text:
# 2021 AIME I Problems/Problem 13

## Problem

Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$, respectively, intersect at distinct points $A$ and $B$. A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$. Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$. Find the distance between the centers of $\omega_1$ and $\omega_2$.

Let $D\equiv\omega\cap\omega_{1}$ and $E\equiv\omega\cap\omega_{2}$. The solution relies on the following key claim:

Claim. $DPEQ$ is a harmonic quadrilateral.

Proof. Using the radical axis theorem, the tangents to circle $\omega$ at $D$ and $E$ are concurrent with line $\overline{ABPQ}$ at the radical center, implying that our claim is true by harmonic quadrilateral properties. $\blacksquare$

Thus, we deduce the tangents to $\omega$ at $P$ and $Q$ are concurrent at $F$ with line $\overline{DE}$. Denote by $O_{1}, O_{2}, O$ the centers of $\omega_{1}, \omega_{2}, \omega$ respectively.

Now suppose $G\equiv\overline{FDE}\cap\overline{O_{1}O_{2}}$. Note that $\overline{OF}\parallel\overline{O_{1}O_{2}}$ giving us the pairs of similar triangles $$\triangle O_{1}GD\sim\triangle OFD~\text{and}~\triangle O_{2}GE\sim\triangle OFE.$$ We thereby obtain $$\dfrac{O_{1}G}{O_{1}D}=\dfrac{OF}{OD}=\dfrac{r\sec\tfrac{120^{\circ}}{2}}{r}=\sec 60^{\circ}=2$$ since $\widehat{PQ}=120^{\circ}$, where $r$ denotes the radius of $\omega$, and $$\dfrac{O_{2}G}{O_{2}E}=\dfrac{OF}{OE}=\sec 60^{\circ}=2$$ as well. It follows that $$O_{1}O_{2}=O_{1}G-O_{2}G=2(O_{1}D-O_{2}E)=2(961-625)=\textbf{672}.$$

Let $O_i$ and $r_i$ be the center and radius of $\omega_i$, and let $O$ and $r$ be the center and radius of $\omega$.

Since $\overline{AB}$ extends to an arc with arc $120^\circ$, the distance from $O$ to $\overline{AB}$ is $r/2$. Let $X=\overline{AB}\cap \overline{O_1O_2}$. Consider $\triangle OO_1O_2$. The line $\overline{AB}$ is perpendicular to $\overline{O_1O_2}$ and passes through $X$. Let $H$ be the foot from $O$ to $\overline{O_1O_2}$; so $HX=r/2$. We have by tangency $OO_1=r+r_1$ and $OO_2=r+r_2$. Let $O_1O_2=d$. $[asy] unitsize(3cm); pointpen=black; pointfontpen=fontsize(9); pair A=dir(110), B=dir(230), C=dir(310); DPA(A--B--C--A); pair H = foot(A, B, C); draw(A--H); pair X = 0.3*B + 0.7*C; pair Y = A+X-H; draw(X--1.3*Y-0.3*X); draw(A--Y, dotted); pair R1 = 1.3*X-0.3*Y; pair R2 = 0.7*X+0.3*Y; draw(R1--X); D("O",A,dir(A)); D("O_1",B,dir(B)); D("O_2",C,dir(C)); D("H",H,dir(270)); D("X",X,dir(225)); D("A",R1,dir(180)); D("B",R2,dir(180)); draw(rightanglemark(Y,X,C,3)); [/asy]$ Since $X$ is on the radical axis of $\omega_1$ and $\omega_2$, it has equal power with respect to both circles, so $$O_1X^2 - r_1^2 = O_2X^2-r_2^2 \implies O_1X-O_2X = \frac{r_1^2-r_2^2}{d}$$since $O_1X+O_2X=d$. Now we can solve for $O_1X$ and $O_2X$, and in particular, \begin{align*} O_1H &= O_1X - HX = \frac{d+\frac{r_1^2-r_2^2}{d}}{2} - \frac{r}{2} \\ O_2H &= O_2X + HX = \frac{d-\frac{r_1^2-r_2^2}{d}}{2} + \frac{r}{2}. \end{align*} We want to solve for $d$. By the Pythagorean Theorem (twice): \begin{align*} &\qquad -OH^2 = O_2H^2 - (r+r_2)^2 = O_1H^2 - (r+r_1)^2 \\ &\implies \left(d+r-\tfrac{r_1^2-r_2^2}{d}\right)^2 - 4(r+r_2)^2 = \left(d-r+\tfrac{r_1^2-r_2^2}{d}\right)^2 - 4(r+r_1)^2 \\ &\implies 2dr - 2(r_1^2-r_2)^2-8rr_2-4r_2^2 = -2dr+2(r_1^2-r_2^2)-8rr_1-4r_1^2 \\ &\implies 4dr = 8rr_2-8rr_1 \\ &\implies d=2r_2-2r_1 \end{align*} Therefore, $d=2(r_2-r_1) = 2(961-625)=\boxed{672}$.

