text:
# compare 2 different sized multivariate samples

I have a two data tables like below. The dataset1 represents failed candidates. The dataset2 represents the successful candidates.

I want to know, by applying some inference statistics which var (variable) has significantly changed the outcome (success and fail).

I have performed descriptive statistics and calculated centrality measures for both data sets column wise. I have also applied Machine-learning approaches for getting feature importance (Decision trees). But is there any way to statistically get that.

Dataset1 (2310x500)

    var1        var2        var3    var4  var500
U1  29.64611415 1576    427.4145    227  532
U2  28.45677687 1902    389.0615    225  484.88
U3  25.58181295 1376    379.1692    200  462
U4  27.02453608 1652    373.9734    210  457.60
U5  24.92124297 1790    330.8919    191  494.8
U6  17.00408003 1272    268.7932    131  409.30
U7  20.37879655 1088    340.4145    171  322
U2310 29.49205  1606    387.9543    222  642

Dataset2 (310x500)

          var1     var2 var3    var4  var500
S1  116.6229945 6954    1910.8676   890 2408
S2  77.54127969 4630    1222.5122   596 1421
S3  85.92342243 5190    1417.5633   655 1818
S4  53.90084493 3472    854.3462    418 939
S5  8.761080115 608     149.72      66  242
S6  4.530369972 356     98.8049     33  124
S7  5.944583534 624     124.3685    45  104
S310    23.41567    1208    418.46   204    266

• As long as var1, 2 etc are the same in the two data sets, then you can merge them and use e.g. logistic regression. – Peter Flom Sep 14 at 14:55
• Yes the var1, var2 etc are same for both data sets. But each data set represents independent group. I will follow your instruction and implement logistic regression. Also would like to see what LDA has to say for my data. Thank you. – Biofreek Sep 16 at 1:49
• LDA could be useful but it makes some pretty stong assumptions – Peter Flom Sep 16 at 10:50


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# 4.1 Linear functions  (Page 2/27)

 Page 2 / 27
$D\left(t\right)=83t+250$

## Representing a linear function in tabular form

A third method of representing a linear function is through the use of a table. The relationship between the distance from the station and the time is represented in [link] . From the table, we can see that the distance changes by 83 meters for every 1 second increase in time.

Can the input in the previous example be any real number?

No. The input represents time so while nonnegative rational and irrational numbers are possible, negative real numbers are not possible for this example. The input consists of non-negative real numbers.

## Representing a linear function in graphical form

Another way to represent linear functions is visually, using a graph. We can use the function relationship from above, $\text{\hspace{0.17em}}D\left(t\right)=83t+250,\text{\hspace{0.17em}}$ to draw a graph as represented in [link] . Notice the graph is a line . When we plot a linear function, the graph is always a line.

The rate of change, which is constant, determines the slant, or slope    of the line. The point at which the input value is zero is the vertical intercept, or y -intercept    , of the line. We can see from the graph that the y -intercept in the train example we just saw is $\text{\hspace{0.17em}}\left(0,250\right)\text{\hspace{0.17em}}$ and represents the distance of the train from the station when it began moving at a constant speed.

Notice that the graph of the train example is restricted, but this is not always the case. Consider the graph of the line $\text{\hspace{0.17em}}f\left(x\right)=2x+1.\text{\hspace{0.17em}}$ Ask yourself what numbers can be input to the function. In other words, what is the domain of the function? The domain is comprised of all real numbers because any number may be doubled, and then have one added to the product.

## Linear function

A linear function    is a function whose graph is a line. Linear functions can be written in the slope-intercept form    of a line

$f\left(x\right)=mx+b$

where $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ is the initial or starting value of the function (when input, $\text{\hspace{0.17em}}x=0\text{\hspace{0.17em}}$ ), and $\text{\hspace{0.17em}}m\text{\hspace{0.17em}}$ is the constant rate of change, or slope of the function. The y -intercept is at $\text{\hspace{0.17em}}\left(0,b\right).$

## Using a linear function to find the pressure on a diver

The pressure, $\text{\hspace{0.17em}}P,$ in pounds per square inch (PSI) on the diver in [link] depends upon her depth below the water surface, $\text{\hspace{0.17em}}d,$ in feet. This relationship may be modeled by the equation, $\text{\hspace{0.17em}}P\left(d\right)=0.434d+14.696.\text{\hspace{0.17em}}$ Restate this function in words.

To restate the function in words, we need to describe each part of the equation. The pressure as a function of depth equals four hundred thirty-four thousandths times depth plus fourteen and six hundred ninety-six thousandths.

## Determining whether a linear function is increasing, decreasing, or constant

The linear functions we used in the two previous examples increased over time, but not every linear function does. A linear function may be increasing, decreasing, or constant. For an increasing function    , as with the train example, the output values increase as the input values increase. The graph of an increasing function has a positive slope. A line with a positive slope slants upward from left to right as in [link] (a) . For a decreasing function    , the slope is negative. The output values decrease as the input values increase. A line with a negative slope slants downward from left to right as in [link] (b) . If the function is constant, the output values are the same for all input values so the slope is zero. A line with a slope of zero is horizontal as in [link] (c) .

#### Questions & Answers

The sequence is {1,-1,1-1.....} has
amit Reply
circular region of radious
Kainat Reply
how can we solve this problem
Joel Reply
Sin(A+B) = sinBcosA+cosBsinA
Eseka Reply
Prove it
Eseka
Please prove it
Eseka
hi
Joel
June needs 45 gallons of punch. 2 different coolers. Bigger cooler is 5 times as large as smaller cooler. How many gallons in each cooler?
Arleathia Reply
find the sum of 28th term of the AP 3+10+17+---------
Prince Reply
I think you should say "28 terms" instead of "28th term"
Vedant
if sequence sn is a such that sn>0 for all n and lim sn=0than prove that lim (s1 s2............ sn) ke hole power n =n
SANDESH Reply
write down the polynomial function with root 1/3,2,-3 with solution
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if A and B are subspaces of V prove that (A+B)/B=A/(A-B)
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write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°)
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Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4
kiruba Reply
what is the answer to dividing negative index
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In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
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give me the waec 2019 questions
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the polar co-ordinate of the point (-1, -1)
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## anonymous 4 years ago Will somebody PLEASE help me? What do you get when you cross cancell 23 and 45? Will somebody PLEASE help me? What do you get when you cross cancell 23 and 45? @Mathematics

1. alfie

Please write down the whole thing, not just the question, because it's not quite easy to understand this way. Anyhow, you cannot cross cancel these two. 23 is a prime number.

2. anonymous

oh ok. thx. I will draw it the best I can. Thank you so much.|dw:1319835048549:dw|

3. anonymous

|dw:1319835171467:dw|

4. alfie

You can't cross cancel these two, but you can cross cancel four and six, so that you have $\frac{45*2}{23*3}$ So you have $\frac{90}{69}$ Now you can divide both denominator and numerator by 3 getting 30/23

5. alfie

You can't cross cancel these two, but you can cross cancel four and six, so that you have $\frac{45*2}{23*3}$ So you have $\frac{90}{69}$ Now you can divide both denominator and numerator by 3 getting 30/23

6. anonymous

Ok. Thank you so much.


text:
# Second-order inhomogeneous differential equation $y''\:-\:4y'\:+\:2y\:=\:2x^2$

I'm trying to solving a 2nd-order inhomogeneous differential equation, but I'm not sure with my answer since I'm only learned it by myself & it has nothing to do with school or homework, so I have no one to ask or correct my answer. Please if you want to help me, i would be really happy & appreciate your help & advice.

