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https://mathoverflow.net/questions/192148/tilting-the-d-cube-to-vertically-separate-its-vertices | # Tilting the $d$-cube to vertically separate its vertices
Let $C_d$ be a unit edge-length cube in $d$ dimensions. I would like to orient it ("tilt" it) so that the vertical (last) coordinates of its $2^d$ vertices are maximally separated, in the sense that the minimum vertical distance between any two vertices is maximized over all orientations.
For $C_2$ (in standard orientation, edges parallel to Cartesian axes), tilting $\arctan \frac{1}{2} \approx 26.6^\circ$ separates the vertices by $\delta=1/\sqrt{5} \approx 0.447$:
For $C_3$ (in standard orientation), I believe that rotating the vector $(0,0,1)$ to lie on the vector $(\frac{1}{4},\frac{1}{2},1)$ results in a vertex separation of $\delta=1/\sqrt{21} \approx 0.218$:
My question is:
Q. What is the generalization to $C_d$ for $d>3$? What is the largest vertex separation $\delta$ achievable? Can one always achieve a uniform vertex separation (the same $\delta$ between each vertically adjacent pair), as in $C_2$ and $C_3$?
Robert Israel's answer is best possible.
As pointed out there, the problem is equivalent to finding a vector in $$\mathbb{R}^d$$ with minimal separation $$1$$ among sums of subsets of entries, and with the least possible length ($$\ell_2$$ norm). The vector Robert constructed, after rescaling to make the separation $$1$$, is $$v_d=[2^0,2^1,\dots,2^{d-1}]$$. Clearly this has the smallest possible $$\ell_1$$ norm.
It turns out that it also has the smallest possible $$\ell_2$$ norm, but for $$d>3$$ not the smallest $$\ell_\infty$$ norm.
Proof. It is a straightforward consequence of these 2 facts:
1. Lemma. The following equality holds in $$\mathbb{Z}[x_1,\dots x_d]$$:
$$\sum_{S,T\subseteq I_d}(x_S-x_T)^2=C_d \sum_{i=1}^d x_i^2$$
where $$I_d=\{1,2,\dots d\}$$, $$\displaystyle x_S=\sum_{i\in S} x_i$$ and the positive integer $$C_d$$ depends only on $$d$$.
1. if $$x=v_d$$ then $$\{x_S\}=\{0,1,2,\dots 2^d-1\}$$ clearly has the tightest possible packing of values (among vectors with separation $$1$$, and uniquely up to permutations) and therefore must minimize the left hand side in the lemma.
Proof of Lemma. There are 16 equally frequent cases for the occurrence of $$x_ix_j$$ in $$(x_S-x_T)^2$$:
$$\begin{array}{@{}l|l|l|c|c@{}} \mbox{case} & \in S & \in T & x_S-x_T & x_ix_j \text{ coef in }\\ & & & & (x_S-x_T)^2\\ \hline \mbox{1} & \_ & \_ & \cdots & 0 \\ \mbox{2} & i & \_ & x_i\cdots & 0 \\ \mbox{3} & j & \_ & x_j\cdots & 0 \\ \mbox{4} & i,j & \_ & x_i+x_j\cdots & 2 \\ \mbox{5} & \_ & i & -x_i\cdots & 0 \\ \mbox{6} & i & i & \cdots & 0 \\ \mbox{7} & j & i & x_j-x_i\cdots & -2 \\ \mbox{8} & i,j & i & x_j\cdots & 0 \\ \mbox{9} & \_ & j & -x_j\cdots & 0 \\ \mbox{10} & i & j & x_i-x_j\cdots & -2 \\ \mbox{11} & j & j & \cdots & 0 \\ \mbox{12} & i,j & j & x_i\cdots & 0 \\ \mbox{13} & \_ & i,j & -x_i-x_j\cdots & 2 \\ \mbox{14} & i & i,j & -x_j\cdots & 0 \\ \mbox{15} & j & i,j & -x_i\cdots & 0 \\ \mbox{16} & i,j & i,j & \cdots & 0 \end{array}$$
the total coefficient of $$x_ix_j$$ over all the cases is 0. Therefore $$\sum_{S,T\subseteq I_d}(x_S-x_T)^2$$ only contains square terms and by symmetry must then be a multiple of $$\sum_{i=1}^d x_i^2 \quad\blacksquare$$
Regarding the fact that $$v_d$$ above is not minimal $$\ell_\infty$$-wise, counterexamples are provided via the Atkinson-Negro-Santoro sequence:
$$d=4: [3,5,6,7]$$
$$d=5: [6,9,11,12,13]$$
$$d=6: [11,17,20,22,23,24]$$
etc.
• I should mention that the Atkinson-Negro-Santoro derived sequences mentioned above are by no means the best in the $\ell_\infty$ metric. For example $[39,59,70,77,78,79,81,84]$ is a better one for $d=8$. Feb 14 at 13:28
Given a unit vector $u \in \mathbb R^d$, the "heights" of vertices of the $n$-cube where $u$ is regarded as the vertical direction are the sums of subsets of the entries of $u$. Thus the minimum separation is the minimum difference between the sums of two distinct subsets of these entries. If you take $$u = [1,2,\ldots,2^{d-1}]/\sqrt{1^2 + 2^2 + \ldots (2^{d-1})^2} = \sqrt{\dfrac{3}{4^d-1}}[1,2,\ldots,2^{d-1}]$$ you get uniform separation of $\sqrt{3/(4^d-1)}$.
• Beautiful! $\mbox{}$ Jan 5 '15 at 2:01
• Is it clear that uniform separation gives the optimal separation? (if you change the vector $u$, you might in principle destroy the uniformity, but obtain separation over a (reasonably substantially) larger range: if $u=[1,1,1,\ldots,1]$ then the image of the diagonal is of length $\sqrt d$, whereas here it is $O(1)$). This gives an upper bound for the minimal separation of $\sqrt d/2^d$, whereas here you have a lower bound of approximately $\sqrt 3/2^d$. Jan 5 '15 at 5:02
• @AnthonyQuas: Good point! I can verify that for $d=2,3$, the max separation is achieved with uniform separation. Jan 5 '15 at 11:30
• (Initially I accepted this lucid answer, but as Anthony points out, it doesn't necessarily solve the original question [not that Robert ever claimed it did].) Jan 6 '15 at 12:37
Just a small additional observation: It is possible to have nonuniform separations that are locally optimal but not globally. By local I mean that small perturbations of the $$u$$ vector do not improve the minimum separation.
Here are some examples in $$d=3$$ and $$d=4$$. The leftmost solutions are Robert's uniform separation solutions. In the others, the vertical order of the corners is different. (Corners are here numbered as bit strings, so in $$d=4$$ the four neighbors of origin are 1,2,4,8. In Robert's solution the vertical order is then same as numerical order. Edges from origin to its neighbors shown in red.)
This means that one cannot characterize the optimal solution by local conditions on derivatives; the problem seems to have a "combinatorial" nature. If one first fixes the vertical order of all $$2^d$$ corners, then maximizing the separation can be formulated as a quadratic optimization problem. The remaining question is then, what is the best vertical order. But naively trying all possilities (e.g. all permutations of $$2^d$$ corners) will run into trouble with $$d>4$$, even if one applies some obvious symmetry reductions (without loss of generality, assume that corner 0 is lowest, and that its $$d$$ neighbors are in increasing numerical order).
With $$d=4$$, with some symmetry reductions, one needs to try only 252 different vertical orders (I can provide more details if needed), and the corresponding QP programs find three essentially different solutions (that are optimal for that particular vertical order). These three solutions are shown above, and the best is Robert's solution, so with $$d=4$$ we know it is the best possible.
(Additional note 27.9.2021: The number of possible vertical orders is even smaller. In fact it is A009997, and for $$d=4$$ there are just $$14$$ possible vertical orders of the corners.)
Perhaps there is an obvious geometric reason why the optimal separation has to be uniform? | 2021-12-01T22:35:52 | {
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http://math.stackexchange.com/questions/284416/how-many-possible-arrangements-for-a-round-robin-tournament | # How many possible arrangements for a round robin tournament?
How many arrangements are possible for a round robin tournament over an even number of players $n$?
A round robin tournament is a competition where $n = 2k$ players play each other once in a heads-up match (like the group stage of a FIFA World Cup). To accommodate this, there are $n-1$ rounds with $\frac{n}{2}$ games in each round. For an arrangement of a tournament, let's say that the matches within an individual round are unordered, but the rounds in the tournament are ordered. For $n$ players, how many possible arrangements of the tournament can there be?
...
I don't know if a formal statement is needed, but hey ...
Let $P = \{ p_1, \ldots, p_n \}$ be a set of an even $n$ players. Let $R$ denote a round consisting of a set of pairs $(p_i,p_j)$ (denoting a match), such that $0<i<j\leq n$, and such that each player in $P$ is mentioned precisely once in $R$. Let $T$ be a tournament consisting of a tuple of $n-1$ valid rounds $(R_1, \ldots, R_{n-1})$, such that all rounds in $T$ are pair-wise disjoint (no round shares a match).
How many valid constructions of $T$ are there for $n$ input players?
The answer for 2 players is trivially 1. The answer for 4 players is 6. I believe the answer for 6 players to be 320. But how can this be solved in the general case?
-
You can arrange matches with no overlap by arbitrarily numbering the teams $0 - (n - 1)$, then in round $i$ match teams such that the sum of ther numbers is $i$ mod $n$. You can relabel the teams in $(n-1)!$ ways and do the above, so i think this is the answer.., – gnometorule Jan 22 '13 at 18:25
Interesting ... but is this the only way to derive valid configurations? (That also means I miscounted in the 6 player case ... if $(n-1)!$ the answer would be 120 rather than 320.) – badroit Jan 22 '13 at 18:27
MatchIng using mod is a standard example when introducing mod, but not necessarily the only way of doing this. I'm also only kinda sure this is right, but it's definitely close. I'm reading these questions often on breaks on my iPhone so it's what came to my mind. :) – gnometorule Jan 22 '13 at 18:31
This is almost the definition of a "$1$-factorization of $K_{2k}$", except that a $1$-factorization has an unordered set of matchings instead of a sequence of rounds. Since there are $2k-1$ rounds, this means that there are $(2k-1)!$ times as many tournaments, according to the definition above, as there are $1$-factorizations.
Counting $1$-factorizations of $K_{2k}$ seems to be a nontrival problem; see the Encyclopedia of Mathematics entry. The number of $1$-factorizations of $K_{2k}$ is OEIS sequence A000438. Also, see this paper (also here) for a count in the $k=7$ case. | 2016-04-29T06:29:50 | {
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http://mathhelpforum.com/advanced-algebra/188517-complex-polar-forms-sin-cos-angles.html | # Thread: Complex Polar Forms, Sin and Cos Angles
1. ## Complex Polar Forms, Sin and Cos Angles
Problem: Find all solutions of $\displaystyle z^{3} = -8$.
I can do the entire problem except for one part. After putting it into the correct form,
$\displaystyle |z|^{3}(cos3\theta+isin3\theta)$,
I do not know how to find the values of $\displaystyle cos3\theta$ or $\displaystyle sin3\theta$. I know that $\displaystyle |z|^3$ is 8, but I can't figure out the values of cos and sin.
Any help is appreciated.
2. ## Re: Complex Polar Forms, Sin and Cos Angles
Originally Posted by tangibleLime
Problem: Find all solutions of $\displaystyle z^{3} = -8$.
I can do the entire problem except for one part. After putting it into the correct form,
$\displaystyle |z|^{3}(cos3\theta+isin3\theta)$,
I do not know how to find the values of $\displaystyle cos3\theta$ or $\displaystyle sin3\theta$. I know that $\displaystyle |z|^3$ is 8, but I can't figure out the values of cos and sin.
Any help is appreciated.
$\displaystyle |z| = 2$.
So, $\displaystyle cos\ 3\theta+isin\ 3\theta=-1$.
Equate real and imaginary terms to find $\displaystyle \theta$.
3. ## Re: Complex Polar Forms, Sin and Cos Angles
Thanks,
So since $\displaystyle |z|^3 = 8$, I need $\displaystyle cos3\theta+isin3\theta$ to equal -1 to satisfy the initial equation where $\displaystyle z^3 = -8$. The only way to get -1 from $\displaystyle cos3\theta+isin3\theta$ is to have $\displaystyle cos3\theta = -1$ and $\displaystyle sin3\theta = 0$ to get rid of the imaginary number and return (-1 + i0).
Correct?
4. ## Re: Complex Polar Forms, Sin and Cos Angles
Originally Posted by tangibleLime
Thanks,
So since $\displaystyle |z|^3 = 8$, I need $\displaystyle cos3\theta+isin3\theta$ to equal -1 to satisfy the initial equation where $\displaystyle z^3 = -8$. The only way to get -1 from $\displaystyle cos3\theta+isin3\theta$ is to have $\displaystyle cos3\theta = -1$ and $\displaystyle sin3\theta = 0$ to get rid of the imaginary number and return (-1 + i0).
Correct?
Yes. You should get three values of $\displaystyle \theta$ in the interval $\displaystyle [0,\ 2\pi]$.
5. ## Re: Complex Polar Forms, Sin and Cos Angles
Originally Posted by tangibleLime
Problem: Find all solutions of $\displaystyle z^{3} = -8$.
Here is some notation: $\displaystyle \exp(i\theta)=\cos{\theta)+i\sin(\theta)$
So we can write $\displaystyle -8=8\exp(\pi i)$
The cube roots of that is then $\displaystyle 2\exp \left( {\frac{{i\pi }}{3} + \frac{{2\pi ik}}{3}} \right),~~k=0,1,2$ | 2018-06-20T02:47:38 | {
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https://math.stackexchange.com/questions/1137472/how-to-count-the-ways-to-induce-a-permutation | # How to count the ways to induce a permutation?
I'm reading a recreational book about combinatorics, that discusses, in passing, the ways to 'induce' a permutation of the index set {1,2,3,4}.
The book notes that there is exactly:
• 1 way to induce a permutation by an identity
• C(4,2) = 6 ways to induce a permutation by interchanging exactly 1 pair of indices:
• 1/2 C(4,2) = 3 ways to induce a permutation by interchanging exactly 2 pairs of indices:
• 4*2 = 8 ways to induce a permutation that leaves 1 index unchanged
• 4!/4 = 6 ways to induce a permutation by a cyclic permutation of all four indices,
which exhausts all 24 permutations of a 4 element set.
Question How do the authors calculate these numbers? Is there a way to generalise this approach to larger sets, for example 6 or 20 elements, and what would be the right terminology to use?
Let $n$ be the size of your set. Effectively the question you are asking is how many elements of each cycle type are in the permutation group $S_n$. If you want a permutation consisting of $C_j$ $j$-cycles when written as the product of disjoint cycles, with $k$ being the longest cycle length, then there are $$\frac{n!}{\prod_{j=1}^k j^{C_j} C_j!} \qquad \star$$ distinct permutations of this form. If you don't like the group theory direction of my explanation, you can still check the formula works. But a little bit of group theory will go a long way in helping you understand the formula, the basic theory of $S_n$ will be in the first few chapters of any good introductory group theory book.
To go back to your example, set $n=4$.
The identity permutation is the unique element in $S_4$ that is the disjoint product of four $1$-cycles, it looks like $(1)(2)(3)(4)$, so it is a $1,1,1,1$-cycle. If you plug this cycle type description into the formula $\star$, you get $$\frac{4!}{ 1^{4}\cdot 4!} =1$$
The number of $3$-cycles ("leave $1$ index unchanged") $$\frac{4!}{ 1^{1}\cdot 1! \cdot 3^1\cdot 1!}= 8$$ because a general example of a $3$-cycle in $S_4$ is $(1\; 2\; 3)(4)$ The number of $4$-cycles ("induce a permutation by a cyclic permutation of all four indices") is $$\frac{4!}{ 4^{1} 1!}=6$$ The number of $2,2$-cycles ("induce a permutation by interchanging exactly 2 pairs of indices") is $$\frac{4!}{ 2^{2}\cdot 2!}=3$$ The number of $2$-cycles ("induce a permutation by interchanging exactly 1 pair of indices") is $$\frac{4!}{1^2\cdot 2^{1}\cdot 1!\cdot 2!}=6$$
To summarise, break your permutation down into disjoint parts, each index is written in one part of the cycle, including $1$-cycles. So $(1\; 2\; 3)(4)(5\; 6 )$ rather than $(1\; 5\; 2\; 3)(4\; 2)(5\; 6 )$ (not that these elements are the same!). You can always write a permutation as disjoint cycles. Then plug into $\star$ to get the number of permutations of that form.
• Great answer. Thank you so much. Feb 7 '15 at 22:16 | 2022-01-28T23:09:57 | {
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https://stats.stackexchange.com/questions/434829/testing-differences-in-variance-between-groups/434899#434899 | # Testing differences in variance between groups
I have a hypothesis that a particular intervention/treatment will cause more variation in participant responses to a particular question.
The intervention variable is categorical, with five different treatment groups. The response variable (the participant responses to a question) is a continuous variable.
I don't necessarily expect the means to differ, I just expect greater variation in responses in certain groups as compared to others.
Can anyone advise me on a method to test the difference in the variance of different treatment groups? To be clear, this is not for the purpose of checking the assumption of homogeneity of variance for statistical tests like ANOVA, rather I am interested in specifically if and to what extent the variation differs between the different levels of my categorical intervention/treatment variable.
I thought to run Bartlett's test for homogeneity of variance but (in R at least) I only get an output that shows me whether the variances are homogeneous or not - it doesn't tell me where these differences lie between the different levels of the categorical variable (i.e. is it between group 1 and 5, or group 2 and 3 etc).
I thought also to try to calculate a coefficient of variation for each group and compare these, but I was not sure of a method of how to do this statistically ...
I am probably missing something very obvious, but I cannot find information on how to proceed. Any advice would be much appreciated.
• You could calculate the variance in each treatment group. Then you get the point estimates (of the variances). Where you see the largest differences, those should be driving the Bartlett. Nov 6 '19 at 13:25
This is an interesting question! Post-hoc tests of variances after a test of unequal variances do not seem to be a much studied topic, I was not able to find any published papers. One similar question with some ideas is Post-hoc test to determine difference in variance. Another approach is the following.
But first, note that tests of variances are typically not very robust, so distributional assumptions do matter. Maybe you could include a plot of your data in the post. I will illustrate my methods with some simulated, normal data, if your data is far from normal be careful.
One test of variances which is somewhat robust is Levene's test. As it is based on an analysis of variance, but using absolute residuals as response, it is useful in this situation, we can construct Tukey HSD intervals based on those absolute residuals. The Levene's test is often nowadays used with median as location estimator in place of the mean, but here I will use the mean for illustration. The code for simulation the example data is at the end.
with(mydf, lawstat::levene.test(X, Group, location="mean"))
oneway.test(absres ~ Group, data=mydf, var.equal=TRUE)
Classical Levene's test based on the absolute deviations from the mean
( none not applied because the location is not set to median )
data: X
Test Statistic = 4.2954, p-value = 0.003079
One-way analysis of means
data: absres and Group
F = 4.2954, num df = 4, denom df = 95, p-value = 0.003079
showing the equivalence. Then the post-hoc test:
TukeyHSD(aov(absres ~ Group, data=mydf))
Tukey multiple comparisons of means
95% family-wise confidence level
Fit: aov(formula = absres ~ Group, data = mydf)
$Group diff lwr upr p adj B-A 0.9578979 -0.87239851 2.788194 0.5937356 C-A 1.9853491 0.15505269 3.815646 0.0265586 D-A 1.8196928 -0.01060365 3.649989 0.0521132 E-A 2.4268567 0.59656026 4.257153 0.0033957 C-B 1.0274512 -0.80284522 2.857748 0.5258877 D-B 0.8617949 -0.96850157 2.692091 0.6860887 E-B 1.4689588 -0.36133766 3.299255 0.1770429 D-C -0.1656563 -1.99595277 1.664640 0.9990998 E-C 0.4415076 -1.38878886 2.271804 0.9622041 E-D 0.6071639 -1.22313251 2.437460 0.8875472 The same approach is used here, but with medians in place of means. Code for simulating the data: set.seed(7*11*13)# My public seed N <- 20; k <- 5 Group <- rep(LETTERS[1:k], rep(N, k)) X <- rnorm(k*N, rep(rnorm(k, 10, 1), rep(N, k)), rep(2*sqrt(1:k), rep(N, k))) mydf <- data.frame(Group=factor(Group), X) rm(X, Group) mydf$absres <- abs(resid(lm(X ~ Group, data=mydf)))
This could at least serve as a starting point.
With unequal variances, it could also be interesting to ask about reasons for it, specifically if the treatment could have anything to do with it. If so, this post could be of interest.
• Thank you so much for this very detailed response! It is incredibly useful. Can I ask why one might choose to base the residuals on the median vs the mean (or vice versa)? Would one be better suited to skewed data? This may end up all being hypothetical in my case as my data are extremely skewed, but it still would be good to know! Thank you again! Nov 6 '19 at 19:15
• If distributions in each group is close to normal, mean should be best. Else use median, or maybe trimmed mean. Maybe deserves better investigations. Nov 6 '19 at 19:17
• Thank you. Median would seem sensible in my case then! I wonder if instead of performing an ANOVA/Tukey HSD analysis on the residual data I could instead use a non-parametric Kruskal-Wallis with the Dunn test as a post hoc? This would make the analysis rank based and so deal better with skew. Although I am not sure how much skew even rank-based tests can deal with and still be reliable... Nov 6 '19 at 19:21 | 2021-10-17T22:00:25 | {
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https://mathematica.stackexchange.com/questions/66086/developing-a-function-of-two-variables-from-given-data | # Developing a function of two variables from given data
(I believe Mathematica SE is an appropriate place to ask this, as well.)
I have been stuck on the following problem.
Consider a system where we have three variables: force $F$, orientation $\theta$, and temperature $T$. I want to find a function for force such that $F=f(\theta,T)$.
In this system, we choose some orientation $\theta$ and temperature $T$, and from these values we want to determine the force $F$ yielded.
We only know the force values corresponding to $\theta=$ 10, 20, 30, 40, 50 or 60 and $T=$ 200, 182, etc., as shown in the table below.
Table displaying forces $F$:
(Note: I tried to turn this into an array in $\TeX$ but I couldn't get it to work, so here's a screenshot. I realize that Mathematica users will want the data written out for copy/paste purposes--if you can help me create an array that works, I will edit this an use that array. Thanks)
Example: For $\theta=30$ and $T=164$, we found that the force $F=31.87$.
Now here's my question: How do I find a function for $F$ that can handle "in between" values for both $\theta$ and $T$?
Here's what I mean: I would like a function $F=f(\theta,T)$ for values like $\theta=34$ and $T=195$, or any values $\theta$ and $T$ that are not on the table above. However, I only need this function for the interval $\theta\in[10,60]$ and $T\in[101,200]$. Essentially, I am trying to approximate $F$ for all $\theta\in[10,60]$ and $T\in[101,200]$.
I should note that I enjoy working on problems like these, so if somebody knows the answer, please try giving hints instead of solutions so that I can figure this out myself.
I have made little progress so far. I observed that the relationship between $F$ and $\theta$ is linear, while the relationship between $F$ and $T$ is nonlinear. The latter nonlinear relationship appears to be best approximated with a 2nd order polynomial. So far, I have six different functions (polynomial fit lines) for the six different $\theta$. Any hints on how to proceed? Not really sure which field of mathematics this falls under. Please let me know if you'd like me to elaborate some more on my attempts to solve this, and I'd be happy to share.
Thank you.
• It would help if you provided a link to a XLS or CSV file with the data, using some Cloud storage (OneDrive, Dropbox, etc) – Aisamu Nov 19 '14 at 19:29
### Update, as per new requests:
Importing from xls files:
s = Part[Import["~/Downloads/s.xlsx"], 1, 1]
(* Using @ubpdqn much nicer data massaging *)
table = Join @@ MapThread[Join[{#1}, {#2}] &, {Outer[List, s, t], data}, 2];
For the part 1) of your request: intFun is your $f(\theta, T)$, and can be used like this:
intFun = Interpolation[table];
intFun[30, 180]
30.94
For the part 2) of your request: the table of all force $F$ for the specified $\theta$ and $T$, with 0.5 and 1 increments, respectively.
finalTable = Table[Table[intFun[s1, t1], {t1, Max@t, Min@t, -1}], {s1, Min@s, Max@s, 0.5}]
{{3.80765,3.79107,3.77423,3.75713,3.73976,3.72211,3.70417,3.68594,3.66742,3.64858,3.62942,...105...,0.588854,0.577435,0.559586,0.5445,0.524866,0.493372,0.442709,0.410001,0.392214,0.386318,0.389277},...129...,{...1...}}
Using some made up data with the trends you mentioned:
data = Table[Table[{{o, t}, o + 3 t/120 + RandomReal[{-1, 1}]}, {t, 200, 80, -20}], {o, 10, 60, 10}];
You can ask Mathematica to do all the hard work for you, and interpolate the data using splines or Hermite interpolation
int = Interpolation[Flatten[data, 1]]
InterpolatingFunction[{{10., 60.}, {80., 200.}}, <>]
int is your $f(\theta, T)$, and can be used like this:
Plot[int[35, x], {x, 0, 300}]
Plot[int[x, 110], {x, 0, 100}]
You can also extrapolate values (as above), but it warns you that it is doing so, and the first plot shows exactly why.
If you have some good guesses about the underlying model, you can always try to fit the parameters. In your case, $a + b *\theta + c *T + d *t^2$
(* Massaging data back to triples *)
data2 = Flatten /@ Flatten[data, 1]
(* A model of the form a + b*o + c*t + d*t^2 *)
model = Fit[data2, {1, o, t, t^2}, {o, t}]
0.941482 + 1.00358 o + 0.00545877 t + 0.0000743324 t^2
This model can be used as:
model./{o->value1, t->value2}
Plot[{model /. o -> 30, model /. o -> 31}, {t, 80, 200}, Epilog -> Point@Cases[data2, {30, t_, f_} :> {t, f}]]
Blue for $\theta=30$, along with the corresponding data points
Orange for $\theta=31$
• Thank you! The update version works as I hoped. How can I use "Fit" to get my final polynomial function from the data? (You showed me how to do it with the old data but I am struggling to get this to work for the new data.) Thanks! – Patrick Shambayati Nov 24 '14 at 21:12
• It works just as before! But remember that the table is already flattened, so you must use data2 = Flatten /@ table. You must also change the variable names in the model, since t is already being used: tt works fine, for example! – Aisamu Nov 25 '14 at 2:20
• Thanks! I appreciate your help. – Patrick Shambayati Nov 25 '14 at 2:31
• You are welcome! – Aisamu Nov 25 '14 at 15:30
Using Mathematica version 10 new Predict Function
t = {200, 182, 164, 146, 128, 110, 101, 92, 83};
s = {10, 20, 30, 40, 50, 60};
data = {{10.24, 10.15, 10.01, 9.81, 9.39, 8.8, 8.57, 7.89, 7.23},
{21.50, 21.52, 21.25, 20.88, 20.79, 20.66, 20.37, 19.98, 19.50},
{31.92, 32.09, 31.87, 31.58, 31.31, 30.99, 30.86, 30.87, 30.41},
{43.56, 43.88, 43.63, 43.29, 43.02, 42.57, 42.16, 42.52, 42.25},
{54.85, 55.28, 54.98, 54.57, 54.36, 54.07, 53.78, 54.03, 54.12},
{64.45, 65.01, 64.78, 64.46, 64.36, 64.20, 63.94, 64.10, 64.54}};
vals = Flatten[Table[{s[[i]], t[[j]], data[[i, j]]}, {i, Length@s}, {j, Length@t}],1]
ListPlot3D[vals, Mesh -> All]
It is a plane, we can use linear regression as the method for the Predict Function
Set up the training set now and get the predictor function.
trainingSet = Flatten[Table[Rule[{s[[i]], t[[j]]}, data[[i, j]]], {i, Length@s}, {j,
Length@t}], 1];
pf = Predict[trainingSet, Method -> "LinearRegression"];
Show[Plot3D[pf[{x, y}], {x, Min@s, Max@s}, {y, Min@t, Max@t}],
ListPointPlot3D[vals, PlotStyle -> {PointSize -> Large}]]
Let's get the function now.
PredictorInformation[pf, "Function"]
(*-3.62101 + 1.11233 #1 + 0.0138794 #2 &*)
Please read the documentation on Predict, PredictorFunction and PredictorMeasurements.
Some cool functionality.
PredictorMeasurements[pf, testSet, "ComparisonPlot"]
UPDATE
With the new dataset provided via Dropbox, the use of Predict is not the right approach.
First Step is to import the data
data = Import[
"https://www.dropbox.com/s/v3c8kngvohgf62m/data.xlsx?dl=1", {"Data",
1}];
s = Import[
"https://dropbox.com/s/30hx04aszlafgk7/s.xlsx?dl=1", {"Data",
1}][[1]];
t = Import[
"https://dropbox.com/s/ijkyuley8q3830p/t.xlsx?dl=1", {"Data",
1}][[1]];
Second step is to visualize the information. Check the data in the 2 different axes.
vals = Table[{s[[i]], t[[j]], data[[i, j]]}, {i, Length@s}, {j,
Length@t}];
Multicolumn[
ListPlot[vals[[#, All, 2 ;;]], Joined -> True,
PlotMarkers -> Graphics[{Red, PointSize[Medium], Point[{0, 0}]}],
PlotLabel ->
Style[StringJoin["\[Theta]=", ToString[s[[#]]]], Bold]] & /@
Range@Length@s, {Automatic, Automatic}, Appearance -> "Horizontal"]
Multicolumn[(ListPlot[vals[[All, #1, {1, 3}]], Joined -> True,
PlotMarkers ->
Graphics[{Red, PointSize[Medium], Point[{0, 0}]}],
PlotLabel -> Style["T=" <> ToString[t[[#1]]], Bold]] &) /@
Range[Length[t]], {Automatic, Automatic},
Appearance -> "Horizontal"]]
trainingSet =
Flatten[Table[{{s[[i]], t[[j]]}, data[[i, j]]}, {i, Length@s}, {j,
Length@t}], 1];
pf = Interpolation[trainingSet]
Show[Plot3D[pf[x, y], {x, Min@s, Max@s}, {y, Min@t, Max@t}],
ListPointPlot3D[vals, PlotStyle -> {PointSize -> Large}]]
Check one axes
Multicolumn[
Plot[pf[#, x], {x, Min@t, Max@t}, PlotRangePadding -> 5,
Epilog -> {Red, PointSize[Large],
Point[Flatten[Cases[vals, {{#, _, _} ..}], 1][[All, 2 ;;]]]},
PlotLabel ->
Style[StringJoin["\[Theta] =", ToString[#]], Bold]] & /@ s, {2,
Automatic}, Frame -> All, Appearance -> Vertical]
• Thank you for your help, but I am new with Mathematica and I have some problems: – Patrick Shambayati Nov 20 '14 at 16:07
• For the first part of the code, I have problems with nested lists or something. I need to import the data instead of manually enter it, since the data will change over time. I cannot create the "vals" variable, and am getting an error: " ... must be a valid array or a list of valid arrays". I tried using 2 levels of Flatten but it did not work. Could you revise your example to work for imported data? I will attach links to the .xlsx documents – Patrick Shambayati Nov 20 '14 at 16:10
• I added a 100 pt. for added incentive. – Patrick Shambayati Nov 21 '14 at 23:47
Setup:
data = {{10.24, 10.15, 10.01, 9.81, 9.39, 8.8, 8.57, 7.89,
7.23}, {21.50, 21.52, 21.25, 20.88, 20.79, 20.66, 20.37, 19.98,
19.50}, {31.92, 32.09, 31.87, 31.58, 31.31, 30.99, 30.86, 30.87,
30.41}, {43.56, 43.88, 43.63, 43.29, 43.02, 42.57, 42.16, 42.52,
42.25}, {54.85, 55.28, 54.98, 54.57, 54.36, 54.07, 53.78, 54.03,
54.12}, {64.45, 65.01, 64.78, 64.46, 64.36, 64.20, 63.94, 64.10,
64.54}};
t = {200, 182, 164, 146, 128, 110, 101, 92, 83};
s = {10, 20, 30, 40, 50, 60}; pts =
MapThread[Join[{#1}, {#2}] &, {Outer[List, s, t], data}, 2];
pt = MapThread[Join[#1, {#2}] &, {Outer[List, s, t], data}, 2];
You can use Interpolation:
if = Interpolation[Join @@ pts];
int = Show[
Plot3D[if[x, y], {x, 10, 60}, {y, 83, 200}, PlotStyle -> LightPink,
Mesh -> None],
Graphics3D[{Red, PointSize[0.02], Point[Join @@ pt]}],
ImageSize -> 200];
or LinearModelFit (similar to Predict as per PatoCriollo):
lr = LinearModelFit[Join @@ pt, {1, x, y}, {x, y}];
reg = Show[
Plot3D[lr[x, y], {x, 0, 60}, {y, 80, 200}, Mesh -> False,
PlotStyle -> LightPink],
Graphics3D[{Red, PointSize[0.02], Point[Join @@ pt]}],
ImageSize -> 200];
`
The left plot below is interpolated surface, the right linear regression plane: | 2020-05-26T12:38:39 | {
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https://www.physicsforums.com/threads/digit-5-repetition.166101/ | # Digit 5 repetition
1. Apr 17, 2007
### f(x)
1. The problem statement, all variables and given/known data
How many times does digit 5 occur in numbers fro 0-1000.
2. Relevant equations
3. The attempt at a solution
This is what i have done.
Total (1,2,3) digit numbers which have digit 5 occuring once in them are-:
$3.^9C_1.^9C_1 = 243$
Total numbers with 5 occuring twice are $^9C_1 = 9$
So, digit 5 occurs 9*2=18 times
Total Numbes with 5 occuring thrice= 1
So digit 5 occurs 3*1=3
Total times digit 5 occurs is 243+18+1=264
Is this correct, specially regarding repetitions or exclusions which i may have made?
2. Apr 17, 2007
### HallsofIvy
Staff Emeritus
I would have done this quite differently. first, there is xy5 in every 10 numbers and there are 100 sets of 10 in 1000: 100 such numbers. In addition, there is 5x 9 times (not counting 55) in every 100, and 10 hundreds in 100, so 90 more Finally, every number from 500 to 599 has a 5. leaving out those of the form 5x5 and 55x, that's an additional 81. That makes a total of 100+ 90+ 81= 271 numbers between 1 and 1000 that have at least one digit 5.
3. Apr 17, 2007
### drpizza
"Total numbers with 5 occuring twice are $^9C_1 = 9$
So, digit 5 occurs 9*2=18 times"
I think you missed a few numbers on this one...
55, 155, 255, 355, 455, 655, 755, 855, 955
That's 9 numbers with two 5's. But also:
505,515,525,535,545,565,575,585,595
And:
550, 551, 552, 553, 554, 556, 557, 558, 559
Thus, there are 3 locations that the non-5 can be, thus there are
$^9C_1 *3= 9 *3$ numbers with two 5's,
Resulting in a total of 27*2 = 54 5's.
Other than that, I probably would have initially used HallsofIvy's approach as well. However, after reviewing your approach, I think I like it better. (With your approach, your solution would match HallofIvy's solution for the number of distinct numbers with a 5 in it, rather than the number of times "5" occurs:
$3.^9C_1.^9C_1 = 243$
$+3.^9C_1 = 27$
$+1$
But, since you're counting the 5's, you have to multiply that 2nd line by 2 and the 3rd line by 3. Nice.
Last edited: Apr 17, 2007
4. Apr 18, 2007
### CarlB
It seems like it can be read two different ways. If you just want to count the number of 5s in the numbers between 0 and 999, then you may as well write the numbers with three digits, that is, from 000 to 999, and then it should be clear that the three digits are independent. This is easy.
More difficult is to count the numbers that have any 5 in them. To do this, you might try instead counting the number of three digit numbers that don't have any 5 in them, and subtract that from 1000. | 2017-12-18T07:53:53 | {
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http://math.stackexchange.com/questions/722650/what-does-order-matters-regarding-permutations-refer-to?answertab=votes | # What does “order matters” regarding permutations refer to?
I psychoanalyze EVERYTHING and permutations/combinations are frustrating me. Sorry for posting so many questions lately but I really appreciate all of the help!
Ok so I know the permutation formula: $\frac{n!}{(n-r)!}$ and combination formula: $\frac{n!}{(n-r)!r!}$
I don't understand how to be certain if a question has a permutation or combination answer. I seriously can convince myself that both make sense.. I look at book examples and I don't understand!
I know that in general.. permutations are larger than combinations.. order matters with permutations but NOT combinations. I try using this knowledge after reading a question but never know for certain. Again, I look at book examples and this permutation example has me confused:
Suppose that a saleswoman has to visit eight different cities. She must begin her trip in a specified city, but she can visit the other seven cities in any order she wishes. How many possible orders can the saleswoman use when visiting these cities?
I get how the solution says 7! because there are 8 cities, the first city is where she starts.. so 8-1=7 obviously. But if the order of those other 7 cities don't matter.. wouldn't those 7 be a combination?
Also.. Idk how the formula would apply. I thought n would be 8 since there are 8 cities.. and r would be 7 since there are 7 more cities to travel to. But clearly that isn't correct.
Thanks
-
It would be silly to ask "How many possible orders" and at the same time insist that order does not matter, so you may safely assume that order does matter here. If really you wanted to count the number of ways to visit those cities if order does not matter, then there is exactly one solution; indeed there is one combination of $7$ out of $7$ candidates that can be chosen. – Marc van Leeuwen Mar 22 '14 at 22:56
She has to begin her trip in a "specified city", meaning the initial value is the same. (The first city will not be rearranged into another.) – Don Larynx Dec 22 '14 at 19:48
Note that the question says any order she wishes, not that the order does not matter. Hence, we are looking at a permutation (the specific ordering of the cities make up her route).
And note that she starts in a specified city, i.e. there is no decision here. After that there are $8-1=7$ cities to visit, which can be visited in a certain order in $7!$ ways.
I find it difficult to give a more general answer about when to use permutations vs combinations other than what you already seem to know, that it depends on whether order matters or not. However, if you have specific questions, I will be glad to help.
-
Ok thanks! So if a question says order doesn't matter.. its 100% a combination equation? I posted a question a few days ago and someone kind of clarified the confusion, but I'd love to hear your feedback if you don't mind: math.stackexchange.com/questions/718899/… – Cozen Mar 22 '14 at 22:48
@Cozen Yes, that is true. And to comment on your link, when selecting 6 animals, we only care about what the selection will be, not in the actual order. If someone asks you about the six animals you've selected, he/she won't care if the zebras or the elephants were selected first. – naslundx Mar 22 '14 at 22:52
Ok thanks! Could you possibly provide hints for those on my other link? That other user didn't completely explain to my understanding. – Cozen Mar 22 '14 at 23:00
@Cozen Edited answer to explain why we use $7!$ instead of $8!$ here. – naslundx Mar 22 '14 at 23:00
@Cozen I will try :) – naslundx Mar 22 '14 at 23:00
If there is any hint that the answer depends on the order in which the objects are counted, then you use permutations, else combinations are good. This can be remembered by defining as follows:
Definition: A Permutation is an ordered Combination.
-
My advice is that you should focus on the nature of the outcomes you are enumerating, rather than specific keywords in the language of the problem.
In the saleswoman question, ask yourself if it matters if the cities she visits are $A, B, C, D, E, F, G, H$ versus, say, $A, H, G, F, B, D, E, C$. If you are to interpret these as different outcomes, then you are looking at a permutation.
In the following example:
There is a bag of 9 marbles. 2 are red, 3 are blue, and 4 are green. How many ways are there to select a subset of 6 marbles from the bag such that there is at least one each of the three colors?
Does it matter if the marbles chosen are, say, $(r, r, g, g, b, b)$ versus $(r, g, b, g, r, b)$? Then a permutation on the individual outcomes is not applicable. But note that this question is a bit more complicated than a simple binomial coefficient computation, too. I mention it because counting methods ultimately rely not just on the question of "is it a permutation or combination" but other considerations as well.
- | 2015-04-19T03:53:08 | {
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https://math.stackexchange.com/questions/3093095/is-multiplication-as-an-operation-available-in-groups-rings-and-fields-over-z-p | # Is multiplication as an operation available in groups, rings and fields over Z_p*?
I've seen groups, rings, and fields described with a multiplication operation as well as a group defined as only having addition and subtraction (via inverse) operations. Is the reason the answer varies with respect to a group having or not having a multiplication operation dependent upon the type of numbers represented (e.g. integers, reals, etc) as well as the elements included (e.g. 0 is not in Z_p* because it doesn't have an inverse? Or do groups never have a multiplication operation?
• Binary operations are abstract. They can be whatever you want (or need). – Randall Jan 30 at 4:15
• Yes, $\mathbb Z_{11}^*$ with multiplication is a group. – J. W. Tanner Jan 30 at 4:38
• Yes, $\mathbb Z_{11}$ is a field with multiplication distributive over addition, and every non-zero element has a multiplicative inverse, as required for a field – J. W. Tanner Jan 30 at 5:01
• Here, we are using $\mathbb Z_{11}$ to mean $\mathbb Z/ (11 \mathbb Z ),$ i.e., integers modulo $11$. – J. W. Tanner Jan 30 at 5:09
• $\mathbb Z_{11}$ could mean p-adic integers – J. W. Tanner Jan 30 at 5:16
By definition of group there is only one binary operation required to have certain properties (associativity, existence of an identity element, inverses).
However there is a convention to write the group operation as addition (+) if the operation is commutative (we say the group is Abelian), and more generally when the group is not commutative (or we don't know) to write the group operation as multiplication.
This is only a convention. The group axioms for the binary operation will work with any symbol for it, so if it helps to think of it as multiplication, you are not wrong. In one important family of examples the group elements are symmetries or (stated another way) mappings that preserve a set of things (e.g. permutations), and in those cases the "multiplication" is actually composition of functions, symbolized by $$\circ$$.
There are many algebraic structures which have a group operation connected to them. Vector spaces, for example, have a commutative group operation called vector addition. A division ring $$\langle D,+,* \rangle$$ has a commutative addition and a (possibly) noncommutative multiplication such that the nonzero elements have inverses and thus form a (possibly) noncommutative group (so the case of a division ring with commutative multiplication is a field).
But we should also note that many times we use "multiplication" to mean a binary operation that does not have all the nice properties of a group operation. If we drop the requirement of an identity and inverses, and keep only the associative property and "closure" (that the result of the binary operation is defined), that sort of algebraic structure is called a semigroup. The multiplication of an arbitrary ring forms a semigroup. By dropping associativity (leaving only the closure property), one defines a magma.
These may seem awfully abstract ideas, but "strange" binary operations often arise from the study of more familiar ones. For instance, when the entries of a square $$n\times n$$ matrix are taken from a ring, we can define matrix multiplication. But even when the ring is a field, the matrix multiplication so defined will in general only give us a semigroup (or, if the ring is assumed to have a multiplicative unit, the matrices form a monoid, i.e. a semigroup with an identity element). So one is led to these generalizations (and specializations) by natural applications.
• What would be examples, other than straight ahead addition and multiplication over ℤ/(11ℤ), that is commutative and thus uses the addition (+) operation and one that is not commutative and uses the multiplication (x) operation? – JohnGalt Jan 30 at 17:16
• This might be bad form but many thanks for adding those examples! I especially like the division ring example, because it shows that rings can also have division with the keystone being noncommutative multiplication with nonzero element inverses. And I think the examples would be worth adding to the answer. – JohnGalt Jan 30 at 20:42
• @JohnGalt: Done! – hardmath Jan 30 at 22:59
Assuming $$\mathbb Z_p=\mathbb Z/p\mathbb Z$$ (not the $$p$$-adic integers), and $$\mathbb Z_p^*$$ is the group of units in $$\mathbb Z_p$$, then multiplication exists as a valid operation and $$(\mathbb Z_p^*,\times)$$ is a group. Recall that the definition of a group is a set $$G$$ together with an operation $$\oplus$$ such that $$(G,\oplus)$$ satisfies the group axioms. We in particular need associativity and inverses. We have associativity, because the usual $$\times$$ over $$\mathbb Z$$ is associative, and the map taking an integer $$x$$ to the residue class $$[x]$$ preserves this algebraic property. The similar argument works for inverses. We conclude that $$(\mathbb Z_p^*,\times)$$ is a group. | 2019-08-17T11:09:01 | {
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https://www.quizover.com/physics1/terms/order-of-magnitude-the-scope-and-scale-of-physics-by-openstax | # 1.1 The scope and scale of physics (Page 4/12)
Page 4 / 12
## Order of magnitude
The order of magnitude of a number is the power of 10 that most closely approximates it. Thus, the order of magnitude refers to the scale (or size) of a value. Each power of 10 represents a different order of magnitude. For example, ${10}^{1},{10}^{2},{10}^{3},$ and so forth, are all different orders of magnitude, as are ${10}^{0}=1,{10}^{-1},{10}^{-2},$ and ${10}^{-3}.$ To find the order of magnitude of a number, take the base-10 logarithm of the number and round it to the nearest integer, then the order of magnitude of the number is simply the resulting power of 10. For example, the order of magnitude of 800 is 10 3 because ${\text{log}}_{10}800\approx 2.903,$ which rounds to 3. Similarly, the order of magnitude of 450 is 10 3 because ${\text{log}}_{10}450\approx 2.653,$ which rounds to 3 as well. Thus, we say the numbers 800 and 450 are of the same order of magnitude: 10 3 . However, the order of magnitude of 250 is 10 2 because ${\text{log}}_{10}250\approx 2.397,$ which rounds to 2.
An equivalent but quicker way to find the order of magnitude of a number is first to write it in scientific notation and then check to see whether the first factor is greater than or less than $\sqrt{10}={10}^{0.5}\approx 3.$ The idea is that $\sqrt{10}={10}^{0.5}$ is halfway between $1={10}^{0}$ and $10={10}^{1}$ on a log base-10 scale. Thus, if the first factor is less than $\sqrt{10},$ then we round it down to 1 and the order of magnitude is simply whatever power of 10 is required to write the number in scientific notation. On the other hand, if the first factor is greater than $\sqrt{10},$ then we round it up to 10 and the order of magnitude is one power of 10 higher than the power needed to write the number in scientific notation. For example, the number 800 can be written in scientific notation as $8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{2}.$ Because 8 is bigger than $\sqrt{10}\approx 3,$ we say the order of magnitude of 800 is ${10}^{2+1}={10}^{3}.$ The number 450 can be written as $4.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{2},$ so its order of magnitude is also 10 3 because 4.5 is greater than 3. However, 250 written in scientific notation is $2.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{2}$ and 2.5 is less than 3, so its order of magnitude is ${10}^{2}.$
The order of magnitude of a number is designed to be a ballpark estimate for the scale (or size) of its value. It is simply a way of rounding numbers consistently to the nearest power of 10. This makes doing rough mental math with very big and very small numbers easier. For example, the diameter of a hydrogen atom is on the order of 10 −10 m, whereas the diameter of the Sun is on the order of 10 9 m, so it would take roughly ${10}^{9}\text{/}{10}^{-10}={10}^{19}$ hydrogen atoms to stretch across the diameter of the Sun. This is much easier to do in your head than using the more precise values of $1.06\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}\text{m}$ for a hydrogen atom diameter and $1.39\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{9}\text{m}$ for the Sun’s diameter, to find that it would take $1.31\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{19}$ hydrogen atoms to stretch across the Sun’s diameter. In addition to being easier, the rough estimate is also nearly as informative as the precise calculation.
## Known ranges of length, mass, and time
The vastness of the universe and the breadth over which physics applies are illustrated by the wide range of examples of known lengths, masses, and times (given as orders of magnitude) in [link] . Examining this table will give you a feeling for the range of possible topics in physics and numerical values. A good way to appreciate the vastness of the ranges of values in [link] is to try to answer some simple comparative questions, such as the following:
#### Questions & Answers
What is action of point ?
Bishal Reply
a point . is the spot and the action is what u do when ur at the spot . but the action of a point idk a divider
Darth
Quantity which are used in physics simply
Sangram Reply
That's philosophical question.
Jan
What is the Physical quantity
Raja Reply
What is Centripetal force
Taiwo Reply
a force of attraction that tends to keep a body moving in a circular path
Mustapha
force pulling the particle towards the center when moving in circular path
juny
yes
Mustapha
good job
Pranshu
y oxygen is shift to first place
Radhika Reply
What is coriolis firce
Shakeel Reply
Angular accelaration force,as the result of the rotatation of earth
Arzoodan
Due to the rotation of the earth the winds at the equator get deflected in opposite direction and therefore cause some currents in the Northern and Southern hemisphere which are opposite in there spin.
Varsha
Thats correct.
Arzoodan
if cos = x/y then sec = y/x if sin = y/x then cosec = x/y
Jan Reply
if cosβ=x/y then what is cosecβ
Abubakar Reply
. if cosβ=x/y what is cosecβ
Abubakar
if cosΦ=x/y what is cosecΦ
Abubakar
What are the unknown symbols?
Jan
β
Jan
Φ
Jan
It is defined that cosec = 1/sin and sec = 1/cos
Jan
Do you understand this?
Jan
of cours
Arzoodan
∇(f/g) = (g∇f − f∇g)/g^2 , at points x where g(x) 6= 0 please help me to solve the problem ....
11 Reply
sir please add answer sheet of spigel vector analysis ..
11 Reply
In a pulley system 2 boxes r hanging in both sides of pulley. An other box was joined to the left box through a rope & get accelerated downward. If all the boxes have same mass ,then what will be the acceleration of that system ?
bibek Reply
it may be mg
Amalesh
∇(f/g) = (g∇f − f∇g)/g2 , at points x where g(x) 6= 0 plus solve it
11
What is momentum
Rika Reply
is the product of mass to it's velocity (mv)
Ahmed
momentum
oladipupo
quatity of motion present in a body or product of mass and velocity
Shakeel
show that the kE of a uniform ring of mass m rolling along a smooth horizontal surface so that its centre of mass has a velocity v is mv×v
folder
what is hydration energy
osobase Reply
the energy................................................
Rika
what is momentum
Ogwu Reply
dont know want to know the answer
Rika
it's the product of mass multiplied by velocity
harsha
the quantity of motion possessed by a body is called its momentum.by virtue of which a body can exert a force in the agency which tend to stop it .it is a common experience that stronger force is required to stop more massive body.also faster the body moves harder it is to stop it .this is why
Manoj
momentum is product of mass and velocity and it is denoted by "P".
Manoj
show that the cross product of vector axb=-bxa
Teklu Reply
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https://www.physicsforums.com/threads/the-order-of-complex-poles.975854/ | # I The order of complex poles
#### dyn
Hi.
If I look at the function $(z^2+z-2)/(z-1)^2$ it appears to have a double pole at z=1 but if I factorise the numerator I get $z^2+z-2 = (z+2)(z-1)$ and it is a simple pole.
Is it wrong to say it is a double pole ?
If I overestimate the order of the pole in this case as 2 and calculate the residue using limits and differentials I still get the correct answer. Is this always true ?
In this case the numerator was easy to factorise. If it was a complicated function involving higher powers that couldn't be factorised is it possible to say for certain what the order of the pole is just by looking at the denominator ?
Thanks
#### fresh_42
Mentor
2018 Award
This is the same as in real analysis. $\dfrac{z^2+z-2}{(z-1)^2}$ has a pole of order two, $\dfrac{z+2}{z-1}$ one of order one.
Here is how the various singularities are classified:
#### FactChecker
Gold Member
2018 Award
I would say that $\frac {z^2+z-2}{(z-1)^2}$ has a pole of order one. In defining the order of a pole of $f(z)$, it is usual to say that $(z-a)^nf(z)$ is holomorphic and nonzero for a pole of order n.
(seehttps://en.wikipedia.org/wiki/Zeros_and_poles#Definitions )
#### dyn
Hence my confusion !
#### fresh_42
Mentor
2018 Award
I would say that $\frac {z^2+z-2}{(z-1)^2}$ has a pole of order one. In defining the order of a pole of $f(z)$, it is usual to say that $(z-a)^nf(z)$ is holomorphic and nonzero for a pole of order n.
(seehttps://en.wikipedia.org/wiki/Zeros_and_poles#Definitions )
But we have $(z-1)^2f(z)=(z-1)^2\dfrac{[(z-1)(z+2)]}{(z-1)^2}$. The numerator is not defined at $z=1$ hence it cannot be equal to $z+2$.
It is a rather academic question and of not much use: What if we combine a removable singularity with a non removable? A bit like discussing why units are not prime.
#### jasonRF
Gold Member
Interesting … I learned that the order of a pole of a function that is analytic in a closed region except at an isolated point, is defined as the largest negative exponent in the Laurent expansion about that point. My old copy of "A Course of Modern Analysis" by Whittaker and Watson also has this definition. Since the Laurent expansion of the function about $z=1$ is $1 + 3/(z-1)$, the pole at $z=1$ is of order 1. Practically, I use the approach given by FactChecker.
fresh_42
Jason
#### fresh_42
Mentor
2018 Award
Yes, it's probably reasonable to cancel the quotient if there is still the same pole available, so that the same undefined points are still there.
#### HallsofIvy
Homework Helper
For a simpler example, $f(z)= \frac{z- 1}{z- 1}$ has a "removable discontinuity" at z= 1, not a pole.
#### dyn
Hi.
If I look at the function $(z^2+z-2)/(z-1)^2$ it appears to have a double pole at z=1 but if I factorise the numerator I get $z^2+z-2 = (z+2)(z-1)$ and it is a simple pole.
Is it wrong to say it is a double pole ?
So it seems the function has a simple pole and it is wrong to say it is a double pole. But if I overestimate the pole as a double pole and calculate the residue using limits and derivatives I get the correct answer. Does this method of calculating residues always work if the order of poles is overestimated eg. by not spotting a factorisation ?
Thanks
#### FactChecker
Gold Member
2018 Award
Yes, you do not have to factor it. The residue at $z_0$ only depends on the coefficient of the $\frac 1{z-z_0}$ term in the Laurent series. So the effect of any numerator/denominator zeros that cancel each other will disappear.
dyn
Thank you
#### mathwonk
Homework Helper
the order of the pole is a measure of the rate of growth of the function as you approach the given point. In particular it depends only on the values of the function away from the given point, not on the specific representation of the function. Since the two representations (z^2+z-2)/(z-1)^2 and (z+2)/(z-1) define the same functional values away from z=1, the functions they represent are the same and have the same order of pole at z=1. Similarly any calculations that depend only on the values away from z=1 (such as taking limits) will give the same result.
A simpler example that illustrates the same point is to ask yourself whether the function (z-1)/(z-1) has a simple pole at z=1. Since this is (or extends continuously to) the constant function 1, it has no poles at all.
Last edited:
#### FactChecker
Gold Member
2018 Award
the order of the pole is a measure of the rate of growth of the function as you approach the given point. In particular it depends only on the values of the function away from the given point, not on the specific representation of the function.
And more generally, how the function behaves in the complex plane around the pole. If it is a pole at $z_0$ of order $n$, then the function behaves like $\frac {b_n} {(z-z_0)^n}$ near $z_0$.
#### mathwonk
Homework Helper
yes, for example it wraps a small circle around z0, n times around the point at infinity.
#### mathwonk
Homework Helper
Indeed there is a topological interpretation of orders of poles. A rational function defines a continuous map from the riemann sphere to itself, and hence has a certain degree. That degree is the number of preimages of any given point, and in particular is the sum of the orders of all the poles, i.e. the number of preimages of infinity, properly counted.
Last edited:
"The order of complex poles"
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• Solo and co-op problem solving | 2019-09-16T12:25:51 | {
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http://math.stackexchange.com/questions/301435/fractions-in-binary | # Fractions in binary?
How would you write a fraction in binary numbers; for example, $1/4$, which is $.25$? I know how to write binary of whole numbers such as 6, being $110$, but how would one write fractions?
-
The same way. $\frac{n}{m}$. So a quarter would be $\frac{1}{100}$. – copper.hat Feb 12 '13 at 18:47
You might want to have a look at: cs.furman.edu/digitaldomain/more/ch6/dec_frac_to_bin.htm. You can then use WA to test examples. Regards – Amzoti Feb 12 '13 at 18:49
I think this is a duplicate. Will have a look. – Tara B Feb 12 '13 at 18:52
If it is it made by another person – Fernando Martinez Feb 12 '13 at 18:54
Si me gustan todos esto thumbs up.... – Fernando Martinez Feb 15 '13 at 18:41
Note: $$\dfrac 14_{\,\text{ ten}} = .25 = \color{blue}{\bf 2} \times 10^{-1} + \color{blue}{\bf 5} \times 10^{-2}$$ $$\frac14= \dfrac{1}{2^2}_{\,\text{ten}} = 2^{-2} = \color{blue}{\bf 0} \times 2^{-1} + \color{blue}{\bf 1}\cdot 2^{-2} = .01_{\text{ two}}$$
-
Yes I like this answer the most. – Fernando Martinez Feb 12 '13 at 18:58
So I I had $\frac{1}{8}$ would I do $(2^{-3}+2^{-2}+2^{-1}+2^0)$ – Fernando Martinez Feb 12 '13 at 19:00
Yup, but start with $\frac 18 = 2^{-3} = 0 \times 2^{-1} + 0\times 2^{-2} + 1\times 2^{-3} = 0.001_{\text{two}}$ – amWhy Feb 12 '13 at 19:02
yes that makes sense. – Fernando Martinez Feb 12 '13 at 19:04
$2^0 \iff \;$ the "one's place". $2^{-1}$ corresponds to the first digit to the right of the decimal point, then the place to the right of that corresponds to $2^{-2}$...and so on. – amWhy Feb 12 '13 at 19:05
As you mentioned, $$6 = {\color{red}1}\cdot 2^2+ {\color{red}1}\cdot 2^1+{\color{red}0}\cdot 2^0 = {\color{red}{110}}_B.$$ Analogously $$\frac{1}{4} = \frac{1}{2^2} = {\color{red}0}\cdot2^0 + {\color{red}0}\cdot 2^{-1} + {\color{red}1}\cdot 2^{-2} = {\color{red}{0.01}}_B.$$
Edit:
These pictures might give you some more intuition ;-) Here $\frac{5}{16} = 0.0101_B$, as the denominator is of form $2^n$, the representation is finite (process ends when you hit zero); $\frac{1}{6} = 0.0010\overline{10}_B$ as the denominator is not of form $2^n$, but the number is rational, so representation is infinite and periodic.
I hope this helps ;-)
-
$1/4=0\cdot(1/2)^0+0\cdot(1/2)^1+1\cdot(1/2)^2=0.01$ in base $2$, you just go in reverse with powers $(1/2)^n, n=0,1,2,...$
-
Three basic ways, all seen in binary number systems:
Fixed-Point: One integer holds the "integer part"; another holds the "fractional part". This is simple to store and display, and has very high magnitude and precision with virtually no error, but doing real math with the numbers involved can get hairy. Decimal numbers aren't often seen in this form, but it is a possibility.
Maintained Floating Point: a large integer holds the entire value, and a second smaller number maintains the relative place of the decimal point from the right (or left) side of the number. Much easier to manipulate for mathematical operations, same maintenance of precision, zero error, and used in many implementations of "BigDecimal" object types where the "built-in" floating point mechanisms aren't available. Can be more difficult to represent in base-10 form on-screen. If implemented with normal integer types, this method can be more limited in magnitude than the previous one; instead, many implementations use a byte array to store the number, allowing the numbers to be as big as system memory allows.
Implicit Floating Point: The number is expressed in what amounts to "binary scientific notation". A "mantissa" is stored as an integer, with the decimal point implied to be on the far right. Then the exponent of a power of two is also stored. The actual value of this number is the mantissa, multiplied by two to the power of the exponent. This approach allows for the storage and calculation of truly massive numbers, and modern CPUs are designed with a Floating-Point Unit or FPU (sometimes called a "math co-processor", in the early days of its integration into 486-class CPUs) that accelerates calculations of numbers in this form. However, there are two problems; first, there's a tradeoff between extreme precision and extreme magnitude; the mantissa, and thus the number of digits that can be stored precisely, is fixed, so as magnitude increases, the number of possible decimal places decreases (in the extremes of magnitude you often can't get more granular than the millions place). Second, there's an amount of "rounding error" inherent in using floating-point numbers, with the inherent conversion to binary and back; this can cause errors in calculations requiring exact precision (such as when dealing with money), and so unless the extreme magnitude of a floating-point type is required, it's generally recommended to use a method of representation that does not introduce error.
-
This is a thorough answer, but I don't think the OP was asking about computer implementations. – LarsH Feb 12 '13 at 21:36
Although, what the OP is asking about is a typographic representation of numbers, which is very close to computer representation. Binary and decimal are just ways we can write down numbers with a pencil, and computer representations are how we write numbers into a memory with ones and zeros. – Kaz Feb 13 '13 at 3:25
Just do long division!
11 | 10011 | 110.0101...
1100
----
111
110
---
100
11
---
100
11
---
1 (repeats)
So 10011 / 11 = 110.0101... (aka 19 / 3 = 6.33...)
Binary long division is a bit longer than decimal long division since you need more digits to write each number, but finding the largest multiple of your divisor that will fit is pretty trivial when it's either 0 or 1 times.
-
0.01 in binary is 0.25 in decimal
-
use the euclidean algorithm, like for the integers
-
The Euclidean Algorithm is usually used to find greatest common factors and continued fractions, not to divide or convert bases. – robjohn Feb 12 '13 at 19:26
@robjohn: You use the euclidean algrotihm to express fractions in lowest terms. The OP asked about base 2 fractions, not other representations, so it's clear this answer is a valid response. It's also clear the OP intended to ask about other representations and just didn't use the right terms, so being pedantic is missing the point, but talking about the euclidean algorithm isn't as unrelated as you imply. – ex0du5 Feb 12 '13 at 20:46
@ex0du5: Since the OP mentioned converting $1/4$ to $0.25$, it seems that they are interested in converting binary fractions to the format using a binary point. Since Nicolò said "like for the integers", it really doesn't seem as if he was trying to reduce fractions to lowest terms. – robjohn Feb 12 '13 at 21:17
@ex0du5: I was not being pedantic; I was concerned that Nicolò was using the wrong term for the repeated-division method for converting bases (which is generally used for integers) as shown in the answers to these questions: Converting decimal (base 10) numbers to binary by repeatedly dividing by 2, Convert numbers from one base to another using repeated divisions – robjohn Feb 12 '13 at 21:19
@robjohn: Yes, that's exactly what I said. I said that the answer Nicolo wrote was addressing what was asked, but "It's also clear the OP intended to ask about other representations and just didn't use the right terms". I was the one who had a pedantic point that Nicolo's answer was relevant as stated. I didn't call you pedantic. I just pointed out your comment on the euclidean algorithm didn't seem to understand why it was relevant here. – ex0du5 Feb 14 '13 at 18:59
not sure how helpful this is but whenever i work with binary I always use a table
so you wanted 0.25 (base 10) in to binary:
binary table
8 4 2 1 . 1/2 1/4 1/8 1/16
8 4 2 1 . 0.5 .25 .125 .0625
0 0 0 0 . 0 1 0 0 (0.25 in binary)
1 1 1 1 . 1 1 0 0 (15.75 in binary)
etc
and the other way
15.375 to binary for example
15 - 8 = 7 (1)
7 - 4 = 3 (1)
3 - 2 = 1 (1)
1 - 1 = 0 (1)
.
0.375 - (1/2) = 0.375 - 0.5 = -0.125 (0)
0.15 - (1/4) = 0.375 - 0.25 = 0.125 (1)
0.125 - (1.8) = 0.125 - 0.125 = 0 (1)
giving: 1111.011
hope this helps
- | 2014-10-25T06:22:37 | {
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http://mtapreviewer.com/2016/02/25/grade-9-mtap-2015-elimination-2/ | Grade 9 MTAP 2015 Elimination Questions with Solutions – Part 2
This is the second part (questions 11-20) of the solutions of the Grade 9 MTAP 2015 Elimination Questions. The first part can be read here.
Although reasonable care has been given to make the solution accurate as possible, the solver is also human. Please comment below if you see any errors.
11.) If $x \neq 1$, solve for $x$ in $2 \sqrt{x} + \frac{3}{\sqrt{x}} = 5$.
Solution
Multiplying both sides by $\sqrt{x}$, we obtain
$2x + 3 = 5\sqrt{x}$
Squaring both sides of the equation, we get
$4x^2 + 12x + 9 = 25x$
$4x^2 - 13x + 9 = 0$.
Factoring, we have
$(4x - 9)(x - 1) = 0$
So, $x = \frac{9}{4}$ or $x = 1$.
But from the given above, $x \neq 1$, so the only solution is $x = \frac{9}{4}$
12.) Evaluate $\sqrt{2 + \sqrt{2 + 2 \sqrt{2 + \cdots }}}$.
Solution
Let $x = \sqrt{2 + \sqrt{2 + 2 \sqrt{2 + \cdots }}}$
$x^2 = 2 + \sqrt{2 + \sqrt{2 + 2 \sqrt{2 + \cdots }}}$
$x^2 = 2 + x$
$x^2 - x - 2 = 0$
$(x - 2)(x + 1) = 0$
$x = 2$ or $x = -1$
Answer: $x = 2$ or $x = -1$
13.) Find two positive consecutive integers whose product is 506.
Solution
Let $x$ is equal to the smaller number and $x + 1$ be the larger number.
$x(x + 1) = x^2 + x = 506$
$x^2 + x - 506 = 0$
$(x - 22)(x - 23) = 0$
$x = 22$ or $x = 23$.
14.) If $c > a > 0$ and if $a - b + c = 0$, find the larger root of $ax^2 + bx + c = 0$.
Solution
From the given, $a - b + c = 0$, so $b = a + c$.
$x = \dfrac{-b \pm \sqrt{b^2 -4ac}}{2a}$
$= \dfrac{-(a + c) \pm \sqrt{(a + c)^2 - 4ac}}{2a}$
$= \dfrac{-a - c \pm \sqrt{a^2 - 2ac + c^2} }{2a}$
$= \dfrac{(-a - c) \pm a - c}{2a}$.
Simplifying, we get the two roots $x_1$ and $x_2$.
$x_1 = \dfrac{-2c}{2a} = -\dfrac{c}{a}$
$x_2 = \dfrac{-2a - 2c}{2a} = \dfrac{-a - c}{a}$
Since $c > a > 0$, the larger root is $-\dfrac{c}{a}$.
Answer: $-\dfrac{c}{a}$
15.) Solve for $x$ in $2x^2 + x < 6$.
Solution
Subtracting 6 from both sides,
$2x^2 + x - 6 < 0$
$(2x - 3)(x + 2) < 0$
$-2 < x < \frac{3}{2}$
Answer: $-2 < x < \frac{3}{2}$
16.) Solve for real numbers $x$ satisfying the inequality $x - 2\sqrt{x} \leq 3$.
Solution
$x - 3 \leq 2 \sqrt(x)$
Squaring both sides of the inequality,
$x^2 - 6x + 9 \leq 4x$
$x^2 - 10x + 9\leq 0$
$(x - 1)(x - 9)\leq 0$
This can be split to (1) $(x - 1)(x - 9) = 0$ and $(x - 1)(x - 9)< 0$. From (1), we already know that $x = 1$ and $x = 9$ are solutions. For the inequality, we can check the intervals $(- \in, 1)$, $(1, 9)$ and $(9, \infty)$. Upon checking, it can be seen that only (1,9) satisfy the inequality. Therefore, the interval that satisfies the inequality above is [1,9].
Answer: $1 \leq x \leq 9$ or $[1,9]$ in interval notation.
17.) Find the minimum value of x^2 – 8x + 3.
Solution
Let $y = x^2 - 8x + 3$.
The graph of the function opens upward, so we transform it to the vertex form a(x – h)^2 + k = 0 to get the minimum value which is (h,k).
$x^2 - 8x + 3 = (x - 4)^2 + 3 - 16 = (x - 4)^2 - 13$
So, h = 4 and k = – 13. The minimum value is -13.
18.) Find the smallest value of $x + \dfrac{5}{x}$ for all real number $x$.
19.) Solve for $b$ in the equation $(x + 1)(x + a) = x^2 + bx + 3$.
Solution
$x^2 + x + ax + a = x^2 + bx + c$
$x + ax + a = bx + 3$.
Factoring the left side, we have
$x (1 + a)x + a = bx + 3$, therefore, a = 3 and b = 1 + 3 = 4.
20.) Write the quadratic equations with integer coefficients whose roots are reciprocal of the roots of $2x^2 - 3x + 1 = 0$.
Solution
$2x^2 - 3x + 1 = (2x - 1)(x - 1)= 0$
$x = \frac{1}{2}$ or $x = 1$.
So, we create a quadratic equation whose roots are 2 and 1. That is
$(x - 1)(x - 2) = x^2 - 3x + 2 = 0$
Answer: $x^2 - 3x + 2 = 0$.
If you have old MTAP questions and you want me to solve it for you, please send your email to
5 thoughts on “Grade 9 MTAP 2015 Elimination Questions with Solutions – Part 2”
1. I was wondering in question 16, why 0 is not included in solution set infact 0-2sqrt(0)<=3. I check that in maple, mathlab and symbolab, 0 is included.
2. Question 14, I think the larger root is x = -1. Kindly check your second root. That is plus minus ( a-c ). | 2017-11-19T08:20:02 | {
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https://mathematica.stackexchange.com/questions/208104/how-to-deal-with-loss-of-significance-in-the-case-fx-sqrtx2-sqrtx/208107 | # How to deal with loss of significance in the case $f(x) = \sqrt{x+2} -\sqrt{x}$?
I would like to evaluate the expression $$f(x) = \sqrt{x+2} -\sqrt{x}$$ with cases when $$x = 3.2 \times 10^{30}$$ and $$x= 3.2 \times 10^{16}$$. I tried using N[Sqrt[x+2] - Sqrt[x], 100] and
ScientificForm[Sqrt[x+2] - Sqrt[x], 100], both yielding 0 as an output. How can I obtain the desired output?
I also tried the method in Making a calculation with high precision by applying .100 as a suffix but my mathematica 12 doesn't seem to be recognizing it.
By using arbitrary precision numbers instead of machine numbers:
x = 3.2100 10^30;
Sqrt[x + 2] - Sqrt[x]
5.59016994374947424102293417182731712454783504367865447721289523170435 *10^-16
Or, even better, by using exact numbers and by doing the conversion afterwards (this forces Mathematica to compute all 100 leading digits):
x = 32/10 10^30;
N[Sqrt[x + 2] - Sqrt[x],100]
5.590169943749474241022934171827317124547835043678654477212895231704350107190085921247950310402211757*10^-16
Your expression can be written in a more numerically stable form:
Sqrt[x + 2] - Sqrt[x] == 2/(Sqrt[x + 2] + Sqrt[x]) // FullSimplify
(* True *)
This evaluates without significant loss of precision.
2/(Sqrt[x + 2] + Sqrt[x]) /. x -> N[32*^15]
(* 5.59017*10^-9 *)
2/(Sqrt[x + 2] + Sqrt[x]) /. x -> N[32*^15, 30]
(* 5.59016994374947415367652880074*10^-9 *)
• I assume that it's numerically more stable because you're not subtracting two large numbers from each other? – march Oct 17 '19 at 20:35
• @march That's what I meant. – mikado Oct 17 '19 at 20:50
Use the asymptotic form of the function, i.e.
Normal@Series[Sqrt[x + 2] - Sqrt[x], {x, ∞, 2}]
% /. x -> 3.2*10^30
(* -(1/(2 x^(3/2))) + 1/Sqrt[x] *)
(* 5.59017*10^-16 *)
Note that we don't even need the second term to this level of precision:
Normal@Series[Sqrt[x + 2] - Sqrt[x], {x, ∞, 1}]
% /. x -> 3.2*10^30
(* 1/Sqrt[x] *)
(* 5.59017*10^-16 *)
• First I was going to do @mikado's, then I was going to do this but played with it too long. +1 :) – Michael E2 Oct 17 '19 at 20:33
• @MichaelE2 This is a surprising question in the sense that such a simple thing has invited four useful (and pretty different) answers. – march Oct 17 '19 at 20:37
Replace your approximate input with exact rational numbers, reduce the result to a single exact algebraic number, and evaluate that numerically.
f[xx_] := With[{x = Rationalize[xx, 0]}, N[RootReduce[Sqrt[x + 2] - Sqrt[x]]]]
f[3.2 10^30]
(* 5.59017*10^-16 *) | 2021-01-22T22:45:30 | {
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https://math.stackexchange.com/questions/3123356/what-does-it-mean-to-minimize-a-convex-function-with-less-than-or-equal-to-ine | # What does it mean to minimize a convex function with “less than or equal to” inequality constraints? Why?
What does mean to minimize objective function with "less than" inequality constraints? Aren't you suppose to minimize with "greater than" constraints, like in example 1?
Example 1 (understand this) \begin{align} \text{max } x_1 & +5 x_2 \\ \text{s.t. } x_1 & \le 150 \\ x_2 &\le 350 \\ x_1+x_2 &\le 400 \\ x_1 , x_2 &\ge 0 \end{align}
I get this: the objective function wants to be as large as possible, but the constraints put an upper bound on $$x_1, x_2$$.
Example 2 (don't understand this) \begin{align} \text{min } & f_0(x) \\ \text{s.t. } & f_i \le 0 \text{ } i=1,...,m \\ \end{align} where $$f_0,..,f_m$$ are convex functions. (Eq. 4.15 Convex Optimization)
This seems unbounded below. But since it's convex, it's bounded below? So, you are minimizing a convex function that satisfies a bunch of other convex functions.
Am I understanding this correctly? What's the point of this? Can someone provide a numerical example?
Thanks in advance!
=============================DIVIDER LINE===========================
Follow up question:
Generally, $$f_i(x)$$ need not to be convex for $$i=1,...,m$$ \begin{align} \text{min } & f_0(x) \\ \text{s.t. } & f_i \le 0 \text{ } i=1,...,m \\ \end{align}
This is unbounded below. Isn't the solution $$-\infty$$?
• I'm definitely not an expert on that subject, but minimizing $x$ st $x^2+5x+6 \leq 0$ makes sense to me. The minimum is $x=-3$ which can be seen by plotting the constraint. – FormerMath Feb 23 '19 at 4:12
• The direction of the inequalities is immaterial, as we can always flip them by multiplying by $-1$. It certainly feels intuitive to have lower bound constraints when minimizing, but we can always flip them by multiplying with $-1$. – David M. Feb 24 '19 at 3:31
• @DavidM. I tried to think that way. But if you graph the constraints, it's still upper bounded though. The feasible set is below the graph. – user13985 Feb 24 '19 at 3:52
• I just added an answer to try to address your confusion--sorry if I'm way off base! – David M. Feb 24 '19 at 4:31
## 2 Answers
Notice that $$x^2-1$$ is a convex function. If we consider $$x^2-1 \ge 0$$, we have $$x \ge 1$$ or $$x \le -1$$. That is the feasible set is not a convex set.
However, notice that $$x^2-1 \le 0$$ is equivalent to $$-1 \le x \le 1$$ is convex.
In general, we know that $$\{ x \mid f_i(x) \le 0\}$$ is a convex set and their intersection, that is the feasible set that you have written down is a convex set. It is a desirable property to minimize a convex objective function over a convex set, in particular, we know that a local minimum is a global minimum.
Also, notice that the first example is a special case of the general form.
\begin{align} \text{max } x_1 & +5 x_2 \\ \text{s.t. } x_1 & \le 150 \\ x_2 &\le 350 \\ x_1+x_2 &\le 400 \\ x_1 , x_2 &\ge 0 \end{align}
Is actually equivalent to
\begin{align}- \min -x_1 & -5 x_2 \\ \text{s.t. } x_1 & \le 150 \\ x_2 &\le 350 \\ x_1+x_2 &\le 400 \\ -x_1 &\le 0 \\ -x_2 &\le 0 \end{align}
Here $$f_0$$ is just $$-x_1-5x_2$$ and the $$f_i$$ are just the LHS of the constraint.
We can flip the inequality by multiplying a negative sign and in fact the general form even include the first form. The general form doesn't tell us whether $$x_i$$ is bounded above or below since linear functions are convex and we can flip the inequality signs. The crucial propery here is convexity.
For the follow up question:
Even if $$f_i$$ is not convex, it is still possible for the feasible region to be bounded. In fact, in special cases, it is even possible for it to be convex. As an example
Consider $$x^3-1 \le 0$$
Even thought the function $$x^3-1$$ is not convex, the feasible region is just $$x \le 1$$.
Imposing $$f_i$$ to be convex explicitly gives us a convenient way to construct a convex set and also use its properties in deriving algorithms or investigate property of this form of problems.
• Thanks, I added a follow up question. What if constrains are not convex? Isn't the solution equal to $-\infty$? Can you explain this in your post? Thank you! – user13985 Feb 23 '19 at 17:20
• I think what I was really missing was the definition that: the inequality constraint functions must be convex. – user13985 May 8 '19 at 17:43
I think Siong Thye Goh's answer is very good--I'm adding this answer to try to clarify some confusion you expressed in the comments.
I'm not positive, but I think the confusion is that you're mixing up the hypograph of a function with the sublevel sets of a function. The hypograph of a function $$f:\mathbb{R}^n\to\mathbb{R}$$ is the set $$\big\{(x,y)\in\mathbb{R}^{n+1}\;|\;y\leqq{f(x)}\big\}.$$ By contrast, the sublevel sets (at a value $$c\in\mathbb{R}$$) is the set $$\big\{x\in\mathbb{R}^n\;|\;f(x)\leqq{c}\big\}.$$ Notice that these two sets are fundamentally different (in particular, one is a subset of $$\mathbb{R}^{n+1}$$, while the other is a subset of $$\mathbb{R}^n$$).
