option.... – Lord Shark the Unknown Feb 28 '18 at 3:53
• And also, the question asks about the remainder modulo $12$, not modulo $4$. – Lord Shark the Unknown Feb 28 '18 at 3:54
• Hint: $\,3^2 = 9 \equiv -3 \pmod{12}\,$. – dxiv Feb 28 '18 at 4:01
• $1/4=3/12{}{}$. – Lord Shark the Unknown Feb 28 '18 at 4:10
Notice that we have
• $3^1 \equiv 3 \mod 12$
• $3^2 \equiv 9 \mod 12$
• $3^3 \equiv 3 \mod 12$
• $3^4 \equiv 9 \mod 12$
• ...
So it goes like in this pattern. In general we have
• $3^{2k} \equiv 9 \mod 12,\ \ k > 0$
• $3^{2k+1} \equiv 3 \mod 12,\ \ k > 0$
And since $29$ is an odd number, the answer should be $3$.
Modular arithmetic does not work well with division. So you cannot say that $\frac{3^{29}}{12} = \frac{3^{28}}{4}$ above, and a collection of similar arguments means your attempt is incorrect.
For the answer, try to find ways to simplify your calculation. For example, note that $3^3 = 27 \equiv 3 \mod 12$, so $$3^{29} \equiv (3^{3})^9 \times 3^2 \equiv 3^9 \times 3^2 \equiv (3^{3})^{3} \times 3^2 \equiv 3^3 \times 3^2 \equiv 3 \times 3^2 \equiv 27 \equiv 3 \mod 12$$
Where I basically used the fact noted many times in a row. Alternately, you could have also said that $3^k$ alternates modulo $27$, and this comes down to the same noted fact.
$$3^2\equiv1\pmod4,3^{2n}=(3^2)^n\equiv1^n$$
$$3^{2n+1}\equiv3\pmod{4\cdot3}$$ | 2020-02-22T15:23:14 | {
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https://classes.areteem.org/mod/forum/discuss.php?d=986 | ## Online Course Discussion Forum
### Math Challenge II-A Combinatorics
Math Challenge II-A Combinatorics
Question 3.26: How many different ways are there to represent 810000 810000 as the product of 3 3 factors if we consider products that differ in the order of factors to be different?
It says that the answer is "We have 810000=243454810000=24⋅34⋅54. Thus, we need to solve a+b+c=4a+b+c=4 a total of three times (once each for 2,3,52,3,5). Thus there are (4+314)3=3375(4+3−14)3=3375 different products.".
However, Shouldn't it be "5+3-1 choose 5" then quantity cubed, which is equal to the cube of 21 which is 9261? Because each exponent can have 5 different results, which are 0,1,2,3,4. There are five, not four numbers, between 0 to 4. Am I wrong or is the answer wrong?
Thanks!
Re: Math Challenge II-A Combinatorics
$$810000=2^4\cdot 3^4 \cdot 5^4$$
As we split it into $3$ factors, each factor is of the form $2^? \cdot 3^?\cdot 5^?$, where the $?$ at the exponents may be from $0$ to $4$. However, we are not just counting one factor; the three factors should look like the following:
$$810000 = (2^a \cdot 3^m \cdot 5^x) (2^b\cdot 3^n\cdot 5^y) (2^c \cdot 3^p \cdot 5^z)$$
For the prime factor $2$, we need $a+b+c=4$. For $3$, we need $m+n+p=4$. For $5$, we need $x+y+z=4$. These are three Stars-and-Bars problems. The $4$ here is the total exponent itself, instead of the number of possible ways. | 2021-09-18T07:08:51 | {
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https://math.stackexchange.com/questions/3019257/how-many-hamiltonian-cycles-are-there-in-a-complete-graph-if-we-discount-the-cyc | # How many Hamiltonian cycles are there in a complete graph if we discount the cycle's orientation or starting point?
Consider a complete graph G with n vertices.
Each vertex is indexed by [n] = {1,2,3...n} where n >= 4.
In this case, a Hamiltonian cycle is determined only by the collection of edges it contains, and we do not need to consider its orientation or starting point.
Question:
1. How many Hamiltonian cycles are there in G?
2. How many Hamiltonian cycles in G contains the edge {1,2}?
3. How many Hamiltonian cycles in G contains the edge {1,2} and {2,3}?
4. How many Hamiltonian cycles in G contains the edge {1,2} and {3,4}?
5. Suppose that M is a set of k <= (n/2) edges, in which no two edges in M share a vertex. How many Hamiltonian cycles contain all edges in M? Give answer in terms of k and n
6. How many Hamiltonian cycles in G do not contain the edge {1,2}, {2,3} and {3,4}?
The question really clusters into two parts.
PART 1: How do I discount "orientation" and "starting point"?
This has to do with 1, 2, 3, 4, and 6.
I can calculate the combinations of edges there can be, but that's not what they're asking. They only want the combinations that form a Hamiltonian cycle.
Additionally, I don't see how you can just know whether a Hamiltonian cycle has crossed through a certain edge.
The more I think about it, the more I feel this is about combinatorial numbers as opposed to graph theory. Are they trying to trick me?
PART 2: How many cycles contain a set of edges that do not share a vertex.
This has to do with question 5, specifically.
My first response is "none..."?
If the graph is a complete graph, then all edges share a vertex at some point right? In that case, M seems to be an empty set and there are no Hamiltonian cycles that cover it. But that doesn't feel right at all...