## Solution 2 (Linearity)

Let $O_{1}$, $O_{2}$, and $O$ be the centers of $\omega_{1}$, $\omega_{2}$, and $\omega$ with $r_{1}$, $r_{2}$, and $r$ their radii, respectively. Then, the distance from $O$ to the radical axis $\ell\equiv\overline{AB}$ of $\omega_{1}, \omega_{2}$ is equal to $\frac{1}{2}r$. Let $x=O_{1}O_{2}$ and $O^{\prime}$ the orthogonal projection of $O$ onto line $\ell$. Define the function $f:\mathbb{R}^{2}\rightarrow\mathbb{R}$ by $$f(X)=\text{Pow}_{\omega_{1}}(X)-\text{Pow}_{\omega_{2}}(X).$$ Then \begin{align*} f(O_{1})=-r_{1}^{2}-(x-r_{2})(x+r_{2})&=-x^{2}+r_{2}^{2}-r_{1}^{2}, \\ f(O_{2})=(x-r_{1})(x+r_{1})-(-r_{2}^{2})&=x^{2}+r_{2}^{2}-r_{1}^{2}, \\ f(O)=r(r+2r_{1})-r(r+2r_{2})&=2r(r_{1}-r_{2}), \\ f(O^{\prime})&=0. \end{align*} By linearity, $$\frac{f(O_{2})-f(O_{1})}{f(O)-f(O^{\prime})}=\frac{O_{1}O_{2}}{OO^{\prime}}=\frac{x}{\tfrac{1}{2}r}=\frac{2x}{r}.$$ Notice that $f(O_{2})-f(O_{1})=x^{2}-(-x^{2})=2x^{2}$ and $f(O)-f(O^{\prime})=2r(r_{1}-r_{2})$, thus \begin{align*}\frac{2x^{2}}{2r(r_{1}-r_{2})}&=\frac{2x}{r}\end{align*} Dividing both sides by $\frac{2x}{r}$ (which is obviously nonzero as $x$ is nonzero) gives us \begin{align*}\frac{x}{2(r_{1}-r_{2})}&=1\end{align*} so $x=2(r_{1}-r_{2})$. Since $r_{1}=961$ and $r_{2}=625$, the answer is $x=2\cdot(961-625)=\boxed{672}$.