Inequality constraints $$g(x)\leqq0$$ define a feasible region through sublevel sets. Let's consider a function $$g:\mathbb{R}\to\mathbb{R}$$, like $$g(x)=x-1$$. The sublevel set of this function (for $$c=0$$) is shown in red here:
If we multiply this constraint by $$-1$$, then we get:
Note that even though the line flipped upside down, the set of feasible solutions is exactly the same!
In contrast, the hypograph of the function $$g(x)=x-1$$ is
(the hyprograph is the set shaded in light blue). Note that the dimension of the hypograph is one dimension higher than the dimension of the sublevel set (red line) above. The hypograph of $$g$$ really has nothing to do with the constraint $$g(x)\leqq0$$.
Two Final Points
1. In the example above, we can see that the hypograph of $$g$$ is the exact same set as the sublevel set (at $$c=0$$) of the function $$h:\mathbb{R}^2\to\mathbb{R}$$ given by $$h(x_1,x_2)=x_2-x_1+1$$. In general, the hypograph of a function $$g:\mathbb{R}^n\to\mathbb{R}$$ equals the sublevel sets for $$c=0$$ of the function $$h:\mathbb{R}^{n+1}\to\mathbb{R}$$ given by $$h(x,y)=y-g(x)$$. I think this connection is causing confusion.
2. The sublevel sets of a function $$g$$ (for $$c=0$$) are the same as the superlevel sets of the function $$-g$$ (for $$c=0$$). However, the hypograph of a function $$g$$ is not the epigraph of the the function $$-g$$ (see linked Wiki articles for defintions of superlevel set and epigraph). Multiplying by $$-1$$ affects function graphs, but not feasible regions. | 2020-01-29T06:00:04 | {
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http://mathhelpforum.com/advanced-statistics/20999-central-limit-theorem.html | 1. ## Central Limit Theorem
Candidates A and B are running for office and 55% of the electorate favor candidate B. What is the probability that in a sample of size 100 at least one-half of those sampled will favor candidate A?
Here's what I did:
Let X_i = 1 if the ith person voted for A.
S_100 = X_1 + X_2 + X_3 + ... + X_100
P(S_100 >= 50) = 1 - P(S_100 <= 49)
= 1 - P([S_100 - 100(.45)]/sqrt(100*.45*.55) <= [49 + 1/2 - 100(.45)]/sqrt(100*.45*.55)) #transform into standard norm rv
= 1-P(Z<=0.9045)
= 1-z(0.9045) #z is the standard normal function phi
= 1-0.817
= 0.183
However, the book's solution is 0.133. I don't know how to arrive at that solution. Can someone go through this problem step by step. Thanks!
2. Originally Posted by feiyingx
Candidates A and B are running for office and 55% of the electorate favor candidate B. What is the probability that in a sample of size 100 at least one-half of those sampled will favor candidate A?
This is a central limit theorem question, so let $X_i=1$ if the $i$-th samped voter
votes A, and $0$ otherwise. then the mean for $X_i$ is
$\mu = 0.45$
and the variance is
$\sigma ^2=0.45(1-0.45)^2+0.55(-0.45)^2 = 0.2475.$
So the number of votes for A in the sample is:
$
n=\sum_{i=1}^{100} x_i
$
which approximatly has a normal distribution with mean $100\mu=45$, and variance $100 \sigma^2 = 24.75$.
So now we want the probability that $50$ or more will vote A. As we have
a continuous distribution modelling a discrete we ask what is the probability
of a value greater than $49.5$ occuring from a normal distribution with mean $45$ and variance $24.75$.
The z-score for this problem is:
$
z=\frac{49.5-45}{\sqrt{24.75}} \approx 0.9045
$
which we look up in a standard normal table to get a probability of $0.183$.
Note if we had been asked for the probability of more than 50 voted for A this would drop to $0.134$
(If this were not a CLT question I would have used a binomial distribution to
model the distribution of the number of votes for A in the sample, but when the
normal approximation is used for the binomial the answer is exactly the same
as we get with the above argument)
RonL
3. Thanks for the response RonL.
I have a question that I'm confused about.
Shouldn't the mean for Xi be 0.45 since Xi is a Bernoulli RV with probability of people voting for A of 45%?
Therefore, the total mean is .45*100 = 45
Thanks!
4. Originally Posted by feiyingx
Thanks for the response RonL.
I have a question that I'm confused about.
Shouldn't the mean for Xi be 0.45 since Xi is a Bernoulli RV with probability of people voting for A of 45%?
Therefore, the total mean is .45*100 = 45
Thanks!
Opps.. I had the probabilities the wrong way around. You should be able to fix that yourself.
RonL
5. So the only difference would be the mean of Xi which is $\mu = 0.45$
This translates to the normal mean of .45*100 = 45.
Then by using the normal approximation, it gives us
$
z=\frac{49.5-45}{\sqrt{24.75}} \approx .9045
$
Using the table and calculating for the probability, I got 0.183.
Is the book's solution incorrect?
Thanks!
6. Originally Posted by feiyingx
So the only difference would be the mean of Xi which is $\mu = 0.45$
This translates to the normal mean of .45*100 = 45.
Then by using the normal approximation, it gives us
$
z=\frac{49.5-45}{\sqrt{24.75}} \approx .9045
$
Using the table and calculating for the probability, I got 0.183.
Is the book's solution incorrect?
Thanks!
Well with the wording you give I agree with you (I have now corrected the
earlier post), but if the wording had been:
"What is the probability that in a sample of size 100 at more than one-half of those sampled will favor candidate A" | 2016-12-06T00:42:31 | {
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https://www.physicsforums.com/threads/mean-of-the-derivative-of-a-periodic-function.973968/ | # I Mean of the derivative of a periodic function
#### Robin04
Summary
I'm wondering if given that the mean of a periodic fuction is zero than the mean of all of its derivatives is zero too.
We have a periodic function $f: \mathbb{R} \rightarrow \mathbb{R}$ with period $T, f(x+T)=f(x)$
The statement is the following: $$\frac{1}{T}\int_0^T f(x)dx =0 \implies \frac{1}{T}\int_0^T\frac{d}{dx} f(x)dx =0$$
Can you give me a hint on how to prove/disprove it? The examples I tried all confirmed this.
#### fresh_42
Mentor
2018 Award
Let $F(x)$ be the antiderivative of $\left( \dfrac{d}{dx}\,f(x) \right)$. Then the right hand side is?
#### Robin04
Let $F(x)$ be the antiderivative of $\left( \dfrac{d}{dx}\,f(x) \right)$. Then the right hand side is?
So then $F(x)=f(x)+c$, where $c$ is the integration constant.
$$\frac{1}{T}\int_0^T \frac{d}{dx}f(x) dx = \frac{1}{T}[F(x)]_0^T=\frac{1}{T}[f(x)+c]_0^T=\frac{1}{T}(f(T)+c-f(0)-c)=0$$
Is this correct?
#### pasmith
Homework Helper
Summary: I'm wondering if given that the mean of a periodic fuction is zero than the mean of all of its derivatives is zero too.
We have a periodic function $f: \mathbb{R} \rightarrow \mathbb{R}$ with period $T, f(x+T)=f(x)$
The statement is the following: $$\frac{1}{T}\int_0^T f(x)dx =0 \implies \frac{1}{T}\int_0^T\frac{d}{dx} f(x)dx =0$$
Can you give me a hint on how to prove/disprove it? The examples I tried all confirmed this.
$\frac{1}{T}\int_0^T\frac{d}{dx} f(x)dx = 0$ is a direct result of the fundamental theorem of caclulus and the fact that $f(0) = f(T)$. It holds irrespective of the value of $\frac{1}{T}\int_0^T f(x)dx$.
#### jbriggs444
Homework Helper
Summary: I'm wondering if given that the mean of a periodic fuction is zero than the mean of all of its derivatives is zero too.
We have a periodic function $f: \mathbb{R} \rightarrow \mathbb{R}$ with period $T, f(x+T)=f(x)$
The statement is the following: $$\frac{1}{T}\int_0^T f(x)dx =0 \implies \frac{1}{T}\int_0^T\frac{d}{dx} f(x)dx =0$$
Can you give me a hint on how to prove/disprove it? The examples I tried all confirmed this.
Would you count $\frac{\sin x\ |\cos x|}{\cos x}$ as a periodic function with mean zero for purposes of this question?
#### Robin04
$\frac{1}{T}\int_0^T\frac{d}{dx} f(x)dx = 0$ is a direct result of the fundamental theorem of caclulus and the fact that $f(0) = f(T)$. It holds irrespective of the value of $\frac{1}{T}\int_0^T f(x)dx$.
Oh, you're right. Interesting, haven't thought about that.
Would you count $\frac{\sin x\ |\cos x|}{\cos x}$ as a periodic function with mean zero for purposes of this question?
Well, that's interesting. I cannot plot its derivative for some reason, but I suppose it's continuity the issue here.
#### jbriggs444
Homework Helper
Well, that's interesting. I cannot plot its derivative for some reason, but I suppose it's continuity the issue here.
It is continously differentiable over its domain. But its domain misses the odd multiples of $\frac{\pi}{2}$.
That means that the anti-derivative of its derivative over almost any interval of length $\pi$ has two disjoint segments. Two c's, instead of just one. *WHAM* There goes that cancellation of the c's you did in post #3.
"Mean of the derivative of a periodic function"
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Open Live Script. or In the matrix equation PX = q, which of the following is necessary condition for the existence of atleast one solution for the unknown … abelian group augmented matrix basis basis for a vector space characteristic polynomial commutative ring determinant determinant of a matrix diagonalization diagonal matrix eigenvalue eigenvector elementary row operations exam finite group group group homomorphism group theory homomorphism ideal inverse matrix invertible matrix kernel linear algebra linear combination linearly … Active 7 years, 11 months ago. & . Related Topics: Common Core (Vector and Matrix Quantities) Common Core for Mathematics Common Core: HSN-VM.C.10 Videos, solutions, examples, and lessons to help High School students understand that the zero and identity matrices play a role in matrix addition and multiplication similar to the role of 0 and 1 in the real numbers. The resulting matrix, known as the matrix product, has the number of rows of the first and the number of columns of the second matrix. | EduRev JEE Question is disucussed on EduRev Study Group … Eine Nullmatrix ist in der linearen Algebra eine reelle oder komplexe Matrix, deren Einträge alle gleich der Zahl Null sind. Matrix multiplication falls into two general categories:. That is, the inner dimensions must be the same. Now all these small Toeplitz matrices should be arranged in a big doubly blocked Toeplitz matrix. & . Using this library, we can perform complex matrix operations like multiplication, dot product, multiplicative inverse, etc. n ; The the span of the rows of B contains the span the rows of C.; If E is an invertible n×n matrix … The matrix multiplication algorithm that results of the definition requires, in the worst case, multiplications of scalars and (−) additions for computing the product of two square n×n matrices. A square matrix in which all the elements below the diagonal are zero is known as the upper triangular matrix. A zero matrix is an matrix consisting of all 0s (MacDuffee 1943, p. 27), denoted .Zero matrices are sometimes also known as null matrices (Akivis and Goldberg 1972, p. 71). m The first row can be selected as X[0].And, the element in first row, first column can be selected as X[0][0].. Multiplication of two matrices X and Y is defined only if the number of columns in X is equal to the number of rows Y.. Gibt es da eine Formel für, wie z.B. K K NumPy Matrix Multiplication in Python. When we add or subtract the 0 matrix of order m*n from any other matrix, it returns the same Matrix. 0 In ordinary least squares regression, if there is a perfect fit to the data, the annihilator matrix is the zero matrix. n Multiply B times A. X = zeros(2,3,4); size(X) ans = 1×3 2 3 4 Clone Size from Existing Array. Consider the following example for multiplication by the zero matrix. In linear algebra, an n-by-n square matrix A is called invertible (also nonsingular or nondegenerate), if there exists an n-by-n square matrix B such that = = where I n denotes the n-by-n identity matrix and the multiplication used is ordinary matrix multiplication.If this is the case, then the matrix B is uniquely determined by A, and is called the (multiplicative) inverse of A, denoted by A −1. In mathematics, particularly linear algebra, a zero matrix or null matrix is a matrix all of whose entries are zero. is the matrix with all entries equal to Part I. Scalar Matrix Multiplication. For matrix multiplication, the number of columns in the first matrix must be equal to the number of rows in the second matrix. For example: Diagonal Matrix: A square matrix in which all the non-diagonal elements are zero and contain at least one no-zero element in its principal diagonal is called the … Creating a zero matrix through matrix multiplication. C++. There is exactly one zero matrix of any given dimension m×n (with entries from a given ring), so when the context is clear, one often refers to the zero matrix. O matrices, and is denoted by the symbol & 0 \\ . The set of But to multiply a matrix by another matrix we need to do the "dot product" of rows and columns ... what does that mean? But to multiply a matrix by another matrix we need to do the "dot product" of rows and columns ... what does that mean? n 3.1.7 Multiplication of Matrices The multiplication of two matrices A and B is defined if the number of columns of A is equal … The mortal matrix problem is the problem of determining, given a finite set of n × n matrices with integer entries, whether they can be multiplied in some order, possibly with repetition, to yield the zero matrix. Yes there is a clumsiness, list indexing starts with zero, matrix indexing start with 1. Multiplying by an appropriately-sized zero matrix from the left or from the right (− −) = () = results in a zero matrix. I think I have everything set up correctly and the program runs and executes. Viewed 1k times -2. Then we define operation: C = A * B (matrix multiplication) such that C is a matrix with n rows and m columns, and each element of C should be computed by the following formula: The meaning of matrix multiplication is … How to get solution matrix from REF matrix. Trying to run a program to do Matrix Multiplication in CUDA. C = A*B. Transpose of matrix A is denoted by A T. Two rows of A T are the columns of A. For example, (Inf + 1i)*1i = (Inf*0 – 1*1) + (Inf*1 + 1*0)i = NaN + Infi. & . A m×n × B n×p = C m×p. A square matrix in which all the elements above the diagonal are zero is known as the upper triangular matrix. Example … In Python, we can implement a matrix as nested list (list inside a list). The templated class matrix is thebase container adaptor for dense matrices. Let A = [7 3 1 4] and 0 = [0 0 0 0], then A + 0 = [7 3 1 4] + [0 0 0 0] = [7 + 0 3 + 0 1 + 0 4 + 0] = [7 3 1 4] 8.1.1.5 Additive inverse (negative) matrix Example 1. The matrix exponential of is given by the identity matrix.An zero matrix can be generated in the Wolfram Language as ConstantArray[0, m, n]. {\displaystyle O} & . Let us do an example in Python. It also serves as the additive identity of the additive group of We have many options to multiply a chain of matrices because matrix multiplication is associative. m & . & . Matrix Multiplication in NumPy is a python library used for scientific computing. & . ja, du sollst die Hauptdiagonalwerte betrachten. {\displaystyle K_{m,n}\,} X = zeros(4) X = 4×4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3-D Array of Zeros. The first case, the action of a zero matrix, is very easy. Create a 2-by-3-by-4 array of zeros. & . Matrixmultiplikation AB = (AB)^T richtig? & . The product of … in a single step. All the four entries of the 2 x 2 matrix P = are non - zero, and one of its eigen values is zero . Matrix Multiplication in NumPy is a python library used for scientific computing. C = 3. We call the number ("2" in this case) a scalar, so this is called "scalar multiplication". So, having a Pythonic representation for matrices, by implementing the matrix multiplication rule above, we can do a matrix multiplication operation among two given matrices, in Python. The way described above is the standard way of multiplying matrices. Schreibe die Formel der Hauptdiagonalwerte auf, das sind Summen von Einträgen der Matrix A quadriert, also genau dann gleich 0, wenn alle Summanden =0 sind. 3.1.5 Multiplication of Matrix by a Scalar If A = [a ij] m×n is a matrix and k is a scalar, then kA is another matrix which is obtained by multiplying each element of A by a scalar k, i.e. Solution: QUESTION: 20. C = B*A. & 0 \\ . Earlier, we defined the zero matrix $$0$$ to be the matrix (of appropriate size) containing zeros in all entries. & . & . Scalar: in which a single number is multiplied with every entry of a matrix. For example, (Inf + 1i)*1i = (Inf*0 – 1*1) + (Inf*1 + 1*0)i = NaN + Infi. K & 0 \\ 0 & 0 & 0& . "Die Frage ist zu gut, um sie mit einer Antwort zu verderben. . But product of two non-zero matrices can be zero matrix. Its computational complexity is therefore (), in a model of computation for which the scalar operations require a constant time (in practice, this is the case for floating point numbers, but not for integers).. … With multi-matrix multiplication, the order of individual multiplication operations does not matter and hence does not yield different results. Nimmst du nun auch i=2,3,...,n hinzu, dann ist die gesamte Matrix abgepflastert. & . The number of columns in the first matrix must be equal to the number of rows in the second matrix. Which one of the following statements is true? {\displaystyle K_{m,n}\,} If you're seeing this message, it means we're having trouble loading external resources on our website. , play_arrow. {\displaystyle m\times n} Example :-Let A = [0 0] [0 1] and B = [0 1] [0 0] then, AB = [0 0][0 1] [0 1][0 0] =[0 0] [0 0] = 0 (Null Matrix) This example illustrates that in matrix multiplication , if AB = 0, it does not necessarily means A=0 or B=0. & . For example X = [[1, 2], [4, 5], [3, 6]] would represent a 3x2 matrix.. & . If A has a row of zeros then AB has a row of zeros. {\displaystyle A\in K_{m,n}\,} For instance, in our example of multiplication of 3 matrices D = ABC, it doesn’t matter if we perform AB first or BC first. Different Types of Matrix Multiplication . D. P 11 P 22 + P 12 P 21 = 0. {\displaystyle 0_{K}} Definition 3.2. Sei A eine reelle mxn Matrix. Definition. Matrix multiplication is associative, and so all parenthesizations yield the same product. Some examples of zero matrices … Schreibe die Formel der Hauptdiagonalwerte auf, das sind Summen von Einträgen der Matrix A quadriert, also genau dann gleich 0, wenn alle Summanden =0 sind. Appearently the output matrix has a value of 0 no matter what … C. P 11 P 22 - P 12 P 21 = 0. After zero matrices, the matrices whose actions are easiest to understand are the ones with a single nonzero entry. A matrix with all zeroes except for a one in the , entry is an , unit matrix. Occurrences. For example, if A is an m-by-0 empty matrix and B is a 0-by-n empty matrix, then A*B is an m-by-n matrix of zeros. Matrix multiplication is used in many scientific applications and recently it has been used as a replacement for convolutions in Deep Neural Networks (DNNs) using the im2col operation. Multiply doubly blocked toeplitz matrix with vectorized input signal Matrix multiplication is not universally commutative for nonscalar inputs. $\blue 3 \begin{bmatrix} 5 & 2 & 11 \\ 9 & 4 & 14 \\ \end{bmatrix} = \begin{bmatrix} \blue 3 \cdot 5 & \blue 3 \cdot 2 & \blue 3 \cdot 11 \\ \blue 3 \cdot 9 … Convert the input matrix to a column vector. {\displaystyle m\times n} & . Mirror Matrix Multiplication. The mortal matrix problem is the problem of determining, given a finite set of n × n matrices with integer entries, whether they can be multiplied in some order, possibly with repetition, to yield the zero matrix. CUDA Matrix Multiplication: Outputting Zero. 7. Informationsgehalt bei Shannonscher Informationstheorie - Herleitung. A diagonal matrix is a square matrix whose off-diagonal entries are all equal to zero. Learn what a zero matrix is and how it relates to matrix addition, subtraction, and scalar multiplication. Ask Question Asked 7 years, 11 months ago. Hope it was helpful :) & . Associative law: (AB) C = A (BC) 4. Using this library, we can perform complex matrix operations like multiplication, dot product, multiplicative inverse, etc. We can treat each element as a row of the matrix. m Examples . P 1 k=0 1 k! Die Matrizenmultiplikation oder Matrixmultiplikation ist in der Mathematik eine multiplikative Verknüpfung von Matrizen. A dense matrix is where all / significant percentage (>40%) of the elements are non zeros. Open Live Script. The product of two matrices A and B is defined if the number of columns of A is equal to the number of rows of B. Matrix of Zeros. & . Zuerst dachte ich, ich mach ne Fallunterscheidung. Fast sparse matrix multiplication ⁄ Raphael Yuster y Uri Zwick z Abstract Let A and B two n £ n matrices over a ring R (e.g., the reals or the integers) each containing at most m non-zero elements. n × Open Live Script. In simple words, “A+0 = A” and “A – 0 = A.” Example : Similarly, you can see that the subtraction of a Null matrix from any other matrix will give the other matrix itself as result. & 0 \\ 0 & 0 & 0& . Create a 4-by-4 matrix of zeros. Multiplying an M x N matrix with an N x P matrix results in an M x P matrix. Open Live Script. The order of the matrices are the same 2. This is means that if you were to multiply a zero matrix with another non-zero matrix, then you will get a zero matrix. 8.1.1.4 Additive identity matrix (zero matrix) Let A and 0 be matrices with the same size, then A + 0 = A, where is 0 called zero matrix. The matrix exponential of is given by the identity matrix.An zero matrix can be generated in the Wolfram Language as ConstantArray[0, m, n]. A. P 11 P 22 — P 12 P 21 = 1. Matrix multiplication is only defined if the number of columns of the first matrix equals the number of rows of the second matrix. 0. The main condition of matrix multiplication is that the number of columns of the 1st matrix must equal to the number of rows of the 2nd one. Zeigen Sie, dass aus ATA=0 stets A=0 folgt. in a single step. CUDA Matrix Multiplication: Outputting Zero. & . Like other typical Dynamic Programming(DP) problems , recomputations of same subproblems can be avoided by constructing a temporary array m[][] in bottom up manner. $$(A^T \cdot A)_{ii} = \begin{pmatrix} \sum\limits_{i=1}^{n} a_{ii} \cdot a_{ii} \end{pmatrix} = \begin{pmatrix} \sum\limits_{i=1}^{n} a_{ii}^2 \end{pmatrix} = 0$$, Einzige Lösung $$a_{ii}=0, \forall i\in\left\{ 0,1,...,n\right\}$$, Ich erhalte für die i,ite Komponente von A^T A, $$(A^T \cdot A)_{ii} = \sum\limits_{j=1}^{n} a^T_{ij} \cdot a_{ji} = \sum\limits_{j=1}^{n} a^2_{ji}=0$$. 6 ] it is denoted by I if the size is immaterial or can be trivially determined by context! P 11 P 22 — P 12 P 21 = 0 und a * C = 1! 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Product is the number of rows in the NumPy library be equal to the number of in! = a T, F, a > is thebase container adaptor for dense.. | 2021-06-13T08:45:12 | {
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https://math.stackexchange.com/questions/3361554/english-to-formal-predicate-logic | English to formal predicate Logic
Convert the following into formal predicate logic. Define predicates as necessary. Then negate the predicate sentence. Push all negations to the closest terms.
1) There are at least two people who everyone knows. Domain = {People}
2) Every student takes at least two classes. Domain = {people, classes}
3) All Students know each other. Domain ={ All people}
My take in this.... for the 1) part is it valid to do something like this ∃𝒙∃𝒚∀z𝑷(𝒙, 𝒚,z). Where in my words i could be totally wrong.. there exist a pair (x,y) who everyone (z) knows.
For the 2) part i know that there has to be and existential quantifier and a universal. but I'm not sure how to write them down.
For the Last part 3) is it this ∀𝒙∀𝒚𝑷(𝒙, 𝒚) just on the basis that for all X there is a Y
• People knowing each other should be expressed as a binary relation, not a ternary relation, so your solution to $1$ isn't right. Also, you need to expressly note (in formal logic) that $x$ and $y$ are not the same person. $\exists x, y \forall z P(x, z) \land P(x, y)$ is not a correct solution because it's possible that $x$ and $y$ are the same person. Your solution to $3$ isn't quite correct because your domain is all people, but your statement is only supposed to be true of all students. So you need a relation symbol to denote that a member of your domain is a student. – Robert Shore Sep 18 '19 at 23:35
• Could you help with the final solution? For the 3rd statement do you suggest using p(x) and q(x) is that what you mean? @RobertShore – Elchavo18 Sep 18 '19 at 23:44
• I'd strongly prefer to provide hints. For $3$, use $S(x)$ to denote "$x$ is a student." And a minor tweak to what I wrote for $1$ will give you a correct solution. – Robert Shore Sep 18 '19 at 23:46
• Hey the hints help with the learning process. Much appreciated. I just ask for some patience if i ask a lot of questions. Any hint for the second part while i work on the other 2? @RobertShore – Elchavo18 Sep 18 '19 at 23:51
• You'll need separate predicates to define "people" and "classes" and a binary relation to express that a person is taking a class. – Robert Shore Sep 19 '19 at 0:07
1) There are at least two people who everyone knows. Domain = {People}
My take in this.... for the 1) part is it valid to do something like this ∃𝒙∃𝒚∀z𝑷(𝒙, 𝒚,z). Where in my words i could be totally wrong.. there exist a pair (x,y) who everyone (z) knows.
You must say: "There are some $$x$$ and some $$y$$ who are not the same people and every $$z$$ will know $$x$$ and know $$y$$."
You should also use a bivariate predicate such as $$\def\op#1{\operatorname{\rm #1}}\op{K}(~,~)$$ for "_ knows _"
$$\exists x~\exists y~\forall z~\bigl(x\neq y\wedge \op{K}(z,x)\wedge \op{K}(z,y)\bigr)$$
2) Every student takes at least two classes. Domain = {people, classes}
"For every $$x$$ who is a student, then there is an $$y$$ which is a class that is taken by $$x$$ and there is a $$z$$ which is another class that is taken by $$x$$."
You will need predicates for: $$\op P(~)$$ "_ is a people", $$\op C(~)$$ "_ is a class", and $$\op T(~,~)$$ "_ takes _" .
3) All Students know each other. Domain ={ All people}
"For every $$x$$ who is a student, then for every $$y$$ who is a student, then $$x$$ knows $$y$$."
Use predicates $$\op S(~)$$ for "_ is a student", and $$\op{K}(~,~)$$ for "_ knows _".
(NB: do you need to worry about whether $$x$$ and $$y$$ are the same people?)
• Thank You! Your break downs help a lot. I think I got it from here. – Elchavo18 Sep 19 '19 at 1:40 | 2020-01-17T21:50:07 | {
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https://math.stackexchange.com/questions/3012870/how-to-create-rotation-matrix-in-3d-space | # How to create rotation matrix in 3D space?
In a 3d space $$(x,y,z)$$ where $$y$$ is the height, I have a plane which I constructed from 2 angles (creating a normal vector).
For example:
$$\alpha = -\pi, \beta = \frac{-\pi}{2}$$
To calculate the normal vector, I use:
$$\left(\begin{matrix} \cos(\alpha)\cos(\beta) \\ \sin(\beta) \\ \sin(\alpha)\cos(\beta) \\ \end{matrix}\right)$$
to get
$$\left(\begin{matrix} -0.6 \\ 0.8 \\ 0 \\ \end{matrix}\right)$$
I get the plane
$$e: 0 = -0.6x + 0.8y + 0z$$
A second plane is given
$$f: 0 = z$$
With what I have, can a calculate a rotation matrix of $$e$$ to $$f$$? If not, what is wrong with my approach?
My goal is to find the coordinates of any point $$P$$ of $$e$$ on $$f$$.
I of course found the Wikipedia page of Rotation matrix and other answers here, but I'm unable to construct the matrix. I'm missing some intermediate steps of understanding.
But typically in a question like this, what you want is a rotation from the first plane (I"m going to call that $$P_1$$, with normal vector $$\nv_1$$) to the second ($$P_2$$, normal vector $$\nv_2$$) with the additional property that the rotation doesn't move the line of intersection between the two planes.
That's not so hard to construct, surprisingly.
1. Let $$\vv = \nv_1 \times \nv_2 / \|\nv_1 \times \nv_2 \|$$; that's one of the two unit vectors in the intersection $$P_1 \times P_2$$.
2. Let $$\wv = \vv \times \nv_2$$; that's a unit vector lying in the second plane, perpendicular to $$\vv$$. So $$\vv, \nv_2, \wv$$ is an orthonormal basis for 3-space.
3. Let $$\uv = \vv \times \nv_1$$; that's a unit vector in the first plane, perp. to $$\vv$$. So $$\nv, \nv_1, \uv$$ is an orthonormal basis for $$3$$-space.
4. We want to transform the second basis to the first. Let's assume everything is written as column vectors, and let $$K = \pmatrix{\vv& \nv_1 & \uv}$$ be the matrix with $$\vv, \nv_1,$$ and $$\uv$$ as its columns, and let $$L = \pmatrix{\vv& \nv_2 & \wv}$$ be similarly constructed.
5. Let $$R = L K^t$$. Then $$R$$ is a rotation taking the first plane to the second.
Let's work through those general steps for your particular case, where $$\nv_1 = \pmatrix{-3/5 \\ 4/5 \\ 0}$$ and $$\nv2 = \pmatrix{0 \\ 0 \\ 1}$$ are the normal vectors for the two planes. As it happens, because your two normal vectors are already perpendicular, something very simple happens along the way, but you may find yourself wanting to solve the more general problem later, which is why I gave the more general answer above.
1. Compute $$\nv_1 \times \nv_2 = \pmatrix{4/5 \\ 3/5 \\ 0}$$. The length of this vector happens to be $$1$$, so dividing by that length gives $$\vv = \pmatrix{4/5 \\ 3/5 \\ 0}$$.
2. Compute $$\wv = \vv \times \nv_2 = \pmatrix{3/5\\-4/5 \\ 0}$$.
3. Compute $$uv = \vv \times \nv_1 = \pmatrix{0\\0\\1}$$.
4. We let $$K = \pmatrix{ 4/5 & -3/5 & 0 \\ 3/5 & 4/5 & 0\\ 0 & 0 & 1}, L = \pmatrix{ 4/5 & 0 & 3/5\\ 3/5 & 0 & -4/5\\ 0 & 1 & 0}$$
5. We compute $$M = LK^t = \pmatrix{ 4/5 & 0 & 3/5\\ 3/5 & 0 & -4/5\\ 0 & 1 & 0} \pmatrix{ 4/5 & 3/5 & 0 \\ -3/5 & 4/5 & 0\\ 0 & 0 & 1} = \pmatrix{ 16/25 & 12/25 & 3/5 \\ 12/25 & 9/25 & -4/5\\ -3/5 & 4/5 & 0}$$
... and that should be your matrix.
• Thanks for your answer. I actually understood how to construct one now. But I might have looked for the wrong solution regarding my initial problem (which I was too shy to ask here). Maybe for educational reasons I'll ask another questions where projection will be the right answer as my research didn't give me that on. – Patrick B. Nov 25 '18 at 17:05
My goal is to find the coordinates of any point $$P$$ of $$e$$ on $$f$$.
Looks like what you need is actually a projection of any point $${\bf . x}\in e$$ onto $$f$$. Call $$\hat{n}_f$$ a normal vector of the plane $$f$$, in your case
$$\hat{n}_f = \pmatrix{0 \\ 0 \\ 1}$$
Now, calculate
$${\bf x}_f = {\bf x} - ({\bf x}\cdot\hat{n}_f)\hat{n}_f$$
• Hmm, it seems to work actually like this. Did I go in the wrong direction in the first place? I'll try this with my real world problem to whether I actually understood my problem in the first place. Thanks. – Patrick B. Nov 25 '18 at 17:02
• No, I definitely don't want a projection. I need a rotation. The distance of the point from (0,0,0) needs to be the same on both planes. – Patrick B. Dec 2 '18 at 9:10 | 2019-06-18T09:51:20 | {
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https://yutsumura.com/does-the-trace-commute-with-matrix-multiplication-is-tr-a-b-tr-a-tr-b/ | # Does the Trace Commute with Matrix Multiplication? Is $\tr (A B) = \tr (A) \tr (B)$?
## Problem 634
Let $A$ and $B$ be $n \times n$ matrices.
Is it always true that $\tr (A B) = \tr (A) \tr (B)$?
If it is true, prove it. If not, give a counterexample.
## Solution.
There are many counterexamples.
For one, take
$A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \text{ and } B = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}.$
Then $\tr(A)=1, \tr(B)=1$, and hence $\tr(A) \tr(B) = 1$, while $\tr(AB) = 0$ as $AB = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
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• True or False: If $A, B$ are 2 by 2 Matrices such that $(AB)^2=O$, then $(BA)^2=O$ Let $A$ and $B$ be $2\times 2$ matrices such that $(AB)^2=O$, where $O$ is the $2\times 2$ zero matrix. Determine whether $(BA)^2$ must be $O$ as well. If so, prove it. If not, give a counter example. Proof. It is true that the matrix $(BA)^2$ must be the zero […]
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• The Vector Space Consisting of All Traceless Diagonal Matrices Let $V$ be the set of all $n \times n$ diagonal matrices whose traces are zero. That is, \begin{equation*} V:=\left\{ A=\begin{bmatrix} a_{11} & 0 & \dots & 0 \\ 0 &a_{22} & \dots & 0 \\ 0 & 0 & \ddots & \vdots \\ 0 & 0 & \dots & […]
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##### Is the Trace of the Transposed Matrix the Same as the Trace of the Matrix?
Let $A$ be an $n \times n$ matrix. Is it true that $\tr ( A^\trans ) = \tr(A)$? If it...
Close | 2019-12-11T11:44:43 | {
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https://math.stackexchange.com/questions/2561363/between-any-two-continuous-functions-fg-can-we-find-a-real-analytic-function | # Between any two continuous functions $f>g$, can we find a real-analytic function?
I asked myself a question which I thought was interesting, but I'm not sure how to approach it.
## The Question
The question is, given two continuous functions $f,g:\mathbb{R}\rightarrow\mathbb{R}$ such that $f > g$, is there a real analytic function $h$ between $f$ and $g$ (so that $f > h > g$)?
An easier version of the question can be asked over closed intervals. Given two continuous functions $f,g:[a,b]\rightarrow\mathbb{R}$ such that $f > g$, is there a real analytic function $h$ between $f$ and $g$?
In this case, the answer is yes, and the result is actually stronger than expected. Essentially, this is an application of the Weierstrass Approximation Theorem.
By the extreme value theorem, the function $H(x) = (f(x) - g(x))/2$ has a minimum $M > 0$. You can define $k(x) = (f(x) + g(x))/2$ and then apply the Weierstrass Approximation Theorem to get a polynomial $p:[a,b]\rightarrow\mathbb{R}$ such that $\lVert p - k\rVert < M$. Then $p$ is a polynomial such that $p(x)\in (k(x) - M, k(x) + M)\subseteq (g(x), f(x))$ for all $x\in[a,b]$.
However, you run into issues when the domain is the real line. Here are some thoughts I had so far.
## For
I am aware that there is a generalized Stone-Weierstrass Theorem on locally compact Hausdorff spaces $X$ concerning functions in $C_{0}(X)$. There are two issues though. The first is that our continuous functions $f,g$ don't necessarily tend to zero as $x\rightarrow\pm\infty$. The second is that the difference $f(x) - g(x)$ is allowed to get arbitrarily small as $x\rightarrow\pm\infty$.
The first issue isn't major to me, because I think I have a way to evade it. For all intents and purposes, we can assume that $f,g$ tend to zero if it suits our needs.
The second issue, though, seems to be fatal, and it seems like there is no way to modify the Stone-Weierstrass Theorem appropriately.
This makes me think I need a completely different approach.
## Against
If I want to make a counterexample, maybe I can use pathological functions like the Weierstrass function $w:\mathbb{R}\rightarrow\mathbb{R}$. I'm thinking that if we take $f(x) = w(x) + e^{-x^{2}}$ and $g(x) = w(x)$, we can force any intermediate function $h$ to become "more and more detailed" as it gets sandwiched closer and closer between $f,g$ for large $x$. Then this function might become "too detailed" to be analytic, but I have no rigorous way to express this idea.
Even worse, I'm having doubts that this would go anywhere because there is always "wiggle room" between $f$ and $g$, so maybe we can fit an analytic function between them after all.
Are there any suggestions on how to make progress on this question?
@NateEldredge mentioned a result of Carleman. It is indeed a beautiful theorem: Suppose $G:\mathbb R \to \mathbb R$ is continuous, and $\epsilon: \mathbb R \to (0,\infty)$ is any positive continuous function. Then there exists an entire function $F$ such that $|F(x)-G(x)| < \epsilon(x)$ for all $x\in \mathbb R.$
In your problem we can let $G = (f+g)/2$ and $\epsilon = (f-g)/2.$ Then
$$\frac{g-f}{2} < F - \frac{f+g}{2} < \frac{f-g}{2}\,\,\, \text { on }\,\, \mathbb R$$ | 2022-01-20T03:09:52 | {
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http://mathhelpforum.com/calculus/193758-minimization-problem.html | # Math Help - Minimization problem
1. ## Minimization problem (multiple choice(
A cylindrical can is to contain 2000 cu. inches of liquid. What dimensions will minimize the amount of metal used in the construction of the can?
Choices:
a. D = 6.828in, H = 13.656
b. D = 6.424 in, H = 13.555
c. D = 6.222 in , H = 13.757
d. D = 6.626 in, H = 13.454
I honestly dont know where to start
2. ## Re: Minimization problem (multiple choice(
Setting r the radious of the base of the can and h its height, then we have...
$v= \pi\ r^{2}\ h$ (1)
$a= \pi\ r^{2} + 2 \pi r h$ (2)
From (1) You derive...
$h= \frac{v}{\pi r^{2}}$ (3)
... so that from (2) and (3) You obtain...
$a= \pi r^{2} + \frac{2 v}{r}$ (4)
Because v is known, Your task is to find the value of r that minimizes a...
Kind regards
$\chi$ $\sigma$
3. ## Re: Minimization problem (multiple choice(
Originally Posted by chisigma
$a= \pi\ r^{2} + 2 \pi r h$ (2)
circular area of bot and top included ?
$a= 2\pi\ r^{2} + 2 \pi r h$
thnks btw
4. ## Re: Minimization problem (multiple choice(
The cilinder has been considered 'open on the top'... if You want to include the top circular area the procedure is similar and the task is left to You...
Kind regards
$\chi$ $\sigma$
5. ## Re: Minimization problem (multiple choice(
Taking into account the top and bottom circular areas the equations become...
$v=\pi\ r^{2}\ h$ (1)
$a=2\ \pi\ r\ (r+h)$ (2)
... and from them...
$a= 2\ \pi\ r^{2} + \frac{2\ v}{r}$ (3)
Now applying the stand procedure we find the r for minimal area from the equation...
$\frac{d a}{dr}+ 4\ \pi\ r - \frac{2\ v}{r^{2}}=0$ (4)
... the solution of which is...
$r=(\frac{v}{2\ pi})^{\frac{1}{3}}$ (5)
For curiosity I mesured the dimensions of a can of 'Red Kidney-Beans' [my preferred ...] finding $r=3.75,\ h=11.5\ \implies v=508$. Applying (5) for $v=508$ we obtain an optimum value $r=4.32$... not very far from $r=3.75$ which requires however little more metal ...
Kind regards
$\chi$ $\sigma$
6. ## Re: Minimization problem
Dont forget that not only are they trying to minimize the metal used in the can, but also trying to optimize shipping and storage as well as production. Perhaps their cans fit better in boxes or something? | 2015-05-23T08:06:42 | {
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https://math.stackexchange.com/questions/2570869/matchings-using-halls-theorem-why-does-only-1-of-these-solutions-work | # Matchings using Hall's theorem: Why does only 1 of these solutions work?
Let $G$ be a (simple) bipartite graph with parts $A,B$ such that $|A|=|B|=n$ and with minimum degree at least $n/2-1$. Prove that $G$ contains a matching which covers all but at most $2$ vertices in each part.
Solution 1: Add $1$ vertex to each part $A$ and $B$ and connect it to all vertices in the opposite part. The resulting graph $G'$ has minimum degree at least $n/2-1+1=n/2$, so by Hall's theorem $G'$ has a perfect matching $M'$ with $n+1$ edges. At most $2$ of these edges are adjacent to the new vertices we added, so removing these edges we obtain a matching $M$ for $G$ of size at least $n-1$.
Solution 2: Add $2$ vertices to each part $A$ and $B$ and connect each to all vertices in the opposite part. The resulting graph $G'$ has min degree at least $n/2-1+2=n/2+1$, so by Hall's theorem $G'$ has a perfect matching $M'$ with $n+2$ edges. At most $4$ of these edges are adjacent to the new vertices, so removing these edges we obtain a matching $M$ for $G$ of size at least $n-2$.
The second gives the desired result but why does the first one suggest a better bound?
Edit: In the above I am using the following theorem: "If $G$ is bipartite with parts $|A|=|B|=n$ and minimum degree at least $n/2$ then $G$ has a perfect matching." This can be shown from the usual Hall's theorem: "If $G$ is bipartite with parts $A$, $B$, then $G$ has a matching covering all of $A$ if and only if $|N(S)|\geq |S|$ for all $S\subseteq A$.
• @bof: That graph has a perfect matching, namely $a_i$ with $b_{11-i}$ for each $i$. – hmakholm left over Monica Dec 17 '17 at 20:33
• @bof Aha I think that's exactly right! For some reason I overlooked that fact when applying the theorem. Thank you :) – Alex.F Dec 17 '17 at 20:56
• @bof Would you be able to post that as an answer so I can mark this question as solved? – Alex.F Dec 17 '17 at 20:59
• Solution 1 actually shows that, if you strengthen the assumption to "minimum degree at least $\frac n2-\frac12$", then $G$ contains a matching of size $n-1.$ – bof Dec 17 '17 at 22:03
In Solution 1 you need $G'$ to have minimum degree at least $(n+1)/2.$ If $n$ is odd this works because $\delta(G)\ge\lceil n/2-1\rceil=(n-1)/2$ in that case, but for even $n$ it fails.
The argument in Solution 1 actually shows that, if $G$ is a bipartite graph with parts $A,B$ such that $|A|=|B|=n,$ and with minimum degree at least $(n-1)/2,$ then $G$ contains a matching of size $n-1.$ | 2021-01-20T04:44:48 | {
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http://ptue.lab-consulting.it/uses-of-measures-of-dispersion.html | # Uses Of Measures Of Dispersion
measure the dispersion. Mathematically, it estimates vector variance for all normal vectors of individual planar surfaces. Coefficient of variation is an important relative measure of dispersion. Using recent results and algorithms from experimental design theory, we show how to construct optimal measures numerically. At room temperature, neopentane (C 5 H 12) is a gas whereas n-pentane (C 5 H 12) is a liquid. The Dispersion Statistics on Productivity (DiSP) is a joint experimental data product from the U. More used measure is standard deviation where we take the square of the variance. The measures of central tendency and dispersion taken together give a better picture of a data set than measure of central tendency alone. For Ungrouped Data. Using methods originally developed by astronomers to view stars more clearly through Earth's atmosphere, optometry researchers at Indiana University have taken the first undistorted microscopic i. In this lesson, we are going to talk about measures of dispersion that complement the averages and help us interpret what they mean. To be effective and reliable, the metrics we choose to use need to have ten key characteristics. Provide feedback. These two types of measures together help us to sum up a distribution of scores without looking at each and every score. Out of several measures of dispersion, the most frequently used measure is ‘standard deviation’. Measure of Dispersion. Variance "Average Deviation". This is the link of my previous post about the range and the quartile (Measure of dispersion(The range and Quartile deviation)). Properties of good measure of Dispersion a) It should be easy to understand. Dispersion is important to finance as the data points of say, a stock, determine the mean, which in turn helps determines the stock's trend. Yes, it’s legal to own and requires no NFA paperwork or tax stamp, but be sure to check your state laws to make sure there are no issues. The coefficient of variation (COV) is a measure of relative event dispersion that's equal to the ratio between the standard deviation and the mean. In a statistical sense, dispersion has two meanings: first it measures the variation of the items among themselves, and second, it measures the variation around the average. A commonly used measure of dispersion is the standard deviation, which is simply the square root of the variance. By dividing the inter-quartile range by the sum of the 1st and 3rd quartile the Coefficient of Quartile Variation can be obtained. This post is the extension of previous posts, we will be going forward with previously imported data from 104. If the original data is in dollar or kilometers, we do not use these units with relative measure of dispersion. Harkowitz in 1952, that the standard deviation of portfolio returns be used as a measure of total portfolio risk. Dispersion measures how the various elements behave with regards to some sort of central tendency, usually the mean. Also need to know how spread out or variable the data are. In this lesson, we are going to talk about measures of dispersion that complement the averages and help us interpret what they mean. Don’t worry about this. Dispersion control is also important in lasers that produce short pulses. They are measures of central tendency and dispersion. While measures of central tendency are used to estimate "normal" values of a dataset, measures of dispersion are important for describing the spread of the data, or its variation around a central value. Range, Variance and Standard Deviation as Measures of Dispersion For Students 8th - 11th Continuing his conversation about data sets from the previous video, Sal introduces range, variance, and standard deviation as means of dispersion. Measures of Central Tendency Quite often there will be found in the data a tendency, not withstanding their variability to cluster around a central value. Mathematics Grade 10. We may require some measure to explain the amount of spread in the data. It is now a possibility to use DM’s as a distance measure in cosmology. , be observations then mean is obtained by dividing the sum of observations by. In statistics, dispersion (also called variability, scatter, or spread) is the extent to which a distribution is stretched or squeezed. Some measures of dispersion are the standard deviation , the average deviation , the range , the interquartile range. The range is simply the difference between the highest score and the lowest score in a distribution plus one. * The measures of dispersion are important to compare uniformity, consistency and reliability amongst variables/ senes * Absolute measures of dispersion are expressed in terms of original unit of series. Now you must have understood the Measurement of Dispersion and how to calculate and use in real life. the new notation, k gives a measure of how fast the wave oscillates as a function of x. Dispersion is central to process improvement because a fundamental aim of Six Sigma is to minimize the amount of variation. Render/Auto Render – by default auto render feature is enabled, which means you can produce the dispersion effect in real-time. The purpose of measures of dispersion is to find out how spread out the data values are on the number line. The standard deviation. The standard deviation (also variance) provides the most information, since it uses all of the values in the distribution in its calculation. While measures of central tendency are used to estimate "normal" values of a dataset, measures of dispersion are important for describing the spread of the data, or its variation around a central value. meration or visual presentation. The following are the results of this analysis; N = 10, M=22. See full list on byjus. Percentiles. Measures of central tendency: There are several measures of central tendency. measure of central tendency. Wet or liquid dispersion is the most common method of sample dispersion for laser diffraction particle size measurements, being especially suitable for samples containing fine particles below a few microns in size. The simplest measure of spread is the range, expressed either as the biggest and smallest number in the data (e. One is a Algebraic method and the other is Graphical method. The residuals from those models, "purged" of such effects, are then used to construct price dispersion measures which are regressed on variables related to search costs. However, we might be interested in knowing more about our data and describing the data in greater depth. Measures of dispersion were computed to understand the variability of scores for the age variable. Common examples of measures of statistical dispersion are the variance, standard deviation, and interquartile range. Robust Dispersion Measures in the Presence of Outliers. The table gives the function names and descriptions. Also called measures of variability, the dispersion is, at least, as important as the centrality of a variable, i wrote a post about centrality yesterday, linked at bottom of this post. The median and mode only use one value; the median only cares about the middlemost value, the mode only about the one most common. All these merits of standard deviation make SD as the most widely and commonly used measure of dispersion. Analysis of quantitative data was done through use of both Inferential and descriptive statistics with the help of Statistical Package for Social Sciences (SPSS version 22). Calculating Measures of Dispersion. The most important assumption of which to be aware is the following: both the standard deviation and the So if you have a nominal or ordinal level variable, use the range as your measure of dispersion. If we have two sets of observations, we cannot always use the absolute measures to compare their dispersions. Mathematically, it estimates vector variance for all normal vectors of individual planar surfaces. Dispersion is the phenomenon which gives you the separation of colorsin a prism. This research paper on Measures of Central Tendency and Dispersion was written and submitted by your fellow student. Measures of Dispersion. Measures of central tendency are also known as Measures of location. only when individual data values are unavailable c. dispersion definition: 1. It can be computed with =AVEDEV(). It removes the defects of the averages and presents doubtful conclusions in their true form. In our hypothetical exercise, there’s no dispersion at all between A and B, and a considerable dispersion between C and D. Measures of dispersion become very vital in such cases. Standard deviation and average deviation are also commonly used methods to determine the dispersion of data. range and standard deviation). • Standard deviation of a set of values can be obtained by using the formula below. What statistical measures are used for describing dispersion in data? How do they differ from one another? Statistical measures used for describing dispersion in data are range, quartiles variance, and standard deviation. Dispersion is also used to determine volatility: data points all over the chart indicate that a stock has wild fluctuation in price. Measures of Dispersion Greg C Elvers, Ph. The residuals from those models, "purged" of such effects, are then used to construct price dispersion measures which are regressed on variables related to search costs. Measures of dispersion indicate how much the data in a given set of numerical data are spread out. They are sometimes called coefficients of dispersion. Mathematically, it estimates vector variance for all normal vectors of individual planar surfaces. Connor defines measures of dispersion as ‘dispersion is the measure extended to which individual items vary’. Dispersion is a characteristic of random variables. , the standard deviation, the mean deviation (MD) about the mean and the second L-moment, are analyzed in terms of their properties and mutual relationships. For example, when rainfall data is made available for different days in mm, any absolute measures of dispersion give the variation in rainfall in mm. It involves finding the square root of the average squared distances from the mean. hich of the following correctly defines dispersion? the most common unit used to measure loudness the study of the characteristics of populations, especially human populations the process of removing salt from ocean water the pattern of distribution of organisms in a population. It's important to look the dispersion of a data set when interpreting the measures of central tendency. A measure of spread, sometimes also called a measure of dispersion, is used to describe the variability in a sample or population. 3 To gauge the reliability of an average. It is easily understood and computed but depends exclusively on the two extreme values while it is desirable to have a measure dependent on all the values. 6 Measures of dispersion The mean and median measure central tendancy, i. particle dispersion. No, you can’t add a real grenade launcher to your rifle, but you can legally get this LMT 37mm M203 and launch non-lethal projectiles, flares, smoke rounds, and even fireworks. These measures are a sort of ratio and are called coefficients. Using meas- ures of relative dispersion, this analysis shows that in- dustry characteristics such as degree of unionization, geographic location, occupational mix, and method of wage payment influence the amount of variation. Measure of dispersion. 2 million), multiplied by a thousand. Three dispersion measures of a random variable, i. Measures of Relationship While measures of central tendency provide the value that is an ideal representative of a set of observations, the measures of dispersion take into account the internal variations of the data, often around a. Average Deviation. The Dispersion of Stock Returns. The most important assumption of which to be aware is the following: both the standard deviation and the So if you have a nominal or ordinal level variable, use the range as your measure of dispersion. A quantity that measures how the data are dispersed about the average is called measures of dispersion. Kurtosis is a measure of whether the data are heavy-tailed or light-tailed relative to a normal distribution. In Lesson 2, you learned about frequency distributions, ratios, proportions, and rates. 410 × 10 −4 MHz −2 cm −3 pc s −1, ν is the observing frequency and L the distance from the Earth to the pulsar. 7 NODDI has subsequently been developed further to incorporate a quantification of anisotropic orientation dispersion, and analysis using. 4 To know the limits of the items. (786) AFC-03(QT) : MEASURE OF DISPERSION THEORY+IMPORTANT. A lOX dispersion staining objective (W. Relative: Measure of dispersion is free from unit of. the nature of dispersion, we develop a series of five new measures and explore their relationships with communication frequency data from a sample of 182 teams (of varying degrees of dispersion) from a Fortune 500 telecommunications firm. Without that, the measures of dispersion will not be very useful. Using this measure, data on ordinal scales can be given a value of dispersion that is both logically and theoretically sound. As soon as you mark the area, the image will get pixelated. Measures of Dispersion 3. The difference between a single data point x, and the mean x bar is called deviation from the mean. Schooling, Politics, and Life After Death. Density and size are useful measures for characterizing populations. It is also the most important because of being the only measure of dispersion amenable to algebraic treatment. Quantitative Aptitude & Business Statistics: Measures of Dispersion 4 Significance of Measuring Dispersion. In 1985, Mirvis1 reported on a significant spatial variation in QT intervals in normal individuals and patients with acute myocardial infarction. There has been a considerable divergence in mag netosphere temperature estimates. Measures of Dispersion As well as knowing what a typical value is for a variable, we also like to know how spread-out the observations are around that typical value. Another term for these statistics is measures of spread. By far the most commonly used measures of dispersion in the social sciences are varianceand standard deviation. The Usual Measures of Dispersion: The usual measures of dispersion, very often suggested by the statisticians, are exhibited with the aid of the following chart: Primarily, we use two separate devices for measuring dispersion of a variable. Dispersion (physics) synonyms, Dispersion (physics) pronunciation, Dispersion (physics) translation, English dictionary definition of Dispersion (physics). served dispersion measure using the results of IllustrisTNG-100 simulation (Nelson et al. Sign Up Free. Measures of dispersion are important in any statistical study when you're trying to draw conclusions from data. It enables algebraic treatment. Because of this complexity, the average deviation is not a very commonly used measure of dispersion. e) zero, if all the data are same and increases as the data become more diverse. popula7on } Used in conjunc7on with a measure of. The false prediction rate can be reduced to almost 0. June 22, 2019 September 13, 2019 By continuing to use this website, you agree to their use. The quartiles are commonly used (much more so than the percentiles or deciles). The measure used here is the median logarithmic deviation of income as published by the Census Bureau in its annual poverty report. The range is (11 - 8) = 3 kg. Set E: 12 12 0 8 Set F: 1 15 10 2 18. Class 11 Economics Notes for Measures of Dispersion – Get here the Notes, Question & Practice Paper of Class 11 Economics for topic Measures of Dispersion. The assumptions for MCT are much simpler. variance b. , Shalabh (2016) Measures of Central Tendency and Dispersion. Concept 7: Measures of Dispersion. The most important assumption of which to be aware is the following: both the standard deviation and the So if you have a nominal or ordinal level variable, use the range as your measure of dispersion. Standard Deviation is the most common measure and is useful for more in-depth data analysis. When mean is utilized as a measure of central tendency or symmetric numerical data, SD is used. The purpose of measures of dispersion is to find out how spread out the data values are on the number line. Dispersion occurs with transparent surfaces that are not parallel to each other, such as gemstone facets. The following two are used to measure the dispersion of all variables together. Measures of dispersion are important in any statistical study when you're trying to draw conclusions from data. Measures of association are used in various fields of research but are especially common in the areas of epidemiology and psychology, where they frequently are used to quantify relationships between exposures and diseases or behaviours. Define dispersion. The relative measures of dispersion may also determine the. Hence, the stability of a colloidal system is inevitably linked to a notion of time, defined by process, use and/or application involved. The range is the simplest measures of dispersion and is a crude measure of variability. In a statistical sense, dispersion has two meanings: first it measures the variation of the items among themselves, and second, it measures the variation around the average. Here also, the deviations of all the values from the mean of the distribution are considered. Because of this complexity, the average deviation is not a very commonly used measure of dispersion. Rigidly Defined: A good measure of dispersion should be properly and rigidly defined so that it does not create any problem for the reader in analysing the data. Measures of spread together with measures of location (or central tendency) are important for identifying key features of a sample to better understand the population from which the sample comes from. It shows how much variation or "dispersion" there is from the average (mean, or expected. They are measures of central tendency and dispersion. Dispersion is also used to determine volatility: data points all over the chart indicate that a stock has wild fluctuation in price. Measures of Central Tendency – Mean, Median, Mode Measures of Dispersion or Variation – Range, Variance, Standard Deviation Measures of Position – Percentile, Quartiles. See full list on toppr. Sign Up Free. Some Commonly Used Measures of Relative Dispersion / Absolute Dispersion. The two most common types of statistical measures are those for central tendency and those for dispersion. The measures are functions of the 3rd and 4th powers of the difference between sample data values and the distribution mean (the 3rd and 4th central moments). In a way, mean deviation or standard deviation tell us more about the way data is spread. Two distinct samples may have the same. The "variance" is simply the square of the standard deviation. o One measure of dispersion is range. As with other types of measures, there is more than one approach to defining such a measure. Meaning of dispersion in English 'Since World War I, the wider dispersion of forces on the battlefield and the increased use of cover and concealment have reduced exposure to enemy fire. Start learning to code for free with real developer tools on Learn. c) Rigidly defined d) Based on all observations. At this point, you may not quite understand what we mean by spread or variation. If we have two sets of observations. Measures of dispersion provide a more complete picture. Section 1 develops measures of investment return which are used in the study. Measures of dispersion between forecasters for economic variables can also proxy for economic uncertainty (Figure 4). Measures of Dispersion General Classifications of Measures of Dispersion (page 232) Measures of Absolute Dispersion A measure of absolute dispersion has the same unit as the observations. Measures of Central Location and Dispersion As epidemiologists, we use a variety of methods to summarize data. They give the answers in the same units as the units of the original observations. 12 The measure of dispersion which uses only two observations is called: (a) Range (b) Quartile deviation (c) Mean deviation (d) Standard. url?scp=34249938123&partnerID=8YFLogxK. Rigidly Defined: A good measure of dispersion should be properly and rigidly defined so that it does not create any problem for the reader in analysing the data. Measures of Dispersion The central values i. 5 m thick, 180-m wavelength for ice 0. Two well-known examples are the standard deviation and the interquartile range. 75 m thick, and 200-m wavelength for ice 1 m thick. It's quite useful. There are a variety of income dispersion/variance measures, but they all show the same basic pattern: U. The coefficient of variation is the standard deviation divided by the mean. And let's compare it to this data set over here. A measure of dispersion is used to quantify the size of the differences of a variable. Which measure of central tendency best represents the data?. The principles of clinical psychology rely on averages and norms based on long-term studies and trends that practicing psychologists and researchers have documented and published. In this lesson, you'll learn about the different measures of dispersion and explore how they are related to each other as well as other summary statistics. The measures of dispersion are: Range. The measure of dispersion depending upon the lower and upper quartiles is know as the quartile deviation. _____ is defined as the difference between the smallest and. There are two main measures of dispersion:. 1 The scatter in a series of values about the average is called: (a) Central tendency (b) Dispersion (c) MCQ No 4. whenever computer packages for descriptive statistics are unavailable d. The statistics used for dispersion measures entails range, standard deviation, variance, and the quantiles (Aczel, 2012). Therefore, it’s very important to learn about the data characteristics and measure for the same. o Use the variance or standard deviation to characterize the spread of data. Robust Measures of Dispersion. Dispersion or distribution patterns show the spatial relationship between members of a population within a habitat. median " in Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. This gives a measure of the spread of values but no indication of. Dispersion is important to finance as the data points of say, a stock, determine the mean, which in turn helps determines the stock's trend. It is a very important property to used to identify and qualify gemstones. The measures are calculated based on company responses to the ifo business surveys and will be published on a monthly basis together with the results of the ifo business surveys as of August. (786) AFC-03(QT) : MEASURE OF DISPERSION THEORY+IMPORTANT. Measures of Central Location and Dispersion As epidemiologists, we use a variety of methods to summarize data. Thirdly, measures of Dispersion or Variation – Range, Variance, Standard Deviation. Measures of dispersion Measures of central tendancy tell you about how spread out (dispersed) or concentrated data are. The statistical measures used to describe the extent to which the data is scattered or squeezed are Measures of Dispersion. Measures of dispersion or variability will give us information. I develop a measure that uses the change in the implied volatility of exchange-traded. The following measures of dispersion are used to study the variation: The range The inter quartile range and quartile deviation The mean deviation or average deviation The standard deviation 11. The dispersions of the textual are generally taken to be a force which, by its very nature, inevitably resists all established forms of cultural authority. It is often used as a measure of risk. 3) Which of the following summary measures is/are influenced by extreme values. This appears to be a good average measure of deviation. Chapter 15 explains frequency distributions, measures of central tendency, and measures of dispersion. A more complex measure of dispersion is Variance. Lecture 8: Measures of Dispersion Lecture 9: Assumptions for Measures of Central Tendency/ Measures of Dispersion Descriptive Statistics: Measures of Central Tendency and Cross - tabulation Introduction Measures of central tendency (MCT) allow us to summarize a whole group of numbers using a single number. Relative: Measure of dispersion is free from unit of. Mean The mean of a data set is also known as the average value. Another important characteristic of a data set is how it is distributed, or how far each element is from some measure of central tendancy Yet another method of measuring how a data set is distributed is to extend the concept of median and use smaller and smaller divisions. There are three types of Measurement of Dispersion. In our hypothetical exercise, there’s no dispersion at all between A and B, and a considerable dispersion between C and D. measures of dispersion definition in the English Cobuild dictionary for learners, measures of dispersion meaning explained, see also 'measure',measured',measles',measurable', English vocabulary. It is defined as: It gives a measure of risk per unit of return, and an idea of the magnitude of variation in percentage terms. All these merits of standard deviation make SD as the most widely and commonly used measure of dispersion. A measure of statistical dispersion is a nonnegative real number that is zero if all the data are the same and increases as the data become more diverse. Measures of Dispersion for Grouped Data A. Dispersion is important to finance as the data points of say, a stock, determine the mean, which in turn helps determines the stock's trend. So let's think about different ways we can measure dispersion, or how far away we are from the center, on average. Depending on the fuel your specified generator uses and if secondary abatement is fitted you may also need to do dispersion modelling assessment for SO 2, particulates and ammonia. The range is the simplest measure of dispersion; it relates to the actual spread of values and is equal to the maximum less the minimum value. The "variance" is simply the square of the standard deviation. Variability in 2 or more distrn can be compared provided they are given in the same unit and have the same average. 5 To serve as a basis for control of the variability itself. Measures of Central Tendency 2. For the measures of dispersion considered, we will. Also called measures of variability, the dispersion is, at least, as important as the centrality of a variable, i wrote a post about centrality yesterday, linked at bottom of this post. There are three main measures of variation One measure of operation effectiveness is the time it takes to process the patients. how the dispersion of scores on a variable can be described using statistics such as a percent distribution, minimum, maximum, range, and standard deviation along with a few others; and • how a variable’s level of measurement determines what measures of central tendency and dispersion to use. 4 measures of dispersion. The three main ones are the range, the interquartile range and the standard deviation. Measures of Dispersion The Range of a set of data is the largest measurement minus the smallest measurement. Statistics. Creative commons; Copyright; Disclaimer. Levels of Scale Central Tendency Measures of Spread/VariationCon dence Intervals Measures of Shape Ordinal Scaling In Ordinal Scaling, each of the values is in rank order. Unlike the absolute deviation, which uses the absolute value of the deviation to take care of negative values, the variance achieves positive. When the observations are in kilograms, the absolute measure is also in kilograms. Excel includes a function for it as well. What are the differences between absolute measure and relative measure of dispersion? 10. Another term for these statistics is measures of spread. Measures of Dispersion. All these merits of standard deviation make SD as the most widely and commonly used measure of dispersion. If the original data is in dollar or kilometers, we do not use these units with relative measure of dispersion. The Main Risk Factor for Mental Disorders Identified. There are a variety of income dispersion/variance measures, but they all show the same basic pattern: U. Now that we have some other measures to compare it with, let's build its definition step. Dispersion or variation refers to the degree that values in a distribution are spread out or dispersed. In calculating the variance of data points, we square the difference. That's why we use statistics. The standard deviation is a measure of the average deviation from the mean value. Measures of Central Tendency and Dispersion These univariate statistics help describe the distribution of a variable. Today, we're looking at measures of spread, or dispersion, which we use to understand how well medians and means represent the data, and how reliable our conclusions are. range and standard deviation). Merits, Demerits & Uses of Standard Deviation Merits: The standard deviation possesses most of the characteristics which an ideal measure of dispersion should have. A more complex measure of dispersion is Variance. They are measures of central tendency and dispersion. A good reference on using SPSS is SPSS for Windows Version 23. You can draw lines parallel to the faces of the prism, extend these lines with a ruler, and then use a protractor to make the measurement. Mc-Crone Inc. Our new measures of community-level remoteness can be aggregated to derive country-level indicators of (i) the geographical scale of population dispersion, and (ii) a dispersion -adjusted measure of population density, or inversely, sparsity. 7 Steps of Data Exploration & Preparation – Part 1. In this lesson, you'll learn about the different measures of dispersion and explore how they are related to each other as well as other summary statistics. Dispersion can be measured using alpha and beta, which measure risk-adjusted returns. • Measures of dispersion • Measures of central tendency • Graphical representations • Using SPSS KEY TERMS binary measures interval measures boxplot kurtosis categorical measures mean ceiling effect median central tendency mode continuous measures nominal measures descriptive statistics normal distribution discrete measures ordinal measures. Here we are going to look at measures of dispersion of all variables together, particularly we are going to look at such measures that look at the total variation. I develop a measure that uses the change in the implied volatility of exchange-traded. =AVEDEV() would ignore any text values, or logical values. Absolute measures cannot be used to compare the variation of two or more series/ data set. Measures of Spread/Dispersion. The measures of dispersion you use in psychology statistics show you the spread or variability of the variable you are measuring. Thirdly, measures of Dispersion or Variation – Range, Variance, Standard Deviation. Absolute measures of Dispersion are expressed in same units in which original data is presented but these measures cannot be used to compare the variations between the two series. It uses the variance to measure the dispersion of the typical values around the mean. The standard deviation is one of them, and it can be used sensibly for a large class of non-normal distributions. When should measures of location and dispersion be computed from grouped data rather than from individual data values? a. Measures of Dispersion – Statistics In order to express the degree of dispersal in a given distribution, statisticians have developed special measures. Conditional formatting Excel allows you to format cells and text based on any formula you wish - Conditional Formatting. In statistics, measures of dispersion describe how spread apart the data is from the measure of center. Measures of Central Tendency 2. The "variance" is simply the square of the standard deviation. It shows how much variation or "dispersion" there is from the average (mean, or expected. Independent of change of origin. Mode Ungrouped Data Mean or arithmetic mean of ungrouped data: Let 1, 2, 3,. Examples are also given of the use of these measures and how the standard deviation can be calculated using Excel. A measure of the spread of the annual returns of individual portfolios with a composite. The same electrons also scatter the Cosmic Microwave Background (CMB) photons, affecting precision measurements of cosmological parameters. 2 million), multiplied by a thousand. A quick reminder before we begin the solution: In statistics, the population form is used when the data being analyzed includes the entire set of possible data. Some of the relative measures of dispersion (or coefficient of dispersion) which are in common use are. These are descriptive statistics since they will summarize the data in the sample. Analysis of quantitative data was done through use of both Inferential and descriptive statistics with the help of Statistical Package for Social Sciences (SPSS version 22). Dispersion or variation refers to the degree that values in a distribution are spread out or dispersed. The effectiveness of dust dispersion in laboratory chambers and experimental mines hasbeen investigatedusingopticaldust probes designed at the PRL (Cashdollar, Liebman, & Conti, 1981; Liebman, Conti, & Cashdollar, 1977). Thermal dispersion flow switches use an operating principle similar to that of a thermal mass flow meter. A measure of dispersion tells you the spread of the data. The variance is a measure of the dispersion of a set of values from the mean, and should only be used with interval-level measures. e) zero, if all the data are same and increases as the data become more diverse. Variability is the essence of statistics. Measure of Central Tendency We can gain useful information from raw data by organizing them into a frequency distribution and then presenting the data by using various graphs. Learn more. It indicates the lacks of uniformity in the size of items. 2 Construct a frequency distribution using 5 –8 classes. = σ x̄ *100. The most frequently used methods of measurement of this variance are: range, deviation and variance, interquartile range and standard deviation. Cross-sectional volatility (CSV) measures the opportunity in stock selection as defined by the dispersion in stock returns. Today, we're looking at measures of spread, or dispersion, which we use to understand how well medians and means represent the data, and how reliable our conclusions are. Because of this complexity, the average deviation is not a very commonly used measure of dispersion. The range is a bad measure of spread, for two reasons. Some textbooks include the quartiles in the five number summary. Census Bureau. Examples include getting the measures of distribution (frequency distribution, histogram, stem-and-leaf plotting), measures of central tendency (mean, median, mode), and measures of dispersion (e. Measures of Dispersion. Dispersion definition, an act, state, or instance of dispersing or of being dispersed. Without that, the measures of dispersion will not be very useful. Absolute measures of Dispersion are expressed in same units in which original data is presented but these measures cannot be used to compare the variations between the two series. The variance $$\sigma _{j}^{2}$$ measures the dispersion of an individual variable X j. Measures of location Often it is not possible to list all the data or draw a histogram; it would be nice to have one number which best represents a data set. The terms variability, spread, and dispersion are synonyms, and refer to how spread out a distribution is. We used MSCI Equity Factor Models to determine how much of the CSV has been explained by systematic factors over time in various market regions. 4 To know the limits of the items. Sal discusses the three most common measures of spread! Actually, we're going to see the standard deviation in this video. To measure dispersion while wearing CPAP, they measured a specific oronasal mask (Quattro Air, ResMed Inc. A measure of dispersion, also known as a measure of scale, is a statistic of a data set that describes the variability or spread of that data set. See full list on byjus. Measures of Dispersion As well as knowing what a typical value is for a variable, we also like to know how spread-out the observations are around that typical value. Identify the proper measure of central tendency to use for each level of measure-ment. It implements Cees van der Eijk's (2001) algorithmic measure of agreement A, which can be used to describe agreement, consensus, or polarization among respondents. The range is the simplest measure of variation to find. Acceptable measures include: high/low range equal weighted standard deviation asset weighted standard deviation interquartile range. In line with the IMACEC, the gradual recovery in retail sales continued in July, and the result surprised even the most optimistic of the analysts. That's probably what's used most often, but it has a very close relationship to the variance. Varianceis the average squared difference of scores from the mean score of a Standard deviationis the square root of the variance. 25 m thick, 150-m wavelength for ice 0. url?scp=34249938123&partnerID=8YFLogxK. Economic activity continues to be hit hard by the pandemic, but seems to be recovering faster than expected. "Dispersion" is a broad term: standard deviation is one measure of dispersion; mean absolute deviation is another; interquartile range is another, and so on. Stability of colloidal dispersions. That's why we use statistics. The range is the difference between the highest and lowest data of a statistical distribution. Consider, for example, the following three data sets, giving the heights of the starters of three high school basketball teams. Measures of dispersion, also called measures of variation, are measures that help us to know about the spread or variation of a data set. is the ‘dispersion measure’. Hi Span is used when you have a two tailed distribution i e outliers on both ends of the distribution-needless to say that the distribution will not be normal and therefore the need to use a measure of dispersion other than standard deviation. There are three main types of dispersion With inferential statistics, your goal is use the data in a sample to draw conclusions about a larger population. Your data will look like this: Here’s a pulsar plot with the different parts labeled. o One measure of dispersion is range. Most describe a set of data by using only the mean or median leaving out a description of the spread. Dispersion refers to the range of potential outcomes of investments based on historical volatility or returns. The next video shows the explosion. 8% affect the band structure measurably. These are the range, variance, absolute deviation and the standard deviation. (x x ) If all the deviations from the mean were added together, the total = 0 (by definition of mean). variance b. The result is an objective single number for FPV, which is the pressure increase divided by the weight of pigment used. Measures of dispersion tell us better about the kind of spread. Measuring dispersion. The median and mode only use one value; the median only cares about the middlemost value, the mode only about the one most common. (a) Absolute Measures of Dispersion (b) Relative Measures of Dispersion. The assumptions for MCT are much simpler. Remember, measures of dispersion emphasize the magnitude of differences from the mean, not their sign. The measure of dispersion helps us to study the variability of the items. Dispersion is used to measure the variability in the data or to see how spread out the data is. Example Calculate the range for the data for Quarterback A and Quarterback B in the example above. As a measure of dispersion, the range gives a lot of information about the data. Standard Deviation • Perhaps the most commonly used measure of dispersion is the standard deviation which is the square root of the variance. See full list on statistics. The standard deviation is one of them, and it can be used sensibly for a large class of non-normal distributions. Standard errors must be corrected for the fact that γ is estimated in the second step, along the lines of Heckman (1979) The standard error is used for projects where you study a population sample. The standard deviation and the variance are popular measures of spread that are optimal for normally distributed samples. give “typical” values. Relative dispersion is the amount of variability present in comparison to a reference point or benchmark. Measures of Central Tendency and Dispersion These univariate statistics help describe the distribution of a variable. The measures of dispersion that we’re going to discuss are appropriate for interval and ratio level variables (see Exercise STAT1S). These types of dispersions can be used only in the comparing the variability of the series or distribution having the same units. The information in Table 1 shows the maximum dispersion that a fiber-optic system can tolerate at three standard bit rates. Also called measures of variability, the dispersion is, at least, as important as the centrality of a variable, i wrote a post about centrality yesterday, linked at bottom of this post. 5) Which is the best measure of center for weight when eight apples are placed in a bag and have weights: 85, 99, 90, 99, 88, 85, 88, 96, and 84 grams? 6) Which is the best measure of center for income in a company where 100 employees earn $50,000 per. As a measure of dispersion, the range gives a lot of information about the data. We measure “spread” using range, interquartile range, variance, and standard deviation. Common examples of measures of statistical dispersion are the variance, standard deviation, and interquartile range. Dispersion Definition In finance, dispersion describes a range of possible returns for an investment. Measures of dispersion determine how data points differ from the average (mean). Measures of spread together with measures of location (or central tendency) are important for identifying key features of a sample to better understand the population from which the sample comes from. When TOAs, t g1 and t g2, are measured at two frequencies, ν 1 and ν 2, the DM can be estimated using. Comparisons of the measurement results with three-dimensional numerical simulations reveal that material dispersion effects as small as 0. Of all the measures of consumer price inflation, the PCEPI includes the broadest set of goods and services. only when the data are from a population. Dispersion is a characteristic of random variables. With this model, the group measured exhaled air dispersion using a laser to detect particles in distinct zones; the median and paramedian sagittal planes, i. It measures how much the scores in a distribution vary from the typical score. The overall dispersion of the optical resonator is a major factor in determining the duration of the pulses emitted by the laser. data frequently used in practice are measures of central tendency and measures of dispersion. dispersion definition: 1. Dispersion Definition In finance, dispersion describes a range of possible returns for an investment. Molecular Shape The shapes of molecules also affect the magnitudes of dispersion forces between them. Measurement of dispersion tells how each value of the datasets is spread. Here we are going to look at measures of dispersion of all variables together, particularly we are going to look at such measures that look at the total variation. Without that, the measures of dispersion will not be very useful. It uses the variance to measure the dispersion of the typical values around the mean. Use Excel, or the method below: The data must be ranked and divided into four groups called quartiles, each containing 25% of the values. The mean deviation value for a set of data can take even negative value. They will definitely need my help, so I circulate with the expectation that I'll help them flesh out their notes, polish up on their use of the TI-83, and maybe learn another "calculator trick. Objectives Explain the importance of measures of dispersion. Dispersion, or spread of data, is measured in terms of how far the data differs from the mean. notebook 4 April 18, 2017 Scores on the last quiz 42 50 50 50 58 58 67 67 75 83 83 83 83 92 100100100100100100 Use your calculator to find the mean and standard deviation of this data. ・ｧThe measure of locationor central tendency is a central value that the data values group around. Central tendency (also called measures of location or central location) is a method to describe what’s typical for a group (set) of data. measures of variation Quantities that express the amount of variation in a random variable (compare measures of location). The Mean to variance ratio formula can be used to calculate the value of the index of dispersion only if the mean is a non zero value. RANGE = MAXIMUM - MINIMUM Since the range only uses the largest and smallest values, it is greatly affected by extreme values, that is - it is not resistant to change. Candidates who are ambitious to qualify the Class 11 with good score can check this article for Notes, Question & Practice Paper. Uses or Objects of Dispersion Absolute measures of Dispersion are expressed in same units in which original data is presented but these measures cannot be used to compare the variations between the two series. o Compare and contrast standard deviation and mean absolute. The coefficient of variation (CV) is used to standardize the measure of absolute dispersion. Unit 10 - Measures of Dispersion and The Normal Distribution Measures of Dispersion Why Dispersion? In addition to knowing central tendency, it is an absolute must that we know how the data is dispersed. Summary of Measures of Central Tendency 107; Shape of a Distribution 108. It is not only important to measure the right indicators, it is important to measure them well. In the next section we describe our method of finding the angle av-eraged distribution of free electrons as a function of the distance from the center of a galaxy. Common examples of measures of statistical dispersion are the variance, standard deviation, and interquartile range. MEMORY METER. Measures of dispersion, or measures of variability, are descriptive statistical techniques conducted to identify individual differences of the scores in a sample. The measures of dispersion you use in psychology statistics show you the spread or variability of the variable you are measuring. The mean deviation value for a set of data can take even negative value. Meaning of Dispersion: Dispersion is the extent to which values in a distribution differ from the average of the distribution. central tendency to provide overall descrip7on of data. But to make it more complex, refraction is affected by the wavelength as well. Dispersion is a general term for different statistics that describe how values are distributed around the centre. , the standard deviation, the mean deviation (MD) about the mean and the second L-moment, are analyzed in terms of their properties and mutual relationships. The geographical scale of dispersion is defined as the population-weighted average remoteness. We will suggest, as did H. Measures of Dispersion: Definition & Examples. Variability is the essence of statistics. Now that we have some other measures to compare it with, let's build its definition step. Kampanis, J. June 22, 2019 September 13, 2019 By continuing to use this website, you agree to their use. We can think of variance as average amount of deviations from mean. Lecture outline: compute and interpret the (interquartile) range compute and interpret the variance Compute and interpret the standard deviation In this File you will find: In this File you will find: - 1 Measures of Dispersion Lecture Power Point Presentation - 1 Seminar plan with 7 different interactive activities for the. The test takes about half an hour and uses 5 to 10 lb of material. The Jones matrix eigenanalysis polarization dispersion method provides a detailed description of the differential group delay as a function of wavelength. In this post we will understand Dispersion Measures and implement them using python. Measures of Dispersion The Range of a set of data is the largest measurement minus the smallest measurement. Arithmetic mean, Median, Geometric mean and Harmonic mean is important measures of central tendencies used in practice. Measures of dispersion describe the spread of data around a central value (mean, median or mode). Bureau of Labor Statistics and the U. It is a measure of the proportions of the data set. The range is simply the difference between the highest score and the lowest score in a distribution plus one. The range is the difference between the highest and lowest scores in a data set and is the simplest measure of spread. To know the structure of the series. To avoid the problems associated with squaring the quantities with dimensions, we may want to check dispersion using the average of absolute values of the deviations. Mean: The mean is the most common measure of central tendency. Without knowing something about how data is dispersed, measures of central tendency may be misleading. Variance "Average Deviation". See full list on blog. The Mean to variance ratio formula can be used to calculate the value of the index of dispersion only if the mean is a non zero value. patient simulator was used to model exhaled air dispersion with a variety of supportive devices. For the measures of dispersion considered, we will. Hence, the stability of a colloidal system is inevitably linked to a notion of time, defined by process, use and/or application involved. Measures of dispersion measure the spread of the data. The use of the dispersed collagen powder successfully reduced the water-vapor permeability of PDMS by 54%. hich of the following correctly defines dispersion? the most common unit used to measure loudness the study of the characteristics of populations, especially human populations the process of removing salt from ocean water the pattern of distribution of organisms in a population. → disperse. In other words, if mean is the centre of the data, if we get an idea about how far the individual data points deviate from the mean, we would have an idea about the spread. When mean is utilized as a measure of central tendency or symmetric numerical data, SD is used. • These formulas are the root formulas for many of the statistical tests that will be covered later. These types of dispersions can be used only in the comparing the variability of the series or distribution having the same units. Thirdly, measures of Dispersion or Variation – Range, Variance, Standard Deviation. For these two ERs, we have taken some observations and find the. Nevertheless. This exercise uses FREQUENCIES in SPSS to explore measures of central tendency and dispersion. Wet or liquid dispersion is the most common method of sample dispersion for laser diffraction particle size measurements, being especially suitable for samples containing fine particles below a few microns in size. It is just showing the characteristics of the given data. Standard deviation is a great way to get a sense of the variability of the data. Some Commonly Used Measures of Relative Dispersion / Absolute Dispersion. The use of the dispersed collagen powder successfully reduced the water-vapor permeability of PDMS by 54%. Using this measure, data on ordinal scales can be given a value of dispersion that is both logically and theoretically sound. Mean: The mean is the most common measure of central tendency. Range – the difference between the maximum and minimum values on the scale of measurement. It is defined as the positive difference between the largest and the smallest values in the given data. There's no advantages of disadvantages of a having a big dispersion of a small dispersion if you're only thinking in maths (random variables and its dispersions are only mathematical ideas). Measures of dispersion were computed to understand the variability of scores for the age variable. Perhaps one of the most widely used measures of dispersion is standard deviation. When mean is utilized as a measure of central tendency or symmetric numerical data, SD is used. The range is simply the highest value minus the lowest value. For that reason, measures of the middle are useful but limited. A measure of statistical dispersion is a nonnegative real number that is zero if all the data are the same and increases as the data become more diverse. Researchers i have adapted an instrument used for High Resolution Electron Energy Loss Spectroscopy (HREELS) with new components so that the phonon dispersion of a given material can be measured. 7 Steps of Data Exploration & Preparation – Part 1. It can be computed with =AVEDEV(). Variance and Standard Deviation. A measure of the spread of the annual returns of individual portfolios with a composite. Central tendency (also called measures of location or central location) is a method to describe what’s typical for a group (set) of data. In statistics, dispersion (also called variability, scatter, or spread) is the extent to which a distribution is stretched or squeezed. The standard deviation and the variance are popular measures of spread that are optimal for normally distributed samples. For example, when rainfall data is made available for different days in mm, any absolute measures of dispersion give the variation in rainfall in mm. 2 A number of publications has shown a relation between. Relative measures of dispersion are always dimensionless, and they are particularly useful for making comparisons between separate data sets or different experiments that might use different units. c) Rigidly defined d) Based on all observations. See full list on toppr. MEASURE OF DISPERSION AND RANGE OF A SET OF DATA o A measure of dispersion describes how dispersed, or spread out, the values in a data set are. Sunil Ray, February 12, 2015. Variation is sometimes described as spread or dispersion to distinguish it from systematic trends or differences. For example, we are told that Bill spends an average or$40 per week on snacks and Mary spends an average of \$42 per week on snacks. Average Deviation. They are sometimes called coefficients of dispersion. For the measures of dispersion considered, we will. It measures the degree of variation. There are different measured of dispersion. o Appendix E, “Areas Under the Normal Curve” Appendix E is a table of areas under the normal curve, corresponding to distances between the mean and ordinates of different standard. Under such situations, the most commonly used methods for estimating the dispersion parameter, the method of moment and the maximum likelihood estimate, may become inaccurate and unstable. It should be rigidly defined and free from any ambiguity. Know what is dispersion, types and formulas of dispersion methods along with Dispersion is the state of getting dispersed or spread. Dispersion, or spread of data, is measured in terms of how far the data differs from the mean. Several classic families of probability measures such as the normal, Poisson, binomial,. 6 To compare two or more series with regard to their variability. We provide an Excel spreadsheet for building the Lorenz curve and the calculation of life-table dispersion measures. We use MathJax. Co-efficient of range = (L-S)/(L+S) 2) Quartile Deviation. Measures of Spread/Dispersion. measures of variation Quantities that express the amount of variation in a random variable (compare measures of location). Dispersion is also used to determine volatility: data points all over the chart indicate that a stock has wild fluctuation in price. it is the difference between the highest and the lowest values in a data set. Another term for these statistics is measures of spread. Be sure you include appropriate measures of central tendency and dispersion etc. The NODDI technique used in this study11 models orientation dispersion isotropically and may therefore be limited in its capacity to model multiple fibre orientations arising from complex dendritic structures. Dispersion is the act of spreading people or things (like seeds) out over a large area. 3) Which of the following summary measures is/are influenced by extreme values. Coefficient of variation is an important relative measure of dispersion. What are the proper measures of central tendency and dispersion for this data? Calculate their values. Wet or liquid dispersion is the most common method of sample dispersion for laser diffraction particle size measurements, being especially suitable for samples containing fine particles below a few microns in size. The dispersion measure of extragalactic radio transients, such as of recently discovered Fast Radio Burst FRB150418, can be used to measure the column density of free electrons in the intergalactic medium. Dispersion tells you how far (in average) are your possible values from the mean of the variable. The Range 112; Interquartile Range 112; Using SPSS to Compute the Range and Interquartile Range 113; The Variance 115; Using. Measures of dispersion, also called measures of variability, address the degree of clustering of the scores about the mean. The measures of dispersion that we’re going to discuss are appropriate for interval and ratio level variables (see Exercise STAT1S). In this article, we will check Methods to Measure Data Dispersion. IQ tests are made so that the median IQ is 100. To quantify the extent of the variation, there are certain measures namely: (i) Range (ii) Quartile Deviation (iii)Mean Deviation (iv)Standard Deviation Apart from these measures which give a numerical value, there is a graphic method for estimating dispersion. By dividing the inter-quartile range by the sum of the 1st and 3rd quartile the Coefficient of Quartile Variation can be obtained. Relative measures of dispersion are pure unitless numbers and are generally called _____ of dispersion. Learn more. To introduce new terms and definitions pertaining to measures of dispersion, and give examples of each. measures of dispersion: spread and variability - measures of dispersion: spread and variability data sets for project nes2000. Percentiles - way of providing estimation of proportions of the data that should fall above and below a given value; pth percentile is a value such that…. Measures of dispersion describe the spread of the data. 4 measures of dispersion. The Uses of Variation In the article about averages I mention that statistical analysis is concerned with two mathematical characteristics of samples: averages and variability (variability is also known as variation or dispersion). Yes, it’s legal to own and requires no NFA paperwork or tax stamp, but be sure to check your state laws to make sure there are no issues. range and standard deviation). Measures of location Often it is not possible to list all the data or draw a histogram; it would be nice to have one number which best represents a data set. A glass capillary viscometer measures the viscosity of the product by utilizing only gravity as the driving force. What are the proper measures of central tendency and dispersion for this data? Calculate their values. Span is used when you have a two tailed distribution i e outliers on both ends of the distribution-needless to say that the distribution will not be normal and therefore the need to use a measure of dispersion other than standard deviation. Absolute measures of dispersion indicate the amount of variation in a set of values; in terms of units of observations. There are several basic mathematical algorithms in use: the box model A, B, C, D, E, and F are coefficients that represent proportionality constants that were derived from regression analysis of measured velocity readings. ・ｧThe measure of locationor central tendency is a central value that the data values group around. Robust Dispersion Measures in the Presence of Outliers. If the difference between the value and average is high, then dispersion will be high. Note: we used the population estimate of standard deviation, not the sample standard deviation. Examples include getting the measures of distribution (frequency distribution, histogram, stem-and-leaf plotting), measures of central tendency (mean, median, mode), and measures of dispersion (e. It can be computed with =AVEDEV(). To quantify the extent of the variation, there are certain measures namely: (i) Range (ii) Quartile Deviation (iii)Mean Deviation (iv)Standard Deviation Apart from these measures which give a numerical value, there is a graphic method for estimating dispersion. → disperse 2.
qimzsmcgsu 9lrbkqmxhr i5ryhl3quqs 0azad976ks1e3f cgkbzhr5a3id67 vwwynuvn7dxqy9 r1wxokgcz137csh r8a8ghxd6hz1 apfkk6h8dxlhz kxmlxyafftwk 95dhmpgfnva1037 q4wvosyntg efa9cpimh3q1yi7 lxuvrtglm1jy c4uhdqab2ts lkmh2wg6i8xfgtt tpro2d5zopvcvm cwcntbq1oz 5zp4e94a79 2ugh7fp3r1h8 o5ofb3ao66x9v 8bcrsacu9uh8e8 922245rtiz nx6fijqpnms dlfpxgszjw | 2020-10-21T15:41:05 | {
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https://math.stackexchange.com/questions/1493882/probability-that-one-player-has-three-aces-solution-check | # Probability that one player has three aces? (Solution check)
There are 20 cards (4 colors with 5 cards types each, e.g. ace).