Hint: if we do consider starting point and orientation, then the number of Hamiltonian cycles is the number of ways that we can order $$[n]$$, i.e. the number of permutations. If you know the order in which to visit the vertices, this tells you exactly the cycle. Each cycle is then counted $$n$$ times for each possible starting point, and twice for each direction around the cycle.
Hint for part 2: A cycle can contain $$\{ 1,2 \}$$ and $$\{ 3,4 \}$$ if it (for example) also contains edge $$\{ 2,3 \}$$. | 2019-05-21T14:35:31 | {
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http://www.math.uah.edu/stat/buffon/Buffon.html |
## 1. Buffon's Problems
Buffon's experiments are very old and famous random experiments, named after comte de Buffon. These experiments are considered to be among the first problems in geometric probability.
### Buffon's Coin Experiment
Buffon's coin experiment consists of dropping a coin randomly on a floor covered with identically shaped tiles. The event of interest is that the coin crosses a crack between tiles. We will model Buffon's coin problem with square tiles of side length 1—assuming the side length is 1 is equivalent to taking the side length as the unit of measurement.
#### Assumptions
First, let us define the experiment mathematically. As usual, we will idealize the physical objects by assuming that the coin is a perfect circle with radius $$r$$ and that the cracks between tiles are line segments. A natural way to describe the outcome of the experiment is to record the center of the coin relative to the center of the tile where the coin happens to fall. More precisely, we will construct coordinate axes so that the tile where the coin falls occupies the square $$S = \left[ -\frac{1}{2}, \frac{1}{2} \right]^2$$.
Now when the coin is tossed, we will denote the center of the coin by $$(X, Y) \in S$$ so that $$S$$ is our sample space and $$X$$ and $$Y$$ are our basic random variables. Finally, we will assume that $$r \lt \frac{1}{2}$$ so that it is at least possible for the coin to fall inside the square without touching a crack.
Next, we need to define an appropriate probability measure that describes our basic random vector $$(X, Y)$$. If the coin falls randomly on the floor, then it is natural to assume that $$(X, Y)$$ is uniformly distributed on $$S$$. By definition, this means that
$\P[(X, Y) \in A] = \frac{\area(A)}{\area(S)}, \quad A \subseteq S$
Run Buffon's coin experiment with the default settings. Watch how the points seem to fill the sample space $$S$$ in a uniform manner.
#### The Probability of a Crack Crossing
Our interest is in the probability of the event $$C$$ that the coin crosses a crack.
The probability of a crack crossing is $$\P(C) - 1 - (1 - 2 r)^2$$.
Proof:
Note that in terms of the basic variables, $$C^c = \left\{r - \frac{1}{2} \lt X \lt \frac{1}{2} - r, r - \frac{1}{2} \lt Y \lt \frac{1}{2} - r\right\}$$.
In Buffon's coin experiment, change the radius with the scroll bar and watch how the events $$C$$ and $$C^c$$ and change. Run the experiment with various values of $$r$$ and compare the physical experiment with the points in the scatterplot. Compare the relative frequency of $$C$$ to the probability of $$C$$.
The convergence of the relative frequency of an event (as the experiment is repeated) to the probability of the event is a special case of the law of large numbers.
Solve Buffon's coin problem with rectangular tiles that have height $$h$$ and width $$w$$.
$1 - \frac{(h - 2 \, r)(w - 2 \, r)}{h \, w}, \quad r \lt \min \left\{ \frac{h}{2}, \frac{w}{2} \right\}$
Solve Buffon's coin problem with equilateral triangular tiles that have side length 1.
Recall that random numbers are simulation of independent random variables, each with the standard uniform distribution, that is, the continuous uniform distribution on the interval $$(0, 1)$$.
Show how to simulate the center of the coin $$(X, Y)$$ in Buffon's coin experiment using random numbers.
$$X = U - \frac{1}{2}$$, $$Y = V - \frac{1}{2}$$, where $$U$$ and $$V$$ are random numbers.
### Buffon's Needle Problem
Buffon's needle experiment consists of dropping a needle on a hardwood floor. The main event of interest is that the needle crosses a crack between floorboards. Strangely enough, the probability of this event leads to a statistical estimate of the number $$\pi$$!
#### Assumptions
Our first step is to define the experiment mathematically. Again we idealize the physical objects by assuming that the floorboards are uniform and that each has width 1. We will also assume that the needle has length $$L \lt 1$$ so that the needle cannot cross more than one crack. Finally, we assume that the cracks between the floorboards and the needle are line segments.
When the needle is dropped, we want to record its orientation relative to the floorboard cracks. One way to do this is to record the angle $$X$$ that the top half of the needle makes with the line through the center of the needle, parallel to the floorboards, and the distance $$Y$$ from the center of the needle to the bottom crack. These will be the basic random variables of our experiment, and thus the sample space of the experiment is $S = [0, \pi) \times [0, 1) = \{(x, y): 0 \le x \lt \pi, \; 0 \le y \lt 1\}$
Again, our main modeling assumption is that the needle is tossed randomly on the floor. Thus, a reasonable mathematical assumption might be that the basic random vector $$(X, Y)$$ is uniformly distributed over the sample space. By definition, this means that $\P[(X, Y) \in A] = \frac{\area(A)}{\area(S)}, \quad A \subseteq S$
Run Buffon's needle experiment with the default settings and watch the outcomes being plotted in the sample space. Note how the points in the scatterplot seem to fill the sample space $$S$$ in a uniform way.
#### The Probability of a Crack Crossing
Our main interest is in the event $$C$$ that the needle crosses a crack between the floorboards.