## Solution 3

Denote by $O_1$, $O_2$, and $O$ the centers of $\omega_1$, $\omega_2$, and $\omega$, respectively. Let $R_1 = 961$ and $R_2 = 625$ denote the radii of $\omega_1$ and $\omega_2$ respectively, $r$ be the radius of $\omega$, and $\ell$ the distance from $O$ to the line $AB$. We claim that$$\dfrac{\ell}{r} = \dfrac{R_2-R_1}{d},$$where $d = O_1O_2$. This solves the problem, for then the $\widehat{PQ} = 120^\circ$ condition implies $\tfrac{\ell}r = \cos 60^\circ = \tfrac{1}{2}$, and then we can solve to get $d = \boxed{672}$. $[asy] import olympiad; size(230pt); defaultpen(linewidth(0.8)+fontsize(10pt)); real r1 = 17, r2 = 27, d = 35, r = 18; pair O1 = origin, O2 = (d,0); path w1 = circle(origin,r1), w2 = circle((d,0),r2), w1p = circle(origin,r1+r), w2p = circle((d,0), r2 + r); pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); pair O = Y[1]; path w = circle(Y[1],r); pair Xp = 5 * X[1] - 4 * X[0]; pair[] P = intersectionpoints(Xp--X[0],w); label("O_1",origin,N); label("O_2",(d,0),N); label("O",Y[1],SW); draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); draw(w^^w1^^w2^^P[0]--X[0]); dot(Y[1]^^origin^^(d,0)); label("X",T,N,gray(0.6)); label("Y",foot(X[0],O1,O2),NE,gray(0.6)); label("\ell",(O+Tp)/2,S,gray(0.6)); [/asy]$

Denote by $O_1$ and $O_2$ the centers of $\omega_1$ and $\omega_2$ respectively. Set $X$ as the projection of $O$ onto $O_1O_2$, and denote by $Y$ the intersection of $AB$ with $O_1O_2$. Note that $\ell = XY$. Now recall that$$d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.$$Furthermore, note that\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\end{align*}Substituting the first equality into the second one and subtracting yields$$2r(R_2 - R_1) = d(O_2X - O_1X) - d(O_2Y - O_1Y) = 2dXY,$$which rearranges to the desired.

## Solution 4 (Quick)

Suppose we label the points as shown below. $[asy] defaultpen(fontsize(12)+0.6); size(300); pen p=fontsize(10)+royalblue+0.4; var r=1200; pair O1=origin, O2=(672,0), O=OP(CR(O1,961+r),CR(O2,625+r)); path c1=CR(O1,961), c2=CR(O2,625), c=CR(O,r); pair A=IP(CR(O1,961),CR(O2,625)), B=OP(CR(O1,961),CR(O2,625)), P=IP(L(A,B,0,0.2),c), Q=IP(L(A,B,0,200),c), F=IP(CR(O,625+r),O--O1), M=(F+O2)/2, D=IP(CR(O,r),O--O1), E=IP(CR(O,r),O--O2), X=extension(E,D,O,O+O1-O2), Y=extension(D,E,O1,O2); draw(c1^^c2); draw(c,blue+0.6); draw(O1--O2--O--cycle,black+0.6); draw(O--X^^Y--O2,black+0.6); draw(X--Y,heavygreen+0.6); draw((X+O)/2--O,MidArrow); draw(O2--Y-(300,0),MidArrow); dot("A",A,dir(A-O2/2)); dot("B",B,dir(B-O2/2)); dot("O_2",O2,right+up); dot("O_1",O1,left+up); dot("O",O,dir(O-O2)); dot("D",D,dir(170)); dot("E",E,dir(E-O1)); dot("X",X,dir(X-D)); dot("Y",Y,dir(Y-D)); label("R",O--E,right+up,p); label("R",O--D,left+down,p); label("2R",(X+O)/2-(150,0),down,p); label("961",O1--D,2*(left+down),p); label("625",O2--E,2*(right+up),p); MA("",E,D,O1,100,fuchsia+linewidth(1)); MA("",X,D,O,100,fuchsia+linewidth(1)); MA("",Y,E,O2,100,orange+linewidth(1)); MA("",D,E,O,100,orange+linewidth(1)); [/asy]$ By radical axis, the tangents to $\omega$ at $D$ and $E$ intersect on $AB$. Thus $PDQE$ is harmonic, so the tangents to $\omega$ at $P$ and $Q$ intersect at $X \in DE$. Moreover, $OX \parallel O_1O_2$ because both $OX$ and $O_1O_2$ are perpendicular to $AB$, and $OX = 2OP$ because $\angle POQ = 120^{\circ}$. Thus$$O_1O_2 = O_1Y - O_2Y = 2 \cdot 961 - 2\cdot 625 = \boxed{672}$$by similar triangles.

