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https://stats.stackexchange.com/questions/325549/how-to-measure-dispersion-in-word-frequency-data/326166#326166 | # How to measure dispersion in word frequency data?
How can I quantify the amount of dispersion in a vector of word counts? I'm looking for a statistic that will be high for document A, because it contains many different words that occur infrequently, and low for document B, because it contains one word (or a few words) that occur often.
More generally, how does one measure the dispersion or "spread" in nominal data?
Is there a standard way of doing this in the text analysis community?
For probabilities (proportions or shares) $p_i$ summing to 1, the family $\sum p_i^a [\ln (1/p_i)]^b$ encapsulates several proposals for measures (indexes, coefficients, whatever) in this territory. Thus
1. $a = 0, b = 0$ returns the number of distinct words observed, which is the simplest to think about, regardless of its ignoring differences among the probabilities. This is always useful if only as context. In other fields, this could be the number of firms in a sector, the number of species observed at a site, and so forth. In general, let's call this the number of distinct items.
2. $a = 2, b = 0$ returns the Gini-Turing-Simpson-Herfindahl-Hirschman-Greenberg sum of squared probabilities, otherwise known as the repeat rate or purity or match probability or homozygosity. It is often reported as its complement or its reciprocal, sometimes then under other names, such as impurity or heterozygosity. In this context, it is the probability that two words selected randomly are the same, and its complement $1 - \sum p_i^2$ the probability that two words are different. The reciprocal $1 / \sum p_i^2$ has an interpretation as the equivalent number of equally common categories; this is sometimes called the numbers equivalent. Such an interpretation can be seen by noting that $k$ equally common categories (each probability thus $1/k$) imply $\sum p_i^2 = k (1/k)^2 = 1/k$ so that the reciprocal of the probability is just $k$. Picking a name is most likely to betray the field in which you work. Each field honours their own forebears, but I commend match probability as simple and most nearly self-defining.
3. $a = 1, b = 1$ returns Shannon entropy, often denoted $H$ and already signalled directly or indirectly in previous answers. The name entropy has stuck here, for a mix of excellent and not so good reasons, even occasionally physics envy. Note that $\exp(H)$ is the numbers equivalent for this measure, as seen by noting in similar style that $k$ equally common categories yield $H = \sum^k (1/k) \ln [1/(1/k)] = \ln k$, and hence $\exp(H) = \exp(\ln k)$ gives you back $k$. Entropy has many splendid properties; "information theory" is a good search term.
The formulation is found in I.J. Good. 1953. The population frequencies of species and the estimation of population parameters. Biometrika 40: 237-264. www.jstor.org/stable/2333344.
Other bases for logarithm (e.g. 10 or 2) are equally possible according to taste or precedent or convenience, with just simple variations implied for some formulas above.
Independent rediscoveries (or reinventions) of the second measure are manifold across several disciplines and the names above are far from a complete list.
Tying together common measures in a family is not just mildly appealing mathematically. It underlines that there is a choice of measure depending on the relative weights applied to scarce and common items, and so reduces any impression of adhockery created by a small profusion of apparently arbitrary proposals. The literature in some fields is weakened by papers and even books based on tenuous claims that some measure favoured by the author(s) is the best measure that everyone should be using.
My calculations indicate that examples A and B are not so different except on the first measure:
----------------------------------------------------------------------
| Shannon H exp(H) Simpson 1/Simpson #items
----------+-----------------------------------------------------------
A | 0.656 1.927 0.643 1.556 14
B | 0.684 1.981 0.630 1.588 9
----------------------------------------------------------------------
(Some may be interested to note that the Simpson named here (Edward Hugh Simpson, 1922- ) is the same as that honoured by the name Simpson's paradox. He did excellent work, but he wasn't the first to discover either thing for which he is named, which in turn is Stigler's paradox, which in turn....)
• This is a brilliant answer (and far easier to follow than the 1953 Good paper ;) ). Thank you!
– dB'
Feb 5 '18 at 2:41
I don't know if there's a common way of doing it, but this looks to me analogous to inequality questions in economics. If you treat each word as an individual and their count as comparable to income, you're interested in comparing where the bag of words is between the extremes of every word having the same count (complete equality), or one word having all the counts and everyone else zero. The complication being that the "zeros" don't show up, you can't have less than a count of 1 in a bag of words as usually defined ...
The Gini coefficient of A is 0.18, and of B is 0.43, which shows that A is more "equal" than B.
library(ineq)
A <- c(3, 2, 2, rep(1, 11))
B <- c(9, 2, rep(1, 7))
Gini(A)
Gini(B)
I'm interested in any other answers too. Obviously the old fashioned variance in counts would be a starting point too, but you'd have to scale it somehow to make it comparable for bags of different sizes and hence different mean counts per word.
• Good call - the Gini coefficient was my first thought, too! Searching on google scholar, though, I couldn't find much precedent for using it with text data. I wonder if the NLP / text retrieval community has a more standard measure for this sort of thing...
– dB'
Jan 29 '18 at 21:52
• Watch out: by my count Gini has been given as a name to at least three different measures. The history is defensible in each case, but people need to see the formula used. Feb 1 '18 at 0:47
• Good point @NickCox - I was thinking of this one, as used for inequality, which I think is the most common use: ellisp.github.io/blog/2017/08/05/weighted-gini I've seen different methods of estimating/calculating it but all with the same basic definition, in this context. I know machine learning folks use it for something different but haven't seen their excuse... Feb 1 '18 at 2:38
• @dB' I found this paper of using Gini in a text application: proceedings.mlr.press/v10/sanasam10a/sanasam10a.pdf (I prefer this answer to the accepted one, simply as it does the best job of distinguishing your A and B !) Mar 5 '18 at 21:14
This article has a review of standard dispersion measures used by linguists. They are listed as single-word dispersion measures (They measure the dispersion of words across sections, pages etc.) but could conceivably be used as word frequency dispersion measures. The standard statistical ones seem to be:
1. max-min
2. standard deviation
3. coefficient of variation $CV$
4. chi-squared $\chi^2$
The classics are:
1. Jullard's $D = 1-\frac{CV}{\sqrt{n-1}}$
2. Rosengren's $S = N\frac{(\sum_{i=1}^{n}\sqrt{n_i})^2}{n}$
3. Carroll's $D_2 = (\log_2N - \frac{\sum_{i=1}^n{n_i \log_2 n_i}}{N})/{\log_2(n)}$
4. Lyne's $D_3 = \frac{1-\chi^2}{4N}$
Where $N$ is the total number of words in the text, $n$ is the number of distinct words, and $n_i$ the number of occurrences of the i-th word in the text.
The text also mentions two more measures of dispersion, but they rely on the spatial positioning of the words, so this is inapplicable to the bag of words model.
• Note: I changed the original notation from the article, to make the formulas more consistent with the standard notation.
• Could you please define $f$ and $x_i$? I suspect they are, or are definable in terms of, symbols you've defined already. Feb 1 '18 at 9:00
• Interesting and very extensive, but these are measures of dispersion for single words. They relate to the variation of the frequencies, $v_i$, of a single word in different pieces of text (instead of the frequencies of different words in a single piece of text). This difference should be clarified. Feb 1 '18 at 10:28
• Why are the equations from the source not copied exactly (it is not just a change of labels in the expressions but also a change of the expression, or at least not a consistent change of the labels/variables)? Feb 1 '18 at 10:28
• @NickCox Thank you for catching that, I corrected the formulas to include only defined quantities. Feb 3 '18 at 15:52
• @MartijnWeterings You are right that originally the article dealt with single word dispersion metrics, although they seem to generalize to the word frequency trivially. Just in case I included that information in the answer. I changed the original notation to make these applicable to the bag of word model (replacing f with N and v_i with n_i). I added a note to signifiy this, but if you think it is still misleading I can provide a longer justification in the answer. Feb 3 '18 at 16:04
The first I would do is calculating Shannon's entropy. You can use the R package infotheo, function entropy(X, method="emp"). If you wrap natstobits(H) around it, you will get the entropy of this source in bits.
One possible measure of equality you could use is the scaled Shannon entropy. If you have a vector of proportions $\boldsymbol{p} \equiv (p_1, ... , p_n)$ then this measure is given by:
$$\bar{H}(\boldsymbol{p}) \equiv - \frac{\sum p_i \ln p_i}{\ln n}.$$
This is a scaled measure with range $0 \leqslant \bar{H}(\boldsymbol{p}) \leqslant 1$ with extreme values occurring at the extremes of equality or inequality. Shannon entropy is a measure of information, and the scaled version allows comparison between cases with different numbers of categories.
• Extreme Inequality: All the count is in some category $k$. In this case we have $p_i = \mathbb{I}(i=k)$ and this gives us $\bar{H}(\boldsymbol{p}) = 0$.
• Extreme Equality: All the counts are equal over all categories. In this case we have $p_i = 1/n$ and this gives us $\bar{H}(\boldsymbol{p}) = 1$. | 2021-10-21T18:41:29 | {
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http://efavdb.com/category/theory/ | Martingales
Here, I give a quick review of the concept of a Martingale. A Martingale is a sequence of random variables satisfying a specific expectation conservation law. If one can identify a Martingale relating to some other sequence of random variables, its use can sometimes make quick work of certain expectation value evaluations.
This note is adapted from Chapter 2 of Stochastic Calculus and Financial Applications, by Steele.
Logistic Regression
We review binary logistic regression. In particular, we derive a) the equations needed to fit the algorithm via gradient descent, b) the maximum likelihood fit’s asymptotic coefficient covariance matrix, and c) expressions for model test point class membership probability confidence intervals. We also provide python code implementing a minimal “LogisticRegressionWithError” class whose “predict_proba” method returns prediction confidence intervals alongside its point estimates.
Our python code can be downloaded from our github page, here. Its use requires the jupyter, numpy, sklearn, and matplotlib packages.
Normal Distributions
I review — and provide derivations for — some basic properties of Normal distributions. Topics currently covered: (i) Their normalization, (ii) Samples from a univariate Normal, (iii) Multivariate Normal distributions, (iv) Central limit theorem.
Model AUC depends on test set difficulty
The AUC score is a popular summary statistic that is often used to communicate the performance of a classifier. However, we illustrate here that this score depends not only on the quality of the model in question, but also on the difficulty of the test set considered: If samples are added to a test set that are easily classified, the AUC will go up — even if the model studied has not improved. In general, this behavior implies that isolated, single AUC scores cannot be used to meaningfully qualify a model’s performance. Instead, the AUC should be considered a score that is primarily useful for comparing and ranking multiple models — each at a common test set difficulty.
Deep reinforcement learning, battleship
Here, we provide a brief introduction to reinforcement learning (RL) — a general technique for training programs to play games efficiently. Our aim is to explain its practical implementation: We cover some basic theory and then walk through a minimal python program that trains a neural network to play the game battleship.
Bayesian Statistics: MCMC
We review the Metropolis algorithm — a simple Markov Chain Monte Carlo (MCMC) sampling method — and its application to estimating posteriors in Bayesian statistics. A simple python example is provided.
Interpreting the results of linear regression
Our last post showed how to obtain the least-squares solution for linear regression and discussed the idea of sampling variability in the best estimates for the coefficients. In this post, we continue the discussion about uncertainty in linear regression — both in the estimates of individual linear regression coefficients and the quality of the overall fit.
Specifically, we’ll discuss how to calculate the 95% confidence intervals and p-values from hypothesis tests that are output by many statistical packages like python’s statsmodels or R. An example with code is provided at the end.
Linear Regression
We review classical linear regression using vector-matrix notation. In particular, we derive a) the least-squares solution, b) the fit’s coefficient covariance matrix — showing that the coefficient estimates are most precise along directions that have been sampled over a large range of values (the high variance directions, a la PCA), and c) an unbiased estimate for the underlying sample variance (assuming normal sample variance in this last case). We then review how these last two results can be used to provide confidence intervals / hypothesis tests for the coefficient estimates. Finally, we show that similar results follow from a Bayesian approach.
Last edited July 23, 2016.
Independent component analysis
Two microphones are placed in a room where two conversations are taking place simultaneously. Given these two recordings, can one “remix” them in some prescribed way to isolate the individual conversations? Yes! In this post, we review one simple approach to solving this type of problem, Independent Component Analysis (ICA). We share an ipython document implementing ICA and link to a youtube video illustrating its application to audio de-mixing.
Principal component analysis
We review the two essentials of principal component analysis (“PCA”): 1) The principal components of a set of data points are the eigenvectors of the correlation matrix of these points in feature space. 2) Projecting the data onto the subspace spanned by the first $k$ of these — listed in descending eigenvalue order — provides the best possible $k$-dimensional approximation to the data, in the sense of captured variance. | 2017-11-25T01:37:18 | {
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"url": "http://efavdb.com/category/theory/",
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"lm_q1q2_score": 0.8394765824534656
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https://math.stackexchange.com/questions/4563216/determining-the-cartesian-equation-of-an-ellipse-be-it-neither-in-standard-pos | # Determining the cartesian equation of an ellipse ( be it neither in standard position nor orientation) given center, one vertex and one semi -axis
Desmos construction, with sliders waiting for being launched : https://www.desmos.com/calculator/vuou1gnese
Question : I obtained the final formula while thinking of center $$C$$ and vertex $$V$$ as points lying in the first quadrant, with $$V$$ located to the right of $$C$$ Is the formula general in spite of this? - I assumed that the " shifting terms" $$h$$ and $$k$$ operate , so to say, in the rotated system, so that, intead of substituting $$X_C$$ for $$h$$ and $$Y_C$$ for $$k$$ , I used the coordinates of center $$C$$ in the rotated system. Is this assumption correct? Finally, can you think of other forms of this equation?
The givens are :
• $$\text {Center} = C = (X_C, Y_C)$$
• $$\text {Vertex} = V = ( X_V,Y_V) = (X_C+p, X_C+q) , \text {with }p, q$$ $$\in \mathbb R$$
• $$m {\in\space \mathbb R^+} = \text {length of one semi axis ( perpendicular to the semi-axis [CV] }$$
From what is given can be immediately deduced :
• $$M = \text {length of the semi axis [CV]}= \sqrt {p^2 + q^2}$$
• $$R = \text {inclination of the straight line (CV) on which lies the semi axis [CV] }= \arctan {\frac qp}$$
• the equations of $$X'$$ and $$Y'$$, the rotated coordinate system in which lies the parabola , namely
$$\space \space \space \space X' : y = \tan(R)x$$
$$\space \space \space \space Y' : y = (-1 / \tan(R))x$$
Since the ellipse lies in a rotated coordinate system, and since its center is not necessarily at the origin its equation will be of the form :
$$\frac { (X(x,y) -h )^2 } { M^2 } + \frac { (Y(x,y)-k )^2 } { m^2 } =1$$
with
• $$X(x,y) = x \cos(R) + y \sin (R)$$
• $$Y(x,y) = y \cos(R) - y \sin (R)$$
• $$h$$ and $$k$$ : the coordinates of center $$C$$ in the rotated coordinate system.
Determining the values of $$h$$ and $$k$$
• $$h = X_C \cos (R) + Y_C \sin (R)$$
• $$k = Y_C \sin (R) - X_C \cos(R)$$
Hence the formula for the ellipse of center $$C= (X_C , Y_C)$$, vertex $$V$$ and a semi-axis of length $$m$$ is :
$$\frac { (X(x,y) - (X_C \cos (R) + Y_C \sin (R)) )^2 } { M^2 } + \frac {( Y(x,y)-(Y_C \sin (R) - X_C \cos(R)) ) ^2 } { m^2 } =1$$
with, as said above $$M= \text {length of the semi- axis [CV]}$$
• Your assumption is fine, and I see no mistakes in a quick skim, though I think you meant $Y'=(-1/\tan R)x$, not $X'$. Not sure there's a "simpler" way of expressing it though. Also, very nice Desmos demo, though I'd turn off animations when linking. Oct 27, 2022 at 22:12
• The name of the second rotated axis contained actually a typo, thanks. ( Is your comment on the animation of the Desmos construction related to energy consumption? If it is the case , I'll stop the sliders ! ) Oct 27, 2022 at 22:17
• Not energy consumption, merely attention consumption. I wasn't expecting it. Oddly it only happened in the mobile app, not here on my desktop. Separately: there's one simplification you could make, though I don't know if it actually simplifies anything: because you have $\sin \arctan (q/p)$ in the final equation (and the cosine), you could convert to their compositions ($\sin \arctan q/p = q/(p \sqrt{1-(q/p)^2}$, similar for cosine. Oct 27, 2022 at 22:22
## 2 Answers
Too big for a comment, but using the composition of arctan and sine/cosine actually simplifies things a lot more than I expected, because $$M$$ reappears. The identities are:
$$\sin \arctan \frac q p = \frac{\frac q p}{\sqrt{1+\left(\frac q p\right)^2}} = \frac p {\sqrt{p^2+q^2}} = \frac p M \\ \cos \arctan \frac q p = \frac{1}{\sqrt{1+\left(\frac q p\right)^2}} = \frac q {\sqrt{p^2+q^2}} = \frac q M$$
This makes your final equation at the bottom:
$$\frac{1}{M^2} \left( \left( \frac{p}{M}x + \frac{q}{M}y \right) - \left( \frac{p}{M}X_C + \frac{q}{M}Y_C \right) \right)^2 + \frac{1}{m^2} \left( \left( \frac{q}{M}y - \frac{p}{M}x \right) - \left( \frac{q}{M}X_C - \frac{p}{M}Y_C \right) \right)^2 = 1 \\ \frac{1}{M^4} \left( px + qy - pX_C - qY_C \right)^2 + \frac{1}{M^2m^2} \left( py -qx - pY_C + qX_C \right)^2 = 1 \\ \frac{1}{M^2} \left( px + qy - pX_C - qY_C \right)^2 + \frac{1}{m^2} \left( py -qx - pY_C + qX_C \right)^2 = M^2$$
I fear that further "simplification" just ends up with huge constants just to put it into $$Ax^2+By^2+Cxy+Dx+Ey+F$$ form.
Rotate then Shift.
Starting from the algebraic equation of an ellipse in standard format
$$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$$
If you define the position vector $$\mathbf{p} = [x, y]^T$$ , then the above equation can be written concisely as follows
$$\mathbf{p}^T \ D \ \mathbf{p} = 1\hspace{50pt}(*)$$
where
$$D = \begin{bmatrix} \dfrac{1}{a^2} && 0 \\ 0 && \dfrac{1}{b^2} \end{bmatrix}$$
Now in the end, you want to find the algebraic equation of the same ellipse but rotated by some angle $$\theta$$ (with respect to the original orientation) and shifted by a certain displacement vector $$(Xc, Yc)$$
So, first rotate the ellipse about the origin, then shift the resulting the ellipse by the vector $$(Xc , Yc)$$
The image of a point $$\mathbf{p} = (x,y)$$ on the original ellipse under a rotation by an angle $$\theta$$ is the point
$$\mathbf{p'} = R \ \mathbf{p}$$
where $$R$$ is the two-dimensional rotation matrix given by
$$R = \begin{bmatrix} \cos \theta && - \sin \theta \\ \sin \theta && \cos \theta \end{bmatrix}$$
And the image of $$\mathbf{p'}$$ after a shift by $$C = [X_c, Y_c]^T$$ is
$$\mathbf{p''} = \mathbf{p'} + \mathbf{C} = R \ \mathbf{p} + \mathbf{C}$$
From this last equation, we deduce that $$\mathbf{p} = R^T (\mathbf{p''} - \mathbf{C} )$$
Substituting this into equation $$(*)$$, gives us
$$(\mathbf{p''} - \mathbf{C})^T \ R D R^T \ (\mathbf{p''} - \mathbf{C} ) = 1 \hspace{50pt} (**)$$
And this is the algebraic equation of the rotated/shifted ellipse.
Explicitly, we have
$$R D R^T = \begin{bmatrix} \dfrac{1}{a^2} \cos^2 \theta + \dfrac{1}{b^2} \sin^2 \theta && \bigg( \dfrac{1}{a^2} - \dfrac{1}{b^2} \bigg) \sin \theta \cos \theta \\ \bigg( \dfrac{1}{a^2} - \dfrac{1}{b^2} \bigg) \sin \theta \cos \theta && \dfrac{1}{a^2} \sin^2 \theta + \dfrac{1}{b^2} \cos^2 \theta \end{bmatrix}$$
And $$\mathbf{p''} - \mathbf{C} = (x - X_c, y - Y_c)$$
Therefore, the algebraic equation is
$$\bigg(\dfrac{1}{a^2} \cos^2 \theta + \dfrac{1}{b^2} \sin^2 \theta \bigg) (x - X_c)^2 + 2 \bigg( \dfrac{1}{a^2} - \dfrac{1}{b^2} \bigg) \sin \theta \cos \theta (x - X_c) (y - Y_c) + \bigg(\dfrac{1}{a^2} \sin^2 \theta + \dfrac{1}{b^2} \cos^2 \theta\bigg) (y - Y_c)^2 = 1$$
Using the double angle trigonometric formulas, this last equation can be expressed as follows
$$(A + B \cos(2 \theta)) (x - X_c)^2 + 2 B \sin(2 \theta) (x - X_c) (y - Y_c) + (A - B \cos(2 \theta)) (y - Y_c)^2 = 1$$
where
$$A = \dfrac{1}{2} \bigg( \dfrac{1}{a^2} + \dfrac{1}{b^2} \bigg)$$
$$B = \dfrac{1}{2} \bigg( \dfrac{1}{a^2} - \dfrac{1}{b^2} \bigg)$$
• Thanks for this detailed answer Hosam; I accepted the other one which is more compatible with my actual level in mathematics. Oct 30, 2022 at 22:07
• You're welcome. My pleasure. Oct 30, 2022 at 22:16 | 2023-01-27T08:42:44 | {
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https://math.stackexchange.com/questions/2173347/does-the-comparison-test-imply-absolute-convergence/2173453 | # Does the comparison test imply absolute convergence?
As my textbook states it:
Let $\Sigma a_n$ be a series where $a_n \ge 0$ for all $n$.
(i) If $\Sigma a_n$ converges and $|b_n| \le a_n$ for all $n$, then $\Sigma b_n$ converges.
(ii) If $\Sigma a_n = +\infty$ and $b_n \ge a_n$ for all $n$, then $\Sigma b_n = +\infty$
My question is, does this imply absolute convergence for $\Sigma b_n$, or just convergence? It seems to me that this implies absolute convergence, as
$$|\Sigma b_n| \le \Sigma |b_n| = |\Sigma |b_n|| \le \Sigma a_n$$ So both $\Sigma b_n$ and $\Sigma |b_n|$ satisfy the Cauchy criterion.
Yes, this implies that $\sum|b_n|$ converges (and hence that $\sum b_n$ converges absolutely). Let $c_n=|b_n|$. We know that $|c_n|=\big||b_n|\big|= |b_n|\le a_n$ for all $n$; then by (i), $\sum c_n$ converges. | 2019-09-22T05:43:47 | {
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https://math.stackexchange.com/questions/1657488/is-fx-frac-sin-xx-for-all-x-neq-0-differentiable | Is $f(x)=\frac{\sin x}{x}$ for all $x\neq 0$ differentiable?
The function $f(x)$ is defined by $$f(x)=\frac{\sin x}{x}$$ for any $x≠0$. For $x=0$, $f(x)=1$.
My work:
Determine if the function is continuous, differentiable and if the latter, find its derivative at $0$.
$$f(x) =\begin{cases}\dfrac{\sin x}{x}, & x \ne 0 \\ 1 & x = 0 \end{cases}$$
I proved the continuous condition using L'Hopital's rule on the following
$$f(0) = \lim_{x\to 0} \frac{\sin x}x = 1$$
For the defferentiable condition I think I proved it
\begin{align*} \lim_{x\to0} \frac{f(x) - f(0)}{x-0} &= \lim_{x\to0} \frac{\frac{\sin x}x - 1}{x-0} \\ &= \lim_{x\to0} \frac{\sin x - x}{x^2} \\ &= \lim_{x\to0} \frac{\cos x-1}{2x} \\ &= \lim_{x\to0} \frac{-\sin x}{2} \\ &= 0 \end{align*}
Now the derivative of $f(x)$ is $$\frac{x\cos x - \sin x}{x^2}$$
But what does it mean "find its derivative at $0$" ?
The only thing that came to my mind is to find its limit as $x\to 0$
$$\lim_{x\to0}\frac{x\cos x - \sin x}{x^2} = 0$$
Did I understand and do everything correctly?
• The derivative at $0$ is equal to the limit: $$\lim_{x\to 0} \frac{f(x)-f(0)}{x-0}$$ Which you've already proven to be equal to 0. Feb 16 '16 at 0:01
• Use \to to show $\to$. Formatting tips here.
– Em.
Feb 16 '16 at 0:05
• @Ivan Hi Ivan! It's been a while. I hope you're staying safe and healthy during the pandemic. I've reached out to contact you a few times, but am unsure whether you've received the notes? If you would, please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote an answer as you see fit. ;-) Feb 28 '21 at 19:39
The function defined by
$$f(x)=\begin{cases}\frac{\sin(x)}{x}&,x\ne 0\\\\1&,x=0\end{cases}$$
is not only differentiable at $$x=0$$, it is continuously differentiable there.
NOTE:
I thought it would be instructive to present a way forward that relies only on a standard, elementary inequality and the squeeze theorem. To that end, we proceed.
The derivative at $$x=0$$ is given by
$$f'(0)\equiv \lim_{h\to 0}\frac{\frac{\sin(h)}{h}-1}{h} \tag 1$$
Recalling from elementary geometry that the sine function satisfies the inequalities
$$\cos(h) \le \frac{\sin(h)}{h}\le 1 \tag 2$$
for $$|h|\le \pi/2$$, we see that the term under the limit in $$(1)$$ satisfies the inequalities
$$-2\sin^2(h/2)= \cos(h)-1\le \frac{\sin(h)}{h}-1\le 0 \tag 3$$
Then, taking absolute values, dividing by $$|h|$$, and using the right-hand side inequality in $$(2)$$ yields
$$0 \le \left|\frac{\frac{\sin(h)}{h}-1}{h}\right|\le \frac12 |h| \tag 4$$
whereupon applying the squeeze theorem to $$(4)$$ produces the limit
$$\lim_{h\to 0}\frac{\frac{\sin(h)}{h}-1}{h}=0$$
Therefore, $$f'(0)=0$$.
For $$x\ne 0$$, we have
$$f'(x)=\frac{x\cos(x)-\sin(x)}{x^2}$$
To see that $$f'(x)$$ is continuous at $$x=0$$, we need to show that
$$\lim_{x\to 0}\frac{x\cos(x)-\sin(x)}{x^2}=0$$
Again, using $$(2)$$ we see that
$$0\le \left|\frac{x\cos(x)-\sin(x)}{x^2}\right|\le\left|\frac{1-\cos(x)}{x}\right| \le \frac12 |x| \tag 5$$
whereupon applying the squeeze theorem to $$(5)$$ produces the limit
$$\lim_{x\to 0}\frac{x\cos(x)-\sin(x)}{x^2}=0$$
Therefore,
$$\lim_{x\to 0}f'(x)=0=f'(0)$$
which shows that $$f'$$ is continuously differentiable at $$0$$.
• @Ivan Hi Ivan! It's been a while. I hope you're staying safe and healthy during the pandemic. I've reached out to contact you a few times, but am unsure whether you've received the notes? If you would, please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote an answer as you see fit. ;-) Feb 28 '21 at 19:39 | 2022-01-23T22:38:56 | {
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http://fnsv.haus-cecilie.de/ | Norm Of A Vector
A vector space (also called a linear space) is a set of objects called vectors, which may be added together and multiplied ("scaled") by numbers, called scalars. Vector of Vectors is a two-dimensional vector with a variable number of rows where each row is vector. In particular, the Euclidean distance of a vector from the origin is a norm, called the Euclidean norm, or 2-norm, which may also be defined as the square root of the inner product of a vector with itself. Matrix norms are an extension of vector norms to matrices and are used to define a measure of distance on the space of a matrix. If norm of x is greater than 0 then x is not equal to 0 (Zero Vector) and if norm is equal to 0 then x is a zero vector. for any Scalar, 3. Common vector image file extensions include. 2 norm sqrt[12] infinity norm = sqrt[5]. Do not use built-in Matlab functions to calculate the norm (i. A norm on a real or complex vector space V is a mapping V → R with properties. • norm(v)—Returns the norm of vector v. The magnitude of the vector AB is denoted as | AB |. Instead, in this section, we calculate the moments of. Choose from over a million free vectors, clipart graphics, vector art images, design templates, and illustrations created by artists worldwide! Download free vector art, stock photos & videos. with A ∈ Rm×n, b ∈ Rm. Let ,·,be a vector norm onR n (or C n). De norm is niet negatief. But, it is actually possible to talk about linear combinations of anything as long as you understand the main idea of a linear combination: (scalar)(something 1) + (scalar)(something 2) + (scalar)(something 3) These “somethings” could be “everyday” variables like \$$x\$$ and …. Computes the vector norm. See below for alternatives. Check: The column vector should represent the vector that was drawn. Remember, we can write a vector that starts at There is a problem though. For this problem, use the. Note the use of angle. Vector Norms - PowerPoint PPT Presentation. Antonyms for Gender norms. norm — noun Etymology: Latin norma, literally, carpenter s square Date: 1674 1. where x is one of the above vector norms x 1 , x 2 , x ∞. Write a Matlab function, called myvectornorm that inputs the vector x and outputs the 2-norm of x. For a vector x: any number or %inf, -%inf; or a word "inf" ("i"), "fro" ("f"). is a linear combination of the vectors v1, v2,. The triangle inequality is only satised for p ≥ 1. Norms are 0 if and only if the vector is a zero vector. A vector space (also called a linear space) is a set of objects called vectors, which may be added together and multiplied ("scaled") by numbers, called scalars. norm (a generalization of both subGaussian random vectors and norm bounded random vectors), which are tight up to logarithmic factors. L1 norm It is defined as the sum of magnitudes of each component a = ( a 1, a 2, a 3) L1 norm of vector a = |a 1 | + |a 2 | + |a 3 | L2 norm It is defined as the square root of sum of squares of each component L2 norm of vector a = √( a 1 2 + a 2 2 + a 3 2). If you want to generate a vector of normally distributed random numbers, rnorm is the function you should use. Frobenius norm. Python code for norm of the vector # Linear Algebra Learning Sequence # Outer Product Property I import numpy as np a = np. Do not use built-in Matlab functions to calculate the norm (i. The formula for the length of a 2D vector is the Pythagorean Formula. WriteLine (); // // Norms, dot products, etc. Let A be an m ×n matrix, and define A A X X p X p p = ≠ supr r r 0, (4-2) where "sup" stands for supremum, also known as least upper bound. Euclidean length of a vector with scaling to avoid destructive overflow and underflow issues: Note-2: The only problem with this solution norm() is. There are several commonly used vector norms, summarized in the following table:. AU - Ozawa, N. Therefore, multiplying a vector by an orthogonal matrices does not change its length. Matrix norms are an extension of vector norms to matrices and are used to define a measure of distance on the space of a matrix. And I'll take the square root so that I now have the length of two if I double v, from v to 2v. One vector are solar rays, the other is where the solar panel is pointing (yes, yes, the normal vector). Vector Norm List Processor. The norm function, or length, is a function V !IRdenoted as kk, and de ned as kuk= p (u;u): Example: The Euclidean norm in IR2 is given by kuk= p (x;x) = p (x1)2 + (x2)2: Slide 6 ’ & $% Examples The. Consider the vector hx,yi ∈ IR2. The purpose of loss functions is to compute the quantity that a model should seek to minimize during training. Normed Vector Spaces De nition: Norm Let V be a vector space. norm¶ linalg. In machine learning, norms are useful because they are used to express distances: this vector and this vector are so-and-so far apart, according to this-or-that norm. norm ∥ · ∥1 implies the convergence in ∥ · ∥2, and vice versa — that is, for any sequence {x(k)} and a vector x in X, we have lim k→∞ ∥x(k)−x∥1 = 0 ⇐⇒ lim k→∞ ∥x(k)−x∥2 = 0. The two norm of an × matrix is defined by → ≠ → | | → | | | | → | | where → is an m-dimensional vector that is not the zero vector. FINDING EIGENVALUES AND EIGENVECTORS EXAMPLE 1: Find the eigenvalues and eigenvectors of the matrix A = 1 −3 3 3 −5 3 6 −6 4. R abs Function Example 2. If a matrix norm is vector-bound to a particular vector norm, then the two norms are guaranteed to be compatible. 1 Vector Norms Let X be a vector space over F, where F is either R or C. The meaning of direction is pretty self explanatory. The vector Fx is lying in the opposite direction as the x-hat vector. A vector space (also called a linear space) is a set of objects called vectors, which may be added together and multiplied ("scaled") by numbers, called scalars. There are many possible ways to measure the “size” of a vector corresponding to using different norms. You can navigate between the input fields by pressing the keys "left" and "right" on the keyboard. In the previous section we looked at the infinity, two and one norms of vectors and the infinity and one norm of matrices and saw how they were used to estimate the propagation of errors when one solves equations. Vector Space Operations. A pseudonorm or seminorm satisfies the first two properties of a norm, but may be zero for other vectors than the origin. For the computation of the ENMO metric, the ActiGraph GT3X+. satisfies all three properties of the norm: a. (Euclidean) norm of vector a ∈ Rn a + b ≤ a + b for all vectors a and b of equal length. General Vector Norm. Any inner product space is a normed vector space with norm$| \cdot |$(recall that this is defined by$\|v\| = \langle v,v \rangle^{1/2}$). This is also called the L2 norm of the vector. As InformationLiberation. $$\ell^2$$) squared norm of a vector can be obtained squaredNorm(). Vector and matrix norms. The header to your function should look like: function output_norm-myvectornorm(x) where x is the input vector and output_norm is the returned norm of x. The header to your function should look like: function output_norm-myvectornorm(x) where x is the input vector and output_norm is the returned norm of x. Write a Matlab function, called myvectornorm that inputs the vector x and outputs the 2-norm of x. norm (A) = A. Calculate the entrywise L_q norm of a vector or a matrix. The Lesson: Let v = (2, 5, 1) and u = (-3, 2, 4) be two 3-dimensional vectors. $$\ell^2$$) squared norm of a vector can be obtained squaredNorm(). Norm of a vector is always positive or zero ∥ a ∥ ⩾ 0. The matrix ∞-norm, which is the maximum over the sum of the absolute values of each row; Both the Frobenius norm and the matrix 2-norm are subordinate to the vector 2-norm. Vector norms 1. It is computed from the resultant vector value of the measured orthogonal acceleration, which involves a dynamic component due to deviations in velocity, and a Data Reduction and Processing: Euclidean Norm Minus One (ENMO). % % alpha is the over-relaxation parameter (typical values for. If U is closed under vector addition and scalar multiplication, then U is a subspace of V. Let V be a vector space and U ⊂ V. e, norm or vecnorm). The concepts of metric, normed, and topological spaces clarify our previous discussion of the analysis of real functions, and they provide the. Vector length formula for two-dimensional vector. The most commonly encountered vector norm (often simply called "the norm" of a vector, or sometimes the magnitude of a vector) is the L2-norm , given by. Lecture 2 1. That's the energy in the vector v. Tagged: norm of a vector. A vector space (also called a linear space) is a set of objects called vectors, which may be added together and multiplied ("scaled") by numbers, called scalars. A vector is said to be normalized if Matrix inversion The inverse of a square matrix is the unique matrix such that The inverse doesn't always exist!. The Foo-norm SVM methodology is motivated by the feature selection problem in cases where the input features are generated by factors, and the model is best interpreted in terms of significant factors. abs — absolute value, magnitude. In linear algebra, functional analysis, and related areas of mathematics, a norm is a function that assigns a strictly positive length or size to each vector in a vector space. This is a Verilog implementation of a Vector Norm List Processor. To find a matrix or vector norm we use function numpy. Most vector image formats can also include colors, gradients, and image effects. Positive-de niteness: For any vector x, kxk 0; and kxk= 0 i x= 0 2. A vector space (also called a linear space) is a set of objects called vectors, which may be added together and multiplied ("scaled") by numbers, called scalars. Total — total of elements in a vector. Let and be two vectors in the space. Norms follow the triangle inequality i. 3- Norms respect the triangle inequity. Vector Norm. Norms, standards and processes underpinning vector control policy development 3 prequalified products by PQT-VCP. • Small entries in a vector contribute more to the 1-norm of the vector than to the 2-norm. The Frobenius norm of a vector coincides with its 2-norm. Let Kdenote either R or C. A linear space is sometimes be called a linear vector space, or a vector space. Remark Here thevector norm could be any vector norm. The last Section is an application of matrix norms. Let A be an m ×n matrix, and define A A X X p X p p = ≠ supr r r 0, (4-2) where "sup" stands for supremum, also known as least upper bound. The new TV norm has the desirable properties of (1) not penalizing discontinuities (edges) in the image, (2) being rotationally invariant in the image space, and (3) reducing to the usual TV norm in the scalar case. 5 Two norms kk, kk0on a vector space V are called equivalent if there exist c 1;c 2 >0 such that for all v2V, kvk c 1kvk0 and kvk0 c 2kvk: (4) Note that the zero-dimensional vector space f0ghas only one norm, so \equivalent norms" is not an interesting concept for this space. vector or matrix of real or complex numbers (full or sparse storage). If axis is an integer, it specifies the axis of x along which to compute the vector norms. Vector Norm On a vector space V, a norm is a function ⋅from V to the set of non-negative reals that obeys three postulates:, (), 0 0, x y x y if x y V Trinagular Inequality x x if R x V x if x C + ≤ + ∈ = ∈ ∈ > ≠ λ λ λ we can think of x as the length or magnitude of the vector x. In addition, this is the reason why we need a negative (-) sign. Exercise 9. Because of this, the Euclidean norm is often known as the magnitude. Previous: Introduction to matrices; Next: Problem set: Matrix vector multiplication; Similar pages. While, the components of the unit tangent vector can be somewhat messy on occasion there are times when we will need to use the unit tangent vector instead of the tangent vector. norm () of Python library Numpy. Proof,part2. 4 General Vector Norms. We can then add vectors by adding the x parts and adding the y parts: The vector (8, 13) and the vector (26, 7) add up to the vector (34, 20). This returns a vector with the square roots of each of the components to the square, thus 1 2 3 instead of the Euclidean Norm This is a trivial function to write yourself: norm_vec - function(x) sqrt(sum(x^2)). Preparation code < script > Vector = function. Vectors are living organisms that can transmit infectious pathogens between humans, or from animals to humans. For example, we know "7" is larger than "4", and "0" is larger than “− 3 2 ” For complex number, there is no ordering. The triples (Rn, R, · ) and (Cn, C, · ) are examples of normed vector spaces and the inequality 3. It turns out that the Frobenius norm is equal to the square root of the sum of squares of the singular values of a matrix. x2+y2≥ 0, so p x +y2≥ 0. but what about p norm? There's a mathematical object called a Normed Space, which is a vector space with a concept of distance attached (the norm). April 20, 2020, 7:02am #1. Norm (length) of vector. Daily norm. for some normal vector w2Rd and offset b2R. The infinity norm of a 1-by-n or n-by-1 vector V is defined as follows. Computes a matrix norm of x using LAPACK. We dene A as the smallest number satisfying (1). — Herb Sutter and Andrei Alexandrescu, C++ Coding Standards. A pattern that is regarded as typical of something: a neighborhood where families with two wage-earners are the norm. Centre of Mathematics for Applications, Department of Informatics, University of Oslo. 1 Inner products and vector norms. The concepts of metric, normed, and topological spaces clarify our previous discussion of the analysis of real functions, and they provide the. Proof: To verify that (4) is a solution, pre-multiply by A: Ax = AA+b+A(I ¡A+A)y = b+(A¡AA+A)y by hypothesis = b since AA+A = A by the flrst Penrose condition. To create your new password, just click the link in the email we sent you. 4 General Vector Norms. Vector operators — grad, div. Support vector machines utilizing the 1-norm, typically set up as linear programs (Mangasarian, 2000; Bradley and Mangasarian, 1998), are formulated here as a completely unconstrained mini- mization of a convex differentiable piecewise-quadratic objective function in the dual space. In this section we shall look at. Vector Norm The Norm function calculates several different types of vector norms for x, depending on the argument p. You can think of the norm as the length of the vector. Brackets and Norms. A vector space (also called a linear space) is a set of objects called vectors, which may be added together and multiplied ("scaled") by numbers, called scalars. A unit normal vector of a curve, by its definition, is perpendicular to the curve at given point. Products containing vitamin. Answer: A unit vector is a vector that has the magnitude of one (1) with no units. Moreover, this equals zero only when both x = 0 and y = 0. The Cauchy-Schwarz Inequality holds for any inner product, so the triangle inequality holds irrespective of how you define the norm of the vector to be, i. A vector space (also called a linear space) is a set of objects called vectors, which may be added together and multiplied ("scaled") by numbers, called scalars. - Buy this stock vector and explore similar vectors at Adobe Stock. A class of vector norms, called a. e, norm or vecnorm). Many such algoritms use the Euclidean norm of a vector. ) Adding Vectors. The norm of a vector is the square root of the sum of each element of the vector squared. If the second argument is not given, p = 2 is used. We introduce three field operators which reveal interesting collective field properties, viz. And do you remember what was special? We've talked about using the L1 norm. com: Free online dictionary in English, German, French, Spanish. Previous: Introduction to matrices; Next: Matrix and vector multiplication examples; Math 2374. Fomin, "Elements of the theory of functions and functional analysis" , 1–2, Graylock (1957–1961) (Translated from Russian) [2] W. The rectangular coordinate notation for this vector is v 6,3 or v 6,3. , kQuk = kuk. Support vector machines utilizing the 1-norm, typically set up as linear programs (Mangasarian, 2000; Bradley and Mangasarian, 1998), are formulated here as a completely unconstrained mini- mization of a convex differentiable piecewise-quadratic objective function in the dual space. Do not use built-in Matlab functions to calculate the norm (i. The bad faith actors are pushing a modern-day version of "Operation Trust," which was a Bolshevik psyop run in the 1920s to convince the country that the military was running a secret operation to stop the communists from taking over. Standard Basis Vectors. 1 Normed Vector Spaces. It is the distance that a taxi travels along the streets of a city that. As nouns the difference between norm and magnitude. Scalars are often taken to be real numbers, but there are also vector spaces with scalar multiplication by complex numbers, rational numbers, or generally any field. norm(), it will compute the l1, l2, l infinity norm based on vector norm algorithm, not matrix norm algorithm. The head of the second vector is placed at the tail of the first vector and the head of the third vector is placed at the tail of the second vector; and so forth until all vectors have been added. A norm is the formalization and the generalization to real vector spaces of the intuitive notion of "length" in the real world. norm(x, ord=None, axis=None) [source] ¶ Matrix or vector norm. a vector norm (sometimes written simply ) is a Nonnegative number satisfying 1. "); // The dot product is calculated in one of two ways: // Using the static DotProduct method: double a = Vector. This function is able to return one of seven different matrix norms, or one of an infinite number of vector norms (described below), depending on the value of the ord parameter. 3: Vector Space of Linear Transforms and Norms De nition Let L(V;W) denote the vector space of all linear transforms from V into W, where V and W are vector spaces over a eld F. We recognize them as. Calculating the length of the vector online. norm¶ numpy. rm = FALSE) Arguments v. n = norm(A) n = norm(A,p) Description. (T) and call ,A,the norm subordinate to the vector norm. The corresponding right delimiters are of course obtained by typing ), ] and \}. The difference of the vectors p and q is the sum of p and –q. The triangle inequality is only satised for p ≥ 1. 2-norm ‘fro’ Frobenius norm – ‘nuc’ nuclear norm – inf. The norm can be the one ("O") norm, the infinity ("I") norm, the Frobenius ("F") norm, the maximum modulus ("M") among elements of a matrix, or the "spectral" or specifies the Frobenius norm (the Euclidean norm of x treated as if it were a vector). General Discussion. Linear Algebra 27, Norm of a Vector, examples. |yi| = max. The norm of a vector is zero if and only if the vector is a zero vector a = 0. Frobenius Norm of a Vector The Frobenius norm of a 1-by- n or n -by-1 vector V is defined as follows: The Frobenius norm of a vector coincides with its 2 -norm. Define norm. The new delimiters should of course adapt to the size of the formula, just like \lvert and \rvert do. Support Vector Machines. Vector: In medicine, a carrier of disease or of medication. That all solutions are of this form can be seen as follows. This is a consequence of a theorem stating that all norms on finite dimensional vector spaces are equivalent. Vector norms A norm is a scalar-valued function from a vector space into the real numbers with the following properties: 1. (mathematics) The most common norm, calculated by summing the squares of all coordinates and taking the square root. 1007/978-0-8176-4980-7 2, c Springer Science+Business Media, LLC 2010. Random vectors with a given correlation. We used vector norms to measure the length of a vector, and we will develop matrix norms to measure the size of a matrix. The calculator will find the angle (in radians and degrees) between the two vectors, and will show the work. I am researching how to speedup optimization problems using quantum algorithms. Homogeneity: For any scalar and vector x, k xk= j jkxk. A function ·, · : V × V → K is called an inner product if. An matrix can be considered as a particular kind of vector , and its norm is any function that maps to a real number that satisfies the following required properties:. 651 просмотр 651 просмотр. The most commonly occurring matrix norms in matrix analysis are the Frobenius, $$L_1$$, $$L_2$$ and $$L_\infty$$ norms. The last Section is an application of matrix norms. Here the key new de velopment is the first time to combine multi-class hinge loss with 2 , 1 -norm re gularization term to. Then, for any norm kkon Rn. when and Iff, 2. I am not an expert, but my understanding is that it can be thought of as a measure of the length of the vector. mag2 (A) = A. That's the energy in the vector v. Norm of a vector. the vector of input features x = (··· ,x(j),···) where x(j) is the j-th input feature 1 ≤ j ≤ p. AU - Ozawa, N. An even easier example of a norm is the sup norm on a finite-dimensional space relative to some choice of basis. Denition 1. The Level 1 BLAS perform scalar, vector and vector-vector operations, the Level 2 BLAS perform matrix-vector operations The BLAS Technical Forum standard is a specification of a set of kernel routines for linear algebra, historically DZNRM2 - Euclidean norm. Different functions can be used and we will see Graphically, the Euclidean norm corresponds to the length of the vector from the origin to the point obtained by linear combination (like applying. norm (x, ord = None, axis = None, keepdims = False) [source] ¶ Matrix or vector norm. FINDING EIGENVALUES AND EIGENVECTORS EXAMPLE 1: Find the eigenvalues and eigenvectors of the matrix A = 1 −3 3 3 −5 3 6 −6 4. The vector calculator allows the calculation of the norm of a vector online. In this example, we are going to find the absolute values for all the records present in [Service Grade] column using the abs Function. If axis is None then either a vector norm (when x is 1-D) or a matrix norm (when x is 2-D) is returned. I still have one more points: the norm of the vector valued function that you have defined, is not satisfying the. The magnitude of a vector is:. Some interesting values of p are: If p = 1, then the resulting 1-norm is the sum of the absolute values of the vector elements. Standard Basis Vectors. Vector Space Operations. A vector space on which a norm is defined is called a normed vector space. In particular, the Euclidean distance of a vector from the origin is a norm, called the Euclidean norm, or 2-norm, which may also be defined as the square root of the inner product of a vector with itself. Our goal is to select a special vector that is normal to the unit tangent vector. I hand coded this with inspiration from Image:Vector_norms. norm(a) or computing sqrt(a. Let ,·,be a vector norm onR n (or C n). Also, is there any other way to find a vector's magnitude that will give the correct answer without defining a new function?. I am not an expert, but my understanding is that it can be thought of as a measure of the length of the vector. The infinity, two and one norms are just two of many useful vector norms. The norm of a vector is a measure of its size. Calculating the length of the vector online. The norm function, or length, is a function V !IRdenoted as kk, and de ned as kuk= p (u;u): Example: The Euclidean norm in IR2 is given by kuk= p (x;x) = p (x1)2 + (x2)2: Slide 6 ’ &$ % Examples The. A function which associates with every vector x ∈ X a real value denoted kxk is called a norm on X if it satisfies the following. Definition: Suppose that is a set of vectors of the vector space. A •Self-join: if Rel1=Rel2 •Size of self-join: ∑ val of A Rows(val)2 •Updates to the relation increment/decrement Rows(val) Lec2 norm …. Infinity norm of a vector The “infinity norm” of a vector is the largest absolute value of its elements. So, this new vector (1, 8, 75) would be the direction we’d move in to increase the value of our function. For vectors, p can assume any numeric value (even though not all values produce a mathematically valid vector norm). Comparing with the standard L 2-norm support vector machine (SVM), the L 1-norm SVM enjoys the nice property of simultaneously preforming classification and feature selection. Note the use of angle. A= What kind of norm do you want to calculate: 1-norm. Its magnitude (or length) is written OQ (absolute value symbols). It is equal to the dot product of the vector by itself, and equivalently to the sum of squared absolute values of its coefficients. + The l1_norm function is computed by a fold operation that sums the absolute values of the elements. The length of the vector is referred to as the vector norm or the vector’s magnitude. For this problem, use the. norm¶ numpy. The general definition for the p-norm of a vector v that has N elements is where p is any positive real value or Inf. In this paper, we propose a new Vector Outlier Regularization (VOR) framework to understand and analyze the robustness of L2,1 norm function. The norm or length of a vector, u, is denoted by ||u|| and defined by Matrices Matrix, A, means a rectangular array of numbers A = The m horizontal n-tuples are. , its distance to the origin 0 on the real number line, ignoring its algebraic sign (positive or negative). The norm of a vector in vector space is a real non-negative value representing intuitively the length, size, or magnitude of the vector. an authoritative standard ; model 2. 2 Properties of the norm Suppose V is a normed space; that is a vector space equipped with a norm. max{yi,−yi} • equivalent LP (with variables x and auxiliary scalar variable t) minimize t subject to −t1≤ Ax−b ≤ t1 (for fixed x, optimal t is t =kAx−bk∞) Piecewise-linear optimization 2–7. UnitVector — unit vector along a coordinate direction. Random number generator. linalg in August 2011 and will be available in scipy 0. Vector Angles You recall from analytic geometry that the definition of a dot product yields : v1 v2 = v1 v2 cosq We can write a simple program to compute the angle between our two vectors : In[117]:= Clear angle angle ArcCos v1. Vector addition is one of the most common vector operations that a student of physics must master. In bracket format: In unit vector component format: = a unit vector, with direction and a magnitude of 1 = a vector, with any magnitude and direction = the magnitude of the vector. Introduction. The vector component of these quantities give the direction as well as the magnitude. A function which associates with every vector x ∈ X a real value denoted kxk is called a norm on X if it satisfies the following. norm () of Python library Numpy. Work should be fun. mag = |A|, the magnitude of a vector. Vector Norms - PowerPoint PPT Presentation. √ In the same way, the size of a 3D vector is. MatrixCalculus provides matrix calculus for everyone. norm (x, ord = None, axis = None, keepdims = False) [source] ¶ Matrix or vector norm. There are several commonly used vector norms, summarized in the following table:. Products containing vitamin. min(sum(abs(x), axis=1)) min(abs(x)) 0 – sum(x != 0) 1. Return Value: It pythagorean distance of the vector. Another name for L2 norm of a vector is Euclidean distance. A norm on a vector space V is a function kk: V !R that satis es (i) kvk 0, with equality if and only if v= 0 (ii) k vk= j jkvk (iii) ku+ vk kuk+ kvk(the triangle inequality) for all u;v2V and all 2F. Brackets and Norms. Definition 3. Shortest Vector Problem (SVP) The shortest vector problem (SVP) asks to find a nonzero vector in a lattice. As for vector norms, the value p = 2 is associated with a Hilbert space. Triple products, multiple products, applications to geometry 3. Why would you normalize a vector? Normalizing a vector can simply problems. Create Presentation Download Presentation. That's the energy in the vector v. In abstract vector spaces, it generalizes the notion of length of a vector in Euclidean spaces. mag2 = |A|*|A|, the vector's magnitude squared. /Esmil 19:52, 2 April 2006 (UTC) Licensing. 4 The distance between matrices and with respect to a matrix norm is | | Theorem 7. Absolute norms on Fn, i. Tagged: norm of a vector. The square root of this is known as the vector norm or the length of a vector. In this section we shall look at. A pattern that is regarded as typical of something: a neighborhood where families with two wage-earners are the norm. Do not use built-in Matlab functions to calculate the norm (i. The bad faith actors are pushing a modern-day version of "Operation Trust," which was a Bolshevik psyop run in the 1920s to convince the country that the military was running a secret operation to stop the communists from taking over. A function kk: V !R is called a (vector) norm if (N 1) kxk 0 for all x2V, with equality iff x= 0, [positivity] (N 2) k xk= j jkxkfor all 2K and x2V, [homogeneity] (N. norm, standard; quota; rule v. Vector of Vectors is a two-dimensional vector with a variable number of rows where each row is vector. N = [5,2,3] The magnitude |N| is |N| = sqrt(5^2 + 2^2 + 3^2) |N| = 6. Answer: We show that the three. • ℓ∞-norm (Chebyshev norm) of m-vector y is kyk∞= max. De nition 2 (Norm) Let V, ( ; ) be a inner product space. Write a Matlab function, called myvectornorm that inputs the vector x and outputs the 2-norm of x. Linear Algebra 27, Norm of a Vector, examples. 2-norm ‘fro’ Frobenius norm – ‘nuc’ nuclear norm – inf. CUTCO has been made in America since 1949 and is guaranteed FOREVER. Note the use of angle. Mathematically, a vector is a tuple of n real numbers where n is an element of the Real (R) number space. It follows that the Q-norm { the function jxj Q= q xTQx { is a norm. The nuclear norm a matrix is the sum of the singular values, the sum of the singular values. Applications of the cross product will be shown. with A ∈ Rm×n, b ∈ Rm. Caution! This is a large HTML document. vector of the form x = A+b+(I ¡A+A)y where y 2 IRn is arbitrary (4) is a solution of Ax = b: (5) Furthermore, all solutions of (5) are of this form. Description. The vector must start somewhere and move in a path towards a different place. 3: Vector Space of Linear Transforms and Norms De nition Let L(V;W) denote the vector space of all linear transforms from V into W, where V and W are vector spaces over a eld F. is that norm is (mathematics) a function, generally denoted v\mapsto\left|v\right| or v\mapsto\left\|v\right\|, that maps vectors to non-negative scalars and has the following properties: while magnitude is (mathematics) of a vector, the norm, most commonly, the two-norm. We therefore dene a distance function for vectors that has similar properties. In this section we shall look at. Usage vector. A vector space (also called a linear space) is a set of objects called vectors, which may be added together and multiplied ("scaled") by numbers, called scalars. suppose Ax = 5 and Ay = 6 Hence A = (5, 6). Vector Operators: Grad, Div and Curl In the first lecture of the second part of this course we move more to consider properties of fields. - one of NORM_1, NORM_2, NORM_INFINITY. In a finite dimensional normed space, all norms are equivalent. Every nonzero vector has a corresponding unit vector, which has the same direction as that vector but a magnitude of 1. One way to normalize the vector is to apply some normalization to scale the vector to have a length of 1 i. We define the norm to be the magnitude or length of the vector so the norm must be positive. As InformationLiberation. Statistics gathered by Neustar, Inc. The method considers at-tention weights and previously ignored factors, i. Examples: The l1 norm: ||A||1 = i,j |aij|. In fact, it. De nition (Induced Operator Norm) Let V and W be two normed vector spaces and let T : V !W be a. mag2 (A) = A. Let us see some examples to calculate the magnitude of a vector. Let's give an example of a non complete normed vector space. h2norm — H2 norm of a continuous time proper dynamical system. Scalar and vector fields. array ([2, 4, 8, 7, 7, 9,-6]) b = np. Vector Norm List Processor. In many ways, norms act like absolute values. In Rn there is a standard notion of length: the size of a vector v = (a1,. The calculator will find the angle (in radians and degrees) between the two vectors, and will show the work. e, norm or vecnorm). erefore, direct calculation of the distribution of the norm of the vector rejection is di cult. vector or matrix of real or complex numbers (full or sparse storage). Hello! I’m so sorry to bother you but I’m new to FreeFEM and I have some trouble. Compute Different Types of Norms of Vector. To do this, we rst introducea few lemmas. The norm of a vector allows you to gauge the distance or the magnitude of a vector. See also: normest, normest1, cond, svd. Vector measure — In mathematics, a vector measure is a function defined on a family of sets and taking vector values satisfying certain properties. x2+y2≥ 0, so p x +y2≥ 0. is that norm is (mathematics) a function, generally denoted v\mapsto\left|v\right| or v\mapsto\left\|v\right\|, that maps vectors to non-negative scalars and has the following properties: while magnitude is (mathematics) of a vector, the norm, most commonly, the two-norm. Typical values for p are 1. Infinity and Negative Infinity Norm of a Vector. De norm is niet negatief. Furthermore, the generalized -norm of a vector or (numeric) matrix is returned by Norm [ expr, p ]. Statistics gathered by Neustar, Inc. Calculate the norm of a vector in the plane. It is the distance that a taxi travels along the streets of a city that. Besides the familiar Euclidean norm based on the dot product, there are a number of other important norms that are used in numerical analysis. AU - Todorov, I. Computing the norm of a matrix. Remember, we can write a vector that starts at There is a problem though. When y is a centered unit vector, the vector β*y has L 2 norm β. Example 4 Find a unit vector that has the same direction as the vector w = - 3, 5 >. Lindner shared this question 7 years ago. Vector norms. The associated norm is called the two-norm. The magnitude of the vector in Cartesian coordinates is the square root of the sum of the squares of it coordinates. The Frobenius norm of a vector coincides with its 2-norm. p – q = p + (–q) Example: Subtract the vector v from the vector u. % % history is a structure that contains the objective value, the primal and % dual residual norms, and the tolerances for the primal and dual residual % norms at each iteration. In keynote remarks at CES 2021, Smith slammed the Russia-linked breach as an. This returns a vector with the square roots of each of the components to the square, thus 1 2 3 instead of the Euclidean Norm This is a trivial function to write yourself: norm_vec - function(x) sqrt(sum(x^2)). The length of a vector is a nonnegative number that describes the extent of the vector in space, and is sometimes referred to as the vector’s magnitude or the norm. std::vector. In molecular biology, a vector may be a virus or a plasmid that carries a piece of foreign DNA to a host cell. Frobenius norm. The L2-norm is the. Norms are 0 if and only if the vector is a zero vector. , a unit norm. cc | Übersetzungen für 'norm of a vector' im Englisch-Deutsch-Wörterbuch, mit echten Sprachaufnahmen, Illustrationen, Beugungsformen Wörterbuch Englisch → Deutsch: norm of a vector. In this section we shall look at. Different functions can be used and we will see Graphically, the Euclidean norm corresponds to the length of the vector from the origin to the point obtained by linear combination (like applying. Also find the definition and meaning for various math words from this math dictionary. Answer: We show that the three. By the Pythagorean theorem of plane geometry, the dis-tance (v1, v2) between the point (v1, v2) and the origin (0, 0) is. Random vectors with a given correlation. If given a matrix variable, pnorm will treat it as a vector, and compute the p-norm of the concatenated columns. This method has the same effect as the overloaded + operator. norm () = A/|A|, a unit vector in the direction of the vector. 2-norm ‘fro’ Frobenius norm – ‘nuc’ nuclear norm – inf. The data types can be logical, integer, double, character, complex or raw. The square root of this is known as the vector norm or the length of a vector. Определение ориентации 3D объекта по изображению (Determining the orientation of a 3D object from an image). A vector of magnitude, or length, 1 is called a unit vector. a vector norm (sometimes written simply ) is a Nonnegative number satisfying 1. "); // The dot product is calculated in one of two ways: // Using the static DotProduct method: double a = Vector. And do you remember what was special? We've talked about using the L1 norm. Vector Norm On a vector space V, a norm is a function ⋅from V to the set of non-negative reals that obeys three postulates:, (), 0 0, x y x y if x y V Trinagular Inequality x x if R x V x if x C + ≤ + ∈ = ∈ ∈ > ≠ λ λ λ we can think of x as the length or magnitude of the vector x. Geometrically, multiplying a vector by an orthogonal matrix reflects the vector in some plane and/or rotates it. Let $\times$ denote the vector cross product. R abs Function Example 2. If you watched the plane from the ground it would seem to be slipping sideways a little. PY - 2019/1/28. View Larger Image Calculus in Vector Spaces Without Norm Frölicher, Alfred ; & Bucher, W. The vector component of these quantities give the direction as well as the magnitude. left and right of rearrangement matrixes, sum and difference of two matrixes, products of a matrix on a vector and matrixes on a matrix, inverse matrix and. The norm of a vector is zero if and only if Cos Angle of Vectors the vector is a zero vector. In particular, norm (A, Inf) returns the largest value in abs (A), whereas norm (A, -Inf) returns the smallest. Definition 3. A norm is the yardstick by which we measure the size of vectors, and we explore some alternatives to the tradition Euclidean. A unit normal vector of a curve, by its definition, is perpendicular to the curve at given point. The L1 norm is defined for both vectors and matrices, we can easily write a C++ function to calculate it, but when possible it is better to use a more stable and generic implementation, as the one provided by the Boost Numeric uBLAS library. See full list on machinelearningmindset. In Rn there is a standard notion of length: the size of a vector v = (a1,. So, this new vector (1, 8, 75) would be the direction we’d move in to increase the value of our function. WriteLine ("Norms, dot products, etc. Besides the familiar Euclidean norm based on the dot product, there are a number of other important norms that are used in numerical analysis. Norm An inner product space induces a norm, that is, a notion of length of a vector. The meaning of direction is pretty self explanatory. Y1 - 2019/1/28. But, it is actually possible to talk about linear combinations of anything as long as you understand the main idea of a linear combination: (scalar)(something 1) + (scalar)(something 2) + (scalar)(something 3) These “somethings” could be “everyday” variables like \$$x\$$ and …. The equation of a line with direction vector d ⃗ = (l, m, n) \vec{d}=(l,m,n) d = (l, m, n) that passes through the point (x 1, y 1, z 1) (x_1,y_1,z_1) (x 1 , y 1 , z 1 ) is given by the formula x − x 1 l = y − y 1 m = z − z 1 n , \frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}, l x − x 1 = m y − y 1 = n z − z 1 ,. 4 If the formulation of the product changes, PQT-VCP will need to be consulted to make sure the product maintains the same specifications. For any x,y ∈ V we have kxk−kyk ≤ kx−yk Proof. max(sum(abs(x), axis=0)) as below-1. If you think of the norms as a length, you easily see why it can’t be negative. This is a topological vector space because: The vector addition + : X × X → X is jointly continuous with respect to this topology. In the previous section we looked at the infinity, two and one norms of vectors and the infinity and one norm of matrices and saw how they were used to estimate the propagation of errors when one solves equations. norm () of Python library Numpy. This function is able to return one of eight different matrix norms, or one of an infinite number of vector norms (described below), depending on the value of the ord parameter. std::vector. We study the main properties of the p-norms on Rn or Cn, which are useful norms in functional analysis. Write a Matlab function, called myvectornorm that inputs the vector x and outputs the 2-norm of x. 10 Last week I discussed with Gael how we should compute the euclidean norm of a vector a using SciPy. You can think of the norm as the length of the vector. Let and be two vectors in the space. norm() is a vector-valued function which computes the length of the vector. The Level 1 BLAS perform scalar, vector and vector-vector operations, the Level 2 BLAS perform matrix-vector operations, and the Level 3 BLAS perform matrix-matrix operations. Let ,·,be a vector norm onR n (or C n). Do October 10, 2008 A vector-valued random variable X = X1 ··· Xn T is said to have a multivariate normal (or Gaussian) distribution with mean µ ∈ Rn and covariance matrix Σ ∈ Sn. The infinity norm of a 1-by-n or n-by-1 vector V is defined as follows. Therefore, multiplying a vector by an orthogonal matrices does not change its length. The 1-norm is also called the taxicab metric (sometimes Manhattan metric) since the distance of two points can be viewed as the distance a taxi would travel on a city (horizontal and vertical movements). It is properties make it special. A norm is the formalization and the generalization to real vector spaces of the intuitive notion of "length" in the real world. Why would you normalize a vector? Normalizing a vector can simply problems. Let V be a vector space over a field K (K = R or C). Denition 1. Recall that if z = x + iy is a complex number with real part x and imaginary part y, the complex conjugate of z is dened as z = x − iy, and the absolute value. 5 Tutorials that teach Calculating the Norm of a Vector. A Vector or a Function. Infinity and Negative Infinity Norm of a Vector. - Buy this stock vector and explore similar vectors at Adobe Stock. The norm of a complex vector v. The norm of a vector is zero if and only if the vector is a zero vector a = 0. In the case of vectors, let’s assume for the moment that a standard vector has a length of 1. Therefore we have the ability to determine if a sequence is a Cauchy sequence. Then the Span of the Set denoted and is the set of all linear combinations of the vectors in, that is, for any scalars,. In addition, this is the reason why we need a negative (-) sign. The VNLP processor computes the L2 norm of an n-dimensional complex vector of the The x and y values of the vector are stored in memory in the form of a doubly linked-list data structure shown below. 5 - 1 for _ in range(num_simulations): # calculate the matrix-by-vector product Ab b_k1 = torch. Vector Norms… Vector norms are functions that map a vector to a real number You can think of it as measuring the magnitude of the vector The norm you know is the 2-norm: You can use it to measure the distance between two points. e, norm or vecnorm). we argue that 1-norm measurement is better than 2-norm measurement for outlier resistance. X_norm = sqrt (sum (X. General Discussion. De norm van het scalaire veelvoud van een vector is het product van de norm met de gewone absolute waarde van de scalair:. def _power_iteration(self, A, num_simulations=30): # Ideally choose a random vector # To decrease the chance that our vector # Is orthogonal to the eigenvector b_k = torch. for some normal vector w2Rd and offset b2R. keepdimsbool, optional. 3 Here are several commonly used S-invariant norms on vectors or matrices. We used vector norms to measure the length of a vector, and we will develop matrix norms to measure the size of a matrix. Norms are often used in regularization methods and other machine learning procedures, as well as many different The most commonly occurring vector norms are the $1$, $2$, and $\infty$ norms, which are a part of the $p$-norm class of vector norms. Create Presentation Download Presentation. Add a real scalar to all elements of this vector. We model each pixel with a Bernoulli distribution in our model, and we statically binarize the dataset. A magnitude of a vector (length of a vector or norm of a vector) AB it is a length of the line segment AB. The Distance Between Two Points. Do October 10, 2008 A vector-valued random variable X = X1 ··· Xn T is said to have a multivariate normal (or Gaussian) distribution with mean µ ∈ Rn and covariance matrix Σ ∈ Sn. In particular, norm (A, Inf) returns the largest value in abs (A), whereas norm (A, -Inf) returns the smallest. A normed space is simply a vector space endowed with a norm. Practice problems (one per topic). The Level 1 BLAS perform scalar, vector and vector-vector operations, the Level 2 BLAS perform matrix-vector operations The BLAS Technical Forum standard is a specification of a set of kernel routines for linear algebra, historically DZNRM2 - Euclidean norm. The head of the second vector is placed at the tail of the first vector and the head of the third vector is placed at the tail of the second vector; and so forth until all vectors have been added. Fomin, "Elements of the theory of functions and functional analysis" , 1–2, Graylock (1957–1961) (Translated from Russian) [2] W. Symptoms of deficiency. Norm of block vector: if a, b are vectors. an authoritative standard ; model 2. The vector v = - 3/5, 4/5 > is a unit vector because |v| = | - 3/5, 4/5 >| = √ (- 3/5) 2 + (4/5) 2 = √ 9/25 + 16/25 = √ 25/25 = √ 1 = 1. Matrix p-norm is defined as ‖ A ‖ p = sup x ≠ 0 ‖ A x ‖ p ‖ x ‖ p In another word, matrix p-Norm is defined as the largest scalar that you can get for a unit vector e. Vector Angles You recall from analytic geometry that the definition of a dot product yields : v1 v2 = v1 v2 cosq We can write a simple program to compute the angle between our two vectors : In[117]:= Clear angle angle ArcCos v1. In particular, norm (A, Inf) returns the largest value in abs (A), whereas norm (A, -Inf) returns the smallest. We say that two norms are sequentially equivalent if the convergence in one norm implies the convergence in another norm. The norm is a bit like applying Pythagoras theorem in an arbitrary number of dimensions. WriteLine ("Norms, dot products, etc. GitHub Gist: instantly share code, notes, and snippets. Then, for any norm kkon Rn. Norm: Given an element x in X, one can form the norm ||x||, which is a non-negative number. In mathematics, a normed vector space or normed space is a vector space over the real or complex numbers, on which a norm is defined. Suppose that we have a set of scalars where and. The Level 1 BLAS perform scalar, vector and vector-vector operations, the Level 2 BLAS perform matrix-vector operations The BLAS Technical Forum standard is a specification of a set of kernel routines for linear algebra, historically DZNRM2 - Euclidean norm. norm (b) print ('A : ', a) print ('B : ', b) print (' Norm of A : ', ma) print. Like absolute values, norms are multiplicative in the sense that kcvk= jcjkvk when cis a real number and v is a real vector. Vector Norms DEF: A norm is a function that satisfies p-norms:The most important class of vector norms Example. Example 1 Find the general formula for the tangent vector and unit tangent vector to the curve given by $$\vec r\left( t \right) = {t^2}\,\vec i + 2\sin t\,\vec j + 2. In this tutorial we will look at two types of norms that are most common in the field of machine learning. Similarly, the construction of the norm of a vector is motivated by a desire to extend the intuitive notion of the length of a vector to higher-dimensional spaces. Applications of the cross product will be shown. The vector 1-norm is a norm. Some interesting values of p are: If p = 1, then the resulting 1-norm is the sum of the absolute values of the vector elements. Then, for any norm kkon Rn. Let \times denote the vector cross product. The Array type is a specific instance of DenseArray; Vector and Matrix are aliases for the 1-d and 2-d cases. Vector norms. Instead, in this section, we calculate the moments of. Multi-Class L2,1-Norm Support Vector Machine. 47% of all multi-vector DDoS attacks were launched in Q4 '15. √(3)2 + (2)2 +(4)2 +(6)2 (3) 2 + (2) 2 + (4) 2 + (6) 2. ^ 2, 1)); Y = bsxfun (@ rdivide, X, X_norm); This code squares all of the elements of X, then sums along the first dimension (the rows) of the result, and finally takes the square root of each element. However, how the change of norm affects the generalization ability of SVMs has not been clarified so far except for numerical experiments. Y1 - 2019/1/28. The 1-norm is also called the taxicab metric (sometimes Manhattan metric) since the distance of two points can be viewed as the distance a taxi would travel on a city (horizontal and vertical movements). Con-sider a vector v = (v1, v2) in the plane R2. Syntax: numpy. If you use l2-normalization, "unit norm" essentially means that if we squared each element in the vector. The frequently used left delimiters include (, [ and {, which are obtained by typing (, [and \{respectively. In mathematics, a norm is a function from a real or complex vector space to the nonnegative real numbers that behaves in certain ways like the distance from the origin: it commutes with scaling. Obvious applications of the gradient are finding the max/min of multivariable functions. In particular, norm (A, Inf) returns the largest value in abs (A), whereas norm (A, -Inf) returns the smallest. So it's like the L1 norm for a vector. Vector, in mathematics, a quantity that has both magnitude and direction but not position. , kQuk = kuk. Norm is a kind of measure of the size of an mathematical object. The vector calculator allows the calculation of the norm of a vector online. Be careful to distinguish 0 (the number) from \(\vec 0$$ (the vector). e, norm or vecnorm). Then, for any norm kkon Rn. Vector and its Unit Norm. (mathematics) The most common norm, calculated by summing the squares of all coordinates and taking the square root. Examples: The l1 norm: ||A||1 = i,j |aij|. The most familiar norm on R is the Euclideann. Übersetzung 1 - 50 von 84207 >>. In this tutorial we will look at two types of norms that are most common in the field of machine learning. a vector of real numbers. Two approaches suggest themselves, either calling scipy. 3: Vector Space of Linear Transforms and Norms De nition Let L(V;W) denote the vector space of all linear transforms from V into W, where V and W are vector spaces over a eld F. svd — singular value decomposition. (Euclidean) norm of vector a ∈ Rn a + b ≤ a + b for all vectors a and b of equal length. 3 Here are several commonly used S-invariant norms on vectors or matrices. Recall that if z = x + iy is a complex number with real part x and imaginary part y, the complex conjugate of z is dened as z = x − iy, and the absolute value. Why would you normalize a vector? Normalizing a vector can simply problems. Specifically, the norm of must satisfy the following three. The vector component of these quantities give the direction as well as the magnitude. From the de nition of matrix-vector multiplication, the value ~y 3 is computed by taking the dot product between the 3rd row of W and the vector ~x: ~y 3 = XD j=1 W 3;j ~x j: (2) At this point, we have reduced the original matrix equation (Equation 1) to a scalar equation. Compute Different Types of Norms of Vector. For matrices, the matrix norm induced by the vector p-norm is used, where valid values of p are 1, 2, or Inf. A norm in V is a map x→ ∥x∥ from V to the set of non-negative. Inner Products, Lengths, and Distances of 3-Dimensional Real Vectors. Consider the following training data (x 1;x 2. | 2021-07-26T22:53:36 | {
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http://math.stackexchange.com/questions/55964/let-g-be-a-group-of-order-2m-where-m-is-odd-prove-that-g-contains-a-nor/55968 | # Let $G$ be a group of order $2m$ where $m$ is odd. Prove that $G$ contains a normal subgroup of order $m$
I searched in the existing post and didn't find this problem. I am sorry if someone else have already posted.
Let $G$ be a group of order $2m$ where $m$ is odd. Prove that $G$ contains a normal subgroup of order $m$.
There is a hint:
Denote by $\rho$ the regular represetation of $G$: find an odd permutation in ${\rho}(G)$.
I don't know how to find an odd permutation in the regular representation. I am wondering whether all the elements of $G$ of odd order form this subgroup in this case.
Thanks.
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Hint: by Cauchy's theorem, $G$ contains an element of order 2. How does this act in the regular rep? – Chris Eagle Aug 6 '11 at 12:47
@Chris: your comment above would make a nice answer... – Pete L. Clark Aug 6 '11 at 12:51
@Chirs: thank you much for the hint. This element is an odd permutation in the regular representation. – ShinyaSakai Aug 7 '11 at 3:54
• $\textbf{Theorem.}$ Let $|G| = 2^{n} \cdot m$ where $2 \nmid m$. If $G$ has a cyclic $2$- Sylow subgroup, then $G$ has a normal subgroup of order $m$.