So here's my problem

$y''\:-\:4y'\:+\:2y\:=\:2x^2$

and here's my step to answer

I know that to solve this 2nd-order inhomogeneous differential equation the general solution is

$y\:=\:y_c\:+\:y_p$

where $y_c$ is the general solution of the homogeneous equation. And $y_p$ is a particular solution

First, I find for $y_c$ through

$y''\:-\:4y'\:+\:2y\:=\:0$

since it's in the form of different roots of the auxiliary equation, i found that $m_1\:=\:2+\sqrt{2}$ and $m_2\:=\:2-\sqrt{2}$ , so the general solution of the homogeneous equation is

$y_c=c_1e^{\left(2+\sqrt{2}\right)x}+c_2e^{\left(2-\sqrt{2}\right)x}$

Second, since it's a polynomial problem so i find for $y_p$ through

$y_p=Ax^2+Bx+C$

first and second derivative of $y_p$ are

$y'_p=2Ax\:+\:B\:\:\:\:\:\:\:and\:\:\:\:\:\:\:\:y"_p=2A$

so now i have to find the undetermined coefficients $A$, $B$, and $C$. I find these coefficients through substitute $y_p$ into $y$, $y'_p$ into $y'$, and $y''_p$ into $y"$, thus

$y''\:-\:4y'\:+\:2y\:=\:\left(2A\right)x^2+\left(2B-8A\right)x\:+\:\left(2A-4B+2C\right)=2x^2$

So, $A\:=\:\frac{1}{2}\:,\:B\:=\:0,\:C\:=\:\frac{1}{2}$ and then substitute this into $y_p$, I find the particular solution which is

$y_p=\frac{1}{2}\left(x^2+1\right)$

and finally I find the general solution in the form of

$y\:=\:c_1e^{\left(2+\sqrt{2}\right)x}+c_2e^{\left(2-\sqrt{2}\right)x}+\frac{1}{2}\left(x^2+1\right)$

This is my answer, am I correct or not? Thank you for your time!

• No, can't be good as the only $x^2$ term in the LHS is $x^2$, while it is $2x^2$ in the RHS. Easier to start with $y=x^2+Ax+B$.
– user65203
Sep 17, 2014 at 13:33
• The second derivative should be typed as y'', not as y". (Compare the outputs: $y''$ and $y"$.) I also suggest cutting down the spending on emoticons; the freed up funds can be used to buy some uppercase letters.
– user147263
Sep 20, 2014 at 6:23

The equation is $$(D^{2}-4D+2)y=2x^{2}$$ Apply the annihilator $D^{3}$ of $x^{2}$ to obtain $$D^{3}(D^{2}-4D+2)y=0.$$ The general solution of this equation is $$f=e^{2x}(Ae^{\sqrt{2}x}+Be^{-\sqrt{2}x})+F+Gx+Hx^{2}.$$ Plugging back into the original equation eliminates the exponential terms and gives \begin{align} (D^{2}-4D+2)f & =(D^{2}-4D+2)(F+Gx+Hx^{2}) \\ & =(2H)-4(G+2Hx)+2(F+Gx+Hx^{2}) \\ & =(2H-4G+2F)+(-8H+2G)x+2Hx^{2}= x^{2}. \end{align} Therefore $$\begin{array}{l} H=1/2\\ -8H+2G=0 \implies G=4H=2 \\ 2H-4G+2F=0 \implies F=2G-H=4-1/2=7/2. \end{array}$$ The final solution is $$f = e^{2x}(Ae^{\sqrt{2}x}+Be^{-\sqrt{2}x})+\frac{7}{2}+2x+\frac{1}{2}x^{2}.$$ $A$ and $B$ are arbitrary constants.

Your steps look good. If you want to check the solution, plug the solution and its derivatives back into the original equation to see if it works.

Your $y_c$ part is fine but there is an error in $y_p$ part. Coefficients should be $A=1$, $B=4$ and $C=7$.

I would use various differential eq calculator to help, e.g,

The results show that the answer should be: yp = x^2+4x+7 enter image description here


text:
# The two consecutive positive integers has a product of 272? What are the 4 integers?

May 12, 2018

$\left(- 17 , - 16\right)$ and $\left(16 , 17\right)$

#### Explanation:

Let a be the smaller of the two integers and let a+1 be the bigger of the two integers:
$\left(a\right) \left(a + 1\right) = 272$, easiest way to solve this is to take the square root of 272 and round down:
$\sqrt{272} = \setminus \pm 16. . .$
16*17 = 272
Thus, the integers are -17,-16 and 16,17

May 12, 2018

16 17

#### Explanation:

If we multiply two consecutive numbers, $n \mathmr{and} n + 1$
we get ${n}^{2} + n$. That is we square a number and add one more on.

${16}^{2} = 256$

256+16=272

So our two numbers are 16 and 17

May 12, 2018

16 and 17

#### Explanation:

$\textcolor{b l u e}{\text{A sort of cheat way}}$

The two number are very close to each other so lets 'fudge' it

$\sqrt{272} = 16.49 \ldots$ so the first number is close to 16

Test $16 \times 17 = 272 \textcolor{red}{\leftarrow \text{First guess gets the prize!}}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{The systematic way}}$

Let the first value be $n$ then the next value is $n + 1$

The product is $n \left(n + 1\right) = 272$

${n}^{2} + n - 272 = 0$

Compare to: $a {x}^{2} + b x + c = 0 \textcolor{w h i t e}{\text{ddd") -> color(white)("ddd}} x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
In this case x->n; color(white)("d")a=1; color(white)("d")b=1 and c=-272

$n = \frac{- 1 \pm \sqrt{1 - 4 \left(1\right) \left(- 272\right)}}{2 \left(1\right)}$

$n = - \frac{1}{2} \pm \frac{\sqrt{1089}}{2}$

$n = - \frac{1}{2} \pm \frac{33}{2}$ The negative is not logical so discard it

$n = - \frac{1}{2} + \frac{33}{2} = 16$

The first number is 16 the second is 17


text:
# Proving double-angle trig identities

I'm having some trouble proving trig identities, this time with the double-angle formula. I want to prove that:

$$\frac{1 + \tan^2 A}{1 - \tan^2 A}=\sec 2A$$

I know that:

$$1 - \tan^2 A = \frac{2 \tan A}{\tan 2A}$$

But I don't know how to get the numerator of the LHS. I also know that:

$$\sec 2A = \frac{1}{2\sin A \cos A}$$

But I can't marry them together. Can anybody point me in the right direction?

-
Did you mean $\dfrac{1+\tan^2 A}{1-\tan^2 A}=\sec 2A$ – Hawk Feb 9 '14 at 15:03

First of all, $\displaystyle2\sin A\cos A=\sin2A$

If $\displaystyle\sec2A=\frac1{2\sin A\cos A}$

we need $\displaystyle\sec2A=\frac1{\sin2A}\iff1=\frac{\sin2A}{\cos2A}$

$\displaystyle\implies\tan2A=1\iff2A=n\pi+\frac\pi4$ where $n$ is any integer

Now, $$\frac{1+\tan^2A}{1-\tan^2A}=\frac{1+\dfrac{\sin^2A}{\cos^2A}}{1-\dfrac{\sin^2A}{\cos^2A}}=\frac{\cos^2A+\sin^2A}{\cos^2A-\sin^2A}=\frac1{\cos2A}$$

Reference : Double-Angle Formulas

-
I liked the previous formatting of your answer more...if you don't mind – Hawk Feb 9 '14 at 15:05
@Hawk, please find the edited version – lab bhattacharjee Feb 9 '14 at 15:07
How did you get from $\frac{1+\tan^2A}{1-\tan^2A}$ to $\frac{\cos^2A+\sin^2A}{\cos^2A-\sin^2A}$? – hohner Feb 9 '14 at 15:08
Yes, this is fine...Your solution is good as always – Hawk Feb 9 '14 at 15:09
@hohner, please find the edited version – lab bhattacharjee Feb 9 '14 at 15:11


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## [1109.2284] Triple Disruptions in The Galactic Centre: Captured and Ejected Binaries, Rejuvenated Stars, and Correlated Orbits

Authors: Idan Ginsburg, Hagai B. Perets

Date: 11 Sep 2011

Abstract: The disruption of a binary star by a massive black hole (MBH) typically leads to the capture of one component around the MBH and the ejection of its companion at a high velocity, possibly producing a hypervelocity star. The high fraction of observed triples ($\sim10$% for F/G/K stars and $\sim50$% for OB stars) give rise to the possibility of the disruption of triples by a MBH. Here we study this scenario, and use direct $N$-body integrations to follow the orbits of thousands of triples, during and following their disruption by a MBH (of $4\times10ˆ6$ M$_\odot$, similar to the MBH existing in the Galactic Centre; SgrA$ˆ*$). We find that triple disruption can lead to several outcomes and we discuss their relative frequency. Beside the ejection/capture of single stars, similar to the binary disruption case, the outcomes of triple disruption include the ejection of hypervelocity binaries; capture of binaries around the MBH; collisions between two or all of the triple components (with low enough velocities that could lead to their merger); and the capture of two or even three stars at close orbits around the MBH. The orbits of single stars captured in a single disruption event are found to be correlated. The eccentricity of the mutual orbits of captured/ejected binaries is typically excited to higher values. Stellar evolution of captured/ejected binaries may later result in their coalescence/strong interaction and the formation of hypervelocity blue stragglers or merger remnants in orbits around SgrA*. Finally, the capture of binaries close to the MBH can replenish and increase the binary frequency near the MBH, which is otherwise very low.