~mathman3880

## Solution 5 (Official MAA)

Like in other solutions, let $O$ be the center of $\omega$ with $r$ its radius; also, let $O_{1}$ and $O_{2}$ be the centers of $\omega_{1}$ and $\omega_{2}$ with $R_{1}$ and $R_{2}$ their radii, respectively. Let line $OP$ intersect line $O_{1}O_{2}$ at $T$, and let $u=TO_{2}$, $v=TO_{1}$, $x=PT$, where the length $O_{1}O_{2}$ splits as $u+v$. Because the lines $PQ$ and $O_{1}O_{2}$ are perpendicular, lines $OT$ and $O_{1}O_{2}$ meet at a $60^{\circ}$ angle.

Applying the Law of Cosines to $\triangle O_{2}PT$, $\triangle O_{1}PT$, $\triangle O_{2}OT$, and $\triangle O_{1}OT$ gives \begin{align*}\triangle O_{2}PT&:O_{2}P^{2}=u^{2}+x^{2}-ux \\ \triangle O_{1}PT&:O_{1}P^{2}=v^{2}+x^{2}+vx \\ \triangle O_{2}OT&:(r+R_{2})^{2}=u^{2}+(r+x)^{2}-u(r+x) \\ \triangle O_{1}OT&:(r+R_{1})^{2}=v^{2}+(r+x)^{2}+v(r+x)\end{align*}

Adding the first and fourth equations, then subtracting the second and third equations gives us $$\left(O_{2}P^{2}-O_{1}P^{2}\right)+\left(R_{1}^{2}-R_{2}^{2}\right)+2r(R_{1}-R_{2})=r(u+v)$$

Since $P$ lies on the radical axis of $\omega_{1}$ and $\omega_{2}$, the power of point $P$ with respect to either circle is $$O_{2}P^{2}-R_{2}^{2}=O_{1}P^{2}-R_{1}^{2}.$$

Hence $2r(R_{1}-R_{2})=r(u+v)$ which simplifies to $$u+v=2(R_{1}-R_{2}).$$

The requested distance $$O_{1}O_{2}=O_{1}T+O_{2}T=u+v$$ is therefore equal to $2\cdot(961-625)=\boxed{672}$.

## Solution 6 (Geometry)

Let circle $\omega$ tangent circles $\omega_1$ and $\omega_2,$ respectively at distinct points $C$ and $D$. Let $O, O_1, O_2 (r, r_1, r_2)$ be the centers (the radii) of $\omega, \omega_1$ and $\omega_2,$ respectively. WLOG $r_1 < r_2.$ Let $F$ be the point of $OO_2$ such, that $OO_1 =OF.$ Let $M$ be the midpoint $FO_1, OE \perp AB, CT$ be the radical axes of $\omega_1$ and $\omega, T \in AB.$

Then $T$ is radical center of $\omega, \omega_1$ and $\omega_2, TD = CT.$

In $\triangle OFO_1 (OF = OO_1) OT$ is bisector of $\angle O, OM$ is median

$\hspace{10mm} \implies O,T,$ and $M$ are collinear.

$$\angle OCT = \angle ODT = \angle OET = 90^\circ \implies$$

$OCTDE$ is cyclic (in $\Omega), OT$ is diameter $\Omega.$ $O_1O_2 \perp AB, OM \perp FO_1 \implies \angle FO_1O_2 = \angle OTE$ $\angle OTE = \angle ODE$ as they intercept the same arc $\overset{\Large\frown}{OE}$ in $\Omega.$ $$OE \perp AB, O_1O_2 \perp AB \implies O_1 O_2 || OE \implies$$ $$\angle OO_2O_1 = \angle O_2 OE \implies \triangle ODE \sim \triangle O_2 O_1 F \implies$$ $$\frac {OE}{OD} = \frac {O_2F}{O_1O_2} \implies \cos \frac {120^\circ}{2} = \frac{r_2 + r - r_1 -r} {O_1O_2}\implies {O_1O_2}= 2|r_2 – r_1|.$$