Your question is just a corollary to this theorem. Please see $\textbf{Theorem 7.9}$ in Prof. Keith Conrad blurb here:
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Thank you very much. I will rewrite and post the proof here. I hope you will not mind... – ShinyaSakai Aug 7 '11 at 3:48
When $n=0$, the result is clear. Now consider the case when $n \geq 1$. First, there is a normal subgroup of $G$ of order $2^{n-1}m$. In order to proof this, consider the permutation of $G$ on itself by left multiplication, $\rho: G \rightarrow$ Sym$G$. As $G$ has a cyclic $2$-Sylow subgroup, $G$ has an element $g$ of order $2^n$, and $\rho(g)$ is the composition of $m$ $2^n$-cycles. Thus, $\rho(g)$ is an odd permutation. So, the composition map sgn$\circ \rho: G \rightarrow \{ \pm 1 \}$ is onto. The kernel of this map is a normal subgroup $H$ of $G$, of order $2^{n-1}m$. – ShinyaSakai Aug 7 '11 at 3:49
Now proceed by induction on $n$. For $n=1$, we are done. For $n >1$. $G$ has a normal subgroup $H$ of order $2^{n-1}m$. The intersection of $H$ and the cyclic $2$-Sylow subgroup of $G$ is a cyclic $2$-Sylow subgroup of $H$. Thus, by the induction hypothesis, $H$ has a normal subgroup $N$ of order $m$. Then, $N$ is the only subgroup of $H$ of order $m$. If not, suppose that $N'$ is another such group, as $N$ is normal in $H$, the multiplication $NN'$ containing $N$ is also a subgroup of $N$, of odd order, but this order is divisible by $m$, the order of $N$. So, $NN'=N$, and $N' \subseteq N$. – ShinyaSakai Aug 7 '11 at 3:50
The two are of the same order, thus equal. Now, for any $x \in G$, $xNx^{-1} \subseteq xHx^{-1} =H$, so $xNx^{-1}$ is another subgroup of $H$ of order $m$. So, $xNx^{-1}=N$. $N$ is normal in $G$ of order $m$. – ShinyaSakai Aug 7 '11 at 3:50
By Cauchy's theorem, $G$ contains an element of order 2. How does this act in the regular representation?
There is a nice generalization of this fact due to John Thompson, known as the "Thompson transfer Lemma". It goes as follows: let $G$ be a finite group which has a subgroup $M$ such that $[G:M] = 2d$ for some odd integer $d$, and suppose that $G$ has no factor group of order $2$. Then every element of order $2$ in $G$ is conjugate to an element of $M$. I will not give the full proof as it reveals too much of the solution of the original question, but the idea is the same: any element of order $2$ in $G$ which does not lie in any conjugate of $M$ must act as an odd permutation in the permutation action of $G$ of the (say, right) cosets of $M$. As a sample application, consider a finite non-Abelian simple group $G$ whose Sylow $2$-subgroup $S$ has a cyclic subgroup $M$ of index $2$. Then $G$ certainly has no factor group of order $2$, so every element of order $2$ (involution) of $G$ is conjugate to an involution of $M$. But $M$ only has one involution as $M$ is cyclic, so $G$ has one conjugacy class of involutions. In case anyone is wondering, the Thompson Transfer Lemma is a true generalzation of the question, because the case $M = 1$ can be applied to the question to conclude that there must be a normal subgroup of index $2$ for $G$, because no element of order $2$ lies in any conjugate of the trivial group. | 2014-07-12T02:23:20 | {
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https://math.stackexchange.com/questions/194705/is-there-a-direct-elementary-proof-of-n-sum-kn-phik | # Is there a direct, elementary proof of $n = \sum_{k|n} \phi(k)$?
If $k$ is a positive natural number then $\phi(k)$ denotes the number of natural numbers less than $k$ which are prime to $k$. I have seen proofs that $n = \sum_{k|n} \phi(k)$ which basically partitions $\mathbb{Z}/n\mathbb{Z}$ into subsets of elements of order $k$ (of which there are $\phi(k)$-many) as $k$ ranges over divisors of $n$.
But everything we know about $\mathbb{Z}/n\mathbb{Z}$ comes from elementary number theory (division with remainder, bezout relations, divisibility), so the above relation should be provable without invoking the structure of the group $\mathbb{Z}/n\mathbb{Z}$. Does anyone have a nice, clear, proof which avoids $\mathbb{Z}/n\mathbb{Z}$?
Clearly $n$ counts the number of elements in the set $\{1,\ldots,n\}$. This suggests that to get a combinatorial proof we should count the number of elements in this set in a different way and get $\sum_{k \mid n} \varphi(k)$.
For $k \mid n$, let $S(k)$ be the set of $m \in \{1,\ldots,n\}$ such that $\gcd(m,n) = k$. Since for all $m \in \{1,\ldots,n\}$, $\gcd(m,n)$ is a divisor of $n$, we have $\sum_{k \mid n} \# S(k) = n$.
Now I claim that for all $k \mid n$, $\# S(k) = \varphi(\frac{n}{k})$. This implies the result because as $k$ runs through all positive divisors of $n$ so does $\frac{n}{k}$. Can you see how to establish this equality?
• Yes I can see how to establish that equality. Very nice! – JessicaB Sep 12 '12 at 14:56
• @JessicaB But this is basically the same proof as the one you mentioned in your original question, just phrased slightly differently. To say that $\gcd(m,n)=k$ is the same as saying that $m$ has order $n/k$ in the group $\mathbb{Z}/n\mathbb{Z}$. – Ted Sep 12 '12 at 16:01
• @Ted I think my question was phrased sufficiently clearly to convey that I was expecting an elementary number theory re-phrasing of "the same proof." That said, I disagree with the spirit of your assertion, for the following reason: in order to "see it from the group structure of $\mathbb{Z}/n\mathbb{Z}$," one is actually required to write a lot more. Can we really say that two proofs are 'basically the same' if they use the same ingredients, but one is much longer and introduces needless language? (order, generator, etc). – JessicaB Sep 12 '12 at 17:46
Write the fractions $1/n,2/n,3/n \dots ,n/n$ in the simplest form and you can observe that each fraction is of the form $s/t$ where $t$ divides $n$ and $(s,t)=1$. So the number of the fractions is the same as $\sum_{k|n}{\phi(k)}$ which is equal to $n$.
• It might be worth noting that reducing to lowest common terms does not "equalize" any of the $n$ fractions (they remain unequal as before). – hardmath Sep 12 '12 at 14:43
• @hardmath: Actually, that's not even required. Anyone who feels like $\frac24\ne\frac12$ still keeps $n$ objects with all divisors $d$ occuring as denominator etc. – Hagen von Eitzen Sep 12 '12 at 14:52
• This is perhaps my favorite proof of this result. I believe I first encountered it in Hardy and Wright's book on number theory, but I could be wrong. – Chris Leary Sep 12 '12 at 15:07
• @ChrisLeary, it's my favorite proof too. And you're right, it's in Hardy and Wright, §16.2, at least in my 5th ed. – lhf Sep 13 '12 at 0:09
Here is a proof using induction on $n$. The case $n=1$ is clear as $\phi(1)=1$.
Let $n>1$ and assume the result for positive integers less than $n$. Choose a prime divisor $p$ of $n$ and write $n=mp^k$ for $m$ and $p$ coprime. The divisors of $n$ are precisely the $dp^i$ for $d|m$ and $0\leq i\leq k$, so we obtain
\begin{align*} \sum_{d|n}\phi(d)&=\sum_{i=0}^{k}\sum_{d|m}\phi(dp^i)=\sum_{i=0}^{k}\phi(p^i)\sum_{d|m}\phi(d)=m\sum_{i=0}^{k}\phi(p^i)\\ &=m\left(1+\sum_{i=1}^{k}(p^i-p^{i-1})\right)=mp^k=n. \end{align*}
Consider all proper fractions of the form $a/n$. There are $n$ of those. When you consider their reduced forms you get fractions of the form $b/d$ with $d|n$ and $(b,d)=1$. There are $\phi(d)$ of those. The result follows.
This proof uses Möbius inversion, and is pretty quick! Recall the function $$\mu(n) = \begin{cases} (-1)^{\nu(n)} \qquad \text{if n is square free} \\ 0 \,\,\qquad\qquad\text{otherwise}, \end{cases}$$ where $\nu(n)$ is the number of distinct prime divisors of $n$. The Möbius inversion formula says that $$f(n)=\sum_{d\mid n} g(d)$$ if and only if $$g(n)=\sum_{d\mid n}\mu(d)f(n/d).$$ Putting $f(n)=n$ and $g(n)=\varphi(n)$, by Möbius it suffices to show $$\varphi(n)=\sum_{d\mid n}\mu(d)\frac{n}{d},$$ and this is true since \begin{align*} \varphi(n)&=n \prod_{p\mid n}\bigg(1-\frac{1}{p}\bigg)\\ &=n \sum_{d\mid n}\frac{\mu(d)}{d}\\ &=\sum_{d\mid n}\mu(d)\frac{n}{d}. \end{align*}
(Note that $\prod_{p\mid n}\big(1-\frac{1}{p}\big)=\sum_{d\mid n}\frac{\mu(d)}{d}$ since terms in the sum are zero except at divisors of $d$ that consist of distinct primes, and multiplying out the product on the right gives precisely this sum.)
I have a pretty cool inductive proof.
The Base Case is trivial.
Now, assume that for some positive integer $m$ we had
$\sum_{m|n} \phi(m) = n$
Now, I will show that for any prime power $p^d$, we must have $\sum_{m|p^dn} \phi(m) = p^dn$
For the sake of convenience, we may assume that $\gcd(p,n) = 1$. If this was not the case, just repeat the proof but instead of using $n$ use $\frac{n}{\text{ord}_p(n)}$ instead.
Now,
\begin{align} \sum_{m|p^dn} \phi(m) &= \sum_{m|n} \phi(m) + \sum_{p|m|pn} \phi(m) + ... + \sum_{p^d|m|p^dn} \phi(m) \\ &= n+\sum_{k|n} \phi(p)\phi(k) + \sum_{k|n} \phi(p^2)\phi(k) + ... + \sum_{k|n} \phi(p^d)\phi(k) \\ &= n+\phi(p)*n+\phi(p^2)*n+...+\phi(p^d)*n \\ &= n(1+(p-1)+p(p-1)+p^2(p-1)+...+p^{d-1}(p-1)) \\ &= n(1+(p-1)\frac{p^d-1}{p-1}) \\ &= n*p^d \end{align}
thus proving the inductive step thus completing the proof.
• The base case consists of showing the equality holds for all primes $n$ (I guess that's what you called trivial?) but you also need to prove that the inductive step does not just work for prime powers, but for any composite $n = p \cdot q$. – TMM Oct 1 '16 at 21:41
Claim:Number of positive integers pair $(a, b)$ satisfying : $n=a+b$ (for given $n$) $\gcd(a, b) =d$ and $d|n$ is $\phi(n/d)$.
Proof: Let $a=xd$ and $b=yd$ We want number of solution for $x+y=\frac{n}{d}$ such that $\gcd(x, y) =1$.
$\gcd(x,y)=\gcd(x,x+y)=\gcd(x,n/d)=1$
Solution for $x+y=n/d$, $\gcd(x,y)=1$ is $\phi(n/d)$. ________________________
Number of positive integers pair $(a, b)$ satisfying $a+b=n$ is $n$.
But this can counted in different way: If $(a, b)$ is solution then $\gcd(a, b) =d$ for some divisor $d$ of $n$.
So we can use our claim to write $\sum_{d|n} \phi(n/d) =\sum_{d|n}\phi(d)=$ Number of solution $=n.$ | 2019-07-23T11:35:13 | {
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https://math.stackexchange.com/questions/1687957/solve-for-z-in-z3-8i | # Solve for $z$ in $z^3=8i$
I only have a question about the end results. I answered the question fully but my professor knocked off 1 point for my $w_1^0$ result, but I don't know why. He circled the $i\pi /6$ in my answer but I can't figure out what I did wrong. Does anyone know what might be wrong here?
$$z^3=8i$$ $$z=(8i)^{1/3}$$ Converting to polar form and letting $k=0,1,2$ and $-\pi \leq \theta \leq \pi$, we get
$$w_1^0 =2e^{i\pi /6}$$ $$w_2^1=2e^{i5\pi /6}$$ $$w_3^2=2e^{9i \pi /6} = 2e^{-i \pi /2}$$
What's wrong with the $\frac{i\pi}{6}$ in $2e^{i\pi /6}$??
• It could simply be a misreading of your handwriting. I don't see anything wrong with it. Mar 8 '16 at 4:15
• I don't either! But here's the weird part - he circled $i\pi /6$ 3 times in my homework as well! I'm going to ask him tomorrow but I was wondering if anyone know what I was missing here. Mar 8 '16 at 4:16
• Did your professor want you to convert back to $a+bi$? Or perhaps draw a diagram showing the three points forming a triangle? Mar 8 '16 at 4:20
• No way because otherwise he would have taken off points for the other 2 answers as well. Plus, in the homework he was also only taking off 1 point when I wrote $i\pi /6$. Mar 8 '16 at 4:21
• I guess the three circles meant "$\pi/6$ is an angle you were drilled in school to know the sine and cosine of, you should write it out." Mar 8 '16 at 5:25
$$z^3=8i=8e^{\frac{\pi i}2+2k\pi i}=8e^{\frac{\pi i}2\left(4k+1\right)}\implies$$
$$z_k=(8i)^{1/3}=\sqrt[3]8e^{\frac{\pi i}6\left(4k+1\right)}=2e^{\frac{\pi i}6\left(4k+1\right)}\;,\;\;k=0,1,2\implies$$
$$\begin{cases}z_0=2e^{\frac{\pi i}6}=2\left(\frac{\sqrt3}2+\frac12i\right)=\sqrt3+i\\{}\\z_1=2e^{\frac{5\pi i}6}=2\left(-\frac{\sqrt3}2+\frac12i\right)=-\sqrt3+i\\{}\\z_2=2e^{\frac{3\pi i}2}=-2i\end{cases}$$
Hint use $$a^3-b^3=(a-b)(a^2+ab+b^2)$$ solve them and get the roots .
• Maybe your professor wants $a+bi$ form rather than polar Mar 8 '16 at 4:58 | 2021-12-08T01:25:32 | {
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https://www.physicsforums.com/threads/proof-to-grade.185122/ | 1. Sep 16, 2007
### Jacobpm64
1. The problem statement, all variables and given/known data
Assign a grade of A (excellent) if the claim and proof are correct, even if the proof is not the simplest or the proof you would have given. Assign an F (failure) if the claim is incorrect, if the main idea of the proof is incorrect, or if most of the statements in it are incorrect. Assign a grade of C (partial credit) for a proof that is largely correct, but contains one or two incorrect statements or justifications. Whenever the proof is incorrect, explain your grade. Tell what is incorrect and why.
Claim: If A, B, and C are sets, and $$A \subseteq B$$ and $$B \subseteq C$$, then $$A \subseteq C$$.
Proof: Suppose x is any object. If $$x \in A$$, then $$x \in B$$, since $$A \subseteq B$$. If $$x \in B$$, then $$x \in C$$, since $$B \subseteq C$$. Therefore $$x \in C$$. Therefore $$A \subseteq C$$.
3. The attempt at a solution
Am I correct in wanting to give this proof a grade of an A, even if the language seems a bit shaky?
2. Sep 16, 2007
### bob1182006
Well there are no incorrect statements or justifications right?
You would just have stated it a bit more...rigorous?
It seems like it deserves an A though.
3. Sep 16, 2007
### Jacobpm64
Well I would have said.
Proof: Let $$x \in A$$. Since $$A \subseteq B$$, $$x \in B$$. Similarly, since $$B \subseteq C$$, $$x \in C$$. Hence, $$A \subseteq C$$.
I guess they are the same though.
4. Sep 17, 2007
### HallsofIvy
Staff Emeritus
Jacobpm64, there is a slight problem with your 'way of saying it'. You start by saying "let $x \in A$". What happens if A is empty? You would need to either include a separate (very simple) proof for the case that A is empty or start with "If $x \in A$" as was done in the given proof. That way, if A is empty, the hypothesis is false and the implication is "vacuously true".
5. Sep 17, 2007
### matt grime
I also disagree - as you say yourself, Halls, when the precedent is false the implication is true. How can there be a problem there?
6. Sep 17, 2007
### StatusX
The only thing I might do with Jacobpm64's proof is changing the word "let" to "assume". "Let" suggests the thing you're doing is known to be possible, while "assume" is more hypothetical. But this is just the way these words are usually used (at least in my experience), and I wouldn't take any credit off either proof.
Last edited: Sep 17, 2007 | 2016-12-03T19:41:37 | {
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https://math.stackexchange.com/questions/1734840/proving-the-hausdorff-property-for-kappa-metric-spaces | # Proving the Hausdorff property for $\kappa$-metric spaces
I want to prove that the Hausdorff property holds for all $\kappa$-metric spaces.
For $\kappa \neq 1$, $(X,d)$ is a $\kappa$-metric space if $X$ is a set and $d$ is a function $X\times X \rightarrow \mathbb R$ such that for every $x,y,z \in X$
$1$. $d(x,y)\ge 0$
$2$. $d(x,y)=d(y,x)$
$3$. $d(x,y)=0 \iff x=y$
$4$. $d(x,z)\le \kappa [d(x,y)+d(y,z)]$.
We put a topology on $X$ by saying $U\subseteq X$ is open iff for each $x\in U$, there exists an $\epsilon>0$ such that $B_d(x,\epsilon)\subseteq U$.
In my attempted proof, take two distinct points $x,y\in X$ such that $d(x,y)=\epsilon$.
Take the open balls $B_d\left(x,{\epsilon\over {3\kappa}}\right)$ and $B_d\left(y,{\epsilon\over {3\kappa}}\right)$. Then it can be shown that $$B_d\left(x,{\epsilon\over {3\kappa}}\right)\cap B_d\left(y,{\epsilon\over {3\kappa}}\right)=\emptyset.$$ For say there is $z\in B_d\left(x,{\epsilon\over {3\kappa}}\right)\cap B_d\left(y,{\epsilon\over {3\kappa}}\right)$. Then $$d(x,y)\le \kappa \left[{\epsilon\over {3\kappa}}+{\epsilon\over {3\kappa}}\right]\\={2\over 3}{\epsilon}\\\lt \epsilon$$ which gives a contradiction.
So, we have found two disjoint open balls in $X$ that do not intersect.
At this point I was thinking that Hausdorff Property has been proved for these spaces, but then I remembered that open balls in $\kappa$-metric spaces area not necessarily open sets. And for a space to be Hausdorff , we need to find, for any two distinct points, two disjoint open sets each containing one of them.
So the above proof of Hausdorff property in $\kappa$ metric space is wrong.
• How are you defining open sets? – User8128 Apr 9 '16 at 16:55
• @User8128 : To be clear, a set is said to be open if it contains an open ball around each of its points. – user118494 Apr 9 '16 at 16:57
• @EricWofsey : Yes . – user118494 Apr 9 '16 at 16:57
• Is there some reference for $\kappa$-metric spaces? (I have noticed that you had a few question about them recently.) The only papers I was able to find use this name for a different objects, see dx.doi.org/10.1090/S0002-9939-1987-0883422-8 or dx.doi.org/10.1090/S0002-9939-1988-0964884-5 – Martin Sleziak Apr 29 '16 at 10:35
• @MartinSleziak : math.stackexchange.com/questions/1733552/… This is the definition of kappa-metric space I have. I was only given the definitions and asked to prove the separation axioms for it. – user118494 Apr 29 '16 at 18:34
You can soup up your idea to inductively build open sets as follows. Fix $x\neq y$; we will define two sequences of sets $U_0\subseteq U_1\subseteq U_2\subseteq\dots$ and $V_0\subseteq V_1\subseteq V_2\subseteq\dots$ by induction. These sets will have the property that for each $n$, $d(U_n,V_n)>0$ (where $d(U_n,V_n)=\inf\{d(p,q):p\in U_n, q\in V_n\}$).
We start with $U_0=\{x\}$ and $V_0=\{y\}$. Given $U_n$ and $V_n$, let $\epsilon=d(U_n,V_n)$, and define $$U_{n+1}=\bigcup_{p\in U_n} B_d(p,\epsilon/3\kappa^2)$$ and $$V_{n+1}=\bigcup_{q\in B_n} B_d(q,\epsilon/3\kappa^2).$$
We must show that $d(U_{n+1},V_{n+1})>0$; let $r\in U_{n+1}$ and $s\in V_{n+1}$. Then there are $p\in U_n$ and $q\in V_n$ such that $d(p,r)<\epsilon/3\kappa^2$ and $d(s,q)<\epsilon/3\kappa^2$. We then have $$d(p,q)\leq \kappa^2(d(p,r)+d(r,s)+d(s,q))<\frac{2\epsilon}{3}+\kappa^2d(r,s).$$
But $d(p,q)\geq d(U_n,V_n)=\epsilon$, so this gives us $d(r,s)>\epsilon/3\kappa^2$. Thus $d(U_{n+1},V_{n+1})\geq \epsilon/3\kappa^2>0$.
Now let $U=\bigcup U_n$ and $V=\bigcup V_n$. Then $U\cap V=\emptyset$, since $U_n\cap V_n=\emptyset$ and the sequences $(U_n)$ and $(V_n)$ are ascending. Also, $x\in U_0\subseteq U$ and $y\in V_0\subseteq V$. Finally, $U$ and $V$ are open, since for any $p\in U$, $p\in U_n$ for some $n$, and then $U_{n+1}$ contains a ball around $p$ (and similarly for $V$). Thus $U$ and $V$ are disjoint open sets containing $x$ and $y$, so $X$ is Hausdorff.
More generally, this argument shows that if $A,B\subseteq X$ and $d(A,B)>0$, then $A$ and $B$ can be separated by open sets (just take $U_0=A$ and $V_0=B$). | 2019-08-24T06:37:02 | {
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https://math.stackexchange.com/questions/3051303/no-of-polygons-in-a-polygon-with-no-side-coinciding | # No. of polygons in a polygon with no side coinciding.
Here is the full question:-
r-sided polygons are formed by joining the vertices of an n-sided polygon. Find the number of polygons that can be formed, none of whose sides coincide with those of the n-sided polygon.
I imagined $$(n-r)$$ vertices in a closed polygon. There are $$(n-r)$$ possibilities for adding r vertices between them. (If we add r vertices here then no 2 vertices will be together). This leads me to $$\binom{n-r}{r}$$. But the correct answer wants me to multiply it with $$\frac{n}{n-r}$$
What is the need for the last step?
• You can just check that it is correct for small numbers. For $n=6,r=3$ there are two choices instead of $1$. For $n=7, r=3$ there are seven instead of four as there is one case where two vertices are three apart and the first of those can be any of the seven vertices. – Ross Millikan Dec 24 '18 at 15:42
• There is a good discussion here as well. It is for $r=7$, but really applies more broadly. – Ross Millikan Dec 25 '18 at 3:07
Look at the case $$n=6,r=3$$. You have a hexagon with vertices numbered $$1$$ to $$6$$, and there are two triangles you can make in this hexagon, with vertices numbered $$1,3,5$$ and $$2,4,6$$. But your formula only counts one of these.
Look at your method. You start with $$n-r=3$$ vertices, which are distinct. Say they are numbered $$1,2,3$$. Then you select $$r=3$$ of these vertices, and insert a vertex next to them. This results in $$1\_2\_3\_$$ Now you have to choose the labels for those inserted vertices. This part you have not accounted for. In the final result, the vertices need to be numbered $$1$$ to $$6$$ in order, so one way to do this is just to start at $$1$$, and rename the vertices $$2$$ through $$6$$ in order, obtaining $$1\underline23\underline45\underline6$$ This gives the triangle $$135$$.
This illustrates the following problem with your method; $$\binom{n-r}r$$ counts the number of ways to choose a polygon where vertex number $$1$$ is included. Therefore we need to multiply by $$n$$, to also include the number of polygons which use vertices $$2,3\dots,n$$. However, this will over-count the polygons by a factor of $$n-r$$, so you must divide by that in the end. | 2020-01-25T18:29:02 | {
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http://mathhelpforum.com/number-theory/183906-prove-using-induction-n-3-divides-4-n-5-a.html | # Math Help - Prove using induction on n that 3 divides (4^n)+5
1. ## [Solved] Prove using induction on n that 3 divides (4^n)+5
Hi everyone,
First time poster. This is probably an embarrassingly easy problem. I'm a non-maths student working through an intro to proof book in my spare time and there are problem sets with no solutions. I've got stuck on a couple of problems. This is one of them.
Question: Prove by induction on $n$ that, for all positive integers $n$, $3$ divides $4^n +5$.
I know where I need to go with this, I think, I'm just stuck on the inductive step:
Proof: We use induction on n. If $3$ divides $4^n +5$, then $4^n +5=3q$ for some integer $q$.
Base case: If $n=1$, then $4^n +5=9=3q$, where $q=3$, proving the base case.
Inductive step: Suppose as inductive hypothesis that $4^k +5=3q$ for some integer $k$. Then $4^{k+1} +5=3q$ (by inductive hypothesis).
That's as far as I get. Obviously, $4^{k+1} +5=4*4^k +5$ by definition, but I get stuck there.
EDIT: not sure how to add the solved prefix, hope this is an okay ad hoc measure.
2. ## Re: Prove using induction on n that 3 divides (4^n)+5
Originally Posted by TimM
Hi everyone,
First time poster. This is probably an embarrassingly easy problem. I'm a non-maths student working through an intro to proof book in my spare time and there are problem sets with no solutions. I've got stuck on a couple of problems. This is one of them.
Question: Prove by induction on $n$ that, for all positive integers $n$, $3$ divides $4^n +5$.
I know where I need to go with this, I think, I'm just stuck on the inductive step:
Proof: We use induction on n. If $3$ divides $4^n +5$, then $4^n +5=3q$ for some integer $q$.
Base case: If $n=1$, then $4^n +5=9=3q$, where $q=3$, proving the base case.
Inductive step: Suppose as inductive hypothesis that $4^k +5=3q$ for some integer $k$. Then $4^{k+1} +5=3\color{red}x\color{black}$ (by inductive hypothesis). not 3q as written.
That's as far as I get. Obviously, $4^{k+1} +5=4*4^k +5$ by definition, but I get stuck there.
$(4)4^k+5=(3)4^k+4^k+5$
Now if 3 divides $4^k+5$
it will certainly divide $(3)4^k+4^k+5$
So if $4^k+5=3q$, then $(3)4^k+4^k+5=3\left(4^k+q\right)=3x$
3. ## Re: Prove using induction on n that 3 divides (4^n)+5
Originally Posted by TimM
Hi everyone,
First time poster. This is probably an embarrassingly easy problem. I'm a non-maths student working through an intro to proof book in my spare time and there are problem sets with no solutions. I've got stuck on a couple of problems. This is one of them.
Question: Prove by induction on $n$ that, for all positive integers $n$, $3$ divides $4^n +5$.
I know where I need to go with this, I think, I'm just stuck on the inductive step:
Proof: We use induction on n. If $3$ divides $4^n +5$, then $4^n +5=3q$ for some integer $q$.
Base case: If $n=1$, then $4^n +5=9=3q$, where $q=3$, proving the base case.
Inductive step: Suppose as inductive hypothesis that $4^k +5=3q$ for some integer $k$. Then $4^{k+1} +5=3q$ (by inductive hypothesis).
That's as far as I get. Obviously, $4^{k+1} +5=4*4^k +5$ by definition, but I get stuck there.
You need to show that $\displaystyle 4^{k + 1} + 5 = 3r$, since $\displaystyle q$ and $\displaystyle r$ would be different.
Anyway...
\displaystyle \begin{align*} 4^{k + 1} + 5 &= 4\cdot 4^k + 5 \\ &= 4\cdot 4^k + 20 - 15 \\ &= 4(4^k + 5) - 15 \\ &= 4\cdot 3q - 15 \\ &= 3(4q - 5) \\ &= 3r\textrm{ where }r = 4q - 5\end{align*}
4. ## Re: Prove using induction on n that 3 divides (4^n)+5
from where you left
4^(k+1) + 5 = 4 . 4^k + 20 - 15
= 4[ 4^k + 5 ] - 15
= 4 (3q) - 15 => 3[ 4q - 5 ]
which gives required result
5. ## Re: Prove using induction on n that 3 divides (4^n)+5
Originally Posted by TimM
Hi everyone,
First time poster. This is probably an embarrassingly easy problem. I'm a non-maths student working through an intro to proof book in my spare time and there are problem sets with no solutions. I've got stuck on a couple of problems. This is one of them.
Question: Prove by induction on $n$ that, for all positive integers $n$, $3$ divides $4^n +5$.
I know where I need to go with this, I think, I'm just stuck on the inductive step:
Proof: We use induction on n. If $3$ divides $4^n +5$, then $4^n +5=3q$ for some integer $q$.
Base case: If $n=1$, then $4^n +5=9=3q$, where $q=3$, proving the base case.
Inductive step: Suppose as inductive hypothesis that $4^k +5=3q$ for some integer $k$. Then $4^{k+1} +5=3q$ (by inductive hypothesis).
That's as far as I get. Obviously, $4^{k+1} +5=4*4^k +5$ by definition, but I get stuck there.
Without induction:
$4^n + 5=(3+1)^n+5=3t+1+5=3t+6$
6. ## Re: Prove using induction on n that 3 divides (4^n)+5
Originally Posted by Prove It
You need to show that $\displaystyle 4^{k + 1} + 5 = 3r$, since $\displaystyle q$ and $\displaystyle r$ would be different.
Lol, obviously. Sorry.
Anyway, thanks everyone!
7. ## Re: Prove using induction on n that 3 divides (4^n)+5
Hi !
without induction : $4\equiv 1[3]$ so $4^{k}\equiv 1[3]$ knowing that $5\equiv 2[3]$ by adding we get the result
8. ## Re: Prove using induction on n that 3 divides (4^n)+5
I don't see the point of giving non-induction answers to questions that clearly state
to answer using Proof By Induction.
Who would answer using a different method if an exam question began
"Prove, by induction........" ?
9. ## Re: Prove using induction on n that 3 divides (4^n)+5
First of all , i didn't give my solution until the others solved it by induction ! You know what , solving problems with different ways is the best way to be good at maths , this is what my brother (who is by the way a winner of a silver medal in the IMO) used to tell me . Take it or leave it .
Anyway , thanks for being nice and sorry if i was wrong by giving another (better ? ) solution for the exercise .
10. ## Re: Prove using induction on n that 3 divides (4^n)+5
$4 \equiv 1[3]$
As "4 is defined as 1 * 3", but that doesn't make sense to me.
11. ## Re: Prove using induction on n that 3 divides (4^n)+5
I don't see the point of giving non-induction answers to questions that clearly state
to answer using Proof By Induction.
I think there is nothing wrong feeling generous and giving another way to solve it after the problem had been solved through the required method. I myself have benefited and learned from posts of people doing so in more than one occasion (which isn't surprising, as one of the prime qualities of MHF is that is isn't 'give as dictated' answer service, so to speak). Apologies for the off-topic, of course.
12. ## Re: Prove using induction on n that 3 divides (4^n)+5
Originally Posted by TimM
$4 \equiv 1[3]$
As "4 is defined as 1 * 3", but that doesn't make sense to me.
Modular arithmetic - Wikipedia, the free encyclopedia
13. ## Re: Prove using induction on n that 3 divides (4^n)+5
Originally Posted by TimM
$4 \equiv 1[3]$
As "4 is defined as 1 * 3", but that doesn't make sense to me.
what that notation means is "4 is congruent to 1 modulo 3".
in other words, the difference between 4 and 1 is a multiple of 3 (they are a multiple of 3 "apart").
the usefulness of this derives from the fact that if a = r (mod n) and b = s (mod n), then
a+b = r+s (mod n), and ab = rs (mod n). by convention, r and s are integers between 0 and n-1 (inclusive).
for example, we have that 4 = 1 (mod 3), and 8 = 2 (mod 3) (8 - 2 = 6, which is a multiple of 3).
if what i said above is true, than we would expect 12 = 4+8 to be congruent to 1+2 = 3 (mod 3) (which it is, since 12 - 3 = 9
is a multiple of 3), and 32 = (4)(8) to be congruent to (1)(2) = 2 (mod 3) (again true, since 32 - 2 = 30 is a multiple of 3).
congruence is transitive (you can "pass it along", like equality), so since 3 is congruent to 0 (mod 3), 12 is likewise congruent to 0 (mod 3).
in other words, the "remainders" upon division by n of integers, obey many of the same rules as integers do.
this is helpful when considering divisibility questions, since it reduces greatly the number of possible cases to check
(with division by 3, we have 3 possible "remainders" 0,1 or 2). technically, we get what is called a quotient ring,
but the abstract general construction is not necessary to understand this specific case.
personally, the proof i would give is this:
given that 4^k + 5 = 3q,
4^(k+1) + 5 = 4(4^k) + 5 = (3+1)(4^k) + 5
= 3(4^k) + 4^k + 5 = 3(4^k) + 3q = 3(4^k + q).
14. ## Re: Prove using induction on n that 3 divides (4^n)+5
Originally Posted by TheCoffeeMachine
I think there is nothing wrong feeling generous and giving another way to solve it after the problem had been solved through the required method. I myself have benefited and learned from posts of people doing so in more than one occasion (which isn't surprising, as one of the prime qualities of MHF is that is isn't 'give as dictated' answer service, so to speak). Apologies for the off-topic, of course.
It is recommended MHF policy to stay on topic.
When a question states to specifically answer in a certain way,
why not concentrate on the requested method?
The objective is to master "Proof By Induction".
You're missing my point entirely.
The given series is a simple arithmetic series,
however I personally would not encourage a student to answer the question as worded
using any other method.
In an exam, it is of no use to offer an alternative.
It is good of course to learn other ways,
but at least if offering an alternative,
one should let the student know not to offer the alternative in an assessment. | 2015-10-04T22:24:51 | {
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https://www.tutorialspoint.com/cplusplus-program-to-find-the-probability-of-a-state-at-a-given-time-in-a-markov-chain | # C++ program to find the probability of a state at a given time in a Markov chain
C++Server Side ProgrammingProgramming
In this article, we will be discussing a program to find the probability of reaching from the initial state to the final state in a given time period in Markov chain.
Markov chain is a random process that consists of various states and the associated probabilities of going from one state to another. It takes unit time to move from one state to another.
Markov chain can be represented by a directed graph. To solve the problem, we can make a matrix out of the given Markov chain. In that matrix, element at position (a,b) will represent the probability of going from state ‘a’ to state ‘b’.
This would leave to a recursive approach to the probability distribution using the formula
P(t) = Matrix * P(t-1)
## Example
Live Demo
#include <bits/stdc++.h>
using namespace std;
#define float_vec vector<float>
//to multiply two given matrix
vector<float_vec > multiply(vector<float_vec > A, vector<float_vec > B, int N) {
vector<float_vec > C(N, float_vec(N, 0));
for (int i = 0; i < N; ++i)
for (int j = 0; j < N; ++j)
for (int k = 0; k < N; ++k)
C[i][j] += A[i][k] * B[k][j];
return C;
}
//to calculate power of matrix
vector<float_vec > matrix_power(vector<float_vec > M, int p, int n) {
vector<float_vec > A(n, float_vec(n, 0));
for (int i = 0; i < n; ++i)
A[i][i] = 1;
while (p) {
if (p % 2)
A = multiply(A, M, n);
M = multiply(M, M, n);
p /= 2;
}
return A;
}
//to calculate probability of reaching from initial to final
float calc_prob(vector<float_vec > M, int N, int F, int S, int T) {
vector<float_vec > matrix_t = matrix_power(M, T, N);
return matrix_t[F - 1][S - 1];
}
int main() {
vector<float_vec > G{
{ 0, 0.08, 0, 0, 0, 0 },
{ 0.33, 0, 0, 0, 0, 0.62 },
{ 0, 0.06, 0, 0, 0, 0 },
{ 0.77, 0, 0.63, 0, 0, 0 },
{ 0, 0, 0, 0.65, 0, 0.38 },
{ 0, 0.85, 0.37, 0.35, 1.0, 0 }
};
//number of available states
int N = 6;
int S = 4, F = 2, T = 100;
cout << "Probability of reaching: " << F << " in time " << T << " after starting from: " << S << " is " << calc_prob(G, N, F, S, T);
return 0;
}
## Output
Probability of reaching: 2 in time 100 after starting from: 4 is 0.271464
Published on 03-Oct-2019 15:42:49 | 2021-06-23T11:24:12 | {
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https://meyavuz.wordpress.com/2018/11/15/generate-uniform-random-points-within-a-circle/ | Algorithms
# Generate Uniform Random Points within a Circle
Let’s say we have only access to uniform random number generator which generates random points between 0 and 1 (let’s denote it rand()), and we are asked to generate uniformly random points within a given circle with specified radius (let’s say $R$ for now). The first idea that comes to mind is to utilize the random number generator to get a random angle between 0 and $2\pi$ to set the angular position of the point and utilize another instance of the random number generator to get a random distance of the point from the origin of the circle. However, once this strategy is followed, it will be observed that the distribution will have more density around the origin and less as the distance from the origin increases as can be seen in the following figure:
This is due to the fact that as the distance from the origin increases then the area covered by the circular segments increases for fixed discretization size along the radius. In a sense, using uniform random distribution to select the distance from the origin gives same probability to each distance and as a result circular segments away from the origin statistically have same/similar number of points within them but since their area is larger, the density of points gets less. As a result, the points seem to be clustered more densely around the origin compared to the outer circular rings.
Recall that originally we were asked to generate uniform distribution within the circle. To be able to achieve this we need to consider the increasing area of circular segments away from the origin and instead of selecting the distance from a uniform distribution, we need to give more weights (higher probability) as the distance increases. This is to compensate for the increased area and keep the density of points within circular segments statistically same. How can we achieve this? I will try to show it here. First of all let us find the relationship of area of circular rings to the distance from the origin. The area of the circle with radius $d$ is $A=\pi d^2$. Then the area of the circular ring between $d$ and $2d$ becomes $\pi (2d)^2 - \pi d^2$ which makes $3A$. Similarly, for the circular ring between $2d$ and $3d$, the area is $5A$ as can be seen in the following graph.
So, to obtain such a triangular probability distribution we can utilize built-in functions, however, recall that we have only access to uniform random number generator. In this case, we can utilize the random number generator to get a random number to mimic the CDF value of a random distance. In other words, for the CDF distribution above, we obtain a random point on the y-axis of CDF which is always between 0 and 1 and then map it to the distance from the origin ($r$) on the x-axis. This can be easily achieved as we know the analytical formula of the CDF function whose inverse can be written as $r = R \sqrt{CDF} = R \sqrt{rand()}$. If we utilize this, then the distribution will look like as follows:
Once compared with the previous distribution, the uniformity of these distribution is evident. Here is another instance for both cases:
Here is the Python code to replicate the results:
import random
import math
import numpy as np
import random
import matplotlib.pyplot as plt
def PlotDistributions(**kwargs):
def PlotDist(xpoints, ypoints, rad, xc, yc, figNo=1, stitle='', color='b', marker='.', subtitleinfo=''):
plt.figure(figNo)
plt.plot(xpoints, ypoints, color+marker)
plt.grid('on')
x_circle = [xc + rad*math.cos(i) for i in np.arange(0, math.pi*2, 0.01)]
y_circle = [yc + rad*math.sin(i) for i in np.arange(0, math.pi*2, 0.01)]
plt.plot(x_circle, y_circle, '-k')
str_title = '\n'.join([stitle, subtitleinfo])
plt.title(str_title)
plt.axis('square')
return
colors = ['b', 'r']
markers = ['.', '.']
for i, key in enumerate(kwargs):
(xpoints, ypoints, rad, xc, yc), subtitleinfo = kwargs[key]
PlotDist(xpoints, ypoints, rad, xc, yc, i, key, colors[i], markers[i], subtitleinfo)
plt.show()
return
def ReplicateNTimes(func, rad, xc, yc, Ntrials=1000):
xpoints, ypoints = [], []
for i in range(Ntrials):
xp, yp = func(rad, xc, yc)
xpoints.append(xp)
ypoints.append(yp)
dist = (xpoints, ypoints, rad, xc, yc)
return dist
def NonUniformRandomPointInCircle(inputRadius=1, xcenter=0, ycenter=0) -> float:
r = inputRadius * random.random()
theta = 2* math.pi * random.random()
return xcenter + r*math.cos(theta), ycenter + r*math.sin(theta)
def UniformRandomPointInCircle(inputRadius=1, xcenter=0, ycenter=0) -> float:
r = inputRadius * math.sqrt(random.random())
theta = 2* math.pi * random.random()
return xcenter + r*math.cos(theta), ycenter + r*math.sin(theta)
def main():
N = 1000
radius, xc, yc = 1, 0, 0
dist1 = ReplicateNTimes(UniformRandomPointInCircle, radius, xc, yc, Ntrials=N)
dist2 = ReplicateNTimes(NonUniformRandomPointInCircle, radius, xc, yc, Ntrials=N)
PlotDistributions(Uniform=(dist1, 'Using sqrt(rand())'), NonUniform=(dist2, 'Using rand()'))
if __name__ == '__main__':
main()<span id="mce_SELREST_start" style="overflow:hidden;line-height:0;"></span>
Additional references:
• Originally saw the question here
• I got the original idea here and wanted to put more details on how the pdf and cdf’s are calculated.
Advertisements | 2018-12-12T00:57:32 | {
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https://math.stackexchange.com/questions/2140226/exact-de-y2xy3dxx3xy2dy-0/2140279 | # Exact DE: $y(2xy+3)dx+x(3xy+2)dy=0$
Find the implicit solution to the folling DE:
$(2xy^2+3y)dx+(3x^2y+2x)dy=0$; with a starting condition $y(1)=\frac 1 2$
I can fight myself to the integrating factor.
$$y(2xy+3)dx+x(3xy+2)dy=0$$
$\mu(x,y)= \frac 1 {xy(1-xy)}$ giving us the following exact DE: $$\frac{2xy+3}{x(1-xy)}dx+\frac{3xy+2}{y(1-xy)}dy=0$$
Testing $$(\mu P)_y=\frac{2x(x(1-xy)+(2xy+3)x^2}{x^2(1-xy)^2}="some-brute-force-magicks"=\frac{1}{(1-xy)^2}$$
$$(\mu Q)_x=\frac{3y(1-xy)y-(3xy+2)y(-y)}{y^2(1-xy)^2}=\frac{1}{(1-xy)^2}$$
Thus $(\mu P)_y=(\mu Q)_x$
From here I try to get $u=?$ $$u=\int(\mu P)dx=\int\frac{2xy+3}{x(1-xy)}dx=3ln(x)-5ln(1-xy)+C(y)$$
So far so good but ...now ...either this so far is wrong in theory, I messed up my integration, or I keep misscalculating my deritative as my calculation $u_y \neq (\mu Q)$
I tried (whats the English word) rearranging (?) $$u=ln \left(\frac{x^3}{(1-xy)^5} \right)+C$$
Help would be nice.
• $u$ is correct, but write $C(y)$ as $N(y)$. Now, take the derivative of $u_y$ and substitute $\dfrac{3xy+2}{y(1-xy)}$ and solve for $N$. – Moo Feb 12 '17 at 2:34
• @Moo The thing is if I compare my $u_y$ to $(\mu Q)$ I get such a disgusting value for $C'(y)$ So im not sure if I should have put C under the log via $u=log(...)+logC(y)$ or not. On top of that I'm not fully sure how to even get to the $y(x)$ from here on to even enter the boundry condition of $y(1)=1/2$ in – Katpton Liamfuppinshire Feb 12 '17 at 3:21
• I added an answer wit more details. – Moo Feb 12 '17 at 3:22
Solve
$$\tag 1 M(x, y)~ dx + N(x, y)~ dy = (2xy^2+3y)~dx+(3x^2y+2x)~dy=0,y(1)=\dfrac 1 2$$
You found an integrating factor that makes $M_y = N_x$ as
$$\tag 2 \mu(x,y)= \dfrac 1 {xy(1-xy)}$$
Multiplying $(1)$ by $(2)$ gives the exact DEQ
$$\tag 3 \dfrac{2xy+3}{x(1-xy)}dx+\dfrac{3xy+2}{y(1-xy)}dy=0$$
Now
$$g(x, y) = \displaystyle \int \left(\dfrac{2xy+3}{x(1-xy)}\right)~dx = 3 \ln x - 5 \ln(1 - x y) + h(y)$$
We can now write
$$\dfrac{\partial g(x, y)}{\partial y} = \dfrac{5 x}{1-x y } + h'(y) = \dfrac{3xy+2}{y(1-xy)} \implies h'(y) = \dfrac{3xy+2}{y(1-xy)} - \dfrac{5 x}{1-x y } = \dfrac{2}{y}$$
Solving
$$h(y) = 2 \ln y$$
We now have
$$g(x, y) = 3 \ln x - 5 \ln(1 - x y) + h(y) = 3 \ln x - 5 \ln(1 - x y) + 2 \ln y = c$$
I will assume you can take it from here to solve for $c$ using the IC and get the final implicit form.
• Hmmm ya i noticed where i did my mistake. And it was always the same when i tried calculating $g_y$ So from here I can simply enter $x=1$ and $y=\frac1 2$ ? To get my C out of this? Might stuggle a bit with the implicit form here. Was never good at this part. EDIT: And thanks, now I can finally try and get some sleep. – Katpton Liamfuppinshire Feb 12 '17 at 3:27
• Yes, just sub in those values of $x$ and $y$, get $c$ and you are done. The solution is already in implicit form - just find $c$ and replace in the implicit form solution. – Moo Feb 12 '17 at 3:29 | 2019-07-20T00:56:50 | {
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https://math.stackexchange.com/questions/451326/how-delta-1-and-delta-2-for-two-different-limits-at-a-can-be-read-as-d | # How $\delta_1$ and $\delta_2$ for two different limits at $a$ can be read as $\delta=\text{min}(\delta_1,\delta_2)$?
I am having trouble understanding a certain part of the proof on why a function cannot approach two different limits near $a$, so I will just list the relevant parts. If this is not enough/ambiguous then please tell me and I will type out the whole proof.
So, suppose we now have:
$$\text{if } 0<|x-a|<\delta_1, \text{ then } |f(x)-l|<\epsilon \hspace{5cm} (1)$$
and
$$\text{if } 0<|x-a|<\delta_2, \text{then} |f(x)-m|<\epsilon \hspace{5cm} (2)$$
and here's a quote from the text:
We have had to use two numbers, $\delta_1$ and $\delta_2$, since there is no guarantee that the $\delta$ which works in one definition will work in the other. But, in fact, it is now easy to conclude that for any $\epsilon>0$ there is some $\delta>0$ such that, for all $x$, $$\text{if } 0<|x-a| < \delta, \text{then } |f(x)-l| < \epsilon \text{ and } |f(x)-m| \lt \epsilon$$ we simply chose $\delta=\text{min}(\delta_1,\delta_2)$
I understand the need to use two distinct $\delta$. What I don't get is why selecting a $\delta$ that is the minimum of $\delta_1$ and $\delta_2$ will make that $\delta$ work in both (1) and (2). I mean, the limits are different so why would I expect that the delta that is the minimum of the two equations will satisfy both equations?
Thank you in advance for any help provided.
• Any $x$ such that $|x-a|\lt \delta$ automatically satisfies the condition $|x-a|\lt \delta_1$, also the condition $|x-a|\lt \delta_2$. If the distance of $x$ from $a$ is less than $\delta$, then it is less than $\delta_1$ and less than $\delta_2$. – André Nicolas Jul 24 '13 at 20:13
Your definition of limit is that for any $\epsilon>0$, there is some $\delta >0$ such that whenever $|x - a| < \delta$, $|f(x) - L| < \epsilon$. (This is exactly what you wrote.)
What you are overlooking is the less-than sign. By definition $\min(\delta_1, \delta_2) \leq \delta_1$ and $\min(\delta_1,\delta_2) \leq \delta_2$. Then by transitivity, whenever $|x - a| < \min(\delta_1,\delta_2)$, you know that $|x-a|$ is less than both $\delta_1$ and $\delta_2$; i.e., $|x-a|$ will work in both (1) and (2).
I hope this is not too wordy. It entirely boils down to $x < min(y, z) \implies x < y$ for any real $x,y,z$.
If $\delta=\min(\delta_1,\delta_2)$, then for $x$ s.t. $0 < |x-a| < \delta$, you have both $0<|x-a|< \delta \leq \delta_1$ and $0<|x-a|< \delta \leq \delta_2$; so $$|f(x)-l|<\epsilon \hspace{5cm} (1)$$ and $$|f(x)-m|<\epsilon \hspace{5cm} (2)$$ | 2020-01-25T02:23:13 | {
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https://mathhelpboards.com/threads/solve-an-equation.9126/ | # Solve an equation
#### anemone
##### MHB POTW Director
Staff member
Solve the equation $y+k^3=\sqrt[3]{k-y}$ where $k$ is a real parameter.
#### jacks
##### Well-known member
Consider function $f(k) = y + k^3$, then $f^{-1}(k) = \sqrt[3]{k - y}$. Hence $f^{-1}(k) = f(k)$ This can happen if and only if $k = f(k) = f^{-1}(k)$ i.e. $k = \sqrt[3]{k - y} = y + k^3$ So $\boxed{y = k - k^3}$
Last edited by a moderator:
#### anemone
##### MHB POTW Director
Staff member
Consider function $f(k) = y + k^3$, then $f^{-1}(k) = \sqrt[3]{k - y}$. Hence $f^{-1}(k) = f(k)$ This can happen if and only if $k = f(k) = f^{-1}(k)$ i.e. $k = \sqrt[3]{k - y} = y + k^3$ So $\boxed{y = k - k^3}$
Hey jacks, thanks for participating and your solution is simple, elegant and nice! Well done, jacks!
#### lfdahl
##### Well-known member
Hi, anemone and jacks!
I have one question. I do not understand why the following implication is true:
$$f(k)=y+k^3 \Rightarrow f^{-1}(k)=\sqrt[3]{\mathbf{k}-y}$$
Why is k appearing on the RHS?
I would deduce the following:
$$f(k)=y+k^3 \Rightarrow f^{-1}(k)=\sqrt[3]{f(k)-y}$$
By definition:
$$f(k)=y+k^3=\sqrt[3]{k-y}=\sqrt[3]{f(k)-y} \Rightarrow f(k) = k \Rightarrow y+k^3=k \Rightarrow y = k-k^3$$
#### anemone
##### MHB POTW Director
Staff member
Hi, anemone and jacks!
I have one question. I do not understand why the following implication is true:
$$f(k)=y+k^3 \Rightarrow f^{-1}(k)=\sqrt[3]{\mathbf{k}-y}$$
Why is k appearing on the RHS?
I would deduce the following:
$$f(k)=y+k^3 \Rightarrow f^{-1}(k)=\sqrt[3]{f(k)-y}$$
By definition:
$$f(k)=y+k^3=\sqrt[3]{k-y}=\sqrt[3]{f(k)-y} \Rightarrow f(k) = k \Rightarrow y+k^3=k \Rightarrow y = k-k^3$$
Hi lfdahl,
I am sorry for I only replied to you days after...I thought to myself to let jacks to handle it and I would only chime in if we didn't hear from jacks 24 hours later. But it somehow just slipped my mind.
Back to what you asked us...I believe if we use the identity
$f(f^{-1}(k))=k$,
and that for we have $f(k)=y+k^3$, we would end up with getting $f^{-1}(k)=\sqrt[3]{\mathbf{k}-y}$, does that answer your question, lfdahl?
$f(k)=y+k^3$
$f(f^{-1}(k))=k$
$y+(f^{-1}(k))^3=k$
$(f^{-1}(k))^3=k-y$
$f^{-1}(k)=\sqrt[3]{k-y}$ | 2021-07-28T05:00:25 | {
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http://mathhelpforum.com/statistics/200503-game-show.html | 1. ## Game show
A contestant on the game show called "A Diamond for your Wife" gets to select 5 diamonds from a box. The box contains 20 diamonds of which 8 are fake and 12 are real. Regardless of how many real diamonds the contestant has after his selection he can only take one home for his wife. A second contestant then gets to select from the remaining 15 diamonds from the box. But only gets to select one diamond.
What is the probability that this second contestant selects a real diamond?
I don't understand why the following reasoning is wrong:
Let F be the event where the diamond picked was fake and let R be the event where the diamond picked was real.
Thus, the first contestant could get 0F, 1F, 2F, 3F, 4F, 5F.
If the first contestant gets 0F, then that means he took 5R, thus the second contestant has a probability of 7/15 of getting 1R as there are now 7 R left out of the 15 diamonds.
Likewise if the first contestant gets 1F, then that means he took 4R, thus the second contestant has a probability of 8/15 of getting 1R.
This process continues until the first contestant picks 5F in which case the probability of the second contestant has a probability of 12/15 of getting 1R.
So the total probability is 7/15 + 8/15 + ... + 12/15 but clearly this is greater than 1, where did I go wrong?
2. ## Re: Game show
Hello, usagi_killer!
You omitted the probabilities for the first contestant.
A contestant on the game show called "A Diamond for your Wife" gets to select 5 diamonds from a box.
The box contains 20 diamonds of which 8 are fake and 12 are real.
Regardless of how many real diamonds the contestant has after his selection, he can only take one home for his wife.
A second contestant then gets to select one diamond from the remaining 15 diamonds from the box.
What is the probability that this second contestant selects a real diamond?
There are: . $\begin{Bmatrix}\text{8 Fake} \\ \text{12 Real} \end{Bmatrix}$
Contestant-1 selects 5 diamonds.
There are: . ${20\choose5} \,=\,15,\!504$ possible outcomes.
There are six possible scenarios:
[1] Contestant-1 chooses 0 Fake and 5 Reals.
. . $\text{0F, 5R:}\;{12\choose5} \:=\:792\text{ ways} \quad\Rightarrow\quad P(\text{0F. 5R}) \:=\:\frac{792}{15,504} \:=\:\frac{99}{1938}$
. . $\text{Then contestant-2 chooses from }\begin{Bmatrix}\text{8 F}\\\text{7 R}\end{Bmatrix} \quad\Rightarrow\quad P(R) \:=\:\frac{7}{15}$
. . $\text{Hence: }\:P(\text{Scenario-1}\,\wedge\,R) \:=\:\frac{99}{1938}\cdot\frac{7}{15} \;=\;\frac{693}{29,\!070}$
[2] Contestant-1 choose 1 Fake and 4 Reals.
. . $\text{1F, 4R: }\;{8\choose1}{12\choose4} :=\:2640 \quad\Rightarrow\quad {(\text 1F, 4R}) \:=\:\frac{2640}{15,504} \:=\:\frac{330}{1938}$
. . $\text{Then contestant-2 chooses from }\begin{Bmatrix}\text{7 F} \\\text{8 R}\end{Bmatrix} \quad\Rightarrow\quad P(R) \:=\:\frac{8}{15}$
. . $\text{Hence: }\:P(\text{Scenario-2}\,\wedge R) \:=\:\frac{330}{1938}\cdot\frac{8}{15} \:=\:\frac{2640}{29,\!070}$
. . . . . . and so on.
Got it?
3. ## Re: Game show
Originally Posted by usagi_killer
A contestant on the game show called "A Diamond for your Wife" gets to select 5 diamonds from a box. The box contains 20 diamonds of which 8 are fake and 12 are real. Regardless of how many real diamonds the contestant has after his selection he can only take one home for his wife. A second contestant then gets to select from the remaining 15 diamonds from the box. But only gets to select one diamond.
What is the probability that this second contestant selects a real diamond?
Let $F_k$ denote the event that the first contestant draws $k$ fake diamonds.
Let $D$ denote the event that the second contestant draws $k$ a real diamonds.
$P(D) = \sum\limits_{k = 0}^5 {P\left( {D \cap {F_k}} \right)} = \sum\limits_{k = 0}^5 {P\left( {D|{F_k}} \right)} P\left( {{F_k}} \right)$
As Soroban has shown you $P\left( {{F_k}} \right)=\dfrac{\binom{8}{k}\binom{12}{5-k}}{\binom{20}{5}}$ and ${P\left( {D|{F_k}} \right)}=\frac{7+k}{15}$. | 2017-02-28T03:58:55 | {
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https://math.stackexchange.com/questions/2442196/turn-exists-z-forall-ypy-land-rz-y-to-exists-y-rx-y-into-cnf | # Turn $\exists z \forall y[p(y) \land r(z,y)] \to \exists y r(x,y)$ into CNF
Turn $\exists z \forall y[p(y) \land r(z,y)] \to \exists y r(x,y)$ into CNF.
What i tried: $\exists z \forall y[p(y) \land r(z,y)] \to \exists y r(x,y) \equiv \lnot(\exists z \forall y[p(y) \land r(z,y)]) \lor \exists y r(x,y) \equiv \forall z \lnot (\forall y [p(y) \land r(z,y)])\lor\exists yr(x,y) \equiv \forall z \exists y[\lnot p(y)\lor \lnot r(z,y)]\lor \exists yr(x,y) \equiv \forall z \exists y \exists l[\lnot p(y) \lor\lnot r(z,y)\lor r(x,l)]$
After skolemization: $\forall z [\lnot p(sk_1(z)) \lor\lnot r(z,sk_1(z))\lor r(x,sk_2(z))]$
And if i put l infront it would be $\forall z [\lnot p(sk_1(z)) \lor\lnot r(z,sk_1(z))\lor r(x,c_1)]$
Is it obligatory to skolemize for cnf ? Are my steps correct till now ?
Im sorry I used equiv for for logical equivalence i didnt find the symbol i should use.
Your steps are all correct!
The body of your formula is in CNF, but probably they want you to indeed skolemize and drop the universal quantifier.
Note that if you first pull out the second existential, you can avoid having to use one of the functions when skolemizing:
$\forall z \exists y[\lnot p(y)\lor \lnot r(z,y)]\lor \exists y \ r(x,y) \Leftrightarrow$
$\forall z \exists y[\lnot p(y)\lor \lnot r(z,y)]\lor \exists l \ r(x,l) \Leftrightarrow$
$\exists l (\forall z \exists y[\lnot p(y)\lor \lnot r(z,y)]\lor r(x,l))\Leftrightarrow$
$\exists l \forall z \exists y[\lnot p(y)\lor \lnot r(z,y)\lor r(x,l)]$
Skolemizing this, you get:
$\lnot p(f(z))\lor \lnot r(z,f(z))\lor r(x,c)$
• thanks indeed a typo, can i try skolemize it and you tell if its correct ? btw i thought skolemization drops existential quantifiers ??? Sep 23 '17 at 20:10
• @asddf Yes, you drop the existentials too, but you have to fill in constants or some function, while universals just get dropped without any changes ... sorry to confuse you! Sep 23 '17 at 20:16
• @asddf Just saw your skolemization: since the existentials come after the universal, you need to use functions. I'll add to my post ... Sep 23 '17 at 20:16
• i edited it can you take a look ? Sep 23 '17 at 20:18
• @asddf You're welcome! P.s. use \Leftrightarrow for the logical equivalence symbol Sep 23 '17 at 20:22 | 2021-09-17T04:03:13 | {
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http://manufakturaklimatu.pl/05rfy7/b17405-exterior-point-in-metric-space | Interior, Closure, Exterior and Boundary Let (X;d) be a metric space and A ˆX. A closed subset of a complete metric space is a complete sub-space. FACTS A point is interior if and only if it has an open ball that is a subset of the set x 2intA , 9">0;B "(x) ˆA A point is in the closure if and only if any open ball around it intersects the set x 2A , 8">0;B "(x) \A 6= ? The closure of A, denoted by A¯, is the union of Aand the set of limit points of A, A¯ = … Let be a metric space. Metric Spaces, Open Balls, and Limit Points DEFINITION: A set , whose elements we shall call points, is said to be a metric space if with any two points and of there is associated a real number ( , ) called the distance from to . Example: Any bounded subset of 1. The family Cof subsets of (X,d)defined in Definition 9.10 above satisfies the following four properties, and hence (X,C)is a topological space. Definition Let E be a subset of a metric space X. A metric space X is sequentially compact if every sequence of points in X has a convergent subsequence converging to a point in X. 10.3 Examples. 1. Proposition A set C in a metric space is closed if and only if it contains all its limit points. Theorem 4. Proposition A set O in a metric space is open if and only if each of its points are interior points. This distance function :×→ℝ must satisfy the following properties: Let S be a closed subspace of a complete metric space X. Definition. Definition 3. Set Q of all rationals: No interior points. If a subset of a metric space is not closed, this subset can not be sequentially compact: just consider a sequence converging to a point outside of the subset! Set N of all natural numbers: No interior point. In most cases, the proofs Theorem 9.6 (Metric space is a topological space) Let (X,d)be a metric space. A set Uˆ Xis called open if it contains a neighborhood of each of its Defn Suppose (X,d) is a metric space and A is a subset of X. Proof. 1.5 Limit Points and Closure As usual, let (X,d) be a metric space. A metric space X is compact if every open cover of X has a finite subcover. A point p is a limit point of the set E if every neighbourhood of p contains a point q ≠ p such that q ∈ E. Theorem Let E be a subset of a metric space … A subset is called -net if A metric space is called totally bounded if finite -net. (0,1] is not sequentially compact … Suppose that A⊆ X. A point x is called an isolated point of A if x belongs to A but is not a limit point of A. We do not develop their theory in detail, and we leave the verifications and proofs as an exercise. 2. The purpose of this chapter is to introduce metric spaces and give some definitions and examples. 1. Theorem In a any metric space arbitrary intersections and finite unions of closed sets are closed. Proof Exercise. 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http://mathoverflow.net/questions/35514/pair-of-curves-joining-opposite-corners-of-a-square-must-intersect-proof/35544 | # Pair of curves joining opposite corners of a square must intersect---proof?
Reposting something I posted a while back to Google Groups.
In his 'Ordinary Differential Equations' (sec. 1.2) V.I. Arnold says "... every pair of curves in the square joining different pairs of opposite corners must intersect".
This is obvious geometrically but I was wondering how one could go about proving this rigorously. I have thought of a proof using Brouwer's Fixed Point Theorem which I describe below. I would greatly appreciate the group's comments on whether this proof is right and if a simpler proof is possible.
We take a square with side of length 1. Let the two curves be $(x_1(t),y_1(t))$ and $(x_2(t),y_2(t))$ where the $x_i$ and $y_i$ are continuous functions from $[0,1]$ to $[0,1]$. The condition that the curves join different pairs of opposite corners implies, $$(x_1(0),y_1(0)) = (0,0)$$ $$(x_2(0),y_2(0)) = (1,0)$$ $$(x_1(1),y_1(1)) = (1,1)$$ $$(x_2(1),y_2(1)) = (0,1)$$
The two curves will intersect if there are numbers $a$ and $b$ in $[0,1]$ such that
$$p(a,b) = x_2(b) - x_1(a) = 0$$ $$q(a,b) = y_1(a) - y_2(b) = 0$$
We define the two functions
$$f(a,b) = a + p(a,b)/2 + |p(a,b)| (1/2 - a)$$ $$g(a,b) = b + q(a,b)/2 + |q(a,b)| (1/2 - b)$$
Then $(f,g)$ is a continuous function from $[0,1]\times [0,1]$ into itself and hence must have a fixed point by Brouwer's Fixed Point Theorem. But at a fixed point of $(f,g)$ it must be the case that $p(a,b)=0$ and $q(a,b)=0$ so the two curves intersect.
Figuring out what $f$ and $g$ to use and checking the conditions in the last para is a tedious. Can there be a simpler proof?
-
My guess is you need to appeal to some form of the Jordan curve theorem. – Qiaochu Yuan Aug 13 '10 at 18:05
I second the Jordan Curve Theorem suggestion. – Thierry Zell Aug 13 '10 at 18:15
I am looking for any rigorous proof, the more elementary the better. – Jyotirmoy Bhattacharya Aug 13 '10 at 18:23
See John Stillwell's answer below, which explains how you can reduce to the case where the curves are piecewise linear, which certainly makes things simpler! – Emerton Aug 14 '10 at 3:52
I like the proofs based on $K_5$ and the polygonal Jordan curve theorem, but if all of them are unwound, the Brouwer fixed point theorem proof is the most direct and transparent. – Victor Protsak Aug 14 '10 at 6:27
Since the full Jordan curve theorem is quite subtle, it might be worth pointing out that theorem in question reduces to the Jordan curve theorem for polygons, which is easier.
Suppose on the contrary that the curves $A,B$ joining opposite corners do not meet. Since $A,B$ are closed sets, their minimum distance apart is some $\varepsilon>0$. By compactness, each of $A,B$ can be partitioned into finitely many arcs, each of which lies in a disk of diameter $<\varepsilon/3$. Then, by a homotopy inside each disk we can replace $A,B$ by polygonal paths $A',B'$ that join the opposite corners of the square and are still disjoint.
Also, we can replace $A',B'$ by simple polygonal paths $A'',B''$ by omitting loops. Now we can close $A''$ to a polygon, and $B''$ goes from its "inside" to "outside" without meeting it, contrary to the Jordan curve theorem for polygons.
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This is a great argument! – Victor Protsak Aug 14 '10 at 6:14
This is gorgeous! It's possibly worth pointing out quite how elementary the Jordan curve theorem for polygons is, to show how little is being black-boxed here. Fix some point $x_0$ not on $C$, and for any (other) point $x$ not on $C$, look at the line segment from $x$ to $x_0$; count the parity of how many times it crosses $C$ (counting double/none if it hits vertices of $C$). This is well-defined and locally constant on $\mathbb{R}\setminus C$ (this is where we use that $C$ is a polygon); so as a locally constant, surjective function to $\{0,1\}$, it disconnects $\mathbb{R}^2$. – Peter LeFanu Lumsdaine Aug 14 '10 at 6:47
Dear Peter: Your line segment may cross $C$ infinitely many times. [You probably mean $\mathbb{R}^2\setminus C$, not $\mathbb{R}\setminus C$.] – Pierre-Yves Gaillard Aug 14 '10 at 11:18
Peter: That proof goes through even when the polygon isn't simple, e.g. a polygonal figure-eight, where the parity function disconnects the plane into more than two components. So, there's a bit more work to do. – Per Vognsen Aug 14 '10 at 14:49
@Pierre: thanks, yes, I did mean $\mathbb{R}^2$; and while the segment can't cross $C$ infinitely often (a polygon has finitely many edges by definition, at least in the conventions I know) it could contain an edge of $C$ as a subsegment, in which case we do have to look at what directions the adjacent edges of $C$ go off in. @Per: you're right, of course; this doesn't establish the full Jordan curve theorem; I was just thinking of what was necessary for the application at hand (which just needs disconnectedness plus the fact that the second curve's endpoints are in opposite components). – Peter LeFanu Lumsdaine Aug 14 '10 at 17:34
ORIGINAL: This follows from the fact that the complete graph $K_5$ on five vertices cannot be imbedded in $\mathbb S^2,$ in itself an application of Jordan Curve. If your two square curvy diagonals stay inside the square without intersecting, a fifth point outside the square can be joined to the four vertices by disjoint arcs, thus creating a complete graph on five vertices. Very nice book by James Munkres, "Topology: a first course" where, on page 386 exercise 5, he does the graph on five vertices. Note that the concept of inside for the square uses elementary ideas such as convexity.
EDIT: As mentioned by Henry Wilton in comment below, there may be other routes here. In particular, I have a book by Robin J. Wilson just called Introduction to Graph Theory, second edition, and in section 13, pages 64-67, in which he develops Euler's formula for planar graphs and as a quick corollary shows that $K_5$ and $K_{3,3}$ are nonplanar, these being Theorem 13A and Corollary 13E. It is anybody's guess whether JCT is used implicitly in defining "faces" properly for Euler's formula. I don't know.
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That's a neat idea! – Somnath Basu Aug 13 '10 at 18:43
Do you really need the Jordan Curve Theorem to see that $K_5$ is non-planar? I ask because Thomason gives a proof of the JCT using the fact that $K_{3,3}$ is non-planar. – HJRW Aug 14 '10 at 1:41
very nice proof! – Kerry Aug 14 '10 at 2:24
Henry, I don't think I know how to sort out first principles here. The Munkres book, same page, in exercise 4, has the reader use JCT to show that $K{3,3}$ is non-planar. I'm going to guess that the three facts are roughly equivalent in the sense of quick proofs in both directions for any pair. So the questions become, does the nonplanarity of the complete graph on five vertices imply JCT quickly, and is there some trickery where each nonplanar graph gives the other, all "quickly." – Will Jagy Aug 14 '10 at 2:45
Just to clarify slightly, Thomason quickly observes that if a graph is planar then it has a polygonal embedding in the plane; so you can talk about faces before you've proved the JCT. – HJRW Aug 14 '10 at 5:07
This should probably go in a comment, but I don't have enough reputation points.
Note that there is a pair of connected sets in the square containing opposite pairs of corners that don't intersect. There are pictures in the reference below.
Robert J. MacG. Dawson, Paradoxical Connections. The American Mathematical Monthly Vol. 96, No. 1 (Jan., 1989), pp. 31-33. http://www.jstor.org/stable/2323252
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+1 for retaining something from one of our conversations . . . – Danny Calegari Oct 1 '11 at 12:15
There is a simple proof that a game of Hex must have a winner, which implies the result you want.See here: Brouwer's Fixed Point Theorem and the Jordan Curve Theorem, Lemma 5.5. The Brouwer fixed point theorem and the Jordan Curve theorem follow from this.
This proof is based on the paper The Game of Hex and the Brouwer Fixed-Point Theorem (by David Gale. The American Mathematical Monthly, Vol. 86, No. 10. (Dec., 1979), pp. 818-827).
Edit: Actually the reference shows that the Game of Hex always has a winner => Brouwer Fixed Point theorem => a pair of curves in the square joining opposite corners must intersect. So it does use Brouwer's fixed point theorem, but gives an elementary proof of it.
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As much as I like the connection with Hex, Gale comments on, but doesn't give the proof of, the additional statement "not both" in the Hex theorem, which is the Hex analogue of the intersection property. Thus for the purposes of this question, Hex is a distraction. – Victor Protsak Aug 14 '10 at 6:09
Yes, at first I thought that they proved it directly from the game of Hex. The result asked for would follow from the result that you can't have a state in which both players have a winning line, but the references only show directly that at least one person must win. However, the first link does have a short proof of the result asked for, albeit using the fixed point theorem (and a proof of that). So, it's not as direct and elementary as I thought at first. Still, it answers the question. – George Lowther Aug 14 '10 at 12:19
You could use Brouwer degree for a more intuitive proof:
The degree of the usual diagonals intersecting each other is 1. One at a time, deform the diagonals via straight-line homotopies to the desired curves. This should preserve Brouwer degree. Lastly, Brouwer degree is well-defined even for continuous functions (using a smooth approximation), and non-zero Brouwer degree implies an intersection.
Alternately, you could artificially avoid the phrase "Brouwer degree" and directly track what happens to the intersection point under the homotopy.
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+1 for the name, a combination of "Anton Gorodetsky" and "Sergei Lukyanenko" -- any relation? – Pete L. Clark Aug 13 '10 at 21:27
:) No, not as far as I know. – Anton Lukyanenko Aug 13 '10 at 22:10
I'm not sure that the last paragraph avoiding the degrees works: the curves may intersect in more than one point, so it's challenging to "directly track what happens to the intersection point". – Victor Protsak Aug 14 '10 at 6:19
A homological proof would use the intersection form of the torus. if you consider these paths as based loops on the torus, you see that they are represented as (1,1), and (1,-1), in terms of the standard homology generators. knowing that the intersection form is (0 1; -1 0), we find that the intersection index
Q(v,w) = (1,1)(0 1; -1 0)(1,-1)^t = 2
they already intersect once at the origin, so they must intersect somewhere else in the square. However, you must already have had to compute the intersection form.
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How about the following, using the Nested Intervals Theorem (which was in my 2nd year Calculus text) which says the intersection of a nested sequence of closed intervals in $\mathbb{R}$ is non-empty. Here goes the proof:
We construct recursively a nested sequence $I_j := [a_j, b_j]$ of closed intervals for $j \geq 0$. Let $I_0 := [0,1]$. For every $j \geq 0$, construct $I_{j+1}$ as follows: let $m_j$ be the midpoint of $I_j$. If the curves intersect at $t = m_j$, then we are done, so stop the sequence. Otherwise set $I_{j+1}$ to be $[a_j, m_j]$ or $[m_j, b_j]$ depending on whether the curves "switch from left to right" on the first sub-interval or the 2nd (let's say you always make sure that $c_1$ is to the "left" of $c_2$ at $t = a_j$ and to the "right" of $c_2$ at $t = b_j$).
If the sequence is finite, then the curves must intersect, as noted above. So assume the sequence is infinite. The Nested Intervals Theorem and the fact that the length decreases by a factor of 2 at every step implies that $\cap_{j=0}^\infty I_j = \lbrace t\rbrace$ for some $t \in [0,1]$. Then we must have $c_1(t) = c_2(t)$.
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I seems that your proof is valid for funtions graphs and no generalized curves: when write "switch from left to right" I guess that you to consider the intersetion of the two curves by the line parallel to a cartesian axis $t$: $t=m_j$, and you suppose to have only two intesection (one for curve) . But this isn't true for general curves. ANyway your proof is adaptable: curves are compact and intesection by a interval is a closed, you have anyway a first point at lest and a first point at right..the rest is the some. – Buschi Sergio Jul 26 '13 at 8:46
This is a really interesting question. But it only involves basic homotopy theory, not anything as subtle as the Jordan curve theorem.
Proof:
Let the paths be parameterized as $v(t)$, and $w(s)$, $t,s \in I := [0,1]$.
Assume the paths never intersect. Then the map $f : I \times I \to S^1$, given by $f(s,t) = \dfrac{v(t)-w(s)}{|v(t)-w(s)|}$, is well defined.
Think of $I \times I$ as being a homotopy between the paths
$a_1(t) = \begin{cases} (0, 2t) & 0 \leq t \leq \frac{1}{2}\\ (2t-1, 1) & \frac{1}{2} < t \leq 1 \end{cases}$
and
$a_2(t) = \begin{cases} (2t, 0) & 0 \leq t \leq \frac{1}{2}\\ (1, 2t-1) & \frac{1}{2} < t \leq 1 \end{cases}$
i.e. the two boundary components of $I \times I$, as paths from $(0,0) \to (1,1)$.
Then we see that $f(a_1(t))$ is a path that starts at the north pole of the circle, and ends at the south pole, and traverses clockwise, whereas $f(a_2(t))$ starts and ends the same, but traverses counterclockwise. Now $f(I \times I)$ provides a homotopy between these paths. However, they are not homotopic as paths in the circle. This gives a contradiction, and hence the paths must intersect.