#### Sep 14, 2011

1109.2284 (/preprints)
2011-09-14, 13:46


text:
# Tag Info

0

If we remove $e = (x,y)$ from $T_2$, then it breaks into two connected components, say $T_{2,x},T_{2,y}$. There is a unique path in $T_1$ from $x$ to $y$, which together with $e$ forms a fundamental cycle of $T_1$. The path starts at $T_{2,x}$ and ends at $T_{2,y}$, hence at some point it crosses between the two. Adding the corresponding edge to $T_2$ would ...

0

The MST indeed adresses conditions 1 and 3 but not conditions 2. The solution of the global problem (as shown by your example) is not the MST but still a tree. Let's call $T$ the solution for the input graph $G$. Let's also call $T_i$ the solution for the problem $G_i$ which is the subgraph of $G$ containing vertices with range index lower or equal to $i$ (I ...

4

As Yuval pointed out, there is some ambiguity of the original exercise as the enqueueing order of the neighbors of a node is not stipulated by the definition of a breadth-first-search (BFS). Proposition 1: Let $G$ be a simple undirected connected graph and $e$ is one of its edges. If $e$ is a part of a cycle, then there is a BFS of $G$ that produces a BFS ...

3

Let's show first that such an edge might belong to a cycle, by considering the case of a square $ab,bc,cd,da$. We will show that $ab$ could belong to BFS trees starting at arbitrary vertices. This is clear if BFS is started from $a$ or from $b$. If BFS is started from $c$, it could enqueue $a,d$ in this order; dequeue $a$; then enqueue $b$. Similarly, if BFS ...

1

It turns out that there's a linear time algorithm, published in Carlson J., Eppstein D. (2006) The Weighted Maximum-Mean Subtree and Other Bicriterion Subtree Problems. In: Arge L., Freivalds R. (eds) Algorithm Theory – SWAT 2006. SWAT 2006. Lecture Notes in Computer Science, vol 4059. Springer, Berlin, Heidelberg. The paper is paywalled, so I have ...

0

As indicated by the comments, this is a minimum spanning tree problem, which can be solved efficiently by Edmonds' algorithm (or Chu–Liu/Edmonds' algorithm).

Top 50 recent answers are included


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# inverse of a 4x4 matrix using calculator

My calculator only supports a 3x3, 3x2, 3x1, 2x3, 2x2, 2x1, 1x3, 1x2, 1x1 matrices. This website uses cookies to ensure you get the best experience. Inverse of a Matrix using Minors, Cofactors and Adjugate (Note: also check out Matrix Inverse by Row Operations and the Matrix Calculator.). Inverse matrix using determinants Apart from the Gaussian elimination, there is an alternative method to calculate the inverse matrix. It is a matrix when multiplied by the original matrix yields the identity matrix. Why does this movie say a witness can't present a jury with testimony which would assist in making a determination of guilt or innocence? MathJax reference. A faster way to obtain the adjoint, adjugate or adjunct of the transpose of a 4x4 (4 by 4) matrix. The inverse of a 2x2 is easy... compared to larger matrices (such as a 3x3, 4x4, etc). Matrix, the one with numbers, arranged with rows and columns, is extremely useful in most scientific fields. Is it illegal to carry someone else's ID or credit card? In this section we see how Gauss-Jordan Elimination works using examples. It only takes a minute to sign up. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. As a result you will get the inverse calculated on the right. By using this website, you agree to our Cookie Policy. Inverse Matrix Calculator Inverse of a matrix is similar to inverse or reciprocal of a number. Are there minimal pairs between vowels and semivowels? By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. We employ the latter, here. Learn more Hire us: How much did the first hard drives for PCs cost? To create your new password, just click the link in the email we sent you. My calculator only supports a 3x3, 3x2, 3x1, 2x3, 2x2, 2x1, 1x3, 1x2, 1x1 matrices. Trick in using Casio fx-570ES calculator. 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Step 1: calculating the Matrix of Minors, Step 2: then turn that into the Matrix of Cofactors, A simple example of finding the inverse matrix of a 4x4 matrix, using Gauss-Jordan elimination Last updated: Jan. 3, 2019 Find the inverse matrix of a 4x4 matrix, Use MathJax to format equations. By using this website, you agree to our Cookie Policy. 4x4 Multiplication can be done when m x n matrices have 4 rows and 4 columns. Use our below online inverse matrix calculator to solve 2x2, 3x3, 4x4 and 5x5 matrices. How can I confirm the "change screen resolution dialog" in Windows 10 using keyboard only? This inverse matrix calculator help you to find the inverse matrix. The inverse of a square n x n matrix A, is another n x n matrix, denoted as A -1. Are there any contemporary (1990+) examples of appeasement in the diplomatic politics or is this a thing of the past? 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In order for MINVERSE to calculate an inverse matrix, the input array must contain numbers only and be a square matrix, with equal rows and columns. 3x3 identity matrices involves 3 rows and 3 columns. How would I reliably detect the amount of RAM, including Fast RAM? The following formula is used to calculate the inverse matrix value of the original 2×2 matrix. The only difference between a solving a linear equation and a system of equations written in matrix form is that finding the inverse of a matrix is more complicated, and matrix multiplication is a longer process. Multiplying by the inverse... \begin{pmatrix}1 & 2 \\3 & 4\end{pmatrix}^{-1}, inverse\:\begin{pmatrix}1 & 2 & 3 \\4 & 5 & 6 \\7 & 2 & 9\end{pmatrix}. About the 4 x 4 matrix inverse calculator The formula that is used to calculate the inverse of a 4 x 4 matrix (say A) is: A - 1 = 1/ det (A) × adj (A) There... For matrices there is no such thing as division, you can multiply but can’t divide. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Matrices are array of numbers or values represented in rows and columns. ... Answer There are mainly two ways to obtain the inverse matrix. Free Matrix Adjoint calculator - find Matrix Adjoint step-by-step This website uses cookies to ensure you get the best experience. We start with the matrix A, and write it down with an Identity Matrix I next to it: (This is called the \"Augmented Matrix\") Now we do our best to turn \"A\" (the Matrix on the left) into an Identity Matrix. MAINTENANCE WARNING: Possible downtime early morning Dec 2, 4, and 9 UTC…, Matrix multiplication/subtraction problem, Checking if a matrix has a left inverse using matrix dimensions and rank. By browsing this website, you agree to our use of cookies. The calculator will perform symbolic calculations whenever it is possible. Is it more efficient to send a fleet of generation ships or one massive one? Create a random matrix A of order 500 that is constructed so that its condition number, cond(A), is 1e10, and its norm, norm(A), is 1.The exact solution x is a random vector of length 500, and the right side is b = A*x. It does not give only the inverse of a 4x4 matrix and also it gives the determinant and adjoint of the 4x4 matrix that you enter. Which direction should axle lock nuts face? Is there any chance I can get the inverse of the 4x4 using my calculator using the matrix mode? An n × n square matrix with ones on the main diagonal and zeros in every other position. Is there a general solution to the problem of "sudden unexpected bursts of errors" in software? Thanks for the feedback. Something tells me that you are meant to solve this without a calculator (perverse or not). by M. Bourne. However, the goal is the same—to isolate the variable. It is much less intuitive, and may be much longer than the previous one, but we can always use it because it is more direct. The inverse of a matrix is expressed by A-1. Please note that this calculator supports both positive and negative numbers, with or without decimals and even fractions. The calculator will find the inverse of the square matrix using the Gaussian elimination method, with steps shown. The matrix has four rows and columns. And, I don't think if there is any property exist. The adjugate matrix is also used in Jacobi's formula for the derivative of the determinant. What do I do to get my nine-year old boy off books with pictures and onto books with text content? ... Calculator Enter a 4x4 matrix and press "execute button". To calculate inverse matrix you need to do the following steps. But unfortunately the resulting formula is somewhat complicated. In this case, (ad-bc) is also known as the magnitude of the original matrix. Show Instructions. Did they allow smoking in the USA Courts in 1960s? Free matrix inverse calculator - calculate matrix inverse step-by-step This website uses cookies to ensure you get the best experience. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. Get the free "4x4 Determinant calculator" widget for your website, blog, Wordpress, Blogger, or iGoogle. First, the original matrix should be in the form below. Using a Calculator to Find the Inverse Matrix Select a calculator with matrix capabilities. Find more Mathematics widgets in Wolfram|Alpha. You can also choose a different size matrix … Why is Buddhism a venture of limited few? Using row reduction to calculate the inverse and the determinant of a square matrix Notes for MATH 0290 Honors by Prof. Anna Vainchtein 1 Inverse of a square matrix An n×n square matrix A is called invertible if there exists a matrix X such that AX = XA = I, where I is the n × n identity matrix. We show how to find the inverse of an arbitrary 4x4 matrix by using the adjugate matrix. For those larger matrices there are three main methods to work out the inverse: Inverse of a Matrix using Elementary Row Operations (Gauss-Jordan) Inverse of a Matrix using Minors, Cofactors and Adjugate; Use a computer (such as the Matrix Calculator) Conclusion Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. We use this everyday without noticing, but we hate it when we feel it. How to professionally oppose a potential hire that management asked for an opinion on based on prior work experience? Inverse of Matrix Calculator. Panshin's "savage review" of World of Ptavvs. Reduce the left matrix to row echelon form using elementary row operations for the whole matrix (including the right one). However it is instructive to know about the division of a matrix into blocks. LU Decomposition of Matrix calculator - Online matrix calculator for LU Decomposition of Matrix, step-by-step. Typo around 4:15. Is there an "internet anywhere" device I can bring with me to visit the developing world? A piece of wax from a toilet ring fell into the drain, how do I address this? I don't think you can find $4×4$ matrix inverse in your calculator unless you find any property which reduces $4×4$ matrix into $3×3$. Examine why solving a linear system by inverting the matrix using inv(A)*b is inferior to solving it directly using the backslash operator, x = A\b.. Inverse of a Matrix using Gauss-Jordan Elimination. In the cofactor grid, the matrix in the first column, third row, C(3,1) should have bottom row (0, 1, 4), not (2, 1, 4). 'Adjoint' of a matrix refers to the corresponding adjoint operator, which is its conjugate transpose. Usually it is performed by multiplying the first row of first matrix with the first column of second one. \begin{pmatrix}1 & 3 & 5 & 9 \\1 & 3 & 1 & 7 \\4 & 3 & 9 & 7 \\5 & 2 & 0 & 9\end{pmatrix}^{-1}, Please try again using a different payment method. We can calculate the Inverse of a Matrix by:. Thanks for contributing an answer to Mathematics Stack Exchange! Multiplied by the original matrix yields the identity matrix property exist sudden unexpected bursts of errors '' Windows... And press  execute button '' appeasement in the email we sent you supports a 3x3, 3x2 3x1. $2 \times 2$ blocks ( i.e of first matrix with on... When m x n matrices have 4 rows and columns any gambits where I have to decline device! 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text:
# Rhombus