Since $r_{1}=625$ and $r_{2}=961$, the answer is $2\cdot|961-625|=\boxed{672}$.

## Video Solution

https://youtu.be/gN7Ocu3D62M ~Math Problem Solving Skills

## Video Solution

Who wanted to see animated video solutions can see this. I found this really helpful.

P.S: This video is not made by me. And solution is same like below solutions.

≈@rounak138


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+0

# Can you solve?

+1
204
9
+65

Solve $$g(1+{g}^{-1}(a))=f(1+{f}^{-1}(1+f(-1+{f}^{-1}(b))))$$ for $$f(1+{f}^{-1}(b))$$

May 2, 2019

#1
+82
+2

First, I need to find f –1(x), g –1(x), and ( f o g)–1(x):

Inverting  f (x):

Inverting g(x):

Inverting the composed function:

Now I'll compose the inverses of  f(x) and g(x) to find the formula for (g–1 o f –1)(x):

Note that the inverse of the composition (( f o g)–1(x)) gives the same result as does the composition of the inverses ((g–1 o f –1)(x)). So I would conclude that

f (x) = 2x – 1
y = 2x – 1
y + 1 = 2x
(y + 1)/2 = x
(x + 1)/2 = y
(x + 1)/2 =  f –1(x)

g(x) = (1/2)x + 4
y = (1/2)x + 4
y – 4 = (1/2)x
2(y – 4) = x
2y – 8 = x
2x – 8 = y
2x – 8 = g –1(x)

( f o g)(x) = f (g(x)) = f ((1/2)x + 4)
= 2((1/2)x + 4) – 1
= x + 8 – 1
= x + 7

( f o g)(x) = x + 7
y = x + 7
y – 7 = x
x – 7 = y
x – 7 = ( f o g)–1(x)

(g–1 o f –1)(x) = g–1( f –1(x))
= g–1( (x + 1)/2 )
= 2( (x + 1)/2 ) – 8
= (x + 1) – 8
= x – 7 = (g–1 o f –1)(x)

( f o g)–1(x) = (g–1 o f –1)(x)

While it is beyond the scope of this lesson to prove the above equality, I can tell you that this equality is indeed always true, assuming that the inverses and compositions exist — that is, assuming there aren't any problems with the domains and ranges and such.

May 2, 2019
#2
+82
+1

did you get the right answer??

May 2, 2019
#3
+65
+1

I actually don't know the answer (or even how to go about solving it), so I asked the question to see if anyone else could figure it out. However, if it is solved, then the results could be groundbreaking.

creepercraft97T3  May 2, 2019
#4
+1

groundbreaking how?

Guest May 2, 2019
#5
+65
+2

The hyperoperations are basically a sequence that extends the 3 basic functions everyone is familiar with.

Multiplication is hyperoperation 2

Exponentiation is hyperoperation 3

Tetration is hyperoperation 4

and it continues on forever (although hyperoperations greater than or equal to 4 have problems with their domains).

Let f(x) be the 'n'th hyperoperation. You can get from hyperoperation 'n' to hyperoperation 'n+1' by setting the 'x' of f(x) to a certain value and then iterating f(x) 'x' times. That takes a lot of work, but for going from hyperoperation 'n' to hyperoperation 'n-1', you just need to use the formula $$f(1+{f}^{-1}(x))$$, where $$f^{-1}(x)$$ is the functional inverse of $$f(x)$$. Using this, you can find out how to go from hyperoperation 'n' to hyperoperation 'n-2' or other values. Basically, you can go from hyperoperation 'n' to hyperoperation 'n-v', where 'v' is an integer.