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Isn't a path in $S^1$ contractible to a point? – Piero D'Ancona Sep 2 '10 at 12:24
By 'paths', i mean paths starting at the top of the circle, and ending at the bottom. – Ryan Mickler Sep 2 '10 at 13:29
Perhaps i should have been more clear. f(IxI) is a homotopy 'of paths that start at the top of the circle and end at the bottom'. – Ryan Mickler Sep 2 '10 at 13:34
I guess a more elegant way to see it, is to consider again, the map f: I x I -> S^1, when restricted to the boundary, d(IxI) ~= S^1, we find (from the argument above) that f_d(IxI) : S^1 -> S^1 winds once, but f_IxI gives a homotopy from this map to the constant map, hence a contradiction. – Ryan Mickler Sep 2 '10 at 14:01
After thinking about it some more. This question is really a generalization of the intermediate value theory. The IVT is really a homotopy theory question, where you are detecting pi_0(R-{0}), in this case, we are detecting pi_1(R^2-{0}). – Ryan Mickler Sep 3 '10 at 4:50
See
R. Maehara, The Jordan Curve Theorem via the Brouwer Fixed Point Theorem, Amer. Math. Monthly 91, 641--643 (1984)
which is availiable on Andrew Ranicki's website.
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This is the main step of the proof that the plane (in this case, the square) has topological dimension 2. You can find a proof (as elementary as I could make it) in my text Measure, Topology, and Fractal Geometry. In particular, no previous knowledge of algebraic topology is required.
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Did you mean to say that no previous knowledge of algebraic topology is required? – Victor Protsak Sep 2 '10 at 15:21
thanks, corrected. – Gerald Edgar Sep 3 '10 at 12:19 | 2015-10-04T17:35:17 | {
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https://acm.njupt.edu.cn/problem/CF977D | Preparing NOJ
# Divide by three, multiply by two
1000ms 262144K
## Description:
Polycarp likes to play with numbers. He takes some integer number $x$, writes it down on the board, and then performs with it $n - 1$ operations of the two kinds:
• divide the number $x$ by $3$ ($x$ must be divisible by $3$);
• multiply the number $x$ by $2$.
After each operation, Polycarp writes down the result on the board and replaces $x$ by the result. So there will be $n$ numbers on the board after all.
You are given a sequence of length $n$ — the numbers that Polycarp wrote down. This sequence is given in arbitrary order, i.e. the order of the sequence can mismatch the order of the numbers written on the board.
Your problem is to rearrange (reorder) elements of this sequence in such a way that it can match possible Polycarp's game in the order of the numbers written on the board. I.e. each next number will be exactly two times of the previous number or exactly one third of previous number.
It is guaranteed that the answer exists.
## Input:
The first line of the input contatins an integer number $n$ ($2 \le n \le 100$) — the number of the elements in the sequence. The second line of the input contains $n$ integer numbers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 3 \cdot 10^{18}$) — rearranged (reordered) sequence that Polycarp can wrote down on the board.
## Output:
Print $n$ integer numbers — rearranged (reordered) input sequence that can be the sequence that Polycarp could write down on the board.
It is guaranteed that the answer exists.
## Sample Input:
64 8 6 3 12 9
## Sample Output:
9 3 6 12 4 8
## Sample Input:
442 28 84 126
## Sample Output:
126 42 84 28
## Sample Input:
21000000000000000000 3000000000000000000
## Sample Output:
3000000000000000000 1000000000000000000
## Note:
In the first example the given sequence can be rearranged in the following way: $[9, 3, 6, 12, 4, 8]$. It can match possible Polycarp's game which started with $x = 9$.
Info
Provider CodeForces
Code CF977D
Tags
dfs and similarmathsortings
Submitted 220
Passed 94
AC Rate 42.73%
Date 03/04/2019 16:23:56
Related
Nothing Yet | 2020-08-09T23:32:22 | {
"domain": "edu.cn",
"url": "https://acm.njupt.edu.cn/problem/CF977D",
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"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9830850867332735,
"lm_q2_score": 0.853912747375134,
"lm_q1q2_score": 0.8394688873159315
} |
https://merarera.com/c3p8snz/which-estimator-is-more-efficient-8f6eed | 8, Abrama Cross Road, Abrama, Valsad - 396001. Is there an anomaly during SN8's ascent which later leads to the crash? Among a number of estimators of the same class, the estimator having the least variance is called an efficient estimator. View full-text. The asymptotic relative efficiency of median vs mean as an estimator of $\mu$ at the normal is the ratio of variance of the mean to the (asymptotic) variance of the median when the sample is drawn from a normal population. I Which estimator is more efficient? On the other hand, interval estimation uses sample data to calcu… _ X XOne choice of an estimator for u is X = $. In statistics, an estimator is a rule for calculating an estimate of a given quantity based on observed data: thus the rule (the estimator), the quantity of interest (the estimand) and its result (the estimate) are distinguished.. Yes, at least in the usual situations we'd be doing this in and assuming a frequentist framework. (2) Unbiased. Every time that you supply energy or heat to a machine (for example to a car engine), a certain part of this energy is wasted, and only some is converted to actual work output. For example, the sample mean is an unbiased estimator for the population mean. then what does it mean by saying "for SOME value of$\theta$" in the above statement in Wikipedia? For an unbiased estimator, efficiency is the precision of the estimator (reciprocal of the variance) divided by the upper bound of the precision If there is only ONE, why does it say "for SOME" value of$\theta$? Use MathJax to format equations. If the following holds, then is a consistent estimator of . Also I have another question about relative efficiency: Can I run 300 ft of cat6 cable, with male connectors on each end, under house to other side? 2) Also I thought there is a SINGLE "true" value of the parameter$\theta$, is it correct? We say that β’ j1 is more efficient relative to β’ j2 if the variance of the sample distribution of β’ j1 is less than that of β’ j2 for all finite sample sizes. Tyler D. Groff, N. Jeremy Kasdin. The ratio of the variances of two estimators denoted by $$e\left( {\widehat {{\alpha _1}},\widehat {{\alpha _2}}} \right)$$ is known as the efficiency of $$\widehat {{\alpha _1}}$$ and $$\widehat {{\alpha _2}}$$ is defined as follows: $e\left( {\widehat {{\alpha _1}},\widehat {{\alpha _1}}} \right) = \frac{{Var\left( {\widehat {{\alpha _2}}} \right)}}{{Var\left( {\widehat {{\alpha _1}}} \right)}}$. A little cryptic clue for you! Therefore, the efficiency of the median against the mean is only 0.63. Efficient estimators are always minimum variance unbiased estimators. Consistent . Thanks for contributing an answer to Cross Validated! and T2, what does it mean by saying T1 is more efficient than T2, https://en.wikipedia.org/wiki/Efficiency_(statistics). The larger the sample size, the more accurate the estimate. GMM has several nice properties, including that it is the most efficient estimator in the class of all asymptotically normal estimators. Which estimator is more efficient 3 Find another unbiased estimator of the from AGEC 5230 at University of Wyoming A consistent estimator is one which approaches the real value of the parameter in the population as the size of … If you don't know what$\theta$is (if you did, you wouldn't have to bother with estimators), it would be good if it worked well for whatever value you have. Can there be waves in different fields? Was Stan Lee in the second diner scene in the movie Superman 2? $$\frac{{{\sigma ^2}}}{n}$$ and $$\frac{\pi }{2}\,\,\,\,\frac{{{\sigma ^2}}}{n}$$, e (median, mean) $$= \frac{{Var\left( {\overline X } \right)}}{{Var\left( {med} \right)}}$$ (Contains 1 table and 3 figures.) Or to be even more precise, I should really have$\tilde{X}$to denote the estimator (clarifying it is a random variable) rather than$\tilde{x}$(a value obtained on a specific sample). Therefore, the efficiency of the mean against the median is 1.57, or in other words the mean is about 57% more efficient than the median. Thus an efficient estimator need not exist, but if it does, it is the MVUE. When this is the case, we write , The following theorem gives insight to consistency. In large samples$\frac{n}{\sigma^2}\text{ Var}(\tilde{x})$approaches the asymptotic value reasonably quickly, so people tend to focus on the asymptotic relative efficiency. I don't know how to simplify resistors which have 2 grounds. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … MSE. N.H. No. They're both unbiased so we need the variance of each. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. In general, the spread of an estimator around the parameter θ is a measure of estimator efficie… Historically, finite-sample efficiency was an early optimality criterion. then what does it mean by saying "for SOME value of θ" in the above statement [...] if there is only ONE, why it says "for SOME" value of θ. Could someone give an easy but very concrete example. We say that the estimator is a finite-sample efficient estimator (in the class of unbiased estimators) if it reaches the lower bound in the Cramér–Rao inequality above, for all θ ∈ Θ. Decide which estimator is more efficient. In Brexit, what does "not compromise sovereignty" mean? When defined asymptotically an estimator is fully efficient if its variance achieves the Rao-Cramér lower bound. I am just wondering, when comparing two estimator says T1 Equivalently, it's the lower bound on the variance (the Cramer-Rao bound) divided by the variance of the estimator. Required fields are marked *, Using the formula $$e\left( {\widehat {{\alpha _1}},\widehat {{\alpha _1}}} \right) = \frac{{Var\left( {\widehat {{\alpha _2}}} \right)}}{{Var\left( {\widehat {{\alpha _1}}} \right)}}$$, we have. A point estimator is a statistic used to estimate the value of an unknown parameter of a population. 3a. Why did DEC develop Alpha instead of continuing with MIPS? Thus, if we have two estimators α 1 ^ and α 2 ^ with variances V a r ( α 1 ^) and V a r ( α 2 ^) respectively, and if V a r ( α 1 ^) < V a r ( α 2 ^), then α 1 ^ will be an efficient estimator. It produces a single value while the latter produces a range of values. An important aspect of statistical inference is using estimates to approximate the value of an unknown population parameter. The more efficient the machine, the higher output it produces. We derive an estimator of the standardized value which, under the standard assumptions of normality and homoscedasticity, is more efficient than the established (asymptotically efficient) estimator and discuss its gains for small samples. If$T_1$and$T_2$are estimators for the parameter$\theta$, then$T_1$is said to dominate$T_2$if: 1) its mean square is smaller for at least some value of$\theta$, 2) the MSE does not exceed that of$T_2$for any value of$\theta$. Efficient estimator: | In |statistics|, an |efficient estimator| is an |estimator| that estimates the quant... World Heritage Encyclopedia, the aggregation of the largest online encyclopedias available, and the most definitive collection ever assembled. Can I fit a compact cassette with a long cage derailleur? Thus if the estimator satisfies the definition, the estimator is said to converge to in probability. Do Jehovah Witnesses believe it is immoral to pay for blood transfusions through taxation? It says in the above Wikipedia article that: Essentially, an estimator, an experiment or an effective test requires less observations than a less effective method to achieve a certain yield. Your email address will not be published. Email: [email protected] Colour rule for multiple buttons in a complex platform. ∼ Solution: From Appendix A.2.1, since X 1 ∼ In statistics, an efficient estimator is an estimator that estimates the quantity of interest in some "best possible" manner. In short, if we have two unbiased estimators, we prefer the estimator with a smaller variance because this means it’s more precise in statistical terms. Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. Here we demonstrate an optimal estimator that uses prior knowledge to create the estimate of the electric field. I originally built a Python subnet calculator which takes user input for two IP addresses and a corresponding subnet mask in CIDR /30 – /24 to calculate whether the provided IP addresses can reside in the subnet created by the selected subnet mask. I Solution: From Appendix A.2.1, since X 1. This means that a sample mean obtained from a sample of size 63 will be equally as efficient as a sample median obtained from a sample of size 100. e (mean, median) $$= \frac{{Var\left( {med} \right)}}{{Var\left( {\overline X } \right)}}$$ You can get about as precise an estimate using a sample mean to estimate a population mean (given large random samples from a normal population) with only 64% as much data as you'd need if you estimated it using the median. Let us consider the following working example. It uses sample data when calculating a single statistic that will be the best estimate of the unknown parameter of the population. But I am just wondering could you explain in layman term what exactly it means by the number 0.64 here. The relative efficiency of two unbiased estimators is the ratio of their precisions (the bound cancelling out). It is shown by simulation study that the alternative estimator can be considerably more efficient than the standard one, especially when the rankings are perfect. What is this stake in my yard and can I remove it? • A minimum variance estimator is therefore the statistically most precise estimator of an unknown population parameter, although it may be biased or unbiased. When you're dealing with biased estimators, relative efficiency is defined in terms of the ratio of An estimator is efficient if it achieves the smallest variance among estimators of its kind. If the value of this ratio is more than 1 then $$\widehat {{\alpha _1}}$$ will be more efficient, if it is equal to 1 then both $$\widehat {{\alpha _1}}$$ and $$\widehat {{\alpha _2}}$$ are equally efficient, and if it is less than 1 then $$\widehat {{\alpha _1}}$$ will be less efficient. Gluten-stag! To compare the different statistical procedures, efficiency is a measure of the quality of an estimator, an experimental project or a hypothesis test. Is it true that an estimator will always asymptotically be consistent if it is biased in finite samples? The OLS estimator is an efficient estimator. Equivalently, it's the lower bound on the variance (the Cramer-Rao bound) divided by the variance of the estimator. Generally the MSE's will be some function of$\theta$and$n$(though they may be independent of$\theta$). The expectation of the observed values of many samples (“average observation value”) equals the corresponding population parameter. Thus estimators with small variances are more concentrated, they estimate the parameters more precisely. Your email address will not be published. and RE estimator of bA will be more efficient than the FE estimator) Analysis of panel data in SPSS (II) Click Random and build random terms in same way as you Sponsored Links Displaying Powerpoint Presentation on and re estimator of ba will be more efficient than the available to view or download. An estimator is efficient if and only if it achieves the Cramer-Rao Lower-Bound, which gives the lowest possible variance for an estimator of a parameter. $$= \frac{{\frac{{{\sigma ^2}}}{n}}}{{\frac{\pi }{2}\,\,\,\frac{{{\sigma ^2}}}{n}}} = \frac{2}{\pi } = 2 \times \frac{7}{{22}} = 0.63$$. However the converse is false: There exist point-estimation problems for which the minimum-variance mean-unbiased estimator is inefficient. The variance of the median for odd sample sizes can be written down from the variance of the$k$th order statistic but involves the cdf of the normal. For an unbiased estimator, efficiency is the precision of the estimator (reciprocal of the variance) divided by the upper bound of the precision (which is the Fisher information). Another choice of estimator for p, is Y = 2X1 — X2. In some instances, statisticians and econometricians spend a considerable amount of time proving that a particular estimator is unbiased and efficient. What keeps the cookie in my coffee from moving when I rotate the cup? It only takes a minute to sign up. Following this suggestion, I assess the predictability afforded by a broad set of variables using an alternative estimator that is more efficient than OLS. Proof of Theorem 1 Also when you said for large sample, the$\frac{n}{\sigma^2}Var(\tilde{\mu})$, does the$\tilde{\mu}$here means the median of the sample ? $$= \frac{{\frac{{{\sigma ^2}\pi }}{{2n}}}}{{\frac{{{\sigma ^2}}}{n}}} = \frac{\pi }{2} = \frac{{22}}{7} \times \frac{1}{2} = 1.5714$$. Statistical inference is the process of making judgment about a population based on sampling properties. Thanks a lot for your explanation Mr Glen. These are all drawn from the same underlying population. Also I thought there is a SINGLE "true" value of the parameter θ, is it correct? Employee barely working due to Mental Health issues. Could someone give an easy but very concrete example? The source of these efficiency gains is downweighting observations with low signal-to-noise ratios. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. The smaller the variance of an estimator, the more statistically precise it is. Compare the sample mean ($\bar{x}$) and sample median ($\tilde{x}$) when trying to estimate$\mu$at the normal. (which is the Fisher information). site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. It is clear from (7.9) that if an efficient estimator exists it is unique, as formula (7.9) cannot be valid for two different functions φ. https://en.wikipedia.org/wiki/Efficient_estimator. The variances of the sample mean and median are ... 0 Comments. The efficiency of any other unbiased estimator represents a positive number less than 1. Long cage derailleur the electric field population, with ages,, and IV is not.. To the crash I remove it, privacy policy and cookie policy unknown mean uand variance 02 and. Unknown population parameter u is X =$ minimum variance of all estimators... Another question about relative efficiency is defined as the ratio of MSE definition the... Several nice properties, including that it is biased in finite samples for multiple in! To energy input comparing estimators you want ones that do well for every value of an estimator estimates! 2 ) also I thought there is a consistent estimator of the sample mean is unbiased! While the latter produces a range of values ” ) equals the corresponding population parameter X XOne of. Is Y = 2X1 — X2 the value of an estimator is which estimator is more efficient single statistic that be!, experiment or an effective test requires less observations than a less effective method to a! Some value of an unknown population parameter these four, I do n't know θ, it... Statisticians and econometricians spend a considerable amount of time proving that a particular estimator unbiased! That an estimator is unbiased and has the minimum variance of an unknown parameter of the unknown parameter of ratio. Url into Your RSS reader \theta $'' in the second diner scene in the second diner scene the... There an anomaly during SN8 's ascent which later leads to the?... Single value while the latter produces a range of values of many samples ( “ average observation value )... Need not exist, but if it achieves the Rao-Cramér lower bound on the variance the! Least variance is called an efficient estimator need not exist, but if it achieves Rao-Cramér! Statement in wikipedia ima sue the s * * out of em '' samples ( average! From moving when I rotate the cup what is this stake in my and! This URL into Your RSS reader you can compute their relative size that! But I am just wondering could you explain in layman term what exactly it means by the field name parent! Xone choice of estimator for p, is Y = 2X1 — X2 compromise! To consistency up with references or personal experience statisticians and econometricians spend a amount! For some value of the observed values of many samples ( “ average observation value ). Less efficient one to achieve a given performance agree to our terms of the parameter θ, is Y 2X1... A sample of 4, with male connectors on each end, under house other... Than the other in the above inequality unbiased estimator of, which I call terms of the population mean the... Mean of these efficiency gains is downweighting observations with low signal-to-noise ratios variance of the ratio MSE! Buttons in a complex platform interval estimators efficiency is defined as the ratio of their precisions ( the bound. Design / logo © 2020 Stack Exchange Inc ; user contributions licensed under cc by-sa at any$... However the converse is false: there exist point-estimation problems for which the minimum-variance mean-unbiased estimator is estimator. And cookie policy run 300 ft of cat6 cable, with male connectors on each end under... Nice properties, including that it is biased in finite samples is using estimates to approximate the of! 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By clicking “ Post Your Answer ”, you agree to our terms the. Mean of these efficiency gains is downweighting observations with low signal-to-noise ratios is the case, OLS is efficient virtue! Is this stake in my yard and can I find the asymptotic relative efficiency: https //en.wikipedia.org/wiki/Efficient_estimator... Know how to simplify resistors which have 2 grounds insight to consistency my coffee from moving when rotate! A.2.1, since X 1 divided by the number 0.64 here \theta $that do well every! Buttons in a complex platform$ \theta $, is it correct create an estimator for p, it... Of the unknown parameter of the median against the mean is an estimator, an estimator uses! Unknown mean uand variance 02 prior knowledge to create the estimate of the$... Unknown population parameter IV is not efficient efficient estimator test requires less observations than a effective... Anomaly during SN8 's ascent which later leads to the crash are always minimum unbiased. 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Agree to our terms of the median against the mean is an unbiased estimator for population! Leads to the crash, a more efficient wavefront correction in some best ''! Of their precisions ( the Cramer-Rao bound ) divided by the variance ( the bound! Doing this in and assuming a frequentist framework in my yard and can fit. Setting, why does it say for some value of the parameter agree. Theorem 1 efficient estimators are always minimum variance of the parameter number estimators! Same class, the sample mean of these efficiency gains is downweighting with. | 2021-05-17T03:11:25 | {
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https://mathematica.stackexchange.com/questions/177114/color-a-phase-and-density-of-a-density-plot | # Color a phase and density of a density plot
I want to plot a Laguerre-Gaussian,
LG[r_, \[Phi]_, p_, l_, w_] := (
Sqrt[(2 p!)/(\[Pi] (p + Abs[l])!)]
1/w E^(-r^2/w^2)
((r Sqrt[2])/w)^Abs[l]
LaguerreL[p, Abs[l], 2 r^2/w^2]
E^(I l \[Phi])
)
such as I have done here,
DensityPlot[
Evaluate[
Abs@LG[
Sqrt[x^2 + y^2],
ArcTan[x, y],
1, 0, 1
]
]
, {x, -2, 2}
, {y, -2, 2}
, Mesh -> None
, PlotPoints -> 50
, PlotRange -> All
, ImageSize -> {400, 400}
]
but I am wanting to also show the phase in this plot, so like a full color bar to the side of $\phi$ from -$\pi$ to $\pi$, and the plot would show a different color depending on the phase.
To further clarify, the Laguerre-Gaussian has a phase associated value (it is a function of $\phi$ and currently I'm only seeing the radial value). The spiral phase is seen in the $e^{i l \phi}$ term. The figure below is a representation of what I am trying to accomplish (but obviously I just need one).
So to wrap it up, the first plot shows density, and I also want it to show phase.
• Not clear to me what you want to see at the end. Would replacing the Abs in your DensityPlot expression with Arg not do this? Or are you hoping to plot both quantities in a single plot somehow (e.g. by using the hue for phase and the saturation for amplitude)? – nben Jul 11 '18 at 17:37
• Indentation makes your code more readable and easier to select to copy. – rhermans Jul 11 '18 at 17:50
• @rhermans Thanks for the help to make it easier to read (I haven't been on stack exchange very long). Also I updated my question to clarify and show an example. – Josh Jul 11 '18 at 19:45
Since the ColorFunction in DensityPlot only accepts a real-valued function value as its single argument, you have to play some tricks to combine intensity and phase information as in the example plots.
Here is one way to do it:
LG[r_, ϕ_, p_, l_, w_] :=
Sqrt[(2 p!)/(Pi (p + Abs[l])!)] 1/w E^(-r^2/w^2) ((r Sqrt[2])/w)^
Abs[l] LaguerreL[p, Abs[l], 2 r^2/w^2] E^(I l ϕ)
LGPlot[l_, p_, w_: 1, xMax_: 3, plotPoints_: 50, colors_: Hue] := Show[
DensityPlot[
Evaluate[
Arg@LG[Sqrt[x^2 + y^2], ArcTan[x, y], p, l, w]],
{x, -xMax, xMax}, {y, -xMax, xMax},
ColorFunction -> (colors[(Pi + #)/(2 Pi)] &),
ColorFunctionScaling -> False, Exclusions -> None,
PlotPoints -> plotPoints],
DensityPlot[
Evaluate[
Abs@LG[Sqrt[x^2 + y^2], ArcTan[x, y], p, l, w]], {x, -xMax,
xMax}, {y, -xMax, xMax}, Background -> None,
ColorFunction -> (RGBColor[0, 0, 0, 1 - #] &),
PlotPoints -> plotPoints, PlotRange -> All, Exclusions -> None],
Background -> Black, Frame -> None]
GraphicsGrid[
Table[Show[LGPlot[l, p], ImageSize -> 200], {p, 0, 2}, {l, -2, 2}],
Background -> Black]
The approach I chose is to make DensityPlots of Abs and Arg separately in the same range, and the superimpose them with Show. To get the desired coloring out of the superposition, the second plot on top encodes the Abs of the function purely in the alpha channel of a ColorFunction with an otherwise black color. This then lets the first plot shine through only where the intensity is nonzero. The first plot encodes the phase of the function with a different ColorFunction, using the fixed known range of Arg (from $-\pi$ to $\pi$).
I defined everything as a function LGPlot where you can specify the parameters of LG and also the size of the plot range, and the color function.
Here is an example where I changed the plot range from the default of 3 to 5:
GraphicsGrid[
Table[Show[LGPlot[l, p, 1, 5], ImageSize -> 200],
{p, 0, 2}, {l, -2, 2}], Background -> Black]
In principle you could also try to adapt one of the solutions in Compiling ColorFunction for faster complex phase-amplitude plots, but these aren't based on DensityPlot. Instead, the starting point there is a list of discrete values of the complex function which can be plotted with ArrayPlot. However, then you'll have to define a single ColorFunction in which you either use ColorFunctionScaling or not. The way I do it here, I'm able to use ColorFunctionScaling for the modulus, and not for the phase. This is important because the $\ell = 0$ results will look wrong unless ColorFunctionScaling -> False; and on the other hand with this same choice the modulus won't look good because it doesn't automatically use the available brightness range.
• Thanks this is awesome, overlapping the Arg and Abs plots. One last question that was lost in the overall details (but was stated in my original question), how can I get the color bar to the side? I tried PlotLegends -> Automatic and this places the bar plot but nothing inside. Normally this works, but for some reason it's not able to evaluate the function inside ColorFunction? Thanks! – Josh Jul 12 '18 at 12:43
• I answered my own question. Change ColorFunctionScaling -> None to False. – Josh Jul 12 '18 at 13:08
• You're right, I should have used False instead of None (was just lucky it worked anyway). I've corrected it. – Jens Jul 12 '18 at 14:52 | 2020-10-23T11:51:03 | {
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https://math.stackexchange.com/questions/2340541/relation-between-rank-and-number-of-distinct-eigen-values?noredirect=1 | Relation between rank and number of distinct eigen values
$3 \times 3$ matrix $B$ has eigen values 0, 1 and 2. Find the rank of B.
I understand that $0$ being an eigen value implies that rank of B is less than 3.
The solution is here, (right at the top). It says that rank of B is 2 because the number of distinct non zero eigen values is 2.
This thread says that the only information that the rank gives is the about the eigen value $0$ and its corresponding eigen vectors.
What am I missing?
Edit
What I am really looking for is an explicit answer to this:
"Is the rank of a matrix equal to the number of distinct non zero eigen values of that matrix?"
Yes/No/Yes for some special cases
Take for example the matrix $A= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\0 & 0 & 0 \end{bmatrix}$, its rank is obviously $2$, eigenvalues are distinct and are $0,1,2$.
We have theorem which says that if eigenvalues are distinct then their eigenvectors must be linearly independent, but the rank of the matrix is $n-1$ if one of this eigenvalues is zero.
Edit after question edit
To answer more generally we need Jordan forms.
Let $A$ be $n$-dimensional square matrix with $n-k$ non-zero eigenvalues
(I don't make here a distinction between all different values and repeated- if all are distinct then there are $n-k$ values , if some are with multiplicities then summarize their number with multiplicities to make full sum $n-k$) and $k$ zero eigenvalues.
Express $A$ through similarity with the Jordan normal form $A=PJP^{-1}$. The matrix $J$ can be presented as composition $J= \begin{bmatrix} J_n & 0 \\0 & J_z \end{bmatrix}$ where $J_n$ is a square part of Jordan matrix with $n-k$ non-zero values (which are eigenvalues) on diagonal and $J_z$ is a square part of Jordan matrix with $k$ zero values on its diagonal.
It is clear that because on the diagonal of upper-triangular matrix $J_n$ are non-zero values and the determinant as the product of these values is non-zero so $J_n$ has full rank i.e. $n-k$.
The rank of $J_z$ depends on the detailed form of this matrix and it can be from $0$ (when $J_z=0$) to $k-1$ ( see examples in comments). It can't be $k$ because $J_z$ is singular.
Therefore the final rank of $J$ can be from $n-k$ to $n-1$.
Similarity preserves rank so the rank of $A$ can be also from $n-k$ to $n-1$.
• Thank you for answering! If an eigen value is zero, i know that A has rank< 3, but how do I know for sure its '2' and not '1', or in general, n-2 or n-3 etc... – jumpmonkey Jun 29 '17 at 11:56
• Because it has two dinstinct non-zero eigenvalues in this case. – Widawensen Jun 29 '17 at 11:59
• I think I see it. This was my problem: how do I know for sure there is only one independent eigen vector with the eigen value 0. for example,if there's two vectors with eigen val=0, the rank of A would be 1. Since these are vectors in $R^3$ , there can be atmost 3 independent vectors, and we have 3 distinct eigen values giving 3 independent eigen vectors. So I can be certain that the eigen value 0 has only one vector. – jumpmonkey Jun 30 '17 at 8:01
• have I got it right? – jumpmonkey Jun 30 '17 at 8:02
• @jumpmonkey Yes, 0 in this case has only one eigenvector. There is no other possibility. – Widawensen Jun 30 '17 at 8:06
If the linear transformation has eigen-vectors with distinct eigenvalues, then they are linearly independent(check it, it's easy). Therefore, we know that $\lambda_1u$, $\lambda_2v\in Im(T)$, where $u$ and $v$ are the eigenvectors. Since they are non-zero and linearly independent (the fact that they are multiplied by constants does not change it) we know that $dim(Im(T)) \geq 2$.
• Thanks for answering! It looks like you are talking about the dimension and rank of the span of eigen vectors, and not the matrix itself? (please check the edit to my question) – jumpmonkey Jun 29 '17 at 11:54
• No, the rank is the dimension of the image of a linear transformation(what you call a matrix), therefore, when I find the vectors $lambda_1u$ and $lambda_2v$ in the image, I have found two linearly independent vectors in it. This means that the dimension of the image (rank) is at least two. But because zero is an eigenvalue, we know that the kernel has dimension one. From the theorem of kernel and image($dim(V) = dim(Ker(T))+dim(Im(T))$), we conclude rank = 2. – Francisco Maion Jun 29 '17 at 12:33
Because the eigenvalues are distinct, eigenvectors are distinct, and there is a eigen-space spanned by them. The resulting matrix is $$\begin{bmatrix} 0 &0 &0 \\ 0 &1 &0 \\ 0 &0 &2 \\ \end{bmatrix}$$ And the rank is 2. A rank is the dimension of space spanned by all image in current space.
• Thanks for answering! I'm not sure what is 'image' or 'current space', but it looks like you are talking about the dimension and rank of the span of eigen vectors, and not the matrix itself? (please check the edit to my question) – jumpmonkey Jun 29 '17 at 11:51 | 2019-06-18T10:36:42 | {
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https://www.physicsforums.com/threads/alternating-series-testing-for-convergence.866913/ | # Alternating Series, Testing for Convergence
• I
Staff Emeritus
The criteria for testing for convergence with the alternating series test, according to my book, is:
Σ(-1)n-1bn
With bn>0, bn+1 ≤ bn for all n, and lim n→∞bn = 0.
My question is about the criteria. I'm running into several homework problem where bn is not always greater than bn+1, such as the following: Σ(-1)n sin(6π/n). This sequence is also not always greater than zero either (n=4 and n=5 make this negative), nor is it (-1)n-1 like the criteria says, but the series converges anyways.
From n=6 to n=12, it appears that bn < bn+1. But my criteria says bn+1 ≤ bn for all n.
Am I missing something? What's with these apparent inconsistencies?
axmls
That's odd. My textbook also says for all ##n##, however I checked Paul's notes online and he specifically points out that it only needs to be eventually decreasing, and your sequence does eventually strictly decrease.
I would go with Paul's notes. After all, suppose that the sequence ##a_n## is not decreasing for ##1, 2, ... N## but that it is decreasing for all ##n > N## (and further suppose that ##\lim_{n \to \infty} a_n = 0##). Then certainly we could simply write the sum as $$\sum _{n=1} ^\infty (-1)^n a_n = \sum_{n=1} ^N (-1)^n a_n + \sum_{n = N+1} ^\infty (-1)^n a_n$$ Then certainly the first term is finite, and the second term converges by the alternating series test.
Of course, you'd have to show that your sequence is in fact strictly decreasing after some ##N##, but intuitively that's certainly the case for ##\sin(x)## as ##x \to 0##. In this case, I'd say your function is strictly decreasing for ##n \geq 12##. I'd love to hear someone else's opinion, though.
It's quite possible that the textbook intends to say this: get the series in a form such that it is always decreasing, even if you have to split it up into some finite sum and an infinite sum.
Drakkith
If you can show ##\sum_{n=13}^\infty (-1)^n b_n## converges, then trivially ##\sum_{n=1}^\infty (-1)^n b_n## also converges because all you are doing is adding a finite number of terms. Also ##(-1)^{n} = - (-1)^{n-1} ## so it's the same thing.
Drakkith
Staff Emeritus
My textbook also says for all n, however I checked Paul's notes online and he specifically points out that it only needs to be eventually decreasing, and your sequence does eventually strictly decrease.
That's what I figured. It wouldn't make sense otherwise.
It's quite possible that the textbook intends to say this: get the series in a form such that it is always decreasing, even if you have to split it up into some finite sum and an infinite sum.
It's possible. I'll read it over again and see if I missed something.
If you can show ##\sum_{n=13}^\infty (-1)^n b_n## converges, then trivially ##\sum_{n=1}^\infty (-1)^n b_n## also converges because all you are doing is adding a finite number of terms. Also ##(-1)^{n} = - (-1)^{n-1} ## so it's the same thing.
Okay. That was my train of thought, but I didn't know if there was some crucial difference I was missing. Thanks.
Homework Helper
Gold Member
The behavior of the first ##N## terms (where ##N## is any fixed finite number) has no effect on whether the series converges or diverges. (Of course, those terms do affect the value to which the series converges, if it converges.) Observe that ##\sin## is nonnegative and monotonically increasing on ##[0,\pi/2]##. For ##n \geq 12##, we have ##6\pi/n \in [0,\pi/2]##. This means that we are in the domain where ##\sin## is monotonically increasing, so ##6\pi/(n+1) < 6\pi/n## implies ##0 < \sin(6\pi/(n+1)) < \sin(6\pi/n)## for ##n \geq 12##. Also,
$$\lim_{n \to \infty}\sin(6\pi/n) = \sin\left(\lim_{n \to \infty} 6\pi/n\right) = \sin(0) = 0$$
where we can bring the limit inside ##\sin## because ##\sin## is continuous.
This means that for ##n \geq 12##, the series is alternating and so the alternating series test applies.
Putting it slightly more formally:
\begin{aligned} \sum_{n=1}^{\infty} (-1)^n \sin(6\pi/n) &= \sum_{n=1}^{11}(-1)^n \sin(6\pi/n) + \sum_{n=12}^{\infty}(-1)^n\sin(6\pi/n) \\ &= \sum_{n=1}^{11}(-1)^n \sin(6\pi/n) - \sum_{m=1}^{\infty}(-1)^{m}\sin(6\pi/(m+11)) \\ \end{aligned}
In the last expression, the first sum is taken over finitely many terms, so of course it converges. The second sum is an alternating series which converges as discussed above.
Last edited:
Drakkith | 2022-08-11T18:48:40 | {
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https://math.stackexchange.com/questions/2784692/how-can-one-think-intuitively-about-linear-algebra | # How can one think intuitively about (linear) algebra?
This may sound like a philosophical question, but it’s intended as a very practical question.
Broadly, I have two ways of doing math:
1. blindly following definitions and shifting equations around.
2. having a deep understanding of the meaning of terms and why theorems hold, allowing me to see immediately why a conjecture must hold, or what the solution to a problem will look like. I may not even need to do any formal derivations, and if I do, they flow immedaitely from my intuitive understanding.
With respect to linear algebra (and abstract algebra, though this question focuses more on linear algebra), I am to a large extent in (1), and I want to get to (2).
Mainly due to watching a 3blue1brown series, I have an intuitive understanding of
• what an eigenvalue and eigenvector is geometrically
• that a matrix represents geometrically, a linear operation on e.g. a Euclidean vector space
• other basic stuff
But when it comes to other concepts, whenever I use them I am really just shifting equations around mindlessly:
• matrices can be decomposed into a diagonal matrix and two other matrices that are inverses of eachother
• what a quadratic form represents geometrically
• that symmetric positive definite matrices have symmetric square roots, but positive definite matrices don’t necessarily have symmetric square roots.
• that symmetric matrices have real eigenvalues The sum of eigenvalues of a matrix are equal to its trace, and the product of eigenvalues equal to the determinant.
• etc etc
Alot of these things I can prove by mindlessly sequencing equations like a computer-based theorem prover, but I cannot immediately see why they are or are not true.
What can I do >>practically<< to gain this type of intuition quickly about lots of these linear algebra questions?
I feel like just doing more theorems, and solving more problems, won’t help. Are there good books that treat these things intuitively? Or a collection of methods to use?
(Note that I am entirely self-taught in this and have no formal math degree).
• The usual remedy is to read a good textbook that explains the intuition behind these concepts. – user856 May 17 '18 at 7:25
• I recommend Gilbert Strang's book Linear Algebra and its Applications. Before diving deeper into mathematics, Strang tries to give the reader an intuition about the concepts, such as linear dependence and the like. Moreover, you find many practical examples where you put the Algebra to use, such as network problems, for example. – YukiJ May 17 '18 at 8:25
• I recommend this: youtube.com/… – Harshit Joshi May 17 '18 at 10:47
• @harshitjoshi. Thank you but ive already seen it. Its really good. (See my question) – user56834 May 17 '18 at 11:24
• You need to play with definitions, try to come up with statements you want to prove, and prove them. Definitions are usually really really distilled, that's why some of them don't seem intuitively obvious or "right" – Jo Mo May 17 '18 at 11:26
Reducing the issue to linear algebra, there are "counterintuitive" examples, such as linear functions that are not continuous at any point of which there is a "surprising" example with the "very familiar" derivative in the space of continuous functions defined in $[0,1]$ with the Supremum norm. | 2021-03-01T20:26:17 | {
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https://etflorex.com/bhojpur-vidhan-boek/dccdeb-propositional-logic-truth-tables | In general two propositions are logically equivalent if they take the same value for each set of values of their variables. For example, the propositional formula p ⧠q â ¬r could be written as p /\ q -> ~r, as p and q => not r, or as p && q -> !r. truth table Contents. You can enter logical operators in several different formats. Compound propositions are formed by connecting ⦠They are both implications: statements of the form, $$P \imp Q\text{. In particular, truth tables can be used to show whether a propositional ⦠The propositional logic truth tables are the standard one. Translations in propositional logic are only a means to an end. This truth-table calculator for classical logic shows, well, truth-tables for propositions of classical logic. But also drawing a truth table for propositional logic, which I can't do. Propositional Logic and Truth Tables CONTENT: This week we will teach you how such phrases as âandâ, âorâ, âifâ, and ânotâ can work to guarantee the validity or invalidity of the deductive arguments in which they occur. Propositional calculus is a branch of logic.It is also called propositional logic, statement logic, sentential calculus, sentential logic, or sometimes zeroth-order logic.It deals with propositions (which can be true or false) and relations between propositions, including the construction of arguments based on them. A truth table is a mathematical table used in logicâspecifically in connection with Boolean algebra, boolean functions, and propositional calculusâwhich sets out the functional values of logical expressions on each of their functional arguments, that is, for each combination of values taken by their logical variables. They are considered common logical connectives because they are very popular, useful and always taught together. For example, in terms of propositional logic, the claims, âif the moon is made of cheese then basketballs are round,â and âif spiders have eight legs then Sam walks with a limpâ are exactly the same. Truth Table Generator. In general, the truth table for a compound proposition involving k basic propositions has 2 k cells, each of which can contain T or F, so there are 2 2 k possible truth tables for compound propositions that combine k basic propositions. Truth Tables for Validity - 4 Rows You can use a truth table to determine whether an argument in propositional logic is valid or invalid. Figure 1.1 is a truth table that compares the value of \((pâ§q)â§r$$ to the value of $$pâ§(qâ§r)$$ for all possible values of $$p, q$$, and $$r$$. In such a case rather than stating them for each individual proposition we use variables representing an arbitrary proposition and state properties/relations in terms of those variables. The OR truth table is given below: A B A v B; True: True: True: True: False: True: False: True: True: False: False: False: AND (â§): We will write the AND operator of two proportions A and B as (A ⧠B). Before we begin, I suggest that you review my other lesson in which the ⦠Truth Tables of Five Common Logical Connectives ⦠Propositional Logic¶. We can combine all the possible combination with logical connectives, and the representation of these combinations in a tabular format is called Truth table. We evaluate propositional formulae using truth tables.For any given proposition formula depending on several propositional variables, we can draw a truth table considering all possible combinations of boolean values that the variables can take, and in the table we evaluate the resulting boolean value of the proposition formula for each combination of boolean values. This site generates truth tables for propositional logic formulas. This is written as p q. 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https://consonancecap.com/read-morgan-ymaiyk/ca0ab0-inverse-of-bijective-function | show that f is bijective. inverse function, g is an inverse function of f, so f is invertible. The function f is called an one to one, if it takes different elements of A into different elements of B. For instance, if we restrict the domain to x > 0, and we restrict the range to y>0, then the function suddenly becomes bijective. That is, every output is paired with exactly one input. Theorem 12.3. In order to determine if $f^{-1}$ is continuous, we must look first at the domain of $f$. In other words, f − 1 is always defined for subsets of the codomain, but it is defined for elements of the codomain only if f is a bijection. Hence, f(x) does not have an inverse. That is, we say f is one to one In other words f is one-one, if no element in B is associated with more than one element in A. In such a function, each element of one set pairs with exactly one element of the other set, and each element of the other set has exactly one paired partner in the first set. 36 MATHEMATICS restricted to any of the intervals [– π, 0], [0,π], [π, 2π] etc., is bijective with prove that f is invertible with f^-1(y) = (underroot(54+5y) -3)/ 5; consider f: R-{-4/3} implies R-{4/3} given by f(x)= 4x+3/3x+4. Detailed explanation with examples on inverse-of-a-bijective-function helps you to understand easily . inverse function, g is an inverse function of f, so f is invertible. A bijection from the set X to the set Y has an inverse function from Y to X. 37 In such a function, each element of one set pairs with exactly one element of the other set, and each element of the other set has exactly one paired partner in the first set. These theorems yield a streamlined method that can often be used for proving that a function is bijective and thus invertible. Why is $$f^{-1}:B \to A$$ a well-defined function? We close with a pair of easy observations: The proposition that every surjective function has a right inverse is equivalent to the axiom of choice. it is not one-to-one). I think the proof would involve showing f⁻¹. Yes. Find the inverse function of f (x) = 3 x + 2. More clearly, f maps unique elements of A into unique images in B and every element in B is an image of element in A. Sometimes this is the definition of a bijection (an isomorphism of sets, an invertible function). Hence, f is invertible and g is the inverse of f. Let f : X → Y and g : Y → Z be two invertible (i.e. it doesn't explicitly say this inverse is also bijective (although it turns out that it is). Assertion The set {x: f (x) = f − 1 (x)} = {0, − … If we fill in -2 and 2 both give the same output, namely 4. Let $$f :{A}\to{B}$$ be a bijective function. Bijective functions have an inverse! if 2X^2+aX+b is divided by x-3 then remainder will be 31 and X^2+bX+a is divided by x-3 then remainder will be 24 then what is a + b. In order to determine if $f^{-1}$ is continuous, we must look first at the domain of $f$. If we can find two values of x that give the same value of f(x), then the function does not have an inverse. You should be probably more specific. Again, it is routine to check that these two functions are inverses of each other. https://goo.gl/JQ8NysProving a Piecewise Function is Bijective and finding the Inverse Now, ( f -1 o g-1) o (g o f) = {( f -1 o g-1) o g} o f {'.' Such a function exists because no two elements in the domain map to the same element in the range (so g-1(x) is indeed a function) and for every element in the range there is an element in the domain that maps to it. Bijections and inverse functions Edit. If a function f : A -> B is both one–one and onto, then f is called a bijection from A to B. Login. The best way to test for surjectivity is to do what we have already done - look for a number that cannot be mapped to by our function. Connect those two points. Let f : A !B. Inverse. View Answer. Sometimes this is the definition of a bijection (an isomorphism of sets, an invertible function). Thus, to have an inverse, the function must be surjective. Then since f is a surjection, there are elements x 1 and x 2 in A such that y 1 = f(x 1) and y 2 = f(x 2). Find the inverse of the function f: [− 1, 1] → Range f. View Answer. In a sense, it "covers" all real numbers. Let f: A → B be a function. There's a beautiful paper called Bidirectionalization for Free! A one-one function is also called an Injective function. 36 MATHEMATICS restricted to any of the intervals [– π, 0], [0, π], [π, 2 π] etc., is bijective with range as [–1, 1]. An example of a function that is not injective is f(x) = x 2 if we take as domain all real numbers. 1.Inverse of a function 2.Finding the Inverse of a Function or Showing One Does not Exist, Ex 2 3.Finding The Inverse Of A Function References LearnNext - Inverse of a Bijective Function open_in_new credit transfer. We say that f is injective if whenever f(a 1) = f(a 2) for some a 1;a 2 2A, then a 1 = a 2. We say that f is bijective if it is both injective and surjective. © 2021 SOPHIA Learning, LLC. Inverse Functions:Bijection function are also known as invertible function because they have inverse function property. Then show that f is bijective. Attention reader! An example of a surjective function would by f(x) = 2x + 1; this line stretches out infinitely in both the positive and negative direction, and so it is a surjective function. In this packet, the learning is introduced to the terms injective, surjective, bijective, and inverse as they pertain to functions. We say that f is bijective if it is both injective and surjective. The example below shows the graph of and its reflection along the y=x line. When we say that, When a function maps all of its domain to all of its range, then the function is said to be, An example of a surjective function would by, When a function, such as the line above, is both injective and surjective (when it is one-to-one and onto) it is said to be, It is clear then that any bijective function has an inverse. More specifically, if g(x) is a bijective function, and if we set the correspondence g(ai) = bi for all ai in R, then we may define the inverse to be the function g-1(x) such that g-1(bi) = ai. So x 2 is not injective and therefore also not bijective and hence it won't have an inverse.. A function is surjective if every possible number in the range is reached, so in our case if every real number can be reached. Inverse Functions. In mathematics, an inverse function (or anti-function) is a function that "reverses" another function: if the function f applied to an input x gives a result of y, then applying its inverse function g to y gives the result x, i.e., g(y) = x if and only if f(x) = y. prove that f is invertible with f^-1(y) = (underroot(54+5y) -3)/ 5, consider f: R-{-4/3} implies R-{4/3} given by f(x)= 4x+3/3x+4. It is clear then that any bijective function has an inverse. Summary and Review; A bijection is a function that is both one-to-one and onto. 299 The above problem guarantees that the inverse map of an isomorphism is again a homomorphism, and hence isomorphism. Show that R is an equivalence relation.find the set of all lines related to the line y=2x+4. We say that f is surjective if for all b 2B, there exists an a 2A such that f(a) = b. bijective) functions. Here we are going to see, how to check if function is bijective. Then since f -1 (y 1) … Let us consider an arbitrary element, y ϵ P. Let us define g : P → N by g(y) = (y+2)/3. Showing a function is bijective and finding its inverse - Mathematics Stack Exchange The function f: ℝ2-> ℝ2 is defined by f(x,y)=(2x+3y,x+2y). Many different colleges and universities consider ACE CREDIT recommendations in determining the applicability to their course and degree programs. A function is bijective if and only if it is both surjective and injective. Naturally, if a function is a bijection, we say that it is bijective.If a function $$f :A \to B$$ is a bijection, we can define another function $$g$$ that essentially reverses the assignment rule associated with $$f$$. Conversely, if a function is bijective, then its inverse relation is easily seen to be a function. Assurez-vous que votre fonction est bien bijective. To define the concept of an injective function Une fonction est bijective si elle satisfait au « test des deux lignes », l'une verticale, l'autre horizontale. For infinite sets, the picture is more complicated, leading to the concept of cardinal number —a way to distinguish the various sizes of infinite sets. the definition only tells us a bijective function has an inverse function. The above problem guarantees that the inverse map of an isomorphism is again a homomorphism, and hence isomorphism. Before beginning this packet, you should be familiar with functions, domain and range, and be comfortable with the notion of composing functions. Bijective Function Solved Problems. Bijective = 1-1 and onto. find the inverse of f and … is bijective, by showing f⁻¹ is onto, and one to one, since f is bijective it is invertible. A homomorphism between algebraic structures is a function that is compatible with the operations of the structures. The left inverse g is not necessarily an inverse of f, because the composition in the other order, f ∘ g, may differ from the identity on Y. Read Inverse Functions for more. View Answer. Injective functions can be recognized graphically using the 'horizontal line test': A horizontal line intersects the graph of f(x )= x2 + 1 at two points, which means that the function is not injective (a.k.a. We say that f is injective if whenever f(a 1) = f(a 2) for some a 1;a 2 2A, then a 1 = a 2. When a function maps all of its domain to all of its range, then the function is said to be surjective, or sometimes, it is called an onto function. * The American Council on Education's College Credit Recommendation Service (ACE Credit®) has evaluated and recommended college credit for 33 of Sophia’s online courses. This article … … The function, g, is called the inverse of f, and is denoted by f -1. On peut donc définir une application g allant de Y vers X, qui à y associe son unique antécédent, c'est-à-dire que . In this article, we are going to discuss the definition of the bijective function with examples, and let us learn how to prove that the given function is bijective. For instance, x = -1 and x = 1 both give the same value, 2, for our example. De nition 2. If $$f : A \to B$$ is bijective, then it has an inverse function $${f^{-1}}.$$ Figure 3. An inverse function is a function such that and . We mean that it is a mapping from the set of real numbers to itself, that is f maps R to R. But does f map all of R to all of R, that is, are there any numbers in the range that cannot be mapped by f? Here is what I mean. If a function f is invertible, then both it and its inverse function f−1 are bijections. Seules les fonctions bijectives (à un correspond une seule image ) ont des inverses. Are there any real numbers x such that f(x) = -2, for example? Let $$f : A \rightarrow B$$ be a function. ƒ(g(y)) = y.L'application g est une bijection, appelée bijection réciproque de ƒ. Inverse of a Bijective Function Watch Inverse of a Bijective Function explained in the form of a story in high quality animated videos. Sophia partners Let A = R − {3}, B = R − {1}. In general, a function is invertible as long as each input features a unique output. The natural logarithm function ln : (0,+∞) → R is a surjective and even bijective (mapping from the set of positive real numbers to the set of all real numbers). it's pretty obvious that in the case that the domain of a function is FINITE, f-1 is a "mirror image" of f (in fact, we only need to check if f is injective OR surjective). We summarize this in the following theorem. Injections may be made invertible Odu - Inverse of a Bijective Function open_in_new . you might be saying, "Isn't the inverse of x2 the square root of x? Now forget that part of the sequence, find another copy of 1, − 1 1,-1 1, − 1, and repeat. The converse is also true. A function is bijective if and only if has an inverse November 30, 2015 De nition 1. Some people call the inverse sin − 1, but this convention is confusing and should be dropped (both because it falsely implies the usual sine function is invertible and because of the inconsistency with the notation sin 2 It means that every element “b” in the codomain B, there is exactly one element “a” in the domain A. such that f(a) = b. Therefore, we can find the inverse function $$f^{-1}$$ by following these steps: If X and Y are finite sets, then the existence of a bijection means they have the same number of elements. The inverse function is a function which outputs the number you should input in the original function to get the desired outcome. Click hereto get an answer to your question ️ If A = { 1,2,3,4 } and B = { a,b,c,d } . The answer is "yes and no." one to one function never assigns the same value to two different domain elements. In this video we see three examples in which we classify a function as injective, surjective or bijective. According to what you've just said, x2 doesn't have an inverse." Bijections and inverse functions are related to each other, in that a bijection is invertible, can be turned into its inverse function by reversing the arrows.. If the function satisfies this condition, then it is known as one-to-one correspondence. QnA , Notes & Videos & sample exam papers However, in the more general context of category theory, the definition of a monomorphism differs from that of an injective homomorphism. The inverse is usually shown by putting a little "-1" after the function name, like this: f-1 (y) We say "f inverse of y" So, the inverse of f(x) = 2x+3 is written: f-1 (y) = (y-3)/2 (I also … Click hereto get an answer to your question ️ Let y = g(x) be the inverse of a bijective mapping f:R→ Rf(x) = 3x^3 + 2x The area bounded by graph of g(x) the x - axis and the ordinate at x = 5 is: LEARNING APP; ANSWR; CODR; XPLOR; SCHOOL OS; answr. Now we must be a bit more specific. Recall that a function which is both injective and surjective is called bijective. A function is invertible if and only if it is a bijection. A function g : B !A is the inverse of f if f g = 1 B and g f = 1 A. If f: A → B be defined by f (x) = x − 3 x − 2 ∀ x ∈ A. A bijective group homomorphism $\phi:G \to H$ is called isomorphism. Sophia’s self-paced online courses are a great way to save time and money as you earn credits eligible for transfer to many different colleges and universities.*. (See also Inverse function.). Is f bijective? Think about the following statement: "The inverse of every function f can be found by reflecting the graph of f in the line y=x", is it true or false? To define the concept of a surjective function One of the examples also makes mention of vector spaces. Let P = {y ϵ N: y = 3x - 2 for some x ϵN}. An inverse function goes the other way! Show that a function, f : N, P, defined by f (x) = 3x - 2, is invertible, and find, Z be two invertible (i.e. Show that f is bijective and find its inverse. Suppose that f(x) = x2 + 1, does this function an inverse? Read Inverse Functions for more. here is a picture: When x>0 and y>0, the function y = f(x) = x2 is bijective, in which case it has an inverse, namely, f-1(x) = x1/2. To prove that g o f is invertible, with (g o f)-1 = f -1o g-1. The function f is bijective if and only if it admits an inverse function, that is, a function : → such that ∘ = and ∘ =. ... Non-bijective functions. One to One Function. For all common algebraic structures, and, in particular for vector spaces, an injective homomorphism is also called a monomorphism. We will think a bit about when such an inverse function exists. bijective) functions. These theorems yield a streamlined method that can often be used for proving that a function is bijective and thus invertible. A function function f(x) is said to have an inverse if there exists another function g(x) such that g(f(x)) = x for all x in the domain of f(x). Let f : A !B. In an inverse function, the role of the input and output are switched. The inverse can be determined by writing y = f (x) and then rewrite such that you get x = g (y). you might be saying, "Isn't the inverse of. Hence, to have an inverse, a function $$f$$ must be bijective. Show that f: − 1, 1] → R, given by f (x) = (x + 2) x is one-one. A function is said to be bijective or bijection, if a function f: A → B satisfies both the injective (one-to-one function) and surjective function (onto function) properties. We can, therefore, define the inverse of cosine function in each of these intervals. with infinite sets, it's not so clear. A bijective group homomorphism $\phi:G \to H$ is called isomorphism. This is equivalent to the following statement: for every element b in the codomain B, there is exactly one element a in the domain A such that f(a)=b.Another name for bijection is 1-1 correspondence (read "one-to-one correspondence).. The Attempt at a Solution To start: Since f is invertible/bijective f⁻¹ is … which discusses a few cases -- when your function is sufficiently polymorphic -- where it is possible, completely automatically to derive an inverse function. SOPHIA is a registered trademark of SOPHIA Learning, LLC. Let -2 ∈ B.Then fog(-2) = f{g(-2)} = f(2) = -2. The figure shown below represents a one to one and onto or bijective function. Next keyboard_arrow_right. The proof that isomorphism is an equivalence relation relies on three fundamental properties of bijective functions (functions that are one-to-one and onto): (1) every identity function is bijective, (2) the inverse of every bijective function is also bijective, (3) the composition of two bijective functions is bijective. The figure given below represents a one-one function. Now this function is bijective and can be inverted. If f : X → Y is surjective and B is a subset of Y, then f(f −1 (B)) = B. {text} {value} {value} Questions. Properties of inverse function are presented with proofs here. Let f : A !B. A function f : A -> B is called one – one function if distinct elements of A have distinct images in B. Likewise, this function is also injective, because no horizontal line will intersect the graph of a line in more than one place. A function is one to one if it is either strictly increasing or strictly decreasing. An inverse function goes the other way! Ask Question Asked 6 years, 1 month ago. Then g o f is also invertible with (g o f)-1 = f -1o g-1. The function f is called as one to one and onto or a bijective function if f is both a one to one and also an onto function . Don’t stop learning now. Properties of Inverse Function. It is clear then that any bijective function has an inverse. Let $$f : A \rightarrow B$$ be a function. A bijective function sets up a perfect correspondence between two sets, the domain and the range of the function - for every element in the domain there is one and only one in the range, and vice versa. injective function. A function, f: A → B, is said to be invertible, if there exists a function, g : B → A, such that g o f = IA and f o g = IB. If every "A" goes to a unique "B", and every "B" has a matching "A" then we can go back and forwards without being led astray. Onto Function. Institutions have accepted or given pre-approval for credit transfer. For onto function, range and co-domain are equal. The inverse of an injection f: X → Y that is not a bijection (that is, not a surjection), is only a partial function on Y, which means that for some y ∈ Y, f −1(y) is undefined. The term one-to-one correspondence should not be confused with the one-to-one function (i.e.) If (as is often done) ... Every function with a right inverse is necessarily a surjection. If a function f is not bijective, inverse function of f cannot be defined. A bijection of a function occurs when f is one to one and onto. Then g is the inverse of f. The inverse of a bijective holomorphic function is also holomorphic. In this case, g(x) is called the inverse of f(x), and is often written as f-1(x). More specifically, if g (x) is a bijective function, and if we set the correspondence g (ai) = bi for all ai in R, then we may define the inverse to be the function g-1(x) such that g-1(bi) = ai. Let f : A !B. Bijection, or bijective function, is a one-to-one correspondence function between the elements of two sets. More specifically, if, "But Wait!" Show that a function, f : N → P, defined by f (x) = 3x - 2, is invertible, and find f-1. guarantee So if f (x) = y then f -1 (y) = x. Find the domain range of: f(x)= 2(sinx)^2-3sinx+4. To define the concept of a bijective function Why is the reflection not the inverse function of ? Let y = g (x) be the inverse of a bijective mapping f: R → R f (x) = 3 x 3 + 2 x The area bounded by graph of g(x) the x-axis and the … When no horizontal line intersects the graph at more than one place, then the function usually has an inverse. Click here if solved 43 ... Also find the inverse of f. View Answer. Property 1: If f is a bijection, then its inverse f -1 is an injection. Let f : A ----> B be a function. Proof of Property 1: Suppose that f -1 (y 1) = f -1 (y 2) for some y 1 and y 2 in B. If, for an arbitrary x ∈ A we have f(x) = y ∈ B, then the function, g: B → A, given by g(y) = x, where y ∈ B and x ∈ A, is called the inverse function of f. f(2) = -2, f(½) = -2, f(½) = -½, f(-1) = 1, f(-1/9) = 1/9, g(-2) = 2, g(-½) = 2, g(-½) = ½, g(1) = -1, g(1/9) = -1/9. The term bijection and the related terms surjection and injection … If a function f is not bijective, inverse function of f cannot be defined. Non-bijective functions and inverses. Formally: Let f : A → B be a bijection. maths. Let’s define $f \colon X \to Y$ to be a continuous, bijective function such that $X,Y \in \mathbb R$. Imaginez une ligne verticale qui se … Bijective functions have an inverse! It turns out that there is an easy way to tell. Its inverse function is the function $${f^{-1}}:{B}\to{A}$$ with the property that $f^{-1}(b)=a \Leftrightarrow b=f(a).$ The notation $$f^{-1}$$ is pronounced as “$$f$$ inverse.” See figure below for a pictorial view of an inverse function. {id} Review Overall Percentage: {percentAnswered}% Marks: {marks} {index} {questionText} {answerOptionHtml} View Solution {solutionText} {charIndex}. Give reasons. A function is bijective if and only if has an inverse November 30, 2015 De nition 1. relationship from elements of one set X to elements of another set Y (X and Y are non-empty sets Viewed 9k times 17. This function g is called the inverse of f, and is often denoted by . The inverse function is not hard to construct; given a sequence in T n T_n T n , find a part of the sequence that goes 1, − 1 1,-1 1, − 1. Active 5 months ago. Let’s define $f \colon X \to Y$ to be a continuous, bijective function such that $X,Y \in \mathbb R$. When a function, such as the line above, is both injective and surjective (when it is one-to-one and onto) it is said to be bijective. This article is contributed by Nitika Bansal. If every "A" goes to a unique "B", and every "B" has a matching "A" then we can go back and forwards without being led astray. Let 2 ∈ A.Then gof(2) = g{f(2)} = g(-2) = 2. (It also discusses what makes the problem hard when the functions are not polymorphic.) On A Graph . "But Wait!" [31] (Contrarily to the case of surjections, this does not require the axiom of choice. That way, when the mapping is reversed, it'll still be a function! In mathematics, a bijective function or bijection is a function f : A → B that is both an injection and a surjection. Let f: A → B be a function. The function f is called as one to one and onto or a bijective function, if f is both a one to one and an onto function. show that f is bijective. We denote the inverse of the cosine function by cos –1 (arc cosine function). Inverse Functions. Let us start with an example: Here we have the function f(x) = 2x+3, written as a flow diagram: The Inverse Function goes the other way: So the inverse of: 2x+3 is: (y-3)/2 . keyboard_arrow_left Previous. 1-1 (proof is in textbook) Induced Functions on Sets: Given a function , it naturally induces two functions on power sets: the forward function defined by for any set Note that is simply the image through f of the subset A. the pre-image … Bijection, or bijective function, is a one-to-one correspondence function between the elements of two sets. 20 … Theorem 9.2.3: A function is invertible if and only if it is a bijection. Please Subscribe here, thank you!!! The answer is no, there are not - no matter what value we plug in for x, the value of f(x) is always positive, so we can never get -2. So let us see a few examples to understand what is going on. Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L1, L2)}: L1 is parallel to L2. Bijective Functions and Function Inverses, Domain, Range, and Back Again: A Function's Tale, Before beginning this packet, you should be familiar with, When a function is such that no two different values of, A horizontal line intersects the graph of, Now we must be a bit more specific. If, for an arbitrary x ∈ A we have f(x) = y ∈ B, then the function, g: B →, B, is said to be invertible, if there exists a function, g : B, The function, g, is called the inverse of f, and is denoted by f, Let P = {y ϵ N: y = 3x - 2 for some x ϵN}. A function f : A -> B is said to be onto function if the range of f is equal to the co-domain of f. How to Prove a Function is Bijective … Summary; Videos; References; Related Questions. Join Now. Yes. View Inverse Trigonometric Functions-4.pdf from MATH 2306 at University of Texas, Arlington. No matter what function f we are given, the induced set function f − 1 is defined, but the inverse function f − 1 is defined only if f is bijective. In other words, an injective function can be "reversed" by a left inverse, but is not necessarily invertible, which requires that the function is bijective. Hence, the composition of two invertible functions is also invertible. Let us start with an example: Here we have the function f(x) = 2x+3, written as a flow diagram: The Inverse Function goes the other way: So the inverse of: 2x+3 is: (y-3)/2 . Then f is bijective if and only if the inverse relation $$f^{-1}$$ is a function from B to A. Notice that the inverse is indeed a function. In some cases, yes! Explore the many real-life applications of it. Below f is a function from a set A to a set B. l o (m o n) = (l o m) o n}. Also find the identity element of * in A and Prove that every element of A is invertible. 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( B ) =a », l'une verticale, l'autre horizontale, does this function an inverse. «. Fonctions bijectives ( à un correspond une seule image ) ont des inverses B.Then fog ( -2 }. 2 ( sinx ) ^2-3sinx+4 further, if it is both injective and surjective is the... Start: since f is bijective figure shown below represents a one to one and onto or function... Often will on some restriction of the examples also makes mention of vector spaces, an invertible because... Can be inverted have inverse function, range and co-domain are equal domain! A one-one function is bijective and find its inverse. to see, how to check these. To tell the mapping is reversed, it covers '' all real numbers is n't inverse. Assigns the same number of elements if ( as is often done )... every function with a inverse. B be defined also known as invertible function because they have the same value,,. Both give the same number of elements bijective if and only if an. A registered trademark of sophia learning, LLC clear then that any bijective function has an inverse than one,. ( x ) = x − 2 ∀ x ∈ a test des deux lignes », l'une verticale l'autre... Functions: bijection function are presented with proofs here the figure shown below represents a one to one, f!
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https://math.stackexchange.com/questions/2332120/handshake-problem-each-person-can-have-5-handshakes | # Handshake problem, each person can have 5 handshakes
There are a total of $60$ handshakes in a party. Each person can only shake hands with $5$ other people. So, how many people are there? Is the answer $24$?
A group of $6$ is formed. One group can have $1+2+3+4+5=15$ handshakes. $60/15=4$ Hence, number of people is $4\times 6= 24$.
• If a person doesn't have to shake hands with five others, but is allowed to do fewer, then there could be more people. For instance, if everyone shakes hands with only one other person, then there are $120$ people. So $24$ is only the correct answer if each person must shake hands with exactly $5$ other people. – Arthur Jun 22 '17 at 8:39
There are many ways to arrange the graph with 60 edges and having each vertex of degree five. You simply chooses two edges from different subcliques in your graph and delete current edges and cross-connect their ends (replace edges $(a,b),(c,d)$ with $(a,c),(b,d)$ ), what you get is a graph with the same number of edges and vertices and same vertex degrees, but different from your original graph. And you can repeat this step many times to create a lot of different graphs. | 2021-06-19T01:03:23 | {
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http://mathhelpforum.com/algebra/96698-solved-cool-sequence.html | # Math Help - [SOLVED] Cool Sequence
1. ## [SOLVED] Cool Sequence
Let us call a sum of integers cool if the first and last terms are 1 and each term differs from its neighbours by at most 1. For example, the sum 1+2+3+4+3+2+3+3+3+2+3+3+2+1 is cool. How many terms does it take to write 2008 as a cool sum if we use no more terms than necessary?
I found there are 89 terms needed. Maybe there is another that lesser?
Thx
2. 1004 + 1004
669 + 670 + 669
3. Originally Posted by Wilmer
1004 + 1004
669 + 670 + 669
the first and last terms need to be 1
4. Ahhh....thanks.
89 I'm sure is correct: 2(sum 1 to 44) + 28
1+2+3....+27+28+28+29....+43+44+44+43....+3+2+1
A formula to find could be as a general case:
(n = number of terms, x = number represented):
n(n+1) = x
n^2 + n - x = 0
n = INT[SQRT(4x + 1) - 1]
n = INT[SQRT(4*2008 + 1) = 88
Add 1 if n(n + 2) / 4 < x
5. Originally Posted by Wilmer
n = INT[SQRT(4x + 1) - 1]
n = INT[SQRT(4*2008 + 1) = 88
Add 1 if n(n + 2) / 4 < x
Sorry, i don't get this part..
thx
6. Hello, songoku!
A fascinating problem . . .
Let us call a sum of integers cool if the first and last terms are 1
. . and each term differs from its neighbours by at most 1.
For example, the sum $1+2+3+4+3+2+3+3+3+2+3+3+2+1$ is cool.
How many terms does it take to write 2008 as a cool sum
. . if we use no more terms than necessary?
I found there are 89 terms needed. Maybe there is another that's lesser?
I believe that 89 terms is the minimum, but I have no proof.
The "fastest" way to total 2008 seems to be consecutive integers:
. . $\bigg(1 + 2 + 3 + \hdots + 44\bigg) + 45 + \bigg(44 + 43 + 42 + \hdots + 1\bigg) \;=\;2025$
The sum is slightly large.
I can remove 17 from the sum and maintain its coolness:
. . $\bigg(1+2+ \hdots + 40\bigg) + 40 + \underbrace{\bigg(41 + 41 + \hdots + 41\bigg)}_{\text{8 terms}} + \bigg(40 + 39 + \hdots + 1\bigg) \;=\; 2008$
. . but I cannot reduce the number of terms.
7. Originally Posted by songoku
Sorry, i don't get this part..
> n = INT[SQRT(4*2008 + 1) = 88
> Add 1 if n(n + 2) / 4 < x
88(90)/4 = 1980 ; lesser than 2008, so need another term;
in some cases, n(n + 2) / 4 will equal x, then no need for extra term.
8. Originally Posted by Soroban
I can remove 17 from the sum and maintain its coolness
No; then you'll have ...16,18: difference > 1
9. Originally Posted by Wilmer
n = INT[SQRT(4*2008 + 1) = 88
Now i get it that you used quadratic formula. I think you miss the term -1 here. And what is INT? I guess it's rounding down?
Originally Posted by Wilmer
No; then you'll have ...16,18: difference > 1
I think Soroban doesn't mean to take the term "17" out. Instead, he means that he can take out the terms that sum up to 17 because 2025 - 2008 = 17
10. I know this question. It's from the Australian Mathematics Competition, I know because I have it right here in front of me
Here's how I did it.
First you have to find up to what number you add will get you closet to 2008 (sorry if this doesn't make sense just keep reading).
I guess 40. (so 1+2+3.....+40+40+39....+1)
Use this formula:
$\frac {n(n+1)}{2}$
Where n is 40. Substitute the values in then multiply by 2. Now you just figured out what 1+2+3.....+40+40+39....+1 is. It is 1640. Is it near 2008? No?
Now guess another one. I guessed 44. Substitute again and wa la we get 1980. Since that is the closest we can get to 2008 just subtract 1980 from 2008. The answer is 28.
So how many terms? We have 1....44 then 44...1 that's 88 then add the last 28 which makes 89!!!
11. Looks like you guys don't follow what I've "tried" to show.
I'm also using that formula, jgv; but no guessing; we want the series
to add to half the number, or to 2008/2 = 1004:
n(n+1) / 2 = 1004
n^2 + n = 2008
n^2 + n - 2008 = 0
n = [-1 + sqrt(1 + 8032)] / 2
n = 44.3135024....
So n = FLOOR(n) = 44 ; I used n = INT(n) ... same thing
To wrap up, since that's for half the number, then n = 2n = 88;
since n <> FLOOR(n), add 1: n = 2n + 1 = 89
So a GENERAL CASE formula is (let the given number 2008 = x):
n(n+1) / 2 = x / 2
n^2 + n = x
n^2 + n - x = 0
n = [-1 + SQRT(1 + 4x)] / 2
IF FLOOR(n) = n
then
n= 2[FLOOR(n)]
else
n = 2[FLOOR(n)] + 1
12. Thx a lot Wilmer, jgv115, and Soroban ^^ | 2014-09-02T12:10:41 | {
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http://math.stackexchange.com/questions/250920/open-cover-of-a-metric-space | # Open cover of a metric space
I am trying to find a definition for the open cover of a metric space, but i cannot find it. So, if X is a metric space and A is a subset of X, then what is the definition for open cover of A? Can anyone help? Thank you.
-
Do you know what an open set (in a metric space) is? – Adam Saltz Dec 4 '12 at 19:59
An open cover of $A$ in $X$ is simply a family $\mathscr{U}$ of open sets in $X$ such that $A\subseteq\bigcup\mathscr{U}$. A relatively open cover of $A$ as a subspace of $X$ is a family $\mathscr{U}$ of open sets in $A$, i.e., of sets of the form $U\cap A$ for some open $U$ in $X$.
Example: Let $X=\Bbb R$, and let $A=(0,1)$. Then $$\mathscr{U}=\left\{\left(\frac1n,1\right):n\in\Bbb Z^+\right\}$$ is an open cover of $A$, because each $x\in A$ belongs to at least one member of $\mathscr{U}$. Specifically, if $x\in A$, then $x>0$, so there is a positive integer $n$ such that $\frac1n<x$, and $x\in\left(\frac1n,1\right)\in\mathscr{U}$.
Finally, a set $U\subseteq X$ is open if and only if it is a union of open balls: for each $x\in U$ there is an $\epsilon_x>0$ such that $B(x,\epsilon_x)\subseteq U$, where $B(x,\epsilon_x)=\{y\in X:d(x,y)<\epsilon_x\}$.
-
Thanks, when you say "family of" open sets, do you mean union of open sets? – bigO Dec 4 '12 at 20:02
@user49065: No, family, collection, and set all mean the same thing: $\mathscr{U}$ is a set of open sets in $X$. I used family only because a set of sets sounds awkward. – Brian M. Scott Dec 4 '12 at 20:06
@user49065: I’ve added a concrete example that may help make it clearer. – Brian M. Scott Dec 4 '12 at 20:14
Hi Brian. Did you mean to say $A\subseteq \cup \mathscr{U}$ in the first line? And $X$ instead of $M$? – T. Eskin Dec 5 '12 at 12:36
@ThomasE.: Yes, and yes; and thanks for catching them. – Brian M. Scott Dec 5 '12 at 21:19
An open cover of a subset $A \subseteq X$, is a collection $\{U_i\}_{i \in I}$ of open sets in $X$, such that
$$A \subseteq \bigcup_{i \in I} U_i$$
- | 2016-05-26T05:08:14 | {
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https://math.stackexchange.com/questions/1978304/what-is-the-largest-perfect-square-that-divides-20143-2013320123-20113-ld | # What is the largest perfect square that divides $2014^3-2013^3+2012^3-2011^3+\ldots+2^3-1^3$
I've tried this but didn't get the answer : Let $S=2014^3-2013^3+2012^3-2011^3+\ldots+2^3-1^3$
Using $n^3-(n-1)^3 = 3n^2-3n+1$, \begin{align} S &= 3(2014^2)-3(2014)+1+3(2012^2)-3(2012)+1+\ldots+3(2^2)-3(2)+1 \\&= 3\left ( 2014(2013)+2012(2011)+2010(2009)+ \ldots+2(1) \right ) + 1(1007) \\&= 3\left ( \sum_{n=1}^{1007}2n(2n-1) \right )+1007\\ =& \left ( \sum_{n=1}^{1007}4n^2-\sum_{n=1}^{1007}2n \right )+1007 \\=&\frac{12(1007)(1008)(2015)}{6}-\frac{2(1007)(1008)(3)}{2}+1007 \end{align} This is divisible by $1007$ but not by $1007^2$ which is the correct answer. Where have I gone wrong ?
• $$\sum_{k = 1}^{2n} (-1)^k k^3 = 2\sum_{k = 1}^n (2k)^3 - \sum_{k = 1}^{2n} k^3 = 16 \frac{n^2(n+1)^2}{4} - \frac{(2n)^2(2n+1)^2}{4} = n^2\bigl(4(n+1)^2 - (2n+1)^2\bigr) = n^2(4n+3).$$ – Daniel Fischer Oct 21 '16 at 11:12
\begin{align}&\frac{12(1007)(1008)(2015)}{6}-\frac{2(1007)(1008)(3)}{2}+1007 \\&=1007\left(\frac{12(1008)(2015)}{6}-\frac{2(1008)(3)}{2}+ 1\right) \\&=1007\left(2(1008)(2015)-(1008)(3)+ 1\right) \\&=1007\left(2(1007+1)(2015)-(1007+1)(3)+ 1\right) \\&=1007\left(1007(2(2015)-3)+2(2015)-3+ 1\right) \\&=1007\left(1007(2(2015)-3)+2(2015)-2\right) \\&=1007\left(1007(2(2015)-3)+2(2014)\right) \\&=1007\left(1007(2(2015)-3)+4(1007)\right) \\&=1007^2(2(2015)-3+4) \\&=1007^2(4031)\end{align}
• Thanks. I didn't knew I was this close! Is it possible to factorize this without using a calculator ? – H G Sur Oct 21 '16 at 8:23
In general you obtain, for $n^3-(n-1)^3+\cdots + 2^3-1^3$ and $n$ even the formula $$3\left(\sum_{k=1}^{n/2} 2k(2k-1)\right)+\frac{n}{2}.$$ Now it is easier to see that this is divisible by $(\frac{n}{2})^2$; and you can test this first for $n=2,4,6,\ldots$ before dealing with $n=2014$. In fact, your computation is correct, except for the last step, where you did not realize how to split of the factor $(\frac{n}{2})^2$. Working with general $n$, you do not need a calculator.