For other uses, see Rhombus (disambiguation).
rhombus
Two rhombi.
Edges and vertices 4
Schläfli symbol { } + { } or 2{ }
Coxeter diagram
Symmetry group Dih2, [2], (*22), order 4
Area $\tfrac{pq}{2}$
Dual polygon rectangle
Properties convex, isotoxal

In Euclidean geometry, a rhombus (◊), plural rhombi or rhombuses, is a simple (non-self-intersecting) quadrilateral whose four sides all have the same length. Another name is equilateral quadrilateral, since equilateral means that all of its sides are equal in length. The rhombus is often called a diamond, after the diamonds suit in playing cards, or a lozenge, though the former sometimes refers specifically to a rhombus with a 60° angle (see Polyiamond), and the latter sometimes refers specifically to a rhombus with a 45° angle.

Every rhombus is a parallelogram, and a rhombus with right angles is a square.[1][2]

## Etymology

The word "rhombus" comes from Greek ῥόμβος(rhombos), meaning something that spins,[3] which derives from the verb ρέμβω (rhembō), meaning "to turn round and round".[4] The word was used both by Euclid and Archimedes, who used the term "solid rhombus" for two right circular cones sharing a common base.[5]

## Characterizations

A simple (non self-intersecting) quadrilateral is a rhombus if and only if it is any one of the following:[6][7]

• a quadrilateral with four sides of equal length (by definition)
• a quadrilateral in which the diagonals are perpendicular and bisect each other
• a quadrilateral in which each diagonal bisects two opposite interior angles
• a parallelogram in which at least two consecutive sides are equal in length
• a parallelogram in which the diagonals are perpendicular
• a parallelogram in which a diagonal bisects an interior angle

## Basic properties

Every rhombus has two diagonals connecting pairs of opposite vertices, and two pairs of parallel sides. Using congruent triangles, one can prove that the rhombus is symmetric across each of these diagonals. It follows that any rhombus has the following properties:

The first property implies that every rhombus is a parallelogram. A rhombus therefore has all of the properties of a parallelogram: for example, opposite sides are parallel; adjacent angles are supplementary; the two diagonals bisect one another; any line through the midpoint bisects the area; and the sum of the squares of the sides equals the sum of the squares of the diagonals (the parallelogram law). Thus denoting the common side as a and the diagonals as p and q, in every rhombus

$\displaystyle 4a^2=p^2+q^2.$

Not every parallelogram is a rhombus, though any parallelogram with perpendicular diagonals (the second property) is a rhombus. In general, any quadrilateral with perpendicular diagonals, one of which is a line of symmetry, is a kite. Every rhombus is a kite, and any quadrilateral that is both a kite and parallelogram is a rhombus.

A rhombus is a tangential quadrilateral.[8] That is, it has an inscribed circle that is tangent to all four of its sides.

## Area

As for all parallelograms, the area A of a rhombus is the product of its base and its height. The base is simply any side length a, and the height h is the perpendicular distance between any two non-adjacent sides:

$A = a \cdot h .$

The area can also be expressed as the base squared times the sine of any angle:

$A = a^2 \cdot \sin \alpha = a^2 \cdot \sin \beta ,$

or as half the product of the diagonals p, q:

$A = \frac{p \cdot q}{2} ,$

or as the semiperimeter times the radius of the circle inscribed in the rhombus (inradius):

$A = 2a \cdot r .$

Another way, in common with parallelograms, is to consider two adjacent sides as vectors, forming a bivector, so the area is the magnitude of the bivector (the magnitude of the vector product of the two vectors), which is the determinant of the two vectors' Cartesian coordinates ( area = x1*y2-x2*y1 ) [9]

The inradius (the radius of the incircle) can be expressed in terms of the diagonals p and q as[8]

$r = \frac{p \cdot q}{2\sqrt{p^2+q^2}}.$

## Dual properties

The dual polygon of a rhombus is a rectangle:[10]

• A rhombus has all sides equal, while a rectangle has all angles equal.
• A rhombus has opposite angles equal, while a rectangle has opposite sides equal.
• A rhombus has an inscribed circle, while a rectangle has a circumcircle.
• A rhombus has an axis of symmetry through each pair of opposite vertex angles, while a rectangle has an axis of symmetry through each pair of opposite sides.
• The diagonals of a rhombus intersect at equal angles, while the diagonals of a rectangle are equal in length.
• The figure formed by joining, in order, the midpoints of the sides of a rhombus is a rectangle and vice-versa.