The event $$C$$ can be written in terms of the basic angle and distance variables as follows: $C = \left\{ Y \lt \frac{L}{2} \, \sin(X) \right\} \cup \left\{ Y \gt 1 - \frac{L}{2} \, \sin(X) \right\}$
The curves $$y = \frac{L}{2} \, \sin(x)$$ and $$y = 1 - \frac{L}{2} \, \sin(x)$$ on the interval $$0 \le x \lt \pi$$ are shown in blue in the scatterplot of Buffon's needle experiment, and hence event $$C$$ is the union of the regions below the lower curve and above the upper curve. Thus, the needle crosses a crack precisely when a point falls in this region.
The probability of a crack crossing is $$\P(C) = 2 L / \pi$$.
Proof:
From symmetry and simple calculus it follows that $$\area(C) = 2 L$$. Of course $$\area(S) = \pi$$.
In the Buffon's needle experiment, vary the needle length $$L$$ with the scroll bar and watch how the event $$C$$ changes. Run the experiment with various values of $$L$$ and compare the physical experiment with the points in the scatterplot. Compare the relative frequency of $$C$$ to the probability of $$C$$.
The convergence of the relative frequency of an event (as the experiment is repeated) to the probability of the event is a special case of the law of large numbers.
Find the probabilities of the following events in Buffon's needle experiment. In each case, sketch the event as a subset of the sample space.
1. $$\{0 \lt X \lt \pi / 2, \; 0 \lt Y \lt 1 / 3\}$$
2. $$\{1 / 4 \lt Y \lt 2 / 3\}$$
3. $$\{X \lt Y\}$$
4. $$\{X + Y \lt 2\}$$
1. $$\frac{1}{6}$$
2. $$\frac{5}{12}$$
3. $$\frac{1}{2 \pi}$$
4. $$\frac{3}{2 \pi}$$
#### The Estimate of $$\pi$$
Suppose that we run Buffon's needle experiment a large number of times. By the law of large numbers, the proportion of crack crossings should be about the same as the probability of a crack crossing. More precisely, we will denote the number of crack crossings in the first $$n$$ runs by $$N_n$$. Note that $$N_n$$ is a random variable for the compound experiment that consists of $$n$$ replications of the basic needle experiment. Thus, if $$n$$ is large, we should have $$\frac{N_n}{n} \approx \frac{2 L}{\pi}$$ and hence $\pi \approx \frac{2 L n}{N_n}$ This is Buffon's famous estimate of $$\pi$$. In the simulation of Buffon's needle experiment, this estimate is computed on each run and shown numerically in the second table and visually in a graph.
Run the Buffon's needle experiment with needle lengths $$L \in \{0.3, 0.5, 0.7, 1\}$$. In each case, watch the estimate of $$\pi$$ as the simulation runs.
Let us analyze the estimation problem more carefully. On each run $$j$$ we have an indicator variable $$I_j$$, where $$I_j = 1$$ if the needle crosses a crack on run $$j$$ and $$I_j = 0$$ if the needle does not cross a crack on run $$j$$. These indicator variables are independent, and identically distributed, since we are assuming independent replications of the experiment. Thus, the sequence forms a Bernoulli trials process.
The number of crack crossings in the first $$n$$ runs of the experiment is $N_n = \sum_{j=1}^n I_j$ which has the binomial distribution with parameters $$n$$ and $$2 L / \pi$$.
The mean and variance of $$N_n$$ are
1. $$\E(N_n) = n \frac{2 L}{\pi}$$
2. $$\var(N_n) = n \frac{2 L}{\pi} \left(1 - \frac{2 L}{\pi}\right)$$
With probability 1, $$\frac{N_n}{2 L n} \to \frac{1}{\pi}$$ as $$n \to \infty$$ and $$\frac{2 L n}{N_n} \to \pi$$ as $$n \to \infty$$.
Proof: a
These results follow from the strong law of large numbers.
Thus, we have two basic estimators: $$\frac{N_n}{2 L n}$$ as an estimator of $$\frac{1}{\pi}$$ and $$\frac{2 L n}{N_n}$$ as an estimator of $$\pi$$. The estimator of $$\frac{1}{\pi}$$ has several important statistical properties. First, it is unbiased since the expected value of the estimator is the parameter being estimated:
The estimator of $$\frac{1}{\pi}$$ is unbiased: $\E \left( \frac{N_n}{2 L n} \right) = \frac{1}{\pi}$
Proof:
This follows from the results above for the binomial distribution and properties of expected value.
Since this estimator is unbiased, the variance gives the mean square error: $\var \left( \frac{N_n}{2 L n} \right) = \E \left[ \left( \frac{N_n}{2 L n} - \frac{1}{\pi} \right)^2 \right]$
The mean square error of the estimator of $$\frac{1}{\pi}$$ is $\var \left( \frac{N_n}{2 L n} \right) = \frac{\pi - 2 L}{2 L n \pi^2}$
The variance is a decreasing function of the needle length $$L$$.
Thus, the estimator of $$\frac{1}{\pi}$$ improves as the needle length increases. On the other hand, the estimator of $$\pi$$ is biased; it tends to overestimate $$\pi$$:
The estimator of $$\pi$$ is positively biased: $\E \left( \frac{2 L n}{N_n} \right) \ge \pi$
Proof:
The estimator of $$\pi$$ also tends to improve as the needle length increases. This is not easy to see mathematically. However, you can see it empirically.