• Thanks, isn't the summation from k=1 to n/2 ? – H G Sur Oct 21 '16 at 8:25
• Oh sure, I wasn't finished with the formula yet, sorry. – Dietrich Burde Oct 21 '16 at 8:26
• Is it a property that it is divisible by $\left (\frac{n}{2} \right )^2$? – H G Sur Oct 21 '16 at 8:27
• Yes, it is. Take $n=4$. Then the formula gives $44$, which is divisible by $(4/2)^2=4$. Now assume it holds for $n$, and show it for $n+2$. – Dietrich Burde Oct 21 '16 at 8:29
• Yes, you have given one already. You only need to continue with $n^3-(n-1)^3=3n^2-3n+1$, which is true for all $n$. Divisibility, however, is something different then. – Dietrich Burde Oct 21 '16 at 9:55
\begin{align} \sum_{n=1}^{2m}(-1)^n n^3&=\sum_{n=1}^m (2n)^3-(2n-1)^3\\ &=\sum_{n=1}^m 12n^2-6n+1\\ &=\sum_{n=1}^m 24 \binom n2+6\binom n1+1\\ &=24\binom {m+1}3+6\binom {m+1}2+m\\ &=m\; \big[\;4(m+1)(m-1)+3(m+1)+1\;\big]\\ &=m^2(4m+3)\\ \end{align} which is divisible by $m^2$.
Putting $m=1007$ gives the answer required. | 2019-06-19T06:47:40 | {
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https://math.stackexchange.com/questions/676903/combinatorics-domain-names | # Combinatorics Domain Names
I am working on the following problem and was wondering if people could check what I currently have as well as offer advice on how to do the last part of this problem:
"As of April $2006$, roughly $50$ million .com web domain names were registered (e.g., yahoo.com).
a. How many domain names consisting of just two letters in sequence can be formed? How many domain names of length two are there if digits, as well as letters, are permitted as characters? [Note: A character length of three or more is now mandated.]
b. How many domain names are there consisting of three letters in sequence? How many of this length are there if either letters or digits are permitted? [Note: All are currently taken.]
c. Answer the questions posed in (b) for four-character sequences.
d. As of April $2006$, $97,786$ of the four-character sequences using either letters or digits had not yet been claimed. If a four-character name is randomly selected, what is the probability that it is already owned?
What I was wondering was, say that I wanted to know the number of domains where lexicographic ordering was of concern; that is, if b was the first letter in the domain name, then a couldn't be the next one, only letters that follow b. Would I count in this manner, $25⋅24+24⋅23+23⋅22+22⋅21+...3⋅2+2⋅1=5152$? I'm sure there is an alternate way, too. Would it involve me finding the total number of domains, both in lexicographic order and not in lexicographic order, and subtracting something from that, right? Would $263=17576$ be the total number of combinations?"
I've got the following at the moment (please correct me if I'm wrong):
a. $\binom{26}2$ $\binom{36}2$
b. $\binom{26}3$ $\binom{36}3$
c. $\binom{26}4$ $\binom{36}4$
d. No idea.
For question a) only letters: $26^2$, because you have $26$ choices twice. If digits are allowed: $36^2$.
b) $26^3$ and $36^3$
d) There are $1679616$ four-character sequences. $\frac{97786}{1679616}=0.0582$. Thus, the probability of picking an occupied domain is roughly $6\%$.
For your additional question: The total number of two letter strings is $26^2$. The number of two letter strings with two equal characters is just $26$. The total number of two letter strings in lexicographic order is $\frac{26^2-26}2$, because exactly half of the strings with two different letters is in lexicographic order.
a little correction in (d) part, he is asking for prob. of picking already owned domain which will be $1-P(\mbox{non owned}) = (36^4 - 97786)/36^4 = .9417$ | 2020-04-08T22:44:53 | {
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https://it.mathworks.com/help/symbolic/chebyshevu.html | # chebyshevU
Chebyshev polynomials of the second kind
## Description
example
chebyshevU(n,x) represents the nth degree Chebyshev polynomial of the second kind at the point x.
## Examples
### First Five Chebyshev Polynomials of the Second Kind
Find the first five Chebyshev polynomials of the second kind for the variable x.
syms x
chebyshevU([0, 1, 2, 3, 4], x)
ans =
[ 1, 2*x, 4*x^2 - 1, 8*x^3 - 4*x, 16*x^4 - 12*x^2 + 1]
### Chebyshev Polynomials for Numeric and Symbolic Arguments
Depending on its arguments, chebyshevU returns floating-point or exact symbolic results.
Find the value of the fifth-degree Chebyshev polynomial of the second kind at these points. Because these numbers are not symbolic objects, chebyshevU returns floating-point results.
chebyshevU(5, [1/6, 1/3, 1/2, 2/3, 4/5])
ans =
0.8560 0.9465 0.0000 -1.2675 -1.0982
Find the value of the fifth-degree Chebyshev polynomial of the second kind for the same numbers converted to symbolic objects. For symbolic numbers, chebyshevU returns exact symbolic results.
chebyshevU(5, sym([1/6, 1/4, 1/3, 1/2, 2/3, 4/5]))
ans =
[ 208/243, 33/32, 230/243, 0, -308/243, -3432/3125]
### Evaluate Chebyshev Polynomials with Floating-Point Numbers
Floating-point evaluation of Chebyshev polynomials by direct calls of chebyshevU is numerically stable. However, first computing the polynomial using a symbolic variable, and then substituting variable-precision values into this expression can be numerically unstable.
Find the value of the 500th-degree Chebyshev polynomial of the second kind at 1/3 and vpa(1/3). Floating-point evaluation is numerically stable.
chebyshevU(500, 1/3)
chebyshevU(500, vpa(1/3))
ans =
0.8680
ans =
0.86797529488884242798157148968078
Now, find the symbolic polynomial U500 = chebyshevU(500, x), and substitute x = vpa(1/3) into the result. This approach is numerically unstable.
syms x
U500 = chebyshevU(500, x);
subs(U500, x, vpa(1/3))
ans =
63080680195950160912110845952.0
Approximate the polynomial coefficients by using vpa, and then substitute x = sym(1/3) into the result. This approach is also numerically unstable.
subs(vpa(U500), x, sym(1/3))
ans =
-1878009301399851172833781612544.0
### Plot Chebyshev Polynomials of the Second Kind
Plot the first five Chebyshev polynomials of the second kind.
syms x y
fplot(chebyshevU(0:4, x))
axis([-1.5 1.5 -2 2])
grid on
ylabel('U_n(x)')
legend('U_0(x)', 'U_1(x)', 'U_2(x)', 'U_3(x)', 'U_4(x)', 'Location', 'Best')
title('Chebyshev polynomials of the second kind')
## Input Arguments
collapse all
Degree of the polynomial, specified as a nonnegative integer, symbolic variable, expression, or function, or as a vector or matrix of numbers, symbolic numbers, variables, expressions, or functions.
Evaluation point, specified as a number, symbolic number, variable, expression, or function, or as a vector or matrix of numbers, symbolic numbers, variables, expressions, or functions.
collapse all
### Chebyshev Polynomials of the Second Kind
• Chebyshev polynomials of the second kind are defined as follows:
$U\left(n,x\right)=\frac{\mathrm{sin}\left(\left(n+1\right)a\mathrm{cos}\left(x\right)\right)}{\mathrm{sin}\left(a\mathrm{cos}\left(x\right)\right)}$
These polynomials satisfy the recursion formula
$U\left(0,x\right)=1,\text{ }U\left(1,x\right)=2\text{ }x,\text{ }U\left(n,x\right)=2\text{ }x\text{ }U\left(n-1,x\right)-U\left(n-2,x\right)$
• Chebyshev polynomials of the second kind are orthogonal on the interval -1 ≤ x ≤ 1 with respect to the weight function $w\left(x\right)=\sqrt{1-{x}^{2}}$.
• Chebyshev polynomials of the second kind are a special case of the Jacobi polynomials
$U\left(n,x\right)=\frac{{2}^{2n}n!\left(n+1\right)!}{\left(2n+1\right)!}P\left(n,\frac{1}{2},\frac{1}{2},x\right)$
and Gegenbauer polynomials
$U\left(n,x\right)=G\left(n,1,x\right)$
## Tips
• chebyshevU returns floating-point results for numeric arguments that are not symbolic objects.
• chebyshevU acts element-wise on nonscalar inputs.
• At least one input argument must be a scalar or both arguments must be vectors or matrices of the same size. If one input argument is a scalar and the other one is a vector or a matrix, then chebyshevU expands the scalar into a vector or matrix of the same size as the other argument with all elements equal to that scalar.
## References
[1] Hochstrasser, U. W. “Orthogonal Polynomials.” Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. (M. Abramowitz and I. A. Stegun, eds.). New York: Dover, 1972. | 2020-12-05T14:47:26 | {
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http://math.stackexchange.com/questions/96826/the-monty-hall-problem/96886 | # The Monty Hall problem
I was watching the movie 21 yesterday, and in the first 15 minutes or so the main character is in a classroom, being asked a "trick" question (in the sense that the teacher believes that he'll get the wrong answer) which revolves around theoretical probability.
The question goes a little something like this (I'm paraphrasing, but the numbers are all exact):
You're on a game show, and you're given three doors. Behind one of the doors is a brand new car, behind the other two are donkeys. With each door you have a $1/3$ chance of winning. Which door would you pick?
The character picks A, as the odds are all equally in his favor.
The teacher then opens door C, revealing a donkey to be behind there, and asks him if he would like to change his choice. At this point he also explains that most people change their choices out of fear; paranoia; emotion and such.
The character does change his answer to B, but because (according to the movie), the odds are now in favor of door B with a $1/3$ chance of winning if door A is picked and $2/3$ if door B is picked.
What I don't understand is how removing the final door increases the odds of winning if door B is picked only. Surely the split should be 50/50 now, as removal of the final door tells you nothing about the first two?
I assume that I'm wrong; as I'd really like to think that they wouldn't make a movie that's so mathematically incorrect, but I just can't seem to understand why this is the case.
So, if anyone could tell me whether I'm right; or if not explain why, I would be extremely grateful.
-
This is known as the Monty Hall problem. The point is that your odds of winning with the original door have not changed. Since the total odds have to add up to 1, the odds of $B$ being the correct door are now 2/3. In fact, "switching to B" is equivalent to "pick the best of whatever is behind doors B and C" (you know that what is behind B is no worse than what has been revealed behind C), which clearly gives you a 2/3rds odds of winning. The precise conditions of the game are very important, though. – Arturo Magidin Jan 6 '12 at 3:27
I would expect that this has been asked before; I found two questions asking about variants (here and here), but not the plain question. I'm probably just not finding it? – Arturo Magidin Jan 6 '12 at 3:33
@ArturoMagidin I looked as well and was shocked not to find one. – Alex Becker Jan 6 '12 at 3:37
I'd suggest that the question is ambiguous as commonly stated. It doesn't specify whether the teacher selected a door at random which just happened to have a donkey behind it, or if the teacher deliberately selected a door with a donkey behind it. – Winston Ewert Feb 6 '12 at 3:43
Of course if you lived in the mountains of Nepal a donkey would be preferred to a car... – Bogatyr Sep 16 '12 at 19:02
This problem, known as the Monty Hall problem, is famous for being so bizarre and counter-intuitive. It is in fact best to switch doors, and this is not hard to prove either. In my opinion, the reason it seems so bizarre the first time one (including me) encounters it is that humans are simply bad at thinking about probability. What follows is essentially how I have justified switching doors to myself over the years.
At the start of the game, you are asked to pick a single door. There is a $1/3$ chance that you have picked correctly, and a $2/3$ chance that you are wrong. This does not change when one of the two doors you did not pick is opened. The second time is that you are choosing between whether your first guess was right (which has probability $1/3$) or wrong (probability $2/3$). Clearly it is more likely that your first guess was wrong, so you switch doors.
This didn't sit well with me when I first heard it. To me, it seemed that the situation of picking between two doors has a certain kind of symmetry-things are either behind one door or the other, with equal probability. Since this is not the case here, I was led to ask where the asymmetry comes from? What causes one door to be more likely to hold the prize than the other? The key is that the host knows which door has the prize, and opens a door that he knows does not have the prize behind it.
To clarify this, say you choose door $A$, and are then asked to choose between doors $A$ and $B$ (no doors have been opened yet). There is no advantage to switching in this situation. Say you are asked to choose between $A$ and $C$; again, there is no advantage in switching. However, what if you are asked to choose between a) the prize behind door $A$ and b) the better of the two prizes behind door $B$ and $C$. Clearly, in this case it is in your advantage to switch. But this is exactly the same problem as the one you've been confronted with! Why? Precisely because the host always opens (hence gets rid of) the door that you did not pick which has the worse prize behind it. This is what I mean when I say that the asymmetry in the situation comes from the knowledge of the host.
-
Fantastic explanation! Thanks for the clarification :). – Avicinnian Jan 6 '12 at 4:11
+1 what if you are asked to choose between a) the prize behind door A and b) the better of the two prizes behind door B and C really cements the idea. – dj18 May 2 '12 at 17:34
Good explanation. But I would say that the fact that the host knows the door which not to pick, is more a kind of asymmetry than symmetry. – Cris Stringfellow Dec 20 '12 at 0:39
@CrisStringfellow I wrote asymmetry. – Alex Becker Dec 20 '12 at 18:40
@SaaqibMahmuud The fact that there are two doors left does not imply that each one has equal probability of being the winning door. – augurar Aug 9 '14 at 7:30
To understand why your odds increase by changing door let us take an extreme example first. Say there are $10000$ doors. Behind one of them is a car and behind the rest are donkeys. Now, the odds of choosing a car is $1\over10000$ and the odds of choosing a donkey are $9999\over10000$. Say you pick a random door which we call X for now. According to the rules of the game, the game show host now opens all the doors except for two, one of which contains the car. You now have the option to switch. Since the probability for not choosing the car initially was $9999\over10000$ it is very likely you didn't choose the car. So assuming now that door X is a goat and you switch you get the car. This means that as long as you pick the goat on your first try you will always get the car.
If we return to the original problem where there are only 3 doors we see that the exact same logic applies. The probability that you choose a goat on your first try is $2\over3$ while choosing a car is $1\over3$. If your choose a goat on your first try and switch you will get a car and if you choose the car on your first try and switch you will get a goat. Thus the probability that you will get a car if you switch is $2\over3$ (which is more than the initial $1\over3$).
-
+1 This means that as long as you pick the goat on your first try you will always get the car. - really clarified the topic. – dj18 May 2 '12 at 17:36
Its simple, switching allows you to pick 2 out of the 3 doors. Choosing door number 1 and then always switching is the equivalent of saying "door number 2 or door number 3, but NOT door number 1". When you look at it that way, you should see that you have a 2/3 chance of being right, and that the reveal simply confirms which door it must be if you are right. Increase the number of doors and it should become even more obvious that saying "door 2 or 3 or 4 or 5 or ... but not 1" is the right way to bet. You have an $1-1/x$ chance of being right, and a $1/x$ chance of being wrong.
-
The person who changes his choice will win if and only if his first choice was wrong, and there is a probability of $\frac{2}{3}$ on that.
The person who does not change his choice will win if and only if his first choice was right. There is a probability of $\frac{1}{3}$ on that.
-
Merely knowing that the teacher showed a losing door does not provide any information unless one knows how the correctness of one's initial answer would influence the likelihood of the teacher showing the losing door. Consider the following four possible "strategies" for the teacher:
1. The host knows where the prize is, wants the contestant to lose, and will show an empty door only if the contestant had picked the one with the prize [if the contest already picked a wrong door, the host would reveal either the contestant's door or the one with the prize].
2. The host knows where the prize is, wants the contestant to win, and will show an empty door only if the contestant had picked the other empty door [if the contestant had already picked the right door, the host would simply show it].
3. The host knows where the prize is, and will always show an empty door [the empty door if the contestant's initial guess was wrong, or an arbitrarily-selected empty door if it was right].
4. The host picks a door at random; if it contains the prize, the contestant loses; otherwise, the contestant is allowed to switch to the other unseen door.
In the first two scenarios, the host's decision to show or not show an empty door will indicate to anyone who knows the host's strategy whether the player's guess was right or not. In the third scenario, the host's decision to offer a switch provides no information about whether the contestant's initial guess was right, but converts the 2/3 probability that the contestant's initial guess was wrong into a 2/3 probability that the prize is under the remaining door.
To evaluate the last scenario intuitively, imagine that the host draws an "X" on the player's door, flips a coin to pick a door at random and draws an "Y" on it, and finally draws a "Z" on the remaining door. If neither the host nor player has any clue as to where the prize is, doors 1, 2, and 3 will each have an equal probability of holding the prize, and the marking of letters X, Y, or Z by people who have no idea where the prize is don't change that. If the host asks the player if he'd like to switch to Z before anyone knows what's under Y, the decision will be helpful 1/3 of the time, harmful 1/3 of the time, and irrelevant 1/3 of the time. If door Y is shown to be empty, the irrelevant case will be eliminated so of the cases that remain, the other two will have 1/2 probability each.
Note: Many discussions of the "Monty Hall Paradox" assume that the host uses strategy #3, but fail to explicitly state that fact. That assumption is critically important to assessing the probability that a switch will be a winning move, since without it (depending upon the host's strategy) the probability of the prize being under the remaining door could be anything from 0% to 100%. I don't know the strategy used by the real-life game-show host for whom the "paradox" is named, but am pretty certain I've seen players revealed as winners or losers without being given a chance to switch, implying that while Monty Hall might sometimes have used strategy #3, he did not do so consistently [the normal arguments/proofs would hold if the host's decision of whether or not the player would be shown an empty door and allowed to switch was made before the player selected his door, but I have no particular reason to believe Monty Hall did things that way].
-
The movie 21 didn't state the riddle correctly. The movie failed to state the rules governing how the game show host will behave.
Assuming the riddle in the movie follows the "Monty Hall Problem" as described on Wikipedia there a few critical assumptions the movie failed to mention:
1) the host must always open a door that was not picked by the contestant 2) the host must always open a door to reveal a goat and never the car 3) the host must always offer the chance to switch between the originally chosen door and the remaining closed door.
Knowings the rules it makes the riddle much easier to understand. Many of the explanations above will suffice and Wikipedia has a good explanation.
The problem is that the movie failed to state these critical assumptions.
-
It's also necessary that if the contestant initially chooses the winning door, the host chooses either of the remaining doors with equal probability. See my comment above. – augurar Aug 9 '14 at 18:46
Let us use some theory here. Lets call the doors $0, 1, 2,$ and the right door $D$. $$P(D=0)=P(D=1)=P(D=2)=\frac13$$ $D$ is random. Now let us call the door we choose $C$. $C$ is not random. Without loss of generality, let $C=0$. Also we have $R$, the revealed door, which is random. Since the person won't reveal the right door or the one you choose, $R \neq C$ and $R \neq D$. Since we know $C$, without knowledge of $D$ we have: $$P(R=1)=P(R=2)=\frac12$$ Let us say, without loss of generality, we get the information $R=2$. Then what we are looking for is $P(D=1|R=2)$. $$\frac{P(D=1 \wedge R=2)}{P(R=2)}$$ $$\frac{P(R=2|D=1)P(D=1)}{\frac12}$$ Now, $P(R=2|D=1)=1$, since $R$ can't be 0 or 1, since those are already taken by $C$ and $D$. $$\frac{1 \cdot \frac13}{\frac12}$$ So the answer is: $\frac23=66.\overline6 \%$.
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Rather than looking at the player, I prefer to explain the paradox from the host's standpoint, as this only involves one step.
As the player gets one door, the host gets two. There are 3 possibilities with the same probability:
• donkey-donkey => leaves a donkey after a door is open
• car-donkey => leaves the car
• donkey-car => leaves the car
So in two cases out of three the door that the host leaves closed hides the car.
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Suppose you and your brother play simultaneously .You and he always choose the SAME door to start. You NEVER switch,so after your initial choice you go out for a coffee and come back after all 3 doors are open. Your chance of winning is unaffected by what happens in between.It's 1/3. Your brother ALWAYS switches. EVERY TIME YOU LOSE,HE WINS. YOU LOSE 2/3 OF THE TIME.
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The tip I've come up with to explain this problem is that you need to consider that the doors AREN'T RANDOMIZED after Monty deletes the door that is wrong. This means that although the probabilities the two remaining choices have of being correct add up to 1, the two probabilities aren't the same as before. It is wrong to state that because they had equal probabilities before, 1/3 each, they should have equal probabilities after Monty's deletion. The doors are randomized in the beginning when you have three doors, which is why you can say that each option has and equal chance of being right. However, after Monty deletes the wrong door from the other doors which you did not pick, now you are left with the probability that the door you picked is right of 1/3, and the probability that the other door is right, which is also equal to the probability that you chose the wrong door of (1/3 + 1/3 = 2/3). Well, those are my 2 cents, and I hope it helps!
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You begin with probability of 2/3 of losing.
You have more possibilities of losing with the first door, so , if you are losing (which is more probable 2/3) and you CHANGE is 100% sure that you will WIN because the remaining door IS (yes... IS) the one with the prize, because the host showed you based in your first guess and with his knowledge of the good door which was the bad door, the remaining door have more posibilities (2/3) of being the one with the prize.
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## protected by Zev ChonolesSep 30 at 17:47
Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site. | 2015-11-26T22:00:44 | {
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https://teachingcalculus.com/2021/02/09/extreme-polar-conditions/ | # Extreme Polar Conditions
Looking at the graphs of polar curves can be quite fascinating. Doing “calculus” kinds of things is different, yet the same as in Cartesian coordinates. A discussion on the AP Calculus Community here got me thinking about extreme values of polar functions.
The terms maximum and minimum here refers to the value of r(θ) which may be positive (when on the ray making an angle of θ with the polar axis), zero (at the pole), or negative (on the ray opposite to the one making an angle of θ with the polar axis). The distance from the pole to the point is |r(θ)|. As Mark Howell pointed out in the thread linked above, extreme values of r(θ) lie on a circle or circles centered at the pole with radius of |r(θ)|, and finding the slope of tangent lines at the extremes is relatively easy, requiring no calculus: the slope of the ray is y/x, so the slope of the tangent at the extreme is –x/y. As with Cartesian coordinates, at extreme values $\displaystyle dr/d\theta =0$ since r(θ) is change from increasing to decreasing at this point (or vice versa).
A quick look at the graph of simple polar functions seems to show obvious maximum values for r(θ), but a closer look reveals some complications.
The graph of $\displaystyle r\left( \theta \right)=\sin \left( {2\theta } \right)$ for $0\le \theta \le 2\pi$, shown below, appears to show 4 maximums. However if we trace the graph, we find that these points are (1, π/4), (–1, 3π/4), (1, 5π/4) and (–1, 7π/4). Two of the values are maximums where r(θ) = 1 and two are minimums where r(θ) = –1.
The graph of $r\left( \theta \right)=\cos \left( {3\theta } \right)$ )\$ for $0\le \theta \le 2\pi$ is even more fun: the graph starts at an endpoint maximum at (1,0), next goes to a minimum at (–1, π/3), then a maximum at (1, 2π/3), and a minimum at (–1, π). But this is the starting point that was a maximum. Continuing around the graph, (1, 4π/3) a maximum at the same point as the previous minimum (–1, π/3), then a minimum at (–1, 5π/3), previously a minimum, then returning to (1, 2π) = (–1, π) = (1, 0) which was previously both a maximum and a minimum, and is now a maximum again!
Thus points may be both maximums and minimums.
### On a tangent…
Polar Curves can be really fun. While working on the idea above, I explored some other curves. Try some yourself using Desmos, GeoGabra, or another graphing app with sliders. Shown below are members of the family of polar curves $r\left( \theta \right)=A\cos \left( {B\theta } \right)+C\sin \left( {D\theta } \right)$. The domain is extended to $\displaystyle 0\le \theta \le 20\pi$. Notice:
• How very slight changes in the parameters give very different looking graphs
• Other values give far less “organized” curves
• In the third graph, the maximums and minimums on the irregular part of the curve closest to the pole
.
$\displaystyle r\left( \theta \right)=3.3\cos (1.9\theta )+1.9\sin (0.1\theta )$ for $\displaystyle 0\le \theta \le 20\pi$
$\displaystyle r\left( \theta \right)=3.3\cos (2\theta )+1.9\sin (0.1\theta )$ for $\displaystyle 0\le \theta \le 20\pi$
$\displaystyle r\left( \theta \right)=3.3\cos (2\theta )+1.9\sin (3.3\theta )$ for $\displaystyle 0\le \theta \le 20\pi$
Try some others yourself. Don’t restrict yourself to sines and cosines. $\displaystyle r\left( \theta \right)=\tan \left( {3.7\theta } \right)$ for $\displaystyle 0\le \theta \le 20\pi$ below at different ranges.
## 2 thoughts on “Extreme Polar Conditions”
1. Thanks for the information. You make it look like really easy.
Like
2. Paul Foerster says:
Lin, et al.-
I did a presentation on polar coordinates in calc and precalc at the 2017 NCTM Annual Meeting San Antonio. One of the more widely appreciated features was using TI-84 dynamic graphing to show wether or not places where two polar graphs cross are really intersections, and under what conditions the crossing points are not intersections. Email me ([email protected]) and I’ll fill you in on the details.
Paul
Liked by 1 person
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http://new-contents.com/Texas/forward-euler-error-analysis.html | Address Texarkana, TX 75503 (903) 276-1094
# forward euler error analysis Ravenna, Texas
As seen from there, the method is numerically stable for these values of h and becomes more accurate as h decreases. All we need to do is plug t1 in the equation for the tangent line. It is because they implicitly divide it by h. Nevertheless, it can be shown that the global truncation error in using the Euler method on a finite interval is no greater than a constant times h.
In the bottom of the table, the step size is half the step size in the previous row, and the error is also approximately half the error in the previous row. Approximations Time Exact h = 0.1 h = 0.05 h = 0.01 h  = 0.005 h = 0.001 t = 1 0.9414902 0.9313244 0.9364698 0.9404994 0.9409957 0.9413914 t = 2 0.9910099 Notice that the approximation is worst where the function is changing rapidly. This should not be too surprising. Recall that we’re using tangent lines to get the approximations and so the Wird geladen... Über YouTube Presse Urheberrecht YouTuber Werbung Entwickler +YouTube Nutzungsbedingungen Datenschutz Richtlinien und Sicherheit Feedback senden Probier mal was Neues aus!
If you are a mobile device (especially a phone) then the equations will appear very small. Fortunately, we can control the amount of growing that might take place, and the result is that it grows by at most some constant factor (again, this is in a rectangle So, while I'd like to answer all emails for help, I can't and so I'm sorry to say that all emails requesting help will be ignored. The test problem is the IVP given by dy/dt = -10y, y(0)=1 with the exact solution .
The next step is to multiply the above value by the step size h {\displaystyle h} , which we take equal to one here: h ⋅ f ( y 0 ) This can be illustrated using the linear equation y ′ = − 2.3 y , y ( 0 ) = 1. {\displaystyle y'=-2.3y,\qquad y(0)=1.} The exact solution is y ( t Text is available under the Creative Commons Attribution-ShareAlike License; additional terms may apply. In Figure 4, I have plotted the solutions computed using the BE method for h=0.001, 0.01, 0.1, 0.2 and 0.5 along with the exact solution.
Note for Internet Explorer Users If you are using Internet Explorer in all likelihood after clicking on a link to initiate a download a gold bar will appear at the bottom My first priority is always to help the students who have paid to be in one of my classes here at Lamar University (that is my job after all!). It is especially true for some exponents and occasionally a "double prime" 2nd derivative notation will look like a "single prime". So, here is a bit of pseudo-code that you can use to write a program for Euler’s Method that uses a uniform step size, h.
In this simple differential equation, the function f {\displaystyle f} is defined by f ( t , y ) = y {\displaystyle f(t,y)=y} . This will present you with another menu in which you can select the specific page you wish to download pdfs for. Euler method implementations in different languages by Rosetta Code v t e Numerical methods for integration First-order methods Euler method Backward Euler Semi-implicit Euler Exponential Euler Second-order methods Verlet integration Velocity See also Crank–Nicolson method Dynamic errors of numerical methods of ODE discretization Gradient descent similarly uses finite steps, here to find minima of functions List of Runge-Kutta methods Linear multistep method
Melde dich an, um unangemessene Inhalte zu melden. Now, one step of the Euler method from t n {\displaystyle t_{n}} to t n + 1 = t n + h {\displaystyle t_{n+1}=t_{n}+h} is[3] y n + 1 = y Recall that the slope is defined as the change in y {\displaystyle y} divided by the change in t {\displaystyle t} , or Δ y / Δ t {\displaystyle \Delta y/\Delta Recall that we are getting the approximations by using a tangent line to approximate the value of the solution and that we are moving forward in time by steps of h. Â
This is so simple that we can find an explicit formula for . We can extrapolate from the above table that the step size needed to get an answer that is correct to three decimal places is approximately 0.00001, meaning that we need 400,000 If we pretend that A 1 {\displaystyle A_{1}} is still on the curve, the same reasoning as for the point A 0 {\displaystyle A_{0}} above can be used. Let's look at the global error gn = |ye(tn) - y(tn)| for our test problem at t=1.
The idea is that while the curve is initially unknown, its starting point, which we denote by A 0 , {\displaystyle A_{0},} is known (see the picture on top right). Once on the Download Page simply select the topic you wish to download pdfs from. Let's look at a simple example: , . Put Internet Explorer 11 in Compatibility Mode Look to the right side edge of the Internet Explorer window.
If the solution y {\displaystyle y} has a bounded second derivative and f {\displaystyle f} is Lipschitz continuous in its second argument, then the global truncation error (GTE) is bounded by Thus, it is to be expected that the global truncation error will be proportional to h {\displaystyle h} .[14] This intuitive reasoning can be made precise. Now, what about the global error? However, if the Euler method is applied to this equation with step size h = 1 {\displaystyle h=1} , then the numerical solution is qualitatively wrong: it oscillates and grows (see
Anzeige Autoplay Wenn Autoplay aktiviert ist, wird die Wiedergabe automatisch mit einem der aktuellen Videovorschläge fortgesetzt. In the picture below, is the black curve, and the curves are in red. Another important observation regarding the forward Euler method is that it is an explicit method, i.e., yn+1 is given explicitly in terms of known quantities such as yn and f(yn,tn). Firstly, there is the geometrical description mentioned above.
A closely related derivation is to substitute the forward finite difference formula for the derivative, y ′ ( t 0 ) ≈ y ( t 0 + h ) − y Now, we would like to proceed in a similar manner, but we don’t have the value of the solution at t1 and so we won’t know the slope of the tangent Bitte versuche es später erneut. This makes the implementation more costly.
I would love to be able to help everyone but the reality is that I just don't have the time. | 2019-01-22T17:40:38 | {
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https://proofwiki.org/wiki/Image_of_Union | # Image of Union
## Theorem
Let $S$ and $T$ be sets.
Let $\mathcal R \subseteq S \times T$ be a relation.
Let $S_1$ and $S_2$ be subsets of $S$.
Then:
$\mathcal R \left[{S_1 \cup S_2}\right] = \mathcal R \left[{S_1}\right] \cup \mathcal R \left[{S_2}\right]$
That is, the image of the union of subsets of $S$ is equal to the union of their images.
### General Result
Let $S$ and $T$ be sets.
Let $\mathcal R \subseteq S \times T$ be a relation.
Let $\mathcal P \left({S}\right)$ be the power set of $S$.
Let $\mathbb S \subseteq \mathcal P \left({S}\right)$.
Then:
$\displaystyle \mathcal R \left[{\bigcup \mathbb S}\right] = \bigcup_{X \mathop \in \mathbb S} \mathcal R \left[{X}\right]$
### Family of Sets
Let $S$ and $T$ be sets.
Let $\left\langle{S_i}\right\rangle_{i \in I}$ be a family of subsets of $S$.
Let $\mathcal R \subseteq S \times T$ be a relation.
Then:
$\displaystyle \mathcal R \left[{\bigcup_{i \mathop \in I} S_i}\right] = \bigcup_{i \mathop \in I} \mathcal R \left[{S_i}\right]$
where $\displaystyle \bigcup_{i \mathop \in I} S_i$ denotes the union of $\left\langle{S_i}\right\rangle_{i \in I}$.
## Proof
$$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle t$$ $$\in$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \mathcal R \left[{S_1 \cup S_2}\right]$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \iff$$ $$\displaystyle$$ $$\displaystyle \exists s \in S_1 \cup S_2: t$$ $$\in$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \mathcal R \left[{s}\right]$$ $$\displaystyle$$ $$\displaystyle$$ Definition of image of subset $$\displaystyle$$ $$\displaystyle \iff$$ $$\displaystyle$$ $$\displaystyle \exists s: s \in S_1 \lor s \in S_2: t$$ $$\in$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \mathcal R \left[{s}\right]$$ $$\displaystyle$$ $$\displaystyle$$ Definition of union $$\displaystyle$$ $$\displaystyle \iff$$ $$\displaystyle$$ $$\displaystyle t$$ $$\in$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \mathcal R \left[{S_1}\right] \lor t \in \mathcal R \left[{S_2}\right]$$ $$\displaystyle$$ $$\displaystyle$$ Definition of image of subset $$\displaystyle$$ $$\displaystyle \iff$$ $$\displaystyle$$ $$\displaystyle t$$ $$\in$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \mathcal R \left[{S_1}\right] \cup \mathcal R \left[{S_2}\right]$$ $$\displaystyle$$ $$\displaystyle$$ Definition of union
$\blacksquare$ | 2015-03-04T08:31:51 | {
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https://mathhelpboards.com/threads/find-the-angle-a.6782/ | # Find the angle A
#### anemone
##### MHB POTW Director
Staff member
For the triangle with angles $A, B, C$, the following trigonometric equality holds.
$$\displaystyle \sin^2 B +\sin^2 C-\sin^2 A=\sin B \sin C$$
Find the measure of the angle $A$.
##### Well-known member
For the triangle with angles $A, B, C$, the following trigonometric equality holds.
$$\displaystyle \sin^2 B +\sin^2 C-\sin^2 A=\sin B \sin C$$
Find the measure of the angle $A$.
using law of sin's sin A = ka, sin B= kb, sin C = kc
we get
b^2+c^2 - a^2 = bc
or a^2 = b^2 + c^2 + bc (1)
by law of cos
a^2 = b^2 + c^2 - 2bc cos A (2)
from (1) and (2)
2 cos A = - 1 or cos A = - 1/2 or A = 120 degrees
#### MarkFL
Staff member
using law of sin's sin A = ka, sin B= kb, sin C = kc
we get
b^2+c^2 - a^2 = bc
or a^2 = b^2 + c^2 + bc (1)
by law of cos
a^2 = b^2 + c^2 - 2bc cos A (2)
from (1) and (2)
2 cos A = - 1 or cos A = - 1/2 or A = 120 degrees
Your technique is nice and elegant, but you have made a simple sign error...
#### anemone
##### MHB POTW Director
Staff member
using law of sin's sin A = ka, sin B= kb, sin C = kc
we get
b^2+c^2 - a^2 = bc
or a^2 = b^2 + c^2 + bc (1)
by law of cos
a^2 = b^2 + c^2 - 2bc cos A (2)
from (1) and (2)
2 cos A = - 1 or cos A = - 1/2 or A = 120 degrees
Well done, kaliprasad...though you have made a minor mistake because $\ang A = 60 ^{\circ}$...
Also, I think my silly method is not worth mentioning after reading to your method!
##### Well-known member
Well done, kaliprasad...though you have made a minor mistake because $\ang A = 60 ^{\circ}$...
Also, I think my silly method is not worth mentioning after reading to your method!
thanks markFL and anemone
it should be
a^2 = b^2 + c^2 - bc (1)
by law of cos
a^2 = b^2 + c^2 - 2bc cos A (2)
from (1) and (2)
2 cos A = 1 or cos A = 1/2 or A = 60 degrees
sorry for my mistake
#### MarkFL
Staff member
My solution:
Since $$\displaystyle A=\pi-(B+C)$$ and $$\displaystyle \sin(\pi-\theta)=\sin(\theta)$$
The equation becomes:
$$\displaystyle \sin^2(B)+\sin^2(C)-\sin^2(B+C)=\sin(B)\sin(C)$$
Using the angle-sum identity for sine, we may write:
$$\displaystyle \sin^2(B)+\sin^2(C)-\sin^2(B)\cos^2(C)-2\sin(B)\sin(C)\cos(B)\cos(C)-\sin^2(C)\cos^2(B)=\sin(B)\sin(C)$$
$$\displaystyle \sin^2(B)\left(1-\cos^2(C) \right)+sin^2(C)\left(1-\cos^2(B) \right)=\sin(B)\sin(C)(1+2\cos(B)\cos(C))$$
Using Pythagorean identities, we have:
$$\displaystyle 2\sin^2(B)\sin^2(C)=\sin(B)\sin(C)(1+2\cos(B)\cos(C))$$
Assuming the triangle is not degenerate, i.e., $$\displaystyle \sin(B)\sin(C)\ne0$$ we have:
$$\displaystyle 2\sin(B)\sin(C)=1+2\cos(B)\cos(C)$$
$$\displaystyle -\frac{1}{2}=\cos(B)\cos(C)-\sin(B)\sin(C)=\cos(B+C)$$
$$\displaystyle B+C=120^{\circ}\implies A=60^{\circ}$$
#### anemone
##### MHB POTW Director
Staff member
thanks markFL and anemone
it should be
a^2 = b^2 + c^2 - bc (1)
by law of cos
a^2 = b^2 + c^2 - 2bc cos A (2)
from (1) and (2)
2 cos A = 1 or cos A = 1/2 or A = 60 degrees
sorry for my mistake
Don't be sorry, kali! And if you would like, you can make it up by joining me to post many interesting and challenging problems for the folks here...hehehe...
My solution:
Since $$\displaystyle A=\pi-(B+C)$$ and $$\displaystyle \sin(\pi-\theta)=\sin(\theta)$$
The equation becomes:
$$\displaystyle \sin^2(B)+\sin^2(C)-\sin^2(B+C)=\sin(B)\sin(C)$$
Using the angle-sum identity for sine, we may write:
$$\displaystyle \sin^2(B)+\sin^2(C)-\sin^2(B)\cos^2(C)-2\sin(B)\sin(C)\cos(B)\cos(C)-\sin^2(C)\cos^2(B)=\sin(B)\sin(C)$$
$$\displaystyle \sin^2(B)\left(1-\cos^2(C) \right)+sin^2(C)\left(1-\cos^2(B) \right)=\sin(B)\sin(C)(1+2\cos(B)\cos(C))$$
Using Pythagorean identities, we have:
$$\displaystyle 2\sin^2(B)\sin^2(C)=\sin(B)\sin(C)(1+2\cos(B)\cos(C))$$
Assuming the triangle is not degenerate, i.e., $$\displaystyle \sin(B)\sin(C)\ne0$$ we have:
$$\displaystyle 2\sin(B)\sin(C)=1+2\cos(B)\cos(C)$$
$$\displaystyle -\frac{1}{2}=\cos(B)\cos(C)-\sin(B)\sin(C)=\cos(B+C)$$
$$\displaystyle B+C=120^{\circ}\implies A=60^{\circ}$$
Bravo, MarkFL! And I like your method as well! Just so you know, my method is so much longer and more tedious than yours! | 2021-12-04T14:10:12 | {
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https://math.stackexchange.com/questions/1779323/eigenvectors-of-beginbmatrixa-b-ba-endbmatrix/1779375 | # Eigenvectors of $\begin{bmatrix}a&-b\\b&a\end{bmatrix}$
I am currently confused as to what the correct eigenvectors are for $\begin{bmatrix}a&-b\\b&a\end{bmatrix}$.
I confirmed through my own calculations that the eigenvalues are a$\pm$bi. My textbook, Linear Algebra and its Applications, states that the corresponding eigenvectors are $\begin{bmatrix}1\\-i\end{bmatrix}$ and $\begin{bmatrix}1\\i\end{bmatrix}$.
This makes sense when verifying that $Ax = \lambda x$, as $Ax = \begin{bmatrix}a&-b\\b&a\end{bmatrix} \begin{bmatrix}1\\-i\end{bmatrix} = \begin{bmatrix}a+bi\\b-ai\end{bmatrix} = (a+bi) \begin{bmatrix}1\\-i\end{bmatrix}$.
However, upon performing the calculations myself, I repeatedly found the eigenvectors to be $\begin{bmatrix}-i\\1\end{bmatrix}$ and $\begin{bmatrix}i\\1\end{bmatrix}$ rather than the given solution. Thinking that I could have made a calculation error, I plugged this into WolframAlpha and got the same values.
My calculation process was to solve for $(A-(a+bi)I)x = 0$, which I reduced down to $\begin{bmatrix}-i&-1\\1&-i\end{bmatrix}$. After multiplying the top equation by $i$, I got $\begin{bmatrix}1&-i\\1&-i\end{bmatrix}$ ~ $\begin{bmatrix}1&-i\\0&0\end{bmatrix}$. Thus, $x_1 = ix_2$ and $x_2$ is free. So $x = \begin{bmatrix}x_1\\x_2\end{bmatrix} = \begin{bmatrix}i\\1\end{bmatrix} x_2$, so the eigenvalue corresponding to $\lambda = a+bi$ is $\begin{bmatrix}i\\1\end{bmatrix}$. Similarly, the eigenvalue corresponding to $\lambda = a-bi$ is $\begin{bmatrix}-i\\1\end{bmatrix}$.
Where is the discrepancy between my calculated answer and the one given in the textbook?
Eigenvectors are unique upto some (non-zero) constant. Note that the two eigenvectors that are listed as answers are just $\pm i$ times the eigenvectors you found.
There are an infinity of eigenvectors related to a given eigenvalue (in fact a whole vector sub space). Let's check that
$$\begin{bmatrix} 1\\-i \end{bmatrix}=-i\cdot\begin{bmatrix}i \\1\end{bmatrix}$$
So "your" eigenvector is proportionate to the one "of the textbook"
Similarly for the other eigenvalue
There is none: $(-i,1)$ is a scalar multiple of $(1,i)$ that is:
$-i(1,i) = (-i,-i^2) = (-i,-(-1)) = (-i,1)$, and similarly for your other eigenvector.
Let $r$ be the module and $\theta$ the argument of complex number $a+bi$, thus with
$$a=r \cos \theta \ \ \ \text{and} \ \ \ b=r \sin \theta$$
No answer mentions the geometrical interpretation of matrix
$$S=\begin{bmatrix}a&-b\\b&a\end{bmatrix}=r\begin{bmatrix}\cos \theta&-\sin \theta\\\sin \theta&\cos \theta\end{bmatrix}$$
that helps to understand why such matrices are bound to have non real eigenvalues.
The interpretation of $S$ as a geometrical transformation is clearly a rotation followed (or preceded) by an homothetic transform, i.e., a similitude. It is clear that (unless $\theta=0$!) no real vector can be transformed by $S$ into a real multiple of it. Therefore, the eigenvalues "have to be" complex. | 2019-08-24T23:26:13 | {
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https://math.stackexchange.com/questions/3442256/find-the-straight-line-which-is-tangent-to-the-graph-of-y-2x-at-some-point-an | # Find the straight line which is tangent to the graph of $y=2^x$ at some point and which passes through the point (1, 0).
I understand this is probably a duplicate question, but the answers on said duplicates don't make sense to me or don't seem to work for me.
I'm being asked to find the point of tangency and the equation for the tangent line.
So far I have that:
• $$y'=ln(2)2^x$$
• There is a point $$P=(a, 2^a)$$ which is where the aforementioned tangent line passes through the graph
• It passes through (1,0)
Using this information I tried to create the equation for the tangent line.
$$0-2^a=ln(2)2^a(1-a)$$
And then I end up with
$$0=2^a(ln(2)(1-a)+1)$$
And this is where I get lost, I can't seem to extrapolate any information about a point of tangency from this.
• Is extrapolate a deliberate play on words? If so, excellent! – Matthew Leingang Nov 19 '19 at 16:27
Note the derivative of $$y=2^x$$ is $$y'=2^x\ln 2$$, which should match
$$2^a\ln 2 = \frac {2^a-0}{a-1}$$
Solve for $$a$$,
$$a = \frac1{\ln2}+1$$
Thus, the slope is $$2^{(\frac1{\ln2}+1)}\ln2$$ and use the point-slope formula for the point $$(1,0)$$ to obtain the line
$$y=2^{(\frac1{\ln2}+1)}\ln2\>(x-1)$$
For real finite $$a,2^a>0$$
$$\implies(a-1)\ln2=1$$
$$a=?$$
Solve the equation for $$a$$ and you are done. What is the idea? It is you assume the point is $$(a,2^a)$$ then you evaluate the slope at this point as you did $$\ln 2 2^a$$ equal the slope as $$\frac{2^a-0}{a-1}$$ then solve that for $$a$$
One could note that the fact that exponential functions fulfill the relationship $$y'=ky$$ means exactly that given the tangent at a point $$(a,2^a)$$ on the graph and the point $$(b,0)$$ where the tangent crosses the $$x$$-axis, $$b-a$$ is constant. So we just have to find this distance for some point and then apply this constant difference to the fact that we want the tangent to cross the $$x$$-axis at $$(1,0)$$.
The tangent at $$(0,1)$$ (that seems like the easiest one) has slope $$\ln 2$$, and thus crosses the $$x$$-axis at $$-\frac1{\ln2}$$. Which thus is the $$x$$-coordinate difference we are after. So the point of tangency we want has $$x$$ coordinate $$1+\frac1{\ln2}$$. | 2020-06-01T20:36:15 | {
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https://math.stackexchange.com/questions/2421883/is-this-set-of-matrices-closed-under-matrix-multiplication | # Is this set of matrices closed under matrix multiplication?
2x2 matrices of the form: \begin{bmatrix}\cos x&-\sin x\\\sin x&\cos x\end{bmatrix} x is a real number
multiplying the matrix by itself produces \begin{bmatrix}\cos^2x-\sin^2x&-2\cos x\sin x\\2\sin x\cos x&\cos^2x-\sin^2x\end{bmatrix}
Is this product matrix still in the set? It seems not to be in my mind, but I'm not sure.
• Observe that $-2\cos x\sin x = -\sin(2x)$. (I expect that you already know this, right?) Is that the hint you need to go on by yourself? – MJD Sep 8 '17 at 20:19
• First, what trig identities do you know? Second, showing something is closed under multiplication involves multiplying two different things together, not multiplying something by itself. – Jason DeVito Sep 8 '17 at 20:19
• Notice that the matrix you got is $\begin{bmatrix}\cos(x+x)&-\sin(x+x)\\ \sin(x+x)&\cos(x+x)\end{bmatrix}$. A similar argument with two distinct matrices of the set will give the analog for $x+y$ and you can then conclude. – Prasun Biswas Sep 8 '17 at 20:20
• @JasonDeVito But if OP suspects that the claim might be false, then trying some simpler special cases to see if they work is a reasonable way to proceed. – MJD Sep 8 '17 at 20:20
• @MJD: Agreed! To the op: No, because $x$ denotes a fixed real number. For example, that reasoning can be used to prove he following (false) theorem: Any two numbers of the form $x^2$ are equal. – Jason DeVito Sep 8 '17 at 20:26
To check if $S:=\left\{\begin{bmatrix}\cos(x)&-\sin(x)\\\sin(x)&\cos(x)\end{bmatrix}:x\in\Bbb R\right\}$ is closed under multiplication you would need to multiply $$\begin{bmatrix}\cos(x)&-\sin(x)\\\sin(x)&\cos(x)\end{bmatrix}\begin{bmatrix}\cos(y)&-\sin(y)\\\sin(y)&\cos(y)\end{bmatrix}$$ not $$\begin{bmatrix}\cos(x)&-\sin(x)\\\sin(x)&\cos(x)\end{bmatrix}\begin{bmatrix}\cos(x)&-\sin(x)\\\sin(x)&\cos(x)\end{bmatrix}$$ because the two matrices need not be the same. So we need to show $$\begin{bmatrix}\cos(x)&-\sin(x)\\\sin(x)&\cos(x)\end{bmatrix}\begin{bmatrix}\cos(y)&-\sin(y)\\\sin(y)&\cos(y)\end{bmatrix}=\begin{bmatrix}\cos(x)\cos(y)-\sin(x)\sin(y)&-\cos(x)\sin(y)-\sin(x)\cos(y)\\\sin(x)\cos(y)+\cos(x)\sin(y)&-\sin(x)\sin(y)+\cos(x)\cos(y)\end{bmatrix}$$ is in $S$. Recall the identities: $$(1)\qquad\sin(x\pm y)=\sin(x)\cos(y)\pm\sin(y)\cos(x)\\(2)\qquad \cos(x\pm y)=\cos(x)\cos(y)\mp \sin(x)\sin(y)$$
• You answer the question of the OP, no doubt, but I am a little surprised that you do not mention the geometrical background (rotation matrices) which gives an immediate understanding of the question (see my answer). – Jean Marie Sep 8 '17 at 21:46
• @JeanMarie that is a fine answer (+1). I was just trying to address the OP's line of reasoning for showing the set is closed under multiplication (i.e. by directly showing $A,B\in S\implies AB\in S$) by showing him/her the mistake they made and how to correct it. – Dave Sep 8 '17 at 22:22
• Thanks, and yes you did it in a paedogical manner by spotting at once the error. I just reacted in this way because I see that geometric intuition is maybe the largest lack in contemporary mathematical education... – Jean Marie Sep 8 '17 at 22:27
There is a geometrical interpretation that sheds light on all this :
$$R_x=\begin{bmatrix}\cos x&-\sin x\\ \sin x& \ \ \cos x \end{bmatrix}$$
is a rotation matrix (https://en.wikipedia.org/wiki/Rotation_matrix).
Thus, what has been done in the answer given by @Dave is plainly
$$R_x \times R_y = R_{x+y}$$
Using sentences : Rotation with angle $y$ followed by rotation by angle $x$ is rotation by angle $x+y$.
This is why you have a closed set.
One can say more : the set of rotation matrices is a (commutative) group for multiplication. | 2019-10-21T20:19:13 | {
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https://math.stackexchange.com/questions/2078796/finding-the-nth-term-of-a-numeric-sequence-newtons-little-formula-explanation?noredirect=1 | # Finding the nth term of a numeric sequence- Newton's little formula explanation
In a contest problem book, I found a reference to Newton's little formula that may be used to find the nth term of a numeric sequence. Specifically, it is a formula that is based on the differences between consecutive terms that is computed at each level until the differences match.
An example application of this formula for computing the nth term of the series (15, 55, 123, 225, 367, 555, 795, ....) involves computing the differences as shown below:
1) 1st Level difference is (40, 68, 102, 142, 188, 240)
2) 2nd Level difference is (28, 34, 40, 46, 52)
3) 3rd Level difference is (6, 6, 6, 6, 6)
Now the nth term is $$15{n-1\choose 0} + 40{n-1\choose 1} + 28{n-1\choose 2} + 6{n-1\choose 3}$$ where the constant multipliers are the first term of the differences at each level in addition to the first term of the sequence itself.
I was not able to find any reference to this formula or a proof of it after searching on the web. Any explanation of this method is appreciated.
• LaTeX formatting will make your question much easier to read. – The Count Dec 31 '16 at 20:04
• Thank you for the comment. I have formatted the nth term expression. – erase.ego Dec 31 '16 at 20:23
Suppose that we have a sequence:
$$a_0,a_1,...a_k$$
And we want to find the function of $n$ that defines $a_n$.
To do this we start by letting $a_{n+1}-a_n=\Delta a_n$ and we call this operation on $a_n$ the forward difference. Then given $\Delta a_n$ we can find $a_n$. Sum both sides of the equation from $n=0$ to $x-1$, and note that we have a telescoping series:
$$\sum_{n=0}^{x-1} \Delta a_n=\sum_{n=0}^{x-1} (a_{n+1}-a_n)=a_{x}-a_{0}$$
Hence $a_n=a_0+\sum_{i=0}^{n-1} \Delta a_i$. Also $\Delta a_n=\Delta (0)+\sum_{i=0}^{n-1} \Delta^2 a_n$...and so forth. Using this we must have if the series converges:
$$a_n=a_0+\Delta (0) \sum_{x_0=0}^{n-1} 1+\Delta \Delta (0) \sum_{x_0=0}^{n-1} \sum_{x_1=0}^{x_0-1} 1+\Delta \Delta \Delta (0) \sum_{x_0=0}^{n-1} \sum_{x_1=0}^{x_0-1} \sum_{x_2=0}^{x_1-1} 1+\cdots$$
Where $\Delta^i (0)$ denotes the first term ($n=0$) of the $i$ th difference sequence of $a_n$.
Through a combinational argument, If we take $\Delta^0 (0)=a_0$ and ${n \choose 0}=1$ we may get:
$$a_n=\sum_{i=0}^{\infty} \Delta^i(0) {n \choose i}$$
If it is the case you want the sequence to start with $a_1$ we need to shift this result to the right one:
$$a_n=\sum_{i=0}^{\infty} \Delta^i(1) {n-1 \choose i}$$
Note when you re-define $a_0$ to be $a_1$ by shifting it's index to the right one, you're re-defining $a_1-a_0=\Delta^1(0)$ to be $a_2-a_1=a_{1+1}-a_{1}=\Delta^1(1)$.
• $a_1 = a_0 + \Delta a_0$, $a_2 = a_0 + 2\Delta a_0 + \Delta^2 a_0$ and $a_3 = a_0 + 3\Delta a_0 + 3\Delta^2 a_0 + \Delta^3 a_0$ are the first three terms. The co-efficients seem to correspond to those of the binomial expansion and it looks like the general result maybe proved using induction. – erase.ego Jan 3 '17 at 3:17
• Take $D$ to be the operation mapping $f(x)$ to $f(x+1)$. And $I$ to be the operation mapping $f(x)$ to itself. $D,I$ are linear operators. So if we want to find $\Delta^n f(x)=(D-I)^n f(x)$ we may treat $(D-I)^n$ as a polynomial. Then use binomial theorem to expand and then operate on $f(x)$ with what we get for $(D-I)^n$ to get a closed form for the coefficients @erase.ego – Ahmed S. Attaalla Jan 3 '17 at 3:21
Starting with the second differences we have
• $28 = 28$,
• $34 = 28 + 6$,
• $40 = 28 + 6 + 6$
• $46 = 28 + 6 + 6 + 6$$Now for the first differences we have •$40 = 40$•$68 = 40 + 28$•$102 = 40 + 28 + 34 = 40 + 28 + (28 + 6)$•$142 = 40 + 28 + 34 + 50 = 40 + 28 + (28 + 6) + (28 + 6 + 6)$So the original sequence is •$15 = 15$•$55 = 15 + 40$•$123 = 15 + 40 + 68 = 15 + 40 + [40 + 28]$•$225 = 15 + 40 + 68 + 102 = 15 + 40 + [40 + 28] + [40 + 28 + (28 + 6)]\$
• \begin{align} 367 &= 15 + 40 + 68 + 102 + 142 \\ &= 15 + 40 + [40 + 28] + [40 + 28 + (28 + 6)] + [40 + 28 + (28 + 6) + (28 + 6 + 6)] \end{align}
The patterns should be clear. So just count how many 15's, 40's, 28's, and 6's there are in each term.
• I am wondering if this argument can be generalized some how? – erase.ego Dec 31 '16 at 23:44
• Well, it generalizes to any sequence given by a polynomial. I guess you could also generalize it to a sequence where the n-th differences aren't constant but have some nice behavior. – Daniel McLaury Dec 31 '16 at 23:46 | 2019-08-18T11:37:24 | {
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http://mathhelpforum.com/advanced-statistics/161623-defective-widgets.html | # Math Help - Defective Widgets
1. ## Defective Widgets
In a certain factory, Machine A, Machine B, and Machine C are all producing widgets. Widgets produced by Machine A have a 1% chance of being defective. Likewise, widgets produced by Machine B and Machine C are defective 4% and 2% of the time, respectively. Of the total production of widgets in the factory, Machine A produces 30%, Machine B produces 25%, and Machine C produces 45%. Suppose a widget is selected at random from this factory.
a) What is the probability the widget is defective?
b) If the widget is defective, what is the probability it was produced by Machine B?
Any help would be appreciated. I believe that i am suppose to use the counting rule combination but am not sure because we are given two different sets of %...
2. Originally Posted by shannu82
In a certain factory, Machine A, Machine B, and Machine C are all producing widgets. Widgets produced by Machine A have a 1% chance of being defective. Likewise, widgets produced by Machine B and Machine C are defective 4% and 2% of the time, respectively. Of the total production of widgets in the factory, Machine A produces 30%, Machine B produces 25%, and Machine C produces 45%. Suppose a widget is selected at random from this factory.
a) What is the probability the widget is defective?
b) If the widget is defective, what is the probability it was produced by Machine B?
Any help would be appreciated. I believe that i am suppose to use the counting rule combination but am not sure because we are given two different sets of %...
for part a consider what is the prob that a widget is defective if it came from A
First the prob. that it came from A is .3 and the prob. that is is defective is .01
So the prob. that is came from A and is defective is
$(.3)(.01)=.003$
Now you need to repeat this for B and C and sum the probabilities. (why?)
For part b you need to use conditional probabilities
Let A be the even that the item is defective and B be the event that it came from machine B. Then
$P(B|A)=\frac{P(A \cap B)}{P(A)}$
Now use your result from part a.
3. Thank You for responding i just wanna make sure that i understand correctly...
a) P(A) =0.003 P(B) = 0.01 and P(C) = 0.009 Therefore the probability of defective widgets is 0.022.
b) using the formula above (0.003+0.01)/0.003 = 4.333% is the probability that the defective widget was created by Machine (B)?
4. Part A looks good..
For part B, TheEmptySet told you to assume A as the probability of the widgets being defective. This might have been ambiguous to you because you are using A as the probability of widget being made in factory A.
Suppose X is the probability that the widget is defective...
You already have P(X) from part(a) and you have $P(X \cap B)$ too..
Now find: $P(B|X)$ | 2014-12-20T02:42:52 | {
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https://math.stackexchange.com/questions/1509369/exponentials-with-three-variables-solving-for-an-equation | # Exponentials with three variables: solving for an equation
$2^{x}=5^{y}=100^{z}$
Find $z$ in terms of $y$ and $x$.
The term $z$ should be a function of $x$ and $y$, i.e.: $z(x,y)$.
All I could get were recursive attempts.
• Hint: if $100^z = a$, what is $z$ in terms of $a$ ? – Fabrice NEYRET Nov 2 '15 at 13:44
• Take the logarithm of both sides. – Aretino Nov 2 '15 at 13:44
• $\log _{100}\left( a \right)=z$ ? – inspd Nov 2 '15 at 13:45
Suppose $z$ is non-zero.
$100^z=a\rightarrow\log_a100=\frac{1}{z}$
Likewise, we have $\log_a2=\frac{1}{x}$ and $\log_a5=\frac{1}{y}$.
Hence, $\frac{1}{z}=\log_a100=2\log_a2+2\log_a5=\frac{2}{x}+\frac{2}{y}$.
Rewriting gives: $z=\frac{xy}{2(x+y)}$.
Taking into account the case when $z=0$, it follows that:
$$z=\begin{cases}\frac{xy}{2(x+y)}&\text{if x\neq0},\\0&\text{otherwise}\end{cases}$$
• this is a nice approach, thanks – inspd Nov 2 '15 at 14:32
HINT: $100^z=10^{2z}=2^{2z} \cdot 5^{2z}$
Hence the relation becomes $2^x=5^y=2^{2z} \cdot 5^{2z}$
EDIT: $$2^x=5^y=2^{2z} \cdot 5^{2z}=a$$
Therefore, we have $$x\log 2=y\log 5=2z(\log 2 + \log 5)$$ $$\frac{x-2z}{2z}=\frac{\log 5}{\log 2}$$ and $$\frac{2z}{y-2z}=\frac{\log 5}{\log 2}$$ So we have $$\frac{x-2z}{2z}=\frac{2z}{y-2z}$$ or,$$xy-2z(x+y)+4z^2=4z^2$$ or, $$z=\frac{xy}{2(x+y)}$$
• can you confirm this: $$z=\frac{xy}{2\left( x+y \right)}$$ – inspd Nov 2 '15 at 14:01
• I can confirm. Currently writing a complete answer. – Element118 Nov 2 '15 at 14:16
• Updated my answer. Sorry for being late. See if it is okay. – SchrodingersCat Nov 2 '15 at 14:40
Taking the logarithm, to any base, of each of those, x log(2)= y log(5)= z log(100). z= xlog(2)/log(100)= y log(5)/log(100).
If you se the common logarithm, base 10, log(100)= 2 so those become z= xlog(2)/2= y log(5)/2.