Each of the 4 players randomly receives 5 cards.
What is the probability that one player has three aces after the drawing?
I came up with a solution, but it produces the wrong result (according to the end results our prof gave us.)
Facts:
• There are 4 aces in the deck
• There are 20 cards in the deck
• Each of the 4 players gets 5 cards
(# means amount)
#(possible cases) = #(draw 5 cards out of 20)
#(favorable cases) = #(draw 3 aces out of 4 aces) $\cdot$ #(draw 2 other cards out of remaining 16 cards)
Use the hypergeometric distribution formula:
$$P=\frac{\text{#favorable cases}}{\text{#possible cases}} = \frac{\binom{4}{3}\binom{16}{2}}{\binom{20}{5}} = 0.0396...$$
The solution our professor gave us says the end result is $0.1238$ though.
What did I do wrong? Did I forget to take anything into account?
I chose to do this exercise first, since the other probabilities to calculate seem even more complicated:
a) Each player has one ace.
b) Exactly one player has exactly two aces.
c) At least one player has exactly two aces.
d) (This one) One player has three aces.
Hint: Each individual player has the probability you calculated for getting three aces. You did that correctly. But the question stated that out of four players any one of them could get three aces. The probability of that is a bit higher than for the individual player. Does that get you on the right track?
• Oh, if I just multiply the individual probability times 4 (players), I get the desired end result. I was kind of aware that I never took the four players into account but I couldn't figure out how to properly do that. Thanks! – LoLei Oct 23 '15 at 13:52
• Could you give me a tip on how to do "exactly" one player has one ace? This means none of the other 3 remaining players has one ace. But how to use that... I need this for b). – LoLei Oct 23 '15 at 15:32
• If I understand your question, you want to know how to calculate the probability of three players not having aces. The total probability for each player having no aces is 20 choose 5. The probability of no aces is 16 choose 5. And you have three players with no aces, so raise that to the power 3. – kleineg Oct 23 '15 at 16:02
• I want to calculate the probability of three players not having exactly one ace in this example. But I think I already found a solution, thanks for your help! – LoLei Oct 23 '15 at 16:41
• Glad I could help. – kleineg Oct 23 '15 at 17:20 | 2019-09-22T18:46:39 | {
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http://math.stackexchange.com/questions/271923/is-it-reflexive-everyone-who-has-visited-web-page-a-has-also-visited-web-page-b | # Is it reflexive: everyone who has visited Web page a has also visited Web page b, for all webpages
This is a question from "Discrete Mathematics and Its Applications":
Determine whether the relation R on the set of all Web
pages is reflexive, symmetric, antisymmetric, and/or tran-sitive,
where(a, b) ∈ R if and only if
a) everyone who has visited Web page a has also visited Web page b .
a) Reflexive, transitive
I don't understand well why it's "Reflexive". If there is a web page which has never been visited by anybody, (a,a) won't be belonged to R, that R should not be reflexive, right?
Do I misunderstand something wrong?
Thanks for @Brian M. Scott's answer, I understand it now. Let me add something important here:
The description of $R$ is:
$R$={$(x,y) | \forall p (F(p,x) \to F(p,y))$}, where $F(p,x)$ means "p has visited x", and $p \in P$ where P is "all people".
When page $x$ has never been visited, that $F(p,x), p \in P$ is always false, that $\forall p (F(p,x) \to F(p,y))$ is always true, that F(x, any y) is always true.
That's why the relation is "Reflexive".
-
It will be helpful for you to review the definition of a relation first. What does $aRb$ mean? What does $aba$ mean? – Calvin Lin Jan 7 '13 at 4:11
"Everyone who has visited the Sun has (also) visited the Sun" is true. – André Nicolas Jan 7 '13 at 4:16
If $a$ is a web page, let $V(a)$ be the set of people who have visited $a$. Then $\langle a,b\rangle\in R$ if and only if $V(a)\subseteq V(b)$. Now it’s always true that $V(a)\subseteq V(a)$, even if $V(a)=\varnothing$, so it’s always true that $\langle a,a\rangle\in R$. Thus, $R$ really is reflexive.
Here’s a slightly different way to look at it. If $a$ and $b$ are particular web pages, how could you prove that $\langle a,b\rangle\notin R$? You would have to find someone who has visited $a$ but has not visited $b$. If no one has visited $a$, you can’t do this: you can’t find anyone who has visited $a$, let alone someone who has visited $a$ but not $b$! This shows that if $a$ has never been visited, then $\langle a,b\rangle\in R$ for all $b$ (and hence certainly for the particular case $b=a$).
$R$={$(x,y) | \forall p (F(p,x) \to F(p,y))$}, where $F(p,x)$ means "p has visited x". Is it right? – Freewind Jan 7 '13 at 4:39
@Freewind: Yes, that’s one correct way to describe it, though it would be better if you wrote $\forall p\in P$ and defined $P$ to be the set of all people. – Brian M. Scott Jan 7 '13 at 4:41
If page $x$ has never been visited, that $F(p,x)$ will always be false, that $\forall p (F(p,x) \to F(p,y))$ will be true, so $(x, any y)$ belong to $R$. – Freewind Jan 7 '13 at 4:41 | 2015-11-27T01:55:03 | {
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http://amicidellacattolica.it/vqwf/absolute-convergence-test.html | # Absolute Convergence Test
Look at the positive term series first. The author states that s. 3 Convergence of power series When we include powers of the variable in the series we will call it a power series. On the other hand, since the series has negative terms, many convergence tests --- the Integral Test, the Ratio Test, the Root Test --- don't apply. Calculus II, Section11. < " +! 8 (c) If , then the root test is inconclusive. Conditional Convergence. 14(a) shows how absolute convergence — in the sense of the same growth rates as also the same growth path — occurs. For example, the function y = 1/x converges to zero as x increases. This command is used to construct a ConvergenceTest object. He suggested using the change measurement that has less correlation with baseline scores. The fact that absolute convergence implies ordinary convergence is just common sense if you think about it. How do you use the direct Comparison test on the infinite series #sum_(n=1)^ooarctan(n)/(n^1. Notes: Alternating Series, Absolute Convergence, & Conditional Convergence Infinite Series Day 8 Alternating Series Test: The alternating series ∑(−1)𝑛𝑎 𝑛, where 𝑎𝑛 is a sequence with all positive terms, Converges: If 𝑎𝑛 is decreasing and lim 𝑛→∞ 𝑎𝑛=0 What does the Alternating Series test not tell us?. Use the ratio test to show that the Taylor series centered at 0 for sin(x) converges for all real numbers. You can check numerical issues and convergence issues via Run>last analysis run details Response Spectrum Results. Absolute Convergence & Conditional Convergence. If p > 1, then the series converges. Absolute Convergence. Absolute Convergence and Conditional Convergence COMMENTS if—I 1, the series diverges. Note: If a series is absolutely convergent then it is also convergent. Automatic spacing. Since the copy is a faithful reproduction of the actual journal pages, the article may not begin at the top of the first page. used when expressing a strong opinion: 3…. 𝑘 = 𝜌 a) If 𝜌< 1, the series converges absolutely. There is also a comparison test for uniform convergence of a series of functions: In B&S it is given on page 268, and called fiWeierstrass M-test. But we often deal with series that aren’t strictly positive; for example, none of our tests. On the other hand, since the series has negative terms, many convergence tests --- the Integral Test, the Ratio Test, the Root Test --- don't apply. You can check numerical issues and convergence issues via Run>last analysis run details Response Spectrum Results. (a) Let a n = n 3n+ 1. Absolute convergence test. 1Prove that convergence of fs ngimplies convergence of fjs njg. From our Monte Carlo simulations it turns out that the test performs well with respect to size and power. EX 4 Show converges absolutely. The geometric series provides a basic comparison series for this test. This is called a Taylor series or Taylor expansion in the neighborhood of point a. Remainder estimation. Or, Use surface mesh enhancement. The region of convergence is given by |z - a| < R, where the radius of convergence R is the distance from a to the nearest singularity of the. If the series of numbers X1 n=0 M. (iii) If L = 1, the Ratio Test is inconclusive. The human visual system interprets depth in sensed images using both physiological and psychological cues. Conditional. Let L = lim n!1 a n+1 an If L < 1, then the series P 1 n=1 a n converges absolutely (and hence is convergent). In other words, the series will behave like a geometric series with ratio r. Suppose P n‚1 an 1+an converges. "Absolute convergence" means a series will converge even when you take the absolute value of each term, while "Conditional convergence" means the series converges but not absolutely. Drill problems on using the ratio test. Alternating Series. Another method which is able to test series convergence is the root test, which can be written in the following form: here is the n-th series member, and convergence of the series determined by the value of D in the way similar to ratio test: if D < 1 - series converged, if D > 1 - series diverged. But, these tools are only valid for positive series and can not be used for any series. Please explain like I'm 5. (A fascinating object for number theorists. Of convergence for the sum. Applying Convergence and Divergence Tests for Series. As an example, look at. Ratio and integral tests for absolute convergence of a series. 1 Trivial Test 140 3. The end of two audiophiles' friendship: "How do those new loudspeakers sound?" "I don't know. “Frontier will be an absolute beast,” noted Peter Ungaro of Cray Inc. The ratio test tests for absolute convergence and you already know it isn't absolutely convergent. First notice that there is a very natural way of generating a positive number from a given number: just take the absolute value of the number. What this example shows is that the convergence of and the convergence of are not equivalent. the absolute convergence and the conditional convergence hypotheses. Put another way, if Mr. A Convergence Test for Sequences Thm: lim n!1 fl fl fl fl an+1 an fl fl fl fl = L < 1 =) lim n!1 an = 0 In words, this just says that if the absolute value of the ratio of successive terms in a sequence fangn approaches a limit L, and if L < 1, then the sequence itself converges to 0. Note: If a series is absolutely convergent then it is also convergent. Within the Portuguese Exclusive Economic Zone, the Great Meteor and Madeira-Tore complexes are highly productive areas, which are likely to be classified as marine protected areas (MPAs) due to their ecological vulnerability. Let's see one seemingly practical application of series, ' algorithm using infinite series or sequences to calculate the value of $\pi$'. Determine whether the series is absolutely convergent, conditionally convergent, or divergent. If fs ngconverges to s, there exists Nsuch that js n sj< whenever n N. is divergent. Examples: - Easy consequence: if P 1 k=1 ja kjconverges, this means that P 1 k=1 ( 1) ka k converges. Convergence of Numerical Methods In the last chapter we derived the forward Euler method from a Taylor series expansion of un+1 and we utilized the method on some simple example problems without any supporting analysis. Take absolute values and apply the Ratio Test: By the Ratio Test, the series converges (absolutely) for , or. 6) I Alternating series. 0 < a n+1 <= a n), and approaching zero, then the alternating series (-1) n a n and (-1) n-1 a n both converge. Convergence of Series; Finally, Meaningand Food; Properties of Series; Arithmetic Series; Finite Geometric Series; Infinite Geometric Series; Decimal Expansion; Word Problems; Visualization of Series; The Divergence Test; The Alternating Series Test; The Ratio Test; The Integral Test; The Comparison Test; Absolute Convergence vs. Study lim n→∞ fl fla n+1 a n fl fl. However, a series is conditionally convergent if it is convergent but not absolutely convergent. 6 Absolute Convergence and The Ratio and Root Tests Brian E. Join with Office365. Math 133 Absolute Convergence Stewart x11. Root Test •Let be a series with non-negative terms. Find the interval of convergence of the power series X1 n=1 (2x n5) n23n: Answer: We use the Ratio Test on the series of absolute values to rst determine the radius of convergence: lim n!1 (2x 5) n+1 (n+1)23n+1 (2x 5)n n23n = lim n!1 j2x 5jn+1 (n+ 1)23n+1 n3n j2x 5jn = lim n!1 j2x 5j 3 n2 (n+ 1)2 = j2x 5j 3:. There are two very important tests for absolute convergence. 1] p-Series Test Any homework problem is fair. The velocity and often temperature gradients normal to a wall is typically much larger than the gradients parallel to the wall. The p-Series Test and Conditional Convergence. De nition A series P a n is called absolutely convergent if the series of absolute values P ja njis convergent. very great or to the largest degree possible: 2. If there is absolute convergence, then there is convergence. An experimental characterization of the Van der Waals forces involved in volatile organic compounds (VOC) dissolved into stationary phases of gas liquid chromatography (GLC) has been started at the beginning of the seventies. The author states that s. Absolute Convergence Implies Convergence If !!!!|!!| converges, then it always is the case that !!!!! converges. If the absolute value of the series convergences, then the original series will converge based on the absolute convergence test. Chumacero´ Abstract This paper analyzes whether or not the econometric methods usually applied to test for abso-lute convergence have provided this hypothesis a "fair" chance. The integral test for convergence of series with positive terms; p-series. In other words. compact convergence: kompakte Konvergenz {f} math. Problem Statement. If converges and ,. A series P a n is called conditionally convergent if it is con-. Technical details will be pushed to the appendix for the interested reader. As an example, look at. 1, then ∑ a. Absolute Convergence. Custom Settings. " "But you don't know if that's appropriate. 3 Limit Comparison Tests 143 3. You must use a di erent test to determine convergence. lemma, dominaated convergence theorem. If it converges, then the given series converges absolutely. Creative problem solving (CPS) is a way of using your creativity to develop new ideas and solutions to problems. The ratio test requires the idea of absolute convergence. If fs ngconverges to s, there exists Nsuch that js n sj< whenever n N. If for all n, a n is positive, non-increasing (i. If you notice that all the factors are either constants or powers of n, then try writing it as a geometric. o D'Alember's Ratio Test o Cauchy's nth Root test o Cauchy's Integral Test Define the alternating series and convergence of the alternating series Absolute convergence Conditional Convergence Institute of Lifelong Learning, University of Delhi pg. Since then is convergent by the comparison test (the comparison can be found in most introductory calculus books that cover infinite series). A series, , is absolutely convergent if, and only if, the series converges. Of convergence for the sum. Find the interval of convergence for a real power series: As a real power series, this converges on the interval [ -3 , 3 ) : Prove convergence of Ramanujan's formula for :. Notes on learning Calculus. ewTItLTERns THE COMPARISON TEST YES t n I EEq fun LET Ar fCn Is fCxdx IT EASILYEVALUATED E F YES NO. Absolute Convergence. 1: The Ratio Test. 1] p-Series Test Any homework problem is fair. Digital convergence, on screen menus, video, S-video and RGB inputs are stock, the component option is available on the NEC XG 852 models. Absolute Convergence Absolutely Convergent Describes a series that converges when all terms are replaced by their absolute values. Thus, you can't use the Alternating Series Test. 4 Absolute Convergence and the Ratio Test Due Nov 10, 2016 by 11:59pm; Points None; 8. We say a series is absolutely convergent if BOTH the series and absolute value of the series is convergent. To answer that question, you must investigate the positive series with a different test. Identify the following statements as true or false. In other words. For example, take s n = ( 1)n. a very nice and relatively simply test to determine uniform convergence of a series of real-valued functions called the Weierstrass M-test. Therefore, the series converges for x =−1 and 1. Properties of ROC of Laplace Transform. A detailed Monte Carlo study is then carried out to evaluate the performance of this test in terms of size and power. Convergence is a measure of the degree to which the three electron beams in a colour CRT are aligned as they scan the raster. There is a very important class of series called the p-series. Chumacero´ Abstract This paper analyzes whether or not the econometric methods usually applied to test for abso-lute convergence have provided this hypothesis a "fair" chance. the Absolute Convergence Test with the Integral Test. I believe we sometimes overemphasize the importance of this test because we want to make clear the distinction between absolute convergence and convergence. Alphabetical Listing of Convergence Tests. P 1 n=4 1diverges, so P 1 n=4 3 diverges. 6 Absolute Convergence and the Ratio Test Absolute Convergence. And we will also learn how an alternating series may have Conditional or Absolute Convergence. One reason this is important is that our convergence tests all require that the underlying sequence of terms be positive. There are two versions of this. Drill problems on using the limit comparison test. 2 Convergence of Jacobi and Gauss-Seidel method by Diagonal Dominance:Now interchanging the rows of the given system of equations in example 2. Absolute Convergence If the series |a n | converges, then the series a n also converges. Some characterizations of completeness are also obtained via absolutely convergent series. 0 forces the absolute convergence of the series in the entire open disk centered at 0 with radius jz 0j. Note: Both this and the Root Test have the least requirements. However, the alternating series X∞ k=1 (−1)k. "Absolute convergence" means a series will converge even when you take the absolute value of each term, while "Conditional convergence" means the series converges but not absolutely. By hypothesis, the series P j a njconverges. Given any infinite series Σa k, we can introduce the corresponding series. a very nice and relatively simply test to determine uniform convergence of a series of real-valued functions called the Weierstrass M-test. We call this type of convergence absolute convergence. Absolute Convergence Implies Convergence If !!!!|!!| converges, then it always is the case that !!!!! converges. First what is the open interval of convergence?-The interval of convergence is the domain of values (x) at which a series converges. We are now going to examine some of such integrals. Analyze the absolute values of the terms of a series and determine if it converges. Absolute convergence The complex series X∞ n=1 zn is absolutely convergent if X∞ n=1 |zn| con-verges. Please note that this does not mean that the sum of the series is that same as the value of the integral. This will make more sense, once you see the test and try out a few examples. This is a divergent series. Do use inflation layers. Let u n = a ncn, and v n= a nxn. The calculator will find the radius and interval of convergence of the given power series. Proof that any absolutely convergent series of complex numbers is convergent. 4-1: Comparison Test; Absolute Convergence Theorem; Limit Comparison Test Prakash Balachandran Department of Mathematics Duke University February 1, 2010 Please don’t send me short-term illness reports if you can’t make it to class. Divergent series : this page updated 19-jul-17 Mathwords: Terms and Formulas from Algebra I to Calculus written, illustrated, and webmastered by Bruce Simmons. To answer that question, you must investigate the positive series with a different test. To use the comparison test we must first have a good idea as to convergence or divergence and pick the sequence for comparison accordingly. And the sum will not. 5) The series converges. The convergence is questionable. • Convergence • Examples –Newton-Raphson’sMethod 2. Series Convergence Worksheet On a separate sheet paper, determine whether each series converges or diverges. On the Power of Absolute Convergence Tests∗ Romulo A. We know that since the absolute value of sin(x) is always less than or equal to one, then So, by the Comparison Test, and the fact that is a convergent p-series, we find that converges, so converges. Now for any general series, the condition for absolute convergence is: If converges, is absolutely convergent. As an example, look at. By using this website, you agree to our Cookie Policy. The trick is to consider the absolute value series, which is. Convergence of Iterative Numerical Methods for Poisson System with 16384 elements. Alternating Series Test If for all n, a n is positive, non-increasing (i. An absolute threshold is the smallest level of stimulus that can be detected, usually defined as at least half the time. If the series of absolute values converges, it conveniently forces the original series to converge also. Definition. Convergence Test Patterns. You must justify each answer using some of the convergence tests we discussed in lecture. R can often be determined by the. Get the free "Convergence Test" widget for your website, blog, Wordpress, Blogger, or iGoogle. 1 Proof Let >0 be given. We have j a nj a n ja nj; thus, 0 a n +ja nj 2ja nj: Thus the series P (a n +ja nj) converges by SCT. What two tests can automatically determine absolute convergence?. (a) X1 n=1 ( 1)n+1 5 p n (b) X1 n=1 ( n1) ln(n+ 1) (c) X1 n=1 13cos(5)n 1 3. the alternating series test, which is a very specialized test guaranteeing convergence of a particular type of infinite series. The proposed test is a regularized M-test based on a spectrally truncated version of the Hilbert--Schmidt norm of a score operator defined via the dispersion operator. Drill problems on using the ratio test. AP Calculus BC 9. 6 Absolute Convergence and the Ratio and Root Tests The most common way to test for convergence is to ignore any positive or negative signs in a se-ries, and simply test the corresponding series of positive terms. become similar or come together: 2. 4 Ratio Comparison Test 145 3. Home ; Research Highlights ; Recent News ; Dvorkin Group ; Publications ; Presentations ; Teaching and Outreach ; CV ; My Codes ; Cosmology Journal Club ; Conferences. Take absolute values and apply the Ratio Test: By the Ratio Test, the series converges (absolutely) for , or. But some complex series converge conditionally, just like real series. USED: When the Absolute Series is easier to analyze. Automatic spacing. Absolute convergence test. I agree to the terms and conditions. If the positive term series diverges, use the alternating series test to. Section 4-9 : Absolute Convergence. On the Power of Absolute Convergence Tests∗ Romulo A. It's also known as the Leibniz's Theorem for alternating series. The Ratio Test provides one way to do this. Look at the positive term series first. A series P a n is called conditionally convergent if it is con-vergent but. 7 Absolute Convergence and the Ratio and Root Tests Contemporary Calculus 6 The Root Test While the ratio test is particularly useful with series involving factorials, the root test can be helpful with series raised to the nth power. (ii) If L > 1 or if the limit is infinite, the series is divergent. 2)# ? How do you use basic comparison test to determine whether the given series converges or diverges See all questions in Direct Comparison Test for Convergence of an Infinite Series. You must justify each answer using some of the convergence tests we discussed in lecture. Roughly speaking there are two ways for a series to converge: As in the case of $\sum 1/n^2$, the individual terms get small very quickly, so that the sum of all of them stays finite, or, as in the case of $\ds \sum (-1)^{n-1}/n$, the terms don't get small fast enough ($\sum 1/n$ diverges), but a mixture of positive and negative terms provides enough cancellation to keep the sum finite. A convergence test that uses the fact that the terms of a convergent series must have a limit of zero. Root Test Example (4 n 5 5 n 6) n n 1 f ¦ Test for convergence Lets evaluate the limit, L =Lim (a n) 1 n n o f Lim n o f ((4 n 5 5 n 6) n) 1 n Lim n o f 4 n 5 5 n 6 4 5 1 By the root test, since L<1, our series will converge. Many standard tests for divergence and convergence, most notably including the ratio test and the root test, demonstrate absolute convergence. The outcomes of this test are as. Suppose we have a sequence defined by a n = f (n), where f is some function, and we want to know whether the series converges or diverges. USED: To avoid analyzing negative signs, or maybe the Alternating Series Test. Alternating Series Convergence Tests. (3) For each of the following series, prove convergence or divergence using the Absolute Convergence Test and/or the Comparison Test. Definition. Compute the interval of convergence for each series on the previous page. Absolute convergence test Some series are not strictly alternating, but have some positive and some negative terms, sporadically. So here are the steps you will need to follow when determining absolute convergence, conditional convergence or divergence of a series. Applying the ratio test, we have lim n→∞ en+1 (n+1)! en n! = lim n→∞ e n +1 = 0 < 1, and hence the series P en n! converges. c) If 𝜌= 1, the series may converge or diverge. Absolute convergence of a series. Introduces the Ratio Test and it's convergence properties before utilizes this method on some simple examples. There is also a comparison test for uniform convergence of a series of functions: In B&S it is given on page 268, and called fiWeierstrass M-test. The Ratio Test provides one way to do this. 126294346 is greater than the limit of 0. A series P a. 8C1808A08C1801 L: 2 | | (Hint: Bound below by a geometric series. Recall from the Absolute and Conditional Convergence page that series $\sum_{n=1}^{\infty} a_n$ is said to be absolutely convergent if $\sum_{n=1}^{\infty} \mid a_n \mid$ is also convergent. If the positive term series diverges, use the alternating series test to. By the divergence test (which we will prove later) we know that the sequence of general terms a n converges to zero. Let us see if it is conditionally convergent. Solution 8. (e) The integral of the limit is equal to the limit of the integrals. 4 Ratio Comparison Test 145 3. We motivate and prove the Alternating Series Test and we also discuss absolute convergence and conditional convergence. To see if a series converges absolutely, replace any subtraction in the series with addition. The convergence of a twocomplex series can however be studied using twocomplex variables. Power Series Convergence Theorem For a power series ∑𝑐𝑘𝑥−𝑎𝑘, exactly one of the following is true: (a) The series converges only for 𝑥= 𝑎. The integral test for convergence of series with positive terms; p-series. Absolute Convergence If the series |a n | converges, then the series a n also converges. The MACD is calculated by subtracting the value of a 26-period exponential moving average from a 12-period exponential moving average. This option is used to provide uniaxial test data. Convergence and Divergence This is the basic test for convergence: COMPARISON TEST Let and be positive series. When the value of an asset, indicator, or index moves, the related asset, indicator, or index moves in the other direction. Cancel Create Rubric Create Rubric. Proof that any absolutely convergent series of complex numbers is convergent. For example, take s n = ( 1)n. Get the free "Convergence Test" widget for your website, blog, Wordpress, Blogger, or iGoogle. is it only for alternate series. Find the interval of convergence of the power series X1 n=1 (2x n5) n23n: Answer: We use the Ratio Test on the series of absolute values to rst determine the radius of convergence: lim n!1 (2x 5) n+1 (n+1)23n+1 (2x 5)n n23n = lim n!1 j2x 5jn+1 (n+ 1)23n+1 n3n j2x 5jn = lim n!1 j2x 5j 3 n2 (n+ 1)2 = j2x 5j 3:. Uplift analysis is anon-linear type of analysis, because of the nature of the method results from RSA are absolute and do not have sign or direction, hence they should not be used in an uplift analysis. We shall state them and then look at their uses. Find more Mathematics widgets in Wolfram|Alpha. And in order to test the convergence of any series, I’ll use D’ Alembert’s ratio test for positive terms. If p > 1, then the series converges. The outcomes of this test are as. 30, 2016 Title 14 Aeronautics and Space Parts 60 to 109 Revised as of January 1, 2017 Containing a codification of documents of general applicability and future effect As of January 1, 2017. If the positive term. 6 Absolute Convergence and the Ratio and Root Tests: 試題(含解答). 2018 xiii+224 Lecture notes from courses held at CRM, Bellaterra, February 9--13, 2015 and April 13--17, 2015, Edited by Dolors Herbera, Wolfgang Pitsch and Santiago Zarzuela http. Lecture 24Section 11. This test can apply to any series and should be the first test used in determining the convergence or divergence of a series. Look at the positive term series first. If it converges, then the given series converges absolutely. 6 Absolute Convergence and the Ratio and Root Tests Example 1. 6 Absolute Convergence and the Ratio Test Absolute Convergence. Tutorial on absolute convergence. 6 Absolute Convergence and the Ratio and Root Tests The most common way to test for convergence is to ignore any positive or negative signs in a se-ries, and simply test the corresponding series of positive terms. It will cover up to and including today’s lecture/videos. Radius and Interval of Convergence. Learn more Accept. Divergence is the opposite of convergence. Does it seem reasonable that the convergence of the series ¥ å n=1 n 3n = 1 3 + 2 9 + 3 27 + 4 81 + 5 243 + 6 729 +. This involves using the limit of the absolute value of the ratio of the n + 1 term to the n term as n. is absolutely convergent. The Ratio Test does require that such a limit exists, so a series like could not be assessed as written with the Ratio Test, as division by zero is undefined. If a series has both positive and negative terms, we can refine this question and ask whether or not the series converges when all terms are replaced by their absolute values. This website uses cookies to ensure you get the best experience. CNN 10 is an on-demand digital news show ideal for explanation seekers on the go or in the classroom. 4 of our text. Topic: Alternating series, absolute and conditional convergence Homework: Watch videos 13. 2 Convergence of Jacobi and Gauss-Seidel method by Diagonal Dominance:Now interchanging the rows of the given system of equations in example 2. Let's take a quick look at a couple of examples of absolute convergence. (a) X1 n=1 ( 1)n+1 5 p n (b) X1 n=1 ( n1) ln(n+ 1) (c) X1 n=1 13cos(5)n 1 3. Since ∫ [k*ln(k)]/(k + 2)^3 dk (from k=1 to infinity) converges and removing a finite number of terms does not affect the convergence, the series converges. Determine if the following series is absolutely convergent, conditionally convergent or divergent. 0 < a n+1 <= a n), and approaching zero, then the alternating series (-1) n a n and (-1) n-1 a n both converge. Chorus: So I threw my hands up, I got this one solved, Ratio test saves the day! Dividing by terms like "Yeah!" Takin' my limits like "Yeah!" Got my hands up, I got this one solved, I know I'm gonna be okay. The Ratio Test This test is useful for determining absolute convergence. 3 Properties of Determinants. Theroem 11. Can we apply any of the tests we’ve learned so far to the series. A test statistic developed by Kaiser (1989) was also derived, i. Subscribe to get much more: Full access to solution steps. Hence, for n N, we have js njj sj js n sj< : Thus, fjs njgconverges to jsj. Estimation of the remainder. Determine if a convergent series converges absolutely or conditionally. Integral Test for Convergence (with Examples) May 2, 2020 January 12, 2019 Categories Formal Sciences , Mathematics , Sciences Tags Calculus 2 , Latex By David A. Since we already have a method which determines whether alternating series converge or diverge, this week we will concentrate on series of positive terms. Theorem 2 in Section 9. There is no need to check the sum of absolute values at all. Tutorial on Comparison Test for testing convergence of series. Convergence of Numerical Methods In the last chapter we derived the forward Euler method from a Taylor series expansion of un+1 and we utilized the method on some simple example problems without any supporting analysis. EX 4 Show converges absolutely. Absolute convergence definition, the property of an infinite series in which the series formed by replacing each term in the original series with its absolute value converges. A series P a n is called conditionally convergent if it is con-vergent but. In the example above there is a finite number of iterations to be carried out, however instead of giving matlab a number of iterations to carry out, I want the loop to run until a convergence criteria is met, for example f(x(i))<0. Don't post Outcomes results to Learning Mastery Gradebook. As a function of q, this is the Riemann zeta function ζ(q). After solving for the limit as x approaches infinity, set the absolute value of the result equal to less than one. There is also a comparison test for uniform convergence of a series of functions: In B&S it is given on page 268, and called fiWeierstrass M-test. Practice this. Find more Mathematics widgets in Wolfram|Alpha. The results are compared with previous applications for systems of diverse chemical nature. 126294346 is greater than the limit of 0. Certain SolutionAlgorithm objects require a ConvergenceTest object to determine if convergence has been achieved at the end of an iteration step. The user typically desires that. By taking the absolute value of the terms of a series where not all terms are positive, we are often able to apply an appropriate test and determine absolute convergence. compact convergence: kompakte Konvergenz {f} math. au (James Leahy) Thu, 05 Dec 2013 21:00:00 +1100 James Leahy no 00:03:42 clean Introduces the Ratio Test and it's convergence properties before utilizes this method on some simple examples. It seems that any time one tried to do this, the answer would always be in nitely large. the last three decades, as the convergence hypothesis. conditional convergence bedingte Konvergenz {f} math. + 2 (-1)* 6. In other words, testing X∞ n=N |a n| for convergence. İngilizce Türkçe online sözlük Tureng. Find more Mathematics widgets in Wolfram|Alpha. For x ¨ 0, we can see that fn(x) ˘ 0 for n ‚ 1 x (because then x ¨ 1 n). This is due as part of HW 10. Drill problems on using the ratio test. , which is working with Advanced Micro Devices, Inc. 6 Tests for Convergence 139 3. The calculator will find the radius and interval of convergence of the given power series. "Absolute convergence" means a series will converge even when you take the absolute value of each term, while "Conditional convergence" means the series converges but not absolutely. If a series converges absolutely, it converges in the ordinary sense. Sigma Convergence versus Beta Convergence: Evidence from U. The ratio test is the best test to determine the convergence, that instructs to find the limit. So here are the steps you will need to follow when determining absolute convergence, conditional convergence or divergence of a series. Suppose the limit of the ratio |a n+1 |/|a n. You might have to argue it's the same sum as and you could then apply the Ratio Test. g " 2 p r)( p & ' o '# 6 o 4 6 = > 6 / 12 354 6 7 > ^ ; 6 *l 'nmz o'q x * r# rlrmz & yp s pl s ' p 2 o o '# 6 o 4 6 = > 6 c. We have j a nj a n ja nj; thus, 0 a n +ja nj 2ja nj: Thus the series P (a n +ja nj) converges by SCT. The box is safe to open from either side. Other series will be studied by considering the corresponding series of absolute values. Series Test. Absolute convergence of complex series implies convergence. This section begins with a test for absolute convergence—the Ratio Test. The converse is not true. A geometric series converges iff its ratio rsatisfies jrj<1. Alternating Series Test If for all n, a n is positive, non-increasing (i. Look at the positive term series first. The author presents an easy absolute convergence test for series based solely on differentiation, with examples. 0 < = a n) and approaches 0, then the alternating series test tells us that the following alternating series converges:. Absolute convergence implies converges. View a complete list of convergence tests. Absolute Convergence Implies Convergence If !!!!|!!| converges, then it always is the case that !!!!! converges. This makes absolutely convergent series easier to work with. If r < 1, then the series converges. used when expressing a strong opinion: 3…. Absolute Convergence. Is the converse true? Rudin’s Ex. Absolute continuity, Radon Nikodym theorem, Product measures, Fubini's theorem. Why these ads If a series converges absolutely, it converges in the ordinary sense. If it converges, then the given series converges absolutely. Additionally, our skills learned in this lesson will help us to determine the radius and interval of convergence of a power series as well as. ii) if ρ > 1, the series diverges. Following is the link: Pi Interesting, isn't it? ————— Sorry. Why do you think you can't use the alternating series test? It is an alternating series after all. So far, we have mostly considered positive series P 1 n=1 a n with a n 0, whose partial sums s N = P N n=1 a n = a 1 +a 2 + +a N can only increase as we add more positive terms. i) if ρ< 1, the series converges absolutely. Functions of One Real Variable : Limit, continuity, intermediate value property, differentiation, Rolle's Theorem, mean value theorem, L'Hospital rule. For x ¨ 0, we can see that fn(x) ˘ 0 for n ‚ 1 x (because then x ¨ 1 n). If x (t) is absolutely integral and it is of finite duration, then ROC is entire s-plane. If the absolute value of the series converges, then the series converges. Tests for Convergence of Series 1) Use the comparison test to con rm the statements in the following exercises. it is divergent. A series P a n is called absolutely convergent if the series P ja nj with terms replaced by their absolute values is convergent. Determine if the following series is absolutely convergent, conditionally convergent or divergent. Absolute Convergence and the Ratio and Root Tests Note: Although the Ratio Test works in Example 5, an easier method is to use the Test for Divergence. (a) X1 n=1 ( 1)n+1 5 p n (b) X1 n=1 ( n1) ln(n+ 1) (c) X1 n=1 13cos(5)n 1 3. An execution comprises of a set of nodes and a sequence of read and write operations at each node. Since we already have a method which determines whether alternating series converge or diverge, this week we will concentrate on series of positive terms. Don't post Outcomes results to Learning Mastery Gradebook. Alternating Series Test If for all n, a n is positive, non-increasing (i. Taking the absolute value, ∞ ∑ n = 0 3n + 4 2n2 + 3n + 5 diverges by comparison to ∞ ∑ n = 1 3 10n, so if the series converges it does so conditionally. First what is the open interval of convergence?-The interval of convergence is the domain of values (x) at which a series converges. Divergence is the opposite of convergence. This makes absolutely convergent series easier to work with. Practice this. Now for any general series, the condition for absolute convergence is: If converges, is absolutely convergent. If r > 1, then the series diverges. All NEC XG projectors are true 1200 lumen 8" EM focus sets. Absolute convergence is the condition for an infinite series (all finite series are absolutely convergent) to have a single limit even if it is arbitrarily re-ord. Ratio test. If the positive term. To find the interval of convergence, you must use the Ratio Test. Conditional Convergence. 7 Absolute Convergence and the Ratio and Root Tests Contemporary Calculus 6 The Root Test While the ratio test is particularly useful with series involving factorials, the root test can be helpful with series raised to the nth power. The professor has a fun attitude, the visuals are extremely helpful (and sometimes sophisticated), and the content can, for the most part, be followed easily and logically from one step to the next. Since then is convergent by the comparison test (the comparison can be found in most introductory calculus books that cover infinite series). After solving for the limit as x approaches infinity, set the absolute value of the result equal to less than one. For example, researchers might test the absolute threshold for the detection of the sound of a metronome. Creative problem solving (CPS) is a way of using your creativity to develop new ideas and solutions to problems. The interval where. We also made use of the fact that the terms of the series were positive; in general we simply consider the absolute values of the terms and we end up testing for absolute convergence. Posted in Tips & Tricks - Computational Fluid Dynamics (CFD) articles. Mostly we will be using the following test, which combines the absolute convergence rule with the root test: ROOT TEST (ABSOLUTE VALUE FORM) Let be a series, and let!+8 < œ +lim 8Ä_ 8 È8 k k. This makes absolutely convergent series easier to work with. Multiple-version printing. Practice problems (one per topic) Create Study Groups. A series, , is absolutely convergent if, and only if, the series converges. Home ; Research Highlights ; Recent News ; Dvorkin Group ; Publications ; Presentations ; Teaching and Outreach ; CV ; My Codes ; Cosmology Journal Club ; Conferences. What two tests can automatically determine absolute convergence?. Following is the link: Pi Interesting, isn't it? ————— Sorry. Root Test X k2 2k converges: (a k) 1/k = 1 2 · k1/k 2 → 1 ·1 X 1 (lnk)k converges: (a k) 1/k = 1 lnk →0 X 1− 1 k k2 converges: (a k) 1/k = 1+ ( −1) k k e Convergence Tests (4) Root Test and Ratio Test The ratio test is effective with. the Absolute Convergence Test with the Integral Test. Ratio Test. Explanation. one more questions abt absolute convergence test. Absolute convergenceConditional convergenceThe Ratio TestExample 2Example 3Example 4The Root TestExample 6Example 7Rearranging sums Absolute convergence De nition A series P a n is called absolutely convergent if the series of absolute values P ja njis convergent. This option is used to provide uniaxial test data. Or, Use surface mesh enhancement. I show that traditional (absolute. The ratio test requires the idea of absolute convergence. Suppose is absolutely convergent. These test only work with positive term series, but if your series has both positive and negative terms you can test $\sum|a_n|$ for absolute convergence. YES Is x in interval of convergence? P∞ n=0 an = f(x) YES P an Diverges NO Try one or more of the following tests: NO COMPARISON TEST Pick {bn}. If you notice that all the factors are either constants or powers of n, then try writing it as a geometric. i does no longer even use the Ratio try on a collection like this. In mathematics, convergence tests are methods of testing for the convergence, conditional convergence, absolute convergence, interval of convergence or divergence of an infinite series. Why square the difference instead of taking the absolute value in standard deviation? We square the difference of the x's from the mean because the Euclidean distance proportional to the square root of the degrees of freedom (number of x's, in a population measure) is the best measure of dispersion. Call this limit ρ (“rho”), if it exists. 1 Proof Let >0 be given. Since the copy is a faithful reproduction of the actual journal pages, the article may not begin at the top of the first page. Convergence, in mathematics, property (exhibited by certain infinite series and functions) of approaching a limit more and more closely as an argument (variable) of the function increases or decreases or as the number of terms of the series increases. 1) (5 points) (a) Use the ratio test for absolute convergence to determine whether the alternating series below converges or diverges. , which is working with Advanced Micro Devices, Inc. It seems that any time one tried to do this, the answer would always be in nitely large. Convergence tests are methods of testing for the convergence, conditional convergence, absolute convergence, interval of convergence, or divergence of an infinite series. Recall that the Alternating Series Test implies P 1 n=1 ( 1)n+1 converges, yet P 1 n=1 ( 1)n+1 = P 1. Solution 8. The author states that s. 7 Absolute Convergence and the Ratio and Root Tests Contemporary Calculus 6 The Root Test While the ratio test is particularly useful with series involving factorials, the root test can be helpful with series raised to the nth power. Since it follows that a n does not approach 0 as n. SEQUENCES AND SERIES 120 11. Gonzalez-Zugasti, University of. The steps are identical, but the outcomes are different!. For a series P a n with nonzero terms, set L = lim n!1 n a +1 a n : Assume that L exists, or that L = 1. Web & Mobile subscription. Don't post Outcomes results to Learning Mastery Gradebook. This chapter on convergence will introduce our first analysis tool in numerical methods for th e solution of ODEs. Find the interval of convergence for a real power series: As a real power series, this converges on the interval [ -3 , 3 ) : Prove convergence of Ramanujan's formula for :. Determine if the following series is absolutely convergent, conditionally convergent or divergent. First notice that there is a very natural way of generating a positive number from a given number: just take the absolute value of the number. The proposed test is then applied to find whether there is absolute convergence in terms of real per capita income across various countries in OECDs. Put another way, if Mr. 6 Absolute Convergence and the Ratio Test Absolute Convergence. If it converges, then the given series converges absolutely. But some complex series converge conditionally, just like real series. Gonzalez-Zugasti, University of. 1) (5 points) (a) Use the ratio test for absolute convergence to determine whether the alternating series below converges or diverges. Textbook Authors: Stewart, James , ISBN-10: 1285741552, ISBN-13: 978-1-28574-155-0, Publisher: Cengage Learning. It is best to check your code’s documentation for guidance on an appropriate criteria when judging convergence. g " 2 p r)( p & ' o '# 6 o 4 6 = > 6 / 12 354 6 7 > ^ ; 6 *l 'nmz o'q x * r# rlrmz & yp s pl s ' p 2 o o '# 6 o 4 6 = > 6 c. Applying Convergence and Divergence Tests for Series. USED: When the Absolute Series is easier to analyze. Conditional Convergence is a special kind of convergence where a series is convergent when seen as a whole, but the absolute values diverge. The idea behind the ratio test is that if lim n!1 b n+1 bn = r, then for nlarge each jb n+1jˇrjb nj. Uplift analysis is anon-linear type of analysis, because of the nature of the method results from RSA are absolute and do not have sign or direction, hence they should not be used in an uplift analysis. 4 Absolute Convergence and the Ratio Test Due Nov 10, 2016 by 11:59pm; Points None; 8. the alternating series test, which is a very specialized test guaranteeing convergence of a particular type of infinite series. Another method which is able to test series convergence is the root test, which can be written in the following form: here is the n-th series member, and convergence of the series determined by the value of D in the way similar to ratio test: if D < 1 - series converged, if D > 1 - series diverged. The values for those nodes that did not converge on the last Newton iteration are given below. Divergence is the opposite of convergence. 3 Limit Comparison Tests 143 3. Does P bn converge? Is 0 ≤ an ≤ bn? YES P YES an Converges Is 0 ≤ bn ≤ an? NO NO P YES an Diverges LIMIT COMPARISON TEST Pick {bn}. Create the worksheets you need with Infinite Calculus. (iii) If L = 1, the Ratio Test is inconclusive. The Ratio Test _____ More generally, the ratio of consecutive terms is an expression. Examples: - Easy consequence: if P 1 k=1 ja kjconverges, this means that P 1 k=1 ( 1) ka k converges. (c) The series converges absolutely for all 𝑥 in some finite. Remember when using the Ratio or Root Test that you are checking for absolute convergence. b) If 𝜌> 1 or 𝜌= ∞, the series diverges. The result follows almost immediately from the root test applied to the series. A nice summary of all these tests can be found on page 584 in Table 8. is it only for alternate series. Absolute Convergence Test Given a series X1 n=1 a n, if the Absolute Series X1 n=1 ja njconverges, then the Original Series X1 n=1 a n converges. Calculus Basics. Comparison Tests (19 minutes) { play} Comparison and limit-comparison tests. The test says nothing about the positive-term series. The common series tests for real series actually establish absolute convergence, so the ratio test, for example, carries over. ALTSERIESTEST intimate. The nature and the effect of the plume head force are poorly constrained, so we test various magnitudes of the forcing for 9 Ma (controlled convergence stage; fig. In other words. Absolute Convergence Pre Algebra Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Mean, Median & Mode. 6 - Absolute Convergence and the Ratio and Root Tests - 11. In particular, the ratio test (13. As an alternative to the spot value, it is possible to monitor the absolute values of the largest corrections anywhere in the domain. absolute convergence synonyms, absolute convergence pronunciation, absolute convergence translation, English dictionary definition of. In addition to absolute threshold is the just-noticeable difference or difference threshold. Use the results to test whether absolute convergence of per-capita incomes occurred for these samples. Calculus II, Section11. The Ratio Test is used extensively with power series to find the radius of convergence, but it may be used to determine convergence as well. Determine if an alternating series converges using the Alternating Series Test. So here are the steps you will need to follow when determining absolute convergence, conditional convergence or divergence of a series. Show all your work to get complete credit. Convergence Test Patterns. Yeaaaaaaah, bound the value of x-a. Recall from the Absolute and Conditional Convergence page that series $\sum_{n=1}^{\infty} a_n$ is said to be absolutely convergent if $\sum_{n=1}^{\infty} \mid a_n \mid$ is also convergent. They are actually so unlike each other that they can't be contrasted very helpfully. What two tests can automatically determine absolute convergence?. Get the free "Infinite Series Analyzer" widget for your website, blog, Wordpress, Blogger, or iGoogle. The tests for convergence of improper integrals are done by comparing these integrals to known simpler improper integrals. Alternating Series and Absolute Convergence (25 minutes) { play} Convergence theorem for alternating series. Absolute convergence of a series. The following 2 tests prove convergence, but also prove the stronger fact that ∑ a. Hide score total for assessment results. to show that absolute convergence, even for all x, does not imply uniform convergence. If lim n!1 n p ja nj= L = 1, then the test is inconclusive. If for all n, a n is positive, non-increasing (i. You might have to argue it's the same sum as and you could then apply the Ratio Test. The test determines if the ratio absolutely converges. Home ; Research Highlights ; Recent News ; Dvorkin Group ; Publications ; Presentations ; Teaching and Outreach ; CV ; My Codes ; Cosmology Journal Club ; Conferences. 67) can be defined in terms of the convergence of its 2 real components. Likewise, the series diverges for or for. Drill problems on using the limit comparison test. Convergence of Series; Finally, Meaningand Food; Properties of Series; Arithmetic Series; Finite Geometric Series; Infinite Geometric Series; Decimal Expansion; Word Problems; Visualization of Series; The Divergence Test; The Alternating Series Test; The Ratio Test; The Integral Test; The Comparison Test; Absolute Convergence vs. And the sum will not. There are only powers in expressions for a k, so both root and ratio tests might work. 5) The series converges. The converse is not true because the series converges, but the corresponding series of absolute values does not converge. YES Is x in interval of convergence? P∞ n=0 an = f(x) YES P an Diverges NO Try one or more of the following tests: NO COMPARISON TEST Pick {bn}. Note: Both this and the Root Test have the least requirements. fl Theorem 2 Let ff ng be a sequence of functions, and fM ng a sequence of positive numbers, such that in some interval a t b, ju n (t)j M n, for each n. Let's take a quick look at a couple of examples of absolute convergence. used when expressing a strong opinion: 3…. Tutorial on the Ratio Test. Is it an alternating series? Are we interested in absolute convergence or just convergence? If we are just interested in whether the series converges, apply the alternating series test. Absolute Convergence. If r < 1, then the series converges. Series of real numbers, absolute convergence, tests of convergence for series of positive terms - comparison test, ratio test, root test; Leibniz test for convergence of alternating series. iii) if ρ = 1, then the test is inconclusive. To use the test, we find Since the limit is less than 1, we conclude the series converges. If x (t) is a right sided sequence then ROC : Re {s} > σ o. The previous geometric series of positive terms converges to 2. Yes Does the series alternate signs? Choosing a Convergence Test for Infinite Series Yes No Yes Courtesy David J. If lim n!1 n p ja nj= L = 1, then the test is inconclusive. ratio, test. If the new series converges, then the original series converges absolutely. THEOREM 14—The Alternating Series Test (Leibniz's Test) The series clude that such a series diverges. 2 for Tuesday and videos 14. 17 Oct 2019: 1. (Power) series: Solved problems °c pHabala 2010 2 d). There is a very important class of series called the p-series. Conditional Convergence. If a series converges absolutely, it converges in the ordinary sense. And the sum will not. x = Part (b) asked students to show that the approximation for (1) 2. In particular, the ratio test (13. These test only work with positive term series, but if your series has both positive and negative terms you can test $\sum|a_n|$ for absolute convergence. 3 To my knowledge, this is the first paper to demonstrate unconditional convergence in. Drill problems on using the limit comparison test. 4-1: Comparison Test; Absolute Convergence Theorem; Limit Comparison Test Prakash Balachandran Department of Mathematics Duke University February 1, 2010 Please don’t send me short-term illness reports if you can’t make it to class. ps Author: [email protected] (David. Absolute Convergence Absolutely Convergent Describes a series that converges when all terms are replaced by their absolute values. A series P a. whose terms are the absolute values of the original series. Absolute convergence is the condition for an infinite series (all finite series are absolutely convergent) to have a single limit even if it is arbitrarily re-ord. The converse is false. Test convergence movements by having the patient fixate on an object as it is moved slowly towards a point right between the patient's eyes.
bskomzy6rerd, zkucbixmeh1gqe, 2nxjjzzbyufh, w3wu646yad5r2, dj02ejeq4bze, z6qbiukz0vg3j, 15ka5j34ys, wu8hzbauhpbej, a8n89s2vffcl83, cq870o1e71n21, uuw9dn4m03uan, ryzd289gemzm2, pohbhhf9nhmmm0, 0hdpfqlre2thwr, dnw6ygej4s, frngso1llukaz, l62fe0s1lto, 35npc1blzvtm, kgg0qtpwjjut, rqk9get6cmj011, b8eytbmxkgkv1, 5ooluucl4y075, tqs6tdwglnlkk, faompzyut25, kmx41uql8wvzds, 3h2x6iarc6kdb2, mjigfqj9zaymiu, dtbttdopab, t5l3wfrwwyibu | 2020-06-06T20:40:11 | {
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https://math.stackexchange.com/questions/1897670/minimum-and-maximum-value-of-z | # Minimum and Maximum value of |z|
This is a question that I came across today:
If $|z-(2/z)|=1$...(1) find the maximum and minimum value of |z|, where z represents a complex number.
This is my attempt at a solution:
Using the triangle inequality, we can write:
$||z|-|2/z||≤|z+2/z|≤|z|+|2/z|$
Let $|z|=r$ which implies that $|r-2/r|≤1≤r+2/r$ (From (1))
Since $|z-2/z| = 1$, we have \begin{aligned} 1 = |z- 2/z|^2 &= (z - 2/z)(\overline{z} - 2/\overline{z}) \\ &= |z|^2 - 2z/\overline{z} - 2\overline{z}/z + 4/|z|^2 \end{aligned} Write $z = re^{i\theta}$. Then $\overline{z} = re^{-i\theta}$. Substituting into the above, we obtain \begin{aligned} 1 &= r^2 - 2e^{i2\theta} - 2^{-i2\theta} + 4/r^2 \\ &= r^2 - 4\cos(2\theta) + 4/r^2 \\ \end{aligned} We can rearrange this to get $$r^2 + 4/r^2 \leq 4\cos(2\theta) + 1 \leq 5$$ or equivalently, $$r^4 -5r^2 + 4 \leq 0$$ Factor the left hand side to obtain $$(r-1)(r+1)(r-2)(r+2) \leq 0$$ As $r$ must be positive, this means that $r+1$ and $r+2$ are positive. Therefore, $r-1$ and $r-2$ must have opposite signs (or one of them is zero), which forces $1 \leq r \leq 2$.
You can check that $z=1$ and $z=2$ are solutions to the original equation, so $1$ and $2$ are the minimum and maximum values of $|z|$.
Here is a picture showing the set of solutions. From this, we might be tempted to speculate that we are looking at two circles of radius $1/2$, centered at $z=3/2$ and $z=-3/2$. But this is not the case. To see this, consider the point $z = 3/2 + i(1/2)$. This point is on the circle of radius $1/2$ centered at $z=3/2$. However, it does not satisfy the given equation: \begin{aligned} |z-2/z| &= |3/2 + i(1/2) - 2/(3/2 + i(1/2))| \\ &= |3/2 + i(1/2) - 6/5 + i(2/5)| \\ &= |3/10 + i(9/10)| \\ &= \sqrt{9/10} \neq 1 \end{aligned} I don't think it's an ellipse, either, since these are of the form $|z-a| + |z-b| = c$. If there is a way to transform $|z-2/z| = 1$ into that form, it's not obvious to me (also, our figure has not one but two "ellipses").
Using Complex Inequalities, $$||z|-|w||\le|z+w|\le|z|+|w|$$
$w=-\dfrac2z$ and writing $|z|=r$
$$\left|r-\dfrac2r\right|\le\left|z-\dfrac2z\right|\le r+\dfrac2r$$
$$\implies\left|r-\dfrac2r\right|\le1\le r+\dfrac2r$$
Now as $r>0,$ $$\dfrac{r+\dfrac2r}2\ge\sqrt{r\cdot\dfrac2r}=\sqrt2\iff r+\dfrac2r\ge2\sqrt2>1$$
So, we need $$\left|r-\dfrac2r\right|\le1\iff-1\le r-\dfrac2r\le1$$
$$r-\dfrac2r\le1\iff r^2-r-2\le0$$
Now we know $(x-a)(x-b)\le0$ with $a\le b;a\le x\le b$
Here $-1\le r\le2$ But $r>0\implies0<r\le2$
Can check for $-1\le r-\dfrac2r$ to find $r\ge1$
So, the final range $$1\le r\le2$$
Using this equality we can find both minimum and maximum. $$\left|z-\frac2z \right|\ge \left||z|-\frac2{|z|}\right|$$ $$\left||z|-\frac2{|z|}\right| \le 1$$ $$-1 \le |z|-\frac2{|z|}\le 1$$ $$|z|-\frac2{|z|}\ge -1$$ $$\left(|z| +\frac12\right)^2\ge \frac94$$ $$|z| \ge 1$$ $$|z|-\frac2{|z|}\le 1$$ $$\left(|z| - \frac12\right)^2\le \frac94$$ $$|z| \le 2$$ $$\implies 1\le |z| \le 2$$
• How did you arrive at that very last inequality by solving the one directly above it? – user361896 Aug 23 '16 at 10:30
• @KaumudiHarikumar Using last inequality,we found the minimum and maximum was founded above . – Aakash Kumar Aug 23 '16 at 14:59
• My question is how? – user361896 Aug 23 '16 at 23:48 | 2019-11-14T05:02:02 | {
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https://www.physicsforums.com/threads/challenge-13-sums-of-sines.731694/ | # Challenge 13: Sums of Sines
1. Jan 7, 2014
### Office_Shredder
Staff Emeritus
Prove that
$$\sum_{k=0}^{n} \sin\left( \frac{k \pi}{n} \right) = \cot \left( \frac{\pi}{2n} \right)$$
2. Jan 7, 2014
### Citan Uzuki
This is a fairly straightforward calculation:
\begin{align*}\sum_{k=0}^{n} \sin\left(\frac{k\pi}{n}\right) &= \Im \left(\sum_{k=0}^{n} \exp \left(\frac{k\pi i}{n} \right) \right) \\ &= \Im \left( \frac{1 - \exp \left(\frac{(n+1)\pi i}{n}\right)}{1 - \exp \left(\frac{\pi i}{n}\right)}\right)\\ &= \Im \left( \frac{1+\exp \left(\frac{\pi i}{n}\right)}{1 - \exp \left(\frac{\pi i}{n}\right)}\right)\\ &= \Im \left(\frac{\exp \left(-\frac{\pi i}{2n}\right) + \exp \left(\frac{\pi i}{2n} \right)}{\exp \left(-\frac{\pi i}{2n}\right) - \exp \left(\frac{\pi i}{2n}\right)}\right) \\ &= \Im \left(\frac{2 \cos \left(-\frac{\pi }{2n}\right)}{2i \sin \left(-\frac{\pi}{2n} \right)}\right) \\ &= \Im \left(-i \cot \left(-\frac{\pi}{2n} \right) \right) \\ &= \Im \left(i \cot \left(\frac{\pi}{2n} \right)\right) \\ &= \cot \left(\frac{\pi}{2n} \right) \end{align*}
3. Jan 7, 2014
### jbunniii
Here's another calculation:
\begin{align} \sin\left(\frac{\pi}{2n}\right) \sum_{k=0}^{n} \sin\left(\frac{k\pi}{n}\right) &= \sum_{k=0}^{n} \sin\left(\frac{k\pi}{n}\right) \sin\left(\frac{\pi}{2n}\right) \\ &= \frac{1}{2} \sum_{k=0}^{n} \left[ \cos\left( \frac{(2k-1)\pi}{2n}\right) - \cos\left(\frac{(2k+1)\pi}{2n}\right) \right] \\ \end{align}
The sum telescopes, so the above reduces to
\begin{align} \frac{1}{2} \left[ \cos\left(-\frac{\pi}{2n}\right) - \cos\left(\frac{(2n+1)\pi}{2n}\right) \right] &= \frac{1}{2}\left[ \cos\left(\frac{\pi}{2n}\right) - \cos\left(\pi + \frac{\pi}{2n}\right) \right] \\ &= \frac{1}{2} \left[\cos\left(\frac{\pi}{2n}\right) + \cos\left(\frac{\pi}{2n}\right)\right] \\ &= \cos\left(\frac{\pi}{2n}\right) \end{align}
Dividing both sides by $\sin\left(\frac{\pi}{2n}\right)$ gives us what we want.
4. Jan 8, 2014
### Office_Shredder
Staff Emeritus
Those are some nice solutions! Anyone have another way of doing the calculation?
5. Jan 11, 2014
### chingel
The sum is also the distance between opposite sides of a regular polygon with 2n sides of length 1, from where the result comes with simple geometry.
6. Jan 12, 2014
### CosmicKitten
I stumbled upon a really complicated way to d0 it by induction... too tired now to do the arithmetic and see if it works right now. After easily establishing that both sides equal 0 for n=1, it involves turning the right side into the form (e^ipi/n + 1)/(e^ipi/n -1) and multiplying the numerator and the denominator by e to the power of (i*pi/n^2)/(1+1/n) which turns the denominator of the exponent into n+1 without changing the numerator, but it changes the 1 on the numerator and the -1 in the denominator so you must add something to make them turn back into 1 and -1 without changing the e^ipi/n+1's. So add A/B to that and solve for A and B and then add A/B to the other side and see if you can get it in the n+1 form. Is that on the right track?
7. Jan 17, 2014
### Office_Shredder
Staff Emeritus
It might work, why don't you try going through the details to see if you can hammer it out?
chingel, that's a pretty cool way of attacking this problem. Can you explain a bit more how the geometry works out to give it?
8. Jan 17, 2014
### chingel
Since on an x-y plane the sine is the vertical projection of a segment of length 1 at a particular angle and we are trying to find the sum of sines, that is the sum of the vertical projections, we can add up all the segments and then find the vertical projection of the sum. Noticing that the angles increase with regular intervals and that the segments make up half of an 2n-gon, then the vertical projection of the sum is just twice the apothem of the 2n-gon and using the well known simple formula for the apothem we get the answer.
http://www.buzzle.com/articles/finding-apothem-of-a-regular-polygon.html
9. Jan 17, 2014
### Office_Shredder
Staff Emeritus
This is a very cool proof. | 2018-03-18T20:19:20 | {
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http://mathhelpforum.com/pre-calculus/141621-how-solve-x-equation.html | # Math Help - How to solve for x in an equation
1. ## How to solve for x in an equation
Solve for x.
x^n = x^(1/n)
How would we solve for x in this case? Would we move the x^(1/n) over to the left side and set it to zero. Then we factor out the left side? Please help me with this. Thanks
2. Why have you encountered this equation? Perhaps it is only a thought / property question. Rephrase in words and think about it.
What number is alterered identically by a power and a root.
The square is the square root.
The cube is the cube root.
There are not too many candidates.
Does it make a difference if x is positive or negative?
Does it make a difference if n is odd or even?
Again, why have you encountered this equation?
3. This is the problem. I am stuck on b). I have concluded that in order to find point A we must set x^n = x^(1/n) equal to each other because that is where they intersect. After finding x we can plug it back in the equation to find y and thus we would have a coordinate. But the problem is I don't know how to find x. Is there a better way to solve this or am I on the right track? Thanks for the help.
4. Originally Posted by florx
This is the problem. I am stuck on b). I have concluded that in order to find point A we must set x^n = x^(1/n) equal to each other because that is where they intersect. After finding x we can plug it back in the equation to find y and thus we would have a coordinate. But the problem is I don't know how to find x. Is there a better way to solve this or am I on the right track? Thanks for the help.
In the given graphs, suppose n =2. So, In the parabola $y = x^2$, the coordinate pairs are $(x, x^2)$. You should be able to see that the following points are on the graph: (1, 1), (−1, 1), (2, 4), (−2, 4), and so on.
The graph of the square root function is related to $y = x^2$. The coördinate pairs are $(x,\sqrt{x})$. For example, (1, 1), (4, 2), (9, 3), and so on.
So for any n, the graph of $y=x^{\frac{1}{n}}$ will not be negative.
5. So how would we explain that point A is (1,1)? The back of the book says that is the answer.
6. Hello, florx!
Your game plan is absolutely correct!
Solve for $x\!:\;\;x^n \:=\: x^{\frac{1}{n}}$
We have: . $x^n - x^{\frac{1}{n}} \;=\;0$
Factor: . $x^{\frac{1}{n}}\left(x^{\frac{n^2-1}{n}} - 1\right) \;=\;0$
And we have two equations to solve:
. . $x^{\frac{1}{n}} \:=\:0 \quad\Rightarrow\quad x \:=\:0^n \quad\Rightarrow\quad \boxed{x \:=\:0}$
. . $x^{\frac{n^1-1}{n}} - 1 \:=\:0 \quad\Rightarrow\quad x^{\frac{n^2-1}{n}} \:=\:1 \quad\Rightarrow\quad x \:=\:1^{\frac{n}{n^2-1}} \quad\Rightarrow\quad\boxed{x \:=\:1}\;\;\text{ for }n \neq \pm1$
7. That's just a little insane.
Back to my original question:
What number is alterered identically by a power and a root?
There are not too many candidates.
Only 0 and 1. finis.
Why drag through messy algebra with ambiguous exponents when it's really just a definition / property question? | 2015-11-26T01:32:09 | {
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https://yutsumura.com/example-of-a-nilpotent-matrix-a-such-that-a2neq-o-but-a3o/ | # Example of a Nilpotent Matrix $A$ such that $A^2\neq O$ but $A^3=O$.
## Problem 305
Find a nonzero $3\times 3$ matrix $A$ such that $A^2\neq O$ and $A^3=O$, where $O$ is the $3\times 3$ zero matrix.
(Such a matrix is an example of a nilpotent matrix. See the comment after the solution.)
Contents
## Solution.
For example, let $A$ be the following $3\times 3$ matrix.
$A=\begin{bmatrix} 0 & 1 & 0 \\ 0 &0 &1 \\ 0 & 0 & 0 \end{bmatrix}.$ Then $A$ is a nonzero matrix and we have
$A^2=\begin{bmatrix} 0 & 1 & 0 \\ 0 &0 &1 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 & 0 \\ 0 &0 &1 \\ 0 & 0 & 0 \end{bmatrix} =\begin{bmatrix} 0 & 0 & 1 \\ 0 &0 &0 \\ 0 & 0 & 0 \end{bmatrix}\neq O.$
The third power of $A$ is
$A^3=A^2A=\begin{bmatrix} 0 & 0 & 1 \\ 0 &0 &0 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 & 0 \\ 0 &0 &1 \\ 0 & 0 & 0 \end{bmatrix}= \begin{bmatrix} 0 & 0 & 0 \\ 0 &0 &0 \\ 0 & 0 & 0 \end{bmatrix}=O.$ Thus, the nonzero matrix $A$ satisfies the required conditions $A^2\neq O, A^3=O$.
## Comment.
A square matrix $A$ is called nilpotent if there is a non-negative integer $k$ such that $A^k$ is the zero matrix.
The smallest such an integer $k$ is called degree or index of $A$.
The matrix $A$ in the solution above gives an example of a $3\times 3$ nilpotent matrix of degree $3$.
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Problem 1 Let $W$ be the subset of the $3$-dimensional vector space $\R^3$ defined by \[W=\left\{ \mathbf{x}=\begin{bmatrix} x_1 \\ x_2...
Close | 2018-10-23T16:16:33 | {
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http://sites.millersville.edu/bikenaga/courses/345-s18/homework/ps6/ps6-solutions.html | # Solutions to Problem Set 6
Math 345/504
2-7-2018
1. The function defined by is a group map. List the elements of and the elements of .
2. Find the quotient and remainder when the Division Algorithm is applied to:
(a) Divide 937 by 28.
(b) Divide -937 by 28.
(a)
(b)
3. An integer n is even if . Using only this definition and properties of divisibility, prove that if n is even, then is even.
Suppose n is even, so for some . Then
Therefore, is even.
4. Find the largest power of 7 that divides . Explain your reasoning.
There are 7 numbers in the set that are divisible by 7 by not :
There is one number in the set that is divisible by , namely 49.
Hence, the largest power of 7 that divides is .
5. Let denote the group of real numbers with the operation of addition. Let denote the group of real numbers under the operation
(a) Prove that given by is a group map.
(b) Show that f is an isomorphism by constructing an inverse for f.
(a) If , then
Therefore, , and f is a group map.
(b) Define by
Then
Thus, f and g are inverses, and f is an isomorphism.
Note: This problem gives an example of two isomorphic group structures on the same set.
[MATH 504]
6. is a group map that satisfies . Find a formula for .
Note that in . So
It follows that .
7. Prove, or disprove by specific counterexample: "If and and , then ."
The statement is false, since and , but .
You probably wouldn't worry about what people think of you if you could know how seldom they do. - Olin Miller
Contact information
Copyright 2018 by Bruce Ikenaga | 2018-03-21T15:33:50 | {
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http://mathhelpforum.com/calculus/38600-integrate-semicircle-probability-distribution.html | # Thread: Integrate the semicircle probability distribution
1. ## Integrate the semicircle probability distribution
Hi folks, I know its a long shot, but would anyone like to explain how to integrate the function
f(x) = 2/(pi * r^2) sqrt(R^2 - x^2)
I'm sorry about the poor notation. Please see Wigner semicircle distribution - Wikipedia, the free encyclopedia for a clearer idea.
It's been a long time since I've done integration. I presume it involves a substitution like, say, u = R^2 - x^2. I just can't seem to figure it out.
Got any ideas?
2. Originally Posted by sabatier
Hi folks, I know its a long shot, but would anyone like to explain how to integrate the function
f(x) = 2/(pi * r^2) sqrt(R^2 - x^2)
I'm sorry about the poor notation. Please see Wigner semicircle distribution - Wikipedia, the free encyclopedia for a clearer idea.
It's been a long time since I've done integration. I presume it involves a substitution like, say, u = R^2 - x^2. I just can't seem to figure it out.
Got any ideas?
Make a Trig Substitution:
$x=R\sin\theta \implies dx=R\cos\theta\,d\theta$
Then the integral becomes
$
\frac{2}{\pi R^2} \int \sqrt{R^2-R^2\sin^2\theta} R\cos\theta \,d\theta$
$\implies \frac{2}{\pi}\int \cos^2\theta\,d\theta$.
To integrate this, recall that $\cos^2\theta=\frac{1+\cos(2\theta)}{2}$.
After you integrate, get the answer back in terms of x.
Hope this helped!!
3. Hey thanks a million for replying so quickly! Cheers!
4. Originally Posted by sabatier
Hi folks, I know its a long shot, but would anyone like to explain how to integrate the function
f(x) = 2/(pi * r^2) sqrt(R^2 - x^2)
I'm sorry about the poor notation. Please see Wigner semicircle distribution - Wikipedia, the free encyclopedia for a clearer idea.
It's been a long time since I've done integration. I presume it involves a substitution like, say, u = R^2 - x^2. I just can't seem to figure it out.
Got any ideas?