In the Buffon's needle experiment, run the simulation 5000 times each with $$L = 0.3$$, $$L = 0.5$$, $$L = 0.7$$, and $$L = 0.9$$. Note how well the estimator seems to work in each case.
Finally, we should note that as a practical matter, Buffon's needle experiment is not a very efficient method of approximating $$\pi$$. According to Richard Durrett, to estimate $$\pi$$ to four decimal places with $$L = \frac{1}{2}$$ would require about 100 million tosses!
Run the Buffon's needle experiment until the estimates of $$\pi$$ seem to be consistently correct to two decimal places. Note the number of runs required. Try this for needle lengths $$L = 0.3$$, $$L = 0.5$$, $$L = 0.7$$, and $$L = 0.9$$ and compare the results.
Show how to simulate the angle $$X$$ and distance $$Y$$ in Buffon's needle experiment using random numbers.
$$X = \pi U$$, $$Y = V$$, where $$U$$ and $$V$$ are random numbers.
Neil Weiss has pointed out that our computer simulation of Buffon's needle experiment is circular, in the sense the program assumes knowledge of $$\pi$$ (you can see this from the simulation result above).
Try to write a computer algorithm for Buffon's needle problem, without assuming the value of $$\pi$$ or any other transcendental numbers. | 2015-01-28T14:08:36 | {
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https://math.stackexchange.com/questions/3194454/different-ways-of-computing-probability | # Different ways of computing probability
There are twelve unique watches and five men. Each of the five men are asked to choose a watch for themselves. What is the probability that at-least two of them choose the same watch.
I could think of two different approaches of solving above problem.
1st approach: $$\text{answer} = 1 - \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 }{ 12^5} \approx 0.618055$$
2nd approach: Consider $$x$$ such that \begin{aligned} x = &\hphantom{{}+{}} \text{no of ways exactly}~\textbf{two}~\text{men choose same watch} \\ &{}+ \text{no of ways exactly}~\textbf{three}~\text{men choose same watch} \\ &{}+\text{no of ways exactly}~\textbf{four}~\text{men choose same watch} \\ &{}+ \text{no of ways exactly}~\textbf{five}~\text{men choose same watch} \\ = &\hphantom{{}+{}} {5\choose 2} \cdot 12 \cdot {11\choose 3} \cdot 3! \\ &+ {5\choose 3} \cdot 12 \cdot {11\choose 2} \cdot 2! \\ &+ {5\choose 4} \cdot 12 \cdot {11\choose 1} \cdot 1! \\ &+ {5\choose 5} \cdot 12 \cdot {11\choose 0} \cdot 0! \\ ={} &118800 + 13200 + 660 + 12 = 132672 \end{aligned} Then $$\text{answer} = x\, / \,12^5 = 132672\, / \, 12^5 \approx 0.53317$$
The two approaches have different answers.
I believe the first approach has no issues, which means i am counting short in the second approach. But not able to figure out how?
• What about 2 choose watch A and 2 others choose Watch B, etc... – DJohnM Apr 20 at 10:25
Three men choose one model and two other men choose a second model: There are $$\binom{5}{3}$$ ways that three of the men can select the same model and twelve models for them to choose. There are eleven ways for the other two men to pick the same model. Hence, there are $$\binom{5}{3}\binom{12}{1}\binom{2}{2}\binom{11}{1} = 1320$$ such distributions.
Two men choose one model, two other men choose a second model, and the fifth man picks a third model: There are $$\binom{12}{2}$$ ways to select which two models are each picked by two men, $$\binom{5}{2}$$ ways for two of the men to pick the more expensive of those two models (or some other feature that distinguishes them if they are the same price), $$\binom{3}{2}$$ ways for two of the other three men to pick the other of those models, and ten ways for the remaining man to pick one of the remaining models. Hence, there are $$\binom{12}{2}\binom{5}{2}\binom{3}{2}\binom{1}{1}\binom{10}{1} = 19800$$ such distributions.
Adding these to the total gives $$118800 + 13200 + 660 + 12 + 1320 + 19800 = 153792$$ Dividing this amount by $$12^5$$ yields $$\approx 0.6180555556$$, which agrees with the answer obtained from your simpler first approach. | 2019-08-21T13:48:43 | {
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http://wangchaofeng.com/parseNote.php?note=notes_math_elementaryAlgebra.note | All notes
ElementaryAlgebra
# Basics
## Exponents
• Ref: wisc classware.
• Ref: Ismor Fischer from UW-Madison.
### Power function
For any real number base x, the powers of x is defined as: $x^0 = 1, x^1 = x, x^2 = x⋅x, x^3 = x⋅x⋅x,$ etc. The exception is $0^0$, which is considered indeterminate.
Powers are also called exponents.
Fractional exponents is defined in terms of roots:
• $x^{1/2} = \sqrt{x}$ , the square root of x.
• $x^{1/3} = \sqrt[3]{x}$ , the cube root of x, etc.
• In general, we have $x^{m/n} = (\sqrt[n]{x})^m$, i.e., the $n^{th}$ root of x, raised to the $m^{th}$ power.
• Negative exponents: $x^{-1}=\frac{1}{x}$.
Power functions have the general form $y = x^p$, for any real value of p, for $x \gt 0$.
$f(x)=x^2$: quadratic equation forms a curved parabola in the XY-plane. In this case, the curve is said to be concave up, i.e., it “holds water.” Similarly, the graph of $–x^2$ is concave down; it “spills water".