Hyperoperation 'n-2' from hyperoperation 'n': $$f(1+f^{-1}(1+f(-1+f^{-1}(b))))$$

I then remembered something I did about a year ago. I set up this equation $$f(a)=f^2(b)$$ and solved for $$f(b)$$, and I got $$f^{0.5}(a)=f(b)$$. Then I did it for the 3rd iteration, then the 4th, and so on. In other words, I learned how to get $$f^{\frac1n}(x)$$ from $$f^n(x)$$. So I figured that I could apply this to hyperoperations.

Basically, if you set up this equation $$g(1+g^{-1}(a))=f(1+f^{-1}(1+f(-1+f^{-1}(b))))$$ ('n-1' and 'n-2' equations equal to each other) and manage to solve the right side for $$f(1+{f}^{-1}(x))$$ (the 'n-1' equation), the left side will, consequently, turn from an 'n-1' equation into an 'n-0.5' equation. This would give us the ability to define fractional hyperoperations (and not just for 'n-0.5'), and if a useful, usable pattern is found, can lead to the generalization of all hyperoperations.

creepercraft97T3  May 2, 2019
#6
+1

Hyperoperations are binary functions

Guest May 2, 2019
#8
+65
+1

They could also be trinary functions, if you want the base to also change around.

creepercraft97T3  May 6, 2019
#9
+1

Yes but then there is only one such function.

Besides you were referring to some hyperoperations as f(x) despite the fact that none of them are unary.

Guest May 6, 2019
edited by Guest  May 6, 2019
#7
+7725
+2

Presumably, both f(x) and g(x) has inverse functions.

$$g(1+{g}^{-1}(a))=f(1+{f}^{-1}(1+f(-1+{f}^{-1}(b))))\\ f^{-1}(g(1+{g}^{-1}(a))) = 1+{f}^{-1}(1+f(-1+{f}^{-1}(b)))\\ f(f^{-1}(g(1+{g}^{-1}(a)))-1) = 1+f(-1+{f}^{-1}(b))\\ f(-1+f^{-1}(b)) = f(f^{-1}(g(1+{g}^{-1}(a)))-1)-1$$

But I can't solve for f(1 + f^{-1}(b))

May 2, 2019

format: plain


text:
## Reversible vs Irreversible Expansion

$w=-P\Delta V$
and
$w=-\int_{V_{1}}^{V_{2}}PdV=-nRTln\frac{V_{2}}{V_{1}}$

jonathanshi_1A
Posts: 22
Joined: Fri Jul 22, 2016 3:00 am

### Reversible vs Irreversible Expansion

I'm a little confused with what a reversible expansion is. I understand how to do the equations, but conceptually I'm not sure what it means to expand reversibly.

Monica Habashy 3A
Posts: 68
Joined: Wed Sep 21, 2016 2:59 pm
Been upvoted: 3 times

### Re: Reversible vs Irreversible Expansion

Reversible means that the reaction can easily go back to its initial form completely. It's like 100% of the reverse reaction went back to create its initial reactants which is why there is no change in internal energy.


text:
# Free University of Brussels (VUB) Vrije Universiteit Brussel (VUB)

## Belgium

1 June 2019 - 31 May 2020

Region: Global
Subject/journal group: All

The table to the right includes counts of all research outputs for Free University of Brussels (VUB) published between 1 June 2019 - 31 May 2020 which are tracked by the Nature Index.

Hover over the donut graph to view the FC output for each subject. Below, the same research outputs are grouped by subject. Click on the subject to drill-down into a list of articles organized by journal, and then by title.

Note: Articles may be assigned to more than one subject area.