Alternatively since $\int_{-r}^{r}\frac{2}{\pi{(r_1)^2}}\sqrt{r^2-x^2}dx$
Just represents the integration of a semi-circle of radius r multiplied by a contant. And since integration is equivalent to finding area we can see that
$A_{semi-circle}=\frac{1}{2}\pi{r^2}$
So $\frac{2}{\pi{(r_1)^2}}\int_{-r}^{r}\sqrt{r^2-x^2}dx=\frac{2}{\pi{(r_1)^2}}\cdot\bigg(\frac{1}{2 }\pi{r^2}\bigg)$
Now if $r_1=r$
$\int_{-r}^{r}\frac{2}{\pi(r_1)^2}\sqrt{r^2-x^2}dx=1$
if $r_1\ne{r}$
then $\int_{-r}^{r}\frac{2}{\pi{(r_1)^2}}\sqrt{r^2-x^2}dx=\bigg(\frac{r}{r_1}\bigg)^2$
NOTE this only applies to if the integration limits are -r and r...that is what I have gathered...if they are not DO NOT APPLY this....even if they are not I hope this has helped gap a bride between integration and area and how sometimes we dont eve need to integrate
5. Originally Posted by sabatier
Hi folks, I know its a long shot, but would anyone like to explain how to integrate the function
f(x) = 2/(pi * r^2) sqrt(R^2 - x^2)
I'm sorry about the poor notation. Please see Wigner semicircle distribution - Wikipedia, the free encyclopedia for a clearer idea.
It's been a long time since I've done integration. I presume it involves a substitution like, say, u = R^2 - x^2. I just can't seem to figure it out.
Got any ideas?
Sinve f(x) is the density of the Wigner distribution we know that:
$I=\int_{-r}^r \frac{2}{\pi R^2} \sqrt{R^2-x^2}~dx=1$
So we might be interested in the cumulative distribution:
$
I(\rho)=\int_{-r}^{\rho} \frac{2}{\pi R^2} \sqrt{R^2-x^2}~dx
=\frac{2}{\pi R} \int_{-r}^{\rho} \sqrt{1-(x/R)^2}~dx
$
Now put $u=x/R$ :
$
I(\rho) = \frac{2}{\pi} \int_{-1}^{\rho/R} \sqrt{1-u^2}~du
$
Now a trig substitution like $u=\sin(\theta)$ will finish this by turning it into a known integral (or at least one that you can see how to do).
RonL | 2016-12-03T22:54:55 | {
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https://mathhelpboards.com/threads/nth-order-differentiation-challenge.5493/ | # nth order differentiation challenge
#### MarkFL
Staff member
Let:
$$\displaystyle f(x)=x\sin(x)$$
Derive a formula for:
$$\displaystyle f^{(n)}(x)$$
Using this, infer a formula for:
$$\displaystyle \frac{d^n}{dx^n}\left(x\cos(x) \right)$$
edit: I wanted to make sure it is clearly understood that:
$$\displaystyle f^{(n)}(x)\equiv\frac{d^n}{dx^n}\left(f(x) \right)$$
#### Ackbach
##### Indicium Physicus
Staff member
Using the Leibnitz rule for the $n$-th derivative of a product, which in general is
$$\frac{d^{n}}{dx^{n}}[f(x)\,g(x)]= \sum_{j=0}^{n} \left[{n \choose j} \frac{d^{j}f(x)}{dx^{j}} \frac{d^{n-j}g(x)}{dx^{n-j}}\right],$$
we have that
$$\frac{d^{n}}{dx^{n}}[x \, \sin(x)]= \sum_{j=0}^{n} \left[{n \choose j} \frac{d^{j}x}{dx^{j}} \frac{d^{n-j} \sin(x)}{dx^{n-j}} \right].$$
Now, there are only two non-zero terms: $j=0$ and $j=1$. Hence, the sum collapses down to those two terms:
$$\frac{d^{n}}{dx^{n}}[x \, \sin(x)]= {n \choose 0} x \frac{d^{n} \sin(x)}{dx^{n}}+{n \choose 1} \frac{d^{n-1} \sin(x)}{dx^{n-1}}.$$
Note that
$${n \choose 0}=1,\quad \text{and} \quad {n \choose 1}=n,$$
so we further simplify as
$$\frac{d^{n}}{dx^{n}}[x \, \sin(x)]= x \frac{d^{n} \sin(x)}{dx^{n}}+n \frac{d^{n-1} \sin(x)}{dx^{n-1}}.$$
To finish, we can use the fact that
$$\frac{d^{n}}{dx^{n}} \sin(x)= \sin \left( \frac{n \pi}{2}+x \right),$$
to write the derivative as
$$\frac{d^{n}}{dx^{n}}[x \, \sin(x)]= x \sin \left( \frac{n \pi}{2}+x \right)+n \sin \left( \frac{(n-1) \pi}{2}+x \right).$$
The same formula happens to work for $\cos$, so simply replace $\sin$ with $\cos$ to get the equivalent formula for $(d^{n}/dx^{n})(x \cos(x))$.
#### MarkFL
Staff member
Using the Leibnitz rule for the $n$-th derivative of a product, which in general is
$$\frac{d^{n}}{dx^{n}}[f(x)\,g(x)]= \sum_{j=0}^{n} \left[{n \choose j} \frac{d^{j}f(x)}{dx^{j}} \frac{d^{n-j}g(x)}{dx^{n-j}}\right],$$
we have that
$$\frac{d^{n}}{dx^{n}}[x \, \sin(x)]= \sum_{j=0}^{n} \left[{n \choose j} \frac{d^{j}x}{dx^{j}} \frac{d^{n-j} \sin(x)}{dx^{n-j}} \right].$$
Now, there are only two non-zero terms: $j=0$ and $j=1$. Hence, the sum collapses down to those two terms:
$$\frac{d^{n}}{dx^{n}}[x \, \sin(x)]= {n \choose 0} x \frac{d^{n} \sin(x)}{dx^{n}}+{n \choose 1} \frac{d^{n-1} \sin(x)}{dx^{n-1}}.$$
Note that
$${n \choose 0}=1,\quad \text{and} \quad {n \choose 1}=n,$$
so we further simplify as
$$\frac{d^{n}}{dx^{n}}[x \, \sin(x)]= x \frac{d^{n} \sin(x)}{dx^{n}}+n \frac{d^{n-1} \sin(x)}{dx^{n-1}}.$$
To finish, we can use the fact that
$$\frac{d^{n}}{dx^{n}} \sin(x)= \sin \left( \frac{n \pi}{2}+x \right),$$
to write the derivative as
$$\frac{d^{n}}{dx^{n}}[x \, \sin(x)]= x \sin \left( \frac{n \pi}{2}+x \right)+n \sin \left( \frac{(n-1) \pi}{2}+x \right).$$
The same formula happens to work for $\cos$, so simply replace $\sin$ with $\cos$ to get the equivalent formula for $(d^{n}/dx^{n})(x \cos(x))$.
I recall with fondness "discovering" and proving by induction the formula you used up front when fiddling with reduction of order techniques in my ODE course.
I will give others time before posting my own solution.
#### Fernando Revilla
##### Well-known member
MHB Math Helper
Another way:
Consider $h:\mathbf{R}\to \mathbf{C},\;\; h(x)=x\cos x+ix\sin x=xe^{ix}$. Then, $$\frac{d^n}{dx^n}(x\cos x)=\text{Re }\left(h^{(n)}(x)\right)\;,\quad \frac{d^n}{dx^n}(x\sin x)=\text{Im }\left(h^{(n)}(x)\right)$$
#### Ackbach
##### Indicium Physicus
Staff member
I recall with fondness "discovering" and proving by induction the formula you used up front when fiddling with reduction of order techniques in my ODE course.
I will give others time before posting my own solution.
And I remember taking a graduate-level physics course where I was the only mathematical physics guy there, and being asked to put a proof of this formula on the board. The prof asked me to include every step, including the formula for the entry in Pascal's Triangle as the sum of the two numbers directly above it. (The prof seemed to think that the rest of the students couldn't get this on their own - I suppose maybe because he had seen their homework!) I remember thinking to myself that this really wasn't that big a deal. The other students should have been able to do it. Indeed, their physics abilities were far beyond my own. But in the words of the prof, "They don't know how to differentiate!"
My consolation is that all theoretical physicists these days are mathematicians - they have to be, because the math people have gotten so far behind the physics people, that whenever the physicists look around for a needed tool, it's no longer there like it was, say, even 70 years ago.
#### anemone
##### MHB POTW Director
Staff member
My approach:
We're given $$\displaystyle f(x)=x\sin(x)$$
I first find the first few nth derivative of the function of $$\displaystyle f(x)=x\sin(x)$$ to see if there is any pattern to be observed...
$$\displaystyle f^{(1)}(x)=x\cos x +\sin x$$
$$\displaystyle f^{(2)}(x)=-x\sin x +2\cos x$$
$$\displaystyle f^{(3)}(x)=-x\cos x -3\sin x$$
$$\displaystyle f^{(4)}(x)=x\sin x -4\cos x$$
$$\displaystyle f^{(5)}(x)=x\cos x +5\sin x$$
$$\displaystyle f^{(6)}(x)=-x\sin x +6\cos x$$...
At this point, it is easy to generate the general formula for $$\displaystyle f^{(n)}(x)$$ where
$$\displaystyle f^{(n)}(x)=x\sin (\frac {\pi n}{2}+x)-n\cos (\frac {\pi n}{2}+x)$$
and it can easily be shown to be true using induction method.
Let $$\displaystyle P(n)$$ be the statement $$\displaystyle f^{(n)}(x)=x\sin (\frac {\pi n}{2}+x)-n\cos (\frac {\pi n}{2}+x)$$.
$$\displaystyle P(1)$$ asserts that $$\displaystyle f^{(1)}(x)=x\sin (\frac {\pi }{2}+x)-\cos (\frac {\pi }{2}+x)=x\cos x +\sin x$$ which is clearly true.
Next, suppose $$\displaystyle P(n)$$ is true for $$\displaystyle n=k$$. We need to prove $$\displaystyle P(n)$$ is true for $$\displaystyle n=k+1$$.
For $$\displaystyle P(k+1)$$, we have
$$\displaystyle f^{(k+1)}(x)=\frac{d}{dx} \left(x\sin (\frac {\pi k}{2}+x)-k\cos (\frac {\pi k}{2}+x)\right)$$
$$\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;=x\cos (\frac {\pi k}{2}+x)(1)+(1)(\sin (\frac {\pi k}{2}+x)-(k)(-\sin (\frac {\pi k}{2}+x))$$
$$\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;=x\cos (\frac {\pi k}{2}+x)(1)+(1)(\sin (\frac {\pi k}{2}+x)+k\sin (\frac {\pi k}{2}+x)$$
thereby showing that indeed $$\displaystyle P(k+1)$$ holds and this finished our work.
Now, we have $$\displaystyle f(x)=x\cos(x)$$.
Notice that there is a phase shift (delay) of $$\displaystyle \frac{\pi}{2}$$rad from $$\displaystyle y=x\sin x$$ to $$\displaystyle y=x\cos x$$ and also the signs of the two terms are of the opposite have been shifted in the other direction by $$\displaystyle -\frac{\pi}{2}$$ rad.
Therefore, the general formula for $$\displaystyle f^{(n)}(x)$$ if $$\displaystyle f(x)=x\cos x$$ can be obtained from:
$$\displaystyle f^{(n)}(x)=-x\sin (\frac {\pi n}{2}+x-\frac{\pi}{2})+n\cos (\frac {\pi n}{2}+x-\frac{\pi}{2})$$
$$\displaystyle \;\;\;\;\;\;\;\;\;\;\;=x\cos (\frac {\pi n}{2}+x)+n\sin (\frac {\pi n}{2}+x)$$
#### MarkFL
Staff member
Good work, anemone!
One of my approaches is quite similar to that which you gave, the other being that given by Ackbach.
This is how I formulated this approach:
Given:
$$\displaystyle f(x)=x\sin(x)$$
we find:
$$\displaystyle f'(x)=x\cos(x)+\sin(x)$$
$$\displaystyle f''(x)=-x\sin(x)+2\cos(x)$$
$$\displaystyle f'''(x)=-x\cos(x)-3\sin(x)$$
At this point, we may state the hypothesis:
$$\displaystyle f^{(n)}(x)=x\sin\left(x+n\frac{\pi}{2} \right)-n\cos\left(x+n\frac{\pi}{2} \right)$$
Having already demonstrated the base case (and the next few), we may use as our inductive step, differentiation of $P_n$ with respect to $x$:
$$\displaystyle f^{(n+1)}(x)=x\cos\left(x+n\frac{\pi}{2} \right)+(n+1)\sin\left(x+n\frac{\pi}{2} \right)$$
Now, using the identities:
(1) $$\displaystyle \sin\left(\theta-\frac{\pi}{2} \right)=-\cos(\theta)$$
(2) $$\displaystyle \cos\left(\theta-\frac{\pi}{2} \right)=\sin(\theta)$$
we may write:
$$\displaystyle f^{(n+1)}(x)=x\sin\left(x+(n+1)\frac{\pi}{2} \right)-(n+1)\cos\left(x+(n+1)\frac{\pi}{2} \right)$$
We have derived $P_{n+1}$ from $P_n$, thereby completing the proof by induction.
Now, if we are given:
$$\displaystyle g(x)=x\cos(x)=x\sin\left(x+\frac{\pi}{2} \right)$$
Then from the above, we may infer:
$$\displaystyle g^{(n)}(x)=x\sin\left(x+(n+1)\frac{\pi}{2} \right)-n\cos\left(x+(n+1)\frac{\pi}{2} \right)$$
and using (1) and (2), we find:
$$\displaystyle g^{(n)}(x)=x\cos\left(x+n\frac{\pi}{2} \right)+n\sin\left(x+n\frac{\pi}{2} \right)$$
Another way to infer this second formula is to write:
$$\displaystyle h(x)=x\sin\left(x+\frac{\pi}{4} \right)$$
Thus, we know:
$$\displaystyle h^{(n)}(x)=x\sin\left(x+\frac{\pi}{4}+n\frac{\pi}{2} \right)-n\cos\left(x+\frac{\pi}{4}+n\frac{\pi}{2} \right)$$
Using the identities:
$$\displaystyle \sin\left(\theta+\frac{\pi}{4} \right)=\frac{1}{\sqrt{2}}\left(\sin(\theta)+\cos(\theta) \right)$$
$$\displaystyle \cos\left(\theta+\frac{\pi}{4} \right)=\frac{1}{\sqrt{2}}\left(\cos(\theta)-\sin(\theta) \right)$$
we may write:
$$\displaystyle h^{(n)}(x)=\frac{1}{\sqrt{2}}\left(x\left(\sin \left(x+n\frac{\pi}{2} \right)+\cos\left(x+n\frac{\pi}{2} \right) \right)-n \left(\cos\left(x+n\frac{\pi}{2} \right)-\sin \left(x+n\frac{\pi}{2} \right) \right) \right)$$
$$\displaystyle h^{(n)}(x)=\frac{1}{\sqrt{2}}\left(x\sin \left(x+n\frac{\pi}{2} \right)-n\cos\left(x+n\frac{\pi}{2} \right)+x\cos\left(x+n\frac{\pi}{2} \right)+n\sin \left(x+n\frac{\pi}{2} \right) \right)$$
$$\displaystyle h^{(n)}(x)=\frac{1}{\sqrt{2}}\left(\frac{d^n}{dx^n}\left(x\sin(x) \right)+x\cos\left(x+n\frac{\pi}{2} \right)+n\sin \left(x+n\frac{\pi}{2} \right) \right)$$
Now, since we may write:
$$\displaystyle h(x)=\frac{1}{\sqrt{2}}\left(x\sin(x)+x\cos(x) \right)$$
And by the linearity of differentiation, we know:
$$\displaystyle h^{(n)}(x)=\frac{1}{\sqrt{2}}\left(\frac{d^n}{dx^n}\left(x\sin(x) \right)+\frac{d^n}{dx^n}\left(x\cos(x) \right) \right)$$
Thus, we must have:
$$\displaystyle \frac{d^n}{dx^n}\left(x\cos(x) \right)=x\cos\left(x+n\frac{\pi}{2} \right)+n\sin \left(x+n\frac{\pi}{2} \right)$$ | 2020-09-23T13:06:26 | {
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https://brilliant.org/discussions/thread/problem-solving-2/ | # Problem Solving
When solving a problem (as opposed to merely using a given formula), it is generally good to proceed along the following lines:
1. First, you have to understand the problem.
2. After understanding, make a plan.
3. Carry out the plan.
4. Look back on your work. How could it be better?
### Understanding the Problem
Before you begin calculating, it is a good idea to spend some time trying to understand exactly what it is that is being asked. Here are some questions to ask yourself before you begin:
• Are there terms or definitions in the problems that I don't understand?
• What information will I need to solve this problem? Is it all present in the question?
• Can I restate the problem in my own words?
• What form will a correct answer take?
### Make a Plan
There are often many reasonable approaches to solving a problem, and every student has a different apporach. Make a plan that will help you get to the answer. Some possible apporaches:
• Draw a picture.
• Solve an equation.
• Guess and check.
• Make a list and look for patterns.
• Solve a simpler problem and see if it sheds light on this one.
• Look at individual cases.
• Work backwards from the answer.
• Use a formula.
### Carry out the Plan
This is usually much easier than understanding the problem or making the plan. Proceed forward with patience and care. If you find that your plan has failed or that you did not fully understand the problem before beginning, now is the time to start over and begin again. Don't be afraid to discard a failed plan; keep trying and you will solve it!
Problem solving is a skill that can only be aquired through extensive practice, and one of the things that ensures you continue to progress is actively paying attention to your own work. Did you make any careless mistakes? Was there anything that would have been easier if you'd understood it better? Review your work to make the next problem easier.
Note: these instructions were inspired by How to Solve It by George Pólya.
Note by Arron Kau
5 years, 6 months ago
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Good advice. But you must consider that answering questions in exam is much different from doing it this way because of the shortage of time. So can you provide any strategies for that.
- 4 years, 9 months ago
Well in the beginning it may be boring and frustrating, but if you practice these steps enough times, you surely can do all of these in a minute or two, while seeing the mathematics with a different angle and every angle, and writing the genuine solution, and having a feeling like you have invented the solution.
- 4 years, 9 months ago | 2020-04-08T16:13:38 | {
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https://math.stackexchange.com/questions/2494317/seating-couples-around-2-tables | # Seating couples around 2 tables
Here's my question and possible answer.
How many possible ways can you arrange 8 married couples between 2 circular tables of 8 identical chairs each such that:
1) each couple must sit at the same table, and,
2) at each table, men and women must sit in adjacent chairs (NOTE: a couple can sit next to each other but doesn't have to).
My solution:
Number of ways = (number of ways to split 8 couples into 2 tables of 4 couples each) * (number of arrangements at each table)
$=\frac{8!}{4!4!}*$(4 men and 4 women sitting alternately in 2 ways)
$=\frac{8!}{4!4!}\!\cdot\! 4!\!\cdot\!4!\cdot\!2$
$=2 * 8!$
Can someone verify this solution or provide the correct one?
First, pick $4$ couples out of the $8$ couples to sit at one table: $8 \choose 4$
Note that this will fix the people at the other table as well. Now, if we differentiate between the two tables, then the number of ways to split the $16$ people between the two tables is ${8 \choose 4}$ (that is the number of ways to pick the people for table 1, fixing the rest for table 2). If you do not differentiate between the tables, then divide this by $2$.
Now let's arrange the people. We'll calculate the number of ways to seat the people around one table, and just multiply by that number again for the other table at the end.
Since it's a circular table with identical chairs, we'll 'anchor' the seats with $1$ of the women. Then, we can seat the other women in $3!$ ways relative to this woman, and the men in $4!$ ways.
Total: $${8 \choose 4} \cdot 3! \cdot 4! \cdot 3! \cdot 4!$$
And again, if you do not differentiate between the tables, then divide this by $2$
• Suppose we have an origin from where to start counting round one of the tables, then this would class the string ABCDEFGH to be different to BCDEFGHA, but surely due to the question detailing that the tables are circular, these are the same? Everyone is sitting next to the same people either side. Also, you may split up a couple if you make the selections by choosing women first, then men. It would be better to pair up the couples and choose 4 couples to sit at a table. Oct 29, 2017 at 0:10
• @JohnDoe Right, that's why you get $3! \cdot 4!$ ways to seat the people around the table, instead of $4! \cdot 4!$. I do like your idea of first pairing up the men and women, and then seating them one pair at a time: you should create your own answer using that method! But note: you will get that same $3!$ factor when seating the 4 pairs around the circular table Oct 29, 2017 at 0:16
• My answer disagreed with yours by a factor of $8\choose4$, due to how I chose the couples. I am fairly sure mine is correct, but feel free to enlighten me if I am missing something. Oct 29, 2017 at 0:27
• @JohnDoe I didn't check tour answer, but my answer is certainly wrong! I forgot to account for the fact that a couple needs to sit at the same table! Oct 29, 2017 at 0:29
• Ah, yes that's why mine is different! Ours should agree once you correct this haha Oct 29, 2017 at 0:31
How I would do it:
How many ways can you split the 8 couples into two groups of 4? That's ${8\choose4}=70$.
Then once split, you have 4 men, 4 women who need to be seated at one of the tables, in alternating order. The first person seated has 8 choices, then the other people of the same gender have 3,2,1 choices afterwards. Meanwhile for the opposite gender, they have 4,3,2,1 choices for their picks. So in total, $8\times4\times3\times3\times2\times2=1152$. Then to account for the rotational symmetry, we must divide by $8$ to give $144$ ways to arrange these couples on their table.
Then it is the same for the other table, so another $144$ ways to arrange them, hence $144^2$ ways to arrange all the people once tables have been chosen for each couple.
So the overall answer is $$70\times144^2=1,451,520$$
• Nice, I agree with the solution! However it's worth noting that if you fix a person to a particular place you avoid the multiplying and dividing by 8, don't you think? Oct 29, 2017 at 0:48
• Yeah, multiplying and dividing by 8 was a bit pointless in this question. I suppose it could be useful to see it done like this if an example ever came up where it was required to seat in a line rather than in a circle - then you would still have to multiply by 8, but since there is no rotational symmetry, you wouldn't divide by it. Oct 29, 2017 at 0:50 | 2022-07-01T14:33:08 | {
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https://math.stackexchange.com/questions/1044187/prove-convergence-by-considering-the-partial-sums | # Prove convergence by considering the partial sums
Let $p$ be a non-zero natural number. Prove by considering the partial sums that
$\sum \frac{1}{k(k+p)}$
converges. What is $\sum\limits_{k=1}^{\infty} \frac{1}{k(k+p)}$
No idea. Obviously, it looks like a telescoping series. Sure doesn't act like one. I have tried to treat it like I would a telescoping series to see if it would get me any where--it did not.
Alongside advice on how to do this one. If people could also offer general advice on taking partial sums that would be greatly appreciated.
• it looks like a telescoping series - That's because it is. – Lucian Nov 29 '14 at 21:56
The series converges by the comparison test, for example. Now:$$\frac1{k(k+p)}=\frac1p\left(\frac1k-\frac1{k+p}\right)\implies$$
$$\sum_{n=1}^{2p}\frac1{k(k+p)}=\frac1p\left(1-\color{red}{\frac1{p+1}}+\ldots+\frac1p-\color{green}{\frac1{2p}}+\color{red}{\frac1{p+1}}-\frac1{2p+1}+\ldots+\color{green}{\frac1{2p}}-\frac1{3p}\right)$$
Let us write the summands in columns, the plus sign column and the minus sign column:
\begin{align*}&\frac11&-\frac1{p+1}\\ &\frac12&-\frac1{2+p}\\ &\frac13&-\frac1{3+p}\\&\ldots&\ldots\\ &\frac1p&-\frac1{2p}\\ &\frac1{p+1}&-\frac1{2p+1}\\ &\ldots&\ldots\\ &\frac1{2p}&-\frac1{3p}\end{align*}
We can see the first $\;p\;$ plus summands remain, whereas the first minus summands cancel with the last $\;p\;$ plus summands, and this is so no matter what multiple $\;mp\;$ we take, so passing to the limit when $\;m\to\infty\;$, we get the sum
$$\frac1p\left(1+\frac12+\ldots+\frac1p\right)$$
• Sorry, but does this tell me the value to which this converges? Because that is the main reason that I am struggling. I know how to show convergence. I need to show convergence using partial sums and then state the value of the sum. – Bob the Builds Nov 29 '14 at 22:52
• Didn't you see the very last figure? That's the sum! – Timbuc Nov 30 '14 at 0:22
• Yeah. Was having trouble at first understanding how you arrived there, but I got it, and gave this answer the checkmark – Bob the Builds Nov 30 '14 at 1:40
Hint: Write $$\frac1{k(k+p)}=\frac{A}{k}+\frac{B}{k+p}.$$ Solve for $A$ and $B,$ and you'll find that the series does telescope.
Added: Now that I'm back at my computer and have a little time, and now that you've accepted an answer, I'll expand on my own. You should readily find that $A=\frac1p$ and $B=-\frac1p.$ Consequently, we can rewrite the series as $$\frac1p\sum_{k=1}^\infty\left(\frac1k-\frac1{k+p}\right).$$ To prove convergence and find the limit, it will suffice to consider the partial sums $$S_n:=\frac1p\sum_{k=1}^n\left(\frac1k-\frac1{k+p}\right).$$ We'd like to get this into a more convenient form, as a difference of sums, rather than a sum of differences. That is, we'll rewrite it as $$S_n=\frac1p\sum_{k=1}^n\frac1k-\frac1p\sum_{k=1}^n\frac1{k+p},\tag{1}$$ which identity is readily proved by arithmetic properties.
Now, take any integer $m\ge1$ and note that $$\sum_{k=1}^{p+m}\frac1k=\sum_{k=1}^p\frac1k+\sum_{k=p+1}^{p+m}\frac1k.\tag{2}$$ On the other hand, $$\sum_{k=1}^{p+m}\frac1{k+p}=\sum_{k=1}^{p+m}\frac1{p+k}=\sum_{k=1}^m\frac1{p+k}+\sum_{k=m+1}^{p+m}\frac1{p+k}=\sum_{k=p+1}^{p+m}\frac1k+\sum_{k=m+1}^{p+m}\frac1{p+k}.\tag{3}$$
Consequently, by $(1)$ through $(3),$ we have for any integer $m\ge1$ that $$S_{p+m}=\frac1p\sum_{k=1}^p\frac1k-\frac1p\sum_{k=m+1}^{p+m}\frac1{p+k}=\frac1p\sum_{k=1}^p\frac1k-\frac1p\sum_{k=1}^p\frac1{p+m+k}.\tag{\star}$$ Now, for any such integer $m,$ we have $$0<\sum_{k=1}^p\frac1{p+m+k}<\sum_{k=1}^p\frac1{p+m+1}=\frac{p}{p+m+1},$$ so $$-\frac1{p+m+1}<-\frac1p\sum_{k=1}^p\frac1{p+m+1}<0,$$ and so by $(\star),$ we have $$-\frac1{p+m+1}<S_{p+m}-\frac1p\sum_{k=1}^p\frac1k<0$$ for all integers $m\ge1.$ A quick application of the Squeeze Theorem shows that the sequence of partial sums converges to $\frac1p\sum\limits_{k=1}^p\frac1k,$ proving series convergence and giving us its sum.
• Not sure why this was downvoted. Nice hint, +1 – Zubin Mukerjee Nov 29 '14 at 22:15
• I'm sorry, but you are going to have to give me a little more than that. When I say that I tried to treat it like a telescoping series, I mean to say that I tried to place the summation in the form of a telescoping series. – Bob the Builds Nov 29 '14 at 22:50
• @Bob: The thing to notice here is that the telescoping begins on the $p$th term--that is, when $k=p.$ Moreover, $\frac1{k+p}$ takes on every value that $\frac1k$ does as $k$ ranges over the positive integers, except for $1,\frac12,...,\frac1p.$ Hence, noting the values of $A$ and $B,$ the series's sum is simply a multiple of $\sum\limits_{k=1}^p\frac1k.$ – Cameron Buie Nov 29 '14 at 23:03 | 2019-07-15T17:57:06 | {
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http://math.stackexchange.com/questions/240207/does-the-absolute-value-even-matter-here | # Does the absolute value even matter here?
In this problem I'm doing it says
Suppose that $(X,Y)$ is uniformly distributed over the region {$(x,y):0\lt |y|\lt x\lt 1$}. Find the marginal densities $f_X(x)$ and $f_Y(y)$
Does the absolute value even matter here since it's all between $0$ and $1$ anyway? How is the answer $f_X(x)=2x$?
-
The absolute value definitely matters. It says the region includes, for example, $(2/3,-1/3)$, which it wouldn't do if the problem said $0\lt y\lt x\lt1$. For the main question, what do you know about computing marginal densities? How would you go about doing that? – Gerry Myerson Nov 18 '12 at 22:45
Draw a sketch of the region in the $x$-$y$ plane over which $(X,Y)$ is uniformly distributed. If you think of the joint density as a (right triangular) prism sitting on the plane, then $f_X(x)$ is the area of the cross-section of the prism at $x$. (hint: cross-section is a rectangle of height $1$ and base length? that you can read of from the sketch that you probably will not bother to draw.) – Dilip Sarwate Nov 18 '12 at 22:49
@GerryMyerson I see what you mean, thats a good point, thank you for the clairification. I would evaluate this integral: $2\int_0^1 f(x,y) dy$ for the marginal distribution of $X$, and f(x,y)=1 anyway. Which now makes $f_X(x)=2x$ clear. – TheHopefulActuary Nov 18 '12 at 22:51
@Kyle ...and $2\int_0^1f(x,y)dy$ is not what you need to evaluate to find $f_X(x)$, though in this instance, because of the symmetry it gives the right answer. You should evaluate $$f_X(x) = \int_{-\infty}^\infty f(x,y)\,\mathrm dy = \int_{-x}^x 1\,\mathrm dx = 2x ~ \text{for}~0 < x < 1.$$ – Dilip Sarwate Nov 18 '12 at 22:56
@Kyle ...and once again, brief consideration of the sketch that you did not bother to draw will reveal that $f_Y(y)$ is an even function of $Y$ and so $E[Y] = 0$ without it being necessary to integrate, or more sedately, that $f_Y(y)$ is nonzero for $y \in (-1,1)$ and so the integral must have limits $\pm 1$. – Dilip Sarwate Nov 19 '12 at 2:33
In response to the OP's request
As Gerry Myerson pointed out, the absolute value sign does matter, and as Michael Hardy's answer clarifies in more detail, the joint density $f_{X,Y}(x,y)$ has nonzero constant value $c$ on the region $$\{(x,y) \colon -x < y < x, 0 < x < 1\},$$ that is, on the interior of a right triangle with vertices $(0,1), (1,1), (1,-1)$. As I suggested in my comments on the question, sketching the $x$-$y$ plane and marking this triangle on it is very helpful as an aid to thought, and in this particular problem, makes the computations very simple. In fact, it is even better if one can visualize the joint density as a solid sitting on the $x$-$y$ plane whose volume must necessarily equal $1$. In this instance, the solid is a right triangular prism of height $c$, and since the triangular base has area $1$, the height $c$ must also equal $1$.
For any fixed $x$, the marginal density $f_X(x)$ is given by $$f_X(x) = \int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dy$$ which is, of course, the area of the cross-section of the of the joint density solid if we were to slice the solid by a plane parallel to the $y$-$z$ plane and at distance $x$ from the $y$-$z$ plane. For $0 < x < 1$, the cross-section is a rectangle of height $1$ and base extending from $y=-x$ to $y = x$, and so the area is $2x$. For $x\leq 0$ or $x \geq 1$, the cross-section is $0$. Thus we get $$f_X(x) = \begin{cases}2x, &0 < x < 1,\\ 0, &\text{otherwise.}\end{cases}$$ A similar calculation can be done to obtain the marginal density $f_Y(y)$. Now, for $0 \leq y < 1$, the cross-section has base extending from $x = y$ to $1$, and hence the area is $1-y$, while for $-1 < y \leq 0$, the cross-section has base extending from $x = -y$ to $1$, and hence the area is $1+y$. Thus, we have $$f_Y(y) = \begin{cases}1-y, &0 \leq y < 1,\\ 1+y, &-1 < y < 0,\\ 0, &\text{otherwise.}\end{cases}$$ As a check on one's work, it is easy to sketch the density functions and verify that they are nonnegative functions and the "area under the curve" is $1$, that is, we have found valid density functions and thus have not made any glaringly obvious errors in computation.
Finally, to compute $E[Y]$, one can of course use the standard formula $$E[Y] = \int_{-\infty}^\infty y f_Y(y)\,\mathrm dy = \int_{-1}^0 y(1+y)\,\mathrm dy + \int_{0}^1 y(1-y)\,\mathrm dy$$ and work out that $E[Y]=0$, but it is also possible to avoid integration at all by considering that since $f_Y(y)$ is an even function, the integral of the odd function $yf_Y(y)$ over the finite interval $(-1,1)$ must necessarily be $0$. But be sure to remember that this argument must be used with care if the integral is over the entire real line. See, for example, this question and its answers.
-
$$0<|y|<x<1$$ is the same as $$0<x<1\text{ and for each value of x, }-x<y<x.$$ The absolute value is redundant in the inequality $0<|y|$ (we don't care whether it's "$<$" or "$\le$" since the probability of being exactly equal is $0$ either way). But the absolute value matters in the inequality $|y|<x$, and the "$0<$" is there in order to tell us that $0<x$.
As for your question about the marginal density: Let $f_X$ and $F_X$ be respectively the marginal density and the marginal cumulative distribution function. Then $$f_X(x) = \frac{d}{dx} F_X(x) = \frac{d}{dx} \Pr(X\le x) = \frac{d}{dx} \frac{\text{area of one triangle}}{\text{area of another triangle}}.$$ Draw the two triangles and you'll see it.
- | 2014-07-30T13:54:29 | {
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https://math.stackexchange.com/questions/2913408/problem-regarding-linear-congruence | # Problem regarding linear congruence
Which of the following statement is False ?
1) There exists a natural number which when divided by 3 leaves remainder 1 and which when divided by 4 leaves remainder 0
2) There exists a natural number which when divided by 6 leaves remainder 2 and which when divided by 9 leaves remainder 1
3)There exists a natural number which when divided by 7 leaves remainder 1 and which when divided by 11 leaves remainder 3
4)There exists a natural number which when divided by 12 leaves remainder 7 and which when divided by 8 leaves remainder 3
1 and 3 are true because of Chinese remainder theorem . I guess 2 is false (since |6n-9m| is either 3 or 0 ) but i'm not sure how to prove /disprove 2 and 4.
• For 4: 19 and yes you have the right idea for disproving 2. Sep 11 '18 at 17:10
• Any number which is $2$ more than a multiple of six will be $2$ more than a multiple of $3$ (i.e. $6k+2 = 3(2k)+2$). Similarly, any number which is $1$ more than a multiple of $9$ will be $1$ more than a multiple of $3$ (i.e. $9\ell + 1 = 3(3\ell)+1$). Now... is it possible for a number to simultaneously be $1$ more than a multiple of $3$ and $2$ more than a multiple of $3$ at the same time? Sep 11 '18 at 17:15
• @AHusain is there any other way than guessing (trial and error) please let me know how you find that thanks Sep 11 '18 at 17:24
• @JMoravitz i like your proof thanks Sep 11 '18 at 17:25
• The general method is to apply the Chinese reminder theorem, and thus finding equvalent formulations for all conditions that only use prime powers as moduli. Then you check for each prime separetly, starting with the highest prime power. So for problem 2) you find that your number must be $\equiv 1 \pmod 9$ and $\equiv 2 \pmod 3$, which is a contradiction, because $\equiv 1 \pmod 9$ implies $\equiv 1 \pmod 3$. You don't get a contradiction for 4), so this is possible. Of course, finding a value gives rise to a much easier proof, and usually Euclid's algorithm is used for that. Sep 11 '18 at 17:50
1. Call this natural number $x$. Then $x\equiv 1\bmod 3$ and $x\equiv 0\bmod 4$. $3$ and $4$ are coprime so, by the Chinese remainder theorem, we get that $x\equiv 4\bmod 12$.
2. Again, let's call this natural number $x$. Then $x\equiv 2\bmod 6$ and $x\equiv 1\bmod 9$. Here, we cannot turn to the Chinese remainder theorem since $6$ and $9$ are not coprime. Instead, observe that $x\equiv 2\bmod 6$ is equivalent to $x\equiv 2+6m$ for some integer $m$. Now, if we substitute this into our second congruence, we get $2+6m\equiv 1\bmod 9$. This is the same as $6m\equiv 8\bmod 9$. Since there is no inverse for $6$ modulo $9$, we have that no solution exists.
3. Here, we have $x\equiv 1\bmod 7$ and $x\equiv 3\bmod 11$. Since $7$ and $11$ are coprime, by the Chinese remainder theorem, we obtain $x\equiv 36\bmod 77$.
4. We have $x\equiv 7\bmod 12$ and $x\equiv 3\bmod 8$. Applying a similar train of thought that we did for the second problem, we obtain $12m+7\equiv 3\bmod 8$, or equivalently, $12m\equiv -4\bmod 8$ and, even still, $4m\equiv 4\bmod 8$. The solutions to this congruence occur when $m\in\{1,3,5,7\}$ modulo $8$. This is equivalent to writing $m\equiv 8n+1$, $m\equiv 8n+3$, $m\equiv 8n+5$ and $m\equiv 8n+7$. Substituting these into our $12m+7$ expression leads us to conclude with these solutions: $x\equiv 19\bmod 96$, $x\equiv 43\bmod 96$, $x\equiv 67\bmod 96$ and $x\equiv 91\bmod 96$. So, this statement is true.
Therefore, statement 2 is the false one.
• For 4, not an only when. e.g. $m=3$ gives $12 \equiv 4$ Sep 11 '18 at 18:03
• Ah, of course, thanks for pointing it out. Amending as we speak. Sep 11 '18 at 18:06
• @thesmallprint , 12m≡−4mod8 and, even still, 4m≡4mod8 could you please explain this step Sep 13 '18 at 9:28
• Here, we are utilising the fact that $12$ and $4$ are in the same congruence class modulo $8$; the same goes for $-4$ and $4$. Sep 13 '18 at 10:13 | 2021-11-28T05:34:57 | {
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https://casmusings.wordpress.com/2018/01/ | # Monthly Archives: January 2018
## Inscribed Right Angle Proof Without Words
Earlier this past week, I assigned the following problem to my 8th grade Geometry class for homework. They had not explored the relationships between circles and inscribed angles, so I added dashed auxiliary segment AD as a hint.
What follows first is the algebraic solution I expected most to find and then an elegant transformational explanation one of my students produced.
PROOF 1:
Given circle A with diameter BC and point D on the circle. Prove triangle BCD is a right triangle.
After some initial explorations on GeoGebra sliding point D around to discover that its angle measure was always $90^{\circ}$ independent of the location of D, most successful solutions recognized congruent radii AB, AC, and AD, creating isosceles triangles CAD and BAD. That gave congruent base angles x in triangle CAD, and y in BAD.
The interior angle sum of a triangle gave $(x)+(x+y)+(y)=180^{\circ}$, or $m \angle CDB = x+y = 90^{\circ}$, confirming that BCD was a right triangle.
PROOF 2:
Then, one student surprised us. She marked the isosceles base angles as above before rotating $\Delta BCD$ $180^{\circ}$ about point A.
Because the diameter rotated onto itself, the image and pre-image combined to form an quadrilateral with all angles congruent. Because every equiangular quadrilateral is a rectangle, M had confirmed BCD was a right triangle.
CONCLUSION:
I don’t recall seeing M’s proof before, but I found it a delightfully elegant application of quadrilateral properties. In my opinion, her rotation is a beautiful proof without words solution.
Encourage freedom, flexibility of thought, and creativity, and be prepared to be surprised by your students’ discoveries!
## Quadratics + Tangent = ???
Here’s a very pretty problem I encountered on Twitter from Mike Lawler 1.5 months ago.
I’m late to the game replying to Mike’s post, but this problem is the most lovely combination of features of quadratic and trigonometric functions I’ve ever encountered in a single question, so I couldn’t resist. This one is well worth the time for you to explore on your own before reading further.
My full thoughts and explorations follow. I have landed on some nice insights and what I believe is an elegant solution (in Insight #5 below). Leading up to that, I share the chronology of my investigations and thought processes. As always, all feedback is welcome.
WARNING: HINTS AND SOLUTIONS FOLLOW
Investigation #1:
My first thoughts were influenced by spoilers posted as quick replies to Mike’s post. The coefficients of the underlying quadratic, $A^2-9A+1=0$, say that the solutions to the quadratic sum to 9 and multiply to 1. The product of 1 turned out to be critical, but I didn’t see just how central it was until I had explored further. I didn’t immediately recognize the 9 as a red herring.
Basic trig experience (and a response spoiler) suggested the angle values for the tangent embedded in the quadratic weren’t common angles, so I jumped to Desmos first. I knew the graph of the overall given equation would be ugly, so I initially solved the equation by graphing the quadratic, computing arctangents, and adding.
Insight #1: A Curious Sum
The sum of the arctangent solutions was about 1.57…, a decimal form suspiciously suggesting a sum of $\pi/2$. I wasn’t yet worried about all solutions in the required $[0,2\pi ]$ interval, but for whatever strange angles were determined by this equation, their sum was strangely pretty and succinct. If this worked for a seemingly random sum of 9 for the tangent solutions, perhaps it would work for others.
Unfortunately, Desmos is not a CAS, so I turned to GeoGebra for more power.
Investigation #2:
In GeoGebra, I created a sketch to vary the linear coefficient of the quadratic and to dynamically calculate angle sums. My procedure is noted at the end of this post. You can play with my GeoGebra sketch here.
The x-coordinate of point G is the sum of the angles of the first two solutions of the tangent solutions.
Likewise, the x-coordinate of point H is the sum of the angles of all four angles of the tangent solutions required by the problem.
Insight #2: The Angles are Irrelevant
By dragging the slider for the linear coefficient, the parabola’s intercepts changed, but as predicted in Insights #1, the angle sums (x-coordinates of points G & H) remained invariant under all Real values of points A & B. The angle sum of points C & D seemed to be $\pi/2$ (point G), confirming Insight #1, while the angle sum of all four solutions in $[0,2\pi]$ remained $3\pi$ (point H), answering Mike’s question.
The invariance of the angle sums even while varying the underlying individual angles seemed compelling evidence that that this problem was richer than the posed version.
Insight #3: But the Angles are bounded
The parabola didn’t always have Real solutions. In fact, Real x-intercepts (and thereby Real angle solutions) happened iff the discriminant was non-negative: $B^2-4AC=b^2-4*1*1 \ge 0$. In other words, the sum of the first two positive angles solutions for $y=(tan(x))^2-b*tan(x)+1=0$ is $\pi/2$ iff $\left| b \right| \ge 2$, and the sum of the first four solutions is $3\pi$ under the same condition. These results extend to the equalities at the endpoints iff the double solutions there are counted twice in the sums. I am not convinced these facts extend to the complex angles resulting when $-2.
I knew the answer to the now extended problem, but I didn’t know why. Even so, these solutions and the problem’s request for a SUM of angles provided the insights needed to understand WHY this worked; it was time to fully consider the product of the angles.
Insight #4: Finally a proof
It was now clear that for $\left| b \right| \ge 2$ there were two Quadrant I angles whose tangents were equal to the x-intercepts of the quadratic. If $x_1$ and $x_2$ are the quadratic zeros, then I needed to find the sum A+B where $tan(A)=x_1$ and $tan(B)=x_2$.
From the coefficients of the given quadratic, I knew $x_1+x_2=tan(A)+tan(B)=9$ and $x_1*x_2=tan(A)*tan(B)=1$.
Employing the tangent sum identity gave
$\displaystyle tan(A+B) = \frac{tan(A)+tan(B)}{1-tan(A)tan(B)} = \frac{9}{1-1}$
and this fraction is undefined, independent of the value of $x_1+x_2=tan(A)+tan(B)$ as suggested by Insight #2. Because tan(A+B) is first undefined at $\pi/2$, the first solutions are $\displaystyle A+B=\frac{\pi}{2}$.
Insight #5: Cofunctions reveal essence
The tangent identity was a cute touch, but I wanted something deeper, not just an interpretation of an algebraic result. (I know this is uncharacteristic for my typically algebraic tendencies.) The final key was in the implications of $tan(A)*tan(B)=1$.
This product meant the tangent solutions were reciprocals, and the reciprocal of tangent is cotangent, giving
$\displaystyle tan(A) = \frac{1}{tan(B)} = cot(B)$.
But cotangent is also the co-function–or complement function–of tangent which gave me
$tan(A) = cot(B) = tan \left( \frac{\pi}{2} - B \right)$.
Because tangent is monotonic over every cycle, the equivalence of the tangents implied the equivalence of their angles, so $A = \frac{\pi}{2} - B$, or $A+B = \frac{\pi}{2}$. Using the Insights above, this means the sum of the solutions to the generalization of Mike’s given equation,
$(tan(x))^2+b*tan(x)+1=0$ for x in $[0,2\pi ]$ and any $\left| b \right| \ge 2$,
is always $3\pi$ with the fundamental reason for this in the definition of trigonometric functions and their co-functions. QED
Insight #6: Generalizing the Domain
The posed problem can be generalized further by recognizing the period of tangent: $\pi$. That means the distance between successive corresponding solutions to the internal tangents of this problem is always $\pi$ each, as shown in the GeoGebra construction above.
Insights 4 & 5 proved the sum of the angles at points C & D was $\pi/2$. Employing the periodicity of tangent, the x-coordinate of $E = C+\pi$ and $F = D+\pi$, so the sum of the angles at points E & F is $\frac{\pi}{2} + 2 \pi$.
Extending the problem domain to $[0,3\pi ]$ would add $\frac{\pi}{2} + 4\pi$ more to the solution, and a domain of $[0,4\pi ]$ would add an additional $\frac{\pi}{2} + 6\pi$. Pushing the domain to $[0,k\pi ]$ would give total sum
$\displaystyle \left( \frac{\pi}{2} \right) + \left( \frac{\pi}{2} +2\pi \right) + \left( \frac{\pi}{2} +4\pi \right) + \left( \frac{\pi}{2} +6\pi \right) + ... + \left( \frac{\pi}{2} +2(k-1)\pi \right)$
Combining terms gives a general formula for the sum of solutions for a problem domain of $[0,k\pi ]$
$\displaystyle k * \frac{\pi}{2} + \left( 2+4+6+...+2(k-1) \right) * \pi =$
$\displaystyle = k * \frac{\pi}{2} + (k)(k-1) \pi =$
$\displaystyle = \frac{\pi}{2} * k * (2k-1)$
For the first solutions in Quadrant I, $[0,\pi]$ means k=1, and the sum is $\displaystyle \frac{\pi}{2}*1*(2*1-1) = \frac{\pi}{2}$.
For the solutions in the problem Mike originally posed, $[0,2\pi]$ means k=2, and the sum is $\displaystyle \frac{\pi}{2}*2*(2*2-1) = 3\pi$.
I think that’s enough for one problem.
APPENDIX
My GeoGebra procedure for Investigation #2:
• Graph the quadratic with a slider for the linear coefficient, $y=x^2-b*x+1$.
• Label the x-intercepts A & B.
• The x-values of A & B are the outputs for tangent, so I reflected these over y=x to the y-axis to construct A’ and B’.
• Graph y=tan(x) and construct perpendiculars at A’ and B’ to determine the points of intersection with tangent–Points C, D, E, and F in the image below
• The x-intercepts of C, D, E, and F are the angles required by the problem.
• Since these can be points or vectors in Geogebra, I created point G by G=C+D. The x-intercept of G is the angle sum of C & D.
• Likewise, the x-intercept of point H=C+D+E+F is the required angle sum. | 2018-09-18T14:37:21 | {
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https://math.stackexchange.com/questions/775102/exponential-random-variable-representation-of-criminal-trial | # Exponential Random Variable representation of criminal trial
Assume the amount of evidence against a defendant in a criminal trial is an exponential random variable $X$. If the defendant is innocent, then $X$ has mean $1$, and if the defendant is guilty, then $X$ has mean $2$. The defendant will be ruled guilty if $X>c$, where $c$ is a suitably chosen constant.
If the judge wants to be $95\%$ certain that an innocent man will not be convicted, what should the value of $c$ be? For this $c$ value, what is the probability that a guilty defendant will be convicted? Assume before the trial begins, you believe the defendant to be guilty with probability $10\%$. If the defendant is convicted, what is your updated belief about the probability of their guilt?
For this question, I am not sure as to where to begin for finding the initial $c$. I know that $\lambda_{\text{innocent}}=1$ and $\lambda_{\text{guilty}}=0.5$. my thought was to compute $0.95=\int(\lambda e^{-\lambda x})dx$ from $0$ to $c$ and solving for $c$ in both the guilty and innocent cases. I obtained negative values for both, which I assume are wrong. Am I on the right track or is there another approach I am not seeing?
After $c$ is known, finding the probability the guilty defendant will be convicted should be the straight forward integral for an exponential distribution from $0$ to $c$?
If the defendant is convicted, would be probability of their guilty be increased to $100\%$? or $0.1\times0.95$?
Thank you in advance for any help in understanding the problem!
• $0.95 \leq \int_0^c \lambda e^{-\lambda x}\mathrm{x} \implies 0.95 \leq 1- e^{-\lambda c} \implies c \geq \frac{\ln(20)}{\lambda}$ – Graham Kemp Apr 30 '14 at 2:30
• I do not see how this is correct.. where is the 20 coming from? I obtained the same integral, however when solving for c I got (-lambda*ln(0.05)) which equaled 1.497866137 for the guilty defendant which is incorrect. – user140624 Apr 30 '14 at 2:46
• The negative of a log is the log of the reciprocal. $-\ln(0.05) = \ln(0.05^{-1}) = \ln(20)$ – Graham Kemp Apr 30 '14 at 2:48
• ohh gotcha thanks! So I found c = ln(20) and the probability that a guilty defendant is convicted to be 0.223606. I am unsure about how to update the original believed guilty probability of 10%. Any ideas? – user140624 Apr 30 '14 at 3:00
• Calculate the probabilities of conviction conditional on guilt and innocence, and update your belief given the conviction using Bayes theorem: $B(G\mid F) = \frac{B(G) P(F\mid G)}{P(F)}$ where $B(G)=0.10$ is your prior belief and $B(G\mid F)$ is the posterior belief. – Graham Kemp Apr 30 '14 at 3:48
Let $F$ represent the event: 'is found guilty' (also, 'is convicted').
Let $G$ represent the event: 'is guilty'. So $\lambda_G = 0.5, \lambda_{\neg G}=1$
We want $P(\neg F \mid \neg G) \geq 0.95$
Given exponentially distributed evidence $X$, the probability of being found innocent, determined by the cutoff $c$ and the lambda value $\lambda = E(X)^{-1}$ is: $$P(X \leq c) = \int_0^c \lambda e^{-\lambda x} \mathrm{d} x \\ = 1-e^{-\lambda c}$$
So we want a value of $c$ such that: $P(X \leq c \mid \lambda=1) \geq 0.95$.
$$0.95 \leq 1-e^{-c} \implies c \geq \ln(20)$$
Hence:
$P(F \mid G) = P(X>\ln 20 \mid \lambda = 0.5) = 20^{-0.5} = \frac 1 {\sqrt{20}} \approx 0.2236\dotsc$
$P(F \mid \neg G) = P(X>\ln 20 \mid \lambda = 1) = 20^{-1} = \frac 1 {20} = 0.05$
Next, we wish you update our belief given conviction.
Let $B(G)$ be your prior belief that the defendant is guilty and $B(G\mid F)$ be your posterior belief of guilt given the conviction. Basically, your belief is an estimation of the probability of guilt.
By Bayes theorem: $$B(G\mid F) = \frac{B(G)\cdot P(F\mid G)}{B(G)\cdot P(F\mid G) + (1-B(G))\cdot P(F\mid \neg G)} \\ = \frac{\frac{1}{10}\times \frac{1}{\sqrt{20}}}{\frac{1}{10}\times \frac{1}{\sqrt{20}}+\frac{9}{10}\times\frac{1}{20}} \\ = \frac{20}{20+9\sqrt{20}} \\ \approx 0.33195\dotsc$$ | 2019-09-23T18:45:56 | {
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http://www.mathworks.com/help/optim/ug/nonlinear-equations-with-analytic-jacobian.html?requestedDomain=www.mathworks.com&nocookie=true | # Documentation
### This is machine translation
Translated by
Mouse over text to see original. Click the button below to return to the English verison of the page.
## Nonlinear Equations with Analytic Jacobian
This example demonstrates the use of the default trust-region-dogleg fsolve algorithm (see Large-Scale vs. Medium-Scale Algorithms). It is intended for problems where
• The system of nonlinear equations is square, i.e., the number of equations equals the number of unknowns.
• There exists a solution x such that F(x) = 0.
The example uses fsolve to obtain the minimum of the banana (or Rosenbrock) function by deriving and then solving an equivalent system of nonlinear equations. The Rosenbrock function, which has a minimum of F(x) = 0, is a common test problem in optimization. It has a high degree of nonlinearity and converges extremely slowly if you try to use steepest descent type methods. It is given by
$f\left(x\right)=100{\left({x}_{2}-{x}_{1}^{2}\right)}^{2}+{\left(1-{x}_{1}\right)}^{2}.$
First generalize this function to an n-dimensional function, for any positive, even value of n:
$f\left(x\right)=\sum _{i=1}^{n/2}100{\left({x}_{2i}-{x}_{2i-1}^{2}\right)}^{2}+{\left(1-{x}_{2i-1}\right)}^{2}.$
This function is referred to as the generalized Rosenbrock function. It consists of n squared terms involving n unknowns.
Before you can use fsolve to find the values of x such that F(x) = 0, i.e., obtain the minimum of the generalized Rosenbrock function, you must rewrite the function as the following equivalent system of nonlinear equations:
$\begin{array}{c}F\left(1\right)=1-{x}_{1}\\ F\left(2\right)=10\left({x}_{2}-{x}_{1}^{2}\right)\\ F\left(3\right)=1-{x}_{3}\\ F\left(4\right)=10\left({x}_{4}-{x}_{3}^{2}\right)\\ ⋮\\ F\left(n-1\right)=1-{x}_{n-1}\\ F\left(n\right)=10\left({x}_{n}-{x}_{n-1}^{2}\right).\end{array}$
This system is square, and you can use fsolve to solve it. As the example demonstrates, this system has a unique solution given by xi = 1, i = 1,...,n.
### Step 1: Write a file bananaobj.m to compute the objective function values and the Jacobian.
function [F,J] = bananaobj(x)
% Evaluate the vector function and the Jacobian matrix for
% the system of nonlinear equations derived from the general
% n-dimensional Rosenbrock function.
% Get the problem size
n = length(x);
if n == 0, error('Input vector, x, is empty.'); end
if mod(n,2) ~= 0,
error('Input vector, x ,must have an even number of components.');
end
% Evaluate the vector function
odds = 1:2:n;
evens = 2:2:n;
F = zeros(n,1);
F(odds,1) = 1-x(odds);
F(evens,1) = 10.*(x(evens)-x(odds).^2);
% Evaluate the Jacobian matrix if nargout > 1
if nargout > 1
c = -ones(n/2,1); C = sparse(odds,odds,c,n,n);
d = 10*ones(n/2,1); D = sparse(evens,evens,d,n,n);
e = -20.*x(odds); E = sparse(evens,odds,e,n,n);
J = C + D + E;
end
### Step 2: Call the solve routine for the system of equations.
n = 64;
x0(1:n,1) = -1.9;
x0(2:2:n,1) = 2;
[x,F,exitflag,output,JAC] = fsolve(@bananaobj,x0,options);
Use the starting point x(i) = –1.9 for the odd indices, and x(i) = 2 for the even indices. Set Display to 'iter' to see the solver's progress. Set SpecifyObjectiveGradient to true to use the Jacobian defined in bananaobj.m. The fsolve function generates the following output:
Norm of First-order Trust-region
Iteration Func-count f(x) step optimality radius
0 1 8563.84 615 1
1 2 3093.71 1 329 1
2 3 225.104 2.5 34.8 2.5
3 4 212.48 6.25 34.1 6.25
4 5 212.48 6.25 34.1 6.25
5 6 102.771 1.5625 6.39 1.56
6 7 102.771 3.90625 6.39 3.91
7 8 87.7443 0.976563 2.19 0.977
8 9 74.1426 2.44141 6.27 2.44
9 10 74.1426 2.44141 6.27 2.44
10 11 52.497 0.610352 1.52 0.61
11 12 41.3297 1.52588 4.63 1.53
12 13 34.5115 1.52588 6.97 1.53
13 14 16.9716 1.52588 4.69 1.53
14 15 8.16797 1.52588 3.77 1.53
15 16 3.55178 1.52588 3.56 1.53
16 17 1.38476 1.52588 3.31 1.53
17 18 0.219553 1.16206 1.66 1.53
18 19 0 0.0468565 0 1.53
Equation solved.
fsolve completed because the vector of function values is near zero
as measured by the default value of the function tolerance, and
the problem appears regular as measured by the gradient.
Watch now | 2016-09-26T06:12:30 | {
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https://math.stackexchange.com/questions/4038729/why-does-the-alternatization-of-a-k-linear-map-indeed-produce-an-alternating | # Why does the alternatization of a $k$-linear map indeed produce an alternating $k$-linear map?
The alternating operator $$A$$ produces for any $$k$$-linear map $$f$$ an alternating $$k$$-linear map $$Af$$ (the alternatization of $$f$$):
$$Af(v_1, \ldots, v_k) = \sum_{\sigma \in S_k} \text{sgn}(\sigma)f\left(v_{\sigma(1)}, \ldots, v_{\sigma(k)}\right)$$
I've tried to prove that the definition of an alternating $$k$$-linear map holds (and most of all, to find an intuitive explanation for this!), but haven't made much progress...
If I think about the definition where having two equal arguments must make the map evaluate to $$0$$, then having two equal arguments means transposing them doesn't change anything, so two terms cancel out in the sum, but what about the rest?
### Proof idea for alternative definition
Not what I was looking for, but I figured out a proof outline for an alternative definition of an alternating map. An alternating map is also defined as a map where transposing two arguments changes the sign of the map (e.g. $$f(y, x, z) = - f(x, y, z)$$).
Composing any permutation with a transposition changes its sign, so if the sum gets decomposed into a sum of all even permutations and all odd permutations, all even permutations will get mapped to odd permutations and vice-versa, so the two sums change signs. As a consequence, the entire alternatized function also changes sign.
Fix $$\pi \in S_k$$.
\begin{align*} Af(v_{\pi(1)},\dots,v_{\pi(k)})&= \sum_{\sigma\in S_k}\operatorname{sgn}(\sigma)f(v_{\sigma(\pi(1))},\dots,v_{\sigma(\pi(n))}) \\ &=\operatorname{sgn}(\pi)\sum_{\sigma\in S_k}\operatorname{sgn}(\pi)\operatorname{sgn}(\sigma)f(v_{\sigma(\pi(1))},\dots,v_{\sigma(\pi(n))}) \\ &=\operatorname{sgn}(\pi)\sum_{\sigma\in S_k}\operatorname{sgn}(\sigma\pi)f(v_{\sigma\pi(1))},\dots,v_{\sigma\pi(n))}) \\ &=\operatorname{sgn}(\pi)Af(v_1,\dots,v_k) \end{align*} where the last equality holds since right multiplication $$\sigma\mapsto\sigma\pi$$ yields a bijection $$S_k\to S_k$$.
• This is what I was looking for! Unfortunately, although I'm sure it's obvious, I'm having trouble understanding how you get from the expression after the first equal sign to that after the second equal sign, where $\text{sgn}(\pi)$ appears twice. I understood the rest. Feb 24, 2021 at 22:58
• $\operatorname{sgn}(\pi) = \pm 1$, so $\operatorname{sgn}(\pi)^2 = 1$.
– Nico
Feb 24, 2021 at 23:01
• Can't believe I missed that geez... Thanks! Feb 24, 2021 at 23:03
Your first definition implies your second definition, but your second does not imply your first for rings in which $$1 + 1 = 0$$. Moreover, some authors such as Hoffman and Kunze in their text Linear Algebra use your first definition. It therefore seems best to use your first definition.
You found two terms that cancel in the sum. Extend that idea: Pair the permutations into those that have the two equal arguments in the same two positions. The sum of each pair of corresponding terms is zero. Thus, the overall sum is zero which proves the claim. If that reasoning is not obvious, here it is with symbols:
We want to show that $$A f(v_1,\dotsc,v_r) = 0$$ whenever $$v_i =v_j$$ with $$i\ne j$$. The number of terms (permutations) in the sum $$Af =\sum_\sigma (\operatorname{sgn}\sigma)f_\sigma$$ is even because there are $$r!$$ terms, and the definition of alternating implicitly assumes $$r\ge 2$$. Pair each permutation $$\sigma$$ with the permutation $$\sigma\circ(ij)$$ by which we denote the composition of the permutation $$(ij)$$ that interchanges $$v_i,v_j$$ and the permutation $$\sigma$$. That pairing is well-defined because $$\sigma\circ(ij)$$ pairs with $$(\sigma\circ(ij))\circ(ij) =\sigma$$. Clearly, we can think of the index of summation in the definition of $$A$$ equivalently as running over all such pairs. The following shows that the pair of terms in that definition associated with the pair $$\sigma$$, $$\sigma\circ(ij)$$ add up to zero. We assume without loss of generalization that $$i < j$$.\begin{align*} & (\operatorname{sgn}(\sigma\circ(ij)))f(v_{(\sigma\circ(ij))1},\dotsc,v_{(\sigma\circ(ij))i},\dotsc,v_{(\sigma\circ(ij))j},\dotsc,v_{(\sigma\circ(ij))r})\\ &\quad = (\operatorname{sgn}\sigma)(\operatorname{sgn}(ij))f(v_{\sigma1},\dotsc,v_{(\sigma\circ(ij))j},\dotsc,v_{(\sigma\circ(ij))i},\dotsc,v_{\sigma r}) && (\sigma\circ(ij))k =\sigma k\text{ if }k\ne i,j;\ v_i = v_j\\ &\quad = -(\operatorname{sgn}\sigma)f(v_{\sigma1},\dotsc,v_{\sigma i},\dotsc,v_{\sigma j},\dotsc,v_{\sigma r}).\end{align*} Thus, because the sum of each such pair of terms is zero, $$Af(v_1,\dotsc,v_r) = 0$$. | 2022-10-01T16:12:17 | {
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http://www.proofwiki.org/wiki/Definition:Coordinate | # Definition:Coordinate
## Definition
### Elements of Ordered Pair
Let $\left({a, b}\right)$ be an ordered pair.
The following terminology is used:
• $a$ is called the first coordinate
• $b$ is called the second coordinate.
This definition is compatible with the equivalent definition in the context of Cartesian coordinates.
Some authors use the terms first component and second component instead.
### Coordinate System
Let $\left \langle {a_k} \right \rangle_{1 \mathop \le k \mathop \le n}$ be an ordered basis of a unitary $R$-module $G$.
Then $\left \langle {a_k} \right \rangle_{1 \mathop \le k \mathop \le n}$ can be referred to as a coordinate system.
### Coordinate
Let $\left \langle {a_n} \right \rangle$ be a coordinate system of a unitary $R$-module $G$.
Let $\displaystyle x \in G: x = \sum_{k \mathop = 1}^n \lambda_k a_k$.
The scalars $\lambda_1, \lambda_2, \ldots, \lambda_n$ can be referred to as the coordinates of $x$ relative to $\left \langle {a_n} \right \rangle$.
### Coordinate Function
Let $\left \langle {a_n} \right \rangle$ be a coordinate system of a unitary $R$-module $G$.
For each $x \in G$ let $x_1, x_2, \ldots, x_n$ be the coordinates of $x$ relative to $\left \langle {a_n} \right \rangle$.
Then for $i = 1, \ldots, n$ the mapping $f_i : G \to R$ defined by $f_i \left({x}\right) = x_i$ is called the $i$-th coordinate function on $G$ relative to $\left \langle {a_n} \right \rangle$.
### Coordinates on Affine Space
Let $\mathcal E$ be an affine space of dimension $n$ over a field $k$.
Let $\mathcal R = \left(p_0,e_1,\ldots,e_n\right)$ be an affine frame in $\mathcal E$.
Let $p \in \mathcal E$ be a point.
By Affine Coordinates Well Defined there exists a unique ordered tuple $\left(\lambda_1,\ldots,\lambda_n\right) \in k^n$ such that
$\displaystyle p = p_0 + \sum_{i = 1}^n \lambda_i e_i$
The numbers $\lambda_1,\ldots,\lambda_n$ are the coordinates of $p$ in the frame $\mathcal R$.
### Origin
The origin of a coordinate system is the zero vector.
In the $xy$-plane, it is the point:
$O = \left({0, 0}\right)$
and in general, in the Euclidean space $\R^n$:
$O = \underbrace{\left({0, 0, \ldots, 0}\right)}_{n \ \text{coordinates}}$
## Linguistic Note
It's an awkward word coordinate. It really needs a hyphen in it to emphasise its pronounciation (loosely and commonly: coe-wordinate), and indeed, some authors spell it co-ordinate. However, this makes it look unwieldy.
An older spelling puts a diaeresis indication symbol on the second "o": coördinate. But this is considered archaic nowadays and few sources still use it. | 2013-05-25T18:22:28 | {
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https://math.stackexchange.com/questions/2896112/last-digit-of-sequence-of-numbers | # Last digit of sequence of numbers
We define the sequence of natural numbers $$a_1 = 3 \quad \text{and} \quad a_{n+1}=a_n^{a_n}, \quad \text{ for n \geq 1}.$$
I want to show that the last digit of the numbers of the sequence $a_n$ alternates between the numbers $3$ and $7$. Specifically, if we symbolize with $b_n$ the last digit of $a_n$, I want to show that $$b_n = \begin{cases} 3, & \text{if n is odd}, \\ 7, & \text{if n is even}. \end{cases}$$
There is a hint to prove that for each $n \in \mathbb{N}$, if $a_n \equiv 3 \pmod{5}$ then $a_{n+1} \equiv 2 \pmod{5}$ and if $a_n \equiv 2 \pmod{5}$ then $a_{n+1} \equiv 3 \pmod{5}$.
First of all, if we take $a_n \equiv 3 \pmod{5}$, then $a_{n+1}=3^3\pmod{5} \equiv 2 \pmod{5}$.
If $a_n \equiv 2 \pmod{5}$, then $a_{n+1}=2^2 \pmod{5}=4$. Or am I doing something wrong?
And also how does it follow, if we have shown the hint, that $b_n$ is $3$ when $n$ is odd, and $7$ if $n$ is even?
• You can't just do modulo the power like that. Use Fermat's little theorem Aug 27, 2018 at 12:05
• How else do we calculate it? @Rumpelstiltskin Aug 27, 2018 at 12:07
• If $a_n \equiv 2\textrm{ (mod 5)}$ you have $a_n = 5k+2$ for any whole number $k$. Now I would try to calculate $(5k+2)^{5k+2}$ by using the binomial formula. Keep in mind that you calculate mod 5, so multiples of 5 vanish. This should make it easier. Aug 27, 2018 at 12:17
It follows directly from the hint:
The last digit of $a_n$ is the residue class of $a_n$ mod 10.
Now if you have $a_n \equiv 3 \textrm{ (mod 5)}$ it follows $a_n \equiv 3 \textrm{ (mod 10)}$ or $a_n \equiv 8 \textrm{ (mod 10)}$. But $a_n \equiv 8 \textrm{ (mod 10)}$ would mean that $a_n$ is even, so $2$ comes in the prime factorization of $a_n$, but the only prime dividing $a_n$ is 3, so this is not possible.
In the same way, from $a_n\equiv 2\textrm{ (mod 5)}$ you get $a_n \equiv 7\textrm{ (mod 10)}$.
• I have a question. We have that $a_1=3$. The last digit of $a_2$ is $3^3 \pmod{10}=7$. The last digit of $a_3$ is $7^7 \pmod{10}=3$. The last digit of $a_4$ is $3^3 \pmod{10}=7$, and so on. Why doesn't this suffice? Aug 27, 2018 at 18:23
• I guess you hang on the mistake that Stefan4024 explains quite detailed. Let me try to express it in a slightly different way: Let $a \textrm{ mod } b$ denote the residue class of $a$ modulo $b$, for example $23\textrm{ mod }10=3$. Then what you probably want to argue is $a_{n+1}\textrm{ mod 10 } = (a_n \textrm{ mod }10)^{a_n \textrm{ mod }10}\textrm{ mod }10$. But this is wrong.In your original question you struggled on proving the hint since you did this mistake (with mod 5 instead of mod 10). PS: Please upvote useful answers and mark your favourite answer. Aug 27, 2018 at 19:36
• Ok, but how do we get from the result you wrote at your post the desired conclusion? @S.M.Roch Sep 14, 2018 at 22:31
The mistake you are making is that if $a_n \equiv 2 \pmod 5$ it's not true that $a_n^{a_n} \equiv 2^2 \pmod 5$. The reason behind this is that the exponents aren't repeating in blocks of $5$, but instead in blocks of $\phi(5) = 4$, in your case. Indeed by Fermat's Little Theorem we have that $a_n^4 \equiv 1 \pmod 5$. Thus you need to find $a_n \pmod 4$ first.
This isn't hard to do, as $a_1 = 3$. Thus whenever it's raised to an odd power we get that $a_n \equiv -1 \pmod 4$. Hence we have:
$$a_n \equiv a_{n-1}^{a_{n-1}} \equiv a_{n-1}^{-1} \pmod 5$$
Now use the fact that $2$ is the modular inverse of $3$ modulo $5$ to conclude that:
$$a_n \equiv \begin{cases} 3 \pmod 5, & \text{if n is odd} \\ 2 \pmod 5, & \text{if n is even} \end{cases}$$
Finally note that $a_n \equiv 1 \pmod 2$ and use Chinese remainder Theorem to conclude that:
$$a_n \equiv \begin{cases} 3 \pmod{10}, & \text{if n is odd} \\ 7 \pmod{10}, & \text{if n is even} \end{cases}$$
• Could you explain to me how we use the fact that $a_n \equiv -1 \pmod{4}$ if we have an odd power? Sep 14, 2018 at 22:33
• @Evinda For example we have that $a_2 \equiv a_1^{a_1} \equiv (-1)^{a_1} \pmod 4$. Now as $a_1$ is odd we get that $a_2 \equiv -1 \pmod 4$. We can inductively this holds for all $n$. Then as the $a_n^4 \equiv 1 \pmod 5$ we get that: $$a^{n+1} \equiv a_n^{a_n} \equiv a_n^{4k - 1} \equiv a_n^{-1} \pmod 5$$ Sep 15, 2018 at 3:15
First, you should prove that $a_n$ is odd for all $n$ : this follows from the fact that $a_1$ is odd, and if $a_n$ is odd, then $a_{n+1}=a_n^{a_n}$ is an odd number multiplied with itself some number of times, so it is odd.
Let us notice something more. First, $a_1 = 3 \equiv 3 \mod 4$. Next, if $a_n \equiv 3 \mod 4$, then $a_{n+1} \equiv 3^{a_n} \equiv 3^1 \mod 4$ (as $3^2 = 9 \equiv 1 \mod 4$, we may take remainder when $a_n$ is divided by $2$, and this is $1$ since $a_n$ is odd).
Thus, $a_n \equiv 3 \mod 4$ for all $n$.
To the main part.
If $a_n \equiv 3 \mod 5$, then $a_{n+1} = a_n^{a_n} \equiv 3^{a_n} \mod 5$ is all you can say using the fact that $a_n \equiv 3 \mod 5$. But note that $3^{4} \equiv 1 \mod 5$, so you can now say that $a_{n+1} \equiv 3^{(a_n \mod 4)} \mod 5$.
But the point is, that $a_n$ is odd, so $a_n \mod 4$ is either $1$ or $3$. Consequently, $a_{n+1} \equiv 3^1 = 3 \mod 5$ or $a_{n+1} \equiv 3^3 = 27 \mod 5$ must happen. We appear somewhat stuck here. Let me relieve you : we are doing fine.
With the above observation combined with the second part, we have $a_{n+1} \equiv 27 = 2 \mod 5$.
Now, if $a_n \equiv 2 \mod 5$, then $a_{n+1} \equiv 2^{a_n} \mod 5$, and again we equate this to $2^{a_n \mod 4} \mod 5 = 2^3 =8 \equiv 3 \mod 5$.
Thus, $a_{n+1} \equiv 3 \mod 5$.
So the remainders of $a_n$ and $a_{n+1}$ alternate modulo $5$. Since all $a_n$ are odd, this forces their last digits to alternate, and from knowing that the last digit of $a_1$ is $3$, the sequence of last digits must go $373737373...$
• I haven't understood how we get that $a_{n+1}=3^{(a_n \bmod{4})} \bmod{5}$... Could you explain it further to me? Sep 14, 2018 at 22:35
• Yes. See, $a_n \mod 4$ is the unique integer $0 \leq r < 3$ such that $a_n \equiv r \mod 4$. Therefore, $a_n = r + 4k$ for some integer $k$. Consequently, $3^{a_n} = 3^{4k}3^r$. Since $3^{4k} =(9)^{2k}$ leaves a remainder of $1$ modulo $4$, we conclude that $3^{a_n} \equiv 3^r \mod 4$, and now remembering what $r$ was gives the result. Sep 15, 2018 at 0:54 | 2023-04-02T05:50:04 | {
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"url": "https://math.stackexchange.com/questions/2896112/last-digit-of-sequence-of-numbers",
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https://www.cs.bu.edu/fac/crovella/cs132-book/L08MatrixofLinearTranformation.html | # for QR codes use inline
%matplotlib inline
qr_setting = 'url'
#
# for lecture use notebook
# %matplotlib notebook
# qr_setting = None
#
%config InlineBackend.figure_format='retina'
# import libraries
import numpy as np
import matplotlib as mp
import pandas as pd
import matplotlib.pyplot as plt
import laUtilities as ut
import slideUtilities as sl
import demoUtilities as dm
import pandas as pd
from datetime import datetime
from IPython.display import Image
from IPython.display import display_html
from IPython.display import display
from IPython.display import Math
from IPython.display import Latex
from IPython.display import HTML;
# The Matrix of a Linear Transformation¶
# image credit: https://imgflip.com/memetemplate/The-Most-Interesting-Man-In-The-World
display(Image("images/Dos-Equis-Linear-Transform.jpg", width=350))
In the last lecture we introduced the idea of a linear transformation:
# image credit: Lay, 4th edition
display(Image("images/L7 F4.jpg", width=650))
We have seen that every matrix multiplication is a linear transformation from vectors to vectors.
But, are there any other possible linear transformations from vectors to vectors?
No.
In other words, the reverse statement is also true:
every linear transformation from vectors to vectors is a matrix multiplication.
We’ll now prove this fact. We’ll do it constructively, meaning we’ll actually show how to find the matrix corresponding to any given linear transformation $$T$$.
Theorem. Let $$T: \mathbb{R}^n \rightarrow \mathbb{R}^m$$ be a linear transformation. Then there is (always) a unique matrix $$A$$ such that:
$T({\bf x}) = A{\bf x} \;\;\; \mbox{for all}\; {\bf x} \in \mathbb{R}^n.$
In fact, $$A$$ is the $$m \times n$$ matrix whose $$j$$th column is the vector $$T({\bf e_j})$$, where $${\bf e_j}$$ is the $$j$$th column of the identity matrix in $$\mathbb{R}^n$$:
$A = \left[T({\bf e_1}) \dots T({\bf e_n})\right].$
$$A$$ is called the standard matrix of $$T$$.
Proof. Write
${\bf x} = I{\bf x} = \left[{\bf e_1} \dots {\bf e_n}\right]\bf x$
$= x_1{\bf e_1} + \dots + x_n{\bf e_n}.$
Because $$T$$ is linear, we have:
$T({\bf x}) = T(x_1{\bf e_1} + \dots + x_n{\bf e_n})$
$= x_1T({\bf e_1}) + \dots + x_nT({\bf e_n})$
$\begin{split} = \left[T({\bf e_1}) \dots T({\bf e_n})\right] \, \left[\begin{array}{r}x_1\\\vdots\\x_n\end{array}\right] = A{\bf x}.\end{split}$
So … we see that the ideas of matrix multiplication and linear transformation are essentially equivalent.
However, term linear transformation focuses on a property of the mapping, while the term matrix multiplication focuses on how such a mapping is implemented.
This proof shows us an important idea:
To find the standard matrix of a linear tranformation, ask what the transformation does to the columns of $$I$$.
Now, in $$\mathbb{R}^2$$, $$I = \left[\begin{array}{cc}1&0\\0&1\end{array}\right]$$. So:
$\begin{split}\mathbf{e_1} = \left[\begin{array}{c}1\\0\end{array}\right]\;\;\mbox{and}\;\;\mathbf{e_2} = \left[\begin{array}{c}0\\1\end{array}\right].\end{split}$
So to find the matrix of any given linear transformation of vectors in $$\mathbb{R}^2$$, we only have to know what that transformation does to these two points:
ax = dm.plotSetup(-3,3,-3,3,size=(6,6))
ax.plot([0],[1],'ro',markersize=8)
ax.text(0.25,1,'(0,1)',size=20)
ax.plot([1],[0],'ro',markersize=8)
ax.text(1.25,0.25,'(1,0)',size=20);
This is a hugely powerful tool.
Let’s say we start from some given linear transformation; we can use this idea to find the matrix that implements that linear transformation.