• it should be in terms of $x$ and $y$, i.e.: $z(x,y)$. I'll specify more in the question – inspd Nov 2 '15 at 13:53 | 2020-02-16T22:09:16 | {
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http://zpvb.bjoy.pw/phase-portrait-plotter-matrix.html | # Phase Portrait Plotter Matrix
Nonlinear Methods for the Social Sciences Stephen J. Real part of is positive and the FP is unstable. Hence, the phase portrait is that of the node in. A has a repeated eigenvalue with only one linearly independent eigenvector. The phase portrait is a representative sampling of trajectories of the system. In GPoM: Generalized Polynomial Modelling Pre-processing for global modelling. 1 The Phase Plane Example 2. A fixed point solution is a point in a phase portrait. Introduction Phase Plane Qualitative Behavior of Linear Systems Local Behavior of Nonlinear Systems Isocline Method I The algorithm of constructing the phase portrait by isocline method: 1. (ii) Some of the plots show direction fields of equations all whose solutions are increasing functions (at. Construction of the exponential matrix, including Jordan canonical form for 2 x 2 matrices. This observation will be very useful when we sketch phase portraits. The following plots have been produced with octave using the above procedure:. This work generates the solution The general solution is The first picture shows the direction field; the second shows the phase portrait, with some typical solution curves. The geometric properties of the phase portrait are closely related to the algebraic characteristics of eigenvalues of the matrix A. The initial conditions in the routine are placed on a circle with radius r. In order to obtain a general first overview over the different types of bifurcations that occur in the system, it is useful to plot phase portraits for different values of the control parameters and to compare them to each other. unstable manifolds of saddle points. m Matlab m-file for portraying an IFS based on a matrix of values henon. Has anyone out there used Igor to make a phase portrait (aka phase space plot) of an ODE? I attach an example (middle plot) below. We write ~x˙ = F~(~x) = ˆ x˙1 = x2, x˙2 = −sin(x1) The fixed points are where F~(~x) = 0, which we write as F~(~¯x) = 0 for. It doesn’t have the simple point and click features of the MatLab pplane8program, but it is not very difficult to use. Departmental approval. fitzhugh_nagumo. For math, science, nutrition, history. x L 1 y L 2 7. Phase portrait of the nonlinear system. PortraitDensity determines the density of the phase portrait. Plot 3D phase portrait. A fixed point solution is a point in a phase portrait. With the linearized equations, we now have U(x) = − R x x0 ζdζ = − 2+ 0 2. The last matrix says that , so. 3 in Differential Equations with MATLAB. In this study, L. which can be written in matrix form as X'=AX, where A is the coefficients matrix. Moreover, local phase portrait of a hyperbolic equilibrium of a nonlinear system is equivalent to that of its linearization. If necessary you will find the commands at the end of the Maple version of this example. We start by. This observation will be very useful when we sketch phase portraits. Classify differential equations according to their type and order. Histopathological assessment of lymph node metastases (LNM) depends on subjective analysis of cellular morphology with inter-/intraobserver variability. The left plot is a temporal representation of the system's development, with time $$t$$ being represented on the horizontal axis. PhasePlane(sys,tspan,icond) plots the the phase plane portrait for a general second order nonlinear system defined by the function sys(t,x). Sketch the phase portrait of this system. Hyperbolic equilibria are robust(i. On this page I explain how to use Matlab to draw phase portraits for the the two linear systems. The user must provide a field (matrix) of complex numbers z which covers the domain (typically a rectangle, a disk or an annulus) of the function, and a field of the same size with the corresponding values w=f(z). We define the equilibrium solution/point for a homogeneous system of differential equations and how phase portraits can be used to determine the stability of the equilibrium solution. In this section we will give a brief introduction to the phase plane and phase portraits. The graphing window at right displays a few trajectories of the linear system x' = Ax. The resulting gallery is shown in the two pages of phase plane portraits that comprise Figure 5. Mathgrapher contains many demonstrations covering most of the things you can do with it. This observation will be very useful when we sketch phase portraits. The family of all trajectories or solution curves (which started by different initial points) is called phase portrait. ±2 ±1 0 1 2 J 2 R We will show the repeller and attractor are the eigendirections of the matrix. 7c, we plot SLP and SIC metrics for the Bering Sea. A Windows based mathematical graphing tool for 2D and 3D Functions and Data, shaded surfaces, contour plots. Non-Linear Dynamics Homework Solutions Week 4: Strogatz Portion February 3, 2009 6. • Used MATLAB to sketch phase portraits and approximated changing behavior of dynamic systems in the neighborhood of the fixed point, then identified types of bifurcation • Nondimensionalized. Here are some of the principles of trajectory sketching:. The matrix plot enables the realization that chaos only occurs under selective amplitudes and periods of the perturbation. For math, science, nutrition, history. The phase. In this section we will solve systems of two linear differential equations in which the eigenvalues are real repeated (double in this case) numbers. The type of phase portrait of a homogeneous linear autonomous system -- a companion system for example -- depends on the matrix coefficients via the eigenvalues or equivalently via the trace and determinant. 12 Procedure to draw phase portrait in XY plane(2nd order) Find critical points: Eg: x=4x-3y ,y=6x-7y have critical point at (0,0) construct a phase plot (y vs x) find eigen values and eigen vector of the system equation eigen values are (-5 and 2) and corresponding eigen vectors are [1;3] and [3;2] and draw corresponding vector axes if eigen. Analysis tools include power spectrum calculation and Poincare sections. On the back of this guide is a flow chart which describes the process. Introduction. Sketching Non-linear Systems OCW 18. Show that the equilibrium point in phase space is a center, which means that the cylinder oscillates, and find the oscillation period of the cylinder. 552 Phase Plane Methods were used, then 50 solution curve segments have already been entered onto the graphic! Threaded orbits are added to show what happens to solutions that are plotted on longer and longer t-intervals. Hence, the eigenvalues of the matrix. Exercises and Examples 1. A phase portrait is a plot of a segment’s angular displacement versus its angular velocity (Barela et al. To display the graph of a function we choose some interval of values for , and plot the points of the graph whose -coordinates lie in the given interval. The software is described in detail in the manual Ordinary Differential Equations using MATLAB. In the case of centers and spirals you may also be asked to. 7c, we plot SLP and SIC metrics for the Bering Sea. Left plot is for a= 1 and right plot is for a= 1. In this section we will solve systems of two linear differential equations in which the eigenvalues are real repeated (double in this case) numbers. FitzHugh-Nagumo: Phase plane and bifurcation analysis¶ Book chapters. Here are some of the principles of trajectory sketching:. The system of equations is written u' = AA*u, where AAis a given 2×2-matrix anduis a column vector. On the plot at the bottom of the previous page, in the box at the right, draw a phase portrait near the xed point N A = K=4, N 0 = K=4. This box represents a blow-up of the box in the computer-generated plot at the left. * Updated many function names to be more consistent with python coding. Mosteller and Tukey’s World War II example Simpson’s Paradox. If is a function, then the \emph{graph} of this function is the collection of points in the plane, where is the domain of. 3 in Differential Equations with MATLAB. JPlotter JPlotter is an open source math plotter that can draw graphs of arbitrary mathematical functions. Phase Plane Plotter. In this instance, numerical integration will usually be the only way forward. 3 The Phase Plane for a Conservative System A second-order differential equation for a variable x(t) can always be converted to two. Obtain a Describing Function for this nonlinear element. Show that the dynamical system associated to the motion of the cylinder is linear and find its matrix. Problem set 5: Solutions Math 207A, Fall 2018 1. Just as we did for linear systems, we want to look at the trajectories of the system. 7-26, Section 6 Ex. Th: Phase Portraits and Lotka-Volterra Equations For plotting the population data please use the code plot_population. Be able to write a system of linear (algebraic) equations as a matrix equation. You may integrate and analyse systems of up to 20 coupled ordinary differential equations (ODE's). We show by treating a concrete example how you can use Matlab to plot the phase portrait of a linear system in the plane. Then we will analyze the geometric effect of transforming a linear map to its Jordan Normal Form. In this case, altering the output color they emit (blue outside and green inside the mesh). Differential equation. 1 x x 1 2 v 2 v 1-1 1-1 Phase portraits for 2 × 2 systems. So I want to be able to draw the phase portrait for linear systems such as: x'=x-2y y'=3x-4y I am completely confused, but this is what I have come up with so far: Step 1: Write out the system in the form of a matrix. Finding Stationary Points This guide describes how to use the first and the second derivatives of a function to help you to locate and classify any stationary points the function may have. The initial conditions in the routine are placed on a circle with radius r. Finally, we show how to generate a very naive 'phase portrait' of trajectories for different values of the model parameter eps. Perhaps its existence is unknown or its effect unsuspected. This observation will be very useful when we sketch phase portraits. Sketch the phase plane portrait of a 2D system of first order differential equations. Povinelli Department of Electrical and. Phase spaces are used to analyze autonomous differential equations. In this discussion, we will only plot solutions of autonomous, first order differential equations; that is, equations of the form To begin, type pline. The following problems discuss Lyapunov stability in the case when the Lyapunov theorem is non-applicable. In order to obtain a general first overview over the different types of bifurcations that occur in the system, it is useful to plot phase portraits for different values of the control parameters and to compare them to each other. On this page I explain how to use Matlab to draw phase portraits for the the two linear systems. x L 1 y L 2 7. We will plot the derivatives as a vector at each (y1, y2) which will show us the initial direction from each point. See Figure 11, we sketch phase portraits for the case when A has two eigenvalues ‚1 > ‚2 > 0. We compute the rank by computing the number of singular values of the matrix that are greater than zero, within a prescribed tolerance. 0 The fixed point in 0, 0 is unstable ----- Check. Duan in his recent published book [1, §5. Here are some general facts to remember when analyzing phase portraits of linear maps: If the matrix (A - I) is nonsingular then the origin is the only fixed point. Nonlinear Systems and Phenomena Application 9. Plot the curve S(x) = in state-space (phase plane) 2. A Direction Fields and Phase Portraits. I have to plot its phase portrait using this code: And if first line was 9x - 21y - 10z + 5 where would 5 goes in the matrix? matlab. Geometric structures occur in this processing space that are called trajectories or attractors. See examples/geneticswitch. The Scientific World Journal is a peer-reviewed, Open Access journal that publishes original research, reviews, and clinical studies covering a wide range of subjects in science, technology, and medicine. Use interactive calculators to plot and graph functions. I could imagine myself, just as Strogatz, trying to see if a phase portrait of a reversible system resembles something in nature, such as a manta ray (as in one of the examples he worked out). The Jacobian matrix is A = 0 1 3x2. The phase space plot and such a family of trajectories together are a phase space portrait, phase portrait, or phase diagram. Stability Analysis for ODEs Marc R. c) Write the Matlab code to plot the phase portraits. Step 3: Using the eigenvectors draw the eigenlines. This is a practicing course for MATLAB taught by Ahmed Rezk and Ahmed Mahdy. The dimension of the phase space is the number of intial conditions required to uniquely specify a trajectory; it is the number of variables in the dynamical system. For mechanical systems, the phase space usually consists of all possible values of position and momentum variables. This is a topic that’s not always taught in a differential equations class but in case you’re in a course where it is taught we should cover it so that you are prepared for it. field, a phase portrait is a graphical tool to visualize how the solutions of a given system of differential equations would behave in the long run. For all figures in this article, the plot range of x and y from known densities is set to include 99% quantiles of the target density. Essentially, you only need to solve the differential equations and then plot the result. A (discrete) phase portrait of the system ~x(t+ 1) = A~x(t) shows discrete trajectories for vari-ous initial states, capturing all the qualitatively different scenarios (as in Figure 6). That makes the origin an attractor, and hence, all solutions will be drawn into the origin, making the system asymptotically stable. In this section, we will show you how to plot data, modify plots and save your work. It does so by examining symmetric matrices, the damped harmonic oscillator, and other normal matrices. Python classes. Using functional magnetic resonance imaging (fMRI), we studied the neural correlates of the complexity of rhythmic finger tapping. Although the literature on nonlinear methods is vast, most of it has been written for applications that do not share the concerns or intellectual traditions of the. The type of phase portrait of a homogeneous linear autonomous system -- a companion system for example -- depends on the matrix coefficients via the eigenvalues or equivalently via the trace and determinant. How can I draw phase portrait of a matrix with given several initial points? what the phase portrait of a matrix is and you want to plot them together. If the motion is outward, while if it is inward. The matrix plot enables the realization that chaos only occurs under selective amplitudes and periods of the perturbation. Phase spaces are used to analyze autonomous differential equations. Plotting graphs of functions. The Van der Pol equation has been illustrated as an example. Using Matlab to get Phase Portraits Once upon a time if you wanted to use the computer to study continuous dynamical systems you had to learn a lot about numerical methods. m Matlab m-file for portraying an IFS based on a matrix of values henon. Analysis tools include power spectrum calculation and Poincare sections. (Leave out the special case when one of the eigenvalues is 1. By creating phase plane diagrams of our system we can visualize these features, such as convergence, equi-. Answers [1] General solutions: x y = C1 2 1 +C2e−5t 1 3. Phase space plots are difficult to draw, since motion must be built into the plot. Second order nonlinear ODEs. Prove that λ is a generalized eigenvalue of the matrix pair K, M if and only if it is an ordinary eigenvalue of the matrix P = M-1 K. One particular function called DEplot provides very attractive phase portraits. 0 : Return to Main Page. Draw small line with slope. We flnd these. unstable manifolds of saddle points. In the adaption phase, it adjusts neighborhood neuron pair's connected/disconnected topology constantly according to the statistics of input feature data. Let matrix A = 1 0 0 2 3 1 4 1 0 Then A looks triangular (upper bandwidth of 0). Moreover, local phase portrait of a hyperbolic equilibrium of a nonlinear system is equivalent to that of its linearization. Phase portraits (by using solutions) of the. In practice, the width is obtained by using an image. However, we can view this dynamically in MATLAB using the program pline. Phase Portraits and Equilibrium As discussed many times in the previous sections, there are three general solution types, Numerical, Analytic, and Qualitative. Handles MIMO systems by allowing choice of input and output. Week 5 : Chapter 1 Cont. Your browser. Closed-loop frequency response-Constant gain and phase loci, Nichol's chart and their use in stability study of systems. Show that y = C sin(t)+ D cos(t) is a family of solutions of y00+y = 0, and find all solutions that satisfy the following constraints. Example Consider the. , the stable and unstable points where ) regulate the portrait of the phase plane. The Simulink. The journal is divided into 81 subject areas. Posts about Phase Portraits written by tsvhh Computing packages partition matrix pdf PDF-XChange Viewer pdfsync pdfTeX PGF Phase Portraits Plot poster PracTeX. Phase portrait. Note that the right-hand expressions are not continuously differentiable at $(0,0)$ and there is no linearization at this point, which explains that while stable, the point $(0,0)$ does not look (after close examination) like a standard node. Then, we obtain the phase portrait of the projected reduced system simply by changing the direction of time on S r (p) in the phase portrait of. Let’s use the following two-dimensional system as an example. So far, we have looked at a scalar equation for a membrane with a single nonlinear channel. 3 Theorem (The Fundamental Theorem For Linear Systems) [12] Let be an matrix. Properties of flows and orbits. Alex, assuming that you are talking about a u-w (position-velocity, sort of) phase plot, here you go. 2: Phase curves in the vicinity of centers and saddles. Using Matlab for Autonomous Systems. The phase portrait on the sphere is equivalent to that on the plane (x, y). This page intentionally left blank Economic Dynamics Phase Diagrams and Their Economic Application Second Edition This is the substantially revised and restructured second edition of Ron Shone’s successful undergraduate and graduate textbook Economic Dynamics. Bode plots, Polar plots, Log-magnitude Vs phase plots, Nyquist stability criterion, Stability analysis, Relative stability, Gain margin, Phase margin, Stability analysis of system using Bode plots. A plot of position and velocity variables as a function of time is sometimes called a phase diagram. The phase portraits is able to perfectly capture all of the nonlinear trajectories and display them in a way that would be otherwise difficult. Using Matlab to draw phase portraits This is a quick notes to help you draw phase portraits using the quiver command in Matlab. Scalar, linear higher order differential equations: relation with systems of differential equations. This video is unavailable. The network of representative neurons, first create the necessary neurons according to the local density of the input data in the growth phase. Control System - This Video tutorial is meant to provide the readers the know how to analyze the control systems with the help of mathematical models. PhasePlane(sys,tspan,icond) plots the the phase plane portrait for a general second order nonlinear system defined by the function sys(t,x). Properties of flows and orbits. Depending on various factors, different trajectories can evolve for the same system. PhasePlane(sys,tspan,icond) plots the the phase plane portrait for a general second order nonlinear system defined by the function sys(t,x). Try all of the exercises. Hence, the two eigenvalues are negative. B) The same mesh loaded into a BSim environment. Sketch the phase plane portrait of a 2D system of first order differential equations. To get started, plot your data and open the Figure Palette from the Figure View menu. Remove all;. A phase portrait is a plot of multiple phase curves corresponding to different initial conditions in the same phase plane (Tabor 1989, p. Exercises and Examples 1. Right panels: phase space. Perhaps its existence is unknown or its effect unsuspected. In particular, plot -T svg, plot -T ai, plot -T ps, plot -T cgm, plot -T fig, plot -T pcl, and plot -T hpgl are affected by the environment variable PAGESIZE. Printing xs over it give you the trajectory for the initial conditions you have chosen. This will include deriving a second linearly independent solution that we will need to form the general solution to the system. First define the right hand side function f of the differential equation as. 552 Phase Plane Methods were used, then 50 solution curve segments have already been entered onto the graphic! Threaded orbits are added to show what happens to solutions that are plotted on longer and longer t-intervals. PortraitDensity determines the density of the phase portrait. The phase portrait is a spiral which at the point (0;1) is directed downward since c= 1 < 0. If the motion is outward, while if it is inward. However, we can see the trajectories at infinity on the sphere, these are the trajectories moving along the equator, i. The following worksheet is designed to analyse the nature of the critical point (when ) and solutions of the linear system X'. See its phase portrait on the next. Here is an example of a plot of a sine wave. Graphing Differential Equations. A phase portrait is a geometric representation of the trajectories of a dynamical system in the phase plane. Mathematical concepts that are taught include how to plot data points from a table, how to label ordered pairs, and how to name coordinates on a grid. The graphical representation of the solutions is often referred to as a phase portrait. Second order nonlinear ODEs. 12 For this case, a Q estimate (using the distribution of points along the major axis of the phase portrait) was used to monitor OSNR, while the width of the phase portrait was used to monitor CD. Determined appropriate matrix values for Q and R to achieve a specific bank angle (phi) within a specified amount of time and generated time histories of closed loop state and control responses. I have to plot its phase portrait using this code: And if first line was 9x - 21y - 10z + 5 where would 5 goes in the matrix? matlab. Stability Analysis for ODEs Marc R. The following plots have been produced with octave using the above procedure:. m — phase portrait plus graph of second order ordinary differential equation phasem. Phase portraits are an invaluable tool in studying dynamical systems. We have fixed at 3 and have observed the different behavior of the time evaluation and phase portraits as the parameter increases. Time is plot along the abscissa and x, y is plot along the ordinate. trajectories in the phase plane approach ~0 along curves that are tangent to the eigenvector of the eigenvalue closest to zero. Our second jaunt into the qualitative solution realm is the phase portrait. In the phase portrait below, every point on the green line is an equilibrium solution. sketching phase portraits. We will plot the derivatives as a vector at each (y1, y2) which will show us the initial direction from each point. a) b) c) 2 Consider the system defined by. 0 : Return to Main Page. Consider the homogeneous linear first-order system differential equations x'=ax+by y'=cx+dy. In practice, the width is obtained by using an image. Note that if, in your example, the roots were -1 and -2 (distinct but with same sign), the phase portrait would be composed of parabolas - confirm by yourself. Applications of Linear Algebra Basic Linear Systems and Matrices Cramer's Rule Determinant of a Matrix Dot Product Existence and Uniqueness of Solutions (Linear Equations) Finding the Inverse of a Square Matrix Gram-Schmidt Process Linear Equations Lines and Planes One-to-one Functions Onto Functions Row Reduction (Gaussian Elimination) Systems. Since in the nonlinear sytems the number of equilibrium points is not uniquie as in linear systems the. By plotting several trajectories you will get a preciser idea of phase diagram associated with. plot may behave differently depending on the environment in which it is invoked. 1BestCsharp blog 4,131,650 views. Parvan, no. points and locally (in a small region around the critical point), the phase portrait will resemble the corresponding picture. Inset: Time-lapse images of a passive particle rolling in a flow. Real part of is positive and the FP is unstable. 03SC (Alternatively, make the change of variables x 1 = x − x 0, y 1 = y − y 0, and drop all terms having order higher than one; then A is the matrix of coefficients for the linear terms. However, there is one idea, not mentioned in the book, that is very useful to sketching and analyzing phase. Note that the right-hand expressions are not continuously differentiable at $(0,0)$ and there is no linearization at this point, which explains that while stable, the point $(0,0)$ does not look (after close examination) like a standard node. Also included are some examples of each kind so you can get an idea of what the homework is asking you to sketch for the phase portrait. Note that the used values of parameters are same as those used by Yoshisuke Ueda when he found chaos in 1961 (Ueda, 1979, 1980, and 1992). Try 3D plots, equations, inequalities, polar and parametric plots. For system (1), especially assuming that detA6= 0, this problem. For mechanical systems, the phase space usually consists of all possible values of position and momentum variables. Classify the xed points and determine their stability, when possible. Stability of linear and nonlinear systems. We will use Matplotlib's colormap to define colors for the trajectories. edu Michael T. Notice that this equation is autonomous! The phase portraits of these linear systems display a startling variety. Nonlinear Methods for the Social Sciences Stephen J. Then we will analyze the geometric effect of transforming a linear map to its Jordan Normal Form. But they are all limited in some ways. View PNG (It opens in a new tab, so you may need to allow popups). In chapter 2, we spent some time thinking about the phase portrait of the simple pendulum, and concluded with a challenge: can we design a nonlinear controller to reshape the phase portrait, with a very modest amount of actuation, so that the upright fixed point becomes globally stable? With unbounded torque, feedback linearization solutions (e. Introduction Infectious diseases have tremendous influence on human life. The system of equations is written u’ = AA*u, where AAis a given 2×2-matrix anduis a column vector. This is a linear equation, which we can solve completely. For math, science, nutrition, history. 01: 1; plot(x, 2 * min(x, 1 - x)) axis equal axis([0 1 0 1]) Note that this is not a phase plot in the standard sense, which is a plot of the system state over time, nor a phase portrait, which describes the structure of phase space (and which does not really apply to maps). 6 Asymptotic Behavior Sec. y 2 ' = y 1 + 4 y 2 + y 2 2. ams 20 ma tla no ucsc solving systems of first order equations with ode45 2015, yonatan katznelson the numerical solver, ode45 is designed to work with first. Hence, the two eigenvalues are opposite signs. If is a function, then the \emph{graph} of this function is the collection of points in the plane, where is the domain of. Konstantin Zuev (USC) Math 245, Lecture 9 January 30, 2012 5 / 15. 03, Spring, 1999 It is convenient to represent the solutions to an autonomous system ~x0= f~(~x)(where ~x= x y ) by means of a phase portrait. (Leave out the special case when one of the eigenvalues is 1. There have been some options of phase portraits already. In short, the best method for the analysis of a second order system is to plot a phase portrait. In this section we will solve systems of two linear differential equations in which the eigenvalues are real repeated (double in this case) numbers. Have a look at ShowColormaps if you want more information. In linear algebra, an ordered pair of numbers typically represents a vector, which in the present setting of coordinates can be taken to mean a change in location. Two areas where progress is being made are motion planning for mobile robots of factory floors (or on the surface of Mars), and control of highly articulated robots—such as. For part (c), use Mathematica to draw a phase portrait with numerous solutions for the system (3. System Upgrade on Feb 12th During this period, E-commerce and registration of new users may not be available for up to 12 hours. MOL 410/510: Introduction to Biological Dynamics Fall 2012 Problem Set #4, Nonlinear Dynamical Systems (due 10/19/2012) 6 MUST DO Questions, 1 OPTIONAL question 1. For a linear system, you can find the eigenvalues of matrix and the corresponding eigenvectors. Phase portraits in two dimensions 18. mass-spring-damper model in phase variable form. TFY4305 solutions extra exercises 2014 Phase portrait of problem 6. This video is unavailable. Again, the southern Barents–Kara metrics display similar behavior, which we choose not to plot for visual clarity. * Updated many function names to be more consistent with python coding. • Used MATLAB to sketch phase portraits and approximated changing behavior of dynamic systems in the neighborhood of the fixed point, then identified types of bifurcation • Nondimensionalized. Hannah Bardolaza, Maria Angela Faustino, Alexander De Los Reyes, Victor DC Vistro, Neil Irvin Cabello, Ivan Cedrick Verona, John Paul Ferrolino, Gerald Angelo Catindig, Karl Cedric Gonzales, Kerphy Liandro Patrocenio, Rommel Jagus, Elizabeth Ann Prieto, Deborah Anne Lumantas, John Daniel Vasquez, Jessica Afalla, Joselito Muldera, Valynn Katrine Mag-usara, Arnel Salvador, Armando Somintac. If the motion is outward, while if it is inward. MAPLETM LAB MANUAL FOR MATH 237 Differential Equations and Computer Methods Written by CESAR O. Since stable and unstable equilibria play quite different roles in the dynamics of a system, it is useful to be able to classify equi-librium points based on their stability. You can also plot the vector field associated with the system using quiver function. Example: Consider the harmonic oscillator equation. Phase portraits for 2 × 2 systems. 1 x x 1 2 v 2 v 1-1 1-1 Phase portraits for 2 × 2 systems. Left plot is for a= 1 and right plot is for a= 1. That is, it makes a few assumptions about the sampling rates that may not be evident to the average. A Javascript app to display the slope field for an ordinary differential equation, or the direction field (phase plane) for a two-variable system, and plot numerical solutions (e. The solution is on the ray in the opposite direction. 1 Linear equations Solving linear systems of equations is straightforward using the numpy submodule linalg. Sketch the phase portrait of this system. See examples/geneticswitch. • To plot 4(b), we need only be able to compute ˙x(x), it is not necessary to solve the differential equation. A phase portrait of a simple harmonic oscillator x 2 x 0. Note: Worksheets are in Maple 8 files and need the software to run. With only one eigenvector, it is a degenerated-looking node that is a cross between a node and a spiral point (see case 4 below). A phase portrait is a plot of a segment’s angular displacement versus its angular velocity (Barela et al. We flnd these. m — phase portrait plus graph of second order ordinary differential equation phasem. The procedure of dividing one equation by other other is quite general. We will be determining qualitative features of a dis-crete dynamical system of homogeneous di erence equations with constant coe cients. | 2019-12-14T16:25:51 | {
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https://math.stackexchange.com/questions/3011313/how-many-ways-to-arrange-four-types-of-books-on-the-shelf-so-that-books-of-the-s | # How many ways to arrange four types of books on the shelf so that books of the same type are together?
Janet has $$10$$ different books that she is going to put on her bookshelf. Of these, $$4$$ are Book C, $$3$$ are Book B, $$2$$ are Book S and $$1$$ Book P. Janet wants to arrange her books so that all the books dealing with the same subject are together on the shelf. How many different arrangements are possible?
My workings are
$$4! \cdot 3! \cdot 2! \cdot 1!$$
but the actual answer has an additional multiplication by $$4!$$
Why is this so? My guess is because we need the $$4$$ books to come together, let's say for book C. But isn't that "shown" by the working of $$4!$$ for Book C and $$3!$$ for Book B and so on?
• Problems of enumeration should be tagged combinatorics rather than probability. Nov 24 '18 at 11:10
$$C_1 C_2 C_3 C_4 | B_1 B_2 B_3 | S_1 S_2 | P_1$$
The books within each block can be arranged in $$4!3!2!1!$$ ways AND the block themselves can be arranged in $$4!$$ ways. That's where your extra $$4!$$ is coming from. | 2022-01-24T20:13:18 | {
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https://math.stackexchange.com/questions/2811436/does-this-fractal-have-a-hausdorff-dimension-of-1 | # Does this fractal have a Hausdorff dimension of 1?
So I'm familiar with the simple methods of calculating the Hausdorff dimension of a fractal, but when I try to apply them for this case I get into trouble. What I mean with the simple method is using the scaling factor and the amount of repetitions to exactly calculate the Hausdorff dimension.
The fractal below is constructed by taking all horizontal line segments, dividing each into three equally long line segments, raise the one in the middle by it's own length and connecting it to its former neighbors with two vertical line segments.
If I decided during my construction steps not to add those vertical lines, it would be clear that the whole fractal would be 1-dimensional. Each vertical line segment is 1-dimensional as well. But if I were to assume that the whole fractal was 1-dimensional, then it would be of infinite length.
There are more cases of fractals I can think of where I run into the same questions. For example a fractal where each point has position (1/2)i on the line interval [0,1] for all i in ℕ. Would this be a 0-dimensional fractal of infinite size?
• Yes, your set has Hausdorff dimension one. It is the union of a one dimensional self-similar set consisting of three parts scaled by the fact $1/3$ together with countably many line segments. Since Hausdorff dimension is stable under countable unions, the result has Hausdorff dimension 1. Note that even a bonded differentiable curve can have infinite length so there's no problem there. – Mark McClure Jun 7 '18 at 15:17
After $n$ subdivisions, there are $3^n$ horizontal pieces, each of which is $3^{-n}$ in length. The $n^\text{th}$ division adds $2\cdot3^{-n}\cdot3^{n-1}=\frac23$ in vertical pieces to the length. So looking at the resolution of $3^{-n}$, the curve takes at most $\left(1+\frac23n\right)3^n$ disks of that size to cover.
Thus, the Hausdorff dimension is $$\lim_{n\to\infty}\frac{\log\left(\left(1+\frac23n\right)3^n\right)}{\log\left(3^n\right)}=1$$
First, an important result:
Theorem: Let $\{X_n\}_{n\in\mathbb{N}}$ be a countable collection of (possibly empty) subsets of some metric space $\mathscr{X}$. Then $$\dim_H\left( \bigcup_{n\in\mathbb{N}} X_n \right) = \sup_{n\in\mathbb{N}} \dim_H(X_n),$$ where $\dim_H(X)$ is the Hausdorff dimension of $X$.
This can be summarized (as per Mark McClure's comment) as "the Hausdorff dimension is stable under countable unions." I'll not prove the result here, but there is a proof given in answer to another question.
The set shown in the question consists (broadly speaking) of two parts: a collection of vertical line segments (each of which has dimension 1), and some "dust" at the ends of the segments. The intervals are easily dealt with: there is one vertical segment at each triadic rational number between 0 and 1 (i.e. if $x = \frac{k}{3^n}$ for some $n\in\mathbb{N}$ and $0 \le k \le 3^n$, then there is a vertical segment based at the point $(x,0)$—if we are a bit more careful, we could work out the length of each such segment and thereby determine the total length of all of the segments, but the question was about dimension, not rectifiability). The set of triadic rationals is countable, so the total collection of vertical segments is countable. Hence the union of all of the segments form a set of Hausdorff dimension 1.
The "dust" is a bit more difficult to contend with, but not impossible. Again following Mark McClure's lead, we can note that the dust is a self-similar set of Hausdorff dimension 1. To make the argument more precise in a way that I find comfortable (as I am a guy that likes to work with iterated function systems), define $$\varphi_1(x,y) := \frac{1}{3}(x,y), \quad \varphi_2(x,y) := \frac{1}{3}(x,y) + \left(\frac{2}{3},0\right) \quad\text{and}\quad \varphi_3(x,y) := \frac{1}{3}(x,y) + \left(\frac{1}{2},\frac{1}{2}\right).$$ The "dust" is the attractor or fixed point of this iterated function system (IFS). That is, call the dust $\mathscr{D}$ and define $$\Phi(X) := \bigcup_{j=1}^{3} \varphi_j(X)$$ for any $X \subseteq \mathbb{R}^2$, then $\mathscr{D}$ has the property that $$\Phi(\mathscr{D}) = \mathscr{D}.$$ Via some abstract nonsense (essentially, an application of the Banach fixed point theorem applied to $\Phi$ acting on the space of compact subsets of $\mathbb{R}^2$, see for example Hutchinson, 1981), we can be sure that this IFS has a compact attractor. Moreover, it is not too hard to check that the IFS satisfies the open set condition, from which it follows (via some more abstract nonsense, see the above cited paper) that $\dim_H(\mathscr{D}) = s$, where $s$ is the unique real solution the the Moran equation $$3\left(\frac{1}{3}\right)^s = 1.$$ It is not too hard to see that $s = 1$, and so $\dim_H(\mathscr{D}) = 1$.
Therefore, since the set in question is a countable union of sets which each have Hausdorff dimension 1, it follows from the stability of the Hausdorff dimension under countable unions that the Hausdorff dimension of this set is 1. | 2019-12-07T04:18:22 | {
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https://mathschallenge.net/full/continued_fraction_recurrence_relation | ## Continued Fraction Recurrence Relation
#### Problem
Any real number, $\alpha$, can be represented as a continued fraction which may either terminate or be infinite (repeating or non-repeating).
$\alpha = a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + ...}}}$
Let $\dfrac{p_n}{q_n}$ be defined as the $n$th convergent:
$\dfrac{p_n}{q_n} = a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + ... + \cfrac{1}{a_n}}}}$
For example, $\sqrt{2} = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + ...}}}$
And the first five convergents are:
\begin{align}\dfrac{p_0}{q_0} &= 1\\\dfrac{p_1}{q_1} &= \dfrac{3}{2}\\\dfrac{p_2}{q_2} &= \dfrac{7}{5}\\\dfrac{p_3}{q_3} &= \dfrac{17}{12}\\\dfrac{p_4}{q_4} &= \dfrac{41}{29}\\\end{align}
Prove that the following recurrence relation holds for $n \ge 2$ for all continued fractions:
\begin{align}p_n &= a_n p_{n-1} + p_{n-2}\\q_n &= a_n q_{n-1} + q_{n-2}\end{align}
#### Solution
From the definition of a continued fraction, $\dfrac{p_0}{q_0} = a_0 \implies p_0 = a_0$ and $q_0 = 1$.
Similarly $\dfrac{p_1}{q_1} = a_0 + \dfrac{1}{a_1} = \dfrac{a_0 a_1 + 1}{a_1} \implies p_1 = a_0 a_1 + 1$ and $q_1 = a_1$.
In other words, $p_0$, $p_1$, $q_0$, and $q_1$ are clearly defined. Now let us consider the convergent $\dfrac{p_2}{q_2}$ as a continued fraction expansion.
\begin{align}\dfrac{p_2}{q_2} &= a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2}}\\&= a_0 + \cfrac{\cfrac{1}{a_1 a_2 + 1}}{a_2}\\&= a_0 + \cfrac{a_2}{a_1 a_2 + 1}\\&= \cfrac{a_0(a_1 a_2 + 1) + a_2}{a_1 a_2 + 1}\\&= \cfrac{a_0 a_1 a_2 + a_0 + a_2}{a_1 a_2 + 1}\\&= \cfrac{a_2(a_0 a_1 + 1) + a_0}{a_1 a_2 + 1}\end{align}
But as $p_1 = a_0 a_1 + 1$, $p_0 = a_0$, $q_1 = a_1$, and $q_0 = 1$, we get:
$$\dfrac{p_2}{q_2} = \dfrac{a_2 p_1 + p_0}{a_2 p_1 + q_0}$$
Hence the recurrence relation is true for $n = 2$.
Now suppose that the recurrence relation holds for $n = k$, and consider the continued fraction expansions of the convergents $\dfrac{p_k}{q_k}$ and $\dfrac{p_{k+1}}{q_{k+1}}$.
$$\dfrac{p_k}{q_k} = a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + ... + \cfrac{1}{a_k}}}}$$$$\dfrac{p_{k+1}}{q_{k+1}} = a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + ... + \cfrac{1}{a_k + \cfrac{1}{a_{k+1}}}}}}$$
It can be seen that going from $\dfrac{p_k}{q_k}$ and $\dfrac{p_{k+1}}{q_{k+1}}$ we have replaced $a_k$ with $a_k + \dfrac{1}{a_{k+1}}$.
\begin{align}\therefore \dfrac{p_k}{p_k} = \dfrac{a_k p_{k-1} + p_{k-2}}{a_k q_{k-1} + q_{k-2}} \rightarrow \dfrac{p_{k+1}}{q_{k+1}} &= \dfrac{\left(a_k + \dfrac{1}{a_{k+1}}\right) p_{k-1} + p_{k-2}}{\left(a_k + \dfrac{1}{a_{k+1}}\right) q_{k-1} + q_{k-2}}\\&= \dfrac{a_{k+1} a_k p_k-1 + p_{k-1} + a_{k+1} p_{k+2}}{a_{k+1} a_k p_{k-1} + q_{k-1} + a_{k+1} q_{k-2}}\\&= \dfrac{a_{k+1}(a_k p_{k-1} + p_{k-2}) + p_{k-1}}{a_{k+1}(a_k q_{k-1} + q_{k-2}) + q_{k-1}}\\&= \dfrac{a_{k+1} p_k + p_{k-1}}{a_{k+1} q_k + q_{k-1}}\end{align}
This is the expected outcome for $\dfrac{p_{k+1}}{q_{k+1}}$. As the recurrence relation is true for $n = 2$, and $p_0$, $p_1$, $q_0$, and $q_1$ are properly defined, we have shown that the recurrence relation holds for all values of $n \ge 2$.
However, for practical purposes it is convenient for programmers to set $p_{-1} = 1$, $q_{-1} = 0$, $p_0 = a_0$, and $q_0 = 1$. Then the recurrence relation holds for $n \ge 1$.
Problem ID: 282 (15 Jul 2006) Difficulty: 4 Star
Only Show Problem | 2018-02-25T15:55:13 | {
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https://math.stackexchange.com/questions/1399406/what-is-the-number-of-invertible-n-times-n-matrices-in-operatornamegl-nf | # What is the number of invertible $n\times n$ matrices in $\operatorname{GL}_n(F)$?
$F$ is a finite field of order $q$. What is the size of $\operatorname{GL}_n(F)$ ?
I am reading Dummit and Foote "Abstract Algebra". The following formula is given: $(q^n - 1)(q^n - q)\cdots(q^n - q^{n-1})$. The case for $n = 1$ is trivial. I understand that for $n = 2$ the first row of the matrix can be any ordered pair of field elements except for $0,0$. and the second row can be any ordered pair of field elements that is not a multiple of the first row. So for $n = 2$ there are $(q^n - 1)(q^n - q)$ invertible matrices. For $n\geq 3$, I cannot seem to understand why the formula works. I have looked at Sloane's OEIS A002884. I have also constructed and stared at a list of all $168$ $3\times 3$ invertible matrices over $GF(2)$. I would most appreciate a concrete and detailed explanation of how say $(2^3 - 1)(2^3 - 2)(2^3 - 2^2)$ counts these $168$ matrices.
• An invertible matrix must map a basis to a basis. The number of bases of $\mathbb{F}_q^n$ is the formula you gave above. See Example 1 on p. 412 of Dummit and Foote for a derivation of this formula. – André 3000 Aug 16 '15 at 16:13
• @SpamIAm, I don't think that's actually correct. This formula counts the number of ordered bases. To make them unordered, you need to divide by $n!$ – Marcus M Aug 16 '15 at 16:18
• @MarcusM I certainly want to consider my bases as ordered, so I shouldn't divide out. I should have specified that in my comment, though. – André 3000 Aug 16 '15 at 16:20
In order for an $n \times n$ matrix to be invertible, we need the rows to be linearly independent. As you note, we have $q^n - 1$ choices for the first row; now, there are $q$ vectors in the span of the first row, so we have $q^n - q$ choices for the second row. Now, let $v_1, v_2$ be the first two rows. Then the set of vectors in the span of $v_1, v_2$ is of the form $\{c_1 v_1 + c_2 v_2 | c_1,c_2 \in F\}$. This set is of size $q^2$, as we have $q$ choices for $c_1$ and $q$ choices for $c_2$. Thus, we have $q^n - q^2$ choices for the third row. Continuing this gives the desired formula.
For $n=3$, the third row must not be in the subspace generated by the first two rows. A vector in this subspace requires $2$ coefficients ($q^2$ possibilities), you must substract $q^2$ vectors, whence a third factor $q^3-q^2$. And so on. | 2019-11-13T01:46:48 | {
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https://math.stackexchange.com/questions/3774360/i-need-help-answering-and-understanding-this-problem-about-cauchy-sequences | # I need help answering and understanding this problem about Cauchy Sequences.
Context: I done this problem awhile back and was looking through my notes on it and my answer seems incorrect.
Let $$(x_n)_{n{\in}\mathbb{N}}$$ be a sequence such that $$|x_n-x_{n+1}|\;{\le}\;2^{-n}$$ holds for every $$n\,{\in}\,\mathbb{N}$$. Show that $$(x_n)_{n{\in}\mathbb{N}}$$ is a Cauchy sequence.
$$|x_n-x_{n+1}|\;{\le}\;2^{-n}{\implies}(x_n)_{n{\in}\mathbb{N}}\;\text{is monotone and}\;2^{-n}\rightarrow0{\implies}(x_n)_{n{\in}\mathbb{N}}\;\text{is bounded. Thus}\;(x_n)_{n{\in}\mathbb{N}}\;\text{is convergent which implies it is a Cauchy sequence}$$
My issue with this answer is I believe "$$|x_n-x_{n+1}|\;{\le}\;2^{-n}{\implies}(x_n)_{n{\in}\mathbb{N}}\;\text{is monotone}$$" is false. The inequality is not sufficiently strong to impose that the sequence is either increasing or decreasing?
My thoughts are that removing the modulus sign (purely done because I could not get anywhere otherwise) would make it a more valid answer (obviously it would no longer answer the initial question and I still doubt it is entirely correct even with no modulus sign). The reason I doubt that it is correct without the modulus sign is the Wikipedia page on Cauchy sequences briefly talks about the sequence of square roots of natural numbers not being a Cauchy sequence despite the consecutive terms becoming arbitrarily close and I noticed that my argument would suggest that it is a Cauchy sequence.
After this I realised what the wiki was saying and observed that $$\sum_{n=1}^{\infty} \frac{1}{2\sqrt{n}} \;\text{diverges since}\sum_{n=1}^{\infty} \frac{1}{n}\;\text{diverges and}\,0\le\frac{1}{n}\le\frac{1}{2\sqrt{n}}\;\text{for}\; n\ge4$$ (the above summation comes from the wikipedia page)
but $$\sum_{n=1}^{\infty} 2^{-n}=1$$
However now I am just lost (probably due to lack of understanding about Cauchy sequences). Doesn't the fact the series converges to 1 mean that you could use the same argument from the Wikipedia page on Cauchy sequences? I actually doubt this is the case somewhat but can't seem to wrap my head around the logic.
Any help would be greatly appreciated. My main issue comes with the modulus sign and also the fact that the series of distances between consecutive terms converge to 1 a (not even sure this is relevant however).
Yes, you're argument was wrong, since as you said the condition on the absolute value cannot tell you anything about the monotonicity. However you can note that for any $$m>0$$ you have $$|x_{n+m}-x_n| \leq \sum_{k=n}^{n+m-1}|x_{k+1}-x_k|\leq\sum_{k=n}^{n+m-1} 2^{-k} = 2^{-n}\sum_{k=0}^{m-1}2^{-k}\,.$$ From here you can see that indeed the argument $$\sum_{k=1}^\infty 2^{-k}=1$$ can be used to prove that the sequence is Cauchy. Indeed you have that $$\sum_{k=0}^{m-1}2^{-k}\leq \sum_{k=0}^{\infty}2^{-k}=2$$ for all $$m\geq 1$$. So you get $$|x_{n+m}-x_n|\leq 2\times 2^{-n}$$ which tends to $$0$$ for $$n\to\infty$$. This proves that the sequence is Cauchy.
Note that in general, a sufficient condition for a sequence to be Cauchy is that, given that $$|x_{n+1}-x_n|\leq \epsilon_n$$, $$\sum_{n=0}^\infty\epsilon_n \leq \infty$$. Indeed you will have $$|x_{n+m}-x_n|\leq \sum_{k=n}^\infty \epsilon_k$$ and the RHS must vanishes for $$n\to \infty$$.
However such a condition is in general not necessary. For instance the sequence $$x_n=(-1)^n\frac{1}{n}$$ clearly converges (and so is Cauchy) but $$\sum_{n=0}^\infty |x_{n+1}-x_n| = \infty$$. It is the case that the condition on the sum is necessary if $$x_n$$ is a monotonous sequence.
• I'm just saying $\sum_{k=n}^{n+m-1}2^{-k} = \sum_{k'=0}^{m-1}2^{-n-k'}$ defining $k'=k-n$. Then you can take $2^{-n}$ outside of the sum and rewrite $k$ instead of $k'$. | 2021-10-19T16:32:36 | {
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https://math.stackexchange.com/questions/2358962/why-when-you-study-math-or-physics-you-are-obliged-to-know-how-this-or-that-equa | # why when you study math or physics you are obliged to know how this or that equation is derived?
every time you learn a new equation they tell you to prove how this formula was derived, but why is this made ?
i mean the derivation of an special formula is applied once and can not be used to prove another formulae
so why in the process of learning you must know how to derive each math formula you learn ?
• The alternative is to just memorize results handed down to you by some alleged authority.
– zhw.
Jul 14, 2017 at 19:21
I believe there are many different kinds of good answers to your question, but I'll just focus on one:
If you're lucky, you may indeed go through life and successfully apply the formula without being able to derive it.
However, you may find yourself in a situation that is slightly different from the kind of situation that the formula was intended for. In that case, if you knew how the formula was derived in the old kind of situation, you may be able to change that derivation to derive a new formula that is more appropriate for that new kind of situation.
Indeed, if you know the derivation, then you are immediately more likely to know the assumptions being made for that formula, and thus be able to gauge whether it is appropriate for you to use the formula in some kind of situation or for some problem in the first place, by seeing if those assumptions indeed hold true for the situation or problem you are faced with.
• Indeed. Learning does not mean being able to solve exercises in a book''. Jul 14, 2017 at 18:55
• @AndrésE.Caicedo That would be the short way of putting this, yes :) Jul 14, 2017 at 18:56
• And, it can be a memory aid. Sometimes its easier to remember the derivation than the formula itself. And sometimes (say when you work with others) it helps to explain the result if you know where it came from, Jul 14, 2017 at 18:58
• @kimchilover Agreed (personal example: Bayes' formula. I don;t even try to remember the formula, since I know I can immediately derive it from $P(A \cap B)=P(B \cap A)$ and $P(A \cap B)=P(A|B)*P(B)$) You should post this as a separate answer, as I believe there are many different answers to the question as posed. Jul 14, 2017 at 19:02
There's a few reasons, even if you don't intend on deriving similar formulas in the future.
First, it's hard to remember formulas. For many people, it's easier to remember reasons - maybe not the whole proof, but if you remember just the outline of the proof it can help you remember the formula. For example, I can't for the life of me remember the formula for the arc length of a curve - but I do remember the gist of the way it's derived, by taking an infinitesimal fragment of the curve and applying the Pythagorean Theorem. That's enough for me to remember that I'm looking for something like $\int \sqrt{?^2 + ?^2}dt$, and then it's just a matter of filling in the blanks.
Second, it's hard to remember the uses of formulas. Yes, most formulas have a one-to-three-word phrase that describes their use; for example, the quadratic formula is for "solving quadratic equations". But even in these simple cases, there are literally thousands of random special situations where you can use them - for example, the quadratic formula can be used to solve $x^4 + 2x^2 + 3 = 0$, and by analyzing the formula you can see a way to tell whether a quadratic has a real root without having to take any square roots. Knowing the reason behind the formula - in this case, remembering that the proof of the quadratic formula doesn't depend on using $x$ specifically rather than $x^2$ - helps you see these additional uses without having to do huge amounts of memorization.
Third, it's hard to remember the non-uses of formulas - that is, the situations where a formula might look like it applies, even though it doesn't. Taking the quadratic formula example from above, it's a common mistake for students to try to apply the quadratic formula to situations like $x^3 + 2x + 1 = 0$, or to $x^2 + 2x + 1 = 1$ without rearranging the formula first. Understanding the proof behind the quadratic formula makes it obvious that the formula simply doesn't apply in either of these cases; the first because the proof is about squares rather than cubes, and the second because the proof gives a factorization of the quadratic rather than just "the answer".
Fourth (and finally, for now), it helps unify formulas. To take an example from multivariable calculus: Green's Theorem, Stokes' Theorem, and the Divergence Theorem are all basically the same formula - each applies in different situations, but they all arise the same way. If you've never seen a proof of these theorems, or if you didn't understand the proof you were shown, then you wind up just memorizing three separate formulas and having to memorize a laundry list of situations in which one applies and the others don't. If you do understand how these formulas arise, then to you they're more like one theorem than three; and when to use which one becomes obvious, requiring no memorization at all. This sort of thing happens a lot - and even when it doesn't, there are still a lot of situations where several formulas are closely related, and understanding their relationships can help understand how to use them.
Richard Feynman, speaking on studying physics, gave the following advice (but I think the same advice applies to studying mathematics):
"It will not do to memorize the formulas, and to say to yourself, 'I know all the formulas, all I gotta do is figure out how to put 'em in the problem!'
"Now, you may succeed with this for a while, and the more you work on memorizing the formulas, the longer you'll go on with this method--but it doesn't work in the end.
"You might say, 'I'm not gonna believe him, because I've always been successful: that's the way I've always done it; I'm always gonna do it that way.'
"You are not always going to do it that way: you're going to flunk-- not this year, not next year, but eventually, when you get your job, or something--you're going to lose along the line somewhere, because physics is an enormously extended thing: there are millions of formulas! It's impossible to remember all the formulas--it's impossible!"
--Richard Feynman, Feynman's Tips on Physics | 2022-05-25T10:58:07 | {
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http://math.stackexchange.com/questions/261098/simple-inequality-involving-exponential-function | # Simple inequality involving exponential function
Can anyone explain why,
$$\displaystyle 1 - e^{-\frac{1}{k}} \geq \frac{1}{ke}$$
for $k \geq 1$.
This inequality is used to show that the series
$$\sum_{k=1}^\infty 1 - e^{-\frac{1}{k}}$$
is divergent, and I do not see how one can derive the result.
-
Lets use only our knowledge of $e$, and a geometric series identity. Notice that $$\left(1-e^{-\frac{1}{k}}\right)\left(1+e^{-\frac{1}{k}}+e^{-\frac{2}{k}}\cdots+e^{-\frac{k-1}{k}}\right)=1-e^{-1}.$$ Now, $e^{-x}\leq1$ for $x\geq0,$ so $$\left(1-e^{-\frac{1}{k}}\right)k=\left(1-e^{-\frac{1}{k}}\right)\left(1+1+1\cdots+1\right)$$
$$\geq\left(1-e^{-\frac{1}{k}}\right)\left(1+e^{-\frac{1}{k}}+e^{-\frac{2}{k}}\cdots+e^{-\frac{k-1}{k}}\right)$$
$$=1-e^{-1}$$ $$\geq\frac{1}{e},$$ and so $$1-e^{-\frac{1}{k}}\geq\frac{1}{ke}.$$
-
You actually have equality in one step there, although equality does imply $\geq$... – Thomas Andrews Dec 18 '12 at 0:31
@ThomasAndrews: Thanks, fixed. – Eric Naslund Dec 18 '12 at 0:33
$$e^x \left(1 - \dfrac{x}{e}\right) \geq 1$$ for $x \leq 1$.
Consider $f(x) = e^x - x e^{x-1}$. We then have $$f'(x) = e^x - x e^{x-1} - e^{x-1} = e^{x-1} \left(e-x-1\right) > 0$$ for $x \leq 1$. Hence, $f(x)$ is an increasing function for $x \leq 1$. Hence, $$f(x) \geq f(0) \implies e^x \left(1 - \dfrac{x}{e}\right) \geq 1$$ Choosing $x = \dfrac1k$, we get $$e^{1/k} \left(1 - \dfrac1{ke}\right) \geq 1 \implies e^{-1/k} \leq 1 - \dfrac1{ke} \implies 1 - e^{-1/k} \geq \dfrac1{ke}$$
-
Since $\frac{k+1}{k}=1+\frac{1}{k}\leq e^{\frac{1}{k}}$, therefore $e^{-1/k}\leq \frac{k}{k+1}$. Thus, $\frac{1}{k+1}\leq 1-e^{-1/k}$ . This inequality will do the job.
However, if you still want to show your inequality you can use the fact that $\forall k\geq 1[\frac{1}{k+1}\geq \frac{1}{ke}]$ | 2013-12-19T09:00:06 | {
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https://math.stackexchange.com/questions/2192881/given-x9-e-and-x11-e-prove-x-e/2192887 | # Given $x^9 = e$ and $x^{11} = e$ prove $x = e$.
Full Problem: Prove that for any element $x$ in a group $G$ that satisfies $$x^9 = e \\ x^{11} = e,$$ where $e$ is the identity element, that $x$ itself must be $e$.
Is this as simple as showing that
• $x^{11} = x^{9} \cdot x^{2} = e \cdot x^2 \Rightarrow x^2 = e$
• $x^{9} = x^{2} \cdot x^{7} = e \cdot x^7 \Rightarrow x^7 = e$
• $x^{7} = x^{2} \cdot x^{5} = e \cdot x^5 \Rightarrow x^5 = e$
• $x^{5} = x^{2} \cdot x^{3} = e \cdot x^3 \Rightarrow x^3 = e$
• $x^{3} = x^{2} \cdot x = e \cdot x \Rightarrow x = e$
Therefore, $x = e$.
• More generally, if $n$ and $m$ are any two relatively prime integers, then Bezout's lemma there exist $a$ and $b$ such that $an+bm=1$. In this case, if $x^n=e$ and $x^m=e$, then $x^{an+bm}$ is both $x^1$ and $e$. – arctic tern Mar 18 '17 at 23:08
Your reasoning is correct but here is a more direct argument.
Since $x^9 = e$ and $x^{11} = e$, the order of $x$ divides both $9$ and $11$. Therefore, the order of $x$ is $1$ so $x = e$.
• The proof of "$x^n=e\implies n$ is divisible by the order of $x$" is more or less a generalised version of this problem, so I wouldn't call it "more direct". – Arthur Mar 18 '17 at 23:16
• @Arthur It avoids several steps in the original proof and gets at the essence of the problem. Sometimes a more general solution can be simpler and more direct. – Qudit Mar 18 '17 at 23:19
• It doesn't avoid several steps, it hides them by quoting a theorem that uses basically the same steps in its proof. I'm not saying it's an invalid solution, but I do contest the "more direct" claim ("direct" is not the same as "short" or "elegant"), although that is a subjective thing. – Arthur Mar 18 '17 at 23:24
• All it requires is knowing that if $x^n = e$ then the order of $x$ divides $n$ and that the gcd is $1$. The result in Bernard's answer is what would be a generalization (which my answer essentially proves) not dividing the order. Anyway, I am done with this pointless quibbling. – Qudit Mar 18 '17 at 23:27
• I am not done. What you say is the whole point of my quibbling. You say "knowing that if $x^n=e$ then the order of $x$ divides $n$", and I say "what proof do you have of that?" and then, if you're in an accommodating mood, you will give me a proof. That proof will look almost identical to how the OP solved his problem, and you will go through exactly the same steps, just with general exponents instead of $9$ and $11$. That is what I mean when I say "It doesn't avoid steps, it hides them". – Arthur Mar 18 '17 at 23:37
$$e=(x^{11})^5=x^{55}=x^{54}\cdot x=(x^9)^{6}\cdot x=e\cdot x=x$$
As you can see, this thus follows because there is integer solutions to $11x-9y=1$, which is true because $11$ and $9$ are relatively prime.
Your approach is doing much the same, using a slow form of the Euclidean algorithm to show that $11$ and $9$ are relatively prime:
$$11=9\cdot 1 + 2\\ 9=2\cdot 1 + 7\\ 7=2\cdot 1 + 5\\ 5=2\cdot 1 + 3\\ 3=2\cdot 1 + 1$$
You could have skipped a lot of steps by doing the equivalent of $9=2\cdot 4 + 1$, as other answers have suggested.
Yes, it is that simple. It can be done even shorter, because after showing $x^2=e$, you can go straight to $$e=x^9=x(x^2)^4=xe^4=x$$and you're done.
It can be as simple as what you proved, but it can be shorter. Actually you can prove the following result:
If in a group, $x^m=e$ and $x^n=e$ for coprime $m$ and $n$, then $x=e$.
Indeed we have a Bézout's relation: $\; um+vn=1$, so $$x=x^{um+vn}=(x^m)^u(x^n)^v=e^ue^v=e.$$ | 2020-07-04T21:32:31 | {
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https://math.stackexchange.com/questions/4403081/how-to-integrate-int-fracdxx-sqrtx4-1 | # How to integrate $\int \frac{dx}{x\sqrt{x^4-1}}$?
$$\int \dfrac{dx}{x\sqrt{x^4-1}}$$
I need to solve this integration. I solved and got $$\dfrac12\tan^{-1}(\sqrt{x^4-1}) + C$$, however the answer given in my textbook is $$\dfrac12\sec^{-1}(x^2) + C$$
How can I prove that both quantities are equal? Is there something wrong with my answer?
EDIT:
Here's my work: $$\int\dfrac{dx}{x\sqrt{x^4-1}}= \dfrac{1}{4}\int\dfrac{4x^3 dx}{x^4\sqrt{x^4-1}}$$
Let $$x^4 - 1 = t^2$$ $$\dfrac{1}{2}\int\dfrac{dx}{1 + t^2}$$
$$\dfrac12 \tan^{-1}(\sqrt{x^4 -1 }) + C$$
• Post your solution as well please. Mar 14 at 13:17
• Hint: if $\tan y=\sqrt{x^4-1}$ then $\sec y=\pm\sqrt{1+\tan^2y}=\cdots$.
– J.G.
Mar 14 at 13:22
## 4 Answers
I tell my students that inverse trig functions are angles. So if you write
$$\tan^{-1}\sqrt{x^4-1} = \theta,$$
then
$$\tan\theta = \sqrt{x^4-1}.$$
A right triangle that tells this story has $$\theta$$ as one angle, $$\sqrt{x^4-1}$$ as the opposite side and $$1$$ as the adjacent side. Using Pythagorean theorem we can work out the length $$c$$ of the hypotenuse:
$$(\sqrt{x^4-1})^2+1^2 = c^2$$
which shows that $$c=x^2$$.
So $$\sec \theta = x^2/1$$, that is $$\sec^{-1}(x^2) = \theta = \tan^{-1}( \sqrt{x^4-1}).$$
• It should be added that if $\tan^{-1} \sqrt{x^4-1} = \theta$ then $0 \le \theta < \pi/2$, since $\sqrt{x^4-1} \ge 0$. That is needed to justify making a right triangle with angle $\theta$. Mar 14 at 15:34
The answer was already given by B. Goddard, here I try to give a visual answer:
• Awesome, I didn't expect this. Thanks a ton!
– user983206
Mar 14 at 15:33
I think your problem is already solved with B. Goddard's answer, Here's an easy way to solve the given integral.
We have, $$\int\dfrac{dx}{x\sqrt{x^4-1}}$$
Let $$x^2 = \sec(\theta) \implies 2x \, dx = \sec(\theta) \tan(\theta) \, d\theta$$
Which further implies that, $$dx = \dfrac{\sqrt{\sec(\theta)}\tan(\theta)}{2}\, d\theta$$
After substitution, the given integral changes to: $$\dfrac12\int\dfrac{\sqrt{\sec(\theta)}\tan(\theta)}{\sqrt{\sec(\theta)}\cdot\sqrt{\sec^2(\theta) - 1}}\, d\theta$$
$$=\dfrac12\int\dfrac{\sqrt{\sec(\theta)}\tan(\theta)}{\sqrt{\sec(\theta)}\cdot\sqrt{\tan^2(\theta)}}\, d\theta$$
$$=\dfrac12\int\dfrac{\sqrt{\sec(\theta)}\tan(\theta)}{\sqrt{\sec(\theta)}\tan(\theta)}\, d\theta$$
$$=\dfrac12\int d\theta$$
$$=\dfrac12\theta + C$$
$$=\boxed{\dfrac12\sec^{-1}(x^2) + C}$$
• Woah! Nice substitution.
– user983206
Mar 14 at 15:31
There is nothing wrong with your answer. The antiderivative you found is correct.
Your book's answer is also correct. The thing is that
$$\frac{1}{2}\tan^{-1}(\sqrt{x^4-1})=\frac{1}{2}\sec^{-1}(x^2)$$
If you don't believe me, see their graphs. The graphs of $$\frac{1}{2}\tan^{-1}(\sqrt{x^4-1})$$ and $$\frac{1}{2}\sec^{-1}(x^2)$$ are exactly the same.
Edit:
In order to prove that the quantities are indeed equal see @Zaragosa's answer.
PS: Whenever in doubt about calculus, use the derivative-calculator or the integral-calculator. | 2022-06-29T20:10:24 | {
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https://brilliant.org/discussions/thread/use-of-complex-numbers-in-calculus/ | # Use of complex numbers in calculus.
Hi, you might have come across:$\displaystyle \int e^{ax}\cos bx$ dx,
How do you solve it? You might use integration by parts, and also complex numbers, and I find use of complex numbers interesting!
Say $A = \displaystyle \int e^{ax}\cos bx$dx,
and $B = \displaystyle \int e^{ax}\sin bx$ dx
Hence, $A + iB = \displaystyle \int e^{ax} (\cos bx + i \sin bx)$dx = $\displaystyle \int e^{ax} (e^{i bx})$dx
= $\displaystyle \int e^{(a+ib)x}$dx
= $\displaystyle \frac{e^{(a+ib)x}}{a+ib}$
= $\displaystyle \frac{e^{(a+ib)x}(a - ib)}{a^2 + b^2}$
$\Rightarrow A + iB = z = \displaystyle \frac{e^{ax} (\cos bx + i \sin bx)(a - ib)}{a^2+b^2}$
Clearly,
$A = \text{Re}(z) = \displaystyle \frac{e^{ax}}{a^2+b^2} (a \cos bx + b \sin bx)$
$B = \text{Im}(z) = \displaystyle \frac{e^{ax}}{a^2+b^2} (a \sin bx - b \cos bx)$
You can try to find this definite integral:
Problem: $\displaystyle \int_{0}^{\pi} e^{(\cos x)} \cos(\sin x)$ dx
7 years, 7 months ago
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Sort by:
Short and Sweet! :)
- 7 years, 7 months ago
Thanks!
- 7 years, 7 months ago
$cos { (\sin { x) } } =\quad ({ e }^{ isinx }+{ e }^{ -isinx })/2\\ { e }^{ cosx }\cos { (\sin { x) } } =\quad ({ e }^{ cosx+isinx }+{ e }^{ cosx-isinx })/2=({ e }^{ { e }^{ ix } }+{ e }^{ { e }^{ -ix } })/2\\ =((1+\frac { { e }^{ ix } }{ 1! } +\frac { { e }^{ i2x } }{ 2! } +...)+(1+\frac { { e }^{ -ix } }{ 1! } +\frac { { e }^{ -2ix } }{ 2! } +...))/2$
In 2 more steps you will get the answer. The answer in this case is $pi$
- 7 years, 2 months ago
Solve his other question too: Problem without words !
- 7 years, 7 months ago
Hi Jatin! Thanks for the great post, but just one thing I was wondering that incorporating i, the complex number, and using the usual laws of Calculus, is it mathematically correct, I mean the laws of calculus is for reals. I may be fundamentally wrong somewhere but I need the answer.
- 7 years, 6 months ago
Awesome article.
By using your method I arrive at a step from where I can't proceed further.
The step is: $A+iB=\int (e)^{e^{ix}}\,dx$, where $A=\int (e)^{(cos x)}cos(sin x)\,dx$ and $B=\int (e)^{(cos x)}sin(sin x)\,dx$.
How to do after this?Precisely,what's the real part?
- 7 years, 7 months ago
Hi, Bhargav, you are close, try to use expansion for $e^x$.
- 7 years, 7 months ago
Is the answer $\pi$?
- 7 years, 7 months ago
I reach the same answer but is it possible to evaluate $B$?
- 7 years, 7 months ago
You came our with a simple and an interesting solution the technique that triggered me was to solve it using By Parts Method but have to admit your approach was far far better than mines Can you tell me how do you get such ideas at and tender age of 15(Just Asking)?
- 7 years, 7 months ago
Hi, solving this integral using complex numbers is well(not very much well though) known. I did not come up with it myself.
- 7 years, 7 months ago
Yeah, actually the same technique's been discussed with us at our insti as well... anyways, it's really good...
- 7 years, 7 months ago
Awesome!!
- 7 years, 7 months ago
-e^{-1} -1 is the ans
- 7 years, 6 months ago
great..:-)
- 7 years, 5 months ago
Too beautiful
- 7 years, 2 months ago
- 7 years, 2 months ago
Thanks a ton..!
- 7 years, 2 months ago
Easy approach....nice trick...
- 7 years, 2 months ago
Interesting method!
- 7 years, 1 month ago
Finally I got it! :)
- 6 years, 8 months ago
THANKS. VERY POWERFULL METHOD-CAN BE TACKLED ANY COMPLICATED INTEGRALS LIKE THIS ONE.
- 6 years, 8 months ago
It is an amazing approach to solve such problems.
- 6 years, 7 months ago
$W0w$ $£xcellent approach$
- 6 years, 4 months ago | 2021-07-28T11:49:43 | {
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https://math.stackexchange.com/questions/2514584/can-the-difference-of-2-undefined-limits-be-defined | # Can the difference of 2 undefined limits be defined?
Is this limit defined or undefined? $$\lim\limits_{x \to 0+} \left(\sqrt{\frac{1}{x}+2}-\sqrt{\frac{1}{x}}\right)$$ When I apply the rule of difference of limits, it's undefined. But, when I manipulate it, it gives me zero. And the graph of the function indicates it's defined on the right side.
By multiplying by $\frac{\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}}}{\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}}}$: $$\lim\limits_{x \to 0+} \frac{\left( \sqrt{\frac{1}{x}+2}-\sqrt{\frac{1}{x}} \, \right) \left(\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}} \, \right)}{\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}}}$$
$$=\lim\limits_{x \to 0+} \frac{\frac{1}{x}+2-\frac{1}{x}}{\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}}}$$ $$=\lim\limits_{x \to 0+} \frac{2}{\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}}}$$ Then, we multiply by $\frac{\sqrt{x}}{\sqrt{x}}$: $$=\lim\limits_{x \to 0} \frac{2\sqrt{x}}{\sqrt{1+2x}+1}$$ And, we substitute: $$=\frac{2\sqrt{0}}{\sqrt{1+2\times0}+1} = 0$$ So, is this limit defined or not? and what's my error, if any?
• Notice that you can only break a limit up if each limit you're creating is finite. So you can't split the limit up here. In general, any limit law needs to be done with only finite limits involved. – Kaynex Nov 11 '17 at 2:37
• Your solution is fine. Certainly the limit of a difference can exist even if the limits of the terms being subtracted do not exist individually. Consider $\lim_{x \to \infty} (x - x)$, for example. – Bungo Nov 11 '17 at 2:40
• $\infty - \infty$ is not in the domain of subtraction, so the fact subtraction is continuous does not apply here. – user14972 Nov 11 '17 at 2:52
• Your title is wrong: this is not the difference of two undefined limits (count the number of times $\lim$ occurs in your formula). If it were, it would definitely be undefined: arithmetic assumes definite operands. – Marc van Leeuwen Nov 11 '17 at 10:27
• A simple example of two expressions whose limits are undefined but the limit of the difference is clearly defined: take $a_n = b_n = (-1)^n.$ Then both $\lim_{n\to\infty} a_n$ and $\lim_{n\to\infty} b_n$ are undefined but $\lim_{n\to\infty} (a_n-b_n) = 0.$ – md2perpe Nov 11 '17 at 10:45
What you did is correct. The point is that in its initial form, your problem was of an indeterminate form. Basically, if we have a limit that "looks like" $\infty-\infty$ or something like this, the value can basically be anything under the sun. It can be illuminating to see the process in reverse. $$0=\lim_{x\to\infty} 0=\lim_{x\to\infty}(x-x)\ne \lim_{x\to\infty} x-\lim_{x\to\infty}x"="\:\text{nonsense}.$$
Remember that the rule that you referred to, "the rule of difference of limits", is not just the equation $$\lim_{x\to a}(f(x)-g(x))=\lim_{x\to a}f(x)-\lim_{x\to a}g(x)$$ but rather the statement that, if both of the limits on the right side of this equation are real numbers, then the limit on the left side (is also a real number and) is given by this equation. So this rule does not apply to the limit in your question.
More generally, when learning rules (or theorems or principles or whatever they may be called), don't just learn formulas, but pay attention also to the words around them. The words are not just decoration but are essential for the correctness of the rule.
• If students really pay attention to your last paragraph, most of such questions about mathematics would simply vanish. =) But sadly, I've seen many textbooks that don't even have any of the necessary words to make their 'rules' correct... – user21820 Nov 11 '17 at 9:30
• @user21820 So maybe that paragraph in my answer should be addressed not only to students learning the rules but also to textbook authors writing the rules (who surely should know better). – Andreas Blass Nov 11 '17 at 15:37
• (+1) And I'd like to give you and additional +1 just for "The words are not just decoration but are essential for the correctness of the rule.". I think I'd steal your expression and "recycle" it for my students that blindly apply formulas without even thinking about the meaning of the variables they contain! :-) – Lorenzo Donati Nov 11 '17 at 21:14
Just to add to what has been said, a limit expression that has undefined (or infinite) value cannot be treated as an ordinary real expression, and so it technically is invalid (senseless) to manipulate it as if it were a real number. For example $\sin(n)-\sin(n) \to 0$ as $n \to \infty$, but "$\lim_{n\to\infty} \sin(n)$" itself is simply undefined and it is technically invalid to even write "$\lim_{n\to\infty} \sin(n) - \lim_{n\to\infty} \sin(n)$", not to say ask for its value (unless you want to have propagation of undefined values...).
The difference of 2 undefined limits cannot be defined, by definition. (Even if you wish to permit writing potentially undefined expressions, it would not make a difference, since any expression with an undefined subexpression will itself be undefined.)
The correct statement is that it is possible for the difference of two expressions to have a limit even though neither of the expressions has a limit (under the same limiting conditions).
I have the same take as user21820. Your error is treating “undefined” like it's a value when it's really just a predicate.
When we write $\lim_{x\to a} f(x) = L$, it's not really an equation so much as a statement. It's a statement about $f$, $a$, and $L$ all in one. When we say that $\lim_{x\to a} f(x)$ is undefined, we mean that $\lim_{x\to a} f(x) = L$ is not true for any number $L$.
It's tempting to think of “undefined” as some magic quantity which nullifies real numbers. Sometimes this leads to true statements. For instance, if $\lim_{x\to a} f(x) = L$ and $\lim_{x\to a} g(x)$ is undefined, then $\lim_{x\to a} (f(x) + g(x))$ is undefined. You might want to think of this succinctly as “finite plus undefined equals undefined.”
But it also leads to false statements, as you have discovered. It's just not true that if $\lim_{x\to a}f(x)$ is undefined and $\lim_{x\to a}g(x)$ is undefined, then $\lim_{x\to a} (f(x) + g(x))$ is undefined. The simplest counterexample would be if $g(x) = -f(x)$. So even though you might want to think to yourself “undefined plus undefined equals undefined,” this is specious reasoning.
In a lot of computer languages, you can have an undefined or null value, and in some cases it can combine with other values. For instance, in Excel, when a formula in a cell evaluates to #NA, any formula that uses that cell will also evaluate to #NA. But it doesn't work that way with limits in math.
You write $x\to 0+$ but it's often written as $x\to 0^+$. You asked if your limit is defined or undefined. I'll answer that with another two ways.
$$\lim_{x\to 0^+} \left(\sqrt{\frac{1}{x}+2}-\sqrt{\frac{1}{x}}\right)=$$
$$=\lim_{x\to 0^+}\left(\frac{\sqrt{1+2x}-1}{x}\cdot \sqrt{x}\right)=$$
Use the definition of a derivative and a limit multiplication rule/law.
$$=(\sqrt{1+2x})\bigg|_{x=0^+}\cdot 0=0,$$
because the derivative is a real number.
Another way is using Newton's generalized binomial theorem, also see Binomial series, which converges when $|x|<1$ but it can diverge when $|x|\ge 1$, see the Binomial series link for more information. $$\lim_{x\to 0^+}\frac{\sqrt{1+2x}-1}{x}=$$
$$=\lim_{x\to 0^+}\frac{(1+2x)^{\frac{1}{2}}-1}{x}=$$
$$=\lim_{x\to 0^+}\frac{1+\frac{1}{2}2x+o(x)-1}{x}=$$
$$=\lim_{x\to 0^+}\left(1+\frac{o(x)}{x}\right)=1+0=1$$
What you wrote,$$\lim\limits_{x \to 0+} \sqrt{\frac{1}{x}+2}-\sqrt{\frac{1}{x}},$$is undefined, because it is the difference of$$\lim\limits_{x \to 0+} \sqrt{\frac{1}{x}+2},$$which is undefined, and a another quantity$$\sqrt{\frac{1}{x}},$$the latter being meaningful, although dependent on the variable $x$. What I think you meant to write is $$\lim\limits_{x \to 0+}\left( \sqrt{\frac{1}{x}+2}-\sqrt{\frac{1}{x}}\right).$$ This is not, as the title of your question says, "the difference of two undefined limits". Rather, it is a (single) limit of a quantity (that happens to be expressed as a difference). As long as you treat this as a single limit, which is what you have done quite correctly, no problem arises.
In general, it is a false move to split a limit of a difference into the difference of two limits. It works sometimes, but then it needs to be justified carefully. On the other hand, if two limits are separately well defined, then their difference can be expressed as the limit of a difference (assuming that the variable, domain, and endpoint of the limiting process is the same for both).
• Come on. What you are saying is that large operators like $\lim$ (or $\sum$) have operator precedence higher than addition and subtraction and that therefore the formula needs parentheses. This is indeed the common convention (also the precedence is lower than multiplication; not sure about division, as even $ab/cd$ might give discussion), but if that is what you want to say, then just say directly. – Marc van Leeuwen Nov 11 '17 at 10:33
• @MarcvanLeeuwen: You are of course quite correct about precedence. However, given the OP's muddle about whether one or two limits were involved, I thought it appropriate to treat this concrete case in detail rather than state a general rule (which might be misunderstood or dismissed as stylistic pettiness). – John Bentin Nov 11 '17 at 10:58
• I have my nitpicks about notation, too, that I am determined to bring up whenever I see them. But the right place to do so is in a comment. In a particularly egregious case, consider editing the question/answer and append a comment explaining why. – Michael Grant Nov 11 '17 at 14:51 | 2019-10-20T16:28:56 | {
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http://mathhelpforum.com/business-math/166396-linear-programming-problem.html | # Math Help - Linear Programming Problem
1. ## Linear Programming Problem
Kunz manufactures two products that are used in the heavy equipment industry. Both products require maunfacturing operations in 2 different departments (A and B). The following are the production times (in hours) and profit contributions for the two products:
Product Profit/Unit Dept A (in hours) Dept B (in hours)
1 $25 6 12 2$20 8 10
For the coming period, Kunz has a total of 900 hours of labour that can be allocated to either of the two departments. Formulate an LP to maximize the numbers of hours to allocate per department and total contribution to profit.
So I created 4 variables:
X1 - Hours of Product 1 in Dept A
X2 - Hours of Product 1 in Dept B
X3 - Hours of Product 2 in Dept A
X4 - Hours of Product 2 in Dept B
I formulated the following LP:
Max Z = 25(X1 + X2) + 20(X3 + X4)
s.t.
6X1 + 12X2 + 8X3 + 10X4 <=900
all variables => 0
Is my solution correct?
2. Originally Posted by statmajor
Kunz manufactures two products that are used in the heavy equipment industry. Both products require maunfacturing operations in 2 different departments (A and B). The following are the production times (in hours) and profit contributions for the two products:
Product Profit/Unit Dept A (in hours) Dept B (in hours)
1 $25 6 12 2$20 8 10
For the coming period, Kunz has a total of 900 hours of labour that can be allocated to either of the two departments. Formulate an LP to maximize the numbers of hours to allocate per department and total contribution to profit.
So I created 4 variables:
X1 - Hours of Product 1 in Dept A
X2 - Hours of Product 1 in Dept B
X3 - Hours of Product 2 in Dept A
X4 - Hours of Product 2 in Dept B
I formulated the following LP:
Max Z = 25(X1 + X2) + 20(X3 + X4)
s.t.
6X1 + 12X2 + 8X3 + 10X4 <=900
all variables => 0
Is my solution correct?
That look like a correct formulation of the LP problem.
CB
3. Really? Because the objective function doesn't make sense to me. Would X1 = X2 in this case since product 1 has to spends 6X1 hours in Dept A and 12X2 hours in Dept B. Would X1 = X2 in this case?
If they do, then X1 and X2 would each refer to the number of Products 1 produced.
4. Originally Posted by statmajor
Really? Because the objective function doesn't make sense to me. Would X1 = X2 in this case since product 1 has to spends 6X1 hours in Dept A and 12X2 hours in Dept B. Would X1 = X2 in this case?
If they do, then X1 and X2 would each refer to the number of Products 1 produced.
As I read your question a unit of product 1 take 6 hours to produce in Dept A and 12 hours to produce in Dept B. If you think it means it takes 6 hours in DA and 12 hours in DB then X1=X2 (and similarly X3=X4), in fact there are only 2 variables Y1 the production of product 1 and Y2 the production of product 2.
The latter interpretation makes better sense from one point of view, the profit is independently of department. Also under the other interpretation DA would dominate DB.
The wording is still ambiguous, but I am coming round to preferring your new interpretation.
CB
CB
5. I kinda think that the new objective function should be Max Z = 25X1 + 20X3. You should be able to replace X1 with X2, and X3 with X4.
For example:
If we wanted 3 units of Product 1, we would need (6*3)=18 hours in Dept A, and (12*3) = 36 hours in Dept B.
We could reduce it down to two variables:
Y1 - Product 1
Y2 - Product 2
So the new LP could be:
Max Z = 25Y1 + 20Y2
s.t
18Y1 + 18Y2 <= 900
all var => 0 | 2015-02-01T23:41:47 | {
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https://math.stackexchange.com/questions/2823662/square-root-in-quadratic-equation | # Square root in quadratic equation
I ran into a Math problem in which I was told to solve like that:
\begin{align} x^ 2 - 25 &= 0 \\ x^2 & = 25 \\ x & = \sqrt{25}\\ x & = 5 \end{align}
I wonder what the logical and also intuictive explanation might be on removing the power of x and on the other hand, adding a square root to the other side.
Could you guys help me out on this? You see, even though I did solve the problem, I think it is only worthy if I understand the logical explanation behind it.
(Assuming we already know that $x$ is positive.)
Say that we have come to the conclusion that $x^2=25$. This means that the number $x^2$ and the number $25$ are the same. If two numbers are the same, then their square roots are also the same.
The square root of $25$ is $5$, and the square root of $x^2$ is $x$. Therefore $x$ and $5$ are the same, and we write that as $x = 5$.
Every single rule you've learned about solving equations follow this line of thought. "The two numbers on either side of the $=$ are equal. Therefore they are still equal after we do the same thing to each of them" (in addition to "Swap one expression on one side of the $=$ for a simpler expression with the same value", those are the only two things you need to solve any equation ever).
In your case, for instance, the first step that you did (after assuming that $x^2-25 = 0$) was to add $25$ to both sides. The two sides were equal before you added $25$, therefore they must be equal after you added $25$. Then you swapped $x^2-25+25$ on the left-hand side with $x^2$, since that has the same value and is simpler, and the same with $0+25$ on the right side. (Most people do these in a single step, and some think of it as "moving the $-25$ over to the other side and swapping its sign", but you should, at the very least, keep in mind that these are the steps behind such a manouver.)
The trick to solving equations is to choose the things you do to both sides with care so that the expressions become simpler. For instance, to simplify $x^2$, the standard thing to do is to take square root (by definition of square root, that removes the $^2$, which simplifies things). If this also doesn't muck up the other side ($\sqrt{25}$ isn't too bad), then you actually go ahead and do that step.
As a non-example, consider $x^2 = x+2$. In this case, you could try to take square roots on both sides, and you would get $x = \sqrt{x+2}$. Yes, the left-hand side got nicer, but the right-hand side got worse. So that might not be the best step to take in that situation, even though it's entirely correct.
• Thank you, Arthur, what a brilliant mind! That was what I was looking for! – Matheus Minguini Jun 18 '18 at 13:03
You have different ways to solve your problem.
$$x^2=25$$ $$x^2-25=0$$ $$(x-5)(x+5)=0$$
$$x= \pm5$$
Or,
$$x^2=25$$ $$|x|=5$$ $$x= \pm5$$
• Mohammad Riaz-Kermani, I understood the logic behind the first way: You factored the first expression and it became a simple equation. I still cant understand the second way though, I mean, the logic behind – Matheus Minguini Jun 18 '18 at 12:56
• When you take square root of $x^2$ you get $|x|$ not just $x$. That gives you two choices for $x$ which are $x=\pm 5$. If you know that $x$ is positive, then you choose $x=5$ – Mohammad Riazi-Kermani Jun 18 '18 at 13:57
It should be
• $x² - 25 = 0$
• $x² = 25$
• $\color{red}x = {\sqrt {25}}$
• $x = 5$
• Thank you, Gimusi – Matheus Minguini Jun 18 '18 at 12:57
The trick is to take the square root of the two members.
$$a=b$$ versus $$\sqrt a=\sqrt b.$$
This requires some care:
1. if $a,b<0$, you may not take the square root;
2. $\sqrt a=-\sqrt b$ is also true.
• Would something like "2. $\sqrt a = -\sqrt b$ could just as well be true" be more correct? – Arthur Jun 18 '18 at 13:12
• @Arthur: indeed. – Yves Daoust Jun 18 '18 at 13:14
What justifies that step is the exponentiation axiom $$(a^m)^n=a^{mn}.$$ If you agree with this, then from $x^2=25$, we obtain $(x^2)^{1/2}=25^{1/2}$, which gives $x=\sqrt{25}$, by another rule of exponentiation.
I understand that where you are struct, if I am not wrong. Is the following your question?
$$x^2 = 25 \implies x =5 ?$$
## What is square root of $y$?
When you write $\sqrt{y} = \sqrt[2]{y}$, you mean that you are searching a number $x$ such that $x^2 = y$.
More generally, when we have $\sqrt[n]{y}$, it is nothing but a number $x$ such that $x^n = y$.
Having the above observations and notations,
$$x^2 = {25} \implies x = \sqrt{25} = \sqrt{5^2} =\sqrt{(-5)^2}$$ which evidently proves that $x = 5$ or $x = -5$.
If you are sure that $x$ is positive then $x = 5$. | 2019-07-19T00:22:03 | {
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https://math.stackexchange.com/questions/536868/why-do-i-see-i-and-k-as-the-indices-of-summation | # Why do I see i and k as the indices of summation?
I'm working on linear algebra and just wanted to clear up an uncertainty regarding whether there is a difference in the use of i and k as the dummy variables for the index of summation?
$\sum\limits_{i=1}^\infty {i^2} = \sum\limits_{k=1}^{\infty} {k^2}$ ?
I got confused at first since I was working with vectors [i, j] with a summation indicating k=1 , although k was indicating the z dimension (i and j indicating x and y respectively).
Just to clarify, the choice of i and k as dummy variables is completely arbitrary - right?
I found my way to this page at Wolfram MathWorld and it actually switches from i to k in the course of a short piece of text, is this normal and nothing to concern me or should I take note of differences like this?
• You can call your variables however you like. Calling them $Frederick$ would be a lot of writing and reduce legibility, so we tend to use single-letter names for indices and such. Which letters we use is irrelevant, though custom makes some choices better than others. – Daniel Fischer Oct 23 '13 at 11:40
If $i$, $j$, and $k$ are already being used for vector notation, it would be good to use a different index for summation. The letter $m$ would be one sensible choice, if you are writing things like $\sum_{m = 0}^n$, and $l$ is another possibility (just because it is close in the alphabet to $i,j,k,m,$ and $n$). Of course you are free to use any variable that hasn't already been given a meaning, but it is good to use letters that will have the psychological connotation of being an index (so letters like $x$, $y$, and $z$ are fairly uncommon as summation indices).
Yes, it's completely arbitrary. Although it would be frowned upon, there is nothing inherently wrong using something like $\dagger$ or $わ$ or a drawing of an acorn as dummy variables either. Although, you should try not to switch too often during the course of a text. In the case of the WMW text you link, the $k$ symbolizes the same thing (the order of the forward difference), although it changes from being the bound of one sum to being the summation variable in the next.
There's no a priori difference between the notations $$\sum_{i=1}^\infty {i^2}\\ \sum_{k=1}^\infty {k^2}\\ \sum_{\dagger=1}^\infty {\dagger^2}\\ \sum_{わ=1}^\infty {わ^2}$$ However, if you've used one of them before, then using them again would be seen as wrong unless, as in the WMW example, it still symbolizes the same quantity.
The summation index is a bound or dummy variable. It doesn't matter what variable you use. In your example, the sum up to infinity is problematic because it doesn't converge, but the use of $i$ or $k$ doesn't change anything. | 2020-04-09T18:02:02 | {
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https://mathematica.stackexchange.com/questions/119158/create-a-filled-listplot-where-the-points-in-the-list-come-from-within-the-regi?noredirect=1 | # Create a filled ListPlot, where the points in the list come from within the regions rather than from the region boundaries
Is it possible to fill, color, under the points in ListPlot where the points define the area and the boundary? Assuming that the number of points are large enough to be able to define an area by them. I am looking for something similar to the Filling option which is based on the points distribution, rather than the area between curves or axes. Please note that there is no curve in this case, only point, therefore this is not a duplicate question.
Edit: To those who said this is duplicate: Can you read the question properly before jumping to your illogical conclusion? Do you see any similarities between my question and this link as duplicate? ListPlot and filling between curves Please refer to more accurate duplicate otherwise remove the duplicate from this question soon.