## Other properties

As topological square tilings As 30-60 degree rhombille tiling
Some polyhedra with all rhombic faces
Identical rhombi Two types of rhombi
Rhombohedron Rhombic dodecahedron Rhombic triacontahedron Rhombic icosahedron Rhombic enneacontahedron


text:
Under the auspices of the Computational Complexity Foundation (CCF)

REPORTS > DETAIL:

### Paper:

TR99-035 | 6th July 1999 00:00

#### Clustering for Edge-Cost Minimization

TR99-035
Authors: Leonard Schulman
Publication: 15th September 1999 16:34
Keywords:

Abstract:

We address the problem of partitioning a set of $n$ points into
clusters, so as to minimize the sum, over all intracluster pairs of
points, of the cost associated with each pair. We obtain a randomized
approximation algorithm for this problem, for the cost functions
$\ell_2^2,\ell_1$ and $\ell_2$, as well as any cost function
isometrically embeddable in $\ell_2^2$.

The maximization problem (maximize the costs of intercluster edges) is
approximated with high probability to within multiplicative factor
$(1-\ep)$; while the minimization problem either receives a
multiplicative $1+\ep$ approximation, or else the optimal clustering
is correctly identified except for a mislabelling of an $\ep$ fraction
of the point set. Given a fixed approximation parameter $\ep$, the
runtime is linear in $n$ for $\ell_2^2$ problems of dimension
$o(\log n / \log \log n)$; and $n^{O(\log \log n)}$ in the general
case.

The case $\ell_2^2$ is addressed by combining three elements: (a)
Variable-probability sampling of the given points, to reduce the size
of the data set. (b) Near-isometric dimension reduction. (c) A
deterministic exact algorithm which runs in time exponential in the
dimension (rather than the number of points). The remaining cases are
addressed by reduction to $\ell_2^2$.

ISSN 1433-8092 | Imprint


text:
# Confirm delete for GridView's CommandField using JavaScript

Been a while since I have had to do this, and to describe my memory as “sieve-like” is probably being generous. After vainly looking for the non-existent OnClientClick property of the CommandField (still can’t quite believe they don’t have that), I tried to recall the best way of wiring this up.

One way is to use a use a TemplateField instead with an ImageButton control or similar and take advantage of its OnClientClick property. If you give it an appropriate CommandName (i.e. “delete”) the GridView will automatically fire the OnRowDeleting event if you have wired it up:

<asp:TemplateField>
<ItemTemplate>
<asp:ImageButton runat="server" ID="DeleteButton"
CommandName="delete"
ImageUrl="~/Images/icon-delete-12x12.png"
OnClientClick="if (!window.confirm('Are you sure you want to delete this item?')) return false;" />
</ItemTemplate>
</asp:TemplateField>

Another option is to hook into the RowDataBound event in code behind to attach the required JavaScript to the button in the CommandField, as is suggested in one of Scott Mitchell’s great ASP.NET 2.0 data access tutorials (see Step 3):

protected void MyGridView_RowDataBound(object sender, GridViewRowEventArgs e) {  if (e.Row.RowType != DataControlRowType.DataRow) return;  int lastCellIndex = e.Row.Cells.Count - 1;  ImageButton deleteButton = (ImageButton) e.Row.Cells[lastCellIndex].Controls[0];  deleteButton.OnClientClick =     "if (!window.confirm('Are you sure you want to delete this item?')) return false;";}

In this case the last cell is an asp:CommandField with ButtonType="Image". Be wary of the double submit issue if you use this approach. Also note that we only return false if the user cancels (i.e. window.confirm returns false). If we just return the result of window.confirm we might end up skipping the JavaScript-triggered postback even if the user wants to go ahead!


text:
# RKHS concepts. Connection with matrix trace function.

With my basic linear algebraic background, am trying to connect the Reproducing Kernel Hilbert Space (RKHS) concepts to example functions over matrices, as I gradually learn about this new concept.

Question: Can $TrX^TKX$ be connected to a RKHS in any way? $K$ is a p.s.d kernel matrix and $X$ is a matrix of reals.

I know that $TrX^TKX$ can be represented as $Tr[(SX)^T(SX)]=||SX||^2_{HS}$, using the hilbert schmidt norm where $S$ is the p.s.d square root of K.(ex:$S=U\lambda^{1/2}$, where $K=U\lambda U^T$ is the eigen-decomposition of $K$). But am trying really hard with my narrow perspective to see the connections of this matrix function with the concept of RKHS. Do shed some light!

-


text:
Comparison when Control is in differen batch
1
0
Entering edit mode
rina • 0
@rina-16738
Last seen 14 days ago
Toulouse

I am performing a differential expression analysis between treated and untreated cell lines using DESeq2. The samples were sequenced in two different batches, however, there is a perfect confounding of the conditions and the batches. All controlled samples are in one batch and all treated samples are in the second one. The perfect confounding does not allow to remove batches or even account for them in the formula. I was wondering is it still valid to compare all treated conditions to the control and then compare the regulation of different genes in response to treatment? For example, I would like to check if AKT1 is upregulated in one treatment when compared to control and downregulated in another one when compared to control.

Thank you!

DESeq2 DifferentialExpression • 103 views
0
Entering edit mode

There is not much —if anything— that can be done that is statistically sound. You could make an assumption that there is no difference between batch 1 and batch 2, but this is most likely an erroneous assumption. If you must proceed with the current data, then regard it as preliminary / pilot data and process each batch independently. Then, transform the normalised counts via vst() or rlog(), followed by a final transformation to Z scores. At least on the Z scale, qualitative comparisons across batch are possible, if even to say things such as 'GeneX is expressed in condition 1, but not condition 2'. Absolute Z > 1.96 is equivalent to p=0.05 on a two-tailed distribution. This is still not ideal, though.

0
Entering edit mode
ATpoint ▴ 590
@atpoint-13662
Last seen 7 hours ago
Germany

Not sure which answer you are expecting. The experiment is confounded as you correctly say, hence all DEGs you see could potentially come from the batch. The interpretation is now on you. It comes now down to time considerations, money and feasability whether you will repeat the experiment properly, or just go ahead pretending there was no batch effect. If you proceed consider results with care, try to validate findings.


text:
# Overview

Code4rena (C4) is an open organization consisting of security researchers, auditors, developers, and individuals with domain expertise in smart contracts.

A C4 code contest is an event in which community participants, referred to as Wardens, review, audit, or analyze smart contract logic in exchange for a bounty provided by sponsoring projects.

During the code contest outlined in this document, C4 conducted an analysis of Perennial contest smart contract system written in Solidity. The code contest took place between December 9—December 15 2021.

## Wardens

13 Wardens contributed reports to the Perennial contest:

1. leastwood
2. kenzo
3. WatchPug (jtp and ming)
4. 0x0x0x
5. 0x1f8b
6. cmichel
7. robee
8. hubble (ksk2345 and shri4net)
9. defsec
10. ye0lde
11. gzeon
12. pmerkleplant
13. broccolirob

This contest was judged by Alex the Entreprenerd.

Final report assembled by itsmetechjay and CloudEllie.

# Summary

The C4 analysis yielded an aggregated total of 11 unique vulnerabilities and 38 total findings. All of the issues presented here are linked back to their original finding.

Of these vulnerabilities, 2 received a risk rating in the category of HIGH severity, 3 received a risk rating in the category of MEDIUM severity, and 6 received a risk rating in the category of LOW severity.

C4 analysis also identified 5 non-critical recommendations and 22 gas optimizations.

# Scope

The code under review can be found within the C4 Perennial contest repository, and is composed of 38 smart contracts written in the Solidity programming language and includes 3531 lines of Solidity code and 7 lines of JavaScript.

# Severity Criteria

C4 assesses the severity of disclosed vulnerabilities according to a methodology based on OWASP standards.

Vulnerabilities are divided into three primary risk categories: high, medium, and low.

High-level considerations for vulnerabilities span the following key areas when conducting assessments:

• Malicious Input Handling
• Escalation of privileges
• Arithmetic
• Gas use

Further information regarding the severity criteria referenced throughout the submission review process, please refer to the documentation provided on the C4 website.

# High Risk Findings (2)

## [H-01] Wrong shortfall calculation

Submitted by kenzo

Every time an account is settled, if shortfall is created, due to a wrong calculation shortfall will double in size and add the new shortfall.

#### Impact

Loss of funds: users won’t be able to withdraw the correct amount of funds. Somebody would have to donate funds to resolve the wrong shortfall.