Count Share
151 22.76

## Outputs by subject (Share)

Subject Count Share
Chemistry 24 8.77
Physical Sciences 92 9.40
1 1
16 0.20
46 5.07
A light dilaton at the LHC
2020-05-02
0.50
Search for an excited lepton that decays via a contact interaction to a lepton and two jets in proton-proton collisions at √s = 13 TeV
2020-05-02
0.01
Search for direct top squark pair production in events with one lepton, jets, and missing transverse momentum at 13 TeV with the CMS experiment
2020-05-02
0.01
Search for high mass dijet resonances with a new background prediction method in proton-proton collisions at √s = 13 TeV
2020-05-02
0.01
Measurement of properties of B0s→ μ+μ− decays and search for B0→ μ+μ− with the CMS experiment
2020-04-28
0.01
Search for heavy Higgs bosons decaying to a top quark pair in proton-proton collisions at √s = 13 TeV
2020-04-27
0.01
Poisson-Lie U-duality in exceptional field theory
2020-04-09
0.17
A search for the standard model Higgs boson decaying to charm quarks
2020-03-24
0.01
Search for lepton flavour violating decays of a neutral heavy Higgs boson to μτ and eτ in proton-proton collisions at √s = 13 TeV
2020-03-18
0.01
Search for a heavy pseudoscalar Higgs boson decaying into a 125 GeV Higgs boson and a Z boson in final states with two tau and two light leptons at √s = 13 TeV
2020-03-11
0.01
Measurement of top quark pair production in association with a Z boson in proton-proton collisions at √s = 13 TeV
2020-03-10
0.01
Search for new neutral Higgs bosons through the H→ZA→ℓ+ℓ−bb⎯ process in pp collisions at √𝑠 = 13 TeV
2020-03-10
0.01
Search for physics beyond the standard model in multilepton final states in proton-proton collisions at √s = 13 TeV
2020-03-10
0.01
Search for a heavy Higgs boson decaying to a pair of W bosons in proton-proton collisions at √𝑠 = 13 TeV
2020-03-06
0.01
Search for dark matter particles produced in association with a Higgs boson in proton-proton collisions at = 13 TeV
2020-03-04
0.01
Bose-Einstein correlations of charged hadrons in proton-proton collisions at √𝑠 = 13 TeV
2020-03-03
0.01
Measurement of the top quark pair production cross section in dilepton final states containing one τ lepton in pp collisions at = 13 TeV
2020-02-28
0.01
Dilaton portal in strongly interacting twin Higgs models
2020-02-20
0.33
Generation of vortices and stabilization of vortex lattices in holographic superfluids
2020-02-18
0.17
Search for top squark pair production in a final state with two tau leptons in proton-proton collisions at = 13 TeV
2020-02-02
0.01
Chaos in the butterfly cone
2020-01-29
0.17
Higgs Boson studies at future particle colliders
2020-01-21
0.05
Quantum complexity of time evolution with chaotic Hamiltonians
2020-01-21
0.08
Search for a charged Higgs boson decaying into top and bottom quarks in events with electrons or muons in proton-proton collisions at √s = 13 TeV
2020-01-16
0.01
Search for electroweak production of a vector-like T quark using fully hadronic final states
2020-01-02
0.01
Superluminal chaos after a quantum quench
2019-12-17
0.46
Study of the B+→J/ψΛ⎯p decay in proton-proton collisions at √s = 8 TeV
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Measurement of the differential Drell-Yan cross section in proton-proton collisions at √s = 13 TeV
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Measurements of differential Z boson production cross sections in proton-proton collisions at √s = 13 TeV
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Strong cosmic censorship for the Dirac field in the higher dimensional Reissner-Nordstrom-de Sitter black hole
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# Top articles by Altmetric score in current window

### The first day of the Cenozoic

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1 June 2019 - 31 May 2020

## International vs. domestic collaboration by Share

• 12.88% Domestic
• 87.12% International

Note: Hover over the graph to view the percentage of collaboration.