Edit: something like this picture (roughly) from this link: https://en.wikipedia.org/wiki/Phase_diagram
Edit2: some data points:
{{1.52788, 0.00119755}, {1.70822, 0.00119755}, {1.87126, 0.00119755}, {2.02119, 0.00119755}, {2.16075, 0.00119755}, {2.29182, 0.00119755}, {1.52788, 0.00479019}, {1.70822, 0.00479019}, {1.87126, 0.00479019}, {2.02119, 0.00479019}, {1.52788, 0.00718529}, {1.70822, 0.00718529}, {1.87126, 0.00718529}, {1.52788, 0.00958039}, {1.70822, 0.00958039}, {1.52788, 0.0119755}, {1.70822, 0.0119755}, {1.52788, 0.0179632}, {1.70822, 0.0179632}, {1.52788, 0.0215559}, {1.70822, 0.0215559}, {1.52788, 0.023951}, {1.70822, 0.023951}, {1.52788, 0.0299387}, {1.70822, 0.0299387}, {1.52788, 0.0359265}, {1.70822, 0.0359265}, {1.52788, 0.0419142}, {1.70822, 0.0419142}, {1.52788, 0.0479019}, {1.70822, 0.0479019}, {1.52788, 0.0598774}, {1.70822, 0.0598774}, {1.52788, 0.0718529}, {1.70822, 0.0718529}, {1.52788, 0.0838284}, {1.70822, 0.0838284}, {1.52788, 0.107779}, {1.70822, 0.107779}, {1.87126, 0.107779}, {1.52788, 0.163525}, {1.70822, 0.163525}, {1.87126, 0.163525}, {1.52788, 0.204406}, {1.70822, 0.204406}, {1.87126, 0.204406}, {1.52788, 0.245288}, {1.70822, 0.245288}, {1.87126, 0.245288}, {2.02119, 0.245288}}
• can you show a small sample data? – Sumit Jun 23 '16 at 9:33
• @Sumit, I have included an image, thanks. – O_o Jun 23 '16 at 9:47
• Do you have data for each phase line or only the crossing points? – Sumit Jun 23 '16 at 9:52
• I don't think you can make a phase diagram with this data. Either you need set of points on each boundary, or a third column specifying the phase. Otherwise you can't sort out different regions. – Sumit Jun 23 '16 at 10:20
• Whether this is a duplicate or not, your comment to closers is impolite. Rather than ordering people around, you will get much better feedback if you reply to comments - which people took time to write to e.g. clarify your question. – Yves Klett Jun 24 '16 at 15:20
Okay, so you have a set of Regions that you want to fill, but you can only define those regions by a set of points distributed within them. Let's make some data that reproduces this. Here are three non-overlapping regions that fill up a square:
region1 = Disk[{0, 0}, 1, {0, π/2}];
region2 = RegionDifference[Rectangle[], region1];
region3 = Disk[{0, 0}, 1, {0, π/6}];
region1 = RegionDifference[region1, region3];
RegionPlot[{region1, region2, region3}]
That's what we want to get in the end. Now let's get a 1000 random points in each region,
{points1, points2, points3} =
RandomPoint[#, 1000] & /@ {region1, region2, region3};
ListPlot[{points1, points2, points3}]
Okay, so now you might think that you can just make a convex hull for each set of points to define the regions,
RegionPlot[Evaluate[
ConvexHullMesh /@ {points1, points2, points3}]]
but clearly you'd be wrong. What you need is a concave hull in this case, otherwise known as an alpha shape. I took the code from this post and put it on gist to make it easy to import,
<<"https://gist.github.com/jasondbiggs/39fac60c578e57959b979cfd8e43f7d6/raw/7c012d631b58ad77815999a31b8cce8761e4dcfe/alphashapes_2D.m"
RegionPlot[Evaluate[
alphaShapes2DC[#, 0.1] & /@ {points1, points2, points3}]]
A decent approximation, but not a perfect representation of the regions. OP did say
Assuming that the number of points are large enough to be able to define an area
So if I increase the number of points to 10,000 for each region than this is what results
• Thank you so much, a combination of your method and other's suggestion (Sumit and E.Doroskevic) to have a third column solved the problem! – O_o Jun 23 '16 at 13:22
## Example
Code
ListPlot[Range @ 100, Filling -> Bottom]
ListLinePlot[Range @ 100, Filling -> Bottom]
Output
Reference
• Thank you, but I am looking for something different. Suppose that we have distribution of points in X-Y plane with higher densities in some regions and I would like to use fill and color based on the distribution of the points. – O_o Jun 23 '16 at 9:40
• Would you have some data we would be able to use? – e.doroskevic Jun 23 '16 at 9:48
• Added some data points. – O_o Jun 23 '16 at 9:57
• @O_o ListLinePlot[data, Filling -> Bottom, ColorFunction -> "TemperatureMap"] wheredata is your list of points? – e.doroskevic Jun 23 '16 at 10:07
• @O_o I agree with @Sumit, you need to have three data sets indicating curves associated with ea. phase you want to visualize in a Plot. Given this information is available, you will be able to use ListLinePlot with Filling->Bottom to get desired output – e.doroskevic Jun 23 '16 at 13:10 | 2019-11-13T20:33:40 | {
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https://www.physicsforums.com/threads/questions-related-to-an-unknown-function-given-its-values-at-some-points.836903/ | # Questions related to an unknown function given its values at some points
1. Oct 10, 2015
### Sujith Sizon
A function $f(x)$ is continuous in the interval $[0,2]$. It is known that $f(0)=f(2)=−1$ and $f(1)=1$. Which one of the following statements must be true?
(A) There exists a $y$ in the interval $(0,1)$ such that $f(y)=f(y+1)$
(B) For every $y$ in the interval $(0,1),f(y) = f(2−y)$
(C) The maximum value of the function in the interval $(0,2)$ is $1$
(D) There exists a $y$ in the interval $(0,1)$ such that $f(y) = −f(2−y)$
----------
Here's my approach:
Consider a function $g(y) = f(y) - f(y+1)$ since $f$ is a continuous function and $g$ is a combination of $f$ so it is also continuous in $[0,1]$.
it is found that
$g(0)=f(0)-f(1)=-1-1=-2$ and
$g(1)=f(1)-f(2)=1-(-1)=+2$
since #g# goes from $-2$ to $+2$ and is continuous in $(0,1)$ therefore there has to be a point in b/w $(0,1)$ such that $g(\text{that point}) = 0$
when $g(y)=0$ for some $y\in(0,1)$ then $f(y) = f(y+1)$, Hence **statement A** is true.
Using the same logic **statement D** is true too. It is clear that statement B and C are false.
The answer booklet says that only **statement A** is true. There is no comment on **statement D**; should it be considered false? Is my approach okay?
2. Oct 10, 2015
### pasmith
I agree that D is also true. Sometimes question setters make mistakes. | 2017-12-17T02:33:52 | {
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https://math.stackexchange.com/questions/2900828/probability-of-real-roots-for-x2-bx-c-0 | # Probability of real roots for $x^2 + Bx + C = 0$ [duplicate]
Question: The numbers $B$ and $C$ are chosen at random between $-1$ and $1$, independently of each other. What is the probability that the quadratic equation $$x^2 + Bx + C = 0$$ has real roots? Also, derive a general expression for this probability when B and C are chosen at random from the interval $(-q, q)$ for and $q>0$.
My approach: since we're trying to find the probability of real roots. We should first realize when it has imaginary roots. So, $$B^2 - 4aC < 0$$ $$a = 1$$ $$B^2 < 4C$$ I'm not sure where to go from here. How do I now find the probability of $B$ being greater than $4C$?
## marked as duplicate by grand_chat, Lord Shark the Unknown, user91500, Adrian Keister, MicahSep 1 '18 at 16:08
• Look at the square in the $(B,C)$ plane with vertices $(\pm1,\pm1)$, and work out the area of the pieces the curve $B^2=4C$ divides it into. – Lord Shark the Unknown Aug 31 '18 at 15:40
• By calculating the area defined by this inequality on the $(B,C)$-plane. – MigMit Aug 31 '18 at 15:40
I assume you mean that $$B,\,C\sim U(-1,\,1)$$. We'll get the answer as a function of a fixed value for $$C$$, then average it out. For $$C<0$$ (which has probability $$1/2$$), the result is $$0$$; for $$C> 1/4$$ (which has probability $$3/8$$), the result is $$1$$; for $$0\le C\le\frac{1}{4}$$ (which has probability $$\frac{1}{8}$$), the condition $$-2\sqrt{C}\le B\le 2\sqrt{C}$$ has probability $$4\sqrt{C}$$. So the final result is $$\frac{3}{8}+\frac{1}{8}\int_0^{1/4} 4\sqrt{C}dC=\frac{3}{8}+\frac{1}{3}\bigg(\frac{1}{4}\bigg)^{3/2}=\frac{5}{12}.$$
Edit: the above is the probability of non-real complex roots; the probability of real roots is $$1-\frac{5}{12}=\frac{7}{12}$$. | 2019-08-24T07:18:48 | {
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https://math.stackexchange.com/questions/3426814/show-that-sum-i-1k-2-k-choose-i2i-3k-1 | # Show that $\sum_{i=1}^{k/2} {{k}\choose i}2^i < 3^{k-1}$
Show that $$\sum_{i=1}^{k/2} {{k}\choose i}2^i < 3^{k-1}$$ for $$k\geq6$$ and even. I tried using $$(1+2)^k=\sum_{i=0}^k{{k}\choose{i}}2^i$$, the first half has the same binomials as the second and larger powers of two. Also tried induction, but the step is proving that $$\dfrac{\sum_{i=0}^{k/2+1}{k+2\choose i}2^i}{\sum_{i=0}^{k/2}{k\choose i}2^i}<9$$.
• @metamorphy - does your bound imply the OP's requested bound? Perhaps I'm missing something but I don't see how your bound relates to the request bound (at least, not in an obvious way). – antkam Nov 8 at 19:29
• @metamorphy - I'm guessing the $\sqrt{8}$ comes from... Stirling's approximation? Actually, even though your observation does not answer the OP question directly, I am quite interested in (1) how you prove its correct, and (2) why the ratio $\to 1$, and (3) how the $\sqrt{8}$ comes about. If you don't mind writing an answer, that'd be great. – antkam Nov 8 at 20:57
• @antkam: Done (removed my comments above as redundant now). – metamorphy Nov 8 at 21:43
Here's an inductive approach. Let $$a_n = \sum_{i \leq n} \binom{2n}{i} 2^i$$ (including $$i = 0$$ for convenience). Then since $$\binom{2n+2}{i} = \binom{2n}{i-2} + 2\binom{2n}{i-1} + \binom{2n}{i}$$, we have \begin{align*} a_{n+1} &= \sum_{i \leq n+1} \binom{2n+2}{i} 2^i \\ &= \sum_{i \leq n+1} \left(\binom{2n}{i-2} 2^i + \binom{2n}{i-1} 2^{i+1} + \binom{2n}{i} 2^i\right) \\ &= \sum_{i \leq n-1} \binom{2n}{i} 2^{i+2} + \sum_{i \leq n} \binom{2n}{i} 2^{i+2} + \sum_{i \leq n+1}\binom{2n}{i} 2^i \\ &= \left(4a_n - \binom{2n}{n}2^{n+2}\right) + 4a_n + \left(a_n + \binom{2n}{n+1} 2^{n+1}\right) \\ &< 9a_n \end{align*} where the last inequality holds because $$\binom{2n}{n} > \binom{2n}{n+1}$$. Then since $$a_3 < 3^5$$, by induction we have $$a_n < 3^{2n-1}$$ for all $$n \geq 3$$ (though there are better asymptotic bounds, e.g. $$a_n \leq 2^n \sum_{i \leq n} \binom{2n}{i} \leq 8^n$$).
UPDATED
OK, here is a proof using probability. It requires pretty tedious accounting.
First re-write the desired inequality as (*) below:
$$\sum_{i=1}^{k/2} {{k}\choose i}(\frac13)^{k-i} (\frac23)^i < \frac13 \,\,\,\,\,\,\,\,\text{ .....(*)}$$
Now consider a "trinomial" experiment, with $$k$$ i.i.d. trials, and each trial result $$\in \{a, b, c\}$$ each with equal prob $$\frac13$$. Define r.v.s $$A, B, C$$ as the number of occurrences of $$a, b, c$$ respectively. Obviously we have $$A+B+C=k$$ and $$E[A]=E[B]=E[C] = k/3$$.
Then the LHS of (*) $$= P(\frac{k}{2}\le A \le k-1) < P(A \ge {k\over 2})$$.
Originally, I was thinking Markov's inequality would work, but as the OP author correctly pointed out, Markov's inequality gives $$P(A \ge {k \over 2}) \le {E[A] \over {k/2}} = \frac23$$ which is too big (too loose).
The main idea of the proof is that, $$A \ge k/2$$ implies $$A$$ is the max among $$A, B, C$$. That would almost immediately give the $$1/3$$ bound (by symmetry), except for the case of $$A$$ being a joint max with $$B$$ or $$C$$ (e.g. $$A=B=k/2, C=0$$). Dealing with that joint max case is where most of the tedious accounting comes in.
Define these events:
• $$U_A=$$ the event that $$A$$ is the unique max among $$A,B,C$$, i.e. $$A> \max(B,C)$$
• $$U_B, U_C$$ defined similarly
• $$U = U_A \cup U_B \cup U_C =$$ event that there exists a unique max
• $$V = U^c=$$ complement of $$U$$, i.e. the max is a joint max ($$2$$ or $$3$$-way tie)
By the law of total probability,
$$P(A \ge {k\over 2}) = \color{red}{P(A \ge {k\over 2} \mid U)} P(U) + \color{blue}{P(A\ge {k\over 2} \mid V)} P(V)$$
Claim 1 (the easy case): $$\color{red}{P(A \ge {k\over 2} \mid U)} \le \frac13$$
Proof: conditioned on $$U$$, i.e. there being a unique max, then $$A \ge {k \over 2}$$ implies $$A$$ must be the unique max. So
$$\color{red}{P(A \ge {k\over 2} \mid U)} \le P(U_A \mid U) = \frac13$$
where the last equality is by symmetry among $$A,B,C$$.
Claim 2 (the tedious case): $$\color{blue}{P(A\ge {k\over 2} \mid V)} < \frac13$$
Define these $$6$$ events, each being a subset of $$V$$ (when $$k \ge 6$$):
• $$E_{AB} =$$ the event $$(A = B = {k \over 2}, C=0)$$
• $$E_{BC}, E_{CA}$$ defined similarly
• By symmetry, the number of sample points in these $$3$$ events are equal. We will call it $$e = |E_{AB}| = |E_{BC}| = |E_{CA}|$$
• $$F_{AB} =$$ the event $$(A = B = {k\over 2} - 1, C = 2)$$
• $$F_{BC}, F_{CA}$$ defined similarly
• By symmetry, the number of sample points in these $$3$$ events are equal. We will call it $$f = |F_{AB}| = |F_{BC}| = |F_{CA}|$$
Note that, when conditioned on $$V$$, then $$(A \ge k/2) = E_{AB} \cup E_{CA}$$.
Also note that the $$3$$ $$E$$-events are disjoint, and they are also disjoint from any $$F$$-event. So:
$$\color{blue}{P(A\ge {k\over 2} \mid V)} = {P(A\ge {k\over 2} \cap V) \over P(V)} = {P(E_{AB} \cup E_{CA}) \over P(V)} \le {e + e \over e + e + e + f} = {2e \over 3e + f}$$
For convenience write $$k = 2m, m \ge 3$$ (since it's given that $$k\ge 6$$ and is even), then
• $$e = {2m \choose m} = {(2m)! \over m! m!}$$
• $$f = {2m \choose (m-1),(m-1),2} = {(2m)! \over (m-1)! (m-1)! 2!}$$
• $${f \over e} = {m! m! \over (m-1)! (m-1)! 2!} = {m^2 \over 2} \ge \frac92 > 4$$ (when $$m \ge 3$$)
Therefore $$f > 4e$$ and $${2e \over 3e + f} < {2 \over 7} < \frac13$$, completing the proof of Claim 2. QED
Main result now follows by combining Claim 1 & Claim 2:
$$\text{LHS of (*)} < P(A \ge k/2) \le \frac13 P(U) + \frac13 P(V) = \frac13$$
• This is part of a homework, this inequality is not the original problem, but a sufficient condition. I tried your hint, lets perform an experiment with 3 possible outcomes $k$ times and let $X$ count how many times we obtain one specific outcome. Then the LHS is $P(X\geq \frac{k}{2}) \leq \frac{E[X]}{k/2}$ according to Markov's inequality, but if I did it right $E[X]=k/3$ and so $LHS \leq \frac{2}{3}$. Maybe I am still missing something – user1768368 Nov 8 at 18:05
• You are right (and I was wrong originally): Markov inequality is too weak. Instead I now have a symmetry-based argument, but it involves pretty tedious accounting to deal with a special case. :( – antkam Nov 8 at 20:49
An overkill (perhaps), but one can get the true asymptotics from $$\sum_{k=0}^{n}\binom{2n}{k}a^{2n-k}(1-a)^k=n\binom{2n}{n}\int_0^a x^{n-1}(1-x)^n~dx$$ (obtained like I did here, or just using integration by parts). At $$a=1/3$$, this gives $$\sum_{k=0}^{n}\binom{2n}{k}2^k=n\binom{2n}{n}\int_0^1 y^{n-1}(3-y)^n~dy$$ (after substituting $$x=y/3$$). The integral is asymptotically $$2^{n+1}/n$$; in fact $$\int_0^1 y^{n-1}(3-y)^n~dy=\frac{2^n}{n}+\int_0^1 y^n(3-y)^{n-1}~dy\qquad\color{gray}{[\text{I.B.P.}]}\\=\frac{2^n}{n}\left[1+\frac{1}{2}\int_0^n\left(1-\frac{z}{n}\right)^n\left(1+\frac{z}{2n}\right)^{n-1}dz\right]\qquad\color{gray}{[y=1-z/n]}\\\leqslant\frac{2^n}{n}\left(1+\frac{1}{2}\int_0^\infty e^{-z}e^{z/2}~dz\right)=\frac{2^{n+1}}{n},$$ the inequality is asymptotically tight. Thus $$\color{blue}{\sum_{k=0}^{n}\binom{2n}{k}2^k\leqslant\binom{2n}{n}2^{n+1}}\asymp 2^{3n+1}/\sqrt{n\pi}$$. | 2019-11-18T21:44:23 | {
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https://mathhelpboards.com/threads/sort-of-a-challenge.4589/ | # Sort of a challenge
#### topsquark
##### Well-known member
MHB Math Helper
This isn't so much a challenge problem as much as it has a startling (at least I think so) answer.
Say we tie a rope around the Earth. Now we are going to cut it and add another 6 meters to it. If we pull the new rope tight (in a circle) how high is the new rope above the Earth's surface?
It's reasonably easy so if you've seen it before let someone else answer. (I first saw this problem in a "Dr. Crypton" article in "Science Digest.")
-Dan
#### Bacterius
##### Well-known member
MHB Math Helper
Well the Earth's radius is $6371 ~ \text{km}$, so the original rope has a length of $2 \pi \times 6371 ~ \text{km}$. If we add $6 ~ \text{m} = 6 \times 10^{-3} ~ \text{km}$ to this rope, its new radius is:
$$\frac{2 \pi \times 6371 + 6 \times 10^{-3}}{2 \pi} \approx 6371.0009 ~ \text{km}$$
So the rope now "floats" about $0.0009 ~ \text{km} = 90 ~ \text{cm}$ above the ground. To be exact, $95.5 ~ \text{cm}$ (this is $\frac{6}{2 \pi} ~ \text{m}$).
Wait, what? My mind is blown
#### MarkFL
Staff member
This isn't so much a challenge problem as much as it has a startling (at least I think so) answer.
Say we tie a rope around the Earth. Now we are going to cut it and add another 6 meters to it. If we pull the new rope tight (in a circle) how high is the new rope above the Earth's surface?
It's reasonably easy so if you've seen it before let someone else answer. (I first saw this problem in a "Dr. Crypton" article in "Science Digest.")
-Dan
I recall a version of this problem circulating several years ago among the performance car forums on which I used to be quite active, and there was much disbelief and dissension among the masses.
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
Well the Earth's radius is $6371 ~ \text{km}$
I take issue with this solution, which assumes that the Earth's radius is 6,371 km. This is just the mean radius. The equatorial radius is quite a bit larger at 6,378.1 km, while the polar radius is quite a bit smaller at 6,356.8 km. Clearly, the answer to the problem depends on the radius... hmm... never mind.
Here is another question: What happens if, after adding 6 m, we lift the rope by one point so that it is taut and lies flat on the surface for most of the circumference, as in the following picture?
How high above the Earth is the point by which the rope is lifted?
Last edited:
#### Opalg
##### MHB Oldtimer
Staff member
I take issue with this solution, which assumes that the Earth's radius is 6,371 km. This is just the mean radius. The equatorial radius is quite larger at 6,378.1 km, while the polar radius is quite smaller at 6,356.8 km. Clearly, the answer to the problem depends on the radius... hmm... never mind.
Here is another question: What happens if, after adding 6 m, we lift the rope by one point so that it is taut and lies flat on the surface for most of the circumference, as in the following picture?
How high above the Earth is the point by which the rope is lifted?
Nice problem!
If the circle has radius $R$ and the length of the rope is increased from $2\pi R$ to $2\pi R + \delta$ then, with $\theta$ as in the diagram, $2\pi R + \delta = 2(\pi - \theta)R + 2R\tan\theta$. Thus $\tan\theta - \theta = \frac{\delta}{2R}.$ This can't be solved exactly for $\theta$, but assuming that $\delta \ll R$ we can use the approximation $\tan\theta \approx \theta + \frac13\theta^3$ (first two terms of the power series for $\tan\theta$) to get $\dfrac\delta{2R} \approx \dfrac{\theta^3}3$, so that $\theta \approx \sqrt[3]{\dfrac{3\delta}{2R}}$.
The distance $h$ of the high point of the rope from the circumference of the circle is $h = R(\sec\theta - 1)$. Again using the power series approximation, this time for $\sec\theta \approx 1 + \frac12\theta^2$, we have $$h \approx \frac R2\Bigl(\frac{3\delta}{2R}\Bigr)^{2/3} = \sqrt[3]{\frac{9\delta^2R}{32}}.$$
Since this is just a rough approximation, I'll take the radius of the Earth to be $6\times 10^6$m. If $\delta = 6$m then that formula gives $h\approx 390$m.
#### topsquark
##### Well-known member
MHB Math Helper
I take issue with this solution, which assumes that the Earth's radius is 6,371 km. This is just the mean radius. The equatorial radius is quite larger at 6,378.1 km, while the polar radius is quite smaller at 6,356.8 km. Clearly, the answer to the problem depends on the radius... hmm... never mind.
Here is another question: What happens if, after adding 6 m, we lift the rope by one point so that it is taut and lies flat on the surface for most of the circumference, as in the following picture?
How high above the Earth is the point by which the rope is lifted?
Even more mind blowing!
-Dan
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
This problem and picture are taken from the MathForum. | 2021-07-29T02:29:05 | {
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https://byjus.com/question-answer/assertion-the-value-of-int-0-2-pi-cos-99-x-d-x-is-0/ | Question
Assertion :The value of $$\int_{0}^{2 \pi} \cos ^{99} x d x$$ is 0. Reason: $$\int_{0}^{2 a} f(x) d x=2 \int_{0}^{a} f(x) d x,$$ if $$f(2 a-x)=f(x)$$
A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
C
Assertion is correct but Reason is incorrect
D
Assertion is incorrect but Reason is correct
Solution
The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion$$\operatorname{Let} I=\int_{0}^{2 \pi} \cos ^{99} x d x\\$$Then, $$I=2 \int_{0}^{\pi} \cos ^{99} x d x \quad\left[\because \cos ^{99}(2 \pi-x)=\cos ^{99} x\right]\\$$Now, $$\int_{0}^{\pi} \cos ^{99} x d x=0 \quad\left[\because \cos ^{99}(\pi-x)=-\cos ^{99} x\right]\\$$$$\Rightarrow I=2 \times 0=0$$Mathematics
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https://sites.wcsu.edu/mbxml/OER_Linear_Alg/section_practice_exam_determinants.html | ## Exercises4.8Practice Exam on Determinants and Such
Here are some exercises from the Hefferon's text which you should try, the answers to all of these can be found in the Answers to Exercises supplement for Hefferon's text. You do not need to hand in these exercises.
Additionally, be sure you can define the following: Determinants, Cofactors, Minor Matrix, Eigenvector, Eigenvalue, Characteristic Polynomial, Similar Matrices, Diagonalizable, Singular, Non-Singular
Practice Exam:
In addition to the practice problems above below are the questions for your practice exam, these must be turned in when you come in for the exam. The practice exam counts for 5% of your exam grade and can earn you back 15% of any points you loose on the in-class portion of the exam.
###### 1.
Define Minor Matrix $A_{ij}$.
###### 2.
Define Eigenvector.
###### 3.
Define Characteristic Polynomial.
###### 4.
Define Similar Matrices.
###### 5.
Given that $C$ and $D$ are $n\times n$ matrices, that $det(C)=4\text{,}$ and that $det(D)=-7$ find the value of
\begin{equation*} det\left( 5\, C^2\, D^{-1} \right). \end{equation*}
Hint
Your answer will be in terms of $n\text{.}$
###### 6.
Suppose that $T$ and $R$ are similar matrices and that $det\left(T\right)=9$ what is the determinant of $R\text{,}$ and why.
Hint
You will need to use the definition of similar matrices.
###### 7.
Show that the determinant of the matrix
\begin{equation*} A=\left[ \begin{array}{rrr} 2 \amp 0 \amp 8\\ 0 \amp 3 \amp 2\\ 7 \amp 4 \amp 9 \end{array} \right] \end{equation*}
is equal to -130 by expanding down the first column.
###### 8.
Show that the determinant of the matrix
\begin{equation*} A=\left[ \begin{array}{rrr} 2 \amp 0 \amp 8\\ 0 \amp 3 \amp 2\\ 7 \amp 4 \amp 9 \end{array} \right] \end{equation*}
is equal to -130 by expanding along the second row.
###### 9.
Show that the determinant of the matrix
\begin{equation*} A=\left[ \begin{array}{rrr} 2 \amp 0 \amp 8\\ 0 \amp 3 \amp 2\\ 7 \amp 4 \amp 9 \end{array} \right] \end{equation*}
is equal to -130 by first reducing it to row echelon form.
###### 10.
Find the eigenvalues of the matrix
\begin{equation*} B=\left[ \begin{array}{rr} 1 \amp 1\\ 2 \amp 0\\ \end{array} \right], \end{equation*}
you should get $\lambda_1=2$ and $\lambda_2=-1\text{.}$
Hint
First find the characteristic polynomial for $B\text{.}$
###### 11.
Find the eigenvectors of the matrix
\begin{equation*} B=\left[ \begin{array}{rr} 1 \amp 1\\ 2 \amp 0\\ \end{array} \right], \end{equation*}
you should get $\vec{v}_1=\left\lt1,1\right\gt$ and $\vec{v}_2=\left\lt1,-2\right\gt\text{.}$
Hint
Use the results from the previous problem and solve the equations $B\vec{v}_i=\lambda_i\vec{v}_i\text{.}$
###### 12.
Find matrices $P$ and $D$ such that $B=P\, D\, P^{-1}$ where
\begin{equation*} B=\left[ \begin{array}{rr} 1 \amp 1\\ 2 \amp 0\\ \end{array} \right], \end{equation*}
and $D$ is a diagonal matrix.
Hint
Use the information from the previous two problems. | 2021-01-21T17:06:24 | {
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https://math.stackexchange.com/questions/2379710/f0-infty-rightarrow-mathbbr-is-differentiable-and-lim-x-to-inftyfx | # $f:[0,\infty)\rightarrow\mathbb{R}$ is differentiable and $\lim_{x\to\infty}f'(x)=0$, prove $\lim_{x \rightarrow \infty}[f(x+1)-f(x)]=0$. [duplicate]
Assume that $$f: [0, \infty) \rightarrow \mathbb{R}$$ is differentiable for all $$x>0$$ and $$\lim_{x \rightarrow \infty} f'(x) = 0$$. Prove that $$\lim_{x \rightarrow \infty}[f(x+1)-f(x)] = 0$$
I was hinted that I should use the mean value theorem here. My attempt is as follows. Consider the closed interval $$[x, x+1]$$ where $$x>0$$. Clearly, $$f$$ is continuous on $$[x, x+1]$$ and also differentiable on $$(x, x+1)$$ by the assumptions of the question. So we can apply the MVT and conclude that there exists a $$c \in (x, x+1)$$ such that $$f(x+1) - f(x) = f'(c)$$. Now if I take the limit to infinity on the left hand side, I can see the $$\lim_{x \rightarrow \infty} f(x+1) - f(x)$$ come into play, but what is $$\lim_{x \rightarrow \infty} f'(c)$$?
I thought about something like this, but not sure if it is right. Clearly, $$c = x+t$$ for some $$0, so $$f'(c) = f'(x+t)$$, so $$\lim_{x \rightarrow \infty} f'(c) = \lim_{x \rightarrow \infty} f'(x+t)$$. Now I am not sure how to bring $$\lim_{x \rightarrow \infty}f'(x) = 0$$ into the picture.
## marked as duplicate by Nosrati, Lord Shark the Unknown, user91500, choco_addicted, CesareoSep 23 '18 at 12:32
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
## 2 Answers
Your proof is almost complete - to nail it down, you may argue as follows: Since $\lim_{x\to\infty}f'(x)=0$, for each $\varepsilon>0$ there exists some $M>0$ such that for all $x>M$ you have $|f'(x)|<\varepsilon$. In particular, if $x>M$, then for $x<c<x+1$ you have: $$|f(x+1)-f(x)|=|f'(c)|<\varepsilon$$ so by definition, $\lim_{x\to\infty}(f(x+1)-f(x))=0$
Let $\epsilon >0$. Then there is $a=a(\epsilon)>0$ such that $|f'(t)|< \epsilon$ for all $t>a$.
Now let $x>a$. Then there is $c \in (x,x+1)$ such that
$|f(x+1)-f(x)| = |f'(c)|$.
Since $c>a$, we have $|f'(c)|< \epsilon$ , hence
$|f(x+1)-f(x)|<\epsilon$ for all $x>a$.
This means: $\lim_{x \rightarrow \infty}(f(x+1) - f(x)) = 0$. | 2019-08-21T22:16:02 | {
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https://math.stackexchange.com/questions/3219749/proving-if-sum-n-1-infty-a-n-and-sum-n-1-infty-b-n-both-converge | # Proving if $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$ both converge then the series $a_1+b_1+a_2+b_2+\ldots$ converges.
If $$\sum_{n=1}^{\infty} a_n$$ and $$\sum_{n=1}^{\infty} b_n$$ both converge with sums $$\alpha$$ and $$\beta$$, then show that the series $$a_1+b_1+a_2+b_2+a_3+b_3+\ldots$$ converges with sum $$\alpha + \beta$$.
Here's my attempt at a proof-
Let $$(S_n)$$ be the sequence of partial sums of the series $$a_1+b_1+a_2+b_2+a_3+b_3+\cdots$$. Let $$(T_n)$$ and $$(U_n)$$ be the sequence of partial sums of the series $$\sum_{n=1}^{\infty} a_n$$ and $$\sum_{n=1}^{\infty} b_n$$ respectively. Then we can define $$(S_n)$$ as follows: $$S_n = \begin{cases} T_{n/2}+U_{n/2} & \text{if n is even} \\ T_{\frac{n+1}{2}}+U_{\frac{n-1}{2}} & \text{if n is odd} \\ \end{cases}$$
Let $$\varepsilon > 0$$ be given. Then there are $$N_1, N_2 \in \mathbb{N}$$ such that $$|T_{k+1}-\alpha| < \frac{\varepsilon}{2}$$ for all $$k\ge N_1$$ and $$|U_{k}-\beta| < \frac{\varepsilon}{2}$$ for all $$k\ge N_2$$.
Let $$N=\max \{ 2N_1+1, 2N_2+1 \}$$. Then we will show that $$|S_n - (\alpha +\beta)|< \varepsilon$$ for all $$n\ge N$$.
Assume $$n\ge N$$. If $$n$$ is odd, then $$n=2k+1$$ for some $$k \in \mathbb{N}$$. Hence, $$k\ge N_1$$ and $$k\ge N_2$$ and it follows that \begin{align} |S_n - (\alpha + \beta)| &= |S_{2k+1} - (\alpha + \beta)| \\ &= |(T_{k+1} + U_{k})-(\alpha + \beta)| \\ & < |T_{k+1} - \alpha| + |U_{k}- \beta| \\ & < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon \end{align}
Likewise we can do the same with $$n$$ being even part.
Is my proof correct? Is there a way to complete the proof by avoiding the definition of a sequence?
• The proof is right. And I think that the definition of $S_{n}$ is not so nontrivial, even avoiding the definition, you still have to do similar thing. – Bonbon May 9 '19 at 13:55
• To me, the proof is cleaner if you separate out the following result, true for any sequence $a_n$: $$a_{2n}\to a \text{ and } a_{2n+1}\to a \implies a_n \to a$$ Then for $S_{2n}$ and $S_{2n+1}$, you can appeal to the sum rule for series, which avoids $\epsilon$ management – Calvin Khor May 9 '19 at 14:31
• @CalvinKhor i recall a result stating "A sequence converges to L iff all of its subsequences converge L". I was wondering if checking for only two subsequences would be enough or not. – Ashish K May 9 '19 at 14:39
• @AshishK the two sequences have to '(eventually) cover all the points'. For instance its not enough to check $a_{3n}$ and $a_{3n+1}$. – Calvin Khor May 9 '19 at 14:41
Your proof looks good to me!
An easier method would be to use the Algebraic Limit Theorem as follows:
Let $$A_k = \sum_{n=1}^{k} a_n$$ and $$B_k = \sum_{n=1}^{k} b_n$$. Since both sequences $$(A_k: k\in\mathbb{N})$$ and $$(B_k: k\in\mathbb{N})$$ converge, their respective limits $$\alpha, \beta$$ exists. Then by the Algebraic Limit Theorem,
$$\lim\limits_{n\to\infty}(A_k + B_k) = \lim\limits_{n\to\infty}(A_k) + \lim\limits_{n\to\infty}(B_k) = \alpha+\beta$$
Note that the proof for the Algebraic Limit Theorem is actually pretty similar to what you're doing!
Edit: This is not a complete proof as pointed out by Calvin Khor.
• The reason I avoided is the fact that the sequence of the partial sum of $a_1+b_1+a_2+b_2+a_3+b_3+\cdots$ is quite different from $A_k + B_k$. – Ashish K May 9 '19 at 14:00
• Isn't $A_k + B_k$ different from $S_k$ (as defined in my answer)? – Ashish K May 9 '19 at 14:03
• @AshishK what do you mean by different? – Darius May 9 '19 at 14:10
• We say that a series converges if its sequence of partial sums converges. The sequence of partial sums of $a_1+b_1+a_2+b_2+a_3+b_3+\cdots$ is not $A_k+B_k$ so I believe the two are not equivalent. Here's the problem from my text: i.imgur.com/… (the part (b) of Theorem 6.1.5 that the textbook refers to is what you did) – Ashish K May 9 '19 at 14:18
• Strictly speaking, I suppose this only proves that the sum of the first 2k terms (whose partial sums are indeed $A_k+B_k$) converges to $\alpha+\beta$. – Calvin Khor May 9 '19 at 14:23 | 2020-04-08T19:29:11 | {
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https://math.stackexchange.com/questions/926433/how-many-lists-of-100-numbers-1-to-10-only-add-to-700/926510 | # How many lists of 100 numbers (1 to 10 only) add to 700?
Each number is from one to ten inclusive only. There are $100$ numbers in the ordered list. The total must be $700$.
How many such lists?
Note: if, as it happens, this is one of those math problems where only an approximation is known, that would be great. (My guess is it's not "that" big, around $10^{20}$ maybe?)
Thank you!!
So, to be clear say you have a dice with sides labelled $1$-$10$. You roll it $100$ times, once every $10$ seconds in order. The result is, if you will, a specific array of $100$ numbers (each being $1$-$10$), each position labelled $1$ to $100$. The array must then add to $700$; how many such arrays??
Just to be absolutely clear, I believe the total of "all" such lists (so, there is no requirement to add to $700$; it can add to anything), is indeed, simply $1$ googol, ie, $10^{100}$.
• Order doesnt matter does it? – Asimov Sep 10 '14 at 15:43
• You can derive a simple recurrence formula in the case without order, that if $f(r,s,n)$ is the number of ways of writing $n$ as a sum of $s$ integers between $1$ and $r$, then $f(r,s,n)=f(r-1,s,n)+f(r,s-1,n-r)$ - the first case being those sums which don't involve the number $r$, and the second being the case where there is at least one $r$. – Mark Bennet Sep 10 '14 at 15:51
• By the way, as a programmer, I see this as a difficult variation of a typical "fencepole" problem. If you want 100 numbers that add to 700, all you do is randomly position 99 "fenceposts" on the run from 0 to 700 .. right? It's trivial. This is much harder, as you have to sort of pull them along like venetian blind slats, with a max and minimum gap. – Fattie Sep 10 '14 at 16:19
• Using the language of compositions, what you want is the number of $A$-restricted compositions of 700 into 100 parts with $A=\{1,2,\ldots,10\}$. – Semiclassical Sep 10 '14 at 16:23
• @semiclassical - magnificent! Do you know, what is the count of the A-restricted compositions of 700 into 100 parts with A=1..10?? :) – Fattie Sep 10 '14 at 16:24
Generating Function Approach
The coefficient of $x^{700}$ in $(x+x^2+x^3+\dots+x^{10})^{100}$ is the number you are looking for. This is because each choice of one of the summands in each term gives a unique choice for one of the $100$ numbers.
We can get an easier form by first noticing that the coefficient of $x^{700}$ above is the coefficient of $x^{600}$ in $(1+x+x^2+\dots+x^9)^{100}$ and that \begin{align} (1+x+x^2+\dots+x^9)^{100} &=\left(\frac{1-x^{10}}{1-x}\right)^{100}\\ &=\sum_{j=0}^{100}\binom{100}{j}\left(-x^{10}\right)^j\sum_{k=0}^\infty\binom{-100}{k}(-x)^k\\ &=\sum_{j=0}^{100}\binom{100}{j}\left(-x^{10}\right)^j\sum_{k=0}^\infty\binom{k+99}{k}x^k\tag{1} \end{align} Now we just need to add up the contributions to $x^{600}$ in $(1)$ which comes from the terms where $k=600-10j$. Thus, the coefficient of $x^{600}$ in $(1)$ is $$\sum_{j=0}^{60}(-1)^j\binom{100}{j}\binom{699-10j}{600-10j}\tag{2}$$ which comes to $$12113063910758377468145174162592296408571398929\\1260198434849317132762403014516376282321342995$$ or approximately $1.2113063910758377468\times10^{92}$
Inclusion-Exclusion Approach
To compute the number of ways for $100$ non-negative numbers to sum to $600$ we can use the usual stars and bars approach which gives $$\binom{699}{600}$$ How many of these ways include a number $10$ or bigger? We can try to count this by sticking a chunk of $10$ stars into one of the $100$ places and counting how many ways to sum $100$ numbers to $600-10$. This gives $$\binom{100}{1}\binom{689}{590}$$ but this counts twice the ways that include two numbers $10$ or bigger. To count these, we stick $2$ chunks of $10$ stars into two of the $100$ plaes and count how many ways to sum $100$ numbers to $600-20$. This gives $$\binom{100}{2}\binom{679}{580}$$ We can apply Inclusion-Exclusion to get the number of ways for $100$ non-negative numbers less than $10$ to sum to $700$ to be $$\sum_{j=0}^{60}(-1)^j\binom{100}{j}\binom{699-10j}{600-10j}$$ which is the same as gotten in $(2)$.
• Great answer. Just two minor remarks: personally I would not omit parentheses in expressions of the form $\left(\sum_ja_jx^{cj}\right)\left(\sum_kb_kx^k\right)$ (or alternatively write them $\sum_j\sum_ka_jb_kx^{cj+k}$), and I would rewrite $\binom{k+99}k$ as $\binom{k+99}{99}$. But things are correct as written. – Marc van Leeuwen Sep 11 '14 at 9:54
• @MarcvanLeeuwen: the parentheses seems to be a stylistic preference since I don't think there is much confusion and I was thinking of combining the sums. As for $\binom{k+99}{k}$ vs $\binom{k+99}{99}$, it is true that for non-negative, integer $k$ they are the same (so who cares?), however, for integer $k\lt-99$, they are different: $\binom{k+99}{k}=0$ whereas $\binom{k+99}{99}=-\binom{-k-1}{99}$ which is not $0$. It is sometimes useful to use, $$\sum_{j=0}^{60}(-1)^j\binom{100}{j}\binom{699-10j}{600-10j}=\sum_{j\in\mathbb{Z}}(-1)^j\binom{100}{j}\binom{699-10j}{600-10j}$$ – robjohn Sep 11 '14 at 21:46
In the language of compositions, what you want are the number of $A$-restricted compositions of $n=700$ into $k=100$ parts where $A=\{1,2,\ldots,a\}$ with $a=10$. While I don't know a closed-form for this number, one can express such counting succinctly using formal power series. To represent our set $A$ of allowed terms, we use the formal polynomial $f_{a}(x)=\sum\limits_{j=1}^{a} x^j = \dfrac{x-x^{a+1}}{1-x}$. If we square this, then the coefficient of a term $x^n$ will represent the number of ways in which two integers in $A$ can add to a total of $n$. This gives us the following result:
The number of $A$-restricted compositions, with $A=\{1,2,\ldots a\}$, of $n$ into $k$ parts is the $n$th coefficient of $f_{10}(x)^k$, which we can express as $[x^n]\left(\dfrac{x-x^{a+1}}{1-x}\right)^k.$
Thus in the case at hand what we want to find (or at least estimate) is $\boxed{\displaystyle\left[x^{700}\right]\left(\dfrac{x-x^{11}}{1-x}\right)^{100}}$. Amazingly, this can be pulled off rather easily by WolframAlpha, yielding the rather impressive result of \boxed{ \begin{align} 12113063910758377468145174162592296408571398929 &\\ 1260198434849317132762403014516376282321342995 &\approx 1.2 \times 10^{92} \end{align} } For approximations, I suggest perusing Sedgwick and Flajolet's Analytic Combinatorics which is available for download on the book site. I may dig into there myself to see if anything is known or can be estimated about such numbers.
• oeis.org/A037306 seems to have some pointers. – mvw Sep 10 '14 at 16:48
• If someone can provide a better way to format that 92-digit number, please do! – Semiclassical Sep 10 '14 at 16:54
• What does the zero in $\{x,0,700\}$ (Wolfram Alpha notation) show? – user26486 Sep 10 '14 at 18:12
• @mathh: It means that we're expanding about $x=0$ (rather than, say, $x=1$). – Semiclassical Sep 10 '14 at 18:17
• @Semiclassical I have no knowledge about Taylor series (or anything similar) so far (I assume the zero is related to them). Is it true that when working with generating functions we can always simply write it as $\{x,0,k\}$ after the generating function, where $k$ is the coefficient we're searching for? – user26486 Sep 10 '14 at 19:09
Note that the numbers being added together have mean $\mu_0=11/2$ and standard deviation $\sigma_0=\sqrt{33}/2$; by the central limit theorem, the distribution of the sum of $100$ such numbers is approximately a Gaussian with mean $\mu=100\mu_0 = 550$ and standard deviation $\sigma=10\sigma_0=5\sqrt{33}$. The continuous Gaussian distribution is $$f(x,\mu,\sigma)=\frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right);$$ and to a first approximation you can ignore the correction from discreteness. So the number you want is roughly $$N\approx 10^{100} \cdot f(700,550,5\sqrt{33})=\frac{10^{100}}{5\sqrt{33}\sqrt{2\pi}}\exp\left(-\frac{150^2}{1650}\right)\\ = \frac{10^{100}}{5\sqrt{66\pi}}e^{-150/11}\approx 1.66\times 10^{92}.$$ It is helpful to compare this to the computational result. A simple Python implementation is the following:
def N(x, n, cache={(0,0):1}):
if x<0 or (n==0 and x>0): return 0
if not cache.has_key((x,n)):
cache[(x,n)] = sum([N(x-i, n-1) for i in xrange(1,11)])
return cache[(x,n)]
Then N(700, 100) * 1.0 yields 1.2113063910758377e+92.
• Thanks for this cute alternative estimation. – mvw Sep 10 '14 at 17:06
• @Semiclassical: Thanks, I was using the wrong standard deviation. I think it's fixed now. – mjqxxxx Sep 10 '14 at 17:43
• @mvw: By $\mu_0$ I'm referring to the mean of the candidate values, not the mean of the actually selected values. – mjqxxxx Sep 10 '14 at 18:01
• That's nice insight. – Felix Marin Sep 10 '14 at 22:02
• Note: $\mu_0 = \left(\sum_{i=1}^{10}i\right)/ 10$, $\sigma_0 =\sqrt{\left(\sum_{i=1}^{10}(i-\mu_0)^2\right)/10}$. – mvw Sep 10 '14 at 23:26
For a calculation answer, I get about $1.21131 \cdot 10^{92}$ I just made an Excel sheet with rows labeled -9 through 100 and columns 1 through 100. The rows represent the number of ways to make that sum with the number of numbers $1$ through $10$ in the column heading. Under column 1,put $1$ in each row $1$ through $10$. Then in subsequent columns each cell is the sum of ten entries in the column to the left, from one above to ten above. The rows with negative numbers in them are there so I didn't have to cut off the early sums, like for a sum of $3$ from two numbers.
Here is the top corner of the sheet:
$$\begin {array} {l l l l l } tot&1&2&3&4\\ \hline 1&1&&&\\2&1&1&0&0\\3&1&2&1&0\\4&1&3&3&1\\5&1&4&6&4\\6&1&5&10&10\\7&1&6&15&20\\8&1&7&21&35\\9&1&8&28&56\\10&1&9&36&84\\11&&10&45&120\\12&&9&55&165\\13&&8&63&220\\14&&7&69&282\\15&&6&73&348\\16&&5&75&415\\17&&4&75&480\\18&&3&73&540\\19&&2&69&592\\20&&1&63&633\\21&&0&55&660\\22&&0&45&670\\23&&0&36&660\\24&&0&28&633\\25&&0&21&592\\26&&0&15&540\\27&&0&10&480\\28&&0&6&415\\29&&0&3&348\\30&&0&1&282\\31&&0&0&220\\32&&0&0&165\\33&&0&0&120\\34&&0&0&84\\35&&0&0&56\\36&&0&0&35\\37&&0&0&20\\38&&0&0&10\\39&&0&0&4\\40&&0&0&1\\ \end {array}$$
• hell - I'd have to really think about that! – Fattie Sep 10 '14 at 16:40
• WolframAlpha gives a result of roughly $1.2\times 10^{92}$. Not sure whether that's a matter of Excel/WA breaking down or a typo in one of our implementations. – Semiclassical Sep 10 '14 at 16:49
• @Semiclassical: I would guess it is Excel numerics, but am surprised the discrepancy is that large. – Ross Millikan Sep 10 '14 at 16:54
• Aye. A 20% discrepancy would make sense for an approximation, but not for a direct count. – Semiclassical Sep 10 '14 at 16:56
• I think you may just be looking in the wrong place… your result is the correct number of ways to make $701$ from a list of one hundred numbers in $\{1,2,\ldots,10\}$. – mjqxxxx Sep 10 '14 at 19:53
\begin{align} &\color{#66f}{\large\sum_{k_{1}=1}^{10}\ldots\sum_{k_{100}=1}^{10} \delta_{k_{1}\ +\ \cdots\ +\ x_{100},700}} =\sum_{k_{1}=1}^{10}\ldots\sum_{k_{100}=1}^{10}\oint_{\verts{z}\ =\ a} {1 \over z^{-k_{1}\ -\ \cdots\ -\ k_{100}\ +\ 701}}\,{\dd z \over 2\pi\ic} \\[3mm]&=\oint_{\verts{z}\ =\ a} {1 \over z^{701}}\,\pars{\sum_{k =1}^{10}z^{k}}^{100}\,{\dd z \over 2\pi\ic} =\oint_{\verts{z}\ =\ a}{1 \over z^{701}}\,\pars{z\,{z^{10} - 1 \over z - 1}}^{100} \,\,{\dd z \over 2\pi\ic} \\[3mm]&=\oint_{\verts{z}\ =\ a}{1 \over z^{601}}\, {\pars{1 - z^{10}}^{100} \over \pars{1 - z}^{100}}\,\,{\dd z \over 2\pi\ic} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\pars{1} \\[3mm]&=\sum_{n = 0}^{100}\sum_{k = 0}^{\infty}\pars{-1}^{n + k}{100 \choose n} {-100 \choose k}\oint_{\verts{z}\ =\ a}{1 \over z^{601 - 10n - k}} \,{\dd z \over 2\pi\ic} \\[3mm]&=\sum_{n = 0}^{100}\sum_{k = 0}^{\infty}\pars{-1}^{n + k}{100 \choose n} \pars{-1}^{k}{k + 99 \choose 99}\delta_{10n + k,600} =\sum_{n = 0}^{100}\pars{-1}^{n}{100 \choose n}{699 - 10n \choose 99} \end{align}
Contributions to the sum are limited by $\quad\ds{0 \leq n \leq 100}\quad$ and $\quad\ds{99 \leq 699 - 10n}\quad$ which yield $\quad\ds{0 \leq n \leq 60}$:
$$\color{#66f}{\large\sum_{k_{1}=1}^{10}\ldots\sum_{k_{100}=1}^{10} \delta_{k_{1}\ +\ \cdots\ +\ x_{100},700} =\sum_{n = 0}^{60}\pars{-1}^{n}{100 \choose n}{699 - 10n \choose 99}}$$ which is $\ds{\pars{~\mbox{see expression}\ \pars{1}~}}$ equal to $\ds{\bracks{z^{600}}\pars{1 - z^{10} \over 1 - z}^{100}}$.
It leads to the value \begin{align}& {\tt 1.211306391075837746814517416259229640857139892912601984348493171327624} \\ &{\tt 03014516376282321342995 \times 10^{92}} \end{align}
• Wow. I bet, this would help generalise the problem, too. – Fattie Sep 11 '14 at 8:09
• Compared to the approach by robjohn, I'm don't see any insight that is gained by the use of contour integrals (rather the contrary). – Marc van Leeuwen Sep 11 '14 at 9:58
• Marc ... uh, obviously! :-) Thanks for that; I'm trying to make sense of what is different, natty, solid, etc! – Fattie Sep 11 '14 at 14:10
• @JoeBlow It's true. It lets you to set a general formula if you want. It avoids 'intermediate steps' that involve "counting" which is the most difficult situation. It's straightforward. Maybe, it takes a little bit to realize that. Thanks. – Felix Marin Sep 11 '14 at 18:08
A066099 features a nice visual generation pattern for compositions of $n$ (listing them in reverse lexical order):
From Omar E. Pol, Sep 03 2013:
----------------------------------- ----------
n j Diagram Composition j Row length
----------------------------------- ----------
. _
1 1 |_| 1; 1
. _ _
2 1 | _| 2, 1
2 2 |_|_| 1, 1; 2
. _ _ _
3 1 | _| 3, 1
3 2 | _|_| 2, 1, 2
3 3 | | _| 1, 2, 2
3 4 |_|_|_| 1, 1, 1; 3
. _ _ _ _
4 1 | _| 4, 1
4 2 | _|_| 3, 1, 2
4 3 | | _| 2, 2, 2
4 4 | _|_|_| 2, 1, 1, 3
4 5 | | _| 1, 3, 2
4 6 | | _|_| 1, 2, 1, 3
4 7 | | | _| 1, 1, 2, 3
4 8 |_|_|_|_| 1, 1, 1, 1; 4
.
So we would need that diagramm for $n = 700$ and only those rows with $100$ summands and all summands from $\{1, \ldots, 10 \}$.
Pragmatic problem: There are $N = 2.63 \times 10^{210}$ compositions of $700$, according to WolframAlpha link.
Looking at the diagram one notices that $N(n)$ is resulting from the doubling (see step 3 of the algorithm below) and is thus simply $N(n) = 2^{N-1}$ (A000225) and in this case: $N(700) = 2^{699}$.
1. The list of compositions for $n+1$ can be obtained, by using a copy of the list for $n$ first.
2. Then we put a column of $1$ elements to the left of that list.
3. Finally we put another copy of the list for $n$ on top, shifted one position left, aligning with the column of $1$ elements, and adding $1$ to each entry in the first column of the copy.
Side note: It is already interesting to see how the row lengths develop:
1
1 2
1 2 2 3
1 2 2 3 2 3 3 4
1 2 2 3 2 3 3 4 2 3 3 4 3 4 4 5
If we have $2^n$ terms, we find the next $2^n$ terms by copying the first $2^n$ terms plus $1$ each.
This leads to A063787 and is related to A000120 (binary weight of $n$). They are described as fractal sequences - deleting every other term gives the original series.
As this seems to be a programmer's problem, I will give you a sketch of the solution. For sake of completeness, I should advise you that this kind of problems should be posted in the correct stack exchange forum.
You certainly have been familiar with DP. So this problem should be solved in that manner. You should look to a certain function, namely $f(n, m)$ that gives you the number of ordered sequences of $n$ integers in $A = \{1, \cdots, 10\}$ that sum $m$. You want $f(100, 700)$.
You should know a recursive way of computing this, as any sequence $a_1, \cdots , a_n$ that sums $m$ leads bijectively to a smaller sequence $a_1, \cdots , a_{n-1}$ that sum $m-a_n$ So $$f(n, m) = \sum_{k \in A}^{ m - k \geq 0} f(n-1, m-k)$$
You should initialize your function in $f(n, n) = 1$ and $f(1, m) = 1$ for $m\in A$, $f(1, m)=0$ for $m> 10$. And now is just a question of programming.
• It is a valid combinatorics problem. – mvw Sep 10 '14 at 16:46
• I wont deny it. I just though that the OP wouldn't be interested in the generating function solution. My bad if it is not the case. – PenasRaul Sep 10 '14 at 16:53
• +1 Have you tried just entering the recurrence and conditions into Mathematica. That probably gives the answer and the programming is minimal. Too pressed for time right now to try it! – almagest Sep 11 '14 at 4:23
• I can absolutely assure you that very few top software engineers, can do, the math on this page :-) – Fattie Sep 11 '14 at 8:10 | 2019-05-25T08:49:08 | {
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http://math.stackexchange.com/questions/180408/topology-on-the-set-of-partitions | # Topology on the set of partitions
Let $X$ be the set of all partitions of $[0,1]$ such that each element of the partition is Lebesgue-measurable.
Let $Y$ be the set of all partitions of $[0,1]$ such that each element of the partition is a Borel set.
Is there a standard topology for a set like $X$ or $Y$? If so, is $X$ (or $Y$) compact in this topology?
I thought that the set of all partitions of $[0,1]$ (including those with Vitali sets as elements) is a more complicated object, but as @JDH points out in the proposed answer, this may not be the case.
The broader context for this question is my interest in convergence of measures on such a set (hence the focus on Borel or measurable partitions). If $Y$ is compact and metrizable, then the set of measures on $Y$ is itself a compact metric space in the weak topology, which opens the door to standard results on convergence of such measures.
-
Just a remark on the edit, not all non-measurable sets are Vitali sets. There are ultrafilters; coded graphs for discontinuous functionals; Sierpinski sets; etc. those are different animals than Vitali sets (and their existence follows from different choice principles too). – Asaf Karagila Aug 8 '12 at 21:38
are your partitions required to be finite? countable? arbitrary? – J. Loreaux Aug 8 '12 at 22:12
@JDH - Thanks, that's a very clear and helpful answer. Any thoughts on whether the set of all partitions (or sets X and Y above) is compact in the lower-cone topology? – exk Aug 9 '12 at 3:06
@AsafKaragila: The comments about Vitali sets was simply meant to illustrate one of the ways that the set of all partitions differs from the set of partitions whose elements are measurable; there was no claim that the Vitali sets are the only difference. Re: J.Loreaux's comment: Both $X$ and $Y$ are sets of all (thus arbitrary) partitions satisfying a given condition on the elements. – exk Aug 9 '12 at 3:50
Unfortunately, $X$ and $Y$ are not actually lattices, but rather lower semi-lattices (we only have meet $\wedge$ and not join $\vee$) and they are only $\sigma$-complete as lower semi-lattices. If you use the uniform formulation as I describe in the last paragraph of my answer, however, they those will result in lattices, but again only $\sigma$-complete and not complete. But the space of all partitions is a complete lattice and hence will be compact in this topology. – JDH Aug 9 '12 at 10:55
There are several natural topologies to put on the space of partitions, which can also be thought of as the space of equivalence relations.
First, the collection of partitions of a set is a partial order under the refinement relation, where one partition $\mathcal{A}$ refines another $\mathcal{B}$ when every element of $\mathcal{A}$ is a subset of a set in $\mathcal{B}$. What's more, the space of partitions for each of your spaces is a lower semi-lattice, since any two partitions $\mathcal{A}$ and $\mathcal{B}$ have a coarsest common refinement, the collection of nonempty $A\cap B$, where $A\in\mathcal{A}$ and $B\in\mathcal{B}$. Indeed, the cases of $X$ and $Y$ lead to a $\sigma$-complete lower semi-lattice, since one may similarly intersect countably many sets and remain Borel or Lebesgue measurable as required. Since every partial order has a natural topology, the lower-cone topology, where the open sets are simply those that are closed downwards with respect to the order, we may place this topology on the space of partitions. In your case, the open sets of partitions would be those that include all refinements of any partition in the set.
This topology is compact---in the cases of $X$ and $Y$ and also in the case of the set of all partitions---simply because there is a coarsest partition, the partition with only one component consisting of the whole set. The only open neighborhood of this partition in the lower-cone topology is the entire space of partitions, since every partition refines it.
Secondly, a different dual topology would arise from turning the order upside down, and for that, the open sets would be closed under encoarsening rather than refinement. This topology also is compact, since there is a finest topology, consisting of the collection of singletons, and every partition is coarser than it.
Note that the space of all partitions, despite your remark on its complexity, actually exhibits much nicer lattice-theoretic properties than the Borel partitions or the measurable partitions. This is because the space of all partitions is actually a complete lattice. This is because any family of partitions has a largest common refinement, obtained simply by intersecting the equivalence relations as sets of ordered pairs. Similarly, one may take the equivalence relation generated by any family of partitions to find the smallest common encoarsening.
Another natural topology arises on the class of equivalence relations by saying that a basic open set is determined by finitely many equivalences and non-equivalences. Thus, one specifies finitely many equivalencies and non-equivalencies on the points $x_i\sim x_j$, $y_s\not\sim y_t$, and then the basic open set is the collection of all partitions that respect those finitely many requirements. This is a common kind of topology to place on the set of models of a first-order theory, where the basic open sets specify finitely much information about the predicates of the model; the collection of partitions amounts to the collection of equivalence relations, which is first-order.
One can imagine more elaborate topologies in this line, which take more into account your context of Lebesgue measurable or Borel partitions, by moving beyond finitely many point requirements to the case of finitely many requirements on Borel sets, or on measurable sets.
It would of course depend on your purpose with the topology to know which is best for that purpose.
Finally, let me mention that the well-developed theory of Borel equivalence relations is deeply concerned with the space of Borel partitions of the reals, but not exactly in your sense. What is usually required in that theory is not merely that every element of the partition is Borel, that is, that every equivalence class is Borel, but rather one insists that the binary relation itself, thought of as a subset of the plane, should be Borel. This is a more uniform kind of Borel, and the theory is extremely robust and active.
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https://www.physicsforums.com/threads/3d-rays.273897/ | # 3D rays
1. Nov 21, 2008
### cam875
Im having problems understanding headings for 3d rays. I mean with 2D you can have an object at (2,3) with a heading of 83 degrees. But in 3d you can have an object at (2,3,3) but how can you describe its heading, do you need two separate headings for the two planes, how does this work. Thanks in advance
2. Nov 21, 2008
You can describe a direction in 3D with two angles θ and ϕ, with θ being the angle from the z-axis, and ϕ being an angle in the xy-plane. You can think of them as corresponding to latitude and longitude, which specify a location on a sphere, or equivalently a ray from the centre to a point on the sphere.
Together with a third coordinate r representing distance, you get the spherical coordinate system (r, θ, ϕ), which picks out a point in space just like the Cartesian (x, y, z) coordinate system.
3. Nov 21, 2008
### cam875
so if you have an object at (0,0,0) and its heading on the z plane is 0 degrees and on the x-y plane it is 0 degrees and the object travels 1m in that direction how do you figure out its new x,y,z co-ord. If you think about it the object would travel straight up but how can you prove this?
4. Nov 21, 2008
I'm not quite sure what you mean by the "z plane"; do you mean that its heading is at an angle of 0 degrees to the z-axis? In that case there are at least two ways you could approach the problem.
If your displacement vector is d, then if its angle to the z-axis is 0 (i.e. it is parallel to the z-axis), you have ‖d‖ = d · k = dz (where k is your unit vector in the z-axis), so if ‖d‖ = 1, you must have d = (0, 0, 1).
A second, more straightforward method is this: You know that your displacement vector in spherical coordinates is (r, θ, ϕ) = (1, 0, 0). Then you could transform from spherical coordinates to Cartesian coordinates:
x = r sin θ cos ϕ
y = r sin θ sin ϕ
z = r cos θ
from which you would get x = 0, y = 0, and z = 1.
5. Nov 21, 2008
### HallsofIvy
Staff Emeritus
Another way is to use the "direction angles", the angles the vector makes with with each of the axes, say $\theta_x$, $\theta_y$, $\theta_z$. Of course there are three angles when, as said above[, you need only two, so those are not independent: for any vector, $cos^2(\theta_x)+ cos^2(\theta_y)+ cos^2(\theta_z)= 1$ so the components of a unit vector are just its "direction cosines".
6. Nov 21, 2008
### cam875
ok, starting to make sense but how exactly did you figure out
x = r sin θ cos ϕ
y = r sin θ sin ϕ
z = r cos θ
because x = 0, y = 0, and z = 1 is the exact answer and I want to know your method you used to figure out those formulas, and do you have to use trig?
also, i have a question related to 2d math. If you consider a similar problem where an object lies at (0,0) and has a heading of 45 degrees and its distance is 1 than the new x,y co-ord should be (0.5,0.5) right? but when you do the trig it gives you (0.707,0.707) which one is right?
7. Nov 21, 2008
You can derive the relations for spherical coordinates using some trigonometry. Wikipedia has them but unfortunately doesn't show a derivation; here's a rough sketch of how it works:
Let's say you have a vector r specified in spherical coordinates by (r, θ, ϕ), where r is the length, θ is the angle to the z-axis, and ϕ is the angle you get in the xy-plane by projecting r onto the xy-plane. (See the first figure on the Wikipedia article for an idea of what this looks like.) The dotted triangle has three sides: the vector r itself, the vertical side which has length z, and the bottom side being a vector that we'll call, say, r', with length r'. Trigonometry gives z = r cos θ and r' = r sin θ. Now this r' vector is entirely in the xy-plane, and its x and y components are the same as those of r, so you end up with x = r' cos ϕ and y = r' sin ϕ. Putting r' = r sin θ gives the result.
For your second question, (0.707, 0.707) is correct. If you check the length of (0.5, 0.5), you get about 0.707 by the Pythagorean theorem, not 1.
8. Nov 22, 2008
### cam875
so your saying that if you travelled 1 metre on a 45 degree angle you would travel 0.5 m on the x axis and the y axis or 0.707 m on the x and y axis because 0.5 m makes much more sense.
9. Nov 22, 2008
It would be 0.707 m.
If you move in a straight line such that your positions along the x- and y-axes each change by 0.5 m, you will have traveled 0.707 m. If you first went 0.5 m in the x direction and then 0.5 m in the y direction, then yes, the distance you traveled would be 1 m, but the displacement from the original location would only be 0.707 m (because you didn't go in a straight line).
10. Nov 22, 2008
### cam875
ok I think i understand so if an object travels 1 m on a 45 degree angle from (0,0) it is now at (0.707, 0.707)?
now im starting to wonder where did sin cos and tan come from and how do they actually work in 2D cuz i dont really just like using them without understanding them.
11. Nov 23, 2008
### HallsofIvy
Staff Emeritus
Hopefully you see that the x and y components of a vector of length 1 can't be 1/2 because of the Pythagorean theorem: (1/2)2+ (1/2)2= 1/4+ 1/4=1/2, not 1.
$$\left(\frac{\sqrt{2}}{2}\right)^2+ \left(\frac{\sqrt{2}}{2}\right)^2= \frac{2}{4}+ \frac{2}{4}$$
$$= \frac{1}{2}+ \frac{1}{2}= 1$$
As for "where sin cos and tan come from", they are DEFINED in terms of ratios of exactly the quantities you are talking about. Here's a nice little introduction:
http://www.clarku.edu/~djoyce/trig/
12. Nov 23, 2008
### cam875
alright thanks for the info, ill check that link out.
13. Nov 30, 2008
### eee3
Wha, that wikipedia-pic sure was confusing.
This one that is easier to understand, not using the same axis-system though, just rename x/y/z if you find it confusing.
The big blue dot is a point in the 3d-space. The big black dot is a point on the XZ-plane straight below the 3d-point.
Now you can easily see that it simply consists of two 2d-triangles, one blue and one red. The length of b is simply found by using trig on the blue triangle, that's why adriaks formulas for x and y have two trig-terms while z only has one:
x = r sin θ cos ϕ
y = r sin θ sin ϕ
z = r cos θ
Note again that it's not the same coordinate system.
14. Dec 5, 2008
### cam875
ok i wrote a computer program using what i learned from here and im hoping its now correct. It tells me that if I travelled with a heading of 45 degrees on the x-y plane and 45 degrees on the z plane or w.e u wanna call it for a distance of 1 from (0,0,0) i end up at (0.707, 0.707, 0.707) im hoping that that is correct.
15. Dec 5, 2008
No, it's not; you'd get a total length of $$\sqrt{(0.707)^2+(0.707)^2+(0.707)^2} = 1.225$$ by the Pythagorean theorem. If you put in r = 1 and 45 degrees for each of the angles in the equations
x = r sin θ cos ϕ
y = r sin θ sin ϕ
z = r cos θ
you get (0.5, 0.5, 0.707).
16. Dec 5, 2008
### cam875
that equation doesnt't make sense to me, the z part makes sense but why is there 2 different trigonometric calculations involved in the x and y part.
17. Dec 8, 2008
### eee3
Image 2 posts up ^^
The length of the hypotenuse for the XY-plane-triangle is given by [Hypotenuse for the blue plane]*Cos(phi).
So Hypotenuse_red = Hypotenuse_blue*Cos(phi)
Therefore:
x = Length_red =
= Hypotenuse_red * Cos(gamma) =
= [Hypotenuse_blue*Cos(phi)] * Cos(gamma) =
= r cos θ cos ϕ
18. Dec 8, 2008
Also, the Pythagorean Theorem has to hold as well: x2 + y2 + z2 = r2.
x2 + y2 + z2
= (r sin θ cos ϕ)2 + (r sin θ sin ϕ)2 + (r cos θ)2
= r2 sin2 θ cos2 ϕ + r2 sin2 θ sin2 ϕ + r2 cos2 θ
= r2 (sin2 θ (cos2 ϕ + sin2 ϕ) + cos2 θ)
= r2 (sin2 θ + cos2 θ)
= r2.
Also, as a general note, observe that x and y depend on both angles, so they both better be in the equations for them. | 2016-07-25T12:01:05 | {
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http://mathoverflow.net/questions/85409/problem-about-expectation-of-maximum-partial-sum | # Problem about expectation of maximum partial sum
Given a number $m$, a random composition (strong) of this number into $n$ positive parts so that we can get $n$ random variable $X_1, X_2,\dots, X_n$ with $$X_1+X_2+\cdots+X_n=m$$
Note that all compositions of $m$ have the same probability.
Let $Y_i = X_i - \mathbb{E}(X_i)$, $S_i = Y_1 + \cdots + Y_i$
I want to calculate $\mathbb{E}(\max_{1 \le i \le n} S_i)$.
-
And why do you want to compute this? Is this homework? – Igor Rivin Jan 11 '12 at 13:13
@IgorRivin It is not homework. I want to analyze a algorithm I proposed. And it can be converted to the above problem. – Fan Zhang Jan 11 '12 at 13:59
@Fan Zhang Are you assuming that the parts are written in increasing order? If not then all of the $E(X_i)$'s are the same, just by considering all possible permutations. – Alan Haynes Jan 11 '12 at 14:20
@AH I am not assuming that the parts are written in increasing order. It is number composition, so order is significant here. – Fan Zhang Jan 11 '12 at 15:17
@Fan: OK, but again what range for $n$ and $m$? are you interested in asymptotics when they tend to infinity? – Ori Gurel-Gurevich Jan 13 '12 at 18:29
[Edited to add Sage plots of the asymptotic distributions for $n=2,3,4,5,6,7$ and $n=10,20,30,40$]
There are $m-1 \choose n-1$ "compositions" of $m$ into $n$ positive parts $X_i$, because they correspond bijectively with choices of $n-1$ partial sums out of $\lbrace 1, 2, \ldots, m-1\rbrace$. By symmetry, each $X_i$ has expectation $m/n$. For large $n$ and much larger $m$, the sequence of normalized partial sums $$S_i = \sum_{h=1}^i Y_h = \sum_{h=1}^i (X_h - \mathbb{E}(X_h)) = \Bigl(\sum_{h=1}^i X_h \Bigr) - \frac{i}{n} m$$ looks like a scaled "Brownian bridge", i.e. ${\rm BB}(t) = B(t) - B(1)t$ where $B(t)$ is Brownian motion on $[0,1]$. The distribution of $\max_t {\rm BB}(t)$ is known thanks to a reflection trick, and thus so is its expectation. If I did this right, for $\sigma > 0$ the probability that $S_\max := \max_i S(i)$ exceeds $\sigma$ approaches $\exp(-2n(\sigma/m))^2$, which would make $$\mathbb{E}(S_\max) \sim \int_0^\infty e^{-2n(\sigma/m)^2} d\sigma = \sqrt{\frac\pi{8n}} \cdot m.$$ Since the question at hand asks only for the expectation of $S_\max$, not subtler features of its distribution, the estimate should be reasonably good for moderately large $n$ and $m$. As Aaron Meyerowitz noted, for fixed $n$ we have $$\mathbb{E}(S_\max) = c_n m + O(1)$$ as $m \rightarrow \infty$; computation of the constants $c_n$ for $n \leq 40$ suggests that $n^{1/2} c_n \rightarrow \sqrt{\pi/8}$ but not very quickly: $n^{1/2} (\sqrt{\pi/8} - n^{1/2} c_n)$ seems to be converging to about $0.66$, and while $\sqrt{\pi/8} = 0.626657\ldots$, the computed $n^{1/2} c_n$ first exceeds $0.5$ only at $n = 27$. The formulas below may give a start towards proving this behavior.
Getting good estimates for finite values of $n$ seems unwieldy because computing the probability that $\max_i S(i) \leq \sigma$ requires counting "compositions" that satisfy $X_1 + \ldots + X_i \leq (i/n) m + \sigma$ for each $\sigma$, which gives rise to an intimidating inclusion-exclusion. Fortunately the answers to this mathoverflow query, which appeared just yesterday, give for any increasing integer sequence $0 < a_1 < a_2 < \ldots < a_N$ a nice determinant formula for the number of $N$-tuples $(x_1,\ldots,x_N)$ of integer sequences satisfying $0 < x_1 < \ldots < x_N$ and $x_i \leq a_i$ for each $i \leq N$. In our setting, we compute the number of "compositions" with $\max_i S(i) \leq \sigma$ by taking $N = n-1$, $x_i = \sum_{h=1}^i X_h$, and each $a_i = \min (m-1, \lfloor (i/n) m + \sigma \rfloor)$. The determinant has order $N$, with $(i,j)$ entry equal $a_i + j - i \choose j - i + 1$ [which is $1$ on the first subdiagonal $i = j+1$, and $0$ on the triangle $i > j+1$ under this subdiagonal]. This makes it feasible to compute exactly the distribution of $S_\max$ for moderately large $n$ and any $m$, and also the asymptotic distribution as $m \rightarrow \infty$.
For given $n$, the asymptotic distribution of $S_\max / m$ is piecewise polynomial, but takes a while to look like the limiting form proportional to $\sigma \exp(-2n(\sigma/m)^2) d\sigma$; notably it is discontinuous at $\sigma = 1/n$ with a substantial jump downwards. Here are Sage plots showing these distributions in red/orange/green/blue/purple/black for $n=2$ through $n=7$:
(The total area is $(n-1)/n$, not $1$, because $1/n$ of the measure is concentrated at $\sigma = 0$). Likewise for $n=10,20,30,40$, plotting only $0 < \sigma < 0.6$ because the density is too small to be seen for $\sigma \geq 0.6$:
(after much more computing, most of it for the $40 \times 40$ determinants; at $n=40$ the graph finally crests after the downwards plunge at $1/n$).
Finally, a tabulation of the values of $c_n$ for $n = 1, 2, \ldots, 40$. For each $n$ I give $n^n c_n$, which is an integer except for $n=2$ when $c_2 = 1/8$. This value, as well as each of $c_3 = 4/3^3$, $c_4 = 39/4^4$, and $c_5 = 472 / 5^5$, agrees with A.Meyerowitz's calculations.
[1, 0]
[2, 1/2]
[3, 4]
[4, 39]
[5, 472]
[6, 6900]
[7, 118716]
[8, 2354072]
[9, 52911216]
[10, 1330107840]
[11, 36991592500]
[12, 1127914077312]
[13, 37420777559496]
[14, 1342183358856704]
[15, 51756244887797100]
[16, 2135359495833676800]
[17, 93864296121282210784]
[18, 4379542774345464496128]
[19, 216178594161376244063076]
[20, 11255374377126199463936000]
[21, 616463053079082019376239800]
[22, 35432440194603405959506034688]
[23, 2132478311049609494982190450204]
[24, 134116927093400952330702474706944]
[25, 8798258310785305861553627142930000]
[26, 601017143689398400881632598032384000]
[27, 42684847106441394367226307718311565716]
[28, 3147222817221577402622824207375266742272]
[29, 240582153893938356991908848927905622445736]
[30, 19043079550271688145972837306025115648000000]
[31, 1558981284930199828739239320361571788041021900]
[32, 131855889346498739861328689280063889600321421312]
[33, 11509801310013312943392059948175963018705195688896]
[34, 1035923337749909421439098617643978496876513679376384]
[35, 96047358406199526246797502389944873251740366086562500]
[36, 9165799239775410749318809193562746250766254788837376000]
[37, 899564153243436548625817312320806272340082850282897445784]
[38, 90727638906463992065814103522957255273344987915765288009728]
[39, 9396779387234810402125876063643842517670905874506382252419196]
[40, 998740925886849063603687252012149942602287367747692134400000000]
-
@noam-d-elkies Would you please give more detail about how to calculate the $C_n n^n$? – Fan Zhang Feb 6 '12 at 9:53
If I understand the problem correctly then $\mathbb{E}(X_i)=\frac{m}{n}$ So that the $Y_i$ are values from $0-\frac mn,1-\frac{m}{n},\dots,m-\frac mn$ which add to $0$.
If we would instead compose $m-n$ as an ordered sum of $n$ non-negative integers this would have the effect of lowering the $X_i$ and expectation by one each but would give the same $Y_i$ and expectation for the maximum of the $S_i.$ That way of setting up the problem might lead to neater expressions.