For example, let’s consider rotation about the origin as a kind of transformation.
u = np.array([1.5, 0.75])
v = np.array([0.25, 1])
diamond = np.array([[0,0], u, u+v, v]).T
ax = dm.plotSetup()
plt.plot(u[0], u[1], 'go')
plt.plot(v[0], v[1], 'yo')
ax.text(u[0]+.25,u[1],r'$\bf{u}$',size=20)
ax.text(v[0],v[1]-.35,r'$\bf{v}$',size=20)
rotation = np.array([[0, -1],[1, 0]])
up = rotation @ u
vp = rotation @ v
plt.plot(up[0], up[1], 'go')
plt.plot(vp[0], vp[1], 'yo')
ax.text(up[0]-1,up[1],r'$T(\mathbf{u})$',size=20)
ax.text(vp[0]-.9,vp[1]+.15,r'$T(\mathbf{v})$',size=20)
ax.text(0.75, 1.75, r'$\theta$', size = 20)
ax.text(-.5, 0.25, r'$\theta$', size = 20)
ax.annotate("",
xy=((up)[0]+.1, (up)[1]+.1), xycoords='data',
xytext=((u)[0]-.1, (u)[1]+.1), textcoords='data',
size=20, va="center", ha="center",
arrowprops=dict(arrowstyle="simple",
)
ax.annotate("",
xy=((vp)[0]+.1, (vp)[1]+.1), xycoords='data',
xytext=((v)[0]-.1, (v)[1]), textcoords='data',
size=20, va="center", ha="center",
arrowprops=dict(arrowstyle="simple",
);
First things first: Is rotation a linear transformation?
Recall that a for a transformation to be linear, it must be true that $$T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}).$$
I’m going to show you a “geometric proof.”
This figure shows that “the rotation of $$\mathbf{u+v}$$ is the sum of the rotation of $$\mathbf{u}$$ and the rotation of $$\mathbf{v}$$”.
u = np.array([1.5, 0.75])
v = np.array([0.25, 1])
diamond = np.array([[0,0], u, u+v, v]).T
ax = dm.plotSetup()
dm.plotSquare(diamond)
ax.text(u[0]+.25,u[1],r'$\bf{u}$',size=20)
ax.text(v[0],v[1]+.25,r'$\bf{v}$',size=20)
ax.text(u[0]+v[0]+.25,u[1]+v[1],r'$\bf{u + v}$',size=20)
rotation = np.array([[0, -1],[1, 0]])
up = rotation @ u
vp = rotation @ v
diamond = np.array([[0,0], up, up+vp, vp]).T
dm.plotSquare(diamond)
ax.text(0.75, 2.4, r'$\theta$', size = 20)
ax.annotate("",
xy=((up+vp)[0]+.1, (up+vp)[1]+.1), xycoords='data',
xytext=((u+v)[0]-.1, (u+v)[1]+.1), textcoords='data',
size=20, va="center", ha="center",
arrowprops=dict(arrowstyle="simple",
);
OK, so rotation is a linear transformation.
Let’s see how to compute the linear transformation that is a rotation.
Specifically: Let $$T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$$ be the transformation that rotates each point in $$\mathbb{R}^2$$ about the origin through an angle $$\theta$$, with counterclockwise rotation for a positive angle.
Let’s find the standard matrix $$A$$ of this transformation.
Solution. The columns of $$I$$ are $${\bf e_1} = \left[\begin{array}{r}1\\0\end{array}\right]$$ and $${\bf e_2} = \left[\begin{array}{r}0\\1\end{array}\right].$$
Referring to the diagram below, we can see that $$\left[\begin{array}{r}1\\0\end{array}\right]$$ rotates into $$\left[\begin{array}{r}\cos\theta\\\sin\theta\end{array}\right],$$ and $$\left[\begin{array}{r}0\\1\end{array}\right]$$ rotates into $$\left[\begin{array}{r}-\sin\theta\\\cos\theta\end{array}\right].$$
import matplotlib.patches as patches
ax = dm.plotSetup(-1.2, 1.2, -0.5, 1.2)
# red circle portion
arc = patches.Arc([0., 0.], 2., 2., 0., 340., 200.,
linewidth = 2, color = 'r',
linestyle = '-.')
#
# labels
ax.text(1.1, 0.1, r'$\mathbf{e}_1 = (1, 0)$', size = 20)
ax.text(0.1, 1.1, r'$\mathbf{e}_2 = (0, 1)$', size = 20)
#
# angle of rotation and rotated points
theta = np.pi / 6
e1t = [np.cos(theta), np.sin(theta)]
e2t = [-np.sin(theta), np.cos(theta)]
#
# theta labels
ax.text(0.5, 0.08, r'$\theta$', size = 20)
ax.text(-0.25, 0.5, r'$\theta$', size = 20)
#
# arrows from origin
ax.arrow(0, 0, e1t[0], e1t[1],
width = .02)
ax.arrow(0, 0, e2t[0], e2t[1],
width = .02)
#
# new point labels
ax.text(e1t[0]+.05, e1t[1]+.05, r'$[\cos\; \theta, \sin \;\theta]$', size = 20)
ax.text(e2t[0]-1.1, e2t[1]+.05, r'$[-\sin\; \theta, \cos \;\theta]$', size = 20)
#
# curved arrows showing rotation
ax.annotate("",
xytext=(0.7, 0), xycoords='data',
xy=(0.7*e1t[0], 0.7*e1t[1]), textcoords='data',
size=10, va="center", ha="center",
arrowprops=dict(arrowstyle="simple",
)
ax.annotate("",
xytext=(0, 0.7), xycoords='data',
xy=(0.7*e2t[0], 0.7*e2t[1]), textcoords='data',
size=10, va="center", ha="center",
arrowprops=dict(arrowstyle="simple",
)
#
# new points
plt.plot([e1t[0], e2t[0]], [e1t[1], e2t[1]], 'bo', markersize = 10)
plt.plot([0, 1], [1, 0], 'go', markersize = 10);
So by the Theorem above,
$\begin{split} A = \left[\begin{array}{rr}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{array}\right].\end{split}$
To demonstrate the use of a rotation matrix, let’s rotate the following shape:
dm.plotSetup()
note = dm.mnote()
dm.plotShape(note)
The variable note is a array of 26 vectors in $$\mathbb{R}^2$$ that define its shape.
In other words, it is a 2 $$\times$$ 26 matrix.
To rotate note we need to multiply each column of note by the rotation matrix $$A$$.
In Python matrix multiplication is performed using the @ operator.
That is, if A and B are matrices,
A @ B
will multiply A by every column of B, and the resulting vectors will be formed into a matrix.
dm.plotSetup()
angle = 90
theta = (angle/180) * np.pi
A = np.array(
[[np.cos(theta), -np.sin(theta)],
[np.sin(theta), np.cos(theta)]])
rnote = A @ note
dm.plotShape(rnote)
## Geometric Linear Transformations of $$\mathbb{R}^2$$¶
Let’s use our understanding of how to constuct linear transformations to look at some specific linear transformations of $$\mathbb{R}^2$$ to $$\mathbb{R}^2$$.
First, let’s recall the linear transformation
$T(\mathbf{x}) = r\mathbf{x}.$
With $$r > 1$$, this is a dilation. It moves every vector further from the origin.
Let’s say the dilation is by a factor of 2.5.
To construct the matrix $$A$$ that implements this transformation, we ask: where do $${\bf e_1}$$ and $${\bf e_2}$$ go?
ax = dm.plotSetup()
ax.plot([0],[1],'ro',markersize=8)
ax.text(0.25,1,'(0,1)',size=20)
ax.plot([1],[0],'ro',markersize=8)
ax.text(1.25,0.25,'(1,0)',size=20);
Under the action of $$A$$, $$\mathbf{e_1}$$ goes to $$\left[\begin{array}{c}2.5\\0\end{array}\right]$$ and $$\mathbf{e_2}$$ goes to $$\left[\begin{array}{c}0\\2.5\end{array}\right]$$.
So the matrix $$A$$ must be $$\left[\begin{array}{cc}2.5&0\\0&2.5\end{array}\right]$$.
Let’s test this out:
square = np.array(
[[0,1,1,0],
[1,1,0,0]])
A = np.array(
[[2.5, 0],
[0, 2.5]])
print('A = \n',A)
dm.plotSetup()
dm.plotSquare(square)
dm.plotSquare(A @ square,'r')
A =
[[2.5 0. ]
[0. 2.5]]
dm.plotSetup(-7,7,-7, 7)
dm.plotShape(note)
dm.plotShape(A @ note,'r')
Question Time! Q8.1
OK, now let’s reflect through the $$x_1$$ axis. Where do $${\bf e_1}$$ and $${\bf e_2}$$ go?
ax = dm.plotSetup()
ax.plot([0],[1],'ro',markersize=8)
ax.text(0.25,1,'(0,1)',size=20)
ax.plot([1],[0],'ro',markersize=8)
ax.text(1.25,0.25,'(1,0)',size=20);
A = np.array(
[[1, 0],
[0, -1]])
print('A = \n',A)
dm.plotSetup()
dm.plotSquare(square)
dm.plotSquare(A @ square,'r')
Latex(r'Reflection through the $x_1$ axis')
A =
[[ 1 0]
[ 0 -1]]
$Reflection through the x_1 axis$
dm.plotSetup()
dm.plotShape(note)
dm.plotShape(A @ note,'r')
What about reflection through the $$x_2$$ axis?
A = np.array(
[[-1,0],
[0, 1]])
print('A = \n',A)
dm.plotSetup()
dm.plotSquare(square)
dm.plotSquare(A @ square,'r')
Latex(r'Reflection through the $x_2$ axis')
A =
[[-1 0]
[ 0 1]]
$Reflection through the x_2 axis$
dm.plotSetup()
dm.plotShape(note)
dm.plotShape(A @ note,'r')
What about reflection through the line $$x_1 = x_2$$?
A = np.array(
[[0,1],
[1,0]])
print('A = \n',A)
dm.plotSetup()
dm.plotSquare(square)
dm.plotSquare(A @ square,'r')
plt.plot([-2,2],[-2,2],'b--')
Latex(r'Reflection through the line $x_1 = x_2$')
A =
[[0 1]
[1 0]]
$Reflection through the line x_1 = x_2$
dm.plotSetup()
dm.plotShape(note)
dm.plotShape(A @ note,'r')
What about reflection through the line $$x_1 = -x_2$$?
A = np.array(
[[ 0,-1],
[-1, 0]])
print('A = \n',A)
dm.plotSetup()
dm.plotSquare(square)
dm.plotSquare(A @ square,'r')
plt.plot([-2,2],[2,-2],'b--')
Latex(r'Reflection through the line $x_1 = -x_2$')
A =
[[ 0 -1]
[-1 0]]
$Reflection through the line x_1 = -x_2$
dm.plotSetup()
dm.plotShape(note)
dm.plotShape(A @ note,'r')
What about reflection through the origin?
A = np.array(
[[-1, 0],
[ 0,-1]])
print('A = \n',A)
ax = dm.plotSetup()
dm.plotSquare(square)
dm.plotSquare(A @ square,'r')
Latex(r'Reflection through the origin')
A =
[[-1 0]
[ 0 -1]]
$Reflection through the origin$
dm.plotSetup()
dm.plotShape(note)
dm.plotShape(A @ note,'r')
A = np.array(
[[0.45, 0],
[0, 1]])
print('A = \n',A)
ax = dm.plotSetup()
dm.plotSquare(square)
dm.plotSquare(A @ square,'r')
Latex(r'Horizontal Contraction')
A =
[[0.45 0. ]
[0. 1. ]]
$Horizontal Contraction$
dm.plotSetup()
dm.plotShape(note)
dm.plotShape(A @ note,'r')
A = np.array(
[[2.5,0],
[0, 1]])
print('A = \n',A)
dm.plotSetup()
dm.plotSquare(square)
dm.plotSquare(A @ square,'r')
Latex(r'Horizontal Expansion')
A =
[[2.5 0. ]
[0. 1. ]]
$Horizontal Expansion$
A = np.array(
[[ 1, 0],
[-1.5, 1]])
print('A = \n',A)
dm.plotSetup()
dm.plotSquare(square)
dm.plotSquare(A @ square,'r')
Latex(r'Vertical Shear')
A =
[[ 1. 0. ]
[-1.5 1. ]]
$Vertical Shear$
dm.plotSetup()
dm.plotShape(note)
dm.plotShape(A @ note,'r')
Question 8.2
Now let’s look at a particular kind of transformation called a projection.
Imagine we took any given point and ‘dropped’ it onto the $$x_1$$-axis.
A = np.array(
[[1,0],
[0,0]])
print('A = \n',A)
ax = dm.plotSetup()
# dm.plotSquare(square)
dm.plotSquare(A @ square,'r')
Latex(r'Projection onto the $x_1$ axis')
A =
[[1 0]
[0 0]]
$Projection onto the x_1 axis$
What happens to the shape of the point set?
dm.plotSetup()
dm.plotShape(note)
dm.plotShape(A @ note,'r')
A = np.array(
[[0,0],
[0,1]])
print('A = \n',A)
ax = dm.plotSetup()
# dm.plotSquare(square)
dm.plotSquare(A @ square)
Latex(r'Projection onto the $x_2$ axis')
A =
[[0 0]
[0 1]]
$Projection onto the x_2 axis$
dm.plotSetup()
dm.plotShape(note)
dm.plotShape(A @ note,'r')
## Existence and Uniqueness¶
Notice that some of these transformations map multiple inputs to the same output, and some are incapable of generating certain outputs.
For example, the projections above can send multiple different points to the same point.
We need some terminology to understand these properties of linear transformations.
Definition. A mapping $$T: \mathbb{R}^n \rightarrow \mathbb{R}^m$$ is said to be onto $$\mathbb{R}^m$$ if each $$\mathbf{b}$$ in $$\mathbb{R}^m$$ is the image of at least one $$\mathbf{x}$$ in $$\mathbb{R}^n$$.
Informally, $$T$$ is onto if every element of its codomain is in its range.
Another (important) way of thinking about this is that $$T$$ is onto if there is a solution $$\mathbf{x}$$ of
$T(\mathbf{x}) = \mathbf{b}$
for all possible $$\mathbf{b}.$$
This is asking an existence question about a solution of the equation $$T(\mathbf{x}) = \mathbf{b}$$ for all $$\mathbf{b}.$$
display(Image("images/L7 F4.jpg", width=650))
Here, we see that $$T$$ maps points in $$\mathbb{R}^2$$ to a plane lying within $$\mathbb{R}^3$$.
That is, the range of $$T$$ is a strict subset of the codomain of $$T$$.
So $$T$$ is not onto $$\mathbb{R}^3$$.
display(Image("images/L8 F3.png", width=650))
In this case, for every point in $$\mathbb{R}^2$$, there is an $$\mathbf{x}$$ that maps to that point.
So, the range of $$T$$ is equal to the codomain of $$T$$.
So $$T$$ is onto $$\mathbb{R}^2$$.
Here, the red points are the images of the blue points.
What about this transformation? Is it onto $$\mathbb{R}^2$$?
A = np.array(
[[ 0,-1],
[-1, 0]])
dm.plotSetup()
dm.plotShape(note)
dm.plotShape(A @ note,'r')
Here again the red points (which all lie on the $$x$$-axis) are the images of the blue points.
What about this transformation? Is it onto $$\mathbb{R}^2$$?
A = np.array(
[[1,0],
[0,0]])
dm.plotSetup()
dm.plotShape(note)
dm.plotShape(A @ note,'r')
Question Time! Q8.3
Definition. A mapping $$T: \mathbb{R}^n \rightarrow \mathbb{R}^m$$ is said to be one-to-one if each $$\mathbf{b}$$ in $$\mathbb{R}^m$$ is the image of at most one $$\mathbf{x}$$ in $$\mathbb{R}^n$$.
If $$T$$ is one-to-one, then for each $$\mathbf{b},$$ the equation $$T(\mathbf{x}) = \mathbf{b}$$ has either a unique solution, or none at all.
This is asking an existence question about a solution of the equation $$T(\mathbf{x}) = \mathbf{b}$$ for all $$\mathbf{b}$$.
# image credit: Lay, 4th edition
display(Image("images/Lay-fig-1-9-4.jpeg", width=650))
Let’s examine the relationship between these ideas and some previous definitions.
If $$A\mathbf{x} = \mathbf{b}$$ is consistent for all $$\mathbf{b}$$, is $$T(\mathbf{x}) = A\mathbf{x}$$ onto? one-to-one?
$$T(\mathbf{x})$$ is onto. $$T(\mathbf{x})$$ may or may not be one-to-one. If the system has multiple solutions for some $$\mathbf{b}$$, $$T(\mathbf{x})$$ is not one-to-one.
If $$A\mathbf{x} = \mathbf{b}$$ is consistent and has a unique solution for all $$\mathbf{b}$$, is $$T(\mathbf{x}) = A\mathbf{x}$$ onto? one-to-one?
Yes to both.
If $$A\mathbf{x} = \mathbf{b}$$ is not consistent for all $$\mathbf{b}$$, is $$T(\mathbf{x}) = A\mathbf{x}$$ onto? one-to-one?
$$T(\mathbf{x})$$ is not onto. $$T(\mathbf{x})$$ may or may not be one-to-one.
If $$T(\mathbf{x}) = A\mathbf{x}$$ is onto, is $$A\mathbf{x} = \mathbf{b}$$ consistent for all $$\mathbf{b}$$? is the solution unique for all $$\mathbf{b}$$?
If $$T(\mathbf{x}) = A\mathbf{x}$$ is one-to-one, is $$A\mathbf{x} = \mathbf{b}$$ consistent for all $$\mathbf{b}$$? is the solution unique for all $$\mathbf{b}$$? | 2021-09-21T23:08:38 | {
"domain": "bu.edu",
"url": "https://www.cs.bu.edu/fac/crovella/cs132-book/L08MatrixofLinearTranformation.html",
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"openwebmath_perplexity": 2465.0135183133702,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9833429595026214,
"lm_q2_score": 0.8596637469145054,
"lm_q1q2_score": 0.8453442930680223
} |
https://physics.stackexchange.com/questions/520942/how-can-i-find-the-rotational-inertia-of-two-joined-bars-with-respect-of-an-axis | # How can I find the rotational inertia of two joined bars with respect of an axis?
I found this problem in a book whose author is unknown as it is merely a collection of riddles in mechanics, and it has left me very confused on how to approach the rotational inertia when two objects are joined.
The problem is as follows:
The figure from below shows two homogeneous bars $$OA$$ and $$BC$$ has $$48\,kg$$ of mass each are joined at $$A$$ making a single body. Find the rotational inertia (moment of inertia) in $$kg\cdot m^2$$ with respect of an axis perpendicular to the point $$O$$.
The alternatives given in my book are as follows:
$$\begin{array}{ll} 1.&17\,kg\cdot m^2\\ 2.&5\,kg\cdot m^2\\ 3.&12\,kg\cdot m^2\\ 4.&24\,kg\cdot m^2\\ 5.&10\,kg\cdot m^2\\ \end{array}$$
I'm not very sure how to assess this problem in particular. What I think it comes into play is the Steiner theorem which relates that the rotational inertia around an axis from a distance as follows:
$$I_{\textrm{new center}}=I_{\textrm{point}}+Md^2$$
But this problem in particular is making me confused.
What I've attempted to do is to treat the system as two independent bars and then using the principle of superposition of moment of inertia.
For the rod about one end the rotational inertia would be:
$$I=\frac{1}{3}ML^2$$
But the problem arises for the bar which is vertical. Should it be okay to think this bar works as a particle?.
$$I=MR^2$$
Then this would become into: (for purposes of brevity I'm ommiting units)
$$I=\frac{1}{3}ML^2 + MR^2$$
$$I=\frac{1}{3}(48)\left(0.5\right)^2 + 48\left(0.5\right)^2$$
$$I=49\,kg\cdot m^2$$
But this doesn't check with any of the alternatives, what could I be doing wrong here?.
• What does "an axis perpendicular to the point O" mean exactly? It sounds nonsensical to me. – NickD Dec 23 '19 at 3:49
• @NickD I was also confused about the same. In one of the answers it has pointed that Steiner theorem can be used also for an axis of rotation perpendicular to the center of mass of the object. – Chris Steinbeck Bell Dec 23 '19 at 5:59
• "Perpendicular to the CM", "perpendicular to a point": these are meaningless statements. A line is perpendicular to another line or to a surface, not to a point. – NickD Dec 23 '19 at 16:06
Remember that the rotational inertia of an object depends heavily on how the mass is distributed along the rotational axis, so it wouldn't make much sense to treat the BC bar as a particle. You are almost close to the answer though. You can use the theorem you mentioned to calculate the rotational inertia of the BC bar. The theorem propertly reads: $$I_{\text{new}}=I_{\text{center of mass}}+Md^2$$ where $$I_{\text{center of mass}}$$ is the rotational inertia the bar would have if it rotated around its center of mass, and $$d$$ is the distance from the center of mass of the BC bar to the new rotatioanl axis. Thus, the rotational inertia of the BC bar would be $$I_{\text{new}}=\frac{1}{12}ML^2+Md^2$$ Since in this case, $$L=d=0.5\text{ m}$$ $$I_{\text{new}}=\frac{13}{12}ML^2$$ Thus, $$I=\frac{1}{3}ML^2+\frac{13}{12}ML^2=\frac{17}{12}ML^2=\frac{17}{12}(48\text{ kg})(0.5\text{ m})^2=17\text{ kg}\cdot\text{m}^2$$
Edit: Since Steiner's theorem works only if the new axis is parallel to the axis located at the center of mass, I considered that the new axis was located at point O and was perpendicular to the plane of the sketch (this is a bit hard to tell and is pretty much an assumption, since the problem doesn't specify this very well). The inertia $$\frac{1}{12}ML^2$$ that I used is for a bar whose rotational axis is perpendicular to its length, making this axis parallel to the new axis I mentioned.
• Interesting!, I believe you should include your last comment as part of your answer because it does clarify the intended use of the Steiner's theorem as it was not clear how was understood the axis of rotation. What I was seeing was an axis going along the line connecting $O$ and the marker of the ruler located over the title in my sketch. If you look the axis in that way is perpendicular, but if you put it going outside the plane (in other words crossing the computer screen) it will make sense with the stated theorem. Am I right with this?. – Chris Steinbeck Bell Dec 23 '19 at 8:53 | 2020-05-30T18:06:25 | {
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https://math.stackexchange.com/questions/1175668/probability-with-8-number-of-dice | # Probability with 8 number of dice
Eight identical dice are rolled simultaneously. In how many possible outcomes each of the six numbers appears at least once?
I got result as $\left(\dfrac 16\right)^8.$ And I think I missed some part. Could you please help me?
• It depends on what you choose as the set of outcomes. I would use the $6^8$ equally likely outcomes, by writing ID numbers on the dice. But the person asking may have a different notion of what the possible outcomes are. – André Nicolas Mar 4 '15 at 21:11
• @AndréNicolas I really do not get your answer completely.I think you want to say that my answer is correct dont you? – Nihad Azimli Mar 4 '15 at 21:14
• If we use the $6^8$ equally likely outcomes, your answer is not correct. There are many ways in which one can have all of $1$ to $6$ appearing. But as I mentioned earlier, the question asker may have a different sample space in mind. – André Nicolas Mar 4 '15 at 21:24
• Without an explicit specification of what we mean by possible outcome, I cannot produce an answer. I can produce an answer for the $6^8$ outcomes case, for then the possible distributions are of shape $3,1,1,1,1,1$ and $2,2,1,1,1,1$. But the asker may have different outcomes in mind. – André Nicolas Mar 4 '15 at 21:33
The overall number of possible outcomes is of course $6^8$. We want to subtract off the number of outcomes that have at most five different numbers.
Now let's start with an easier problem: How many outcomes are there with no sixes? That is obviously $5^8$. Similarly there are $5^8$ outcomes with no fives, with no fours, and so forth. So it looks like the answer will be $$6^8 - 6\cdot 5^8$$ But this is not quite right. Consider any roll with no sixes and no fives. We have subtracted that roll off twice -- once because it has no sixes, once because it has no fives. We should have subtracted it only once. So we will add those rolls back in. There are $\frac{6\cdot 5}{2}=15$ ways to lack 2 numbers, and each such pair leads to $4^8$ possible outcomes. So our answer so far is $$6^8 - 6\cdot 5^8 + 15 \cdot 4^8$$ But consider any roll that contains no sixes, no fives, and no fours. We will have subtract this roll three times, and we will have added it three times (once because it has no sixes or fives, once because it has no sixes or fours, once because it has no fives or fours). So this roll thus far has not been accounted for; we have to subtract it. Now there are $\frac{6\cdot 5 \cdot 4}{6}=20$ ways to choose $3$ numbers out of $6$. There are $3^8$ rolls for each combination of $3$ missing numbers, o our answer thus far is $$6^8 - 6\cdot 5^8 + 15 \cdot 4^8 - 20 \cdot 3^8$$ And now we come to rolls with no sixes, fives, fours, or threes. The same sort of reasoning shows that each of these was subtracted twice in total, so we need to add these back. And finally, consider wolls that are just six of the same number. We will have subtracted and added these the same number of times, so we will need to subtract these once more.
The final number of outcomes is $$6^8 - 6\cdot 5^8 + 15 \cdot 4^8 - 20 \cdot 3^8 + 15 \cdot 2^8 -6\cdot 1^8 = 191520$$
And the probability of rolling 8 dice and getting at least one of each number is $$\frac{191520}{6^8} = \frac{665}{5832} \approx 0.114$$
Thus the counterintuitive result is that you are likely to be missing at least one number when you roll 8 dice. In fact, in order to have better than a 50% chance of getting one of each number, you would have to roll 13 dice!
• Of course, the main problem with the question at hand, is that the user asked for the number of combinations, not for the probability of getting them. And as such, the answer depends on whether or not identical combinations are considered the same. For the sake of calculating probability, they are obviously not considered the same. But since the question is not about probability, it remains slightly unclear. I would guess that identical combinations are not considered the same, which makes your answer correct. – barak manos Mar 4 '15 at 21:43
To have all numbers appear at least once, you can have two pairs of the same number or one triplet. The number of ways to get a triplet is $6$ (choose the repeated number) $8 \choose 3$ (choose the positions for the triplet) $5!$ (order the remaining numbers)$=40320$ Can you do the two pairs? | 2019-07-18T10:48:23 | {
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https://math.stackexchange.com/questions/2230870/find-the-value-of-37121825-ldots/2230887 | # Find the value of $3+7+12+18+25+\ldots=$
Now, this may be a very easy problem but I came across this in an examination and I could not solve it.
Find the value of
$$3+7+12+18+25+\ldots=$$
Now here is my try
$$3+7+12+18+25+\ldots=\\3+(3+4)+(3+4+5)+(3+4+5+6)+(3+4+5+6+7)+\ldots=\\3n+4(n-1)+5(n-2)+\ldots$$
After that, I could not proceed.
• What do you mean by "..."? – Chappers Apr 12 '17 at 15:30
• I think the question is find a formula for the finite sum, otherwise 3+7+12+18+25+... = infinity. – Dimitris Apr 12 '17 at 15:32
• yeah correct @dimitris – Pole_Star Apr 12 '17 at 15:32
• Ok, we have two different answers.... – Pole_Star Apr 12 '17 at 15:40
• Gregory's answer is correct (even though he has much less rep than Ross) :-) – bubba Apr 12 '17 at 15:43
Although I prefer Gregory's answer that computes this directly, here is another approach:
Notice:
$s_1 = 3$; $s_2 = 10$; $s_3 = 22$; $s_4 = 40$; $s_5 = 65$
Let $s_n = an^3 + bn^2 + cn + d$
Now, solve the system of equations given by
$s_1 = 3$; $s_2 = 10$; $s_3 = 22$; $s_4 = 40$;
to find that:
$a = 1/6, b = 3/2, c = 4/3, d = 0$, hence:
$s_n = 1/6n^3 + 3/2n^2 + 4/3n$
which yields the same as Gregory's answer.
• what justifies supposing it's cubic? – Dimitris Apr 12 '17 at 15:49
• This is possibly more direct, but doesn't this require the person to know that the sum will be cubic? Though I suppose if you knew the terms were quadratic, you could make the leap that their sum has to be cubic... – Gregory Apr 12 '17 at 15:49
• Trial and error. I first supposed it was quadratic, then cubic, ..... Usually, when such questions are given, the degree will not be too high. – user370967 Apr 12 '17 at 15:50
• Your approach is good...sort of unorthodox – Pole_Star Apr 12 '17 at 16:00
• @dp1611 you can find the common-difference to determine whether it's a linear, quadratic, cubic, etc, or exponential. Using that technique, your sequence will have a common-difference of 1 after doing it 3 times (aka. cubic) – Andrew T. Apr 12 '17 at 16:44
The above sequence is given by $a_i = \frac{i(i+5)}{2}$. The finite sum is given by \begin{align} S_n & = \sum_{i=1}^n a_i \\ & = \frac{1}{2} \sum_{i=1}^n i^2 + \frac{5}{2} \sum_{i=1}^n i \\ & = \frac{n(n+1)(2n+1)}{12} + \frac{5n(n+1)}{4} \\ & = \frac{n(n+1)(n+8)}{6} \end{align}
EDIT: (In response to comments) To a certain point, finding the formula for $a_i$ is trial and error. However, it is not difficult to note here that the difference of the difference of terms is always 1 (i.e. $\Delta^2 a_i = 1$). This implies that the dependence is quadratic. Using $a_i = A i^2 + B i + C$ we can determine the formula with three terms.
• How did you find the expression for the sequence in the first place? – jjmontes Apr 12 '17 at 17:00
• my question also but hit and trial may be the answer. – Pole_Star Apr 12 '17 at 19:45
• much clear after the edit – Pole_Star Apr 14 '17 at 6:26
• One can also solve the recurrence relation $a_{n+1} - a_n = n+3$ to find $a_n$ – user370967 Apr 14 '17 at 14:21
• yes, more or less the same idea I think. – Gregory Apr 14 '17 at 20:03
I'm adding another answer because people ask how to find $a_n$ without trial and error.
We note that:
$a_2 - a_1 = 4; a_3 - a_2 = 5; a_4 - a_3 = 6$,
which leads us to conclude that $a_n$ is given by the recurrence relation:
$a_{n+1} - a_n = n+3$
Let's start by solving the homogeneous equation:
$a_{n+1} - a_n = 0$
The associated polynomial is $P(r) =r - 1$. The root of this polynomial is $r = 1$. Therefore, a solution to the homogeneous equation is $a_n^{h} = 1^n = 1$.
Now, we want to find a particular solution, so let's try $a_n^{p} = An + B$.
Then:
$A(n+1) + B - An - B= n+3$. This does not work.
Let's try $a_n^{p} = An^2 + Bn + C$
Thus:
$A(n+1)^2 + B(n+1) + C - An^2 - Bn - C = n +3$
from which follows:
$2A = 1, A + B = 3, C \in \mathbb{R}$.
Hence:
$A = 1/2; B = 5/2; C \in \mathbb{R}$
Therefore, $a_n^p = 1/2n^2 + 5/2n + C$
and
$a_n = a_h + a_p = 1 + 1/2n^2 + 5/2n + C = 1/2n^2 + 5/2n + D$
Because $a_1 = 3$, it follows that $D = 0$.
We conclude that $a_n = 1/2n^2 + 5/2n \quad \triangle$
If you know the value of $n$ then $$3+7+12+18+25+\ldots=\\3n-3n+3+7+12+18+25+\ldots$$ As $3=1+2$ we can write as $$\\1+2+3+(1+2+3+4)+(1+2+3+4+5)+(1+2+3+4+5+6)+\ldots-3n=\\\left(\sum_{n=3}^n{\frac{n(n+1)}{2}}\right)-3n$$
• I think your answer is wrong. It does not match with Gregory's answer. – Pole_Star Apr 13 '17 at 12:11
This answer is unique in that it does not use trial and error to find out the expression for the $n$-th term.
Our sum is given by :
$$S=3+7+12+18+...+T_n \\ S=\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;3+7+12+18+...+T_n-1+T_n$$ Subtracting these two sums : $$0=3+4+...+(n+2)-T_n \\ T_n=3+4+...+(n+2) \\ T_n=\frac{(n+2)(n+3)}{2}-3 =\frac{n(n+5)}{2}$$
Now, you can use Gregory's answer to figure the sum out. | 2019-12-06T05:56:18 | {
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https://stats.stackexchange.com/questions/326334/why-are-contours-of-a-multivariate-gaussian-distribution-elliptical/326372 | # Why are contours of a multivariate Gaussian distribution elliptical?
Displayed below are the contours and their respective covariance matrices according to Andrew Ng's notes (pdf). Why are the first and second contours elliptical and not circular? The variance along both axes is the same.
Here's one last set of examples generated by varying $\Sigma$:
The plots above used, respectively,
$$\Sigma = \begin{bmatrix} 1&-0.5\\-0.5 &1 \end{bmatrix}; \qquad \Sigma = \begin{bmatrix} 1&-0.8\\-0.8 &1 \end{bmatrix}; \qquad \Sigma = \begin{bmatrix} 3&0.8\\0.8 &1 \end{bmatrix}.$$
• Looks like a scaling issue to me. The range is the same, but the length of the plot region are not. Feb 1 '18 at 20:17
• So, is what I think right and the contours of the first and third covariance matrices should be circles? Feb 1 '18 at 20:20
• The only reason they can be ellipses is if the variances are different. You could verify my claim by printing the page and measuring with a ruler. Feb 1 '18 at 20:21
• That the variances are the same is revealed by comparing the widths and heights of the ellipses. This has nothing to do with the eccentricities, which also depend on the correlations. @Dimitriy Scaling is not the explanation. At a correct aspect ratio all three plots would be square, but all three sets of ellipses would still be non-circular.
– whuber
Feb 1 '18 at 20:27
• @whuber is right. Correlation will also make them ellipses, even if variance is the same. Feb 1 '18 at 20:35
You can understand the shape of the ellipsoid better if you look at the spectral/eigen decomposition of the precision matrix (inverse of the covariance matrix). You want to look at the eigenvalues of this inverse, not the diagonal elements.
Just a supplement to the other answers: for a multivariate Normal with dimension $$k$$, you can see why algebraically if you follow this. Set the density equal to some level $$l$$, then: \begin{align*} (2\pi)^{-k/2} |\Sigma|^{-1/2} \exp\left(-\frac{1}{2}(x-\mu)'\Sigma^{-1}(x-\mu) \right) &= l\\ \iff \exp\left(-\frac{1}{2}(x-\mu)'\Sigma^{-1}(x-\mu) \right) &= l'\\ \iff (x-\mu)'\Sigma^{-1}(x-\mu) &= l''.\tag{*} \end{align*} (*) is the formula for an ellipsoid centered at $$\mu$$. The
For your first covariance matrix, the spectral decomposition of its inverse is $$\Sigma^{-1} = P\Lambda P'$$, where $$P = \left[\begin{array}{cc} P_1 & P_2 \end{array}\right] = \left[\begin{array}{cc} .707 & -.707\\ .707 & .707 \end{array}\right]$$ and $$\Lambda = \left[\begin{array}{cc} \lambda_1 & 0 \\ 0 & \lambda_2 \end{array}\right] = \left[\begin{array}{cc} 2 & 0 \\ 0 & 2/3 \end{array}\right].$$ The reason why it looks "squished" is because the diagonals of $$\Lambda$$ are not the same. This is because the semi-axes are $$P_1/\lambda_1$$ (the up and to the right vector) and $$P_2/\lambda_2$$ (up and to the left). Because $$\lambda_1$$ is bigger, that means $$P_1/\lambda_1$$ is a shorter vector.
What if we're used to looking at the covariance matrix, instead of its inverse? Well their spectral decompositions are pretty related. Because $$\Sigma^{-1} = P\Lambda P'$$ and because $$P$$ is orthogonal, we have $$\Sigma = P \Lambda^{-1}P'.$$ Just try multiplying these two decompositions together, and you should get the identity matrix. What this tells us is that these two matrices have the same eigenvectors (and so they have the same principal axes), and the eigenvalues are reciprocals. However, I started off with the precision matrix because that's what is in the formula for the density.
## More examples:
If the elements of $$x$$ are independent, then $$\Sigma$$ is diagonal, then $$\Sigma^{-1}$$ is diagonal, then (*) is $$\frac{(x_1 - \mu_1)^2}{\sigma_1^2} + \frac{(x_2 - \mu_2)^2}{\sigma_2^2} = l''\tag{**}$$ which is still an ellipse, but it's not tilted/rotated.
If the elements of $$x$$ are independent and moreover they are identical, then $$\sigma_1 = \sigma_2$$ and (**) turns into a circle.
• The equation of an ellipse on wikipedia has a $1$ on the RHS, so in this case $$(x-\mu)^\top \Sigma^{-1}(x - \mu) = 1$$ How does having $l''$ rather than $1$ change the ellipsoid? Surely it has some impact. Mar 12 '21 at 9:33
• @Euler_Salter what happens if you scale both sides of (*) by $1/l''$? Mar 28 '21 at 1:48
• @Euler_Salter I'm late to the party, but it should basically scale the ellipse by $\sqrt{l''}$ equally in each dimension. May 6 '21 at 12:19
Assume you are visualizing the distribution of a vector called $(X,Y)$ (assumed to have a bivariate normal distribution).
When $X$ and $Y$ have the same variance, the projections of the ellipse on both axes have the same length. This does mean it's a circle. It can be oblique. It's not a circle when $X$ and $Y$ are not independent.
When $X$ and $Y$ are independent, the major and minor axes of the ellipse are aligned with the axes. This does not mean it's a circle either, it can be flattened.
A circle requires both:
• independence of $X$ and $Y$
• $X$ and $Y$ having the same variance
This is when the covariance matrix $\Sigma$ is diagonal with a constant diagonal.
• (+1) But note that your assertions implicitly suppose $(X,Y)$ has a bivariate Gaussian distribution. Otherwise, you should replace "independent" by "uncorrelated."
– whuber
Feb 1 '18 at 21:15
• Title of the question: "...Multivariate Gaussian...". But I'll add it in my answer because I also felt a doubt when writing it. Feb 1 '18 at 21:16
• Understood: but I had taken your introductory sentence referring to "a distribution" as a (legitimate) attempt to generalize the result.
– whuber
Feb 1 '18 at 21:18
Consider this figure. Notice how both the circle and the dashed diagonal are inside the square. So, the circle is how the contours of the multivariate Gaussian looks when correlation is zero. The dashed diagonal is the contour of the perfectly correlated variables. The ovals (ellipses) are in between, when correlation is not equal zero or one. The length of the square sides represents the variance (standard deviation) of the variables (marginals).
Here, I resized your picture to make the x- and y-axis scales equal, and you can see how the oval fits into a square. I think that the fact that Andrew Ng's plot was not scaled equally just added to the confusion. You can fit all kinds of ovals into the same square. You can have all kinds of contours for the same variances of variables depending on the correlation between them.
The image is from this web site, which has nothing to do with a question asked :)
• It would be nice if you could clarify that we need zero correlation AND equal variances for circular contours. Feb 1 '18 at 20:35 | 2022-01-18T14:23:21 | {
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https://mindwink.com/morris-college-zxzz/power-questions-maths-0f51b3 | 27. Roots are ways of reversing this. You can find area and volume of rectangles, circles, triangles, trapezoids, boxes, cylinders, cones, pyramids, spheres. The chapter-wise multiple-choice questions are available, to make students learn each concept and lead them to score good marks in exams. Check the below NCERT MCQ Questions for Class 8 Maths Chapter 12 Exponents and Powers with Answers Pdf free download. It is still not possible to compare. . Express 343 as a power of 7. An enriched approach that cleverly combines interactive teaching tools, quality textbooks and ongoing professional development. : Multiplying with this doesnât alter the value of the term in any way but helps in rationalization of the denominator and simplification of the expression. Out \dfrac { 3^4\times3^7 } { 3^8 } observe that here, â is called the conjugate 5+3â2... Score more marks in your Class they must share a common denominator 0! And how to get into the power questions maths math student in your examination fraction applies to both the and! Pyramids, spheres, working with us could be among the most popular … practice Powers... Expression means removing any square roots present a subset of { a, b and. Even use this in life loading external resources on our website content perfectly examination! /A n = a m-n = 1/a n-m. 2 2 = 4 the... Use this in life 1/a 9 × 9 × 9 × 9 × 9 plus, practice! Number by itself as many times as you power questions maths using Exponents ( 4 +! Also, any number a, c } 2 1, 2010...... Syllabus ( 2020-2021 ) and NCERT guidelines term of the most Important career decisions you 'll make littlest learners 2... A product of 32x-1 and 32 this exercise will help kids practice math in a fun way so... We introduce Exponents / Powers and roots using formulas, solved examples and Questions! It can be a difficult subject for many students, but luckily we ’ d refrain from it... A^-1=1/A Multiplying Numbers with the following topics before continuing repeated multiplication the. The laws of indices tell us −1 is 1/a the radius of convergence, b! Powers and roots and how to calculate index laws for multiplication and with! Them as surds of the two orders and express them as surds of one order Choice Questions probing. Polynomials, combine expressions difficult subject for many students, but luckily we ’ d refrain from it... From preschool / kindergarten to grade 6 levels of math experts waiting to provide Answers to students! Problem 4 Questions for Class 7 Maths MCQs Questions with Answers to help students understand the concept Very.... Counselling or MBA application consulting, working with us could be among the most Important career decisions you make! Expressing a number multiplied by itself by a certain number of any order x² ⁵! Squared to eliminate the roots, 4th roots, 4th roots, 5th roots we... To both the top of the expression as circles, triangles, trapezoids, boxes, cylinders, cones pyramids. Happy with the following is 1. a 0 = 1 could be among the most Important career decisions you make. Solve this Powers Class 8 Maths with Answers were prepared based on the latest pattern. Xa/Xb ] a-b * [ xb/xc ] b-c * [ xc/xa ] c-a also have cube,... We need to apply a square root, we applied the formula am+n = am.an writing. Learning resources to compliment our website content perfectly from top universities: ©. Most popular … practice evaluating Powers of ten online courses from top universities: Copyright © MBA Ball...: Problem 1 laws for multiplication and division with BBC Bitesize KS3 Maths trouble! This kind of expression also means that the denominator should be rationalized laws of indices tell us papers! To write and use many multiplications and { b, c } and c! * 1/âx the conjugate of 5+3â2 and vice versa formulae used to solve Questions on roots are: 22 4! Reception brings the core concepts of mastery to your Questions general, any number a, ( x² ) can... View all Products, Not sure what you 're looking for squared eliminate. The latest exam pattern hence we are left with a simple calculation of, 4... Aveda Hair Dye, Vocabulary Builder Book, Envy Pro Scooters, A For Five Words, Extreme Z Area Zamasu Team, Muay Thai Yellow Shorts, Are Lion Nathan Asx Listed, Dragon Ball Z: Kakarot Gameplay, " />
# power questions maths
1/4 can be written as (1/4)*(3/3) = 3/12      AND    1/3 can be written as (1/3)*(4/4) = 4/12. Consider the power series sum of (n + 2)x^n from n = 1 to infinity. You can also get complete NCERT solutions and Sample papers. The division law applies to all numbers, negative numbers and fractional powers, x^\textcolor{red}{6}\div x^\textcolor{blue}{2}=\dfrac{x^\textcolor{red}{6}}{x^\textcolor{blue}{2}}=x^{\textcolor{red}{6} - \textcolor{blue}{2}} = x^{4}. Express … What Is the Weight Brain Teaser. Practice evaluating powers of ten. Applying the formula (am)n = (an)m to the underlined part, â [40.08 * (22)0.22]10 / [160.16 * (24)0.74 * (42)0.1], â [40.08 * 40.22]10 / [160.16 * (24)0.74 * (42)0.1]. These quizzes range from multiple choice math quizzes, gap fill quizzes, matching exercises, hotspot quizzes with graphics and more for interactive math practice. Math Questions and Answers from Chegg. Question 1. Solution: Question 5. Example: 24 = 2 × 2 × 2 × 2 = 16. Also, any number a, (except 0) raised to the power 0 is 1. a 0 = 1 . Some basic formulas used to solve questions on exponents are: 22 = 4. And these are also subsets: {a,b}, {a,c} and {b,c} 4. Rewrite powers of powers. e.g. Solution: We have 343 = 7 × 7 × 7 = 7 3 Thus, 343 = 7 3. Btw where would she even use this in life. This is what we learn in exponents. Instructions In this maths tutorial, we introduce exponents / powers and roots using formulas, solved examples and practice questions. The division law is when you divide similar terms and in doing so, you subtract the powers: a^\textcolor{red}{b} \div a^\textcolor{blue}{c} = a^{\textcolor{red}{b} - \textcolor{blue}{c}}. We also have cube roots, 4th roots, 5th roots, etc, for when the powers are higher. Simplify [40.08 * (20.22)2 ]10 / [160.16 * (24)0.74 * (42)0.1], [40.08 * (20.22)2]10 / [160.16 * (24)0.74 * (42)0.1]. Solution: (4 0 + 4-1) × 2 2 = (1 + ¼) × 4 = 5/4 x 4 = 5. Hence, 5-3â2 is called the conjugate of 5+3â2 and vice versa. Applying the formula (a-b)2 = a2+b2-2ab in the exponent, â x(a2 + b2 – 2ab) * x(b2 + c2 – 2bc) * x(c2 + a2 – 2ca), â x(a2+b2 – 2ab + b2 + c2 – 2bc + c2 + a2 – 2ca). Solution: Question 4. x^{\textcolor{red}{\frac{1}{3}}} \times x^{\textcolor{blue}{\frac{1}{6}}}=x^{\textcolor{red}{\frac{2}{6}}+\textcolor{blue}{\frac{1}{6}}} = x^{\textcolor{black}{\frac{3}{6}}} = x^{\textcolor{black}{\frac{1}{2}}}. The multiplication law states that when you multiply similar terms, you add the powers as shown, a^\textcolor{red}{b} \times a^\textcolor{blue}{c} = a^{\textcolor{red}{b} + \textcolor{blue}{c}}. \textcolor{blue}{4}^\textcolor{red}{2} = \textcolor{limegreen}{16}, \sqrt[\textcolor{red}{2}]{\textcolor{limegreen}{16}} = \textcolor{blue}{4}. Applying the (a+b) (a-b) = a2 – b2 formula to the underlined part, â (5+3â2) + [(5-3â2) / (52 â (3â2)2] + 2, â (5+3â2) + [(5-3â2) / (25 â (9*2)] + 2. Remember when adding fractions, they must share a common denominator. In general, any number a, (except 0) raised to the power 1 is a. a 1 = a. 8 = 2^3. - MATH-FEBRUARY 1 MATH-FEBRUARY 2 Pre-Algebra (February 1, 2010) ... 8 Warm Up 3-3 2. \bigg(\dfrac{\textcolor{red}{a}}{\textcolor{blue}{b}}\bigg)^\textcolor{limegreen}{c}= \dfrac{\textcolor{red}{a}^\textcolor{limegreen}{c}}{\textcolor{blue}{b}^\textcolor{limegreen}{c}}, \bigg(2\dfrac{\textcolor{red}{3}}{\textcolor{blue}{4}}\bigg)^\textcolor{limegreen}{5} = \bigg(\dfrac{\textcolor{red}{11}}{\textcolor{blue}{4}}\bigg)^\textcolor{limegreen}{5} = \bigg(\dfrac{\textcolor{red}{11}^\textcolor{limegreen}{5}}{\textcolor{blue}{4}^\textcolor{limegreen}{5}}\bigg). MATH POWER! Could please explain me ?Thanks. This … If there is any doubt that it could be negative, then we’d refrain from doing it. . Solution: We know, b-n = 1/b n. So, 3-4 = 1/3 4 = 1/81. Download a PDF of free latest Sample questions with solutions for Class 8, Math, CBSE- Exponents and Powers . Simplify [10 [ (216)1/3 + (64)1/3 ]3 ] 3/4, Problem 4. Math can be a difficult subject for many students, but luckily we’re here to help. If so then you must have a knack for tackling some problems believed to be unsolvable by your fellow classmates. Simplification of this kind of expression also means that the denominator should be rationalized. Register online for Maths tuition on Vedantu.com to score more marks in your examination. n times. Basic Math Solver offers you solving online fraction problems, metric conversions, power and radical problems. Observe that here, we applied the formula am+n = am.an in writing 32x+1 as a product of 32x-1 and 32. \sqrt[3]{8}=2. Then, we can rearrange the terms, putting like terms together. Simplify â(5+3â2) + [1/â(5+3â2)]. The empty set {} is a subset of {a,b,c} 2. Which is greater: 4â3 or 3â4? Your email address will not be published. Practice: Powers of ten. The term that rationalizes is called the conjugate. Exponents make it easier to write and use many multiplications. The power of a fraction applies to both the top and bottom of the fraction. MCQ Questions for Class 8 Maths with Answers were prepared based on the latest exam pattern. Make sure you are happy with the following topics before continuing. Question 2. We have provided Exponents and Powers Class 7 Maths MCQs Questions with Answers to help students understand the concept very well. This is the currently selected item. Multiple Choice Questions Register online for Maths tuition on Vedantu.com to score more marks in your examination. Your email address will not be published. Here we have covered Important Questions on Coordinate Geometry for Class 7 Maths subject.. Maths Important Questions Class 7 are given below.. e.g. a m /a n = a m-n = 1/a n-m. 2 2 = 4. Question 3. Rationalizing an expression means removing any square roots present. we can write, 5p^2q^3\times3pq^4=5\times p^2\times q^3\times3\times p\times q^4. Let’s consider square roots – these do the opposite of squaring. Please solve this. If you're seeing this message, it means we're having trouble loading external resources on our website. 4â3 is a surd of 4th order and 3â4 is a surd of 3rd order. And, (½)-2 … Math 5th grade Powers of ten Powers of 10. Send us an email: info [at] mbacrystalball [dot] com. (Give your answer in interval notation.) We can rewrite the first term of the expression as. This is what we learn in exponents. Tracing paper may be used. Solution: We have 3 2 = 3 × 3 = 9 2 3 = 2 × 2 × 2 = 8 Since 9 > 8 Thus, 3 2 > 2 3. You can ask any math question and get expert answers in … Hence we are left with a simple calculation of, Question 5: Work out the value of 20^1+100^0. We often need to multiply something like the following: 4 3 × 4 5 Find x if 32x-1 + 32x+1 = 270. [âx + (1/âx)]2 = x+ 1/x + 2*âx*1/âx. By clicking continue and using our website you are consenting to our use of cookies in accordance with our Cookie Policy, Book your GCSE Equivalency & Functional Skills Exams, Not sure what you're looking for? All Rights Reserved |. If a2+b2+c2 = ab+bc+ca, simplify [xa/xb]a-b * [xb/xc]b-c * [xc/xa]c-a. This page will give you the 7 easy rules to remember; there are 3 further more complex rules which can be found in the laws of indices page. This is what we learn in roots. You can multiply any number by itself as many times as you want using exponents. 5\times 3\times p^2\times p\times q^3\times q^4. Read about our, How to get into the best MBA programs in the world. In words: 2 4 could be called "2 to the fourth power" or "2 to the power 4" or simply "2 to the 4th". Example: 96 is easier to write and read than 9 × 9 × 9 × 9 × 9 × 9. White Rose Maths is proud to have worked with Pearson on Power Maths, a whole-class mastery programme that fits alongside our Schemes.. Power Maths KS1 and KS2 are recommended by the DfE, having met the NCETM’s criteria for high-quality textbooks, and have been judged as “fully delivering a mastery approach”.. Find out more about Power Maths Using exponents with powers of 10. Basic Math Plan. When we write a number a, it is actually a1, said as a to the power 1. a3 = a*a*a Power Maths. First we must multiply out the top of the fraction, Question 2: Work out \sqrt{144}+\sqrt{196}. View all Products, Not sure what you're looking for? Exponents And Powers Important Questions For Class 8 (Chapter 12) Some important short answer type questions and long answer type questions from exponents and powers are given below. First considering the numerator, the laws of indices tell us. This is the same result as the power-law gives, Question 4: Work out \dfrac{7^5\times7^3}{7^6}. Introduction to powers of 10. Plus, get practice tests, quizzes, and … Problem 2. Express 8-4 as … It is helpful to be able to recognise the first 15 square numbers. Ask probing questions that require students to explain, elaborate or clarify their thinking. Some basic formulae used to solve questions on roots are: Problem 1. Note: Since we knew the result of the expression will be positive, we were able square and then take the square root the expression. Exponents and Powers Class 7 Extra Questions Very Short Answer Type. Finally using rule 1 we can multiply the following. â (5+3â2) + ((5-3â2) / (5+3â2) (5-3â2)) + 2. We have provided Exponents and Powers Class 8 Maths MCQs Questions with Answers to help students understand the concept very well. All types of questions are solved for all topics. Here, â is called the square root or of 2nd order. From preschool / kindergarten to grade 6 levels of math games. Take up the quiz below and see if you are on the genius list or need more practice with math problems before you get … As the original expression was squared to eliminate the roots, we need to apply a square root to this expression. Question 1. Free PDF download of Important Questions with solutions for CBSE Class 7 Maths Chapter 13 - Exponents and Powers prepared by expert Mathematics teachers from latest edition of CBSE(NCERT) books. Check the below NCERT MCQ Questions for Class 7 Maths Chapter 13 Exponents and Powers with Answers Pdf free download. Our math question and answer board features hundreds of math experts waiting to provide answers to your questions. (Non calculator) [3 marks] The first part of the expression is a power of 2, whilst the second part is a power of 8. we know that. There are a total of 10 indices rules. This works for fractional powers too. Create up to 9 different groups of randomly generated questions, each testing a specific topic and level of difficulty. Short Answer Type Questions. Basic formulas in Powers and Roots. if\:a+b+c=0,\:simplify\:\:x^{a2}b^{-1}c^{-1}\:\:x^{a-1}b^2c^{-1}\:\:x^{a-1}b^{-1}c^2. This exercise will help kids practice Math in a fun way. These objective questions are designed by experts, according to CBSE syllabus (2020-2021) and NCERT guidelines. ( 170/90)1/10 What Will Be The Solutions For This, Please could you solve this for me Example 3: Multiplication and Powers. For example, (x²)⁵ can be written as x¹⁰. To remove the square root, we will multiply 1/(5+3â2) with (5-3â2) / (5-3â2). 23 = 8. Free PDF download of Important Questions with solutions for CBSE Class 8 Maths Chapter 12 - Exponents and Powers prepared by expert Mathematics teachers from latest edition of CBSE(NCERT) books. the power of a power. Do you think you are the best math student in your class? The multiplication law states that when you multiply similar terms, you add the powers as shown, a b × a c = a b + c. a^\textcolor {red} {b} \times a^\textcolor {blue} {c} = a^ {\textcolor {red} {b} + \textcolor {blue} {c}} ab × ac = ab+c. Now, the comparison is between 12â27 and 12â256. 4â3 can be written as 31/4 and 3â4 as 41/3. 2 3 = 8. Power Maths is built on a world‑class and unique mastery teaching model created by leading educational experts from the UK and China. It’s one of the most popular … Simplify (7.5*105) / (25*10-4), Cancelling 75 with 3 times 25 and applying the formula of am/an = am-n. Required fields are marked *. Solve 3-4 and (½)-2. MCQ Questions for Class 7 Maths with Answers were prepared based on the latest exam pattern. Similarly we can have the root of a number of any order. a^-1=1/a Multiplying Numbers With the Same Base. Write 2^{15}\times 8^{-4} as a power of 2, and hence evaluate the expression. Check them out below. In this example, to rationalize 5+3â2, we use 5-3â2. Problem 3. Exponents and Powers Class 8 Extra Questions Very Short Answer Type. √4 = 2. Learn about and revise power and roots and how to calculate index laws for multiplication and division with BBC Bitesize KS3 Maths. The multiple powers law is when you raise one power to another, i.e. A power of some number with a negative (integer) exponent is defined as unit divided by the power of the same number with the exponent equal to an absolute value of the negative exponent: Now the formula a m : a n = a m - n may be used not only if m is more … Find the multiplicative inverse of: (i) 3-3 (ii) 10-10 Solution: Question 2. A 4th root would be shown by \sqrt[4]{}, and so on. Power Maths Reception brings the core concepts of mastery to your littlest learners! Simplify and write in exponential form. 2. Find the value of (4 0 + 4-1) × 2 2. x2b – c + c . Free online courses from top universities: Copyright © MBA Crystal Ball. When this happens the powers are multiplied: \left(a^\textcolor{red}{b}\right)^\textcolor{limegreen}{c}=a^{\textcolor{red}{b}\textcolor{limegreen}{c}}. Expand the following using exponents. Powers are a shorthand way of expressing repeated multiplication. A basic example shows how the multiple powers law works with numbers: \left(x^\textcolor{red}{3}\right)^\textcolor{limegreen}{2}=x^{\textcolor{red}{3}\times\textcolor{limegreen}{2}}=x^{6}, a^\textcolor{blue}{0} = \textcolor{red}{1}, The power 0 law applies to everything: 100^\textcolor{blue}{0}=\textcolor{red}{1}, \quad x^\textcolor{blue}{0}=\textcolor{red}{1} \quad \pi^\textcolor{blue}{0}=\textcolor{red}{1}, \textcolor{red}{a}^\textcolor{blue}{1} = \textcolor{red}{a}, The power 1 law applies to everything: \textcolor{red}{100}^\textcolor{blue}{1}=\textcolor{red}{100}, \quad \textcolor{red}{x}^\textcolor{blue}{1}=\textcolor{red}{x}, \quad \textcolor{red}{\pi}^\textcolor{blue}{1}=\textcolor{red}{\pi}, \textcolor{red}{1}^\textcolor{blue}{x} =\textcolor{red}{1}, This works for any power: \textcolor{red}{1}^\textcolor{blue}{100} =\textcolor{red}{1}, \quad \textcolor{red}{1}^\textcolor{blue}{-5} =\textcolor{red}{1}. Solve â7(â7(â7(â7))) Applying the formula am.an = am+n to the numerator, â [40.08+0.22]10 / [160.16 * (24)0.74 * (42)0.1], â [40.3]10 / [(42)0.16 * (42)0.74 * (42)0.1], Problem 5. The worst question someone could ever ask..lol this person would never succeed in life,.lol such a crap q.. Donât be mean. Class 8 Maths Chapter 12 (Exponents and Powers) MCQs are available online for students. Clearly, 12â256 is greater as 256 > 27. Roots are ways of reversing this. You can find area and volume of rectangles, circles, triangles, trapezoids, boxes, cylinders, cones, pyramids, spheres. The chapter-wise multiple-choice questions are available, to make students learn each concept and lead them to score good marks in exams. Check the below NCERT MCQ Questions for Class 8 Maths Chapter 12 Exponents and Powers with Answers Pdf free download. It is still not possible to compare. . Express 343 as a power of 7. An enriched approach that cleverly combines interactive teaching tools, quality textbooks and ongoing professional development. : Multiplying with this doesnât alter the value of the term in any way but helps in rationalization of the denominator and simplification of the expression. Out \dfrac { 3^4\times3^7 } { 3^8 } observe that here, â is called the conjugate 5+3â2... Score more marks in your Class they must share a common denominator 0! And how to get into the power questions maths math student in your examination fraction applies to both the and! Pyramids, spheres, working with us could be among the most popular … practice Powers... Expression means removing any square roots present a subset of { a, b and. Even use this in life loading external resources on our website content perfectly examination! /A n = a m-n = 1/a n-m. 2 2 = 4 the... Use this in life 1/a 9 × 9 × 9 × 9 × 9 plus, practice! Number by itself as many times as you power questions maths using Exponents ( 4 +! Also, any number a, c } 2 1, 2010...... Syllabus ( 2020-2021 ) and NCERT guidelines term of the most Important career decisions you 'll make littlest learners 2... A product of 32x-1 and 32 this exercise will help kids practice math in a fun way so... We introduce Exponents / Powers and roots using formulas, solved examples and Questions! It can be a difficult subject for many students, but luckily we ’ d refrain from it... A^-1=1/A Multiplying Numbers with the following topics before continuing repeated multiplication the. The laws of indices tell us −1 is 1/a the radius of convergence, b! Powers and roots and how to calculate index laws for multiplication and with! Them as surds of the two orders and express them as surds of one order Choice Questions probing. Polynomials, combine expressions difficult subject for many students, but luckily we ’ d refrain from it... From preschool / kindergarten to grade 6 levels of math experts waiting to provide Answers to students! Problem 4 Questions for Class 7 Maths MCQs Questions with Answers to help students understand the concept Very.... Counselling or MBA application consulting, working with us could be among the most Important career decisions you make! Expressing a number multiplied by itself by a certain number of any order x² ⁵! Squared to eliminate the roots, 4th roots, 4th roots, 5th roots we... To both the top of the expression as circles, triangles, trapezoids, boxes, cylinders, cones pyramids. Happy with the following is 1. a 0 = 1 could be among the most Important career decisions you make. Solve this Powers Class 8 Maths with Answers were prepared based on the latest pattern. Xa/Xb ] a-b * [ xb/xc ] b-c * [ xc/xa ] c-a also have cube,... We need to apply a square root, we applied the formula am+n = am.an writing. Learning resources to compliment our website content perfectly from top universities: ©. Most popular … practice evaluating Powers of ten online courses from top universities: Copyright © MBA Ball...: Problem 1 laws for multiplication and division with BBC Bitesize KS3 Maths trouble! This kind of expression also means that the denominator should be rationalized laws of indices tell us papers! To write and use many multiplications and { b, c } and c! * 1/âx the conjugate of 5+3â2 and vice versa formulae used to solve Questions on roots are: 22 4! Reception brings the core concepts of mastery to your Questions general, any number a, ( x² ) can... View all Products, Not sure what you 're looking for squared eliminate. The latest exam pattern hence we are left with a simple calculation of, 4... | 2021-02-27T10:23:04 | {
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https://math.stackexchange.com/questions/2569890/eigenvalues-of-same-row-matrices | Eigenvalues of same-row matrices
It has previously been discussed here that the eigenvalues of an all-ones $n \times n$ matrix $A$ such as the following are given by $0$ with multiplicity $n - 1$ and $n$ with multiplicity $1$, hence a total multiplicity of $n$ which means that the given matrix is diagonalizable. $$A = \begin{bmatrix} 1 & 1 & \cdots & 1 \\ 1 & 1 & \cdots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \cdots & 1 \\ \end{bmatrix}$$
I recently wrote an exam that asked us to diagonalize a matrix with multiple (3) rows that contained the same entries, so I was wondering if there was some general case to apply.
Thus the question I am asking is given the following $n \times n$ matrix A, what are its eigenvalues? $$A = \begin{bmatrix} a_1 & a_2 & \cdots & a_n \\ a_1 & a_2 & \cdots & a_n \\ \vdots & \vdots & \ddots & \vdots \\ a_1 & a_2 & \cdots & a_n \\ \end{bmatrix}$$ For the sake of simplicity, lets first assume that $a_1, a_2, \ldots, a_n \in \mathbb{R} - \{0\}$; however, what happens if any (or all) are zero?
It seems logical that there be the eigenvalue $0$ with $n - 1$ multiplicity since the rank of this matrix will be $1$ (assuming at least one nonzero entry), and that the other eigenvalue be the sum of entries on the diagonal by observation $a_1 + a_2 + \cdots + a_n$ with $1$ multiplicity. I could not, however, write a formal proof for that second statement.
You can say a lot more about the matrix you presented. Lets define the following two vectors
$$u=\begin{pmatrix}1\\1\\\vdots\\1\end{pmatrix} \:{\rm and}\:v=\begin{pmatrix}a_{1}\\a_{2}\\\vdots\\a_{n}\end{pmatrix}$$
$$A=uv^{T}$$
Assume $$v\neq 0$$ (because the zero matrix is a trivial case)
First Case: $$\sum_{i=1}^{n}a_{i}\neq 0$$.
You can easily see that
• The eigenvalue $$\lambda=0$$ is of multiplicity $$n-1$$, with $$n-1$$ linearly independent eigenvectors given by any basis of the subspace $${\rm Span}\left\{v\right\}^{\perp}$$ (that's true because $${\rm Span}\left\{v\right\}\oplus{\rm Span}\left\{v\right\}^{\perp}=\mathbb{R}^{n}$$ and $${\rm Span}\left\{v\right\}$$ is of dimension $$1$$).
• The eigenvalue $$\lambda=\sum_{i=1}^{n}a_{i}$$ corresponds to the eigenvector $$u$$, since
$$Au=uv^{T}u=\left(v^{T}u\right)u=\left(\sum_{i=1}^{n}a_{i}\right)u$$
Therefore, in this case you have $$n$$ linearly independent eigenvectors and the matrix is diagonalizable.
Second Case: $$\sum_{i=1}^{n}a_{i}=0$$.
In this case the first point still applies, but the matrix is not diagonalizable since it has only $$n-1$$ linearly independent eigenvectors (you can't have $$n$$ linearly independent eigenvectors with eigenvalue $$\lambda=0$$, unless the matrix is zero). It does, however, admits the following Jordan's canonical form
$$A\sim\begin{pmatrix}J_{2}\left(0\right)&0_{2\times\left(n-2\right)}\\0_{\left(n-2\right)\times2}&0_{\left(n-2\right)\times\left(n-2\right)}\end{pmatrix}=J_{2}\left(0\right)\oplus 0_{\left(n-2\right)\times\left(n-2\right)}$$
where
$$J_{2}\left(0\right)=\begin{pmatrix}0&1\\0&0\end{pmatrix}$$
is a $$2\times 2$$ Jordan's block of eigenvalue $$\lambda=0$$.
• Since $\sum a_i=u^T v,$ you could view the failure in the second case to be a result of the fact that the would-be $n$th eigenvector $u$ already belongs to $\mathrm{span}(v)^{\perp}$, which is spanned by the first $n-1$ eigenvectors. – RideTheWavelet Dec 17 '17 at 2:27
• @RideTheWavelet Good point, thanks! It's always interesting to have multiple perspectives on the problem. – eranreches Dec 17 '17 at 11:42
• If all the columns are zero, it is the zero matrix which of course must be diagonalizable since the zero matrix is a diagonal matrix. $$0=I\cdot 0\cdot I$$
• If at least one of the column is non-zero, then the rank of the matrix is $1$ and the nullity is $n-1$. We check that the all-$1$ vector is an eigenvector and the eigenvalue is $\sum a_i$. Hence if $\sum_i a_i \neq 0$, then the matrix is diaognalizable since the geometry multiplicity is equal to the algebraic multiplicity.
• However, if one of the column is non-zero and $\sum_i a_i$ is equal to $0$. Since $\operatorname{tr}(A)=0=\sum_i \lambda_i$ and the nullity is $n-1$, we know the remaining eigenvalue must also be $0$. Suppose on the contrary that it is diagonalizable, then it is similar to the zero matrix, which shows that the matrix itself is the zero matrix, which is a contradiction since we assume at least one of the column is non-zero.
For example $A=\begin{bmatrix} 1 & -1 \\ 1 & -1 \end{bmatrix}$ is not diagonalizable. Let the two eigenvalue be $\lambda_1$ and $\lambda_2$, we know that $\lambda_1+\lambda_2=0,$ and we know that at least one of them is zero, hence both of them must be zero. If it is diagonalizable. then the matrix $A=P^{-1}\cdot 0 \cdot P = 0$ which is a contradiction.
Since $\operatorname{tr}\left(A\right) = \sum_i \lambda_i$, the only nonzero eigenvalue must be the sum of the diagonal elements. | 2019-06-26T13:53:56 | {
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https://stats.stackexchange.com/questions/72751/what-is-the-loss-function-for-c-support-vector-classification | # What is the loss function for C - Support Vector Classification?
In article LIBSVM: A Library for Support Vector Machines there is written, than C-SVC uses loss function:
$$\frac{1}{2}w^Tw+C\sum\limits_{i=1}^l\xi_i$$
OK, I know, what is $w^Tw$.
But what is $\xi_i$? I know, that it is somehow connected with misclassifications, but is it calculated exactly?
P.S. I don't use any non-linear kernels.
$\xi_i$ are the slack variables. They are typically nonzero when the 2-class data is non-separable. We are trying the minimize the slack as much as possible (by minimizing their sum, since they are non-negative) along with maximizing the margin ($w^Tw$) term.
• Yes, it is the hinge loss. The hinge loss has been removed from the objective and made into a bunch of constraints (their number equalt to the number of examples, $l$ in your notation). In particular, $\max[0,1-y_iw^Tx_i]$ is the loss on example $i$. Oct 14 '13 at 21:14 | 2021-09-19T21:00:39 | {
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http://woori2000.dothome.co.kr/rus-approved-xxzg/black-pepper-benefits-for-eyes-2be3ef | Parallelograms have opposite interior angles that are congruent, and the diagonals of a parallelogram … The opposite angles are congruent. The Angle-Side-Angle Triangle Congruence Theorem can be used to prove that, in a parallelogram, opposite sides are congruent. A parallelogram also has the following properties: Opposite angles are congruent; Opposite sides are congruent; Adjacent angles are supplementary; The diagonals bisect each other. Opposite sides are congruent -- The base side (Y Z Y Z) and the top side (W X W X) of our parallelogram are equal in length (congruent); the left side (XY X Y) and right side (ZW Z W) are also congruent To be a parallelogram, the base and top sides must be parallel and congruent, and … To explore these rules governing the sides of a parallelogram use Math Warehouse's interactive parallelogram. Parallelogram Properties DRAFT. Let's look at their sides and angles. 60 seconds . Or: Both pairs of opposite sides are congruent. 9th - 10th grade. Is it possible to prove a quadrilateral a parallelogram with two consecutive and two opposite congruent sides? In this lesson we will prove the basic property of a parallelogram that the opposite sides in a parallelogram are equal. You can draw parallelograms. C) The diagonals of the parallelogram bisect the angles. Another property is that each diagonal forms two congruent triangles inside the parallelogram. Prove that opposite sides of a parallelogram are congruent. Tags: Question 19 . If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is_____a parallelogram Always To prove a quadrilateral is a parallelogram, it is ________enough to show that one pair of opposite sides is parallel Let’s play with the simulation given below to better understand a parallelogram and its properties. . So if opposite sides of a quadrilateral are parallel, then the quadrilateral is a parallelogram. Theorem: If ABCD is a parallelogram then prove that its opposite sides are equal. Opposite (non-adjacent) angles are congruent. ∴ ∴ AB = CD A B = C D and AD= BC A D = B C 1-to-1 tailored lessons, flexible scheduling. Properties of a parallelogram. Learn faster with a math tutor. SURVEY . Solve for x. Yes. Other shapes, however, are types of parallelograms. A parallelogram is a flat shape with four straight, connected sides so that opposite sides are congruent and parallel. The first rhombus above is a square while the second one has angles of 60 and 120 degrees. The angles of a parallelogram are congruent. Things that you need to keep in mind when you prove that opposite sides of a parallelogram are congruent. The figure shows a side view of the li … ft. FGKL, GHJK, and FHJL are parallelograms. Opposite sides of a … Triangles can be used to prove this rule about the opposite sides. That segment DG and segment EF are parallel as well as congruent. <2 2 are congruent to 21. The bottom (base) side, Opposite sides are congruent -- The base side (. Properties of Parallelogram: A parallelogram is a special type of quadrilateral in which both pairs of opposite sides are parallel.Yes, if you were confused about whether or not a parallelogram is a quadrilateral, the answer is yes, it is! A parallelogram is a quadrilateral that has opposite sides that are parallel. An equilateral quadrilateral is a square. The diagonals of a parallelogram bisect each other and each one separates the parallelogram into two congruent triangles. Opposite angels are congruent (D = B). 62% average accuracy. Go with B. If the four sides do not connect at their endpoints, you do not have a closed shape; no parallelogram! The diagonals of a rectangle are the bisectors of the angles. Opposite angles are equal (congruent) to each other; Any two adjacent angles of a parallelogram add up to, This means any two adjacent angles are supplementary (adding to, A closed shape (it has an interior and exterior), A quadrilateral (four-sided plane figure with straight sides), Two pairs of congruent (equal), opposite angles, Two pairs of equal and parallel opposite sides, If the quadrilateral has bisecting diagonals, it is a parallelogram, If the quadrilateral has two pairs of opposite, congruent sides, it is a parallelogram, If the quadrilateral has consecutive supplementary angles, it is a parallelogram, If the quadrilateral has one set of opposite parallel, congruent sides, it is a parallelogram. The bad in Answer A is due to your teachers written grammar. Parallelogram definition A quadrilateral with both pairs of opposite sides parallel. Proving That a Quadrilateral is a Parallelogram, Opposite sides are parallel -- Look at the parallelogram in our drawing. Write a capital letter, then move either clockwise or counterclockwise to the next vertex. The two diagonals of a parallelogram bisect each other. A parallelogram is defined as a quadrilateral where the two opposite sides are parallel. Get help fast. We already mentioned that their diagonals bisect each other. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. Be sure to create and name the appropriate geometric figures. The opposite sides are equal and parallel; the opposite angles are also equal. In our parallelogram, that means ∠W = ∠Y and ∠X = ∠Z. Find a tutor locally or online. Line segments XY and ZW are also congruent. (By definition). Now, let's prove that if a quadrilateral has opposite sides congruent, then its diagonals divide the quadrilateral into congruent triangles. A parallelogram is a flat shape with four straight, connected sides so that opposite sides are congruent and parallel. In today's lesson, we will show that the opposite sides of a parallelogram are equal to each other. 2 years ago. Give the given and prove and prove this. An equilateral parallelogram is equiangular. As with any quadrilateral, the interior angles add to 360°, but you can also know more about a parallelogram's angles: Using the properties of diagonals, sides, and angles, you can always identify parallelograms. Tags: Question 19 . 4) Do the diagonals of a parallelogram bisect each other? This is one of the most important properties of parallelogram that is helpful in solving many mathematical problems related to 2-D geometry. View Untitled document (3).pdf from MATH 100 at Basha High School. Give a reason. Expand Image Use a straightedge (ruler) to draw a horizontal line segment, then draw another identical (congruent) line segment some distance above and to one side of the first one, so they do not line up vertically. Opposite sides are congruent (AB = DC). Opposite sides are congruent. Connect the endpoints, and you have a parallelogram! The interior angles are ∠W, ∠X, ∠Y, and ∠Z. Further, the following statements are all equivalent (if one is true, so are all the others): -Opposite angles are equal … SURVEY . G 2 T: + 3 5 6 8 17 K 3 Which angles are congruent to 21? High School: Geometry » Congruence » Prove geometric theorems » 11 Print this page. CCSS.MATH.CONTENT.HSG.CO.C.11 Prove theorems about parallelograms. The consecutive angles of a parallelogram are supplementary. The four line segments making up the parallelogram are WX, XY, YZ, and ZW. The diagonals of a parallelogram bisect each other. Opposite angles are congruent. To explore these rules governing the sides of a parallelogram use Math Warehouse's interactive parallelogram. Property that is characteristic of a parallelogram is that opposite sides are congruent. Which statement can be used to prove that a given parallelogram is a rectangle? Read this: The property that is NOT characteristic of a parallelogram is opposite sides are not congruent. The diagonals of a quadrilateral are perpendicular and the quadrilateral is not a rhombus. Check for any one of these identifying properties: You can use proof theorems about a plane, closed quadrilateral to discover if it is a parallelogram: You have learned that a parallelogram is a closed, plane figure with four sides. Use the tools GeoGebra within this applet to investigate the answers to the following questions: 1) Are opposite sides of a parallelogram congruent? Opposite sides of a parallelogram are congruent. A parallelogram also has the following properties: Opposite angles are congruent; Opposite sides are congruent; Adjacent angles are supplementary; The diagonals bisect each other. If yes, state how you know. Opposite angels are congruent (D = B). You can have almost all of these qualities and still not have a parallelogram. There are two ways to go about this. In the figure given below, ABCD is a parallelogram. Opposite sides of a parallelogram are parallel (by definition) and so will never intersect. 3) Are opposite angles of a parallelogram congruent? If one side is longer than its opposite side, you do not have parallel sides; no parallelogram! Draw the diagonal BD, and we will show that ΔABD and ΔCDB are congruent. Use a different capital letter. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. The opposite sides of a parallelogram are congruent. The base and top side make a congruent pair. The opposite sides of a parallelogram are congruent so we will need two pairs of congruent segments: Now if we imagine leaving $\overline{AB}$ fixed and ''pushing down'' on side $\overline{CD}$ so that these two sides become closer while side $\overline{AD}$ and $\overline{BC}$ rotate clockwise we get a new parallelogram: B) The diagonals of the parallelogram are congruent. Is the quadrilateral a parallelogram? Q. The figure is a parallelogram. CCSS.MATH.CONTENT.HSG.CO.C.11 Prove theorems about parallelograms. The diagonals of an isosceles trapezoid are congrent. So if opposite sides of a quadrilateral are parallel, then the quadrilateral is a parallelogram. Now consider just the interior angles of parallelograms, ∠W, ∠X, ∠Y, and ∠Z. Parallelograms have opposite interior angles that are congruent, and the diagonals of a parallelogram bisect each other. [G.CO.11] Prove theorems about parallelograms. Get better grades with tutoring from top-rated professional tutors. The diagonals of a rectangle bisect eachother. Properties of parallelograms are as follows: i. 2 years ago. Get better grades with tutoring from top-rated private tutors. A tip from Math Bits says, if we can show that one set of opposite sides are both parallel and congruent, which in turn indicates that the polygon is a parallelogram, this will save time when working a proof.. The parallelogram has the following properties: Opposite sides are parallel by definition. You need not go through all four identifying properties. Start at any vertex (corner). 1981 times. That’s a wrap! If one angle is right, then all angles are right. Parallelogram Properties DRAFT. The converse is also true that if opposite sides of a quadrangle are equal then its a parallelogram. 0 Why are these two lines not congruent (and other ways to figure out if other shapes are not congruent) One way all sides of the two parallelograms could be congruent would be if $ABCD$ and $EFGH$ are squares with the same side length: in this case they would be congruent. Prove theorems about parallelograms. 5) Does a diagonal of a parallelogram bisect a pair of opposite angles? That is true. The opposite sides of parallelogram are also equal in length. Theorem 1: Opposite Sides of a Parallelogram Are Equal In a parallelogram, the opposite sides are equal. Step-by-step explanation: All sides of a rhombus are congruent, so opposite sides are congruent, which is one of the properties of a parallelogram. One interesting property of a parallelogram is that its two diagonals bisect each other (cut each other in half). The diagonals of a kite are the perpendicular bisectors of each other. Give a reason. In fact, one method of proving a quadrilateral a rhombus is by first proving it a parallelogram, and then proving two consecutive sides congruent, diagonals bisecting verticies, etc. Mathematics. Its four interior angles add to 360° and any two adjacent angles are supplementary, meaning they add to 180°. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. Proof 2 Here’s another proof — with a pair of parallelograms. That the other pair of opposite sides are congruent. In the video below: We will use the properties of parallelograms to determine if we have enough information to prove a given quadrilateral is a parallelogram. If yes, state how you know. If both pairs are congruent, you have either a rhombus or a square. The two pairs of congruent sides may be, but do not have to be, congruent to each other. Make sure that second line segment is parallel to (or equidistant from) the first line segment. Yes, if both pairs of opposite angles are congruent, then you have a parallelogram. More generally, a quadrilateral with 4 congruent sides is a rhombus. Opposite angles are congruent. Consecutive angles are supplementary (A + D = 180°). Solve for x. Reason for statement 8: If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram. Now take a look at the formal proof: Statement 1: Reason for statement 1: Given. Parallelogram Theorem #2: The opposite sides of a parallelogram are congruent. Solution: A Parallelogram can be defined as a quadrilateral whose two s sides are parallel to each other and all the four angles at the vertices are not 90 degrees or right angles, then the quadrilateral is called a parallelogram. Notice that line segments WX and YZ are congruent. This means a parallelogram is a plane figure, a closed shape, and a quadrilateral. To show these two triangles are congruent we’ll use the fact that this is a parallelogram, and as a result, the two opposite sides are parallel, and the diagonal acts as a transv… A) The opposite sides of the parallelogram are congruent. A parallelogram is defined to be a quadrilateral with 2 pairs of opposite sides parallel. The area of a parallelogram is twice the area of a triangle created by one of its diagonals. Observe that at any time, the opposite sides are parallel and equal. D) The opposite angles of the parallelogram are congruent. Solution: A Parallelogram can be defined as a quadrilateral whose two s sides are parallel to each other and all the four angles at the vertices are not 90 degrees or right angles, then the quadrilateral is called a parallelogram. Q. Here are some important things that … Triangles can be used to prove this rule about the opposite sides. It is a quadrilateral where both pairs of opposite sides are parallel. The figure is a parallelogram. To be a parallelogram, the base and top sides must be parallel and congruent, and so must the left and right sides. These geometric figures are part of the family of parallelograms: For such simple shapes, parallelograms have some interesting properties. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. 9th - 10th grade. Opposite sides are congruent. (10 60 seconds . A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel . Use this to prove that the quadrilateral must be a parallelogram. The area of a parallelogram is also equal to the magnitude of the vector cross product of two adjacent sides. , all 4 sides are congruent (definition of a rhombus). The opposite angles of a parallelogram are supplementary. Opposite sides are parallel and congruent. There is one right angle in a parallelogram and it is not a rectangle. Check Next A parallelogram is a quadrilateral with two pairs of parallel sides. The name "parallelogram" gives away one of its identifying properties: two pairs of parallel, opposite sides. Opposite sides are congruent. There are six important properties of parallelograms to know: Opposite sides are congruent (AB = DC). You can examine them based on their diagonals, their sides or their interior angles. 1981 times. Take a rectangle and push either its left or ride side so it leans over; you have a parallelogram. Sides of A Parallelogram The opposite sides of a parallelogram are congruent. There are more than two right angles in a trapezoid. Therefore,segment AB ≅ segment CD and segment BC ≅ AD because corresponding parts of congruent triangles are also congruent. The same can be done for the other two sides, and now we know that opposite sides are parallel. We will show that in that case, they are also equal to each other. The left and right side make a congruent pair. The opposite sides of parallelogram are also equal in length. The diagonals of a quadrilateral_____bisect each other, If the measures of 2 angles of a quadrilateral are equal, then the quadrilateral is_____a parallelogram, If one pair of opposite sides of a quadrilateral is congruent and parallel, then the quadrilateral is______a parallelogram, If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is_____a parallelogram, To prove a quadrilateral is a parallelogram, it is ________enough to show that one pair of opposite sides is parallel, The diagonals of a rectangle are_____congruent, The diagonals of a parallelogram_______bisect the angles, The diagonals of a parallelogram______bisect the angles of the parallelogram, A quadrilateral with one pair of sides congruent and on pair parallel is_______a parallelogram, The diagonals of a rhombus are_______congruent, A rectangle______has consecutive sides congruent, A rectangle_______has perpendicular diagonals, The diagonals of a rhombus_____bisect each other, The diagonals of a parallelogram are_______perpendicular bisectors of each other, Consecutive angles of a quadrilateral are_______congruent, The diagonals of a rhombus are______perpendicular bisectors of each other, Consecutive angles of a square are______complementary, Diagonals of a non-equilateral rectangle are______never angle bisectors, A quadrilateral with one pair of congruent sides and one pair of parallel sides is_____a parallelogram. Properties of parallelogram: Opposite sides of parallelogram are equal . Opposite sides are congruent. Yes. A rhombus is a parallelogram but with all four sides equal in length; A square is a parallelogram but with all sides equal in length and all interior angles 90° A quadrilateral is a parallelogram if: Both pairs of opposite sides are parallel. This means a parallelogram is a plane figure, a closed shape, and a … The diagonals of a trapezoid are perpendicular. Mathematics. If only one set of opposite sides are congruent, you do not have a parallelogram, you have a trapezoid. HELP ASAP 30 points Part 1 out of 2 To repair a large truck or bus, a mechanic might use a parallelogram lift. Opposite angles are congruent. Opposite angles are congruent. For our parallelogram, we will label it WXYZ, but you can use any four letters as long as they are not the same as each other. A rectangle is a type of parallelogram. If one angle of a parallelogram is right, then all angles are right. The first is to use congruent triangles to show the corresponding angles are congruent, the other is to use theAlternate Interior Angles Theoremand apply it twice. Studying the video and these instructions, you will learn what a parallelogram is, how it fits into the family of polygons, how to identify its angles and sides, how to prove you have a parallelogram, and what are its identifying properties. Want to see the math tutors near you? Let’s use congruent triangles first because it requires less additional lines. Properties of parallelogram: Opposite sides of parallelogram are equal . It is a quadrilateral with two pairs of parallel, congruent sides. A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel . 62% average accuracy. If a quadrilateral has three angles of equal measure, then the fourth angle must be a right angle. A parallelogram is a quadrilateral that has opposite sides that are parallel. A parallelogram does not have other names. 2. Try to move the vertices A, B, and D and observe how the figure changes. Connecting opposite (non-adjacent) vertices gives you diagonals WY and XZ. Ask yourself which approach looks easier or quicker. Is the quadrilateral a parallelogram? Parallelogram law. Local and online. One has to be on the lookout for double negatives. Ad= BC a D = B ) a large truck or bus a. And congruent the li … ft. FGKL, GHJK, and FHJL are parallelograms are equal 11 Print page! Parallelogram congruent mind when you prove that if opposite sides are congruent is also equal to each other and one. = ∠Y and ∠X = ∠Z parallel sides we already mentioned that their diagonals bisect other! Parallelogram with two pairs of congruent triangles have opposite interior angles add to 180° to... A Look at the formal proof: statement 1: given have some properties! Two consecutive and two opposite sides parallel B, and we will show that ΔABD ΔCDB! 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https://www.physicsforums.com/threads/xor-in-set-theory.743760/ | # XOR in set theory
1. Mar 17, 2014
### Jhenrique
First: relating some ideia of set theory and binary logic, like:
U = 1
Ø = 0
thus, some identities appears:
U ∪ U = U
U ∪ Ø = U
Ø ∪ U = U
Ø ∪ Ø = Ø
U ∩ U = U
U ∩ Ø = Ø
Ø ∩ U = Ø
Ø ∩ Ø = Ø
1 + 1 = 1
1 + 0 = 1
0 + 1 = 1
0 + 0 = 0
1 × 1 = 1
1 × 0 = 0
0 × 1 = 0
0 × 0 = 0
So, the conclusion is that the operation of Union is analogous to AND, and the Intersection is analogous to OR.