$f(x)=x^{-1}$. This is the graph of a hyperbola, which has two branches, one in the first quadrant and the other in the third. As we move to the right, the graph approaches the X-axis as a horizontal asymptote.
The hyperbola is one of the three kinds of conic section, formed by the intersection of a plane and a double cone. The other conic sections are the parabola and the ellipse.
### Exponential function
The var $x$ is an exponent. $f(x) = 2^x$.
The inverse of the exponential function $y = b^x$ is, by definition, the logarithm function $y = log_b (x)$.
#### Lambert W function
quora.com. Solve a hard equations: $2^x = x$.
$$x = 2^x$$ $$x = e^{\ln(2^x)} = e^{x \ln 2}$$
To further on, we needs knowlege of Lambert W function.
The Lambert W function (also called the omega function or product logarithm, is a set of functions, namely the branches of the inverse relation of the function $f(z) = ze^z$ where $z$ is any complex number. In other words $$z=f^{-1}(ze^{z})=W(ze^{z})$$ which can also be expressed as $$f(W) = We^W$$ Wolfram.
By substituting the above equation in $z'=ze^{z}$, we get the defining equation for the W function (and for the W relation in general): $$z'=W(z')e^{W(z')}$$ for any complex number $z'$.
Since the function $f$ is not injective, the relation W is multivalued (except at 0).
##### Examples
\begin{aligned} 2^{t}&=5t\\ 1&={\frac {5t}{2^{t}}}\\ 1&=5t\,e^{-t\ln 2}\\ {\frac {1}{5}}&=t\,e^{-t\ln 2}\\ {\frac {-\ln 2}{5}}&=(-t\ln 2)\,e^{(-t\ln 2)}\\ W\left({\frac {-\ln 2}{5}}\right)&=-t\ln 2\\ t&=-{\frac {W\left({\frac {-\ln 2}{5}}\right)}{\ln 2}} \end{aligned}
Another example: \begin{aligned} x^{x}&=z\\ \Rightarrow x\ln x&=\ln z\\ \Rightarrow e^{\ln x}\cdot \ln x&=\ln z\\ \Rightarrow \ln x&=W(\ln z)\\ \Rightarrow x&=e^{W(\ln z)}\\ \end{aligned}
or, equivalently, $x={\frac {\ln z}{W(\ln z)}}$ since: $\ln z=W(\ln z)e^{W(\ln z)}$ by definition.
For quadratic equation $0 = ax^2 + bx + c$, the quadratic formula is: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
### Formula derivation
For equation $0 = ax^2 + bx + c$. $$0 = x^2 + \frac{b}{a} x + \frac{c}{a} \quad (\forall a \neq 0)$$ Using binomial theorem, $(x+a)^2 = x^2 + 2xa + a^2$, so: $$0 = x^2 + \frac{b}{a} x + (\frac{b}{2a})^2 - (\frac{b}{2a})^2 + \frac{c}{a}$$ $$(x + \frac{b}{2a})^2 = (\frac{b}{2a})^2 - \frac{c}{a}$$ $$x + \frac{b}{2a} = \pm \sqrt{ (\frac{b}{2a})^2 - \frac{c}{a} }$$ In this step, $(\frac{b}{2a})^2 - \frac{c}{a} = \frac{b^2 - 4ac}{(2a)^2} \geq 0 \Rightarrow b^2-4ac \geq 0$ $$x + \frac{b}{2a} = \pm \sqrt{ (\frac{b^2}{(2a)^2}) - \frac{4ac}{(2a)^2} }$$ $$x = -\frac{b}{2a} \pm \frac{1}{2a} \sqrt{b^2 - 4ac}$$ $$x = -\frac{b \pm \sqrt{b^2 - 4ac}}{2a}$$
## Binomial Theorem
$$(a+b)^n = \sum_{i=0}^{n} \binom{n}{i} a^i b^{n-i}$$ where $\binom{n}{i}$ denotes the total number of different combinations of $i$ items chosen from within $n$ items. $$\binom{n}{i} = \frac{n!}{i!(n-i)!}$$
See wikipedia: binomial expansion visualization for geometry interpretation of binomial expansion.