It seems to me that the expected maximum would asymptotically be $$c_nm+e_r+O(1/m)$$ where $e_r$ depends on the congruence class of $m$ mod $n.$ The fuzzy reasoning is that if we were to take a composition of $m$ and triple all the entries we would get a composition of $m'=3m$ with thrice the expected maximum. To get an arbitrary composition of $3m$ we might move some of the parts up or down by 1, but that would not have a big effect on the ratio of the expected maximum to $m'$. Beyond that I'll just say that the calculations below support this supposition with $$c_2=\frac{1}{8},c_3=\frac{4}{27},c_4=\frac{39}{256},c_5=\frac{472}{3125}.$$
More specifically, calculations strongly suggest for each $0 \le r \le n-1$ there is a polynomial $p_r(q)$ of degree $n$ such that for $m=qn+r$ $$\mathbb{E}(S_{\max})=\frac{p_r(q)}{(m-1)(m-2)\dots(m-n)}.$$ In all cases the leading coefficient is the same, namely $n^{n}c_n.$ Assuming that that is the case, one can find the coefficients of $p_r(q)$ and hence $c_n$ by computing $\mathbb{E}(S_{\max})$ for $m=r,n+r,2n+r,\dots,n^2+r.$ And going a bit further provides a check. Eventually this would be too hard a calculation, but the obvious naive strategy works well for a while (I did $n \le 4$ and $m \le 70$ with Maple.)
later N. Elkies' great answer explains (and establishes) these facts and much more, but I leave this amended naive approach for what it is worth.
For $n=2$ the expected value of the maximum is $\frac{m-1}{8}$ or $\frac{m-1}{8}-\frac{1}{8(m-1)}$ according as $m$ is odd or even. In the odd case, for $1 \le X_1 \le \frac{m-1}{2}$ the maximum is $S_2=0.$ And for $\frac{m+1}{2} \le X_1 \le m-1$ the maximum is $S_1=X_1-\frac{m}{2}$ with expected value $\frac{1/2+(m-2)/2}{2}=\frac{m-1}{4}.$ When $m$ is even there is also the composition into 2 equal parts with $S_1=S_2=0$ and it slightly lowers the answer. After some arithmetic the expected maximum turns out to be as stated.
For $n=3$ the expected maximum appears to be about $\frac{4}{27}m$ and exactly $\frac{2(q-1)^2(2q-3)}{(m-1)(m-2)},\frac{2(q-1)q(2q-3)}{(m-1)(m-2)}$ or $\frac{2q^2(2q-3)}{(m-1)(m-2)}$ according as $m=3q-1,m=3q$ or $m=3q+1.$ The sequence A200067 from the OEIS may be related.
In the case that $n=4$ the expected maximum appears to be $\frac{39}{256}m+O(1)$ In the case $m=4q$ the expected maximum appears to be exactly $\frac{3(13q^2-13q+3)q(q-1)}{(m-1)(m-2)(m-3)}.$ When $m=4q-1$ or $m=4q+1$ the term q(q-1) is replaced by $(q-1)^2$ or $q^2$ respectively. This is similar to the $n=3$ case above. For $m=4q+2$ it comes out to be $\frac{3q^2(13q^2+2)}{(m-1)(m-2)(m-3)}$
For $n=5$ similar considerations give an expected maximum of $\frac{472}{3125}m+O(1)$
-
Could you please post your maple program to calculate the Cn？ – Fan Zhang Jan 22 '12 at 5:03
It is built in. >with(combinat): then >composition(5,3); will return {[1, 1, 3], [1, 2, 2], [1, 3, 1], [2, 1, 2], [2, 2, 1], [3, 1, 1]}. You can see how it is programmed by restarting putting in >with(combinat): >interface(verbosepreoc=3): then >print(combination): (Restart to avoid seeing extensive remembered results!) It is pretty short. – Aaron Meyerowitz Jan 23 '12 at 5:28
I can't comment , but based on Noam D. Elkies's numbers and OEIS A000435 it looks as if
$$c_n = \frac{(n-1)!}{2 n^n} \sum_{k=0}^{n-2} \frac{n^k}{k!}.$$
-
@Henry Do you understand Elkie's approach? – Fan Zhang Apr 18 '12 at 1:45 | 2015-04-26T05:08:13 | {
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https://math.stackexchange.com/questions/4128705/help-solving-system-of-differential-equations | # Help Solving System of Differential Equations
I am working through some practice problems, and I am getting a different answer from what the back of my book says:
Express the general solution of the given system of equations in terms of real valued functions:
$$\begin{equation*} x' = \begin{pmatrix} 2 & -5 \\ 1 & -2 \\ \end{pmatrix}x \end{equation*}$$
I get eigenvalues of $$\pm i$$ and eigenvectors
$$\begin{equation*} v_1 = \begin{pmatrix} 2+i \\ 1 \\ \end{pmatrix} \end{equation*}$$
$$\begin{equation*} v_2 = \begin{pmatrix} 2-i \\ 1 \\ \end{pmatrix} \end{equation*}$$
The general solution with complex values is
$$\begin{equation*} x = e^{it} \begin{pmatrix} 2+i \\ 1 \\ \end{pmatrix} \end{equation*}$$
which becomes
$$\begin{equation*} x = (\cos(t)+i\sin(t)) \begin{pmatrix} 2+i \\ 1 \\ \end{pmatrix} \end{equation*}$$
Expanding this and simplifying, I get
$$\begin{equation*} x = \begin{pmatrix} 2\cos(t)-\sin(t) \\ \cos(t) \\ \end{pmatrix} + i \begin{pmatrix} \cos(t)+2\sin(t) \\ \sin(t) \\ \end{pmatrix} \end{equation*}$$
and then adding the constants, I get
$$\begin{equation*} x = c_1 \begin{pmatrix} 2\cos(t)-\sin(t) \\ \cos(t) \\ \end{pmatrix} + c_2 \begin{pmatrix} \cos(t)+2\sin(t) \\ \sin(t) \\ \end{pmatrix} \end{equation*}$$
Looking at the solution at the back of the book, it says the correct answer is
$$\begin{equation*} x = c_1 \begin{pmatrix} 5\cos(t) \\ 2\cos(t)+\sin(t) \\ \end{pmatrix} + c_2 \begin{pmatrix} 5\sin(t) \\ -\cos(t)+2\sin(t) \\ \end{pmatrix} \end{equation*}$$
Where did I go wrong? Thank you.
• What happened to $\pmatrix{2-i\\1}$? May 5, 2021 at 23:52
• Why do you think that your answer is different from the book? May 5, 2021 at 23:52
• change $c_1 \to 2a - b$ and $c_2 \to a + 2b$ in your solution (this is allowed since $c_1$ and $c_2$ are arbitrary) May 5, 2021 at 23:56
• @J.W.Tanner , there is a theorem saying in the system $x'=Ax$, where each value of A is real and continuous, and if $x=u(t)+iv(t)$ is a complex solution of $x'=Ax$, then its real part $u(t)$ and imaginary part $v(t)$ are also solutions to the equation. May 5, 2021 at 23:57
$$\begin{pmatrix} 2+i \\ 1\end{pmatrix}$$ is proportional to $$\begin{pmatrix} 5 \\ 2-i\end{pmatrix}$$, so your solution is equivalent to the one given. You didn't do anything wrong. | 2022-07-05T12:19:24 | {
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http://mathhelpforum.com/calculus/169692-evaluate-integral.html | # Math Help - Evaluate the Integral
1. ## Evaluate the Integral
This one has had me stuck for a good hr.
The integral has a min of 0 and a max of 1.
S (4-x)/(sqrt(4-x^2)) dx
I'm trying to use the formula S (du)/(sqrt(a^2-u^2)) = arcsin u/a + C
I set "u" = (x^2), "du" = 2x, and a = 2.
Typically the problems I get will have the du equal the numerator and it's pretty easy from there. But my du does not equal the numerator and I'm stuck.
I think the answer for the interval (disregarding the min and max) is:
Sqrt(4-x^2) +
4 arcsin(x/2) + C.
And the part in underlined bold I have absolutely no idea where it came from.
-2 + Sqrt(3) + (2pi)/3
Basically I cannot figure a way to get that answer or the steps involved. Can anyone help? Thank you.
2. It is easy to obeserve that is...
$\displaystyle \frac{4-x}{\sqrt{4-x^{2}}}= \frac{2}{\sqrt{1-(\frac{x}{2})^{2}}} - \frac{\frac{x}{2}}{\sqrt{1-(\frac{x}{2})^{2}}}$
Kind regards
$\chi$ $\sigma$
3. $\displaystyle \int\frac{4-x}{\sqrt{4-x^{2}}}\;{dx} = \int \frac{4}{\sqrt{4-x^2}}-\frac{x}{\sqrt{4-x^2}}\;{dx}= 4\int\frac{1}{\sqrt{1-x^2}}\;{dx}+\int\frac{(1-x^2)'}{2\sqrt{1-x^2}}\;{dx}.$
The first one should be easy enough. The second is of the form $\int\frac{f'(x)}{2\sqrt{f(x)}}\;{dx} = \sqrt{f(x)}+k$.
4. I am still not following.
@Chisigma. I see that you broke them up to two separate integrals and divided by 2. But after that I don't see where to go from there.
@TheCoffeeMachine. I see what you did on the first two steps but on the 3rd one when you add them, I am lost. In the front of my calc book I have a sheet with basic integration formulas, it doesn't show anything that relates to yours. I also don't understand where you got the second form. Is their a good site with more formulas?
I also tried to look at how Wolfram Alpha solved it
(4-x)/(sqrt(4-x^2)) - Wolfram|Alpha
And the part that me throws me off on their solution is that they set x = 2sin(u) and u = arcsin (x/2)
I got a good 2+hrs into this one problem and 3 pages of scratch paper. I'm at a loss. Thank you in advance!
Derek Z
5. Originally Posted by DerekZ10
I am still not following.
@Chisigma. I see that you broke them up to two separate integrals and divided by 2. But after that I don't see where to go from there.
Just in case a picture helps...
... where (key in spoiler) ...
Spoiler:
... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this at first case x, then after the 'sub', theta), 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).
The general drift is...
Then back substitute from sin theta to x/2 to find F.
What TCM was hinting towards is kind of like this...
... where (key in spoiler) ...
Spoiler:
The general drift is...
Edit: BTW, using the trig sub for the whole fraction like Wolfram does...
... means that after back-substituting you have an expression
$2 \cos(\arcsin(\frac{x}{2}))$
which with some Pythag will iron out to the same as above.
________________________________________________
Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!
6. Originally Posted by DerekZ10
∫ (4-x)/(sqrt(4-x^2)) dx
I'm trying to use the formula S (du)/(sqrt(a^2-u^2)) = arcsin u/a + C
...
I think the answer for the integral (disregarding the min and max) is:
Sqrt(4-x^2) +
4 arcsin(x/2) + C.
And the part in underlined bold I have absolutely no idea where it came from.
...
Basically I cannot figure a way to get that answer or the steps involved. Can anyone help? Thank you.
Two others have covered this quite well. I'll also give it a try.
$\displaystyle {{4-x}\over{\sqrt{4-x^2}}}={{4}\over{\sqrt{4-x^2}}}+{{-x}\over{\sqrt{4-x^2}}}$
You apparently have no problem with the integral: $\displaystyle \int {{4}\over{\sqrt{4-x^2}}} dx\,.$
For the integral $\displaystyle \int {{-x}\over{\sqrt{4-x^2}}} dx\,,$ use the substitution:
$\displaystyle u = 4-x^2\quad\to\quad du=-2x\,dx\,.$
Can you take it from here?
7. Yes I got it, thank you! | 2015-07-28T01:30:37 | {
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https://math.stackexchange.com/questions/2394582/show-that-fx-is-an-odd-function | # Show that $f(x)$ is an Odd Function
Show that $$f(x) = \ln \left(x+\sqrt{x^2+1}\right)$$ is an odd function.
My attempt:
$$f(-x)=\ln\left(-x+\sqrt{(-x)^2+1}\right)=\ln\left(-x+\sqrt{x^2+1}\right).$$
How should I proceed? I know that if $f(-x)=-f(x)$, the function is odd.
• "I know that et cetera" ... so you must show, and it suffices showing, that $-x+\sqrt{x^2+1}=\frac{1}{x+\sqrt{x^2+1}}$ for all $x$. – user228113 Aug 15 '17 at 16:30
Try to use $$(\sqrt{x^2+1}-x)(\sqrt{x^2+1}+x)=1$$
You are almost done : Simply note that $f(x)+ f(-x) = \ln(\sqrt {x^2 + 1} + x) + \ln(\sqrt{x^2 + 1}-x) = \\ \ln((\sqrt {x^2 + 1} + x)(\sqrt{x^2 + 1}-x)) = \ln (x^2 + 1 - x^2) = \ln 1 = 0$.
Hence, the function is an odd function.
$$\ln(-x+\sqrt{x^2+1})=\ln\left(\frac{1}{\sqrt{x^2+1}+x}\right)=-\ln(\sqrt{x^2+1}+x)$$
• How come you could take its reciprocal? – Andrew Tawfeek Aug 15 '17 at 17:57
• I am just multiplying on the top and bottom by $\sqrt{x^2+1}+x$ – tattwamasi amrutam Aug 15 '17 at 18:41
Note that $\sqrt{1+x^2}$ is even function.
Now, $$\int_{-x}^{x}\frac{1}{\sqrt{1+t^2}}dt=2\int_{0}^{x}\frac{1}{\sqrt{1+t^2}}dt=2\ln(x+\sqrt{1+x^2})$$
Again, $$\int_{-x}^{x}\frac{1}{\sqrt{1+t^2}}dt=2\int_{-x}^{0}\frac{1}{\sqrt{1+t^2}}dt=-2\ln(-x+\sqrt{1+x^2})$$
Hence $\ln(x+\sqrt{1+x^2})=\ln(-x+\sqrt{1+x^2})$ i.e. odd function.
$$f(\sinh\theta) = \theta$$ and both $\theta$ and $\sinh\theta$ are odd functions, hence $f$ is an odd function as well ($f(x)=\text{arcsinh}(x)$).
One more:
$f(x) = \ln (x + \sqrt{x^2+1})$, $x \in \mathbb{R}.$
$\star)$ $f(-x) = \ln(-x + \sqrt{x^2 +1}) =$
$\ln (\frac{1}{x+\sqrt{x^2+1}} )=$
$- \ln (x + \sqrt{x^2+1})$.
Combining:
$f(-x) = - f(x)$.
Used :
$\ln [( -x +\sqrt{x^2+1}) \frac{x + \sqrt{x^2 +1}}{x+\sqrt{x^2+1}}] =$
$\ln ( \frac{1}{x + \sqrt{x^2+1}} )=$ $- \ln (x + \sqrt{x^2+1})$. | 2019-07-16T07:55:45 | {
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https://stats.stackexchange.com/questions/411018/poisson-distribution-why-does-time-between-events-follow-an-exponential-distrib | # Poisson distribution: why does time between events follow an exponential distribution?
I was reading an article, and came across the following:
Purchase count follows a Poisson distribution with rate λ. In other words, the timing of these purchases is somewhat random, but the rate (in counts/unit time) is constant. In turn, this implies that the inter-purchase time at the customer level should follow an exponential distribution.
It's been quite a while since I did any statistics so I am struggling with the definitions of a Poisson distribution. What I understand by the "rate is constant" is that if a customer purchases 1 thing on average in a week, they purchase 4 things on average in a four-week period. Is this correct?
Where I believe I am confused is with the final sentence. Is this saying that the time between a customers purchases would grow exponentially as time goes on? Doesn't this contradict the idea that we have a constant rate of purchase?
• For a full account of this, including definitions, see stats.stackexchange.com/questions/214421. For related posts, see stats.stackexchange.com/… Another useful search (focusing on "memoryless") is stats.stackexchange.com/…. – whuber May 31 at 16:29
• $${}$$ "if a customer purchases 1 thing on average in a week, they purchase 4 things on average in a four-week period. Is this correct?" $\textbf{Yes}.$ $${}$$ "Is this saying that the time between a customers purchases would grow exponentially as time goes on?" $\textbf{No},$ that's not what it means at all. $\qquad$ – Michael Hardy May 31 at 19:17
## 2 Answers
Let $$X_t$$ be the number of arrivals in the Poisson process with rate $$\lambda$$ between time $$0$$ and time $$t\ge0.$$ Then we have $$\Pr(X_t=x) = \frac{(\lambda t)^x e^{-\lambda t}}{x!} \text{ for } x=0,1,2,3,\ldots$$
Let $$T$$ be the time until the first arrival.
Then the following two events are really both the same event: $$\Big[ X_t=0\Big]. \qquad \Big[ T>t \Big].$$ So they both have the same probability. Thus $$\Pr(T>t) = \Pr(X_t=0) = \frac{(\lambda t)^0 e^{-\lambda t}}{0!} = e^{-\lambda t}.$$ So $$\Pr(T>t) = e^{-\lambda t} \text{ for } t\ge0.$$ That says $$T$$ is exponentially distributed.
• That's very clever. Thanks! – J. Stott Jun 3 at 8:45
Exponential distribution doesn't imply that time between events grows exponentially. All it tells you is that probability to wait longer between events declines very quickly with waiting time. The probability density is:$$\lambda e^{-\lambda t}$$ where $$\lambda$$ is Poisson intensity, i.e. average number of events in unit of time, and $$t$$ is the waiting time. The average waiting time is obviously $$1/\lambda$$. | 2019-11-13T09:20:53 | {
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http://mathhelpforum.com/calculus/56143-solved-series.html | 1. ## [SOLVED] Series
I am given:
And it says to find the first few coefficients, c0,c1...,to c4, which I can't find. Then it says to find the radius of convergence R of the power series. I'm not sure why they have the fraction equaling the series. I know I can use the ratio or root test to find the radius of convergence, but what exactly is the above problem stating?
Thanks,
Matt
2. Okay, I got help and figured out that I should use the form:
$\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$
So then I got to:
$\frac{6x}{10}\sum_{n=0}^{\infty}\left(\frac{-x}{10}\right)^n$
and $(-1)^n\left(\frac{1}{10}\right)^n x^n$
Now, I figured out that the radius of convergence is 10, but I'm still unsure for the coefficients.
3. Hello, matt3D!
$\frac{6x}{10+x} \:=\:\sum^{\infty}_{n=0} c_nx^n$
(a) Find the first few coefficients, to $c_4$
(b) Find the radius of convergence R of the power series.
Use long division . . .
$\begin{array}{ccccccc}
& & & \frac{6}{10}x & -\frac{6}{10^2}x^2 & +\frac{6}{10^3}x^3 & - \quad\cdots \\
& & ---&---&---&--- \\
10+x & ) & 6x \\
& & 6x & +\frac{6}{10}x^2 \\
& & --- & --- \\
& & & -\frac{6}{10}x^2 \\
& & & -\frac{6}{10}x^2 & -\frac{6}{10^2}x^2 \\
& & & --- & --- \\
& & & & \frac{6}{10^2}x^3 \\
& & & & \frac{6}{10^2}x^3 & + \frac{6}{10^3}x^4 \\
& & & & ---&---\end{array}$
We see that: . $\frac{6x}{10+x} \:=\:\frac{6}{10}x - \frac{6}{10^2}x^2 + \frac{6}{10^3}x^3 - \frac{6}{10^4}x^4 + \cdots$
(a) The first five coefficients are: . $\begin{Bmatrix}c_o &=& 0 \\ \\[-4mm]c_1 &=&\frac{6}{10} \\ \\[-4mm]c_2 &=& \text{-}\frac{6}{10^2} \\ \\[-4mm]c_3 &=& \frac{6}{10^3} \\ \\[-4mm]c_4 &=& \text{-}\frac{6}{10^4}\end{Bmatrix}$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
The general term is: . $a_n \:=\:(\text{-}1)^{n+1}\frac{6}{10^n}x^n$ . . . for $n \geq 1$
Ratio Test: . $\frac{a_{n+1}}{a_n} \;=\; \left|\frac{6x^{n+1}}{10^{n+1}} \cdot \frac{10^n}{6x^n}\right| \;=\;\left|\frac{x}{10}\right|$
And we have: . $\left|\frac{x}{10}\right| \:<\:1 \quad\Rightarrow\quad |x| \:< \:10$
(b) Therefore, the radius of convergence is: . $R \:=\:10$
4. Originally Posted by matt3D
I am given:
And it says to find the first few coefficients, c0,c1...,to c4, which I can't find. Then it says to find the radius of convergence R of the power series. I'm not sure why they have the fraction equaling the series. I know I can use the ratio or root test to find the radius of convergence, but what exactly is the above problem stating?
Thanks,
Matt
Let's find a more general power series. Consider
$\frac{ax^m}{b+cx^p}$
Simple algebra shows that this is equivalent to
$\frac{ax^m}{b}\cdot\frac{1}{1+\frac{c}{b}x^p}$
$=\frac{ax^m}{b}\cdot\frac{1}{1+\left(\frac{c^{\tfr ac{1}{p}}x}{b^{\tfrac{1}{p}}}\right)^p}$
Now we know that
$\frac{1}{1+x^p}=\sum_{n=0}^{\infty}(-1)^nx^{np}$
$\therefore\quad\frac{1}{1+\left(\frac{c^{\tfrac{1} {p}}x}{b^{\tfrac{1}{p}}}\right)^p}=\sum_{n=0}^{\in fty}(-1)^n\left(\frac{c^{\frac{1}{p}}x}{b^{\frac{1}{p}}} \right)^{np}$
$=\sum_{n=0}^{\infty}\frac{(-1)^nc^nx^{np}}{b^n}$
$\therefore\quad\frac{ax^n}{b+cx^p}=\frac{ax^m}{b}\ cdot\frac{1}{1+\left(\frac{c^{\tfrac{1}{p}}x}{b^{\ tfrac{1}{p}}}\right)^p}$
$=\frac{ax^m}{b}\sum_{n=0}^{\infty}\frac{(-1)^nc^nx^{np}}{b^n}$
$=a\sum_{n=0}^{\infty}\frac{(-1)^nc^nx^{np+m}}{b^{n+1}}$
Using the root test we check for the ROC...to do this we must find all x such that
$\lim_{n\to\infty}\sqrt[n]{\left|\frac{(-1)^nc^nx^{np+m}}{b^{n+1}}\right|}$
$=\frac{c|x|^p}{b}<1$
Solving for x gives
$|x|<\sqrt[p]{\frac{b}{c}}$
I'll leave you the endpoints...they are fairly easy.
5. Wow, thanks Soroban, I didn't know you could use long division. Mathstud28 I'm still trying to figure your post out, but thanks for the info! | 2016-09-26T19:37:24 | {
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https://mathhelpforum.com/threads/graph-the-inequality.283182/ | Graph the Inequality
xyz_1965
See attachment for question and picture of graph.
Cervesa
$xy > 0$ for all points in quad IV ... why?
joshuaa
I understand why you are confused!
you think
xy >= 4 as same as y >= 4/x
They are not the same
here
y >= 4/x
any points on the curves or above will satisfy the condition
here
xy >= 4
any point on the right curve or above it will satisfy the condition
but for the left curve any points on the curve or below it will satisfy the condition
for example
x = -2
y = -1 (above the left curve)
xy = (-2)(-1) = 3 will not satisfy the condition xy >= 4 cuz 3 is less than 4. so we need values equal or below the left curve
Debsta
MHF Helper
If x is pos, xy>4 implies y>4/× (dividing by a positive keeps the inequality sign).
But, if x is neg, xy>4 implies y<4/x (dividing by a negative reverses inequality sign).
So, in the first quadrant, where x is pos, you shade the area above.
In the third quadrant, where x is neg, you shade the area below.
xyz_1965
$xy > 0$ for all points in quad IV ... why?
The product of xy must be positive for xy > 0 to be a true statement.
xyz_1965
I understand why you are confused!
you think
xy >= 4 as same as y >= 4/x
They are not the same
here
y >= 4/x
any points on the curves or above will satisfy the condition
here
xy >= 4
any point on the right curve or above it will satisfy the condition
but for the left curve any points on the curve or below it will satisfy the condition
for example
x = -2
y = -1 (above the left curve)
xy = (-2)(-1) = 3 will not satisfy the condition xy >= 4 cuz 3 is less than 4. so we need values equal or below the left curve
You said:
"any point on the right curve or above it will satisfy the condition
but for the left curve any points on the curve or below it will satisfy the condition"
I assume that by "satisfy the condition" you meant to say the condition of the given inequality or
xy greater than or equal to 4.
xyz_1965
If x is pos, xy>4 implies y>4/× (dividing by a positive keeps the inequality sign).
But, if x is neg, xy>4 implies y<4/x (dividing by a negative reverses inequality sign).
So, in the first quadrant, where x is pos, you shade the area above.
In the third quadrant, where x is neg, you shade the area below.
As joshuaa said:
"any point on the right curve or above it will satisfy the condition
but for the left curve any points on the curve or below it will satisfy the condition"
joshuaa
You said:
"any point on the right curve or above it will satisfy the condition
but for the left curve any points on the curve or below it will satisfy the condition"
I assume that by "satisfy the condition" you meant to say the condition of the given inequality or
xy greater than or equal to 4.
the condition is
xy greater than or equal to 4
i advise you to always Test your final solution whenever you have a problem with inequalities
xyz_1965
xyz_1965
the condition is
xy greater than or equal to 4
i advise you to always Test your final solution whenever you have a problem with inequalities
joshuaa
what you have said correct except
let x = -4, and y = -3
(-4)(-3) >= 4
12 >= 4
True statement again proving that we must shade below the graph of the function in quadrant 3
xyz_1965 | 2020-01-29T20:34:58 | {
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https://math.stackexchange.com/questions/2075949/how-to-draw-diagram-for-line-and-parabola/2076079 | # How to draw diagram for line and parabola.
I am finding the area bounded by parabola and line using definite integral. When its about line y = x and parabola $y^2 = x$. I know the line passes through origin. But when we have given equation of line like x = 4y - 2 and parabola $x^2 = 4y$ I got confused how to draw diagram.
I have no problem in finding points using given equations.
x = 4y - 2 and $x^2 = 4y$
We have when x = 2, y = 1.
And when x = -1, y = -1/4.
But problem comes in drawing sketch for this. Also when we find area why we subtract the integral of line from parabola or parabola from line.
Edit -
In this question I have doubts in drawing graph when the equation of line in the form of ax - by + c = 0.
Second on finding area why we don't add integrals of both?
• $y^x=2$ does not define a parabola, probably there's a typo. However, are you able to find things like the axis and the vertex of a parabola? They should give you hints on how to draw it. – user402433 Dec 29 '16 at 13:14
• With the hope of avoiding duplication of effort, could you please clarify how this differs from Find the area of the region bounded by parabola and line.? In that post as well, it seems you're struggling with how to set up the area of a plane region (bounded by two or more graphs) as an integral. Is that the underlying question? Thank you. :) – Andrew D. Hwang Dec 29 '16 at 13:43
• @Andrew D. Hwang I can't ask many doubts in that one question so I asked different question here. Now I updated it with my doubts. – user404716 Dec 29 '16 at 13:47
Disclaimer: This answer invokes infinitesimals in a squishy, pragmatic way. The results are correct, but justifying their correctness requires work beyond the scope of this question.
The fundamental idiom of integral calculus (a name I'm trying to popularize) states:
To calculate a total quantity (area, length, volume, work...), split the quantity into infinitesimal increments that can be computed using algebra, then add them up (i.e., integrate) to find the total.
Suppose we wish to calculate the area of the plane region $R$ bounded below and above by the graphs $y = f_{1}(x)$ and $y = f_{2}(x)$, and bounded to the left and right by the vertical lines $x = a$, $x = b$. More succintly, $R$ is the set of $(x, y)$ with $$f_{1}(x) \leq y \leq f_{2}(x),\qquad a \leq x \leq b.$$
To apply the fundamental idiom, we slice $R$ into "thin vertical strips": For each $x$ between $a$ and $b$, we have $f_{1}(x) \leq y \leq f_{2}(x)$. The portion of $R$ "at $x$" (or, if you prefer, "between $x$ and $x + dx$") is a rectangle of height $f_{2}(x) - f_{1}(x)$ and width $dx$, and so has area $$dA = (f_{2}(x) - f_{1}(x))\, dx.$$ The total area of $R$ is the sum (i.e., integral) of these increments: $$\text{area} = \int_{a}^{b} (f_{2}(x) - f_{1}(x))\, dx.$$
Entirely similar ideas work if our region $R$ can be sliced easily into horizontal strips, i.e., if $R$ can be conveniently defined by inequalities $$g_{1}(y) \leq x \leq g_{2}(y),\qquad c \leq y \leq d.$$ In this event, $$\text{area} = \int_{c}^{d} (g_{2}(y) - g_{1}(y))\, dy.$$
As for sketching in practice: Draw the specified curves separately, then determine (by inspection or calculation) where they intersect. If necessary, decide whether to calculate the area by slicing the region into vertical or horizontal strips, and express the region accordingly using graphs of functions. The points of intersection give the limits of integration.
In more complicated examples, it may be necessary to split the region into sub-regions, and to use the preceding paragraph separately on each piece. (Commonly, for example, one wants the region between two graphs that cross in the interior of the interval of integration.)
In your specific example, you have the line $x = 4y - 2$ (which can be sketched by finding two points on the line), and the parabola $x^{2} = 4y$, which opens to the right, is symmetric across the $x$-axis, and passes through the origin and the point $(2, 1)$.
Since each curve is given in the form of $x$ as a function of $y$, horizontal slices are easiest. Your sketch will tell you
• Which curve is the left edge, $x = g_{1}(y)$, and which is the right edge, $x = g_{2}(y)$;
• What the limits of integration, $c \leq y \leq d$, are.
• I can't understand still why we subtract two areas to find bounded area. – user404716 Dec 29 '16 at 17:41
• The two curves specify the endpoints of a thin rectangular slice; the relevant dimension of the slice is the difference: top minus bottom, or right minus left. Do the accompanying diagrams not help? – Andrew D. Hwang Dec 29 '16 at 18:00
• No its not helpful. – user404716 Dec 30 '16 at 5:20
• Is there a particular part of the argument you don't understand? In itemized form, the argument boils down to: 1. If a region is cut into thin strips, the total area of the region is equal to the sum of the areas of the strips. 2. Each strip is (approximately, but nearly enough so that the conclusion is correct) a rectangle, with one side very small. 3. The distance between two points on a number line is the larger minus the smaller. – Andrew D. Hwang Dec 30 '16 at 12:55
• Actually the terms and language used by you is very difficult to understand. – user404716 Dec 30 '16 at 18:39 | 2021-08-03T21:39:07 | {
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https://math.stackexchange.com/questions/2685864/slope-of-the-tangent | # Slope of the tangent
I have the following exercise.
$1$. Find the slope of the tangent to the curve $y = x^3 -4x + 1$ at the point where $x = a$.
$2$. Find the equations of the tangent lines in the points $(1, -2)$ and $(2,1)$.
$3$. Graph the curve and the two tangents in a single Cartesian plane.
I understand perfectly how to elaborate the answers for $2$nd and $3$rd part, but I have doubts about how the first part should be answered: Find the slope of the tangent to the curve $y = x^3 -4x + 1$ at the point where $x = a$.
• Take the derivative at $a$? – Andrew Li Mar 11 '18 at 0:42
• I understand perfectly how to elaborate the answers for 2nd and 3th part Use one of those to find the tangent at point $(a, a^3-4a+1)$. – dxiv Mar 11 '18 at 0:43
• Can you get started with the third part at all? Where are you stuck? – saulspatz Mar 11 '18 at 0:51
• Please remember that you can choose an aswer among the given if the OP is solved, more details here meta.stackexchange.com/questions/5234/… – gimusi Mar 12 '18 at 23:55
HINT
1. Remember that the slope in a point x is given by $f'(x)$
2. The line equation in a point $(x_0,y_0)$ is $(y-y_0)=f'(x_0)(x-x_0)$
• the slope of the tangent to the curve y=x slope of the tangent to the curve y=x3−4x+1 at the point where x=a3−4x+1 at the point where x=a is m=3a^2-4. is this correct? – Tulipa Kauffmaniana Mar 11 '18 at 4:28
• Yes exactly! $m=3a^2-4$ is the slope – gimusi Mar 11 '18 at 6:57
(1) Find $\,f'$(x) and then substitute x = a as the derivative of a function at a point is equal to the tangent slope at that point. You're supposed to get 3a$^2$-4.
(2) You can use the point-slope form of a line ie $\,y-y_0 = \frac{dy}{dx}.(x-x_0)$ where $\,(x_o,y_0)$ are the given points.
Here the equations should simplify to $\,y = -(x+1)$ and $\, y= 8x-15$
(3) | 2019-08-19T05:32:18 | {
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https://math.stackexchange.com/questions/2128652/the-matrix-representation-of-a-reflection-operator-across-the-plane-x2y3z-0 | # The matrix representation of a reflection operator across the plane $x+2y+3z=0$
Let $T:\mathbb{R}^3\rightarrow \mathbb{R}^3$ be the reflection across the plane $x+2y+3z=0$, find the matrix of this linear operator $T$ in respect to the basis $B=\left\{v_1,v_2,v_3\right\}$, where we have:
$v_1=\begin{bmatrix}1\\ 1\\ -1\end{bmatrix}$ $v_2=\begin{bmatrix}-1\\ 2\\ -1\end{bmatrix}$ $v_3=\begin{bmatrix}1\\ 2\\ 3\end{bmatrix}$
First of what does it mean that $T$ is a reflection across the given plane? What will hapen to let's say the vector $(1,0,0)$ if it's passed through $T$? In my textbook, there was a short example that mentioned a reflection operator, which transformed a given vector $(a,b,c)$ to the vector $(a,b,-c)$ So that would be the reflection across the standard plane $x+y+z=0$, but here, will it act as if given a vector $(a,b,c)$ you get out a vector $(a,2b,-3c)$?
EDIT: after reading the comments, I got the matrix representation as:
$T=\begin{bmatrix}1&-1&-1\\ 1&2&-2\\ -1&-1&-3\end{bmatrix}$
Now how can I find the representation of this operator in respect to the standard basis vectors $(e_1,e_2,e_3)$?
I read about finding the change of basis matrix so I calculated $T^{-1}$:
$T^{-1}=\begin{bmatrix}\frac{4}{7}&\frac{1}{7}&-\frac{2}{7}\\ -\frac{5}{14}&\frac{2}{7}&-\frac{1}{14}\\ -\frac{1}{14}&-\frac{1}{7}&-\frac{3}{14}\end{bmatrix}$
Is this it or is there more to it?
• The vectors in the plane will be preserved (multiplied by 1) and the vectors perpendicular to the plane will be mirrored (multiplied by -1). Maybe that can help. – mathreadler Feb 4 '17 at 11:13
• @mathreadler so v1 will map into v1,v2 into v2 and v3 into -v3? – adadaae12313412 Feb 4 '17 at 11:17
• In general, there is a householder matrix that describes reflections of this type. – Andres Mejia Feb 4 '17 at 11:28
• v1 map onto v1, v2 onto v2, v3 onto -v3 – mathreadler Feb 4 '17 at 11:36
• @mathreadler can you review mt question, I edited it – adadaae12313412 Feb 4 '17 at 11:42
The person who prepared you this question has made life very easy for you. Verify following facts:
1. $v_1 \in$ the plane (its coordinates verify the equation of the plane)
2. $v_2$ also lies in this plane.
3. $v_3 \perp v_1$ (calculate the dot product)
4. $v_3 \perp v_2$
So the reflection maps: $\begin{cases} v_1 \mapsto v_1 \\ v_2 \mapsto v_2 \\ v_3 \mapsto -v_3 \end{cases}$ And the matrix w.r.t. this basis is $\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0& 0& -1\end{pmatrix}$
There is more to it. You can find a canonical basis (make an eigenvalue decomposition) $${\bf T = S}^{-1}{\bf DS}$$
where $${\bf D} = \begin{bmatrix}1&0&0\\0&1&0\\0&0&-1\end{bmatrix}$$ and the two leftmost columns of $\bf S$ are in the plane, and the rightmost is perpendicular to it. The eigenvalues $1$ means "preserve" vector, which is what happen to the components which lie in the plane. -1 means to flip the direction 180 degrees which is what should happen to the part of the vectors pointing right out of the plane.
In other words, if we
1. write the vector to be as a linear combination of two vectors in the plane and one perpendicular to it.
2. we can just flip the one perpendicular (multiply with -1)
3. reassemble our vector.
That is basically what multiplying with ${\bf S}^{-1}\bf DS$ would mean step-by-step.
• Okay, but can you answer this to me? After finding the matrice $T^{-1}$, we can find the matrice representation in respect to the standard basis by calculating $A=TET^{-1}$, where $E$ is the matrice made up of the vectors e1,e2,e3? – adadaae12313412 Feb 4 '17 at 11:58
• if $T$ is defined as you do, it will be like a $S^{-1}$ in my presentation and $E$ will be a diagonal matrix like $\bf D$. – mathreadler Feb 4 '17 at 12:06
As noted above, vectors $v_1,v_2 \in p$ (plane),and $v_3$ is $\perp$ to $v_1,v_2$. Since $v_1,v_2$ are in the plane their reflectection will be the same vector $v_1,v_2$ and since $v_3$ is the normal vector for this plane it's reflection will be the same vector just in the opposite direction ( $-v_3$). So your reflection matrix in the base $B=\{v_1,v_2,v_3\}$ will look like this:
$R=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & -1 \end{bmatrix}_B$
To get the matrix representation in the standard base you can use the change of basis matrix $T$. Notice that $T^{-1}$ is the matrix which will change $B_{std} \rightarrow B$ and $T$ will change $B\rightarrow B_{std}$.
The matrices will look like this: $T^{-1}=\begin{bmatrix} \frac{4}{7}& \frac{1}{7} & \frac{-2}{7}\\ & & &\\ \frac{-5}{14} & \frac{2}{7} & \frac{-1}{14}\\ & & &\\ \frac{1}{14} & \frac{1}{7} & \frac{3}{14} \end{bmatrix} , T=\begin{bmatrix} 1 & -1 & 1\\ 1 & 2 & 2 \\ -1 & -1 & 3 \end{bmatrix}$
And finally $R'=TRT^{-1}=\begin{bmatrix} \frac{6}{7} & \frac{-2}{7} & \frac{-3}{7}\\ &&&\\ \frac{-2}{7} & \frac{3}{7} & \frac{-6}{7}\\ &&&\\ \frac{-3}{7} & \frac{-6}{7} & \frac{-2}{7} \end{bmatrix}$
Let us call the plane normal as vector $$\vec n=(1,2,3)$$ and let the incident vector be $$\vec r=(x,y,z)$$ and the reflected vector be $$\vec r'=(x',y',z')$$. When the vector $$\vec r$$ is reflected upon the plane, its projection on the plane remains unchanged and only its normal component with respect to the plane will be reversed to become the reflected vector $$\vec r'$$. The normal component of $$\vec r$$ with respect to the plane is: $$\vec r_n=\frac{\vec r\bullet\vec n}{\vec n\bullet \vec n}\vec n$$ And the projection of $$\vec r$$ on the plane is: $$\vec r_p=\vec r-\vec r_n$$ So the reflected vector is: $$\vec r'=-\vec r_n+\vec r_p$$ or: $$\vec r'=-\vec r_n+\vec r-\vec r_n$$ or $$\vec r'=\vec r-2\vec r_n$$ or $$\vec r'=\vec r-2\frac{\vec r\bullet\vec n}{\vec n\bullet \vec n}\vec n$$ If we substitute the numerical value of the plane normal we get: $$\vec r'=\vec r-2\frac{x+2y+3z}{14}(1,2,3)$$ or $$\left\{\begin{array}{c}x'=x-\frac{x+2y+3z}{7}\\y'=y-2\frac{x+2y+3z}{7}\\z'=z-3\frac{x+2y+3z}{7}\end{array}\right.$$ or $$\left\{\begin{array}{c}x'=\frac{6x-2y-3z}{7}\\y'=\frac{-2x+3y-6z}{7}\\z'=\frac{-3x-6y-2z}{7}\end{array}\right.$$ So the matrix reperesentation is: $$R=\begin{bmatrix} \frac{6}{7}& \frac{-2}{7} & \frac{-3}{7}\\ & & &\\ \frac{-2}{7} & \frac{3}{7} & \frac{-6}{7}\\ & & &\\ \frac{-3}{7} & \frac{-6}{7} & \frac{-2}{7} \end{bmatrix}$$ | 2021-06-15T11:01:16 | {
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https://getmedicalwebsite.com/mexican-restaurant-btk/c9742f-convergence-in-probability-and-convergence-in-distribution | Convergence in probability: Intuition: The probability that Xn differs from the X by more than ε (a fixed distance) is 0. Contents . %%EOF The weak law of large numbers (WLLN) tells us that so long as $E(X_1^2)<\infty$, that Under the same distributional assumptions described above, CLT gives us that n (X ¯ n − μ) → D N (0, E (X 1 2)). Yes, you are right. (max 2 MiB). The concept of convergence in distribution is based on the … 6 Convergence of one sequence in distribution and another to … probability zero with respect to the measur We V.e have motivated a definition of weak convergence in terms of convergence of probability measures. We say V n converges weakly to V (writte • Convergence in probability Convergence in probability cannot be stated in terms of realisations Xt(ω) but only in terms of probabilities. It’s clear that $X_n$ must converge in probability to $0$. CONVERGENCE OF RANDOM VARIABLES . 288 0 obj <>stream Suppose B is the Borel σ-algebr n a of R and let V and V be probability measures o B).n (ß Le, t dB denote the boundary of any set BeB. Note that if X is a continuous random variable (in the usual sense), every real number is a continuity point. I will attempt to explain the distinction using the simplest example: the sample mean. This is typically possible when a large number of random effects cancel each other out, so some limit is involved. It is easy to get overwhelmed. This leads to the following definition, which will be very important when we discuss convergence in distribution: Definition 6.2 If X is a random variable with cdf F(x), x 0 is a continuity point of F if P(X = x 0) = 0. Xt is said to converge to µ in probability … Suppose that fn is a probability density function for a discrete distribution Pn on a countable set S ⊆ R for each n ∈ N ∗ +. 5.2. $$Is n the sample size? Convergence in probability. R ANDOM V ECTORS The material here is mostly from • J. Definitions 2. Just hang on and remember this: the two key ideas in what follows are \convergence in probability" and \convergence in distribution." By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy, 2020 Stack Exchange, Inc. user contributions under cc by-sa. Also, Could you please give me some examples of things that are convergent in distribution but not in probability? Types of Convergence Let us start by giving some deflnitions of difierent types of convergence. n!1 0. Convergence in distribution is the weakest form of convergence typically discussed, since it is implied by all other types of convergence mentioned in this article. Note that although we talk of a sequence of random variables converging in distribution, it is really the cdfs that converge, not the random variables. Noting that \bar{X}_n itself is a random variable, we can define a sequence of random variables, where elements of the sequence are indexed by different samples (sample size is growing), i.e. x) = 0. n!1 . This question already has answers here: What is a simple way to create a binary relation symbol on top of another? In econometrics, your Z is usually nonrandom, but it doesn’t have to be in general. 0 1. If it is another random variable, then wouldn't that mean that convergence in probability implies convergence in distribution? Definition B.1.3. X. n 2.1.2 Convergence in Distribution As the name suggests, convergence in distribution has to do with convergence of the distri-bution functions of random variables. dY. The basic idea behind this type of convergence is that the probability of an “unusual” outcome becomes smaller and smaller as the sequence progresses. In other words, for any xed ">0, the probability that the sequence deviates from the supposed limit Xby more than "becomes vanishingly small. Note that the convergence in is completely characterized in terms of the distributions and .Recall that the distributions and are uniquely determined by the respective moment generating functions, say and .Furthermore, we have an equivalent'' version of the convergence in terms of the m.g.f's I posted my answer too quickly and made an error in writing the definition of weak convergence. Then X_n does not converge in probability but X_n converges in distribution to N(0,1) because the distribution of X_n is N(0,1) for all n. 1.1 Almost sure convergence Definition 1. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. (4) The concept of convergence in distribtion involves the distributions of random ari-v ables only, not the random ariablev themselves. Proposition7.1Almost-sure convergence implies convergence in … Convergence in probability gives us confidence our estimators perform well with large samples. Viewed 32k times 5.$$\lim_{n \rightarrow \infty} F_n(x) = F(x),$$1.2 Convergence in distribution and weak convergence p7 De nition 1.10 Let P n;P be probability measures on (S;S).We say P n)P weakly converges as n!1if for any bounded continuous function f: S !R Z S f(x)P n(dx) ! Formally, convergence in probability is defined as And, no, n is not the sample size. h����+�Q��s�,HC�ƌ˄a�%Y�eeŊd뱰�c�BY()Yِ��\J4al�Qc��,��o����;�{9�y_���+�TVĪ:����OZC k��������� ����U\[�ux�e���a;�Z�{�\��T��3�g�������dw����K:{Iz� ��]R�؇=Q��p;���I��bJ%�k�U:"&��M�:��8.jv�Ź��;���w��o1+v�G���Aj��X��菉�̐,�]p^�G�[�a����_������9�F����s�e�i��,uOrJ';I�J�ߤW0 Na�q_���j���=7� �u�)� �?��ٌ�f5�G�N㟚V��ß x�Nk Topic 7. h�ĕKLQ�Ͻ�v�m��*P�*"耀��Q�C��.$$, $$\sqrt{n}(\bar{X}_n-\mu) \rightarrow_D N(0,E(X_1^2)).$$, $$\lim_{n \rightarrow \infty} F_n(x) = F(x),$$, https://economics.stackexchange.com/questions/27300/convergence-in-probability-and-convergence-in-distribution/27302#27302. Convergence in distribution in terms of probability density functions. Convergence in probability and convergence in distribution. And $Z$ is a random variable, whatever it may be. $$\forall \epsilon>0, \lim_{n \rightarrow \infty} P(|\bar{X}_n - \mu| <\epsilon)=1. I have corrected my post. suppose the CLT conditions hold: p n(X n )=˙! Xn p → X. d: Y n! In particular, for a sequence X1, X2, X3, ⋯ to converge to a random variable X, we must have that P( | Xn − X | ≥ ϵ) goes to 0 as n → ∞, for any ϵ > 0. Click here to upload your image Over a period of time, it is safe to say that output is more or less constant and converges in distribution. Under the same distributional assumptions described above, CLT gives us that Convergence in Probability. For example, suppose X_n = 1 with probability 1/n, with X_n = 0 otherwise. It tells us that with high probability, the sample mean falls close to the true mean as n goes to infinity.. We would like to interpret this statement by saying that the sample mean converges to the true mean.$$\forall \epsilon>0, \lim_{n \rightarrow \infty} P(|\bar{X}_n - \mu| <\epsilon)=1. However, $X_n$ does not converge to $0$ according to your definition, because we always have that $P(|X_n| < \varepsilon ) \neq 1$ for $\varepsilon < 1$ and any $n$. where $\mu=E(X_1)$. Convergence of the Binomial Distribution to the Poisson Recall that the binomial distribution with parameters n ∈ ℕ + and p ∈ [0, 1] is the distribution of the number successes in n Bernoulli trials, when p is the probability of success on a trial. %PDF-1.5 %���� $$plim\bar{X}_n = \mu,$$ The answer is that both almost-sure and mean-square convergence imply convergence in probability, which in turn implies convergence in distribution. 5 Convergence in probability to a sequence converging in distribution implies convergence to the same distribution. Knowing the limiting distribution allows us to test hypotheses about the sample mean (or whatever estimate we are generating). We write X n →p X or plimX n = X. Convergence in distribution means that the cdf of the left-hand size converges at all continuity points to the cdf of the right-hand side, i.e. most sure convergence, while the common notation for convergence in probability is X n →p X or plim n→∞X = X. Convergence in distribution and convergence in the rth mean are the easiest to distinguish from the other two. 4 Convergence in distribution to a constant implies convergence in probability. A sequence of random variables {Xn} is said to converge in probability to X if, for any ε>0 (with ε sufficiently small): Or, alternatively: To say that Xn converges in probability to X, we write: Convergence in probability gives us confidence our estimators perform well with large samples. e.g. If fn(x) → f∞(x) as n → ∞ for each x ∈ S then Pn ⇒ P∞ as n → ∞. convergence of random variables. The concept of convergence in probability is based on the following intuition: two random variables are "close to each other" if there is a high probability that their difference will be very small. To say that Xn converges in probability to X, we write. We note that convergence in probability is a stronger property than convergence in distribution. Convergence in distribution of a sequence of random variables. Although convergence in distribution is very frequently used in practice, it only plays a minor role for the purposes of this wiki. P n!1 X, if for every ">0, P(jX n Xj>") ! You can also provide a link from the web. $$\bar{X}_n \rightarrow_P \mu,$$. Your definition of convergence in probability is more demanding than the standard definition. Then define the sample mean as $\bar{X}_n$. A quick example: $X_n = (-1)^n Z$, where $Z \sim N(0,1)$. This is fine, because the definition of convergence in 4 distribution requires only that the distribution functions converge at the continuity points of F, and F is discontinuous at t = 1. I understand that $X_{n} \overset{p}{\to} Z$ if $Pr(|X_{n} - Z|>\epsilon)=0$ for any $\epsilon >0$ when $n \rightarrow \infty$. Convergence in Distribution [duplicate] Ask Question Asked 7 years, 5 months ago. Convergence in probability is stronger than convergence in distribution. 16) Convergence in probability implies convergence in distribution 17) Counterexample showing that convergence in distribution does not imply convergence in probability 18) The Chernoff bound; this is another bound on probability that can be applied if one has knowledge of the characteristic function of a RV; example; 8. Convergence in probability. Download English-US transcript (PDF) We will now take a step towards abstraction, and discuss the issue of convergence of random variables.. Let us look at the weak law of large numbers. dY, we say Y n has an asymptotic/limiting distribution with cdf F Y(y). Active 7 years, 5 months ago. X a.s. n → X, if there is a (measurable) set A ⊂ such that: (a) lim. Z S f(x)P(dx); n!1: Econ 620 Various Modes of Convergence Definitions • (convergence in probability) A sequence of random variables {X n} is said to converge in probability to a random variable X as n →∞if for any ε>0wehave lim n→∞ P [ω: |X n (ω)−X (ω)|≥ε]=0. Convergence in Distribution p 72 Undergraduate version of central limit theorem: Theorem If X 1,...,X n are iid from a population with mean µ and standard deviation σ then n1/2(X¯ −µ)/σ has approximately a normal distribution. • Convergence in mean square We say Xt → µ in mean square (or L2 convergence), if E(Xt −µ)2 → 0 as t → ∞. Constant and converges in distribution is very frequently used in practice, it is safe to say X.! 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Types of convergence in probability to a constant implies convergence in probability gives us confidence our estimators perform well large! Both almost-sure and mean-square convergence imply convergence in probability, which in turn implies convergence in mean... Using the simplest example: $X_n = 0$ otherwise, whatever it may be sample of variables... Click here to upload your image ( max 2 MiB ) distribtion involves the distributions random. Although convergence in probability is a random variable, whatever it may be asymptotic/limiting distribution with cdf Y! Idea is to extricate a simple deterministic component out of a random variable, whatever it may be then... Need some clarification on what the subscript $n$ means number random... Then would n't that mean that convergence in distribution. ( measurable ) set a ⊂ that!, almost-sure and mean-square convergence do not imply each other out, so some limit is involved we X. Unusual outcome keeps … this video explains what is a continuity point to.... Value, or another random variable ( in the usual sense ), and write \bar! Period of time, it only plays a minor role for the purposes of wiki... Weak convergence role for the purposes of this wiki with probability 1, =! Generating ) hold: p n ( 0,1 ) $, p ( n... }$ involves the distributions of random variables $\ { \bar { }! Same distribution. has approximately an ( np, np ( 1 −p ) ) distribution. distribution function X. Probability, which in turn implies convergence to the same distribution. question. Of random variables max 2 MiB ) mean as$ \bar { X } _n ! Typically possible when a large number of random ari-v ables only, not the size... F Y ( Y ) X_i\ } _ { i=1 } ^n $that are convergent in distribution ; ’... Stronger statement provide a link from the web MiB ) with cdf F Y ( Y ) gives! The limiting distribution allows us to test hypotheses about the sample mean as$ \bar { X _n\... Create a binary relation symbol on top of another ( max 2 MiB ) weak.. Question already has answers here: what is meant by convergence in distribution in terms of probability measures is... With probability 1, X = Y. convergence in distribtion involves the distributions of random variables im little... Another random variable, whatever it may be mean ; convergence in probability us. Symbol on top of another { i=1 } ^n $of X as n goes to infinity that... ), and write X_1, X_2, \ldots$, your $Z$.... Distribution allows us to test hypotheses about the sample mean as ${! Is$ Z $a specific value, or another random variable ( in the usual sense ), write... The probability of unusual outcome keeps … this video explains what is meant by convergence in implies.$ with probability $1/n$, where $Z$, where $\sim. Role for the purposes of this wiki where$ Z $a specific value, or another random variable in... Some examples of things that are convergent in distribution but not in probability is a continuous variable!, p ) random variable, whatever it may be test hypotheses about sample... ( X n converges to the distribution function of X as n goes to infinity less constant and converges distribution. N ( 0,1 )$ _ { n=1 } ^ { \infty } $large samples define the mean! The answer is that both almost-sure and mean-square convergence do not imply each other hang on and remember this convergence in probability and convergence in distribution! Involves the distributions of random variables } ^ { \infty }$ $a specific value or. Say that Xn converges in distribution to a sequence of random variables$ \ \bar! Is that both almost-sure and mean-square convergence imply convergence in distribution and another to … convergence in to. A sequence $X_1, X_2, \ldots$ quick example: $X_n =$. That convergence in probability is stronger than convergence in distribution of a sequence of random variables \. Surely ( a.s. ), and write in practice, it is another random.! Xj > '' ) explains what is meant by convergence in distribution tell us something very different is... Much stronger statement probability ; convergence in probability implies convergence in distribution of a variable... That both almost-sure and mean-square convergence do not imply each other mean that convergence in distribution ''... Create a binary relation symbol on top of another some clarification on what the subscript n! Some examples of things that are convergent in convergence in probability and convergence in distribution is very frequently used in practice, it only a... And converges in probability is stronger than convergence in distribution. ari-v ables only, not the random ariablev.! Some deflnitions of difierent types of convergence the subscript $n$ is usually nonrandom but... Possible when a large number of random effects cancel each other constant implies in. Would n't that mean that convergence in probability is stronger than convergence in distribution and another to … in... The convergence of probability measures something very different and is primarily used for hypothesis testing continuous random variable p!. Is that both almost-sure and convergence in probability and convergence in distribution convergence imply convergence in distribution. in Quadratic mean convergence... { i=1 } ^n $deflnitions of difierent types of convergence in probability to X, if for ! 9 convergence in distribution tell us something very different and is primarily used for hypothesis testing, denoted n. N has an asymptotic/limiting distribution with cdf F Y ( Y ) Z s F ( X n converges the... Role for the purposes of this wiki purposes of this wiki almost-sure and mean-square convergence do not each! This question already has answers here: what is meant by convergence in distribution tell us something very different is. Idea is to extricate a simple way to create a binary relation symbol on top of another the we., X_2, \ldots$ more or less constant and converges in probability X!: p n ( X n converges to X almost surely ( a.s.,! Nonrandom, but it doesn ’ t have to be in general n means... Every > 0, p ( jX n Xj > '' ) i just need some clarification on the... → X, if there is a continuous random variable has approximately (... To infinity question already has answers here: what is a continuity point, ( )!
Latest Videoke Machine, Isle Of Man Steam Packet Ships, Loud House Episodes Wiki, Cristine Reyes Netflix, Manhattan College Baseball Coaches, Which Of The Following Statements About Gdp Is True, Miitopia Red Orochi, | 2021-03-04T03:21:12 | {
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https://stats.stackexchange.com/questions/127121/do-logs-modify-the-correlation-between-two-variables | # do logs modify the correlation between two variables?
I am applying logs to two very skewed variables and then doing the correlation. Before logs the correlation is 0.49 and after logs it is 0.9. I thought the logs only change the scale. How is this possible? Here below the graphs for each of them. Perhaps I haven't applied the right transformation?
• Logarithms are manifestly a nonlinear transformation and so in general correlations will change, often substantially. You just found that out. It's not a problem; on the contrary, it is usually desired behaviour. The correlation is a measure of how far data can be approximated by a straight line; far from needing another transformation, you are evidently using the most appropriate transformation possible. – Nick Cox Dec 8 '14 at 9:43
• Note that skewness here is evident and important, but secondary. The most important reason for transformation is that the relationship on the original scale is nonlinear. – Nick Cox Dec 8 '14 at 9:44
• @NickCox: I'd quibble with "the correlation is a measure of how far data can be approximated by a straight line". This is correct for Pearson's correlation, but there are other correlation coefficients without that linearity assumption, and sometimes these are more suitable than finding a transformation and then applying Pearson's correlation. – Stephan Kolassa Dec 8 '14 at 9:46
• You're correct, naturally, and I would make the same correction in reverse. It's too late to edit my comment, but I meant "as used here". I think "correlation" does default to "Pearson correlation", but you are quite right to want explicit statements. (I first calculated Spearman correlations in 1965!) – Nick Cox Dec 8 '14 at 9:50
• Taking logs can even change the sign of the Pearson correlation ... but the rank correlations won't change at all. – Glen_b May 8 '17 at 7:15
There are multiple different types of correlation. The most common one is Pearson's correlation coefficient, which measures the amount of linear dependence between two vectors. That is, it essentially lays a straight line through the scatterplot and calculates its slope. This will of course change if you take logs!
If you are interested in a measure of correlation that is invariant under monotone transformations like the logarithm, use Kendall's rank correlation or Spearman's rank correlation. These only work on ranks, which do not change under monotone transformations.
Here is an example - note how the Pearson correlation changes after logging, while the Kendall and the Spearman ones don't:
> set.seed(1)
> foo <- exp(rnorm(100))
> bar <- exp(rnorm(100))
>
> cor(foo,bar,method="pearson")
[1] -0.08337386
> cor(log(foo),log(bar),method="pearson")
[1] -0.0009943199
>
> cor(foo,bar,method="kendall")
[1] 0.02707071
> cor(log(foo),log(bar),method="kendall")
[1] 0.02707071
>
> cor(foo,bar,method="spearman")
[1] 0.03871587
> cor(log(foo),log(bar),method="spearman")
[1] 0.03871587
The following earlier question discusses Kendall's and Spearman's correlation: Kendall Tau or Spearman's rho?
• Spearman rank correlation is also possible here. – Nick Cox Dec 8 '14 at 9:45
• And if I wanted to put together these two variables in a PCA, how would I do it? With the logs applied? – DroppingOff Dec 8 '14 at 10:30
• And if I want to include these variables in a clustering procedure? What is it better? Can I say that I include only one because they are highly correlated? Do I need to include the logged variable (it is going to be more difficult to interpret it. – DroppingOff Dec 8 '14 at 10:32
• At some point, it makes sense to ask a new question, because by now, we have left the correlation issue far behind. @NickCox wrote some very good comments. Looking at the scatterplot of your logged data, I'd recommend taking logs and including only one of the two variables, since they apparently carry pretty much the same information. – Stephan Kolassa Dec 8 '14 at 10:37
• I agree strongly with @Stephan Kolassa that new questions require a new thread (and much more context on what you are doing). If you have data like this, you do need to think logarithmically. It takes time and experience, but eventually it becomes easier to interpret what is happening when it's expressed on an appropriate scale. – Nick Cox Dec 8 '14 at 10:44 | 2019-10-22T01:50:00 | {
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https://math.stackexchange.com/questions/4249639/do-you-lose-solutions-when-differentiating-to-solve-an-integral-equation | # Do you lose solutions when differentiating to solve an integral equation?
The question is more general but here's the problem that motivated it: I want to find all solutions to the integral equation
$$f(x) + \int_0^x (x-y)f(y)dy = x^3.$$
Differentiating twice with respect to x this yield the second order differential equation
$$f''(x) + f(x) = 6x.$$
The solution to this last equation is something of the form $$f(x) = A \cos(x + \phi) + f_p$$ where $$f_p$$ can be found by variation of parameters or other tools and $$A, \phi$$ are determined by initial values.
However, what I want to know is whether I'm accounting for all solutions here. Do I not "lose information" when differentiating the original equation? If so, how can I construct all solutions from the ones I just found?
• Of course, in this case, we can see by inspection that $f_p(x) = 6x$. Sep 14, 2021 at 19:02
• Using the Laplace transform on $$f(x) + x\circledast f(x) = x^3$$ we have $F(s) +\frac{F(s)}{s^2}=\frac{6}{s^4}$ and deriving twice the full equation we get $$s^2F(s) + F(s) = \frac{6}{s^2}$$ both with solution $$f(x) = 6(x-\sin x)\theta(x)$$ with $\theta$ the Heaviside step function. Sep 14, 2021 at 19:05
It's the other way around: when you differentiate both sides of an equation, you should be concerned about creating solutions. This is because $$f'(x)=g'(x)$$ only implies $$f(x)=g(x)+c$$ for some $$c$$. So any solution to the IE will satisfy the DE but not the other way around.
In effect the IE has the boundary conditions built into it in a way the DE does not. This isn't totally obvious, so let's walk through it in your example.
Substitute $$x=0$$ into the original IE to see $$f(0)=0$$. Differentiate once to get
$$f'(x)+\int_0^x f(y) dy=3x^2$$
and then substitute $$x=0$$ to get $$f'(0)=0$$. Thus the IE* implies the ODE IVP $$f''(x)+f(x)=6x,f(0)=0,f'(0)=0$$, which has a unique solution as you probably already know.
To see the ODE IVP implies the IE, you run the procedure in reverse:
$$\int_0^x f''(y) + f(y) dy = \int_0^x 6y dy \\ f'(x)-f'(0)+\int_0^x f(y) dy = 3x^2.$$
Use the initial condition:
$$f'(x)+\int_0^x f(y) dy = 3x^2.$$
Now you integrate again:
$$\int_0^x f'(y) dy + \int_0^x \int_0^y f(z) dz dy = \int_0^x 3y^2 dy \\ f(x)-f(0) + \int_0^x \int_0^y f(z) dz dy = x^3.$$
Use the other initial condition:
$$f(x)+\int_0^x \int_0^y f(z) dz dy = x^3.$$
This looks different from the original thing, but it is actually the same. One way to make it look the same is to use integration by parts together with the initial conditions on the outer integral of the second term. Another way is to interchange the order of integration; after the interchange you can simply do the inner $$dy$$ integral since $$f(z)$$ doesn't depend on $$y$$. The fact that these both work is a quite general thing, cf. https://en.wikipedia.org/wiki/Order_of_integration_(calculus)#Relation_to_integration_by_parts
* Technically you need the IE and a $$C^2$$ assumption to run this calculation. The shortcut way that I can think of to get this regularity assumption is to just assume it up front, check that the solution you get has the desired regularity (which it does) and then study the general uniqueness theory of the IE (which in this setting is the Fredholm alternative) to conclude that you didn't miss any irregular solutions.
• Looking at this as an outsider, the only thing that is missing seems to be an argument that a solution to the integral equation will indeed have to be two times differentiable. Just for the sake of completeness. Sep 14, 2021 at 8:45
• @BrianBi what if I have the equation $x(x-1)=0$ and I divide both sides by $x$? Sep 15, 2021 at 19:27
Something that I think that high school math doesn't explain very clearly is the theory of extra/missing solutions. Let's say we have equations $$(1)$$ and $$(2)$$. If $$(1)\rightarrow (2)$$, that means $$(\text{x satisfies (1)})\rightarrow(\text{x satisfies (2)})$$, so every solution of $$(1)$$ will also be a solution of $$(2)$$. However, $$(1)\rightarrow (2)$$ does not guarantee that every solution of $$(2)$$ will be a solution of $$(1)$$. So when you have a chain of equivalent equations, every solution of the final equation will be a solution of the original equation. When you have a chain of implication, you can have extraneous solutions. To lose solutions, you have to have a situation where $$(2)\rightarrow (1)$$, i.e. your later equation is sufficient, but not necessary, for the initial equation to be true. For instance, if you go from $$x^2=y^2$$ to $$x=y$$, the first equation is a consequence of the second, but the second does not necessarily follow from the first.
In your case, you are differentiating. If equation $$(1)$$ is $$y=f(x)$$, and equation $$(2)$$ is $$y'=f'(x)$$, we have $$(1)\rightarrow(2)$$ (if two functions are equal, then their derivatives are equal as well), but we don't have $$(2)\rightarrow(1)$$ (that is, if two derivates are equal, that doesn't mean the original functions were equal). So, for instance, if you have $$f(x) = x+1$$ and take the derivative of both sides, you get $$f'(x) = 1$$, and $$f(x)=x$$ is a solution to the second equation, but not the first.
Differentiating does in fact lose information, but losing information means that you are getting more solutions, not fewer. The more information you have, the more possibilities you can eliminate. If you know $$y=x+1$$, there are some ordered pairs that you can eliminate (for instance, you that (1,0) is not a solution). If you also know that $$2y=3x$$, then you can eliminate more ordered pairs. Remember that equations don't rule solutions in, they only rule solutions out. | 2022-05-19T15:45:36 | {
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https://mathematica.stackexchange.com/questions/269120/is-there-any-way-to-get-nthcomposition-of-a-function | # Is there any way to get nthComposition of a function?
I need to find the nthComposition of a function. I thought there could be something like nthComposition[f,x,5] means five times composition of f with x as input parameter. I don't want to repeat f in my code as there is no fixed number for my composition, it could change every time. Assume it is my function:
f [va_] := ( va*2);
Composition[f, f][s]
4 s
Composition[f, f, f][s]
8 s
Depending on exactly how you want to use this, you might want to bundle this up into its own function (or function overload).
f[nestLevel_, va_] := Nest[f, va, nestLevel]
or
compf[nestLevel_] := Composition @@ ConstantArray[f, nestLevel]
Usage:
f[2, x]
(* f[f[x]] *)
(* or 4 x using your sample function *)
compf[3]
(* f@*f@*f *)
compf[3][x]
(* f[f[f[x]]] *)
(* or 8 x using your sample function*)
• thanks but I'm a bit confused about the first solution : f[nestLevel_, va_] := Nest[f, va, nestLevel] . we are defining f on left side but giving it to Nest as input on right side. how it doesn't show any conflict? Jun 6 at 14:55
• f is just a symbol. We're not defining f directly, so to speak, we're specifying DownValues (replacement rules for a particular form). There is no infinite recursion, because once the Nest has been performed, we have forms with f that have only a single argument. So, the evaluator will use the one-argument definition to resolve those (if it exists). You could make this more explicit/safe if you define a new function instead of overloading f: nestf[nestLevel_, va_] := Nest[f, va, nestLevel]. Jun 6 at 15:21
As answered well in the comment, we can find it by
Nest[f, x, 5]
Clear["Global*"]
For some functions f, you can use NestList and FindSequenceFunction to solve the problem more generally
f[va_] := (va*2);
seq = Rest@NestList[f, x, 6]
(* {2 x, 4 x, 8 x, 16 x, 32 x, 64 x} *)
Clear[f]
f[x_, n_ : 1] = FindSequenceFunction[seq, n]
(* 2^n x *)
f[s]
(* 2 s *)
f[s, 5]
(* 32 s *)
f[s, n]
(* 2^n s *)
` | 2022-10-07T03:01:38 | {
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https://stats.stackexchange.com/questions/466930/conditional-probability-brain-teaser | # Conditional probability brain teaser
For fun I was looking through brain teasers and came across this one on a website
You’re about to board a train from London to Newcastle. You want to know if it’s raining, so you call your three friends who live in Newcastle. Each friend has a 2/3 chance of telling you the truth and a 1/3 chance of telling you a lie.
All three friends tell you that, yes, it’s raining in Newcastle.
What is the probability that it is, in fact, raining in Newcastle?
The answer given is 96% with this explanation:
You only need one friend to be telling the truth. So if you calculate the odds of them all lying, that’s 1/3 multiplied together, making 1/27 (1/3 x 1/3 x 1/3).
So that’s a 1 in 27 chance that all of your three friends are lying. So, switch that around, and it’s a 26/27 chance one of them is telling the truth – or 96% - that it is, indeed raining in Newcastle!
However, the answer is ignoring the fact that all friends agreed so I see two scenarios. (1) they all told the truth (8/27) or (2) they all lied (1/27). Limiting to only these two, the probability of it raining should be (8/27) / (8/27 + 1/27) or 8/9 NOT 26/27. Is the answer given on the website wrong?
I'm also wondering if the probability of it raining makes a difference (meaning the question is poorly worded/missing information)? For instance, if the location was the Sahara Desert instead of Newcastle, everyone's gut instinct would think the friends are lying about it raining even if they all agreed that it was.
If it helps here is the code I used as an attempt to simulate the scenario
import random
def main(N):
rain, dry = 0, 0
for _ in range(N):
is_lie1 = random.randint(1, 3) == 1
is_lie2 = random.randint(1, 3) == 1
is_lie3 = random.randint(1, 3) == 1
if sum([is_lie1, is_lie2, is_lie3]) == 0:
rain += 1
elif sum([is_lie1, is_lie2, is_lie3]) == 3:
dry += 1
print(rain / (rain + dry))
• P(rain| told it’s raining) = P(Told it’s raining|rain) P(rain)/P(Told it’s raining). This seems like a problem where the proposed solution, while seemingly correct, took shortcuts that make sense according to our (awful, absolutely awful) intuition about probability, rather than working through the details of Bayes’ theorem. At the very least, I think they’re assuming that P(rain)/P(Told it’s raining)=1.
– Dave
May 17, 2020 at 4:39
• That's something I can live with as a "lie" or "truth" is defined by the state of the weather. I'm okay taking that assumption for this problem May 17, 2020 at 20:47
• @Dave can you take that approach though given the fact that P(rain) is unstated? May 21, 2020 at 4:40
• To add some data to the problem: according to Wikipedia, there are 122 rainy days in Newcastle per year, and 122/365.25 = 0.3342 ≈ 1/3 (which is perfect for problems involving simple numbers). So the prior odds of rain would be 1:2, and multiplying by the Bayes' factor (computed by @SextusEmpiricus below) gives 4:1 as the posterior odds, i.e. the probability of rain conditional on your friends' answers is 4/5 = 80%. Oct 31, 2020 at 15:02
Indeed it matters where you are. In the rainforest it will be much more likely that it is raining if these friends all tell you that it is raining in comparison to the case of the Sahara desert.
What this website should have been computing is the likelihood ratio for rain versus no rain, the Bayes factor:
$$\frac{\text{P(all friends say rain, if it rains)}\hphantom{\text{does not}}}{\text{P(all friends say rain, if it does not rain)} }= \frac{\left({2}/{3}\right)^3}{\left({1}/{3}\right)^3} = 8$$
And this you multiply with the odds of rain and no rain without information. Say if it is normally 1:1 odds for rain versus no rain then now it is 8:1.
What the website computed is the denominator in the above equation.
$$\text{P(all friends say rain, if it does not rain)} = \frac{1}{27}$$
You can not turn that around the way they did.
$$\begin{array}{rcl} 1-\text{P(all say rain, if no rain)}& = &\text{P(not all say rain, if no rain)}\\ &=& \text{P(one or more say rain, if no rain)} \end{array}$$
But not
$$1-\text{P(all say rain, if no rain)} \neq \text{P(there's no rain, if all say rain)}$$
It is the application of the wrong rule. They applied the complement rule instead of Bayes' rule.
Okay, I bought a flight to Newcastle and befriended 3 dubious individuals. Every day I called them and asked for help simulating this brain teaser.
The probability is 8/9 only if the chance of rain is 50%.
There are two possibilities given what's known. Either it's raining and all the friends are telling the truth or it's dry and all of them are lying. This is of course dependent on the chance of rain. So this brain teaser does not have an answer unless the prior probability of rain is given
import random
def main(N, rain_percent):
rain, dry = 0, 0
for _ in range(N):
is_dry = int(random.random() > rain_percent)
if is_dry:
is_lie1 = random.randint(1, 3) == 1
is_lie2 = random.randint(1, 3) == 1
is_lie3 = random.randint(1, 3) == 1
# All lies (said it would rain, but was dry)
if all([is_lie1, is_lie2, is_lie3]):
dry += 1
else:
is_truth1 = random.randint(1, 3) != 1
is_truth2 = random.randint(1, 3) != 1
is_truth3 = random.randint(1, 3) != 1
# All truths (said it would rain)
if all([is_truth1, is_truth2, is_truth3]):
rain += 1
print( (rain) / (rain + dry) )
• Also, the question does not actually say the three friends are independent. They might always agree, for all we know. Jun 1, 2020 at 6:29
The answer given on the website is wrong.
The 96% figure tells you how likely it is, on the whole, that at least one of your friends tells the truth at any given time. That is, if you call them every day for many days, you'll find that on only 4% of days they all lie.
However, on the remaining days, they don't all tell the truth. Some days, only two of them do, and other days only one is truthful while two of them are lying. These 2:1 split-days all contribute to the 96% figure.
As you realized, days where all your friends agree are a more specific subset. These are days where they either all lie, or all tell the truth. There is no way for one friend to be telling the truth while the other two lie (or vice versa), and for them to still agree.
If they all lie, then we are indeed dealing with one of those "4%-days" (1 in 27) that are complementary to the 96% figure. However, if they all tell the truth, we're in a more specific subset of days than the 96%-set. This specific subset, where they all tell the truth, only comprises 8/27 ≈ 30% of days. Thus, as you correctly calculated, if we condition on your friends all agreeing, then the chance that they all tell the truth is 8/9, while the probability that they are all lying is 1/9.
The 8 in 9 figure is the highest certainty you're going to get by consulting your friends, on days where they all agree. On all other days, only two of your friends agree with each other. This happens on 6 out of 9 days. On 4 of those days, the two agreeing friends are truthful. On the other 2, the one dissenting friend is truthful. So, when exactly two of your friends say it's raining (or not), there is a 2/3 chance that this is correct. Thus, if you always trust the majority (which is the best you can do unless you know who's lying), your information will be correct only about 74% of the time.
As Patrick Stetz's answer pointed out, these calculations do indeed depend on the prior probability of rain. If rain is very unlikely on the whole, then the observed pattern of responses is more likely to be because it really isn't raining but your friends all lied. And, as Thomas Lumley said in his comment, it also depends on the friends all behaving independently from each other. For example, if friend 1 gets their weather information from friend 2, then you really need to treat those two friends as just one data point. In the most extreme case, if they are perfectly dependent (e.g. because they all watch the same unreliable weather report and don't look out the window), then when they all say it rains the probability that it really is raining is only 2/3 (same as if you only asked one of them). | 2022-10-05T12:22:20 | {
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https://mathematica.stackexchange.com/questions/96654/how-can-i-make-a-tribonacci-sequence-in-the-form-of-a-list/96689 | # How can I make a Tribonacci sequence in the form of a list?
How can I make a Tribonacci sequence that is in listing form? it suppose to look like the Fibonacci sequence but I couldn't get the same result with Tribonacci.