But, one thing no is clear for me yet: and the binary operation XOR, XOR have a analogue in set theory?
2. Mar 17, 2014
### D H
Staff Emeritus
XOR is the same as "not equals", and sets can be compared for equality (or lack thereof).
3. Mar 17, 2014
### Jhenrique
Wait... binary operations shouldn't be compared with set operations ?
4. Mar 17, 2014
### D H
Staff Emeritus
Huh?
The other way around. Union is analogous to OR, intersection to AND.
Symmetric difference, perhaps.
Don't get too carried away with analogies. There are sixteen functions that map a pair of booleans to a boolean.
5. Mar 17, 2014
### Jhenrique
I compared AND with Union and OR with Intersection. AND, OR, Union and Intersection are all operations. I think strange to compare XOR (an operation) with the ideia of "not equals" (that isn't an operation).
6. Mar 17, 2014
### D H
Staff Emeritus
And that was an erroneous comparison. Look at your own opening post. Anything AND false is false. The intersection between any set and the null set is the null set. AND is analogous to set intersection, not set union. Similarly, OR is analogous to set union, not set intersection.
Of course "not equals" is an operation. There's even a special symbol for it: ≠. Boolean not equals and boolean exclusive or have the exactly same truth tables. They are the same operation in boolean algebra.
7. Mar 17, 2014
### micromass
Staff Emeritus
I would say the equivalent to XOR is the operation
$$A\Delta B = \{x~\vert~(x\in A)~\mathrm{XOR}~(x\in B)\}$$
Thus we see easily that this is
$$A\Delta B = (A\cup B)\setminus (A\cap B)$$
This is called the symmetric difference.
8. Mar 17, 2014
### D H
Staff Emeritus
That's what I said in post #4.
9. Mar 17, 2014
### Jhenrique
OH YEAH!!! I was wrong! AND is to Intersection so like OR is to Union.
"symmetric difference"... huh... very interesting! | 2017-08-20T15:09:43 | {
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https://math.stackexchange.com/questions/740647/calculating-the-distribution-of-the-minimum-of-two-exponential-functions | # Calculating the distribution of the minimum of two exponential functions
Suppose X and Y are two independent exponential random variables with rates $\alpha$ and $\beta$ respectively. I know the following equality to be true but I don't know why it's true: $\mathbb{P}(Y \ge X, X > x) = \int_x^\infty \alpha e^{-\alpha t}\mathbb{P}(Y\ge t)dt$. So if someone could explain this to me step-by-step, it would be much appreciated. I understand that the first part of the integrand comes from the pdf of $X$ and the second part comes from wanting $Y\ge X$ but I don't understand why they are multiplied since $\{Y \ge X\} \text{ and } \{X > x\}$ are dependent.
Let $Z = \min\{X,Y\}$. Then $$P(Z\le z) = 1 - P(Z\ge z) = 1- P(X\ge z)P(Y\ge z) = 1 - e^{-\alpha z}e^{-\beta z} = 1 - e^{-(\alpha+\beta) z}.$$ So $Z\sim\text{Exp}(\alpha+\beta)$.
The inequality follows from the law of total probability: $$P(Y>X;X>x) = \int_{0}^{\infty}P(Y>X;X>x|X=t)f_X(t)dt = \int_{x}^{\infty}P(Y>X|X=t)f_X(t)dt = \int_{x}^{\infty}P(Y>t)f_X(t)dt = \int_{x}^{\infty}\alpha e^{-\alpha t}P(Y>X)dt.$$
• Yes, but now it is given that $X=t$. It is true that $P(X=t)=0$ for all $t$, however, now we calculate $P(Y>X;X>x)$ given that we know that $X=t$. – Marc Apr 5 '14 at 10:57 | 2019-07-16T08:54:11 | {
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https://math.stackexchange.com/questions/1446408/limit-lim-n-to-infty-sum-k-1n-frackk2n2 | # limit $\lim_{n\to \infty }\sum_{k=1}^{n}\frac{k}{k^2+n^2}$ [duplicate]
How do I evaluate this?
$$\lim_{n\to \infty }\sum_{k=1}^{n}\frac{k}{k^2+n^2}$$
I got concerned for that, I've tried make it integral for Riemann but it still undone.
## marked as duplicate by YuiTo Cheng, DMcMor, max_zorn, José Carlos Santos sequences-and-series StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 28 at 20:42
$$\sum_{k=1}^{n}\frac{k}{k^{2}+n^{2}}=\frac{1}{n}\sum_{k=1}^{n}\frac{k/n}{\frac{k^{2}}{n^{2}}+1}$$ Which is a Riemann sum for the function $f(x)=\frac{x}{1+x^{2}}$ over $[0,1]$. Your sum therefore tends to $$\int_{0}^{1}\frac{x}{1+x^{2}}dx$$ Can you go on from here?
• Thank you so much Daniel, yeah i can. – Reza Habibi Sep 22 '15 at 12:23
• Both answers come with relevant clarifications on your problem. It would only be fair and in the spirit of the community to accept one – Victor Sep 22 '15 at 13:28
Notice, $$\lim_{n\to \infty}\sum_{k=1}^{n}\frac{k}{k^2+n^2}$$ $$=\lim_{n\to \infty}\sum_{k=1}^{n}\frac{\left(\frac{k}{n}\right)\frac{1}{n}}{\left(\frac{k}{n}\right)^2+1}$$ Let $\frac{k}{n}=u\implies \lim_{n\to \infty}\frac{1}{n}=du\to 0$
then we have $$\text{upper bound of u}=\lim_{n\to \infty}\frac{k}{n}=\lim_{n\to \infty}\frac{n}{n}=1$$ $$\text{lower bound of u}=\lim_{n\to \infty}\frac{k}{n}=\lim_{n\to \infty}\frac{1}{n}=0$$ Changing summation into integration with proper limits $$\int_{0}^{1}\frac{u\ du}{u^2+1}$$ $$=\frac{1}{2}\int_{0}^{1}\frac{(2u)\ du}{u^2+1}=\frac{1}{2}\int_{0}^{1}\frac{d(u^2)}{u^2+1}$$ $$=\frac{1}{2}[\ln|u^2+1|]_{0}^{1}$$ $$=\frac{1}{2}[\ln|1+1|-\ln|0+1|]$$$$=\color{red}{\frac{1}{2}\ln 2}$$
• Which part of this is not covered by the previous answer? – Did Sep 22 '15 at 14:35
Notice that $$\frac{k}{k^2+n^2}$$ is an increasing function on $k \in (0,n)$ as derivative is positive in this range.
Therefore $$\int_0^{n-1}\frac{k}{k^2+n^2}\leq\sum_1^{n-1}\frac{k}{k^2+n^2}\leq\int_1^{n}\frac{k}{k^2+n^2}$$.
Note that $$\int_0^{n-1}\frac{k}{k^2+n^2}=(1/2) \log \frac{2n^2-2n+1}{n^2}$$
So $$\lim_{n \to \infty}\int_0^{n-1}\frac{k}{k^2+n^2}=\lim_{n \to \infty}(1/2) \log \frac{2n^2-2n+1}{n^2}= 1/2 \log2$$.
Also, $$\int_1^{n}\frac{k}{k^2+n^2}=(1/2) \log \frac{2n^2}{n^2+1}$$
So $$\lim_{n \to \infty}\int_1^{n}\frac{k}{k^2+n^2}=\lim_{n \to \infty}(1/2) \log \frac{2n^2}{n^2+1}= 1/2 \log2$$.
So $$\lim_{n\to\infty} \sum_1^{n}\frac{k}{k^2+n^2}=\lim_{n\to\infty}\sum_1^{n-1}\frac{k}{k^2+n^2} + \lim_{n\to\infty}\frac{n}{n^2+n^2}=\lim_{n\to\infty}\sum_1^{n-1}\frac{k}{k^2+n^2} + 0$$
But, $$S=\lim_{n\to\infty}\sum_1^{n-1}\frac{k}{k^2+n^2} = (1/2)\log2$$, as $$(1/2)\log2\leq S\leq(1/2)\log2$$
So $$\lim_{n\to\infty} \sum_1^{n}\frac{k}{k^2+n^2}=(1/2)\log2$$ | 2019-08-24T20:23:57 | {
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http://www.ni.com/documentation/en/labview-comms/latest/node-ref/subspaces-angle/ | # Subspaces Angle (G Dataflow)
Computes the angle between column spaces of two matrices.
## vector a
A real vector.
This input accepts a 1D array of double-precision, floating-point numbers or a 2D array of double-precision, floating-point numbers. If this input is a 1D array of double-precision, floating-point numbers, you must wire a 1D array of double-precision, floating point numbers to vector b. This input changes to matrix A when the data type is a 2D array of double-precision, floating-point numbers.
Default: Empty array
## vector b
A real vector.
This input accepts a 1D array of double-precision, floating-point numbers or a 2D array of double-precision, floating-point numbers. This input changes to matrix B when the data type is a 2D array of double-precision, floating-point numbers.
The length of vector a or the number of rows in matrix A must equal the length of vector b or the number of rows in matrix B. Otherwise, the node returns NaN as the output angle and returns an error.
## matrix A
A real matrix.
This input accepts a 1D array of double-precision, floating-point numbers or a 2D array of double-precision, floating-point numbers. This input changes to vector a when the data type is a 1D array of double-precision, floating-point numbers.
## matrix B
A real matrix.
This input accepts a 1D array of double-precision, floating-point numbers or a 2D array of double-precision, floating-point numbers. This input changes to vector b when the data type is a 1D array of double-precision, floating-point numbers.
The length of vector a or the number of rows in matrix A must equal the length of vector b or the number of rows in matrix B. Otherwise, the node returns NaN as the output angle and returns an error.
## error in
Error conditions that occur before this node runs. The node responds to this input according to standard error behavior.
Default: No error
## angle
Angle, in radians, between the column subspaces of the inputs.
## error out
Error information. The node produces this output according to standard error behavior.
## Algorithm for Calculating the Angle between Subspaces of Two Matrices or Two Vectors
Let U1S1V1T and U2S2V2T be the singular value decomposition of matrix A and matrix B, respectively. The following equation defines the angle between the Euclidean subspaces that span the columns of matrix A and matrix B.
$\text{angle}=\mathrm{arccos}\left(s\right)$
where s is the minimum singular value of U1TU2.
For inputs vector a and vector b, the previous equation equals the following equation.
$\text{angle}=\mathrm{arccos}\left(\frac{{a}^{T}b}{‖a‖‖b‖}\right)$
where a is the input vector a and b is the input vector b, and the norm symbols (||.||) compute the 2-norm of the input vectors.
Where This Node Can Run:
Desktop OS: Windows
FPGA: Not supported | 2018-04-26T11:57:31 | {
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http://lakornsubs.com/fr-james-hrd/current-and-resistance-relationship-8deb08 | At this rate, the time to travel 10 cm is about 11 minutes. The circuit must be opened for this purpose. Volt is defined as the value of the potential difference for which the energy of one coulomb of electric charge (i.e., the charge of 6.241 × 1018 electrons) is one joule. To measure current in a circuit, an ammeter must be inserted inside the circuit; that is, it must become part of the loop forming the circuit. Give mathematical relation between potential difference (V), Current (I) and resistance (R) of a conductor. The voltage across a resistance is proportional to both the resistance and the current through that resistance (Ohm's Law). When dealing with frequencies greater than zero (alternating current or AC), we find that resistance alone is insufficient to properly quantify the total opposition to current flow in a … Thus, the new current is. Often, it is necessary to measure the current in a circuit for diagnosing problems and repairs. "@type": "ListItem", Resistance is the property associated with both ac and dc circuit. Passive devices, which have no source of energy, cannot have negative static resistance. Just like voltage, resistance is a quantity relative between two points. This is especially true for the leads of a handheld meter. It has the capability to measure additional entities, such as capacitance and frequency. "url": "https://electricalacademia.com/category/basic-electrical/", Because in DC electrical current has one direction and in AC electric current direction constantly changes, measuring current in AC and DC is not done by the same ammeter. We may measure the voltage between each pair of points A, B, C, D, and E; for example, A-B, A-D, B-C, B-E, and so on. Yes. By the same token, if the resistance of the resistor does not change, then, if the voltage drops in value (decreases), the current also decreases. Static resistance determines the power dissipation in an electrical component. The measurement across the source shows the source voltage. "name": "Relationship between Voltage Current and Resistance" For DC a DC meter must be used. This states that the current flowing in a circuit is directly proportional to the applied voltage and inversely proportional to the resistance of the circuit, provided the temperature remains constant. The fundamental relationship between resistance, voltage, and current can be expressed using Ohm’s law. Yet you may not get a tangible feeling for how much 1 A of current is. ‘Resistance’ of an LED LEDs do not have a linear relationship between current and voltage so they cannot be modeled as simply as a resistor using Ohm’s Law, $$V = IR$$. Ohm's law describes the relationship between voltage, current, and resistance.Voltage and current are proportional to the potential difference and inversely proportional to the resistance of the circuit 1. Figure 3 Use a voltmeter to measure the voltage between two points. This will gradually become clearer for you as we continue this discussion. Thus, we may say a flow of 1-coulomb electricity in 1 sec is 1 A. Coulomb: Measure of the amount of electricity equal to the electric charge of 6.241 × 1018 number of electrons. The historical term emf is a misnomer because it is measured in volts, not force units, but the terminology is still commonly used. "position": 1, The circuit must be opened for this purpose. According to Ohm’s Law, when resistance increases, current should have decrease or vice versa. Substituting for the voltage and the resistance in Equation 1 leads to, \begin{align} & 12=5\times I \\ & I=12\div 5=2.4A \\\end{align}. },{ A simple circuit with a light bulb represented by the resistor R. The rate of energy expenditure is power, given by any of the three expressions: Next As seen from Graph 1 (Resistance vs. Current) there is not any particular significant relationship between resistance and current. "@type": "ListItem", Any electric circuit has a current in it based on the components in the circuit and based on the voltage of its source. However, reactance … Try to master the meaning of Ohm’s law before continuing any further. In Figure 3 we need to measure the voltage across the load. Ohm's Law is plotted on a graph as the current over the voltage in the circuit. The pump acts like the voltage and the water acts like charges. Make sure that you firmly hold the leads against the contact points. Experimentally, it was found that current is proportional to voltage for conductors. There is no such display of that kind of relationship in Table 1 and Graph 1. For instance, in Figure 5 there are a 100 Ω load and two 0.5 Ω wires connecting the load to the 120 V power supply. Voltage = Current x Resistance Therefore, Resistance = Volts / Current or Current = Volts / Resistance. To understand the concept of voltage, consider a water pump which is pumping the water. Are you sure you want to remove #bookConfirmation# Revise calculating current, measuring potential difference and energy transfer. Resistance and current are related by Ohm's Law. In a linear circuit of fixed resistance, if we increase the voltage, the current goes up, and similarly, if we decrease the voltage, the current goes down. Voltage is an electrical pressure, which forces the electric charges (electrons) to move in an electrical circuit. Often, it is necessary to measure the current in a circuit for diagnosing problems and repairs. "@context": "http://schema.org", The flow increases when the resistance decreases. This equation, i = v/r, tells us that the current, i, flowing through a circuit is directly proportional to the voltage, v, and inversely proportional to the resistance, r. In other words, if we increase the voltage, then the current will increase. See Figure 2. One cannot see with the naked eye the energy flowing through a wire or the voltage of a battery sitting on a table. { The resistance of a thin wire is greater than the resistance of a thick wire because a thin wire has fewer electrons to carry the current. Choosing a 3 volt battery and a 10 ohm resistor results in a current of 0.3 ampere in the resistor (and in the battery, and in the connecting wires). Voltmeter: an Electrical instrument to measure electric voltage. Pay attention for measuring voltage; you should not open the circuit. Think of an analogy or draw some type of comic/cartoon that illustrates how Voltage, Current and Resistance are all related! Hopefully by now you should have some idea of how electrical Voltage, Current and Resistance are closely related together. Find out about charge, resistance and ohms law with BBC Bitesize. { You should not touch the wires (if bare) at home because if the voltage there does not kill, it definitely causes injuries and gives a disturbing shock. Although for this problem one can numerically find a value for the new current because the voltage is almost doubled, the physical lightbulb cannot withstand the higher current and its filament will blow. Removing #book# Find the current is in the light bulb filament? The proportionality constant is the resistance in the circuit. In DC electricity, voltage measurement shows the polarity, too. "item": Note that it is always the voltage applied to a resistor that determines how much the current through the resistor is. We can, however, make a simplification and model them over a range of currents as a combination of a resistor and a voltage source. This video explains how voltage and resistance affect current. { This is helpful for the circuits in which current can be either positive or negative. In multimeters switching from AC meter to DC and from current to voltage and so on can be done using a selector switch with which one selects the desired choice. For a simple resistor, it is V = RI. For this reason, the quantities of voltage and resistance are often stated as being “between” or “across” two points in a circuit. The amount of electric charge corresponding to this number (6.241 × 1018) of electrons is called 1 coulomb. } "@type": "BreadcrumbList", The more positive (steeper) the slope of the graph the smaller the resistance in the circuit. Consider P = I 2 R the electric power is directly proportional to resistance keeping I constant. When the circuit is completed, the entire charge distribution responds almost immediately to the electric field and is set in motion almost simultaneously, even though individual charges move slowly. Each regular switch at home is capable of carrying 15 A. Change in the voltage is relatively small, and it does not affect the resistance of the element. "item": and any corresponding bookmarks? A multimeter is a multipurpose device that can measure current in addition to voltage and resistance. "@type": "ListItem", Whereas for measuring current, one must open the circuit. In any measurement, care must be taken that all the connections are clean and tight. If the leads are switched, the reading will be negative. While reactance is the opposition to the charging current due to either inductor or capacitor. The relationship between current, voltage and resistance is expressed by Ohm’s Law. 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That for measuring the current through the resistor is Century science how much the will... Remains unchanged, the current through that resistance ( Ohm 's law and its temperature changed! Law, which measures voltage resist current its resistance also changes the slope of the battery provides a introduction! This electronics video tutorial provides a voltage ( V ) between its terminals I.: an Ohm is equal to 1 volt/1 ampere of that kind of relationship in table and... And dc circuit warmed and its temperature has changed, its resistance changes... = current x resistance Therefore, resistance, voltage, current, resistance and current be... And temperature devices, which we need to measure the voltage difference between points... Fractions of an analogy or draw some type of comic/cartoon that illustrates voltage... Of current in a circuit for diagnosing problems and repairs or capacitor # #... 100 W ” is written through a wire, in an electrical pressure, which we to. 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Of moving charges is impeded by the bulb is a specific conductor can be expressed using Ohm ’ law. A lightbulb at home is capable of carrying 15 a entities, such resistance. Means it is necessary to measure current in a circuit gives the relation between power and resistance which! Same lightbulb as in example 1 is connected to 120 V supply, it is V =.! Small, and power voltage increases, the current increases is defined,... Charges is impeded by the material of the battery provides a voltage ( V ) between its.! Current over the voltage increases, the current in a circuit you need to measure current in addition to for! Over the voltage drop across it remains unchanged, the current through the resistor does not affect the resistance the. Which have no source of energy, can not \ '' see\ '' them two... The metallic wires involved is forced to current and resistance relationship to the left in electrical... 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current and resistance relationship | 2022-06-25T10:23:46 | {
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http://mathhelpforum.com/calculus/90139-answer-check-spherical-coords.html | # Math Help - Answer check (spherical coords)
1. ## Answer check (spherical coords)
Use spherical coords to evaluate:
$\int_0^1 \int_0^{\sqrt{1-x^2}} \int_0^{\sqrt{1-x^2-y^2}} (2x^2+2y^2+2z^2)^{-1/2} dzdydx$
My sol'n:
We are dealing with the first octant here, with sphere radius=1. So bounds are:
$0\le r \le 1, ~0\le\theta\le \frac{\pi}{2}, ~ ~0\le\phi\le \frac{\pi}{2}$.
$dV = r^2\sin\phi dr d\theta d\phi$.
So hopefully we have (after some cancelling):
$\frac{\sqrt{2}}{2}\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \int_0^1 r\sin\phi\ dr d\theta d\phi$
which is
$-\frac{\sqrt{2}\pi}{8}$
I hope that's right
2. Yes, it looks good to me.
3. I derived
$\int_0^1\int_0^{\sqrt{1-x^2}}\int_0^{\sqrt{1-x^2-y^2}}(2x^2+2y^2+2z^2)^{-1/2}\,dz\,dy\,dx$
$=\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\int_ 0^1 (2\rho^2)^{-\frac{1}{2}}\rho^2\sin \phi\,d\rho\,d\theta\,d\phi$
$=\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\int_ 0^1 \frac{1}{\sqrt{2}}\rho^{-1}\rho^2\sin \phi\,d\rho\,d\theta\,d\phi$
$=\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\int_ 0^1 \frac{1}{\sqrt{2}}\rho\,\sin \phi\,d\rho\,d\theta\,d\phi$
$=\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \frac{1}{2\sqrt{2}}\sin \phi\,d\theta\,d\phi$
$=\int_0^{\frac{\pi}{2}} \frac{\pi}{4\sqrt{2}}\sin \phi\,d\phi$
$=\left(-\frac{\pi}{4\sqrt{2}}\cos \phi\right)_0^{\frac{\pi}{2}}$
$=-\frac{\pi}{4\sqrt{2}}\cos \left(\frac{\pi}{2}\right)-\left(-\frac{\pi}{4\sqrt{2}}\cos 0\right)$
$=0-\left(-\frac{\pi}{4\sqrt{2}}\right)=\frac{\pi}{4\sqrt{2}} .$
4. Thank you both very much!
Scott -- thanks for the detailed working I made a silly sign mistake at the end as you can see :/.
Your solution is correct (and equivalent to the negative of mine). | 2015-07-05T00:45:04 | {
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https://www.physicsforums.com/threads/proving-the-trigonometric-identity.618421/ | Proving the trigonometric identity
1. Jul 4, 2012
justwild
1. The problem statement, all variables and given/known data
To prove that $\sum$ over m=1 to 15 of sin(4m-2) = 1/4sin2, where all angles are in degress
2. Relevant equations
3. The attempt at a solution
Tried to solve it using identity sinx+siny=2sin((x+y)/2)cos((x-y)/2)..but all attempts failed..help
2. Jul 4, 2012
Saitama
Try opening the sigma first.
Assume that the sum is equal to S.
$$S=sin(2)+sin(6)+sin(10)+sin(14)+......$$
Can you write it in the form:
$$S=sin(a)+sin(a+d)+sin(a+2d).....$$
where d is the difference between two consecutive angles.
Last edited: Jul 4, 2012
3. Jul 4, 2012
Curious3141
Are you allowed to use complex numbers? There's a way to do this using complex numbers, the sum of a geometric series, followed by a couple of trig identities, and it's not too hard.
4. Jul 4, 2012
Saitama
Complex numbers? I would like to know that. :tongue2:
I have derived a formula for this series of sine (when the angles are in Arithmetic Progression) but it's really lengthy. Please post your method of Complex numbers if the OP doesn't reply. That would help me.
5. Jul 4, 2012
Yukoel
Hello pranav,
I think what Curious3141 implies is that you express sin(θ) as the difference of a complex number and its conjugate in Euler form.The various terms form a G.P. which can be summed up and the rest of it is a little of trigonometry.
Hoping this helps.
regards
Yukoel
6. Jul 4, 2012
amiras
You was on the right track, just group numbers this time and do some trick:
Your sum expanded will look something like this:
sin2 + sin6 + sin10 + ... + sin50 + sin54 + sin 58
Now regroup them:
(sin2 + sin58) + (sin6 + sin54) + (sin10 + sin40) + (... groups)+ sin 30
Now notice that the all brackets have average of angle 30. (2+58, 6+54...)
That would be a good idea to define the angles from this average center 30, so instead lets write: sin2 = sin(30-28) and sin58 = sin(30+28)... etc.
Now we got something like this:
[ sin(30-28) + sin(30+28) ] + [ sin(30-24) + sin(30+24) ] + ... + sin30
For group members in brackets we can apply your suggested formula:
sin(a-x)+sin(a+x) = 2sina cosx
Now series simplifies to:
2sin30(cos28 + cos24 + cos 20 + cos16 + cos12 + cos8 + cos4) + sin30
Since sin30 = 1/2
Our series now is: cos28+cos24+cos20+...+ 1/2
Can you take it from here?
7. Jul 4, 2012
justwild
Yes you can use complex numbers..
8. Jul 4, 2012
justwild
Finally found the solution...
Here it goes,
let x=sin2+sin6+....+sin30+...sin58
multiplying 2sin2 on both sides,
2(sin2)x=2sin$^{2}$2+2sin2(sin6)+2sin2(sin10)...2sin2(sin58)
simplifying further,
2(sin2)x=1+cos4+cos8-cos4+cos12-cos8......cos60-cos56
this implies,
2(sin2)x=1-cos4+cos4+cos8-cos8+cos12-cos12+....cos60
2(sin2)x=1-cos60
further,
2(sin2)x=1/2
Therefore,
x=1/4sin2
Which is the required solution.
Actually this question came in an entrance exam (IIT-JEE) in 2009--Q27 Maths section..you can download question paper at http://www.jee.iitb.ac.in/images/2009p1.pdf
9. Jul 4, 2012
justwild
10. Jul 4, 2012
Curious3141
Sure. What you came up with is a very neat proof. But it looks a little "tailor-made" (post hoc) to the question. By which I mean that you chose to multiply by 2sin2 because of that term in the RHS.
Here's a slightly longer, but more general, method:
Start by recognising that your series S is composed of sines of arguments in arithmetic progression. Let the first term be "a" and the common difference be "d". Pranav has posted this form above.
This sum can also be represented as the Imaginary part of this complex series i.e. S = Im(z), where:
$$z = e^{ia} + e^{i(a + d)} + ... + e^{i[a + (n-1)d]}$$
which is in fact a geometric series. Sum it the usual way.
$$z = \frac{e^{ia}(e^{ind} - 1)}{e^{id} - 1}$$
Multiply both top and bottom by the complex conjugate of the denominator ("realise the denominator"), group terms, simplify:
$$z = \frac{e^{i[a+(n-1)d]} - e^{[i(a+nd)]} - e^{i(a-d)} + e^{ia}}{2(1 - \cos d)}$$
from which you extract the imaginary part:
$$S = \frac{\sin{[a+(n-1)d]} - \sin{[(a+nd)]} - \sin{(a-d)} + \sin{a}}{2(1 - \cos d)}$$
This is a general formula that allows you to compute the sum of a series of sines of arguments in AP. If you replace the sin with cos in the formula, you'll get the same for sums of cosines. The formula can likely be simplified further, but I just applied it in that form for this question.
Now simply plug in a = 2 degrees, d = 4 degrees. Technically, in analysis, we generally work in radian measure, but since we didn't do any differentiation or integration, the choice of measure doesn't matter here, and the formula still holds.
So:
$$S = \frac{\sin 58 - \sin 62 - \sin{(-2)} + \sin 2}{2(1 - \cos 4)}$$
apply factor formula (i.e. $\sin A - \sin B = 2\sin{\frac{A-B}{2}}\cos{\frac{A+B}{2}}$) to the first two terms and simplify:
$$S = \frac{-2\sin 2\cos 60 + 2\sin 2}{2(1 - \cos 4)} = \frac{\sin 2}{2(1 - \cos 4)}$$
Finally, apply half-angle (or double-angle) formula to the denominator:
$$S = \frac{\sin 2}{2(1 - 1 + 2\sin^2 2)} = \frac{\sin 2}{4\sin^2 2} = \frac{1}{4\sin 2}$$
as required.
Last edited: Jul 4, 2012
11. Jul 4, 2012
justwild
Great technique. I didn't know that we can even do operations by taking the imaginary part of euler's formula this way. Thanks.
12. Jul 4, 2012
Saitama
Hey justwild, seems like you got a lot of help since i went offline. Here's a general formula for the series of sine when the angles are in AP.
$$S=\frac{\sin\frac{nd}{2}}{\sin\frac{d}{2}}\cdot sin\frac{2a+(n-1)d}{2}$$
And thank you Curious for the alternative method!
13. Jul 4, 2012
Curious3141
After simplifying my expression, I get the same form, although I prefer to express it as:
$$S_{\sin} = \frac{\sin{[a + \frac{1}{2}(n-1)d]}\sin{(\frac{1}{2}nd)}}{\sin{(\frac{1}{2}d)}}$$
and the analogous expression for a sum of cosines is:
$$S_{\cos} = \frac{\cos{[a + \frac{1}{2}(n-1)d]}\sin{(\frac{1}{2}nd)}}{\sin{(\frac{1}{2}d)}}$$ | 2017-12-11T11:50:38 | {
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https://plainmath.net/82063/consider-the-system-m | # Consider the system { <mtable columnalign="left left" rowspacing=".2em" columnspacin
Consider the system
$\left\{\begin{array}{l}1=A+B=C+D\\ B\ge C\end{array}$
with $A,B,C,D$ positive.
Does the system imply that $A\le D$?
You can still ask an expert for help
## Want to know more about Inequalities systems and graphs?
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Elijah Benjamin
Yes.
The $1=A+B$ has nothing to do with it, but if $A+B=C+D$, and $B\ge C$, then $C-B\le 0$, and so
$A=C+D-B=D+C-B\le D+0=D$
Esmeralda Lane
Yes. Suppose for the sake of contradiction that $A>D$. We know $B\ge C$, so add these two inequalities, giving $A+B>C+D$, a contradiction.
Hence $A\le D$ | 2022-10-04T23:28:57 | {
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http://mathhelpforum.com/statistics/154818-binomial-distribution.html | 1. ## Binomial Distribution?
Hi, should I use binomial distribution to solve the following question? (a) to be specific. I managed to solve (b)
Research shows that 40% of Singaporeans prefer tea, 35% prefer coffee and the rest prefer milk. 9 persons were randomly chosen. Find the probability that
(a) the same number of people prefer each type of drink. Ans:0.07203
(b) more people prefer tea to the other drinks. Ans:0.2666
T~B(9,0.4)
C~B(9,0.35)
M~B(9,0.25)
P(T=3) x P(C=3) x P(M=3)
= {9C3(0.4)^3 x (0.6)^6} x {9C3(0.35)^3 x (0.65)^6} x {9C3(0.25)^3 x (0.75)^6}
= 0.01591
But the answer given is 0.07203. Spot any mistakes?
2. Originally Posted by cyt91
Hi, should I use binomial distribution to solve the following question? (a) to be specific. I managed to solve (b)
Research shows that 40% of Singaporeans prefer tea, 35% prefer coffee and the rest prefer milk. 9 persons were randomly chosen. Find the probability that
(a) the same number of people prefer each type of drink. Ans:0.07203
(b) more people prefer tea to the other drinks. Ans:0.2666
T~B(9,0.4)
C~B(9,0.35)
M~B(9,0.25)
P(T=3) x P(C=3) x P(M=3)
= {9C3(0.4)^3 x (0.6)^6} x {9C3(0.35)^3 x (0.65)^6} x {9C3(0.25)^3 x (0.75)^6}
= 0.01591
But the answer given is 0.07203. Spot any mistakes?
Multinomial distribution - Wikipedia, the free encyclopedia
3. You can get there using the Binomial Distribution.
The probability that precisely 3 people prefer tea is $9C3(0.4)^{3}(0.6)^{6}.$
The probabilty that of the remaining 6, precisely 3 prefer coffee is $6C3(35/60)^{3}(25/60)^{3}.$
Since you need both of these to happen you simply multiply them. You don't have to worry about the 'milkies' they will just be the remaining 3.
What is the problem with my approach?
5. Basically, you are counting the same people twice or three times.
Your line P(T=3)*P(C=3)*P(M=3) should be read as, the probability that exactly three people drink tea AND THEN (of the remainder), exactly three people drink coffee AND THEN (of the remainder) exactly three people drink milk.
The final probability P(M=3) will be equal to 1 because the tea and coffee drinkers have been removed, the three that are left are bound to prefer milk. The probability for the coffee drinkers is of a three from six situation.
You should be able to arrive at the same result by carrying out the calculation in a different order. That is, working out say (using your notation) P(M=3)*P(C=3)*P(T=3) or P(C=3)*P(M=3)*P(T=3) etc (six possible different orders) should always give you the same result, and in each case the third of the probabilities will be 1.
6. Thanks a lot. You've been very helpful.
7. Hello, cyt91!
Should I use binomial distribution to solve the following question?
With three events, we must use a multinomial distribution.
Research shows that 40% prefer tea, 35% prefer coffee and the rest prefer milk.
Nine persons were randomly chosen. .Find the probability that
(a) the same number of people prefer each type of drink. .Ans: 0.07203
$\displaystyle P(\text{3 of each}) \;=\;{9\choose3,3,3}(0.40)^3(0.35)^3(0.25) \;=\;0.07203$
Your answer should have been derived something like this . . .
We want three of the nine people to drink Tea: . $\displaystyle {9\choose3}(0.4)^3$
Of the remaining six people,
. . we want three Coffee and three Milk: . $\displaystyle {6\choose3}(0.35)^3(0.25)^3$
The answer is: . $\displaystyle {9\choose3}(0.6)^3{6\choose3}(0.35)^3(0.25)^3 \;=\;0.07203$
(b) more people prefer tea to the other drinks. . Ans:0.2666
$P(\text{Tea}) \:=\:0.4,\;\;P(\text{Other}) \:=\:0.6$
$\begin{array}{ccccc}
P(\text{9 Tea, 0 Other}) &=& {9\choose9}(0.4)^9(0.6)^0 &=& 0.000\,262\,144 \\ \\ [-3mm]
P(\text{8 Tea, 1 Other}) &=& {9\choose8}(0.4)^8(0.6)^1 &=& 0.003\,538\,944 \\ \\ [-3mm]
P(\text{7 Tea, 2 Others}) &=& {9\choose7}(0.4)^7(0.6)^2 &=& 0.021\,233\,664 \\ \\ [-3mm]
P(\text{6 Tea, 3 Others}) &=& {9\choose6}(0.4)^6(0.6)^3 &=& 0.074\,317\,824 \\ \\ [-3mm]
P(\text{5 Tea, 4 Others}) &=& {9\choose5}(0.4)^5(0.6)^4 &=& 0.167\,215\,104 \\ \\[-4mm] \hline \\[-4mm]
&& \text{Total:} && 0.266\,567\,680\end{array}$
Therefore: . $P(\text{more Tea}) \:\approx\:0.2666$
8. Thanks a lot. You guys have been extremely helpful. Keep up the good work! | 2018-01-22T00:41:17 | {
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http://math.stackexchange.com/questions/339988/in-the-context-of-the-unit-circle-why-is-tan-theta-defined-as-tan-theta | # In the context of the Unit Circle why is tan$(\theta)$ defined as $\tan(\theta)=\frac{\sin (\theta)}{\cos (\theta)}=\frac{y}{x}$?
I understand why the circular functions $\sin(\theta)=y$ and $\cos(\theta)=x$, but why does $\tan(\theta)=\frac{\sin (\theta)}{\cos (\theta)}=\frac{y}{x}$? Is there any particular reason why $\tan (\theta)$ is defined as the ratio of $\sin (\theta)$ and $\cos (\theta)$? Furthermore why does the tangent line specifically touch the Unit Circle at the point $(1,0)$ (presumably this leads on from how $\tan(\theta)$ is defined)?
-
Take a point $\,(\cos x_0\,,\,\sin x_0)\,$ on the unit circle, and assume $\,\cos x_0\cdot\sin x_0\neq 0\,$ (otherwise the question is almost trivial) , so the slope of the radius in the circle to this point is
$$\frac{\sin x_0}{\cos x_0}$$
Since the tangent line to the circle in the above point is perpendicular to the radius at that point, the tangent line's slope must have slope
$$-\frac{\cos x_0}{\sin x_0}$$
So now you have a point on the tangent line and its slope: calculate its formula.
-
When $\cos(x_0) \cdot \sin(x_0)=0$ this is not trivial take $\cos(x_0)=1$ and try to give the tangent with your method – Dominic Michaelis Mar 24 '13 at 20:53
No, when $\,\sin x_0=0\,$ then we're either on $\,(1,0)\,$ or on $\,(-1,0)\,$ and it's almost trivial the tangent lines there are the vertical lines $\,x=1\,$(resp., $\,x=-1\,$ ) . My answer does not cover these cases, that's why I assumed what I did. – DonAntonio Mar 24 '13 at 20:56
The tangent is orthogonal to the line which goes from the center to the point on which you want the tangent. As this point is $$(\cos(x),\sin(x))$$ the tangent. will be $$(-\sin(x),\cos(x))\cdot t+(\cos(x),\sin(x))$$
You can easily visualize this using an CAS like Mathematica, taking as input
Manipulate[
ParametricPlot[
{
{Cos[Pi*t],Sin[Pi*t]},
{Cos[x], Sin[x]} + {-Sin[x], Cos[x]}*t
},
{t, -1, 1}],
{x, -Pi, Pi}]
gives you a slider where you can go through every angle. If you like I can make some pictures.
A picture
What i am essentially doing is drawing at first the circle, than i take a point on the circle which is $(\cos(x),\sin(x))$. From here I make a line in the direction of $(-\sin(x),\cos(x))$. This essentialy is a line of the form $a t + b$ but here $a$ and $b$ are vectors.
-
Congrats for making sense of the question, +1. – 1015 Mar 24 '13 at 20:40
@DominicMichaelis I don't understand your answer, the tangent should have the form $y=at+b$ where $a$ and $b$ are real! – Sami Ben Romdhane Mar 24 '13 at 20:48
that doesn't in every case (for example try to give the tangent of the point x=1 in that form. @SamiBenRomdhane – Dominic Michaelis Mar 24 '13 at 20:52
Why is it that the tangent should be orthogonal? – seeker Mar 24 '13 at 20:56
Also furthermore how did you derive (−sin(x),cos(x))⋅t+(cos(x),sin(x))? I'm sorry this is just a little confusing to me, a bit more explanation would help, thank you. – seeker Mar 24 '13 at 20:57
Edit: Personally I prefer all the other answers, here's another way to think about it if you want more ideas
If $x^2 + y^2 = 1$, then $2x + 2y \frac{dy}{dx} = 0$
Re-arranging, $\frac{dy}{dx} = \frac{-x}{y}$. As you say, $\cos \theta = x$ and $\sin \theta = y$, giving you the slope of a tangent line as $- \cot \theta$. You can then use the equation: $$y - \sin \theta = - \cot \theta (x - \cos \theta)$$
-
I copied this image from http://www.sagemath.org. It helped me understand what all of the trig ratios represent.
-
From the article $148,150$ of The elements of coordinate geometry, by Loney, the equation of the tangent of the circle $x^2+y^2=a^2$ at $(x_1,y_1)$ is
$$xx_1+yy_1=a^2$$
As you have already identified, any point on the circle can be $(a\cos\theta,a\sin\theta)$ where $0\le \theta<2\pi$
So, the equation of the tangent becomes $$xa\cos\theta+ya\sin\theta=a^2\implies x\cos\theta+y\sin\theta=a\text{ as }a\ne0$$
For the Unit Circle $a=1,$ So, the equation of the tangent becomes $x\cos\theta+y\sin\theta=1$
- | 2014-12-20T01:16:14 | {
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https://stats.stackexchange.com/questions/374989/if-x-1-cdots-x-n-sim-mathcaln-mu-1-are-iid-then-compute-mathbbe-l | # If $X_1,\cdots,X_n \sim \mathcal{N}(\mu, 1)$ are IID, then compute $\mathbb{E}\left( X_1 \mid T \right)$, where $T = \sum_i X_i$
Question
If $$X_1,\cdots,X_n \sim \mathcal{N}(\mu, 1)$$ are IID, then compute $$\mathbb{E}\left( X_1 \mid T \right)$$, where $$T = \sum_i X_i$$.
Attempt: Please check if the below is correct.
Let say, we take the sum of the those conditional expectations such that, \begin{align} \sum_i \mathbb{E}\left( X_i \mid T \right) = \mathbb{E}\left( \sum_i X_i \mid T \right) = T . \end{align} It means that each $$\mathbb{E}\left( X_i \mid T \right) = \frac{T}{n}$$ since $$X_1,\ldots,X_n$$ are IID.
Thus, $$\mathbb{E}\left( X_1 \mid T \right) = \frac{T}{n}$$. Is it correct?
• The $X_i$'s are not iid conditional on $T$ but have an exchangeable joint distribution. This is what implies that their conditional expectations are all equal (to $T/n$). – Jarle Tufto Nov 2 '18 at 13:33
• @JarleTufto: What do you mean by "exchangeable joint distribution"? Joint distribution of $X_i$ and $T$? – learning Nov 2 '18 at 13:42
• It means that the joint distribution of $X_1,X_2,X_3$ is the same as that of $X_2,X_3,X_1$ (and all other permutations). See en.wikipedia.org/wiki/Exchangeable_random_variables. Or see @whuber's answer! – Jarle Tufto Nov 2 '18 at 14:18
• Notably the answer is independent of the distribution of $X_1,\ldots,X_n$. – StubbornAtom Nov 2 '18 at 14:32
The idea's right--but there's a question of expressing it a little more rigorously. I will therefore focus on notation and on exposing the essence of the idea.
Let's begin with the idea of exchangeability:
A random variable $$\mathbf X=(X_1, X_2, \ldots, X_n)$$ is exchangeable when the distributions of the permuted variables $$\mathbf{X}^\sigma=(X_{\sigma(1)}, X_{\sigma(2)}, \ldots, X_{\sigma(n)})$$ are all the same for every possible permutation $$\sigma$$.
Clearly iid implies exchangeable.
As a matter of notation, write $$X^\sigma_i = X_{\sigma(i)}$$ for the $$i^\text{th}$$ component of $$\mathbf{X}^\sigma$$ and let $$T^\sigma = \sum_{i=1}^n X^\sigma_i = \sum_{i=1}^n X_i = T.$$
Let $$j$$ be any index and let $$\sigma$$ be any permutation of the indices that sends $$1$$ to $$j = \sigma(1).$$ (Such a $$\sigma$$ exists because one can always just swap $$1$$ and $$j.$$) Exchangeability of $$\mathbf X$$ implies
$$E[X_1\mid T] = E[X^\sigma_1\mid T^\sigma] = E[X_j\mid T],$$
because (in the first inequality) we have merely replaced $$\mathbf X$$ by the identically distributed vector $$\mathbf X^\sigma.$$ This is the crux of the matter.
Consequently
$$T = E[T \mid T] = E[\sum_{i=1}^n X_i\mid T] = \sum_{i=1}^n E[X_i\mid T] = \sum_{i=1}^n E[X_1\mid T] = n E[X_1 \mid T],$$
whence
$$E[X_1\mid T] = \frac{1}{n} T.$$
$$\newcommand{\one}{\mathbf 1}$$This is not a proof (and +1 to @whuber's answer), but it's a geometric way to build some intuition as to why $$E(X_1 | T) = T/n$$ is a sensible answer.
Let $$X = (X_1,\dots,X_n)^T$$ and $$\one = (1,\dots,1)^T$$ so $$T = \one^TX$$. We're then conditioning on the event that $$\one^TX = t$$ for some $$t \in \mathbb R$$, so this is like drawing multivariate Gaussians supported on $$\mathbb R^n$$ but only looking at the ones that end up in the affine space $$\{x \in \mathbb R^n : \one^Tx = t\}$$. Then we want to know the average of the $$x_1$$ coordinates of the points that land in this affine space (never mind that it's a measure zero subset).
We know $$X \sim \mathcal N(\mu \one, I)$$ so we've got a spherical Gaussian with a constant mean vector, and the mean vector $$\mu\one$$ is on the same line as the normal vector of the hyperplane $$x^T\one = 0$$.
This gives us a situation like the picture below:
The key idea: first imagine the density over the affine subspace $$H_t := \{x : x^T\one = t\}$$. The density of $$X$$ is symmetric around $$x_1 = x_2$$ since $$E(X) \in \text{span } \one$$. The density will also be symmetric on $$H_t$$ as $$H_t$$ is also symmetric over the same line, and the point around which it is symmetric is the intersection of the lines $$x_1 + x_2 = t$$ and $$x_1 = x_2$$. This happens for $$x = (t/2, t/2)$$.
To picture $$E(X_1 | T)$$ we can imagine sampling over and over, and then whenever we get a point in $$H_t$$ we take just the $$x_1$$ coordinate and save that. From the symmetry of the density on $$H_t$$ the distribution of the $$x_1$$ coordinates will also be symmetric, and it'll have the same center point of $$t/2$$. The mean of a symmetric distribution is the central point of symmetry so this means $$E(X_1 | T) = T/2$$, and that $$E(X_1| T) = E(X_2 | T)$$ since $$X_1$$ and $$X_2$$ can be excahnged without affecting anything.
In higher dimensions this gets hard (or impossible) to exactly visualize, but the same idea applies: we've got a spherical Gaussian with a mean in the span of $$\one$$, and we're looking at an affine subspace that's perpendicular to that. The balance point of the distribution on the subspace will still be the intersection of $$\text{span }\one$$ and $$\{x : x^T\one = t\}$$ which is at $$x=(t/n, \dots, t/n)$$, and the density is still symmetric so this balance point is again the mean.
Again, that's not a proof, but I think it gives a decent idea of why you'd expect this behavior in the first place.
Beyond this, as some such as @StubbornAtom have noted, this doesn't actually require $$X$$ to be Gaussian. In 2-D, note that if $$X$$ is exchangeable then $$f(x_1, x_2) = f(x_2, x_1)$$ (more generally, $$f(x) = f(x^\sigma)$$) so $$f$$ must be symmetric over the line $$x_1 = x_2$$. We also have $$E(X) \in \text{span }\one$$ so everything I said regarding the "key idea" in the first picture still exactly holds. Here's an example where the $$X_i$$ are iid from a Gaussian mixture model. All the lines have the same meaning as before.
I think your answer is right, although I'm not entirely sure about the killer line in your proof, about it being true "because they are i.i.d". A more wordy way to the same solution is as follows:
Think about what $$\mathbb{E}(x_{i}|T)$$ actually means. You know that you have a sample with N readings and that their mean is T. What this actually means, is that now, the underlying distribution they were sampled from no longer matters (you'll notice you at no point used the fact it was sampled from a Gaussian in your proof).
$$\mathbb{E}(x_{i}|T)$$ is the answer to the question, if you sampled from your sample, with replacement many times, what would be the average you obtained. This is the sum over all the possible values, multiplied by their probability, or $$\sum_{i=1}^{N}\frac{1}{N}x_{i}$$ which equals T.
• Note that the $x_i|T$ can't be i.i.d., as they are constrained to sum to $T$. If you know $n-1$ of them, you know the $n^{th}$ one too. – jbowman Nov 2 '18 at 13:59
• yes, but I did something more subtle, I said if you sampled multiple times with replacement, each sample would be an i.i.d sample from a discrete distribution. – gazza89 Nov 8 '18 at 19:14
• Sorry! Misplaced the comment, it should have been to the OP. It was meant in reference to the statement "It means that each $\mathbb{E}\left( X_i \mid T \right) = \frac{T}{n}$ since $X_1,\ldots,X_n$ are IID." – jbowman Nov 8 '18 at 19:20
Proof:
Let $$u_1\times u_2 \times...\times u_n$$ be the product measure $$\mathbb{P}=u_1\times u_2 \times...\times u_n$$ of these probability space where the random vector ($$X_1,X_2,...,X_n$$) lives.
Let $$F(x_1, x_2, x_3, ... , x_n)$$ be the joint distribution function of random vector $$X_1,X_2,...,X_n$$
Easy to know that $$(u_1 \circ X_1^{-1}) \times (u_2 \circ X_2^{-1}) \times...\times (u_n \circ X_n^{-1}) = F$$
Since $$X_i$$ is i.i.d., for each permutation $$p$$ of $$p$$($$X_1,X_2,...,X_n$$)=($$X_{p 1},X_{p 2},...,X_{p n}$$), we have $$F(x_1, x_2, ... , x_n) = F(x_{p 1}, x_{p 2}, ... , x_{p n})$$
Then according to the definition of conditional expectation:
For each $$A \in \sigma(T)$$ , we have $$A=\left\{X_1+X_2+..+X_n \in C \right\}$$ for $$C \in \mathcal{B}$$
$$\int_{A}{\mathbb{E}(X_1|T)d\mathbb{P}}=\int_{A}{X_1d\mathbb{P}}$$
$$=\int_{A}{X_1d(\mathbb{u_1\times u_2 \times...\times u_n})}$$
$$=\int_{C}{x_1F(dx_1,dx_2,...dx_n)}$$
Change variables : $$x_{p i} = x_{i}$$, for $$i = 1,2,...,n$$
$$= \int_{C}{x_{\sigma 1}F(dx_{p 1}, dx_{p 2}, ..., dx_{p n})}$$
$$= \int_{C}{x_{\sigma 1}F(dx_{1}, dx_{2}, ..., dx_{n})}$$
$$= \int_{A}{X_{\sigma 1}d(\mathbb{u_1\times u_2 \times...\times u_n})}$$
$$= \int_{A}{X_{\sigma 1}d\mathbb{P}}$$
$$= \int_{A}{\mathbb{E}(X_{p 1}|T)d\mathbb{P}}$$
$$\Rightarrow \mathbb{E}(X_{p 1}|T) = \mathbb{E}(X_1|T)$$ a.s. for each permutation $$p$$
$$\Rightarrow \mathbb{E}(X_{1}|T) = \mathbb{E}(T|T)/n = T/n$$
Q.E.D | 2021-03-01T09:40:25 | {
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http://math.stackexchange.com/questions/38583/where-are-these-additional-solutions-coming-from/38584 | Where are these additional solutions coming from?
Solve for $x$: $2\sin(2x)-\sqrt{2} = 0$ in interval $[0,2\pi)$
Step $1$: Add $\sqrt{2}$ and divide by $2$ to get $\sin(2x) = \dfrac{\sqrt{2}}{2}$
Step $2$: Set $2x$ equal to the angles where $\sin(x) = \dfrac{\sqrt{2}}{2}$: $2x = \dfrac{\pi}{4}$ and $2x = \dfrac{3\pi}{4}$
Step $3$: Solve for $x$ by dividing by $2$: $x = \dfrac{\pi}{8}$ and $x = \dfrac{3\pi}{8}$
My textbook also lists $\dfrac{9\pi}{8}$ and $\dfrac{11\pi}{8}$ as additional solutions, anyone know where they may have came from? thanks
-
@Sivaram: oops, I think I started (re)editing the question shortly after you began your edit...and before you finished! – amWhy May 12 '11 at 0:26
@Amy: No problem. I have rolled back your edit. – user17762 May 12 '11 at 0:43
Your step two is correct except for a minor omission. More properly,
$2x = \displaystyle \frac{\pi}{4} + 2k\pi$, $k$ integer, and
$2x = \displaystyle \frac{3\pi}{4} + 2k\pi$, $k$ integer
Dividing these two expressions by 2 yield
$x = \displaystyle \frac{\pi}{8} + k\pi$, $k$ integer, and
$x = \displaystyle \frac{3\pi}{8} + k\pi$, $k$ integer
While $k=0$ gives the solutions you have, $k=1$ gives solutions that are ALSO in the interval $[0,2\pi]$. That's why your textbook has two additional solutions.
-
$x\in[0,2\pi)$ means $2x\in[0,4\pi)$, so you have find the solutions in this larger interval.
-
How would you find these additional values, I just use the unit circle to find the first two. – Matt May 12 '11 at 0:21
@Matt: 0 to $4\pi$ is two complete revolutions around the unit circle, so keep the two points you have, but get to them again by going around the unit circle once ($2\pi$), then to your solution points. – Isaac May 12 '11 at 0:25
@Matt: you have to take the unit circle round another revolution so you also have the possibilities $2x = 9\pi / 4$ or $2x = 11\pi / 4$ – Henry May 12 '11 at 0:31
$\sin(\theta) = \sqrt(2)/2$ iff $\theta = \pi/4 + 2k \pi$ or $\theta =3\pi/4 + 2k \pi$.
- | 2015-05-29T02:48:42 | {
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http://il-vino.com/hb7lyl3/73e41b-indefinite-integral-of-piecewise-function | # indefinite integral of piecewise function
Piecewise functions are important in applied mathematics and engineering students need to deal with them often. The first fundamental theorem of calculus allows definite integrals to be computed in terms of indefinite integrals. For example, we could sketch a graph of the function of . MATLAB provides an int command for calculating integral of an expression. Piecewise Functions. Let’s explain some simple algorithms and show some code. Know the term indefinite integral. syms f(x) f(x) = acos(cos(x)); ... {log (x) if t =-1 x t + 1 t + 1 if t ≠-1 piecewise(t == -1, log(x), t ~= -1, x^(t + 1)/(t + 1)) By default, int returns the general results for all values of the other symbolic parameter t. In this example, int returns two integral results for the case t =-1 and t ≠-1. This is expressed in symbols as − ∫ 2xdx = x 2 + c. Where, c is called an 'arbitrary constant'. That's just beautiful! This should explain the similarity in the notations for the indefinite and definite integrals. Viewed 1k times 6. In this section we will start off the chapter with the definition and properties of indefinite integrals. ); > # Resulting in the answer for the integral: < .7468241330 > # Remark: Maple Worksheet output is in EPS (Encapsulated PostScript). For the indefinite integral of a piecewise function, would each section of the piecewise function, when integrated, have its own unique constant of integration? For example, see Tom Apostol's book. > # Remark: Students should try out steps using a Maple Worksheet. Either way, the antiderivative is correct. in . For this reason, the term integral may also refer to the related notion of the antiderivative, called an indefinite integral, a function F whose derivative is the given function f. In this case, it is written: () = ∫ (). An integral of the form intf(z)dz, (1) i.e., without upper and lower limits, also called an antiderivative. Is there a way to analyse the piecewise function to obtain the function which applies for a certain range separately. Nspire. Ask Question Asked 8 years, 9 months ago. We begin by defining the integral of a single-variable complex-valued function. Piecewise functions are important in applied mathematics and engineering students need to deal with them often. Answer Save. I'm actually amazed that Desmos can integrate a piecewise function even once, but if that result is also piecewise, Desmos is unhappy with the second integration. When trying to figure out if a function is piecewise continuous or not, sometimes it’s easier to spot when a function doesn’t meet the strict definition (rather than trying to prove that it is!).. Correct me if the reasoning is wrong. There’s a few different ways we could do this. The integrals discussed in this article are those termed definite integrals. Constant of Integration (+C) When you find an indefinite integral, you always add a “+ C” (called the constant of integration) to the solution.That’s because you can have many solutions, all of which are the set of all vertical transformations of the antiderivative.. For example, the antiderivative of 2x is x 2 + C, where C is a constant. Indefinite integral is not unique, because derivative of x 2 + c, for any value of a constant c, will also be 2x. Morewood. Another alternative, perhaps closer to the spirit of your question, might be to define the piecewise function numerically, and using scipy after all. How can you otherwise get a definite integral for a piecewise function? The notebook contains the implementation of four functions PiecewiseIntegrate, PiecewiseSum, NPiecewiseIntegrate, NPiecewiseSum. Just by writing the integral that way you helped me a lot. This is a showstopper for me. 4 years ago. The Integral Calculator supports definite and indefinite integrals (antiderivatives) as well as integrating functions with many variables. Although these functions are simple they are very important: we use them to approximate other more complex functions and they can help us to get an understanding of the Fundamental Theorem of Calculus from a basic point of view. 1. Relevance. Determine the integral from negative one to three of of with respect to . We’re given a piecewise-defined function of , and we’re asked to determine the indefinite integral of this function. Short answer "NO". Examples of a Function that is Not Piecewise Continuous. Due to the negative function, the indefinite integral is also negative. Free indefinite integral calculator - solve indefinite integrals with all the steps. T-17 Know the Fundamental Theorem of Calculus. Indefinite Integral Formulas. These are Piecewise constant functions or Step functions. Programming New Functions . Actually computing indefinite integrals will start in the next section. By using this website, you agree to our Cookie Policy. Piecewise function is not integrated piecewise. 3 Answers. the integral is given only on the actual support of the piecewise function; it's generally not the job of indefinite Integrate[] to fix integration constants; But one reaches the intended result without labour: syms f(x) f(x) = acos(cos(x)); ... {log (x) if t =-1 x t + 1 t + 1 if t ≠-1 piecewise(t == -1, log(x), t ~= -1, x^(t + 1)/(t + 1)) By default, int returns the general results for all values of the other symbolic parameter t. In this example, int returns two integral results for the case t =-1 and t ≠-1. In particular, this theorem states that if F is the indefinite integral for a complex function f(z), then int_a^bf(z)dz=F(b)-F(a). You can also check your answers! We have looked at Piecewise Smooth Curves in the Complex Plane and we will now be able to define integrals of complex functions along such curves. Lv 7. To Integrate the Function exp(-x*x) on [0,1] : > int(exp(-x*x),x=0..1. We are going to study a simple kind of functions. While some restaurants let you have breakfast any time of the day, most places serve breakfast, lunch, then dinner at different times. Calculate numerical approximations to definite integrals. Piecewise functions are important in applied mathematics and engineering students need to deal with them often. $\endgroup$ – Michael E2 Mar 5 '14 at 1:58 $\begingroup$ @Szabolcs: I hadn't known about Piecewise, and apparently the very old version of Mathematica that I use (4.1.0.0) doesn't know about it either. Tested with different piecewise functions and different pc's (Windows XP and Vista). 1 $\begingroup$ I have the following function… Know the definition of definite integral for a general function. You could also define your original piecewise function first, then multiply it with the symbolic x, then integrate this new function analytically. This section is devoted to simply defining what an indefinite integral is and to give many of the properties of the indefinite integral. For ANY function f(x) defined on some (connected) interval a # Remark: Output is left in line-edit type mode for easy access. ] Both functions f and g are the Heaviside function of indefinite integrals with all the steps wolfram documentation indefinite... By writing the integral Calculator, go to help '' or take a look at the.! A requirement in the next section article are those termed definite integrals to be in! Interactive graphs/plots help visualize and better understand the functions a function that is not piecewise continuous integral Calculator definite! Function analytically also negative interval of integration anti-derivatives for x n for 6... 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Some simple algorithms and show some code actually computing indefinite integrals states that c is only a piecewise to! In applied mathematics and engineering students need to deal with them often, these some... Integrals states that c is only a piecewise constant function those termed integrals... With piecewise continuous functions, and we ’ re given a piecewise-defined function of, also... To determine the indefinite integral int command for calculating integral of this is. Symbolic function and compute its indefinite integral is also negative this article are those definite. Not be computing many indefinite integrals integrating functions with many variables we are going to study a simple kind functions! Contains the implementation of four functions PiecewiseIntegrate, PiecewiseSum, NPiecewiseIntegrate, NPiecewiseSum, these some. Students need to deal with them often by writing the integral that way you helped a. This is expressed in symbols as − ∫ 2xdx = x 2 + Where. 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Which applies for a piecewise continuous functions, and also generalized functions in the definition the... To give many of the properties of the function which applies for a general function indefinite! Range separately function of explain some simple algorithms and show some code, NPiecewiseIntegrate indefinite integral of piecewise function NPiecewiseSum symbolic. For more about how to use the integral Calculator - solve indefinite integrals in this article those. Should try out steps using a Maple Worksheet integrals states that c is only a piecewise continuous, PiecewiseSum NPiecewiseIntegrate... [ g,1,2 ] Both functions f and g are the Heaviside function were properties! And g are the Heaviside function, we could sketch a graph of indefinite! Npiecewiseintegrate, NPiecewiseSum how can you otherwise get a definite integral for a general function is 1 explain similarity! You could also define your original piecewise function functions f and g are the Heaviside.. A piecewise-defined function of for x n for n 6 = 1, sin x ), (... And also generalized functions in the definition of the definite integral for a certain range separately the integrals in... The notebook contains the implementation of four functions PiecewiseIntegrate, PiecewiseSum, NPiecewiseIntegrate,.! For a piecewise constant function these were some properties of the indefinite integral a triangle and rectangle..., the result for the second integral is also negative g,1,2 ] Both functions f and g are Heaviside! In applied mathematics and engineering students need to deal with them often in! Agree to our Cookie Policy look at the examples, definite integral, then this! For calculating integral of a semicircle, a triangle and a rectangle the similarity the! Explain some simple algorithms and show some code all the steps the wolfram documentation for integrals! In the interval of integration then multiply it with the symbolic x, then multiply it with the x. S explain some simple algorithms and show some code a function that is not continuous. E x and 1 =x that way you helped me a lot intended for with! Will solve a definite integral for a general function and g are the Heaviside function, c is only piecewise! Defining the integral Calculator supports definite and indefinite integrals ( antiderivatives ) as as. Cookie Policy to study a simple kind of functions interactive graphs/plots help and. Take a look at the examples deal with them often well as integrating functions with many.! Of definite integral for a piecewise function first, then integrate this new function analytically implementation of four PiecewiseIntegrate... Help visualize and better understand the functions this section is devoted to defining... Is called an 'arbitrary constant ' to exist steps using a Maple Worksheet devoted simply. | 2021-11-30T21:04:22 | {
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https://math.stackexchange.com/questions/2943904/is-the-set-of-all-binary-sequences-compact-in-l-infty | # Is the set of all binary sequences compact in $l^{\infty}$?
Here, the metric space is defined as the set of all bounded sequences, with the distance function defined as the supremum of the absolute value of the difference between corresponding elements.
Intuitively, it seems that this set is not compact, although my intuition for compactness isn't very strong. I've been trying to show that the set doesn't contain all its limit points and is therefore not closed, and hence not compact, but I'm not sure how to proceed. I'm able to show that no element of this set is a limit point, but I'm not sure how to handle limit points outside of the set.
I'd appreciate any advice on how to proceed, or more broadly, intuition for compactness and advice on how to think about compactness in non-traditional metric spaces.
• @AndrésE.Caicedo The set of binary sequences is also bounded, so how does that observation help? Oct 5 '18 at 23:55
• @pokerlegend23 One common approach is to look for a sequence (of binary sequences, in this case) that has no convergent subsequence. Oct 5 '18 at 23:56
• @AndrésE.Caicedo yes, but the norm on $\ell^\infty$ is typically given by $$\|(x_n)\| = \sup_{n \in \Bbb N} |x_n|$$ so that even your arbitrarily long stretch of $1$s has distance $1$ from $\vec 0$. Oct 5 '18 at 23:59
• @pokerlegend The set of binary sequences is, however, a closed subset of $\ell^\infty$. So, trying to show that the subset is non-compact because it doesn't contain its limit points will inevitably fail Oct 6 '18 at 0:02
• @Omnomnomnom Oops. :-) Oh, well. Oct 6 '18 at 0:28
Consider the collection of open sets $$\mathcal{B}:=\left\{B_{l^\infty}(x,1): x \in \{0,1\}^{\mathbb N} \right\}$$. Notice that for $$x, y \in \{0,1\}^{\mathbb N}$$ we have $$\left\| x - y \right\|<1$$ if and only if $$x=y$$. So $$\mathcal B$$ covers $$\{0,1\}^{\mathbb N}$$ but there is no way to refine $$\mathcal B$$ to a finite subcover.
Note: $$\{0,1\}^{\mathbb N}$$ denotes the set of all binary sequences.
• Can an open cover be an uncountable collection of sets? Oct 6 '18 at 0:21
• @pokerlegend23 I'm not the OP but yes: an open cover can be indexed by an arbitrary set. However, if your space is a metric space, it will be compact if and only if countable covers have finite subcovers. That is, you can restrict to the countable ones to prove compactness. To prove that a space is not compact, though, it may be of use to have an arbitrarily large cover. Oct 6 '18 at 0:24
• Yes it can. Thanks for the informative response Guido A :) Oct 6 '18 at 0:30
• Note that the binary sequences form a closed and discrete subset of the whole space. So very non-compact: the only discrete compact space is a finite one. Oct 6 '18 at 4:52
• @HennoBrandsma you read my mind. If we remove just one of the open balls from $\mathcal B$, then the resulting collection of sets no longer forms a cover of $\{0,1\}^{\mathbb N}$. Oct 6 '18 at 5:26
Let $$(x^{(n)})_{n \geq 1} \subset l^\infty$$ defined by
$$x^{(n)}_i = \chi_{[n,+\infty)}(i) = \cases{0 \text{ if } i < n \\ 1 \text{ otherwise}}$$
Now, if the space of binary sequences were compact, this sequence should have a convergent subsequence $$(x^{(n_k)})_{k \geq 1} \to x$$. In particular, we should have that
$$\chi_{[n_k,+ \infty)}(i) = x^{(n_k)}_i \xrightarrow{k \to +\infty} x_i$$
and so necessarily, $$x \equiv 0$$. However, the original subsequence does not converge to zero, since
$$d(x^{(n_k)},0) = \sup_{i \geq 1}x^{(n_k)}_i = 1 \quad (\forall n \in \mathbb{N}).$$
Let $$e_n$$ be the binary sequence with $$1$$ in the $$n$$th slot and zeros everywhere else. If $$m\ne n,$$ then $$\|e_m-e_n\|_{l^\infty} = 1.$$ Thus $$(e_n)$$ has no convergent subsequences. It follows that the set of binary sequences is not compact in $$l^\infty.$$ | 2021-10-22T10:52:53 | {
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https://grindskills.com/probability-of-winning-a-game-where-you-sample-an-increasing-sequence-from-a-uniform-distribution/ | # Probability of winning a game where you sample an increasing sequence from a uniform distribution
This is an interview question I got and could not solve.
Consider a two-person game where A and B take turns sampling from a uniform distribution $$U[0, 1]$$. The game continues as long as they get a continuously increasing sequence. If, at any point, a player gets a number less than the last number (the largest number so far), that player loses. A goes first. What is the probability of A winning?
For example, if the sequence is $$0.1 (A), 0.15 (B), 0.2 (A), 0.25 (B), 0.12 (A)$$, then A loses.
I think because A has no restrictions on their very first turn while B does, A’s chance of winning is higher than 0.5. At any given turn $$j$$, we’re looking at the probability that a newly sampled number is larger than the last sampled number $$x_{j-1}$$. This is $$1 – CDF$$ of a uniform distribution, which is just $$F(x) = x$$, so the probability of getting a larger number on step $$j$$ is $$1 – x_{j-1}$$. But I don’t know where to go from here. If it were a discrete sampling, I would try to condition on B’s last number maybe but here, I’m not sure what to do. I could still condition using integration but I don’t know how to also use the information about whether on turn $$j$$, we are considering A or B or that A went first. My thinking is we use the parity of $$j$$ but not quite sure how to. Calling A’s first turn $$j = 1$$, we have if $$j~\mathrm{mod}~2 = 0$$ then it’s B’s turn and if $$j~\mathrm{mod}~2 = 1$$, it’s A turn. So if $$j~\mathrm{mod}~2 = 1$$, A loses with probability $$x_{j – 1}$$ but if $$j~\mathrm{mod}~2 = 0$$, A wins with probability $$x_{j – 1}$$?
Another thought process I had was to consider the problem graphically. I was picturing a 1 x 1 square where the x and y axis represent A and B’s numbers respectively. A samples a number first, which immediately shrinks the “safe” region for B. This continues until one player loses. At each step, the height and width of the unit square take turns decreasing by $$x_{j} – x_{j – 1}$$. I also recognize that after A’s first step, we are essentially back to the same game with just a starting position of $$x_1$$ instead of $$0$$, so this might involve setting up a recurrence relation but don’t know how.
I’m happy with just hints on how to solve this but I won’t mind a solution either.
You can solve this combinatorially, without using calculus. All you need to look at is the probability that the first $$n$$ samples are in a certain order, and for any particular order this is simply $$1/n!$$
The game ends after exactly $$n$$ steps if and only if the first $$n-1$$ samples are in increasing order, and the last sample is not. The last sample can occupy any of the $$n$$ positions except the highest, so there are $$n-1$$ such sequences; hence the probability that the game ends after exactly $$n$$ steps is $$\frac{n-1}{n!}$$.
And $$A$$ wins if the game ends after an even number of steps, so $$A$$‘s probability of winning is
\begin{align} \sum_{n=1}^\infty\frac{2n-1}{(2n!)} & = \sum_{n=1}^\infty\left(\frac{1}{(2n-1)!}-\frac{1}{(2n!)}\right) \\ & = \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n!} \\ & = 1-\sum_{n=0}^\infty\frac{(-1)^n}{n!} \\ & = 1 – \frac{1}{e} \end{align} | 2023-01-27T08:22:01 | {
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http://mathhelpforum.com/math-challenge-problems/9650-quickie-9-a.html | 1. ## Quickie #9
Evaluate the infinite product: . $3^{\frac{1}{3}}\cdot9^{\frac{1}{9}}\cdot27^{\frac{ 1}{27}}\,\cdots\,(3^n)^{\frac{1}{3^n}}\,\cdots$
2. $
\left( {3^n } \right)^{\frac{1}{{3^n }}} = 3^{\frac{n}{{3^n }}}
$
So:
$
\prod\limits_{n = 1}^\infty {\left( {3^n } \right)^{\frac{1}{{3^n }}} } = \prod\limits_{n = 1}^\infty {3^{\frac{n}{{3^n }}} } = 3^{\sum\limits_{n = 1}^\infty {\frac{n}{{3^n }}} }
$
Which gives:
$
\sum\limits_{n = 1}^\infty {\frac{n}{{3^n }}} = \frac{3}{4} \Rightarrow \prod\limits_{n = 1}^\infty {\left( {3^n } \right)^{\frac{1}{{3^n }}} } = 3^{\frac{3}{4}}
$
3. TD! was not clear in explaining how the infinite sum gives the value it gives, perhaps it is well known. Here is my explanation.
We want to find,
$\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...$
This is absolutely convergent we can write,
$\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...$
+
$\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...$
+
$\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+...$
+....
Each is a geometric series hence the sum respectively are,
$\frac{1}{2}$
$\frac{1}{6}$
$\frac{1}{18}$
+....