https://www.quora.com/?digest_story=4219692 Here is a general approach for solving problems like this. Before getting into the details, let's first consider a simpler version -- if x+y+z=0, proving that x^3+y^3+z^3=3xyz. Now, some people happen to remember the identity that factors x^3+y^3+z^3−3xyz, which makes that problem easy. But the point is that even for that case, the amount of brute-force work it takes to multiply everything out and prove that identity is not negligible, and one can imagine how much work it would be to do this for seventh powers. So that's the motivation of why we need a better approach. Now, if we have symmetric expressions for three variables x,y,z, it is interesting to look at them as roots of a third-degree polynomial. In particular, if x+y+z=0, then these guys are the roots of a polynomial t^3+at+b for some a,b -- and it is not hard to see that a=xy+yz+zx, b=−xyz -- this follows from the definition of a polynomial with particular roots. The nice thing is that this will be the only brute force work that will be needed. With this approach, the simpler version of the problem is a one-line solution: take t^3+at+b=0 for t=x,y,z, add them up, and since any constant times the sum of the three variables vanishes anyway, we get x^3+y^3+z^3+3b=0. Finally turning to the more complex problem given: note that x^2+y^2+z^2=(x+y+z)^2−2(xy+yz+zx), so x^2+y^2+z^2=−2a. As for the higher powers, consider the following: t^5+at^3+bt^2=0 so summing across x,y,z and reusing the results for second and third powers already known yields x^5+y^5+z^5=−a(−3b)−b(−2a)=5ab Finally, t^6=(at+b)2, so t^7=t(at+b)^2=a^2t^3+2abt^2+b^2t Again doing the sum yields x7+y7+z7=a2(−3b)+2ab(−2a)=−7a2b Therefore, we just need to prove that (−2a/2)(5ab/5)=−7a^2b/7, which is obvious. | 2017-12-17T17:45:56 | {
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https://www.coursehero.com/file/p2ievuc/a-Prove-that-the-sequence-defined-by-y-n-sup-a-k-k-n-converges-Proof-For-any-n/ | a Prove that the sequence defined by y n sup a k k n converges Proof For any n
# A prove that the sequence defined by y n sup a k k n
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(a) Prove that the sequence defined byyn= sup{ak:kn}converges.Proof.For anyn∈ {1,2,3, . . .}, define the setAn:={an, an+1, an+2, . . .}.2
In other words,yn= supAn. As defined,An+1An, soyn+1yn.Since this is true for anyn, we see that the sequence (yn) is decreasing.Also, since (an) is bounded, there existsM>0 such that|an| ≤Mfor alln{1,2,3, . . .}. In particular, for anyn,-Manyny1,so we see that the sequence (yn) is bounded.Therefore, since (yn) is a bounded, decreasing sequence, the Monotone ConvergenceTheorem implies that it converges.(b) Thelimit superiorof (an), or lim supan, is defined bylim supan= limyn,whereynis the sequence from part (a) of this exercise. Provide a reasonable definitionfor lim infanand briefly explain why it always exists for any bounded sequence.Answer.Here’s the definition of lim infan:For a bounded sequence (an), define the sequence (zn) byzn= inf{ak:kn}.Then thelimit inferiorof (an) is defined to belim infan:= limzn.The sequence (zn) is increasing and bounded (it’s bounded below byz1and above bythe upper bound for (an)), so the Monotone Convergence Theorem implies it converges,so the above limit definitely exists.(c) Prove that lim infanlim supanfor every bounded sequence, and give an example ofa sequence for which the inequality is strict.Proof.Note that, with (yn) and (zn) defined as above,znanynfor alln∈ {1,2,3, . . .}. In particular,znyn. Therefore, by Theorem 2.3.4(ii),lim infan= limznlimyn= lim supan.Consider the sequence (an) given byan= 2 + (-1)n(1 + 1/n).Then it’s straightforward to show that lim infan= 1 and lim supan= 3, so the inequalityis strict for this sequence.3
(d) Show that lim infan= lim supanif and only if limanexists. In this case, all three sharethe same value.Proof.To show the forward implication, assume lim infan= lim supan. In other words,with (yn) and (zn) defined as above, limzn= limyn. Now, we know thatznanynfor alln∈ {1,2,3, . . .}, so the Squeeze Theorem implies that limanexists and is equalto the common limit of theyn’s andzn’s.Turning to the reverse implication, assume liman=Lfor someLR. Let>0. Since(an)L, there existsNNsuch that, whenevernN,|an-L|</2or, equivalently,-/2< an-L </2.
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• Summer '19
• Math, lim, Mathematical analysis, Limit of a sequence, Limit superior and limit inferior | 2020-10-30T21:30:13 | {
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https://stats.stackexchange.com/questions/447863/log-transforming-target-var-for-training-a-random-forest-regressor/448153 | # Log-Transforming target var for training a Random Forest Regressor
I have a variable that I want to model, which has a skewed distribution. Log transforming the var gives is a normal-like distribution. When training a Random Forest regressor on the non-transformed var, I get worse performance than when I log-tranform the var. I am bit puzzled about whether I should do this knowning that the random forest regressor is predicting the mean of the leafs. If trained on a log tranformed var, that means that the prediction is the mean of the logs of the values in the leafs. Which (when tranformed back) is not equal to the mean of the real values.
Any opinion?
I will be assuming that by "better performance" you mean better CV/validation performance, and not train one.
I want to invite you to think of what the effect of log-transforming the target variable is on single regression trees
Regression trees make splits in a way that minimizes the MSE, which (considering that we predict the mean) means that they minimize the sum of the variances of the target in the children nodes.
What happens if your target is skewed?
If your variable is skewed, high values will affect the variances and push your split points towards higher values - forcing your decision tree to make less balanced splits and trying to "isolate" the tail from the rest of the points.
Example of a single split on non-transformed and transformed data:
As a result overall, your trees (and so on RF) will be more affected by your high-end values if your data is not transformed - which means that they should be more accurate in predicting high values and a bit less on the lower ones.
If you log-transform you reduce the relative importance of these high values, and accept having more error on those while being more accurate on the bulk of your data. This might generalize better, and - in general - also makes sense. Indeed in the same regression, predicting $$\hat{y}=105$$ when $$y=100$$ is better than predicting $$\hat{y}=15$$ when $$y=11$$, because the error in relative terms often matters more than the absolute one.
Hope this was useful!
• This is super useful. idd by performance I am referring to CV. My only concern has to do with with the averaging of the log-transformed values but I do see the overall point of transforming to mitigate your estimator putting more emphasis on the overall performance rather than the densest part of the distribution Feb 6, 2020 at 12:28
• it's the opposite - by logtrasforming you try to have a more general performance and focus "less" on the tails and more on the dense part of the distribution. And it's not just for regression trees, the same works for any regression problem as far as I know. Feb 6, 2020 at 13:13
• @LetsPlayYahtzee - that follows the same problem - the average of the log is not the log of the averages, and happily so: averaging the log values rather than taking the log of the averages is LESS sensitive to tail values, and therefore is in line with the choice that you make of log transforming your target in order to make the skewness less impactful of the results Feb 6, 2020 at 20:49
Tangentially, the marginal distribution (that is, the distribution obtained when plotting a histogram) of the outcome is irrelevant in regression since most regression methods make assumptions about the conditional distribution (that is, the distribution obtained when plotting the histogram of the outcome were I to only observe outcomes which have the same features). Now, on to your question.