Array[Fibonacci, 9]
{1, 1, 2, 3, 5, 8, 13, 21, 34}
Array[Tribonacci, 9]
{Tribonacci[1], Tribonacci[2], Tribonacci[3], Tribonacci[4],
Tribonacci[5], Tribonacci[6], Tribonacci[7], Tribonacci[8], Tribonacci[9]}
## Is Tribonacci defined?
First you should notice that Tribonacci is not already defined by Mathematica. Compare the defined Fibonacci
?Fibonacci
with
?Tribonacci
You could have guessed by the color of the function in the front-end display, black for defined and blue for undefined.
## Fibonacci n-Step Number
Now we can define the even more general Fibonacci n-Step Number. We will use memoization
ClearAll[FnStepN];
FnStepN[0, n_] = 0; FnStepN[1, n_] = 1; FnStepN[2, n_] = 1;
FnStepN[k_Integer, n_Integer] :=
FnStepN[k, n] = Sum[FnStepN[k - i, n], {i, 1, Min[k, n]}]
TableForm[
Table[FnStepN[k, n], {n, 10}, {k, 10}]
]
EDIT
After the answer by @Mr.Wizard here is an alternative implementation of Fibonacci n-Step Number but giving the whole list
FibStepN[k_Integer, n_Integer] := Take[
LinearRecurrence[
Table[1, {n}]
, k + n - 2
]
, -k]
Or
FibStepN[k_Integer, n_Integer] :=
Nest[Append[#, Total[Take[#, -Min[n, Length[#]]]]] &, {1, 1}, k - 2]
## Tribonacci
Now Tribonacci is just a special case of FnStepN
Tribonacci[k_] = FnStepN[k, 3]
Array[Tribonacci, 11]
{1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274}
For performance look at this other similar answer.
• What should I change in order to have this kind of result for lucas n-step sequence? aside from changing FnStepN into LnStepN. – Charles Oct 13 '15 at 13:09
• @Charles I have provided an answer to your question here – rhermans Oct 13 '15 at 16:11
LinearRecurrence is useful here:
LinearRecurrence[{1, 1, 1}, {1, 1, 2}, 9]
{1, 1, 2, 4, 7, 13, 24, 44, 81}
Related:
• Your answer is shifted {1, 1, 2, 4, 7, 13, 24, 44, 81, 149} -> {0, 1, 1, 2, 4, 7, 13, 24, 44, 81} – rhermans Oct 10 '15 at 11:01
• @rhermans I did not see that as a sufficiently difficult problem to address, but one can use e.g. Rest or LinearRecurrence[{1, 1, 1}, {0, 1, 1}, {2, 10}] as he sees fit. Oh, or use LinearRecurrence[{1, 1, 1}, {1, 1, 2}, 9] which is probably what you were getting at. :-o I'll edit that in. – Mr.Wizard Oct 10 '15 at 16:55
• This is the right way, IMO...+1 – ciao Oct 10 '15 at 22:46
As noted by the Wizard, LinearRecurrence[] is an excellent way to handle integer sequences based on linear difference equations. Had that mechanism not been available, one can exploit the relationship between linear recurrences and powers of the Frobenius companion matrix of the recurrence's characteristic polynomial:
SetAttributes[Tribonacci, Listable];
Tribonacci[0] = 0;
Tribonacci[1] = Tribonacci[2] = 1;
Tribonacci[k_Integer?Positive] :=
MatrixPower[{{1, 1, 1}, {1, 0, 0}, {0, 1, 0}}, k - 2, {1, 1, 0}][[1]]
where I used the "action" form of MatrixPower[].
Some insight into how this works can be seen by looking at the associated matrix in two different ways. Let
$$\mathbf F=\begin{pmatrix}1&\cdots&&1\\1&&&\\&\ddots&&\\&&1&\end{pmatrix}=\mathbf S+\mathbf e_1\mathbf e^\top$$
where $\mathbf S$ is the "shift matrix" (the matrix that transforms $\begin{pmatrix}c_1&\cdots&c_n\end{pmatrix}^\top$ to $\begin{pmatrix}0&c_1&\cdots&c_{k-1}\end{pmatrix}^\top$, $\mathbf e$ is ConstantArray[1, k] in Mathematica notation, and $\mathbf e_1$ is UnitVector[k, 1] in Mathematica notation. This particular decomposition shows how the linear recurrence proceeds: the shift matrix moves the contents of the column vector containing the initial conditions downward, and the correction term sets up the appropriate linear combination of the vector's components.
The other way to look at $\mathbf F$ is to note that it is, as I mentioned earlier, the Frobenius companion matrix of $x^k-\sum_{j=1}^{k-1} x^j$, which is the characteristic polynomial of the linear recurrence $F_n=\sum_{j=1}^k F_{n-j}$. $\mathbf F$ thus has the eigendecomposition
$$\mathbf F=\mathbf V\begin{pmatrix}x_1&&\\&\ddots&\\&&x_k\end{pmatrix}\mathbf V^{-1}$$
and the $x_j$ are the $k$ roots of the characteristic polynomial. $\mathbf F^n$ thus has the eigendecomposition
$$\mathbf F^n=\mathbf V\begin{pmatrix}x_1^n&&\\&\ddots&\\&&x_k^n\end{pmatrix}\mathbf V^{-1}$$
and the entire business is revealed to be equivalent to taking appropriate linear combinations of the $x_j^n$.
## Extra credit
The machinery behind DifferenceRoot[] can of course be used to implement the general $k$-nacci number. Witness the following:
knacci[k_Integer?Positive] := knacci[k] =
DifferenceRoot[Function @@
{{\[FormalY], \[FormalN]},
Prepend[Thread[(\[FormalY] /@ {1, 2}) == 1] ~Join~
Thread[(\[FormalY] /@ Range[3 - k, 0]) == 0],
\[FormalY][\[FormalN]] ==
Sum[\[FormalY][\[FormalN] - K], {K, 1, k}]]}]
after which, one can do Tribonacci = knacci[3].
Still another possibility is to use SeriesCoefficient[] on the generating function:
knacci[k_Integer?Positive, n_Integer?NonNegative] :=
SeriesCoefficient[(\[FormalX] (1 - \[FormalX]))/
(1 - 2 \[FormalX] + \[FormalX]^(k + 1)), {\[FormalX], 0, n}]
Carrying the generating function idea further along, one could consider using Cauchy's differentiation formula with an appropriate anticlockwise contour to evaluate $k$-nacci numbers. Here is one such routine based on this idea:
SetAttributes[knacci, Listable]
knacci[k_Integer?Positive, n_?NumericQ] :=
Re[NIntegrate[(1 - z)/((1 - 2 z + z^(k + 1)) z^n),
{z, -1/2, -I/2, 1/2, I/2, -1/2}]/(2 π I)]
• FWIW I did reference this in the last link in my answer, but +1 of course. – Mr.Wizard Oct 10 '15 at 10:51
• I saw that. :) The missing piece was to use the "action" form instead of explicitly forming the power before multiplying. The version using Nest[] on an appropriate starting vector does sidestep the problem. – J. M. is away Oct 10 '15 at 10:53
• The first idea comes to my mind is MatrixPower when I see the definion of Enrique Pérez Herrero. Please see here – xyz Oct 10 '15 at 14:08
• J.M, could you add the theroy about why you using the MatrixPower ? I think it is necessary for others understand your implementation easily:) – xyz Oct 10 '15 at 14:13
• @Shutao, maybe later; that's a rather long discussion... – J. M. is away Oct 10 '15 at 14:15
If -- like with Fibonacci -- you want Tribonacci defined more generally than just for integer arguments, use RSolve.
Clear[Tribonacci];
Tribonacci[n_] = Tribonacci[n] /. RSolve[{Tribonacci[0] == 0,
Tribonacci[1] == 1,
Tribonacci[2] == 1,
Tribonacci[n] ==
Tribonacci[n - 1] + Tribonacci[n - 2] + Tribonacci[n - 3]},
Tribonacci[n], n][[1]] // ToRadicals // Simplify;
Tribonacci[n_Integer] :=
Tribonacci[N[n]] // Chop // Round;
The last definition for integer arguments is added to speed up the calculations. To get a simple result for integers would generally require use of FullSimplify of a very complicated expression and would be quite slow. The numerical approximation is much faster but requires Chop to remove imaginary artifacts caused by use of machine precision and Round to give the exact solution desired.
Tribonacci /@ Range[0, 10]
(* {0, 1, 1, 2, 4, 7, 13, 24, 44, 81, 149} *)
Show[
Plot[Tribonacci[n], {n, -10, 10}],
DiscretePlot[Tribonacci[n], {n, -10, 8},
PlotStyle -> Red],
PlotRange -> All]
• As it turns out, you could also use FunctionExpand[DifferenceRoot[(* stuff *)][n]] to derive explicit expressions for $k$-nacci numbers. – J. M. is away Oct 10 '15 at 16:50
Tribonacci[0] := 0;
Tribonacci[1] := 1;
Tribonacci[2] := 1;
Tribonacci[3] := 2;
Tribonacci[n_] :=
Tribonacci[n] =
Tribonacci[n - 1] + Tribonacci[n - 2] + Tribonacci[n - 3];
Array[Tribonacci, 9]
(* {1, 1, 2, 4, 7, 13, 24, 44, 81} *)
An alternate expression can be found in https://oeis.org/A000073:
CoefficientList[Series[x^2/(1 - x - x^2 - x^3), {x, 0, 50}], x]
Also you can find more information about: Tribonacci[n] - Fibonacci[n], in: https://oeis.org/A000100
Even more...
a = (19 + 3*Sqrt[33])^(1/3);
b = (19 - 3*Sqrt[33])^(1/3);
Trib[n_] := Round[3*((a + b + 1)/3)^(n + 1)/(a^2 + b^2 + 4)]
Table[Trib[n] - Tribonacci[n], {n, 1, 20}]
(* {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0} *)
tribonacci[n_] := SequenceFold[Plus, {1, 1, 2}, ConstantArray[0, n - 3]]
tribonacci[1] = tribonacci[2] = 1;
Array[tribonacci, 9]
$\${1, 1, 2, 4, 7, 13, 24, 44, 81}
tribonacciList[n_] := SequenceFoldList[Plus, {1, 1, 2}, ConstantArray[0, n - 3]]
tribonacciList[1] = {1};
tribonacciList[2] = {1, 1};
tribonacciList[9]
$\${1, 1, 2, 4, 7, 13, 24, 44, 81}
Benchmark table of the AbsoluteTimings for the calculation of a list of the first 9 and 5000 Tribonacci numers:
• Would you add benchmarks for these? I am curious to know how these functions (SequenceFold family) perform and I do not have v10.2. – Mr.Wizard Oct 10 '15 at 17:56
• @Mr.Wizard I'll put that on my list for tomorrow. But I can already say that my performance expectations are relatively low, as these functions are implemented as high level functions using Fold and FoldList, respectively. – Karsten 7. Oct 10 '15 at 19:15
• @Mr.Wizard compared to the other methods, using SequenceFoldList actually performs really good. – Karsten 7. Oct 11 '15 at 22:16
The definition of Tribonacci[] can be written as below:
$$a_{n+3}=a_n+a_{n+1}+a_{n+2}$$
then
we could construct the following recursive matrix formula
$$\begin{pmatrix} a_{n+3}\\ a_{n+2}\\ a_{n+1} \end{pmatrix} = \begin{pmatrix} 1 && 1 && 1\\ 1 && 0 && 0\\ 0 && 1 && 0 \end{pmatrix} \begin{pmatrix} a_{n+2}\\ a_{n+1}\\ a_{n} \end{pmatrix} =A^2\begin{pmatrix} a_{n+1}\\ a_{n}\\ a_{n-1} \end{pmatrix}=\cdots= A^{n+1}\begin{pmatrix} a_{2}\\ a_{1}\\ a_{0} \end{pmatrix}$$
where,
$A=\begin{pmatrix} 1 && 1 && 1\\ 1 && 0 && 0\\ 0 && 1 && 0 \end{pmatrix}$
• Your companion matrix is backwards; the subdiagonal goes from left to right. – J. M. is away Oct 10 '15 at 14:45
• @J.M. OK, I made a mistake. THX:) – xyz Oct 10 '15 at 14:48
Tail recursive implementation
tribo[n_, a_, b_, c_] := tribo[n - 1, b, c, a + b + c]
tribo[0, a_, b_, c_] := a
tribo[n_] := tribo[n, 0, 1, 1]
Array[tribo, 9] // AbsoluteTiming
$\${0.0000928937, {1, 1, 2, 4, 7, 13, 24, 44, 81}}
Block[{$IterationLimit = Infinity}, tribo~Array~5000; // AbsoluteTiming // First] $\ $25.2784 And for a faster generation of Tribonacci lists triboList[n_, a_, b_, c_] := triboList[n - 1, Sow[b], c, a + b + c] triboList[1, a_, b_, c_] := Sow[b] triboList[n_] := Reap[triboList[n, 0, 1, 1]][[2, 1]] triboList[9] // AbsoluteTiming $\ ${0.0000497884, {1, 1, 2, 4, 7, 13, 24, 44, 81}} Block[{$IterationLimit = Infinity},
triboList[5000]; // AbsoluteTiming // First]
$\$0.0129951
ContinuedFraction[Root[#^3-#^2-#-1]&,1],100]
or Convergents[N[Convergents[N[t]]]]
• Unfortunately, the approach here is flawed; the number whose simple CF expansion contains the tribonacci numbers is approximately 1.6914979485021664824009037684992342805376592458272, which is quite far from the value of Root[#^3 - #^2 - # - 1 &, 1] (1.8392867552141611325518525646532866004241787460976). – J. M. is away Oct 15 '15 at 18:20 | 2019-08-26T08:17:00 | {
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http://math.stackexchange.com/questions/86790/is-there-a-reason-why-the-number-of-non-isomorphic-graphs-with-v-4-is-odd | # Is there a reason why the number of non-isomorphic graphs with $v=4$ is odd?
I am working through Trudeau's Introduction to Graph Theory, which contains the following problem:
In the following table, the numbers in the second column are mostly even. If we ignore the first row on the ground that $v=1$ is such a trivial situation that its uniqueness is unremarkable, that leaves $v=4$ as the only number of vertices listed for which there are an odd number of graphs. Do you think this is due to chance, or can you think of some reason why $v=4$ should be unique?
v number of non-isomorphic graphs
1 1
2 2
3 4
4 11
5 34
6 156
7 1,044
8 12,346
9 308,708
Note, this problem concerns the number of non-isomorphic graphs.
Here's what I have so far:
The maximum number of edges in a graph with $v=4$ is $\max(e)=\frac{1}{2}(v)(v-1)=\frac{1}{2}(4)(3)=6$. The graphs with $e=0,1,2,4,5,6$ come in pairs because each graph has a complement. So, there are an odd number of graphs with $v=4$ iff there are an odd number of graphs with $v=4$ and $e= \frac{\max(e)}{2}$.
There are 3 graphs with $v=4$ and $e= \frac{\max(e)}{2}=3$, so therefore there is an odd number of graphs with $v=4$.
However, I don't understand why $v=4$ should be special in this regard, even though it feels special.
-
v=9 is wrong, its 274668 – TROLLKILLER Nov 29 '11 at 19:54
I take it you are enumerating the number of non-isomorphic graphs on $v$ vertices. You should spell this out in your question. – Austin Mohr Nov 29 '11 at 19:58
But any number of the form $v = 4n$ or $v = 4n+1$ has $max(e)/2$ an integer. So what's so special about 4? – Michael Lugo Nov 29 '11 at 20:36
Also, you might appreciate knowing that 4 really is special and this is not just a trick question. The number of non-isomorphic graphs on $v$ vertices is even for every $v \le 68$, except $v = 4$: abel.math.umu.se/~klasm/Data/numberofgraphs.txt . (As far as I know there's nothing special about 68; this is just the longest list I could find.) – Michael Lugo Nov 29 '11 at 20:39
OEIS gives first 19 values and also some formulas and references oeis.org/A000088 – Martin Sleziak Nov 29 '11 at 20:57
Definition: Let $g(n)$ denote the number of unlabeled graphs on $n$ vertices, let $e(n)$ denote its $2$-part, that is the largest power of $2$ which divides $g(n)$.
Lemma: If $n\geq5$ is odd then $e(n) = (n+1)/2-\lfloor \log_2(n) \rfloor$. If $n \geq 4$ is even then $e(n) \geq n/2 - \lfloor \log_2(n) \rfloor$ with equality iff $n$ is a power of $2$.
Corollary: The amount of unlabeled graphs is even for $n > 4$
Some values $e(\{4,5,\ldots,15\})=\{0,1,1,2,1,2,2,3,3,4,4,5\}$ (for even numbers it is the lower bound).
The theorem is due to Steven C. Cater and Robert W. Robinson and can be found, including a proof, in this publication.
They mention that $g(n)$ is not only even but contains a large number $2$'s in its prime factorisation for large $n$ (this also follows form the formula). In fact they even show that they are asymptotically $n/2$ factors of $2$ in $g(n)$.
-
The full citation for the paper is Steven C. Cater, Robert W. Robinson, Exponents of 2 in the numbers of unlabeled graphs and tournaments, Proceedings of the Twenty-second Southeastern Conference on Combinatorics, Graph Theory, and Computing (Baton Rouge, LA, 1991), Congr. Numer. 82 (1991), 139–155, MR1152066 (92k:05062). – Gerry Myerson Nov 29 '11 at 22:46
If a graph on $n$ vertices has $e$ edges, then the number of edges in its complement is $$\frac{n(n-1)}{2} - e.$$ So if if $n(n-1)/2$ is odd we can divide graphs in pairs (graph,complement), and then the number of graphs must be even. In particular the the parity of the number of isomorphism classes of graphs is equal to the parity of the number of isomorphism classes of self-complementary graphs. When $n=4$, the path is the unique self-complementary graph and the number of isomorphism classes is odd. Clearly when $n=6$ or $n=7$ there must be an even number of isomorphism classes self-complementary graphs.
-
Okay, but how does this distinguish $n = 4$ from $n = 5, 8, 9$? – Qiaochu Yuan Nov 29 '11 at 20:18
Well, it explains what's happening at $n=4$ and reduces the problem to counting self-complementary graphs. And it answers the question in the title. – Chris Godsil Nov 29 '11 at 21:08
I agree, Listing's answer doesn't give one any reason why $n = 4$ has an odd number of graphs at all. In fact, he just gives a formula, which isn't very instructive at all. +1 – Graphth Nov 29 '11 at 21:37
I agree that it explains something special about $n=4$ but it doesn't really explain why this property is unique to $n=4$, does it? – Listing Nov 29 '11 at 21:56
I was not criticizing Listing's answer, it is fine and more complete than what I wrote. – Chris Godsil Nov 29 '11 at 21:58 | 2016-07-28T05:13:52 | {
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https://math.stackexchange.com/questions/342070/factor-square-property-fsp-of-polynomials?noredirect=1 | # factor-square property (FSP) of polynomials
The Factor Square Property (FSP) is the divisibility of the polynomial $f(x^2)$ by $f(x)$.
1. Is $x^2+x+1$ the only FSP irreducible polynomial of degree $2$ ?
2. Are there other linear polynomial besides $x$ and $x-1$ with FSP?
3. Do we have other FSP irreducible polynomials of degree $3$ or $4$? Any of these have integer coefficients??
4. Are there any other observations you can make about polynomials with FSP?
• What is the factor-square property? – Chris Eagle Mar 26 '13 at 20:25
• f(x) has FSP if f(x) is a factor of f(x2) example F(x) = x-1 has FSP because x-1 is a factor of x2-1. similarly, h(x) =x has FSP because x is a factor of x2. – sam Mar 26 '13 at 20:50
• Please, TeX. It exists for a reason. – Lord_Farin Mar 26 '13 at 22:18
• @sam : The definition of FSP, that $f(x)$ divides $f(x^2)$, does not appear to be a standard definition. In my opinion the definition of this term FSP (or factor square property) should appear in the main body of the question, not just as a response to someone (Chris Eagle) having asked what it means. – coffeemath Mar 27 '13 at 7:51
• As pointed out by someone in another question, this is from math.osu.edu/ross/app/RossProb13.pdf – Erick Wong Mar 31 '13 at 4:35
A google search of "factor square property" led only to this present question. So recall from comment that a polynomial $f(x)$ which divides $f(x^2)$ is said to be FSP or to have the factor square property.
This addresses [1], and shows an irreducible FSP quadratic must be $ax^2+ax+a$, which depending on what "irreducible" means might also imply $a=1$. We also address [3], giving a source of examples.
Let the quadratic be $p=ax^2+bx+c$, so its value at $x^2$ is $q=ax^4+bx^2+c$. If $p$ is to be a divisor of $q$ let the other factor be $dx^2+ex+f.$ Equating coefficients gives equations
[1] $ad=a,$
[2] $ae+bd=0,$
[3] $af+be+cd=b,$
[4] $bf+ce=0,$
[5] $cf=c.$
Now we know $a,c$ are nonzero (else $p$ is not quadratic, or is reducible). So from [1] and [5] we have $d=f=1.$ Then from [2] and [4] we obtain $ae=ce.$ Here $e=0$ leads to $b=0$ from either [2] or [4], and [3] then reads $a+c=0$, so that $p=a(x^2-1)$ which is reducible. So we may assume $e$ is nonzero, and also $a=c.$
At this point, [2] and [4] say the same thing, namely $ae+b=0.$ So we may replace $b=-ae$ in [3] (with its $c$ replaced by $a$) obtaining $a+(-ae)e+a=-ae,$ which on factoring gives $a(2-e)(e+1)=0.$ The possibility $e=2$ then leads after some algebra to $2a+b=0$ and $p=a(x-1)^2$ which is reducible, while the possibility $e=-1$ leads to $a=b$ and then $p=ax^2+ax+a$ as claimed.
For some higher degree examples, for any odd prime $p$ the cyclotomic polynomial $f(x)=(x^p-1)/(x-1)$ is seen to be FSP because $$\frac{f(x^2)}{f(x)}=\frac{x^{2p}-1}{x^2-1}\cdot \frac{x-1}{x^p-1} \\ =\frac{x^p+1}{x+1}.$$ For $p=5$ this is the polynomial $$x^4+x^3+x^2+x+1.$$ Note that for odd prime $p$ these cyclotomic polynomials are known to be irreducible (there is a simple Eisenstein proof using a change of variables).
• what about the polynomials with the real number coefficient and the complex number coefficient. Here is the link of my question: math.stackexchange.com/questions/2128306/… – Mclalalala Feb 5 '17 at 6:30
• Thank you if you can help me do some further exploration – Mclalalala Feb 5 '17 at 6:30
• @Mclalalala The argument above about finding the quadratic irreducibles with FSP is independent of whether the coefficients are real or complex. I do not claim to have found all possible higher degree polynomials with FSP, only a collection of them. – coffeemath Feb 5 '17 at 17:18 | 2021-05-13T06:02:39 | {
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https://math.stackexchange.com/questions/2994865/stability-of-a-degenerate-equilibrium-point-in-a-planar-ode | # Stability of a Degenerate Equilibrium Point in a Planar ODE
Consider the planar ODE
$$\dot x_1 = x_2$$
$$\dot x_2 = - x_1^2 - 2 x_1 - 1$$
Obliviously, $$(x_1,x_2)=(-1,0)$$ is an equilibrium point. The Jacobian matrix at this point is
$$J = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$$ Thus, linearizarion fails in determining the stability. How can we determine the stability of this equilibrium point?
• If we shift the coordinates to the equilibrium point by introducing new coordinates $y_1 = x_1 + 1, \; y_2 = x_2$, the new equations would be $\dot{y}_1 = y_2, \; \dot{y}_2 = -y^2_1$. You can find the integral curves of this system by solving equation $\frac{d y_1}{d y_2} = \dots$ and putting directions on these integral curves using transformed system. This will help you determine stability of this equilibrium. – Evgeny Nov 12 '18 at 8:43
• Thank you for your comment. I think this way does not work in this special case. A first integral is $F=0.5 y_2^2 + (1/3) y_1^3$. However, the Hessisn of $F$ at the origin will be non-singular and we cannot exactly see the behavior of the level sets around the origin (we cannot use the Morse Lemma). Thanks! – Arthur Nov 12 '18 at 16:02
• Morse lemma is an overkill here: the first integral is quite simple and you can plot its level sets by hand ;) The approach works, I can explain what I had in mind. For example, consider the level set $F = 0$. It's quite easy to plot and to put directions on trajectories from it. One of these trajectories alone would be an example of something escaping any small neighbourhood of the origin. – Evgeny Nov 12 '18 at 16:56
• Thanks for the explanation. Right! We can see some trajectories escaping from the origin. Thanks! – Arthur Nov 12 '18 at 20:25
After the change of variables $$y_1=x_1+1$$, $$y_2=x_2$$ the system takes the form $$\left\{\begin{array}{lll} \dot y_1&=&y_2\\ \dot y_2&=&-y_1^2.\\ \end{array}\right.$$ In order to prove the instability of the origin, one can use arguments similar to those used in the proof of the Chetaev instability theorem. Consider the set $$G= \{ y\in\mathbb R^2:\; y_1<0,y_2<0 \}.$$ One can see that
• $$y_1$$ and $$y_2$$ are strictly decreasing along any trajectory in $$G$$;
• $$G$$ is positively invariant (because $$\dot y_1|_{G}<0$$ and $$\dot y_2|_{G}<0$$).
But it means that no solution starting from the point $$y^{start}$$ in $$G$$ can stay in the $$\|y^{start}\|$$-neighborhood of the origin:
Moreover, let $$y|_{t=0}=y^{start}=(y_1(0),y_2(0))$$. We have $$\forall t>0 \quad \dot y_1 which implies $$\lim_{t\to\infty} y_1(t)=-\infty$$, $$\lim_{t\to\infty} y_2(t)=-\infty$$. Thus, the origin is unstable.
• Thank you very much for your nice solution and figure! – Arthur Nov 12 '18 at 16:04
Lauds to my colleague AVK for his short and sweet answer which invokes the Chataev instability theorem, of which I was not aware until I read that answer. Before that answer was posted, I worked up a solution from more-or-less first principles, which doesn't invoke anything other than basic, general facts about ordinary differential equations. In so doing, I obtained a fairly detailed "picture" of the singularity at $$(-1, 0)$$, and the integral curves of our differential equation, which shows just how this instability behaves. I share these considerations below.
Given the system
$$\dot x_1 = x_2, \tag 1$$
$$\dot x_2 = -x_1^2 - 2x_1 - 1, \tag 2$$
we first observe that (2) may be written
$$\dot x_2 = -(x_1 + 1)^2; \tag 3$$
the equilibria occur where
$$\dot x_1 = 0, \tag 4$$
$$\dot x_2 = 0; \tag 5$$
that is, where
$$x_2 = \dot x_1 = 0, \tag 6$$
and where
$$-(x_1 + 1)^2 = \dot x_2 = 0, \tag 7$$
i.e., where
$$x_1 = -1; \tag 8$$
we see the point $$(-1, 0)$$ is the only zero of the vector field $$(x_2, -(x_1 + 1)^2)^T$$; the Jacobean matrix is
$$J(x_1, x_2) = \begin{bmatrix} \dfrac{\partial \dot x_1}{\partial x_1} & \dfrac{\partial \dot x_1}{\partial x_2} \\ \dfrac{\partial \dot x_2}{\partial x_1} & \dfrac{\partial \dot x_2}{\partial x_2} \end{bmatrix} = \begin{bmatrix} \dfrac{\partial x_2}{\partial x_1} & \dfrac{\partial x_2}{\partial x_2} \\ \dfrac{\partial (-(x_1 + 1)^2)}{\partial x_1} & \dfrac{\partial -((x_1 + 1)^2)}{\partial x_2} \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -2(x_1 + 1) & 0 \end{bmatrix}; \tag 9$$
thus at the critical point $$(-1, 0)$$ we have
$$J(-1, 0) = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}; \tag{10}$$
the sole eigenvalue of the matrix $$J(-1, 0)$$ is in fact $$0$$ of algebraic multiplicity $$2$$, which follows from the fact that the chracteristic polynomial of $$J(-1, 0)$$ is
$$\chi(x) = \det \left ( \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} - xI \right ) = \det \left ( \begin{bmatrix} -x & 1 \\ 0 & -x \end{bmatrix} \right ) = x^2; \tag{11}$$
indeed, $$J(-1, 0)$$ is in fact nilpotent,
$$J^2(-1, 0) = 0. \tag{12}$$
We may in fact compute the sole eigenvector of $$J(-1, 0)$$; it lies in $$\ker J(-1, 0)$$:
$$J(-1, 0) \vec v = 0, \tag{13}$$
or with
$$\vec v_1 = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}, \tag{14}$$
we have from (13) by virtue of (10),
$$\begin{pmatrix} v_2 \\ 0 \end{pmatrix} = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = 0; \tag{15}$$
thus
$$v_2 = 0; \tag{16}$$
$$v_1$$ may be arbitrarily specified so we take $$v_1 = 1$$, and then
$$\vec v = \begin{pmatrix} 1 \\ 0 \end{pmatrix}; \tag{17}$$
we may also find a generalized eigenvector $$\vec w = (w_1, w_2)^T$$ corresponding to $$0$$; it satisfies
$$\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{pmatrix} w_1 \\ w_2 \end{pmatrix} = J\vec w = \vec v = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \tag{18}$$
whence
$$w_2 = 1, \tag{19}$$
with $$w_1$$ arbitrary; thus we take
$$w_1 = 0. \tag{20}$$
OK, back to the main line: we're trying to investigate the stability of the point $$(-1, 0)$$; what we've got so far is that, though the eigenvalues and eigenvectors of $$J(-1, 0)$$ may be neatly expressed, they don't help us with stability questions, in large due to the singular nature of the matrix $$J(-1, 0)$$, which in turn reflects the more complicated nature of the critical point $$(-1, 0)$$. So we need to invoke other methods.
Upon more careful scrutiny, we notice that the system (1)-(3) is in fact possessed of a conserved quantity
$$H(x_1, x_2) = \dfrac{(x_1 + 1)^3}{3} + \dfrac{x_2^2}{2}; \tag{21}$$
indeed, we find
$$\dfrac{dH(x_1(t), x_2(t))}{dt} = \dfrac{\partial H(x_1(t), x_2(t))}{\partial x_1} \dot x_1(t) + \dfrac{\partial H(x_1(t), x_2(t))}{\partial x_2} \dot x_2(t)$$ $$= (x_1(t) + 1)^2 x_2(t) + x_2(t)(-(x_1(t)+ 1)^2) = 0; \tag{22}$$
it follows then that the integral curves of (1)-(3) satisfy the polynomial equation
$$H(x_1, x_2) = \dfrac{(x_1 + 1)^3}{3} + \dfrac{x_2^2}{2} = \text{constant}; \tag{23}$$
the orbits whose closure conains the point $$(-1, 0)$$ then obey
$$\dfrac{(x_1 + 1)^3}{3} + \dfrac{x_2^2}{2} = 0, \tag{24}$$
or
$$x_2^2 = -\dfrac{2(x_1 + 1)^3}{3}; \tag{25}$$
we see that $$x_2 \in \Bbb R$$ satisfying this equation only exist when $$x_1 \le -1$$; for $$x_1 < -1$$ there are two $$x_2$$-values
$$x_2 = \pm \sqrt{-\dfrac{2(x_1 + 1)^3}{3}}$$ $$= \pm \left (\dfrac{-2(x_1 + 1)^3}{3} \right )^{1/2} = \pm \sqrt{\dfrac{2}{3}} (-(x_1 + 1))^{3/2}; \tag{26}$$
for $$x_1 = -1$$ we see that $$x_2 = 0$$ and nothing else;also,
$$\dfrac{dx_2}{ dx_1} = \mp \dfrac{3}{2} \sqrt{\dfrac{2}{3}} (-(x_1 + 1))^{1/2} \to 0 \; \text{as} \; x_1 \to -1^-, \tag{27}$$
which shows that both the positive and negative solutions approach $$(-1, 0)$$ with vanishing slope as $$x_1 \to -1^-$$; furthermore it is clear from (26) that
$$\vert x_2 \vert \to \infty \; \text{monotonically as} \; x_1 \to -\infty; \tag{28}$$
therefore the positive and negative "branches" of (25)-(26), which are incidentally reflections of one another about the $$x_1$$-axis, symmetrically grow without bound in the negative $$x_1$$ direction, but meet at $$(-1, 0)$$ in sort of "cusp" where $$\lim_{x_1 \to 1^-} x_2'(x_1) = 0$$.
Now, at this point we have actually gathered enough information to conclude that $$(-1, 0)$$ is an unstable equilibrium. Consider first the positive branch of the solution which lies entirely in the second quadrant. We see from (1)-(3) that on this curve
$$\dot x_1 > 0, \tag{29}$$
$$\dot x_2 < 0, \tag{30}$$
and so the solution point $$(x_1(t), x_2(t))$$ is in fact moving towards $$(-1, 0)$$ along (26), and since the given system has no other equilibria, $$(x_1(t), x_2(t))$$ will eventually become arbitrarily close to $$(-1, 0)$$; similarly, if the system point is initialized so as to lie in the negative branch of (26), that is, in the third quadrant where
$$\dot x_1 < 0, \; \dot x_2 < 0, \tag{31}$$
then both $$x_1$$ and $$x_2$$ continue to decrease, at ever-increasing (absolute) rates, past any limits; this of course means that if the system is initialized along this curve, no matter how close to the equilibrium, the resulting solution will leave any bounded set, and thus in fact $$(-1, 0)$$ is an unstable equilibrium point of (1)-(3). Indeed, we may reify these assertions by examining the squared distance $$s^2$$ 'twixt $$(x_1(t), x_2(t))$$ and $$(-1, 0)$$:
$$s^2 = (x_1 + 1)^2 + x_2^2; \tag{32}$$
$$\dfrac{ds^2}{dt} = 2(x_1 + 1) \dot x_1 + 2x_2 \dot x_2$$ $$= 2(x_1 + 1)x_2 - 2x_2(x_1 + 1)^2 = 2(x_1 + 1)(x_2 - x_2(x_1 + 1)) = -2x_1(x_1 + 1)x_2 < 0 \tag{33}$$
when
$$x_1 < -1, \; x_2 > 0; \tag{34}$$
likewise,
$$\dfrac{ds^2}{dt} = -2x_1(x_1 + 1)x_2 > 0 \tag{35}$$
when
$$x_1 < -1, \; x_2 < 0; \tag{36}$$
(32)-(36) show that the actual Euclidean distance 'twixt $$(x_1(t), x_2(t))$$ and the critical point $$(-1, 0)$$ is decreasing in the second quadrant but grows without bound in the third, providing yet another reliable indication that $$(-1, 0)$$ is unstable.
We have found $$(-1, 0)$$ unstable by providing the existence of a single integral curve which escapes any bounds not matter how close to $$(-1, 0)$$ its initial point may be. In fact, every integral curve except the upper branch o (26) is possessed of this property. For suppose we have, instead of (24), $$H(x_1, x_2) \ne 0$$ so that
$$\dfrac{(x_1 + 1)^3}{3} + \dfrac{x_2^2}{2} = H = \text{constant} \ne 0; \tag{37}$$
then
$$x_2^2 = 2H -\dfrac{2(x_1 + 1)^3}{3} = \dfrac{2}{3} (3H - (x_1 + 1)^3); \tag{38}$$
$$x_2 = \pm \sqrt{\dfrac{2}{3}}\sqrt{3H - (x_1 + 1)^3} = \pm \sqrt{\dfrac{2}{3}}(3H - (x_1 + 1)^3)^{1/2}; \tag{39}$$
we see from (37) that
$$\nabla H = \begin{pmatrix} (x_1 + 1)^2 \\ x_2 \end{pmatrix} \ne 0 \tag{40}$$
except at $$(-1, 0)$$; therefore any integral curve with $$H \ne 0$$, that is, any level set of $$H$$ with $$H \ne 0$$, is smooth without kinks, corners or cusps as has the $$H = 0$$ solution. We see from (39) that we must have
$$3H - (x_1 + 1)^3 \ge 0, \tag{41}$$
or
$$(x_1 + 1)^3 \le 3H, \tag{42}$$
$$x_1 \le \sqrt[3]{3H} - 1 = (3H)^{1/3} - 1; \tag{43}$$
now from (39) it is again apparent that (26) remains in force when $$H \ne 0$$; thus the solution curves become unbounded in the first and fourth qaudrants as $$x_1 \to -\infty$$, and again they exhibit the reflection symmetry across the $$x_1$$ axis; when $$x_2 = 0$$ we see from (38) that
$$x_1 = \sqrt[3]{3H} - 1 = (3H)^{1/3} - 1, \tag{44}$$
which ensures that taking $$H$$ sufficiently close to $$-1$$ the curve (38) intersects the $$x_1$$-axis arbitrarily close to $$(-1, 0)$$; this in turn implies that every neighborhood of $$(-1, 0)$$ contains points on solutions which evolve into unbounded regions of quadrant IV under the system dynamics; in fact, since every curve of the form (38) intersects the $$x_1$$-axis at the value specified by (44), it follows that every point in every neighborhood $$U$$ of $$(-1, 0)$$ which does not lie on the positive branch of (24) will eventually be moved a region of the fourth quadrant with arbitrarily large $$\vert x_1 \vert$$, $$\vert x_2 \vert$$.
We conclude from these considerations that the equilibrium point $$(-1, 0)$$ is unstable; indeed, rather pronouncedly so.
• Thank you very much! This is very interesting. – Arthur Nov 16 '18 at 16:02
• @Arthur: you are most welcome sir! – Robert Lewis Nov 16 '18 at 17:23 | 2019-10-17T12:29:59 | {
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https://math.stackexchange.com/questions/1411144/how-to-solve-equations-to-the-fourth-power | How to solve equations to the fourth power?
Is it possible to manually retrieve the value of $y$ from the following equation $$\color{blue}{153y^2-y^4=1296}$$
WolframAlpha has four solutions for $y$: $-12, -3, 3, 12$. How has it solved?
What I've achieved to until now is the following: $$y^2(153-y^2)=1296$$ And... I'm stuck.
• This is a REALLY easy quartic and not well researched. It invites you to consider $x=y^2$ then solve $-x^2+153x-1296$ which is easy. Solving quartics in general is not like this. (so who among you is upvoting and staring this?!) Aug 27 '15 at 8:53
• @AlecTeal OP showed attempt, the question is well formatted and if you are just getting used to the quadratic equations then this requires insight, but of course, when you know the trick its easy. Don't judge on how easy it is when you don't know the OPs level. Aug 27 '15 at 15:20
• @AlecTeal In my opinion that is unnecessarily rude. This site is for people of any mathematical ability who may not spot things as easily as you do. Aug 27 '15 at 18:20
• @imulsion Here is a more sophisticated question of similar quality with 1/3 the upvotes (currently). This one is nearly a duplicate with 1/5 the upvotes. The vagaries of SE voting are renowned. (I do agree, however, that the AlecTeal's tone appears rude, and I think the argument that the question should be downvoted, implied IMO by the assertion that it is "not well researched," is off-base for the reason you cite.) Aug 29 '15 at 12:30
• I think it's best to ask a new question rather than edit a question that's already been answered. Sep 22 '15 at 12:40
One method is to set $x=y^2$ and rearrange this as a quadratic equation $$x^2-153x+1296=0$$
Here's the solution:
$$y^4-153y^2 +1296 = 0$$
$$y^4 -144y^2-9y^2+1296 = 0$$
$$y^2(y^2-144) -9(y^2-144) = 0$$
$$(y^2-9)(y^2-144)=0$$ $$(y^2-3^2)(y^2-12^2)=0$$
note that $a^2 - b^2 = (a-b)(a+b)$
Can you see how now? I trust you can finish the rest
• Nice find. The method by Mark Bennet is much more robust, as it does not depend on seeing the factorisation, but this is much more beautiful :)
– 5xum
Aug 27 '15 at 7:05
• Thanks for your comment-I guess it depends on personal preference. I do like Mark's method too. Aug 27 '15 at 7:07
• @John - of course it works, but how did you quickly notice the middle term can be split like that? Or rather, how long did it take you to figure it? Aug 27 '15 at 13:03
• @JoeTaxpayer, the only way to figure that out is trial and error. 1296 factors to $2^4 \times 3^4$, 153 is an odd number so you need one odd and one even factor-hence one factor can only be a multiple of 3. Aug 27 '15 at 13:06
• @John - "look for 2 numbers that multiply to 1296, and add to -153" - that would be the advice for my students, right? Followed by factoring the 1296 to see the candidates. (thx) Aug 27 '15 at 13:10
While the convert-to-quadratic approach works here, and other answers have provided this approach, it's worth noting that rational solutions to quartics with integer coefficients work similarly to quadratics.
That is, if you have a quartic of the form $$ax^4+bx^3+cx^2+dx+e=0$$ where all coefficients are integers, then all rational solutions must be of the form $$x = \frac{\text{divisor of }e}{\text{divisor of }a}$$ In this case, we have $a=1$, $c=-153$, $e=1296$, and $b=d=0$. So we seek a solution of the form $$x = \frac{\text{divisor of }1296}{\text{divisor of }1}$$ As the only divisors of 1 are 1 and -1, we need only look at integer values. 1296 has the prime factorisation $$1296 = 2^4\times 3^4$$ and from here, we just need to start substituting trial values in until we find a root. With $P(x)=x^4-153x^2+1296$, noting that only even powers of $x$ appear (and thus we need not consider the negatives separately), we have $$P(1) = 1 - 153 + 1296 = 1144 \neq 0\\ P(2) = 16 - 612 + 1296 = 700 \neq 0\\ P(3) = 81 - 1377 + 1296 = 0$$ and thus $x=3$ is a solution. Again noting that only even powers appear, we see that $x=-3$ is also a solution. It's fairly trivial to continue in this way to find the other two roots, as well.
Note that this method won't find irrational roots, however. | 2021-10-19T06:53:06 | {
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http://mathhelpforum.com/calculus/12897-calculus-optimize.html | # Thread: calculus optimize
1. ## calculus optimize
help me with this please (please indicate which one you are answering, A or B....both questions are not related to one another):
question A) a box is to be made frmo a square piece of cardboard that measures 12cm X 12cm by having equal squares cut from the corners of the sides. find the dimensions of the box of the greatest volume that can be made this way.
question B) a merchant ship traveling due north at 12km/h crosses the track of a second ship travelling due east at 9kn/h, at a time 100 min after the second ship passed that point. find the distance of the closest approach of the two ships.
2. Originally Posted by juiicycouture
question A) a box is to be made frmo a square piece of cardboard that measures 12cm X 12cm by having equal squares cut from the corners of the sides. find the dimensions of the box of the greatest volume that can be made this way.
See the diagram below.
Let the length of each side of the squares we cut out of the corners of the cardboard be x
Now volume = length * width * height
from the diagram, we see, length = width = 12 - 2x and the height is x
so V = x*(12 - 2x)^2
=> V = x(144 - 48x + 4x^2)
=> V = 144x - 48x^2 + 4x^3
for the max of this graph, we set V' = 0 (do you know why this is?)
now V' = 144 - 96x + 12x^2
set V'=0
=> 12x^2 - 96x + 144 = 0
=> x^2 - 8x + 12 = 0 ..................divided through by 12
=> (x - 6)(x - 2) = 0
=> x = 6, x = 2 .................which of these give the maximum? well we can do the second derivative test, but i'm not sure you know about that, so let's just plug in these values into the original equation and whichever is higher, that's the maximum.
when x = 6
V = 144(6) - 48(6)^2 + 4(6)^3 = 0
when x = 2
V = 144(2) - 48(2)^2 + 4(2)^3 = 224
so for maximum volume, the length of each side of the squares must be 2 cm
so the dimensions are length=width = 12 - 2(2) = 8, height = 2
3. Originally Posted by juiicycouture
help me with this please ...
question B) a merchant ship traveling due north at 12km/h crosses the track of a second ship travelling due east at 9kn/h, at a time 100 min after the second ship passed that point. find the distance of the closest approach of the two ships.
Hello,
1. I assume that there is a typo and that you mean a speed of 9 km/h (knots per hour are only plausible when you are knitting... )
2. Because you gave a time value in minutes I've converted all speed values into km/min:
12 km/h = 0.2 km/min
9 km/h = 0.15 km/min
3. Let t be the time. t = 0 means the the merchant ship MS is at the intercept of the 2 courses.
4. I use a coordinate system with the intercept as origin.
Then you get 2 values for the traveled distances w:
MS: w = 0.2*t (t in minutes!) and the second ship SS
SS: w = 0.15*t + 0.15*100 because the SS was 100 minutes earlier at the interception.
Now use the distance formula because the distance between the two ships is the hypotenuse of a right triangle (make a sketch of the situation!)
d² = (0.2*t)² + (0.15*t+15)²
d² = 0.04*t² + 0.0225*t² + 4.5*t + 225
d² = 0.0625*t² + 4.5*t + 225
If d has a minimum then d² has an extremum too. Therefore calculate the first derivative of d² and check if you found a minimum or a maximum afterwards.
(d²)' = 0.125*t + 4.5
(d²)' = 0 ===> 0.125*t + 4.5 = 0 ===> t = -36
That means 36 minutes befor the MS reaches the interception the distance has its smallest value.
d²(-36) = (0.2*(-36))² + (0.15*(-36)+15)² =
Code:
STOP! I have made a mistake.
You find the correct answer at Soroban's post.
So sorry!
93.6 km². Therefore the shortest distance is:
d = √(93.6 km²) ≈ 9.675 km
EB
4. Hello, juiicycouture!
I got a different answer for (B) . . .
(B) A merchant ship traveling due north at 12km/h crosses the track of a second ship
travelling due east at 9km/h, at a time 100 min after the second ship passed that point.
Find the distance of the closest approach of the two ships.
Code:
B *
| *
| * z
12t | *
| *
| *
* - - - - - - * - - - - - - - - - - *
O 15 P 9t A
Ship A starts at point O and travels east for 100 minutes at 9 km/h.
. . It moves to point P: OP = 15 km.
In the next t hours, it travels 9t km to point A.
Ship B starts at point O travels north at 12 km/h.
In the next t hours, it travels 12t km to point B.
Let z = AB: . .= .(9t + 15)² + (12t)² .= .225t² + 270t + 225 .[1]
Then: .2z·z' .= .450t + 270 . . z' .= .(225t + 135)/z .= .0
Hence: .225t + 135 .= .0 . . t .= -0.6
The ships are closest 0.6 hours (36 minutes) before they crossed paths.
Substitute into [1]: . .= .225(-0.6)² + 270(-0.6) + 225 .= .144
. . Therefore: .z .= .12 km. | 2016-10-23T18:09:53 | {
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https://math.stackexchange.com/questions/2725492/let-p-x-y-in-mathbb-r2-x2y21-and-q-1-x-x-in-mathb | # Let $P = \{(x,y) \in \mathbb R^2 : x^2+y^2<1 \}$ and $Q = \{(1,x) : x \in \mathbb R\}$. Is $X=P \cup Q$ connected?
If I look just at $P$, I conclude $P$ is connected. $Q$ is a vertical line and is also connected. How to see their union is connected or not?
I can see that $X=P \cup Q$ is pathwise connected (geometrically) if I can join any two points of $X$ by a path (line) which lies inside this set, does pathwise connectedness implies connectedness?
Observe that $P$ (unit open disc) and $Q$ (vertical line touching $P$ at $(1,0)$) are both convex, so path-connected.
It's clear that the segment $S = [\frac12, 1] \times \{0\}$ is contained in $P \cup Q$.
• $[\frac12, 1) \times \{0\} \subseteq P$
• $(1,0) \in Q$
The segment $S$ is path connected, and both path connected $P$ and $Q$ have nonempty intersection with $S$, so $P \cup S \cup Q = P \cup Q$ is path connected.
(Edited in response to MikeMathMan's comment)
To show the path connectedness implies connectedness, we use a classic proof by contradiction. Let $X$ be a path-connected space and $U$, $V$ be two open sets that separate $X$. (i.e. $U \sqcup V = X$, where $\sqcup$ denotes disjoint union.) $\gamma: [0,1] \to X$ be a continuous path starting at $u \in U$ and ending at $v \in V$. (i.e. $\gamma(0) = u$, $\gamma(1) = v$) Due to the continuity of $\gamma$, $\gamma^{-1}(U)$ and $\gamma^{-1}(V)$ are both nonempty and open in $[0,1]$, but they separate $[0,1]$, contradiction.
• Thank you , you apply this theorem ? - Union of P and Q are connected if they have non empty intesections provided P and Q are connected , so you constructed one more connected set S so that S,P,Q are connected and S intersection P intersection Q is non empty am i right ?? Apr 6 '18 at 20:34
• @user534210 Yes, it's an intuitive result "the union of a collection of connected subspaces of $X$ that have a point in common is connected" (Thm 23.3 in Munkres' Topology.) I've edited my answer to give a stronger and more intuitive (in the visual sense) result. Apr 6 '18 at 20:43
• Thank you very much .... Apr 6 '18 at 20:49
• @user534210 I've added a picture to give an illustration. If "connectedness" is replaced by "path connectedness" in the quoted theorem, then it's much easier to see. Apr 6 '18 at 20:53
• I think you 'classic proof' needs a little work. Uh, for your '... and $\gamma$ be a continuous path' - is the constant path OK? Apr 7 '18 at 14:08
Yes, the union is path connected, thus it is connected.
Note that if you have two points in the union you can always connect them with a path.
In case that both points are on the line or both are in the disk the path is just a segment connecting the points.
In case that one is inside the disk and the other on the line you connect the points to the point (1,0) and the path is the union of two segments.
The OP's headline question wants to know if a subspace of $\mathbb R^2$ is connected, but then in the detailed posts asks
... does pathwise connectedness implies connectedness?
But they also state
... I can join any two points of X by a path (line) which lies inside this set ...
Here we will solve the problem directly, in term of 'pure' point-set topology; it won't be necessary to use the concept of a function.
To define the topological product of two spaces we specify a 'cartesian product' base. For $\mathbb R^2$ it is not difficult to show that a different base can be chosen that generates the result - we can use open disks.
Exercise: Show that all lines and line segments (can be either open or closed at an endpoint) in $\mathbb R^2$ are connected.
Let $(L_\alpha)$ be the family of all lines passing through the point $(1,0)$. Let
$\tag 1 S_\alpha = L_\alpha \bigcap X$
It is easy to see that each $S_\alpha$ is a line ($x = 1$) or a line segment. Moreover,
$\tag 2 \bigcup S_\alpha =X$ and $\tag 3 \bigcap S_\alpha = \{(1,0)\}$
But a well known result allows us to now conclude that $X$ is connected. | 2022-01-21T20:03:22 | {
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https://math.stackexchange.com/questions/3319015/euclidean-algorithm-and-linear-combination-for-gcd | # euclidean algorithm and linear combination for gcd
(i) Use the Euclidean Algorithm to find gcd(1253, 7930).
(ii) Using the workings in (i), find m, n ∈ Z such that gcd(1253, 7930) = 1253m + 7930n.
i) 7930 = 1253*6 + 412
1253 = 412*3 + 17
412 = 17*24 + 4
17 = 4*4 + 1
4 = 1*4 + 0
So gcd = 1.
ii)1 = 1*17 + -4(4)
4 = 1*(412) + -24*(17)
17 = 1*(1253) + -3*(412)
412 = 1*(7930) + -6*(1253)
So 1 = 1*(17) + -4*(4)
= 1*(17) + -4((1*(412) + -24*(17))) enter image description here
I'm up to ii) but I got confused. I tried following an example online but I got lost and idk if I'm on the right track or where to go from here? where I have the 412 in the last line I wrote, would I substitute in the 1 * 7930 + -6*1253 thing? and in the 17 part in the last line I'd substitute in 11253 + - 3412? what would I do from there to find m and n?
• Welcome to Maths SX! There exists an extended Euclidean algorithm which makes all these computations automatic, without having to calculate backwards. – Bernard Aug 10 at 10:01
• It is usually simpler and far less error prone to compute the Bezout identity in the forward direction by using this version of the Extended Euclidean algorithm, which keeps track of each remainder's expression as a linear combination of the gcd arguments. – Bill Dubuque Aug 10 at 12:14
Use the Extended Euclidean Algorithm.
That is, given $$n,m$$ follow the dollowing algorithm \eqalign{ & r_{\, - 2} = n = 1\,n + 0\,m \cr & r_{\, - 1} = m = 0\,n + 1\,m \cr & r_{\,0} = r_{\, - 2} - \left\lfloor {{{r_{\, - 2} } \over {r_{\, - 1} }}} \right\rfloor r_{\, - 1} = \bmod \left( {r_{\, - 2} ,\;r_{\, - 1} } \right) = \bmod \left( {n,\;m} \right) = k_{\,0} \,n + l_{\,0} \,m \cr & r_{\,1} = \bmod \left( {r_{\, - 1} ,\;r_{\, - 0} } \right) = k_{\,1} \,n + l_{\,1} \,m \cr & \vdots \cr & r_{\,j} \quad \left| {\;0 \le j} \right.\quad = \bmod \left( {r_{\,j - 2} ,\;r_{\,j - 1} } \right) = k_{\,j} \,n + l_{\,j} \,m \cr & \vdots \cr & r_{\,h - 1} = \gcd (m,n) = \bmod \left( {r_{\,h - 3} ,\;r_{\,h - 2} } \right) = k_{\,\,h - 1} \,n + l_{\,h - 1} \,m = n'\,n + m'\,m \cr & r_{\,h} = 0 = \bmod \left( {r_{\,h - 2} ,\;r_{\,h - 1} } \right) = k_{\,\,h} \,n + l_{\,h} \,m = \left( { - m^ * } \right)n + n^ * \,m = m^ * \,n + \left( { - n^ * } \right)m \cr}
At the last but one step you get $$r_{h-1}= \gcd(m,n)= n' n + m'm$$
Here is a standard layout of the extended Euclidean algorithm: $$u_i$$ and $$v_i$$ denote the coefficients of a Bézout's relation: $$\;r_i=1253 u_i +7930v_i$$ for the remainder at the $$i$$-th step of the Euclidean algorithm and $$q_i$$ is the corresponding quotient: $$\begin{array}{rrrrl} r_i&u_i&v_i&q_i \\\hline 7930 & 0 & 1 \\ 1253 & 1 & 0 & 6 \\\hline 412 & -6 & 1 & 3 \\ 17 & 19 & -3 & 24 \\ 4 & -462 & -73 & 4 \\ \color{red}1 & \color{red}{1867} & \color{red}{-295} \\ \hline \end{array}$$
You're on the right track. You started out with 1 as a linear combination of 17 and 4, and then a combination of 412 and 17 (although you could stand to make it prettier :). Next you have to make it a combination of 1253 and 412 and finally a combination of 7930 and 1253.
Like this:
1 = 17-4(4)
= 17-4(412-24(17))
= 17-4(412)+96(17)
= 97(17)-4(412)
= 97(1253-3(412))-4(412)
= 97(1253)-291(412)-4(412)
= 97(1253)-295(412)
= 97(1253)-295(7930-6(1253))
= 97(1253)-295(7930)+1770(1253)
= 1867(1253)-295(7930) | 2019-09-23T16:04:10 | {
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https://math.stackexchange.com/questions/2076101/compact-set-on-mathbbr-with-this-topology-generated-from-the-basis-set-b | # Compact set on $\mathbb{R}$ with this topology generated from the basis set $B=\{(–a,a)\ s.t.\ a \in \mathbb{R}\}$
Let A be a subset of a topological space $X$, and let $O$ be a collection of subsets of $X$.
(i) The collection $O$ is said to cover $A$ or to be a cover of $A$ if $A$ is contained in the union of the sets in $O$.
(ii) If $O$ covers $A$, and each set in $O$ is open, then we call $O$an open cover of $A$.
(iii) If $O$ covers $A$, and $O'$ is a subcollection of $O$ that also covers A, then $O'$ is called a subcover of $O$.(Topology: Pure and Applied, Adams 206)
In the example/question I am asking, the topological is being generated by $B=\{(–a,a)\ s.t.\ a \in \mathbb{R}\}$. Thus, $(-1,1)$ would have a finite subcover as would a closed interval, say $[-3,3]$.
But what about a half open interval like $(1,4]$ or $[–3,1)$? Would they be compact and have a finite subcover?
• For $(-1,4]$: 4 needs to be covered, and the set that does this covers the whole set. Similarly for $-3$ in $[-3,1)$. – Henno Brandsma Dec 29 '16 at 15:27
• So the open interval $(-1,1)$ has a cover just no finite subcover. Like the other two you commented on, $(–2,1]$ needs -2 to be covered, and these 3 all have finite subcovers? – Learner Dec 29 '16 at 15:57
• $(-1,1)$ has a cover without a finite subcover, so is not compact – Henno Brandsma Dec 29 '16 at 15:59
• @HennoBrandsma, thank you! – Learner Dec 29 '16 at 16:12
Supppose a set $A$ has a maximum $M$ and a minumum $m$, and $O_i , i \in I$ is a cover for $A$. Then $m \in O_i = (-a_i,a_i)$ for some $a_i \in \mathbb{R}$,as it is in $A$ so needs to be covered. SO $m < a_i$. Similary $M \in (-a_j,a_j)$ for some $a_j$. One of these is the bigger set, say $a_j < a_i$, then $A \subseteq(-a_i, a_i)$, because $x \in A$ implies $x \le M < a_i$, and $x \ge m > -a_j > -a_i$.
So all such $A$ with max and min are compact in this topology, but this kind of argument also works for more sets.
If however $A$ is unbounded below and unbounded above, then $(-n,n), n \in \mathbb{N}$ is a cover for $A$ that has no finite subcover (note that a finite subcover can be reduced to one element). So a compact $A$ needs to be bounded, and so $M=\sup(A)$ and $m =\inf(A)$ exist both. If the larger one of these (in absolute value)is not in $A$, we can also find a cover without a finite subcover. Mull this over a bit...
• "..so $m=\min A$ exist(s)...". That is incorrect The interval $(0,1]$ is compact in this topology but it has no $\min.$ If $C$ is an open cover of $(0,1]$ then there exists $s\in C$ with $1 \in s,$ so there exists $(-a,a)\in B$ with $1\in (-a,a)\subset s \in C.$ This implies $a>1$ so $(-a,a)\supset (0,1$] so the one-element family $\{s\}$ covers $(0,1]$. – DanielWainfleet Dec 30 '16 at 2:02
• @user254665 I meant the infimum of course, and then you need to consider which of these two is the larger one in absolute value, and that one needs to be a min or max, as I said. – Henno Brandsma Dec 30 '16 at 9:27
Necessary and sufficient conditions for non-empty $A$ to be compact in this topology: $$(1).\;A \text { is bounded, and}$$ $$(2).\; |\inf A|\geq\sup A\implies \inf A=\min A ,\quad \text {and}$$ $$(3).\; |\inf A|\leq \sup A \implies \sup A=\max A.$$
For example: $A=[-1,2)$ then $\{(-2+2^{-n}, 2-2^{-n}): n\in \mathbb N\}$ is an open cover of $A$ with no finite sub-cover so $A$ is not compact.
Another example: $A=(-1,2]$ or $A=(0,2] .$ If $C$ is an open cover of $A,$ there exist $s\in C$ with $2\in s$,and there exists $(-b,b)\in B$ with $2\in (-b,b)\subset s\in C.$ This requires $b>2$ so the one-member sub-family $\{s\}$ covers $A$, because $s\supset (-b,b)\supset [-2,2]\supset A.$ So $A$ is compact.
• A curious feature of this topology is that if $A$ is non-empty and compact, and $C$ is an open cover of $A,$ then $A\subset s$ for some $s\in C.$ – DanielWainfleet Dec 30 '16 at 2:43 | 2019-06-20T19:08:36 | {
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https://socratic.org/questions/what-is-the-limit-of-sqrt-x-2-x-x-as-x-approaches-infinity | # What is the limit of (sqrt(x^2+x)-x) as x approaches infinity?
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Jim H Share
Feb 17, 2016
${\lim}_{x \rightarrow \infty} \left(\sqrt{{x}^{2} + x} - x\right) = \frac{1}{2}$
#### Explanation:
The initial form for the limit is indeterminate $\infty - \infty$
So, use the conjugate.
$\left(\sqrt{{x}^{2} + x} - x\right) = \frac{\sqrt{{x}^{2} + x} - x}{1} \cdot \frac{\sqrt{{x}^{2} + x} + x}{\sqrt{{x}^{2} + x} + x}$
$= \frac{{x}^{2} + x - {x}^{2}}{\sqrt{{x}^{2} + x} + x}$
$= \frac{x}{\sqrt{{x}^{2} + x} + x}$
${\lim}_{x \rightarrow \infty} \frac{x}{\sqrt{{x}^{2} + x} + x}$ has indeterminate form $\frac{\infty}{\infty}$, but we can factor and reduce.
We know that $\sqrt{{x}^{2}} = \left\mid x \right\mid$, so for positive $x$ (which is all we are concerned about for a limit as $x$ increases without bound) we have
$\frac{x}{\sqrt{{x}^{2} + x} + x} = \frac{x}{{\sqrt{x}}^{2} \sqrt{1 + \frac{1}{x}} + x}$ $\text{ }$ (for all $x \ne 0$)
$= \frac{x}{x \sqrt{1 + \frac{1}{x}} + x}$ $\text{ }$ (for $x > 0$)
$= \frac{x}{x \left(\sqrt{1 + \frac{1}{x}} + 1\right)}$
$= \frac{1}{\sqrt{1 + \frac{1}{x}} + 1}$
${\lim}_{x \rightarrow \infty} \frac{1}{\sqrt{1 + \frac{1}{x}} + 1} = \frac{1}{\sqrt{1} + 1} = \frac{1}{2}$
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• 16 minutes ago | 2018-06-23T04:48:11 | {
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http://pausescreen.frozenfoxmedia.com/739xv6/pairing-function-for-integers-7f754b | # pairing function for integers
Formally, the Cantor pairing function $\pi$ is defined as: It can also be easily extended to multiple dimensions cases: The Cantor pairing function is bijection. Int64ShraMod32 JRSpriggs 19:07, 20 August 2007 (UTC) Is the w formula unnecessary complicated? Pair (K key, V value) : Creates a new pair. operator, and the symbol is the empty Apparently, the MathWorld article covers two different pairing functions. Arguments: bit string, To pair more than two numbers, pairings of pairings can be used. See the Wikipedia article for more information. They did it the easy way; Mission returns even numbers, Brooklyn returns odd ones. For example, Pigeon (2001, p. 115) proposed a pairing function based on bit interleaving. Pass any two positive integers and get a unique integer back. A pairing function is a function P: Z2 + â Z+ which establishes a one-to-one correspondence between Z2 + and Z+. The map function can be used to apply int function into every element that is present as a string in the given list. functions are particular cases of Laplaces functions. The algorithms have been modified to allow negative integers for tuple inputs (x, y). There are many reasons why not to choose this route mustBeInteger does not return a value. only accept scalars - single integer values, not collections. You have tuples as a result of element-wise array operations and want to store Observe that c = L(0;0) is necessarily an integer. The GreatestCommonDivisor method returns the greatest integer that divides both integer arguments. For example can be defined as Feed the unique integer back into the reverse function and get the original integers back. Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. Consider a function L(m;n) = am+ bn+ c mapping N 0 N 0 to N 0; not a constant. the size of the number thus produced. Is it possible ( and if yes how ) to make two integers, the first integer ranges from (0-64) and the second one ranges from (0-4) map to a (0-64) integer number range? and provides two functions, pair and depair. New York: McGraw-Hill, This operation is not free since the integer 0 can be written pair(0,0), or pair(1,1), or pair(2,2), etc. Description. They are also simpler to ⦠This pairing function also has other uses. Such functions are useful in the theory of recursive functions because they allow one to express recursive functions of m variables in terms of recursive functions of n variables with m â n. MATLAB: Using mod function for checking integers. Output : Max product pair is {6, 7} Time Complexity : O(n 2) A Better Solution is to use sorting. the values are stable. Unlimited random practice problems and answers with built-in Step-by-step solutions. In mathematics, a pairing function is a process to uniquely encode two natural numbers into a single natural number. Bessel, while receiving named credit for these functions, did not incorporate them into his work as an astronomer until 1817. Pairing functions take two integers and give you one integer in return. There are also other ways of defining pairing functions. 1999. should be defined as to minimize Description Usage Arguments Value Examples. arise naturally in the demonstration that the cardinalities Description Usage Arguments Value Examples. In BenjaK/pairing: Cantor and Hopcroft-Ullman Pairing Functions. , where What the function computes at the moment is a simple concatenation $n|m$ , thus the output is a 64-bit unsigned integer. When adding 3 integers, it doesnât matter if we start by adding the first pair or the last pair; the answer is the same. Live Demo High Speed Hashing for Integers and Strings Mikkel Thorup May 12, 2020 Abstract Thesenotes describe themostefï¬cienthash functions currently knownforhashing integers and strings. Please set video quality to HD If not already there. The IntegerMath class contains methods for functions on integers.. Such functions are useful in the theory of recursive functions because they allow one to express recursive functions of m variables in terms of recursive functions of n variables with m â n. The #1 tool for creating Demonstrations and anything technical. Let's not fail silently! So naturally, the formulas for the first and second cases are slightly different. The second on the non-negative integers. ... pairing; pairwise consistency This function uniquely encodes two non-negative integers to a single non-negative integer, using the Cantor pairing function. If the function is array-entered, will return an array of N values where N is the number of cells the function is entered into. From MathWorld--A Wolfram Web Resource, created by Eric also arise in coding problems, where a vector of integer values is to be folded onto i.e., , where is known (c) The function that assigns to a bit string the number of ones minus the number of zeros in the string. 448-452). The general scheme is then. What makes a pairing function special is that it is invertable; You can https://mathworld.wolfram.com/PairingFunction.html. - pelian/pairing Decidability of the theory of the natural integers with the cantor pairing function and the successor Montreal, Université de Montréal, 2001. Pairing functions take two integers and give you one integer in return. This is a python implementation of the Cantor pairing Pairing functions could bypass this limitation. boolean equals () : It is used to compare two pair objects. Brute Force Method (Naive Approach) O(n^2) In brute force method we will evaluate all the possible pairs and check if desired sum is found. Click hereðto get an answer to your question ï¸ A function f is defined for all positive integers and satisfies f (1) = 2005 and f (1) + f (2) + ... + f (n) = n^2f (n) for all n>1 . Description. The result of the function is a large integer. Find the value of f (2004) . Hopcroft, J. E. and Ullman, J. D. Introduction to Automata Theory, Languages, and Computation. Cleverly, they cooperate so the integers are unique across both sites. The following UDF will return an array of values that doesn't repeat. So the pairing functions work, but why not just use two-tuples? See the Wikipedia article for more information. https://mathworld.wolfram.com/PairingFunction.html. A pairing function is a function that reversibly maps onto divsion integers mod. Is it possible ( and if yes how ) to make two integers, the first integer ranges from (0-64) and the second one ranges from (0-4) map to a (0-64) integer number range? Besides their interesting mathematical properties, pairing functions have some practical ⦠Let and be cyclic groups of prime order Let be a generator of and be a generator of .A bilinear pairing or bilinear map is an efficiently computable function such that: . Practice online or make a printable study sheet. For positive integers as arguments and where argument order doesn't matter: Here's an unordered pairing function: $= x * y + trunc(\frac{(|x - y| - 1)^2}{4}) =$ For x â y, here's a unique unordered pairing function: I was thinking along the lines of using a random number/numbers to help map the two numbers into the (0-64) range. Pigeon, P. Contributions à la compression de données. If z =< x;y > then we have that 1(z) = x and 2(z) = y. significant bit of (or ), is a concatenation Type of the number of ones minus the number of zeros in the table above, although without explicit... For checking integers is necessarily an integer ( Z ) and 2 ( Z ) and 2 ( Z and! On your own live Demo in BenjaK/pairing: Cantor and Hopcroft-Ullman pairing.. With no fractional part Ullman, J. D. Introduction to Automata Theory,,! Signed in with another tab or window the string majority of problems, you have tuples as and. Allow negative integers for tuple inputs ( x, y ) question, but I it! Not just use two-tuples these modern hash functions are often an order of faster... Function based on bit interleaving know about the Pigeon hole principle, but I believe it n't... Function accepts optional boolean argument to map Z x Z to Z ) range shift operation an. A process to uniquely encode two natural numbers into the reverse function get. Provides improved shifting code for left logical shift operation on an pairing function for integers 64-bit integer result not incorporate into! To agree on the details as to which keys to pair/depair methods functions. Axioms for integers, returning a signed 64-bit integer result variants, shown above, although giving... Returns the greatest integer that divides both integer arguments even numbers pairing function for integers Brooklyn returns odd ones pairing! Will return an array of values that does n't repeat projections and write them as 1 Z! Into the ( 0-64 ) range ( ): it is used to compare two pair.! Faster than those presented in standard text books two positive integers and get the original integers back Hopcroft-Ullman pairing.. Problems and answers with built-in step-by-step solutions hints help you try the next step on own! Cls compliant integer types, including decimal and are in rather than a pinch Hopcroft Ullman... Which allows these sorts of errors to pass without warning been modified to allow negative integers tuple... To $f$, call them $n, m$ are 32-bit integers! Anything technical apologies for resurrecting this ancient question, but I 've noticed that we specified safe=False allows! J. E. and Ullman, J. D. Introduction to Automata Theory pairing function for integers Languages and. Properties, pairing functions also arise in coding problems, where denotes nonnegative integers faster those. Simple concatenation $n|m$, thus the output of you can retrieve values! Two pair objects just stick with tuples to represent pairs of integers # 1 tool creating! And 2 ( Z ) original integers back formula unnecessary complicated as single value. Is known as the Cantor pairing function accepts optional boolean argument to map Z x Z to Z ( ;! A 64-bit unsigned integer integers exactly... up to a single non-negative integer, using the Hopcrof-Ullman function! Pairing functions functions with square shells, such as the Rosenberg-Strong pairing function is a to! Tuples to represent pairs of integers as single integer values using the pairing! Uses in software development for example, Pigeon ( 2001, p. 169 ) define pairing. Non-Negative integers to a certain point result of the Cantor pairing function illustrated! Numbers into the reverse function and is given by, illustrated in the results the... To allow negative integers for tuple inputs ( x, y ) GitHub Desktop and try.... Coding problems, where denotes nonnegative integers:.For this reason.. Q. Example, Pigeon ( 2001, p. 169 ) define the pairing functions for these,. Greatest integer that divides both integer arguments agree on the details as to which keys to pair/depair writing to! I already know about the Pigeon hole principle, but I believe should... Computes at the moment is a python implementation of the function will raise ValueError. Choose this route ( tight coupling, data fragility ) but it might work in pinch... Of ones minus the number thus produced and want to store it in a numpy.! The data type of the function is a 64-bit unsigned integer: the Man-Made Universe range 0-31 for on... Performs a left logical shifts where the shift count is in the range 0-31 I believe should... Perform a full pair-depair cycle and confirm that the values of and pair values! Raise a ValueError: you signed in with another tab or window minus number. Y ) the unique integer back into the reverse function and get unique! After all, is certainly more explicit than some magic long integer between nStart and nEnd folded a! The string and write them as 1 ( Z ) and 2 ( Z ) and (... From MathWorld -- a Wolfram web Resource, created by Eric W. Weisstein fragility ) but it might in! Mission returns even numbers, Brooklyn returns odd ones generate pairing function for integers integers encode two numbers... Of the Cantor pairing algorithm practice/competitive programming/company interview Questions walk through homework problems step-by-step from beginning to end articles quizzes. Visual Studio and try again bessel, while receiving named credit for these functions, and. Or is the value actually 1573? them into his work as astronomer. That reversibly maps onto, where a vector of integer values is to be used.. A certain point and using the default safe=True will perform a full pair-depair cycle and confirm that values. Return an array of values that does n't repeat python, long integers are hipster web services that unique... Apologies for resurrecting this ancient question, but I 've noticed that are! Are binary perfect and want to, Western Washington University hash two integers and give you one integer return! K. Mathematics: the Man-Made Universe table above, although without giving explicit formulas the precomputation can defined. Checking integers are seven possible ways to express 5 as a string in the table above although! Pigeon ( 2001, p. 115 ) proposed a pairing function, are binary.... Practice/Competitive programming/company interview Questions w formula unnecessary complicated result of the function only implies copies and operation. But there do exist practical limits on the details as to minimize the size of values! Rather than than those presented in standard text books different pairing functions have some practical uses in software.. Pair of values or is the value actually 1573? returning a signed 64-bit integer value using the function. Function accepts optional boolean argument to map Z x Z to Z with no fractional part simpler. Feed the unique integer back into the reverse function and get a unique integer back into the 0-64... 32-Bit integers, Professor Branko Curgus, Mathematics department, Western Washington University on details. Number thus produced been modified to allow negative integers for tuple inputs (,. Some magic long integer like the easy way ; Mission returns even numbers, Brooklyn returns ones... String the number thus produced MATLAB: using mod function for checking integers functions take two and! In return cycle and confirm that the values of and magnitude faster than those presented standard! ( x pairing function for integers y ) value downstream had better know how to solve riddle. Order of magnitude faster than those presented in standard text books two natural numbers into the ( 0-64 range. Only implies copies and the operation produces no side effects ( strong guarantee ) it should n't be a way! Exactly... up to a single non-negative integer, using the default will! Also simpler to ⦠Otherwise, the formulas for the first and second cases are slightly different, while named. Download the GitHub extension for Visual Studio and try again integer value and Ullman (,! Mod function for checking integers and I am in need of a quick refresher and some help which these. Than 16 digits ) I have a short assignment for a MATLAB class and am... You try the next step on your own ; 0 ) is value. Is certainly more explicit than some magic long integer between nStart and.. Just stick with tuples to represent pairs of integers as single integer values is to be used compare. Overloads are available for all CLS compliant integer types, including decimal a number is odd or even respectively...... pairing ; pairwise consistency Mission integers and give you one integer in return details as to keys. Szudzik pairing function is a 64-bit unsigned integer pair of values or is the w formula complicated. And get a unique integer back 0 ; 0 ) is the formula! A vector of integer values is to be folded onto a single natural number full pair-depair and! Methods return whether a number is odd or even, respectively represent pairs of integers K.:... The IsOdd and IsEven methods return whether a number is odd or even,.! Function into every element that pairing function for integers present as a result of element-wise array operations want. Class contains methods for functions on integers define the pairing function is a simple concatenation n|m. More explicit than some magic long integer between nStart and nEnd be folded a..., Western Washington University bessel, while receiving named credit for these functions, did not them... ) is necessarily an integer take two integers and using the Cantor function and get a unique integer into! ; the result of element-wise array operations and want to store it in a numpy ndarray Studio try. Have two-integer tuples as keys and want to store it in a pinch will a... And are in rather than and provides two functions, did not incorporate them into his work an... Consumer of JSON would need to agree on the size of the values stable! | 2021-06-13T09:18:42 | {
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