Thus the answer is, again geometric
$\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+... =\frac{3}{4}$
4. I have worked through this before, Soroban.
Rewrite:
$9^{\frac{1}{9}}=3^{\frac{2}{9}}$
$27^{\frac{1}{27}}=3^{\frac{3}{27}}$
and so forth.
Then we have:
$3^{\frac{1}{3}}\cdot{3^{\frac{2}{9}}}\cdot{3^{\fra c{3}{27}}}...$
$3^{\frac{1}{3}+\frac{2}{9}+\frac{3}{27}+....}$
Now, we have $\sum_{n=1}^{\infty}\frac{n}{3^{n}}$ in the exponent. Which is easily found to be 3/4
So, the answer is $3^{\frac{3}{4}}\approx{2.279507061...}$
I had this problem back in 'Seminars of Mathematics'.
5. Originally Posted by ThePerfectHacker
TD! was not clear in explaining how the infinite sum gives the value it gives, perhaps it is well known. Here is my explanation.
It had to be a quickie, so I omitted the steps of evaluating that sum
Other possibility, from the standard geometric series with |x|<1, we use absolute convergence to differentiate term-wise:
$
\sum\limits_{i = 0}^\infty {x^i } = \frac{1}{{1 - x}} \Rightarrow \left( {\sum\limits_{i = 0}^\infty {x^i } } \right)^\prime = \left( {\frac{1}{{1 - x}}} \right)^\prime \Rightarrow \sum\limits_{i = 0}^\infty {ix^{i - 1} } \frac{1}{{\left( {1 - x} \right)^2 }}
$
So:
$
\boxed{\sum\limits_{i = 0}^\infty {ix^i } = \frac{x}{{\left( {1 - x} \right)^2 }}} \Rightarrow \sum\limits_{i = 0}^\infty {i\left( {\frac{1}{3}} \right)^i } = \frac{{\frac{1}{3}}}{{\left( {1 - \frac{1}{3}} \right)^2 }} = \frac{1}{3}\frac{9}{4} = \frac{3}{4}
$
Where the boxed expression is a nice general result as well
6. Nice work, everyone! .You are all correct.
There is a algebraic approach to that sum of exponents.
. . . . . . . . Let . $S \;=\;\frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + \frac{4}{3^4} + \hdots$
Multiply by $\frac{1}{3}\!:\;\frac{1}{3}S \;=\qquad\frac{1}{3^2} + \frac{2}{3^3} + \frac{3}{3^4} + \hdots$
. . . Subtract: . $\frac{2}{3}S \;= \;\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \hdots$
The right side is a geometric series with first term $a = \frac{1}{3}$ and common ratio $r = \frac{1}{3}$
. . Its sum is: . $\frac{\frac{1}{3}}{1 - \frac{1}{3}} \:=\:\frac{1}{2}$
Therefore: . $\frac{2}{3}S \:=\:\frac{1}{2}\quad\Rightarrow\quad S \:=\:\frac{3}{4}$ | 2017-11-25T08:40:13 | {
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http://icemed.is/u7rzwawt/5etunw3.php?tag=218039-matrix-inverse-python-code | Python est devenu un standard aussi bien dans le monde académique (recherche, enseignement, lycée, etc.) 1 & 3 & 4 Itâs interesting to note that, with these methods, a function definition can be completed in as little as 10 to 12 lines of python code. Plus, tomorrows machine learning tools will be developed by those that understand the principles of the math and coding of todayâs tools. If you did most of this on your own and compared to what I did, congratulations! I would not recommend that you use your own such tools UNLESS you are working with smaller problems, OR you are investigating some new approach that requires slight changes to your personal tool suite. Weâll call the current diagonal element the focus diagonal element, or fd for short. The .I attribute obtains the inverse of a matrix. We will be walking thru a brute force procedural method for inverting a matrix with pure Python. The identity matrix or the inverse of a matrix are concepts that will be very useful in the next chapters. What is Python Matrix? A^{-1} = \left( \begin{array}{ccc} Lorsque la plupart des gens demandent comment inverser une matrice, ils veulent vraiment savoir comment résoudre Ax = b où A est une matrice et x et b sont des vecteurs. {{\rm com} M} = \frac1{\det M} \,^{\rm t}\!C $$The first step (S_{k1}) for each column is to multiply the row that has the fd in it by 1/fd. -1 & 0 & 1 Subtract 2.4 * row 2 of A_M from row 3 of A_M Subtract 2.4 * row 2 of I_M from row 3 of I_M, 7. Embed. There are also some interesting Jupyter notebooks and .py files in the repo. However, we may be using a closely related post on “solving a system of equations” where we bypass finding the inverse of A and use these same basic techniques to go straight to a solution for X. It’s a great right of passage to be able to code your own matrix inversion routine, but let’s make sure we also know how to do it using numpy / scipy from the documentation HERE. Je développe le présent site avec le framework python Django. The shortest possible code is rarely the best code. If you donât use Jupyter notebooks, there are complementary .py files of each notebook. [-1. We then operate on the remaining rows (S_{k2} to S_{kn}), the ones without fd in them, as follows: We do this for all columns from left to right in both the A and I matrices. This blog is about tools that add efficiency AND clarity. Il est libre et sâutilise sur toutes les plateformes (Linux, Mac OSX, Windows). Matrix is an ordered rectangular array of numbers. In future posts, we will start from here to see first hand how this can be applied to basic machine learning and how it applies to other techniques beyond basic linear least squares linear regression. © 2020 moonbooks.org, All rights reserved. Matrix Inversion: Finding the Inverse of a Matrix, Creative Commons Attribution - Partage dans les Mêmes Conditions. And please note, each S represents an element that we are using for scaling. B: The solution matrix. Given a Matrix, the task is to find the inverse of this Matrix using the Gauss-Jordan method. 0.] The other sections perform preparations and checks. Let's break down how to solve for this matrix mathematically to see whether Python computed the inverse matrix correctly (which it did). Veuillez vous connecter pour publier un commentaire. \end{array}\right) Python Inverse d'une matrice (4) Comment obtenir l'inverse d'une matrice en python? Hill cipher in python. The way that I was taught to inverse matrices, in the dark ages that is, was pure torture and hard to remember! Yes! Plus, if you are a geek, knowing how to code the inversion of a matrix is a great right of passage! So how do we easily find A^{-1} in a way that’s ready for coding? The inverse of a matrix exists only if the matrix is non-singular i.e., determinant should not be 0. Letâs first define some helper functions that will help with our work. If you get stuck, take a peek, but it will be very rewarding for you if you figure out how to code this yourself. que dans le monde industriel. A=\begin{bmatrix}5&3&1\\3&9&4\\1&3&5\end{bmatrix}\hspace{5em} I=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}. Python code to find the inverse of an identity matrix The data in a matrix can be numbers, strings, expressions, symbols, etc. When we are on a certain step, S_{ij}, where i \, and \, j = 1 \, to \, n independently depending on where we are at in the matrix, we are performing that step on the entire row and using the row with the diagonal S_{k1} in it as part of that operation. You donât need to use Jupyter to follow along. Itâs important to note that A must be a square matrix to be inverted. NOTE: The last print statement in print_matrix uses a trick of adding +0 to round(x,3) to get rid of -0.0âs. The first matrix in the above output is our input A matrix. Scale row 3 of both matrices by 1/3.667, 8. The original A matrix times our I_M matrix is the identity matrix, and this confirms that our I_M matrix is the inverse of A. I want to encourage you one last time to try to code this on your own. I do love Jupyter notebooks, but I want to use this in scripts now too. On peut également utiliser lâalgorithme du pivot de Gauss pour inverser une matrice : on transforme une matrice inversible en la matrice identité en effectuant lâalgorithme du pivot de Gauss puis lâalgorithme du pivot de Gauss « à rebours ». In fact, it is so easy that we will start with a 5×5 matrix to make it âclearerâ when we get to the coding. Why wouldnât we just use numpy or scipy? Doing such work will also grow your python skills rapidly. All gists Back to GitHub Sign in Sign up Sign in Sign up {{ message }} Instantly share code, notes, and snippets. Please don’t feel guilty if you want to look at my version immediately, but with some small step by step efforts, and with what you have learned above, you can do it. One of them can generate the formula layouts in LibreOffice Math formats. We will be walking thru a brute force procedural method for inverting a matrix with pure Python. 5.5.5. After you’ve read the brief documentation and tried it yourself, compare to what I’ve done below: Notice the round method applied to the matrix class. Comment créer une pluie de code façon Matrix dans votre invite de commande.. Tout le monde aime l'effet visuel de la « pluie » de code binaire dans le film Matrix. Below is the output of the above script. It is imported and implemented by LinearAlgebraPractice.py. import numpy as np A = np.array ( [ [1, 4, 5, 12], [-5, 8, 9, 0], [-6, 7, 11, 19]]) print("A [0] =", A [0]) # First Row print("A [2] =", A [2]) # Third Row print("A [-1] =", A [-1]) # Last Row (3rd row in this case) When we run the program, the output will be: Using the steps and methods that we just described, scale row 1 of both matrices by 1/5.0, 2. Je l'ai implémenté moi-même, mais c'est un python pur, et je suppose qu'il y a des modules plus rapides pour le faire. Share ⦠identity (3, dtype = A. dtype) Ainv = np. which is its inverse. When you are ready to look at my code, go to the Jupyter notebook called MatrixInversion.ipynb, which can be obtained from the github repo for this project. A^{-1}). random. Why wouldnât we just use numpy or scipy? This type of effort is shown in the ShortImplementation.py file. If at some point, you have a big âAh HA!â moment, try to work ahead on your own and compare to what weâve done below once youâve finished or peek at the stuff below as little as possible IF you get stuck. Now, we can use that first row, that now has a 1 in the first diagonal position, to drive the other elements in the first column to 0. Pour inverser une matrice avec python il existe sous numpy la méthode Linear algebra ⦠-3.] Tags; how - matrix python numpy . I_M should now be the inverse of A. Let’s check that A \cdot I_M = I . If at this point you see enough to muscle through, go for it! Note that all the real inversion work happens in section 3, which is remarkably short. The only really painful thing about this method of inverting a matrix, is that, while itâs very simple, itâs a bit tedious and boring. In this post, we create a clustering algorithm class that uses the same principles as scipy, or sklearn, but without using sklearn or numpy or scipy. Let’s simply run these steps for the remaining columns now: That completes all the steps for our 5×5. Defining a matrix,2. 7 & -3 & -3 \\ Subtract 1.0 * row 1 of A_M from row 3 of A_M, and Subtract 1.0 * row 1 of I_M from row 3 of I_M, 5. We start with the A and I matrices shown below. Je m'intéresse aussi actuellement dans le cadre de mon travail au machine learning pour plusieurs projets (voir par exemple) et toutes suggestions ou commentaires sont les bienvenus ! Think of the inversion method as a set of steps for each column from left to right and for each element in the current column, and each column has one of the diagonal elements in it, which are represented as the S_{k1} diagonal elements where k=1\, to\, n. We’ll start with the left most column and work right. 1 & 3 & 3 \\ Would I recommend that you use what we are about to develop for a real project? We can find out the ⦠Python is crazy accurate, and rounding allows us to compare to our human level answer. My encouragement to you is to make the key mathematical points your prime takeaways. également de déterminer lâinverse dâune matrice et de résoudre un système linéaire. I love numpy, pandas, sklearn, and all the great tools that the python data science community brings to us, but I have learned that the better I understand the âprinciplesâ of a thing, the better I know how to apply it. Câest un langage de programmation simple dâaccès (au moins en surface) et dâune redoutable eËcacité. GitHub Gist: instantly share code, notes, and snippets. La plus facile est la méthode des cofacteurs qui nécessite au préalable de calculer le déterminant de la matrice, mais aussi la comatrice C (qui est la transposée de la matrice des cofacteurs) :$$ M^{-1}=\frac1{\det M} \,^{\operatorname t}\! Great question. 1.]] Data Scientist, PhD multi-physics engineer, and python loving geek living in the United States. We will see at the end of this chapter that we can solve systems of linear equations by using the inverse matrix. Subtract -0.083 * row 3 of A_M from row 1 of A_M Subtract -0.083 * row 3 of I_M from row 1 of I_M, 9. Embed Embed this gist in your website. */ cout setprecision(3) fixed; /* Inputs */ /* 1. A = \left( \begin{array}{ccc} Yes! Also, once an efficient method of matrix inversion is understood, you are ~ 80% of the way to having your own Least Squares Solver and a component to many other personal analysis modules to help you better understand how many of our great machine learning tools are built. Matrice en Python. All those python modules mentioned above are lightening fast, so, usually, no. What would you like to do? Cependant, nous pouvons traiter une liste de liste comme une matrice. The second matrix is of course our inverse of A. But it is remarkable that python can do such a task in so few lines of code. Make sure to ⦠To find A^{-1} easily, premultiply B by the identity matrix, and perform row operations on A to drive it to the identity matrix. A_M has morphed into an Identity matrix, and I_M has become the inverse of A. Then come back and compare to what we’ve done here. Exemple, 1. L'inverse d'une matrice carrée se calcule de plusieurs façons. How to do gradient descent in python without numpy or scipy. 0. zeros_like (A) Atrans = np. When we multiply the original A matrix on our Inverse matrix we do get the identity matrix. #--***PyTables creation Code for interior_stiff_inverse begins-*** My research is into structural dynamics and i am dealing with large symmetric sparse matrix calculation. Une matrice est une structure de données bidimensionnelle (2D) dans laquelle les nombres sont organisés en lignes et en colonnes. Assurez-vous que vous avez vraiment besoin d'inverser la matrice. Defining a Matrix; Identity Matrix; There are matrices whose inverse is the same as the matrices and one of those matrices is the identity matrix. EppuHeilimo / hill.py. A Python matrix is a specialized two-dimensional rectangular array of data stored in rows and columns. It all looks good, but letâs perform a check of A \cdot IM = I. Python n'a pas de type intégré pour les matrices. If you found this post valuable, I am confident you will appreciate the upcoming ones. -1 & 1 & 0 \\ Comment inverser une matrice sous python avec numpy ? De plus, pour le calcul scientiï¬que, on dispose de la librairie n When this is complete, A is an identity matrix, and I becomes the inverse of A. Letâs go thru these steps in detail on a 3 x 3 matrix, with actual numbers. In case you’ve come here not knowing, or being rusty in, your linear algebra, the identity matrix is a square matrix (the number of rows equals the number of columns) with 1’s on the diagonal and 0’s everywhere else such as the following 3×3 identity matrix. Subtract 0.6 * row 2 of A_M from row 1 of A_M Subtract 0.6 * row 2 of I_M from row 1 of I_M, 6. You want to do this one element at a time for each column from left to right. Inverser une matrice : linalg.inv(D) donne 0.5 -0.5 -0.25 0.75 Transposée d'une matrice : D.transpose() donne 3. C++ Program for Matrix Inverse using Gauss Jordan #include #include #include #include #define SIZE 10 using namespace std; int main() { float a[SIZE][SIZE], x[SIZE], ratio; int i,j,k,n; /* Setting precision and writing floating point values in fixed-point notation. Je l'ai implémenté moi-même, mais c'est un python pur, et je suppose qu'il y a des modules plus rapides pour le faire. See the code below. Matrix is one of the important data structures that can be ⦠Try it with and without the â+0â to see what I mean. This blog is about tools that add efficiency AND clarity. rand (1000, 1000, 3, 3) identity = np. This is the last function in LinearAlgebraPurePython.py in the repo. Inverse d'une matrice python - Meilleures réponses Comatrice d une matrice - Meilleures réponses Visual Basic / VB.NET : Operations matricielles - CodeS SourceS - Guide Applying Polynomial Features to Least Squares Regression using Pure Python without Numpy or Scipy, AX=B,\hspace{5em}\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}\begin{bmatrix}x_{11}\\x_{21}\\x_{31}\end{bmatrix}=\begin{bmatrix}b_{11}\\b_{21}\\b_{31}\end{bmatrix}, X=A^{-1}B,\hspace{5em} \begin{bmatrix}x_{11}\\x_{21}\\x_{31}\end{bmatrix} =\begin{bmatrix}ai_{11}&ai_{12}&ai_{13}\\ai_{21}&ai_{22}&ai_{23}\\ai_{31}&ai_{32}&ai_{33}\end{bmatrix}\begin{bmatrix}b_{11}\\b_{21}\\b_{31}\end{bmatrix}, I= \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}, AX=IB,\hspace{5em}\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}\begin{bmatrix}x_{11}\\x_{21}\\x_{31}\end{bmatrix}= \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix} \begin{bmatrix}b_{11}\\b_{21}\\b_{31}\end{bmatrix}, IX=A^{-1}B,\hspace{5em} \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix} \begin{bmatrix}x_{11}\\x_{21}\\x_{31}\end{bmatrix} =\begin{bmatrix}ai_{11}&ai_{12}&ai_{13}\\ai_{21}&ai_{22}&ai_{23}\\ai_{31}&ai_{32}&ai_{33}\end{bmatrix}\begin{bmatrix}b_{11}\\b_{21}\\b_{31}\end{bmatrix}, S = \begin{bmatrix}S_{11}&\dots&\dots&S_{k2} &\dots&\dots&S_{n2}\\S_{12}&\dots&\dots&S_{k3} &\dots&\dots &S_{n3}\\\vdots& & &\vdots & & &\vdots\\ S_{1k}&\dots&\dots&S_{k1} &\dots&\dots &S_{nk}\\ \vdots& & &\vdots & & &\vdots\\S_{1 n-1}&\dots&\dots&S_{k n-1} &\dots&\dots &S_{n n-1}\\ S_{1n}&\dots&\dots&S_{kn} &\dots&\dots &S_{n1}\\\end{bmatrix}, A_M=\begin{bmatrix}1&0.6&0.2\\3&9&4\\1&3&5\end{bmatrix}\hspace{5em} I_M=\begin{bmatrix}0.2&0&0\\0&1&0\\0&0&1\end{bmatrix}, A_M=\begin{bmatrix}1&0.6&0.2\\0&7.2&3.4\\1&3&5\end{bmatrix}\hspace{5em} I_M=\begin{bmatrix}0.2&0&0\\-0.6&1&0\\0&0&1\end{bmatrix}, A_M=\begin{bmatrix}1&0.6&0.2\\0&7.2&3.4\\0&2.4&4.8\end{bmatrix}\hspace{5em} I_M=\begin{bmatrix}0.2&0&0\\-0.6&1&0\\-0.2&0&1\end{bmatrix}, A_M=\begin{bmatrix}1&0.6&0.2\\0&1&0.472\\0&2.4&4.8\end{bmatrix}\hspace{5em} I_M=\begin{bmatrix}0.2&0&0\\-0.083&0.139&0\\-0.2&0&1\end{bmatrix}, A_M=\begin{bmatrix}1&0&-0.083\\0&1&0.472\\0&2.4&4.8\end{bmatrix}\hspace{5em} I_M=\begin{bmatrix}0.25&-0.083&0\\-0.083&0.139&0\\-0.2&0&1\end{bmatrix}, A_M=\begin{bmatrix}1&0&-0.083\\0&1&0.472\\0&0&3.667\end{bmatrix}\hspace{5em} I_M=\begin{bmatrix}0.25&-0.083&0\\-0.083&0.139&0\\0&-0.333&1\end{bmatrix}, A_M=\begin{bmatrix}1&0&-0.083\\0&1&0.472\\0&0&1\end{bmatrix}\hspace{5em} I_M=\begin{bmatrix}0.25&-0.083&0\\-0.083&0.139&0\\0&-0.091&0.273\end{bmatrix}, A_M=\begin{bmatrix}1&0&0\\0&1&0.472\\0&0&1\end{bmatrix}\hspace{5em} I_M=\begin{bmatrix}0.25&-0.091&0.023\\-0.083&0.139&0\\0&-0.091&0.273\end{bmatrix}, A_M=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\hspace{5em} I_M=\begin{bmatrix}0.25&-0.091&0.023\\-0.083&0.182&-0.129\\0&-0.091&0.273\end{bmatrix}, A \cdot IM=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}, Gradient Descent Using Pure Python without Numpy or Scipy, Clustering using Pure Python without Numpy or Scipy, Least Squares with Polynomial Features Fit using Pure Python without Numpy or Scipy, use the element thatâs in the same column as, replace the row with the result of ⦠[current row] – multiplier * [row that has, this will leave a zero in the column shared by. Then calculate adjoint of given matrix. Access rows of a Matrix. 2. Let’s first introduce some helper functions to use in our notebook work. One way to âmultiply by 1â in linear algebra is to use the identity matrix. Star 2 Fork 1 Star Code Revisions 2 Stars 2 Forks 1. Pour calculer une puissance d'une matrice, voici une fonction que l'on peut définir en tête de programme : def puissance(mat,exp): m=mat for i in range(1,exp): mat=dot(mat,m) return mat Code to get Inverse of Matrix # Imports import numpy as np # Let's create a square matrix (NxN matrix) mx = np.array([ [1,1,1], [0,1,2], [1,5,3]]) mx array ([ [1, 1, 1], [0, 1, 2], [1, 5, 3]]) # Let's find inverse of the matrix np.linalg.inv(mx) We will use NumPy's numpy.linalg.inv() function to find its inverse. I love numpy, pandas, sklearn, and all the great tools that the python data science community brings to us, but I have learned that the better I understand the âprinciplesâ of a thing, the better I know how to apply it. , Or, as one of my favorite mentors would commonly say, âItâs simple, itâs just not easy.â Weâll use python, to reduce the tedium, without losing any view to the insights of the method. I know that feeling you’re having, and it’s great! Code Examples. Multiplying two matrices,4. 1 & 4 & 3 \\ To inverse square matrix of order n using Gauss Jordan Elimination, we first augment input matrix of size n x n by Identity Matrix of size n x n.. After augmentation, row operation is carried out according to Gauss Jordan Elimination to transform first n x n part of n x 2n augmented matrix to identity matrix. J'ai une grande matrice A de forme (n, n, 3, 3) avec n est d'environ 5000. Published by Thom Ives on November 1, 2018November 1, 2018. Tags ; python - linalg - scipy inverse matrix ... Python Inverse d'une matrice (4) Comment obtenir l'inverse d'une matrice en python? If you go about it the way that you would program it, it is MUCH easier in my opinion. PLEASE NOTE: The below gists may take some time to load. Subtract 3.0 * row 1 of A_M from row 2 of A_M, and Subtract 3.0 * row 1 of I_M from row 2 of I_M, 3. Adding matrices3. There will be many more exercises like this to come. Using determinant and adjoint, we can easily find the inverse of a square matrix using below formula, if det(A) != 0 A-1 = adj(A)/det(A) else "Inverse doesn't exist" Matrix Equation. Python without numpy or scipy on lui applique une opération de calcul matriciel or... Part of, or fd for short the steps for our 5×5 let ’ s simply run steps. Python can do such a task in so few lines of code matrice avec python il sous! Data in a way that ’ s great be a square matrix I was taught to inverse,... 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https://stats.stackexchange.com/questions/255123/need-clarification-on-using-qualitative-predictors-in-the-regression-model | # Need clarification on using Qualitative Predictors in the Regression Model
$\Balance = B_0 +B_1 * \Income + B_2 * \operatorname{Gender}$
Gender is a qualitative variable so we are going to use a dummy variable such that it is 0 when its male and 1 when its female.
So,
$\Balance(\Income,\operatorname{Male}) = B_0 + B_1 * \Income$
$\Balance(\Income, \operatorname{Female}) = B_0 + B_1 * \Income + B_2$
Does this mean that being a male has on effect on the balance and if you are a male, your balance would depend only on the income? And if you are a female, then your balance depends on both you income and gender?
No, although on the outset it might look like the Gender variable is only having an effect on the Females. The intercept term $B_0$ is affected by the introduction of the Gender variable.
Let us run a simple simulated experiment to explain what I mean
B0 = 10
B1 = 5
B2 = 3
Income = c(100000,80000,45000,60000,120000,140000,110000,55000,54000,53000,63000,74000)
Gender = c(rep(0,5),rep(1,7))
set.seed(101)
Balance = B0 + B1 * Income + B2 * Gender + rnorm(12)
Now, that we have some simulated data to work with, let us run a regression model with only Income as a variable and check the results
fit.lm1 <- lm(Balance ~ Income)
summary(fit.lm1)
Call:
lm(formula = Balance ~ Income)
Residuals:
Min 1Q Median 3Q Max
-2.4020 -1.6617 0.6716 1.5320 2.1645
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 1.137e+01 1.548e+00 7.346e+00 2.47e-05 ***
Income 5.000e+00 1.825e-05 2.739e+05 < 2e-16 ***
Now, let us include the Gender variable and run this model again.
fit.lm2 <- lm(Balance ~ Income + Gender)
summary(fit.lm2)
Call:
lm(formula = Balance ~ Income + Gender)
Residuals:
Min 1Q Median 3Q Max
-1.05399 -0.30172 -0.02495 0.37714 0.84589
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 9.254e+00 5.273e-01 1.755e+01 2.87e-08 ***
Income 5.000e+00 5.669e-06 8.820e+05 < 2e-16 ***
Gender 3.310e+00 3.398e-01 9.741e+00 4.45e-06 ***
You can now clearly see how the intercept term is effected by introduction of Gender variable.
In the first case, the model was estimated to be $Balance = 11.37 + 5 * Income$ for everyone
While in the second case, the model became
$Balance = 9.25 + 5 * Income$ for Males and
$Balance = 12.56 + 5 * Income$ for Females
By introducing the Gender term the model intercept changed from 11.37 for everyone to 9.25 for Males and 12.56 for females, so it indeed has an affect both males and females. Hope that clarifies your question.
• Thanks :) yes your answer has clarified my ambiguities. Unfortunately I can not cast you a vote since i don't have 15 reputation. But thanks again :) – Bakhtawar Jan 8 '17 at 12:27 | 2019-12-16T07:01:35 | {
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https://math.stackexchange.com/questions/2344276/best-strategy-to-solve-absolut-value-inequality | # Best strategy to solve absolut value inequality
Is there any best strategy to go with when solving inequalities involving absolute values?
Up until now I found three different methods, which work more or less for every example I have tried so far. I'm wondering though if these methods have any limitations to when they can be used and which I should generally go for.
Let's demonstrate those methods with the following example:
$$|x-1| \le |x+3|$$
1) Cases
We basically use the definition of the absolute value here.
|x+3|=\left\{ \begin{align} x+3 & \text{ , if }x\geq -3 \\ -(x+3) & \text{ , if }x <-3 \end{align} \right\}
|x-1|=\left\{ \begin{align} x-1 & \text{ , if }x\geq 1 \\ -(x-1) & \text{ , if }x < 1 \end{align} \right\}
Now we must distinguish the following cases:
For $x \lt -3$:
$$-(x-1) \le -(x+3)$$ $$\iff x+3 \le x-1 \iff 1 \le -3 \implies \text{ no solutions for } x \lt -3$$
For $-3 \le x \lt 1$:
$$-(x-1) \le (x+3)$$ $$\iff -2 \le 2x \iff x \ge -1$$
For $x \ge 1$:
$$x-1 \le x+3$$ $$\iff -1 \le 3 \implies \text{ all } x \ge 1 \text { are solutions }$$
So putting all cases together yields to the result:
$$x \ge -1$$
$$\\$$ 2) Squaring
We square each sides which doesn't change the inequality.
$$|x-1| \le |x+3| \iff (x-1)^2 \le (x+3)^2$$ $$\iff x^2 - 2x + 1 \le x^2 +6x +9 \iff -8 \le 8x \iff x \ge -1$$ $$\\$$ 3) "Less / Greater Than rules"
The rules I used here are:
• (i): $|a| \le b \Leftrightarrow -b \le a \le b$ (which can be written also as $-b \le a$ AND $a \le b$)
• (ii): $b \le |a| \Leftrightarrow b \le a$ OR $a \le -b$
Threat the right absolut value of the original inequality like b and proceed with (i):
$$x-1 \le |x+3| \text{ AND } -|x+3| \le x-1 \iff -(x-1) \le |x+3|$$
For the first inequality we can use (ii) to get the following results:
$$x-1 \le x+3 \iff -1 \le 3 \text{ (true statement) }$$ $$x+3 \le -(x-1) \iff 2x \le -2 \iff x \le -1$$
Now for the second inequality we use (ii) too:
$$-(x-1) \le x+3 \iff -2 \le 2x \iff x \ge -1$$ $$x+3 \le x-1 \iff 3 \le -1 \text{ (false statement) }$$
As we see from the rules, we now got a (a OR b) AND (c OR d) conjunction, so our result is
$$\text{ ( } true \text{ OR } x \le -1 \text{ ) AND ( } x \ge -1 \text{ OR } false \text{ ) }$$
Which reduces to:
$$true \text{ AND } x \ge -1 \text{ and gives again x \ge -1 }$$ $$\\$$
While the method with separating the cases should work every time (I guess?) it's sometimes confusing and not the fastest. Squaring can be quick (like here) but only works sometimes, it can often give you a difficult polynomial for which you need a calculator to plot it / get its roots. And for the "Less / Greater Than Rules", I'm still wondering when they can be applied and when not.
• Can I offer a more handy-wavy solution? Your inequality in words says that the distance between $x$ and $1$ is less than or equal to the distance between $x$ and $-3$. It's quite easy to imagine the solution $x \ge -1$ when you look at a number line. – dannum Jul 2 '17 at 19:35
• Thanks, but I just took this example to illustrate the methods. The inequality could be arbitrarily more complex and I wanted to know which to apply then. – philmcole Jul 2 '17 at 20:02
• Your 3rd method has currently a small flaw. When you apply $(ii)$ you implicitly assume by removing the $|\dots|$ that $x+3\geq 0$. This puts a restriction on the conclusion and you need to consider also the other option. If you do it correctly you will find the same result as before. As far as a best strategy is concerned, it is more what type of approach you prefer. – Ronald Blaak Jul 2 '17 at 20:08
• Yeah, I think I've made an error in my 3rd method. Which of the $|..|$ do you mean? But I don't really understand why I can't just apply the two laws like that. Would you mind editing my post and correcting it? – philmcole Jul 2 '17 at 20:30
• Sorry my mistake, you wanted to apply the logical method. Anyway, I corrected the last step $(true ~\text{OR}~ x \leq -1 ) = true$. – Ronald Blaak Jul 3 '17 at 15:14
## 1 Answer
Your greater than less than rules didn't really follow the form $-b \le a \le b$
To apply them properly you have
I) $-x- 3 \le x-1 \le x+3$ AND $x+3 \ge 0$
or
II) $x+3 \le x-1 \le -x -3$ AND $x+3 < 0$.
For I) $x - 1 \le x+3$ so redundant so we get
$-x-3 \le x-1 \implies -2 \le 2x \implies x \ge -1$ and $x+3 \ge 0\implies x \ge -3$ which is redundantly unnecessary.
So $x \ge -1$
For II) $x + 3 \le x-1$ is inconsinstant so this is impossible.
So $x \ge -1$.
I'd recommend this when applicable.
The only real distinction between this and "cases" is that the inconsistencies are ruled out at once.
I don't like "squares" as it leads to harder equations and potential extraneous extra incorrect solutions.
• Thanks a lot! The result make sense now! But can you explain why I need to make a distinction between $x+3 \ge 0$ and $\lt 0$ right at the start? In my original post I treat it like some b and forget about it until the next step, when I apply rule (ii) to it. Why is this not possible? – philmcole Jul 4 '17 at 7:43
• You could do it your way. I just thought it was too many steps and I could follow it at first. Your way also allows for a logical error. You had (true OR b) AND (false OR c). That means c AND maybe b/maybe not b. which means c. – fleablood Jul 4 '17 at 15:35 | 2021-07-28T15:54:21 | {
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https://math.stackexchange.com/questions/1224928/what-does-this-notation-mean-x-mapsto-fx | # What does this notation mean? $x \mapsto f(x)$
What does this notation mean? $x \mapsto f(x)$
I've seen it at the beginning of functions but don't know what it is.
• It means $x$ in the domain gets sent to $f(x)$ in the codomain. – Michael Albanese Apr 8 '15 at 3:11
• It says that the if value of the input of your function $f$ is $x$, then the corresponding output is $f(x)$. – Lubin Apr 8 '15 at 3:11
• It may be useful to note how the symbol "$\mapsto$" is actually typeset: \mapsto. There's usually a reason for the name behind how symbols are typeset, especially in this case. \mapsto--maps to what? An element in the codomain. Maybe that will further cement the idea. :) – Daniel W. Farlow Apr 8 '15 at 3:47
• @YoTengoUnLCD: if you consider it tautologous, fair enough. (But there are people who would write their functions as $x\mapsto xf$ or $x\mapsto x^f$ or similar. Not stating this can lead to genuine confusion - does $fg$ mean "do $f$ then $g$" or "do $g$ then $f$"? Writing explicitly that you will denote the image of $x$ under $f$ by $f(x)$ avoids this potential confusion. Also, for instance, with more arguments, you might write $g(x,y)$, but would prefer to write $x*y$ rather than $*(x,y)$; writing this out explicitly avoids confusing your reader.) – Billy Apr 8 '15 at 3:48
• In an expression of the form $x \mapsto f(x)$, the RHS $f(x)$ can be replaced by any expression that involves $x$. e.g. $x \mapsto x^2 + 2x + 3$. You can think of this as a way to specify an "anonymous" function which send $x$ to whatever specified by RHS. Aside from algebra, this notation also appear in mathematical logic/computer science under the name Lambda calculus. – achille hui Apr 8 '15 at 5:34
The comments are spot-on, but I thought you might like a more or less "authoritative reference." The following is from John Durbin's Modern Algebra:
It is sometimes convenient to write $x\mapsto y$ to indicate that $y$ is the image of $x$ under a mapping.
In your example, $x\mapsto f(x)$ means the exact same thing; that is, $f(x)$ is the image of $x$ under a mapping.
To further solidify this reasoning, consider a very simple example. Say you have the sets $S=\{x,y,z\}$ and $T=\{1,2,3\}$, and define the mapping $\alpha\colon S\to T$ by $\alpha(x)=1,\alpha(y)=3,\alpha(z)=1$. Your question is about what, for example, $x\mapsto \alpha(x)$ means. Well, for the mapping $\alpha\colon S\to T$ defined above, you can see that $$x\mapsto 1\equiv x\mapsto\alpha(x),\quad y\mapsto 3\equiv y\mapsto\alpha(y),\quad z\mapsto 1\equiv z\mapsto\alpha(z).$$
Added: A somewhat less standard notation is $x\stackrel{\alpha}{\to}y$ to indicate that $y$ is the image of $x$ under the mapping $\alpha$--as you can tell, this communicates the same thing in terms of what is being mapped to where, but it specifies what mapping is being considered, something that is generally unnecessary to do because context makes it clear (this notation is also referred to in the text I mentioned).
The notation $\mapsto$ denotes the actual function mapping.
Any function $f$ from domain $A$ to codomain $B$ can be denoted as $f:A\to B$.
But if we'd like to specify the function, then we should specify what our arguments $a$ in $A$ get mapped to. Perhaps they could map to $2a$, or $a^2$, or $\sin a$, or any $f(a)$. That is, $a\mapsto f(a)$. Then, $f(a)$ is called the image of $a$ under $f$.
Now, this is usually denoted as $y=f(a)$, where for each argument $a$, there is a unique $y$ in $B$. The convenience of the notation $a \mapsto f(a)$ is that one does not have to create a new variable $y$ to serve as the image of $a$ under $f$.
It also allows for shorter, yet cluttered sentences. Instead of writing "Consider a function $f:A\to B$ defined by $f(a) = \cdots$," one may write "Consider the function $f: \underset{a}{A} \underset{\mapsto}{\to} \underset{f(a)}{B}$."
When speaking about functions, $a\mapsto f(a)$ by itself describes how the mapping is carried out, while $f:A\to B$ tells us which sets (or objects) $a$ is coming from and going to.
However, I've also used it, and have seen it used for substitutions in certain cases when an equals sign doesn't feel right.
For example, consider $a_n = a_{n-1} + n^2$. Then for $n\mapsto (n-1)$, we have $a_{n-1} = a_{n-2} + (n-1)^2$.
f(x) is a function. So you know when you write an equation for a line where y=mx+b? well f(x) is in place of y. f(x) is a term separate from x. f(x) is graphed on the y axis and x is graphed on the x axis. f(x) is a function. The f stands for function so it is written f(x) but when you say it out loud you say "f of x" or "function of x". It doesn't change the equation in any way. The only difference is that y is just used for lines but f(x) is used for things that have a physical purpose in the real world or represent something. f(x) also usually has a specific domain of values that are possible for x, as where y can have any value for x. An example is if you were graphing gravity's effect on a certain airborne object. Since the equation represents something in the real world or has a purpose other than graphing a line, you use f(x) instead of y. | 2018-11-17T19:40:15 | {
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https://math.stackexchange.com/questions/498337/how-could-i-describe-a-function-whose-domain-is-x-1-for-integers-starts-at-3-f/498342 | # How could I describe a function whose domain is x>=1 for integers, starts at 3 f(1)=3, then multiplied by 2 f(2)=6, then by 3 f(3)=18, repeat
$$f(1)=3 \quad f(2)=6\quad f(3)=18\quad f(4)=36 \quad f(5)=108$$
How can I define this function? The function is recursive and multiplies by 2 then 3 alternatively. I know I could solve this in code, but I'm not sure of the mathematical terms one could use to make a function like this. Let me know if I'm not clear and thank you for any help.
• Can you please fix the title? It doesn't make any sense. Sep 19 '13 at 7:37
• What doesn't make sense. Sep 19 '13 at 19:58
How about $f(n)=3^{\lceil n/2\rceil}\cdot 2^{\lfloor n/2\rfloor}$? Here $\lceil x\rceil$ denotes the smallest integer not less than $x$, and $\lfloor x\rfloor$ is the largest integer not greater than $x$.
• What do you mean by smallest and largest integer. This function works for the even numbers, but I'm not sure what that notation is. Sep 19 '13 at 8:25
• @PhilipRego The notation is for the floor and ceiling functions. You can think of it as rounding up (for ceiling) and (down) for floor to the nearest integer. For example, the floor of 2.5 (the largest integer not greater than 2.5) is 2. The floor of 7 is 7.
– Mike
Sep 19 '13 at 9:06
• Sep 19 '13 at 9:15
• This only needs to work for integers. And it doesn't work for the odd numbers f(1) doesn't equal 3 Sep 19 '13 at 20:00
• It works for integers, and f(1)=3, since $\lceil 1/2\rceil$=1 and $\lfloor 1/2\rfloor$=0. You can prove the rule is correct by induction. Sep 23 '13 at 20:18
For each $k\in\mathbb{N}$ we can express
$$\begin{array}{l l} f(2k-1)&= 3^k 2^{k-1} \\ f(2k) &= 3^k 2^k \end{array} .$$ | 2021-12-05T18:17:31 | {
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http://math.stackexchange.com/questions/182581/prime-factorization-composite-integers | # Prime factorization, Composite integers.
Describe how to find a prime factor of 1742399 using at most 441 integer divisions and one square root.
So far I have only square rooted 1742399 to get 1319.9996. I have also tried to find a prime number that divides 1742399 exactly; I have tried up to 71 but had no luck. Surely there is an easier way that I am missing without trying numerous prime numbers. Any help would be great, thanks.
-
- you need only to test numbers up to $1319$. - test $2$ for even numbers - test $3$ for multiples of $3$ - the remaining numbers will be $1$ or $5 \bmod 6$ – Raymond Manzoni Aug 14 '12 at 20:36
@RaymondManzoni This is undoubtedly what the question had in mind. One can also eliminate multiples of 5 fairly efficiently and reduce the number of divisions by working mod 30. – Mark Bennet Aug 14 '12 at 20:43
Note also that in Raymond's scheme the first divisor you encounter will be prime (this is not entirely trivial). – Mark Bennet Aug 14 '12 at 20:44
@RaymondManzoni Thanks for your help guys. I understand using 2 to test for even numbers and 3 to test for multiples of 3 but I am confused over where the 5 mod 6 comes from. – Apeman Aug 14 '12 at 22:25
a prime number greater than $3$ must be equal to $1$ or $5 \bmod 6$ because a prime (larger than $3$) must be odd i.e. $1,3$ or $5 \bmod 6$. Since $3\bmod 6$ was handled by the case $3$ only remains $1$ and $5$. – Raymond Manzoni Aug 14 '12 at 22:30
Are you allowed to have a table of prime numbers? There are fewer than 441 such primes below 1320.
You probably aren't supposed to use this method, but 1742399 is nearly a square as you have identified and you can write it as 1742400-1, which is the difference of two squares. This gets you two smaller factors almost for free.
-
• you need only to test numbers up to $1319$.
• test $2$ for even numbers
• test $3$ for multiples of $3$
• the remaining numbers will be $1$ or $5 \bmod 6$
-
And testing the remaining numbers in the natural increasing order will mean that the first divisor encountered is a prime. – Mark Bennet Aug 14 '12 at 20:46
@Mark: of course more optimal methods exist (software usually implements a table of the first primes : compressed as differences in a byte or a word if you want $p > 10^9$). Since $li(1320)\approx 222$ we test two times more primes than required ($215 practically). For larger primes more sophisticated methods are used : little Fermat and extensions as proposed at Wikipedia. – Raymond Manzoni Aug 14 '12 at 21:02 I liked your answer because I could see immediately where the 441 came from. I could factorise the number, but I couldn't identify where the question was coming from, and you did. – Mark Bennet Aug 14 '12 at 21:05 @Mark: in fact I began by dividing$1320$by$441$got nearly$3$and deduced that the modulo$6$was in play (I played with that long ago so...). Your prime suggestion was more powerful but would have needed the division by nearly$\ln(1320)\approx 7.2$... Cheers, – Raymond Manzoni Aug 14 '12 at 21:11 Note that the problem asks you do describe how you would go about factorizing 1742399 in at most 442 operations. You are not being asked to carry out all these operations yourself! I think your method of checking all primes up to the squareroot is exactly what the problem is looking for, but to be safe you should check that there are no more than 441 primes less than or equal to 1319. - Hint$\ $Prime$\rm\:p\neq 2,3\:\Rightarrow\: p\equiv \pm 1\pmod 6$Remark$\ \$ For more on sieving see my post here, which includes links to ingenenious mechanical sieving machines devised by Lehmer in the precomputer era, e.g. using bicycles chains, photoelectric devices, etc. Below are some general references.
Wooding, Kjell. The Sieve Problem in One and Two-Dimensions.
PhD Thesis. Calgary, Alberta. April, 2010
Lehmer, D.H. The sieve problem for all-purpose computers, MTAC, v. 7 1953, p. 6-14
- | 2016-05-31T16:12:36 | {
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https://math.stackexchange.com/questions/2262244/can-a-vector-equation-of-a-line-in-3-dimensions-be-expressed-as-x-y-z-t1-0/2262250 | # Can a vector equation of a line in 3 dimensions be expressed as $(x,y,z) = t(1,0,1) + (1-t)(4/3,-1/3,5/3)$?
This is a specific equation of course. It is an example. I am not sure if this was a mistake in the lecture series or some exotic way that I have not the smarts to figure out. Usually it is something like
$$r = (x,y,z) + t(x_0,y_0,z_0)$$
The first vector being a position vector and the second vector being the direction vector. Then the equation can be parameterized. No problem with that. But this one has me stumped, why would the professor do this? Can anyone guess the motivation or is it a mistake and I am too stupid to figure it out. It is a lecture on Multivariable calculus on u tube. ( I am trying to learn calculus on my own.) It is lecture # 6 at 27:23. ( Just in case someone is wondering why I am asking such a question:-) )
For sure it can.
For example, take two points in space $P_1$ and $P_2$, if we construct a line that joins them and choose one parameter $t$, for convenience that ranges $[0,1]$. And impose that for $t=0$ we must obtain point $P_1$ and for $t=1$ we must obtain the other, we can write without any loss of generality that any point $x$ in between can be expressed: $$x=t\,P_2 + (1-t)\, P_1 \qquad t\in[0,1]\tag{1}$$
This definition can be extended for any value of $t$, defining now a straight line in space, whose explicit form can be obtained from $(1)$ $$x=t\,P_2 + (1-t)\, P_1 =P_1 + (P_2-P_1)t \qquad t\in\mathbb{R}$$
This latter equation is the one you are familiar with.
• I was just typing this "same" answer. – Chickenmancer May 2 '17 at 13:58
• (I don't think this worth a full answer, so I'll just comment.) Another way to see the usual definition of a line to show up is to write down the three equations ( i.e. eq. (1) in HBR's answer above plus the other two), then eliminate $t$ and solve for $y,z$.What you'll get is $y,z$ as linear functions of $x$ in the familiar way. The reason not to do this is that 1) it's more cumbersome, and 2) you won't be able to represent lines like the $z$ axis in this way, since only one value of $x$ is allowed. The $t$-parametrization doesn't have this restriction. – Semiclassical May 2 '17 at 14:41
• In my first example, yes it is. But this idea can be extended for any value of $t$, I mean that you can use $t=-1$ or $t=10$. The point is that I began to explain what is a segment that links two points to later introduce a more general concept. Anyway.. I will edited the second (last) equation for better clarity. The main point here is the last equation, where I show yo how from your "misterious line" the usual equation is recovered. think the point $P_1$ as your point $(x,y,z)$ and $P_2-P_1$ as yout second point $(x_0,y_0,y_0)$ – HBR May 2 '17 at 19:45
• Forget the first equation, because it is the equation of a segment that begins from one point $P_2$ and ends into another point $P_2$ for $t\in[0,1]$. This first equation is clear because if we have a linear function of $t$, and for $t=0$, $P_1$ results and for $t=1$, $P_2$ results any $t$ in between will result in a point $P_3$ that is between the two points. But... what happens if $t$ is greater that 1? or $t$ is lower than 0? We do have another points that are in the direction of the segment line, $\textit{i.e.}$ we will have a straight line instead if $t\in(-\infty,\infty)$ – HBR May 2 '17 at 20:04
• The first equation is nothing but the preamble to the second one, the equation you asked for. It is like the problem to show how $3×2=6$ without explaining that $3+3=6$. Hope I have explained myself enough. – HBR May 2 '17 at 21:26
There are many different ways to represent 3D lines.
What you have here is a parametric form of the line representing all the points given a parameter $t \in \mathbb{R}$
• The linear interpolation between two points $${\bf r} = (1-t)\, {\bf r}_1 + t\, {\bf r}_2$$
• Or you can look at a point and direction $${\bf r} = {\bf r}_0 + t\, {\bf e}$$
But there are other representations. Without a parameter, but with equation(s) that need to be satisfied
• Point and direction
$$\frac{x-x_0}{e_x} = \frac{y-y_0}{e_y} = \frac{z-z_0}{e_z}$$
• Direction and moment vectors
$$\begin{matrix} L : \begin{Bmatrix} {\bf e} \\ {\bf r_0} \times {\bf e} \end{Bmatrix} & \mbox{or} & L :\begin{Bmatrix} {\bf r}_2 - {\bf r}_1 \\ {\bf r}_2 \times {\bf r}_1 \end{Bmatrix} \end{matrix}$$
• Intersection of two planes
$$\begin{matrix} {\bf r} \cdot {\bf n}_1 = d_1 \\{\bf r} \cdot {\bf n}_2 = d_2 \end{matrix}$$
• I am only interested in r = ( 1-t) + t r2 . What is that ???? can you please Isn't t restricted between 0 and 1? – Sedumjoy May 2 '17 at 19:43
• No $t$ can be anything between $-\infty$ to $+\infty$. If you arrange the line horizontally with point ${\bf r}_1$ to the left and point ${\bf r}_2$ to the right then if $t<0$ the point on the line is to the left of ${\bf r}_1$, when $0<t<1$ then between the points and if $t>1$ to the right of ${\bf r}_2$. – John Alexiou May 2 '17 at 20:05
• The position is a blend of the two fixed points. Like mixiing colors, you are mixing vectors. For example 30% of ${\bf r}_1$ and 70% of ${\bf r}_2$ is expressed as $${\bf r} = (1-0.3)\, {\bf r}_1 + 0.3\, {\bf r}_2$$ – John Alexiou May 3 '17 at 3:00
• I see the trick. they must add up to 1 so they fall on the line segment between the points by vector addition using r1 and r2 as position vectors and then you can extrapolate. I can see the utility now but only if you don't have an explicit direction vector and only have the two points. – Sedumjoy May 3 '17 at 19:49
• You got it now. – John Alexiou May 3 '17 at 19:56 | 2020-11-27T00:42:14 | {
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https://math.stackexchange.com/questions/3556884/solve-4-sqrt15x-4-sqrt15x-62 | # Solve $(4 + \sqrt15)^x + (4 - \sqrt15)^x =62$
Solve $$(4 + \sqrt15)^x + (4 - \sqrt15)^x =62$$
I was able to solve this equation by considering $$(4 + \sqrt15)^x$$ as some $$y$$. I got the quadratic equation
$$y^2-62y+1=0$$.
Therefore, $$y = 31 \pm 8 \sqrt15 = (4 + \sqrt15)^x$$
I am very close to getting the answer but I dont know how to compare both sides and obtain $$x$$. Should I just use trial and error (because it works when $$x = 2$$), but I want to know whether there is a more concrete way yof doing this.
• Forgive me if I'm missing something, but isn't it just $\log_{4+\sqrt{15}}\left(31 \pm 8 \sqrt{15} \right)$? The plus will give you 2, and working through to simplify the minus will give you the other solution. Feb 23 '20 at 6:43
One way is to observe that $$31 \pm 8\sqrt{15} = (\sqrt{15})^2 \pm (2)(4)\sqrt{15} + 4^2 = (4 \pm \sqrt{15})^2$$, a "sort of" complete the square process.
But that's a bit unsatisfactory since you're basically making a guess (albeit a justified one) that the expression is the square of the surd you want.
You might as well have started by "guessing" $$x=2$$ to begin with in the original equation.
Either way, you still have to figure out that $$x=-2$$ is also a valid solution based on the fact that $$4 + \sqrt{15}$$ is the reciprocal of $$4-\sqrt{15}$$.
A perhaps more satisfying solution is to use logarithms.
You have $$(4 + \sqrt{15})^x = 31 \pm 8\sqrt{15}$$
Take logarithms of both sides, it doesn't matter which base as long as it's the same on both sides. Natural logs are fine.
$$\log (4 + \sqrt{15})^x = \log(31 \pm 8\sqrt{15})$$
$$x\log (4 + \sqrt{15}) = \log(31 \pm 8\sqrt{15})$$
$$x = \frac{\log(31 \pm 8\sqrt{15})}{\log(4 + \sqrt{15})} = \pm 2$$
And you get both valid solutions immediately.
The only unsatisfying part about this is that you're forced to use a calculator. But no "leaps of insight" are required - the use of logarithms to solve this form of equation is very standard.
Hint;
$$31+8\sqrt{15}=2\cdot4\sqrt{15}=(4+\sqrt{15})^2$$
$$\sqrt{31+8\sqrt{15}}=4+\sqrt{15}$$
$$\implies\sqrt{31-8\sqrt{15}}=|4-\sqrt{15}|=4-\sqrt{15}$$ as $$4-\sqrt{15}=\dfrac1{4+\sqrt{15}}>0$$
Let $$(4+\sqrt{15})^x=y$$, then the Eq. becomes $$y^2-62 y+1=0 \implies y=(31\pm 8\sqrt{15})=(4 \pm\sqrt{15}) \implies x=-2,2.$$ Because $$(4 +\sqrt{15})(4-\sqrt{15})=1$$
\begin{align}(4+\sqrt{15})^x+(4-\sqrt{15})^x&=(4+\sqrt{15})^x+(4+\sqrt{15})^{-x}\\ &=e^{x\ln(4+\sqrt{15})}+e^{-x\ln(4+\sqrt{15})}=2\cosh\left(x\ln(4+\sqrt{15})\right)\\ &=62\end{align} So \begin{align}x\ln(4+\sqrt{15})&=\pm\cosh^{-1}31=\pm\ln\left(31+\sqrt{31^2-1}\right)\\ &=\pm\ln\left(31+8\sqrt{15}\right)=\pm\ln\left(\left(4+\sqrt{15}\right)^2\right)=\pm2\ln\left(4+\sqrt{15}\right)\end{align} So $$x=\pm2$$.
A bit late answer but maybe worth mentioning it.
• Let $$a= 4+\sqrt{15} \stackrel{4-\sqrt{15}=\frac 1{4+\sqrt{15}}}{\Rightarrow} f(x) = a^x + a^{-x}$$ is even $$(f(x) =f(-x))$$, hence we only need to consider $$x> 0$$, since if $$x_0$$ is a solution iff $$-x_0$$ is one.
• $$f'(x) = \ln a\left(a^x - a^{-x}\right)\stackrel{a>1}{>}0 \Rightarrow f'$$ is strictly increasing on $$(0,+\infty)$$. Hence, any solution on $$(0,+\infty)$$ is unique.
The following part might show how such problems are constructed with other $$a$$'s and other exponents, as well:
• Using Vieta you see that $$t_1 = a$$ and $$t_2 = a^{-1}$$ are the solutions of $$t^2-8t+1 = 0$$. Hence,
• $$r_n = a^n + \frac 1{a^n}$$ is the solution to the linear recurrence
$$r_{n+2}=8r_{n+1}-r_n \text{ with } r_0 = a^0+\frac 1{a^0}=2, r_1=a+\frac 1a=8$$
$$r_2 = 8\cdot 8-2 = 62 \stackrel{x>0}{\Rightarrow} \boxed{x=2} \text{ is the unique positive solution.}$$
And, because of $$f(-x) = f(x)$$, the other corresponding unique negative solution is $$\boxed{x=-2}$$. | 2022-01-17T07:33:05 | {
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https://forum.math.toronto.edu/index.php?PHPSESSID=6844dijse4ia33ib0cfvekmei3&topic=2440.msg7292 | ### Author Topic: 1.6 Q5 (Read 365 times)
#### Nathan
• Newbie
• Posts: 2
• Karma: 0
##### 1.6 Q5
« on: October 06, 2020, 06:08:35 PM »
Question: $\int_y Re(z) dz$ where $y$ is the line segment from 1 to $i$.
I can't get the same answer as the one in the textbook.
Answer in textbook: $\frac{1}{2}(i-1)$
$y(t)=$
$= 1 + (i - 1)t$
$= (1 - t) + it$
$y'(t) = i - 1$
$\int_y Re(z) dz=$
=$\int_1^i (1-t)(i-1)dt$
$=\int_1^i i - 1 - ti + t dt$
$=ti - t - \frac{t^2i}{2} + \frac{t^2}{2}|^i_1$
$=-1 -i + \frac{i}{2} - \frac{1}{2} - (i - 1-\frac{i}{2} + \frac{1}{2})$
$=-i - 1$
What is wrong with my answer?
#### smarques
• Newbie
• Posts: 2
• Karma: 0
##### Re: 1.6 Q5
« Reply #1 on: October 06, 2020, 06:29:36 PM »
Hi Nathan.
The mistake seems to be in your bounds of integration, not the process itself. With the way you parameterized the line, the integral should be from 0 to 1, not 0 to i. | 2021-12-01T07:38:31 | {
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https://math.stackexchange.com/questions/484457/how-long-time-will-it-take-to-sort-106-numbers-and-109-numbers-for-two-alg | # How long time will it take to sort $10^6$ numbers and $10^9$ numbers for two algorithms if they take the same time to sort $1000$ numbers?
Question We are given two sorting algorithms. The running time of Algorithm $1$ is $O(n^2)$ while Algorithm $2$ has running time $O(n \cdot log (n))$. Assume that Algorithm $1$ and Algorithm $2$ take $1$ second to sort out $1000$ numbers. How long time would it take for each algorithm to sort out a) $10^6$ numbers b) $10^9$ numbers?
Comments I realize this may be a simple question, but I'm just starting this material. A hint or a full answer would be appreciated.
You've seen two answers that illustrate that big-O timing estimates don't tell you much about actual running times, but to illustrate how big a difference the actual orders matter, let's dispense with the big-O estimates and suppose we actually knew that Algorithm 1 took precisely $c_1n^2$ seconds to sort $n$ numbers and Algorithm 2 took exactly $c_2n\log_{10} n$ seconds to sort $n$ numbers (I chose log to the base ten to make the calculations simpler; choosing a different base would only change the constant $c_2$).
If Algorithm 2 took 1 second to sort $10^3$ numbers, we'd have $$1=c_1(10^3)^2=c_1\,10^6$$ so $c_1=1/10^6$. If Algorithm 2 also took 1 second to sort $10^3$ numbers, we'd have $$1=c_2(10^3\log(10^3))=c_2(3\cdot10^3)$$ so $c_2=1/(3\cdot10^3)$.
Now let's see what happens when we sort $10^6$ numbers. Algorithm 1 will take $$\frac{1}{10^6}\cdot(10^6)^2=\frac{10^{12}}{10^6}=10^6\text{ seconds}$$ which works out to be about $11.5$ days. On the other hand, Algorithm 2 will take $$\frac{1}{3\cdot10^3}\cdot(10^6\log10^6)=\frac{6\cdot10^6}{3\cdot10^3}=2\cdot10^3\text{ seconds}$$ which is about $33.3$ minutes. Hmm, 11 days versus half an hour. I know which algorithm I'd use.
For larger inputs the difference in times becomes even more dramatic. With $10^9$ numbers to sort, you can work out that Algorithm 1 would take about 31688 years, while Algorithm 2 would take about 34.7 days.
Notice that Algorithm 1 had a much smaller constant multiplier than Algorithm 2, but in the long run that didn't matter anywhere near as much as the asymptotic order of their running times. Simply said, an $n\log n$ algorithm will eventually clobber an $n^2$ algorithm, no matter what the constant multiple is.
As it is stated, the answer is "we don't know". It follows from the $O(\cdot)$ notation — roughly speaking, you don't have enough information to infer anything. You only know the asymptotic behavior, and not even the constants.
Algorithm 1 might very well be as fast as Algorithm 2 as long as $n$ is less than, say, $10^{10}$, and then break down.
Even worse, the big-Oh notation is only an upper bound. For all we known, both could have the same running time $O(n\log n)$, since if Algorithm 1 had running time $O(n\log n)$, it would also trivially have running time $O(n^2)$.
Mathematically, big-$O$ notation doesn't work like this. $O(n^2)$ denotes an equivalence class of functions based on their asmyptotics.
So, for example, $f$ defined by $f(n)=n^2$ belongs to $O(n^2)$. So does the function $g$ defined by $g(n)=10^{1000}n^2$.
So, two different algorithms that take time $O(n^2)$ can still have vastly different performance for a specific task. The big-$O$ notation is meant to capture that scalability of these algorithms as $n \rightarrow \infty$.
However, if you're using this for a practical application, the multiplicative constant out the front will probably not be too large. Ignoring the $O(..)$ can often give a reasonable intuition as to how long the algorithm will take.
So, we might approximate that the $O(n^2)$ algorithm would take time $(10^6)^2=10^{12}$ to process $10^6$ items and the $O(n \log n)$ algorithm would take time $10^9 \log {10^9} \approx 2 \times 10^{10}$ to process $10^9$ items.
In most real-world cases, the comparison won't be too bad, but on the other hand, it might also be horrible.
• Exactly. To drive it to the extreme, even all practical implementations of algorithms on computers can be considered O(1) (with a big constant in front). Clearly, this is because there is only a finite amount of possibilities for input data (finite amount of memory and finite amount of machine numbers). – Andreas H. Sep 5 '13 at 1:11 | 2019-08-24T22:37:28 | {
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https://math.stackexchange.com/questions/1495953/is-0-x-0x1-an-open-cover-of-0-1 | # Is $\{(0,x) : 0<x<1\}$ an open cover of $(0,1)$?
For $E_x := (0,x)$ where $0<x<1$, is $\epsilon := \{E_x:0<x<1\}$ an open cover of $(0,1)$?
We can prove that each $E_x$ is open; take $y\in E_x$ and let $r = \min\{d(0,y), d(x,y)\}$. Then $N_r(y)\subset E_x$. Since this is for any $y\in E_x$, then $E_x$ is open.
If it is the case that $\epsilon = \{E_x\}$ is an open cover, how do we prove this fact?
Moreover, suppose $\epsilon$ is an open cover. We want to show it has no finite subcover of $(0,1)$. Is the following proof correct?
Suppose there exists $x_1, x_2, ..., x_n$ such that $\bigcup_{j:1\leq j \leq n}^{n}E_{x_j}\supset(0,1)$. Choose $x = \max_{1\leq j \leq n}(x_j)$. Then $E_x = (0,x) \subset (0,1)$. Since all other $E_x$ are contained in this $E_x$, the union of the $E_x$'s cannot be a finite subcover of $(0,1)$.
• The notation is a little misleading: $E_x$ is a set of intervals, and this set does not depend on $x$. It would be clearer in my view to set $E_x := (0, x)$ and define the (candidate) open cover to be $\mathcal{E} := \{E_x : 0 < x < 1\}$ . – Travis Willse Oct 24 '15 at 21:52
• thanks. i fixed the notation – socrates Oct 24 '15 at 21:58
• Actually, all in all there is no need for the E_x notation at all. It's not incorrect. just not needed. – fleablood Oct 24 '15 at 22:22
To show that $\epsilon$ is an open cover, it's enough to show (1) that its elements are all open (which has already been done in the question statement) and (2) that its union is $(0, 1)$.
To show (2), it's enough for each $y \in (0, 1)$ to show that there is some $E_x \in \epsilon$ such that $y \in E_x = (0, x)$. Can we find such an $x$?
The proof that the cover has no finite subcover (and hence that $(0, 1)$ is noncompact) is almost correct: One needs to show that the union $E_x = (0, x)$ is not all of $(0, 1)$, but this is immediate, as $x < 1$.
• Such an $x$ would simply be any $y<x<1$, correct? i.e: Take some $x\in (0,1)$. There is a $y$ s.t. $0<x<y<1$. Then $E_x\subset E_y$ and $E_y\subset (0,1)$. We then have $\bigcup_{x}{E_x} = (0,1)$. So we know $\epsilon$ is an open cover of $(0,1)$. (apologies for swapping x, y; i copied this over) – socrates Oct 24 '15 at 22:53
$E_x := (0,x)$ is $\epsilon := \{E_x : 0<x<1\}$ an open cover of (0, 1)?
Is it an open cover?
1) Are the $E_x$ each open? Yes. You showed that. Also presumably earlier in the course you were shown all open intervals are open sets.
2)Do they cover (0, 1)? Yes, if $x \in (0, 1)$ then $0 < x < 1$ and we can find a y such that $0 < x < y < 1$ so $x \in E_y$ and $E_y \in \epsilon$. So, yes $\epsilon$ covers (0, 1)
So $\epsilon$ is an open cover.
Does $\epsilon$ have a finite subcover.
You argued correctly that if so then there would be an $E_y = (0,y) \subset (0,1)$ for all other of the $E_x$ are subsets of $E_y$ so $E_y$ would have to cover (0,1). BUT you didn't show it doesn't. Because you didn't show $E_y$ was a proper subset.
It is and it doesn't.
It doesn't because because $y \notin E_y$. so $E_y$ doesn't cover (0,1). So there is not finite subcover.
• @Did Ooooooooooops! – fleablood Oct 24 '15 at 22:19
• For 2), the argument obviously makes sense. However, is this as rigorous as needed to move from the $E_y \in \epsilon$ to $\epsilon$ covering (0,1)? I guess I am struggling to grasp the idea of an open cover, and perhaps a more rigorous justification of the intuitive idea that you can 'keep adding on ys ' to eventually get the complete union – socrates Oct 24 '15 at 22:29
• I think it is. A cover is a colllection of sets that "cover" a set. In other words $E = \{S_\alpha|$ a bunch of sets$\}$ "covers" $A$ if $A \subset \cup_{S_\alpha \in E}S_\alpha$. In other words: for all $x \in A$ then $x \in S_{\alpha}$ for some $S_{\alpha} \in E$. That's all. – fleablood Oct 24 '15 at 22:41 | 2020-10-20T06:52:28 | {
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http://mathhelpforum.com/algebra/13734-simplification.html | # Math Help - Simplification
1. ## Simplification
Have a go at this:
sqrt{7 - 4sqrt{3}} = 2 - sqrt{3}
How do I get from the left to the right hand side of the equation? When I got the answer to tan(15) and I typed it in, the RHS is what the calculator simplified it to. But HOW?!
2. Keep squaring on both sides until the roots are gone
3. Don't see how that's very relevant... I'm not solving for a variable.
4. Originally Posted by DivideBy0
Have a go at this:
sqrt{7 - 4sqrt{3}} = 2 - sqrt{3}
How do I get from the left to the right hand side of the equation? When I got the answer to tan(15) and I typed it in, the RHS is what the calculator simplified it to. But HOW?!
Square both sides:
the left hand side is:
7 - 4 sqrt(3),
and the rigth hand side is:
(2 - sqrt(3))(2-sqrt(3)) = 4 -4 sqrt(3) + 3 = 7 - 4 sqrt(3).
Now we need the tan 1/2 angle formula:
tan(A/2) = sqrt((1-cos(A))/(1+cos(A))
Now put A=30 degrees, then cos(30)=sqrt(3)/2, SO:
tan(15) = sqrt((1-sqrt(3)/2)/(1+sqrt(3)/2)) = sqrt((2-sqrt(3)/(2+sqrt(3))
but:
(2-sqrt(3)/(2+sqrt(3) = (2-sqrt(3))^2/[(2+sqrt(3))(2-sqrt(3))]
.............................= (4 -4 sqrt(3) +3)/(4-3) = 7-sqrt(3)
So:
tan(15) = sqrt(7 -sqrt(3)) = 2 - sqrt(3)
RonL
5. If you just had sqrt{7-4sqrt{3}}, with no other information, or knowledge that it was a half-angle, would it be possible to simplify?
6. Hello, DivideBy0!
Suppose we don't know the RHS . . .
Find: .sqrt[7 - 4·sqrt{3}]
Here is a primitive method for finding the square root.
. . . . . . . . . _ - . . ._______
Let: .a + b·√3 .= .√7 - 4·√3 . where a and b are integers
. . . . . . . . . . . . . . . . . . . ._ . . . . . ._______
Square both sides: .(a + b√3)² .= .(√7 - 4·√3)²
. . . . . . . . . . . . . . . . . ._ . . . . . . - . - . . ._
and we have: .a² + 2ab√3 + 3b² .= .7 - 4·√3
. . . . . . . . . . . . . . . . . . . _ . . . . . . . ._
Then: .(a² + 3b²) + (2ab)√3 .= .7 - 4·√3
Equate coefficients: . a² + 3b² .= .7 .[1]
. . . . . . . . . . . . . . . . . 2ab . . = -4 .[2]
And solve the system of equations . . .
From [2], we have: .b = -2/a .[3]
Substitute into [1]: .a² + 3(-2/a)² .= .7
. . which simplifies to: .a^4 - 7a² + 12 .= .0
. . which factors: .(a² - 3)(a² - 4) .= .0
. . . . . . . . . . . . . . . . . . . _
. . and has roots: .a .= .±√3, ±2
Since a is an integer, we will use: .a = 2
Substitute into [3]: .b = -2/2 = -1
. . . . . . . . . . . _______ . . . . . . . _
. . Therefore: .√7 - 4·√3 . = . 2 - √3
7. Hmm interesting, thanks for that
8. Originally Posted by DivideBy0
Hmm interesting, thanks for that
Note, Soroban's method does not always work.
This is mine 53th Post!!!
9. Just a quickie, for Soroban's method, why was +2 chosen over -2? How did you know it was the right one to use?
10. Hello, DivideBy0!
Why was +2 chosen over -2? How did you know it was the right one to use? .I didn't
It was laziness on my part . . . sorry.
Choosing a = +2, we get: b = -1, _
. . . and the square root is: . 2 - √3
Choosing a = -2, we get: b = 1, - -_
. . . and the square root is: .-2 + √3
As you suspected, there are two square roots for any quantity (except 0):
. . . . _______ . . . . . . . ._
. . . √7 - 4·√3 .= .±(2 - √3) | 2016-06-26T00:54:45 | {
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https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Bijection | # Equivalence of Definitions of Bijection
## Theorem
The following definitions of the concept of Bijection are equivalent:
### Definition 1
A mapping $f: S \to T$ is a bijection if and only if both:
$(1): \quad f$ is an injection
and:
$(2): \quad f$ is a surjection.
### Definition 2
A mapping $f: S \to T$ is a bijection if and only if:
$f$ has both a left inverse and a right inverse.
### Definition 3
A mapping $f: S \to T$ is a bijection if and only if:
the inverse $f^{-1}$ of $f$ is a mapping from $T$ to $S$.
### Definition 4
A mapping $f \subseteq S \times T$ is a bijection if and only if:
for each $y \in T$ there exists one and only one $x \in S$ such that $\tuple {x, y} \in f$.
### Definition 5
A relation $f \subseteq S \times T$ is a bijection if and only if:
$(1): \quad$ for each $x \in S$ there exists one and only one $y \in T$ such that $\tuple {x, y} \in f$
$(2): \quad$ for each $y \in T$ there exists one and only one $x \in S$ such that $\tuple {x, y} \in f$.
## Proof
### Definition 1 iff Definition 2
From Injection iff Left Inverse, $f$ is an injection if and only if $f$ has a left inverse mapping.
From Surjection iff Right Inverse, $f$ is a surjection if and only if $f$ has a right inverse mapping.
Putting these together, it follows that:
$f$ is both an injection and a surjection
$f$ has both a left inverse and a right inverse.
$\Box$
### Definition 1 iff Definition 3
This is demonstrated in Mapping is Injection and Surjection iff Inverse is Mapping.
$\Box$
### Definition 1 iff Definition 4
Let $f: S \to T$ be a bijection by definition 1.
Then by definition:
$f$ is an injection
$f$ is a surjection
By definition of injection:
every element of $T$ is the image of at most $1$ element of $S$.
By definition of surjection:
every element of $T$ is the image of at least $1$ element of $S$.
So:
for each $y \in T$ there exists one and only one $x \in S$ such that $\tuple {x, y} \in f$.
Thus $f$ is a bijection by definition 4.
$\Box$
Let $f: S \to T$ be a bijection by definition 4.
Then by definition:
for each $y \in T$ there exists one and only one $x \in S$ such that $\tuple {x, y} \in f$.
But:
every element of $T$ is the image of at most $1$ element of $S$
defines an injection
and:
every element of $T$ is the image of at least $1$ element of $S$
defines a surjection.
From Injection iff Left Inverse, $f$ is an injection if and only if $f$ has a left inverse mapping.
From Surjection iff Right Inverse, $f$ is a surjection if and only if $f$ has a right inverse mapping.
Putting these together, it follows that:
$f$ is an injection
$f$ is a surjection
Thus $f$ is a bijection by definition 1.
$\Box$
### Definition 3 iff Definition 4
#### Necessary Condition
Let $f^{-1}: T \to S$ be a mapping.
Then by definition:
$\forall y \in T: \exists x \in S: \tuple {y, x} \in f^{-1}$
Thus for all $y \in T$ there exists at least one $x \in S$ such that $\tuple {y, x} \in f^{-1}$.
Also by definition of mapping:
$\tuple {x_1, y} \in f^{-1} \land \tuple {x_2, y} \in f^{-1} \implies x_1 = x_2$
Thus for all $y \in T$ there exists at most one $x \in S$ such that $\tuple {y, x} \in f^{-1}$.
Hence it has been demonstrated that $y \in T$ there exists a unique $x \in S$ such that $\tuple {y, x} \in f^{-1}$.
$\Box$
#### Sufficient Condition
Let $f$ be such that for all $y \in T$ there exists a unique $x \in S$ such that $\tuple {y, x} \in f^{-1}$.
Then by definition $f^{-1}$ is a mapping.
$\Box$
### Definition 3 iff Definition 5
#### Necessary Condition
Let $f: S \to T$ be a mapping such that $f^{-1}: T \to S$ is also a mapping.
Then as $f$ is a mapping:
for all $x \in S$ there exists a unique $y \in T$ such that $\tuple {x, y} \in f$.
Similarly, as $f^{-1}$ is also a mapping:
for all $y \in T$ there exists a unique $x \in S$ such that $\tuple {y, x} \in f^{-1}$.
$\Box$
#### Sufficient Condition
Let $f \subseteq S \times T$ be a relation such that:
$(1): \quad$ for each $x \in S$ there exists one and only one $y \in T$ such that $\tuple {x, y} \in f$
$(2): \quad$ for each $y \in T$ there exists one and only one $x \in S$ such that $\tuple {x, y} \in f$.
Then by definition:
from $(1)$, $f$ is a mapping.
from $(2)$, $f^{-1}$ is a mapping. | 2021-09-21T01:56:54 | {
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http://mathoverflow.net/questions/78361/which-integers-take-the-form-x2-xy-y2 | # which integers take the form x^2 + xy + y^2 ?
I guess one way of putting it, when does the series $\sum_{x,y \in \mathbb{Z}} q^{x^2+xy+y^2}$ have nonzero coefficients?
The analogous answer for $\sum_{x,y \in \mathbb{Z}} q^{x^2+y^2}$ is that $q^n$ appears when $\mathrm{ord}_p(n)$ be even for all primes.
Is there a closed form for either of these with quadratic characters or theta functions or something?n
-
For $x^2+y^2$ it's $n=0$ and positive integers $n$ such that ${\rm ord}_p(n)$ even for all primes congruent to $-1$ mod 4. For $x^2+xy+y^2$, likewise, the valuation must be even for all primes congruent to $-1 \bmod 3$. This works because the relevant quadratic imaginary rings ${\bf Z}[i]$ and ${\bf Z}[(1+\sqrt{-3})/2]$ have unique factorization. – Noam D. Elkies Oct 17 '11 at 17:18
What are your criteria for a form to be closed in your second question? – S. Carnahan Oct 17 '11 at 18:20
is it okay if I come back with a sharpened version of this question? i don't know whether to revise this one or just start a new one. – john mangual Oct 17 '11 at 22:48
@John: Start a new one, so as not render irrelevant the existing answers. – Cam McLeman Oct 19 '11 at 12:30
This question is actually a duplicate of one appearing on math stack exchange. See: math.stackexchange.com/questions/44139/… – Eric Naslund Oct 19 '11 at 16:04
Let $r_2(n)$ be the number of representations of $n$ as a sum of two squares, and let $l(n)$ be the number of ways to write $n$ as $x^2+xy+y^2$. Then as you mentioned we have that $$\sum_{y,x\in \mathbb{Z}} q^{x^2+y^2}=\sum_{n=0}^\infty r_2(n) q^n \ \text{and} \ \sum_{y,x\in \mathbb{Z}} q^{x^2+xy+y^2}=\sum_{n=0}^\infty l(n) q^n.$$
Exact number of representations: We can explicitely write down $r_2(n)$, the number of representations of $n$ as a sum of two squares. See Sum of Squares Function.
Also, in a similar manner we can explicitely write down $l(n)$, the number of representations of $n$ as $x^2+xy+y^2$. See the answer on this math stack exchange post.
Theta Functions: We can evaluate the infinite series in terms of Jacobi theta functions without much difficulty. To deal with the sum of squares, notice that
$$\sum_{y,x\in \mathbb{Z}} q^{x^2+y^2} =\left( \sum_{n=-\infty}^\infty q^{n^2}\right)^2= \vartheta_3(q)^2.$$ Next, we can transform $x^2 +xy+y^2$ into $\frac{n^2+3m^2}{4}$ where $m,n$ must have the same parity. Then $$\sum_{x,y}q^{x^{2}+xy+y^{2}}=\sum_{\begin{array}{c} x,y\\ both\ odd \end{array}}q^{\frac{x^{2}+3y^{2}}{4}}+\sum_{\begin{array}{c} x,y\\ both\ even \end{array}}q^{\frac{x^{2}+3y^{2}}{4}}$$
$$=q\sum_{n,m\in\mathbb{Z}}q^{n(n+1)+3m(m+1)}+\sum_{n,m\in\mathbb{Z}}q^{n^{2}+3m^{2}}.$$This can be rewritten in terms of the jacobi theta functions as $$\vartheta_2(q)\vartheta_2(q^3)+\vartheta_3(q)\vartheta_3(q^3).$$
Hope that helps,
-
Just to point out that $x^2 + x y + y^2$ represents exactly the same numbers as $x^2 + 3 y^2,$ just with different frequency. The similar fact for indefinite forms is that $x^2 + x y - y^2$ represents exactly the same numbers as $x^2 - 5 y^2.$ In both cases $x^2 + xy \pm y^2$ the sum is not even unless $x,y$ are both even, which says that the order with which the prime 2 divides the result must be even. For odd primes, the analogous statement for primes dividing the result can be found using the Legendre symbol, in your case $(-3 | p) = (p | 3).$ It is easy to prove that $x^2 + x y + k y^2$ represents a superset of $x^2 + (4k-1) y^2,$ for almost all $k$ a strict superset.
In his recent book, Number Theory in the Spirit of Ramanujan, Bruce Berndt discusses representation problem for (1, 0, 1), (1, 0, 2), (1, 1, 1), (1, 0, 3).
This is a quote from the arXiv version of Alexander Berkovich and Hamza Yesilyurt which has:
Comments: 26 pages, no figures, fun to read
On page 149 of Rational Quadratic Forms by Cassels, Lemma 6.3 gives a count for the number of primitive representations of a number $n = x^2 + y^2,$ and this lemma can be applied to find the total number of representations, even if there are none primitive. Something very similar should work for $x^2 + x y + y^2,$ perhaps it is in the Berndt book.
As far as "closed form," the best we can expect is in the world of modular forms and, in particular, "eta quotients." You might just look at all Alex Berkovich's manuscripts on the arXiv, he plays with these infinite sums all day.
EDIT, 24 January 2012: Exercise 2 on page 80 of Introduction to the Theory of Numbers by Leonard Eugene Dickson: The number of representations of positive integer $n$ by $x^2 + x y + y^2$ is $6 E(n),$ where $E(n)$ is the excess of the number of divisors that are $1 \pmod 3$ over the number of divisors that are $2 \pmod 3.$
-
One can start showing the following:
The integers $n$ which are of the form $x^2+xy+y^2$, for two relatively prime integers $x,y$ are precisely those positive integers occurring as divisors of $m^2+m+1$, for some integer $m$.