#### Proof of Concept

We can see in the settleAccount of OptimisticLedger that self.shortfall ends up being self.shortfall+self.shortfall+newShortfall: (Code ref)

function settleAccount(OptimisticLedger storage self, address account, Fixed18 amount)
internal returns (UFixed18 shortfall) {

if (newBalance.sign() == -1) {
newBalance = Fixed18Lib.ZERO;
}

self.balances[account] = newBalance.abs();
}

Additionally, you can add the following line to the “shortfall reverts if depleted” test in Collateral.test.js, line 190:

await collateral.connect(productSigner).settleAccount(userB.address, -50)

Previously the test product had 50 shortfall. Now we added 50 more, but the test will print that the actual shortfall is 150, and not 100 as it should be.

Move the setting of self.shortfall to inside the if function and change the line to:

    self.shortfall = shortfall

kbrizzle (Perennial) confirmed:

Excellent find 🙏

Alex the Entreprenerd (judge) commented:

Agree with the finding shortfall = self.shortfall.add(newBalance.abs()); is already shortfal + newBalance.abs() So performing line 73 self.shortfall = self.shortfall.add(shortfall); is adding shortfall again

## [H-02] withdrawTo Does Not Sync Before Checking A Position’s Margin Requirements

Submitted by leastwood

#### Impact

The maintenanceInvariant modifier in Collateral aims to check if a user meets the margin requirements to withdraw collateral by checking its current and next maintenance. maintenanceInvariant inevitably calls AccountPosition.maintenance which uses the oracle’s price to calculate the margin requirements for a given position. Hence, if the oracle has not synced in a long time, maintenanceInvariant may end up utilising an outdated price for a withdrawal. This may allow a user to withdraw collateral on an undercollaterized position.

#### Proof of Concept

https://github.com/code-423n4/2021-12-perennial/blob/main/protocol/contracts/collateral/Collateral.sol#L67-L76

function withdrawTo(address account, IProduct product, UFixed18 amount)
notPaused
collateralInvariant(msg.sender, product)
maintenanceInvariant(msg.sender, product)
external {
_products[product].debitAccount(msg.sender, amount);
token.push(account, amount);

emit Withdrawal(msg.sender, product, amount);
}

https://github.com/code-423n4/2021-12-perennial/blob/main/protocol/contracts/collateral/Collateral.sol#L233-L241

modifier maintenanceInvariant(address account, IProduct product) {
_;

UFixed18 maintenance = product.maintenance(account);
UFixed18 maintenanceNext = product.maintenanceNext(account);

if (UFixed18Lib.max(maintenance, maintenanceNext).gt(collateral(account, product)))
revert CollateralInsufficientCollateralError();
}

https://github.com/code-423n4/2021-12-perennial/blob/main/protocol/contracts/product/types/position/AccountPosition.sol#L71-L75

function maintenanceInternal(Position memory position, IProductProvider provider) private view returns (UFixed18) {
Fixed18 oraclePrice = provider.priceAtVersion(provider.currentVersion());
UFixed18 notionalMax = Fixed18Lib.from(position.max()).mul(oraclePrice).abs();
return notionalMax.mul(provider.maintenance());
}

#### Tools Used

Manual code review.

Consider adding settleForAccount(msg.sender) to the Collateral.withdrawTo function to ensure the most up to date oracle price is used when assessing an account’s margin requirements.

kbrizzle (Perennial) confirmed:

Great catch 🙏

Alex the Entreprenerd (judge) commented:

With most onChain protocols where there is potential for undercollateralized positions and liquidations, it is very important to accrue a user position before making any changes to their balance.

The warden identified a potential way for a user to withdraw funds while their account is below margin requirements.

Because this impacts the core functionality functionality of the protocol (accounting), I’m raising the severity to high

Mitigation seems to be straightforward

# Medium Risk Findings (3)

## [M-01] No checks if given product is created by the factory

Submitted by 0x0x0x

An attacker can create a fake product. Collateral contract does not check whether the given product is created by the factory. A malicious product can return arbitrary maintenance amounts, therefore they can make any deposit to fake product stuck (simply return collateral - 1 as maintenance) and fake product owner can change the maintenance and liquidate whenever wanted and claim full collateral.

This is a serious attack vector, since by only interacting with Collateral contract the user can lose funds. Furthermore, this vulnerability poses big risks in combination web vulnerabilities. Users always have to check the product address to avoid getting scammed, but likely most users will only check the contract address.

Furthermore, if another protocol would want to add perennial protocol for its use case, the other protocol has to be careful with this behaviour and keep track of whitelisted products. This complicates adoption of perennial protocol, since whitelist has to be managed manually or else this vulnerability will likely be exploitable.

#### Mitigation step

Add a mapping inside Collateral, which verifies whether a product is created by factory or not. This mapping should get updated by the factory. This will add a little bit gas cost, but will eliminate a serious attack vector.

Or less gas efficient option is directly call a function from factory to verify.

kbrizzle (Perennial) disagreed with severity:

Good find 🙏

On severity: In general there’s a whole plethora of ways a Product owner could create a product that harmful a user which choses to take part in it. The design model of Perennial is to give the Product owners the freedom to create risky products, but to segregate that risk to only those products, and this issue does not break this segregation of risk. This is especially important during our gated-beta phase while we learn which levers are worth it to lock down.

That being said this is low hanging fruit to correct, so we think it should still be 2 (Medium) severity.

Alex the Entreprenerd (judge) commented:

Given the extensible nature of the Collateral contract, users can call the contract with any user input for product this could be used maliciously against the users as the warden highlighted.

This can definitely happen, but is contingent on a set of external factors, so I think medium severity is more appropriate

## [M-02] Multiple initialization of Collateral contract

Submitted by 0x1f8b

#### Impact

The attacker can initialize the contract, take malicious actions, and allow it to be re-initialized by the project without any error being noticed.

#### Proof of Concept

The initialize method of theCollateral contract does not contain the initializer modifier, it delegates this process to check if the factory is different from address(0) in theUFactoryProvider__initialize call, however the factory_is other than address(0), so an attacker could use front-running by listening to the memory pool, initialize the contract with a factory=address(0), perform malicious actions, and still allow it to be started later by the I draft the contract without noticing any error.

Source reference:

• Collateral.initialize

#### Tools Used

Manual review

Use initializer modifier

kbrizzle (Perennial) marked as duplicate:

Alex the Entreprenerd (judge) commented:

Disagree with sponsor that this is duplicate of #73

Alex the Entreprenerd (judge) commented:

The finding mentions that initialize can be called multiple times, specifically by first having factory set to address(0) and then setting it to it’s final value.

This is true.

Calling initialize multiple times could allow to set certain values up with a different token, then change to another token and set the factory. This could be used to trick the protocol accounting to later take advantage of other depositors.

Alex the Entreprenerd (judge) commented:

The specific vulnerability is the lack of the initializer modifier on initialize, the consequences can be problematic only if a malicious actor were to set the contracts up.

Because this can be done, and it’s reliant on a specific setup, I’ll mark the finding as medium severity, mitigation is straightforward

kbrizzle (Perennial) commented:

My understanding of these is that:

• # 13 would allow an attacker to front-run the initialization of a contract to gain admin, and the subsequent legitimate attempt to initialize would still work (thereby possibly going unnoticed).

If we want to tag these as separate issues, I think that makes sense as the effects are slightly different. Both are ultimately a documentation issue though solved by this remediation, so I’m wondering why one is tagged as 0 (Non-critical) and the other 2 (Medium).

Is that correct, or is there an issue here even with proper atomic deploy-and-initialize usage with OZ’s upgrades plugin? Just want to make sure we’re not missing something else here. Also - we do not have an initializer modifier available in the code base, so the remediation suggestion was a little confusing.