If you are evaluating the performance of on the transformed outcome, the results can be misleading. Because the log essentially squeezes the outcomes, the variance is also shrunk meaning predictions will be closer to the observations. This shrinks the loss and appears to make your model better. Try doing this
from sklearn.dummy import DummyRegressor
from sklearn.model_selection import cross_val_score
cross_val_score(DummyRegressor(), X, y, scoring = 'neg_mean_squared_error')
cross_val_score(DummyRegressor(), X, np.log(y), scoring = 'neg_mean_squared_error')
Same data, but the scores are immensely different. Why? Because the log shrinks the variance of the outcomes making the model appear better even though it does nothing different.
If you want to transform your outcome, you can:
• Train the model on the transformed outcomes
• Predict on a held out set
• Re-transform the predictions to the original space
• Evaluate the prediction quality in the original space
Sklearn makes this very easy with their TransformedTargetRegressor.
from sklearn.ensemble import RandomForestRegressor
from sklearn.compose import TransformedTargetRegressor
from sklearn.model_selection import GridSearchCV, train_test_split
from sklearn.pipeline import Pipeline
from sklearn.datasets import make_regression
import numpy as np
rf = RandomForestRegressor()
log_rf = TransformedTargetRegressor(rf, func = np.log, inverse_func=np.exp)
params = {'regressor__n_estimators': [10,100,1000]}
gscv = GridSearchCV(log_rf, param_grid=params,refit = True)
X,y = make_regression(n_samples = 10_000, n_features=50, n_informative=5)
y -= y.min()-1 #Make the outcome positive.
Xtrain, Xtest, ytrain, ytest = train_test_split(X,y, test_size = 0.25)
gscv.fit(Xtrain, ytrain)
This will ensure that the model is trained on the log-transformed outcomes, back transforms into the original space, and evaluates the loss in the original space.
• that's exactly what I am doing! My question has to do with the fact that the RF regressor will average log-transformed values, that I then convert back to the original space. But the mean of the transformation when (converted back to the original space) is not equal to the mean in the original space. Feb 4, 2020 at 18:31
• I would encourage you to predict, transform, then compute the loss rather than predict, compute the loss, then transform. Does that make sense? Feb 4, 2020 at 18:33
• That's what I am doing :) The performance (on the original var) after training on the log-transformed var and projecting back, is better. But I can't tell if this is by accident - since as I mentioned above the RF regressor is applying a mean on the leaf values. Feb 4, 2020 at 18:35
• Great answer and example! Thanks :) Jul 15, 2020 at 10:08
• A warning from ML Engineering perspective is due here: such inverse transforms that influence prediction distributions have to be carried over with the model everywhere, so unless your organization versions every model and its metadata (such as required transforms) and controls every line of the validation and inference code (to apply the transforms consistently both for offline and online) as opposed to using canned libraries and software, you are advised against transforms that are external to the model object Jan 29 at 13:04 | 2022-05-27T06:01:14 | {
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https://www.physicsforums.com/threads/limit-with-trigonometric-and-polynomial-function.839248/ | # Limit with trigonometric and polynomial function.
1. Oct 23, 2015
### FaroukYasser
1. The problem statement, all variables and given/known data
For $$\lim _{ x\rightarrow \infty }{ \frac { { x }^{ 2 }+{ e }^{ -{ x }^{ 2 }\sin ^{ 2 }{ x } } }{ \sqrt { { x }^{ 4 }+1 } } }$$, determine whether it exists. If it does, find its value. if it doesn't, explain.
2. Relevant equations
Sand witch theorem and arithmetic rule
3. The attempt at a solution
I reached that the limit is 1 using the following:
$$\lim _{ x\rightarrow \infty }{ \frac { { x }^{ 2 }+{ e }^{ -{ x }^{ 2 }\sin ^{ 2 }{ x } } }{ \sqrt { { x }^{ 4 }+1 } } } =\lim _{ x\rightarrow \infty }{ \frac { { x }^{ 2 } }{ \sqrt { { x }^{ 4 }+1 } } +\frac { 1 }{ { e }^{ { x }^{ 2 }\sin ^{ 2 }{ x } }\sqrt { { x }^{ 4 }+1 } } } \\ \frac { 1 }{ { e }^{ x } } <\frac { 1 }{ { e }^{ { x }^{ 2 } }\sqrt { { x }^{ 4 }+1 } } <\frac { 1 }{ { e }^{ { x }^{ 2 }\sin ^{ 2 }{ x } }\sqrt { { x }^{ 4 }+1 } } <\frac { 1 }{ \sqrt { { x }^{ 4 }+1 } } \\ \\ \lim _{ x\rightarrow \infty }{ \frac { 1 }{ { e }^{ x } } } =\lim _{ x\rightarrow \infty }{ \frac { 1 }{ \sqrt { { x }^{ 4 }+1 } } } =0$$
Therefore by the sandwich theorem:
$$\lim _{ x\rightarrow \infty }{ \frac { 1 }{ { e }^{ { x }^{ 2 }\sin ^{ 2 }{ x } }\sqrt { { x }^{ 4 }+1 } } } =0$$
Also,
$$\lim _{ x\rightarrow \infty }{ \frac { { x }^{ 2 } }{ \sqrt { { x }^{ 4 }+1 } } } =1$$
Hence:
$$\lim _{ x\rightarrow \infty }{ \frac { { x }^{ 2 }+{ e }^{ -{ x }^{ 2 }\sin ^{ 2 }{ x } } }{ \sqrt { { x }^{ 4 }+1 } } } =1+0=1$$
I tried to put the limit into wolfram but it gave me a time limit exceeded. Is there a reason for this? Does the limit really not exist? And is there anything wrong in my argument?