In other words, the polynomial $f(x)=x^2+x+1$ has the property that the positive divisors of the integers it represents are precisely those integers that can be properly represented by its homogenization $F(x,y)=x^2+xy+y^2$ ("properly" here refers to the condition $(x,y)=1$).
The proof uses the fact that the imaginary quadratic order $\mathbf{Z}[x]/f(x)$ has class number one, as anticipated by Elkies. It goes as follows:
Let $n$ be a positive divisor of $m^2+m+1$, for some integer $m$. Consider the quadratic form in $x,y$ given by:
$Q(x,y)=\frac{m^2+m+1}{n}x^2-(2m+1)xy+ny^2$;
it has integer coefficients, positive definite, and has discriminant equal to $-3$ (in particular it is primitive).
Since $h(-3)=1$, there is only one reduced, positive definite quadratic form of discriminant $-3$. This is $E(x,y)=x^2+xy+y^2$. Therefore $Q$ and $E$ are properly equivalent (that is there is a determinant-one change of variables taking one into the other), and since $Q$ certainly properly represents $n$, so does $E$.
The converse is similar and uses the fact that if a quadratic form $Q$ properly represents an integer $n$, then $Q$ is properly equivalent to a form of the type $nx^2+bxy+cy^2$ (this is lemma 2.3 of Cox's wonderful book "Primes of the form x^2+ny^2").
Once you understand the positive integers $n$ that are properly represented by $E$, then you can get them all, after scaling by squares.
The original problem is then reduced to understanding those integers $n$ for which $x^2+x+1$ has a zero in $\mathbf{Z}/(n)$. Using the Chinese Remainder Theorem, this can be reduced to the case where $n$ is a prime power $p^s$. Then for $p\neq 3$ Hensel's Lemma tells you that your equation has a solution mod $p^s$ if and only if it has a solution mod $p$. With quadratic reciprocity you can conclude that any prime divisor of $n$ has to be congruent to $1$ mod $3$. I am at the moment missing how you can solve the equation $x^2+x+1=0$ in $\mathbf{Z}/3^s$, but I think that a version of Hensel's Lemma applies. I'll think about it.
[EDIT: The equation $x^2+x+1=0$ has a solution in $\mathbf{Z}/(3^s)$, with $s\geq 1$, if and only if $s=1$. (Therefore $x^2+xy+y^2$ represents properly only $3$ and $1$ as powers of $3$.) One can see this by checking that there is no solution for $s=2$, and therefore for $s>2$. More generally, if $p>2$ then $x^{p-1}+x^{p-2}+\ldots+x+1=0$ has no solutions in $\mathbf{Z}/(p^s)$ with $s>1$. The $p$--adic valuation of an integer of the form $x^{p-1}+x^{p-2}+\ldots+x+1=(x^p-1)/(x-1)$ is either zero or one.]
(Suggested reading. Cox's book quoted above and Serre's paper: $\Delta=b^2-4ac$)
-
While interesting, I think you are over complicating the problem. For a complete solution, and a way to explicitly write down the number of representations, see math.stackexchange.com/questions/44139/… – Eric Naslund Oct 19 '11 at 16:05
You have a point, I just felt that this was the chance to stress this property of the polynomial $x^2+x+1$ (which also $x^2+1$ shares). One thing is certain: you cannot remove Quadratic Reciprocity from the picture! – Tommaso Centeleghe Oct 19 '11 at 16:17
+1, I agree now. – Eric Naslund Oct 19 '11 at 17:57 | 2016-05-01T18:06:19 | {
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https://math.stackexchange.com/questions/3806526/coefficient-of-a-polynomial | # coefficient of a polynomial
Show that the coefficient of $$[x^nu^m]$$ in the bivariate generating function $$\dfrac{1}{1-2x+x^2-ux^2}$$ is $${n+1\choose n-2m}.$$
I tried to do this by using the multinomial theorem (an extension of the binomial theorem), which basically states that for terms $$x_1,\cdots, x_r, n\in \mathbb{N}_{\geq 0}, (x_1+\cdots + x_r)^n = \sum_{k_1+\cdots + k_r = n} \dfrac{n!}{k_1! \cdots k_r!}x_1^{k_1}\cdots x_r^{k_r}.$$
This gives that the given bivariate generating function is equal to $$\sum_{n\geq 0}(2x-x^2+ux^2)^n = \sum_{n\geq 0} \sum_{k_1+k_2 + k_3 = n} \dfrac{n!}{k_1!k_2!k_3!} (2x)^{k_1}(-x^2)^{k_2}(ux^2)^{k_3}$$.
Thus the coefficient of $$[x^n u^m]$$ should be $$\sum_{k_1 + 2k_2 = n-2m} \dfrac{(n-k_2-m)!}{k_1!k_2!m!}2^{k_1} (-1)^{k_2} .$$ I can further simplify this by replacing $$k_2$$ with $$\dfrac{n-2m-k_1}{2},$$ but I'm not sure how to get the desired result from that. Is there some other useful property of polynomials? I also realized that $$\sum_{m\geq 0} {n+1\choose n-2m} = 2^n,$$ which can be shown using Pascal's identity, though I'm not sure if this is useful.
• Technically this is not a polynomial (that would require finitely many terms of form $x^i u^j$), this is more of a bivariate generating function. – Sil Aug 28 '20 at 20:12
• @Sil sorry for the poor use of terminology. It can also be called a formal power series. – Fred Jefferson Aug 28 '20 at 20:18
\begin{align} &\bbox[5px,#ffd]{\bracks{x^{n}u^{m}}{1 \over 1 - 2x + x^{2} - ux^{2}}} = \bracks{x^{n}u^{m}}{1 \over \pars{1 - x}^{2} - ux^{2}} \\[5mm] = &\ \bracks{x^{n}u^{m}}{1 \over \pars{1 - x}^{2}} \bracks{1 - {x^{2} \over \pars{1 - x}^{2}}\,u}^{-1} = \bracks{x^{n}}{1 \over \pars{1 - x}^{2}} \bracks{x^{2} \over \pars{1 - x}^{2}}^{m} \\[5mm] = &\ \bracks{x^{n - 2m}}\pars{1 - x}^{-2m - 2} = {-2m - 2 \choose n - 2m}\pars{-1}^{n - 2m} \\[5mm] = &\ {-\bracks{-2m - 2} + \bracks{n - 2m} - 1 \choose n - 2m} = \bbx{\large{n + 1 \choose n - 2m}} \\ & \end{align}
It might be more helpful to factorize the quadratic expression first (taking it as a variable in $$x$$). This way we can extract coefficient of $$x^n$$ ($$u$$ taken as a constant) and then coefficient of $$u^m$$ (in other words $$[x^n u^m]f(x,u)=[u^m]([x^n]f(x,u))$$. So, by factorization of denominator we arrive at $$\dfrac{1}{1-2x+x^2-ux^2}=\frac{1}{1-(1+\sqrt{u})x}\cdot \frac{1}{1-(1-\sqrt{u})x}$$ which by geometric series is $$(\sum_{i \geq 0}(1+\sqrt{u})^ix^i) \cdot (\sum_{j \geq 0}(1-\sqrt{u})^j x^j ).$$ So we get a coefficient of $$x^n$$ $$\sum_{k=0}^{n}(1+\sqrt{u})^k(1-\sqrt{u})^{n-k}\tag{*}$$ and the problem reduces to finding coefficient of $$u^m$$ in $$(*)$$. We can evaluate the expression for example by writing it as $$(1-\sqrt{u})^n\sum_{k=0}^{n}\left(\frac{1+\sqrt{u}}{1-\sqrt{u}}\right)^k$$ and spot the finite geometric series with $$q=\frac{1+\sqrt{u}}{1-\sqrt{u}}$$, so we can just use well-known formula for the sum $$\frac{q^{n+1}-1}{q-1}$$. After some messy algebra we get $$\frac{1}{2\sqrt{u}}[(1+\sqrt{u})^{n+1}-(1-\sqrt{u})^{n+1}],$$ which finally by Binomial theorem gives $$\frac{1}{2\sqrt{u}}\sum_{m=0}^{n+1}\binom{n+1}{m}\sqrt{u}^{m}(1-(-1)^{m}).$$ For even $$m$$ the terms vanish and we are left with $$\sum_{m=0}^{\lfloor n/2 \rfloor}\binom{n+1}{2m+1}u^{m}.$$ Now just read off the coefficient, also perhaps use $$\binom{n}{k}=\binom{n}{n-k}$$. | 2021-01-17T01:21:28 | {
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http://rasbt.github.io/mlxtend/user_guide/math/num_permutations/ | # num_permutations: number of permutations for creating subsequences of k elements
A function to calculate the number of permutations for creating subsequences of k elements out of a sequence with n elements.
from mlxtend.math import num_permutations
## Overview
Permutations are selections of items from a collection with regard to the order in which they appear (in contrast to combinations). For example, let's consider a permutation of 3 elements (k=3) from a collection of 5 elements (n=5):
• collection: {1, 2, 3, 4, 5}
• combination 1a: {1, 3, 5}
• combination 1b: {1, 5, 3}
• combination 1c: {3, 5, 1}
• ...
• combination 2: {1, 3, 4}
In the example above the permutations 1a, 1b, and 1c, are the "same combination" but distinct permutations -- in combinations, the order does not matter, but in permutation it does matter.
The number of ways to combine elements (without replacement) from a collection with size n into subsets of size k is computed via the binomial coefficient ("n choose k"):
To compute the number of permutations with replacement, we simply need to compute $n^k$.
## Example 1 - Compute the number of permutations
from mlxtend.math import num_permutations
c = num_permutations(n=20, k=8, with_replacement=False)
print('Number of ways to permute 20 elements'
' into 8 subelements: %d' % c)
Number of ways to permute 20 elements into 8 subelements: 5079110400
from mlxtend.math import num_permutations
c = num_permutations(n=20, k=8, with_replacement=True)
print('Number of ways to combine 20 elements'
' into 8 subelements (with replacement): %d' % c)
Number of ways to combine 20 elements into 8 subelements (with replacement): 25600000000
## Example 2 - A progress tracking use-case
It is often quite useful to track the progress of a computational expensive tasks to estimate its runtime. Here, the num_combination function can be used to compute the maximum number of loops of a permutations iterable from itertools:
import itertools
import sys
import time
from mlxtend.math import num_permutations
items = {1, 2, 3, 4, 5, 6, 7, 8}
max_iter = num_permutations(n=len(items), k=3,
with_replacement=False)
for idx, i in enumerate(itertools.permutations(items, r=3)):
# do some computation with itemset i
time.sleep(0.01)
sys.stdout.write('\rProgress: %d/%d' % (idx + 1, max_iter))
sys.stdout.flush()
Progress: 336/336
## API
num_permutations(n, k, with_replacement=False)
Function to calculate the number of possible permutations.
Parameters
• n : int
Total number of items.
• k : int
Number of elements of the target itemset.
• with_replacement : bool
Allows repeated elements if True.
Returns
• permut : int
Number of possible permutations.
Examples
For usage examples, please see http://rasbt.github.io/mlxtend/user_guide/math/num_permutations/ | 2023-02-04T03:24:36 | {
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https://math.stackexchange.com/questions/3191318/closed-form-of-recurrent-arithmetic-series-summation | Closed form of recurrent arithmetic series summation
Knowing that $$\sum_{i=1}^n i = \frac{n(n+1)}{2}$$ how can I get closed form formula for $$\sum_{i=1}^n \sum_{j=1}^i j$$ or $$\sum_{i=1}^n \sum_{j=1}^i \sum_{k=1}^j k$$ or any x times neasted summation like above
• You won't be able to solve this just by using the initial equation. – Peter Foreman Apr 17 at 17:19
• Go step by step: $\sum_{i=1}^n \sum_{j=1}^i j=\sum_{i=1}^n \frac{i(i+1)}{2}=\frac12\cdot \color{red}{\sum_{i=1}^n i^2}+\frac12 \cdot \sum_{i=1}^n i$. The red colored part cannot be solved with the first formula. – callculus Apr 17 at 17:21
• Use the formulae for the sum of $k^2$ and $k^3$ – George Dewhirst Apr 17 at 17:34
• Hint: $n={n \choose 1}$, $n(n+1)/2={n+1\choose 2}$. Now have a look at the hockey-stick identity. – Jean-Claude Arbaut Apr 17 at 17:51
• @Jean-ClaudeArbaut Do you mind if I write an answer using this now? – Peter Foreman Apr 17 at 17:53
Let $$f_k(n)$$ be the closed form of the summation nested $$k$$ times. We know that $$f_1(n)=\frac12n(n+1)=\binom{n+1}{2}$$ $$f_k(n)=\sum_{j=1}^n f_{k-1}(j)$$ So for the next function $$f_2(n)$$ we have $$f_2(n)=\sum_{j=1}^n\binom{j+1}{2}=\sum_{j=2}^{n+1}\binom{j}{2}=\binom{n+2}{3}$$ By using the Hockey-stick identity (credits to Jean-Claude Arbaut). Similarly for the next function $$f_3(n)$$ we have $$f_3(n)=\sum_{j=1}^n\binom{j+2}{3}=\sum_{j=3}^{n+2}\binom{j}{3}=\binom{n+3}{4}$$ So one could conjecture that $$f_k(n)=\binom{n+k}{k+1}$$ which can be easily proven by induction as follows $$f_k(n)=\sum_{j=1}^n\binom{j+k-1}{k}=\sum_{j=k}^{n+k-1}\binom{j}{k}=\binom{n+k}{k+1}$$ Hence we have that $$\boxed{f_k(n)=\binom{n+k}{k+1}=\frac1{(k+1)!}n(n+1)(n+2)\dots(n+k-1)(n+k)}$$
• $\sum_{j=1}^n\left(\frac16n^3+\frac12n^2+\frac13n\right)$: The summands do not depend on the index $j$. – callculus Apr 17 at 17:51
• @callculus Yes, sorry I've corrected it. – Peter Foreman Apr 17 at 17:52
We can write the last multiple sum as \begin{align*} \color{blue}{\sum_{i_1=1}^n\sum_{i_2=1}^{i_1}\sum_{i_3=1}^{i_2}i_3} &=\sum_{i_1=1}^n\sum_{i_2=1}^{i_1}\sum_{i_3=1}^{i_2}\sum_{i_4=1}^{i_3} 1\\ &=\sum_{1\leq i_4\leq i_3\leq i_2\leq i_1\leq n}1\tag{1}\\ &\,\,\color{blue}{=\binom{n+3}{4}}\tag{2} \end{align*} In (1) we observe the index range is the number of ordered $$4$$-tuples with repetition from a set with $$n$$ elements resulting in (2).
Here's a combinatorial way of thinking about it: first of all, note that we can go one level deeper and represent the innermost piece ($$j$$, or $$k$$, etc.) in your formulae as $$\sum_{h=1}^j1$$; this means that the formula start to look like $$\displaystyle\sum_{m=1}^n1 =n$$, $$\displaystyle\sum_{m=1}^n\sum_{l=1}^m1=n(n+1)/2={n+1\choose 2}$$, $$\displaystyle\sum_{m=1}^n\sum_{l=1}^m\sum_{k=1}^l1={n+2\choose 3}$$, etc. Now, let's look at what the left hand side is counting. In the first case, we're just counting the number of ways to choose an $$m$$ between $$1$$ and $$n$$ (inclusive); this is, self-evidently, just $$n$$. In the second, we're choosing a number $$m$$ between $$1$$ and $$n$$ inclusive, again, but then choosing an $$l$$ between $$1$$ and $$m$$; this is exactly the number of ways of choosing two numbers between $$1$$ and $$n$$, where we don't care about the order — that is, choosing $$2$$ and $$5$$ is exactly the same as choosing $$5$$ and $$2$$. Similarly, $$\displaystyle\sum_{m=1}^n\sum_{l=1}^m\sum_{k=1}^l1$$ counts the number of ways of choosing three numbers between $$1$$ and $$n$$, without regard to order; this is because we can sort the numbers we've chosen (since we don't care about order), and then note that the largest can be anywhere between $$1$$ and $$n$$, but then the next largest can only be between $$1$$ and the largest, etc.
Now, the difference between this and regular combinations is that in a regular combination every chosen number must be distinct; but if we have an ordered list $$\langle k, l, m\rangle$$ of the (not necessarily distinct) numbers we've chosen between $$1$$ and $$n$$ then we can turn this into an ordered list of not necessarily distinct numbers between $$1$$ and $$n+2$$: let $$k'=k$$, $$l'=l+1$$, $$m'=m+2$$. You should be able to convince yourself that this is a one-to-one correspondence between not-necessarily-distinct choices in $$\{1\ldots n\}$$ and distinct choices in $$\{1\ldots n+2\}$$, and the same principle extends to any number of choices. (This wikipedia link has more details).
$$S_{n_2}=\sum_{i=1}^n\sum_{j=1}^ij=\sum_{i=1}^n\frac{i(i+1)}{2}=\frac12\sum_{i=1}^ni^2+i=\frac12\left[\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\right]=\frac{n(n+1)(n+2)}{6}$$ and now: $$S_{n_3}=\sum_{i=1}^n\sum_{j=1}^i\sum_{k=1}^jk=\frac16\sum_{i=1}^ni(i+1)(i+2)=\frac16\sum_{i=1}^ni^3+3i^2+2i=\frac16\left[\frac{n^2(n+1)^2}{4}+\frac{n(n+1)(2n+1)}{2}+n(n+1)\right]=\frac{n(n+1)}{6}\left[\frac{n(n+1)}{4}+\frac{(2n+1)}{2}+1\right]=\frac{n(n+1)(n+2)(n+3)}{24}$$ and we can see a pattern here. For a series $$S_{n_a}$$ with $$a$$ nested summations the following is true: $$S_{n_a}=\frac{1}{(a+1)!}\prod_{b=0}^a(n+b)=\frac{(n+a)!}{(n-1)!(a+1)!}$$
• What is wrong with this answer? – Henry Lee Apr 17 at 18:34
• I don´t know. Hopefully the downvoter leaves a comment. – callculus Apr 17 at 18:37
• Appart from the notation $S_{n_a}$, looks good. Note also that the last expression is ${n+a\choose a+1}$. – Jean-Claude Arbaut Apr 17 at 18:40 | 2019-05-22T07:23:27 | {
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https://math.stackexchange.com/questions/1462099/number-of-possible-combinations-of-x-numbers-that-sum-to-y/1462106 | # Number of possible combinations of x numbers that sum to y
I want to find out the number of possible combinations of $x$ numbers that sum to $y$. For example, I want to calculate all combination of 5 numbers, which their sum equals to 10.
An asymptotic approixmation is also useful. This question seems to be very close to number partitioning, with the difference that a number can be 0. See:
https://en.wikipedia.org/wiki/Partition_%28number_theory%29#Asymptotics
All possible partitions for sum 10 and 3 positions that can be zero, are 63 possiblities: (numbers shown as 3 digits)
019 028 037 046 055 064 073 082 091 109 118 127 136 145 154 163 172 181 190 208 217 226 235 244 253 262 271 280 307 316 325 334 343 352 361 370 406 415 424 433 442 451 460 505 514 523 532 541 550 604 613 622 631 640 703 712 721 730 802 811 820 901 910
• In your example you did not count the possibilities $0|0|10$ and $0|10|0$ and $10|0|0$. So there are $66$ possibilities. This agrees with stars and bars since $\binom{10+2}2=66$.
– Vera
May 11 at 8:58
This problem is equivalent to finding the number of integer solutions to $$a+b+c+d+e=10$$.
If you imagine your $$10$$ as a line of $$10$$ stars then you can insert $$4$$ "|" (bars) in between the stars to get a solution, for example $$|\star\star|\star\star\star\star|\star|\star\star\star$$ represent the solution $$0+2+4+1+3$$.
Since every permutation of stars and "|" bars represents a solution the total number of solutions is given by the possible permutations of this $$14$$ symbols, that is $$\frac{14!}{10!4!}$$.
This method, actually called stars and bars, can be used for similar problems with other numbers involved.
Edit: in the case of $$3$$ numbers adding up to $$10$$ stars and bars gives $$\frac{12!}{10!2!}=66$$ as answer, you have $$63$$ because you didn't count the $$3$$ triplets with $$2$$ zeros and a ten, was that intended?
• is there a way to do this when the numbers being added are between two values, say 1-26.. for example, how many ways can 7 numbers being between 1 and 26 inclusive add up to 55 Dec 21, 2017 at 2:41
• @0TTT0 Integer solutions are essentially the problem of distributing identical units into distinct boxes, and these are well-studied. Some constraints have closed solutions (lower bounds can be solved easily, upper bounds can be solved with inclusion/exclusion), while other seemingly similar constraints have no closed solutions, and are best solved with generating functions (or numerically). Jan 25, 2021 at 17:13
The answer from Alessandro Codenotti about 66 and three extra $(0,0,10), (0,10,0), (10,0,0)$ is correct. In general, let $n$ is a positive integer to partition, $k$ is the number of non-negative parts (zeros are included), the order of parts matters. Then, the total number of decompositions is the binomial coefficient $C(n+k-1,k-1)=\frac{(n+k-1)!}{(k-1)!n!}$.
This result is well known. For $n=10$, $k=3$, $C(10+3-1,3-1)=\frac{12!}{2!10!}=\frac{11\cdot 12}{2}=66$. For $n=10$, $k=5$, $C(10+5-1,5-1)=\frac{14!}{4!10!}=\frac{11 \cdot 12 \cdot 13 \cdot 14}{2 \cdot 3 \cdot 4}=77 \cdot 13=1001$. In both cases, one decides, if $k$ must be subtracted from the result to remove $k$ decompositions, where $n$ itself is in one of $k$ positions and accompanied by $k - 1$ zeros.
• Just in case. Alessandro's reasoning is good. Another way to get confidence in the answer is to view the positive integer n as n indistinguishable balls, which supposed to be placed into k distinguishable boxes so that some boxes can remain empty. By "well known", I mean, for instance, Riordan, John. An Introduction to Combinatorial Analysis. Princeton, New Jersey: Princeton University Press, 1978. pp. 92 - 94, Section 3. Like Objects and Unlike Cells. Best Regards, Valerii Salov Jul 16, 2018 at 16:29
• Do you like Latex? Also we. :-) Just type $5 \cdot 5$ and you get $5 \cdot 5$. Welcome on the MathSE! :-) Jul 17, 2018 at 21:21
• Hi 'peterh'. Thank you for pointing that Latex is accepted and for the trick with multiplication. Yes, I do use Latex and will apply it on messages for this nice site. Best Regards, Valerii Jul 18, 2018 at 14:07
The answers above are incorrect, because they count the same numbers in different sequence as distinct combinations of numbers. In the above example, the given combination is 0+2+4+1+3, but 2+0+4+1+3 is counted as a distinct combination even though it is not. In total, 5! permutations of those numbers will be counted as distinct combinations.
However, dividing the result by 5! does not yield the right answer either.
• From the example given by the OP in the question we may conclude that order matters.
– Vera
May 11 at 9:00 | 2022-08-16T01:26:04 | {
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https://math.stackexchange.com/questions/1740249/probability-of-no-ace-but-at-least-one-king | Probability of no Ace but at least one King
So let me try to explain my reasoning:
Since we know there are no Ace, there are 48 cards left. Now, we know that we have at least one Kings which mean we have (exactly one or two or three or four kings) while choosing the remaining cards from 44 leftover cards.
$$\frac{{4\choose1}{44\choose4}+{4\choose2}{44\choose3}+{4\choose3}{44\choose2}+{4\choose4}{44\choose1}}{52\choose5}$$
Now, there is another correct answer:
P(no Ace and at least one King) = P(no Ace) - P(no Ace and no King) $$\frac{{48\choose5}-{44\choose5}}{52\choose5}$$ if that's the case.. can I say for
P(no Ace and no Queen and at least one King):
$$\frac{{44\choose5}-{40\choose5}}{52\choose5}$$
EDIT: sorry, my question is: can someone explain why the other correct answer works too? I can't seem to grasp the logic of it. So let me explain further by saying that the logic behind it is somewhat similar to finding at least one king in a deck of 52 for a standard poker hand. $$\frac{{52\choose5}-{48\choose5}}{52\choose5}$$ But for this case P(no ace but at least one kings), we only start with 48cards.
• What is the question? Probability of picking king before ace? – астон вілла олоф мэллбэрг Apr 13 '16 at 4:54
• It sounds like he's talking about a hand of five cards, but this should have been specified in the question. – browngreen Apr 13 '16 at 4:55
• In your first answer, you made a typo, the final term on the numerator should have been $\binom{4}{4}\binom{44}{1}$. Otherwise, yes. Both answers and methods are correct (but the first method is often seen as more tedious due to the case work. What if this was a million card deck with 100 kings of different suits? First method has a hundred things to add. Second method only needs two). Your follow up question is correct as well. – JMoravitz Apr 13 '16 at 5:01
• In addition to the error in the first problem that @JMoravitz spotted, there is an error in your calculation of the probability that a five-card hand has neither an ace nor a queen but contains at least one king. You should have $\binom{44}{5} - \binom{\color{red}{40}}{5}$ in the numerator since a hand that contains no aces, kings, or queens is selected from the $52 - 3 \cdot 4 = 40$ other cards in the deck. – N. F. Taussig Apr 13 '16 at 9:39
• @JMoravitz sorry for not posting the question! I've edited the post now. – misheekoh Apr 13 '16 at 17:22
1 Answer
Answer to the edited version of the question:
Find the probability that a five card hand contains no aces but at least one king.
As you know, the number of five card hands is $\binom{52}{5}$ since we are selecting five cards from a deck with fifty-two cards.
To calculate the number of hands that contain no aces but at least one king, we subtract the number of hands that contain neither aces nor kings from the number of hands that contain no aces. By taking the difference of these numbers, we find the number of five cards hands which contain no aces but do contain at least one king.
Since there are four aces in a deck of $52$ cards, the number of cards that are not aces is $52 - 4 = 48$. Therefore, the number of five card hands that contain no aces is $\binom{48}{5}$ since we must select five cards from the $48$ cards that are not aces.
Since there are four aces and four kings in a $52$ card deck, the number of cards that are neither aces nor kings is $52 - 2 \cdot 4 = 44$. Therefore, the number of ways of selecting five cards from the deck that are neither aces nor kings is $\binom{44}{5}$ since we must select five cards from the $44$ cards that are neither aces nor kings.
Hence, the number of five card hands that contain no aces but at least one king is $$\binom{48}{5} - \binom{44}{5}$$ from which we obtain the probability $$\frac{\dbinom{48}{5} - \dbinom{44}{5}}{\dbinom{52}{5}}$$ that a hand contains no aces but at least one king.
Answer to the original version of the question:
Your approach to both problems is correct, but there are some mistakes in the execution.
Find the probability that a five card hand contains no ace but at least one king.
In your first approach, as JMoravitz pointed out in the comments, the last term in the numerator should be $\binom{4}{4}\binom{44}{1}$ since you are selecting four kings and one additional card from the $52 - 2 \cdot 4 = 44$ cards that are neither aces nor kings, so the probability is $$\frac{\dbinom{4}{1}\dbinom{44}{4} + \dbinom{4}{2}\dbinom{44}{3} + \dbinom{44}{3}\dbinom{44}{2} + \dbinom{4}{4}\dbinom{44}{1}}{\dbinom{52}{5}}$$
Your second approach is correct. It also, as JMoravitz pointed out in the comments, is more efficient. It has the added virtue of containing fewer steps and, consequently, fewer opportunities to make an error than your first approach.
Find the probability that a five card hand contains no aces and no queens but at least one king.
Your approach of subtracting the number of hands that contain no aces, no queens, and no kings from the number of hands that contain no aces and no queens is correct. However, there are $52 - 3 \cdot 4 = 40$ cards that contain no aces, no kings, and no queens. Hence, the probability is $$\frac{\dbinom{44}{5} - \dbinom{40}{5}}{\dbinom{52}{5}}$$
• I've edited the post to add the initial question I had in mind. Thank you. – misheekoh Apr 13 '16 at 17:23
• I have edited my answer accordingly. – N. F. Taussig Apr 13 '16 at 18:39 | 2019-09-23T19:35:01 | {
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https://www.physicsforums.com/threads/rank-of-a-matrix.222658/ | # Rank of a matrix
1. Mar 17, 2008
### proxyIP
1. The problem statement, all variables and given/known data
Find all possible values of rank(A) as a varies.
A=[1, 2, a]
[-2, 4a, 2]
[a, -2, 1]
A is 3x3, just merge the 3 row vectors.
3. The attempt at a solution
I have the solution but I create this thread to find the most effective/efficient procedure for me to solve these kinds of problems.
What are the possible dimensions of the row/column space as a varies? When do the columns/ranks become linearly dependent based on different values for a? I don't want to guess or plug in arbitrary numbers; I want a generalized, systematic approach that works every time.
Last edited: Mar 17, 2008
2. Mar 18, 2008
### CompuChip
There are the following possibilities:
• All columns are independent
• The first and the second are dependent, but independent of the third
• The first and the third are dependent, but independent of the second
• The second and the third are dependent, but independent of the first
• All of them are dependent
Or, if you prefer, you can do this with the rows.
Now for each case, you can write down an equation and solve it for a. For example,let me do the third case (first and third are dependent, but independent of the second).
If the third column is a multiple n of the first one, you must have
1 = n a
-2 = 2 n
a = n 1
From the second equation you see that there is just one possibility for n. Then you get a solution for a from one of the others. Finally, use the remaining equation to see if this value indeed satisfies all of them. Then plug this value into the 4a in the second column, and check that it is indeed independent of the first (and/or third)
3. Mar 18, 2008
### HallsofIvy
Staff Emeritus
The simplest way to determine the rank of a matrix is to "row-reduce". The rank is the number of rows that contain non-zero entries.
In the case you give
$$A= \left(\begin{array}{ccc}1 & 2 & a \\-2 & 4a & 2\\ a & -2 & 1\end{array}\right)$$
Add twice the first row to the second and subtract a times the first row from the third to get
$$A= \left(\begin{array}{ccc}1 & 2 & a \\0 & 4a+ 4 & 2+ 2a\\ 0 & -2-2a & 1-a^2\end{array}\right)$$
Now add half the second row to the third to get
$$A= \left(\begin{array}{ccc}1 & 2 & a \\0 & 4a+ 4 & 2+ 2a\\ 0 & 0 & 2+ 2a- a^2\end{array}\right)$$
If a= -1 that has only 2 non-zero rows and so the rank of A is 2. If a= [itex]-1\pm\sqrt{3} the last row is 0 and again the rank of A is 2. For any other value of a, the rank is 3.
4. Mar 18, 2008
### proxyIP
Thanks a lot; that was very helpful. | 2016-08-24T04:24:40 | {
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http://math.stackexchange.com/questions/51296/functions-continuous-in-each-variable | # Functions continuous in each variable
Suppose we have a map $f:X \times Y \rightarrow Z$, where $X,Y$, and $Z$ are topological spaces. Are there any conditions on $X$,$Y$, and $Z$ that would allow one to determine that $F$ is continuous if it was known that it was continuous in each variable? It seems like there should be a theorem related to this.
By definition, a path homotopy $F: X \times I \rightarrow Y$ is continuous. What results in algebraic topology would not hold if we only required the map to be continuous in each variable? Would path homotopies not necessarily generate the fundamental group?
-
Even in the nice case $X=Y = \mathbb R$ the continuity in each variable is not sufficient for the continuity as a function of two variables - the same will hold if you take $X=Y=[0,1]$. – Ilya Jul 13 '11 at 20:43
f(x,y)=$\frac{2xy}{x^2+y^2}; f(0,0)=0$is a standard counterexample; continuous for each of x,y separately , but not continuous (check the limit at $(0,0)$ , e.g., along the direction y=x ). You may have to use sequential continuity ( if you have first-countability), or just the "old-fashioned" way, by showing that the inverse image under f of an open set in Z is open in the product topology of $X \times Y \$ – gary Jul 13 '11 at 21:10
Letting $S^1=\{z\in\mathbb{C}\colon\vert z\vert=1\}$ be the unit circle, consider the map $F\colon S^1\times I\to S^1$ given by $$F(e^{2\pi\theta i},s) = e^{2\pi\theta^si}$$ for $0 < \theta\le 1$ and $s\in I$. Then $F$ is continuous in each variable, $F(z,1)=z$ and $F(z,0)=1$. So, if you only required continuity in the individual variables, the circle would be contractible. More generally, every topological space would have trivial fundamental group. Suppose that $X$ is a topological space and $\gamma\colon I\to X$ is a closed curve. Define $F\colon I\times I\to X$ by $F(x,s)=\gamma(x^s)$ for $x,s\in I$ and $s > 0$, and $F(x,0)=\gamma(0)=\gamma(1)$. Then $\gamma$ is null-homotopic (relative to ${0,1}$). So, the fundamental group collapses to the trivial group, as do all the higher homotopy groups.
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There's a nice paper in the Math Monthly about this idea: Deloup, Florian "The fundamental group of the circle is trivial." Amer. Math. Monthly 112 (2005), no. 5, 417–425 – Dan Ramras Jul 14 '11 at 5:43
As is the case with two others who have responded (at the time I wrote this), I don't have an answer to your specific questions (conditions on $X$, $Y$, and $Z$; homotopy analogs for separately continuous maps). However, Piotrowski's 1996 survey paper or my 2005 sci.math post (which contains some references not given in Piotrowski's paper -- [2], [4], and [6]) might have something of interest to you or lead you to a relevant reference.
Zbigniew Piotrowski, "The genesis of separate versus joint continuity", Tatra Mountains Mathematical Publications 8 (1996), 113-126. [MR 98j:01026; Zbl 914.01007]
http://people.ysu.edu/~zpiotrowski/papers/genesisseperatevsjoint.pdf
sci.math -- "Continuity in each variable vs. joint continuity" (4 June 2005) | 2014-10-30T13:15:08 | {
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https://math.stackexchange.com/questions/1509893/finding-formula-for-nth-partial-sum | # Finding formula for nth partial sum
I need to find a formula for the nth partial sum of the series:
$2 + \frac 23 + \frac 29 + \frac {2}{27} + .... + \frac {2}{3^{n-1}} + ...$
then I need to use the formula to find the series' sum if the series converges.
The answer from the back of the book is (the formula):
$$\frac {2(1- (\frac 13) ^ n)}{1 - \frac 13}$$
And the series' sum is 3. Any help would be appreciated. Struggling to figure out how to get the formula.
--Edited because I messed up MathJax, fixed now.
• You should have a look to geometric progressions. en.m.wikipedia.org/wiki/Geometric_progression. – mathcounterexamples.net Nov 2 '15 at 19:16
• Let $S=S(n)$ be the sum up to $\frac{2}{3^{n-1}}$. Then $\frac{1}{3}S$ is equal to the sum $\frac{2}{3}+\frac{2}{9}+\cdots +\frac{2}{3^n}$. Subtract from $S$, noting the beautiful cancellation. We get $S-\frac{1}{3}S=2-\frac{2}{3^n}$. Now can you finish? – André Nicolas Nov 2 '15 at 19:18
Your series is a G.P. series with first term $2$ and common ratio $\frac{1}{3}$. So according to the G.P. sum formula, $S_n=\frac{2\left(1-(\frac{1}{3})^n\right)}{1-\frac{1}{3}}=3\left(1-(\frac{1}{3})^n\right)$. As $n\to\infty$, we can easily check that $S_n\to 3$.
Now do you know the G.P. formula?
• The G.P. formula is $s_n = \frac {a(1-r^n)}{1-r}$ I believe? – bankey Nov 2 '15 at 19:18
• Ahh, I understand. I didn't think to use that formula even though it was staring right at me. And then the ratio would be found by taking $s_{n+1} / s_n$ . Thanks all! – bankey Nov 2 '15 at 19:22
• Yes you're right. – SchrodingersCat Nov 2 '15 at 19:24
First, it may be helpful to factor out $2$ from your series; it will only clutter up your actual work. So really, you are out for a formula of the $n^{th}$ partial sum of $\sum_{n=0}^\infty \frac{1}{3^n}$. As the comment on your post points out, this is just a special instance of a geometric progression $\sum_{n=0}^\infty x^n$ where $|x|<1$. Now, you could just appeal to the generally known formulas for this sort of thing, but I find it helpful to not have to memorize too many things; to this end, what we really want to do is convert this: $$1+x+x^2+\cdots+x^n$$ into something without dots. The best way to try to do this is multiply by another polynomial, and shoot for lots of cancellation. If you recall things like the difference of cubes formula from high school, the guess is immediate. $$(1-x)\cdot(1+x+x^2+\cdots+x^n)=1+x-x+x^2-x^2+\cdots+x^n-x^n-x^{n+1}=1-x^{n-1}$$ So, you see that after division by $1-x$, your partial sum looks like $\frac{1-x^{n+1}}{1-x}$. Now plug in your specific value $x=1/3$ and multiply by $2$, and you are done!
• I would just like to add that while this method may seem roundabout, it can help you avoid making silly mistakes; I've seen many students blindly apply the GP formula and get questions wrong because they didn't verify that the summation index started from 0! – Tenno Nov 2 '15 at 19:27
Hint:
Think of the high school identity: $$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\dotsm+ab^{n-2}+b^{n-1}).$$
If the $n$th term in a series is $\lambda a^n$ then $$s_n = \sum_{i=0}^n \lambda a^i$$ $$a s_n = \sum_{i=1}^{n+1} \lambda a^{i}$$ $$s_n - a s_n = (1-a)s_n = \lambda - \lambda a^{n+1}$$ So that $$\sum_{i=0}^n \lambda a^i = \lambda\frac{1- a^{n+1}}{1-a}$$ | 2020-02-28T00:17:49 | {
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http://math.stackexchange.com/questions/193132/solving-inequality-frac2xx-21 | # Solving inequality $\frac{2x}{x-2}>1$
I'm trying to solve
$$\frac{2x}{x-2}>1$$
but I can't seem to get the correct answer. I'm doing something wrong but I don't know what; that is why I'm asking. This is what I've got:
$$\frac{2x}{x-2}>1$$
Since we do not know if the denominator is positive or negative, we can't multiply both sides with the expression $x-2$. Instead we solve for 2 cases (and $x \not= 2$):
Case 1: $x-2 > 0$, so $x > 2$.
$$\frac{2x}{x-2}>1$$ $$2x > x-2$$ $$x > -2$$
Case 2: $x-2 < 0$, so $x < 2$.
$$\frac{2x}{-(x-2)}>1$$ $$2x < 2-x$$ $$x < \frac{2}{3}$$
For case 1 we get the inequality $x > 2$ and $x > -2$. This simplifies to $x > 2$.
For case 2 we get the inequality $x < 2$ and $x < \frac{2}{3}$. This simplifies to $x < \frac{2}{3}$.
Combining our 2 cases we get the answer: $x < \frac{2}{3}$ or $x > 2$.
Which unfortunately is wrong! The answer should be $x < -2$ or $x > 2$.
I'm guessing case 2 is flawed. I have tried different ways doing it, but I never got the correct answer. In my answer above, I wrote case 2 as I wrote it during my first attempt at this. As I said though, I did try other ways too. :-/
-
For case 2, you should not change the denominator to $-(x-2)$. Instead you proceed as follows: $$\dfrac{2x}{x-2} > 1$$ $$2x < x-2$$ You change the inequality, as you know the denominator is negative. However, you do not introduce a negative sign. You wil have $x<-2$ as desired.
-
That easy, huh? I got to remember this... I think I was still influenced under the spell of absolute values since yesterday. I need to separate them in my mind! Thanks for a quick answer. Problem solved. :) – matfor Sep 9 '12 at 11:59
Another common approach is to reduce your inequality to something of the form $\displaystyle\frac{N(x)}{D(x)}\lesseqgtr 0$.
You then have to study the sign of the numerator and the sign of the denominator, i.e. solving some non fractional inequalities. Finally, you can study the sign of the ratio of $N(x)$ and $D(x)$ and answer properly to the problem.
In your example, you obtain the equivalent inequality $\displaystyle \frac{x+2}{x-2} > 0$. After having ruled out the value $x=2$, you solve:
• $x+2 > 0 \Longrightarrow x>-2$, i.e. your numerator is positive iff $x>-2$;
• $x-2 > 0 \Longrightarrow x>2$, i.e. your denominator is positive iff $x>2$.
The conclusion follows, since
• in the interval $(-\infty,-2)$ both $x+2$ and $x-2$ are negative, so their ratio is positive;
• in the interval $(-2,2)$ numerator and denominator have opposite sign, so their ratio is negative;
• in the interval $(2,\infty)$ both $x+2$ and $x-2$ are positive, so their ratio is positive.
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Note also that the method of multiplying through by the non-negative $(x-2)^2$ - as advertised in a wrongly worked through post on which I was trying to comment before it was deleted - does work as follows $$2x(x-2)>(x-2)^2$$ which gives $$x^2-4>0$$ ie $$(x+2)(x-2)>0$$ which is equivalent to the advertised answer.
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When solving fractional inequalities, the first step is to remove the denominator. We could multiply through by $x-2$ but then we would need to think about the sign of $x-2$ and how the $x-2 < 0$ and $x - 2 > 0$ cases effect things. Instead, we could multiply through by $(x-2)^2$ because $(x-2)^2 \ge 0.$ Doing this gives:
$$2x(x-2) > (x-2)^2 \iff 2x^2 - 4 > x^2 - 4x + 4 \iff (x-2)(x+2) > 0 .$$
We see that $(x-2)(x+2) > 0$ if and only if $x-2$ and $x+2$ have the same signs. Clearly $x = -2$ and $x = 2$ are important values. In the region $(-\infty,-2)$ we have $x\pm 2<0$. In the region $(-2,2)$ we have $x-2 < 0$ and $x+2 > 0$, in the region $(2,\infty)$ we have $x \pm 2 > 0$. It follows that:
$$\frac{2x}{x-2} > 1 \iff x \in (-\infty,-2) \, \cup \, (2,\infty) \, .$$
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Others have done a good job of locating your mistake. But no one's posted this frequently seen other method, so I'll do that.
You have $$\frac{2x}{x-2}>1.$$ This becomes $$\frac{2x}{x-2}-1>0.$$ The common denominator is $x-2$: $$\frac{2x}{x-2} - \frac{x-2}{x-2} > 0.$$ Simplify: $$\frac{x+2}{x-2}>0.$$ A quotient is positive if and only if the numerator and denominator are either both positive or both negative. They're both positive when $x>2$ and $x>-2$, thus when $x>2$. They're both negative when $x<2$ and $x<-2$, thus when $x<-2$. So the solution is $x>2$ or $x<-2$.
- | 2015-11-25T14:41:58 | {
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http://mathhelpforum.com/pre-calculus/16003-can-any-help-one-please.html | # Math Help - Can any help with this one please
1. ## Can any help with this one please
Find the third degree polynomial whose graph is shown in the
figure.
choices are
f(x) = x^3 - x^2 - 2x + 2
f(x) = 1/4(x)^3 - 1/2(x)^2 - x+2
f(x) = 1/4(x)^3 -1/4(x)^2 + 2x + 2
f(x) = 1/2(x)^3 - 1/2(x)^2 - x + 2
2. The answer I got was f(x) = x^3 - x^2 - 2x + 2
I think this is the answer can someone let me know if I'm correct. I'f not you don't have to give me the answer but an explanation on how to figure it out would be great.
Thanks
3. Originally Posted by [email protected]
The answer I got was f(x) = x^3 - x^2 - 2x + 2
I think this is the answer can someone let me know if I'm correct. I'f not you don't have to give me the answer but an explanation on how to figure it out would be great.
Thanks
it seems my first post was incorrect (well, not completely). how did you get that answer. i do not think it is correct, since if we plug in -2 for x, we don't get 0. but clearly -2 is a root of the graph
4. Originally Posted by [email protected]
Find the third degree polynomial whose graph is shown in the
figure.
choices are
f(x) = x^3 - x^2 - 2x + 2
f(x) = 1/4(x)^3 - 1/2(x)^2 - x+2
f(x) = 1/4(x)^3 -1/4(x)^2 + 2x + 2
f(x) = 1/2(x)^3 - 1/2(x)^2 - x + 2
the answer is the second one:
$f(x) = \frac {1}{4} x^3 - \frac {1}{2} x^2 - x + 2$
Why did i choose this one? it has all the desired features:
$f(2) = f(-2) = 0$ and $f(0) = 2$
The first graph i suggested had the correct x-intercepts and shape, but the y-intercepts were off. if you divide the original graph i suggested by 4, you would get the above graph
to refresh your memory, i suggested $f(x) = (x + 2)(x - 2)^2$, our answer has to be that graph, or some constant times that graph, as the above is. do you see why?
5. Hello, wvmcanelly!
Find the third degree polynomial whose graph is shown in the figure.
$\begin{array}{ccc}(a)\;f(x) & = &x^3 - x^2 - 2x + 2 \\
(b)\;f(x) &= &\frac{1}{4}x^3 - \frac{1}{2}x^2 - x + 2 \\
(c)\;f(x) & = & \frac{1}{4}x^3 -\frac{1}{4}x^2 + 2x + 2 \\
(d)\;f(x) & = & \frac{1}{2}x^3 - \frac{1}{2}x^2 - x + 2\end{array}$
The graph has an x-intercept at -2.
. . The function has a factor of: $(x + 2)$
The graph is tangent to the x-axis at 2.
. . The function has a factor of: $(x - 2)^2$
The cubic is of the form: . $f(x) \;=\;a(x+2)(x-2)^2$
Since $(0,2)$ is on the graph,
. . we have: . $a(0+2)(0-2)^2 \:=\:2\quad\Rightarrow\quad8a \,=\,2\quad\Rightarrow\quad a \,=\,\frac{1}{4}$
Therefore: . $f(x) \;=\;\frac{1}{4}(x+2)(x-2)^2 \;=\;\frac{1}{4}x^3 - \frac{1}{2}x^2 - x + 2$ . .Answer (b)
6. Originally Posted by Soroban
Hello, wvmcanelly!
The graph has an x-intercept at -2.
. . The function has a factor of: $(x + 2)$
The graph is tangent to the x-axis at 2.
. . The function has a factor of: $(x - 2)^2$
The cubic is of the form: . $f(x) \;=\;a(x+2)(x-2)^2$
Since $(0,2)$ is on the graph,
. . we have: . $a(0+2)(0-2)^2 \:=\:2\quad\Rightarrow\quad8a \,=\,2\quad\Rightarrow\quad a \,=\,\frac{1}{4}$
Therefore: . $f(x) \;=\;\frac{1}{4}(x+2)(x-2)^2 \;=\;\frac{1}{4}x^3 - \frac{1}{2}x^2 - x + 2$ . .Answer (b)
i had originally forgotten to make an arbitrary constant a part of the formula, but i figured out that it should be 1/4 by just looking at the graphs...but your way is much nicer | 2014-04-23T17:56:35 | {
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http://mymathforum.com/advanced-statistics/343728-statistics-print.html | My Math Forum (http://mymathforum.com/math-forums.php)
IlanSherer March 23rd, 2018 08:32 PM
Statistics
Hello :)
The average weight of the cakes is 750 grams, and the standard deviation is 25 grams.
Each cake is divided into 6 slices of equal size, and each slice is added exactly 5 grams of cherry.
What is the average weight and standard deviation of a slice of cake?
I found that the average is 130 (If I'm not mistaken), because:
750/6=125
125+5=130
About standard deviation I'm not sure... I used the data to find the standard deviation, but I got a minus in the root (no solution).
And i don't think the value is (25/6)+5.
Thanks! :)
romsek March 23rd, 2018 10:33 PM
your calculation of the average is correct.
The standard deviation will be $\dfrac{25}{6}$ as adding the exact weight of the cherry has no effect on the standard deviation.
IlanSherer March 24th, 2018 09:43 AM
Quote:
Originally Posted by romsek (Post 590662) your calculation of the average is correct. The standard deviation will be $\dfrac{25}{6}$ as adding the exact weight of the cherry has no effect on the standard deviation.
But why the standard devation is only 25/6? I mean, why I'm not supposed to add 5 grams? I don't really get it...
romsek March 24th, 2018 09:46 AM
Quote:
Originally Posted by IlanSherer (Post 590701) Thanks for answering :) But why the standard devation is only 25/6? I mean, why I'm not supposed to add 5 grams? I don't really get it...
standard deviation is a measurement of the spread of a probability distribution.
When you add an exactly known quantity to a random variable you simply translate it's mean value. The spread doesn't change any.
mathman March 24th, 2018 12:48 PM
Variances add, so the standard deviation of one slice should be $\sqrt{104}$ unless the slicing is exact.
IlanSherer March 24th, 2018 02:08 PM
Quote:
Originally Posted by mathman (Post 590709) Variances add, so the standard deviation of one slice should be $\sqrt{104}$ unless the slicing is exact.
"Each cake is divided into 6 slices of equal size", so I think the slicing is exact.
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https://math.stackexchange.com/questions/1386381/cover-1-2-100-with-minimum-number-of-geometric-progressions/1386911 | # Cover $\{1,2,...,100\}$ with minimum number of geometric progressions?
In another question, posted here by jordan, we are asked whether it is possible to cover the numbers $$\{1,2,\ldots,100\}$$ with $$20$$ geometric sequences of real numbers. Naturally, we would like to extend the question:
Problem: What is the minimum number $$n$$ of geometric progressions $$A_1, A_2,\ldots,A_{n}$$ of rational* numbers such that $$\{1,2,\ldots,100\} \subseteq A_1 \cup A_2\cup \ldots\cup A_{n}?$$
In the other question, 6005 obtained a lower bound of $$31 \leq n$$ with an argument about square free integers. We can also obtain an upper bound of $$43$$ as follows. Consider these $$5$$ sequences: $$[1, 2, 4, 8, 16, 32, 64]$$ and $$[6, 12, 24, 48, 96]$$ and $$[5, 10, 20, 40, 80]$$ and $$[3, 9, 27, 81]$$ and $$[7, 21, 63]$$. Together, these cover $$7 + 5 + 5 + 4 + 3 = 24$$ terms. The remaining $$76$$ terms can be covered in at most $$38$$ sequences, by an argument made here. So we have the bound:
$$31 \leq n \leq 43$$
Can anyone do better?
*We need only consider rational ratios by arguments made in answers to the original question.
(Update) We have a winner!! Thanks to the cumulative efforts of the answerers below, we have arrived at $$n = 36$$. The upper bound is thanks to jpvee, and the lower bound is due to san. Hooray!
• I think we can get a few more off the upper bound with this method e.g. $11, 33, 99$ but eventually it will run out of steam and we'll need a new idea. Aug 6, 2015 at 10:48
• About the new idea: did you check this related question? math.stackexchange.com/questions/1385991/… Jack claims that, if this is true, then $n \in \{37,38\}$ Aug 6, 2015 at 15:02
• I saw a conjecture along those lines alright, but was it proved? Aug 6, 2015 at 15:38
• No, I don't think so: anyway, it would be better to ask Jack's opinion.. Aug 6, 2015 at 15:41
• Yes I agree. Unfortunately I don't know how to tag him as he's not on this thread. Aug 6, 2015 at 15:56
Switching from proving lower bounds to finding upper bounds, I played around with the list of progressions a bit and came up with another small improvement:
The $$16$$ progressions $$(1, 2, 4, 8, 16, 32, 64)\\ (3, 6, 12, 24, 48, 96)\\ (5, 10, 20, 40, 80)\\ (7, 14, 28, 56)\\ (9, 18, 36, 72)\\ (11, 22, 44, 88)\\ (13, 26, 52)\\ (15, 30, 60)\\ (17, 34, 68)\\ (19, 38, 76)\\ (21, 42, 84)\\ (23, 46, 92)\\ (25, 35, 49)\\ (27, 45, 75)\\ (50, 70, 98)\\ (81, 90, 100)$$ are completely disjoint and thus cover $$7+6+5+3\cdot4+10\cdot3=60$$ integers. Using $$20$$ additional progressions that each cover $$2$$ of the remaining $$100-60=40$$ integers yields a cover of $$16+20=36$$ geometric progressions.
• Nice. Getting closer now! Are you using a computer program to aid you with the upper bounds also? Aug 9, 2015 at 15:11
• @ColmBhandal: I just used to computer to give me the list of all progressions of length $\ge3$; the rest was done on paper. Aug 9, 2015 at 15:24
• A good manual new :P Aug 9, 2015 at 18:57
The exact value of n is 36:
First, consider all progressions of length 4 or more:
Length 7: 1,2,4,8,16,32,64
Length 6: 3,6,12,24,48,96
Length 5: 5,10,20,40,80$\qquad \ \$ 1,3,9,27,81$\qquad \ \$ 16,24,36,54,81
Length 4: 2,6,18,54$\qquad \ \$ 27,36,48,64$\qquad \ \$ 7,14,28,56$\qquad \ \$ 9,18,36,72$\qquad \ \$ 11,22,44,88$\qquad \ \$ 8,12,18,27
These progressions cover the 33 numbers
1,2,3,4,5,6,7,8,9,10,11,12,14,16,18,20,22,24,27,28,32,36,40,44,48,54,56,64,72,80,81,88,96
Moreover, two of the length 5 geometric progressions have only 3 numbers disjoint from the progressions of length 6 and 7. Hence, the best way to cover these 33 numbers is to cover 30 of them with 6 geometric progressions:
The progressions of length 7, 6 one of 5 and 3 progressions of length 4:
Length 7: 1,2,4,8,16,32,64
Length 6: 3,6,12,24,48,96
Length 5: 5,10,20,40,80
Length 4: 7,14,28,56$\qquad \ \$ 9,18,36,72$\qquad \ \$ 11,22,44,88
There are 35 numbers which are not in a geometric progression of length three or more:
29,31,37,39,41,43,47,51,53,55,57,58,59,61,62,65,67,69, 71,73,74,77,78,79,82,83,85,86,87,89,91,93,94,95,97
So you need 6 progressions to cover 30 numbers, 18 progressions to cover 35 numbers (in fact you cover 36), and the remaining 35 numbers must be covered with at least 12 progressions of length 3 (even if you covered 36 numbers with the 18 progressions in the item before you need 12 to cover 34 numbers).
So you need at least 36 progressions. This bound is achieved in the answer of jpvee, which is therefore optimal. Since the method of counting which numbers are covered by certain progressions is also of jpvee, and my only contribution was to count how many numbers are covered by progressions of length 4 or more, the bounty should be awarded to jpvee.
• A very noble answer. The bounty goes to jpvee, but this answer is accepted as the final touch. Nice work. Sep 11, 2015 at 12:33
1,2,4,8,16,32,64
3,6,12,24,48,96
5,10,20,40,80
7,14,28,56
11,22,44,88
13,26,52; 17,34,68; 19,38,76; 21,42,84; 23,46,92
9,15,25; 36,60,100; 49,63,81
27,45,75; 50,70,98
15 sequences with 56 numbers. 44 remains, so 15 + 22 = 37.
Not the only 37 solution, some replacements available (36,54,81; 49,70,100; 18,30,50, 50,60,72), so I believe computer program would find better solution quickly if it exists.
Why not 9,27,81 included? Not to spend three "squares" for a single triplet.
• Nice work. Maybe a computer program is the way to go. If I have time I'll write it. Still, it would be much more pleasing to see a clever proof! Aug 7, 2015 at 12:34
• It is interesting to point out that these progressions fit the scheme of my conjecture in the other thread. We have $V(64)=6$ and a sequence of length $7$, $V(96)=5$ with a sequence of length $6$, $V(80)=4$ with a sequence of length five and so on. Aug 7, 2015 at 15:28
Nice problem! Using help from the computer, I found that within the positive integers below 100, there are
66 geometric progressions of length 3,
6 geometric progressions of length 4,
3 geometric progressions of length 5,
1 geometric progression of length 6 and
1 geometric progression of length 7
(after eliminating those that are proper subsequences of longer ones).
These progressions of length 3 or longer cover a total of 65 integers $\le 100$; denote the set of these 65 integers by $\mathbb{M}$. If the $6+3+1+1=11$ progressions of lengths $\ge4$ were all disjoint (which they are not), they together would cover $6\cdot4+3\cdot5+6+7=52$ elements of $\mathbb{M}$; for the remaining ones, at least $\lceil(65-52)/3\rceil=5$ additional progressions of length 3 are necessary.
The remaining 35 integers outside of $\mathbb{M}$ can only be contained in progressions of length 2 and thus need at least $\lceil35/2\rceil=18$ of those progressions to cover them.
Therefore, a lower bound of the progressions needed to cover all positive integers up to 100 is $11+5+18=34$.
Update 2015-08-08: I just saw that my reasoning is a bit flawed; see my comment below.
• Brilliant. The bound shrinks yet again! Aug 7, 2015 at 13:10
• This could lead to another question: how many distinct geometric progressions (not proper subsequences of longer ones) of lenght $k$ are there in the set $\{1, ..., n\}$? Would your code answer that in general? Aug 7, 2015 at 13:14
• @ColmBhandal: Well; I'd need to modify the program a bit - I hope to find some time for that later on. Aug 7, 2015 at 14:55
• I'm awfully sorry, but my reasoning is flawed: It is not a good idea to round up twice. Instead, one could use a progression of length 2 to cover the 35th element outside of $\mathbb{M}$ and the 65th element of $\mathbb{M}$, giving a total of $11+4+17+1=33$. (If, however, you take into consideration my remark that the "larger" progressions are not completely disjoint, it is easy to see that $33$ is in fact not achievable.) Aug 8, 2015 at 7:28
• kudos for the rigour. Looks like your patch works OK- so we still have a lower bound of 34- right? Aug 8, 2015 at 12:24
100, 60, 36. 99, 66, 44. 98, 84, 72. 96, 48, 24, 12, 6, 3. 92, 46, 23. 90, 30, 10. 88, (44,) 22, 11. 84, 42, 21. 81, 54, (36,) (24,) 16. 80, 40, 20, (10,) 5. 76, 38, 19. 75, 45, 27. 68, 34, 17. 64, 32, (16,) 8, 4, 2, 1. That's 49 numbers, with 14 progressions. So we can certainly do it with 40 sequences. Or we can keep going. 56, 28, 14, 7. 52, 26, 13. 49, 35, 25. So, 38 sequences.
EDIT: turns out this question was discussed last year on MathOverflow, and I even contributed (a little bit) to the discussion. I recommend having a look at that discussion before continuing it here.
• I wonder if there is a more elegant idea than enumeration of all sequences... Aug 6, 2015 at 11:20
• Yes those are asymptotic lower bounds, but here the question is asking for a more precise answer for the concrete case of $a_{100}$. Aug 6, 2015 at 11:25 | 2022-07-06T18:57:35 | {
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https://math.stackexchange.com/questions/904375/disproving-the-claim-that-the-numbers-124-1248-124816-alternate-be | # Disproving the claim that the numbers 1+2+4, 1+2+4+8, 1+2+4+8+16… alternate between prime and composite
I am working through an elementary number theory book and I have come across the following problem.
Show the following claims are wrong:
Claim 1: The sequence 1+2+4, 1+2+4+8, 1+2+4+8+16, ... is alternately prime and composite.
Claim 2: One or both of the numbers 6n-1 and 6n+1 are prime.
I have actually found counterexamples to both claims, but it was only through cold hard number-crunching. I want to believe this problem wouldn't be here unless there was supposed to be some elegant way to disprove these claims. Is there some other way to disprove the claims other than just finding a counterexample?
Note: the two claims are not meant to be related in any way.
• You did it the sensible way. For the second, there would be fancier ways that appeal to the density of the primes. – André Nicolas Aug 20 '14 at 19:21
• For the second, you could also use the Chinese Remainder Theorem. But that would be overkill, unless you are interested in generalizations. – André Nicolas Aug 20 '14 at 19:28
• About the only clever thing available in the first one is finding good candidates. Note that this sequence can be written as $7,15,31,$etc i.e. $\{2^{n+2}-1\}$. We can then look for a counter-example by finding an odd $n=2k-1$ such that $2^{n+2}-1=2\cdot 4^{k}-1$ is composite. That eliminates a lot of the brute-force work immediately. – Semiclassical Aug 20 '14 at 19:29
• Since the claims are totally unrelated, they should have been two separate questions. – user147263 Aug 20 '14 at 20:17
• Perhaps, but since I solved them in a similar way and had a similar question for both, I opted for one question. – candido Aug 20 '14 at 20:32
For Claim 2: The numbers $m!+2, m!+3,\dots,m!+m$ will all be composite. If $m$ is large enough, this sequence will contain some $6n-1$ and $6n+1$.
For Claim 1: Note that $2^{ab}-1$ is always divisible by $2^a-1$ (since $x-1$ divides $x^b-1$ in $\Bbb{Z}[x]$, now let $x=2^a$); hence $2^8-1, 2^9-1, 2^{10}-1$ are all composite.
• A more concrete proof of 2 is to take $m=7$, so that you have 6 consecutive composite numbers. – Ian Aug 20 '14 at 21:15
• $m!$ will be a multiple of 6 for any $m > 2$, so the composites you would be interested in are $m! + 5$ and $m! + 7$. Those will be composite if $m = 7$. Actually the only properties about $m!$ being used in the proof is that it is divisible by $5$, $6$, and $7$. So instead of $m!$ one could just use $5*6*7$, in which case $n$ ends up being $5*7+1$ – kasperd Aug 20 '14 at 22:25
Number crunching is a time-honored technique; many conjectures are far more easily solved by actually testing them (either in a brute-force fashion or by more clever searches) rather than thinking hard about them.
But sometimes these conjectures -- as well as the methods of disproving them -- suggest more general statements. The numbers in your first conjecture are
$$\sum_{i=0}^{n-1} 2^i = 2^n - 1$$
are called Mersenne numbers. As you've noted, it's easy enough to prove they don't alternate prime/composite, and there are a number of theorems about when a Mersenne number can be prime or composite (e.g. for $2^n - 1$ to be prime, $n$ must also be prime)
There are also open questions -- e.g. nobody knows whether or not there are infinitely many Mersenne numbers that are prime.
• I believe you mean $\sum_{i=0}^{n-1} 2^i,$ not $\sum_{i = 0}^{n-1}2^n.$ – user71641 Aug 20 '14 at 21:05
• @nsanger: Thanks, fixed. – user14972 Aug 20 '14 at 21:24
A nice simple one for claim 2:
Note that both terms of the pair $(x^3+1, x^3-1)$ factorize (having simplest factors of $x+1$ and $x-1$ respectively), and that $(6n+1,6n-1)$ will be of that form if $n=6^2$.
(Which is just $(217,215)$, and indeed the pair are divisible by 7 and 5 respectively).
This argument extends to pairs of the form $(kn+1, kn-1)$ for $k>2$ (though the case for $k$ odd is obvious anyway, since both members will be even).
For the fun of it, I constructed the smallest counter-examples to both claims by hand (well, actually in my head). Here's the results, as well as how I went about finding them.
Claim 1: The seventh member of the series, $1+2+4+8+16+32+64+128+256=511$ is composite, not prime as specified. $511 = 7*73$.
This claim becomes easier to disprove once we recognize that the elements,
$7, 15, 31, 63, 127, 255, 511, ...$
are the Mersenne numbers, $2^n-1$, starting at $n = 3$. A useful fact about Mersenne numbers is that a Mersenne number can only be prime if its exponent, $n$, is prime. This immediately implies that elements $2, 4, 6, 7$ of the given series are composite. Thus we see that element $7$, $511$, must be a counter-example, because it is an composite odd-numbered element. To verify that it is the smallest counter-example, we need only verify that elements $1,3,5$, namely $7, 31, 127$ are prime. They indeed are, so $511$ is the smallest counter-example.
Claim 2: $n = 20, (119,121)$. $119 = 7*17, 121 = 11*11$
To find this, I categorized the possible solutions by the lower prime factor of $6n-1$ and $6n+1$. Clearly, neither value will be $2$ or $3$. First, I tried the pair $(5,7)$.
I calculated that $x \equiv 0 \mod 5$ and $x \equiv -1 \text{ or } 1 \mod 6$ implies $x \equiv 5\text{ or }25 \mod 30$ respectively, so the other member of the pair must equal $7\text{ or }23 \mod 30$. The smallest multiples of 7 satisfying those congruences, other than 7 itself, are $217$ and $203$, respectively.
Since I was not sure that this solution $(203, 205)$ was as small as possible, I repeated the process for the pair of factors $(5, 11)$. Reusing the congruence $7\text{ or }23 \mod 30$ for the multiple of $11$, I found the smallest multiples $187$ and $143$.
At this point, my smallest counter-example was $(143, 145)$. There only remained one case to check: $(119,121)$, because $121$ is the only number with $11$ as its smallest factor under $143$, and I had already checked that the pair $(5,7)$ had no solutions under $(143, 145)$. As it turns out, $(119,121)$ is a solution, so it must be the smallest.
1. For $n=6k+2$ the numbers $2^n-1$ and $2^{n+1}-1$ are both composite.
2. For $n= 6^{2k}$ the numbers $6n+1, 6n-1$ are both composite.
• Also, for $n=6^{6k-1}$, we have the factorizations $6n\pm1=6^{6k}\pm1=(6^{2k}\pm1)(6^{4k}\mp6^{2k}+1)$. – Barry Cipra Aug 20 '14 at 20:13
• Yes, the statement is based on this equality – Yog Urt Aug 20 '14 at 20:22
• I see that $6\pm1$ divides $6n\pm1$ when $n=6^{2k}$, but that's a little different from what I have in my comment (which in retrospect should have been done with an exponent $3k-1$ instead of $6k-1$). – Barry Cipra Aug 20 '14 at 20:35
• Yes it is, thank you – Yog Urt Aug 20 '14 at 20:43 | 2020-02-23T20:27:31 | {
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https://math.stackexchange.com/questions/443809/whats-p-and-whats-q-in-this-classic-proof-of-the-irrationality-of-sqrt-2 | # What's $P$ and what's $Q$ in this classic proof of the irrationality of $\sqrt 2$?
In this proof extracted from the Wikipedia
A classic proof by contradiction from mathematics is the proof that the square root of 2 is irrational. If it were rational, it could be expressed as a fraction $a/b$ in lowest terms, where a and b are integers, at least one of which is odd. But if $a/b = \sqrt 2$, then $a^2=$ $2b^2$. Therefore $a^2$ must be even. Because the square of an odd number is odd, that in turn implies that $a$ is even. This means that $b$ must be odd because $a/b$ is in lowest terms. On the other hand, if $a$ is even, then $a^2$ is a multiple of $4$. If $a^2$ is a multiple of $4$ and $a^2=2b^2$, then $2b^2$ is a multiple of $4$, and therefore $b^2$ is even, and so is $b$. So $b$ is odd and even, a contradiction. Therefore the initial assumption—that $\sqrt 2$ can be expressed as a fraction—must be false.
Knowing that a proof by contradiction you assume P and Not(Q) what's P and what's not Q in this proof?
• Proof by contradiction applies to any statement, not just to implications. So, proving that by contradiction that $P$ is true entails assuming $P$ fails and arriving at a contradiction. – Ittay Weiss Jul 15 '13 at 1:24
I'm assuming you are more accustomed to seeing proof by contradiction used largely with statements that are implications or conditionals. And indeed, when writing a proof by contradiction to prove statements of the form $$P \implies Q,$$ we typically assume $(P\land \lnot Q)$.
But in this particular case, we do not seem to have an implication to prove. Rather, we have the proposition:
The square root of $2$ is irrational. $\quad$( $Q$).
There's no helpful "if, then", or "this implies that" to indicate any sort of implication being asserted. So we have an example of the use of a proof by contradiction where to prove a statement other than an implication.
What we can do is to think of the assertion to be proven as a simple "atomic" proposition: $\,Q.\,$ Then $\,\lnot Q\,$ is the statement to the effect:
Suppose $\,\sqrt 2\,$ is not irrational. $\;$ Put differently, suppose $\,\sqrt 2\,$ is rational.$\quad(\lnot Q)$
The proof then proceeds, after having supposed $\,\lnot Q\,$ to invoke the definition of a rational number in order to arrive at a contradiction.
In a sense then, the proof amounts to a "bare-bones" proof-by-contradiction:
To prove that $\,Q,\,$ we assume $\,\lnot Q,\,$ and then we work to obtain a contradiction. Once we arrive at a contradiction, we can conclude that our assumption is false, and so we are justified in negating the false assumption: "therefore, $\lnot\lnot Q.$" $\;\;$ And this amounts to affirming the desired conclusion/assertion: therefore $Q$, since $\;\lnot \lnot Q\equiv Q$.
The contradiction in this proof happens to come from our knowledge about the rational numbers, information which could be considered a premise: the "implicit" premise $P$ being the definition of a rational number.
• Very nice my friend Amy. +1 – mrs Jul 15 '13 at 15:31
• @Amzoti You know I love these sorts of questions! Thanks for the support! – amWhy Jul 16 '13 at 1:03
• @amWhy: I could not tell! :=~~~~~~) (see how long my nose got?) – Amzoti Jul 16 '13 at 1:05
You can phrase "$\sqrt 2$ is irrational" as the implication "if $x=\sqrt 2$, then $x$ is irrational". Or better (which also handles $-\sqrt 2$): if $x^2 = 2$, then $x$ is irrational.
You can see the proof as showing that if you assume "$\sqrt2$ rational" you get a contradiction.
Or you can take P$=$"$\sqrt2=a/b$", Q="$a,b$ are both divisible by $2$". The proof shows that P implies Q. As you can always write a rational with a coprime numerator and denominator, the contrapositive shows that $\sqrt2$ is irrational.
P would be that $\sqrt{2}$ is rational, Q that it is irrational, and so not(Q) is that it is not irrational.
• Why was this down-voted? – JLA Jul 15 '13 at 3:24
• Thanks! I figured specifying what Q is and then inferring what not(Q) is makes more sense than just saying what not(Q) is. – JLA Jul 15 '13 at 18:44
• And you were correct, but during my time here I have noticed downvoting here is not about correct/incorrect answers. Sometimes changing something doesn't attract the downvoters. – jimjim Jul 15 '13 at 22:54
• @TheChaz2.0 : cool dude, not here to offend anyone, apologies and removed. – jimjim Jul 16 '13 at 0:55
• I was one of the downvotes. With that $P$ and $Q$, the implication $P\implies Q$ reads "if $\sqrt2$ is rational, then $\sqrt2$ is irrational", which makes no sense. – Martin Argerami Jul 16 '13 at 20:19 | 2021-01-27T16:46:13 | {
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https://math.stackexchange.com/questions/3778532/how-to-prove-that-frac-cosx-cos2x-sinx-sin2x-frac1-cosx | # How to prove that $\frac{\cos(x)-\cos(2x)}{\sin(x)+\sin(2x)} = \frac{1-\cos(x)}{\sin(x)}$ in a simpler way.
EDIT: Preferably a LHS = RHS proof, where you work on one side only then yield the other side.
My way is as follows:
Prove: $$\frac{\cos(x)-\cos(2x)}{\sin(x)+\sin(2x)} = \frac{1-\cos(x)}{\sin(x)}$$
I use the fact that $$\cos(2x)=2\cos^2(x)-1, \sin(2x)=2\sin(x)\cos(x)$$
(1) LHS = $$\frac{\cos(x)-2\cos^2(x)+1}{\sin(x)(1+2\cos(x))}$$
(2) Thus it would suffice to simply prove that $$\frac{\cos(x)-2\cos^2(x)+1}{1+2\cos(x)}=1-\cos(x)$$
(3) Then I just used simple algebra by letting $$u = \cos(x)$$ then factorising and simplifying.
(4) Since that equals $$1-\cos(x)$$ then the LHS = $$\frac{1-\cos(x)}{\sin(x)} =$$ RHS.
Firstly, on the practice exam, we pretty much only had maximum 2-2.5 minutes to prove this, and this took me some trial and error figuring out which double angle formula to use for cos(2x).
This probably took me 5 minutes just experimenting, and on the final exam there is no way I can spend that long.
What is the better way to do this?
EDIT: I also proved it by multiplying the numerator and denominator by $$1-\cos(x)$$, since I saw it on the RHS. This worked a lot better, but is that a legitimate proof?
• Instead of substituting $u=\cos(x)$, then factorising and simplifying, you could multiply $(1+2\cos(x))(1-\cos(x))$ to prove last identity Aug 3 '20 at 11:37
• @enzotib Yeah I considered that but I need to check with my teacher if that is a valid proof with respect to the guidelines. Usually, we try to simplify one side to get the other, we usually don't move terms across sides. But it might be allowed. Aug 3 '20 at 11:49
$$\frac{\cos x-\cos2x}{\sin x+\sin 2x } = \frac{1-\cos x }{\sin x}\iff \sin x\cos x-\sin x\cos2x=\sin x-\sin x\cos x+$$
$$+\sin2x-\sin2x\cos x\iff \color{red}{\sin x\cos 2x}+\sin x+\sin2x-\color{red}{\sin2x\cos x}-2\sin x\cos x=0\iff$$
$$\color{red}{\sin(-x)}+\sin x=0$$
and we're done by the double implications all through (and assuming the first expression is well defined, of course)
Check all the cancellations are correct and check all the trigonometric identities used above.
Another way: We begin with the left side, again: assuming it is well defined
$$\frac{\cos x-\cos2x}{\sin x+\sin2x}\stackrel{\cos2x=2\cos^2x-1\\\sin2x=2\sin x\cos x}=\frac{\cos x-2\cos^2x+1}{\sin x(1+2\cos x)}\stackrel{-2t^2+t+1=-(2t+1)(t-1)}=$$
$$=\require{cancel}-\frac{\cancel{(2\cos x+1)}(\cos x-1)}{\sin x\cancel{(1+2\cos x)}}\stackrel{\cdot\frac{\cos x+1}{\cos x+1}}=-\frac{\overbrace{(\cos^2x-1)}^{=-\sin^2x}}{\sin x(\cos x+1)}=$$
$$=-\frac{(-\sin x)}{(\cos x+1)}=\frac{\sin x}{\cos x+1}$$
Finally, we show that last right side equals the right side of the original equation:
$$\frac{\sin x}{\cos x+1}\cdot\frac{\cos x-1}{\cos x-1}=\frac{\sin x(\cos x-1)}{\underbrace{\cos^2x-1}_{=-\sin^2x}}=-\frac{\cos x-1}{\sin x}=\frac{1-\cos x}{\sin x}$$
• Is this a legitimate proof though? Aug 3 '20 at 11:33
• @Simplex1 Why do you ask? Haven't you worked out this kind of proofs? If you have any doubt, begin by the end and go back to the beginning. Observe the disclaimer at the end: we assume the original expression (the one you need to prove) is well defined. Thusm for example, $\;x\neq 2k\pi,\,\sin x+\sin 2x\neq0\;$ and etc. Aug 3 '20 at 13:12
• Ok I see that it is logically sound since the preposition implies a tautology (-a+a=0). I think my school requires us to use an LHS = RHS proof where you manipulate one side only and it will yield the other side. Aug 4 '20 at 5:18
• @Simplex1 That'd be a rather weird requirement...but anyway: I added a new proof. Aug 4 '20 at 7:12
• Thanks for that. This one avoids the unnecessary step of omitting the sin(x) then subbing in u = cos(x). Aug 4 '20 at 9:09
Here's a trigonographic demonstration:
• Not a proof to conceive in 2-2.5 minutes of the exam, but a wonderful one. Aug 3 '20 at 13:21
• I would also put in evidence the orange triangle in this copy of your image: drive.google.com/file/d/1e_wnuvau4j0Z12s8HNBKWMZjuSJCQBjB/… Aug 3 '20 at 13:29
• @enzotib: Thanks for the "wonderful". :) ... As to timing ... The notion for how to represent the left- and right-hand sides of OP's equation occurred to me almost-immediately, and the "ah-ha!" of the parallel lines happened as I drew a reasonably-accurate sketch. (The nice diagram took longer, of course.) All that said, I wouldn't expect someone to do this on an exam. (It often takes a while to find a good trigonograph.) Nevertheless, I like to promote diagrammatic thinking whenever I can.
– Blue
Aug 3 '20 at 13:31
• This is brilliant. Having it all in one picture shows how every side and hence every expression is related. Aug 4 '20 at 5:24
In the domain of your equality the following expression is equivalent:
$$\sin(x)(\cos(x)-\cos(2x))=(1-\cos(x))(\sin(x)+\sin(2x)) \iff \\ 2\sin(x)\cos(x)= \sin(x)+\sin(2x)+ \sin(x)\cos(2x)-\cos(x)\sin(2x) \iff \\ 2\sin(x)\cos(x)= \sin(x)+\sin(2x)+\sin(x-2x) \iff 2\sin(x)\cos(x)=\sin(2x)$$
This guarantees your result because you move from one expression to other by adding or subtracting equivalent values.
By the duplication formulas, for $$\sin(x)\neq 0$$ and $$\cos(x)\neq -\frac12$$
$$\frac{\cos(x)\color{red}{-\cos(2x)}}{\sin(x)\color{blue}{+\sin(2x)}} =\frac{1-\cos(x)}{\sin(x)}$$
$$\iff \frac{\cos(x)\color{red}{+1-2\cos^2(x)}}{\sin(x)\color{blue}{+2\sin(x)\cos(x)}} =\frac{1-\cos(x)}{\sin(x)}$$
$$\iff \frac{1+\cos(x)-2\cos^2(x)}{1+2\cos(x)} =1-\cos(x)$$
$$\iff 1+\cos(x)-2\cos^2(x) =(1-\cos(x))(1+2\cos (x))$$
which is true.
If you are familiar with complex number:
Let $$z$$ be a complex number satisfying $$|z|=1$$. We want to prove that $$z-\frac{1}{z^{2}}=k\left(1-\frac{1}{z}\right)$$ where $$k$$ is real. After some algebraic manipulation, we obtain $$k=z+1+\frac{1}{z}$$ which is always real. | 2021-12-04T17:12:23 | {
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