Alex the Entreprenerd (judge) commented:

@kbrizzle Yes, this is something that can still be done, arguably only by your deployer

You could deploy and initialize the contract with factory set to 0, use that to perform operations with a token you created (so it has no cost to you), then initialize again with factory set to the proper value

Because certain functions do not check for the product being created by the factory, this can be done as a way to lay the ground for a rug against future depositors

So to my eyes this finding highlights the consequences of a lack of the initializer modifier, while the other is the usual initializer can be frontrun non critical finding

Alex the Entreprenerd (judge) commented:

@kbrizzle to clarify: you won’t be frontrun with your deployment as the deploy + initialize is done in one tx. But the Collateral  not having initializer can be used by the deployer for malicious purposes because what was perceived as an unchangeable token can actually be changed if you set up initialize with the factory address set to 0

kbrizzle (Perennial) commented:

@Alex the Entreprenerd I see, that makes sense from the perspective of the token seeming like it should be immutable 👍

FWIW in this scenario the deployer would have to run the risk of someone else initializing and taking admin for the duration that the malicious initial token was active - so I’m not sure how feasible that would be.

That being said, I think a UInitializable would be a great addition to our lib anyways, so let’s go this route since this does seem like a solution with better guarantees. Thanks for the explanation!

Alex the Entreprenerd (judge) commented:

The attack is dependent upon external conditions (malicious deployer, nobody else calling initialize, setting up malicious accounting), for that reason (as per the docs) I believe medium severity to be proper.

Submitted by WatchPug, also found by cmichel, defsec, and ye0lde

function sync() public {
(, int256 feedPrice, , uint256 timestamp, ) = feed.latestRoundData();
Fixed18 price = Fixed18Lib.ratio(feedPrice, SafeCast.toInt256(_decimalOffset));

if (priceAtVersion.length == 0 || timestamp > timestampAtVersion[currentVersion()] + minDelay) {
priceAtVersion.push(price);
timestampAtVersion.push(timestamp);

emit Version(currentVersion(), timestamp, price);
}
}

On ChainlinkOracle.sol, we are using latestRoundData, but there is no check if the return value indicates stale data. This could lead to stale prices according to the Chainlink documentation:

#### Recommendation

Consider adding missing checks for stale data.

For example:

(uint80 roundID, int256 feedPrice, , uint256 timestamp, uint80 answeredInRound) = feed.latestRoundData();
require(feedPrice > 0, "Chainlink price <= 0");
require(timestamp != 0, "Round not complete");

kbrizzle (Perennial) confirmed

Alex the Entreprenerd (judge) commented:

Agree with the finding, while you can get started with Chainlink’s feed can be used with one line of code, in a production environment it is best to ensure that the data is fresh and within rational bounds

# Disclosures

C4 is an open organization governed by participants in the community.

C4 Contests incentivize the discovery of exploits, vulnerabilities, and bugs in smart contracts. Security researchers are rewarded at an increasing rate for finding higher-risk issues. Contest submissions are judged by a knowledgeable security researcher and solidity developer and disclosed to sponsoring developers. C4 does not conduct formal verification regarding the provided code but instead provides final verification.

C4 does not provide any guarantee or warranty regarding the security of this project. All smart contract software should be used at the sole risk and responsibility of users.


text:
# AREA OF A TRIANGLE

• PRACTICE (online exercises and printable worksheets)

## AREA OF A TRIANGLE USING A BASE/HEIGHT PAIR

An earlier section, Area Formulas: Triangle, Parallelogram, Trapezoid,
showed how to find the area of a triangle with a base/height pair.
The key results are summarized here for convenience:

 Take a vertex of a triangle. Drop a perpendicular to the opposite side or (if needed) to an extension of the opposite side. The segment just created is called a height or altitude of the triangle. The side opposite the vertex is called the corresponding base. Since every triangle has three vertices, every triangle has three base/height pairs. Area is half base times height: $$\begin{gather} \text{area of triangle}\cr \text{with base \,b\, and height \,h\,}\cr = \frac 12 bh \end{gather}$$ Drop a perpendicular to the opposite side ... ... or, to an extension of the opposite side.

 A derivation of the triangle area formula ($\,bh/2\,$) was given in Area Formulas: Triangle, Parallelogram, Trapezoid for the situation where the altitude ‘hits’ the base (left-most picture above). The formula also works when the altitude doesn't ‘hit’ the base, but instead hits an extension of the base, as shown at right. Here's how: Suppose you want the area of the blue triangle at right ($\,\triangle ABC\,$). This triangle has base $\,b\,$ and height $\,h\,$. Let $\,x\,$ denote the length of the red-dashed segment. Then: \begin{align} &\text{area of blue triangle}\cr\cr &\qquad = \quad \text{area of big right triangle } (\triangle ABD) - \text{area of small right triangle } (\triangle CBD)\cr\cr &\qquad = \quad \frac 12(b+x)h - \frac 12 xh\cr\cr &\qquad = \quad \frac 12bh + \frac 12xh - \frac 12xh\cr\cr &\qquad = \quad \frac 12bh \end{align} area of blue triangle is $\,\frac 12 bh\,$

Note that there are infinitely many triangles with a given base/height pair.
For example, all the triangles below have the same base/height pair.
Same base/height pair—same area!
A base/height pair uniquely determines the area of a triangle, but not the shape of the triangle.

 All these triangles have the same base/height pair, hence the same area.

## AREA OF A TRIANGLE USING TWO SIDES AND AN INCLUDED ANGLE

By the SAS (side-angle-side) triangle congruence theorem,
two sides and an included angle uniquely defines a triangle.

Let $\,a\,$ and $\,b\,$ denote two sides of a triangle,
with included angle $\,\theta\,$.

The sketches at right and below show the two situations that can occur when finding the altitude that has corresponding base $\,b\,$. In both cases, the altitude length is found using the yellow triangle (which is a right triangle with hypotenuse $\,a\,$).

In both cases, the altitude length is $\,h = a\sin\theta\,$.

Thus: \begin{align} &\text{area of triangle with sides a and b and included angle \theta}\cr\cr &\qquad = \frac 12(\text{base})(\text{height})\cr\cr &\qquad = \frac 12 b(a\sin\theta)\cr\cr &\qquad = \frac 12 ab\sin\theta \end{align} In words, the area is:

half the product of the two sides,
times the sine of the included angle

 (When we write $\,180^\circ - \theta\,$, it is assumed that $\,\theta\,$ is measured in degrees.)
Master the ideas from this section


text:
# what does csnumgdef do?

What does \csnumgdef do? From code examples, it appears to take two arguments, and one is led to believe that it defines a number var and assigns to it the value val, as in \csnumgdef{var}{val}. If var is being used as a simple counter, what are the advantages/disadvantages of this syntax over something like:

\newcounter{var}
\setcounter{var}{val}

Lastly, what is the etymology of csnumgdef, in particular the cs and the g? I have noticed other commands that appear to be in the cs family, such as \csname, \csuse, and \csdef, but am not familiar with what unites them.

• What package is \csnumgdef from? (cs stands for "control sequence". \csname, \csnumgdef, \documentclass, \begin, are all examples of control sequences. g possibly means "global", for a global assignment) – Phelype Oleinik Nov 19 at 23:44
• @PhelypeOleinik: I believe it's not part of any package, but rather is part of the TeX core. For example, it is used here, which uses only two packages (biblatex, filecontents), neither of which mentions csnumgdef in their manuals. Thanks for the explanation of cs and g. – user001 Nov 19 at 23:48
• no it is a macro from a package not anything built in to tex – David Carlisle Nov 20 at 0:02
• From the three you mentioned (\csname, \csuse, and \csdef), only \csname is a TeX primitive (\csname something\endcsname expands to \something). The other two (and possibly more \cs<whatever> are macros which (usually) use \csname to make a control sequence of the argument. \csuse{<arg>} is just \csname <arg>\endcsname. \csdef{<name>}{<definition>} does \def\<name>{<definition>}. – Phelype Oleinik Nov 20 at 0:11
• One more note: If you are not sure if something is a TeX primitive or a macro, you can use \show. If you do \show\csname, TeX will print > \csname=\csname to the terminal, meaning it's a primitive. If you do \show\csdef, TeX will print: > \csdef=\protected macro: #1->\expandafter \def \csname #1\endcsname. – Phelype Oleinik Nov 20 at 0:40

\csnumgdef, from the etoolbox package (for you, loaded by biblatex) is defined as:

\newrobustcmd*{\csnumgdef}[1]{%
\expandafter\numgdef\csname#1\endcsname}

in terms of \numgdef which, in turn, is defined as:

\newrobustcmd*{\numgdef}[2]{%
\ifundef#1{\let#1\z@}{}%
\xdef#1{\the\numexpr#2}}

Let's walk an example step by step. Suppose you do \csnumgdef{foo}{1+1}. First, \csnumgdef takes one argument, foo, and does (the {1+1} is still there from the previous step, untouched):

\expandafter\numgdef\csname foo\endcsname{1+1}

The \expandafter skips over \numgdef and expands \csname. \csname builds a control sequence (thus the cs in the name of \csnumgdef) out of everything until the matching \endcsname, which is the argument foo. This becomes:

\numgdef\foo{1+1}

so you can conclude that \csnumgdef{foo} and \numgdef\foo are equivalent.