Thank you
2. Oct 23, 2015
### BvU
I think you're doing just fine. which of the two did wolfie suffocate on ?
3. Oct 23, 2015
### FaroukYasser
Wolfram exceeded time on the original expression without breaking it down. (Although I have no idea why it did) | 2017-08-20T15:49:17 | {
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http://math.stackexchange.com/questions/113120/finding-lim-n-to-infty-phin-n2-when-phin-sum-i-1n-phin?answertab=votes | # Finding $\lim_{n\to\infty}\Phi(n)/n^2$, When $\Phi(n)=\sum_{i=1}^{n}\phi(n)$
This exercise is meant to be 'explored' computationally. However, I implemented it in C++ and did not get anything better than a sequence of pseudo-random numbers.
Let $\Phi(n)=\sum_{i=1}^{n}\phi(n)$. Investigate the value of $\Phi(n)/n^2$ for increasingly large values of $n$, such as $n=100$, $n=1000$, and $n=10000$. Can you make a conjecture about the limit of this ratio as $n$ grows large without bound?
Notice that $\Phi(n)=n\phi(n)$. Hence, $\Phi(n)/n^2=\phi(n)/n$. Moreover, the largest value $\phi(n)/n$ ever attains is $1$ at $n=1$; everything else falls within the interval $(0,1)$, and the closest it gets to $1$ again is when $n$ is prime (since $\phi(p)=p-1$, and $(p-1)/p\approx1$ for very large primes $p$).
However, I am tempted to say that this function diverges, and that no conjecture about its limit can be concluded as a result.
What do you guys think?
-
Since $\phi(n)$ is asymptotically $n$, I'm tempted to say $\Phi(n)/n^2$ approaches $1/2$, as $\sum\limits_{i=1}^ni=\frac{i^2+i}{2}$. – Alex Becker Feb 25 '12 at 3:20
I think the question is supposed to be about $\sum_{i=1}^n\phi(i)$. – Gerry Myerson Feb 25 '12 at 3:22
Gerry, I thought the same. However, I copied the book's question character by character (Elementary Number Theory and Its Applications 5E. by Kenneth H. Rosen, page 237). Could he have made a typo? – EsX_Raptor Feb 25 '12 at 3:27
I think the function $f(n) = \phi(n)/n$ has no limit, because for any value taken by $f$, say $f(n_0)$, then $(n_0^k)_{k>0}$ defines a subsequence that converges to this value (using the fact that $f(n_0^k) = f(n_0)$). – Joel Cohen Feb 25 '12 at 3:36
Hey guys, look at what I found. en.wikipedia.org/wiki/… Is that corroborating that the limit indeed does not exist? The mathematics are too convoluted for me to understand at this point. :/ – EsX_Raptor Feb 25 '12 at 3:46
There is a very old result that says $$\lim_{n\to\infty}\frac{\sum_{k=1}^n \varphi(k)}{n^2}=\frac{3}{\pi^2}.$$
The error term I have in notes is $O(x(\log x)^{2/3}(\log\log x)^{4/3})$, but undoubtedly there have been improvements on that. There is a large literature.
Added: The OP quoted correctly the textbook source of the problem, which asks about the behaviour of $(\sum_{i=1}^n\varphi(n))/n^2$. This is undoubtedly a typo, since $\sum_{i=1}^n\varphi(n)=n\varphi(n)$.
The ratio $\dfrac{\varphi(n)}{n}$ certainly bounces around a lot, and can be made arbitrarily close to $0$, and, much more easily, arbitrarily close to $1$.
-
Is that considering $\Phi(n)=\sum_{i=1}^{n}\phi(n)$ or $\Phi(n)=\sum_{i=1}^{n}\phi(i)$? – EsX_Raptor Feb 25 '12 at 4:50
What you wrote, both in the post and in the comment, cannot be right, since you are writing $\sum_{i=1}^n \varphi(n)$. I assumed that you mean $\sum_{i=1}^n\varphi(i)$. If we are summing over the divisors of $n$, that is an entirely different function, easy to get explicit formulas for, but somewhat chaotic. – André Nicolas Feb 25 '12 at 4:58
The typo is fixed in the sixth edition, which defines $\Phi(n)$ to be $\sum_{i=1}^n \phi(i)$. – David Moews Feb 25 '12 at 5:51
@AndréNicolas: "Undoubtedly there have been improvements on that." Surprisingly, there haven't been any. That result is due to Walfisz, and it remains the best. See this blog post for details: enaslund.wordpress.com/2012/01/15/… – Eric Naslund Feb 25 '12 at 13:44
Here is a detailed note regarding the Totient Summatory function. Part 1 and 2 should be of interest, and in part 2 there is a short proof.
Also see this Math Stack Exchange question and answer.
- | 2016-05-06T10:04:42 | {
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