Now \numgdef expands, grabbing two arguments, \foo and 1+1:

\ifundef\foo{\let\foo\z@}{}%
\xdef\foo{\the\numexpr 1+1}

The \ifundef test checks if \foo is undefined, and does \let\foo\z@ if so, otherwise it does nothing.

Now for the good part: \xdef is equivalent to \global\edef (and now you see that the g in (cs)numgdef stands for \global), which will make a global assignment, and \edef will expand everything inside the definition. It will expand \the, which will trigger \numexpr. \numexpr will then evaluate the integer expression following it (1+1) and leave the result there. After the expansion, you have (sort of):

\global\def\foo{2}

and now things go as usual, and now \foo expands to 2.

In short, \csnumgdef{<name>}{<intexpr>} will evaluate the integer expression <intexpr> and will store it in \<name>. The version without cs in the name takes a control sequence as argument (as opposed to a control sequence name): \numgdef\<name>{<intexpr>}, and the version without g does a local assignment.

These naming conventions are useful (especially when followed correctly ;-) because you don't need to look up the definition of some command to find ouy what it does. If you know that \numdef\foo{1+1} does, then it's not hard to deduce what \csnumdef or \numgdef do (given, of course, you know these conventions).

expl3's conventions define that a function that contains set in its name does an assignment (much like the def in \numdef). If the function is gset then such assignment is global. Knowing that, it's fairly easy to guess what \int_set:Nn do: it's (almost1) the same as \numdef :-)
If you know that, then \int_gset:Nn is easy.

But what about the cs version, you ask. This one is denoted in the argument specification. Instead of a single Normal argument, the argument is a csname. \int_set:cn (equivalent of \csnumdef) and \int_gset:cn (equivalent of \csnumgdef) do exactly that :-)

1 \numdef stores the number in a macro, while \int_set:Nn stores it in a count register.

• The walkthrough and especially the comparison between \csnumgdef{foo}{val} and \numgdef\foo{val} were most illuminating. – user001 Nov 20 at 0:24
• @user001 Glad it helped :-) See the note I added a note on naming conventions – Phelype Oleinik Nov 20 at 0:30
• Let's mention that \let#1\z@ within the definition of \numgdef makes it possible to do \csnumgdef{foo}{\csuse{foo}+1} without defining/initializing \foo as in case \foo is undefined, this will define/initialize \foo to be equal to 0. – Ulrich Diez Nov 20 at 11:35

\csnumgdef is defined within the etoolbox package. This package "is a toolbox of programming tools geared primarily towards LaTeX class and package authors. It provides LaTeX frontends to some of the new primitives provided by e-TeX as well as some generic tools which are not related to e-TeX but match the profile of this package." (from the etoolbox documentation.

The cs, g and def nomenclature is used through the documentation, roughly defined as

• cs = <csname>
• g = global
• def = \def

Another interpretation is that gdef is similar to \gdef or \global\def (or, possible \xdef - an expanded version of \gdef).

In some contexts, its convenient to just consider the control sequence name (say, foo), rather than the control sequence itself (say, \foo). For example, perhaps you want to define an auxiliary control sequence that should accompany some definition and, for the sake of coding, you use the same basic definition. Here's an example:

\makeatletter
\newcommand{\foo}{foo}
\newcommand{\foo@bar}{bar}
\newcommand{\print}[1]{\csname #1\endcsname: \csname #1@bar\endcsname}
\makeatother

\print{foo}

which prints

foo: bar

It's easy to pass foo to \print and then have access to \foo@bar. However, if you have passed \foo to \print, you'd have to remove the backslash \ in order to string add the suffix @bar and have access to \foo@bar. So, depending on the user (package author), it might be more convenient to work with control sequence names rather than the control sequences themselves.

etoolbox explicitly defines some of these functions in order to force a specific use. \csnumgdef{<var>}{<val>} is not the same a

\newcounter{<var>}
\setcounter{<var>}{<val>}

since <var> in \csnumgdef is not a counter. Instead, it's a regular control sequence. However, since <val> is evaluated using \numexpr, only arguments that are valid in that context is accepted. So, it may seem similar to using a counter, but internally it's different. Since it uses \numexpr, you can do something like \csnumgdef{foo}{5+2} and have 7 stored in \foo. You don't need \thefoo to print the number if you'd gone the route of defining a counter.

• Thanks for explaining the nomenclature as well as for clarifying that the first argument of csnumgdef is a control sequence name rather than a variable. – user001 Nov 20 at 0:18


text:
## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 3 - Derivatives - 3.1 Introducing the Derivative - 3.1 Execises: 9

#### Answer

$a.$ The slope is $m_{tan} = 6$. $b.$ The equation is $y=6x-14.$ $c.$ The graph is on the figure below.

#### Work Step by Step

By definition (1) we have $a.$ The slope is given by $$m_{tan}=\lim_{x\to 3}\frac{f(x)-f(3)}{x-3} =\lim_{x\to 3}\frac{x^2-5-(3^2-5)}{x-3}=\lim_{x\to 3}\frac{x^2-9}{x-3} = \lim_{x\to 3}\frac{(x-3)(x+3)}{x-3} = \lim_{x\to 3}(x+3) = 6$$ where in the last step we used the substitution to evaluate the limit. $b.$ Applying formula for tangent line $y-f(a) = m_{tan}(x-a)$ with $a=3$ and $m_{tan} = 6$ we have $$y-(3^2-5)=6(x-3)\Rightarrow y-4=6x-18$$ which gives $$y=6x-14.$$ $c.$ The graph is on the figure below. The function is solid and the tangent is dashed

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
How do you express (x^2-2x-1) / ((x-1)^2 (x^2+1)) in partial fractions?

Nov 15, 2017

The answer is $= - \frac{1}{x - 1} ^ 2 + \frac{1}{x - 1} - \frac{\left(x - 1\right)}{{x}^{2} + 1}$

Explanation:

Perform the decomposition into partial fractions

$\frac{{x}^{2} - 2 x - 1}{{\left(x - 1\right)}^{2} \left({x}^{2} + 1\right)} = \frac{A}{x - 1} ^ 2 + \frac{B}{x - 1} + \frac{C x + D}{{x}^{2} + 1}$

$= \frac{A \left({x}^{2} + 1\right) + B \left(x - 1\right) \left({x}^{2} + 1\right) + \left(C x + D\right) {\left(x - 1\right)}^{2}}{{\left(x - 1\right)}^{2} \left({x}^{2} + 1\right)}$

The denominators are the same, compare the numerators

$\left({x}^{2} - 2 x - 1\right) = A \left({x}^{2} + 1\right) + B \left(x - 1\right) \left({x}^{2} + 1\right) + \left(C x + D\right) {\left(x - 1\right)}^{2}$

Let $x = 1$

$- 2 = 2 A$, $A = - 1$

Coefficients of ${x}^{2}$

$1 = A - B - 2 C + D$

Coefficients of $x$

$- 2 = B + C - 2 D$

and

$- 1 = A - B + D$, $\implies$, $- B + D = 0$, $B = D$

$1 = - 1 - B - 2 C + B$, $\implies$, $2 C = - 2$, $\implies$, $C = - 1$

$- 2 = B - 1 - 2 B$, $\implies$, $B = 1$

$\implies$, $D = 1$

Finally,

$\frac{{x}^{2} - 2 x - 1}{{\left(x - 1\right)}^{2} \left({x}^{2} + 1\right)} = - \frac{1}{x - 1} ^ 2 + \frac{1}{x - 1} - \frac{\left(x - 1\right)}{{x}^{2} + 1}$