Note: Collaboration is determined by the fractional count (Share), which is listed in parentheses.


text:
# High count clusters

### Help Support The Rocketry Forum:

#### KilroySmith

A large shipping tube (~6 inch diameter and 4 feet long) fell into my lap at work, and I'd like to mount about 19 motors (18 C's and 1 E) in it. According to OpenRocket, it won't go very high (about 600 feet), but it should be entertaining for a $50 rocket. Any great things to NOT do? I'm figuring a 3mm or 5mm plywood thrust plate for the fore ends of the motors with holes drilled for ejection charges, and a 3mm aft plate that'll be 90% empty space with all the 18mm holes drilled in it. I'll use electronic deployment. And, yes, I have a launcher capable of lighting 19 igniters. No worries on that front. Thoughts? /frank #### rharshberger ##### Well-Known Member A large shipping tube (~6 inch diameter and 4 feet long) fell into my lap at work, and I'd like to mount about 19 motors (18 C's and 1 E) in it. According to OpenRocket, it won't go very high (about 600 feet), but it should be entertaining for a$50 rocket.

Any great things to NOT do? I'm figuring a 3mm or 5mm plywood thrust plate for the fore ends of the motors with holes drilled for ejection charges, and a 3mm aft plate that'll be 90% empty space with all the 18mm holes drilled in it.

I'll use electronic deployment.

And, yes, I have a launcher capable of lighting 19 igniters. No worries on that front.

Thoughts?

/frank

Good Luck, and build light.

18mm e?

#### Rex R

##### LV2
you will need an FAA waiver to fly this beast (over 4.4 oz. propellant weight).
Rex

#### KennB

##### I-95 Envoy
TRF Supporter
Check out the Hellboy at this site. This is a very successful 19 motor cluster (the picture on the home page shows the two stage version with a booster; click on the Hellboy tab at the top).

You'll have about 190 total Ns in your cluster, equivalent to a small H-impulse motor, so HPR rules will apply.

#### Micromeister

##### Micro Craftman/ClusterNut
TRF Supporter
As others have posted even using 18mm motors and a core E your cluster will be well over the 4.4oz of propellent limit requiring an HPR waiver from the FAA.

Large cluster flying isn't unheard of but the waiver requirment does throw another element into the mix.

Do keep in mind average igniters require about 2amps each so with 19 motors your going to need a power supply capable of suppling about 40amps.
Good luck and be sure to fly safe.

#### KilroySmith

##### Well-Known Member
Thanks for the responses. I'd forgotten about the 4.4 oz limit, I'll have to go review those restrictions and modify my plans against that. I have an 8-site launcher that's capable of supplying about 10 amps per site, I've driven 50 amps through it in testing so I think I'm OK. I'd probably use Q2G2 igniters though, to keep the current down and ignition variability to a minimum.

#### 75Grandville

##### Well-Known Member
Search on "Boris Katan", user ID delta22

He regularly does all sorts of high count clusters. He has tutorials on making your own ignitors, dos and don'ts, etc. Would suggest that as some preliminary reading.

I've read some of his stuff with an eye to high-count clusters, but haven't had a chance to try yet.

#### KennB

##### I-95 Envoy
TRF Supporter
Search on "Boris Katan", user ID delta22

He regularly does all sorts of high count clusters. He has tutorials on making your own ignitors, dos and don'ts, etc. Would suggest that as some preliminary reading.

I've read some of his stuff with an eye to high-count clusters, but haven't had a chance to try yet.
The link in Post #5 above will bring you to Boris' site.

#### marcusSRG

##### Well-Known Member
I'd do a flash pan instead of worrying about all of those igniters. Boris ignites his Hellboy with a variant of the flashpan...BP, quickmatch and lots of tape.


text:
What is FEA | Finite Element Analysis?

The Finite Element Analysis (FEA) is the simulation of any given physical phenomenon using the numerical technique called Finite Element Method (FEM). Engineers use it to reduce the number of physical prototypes and experiments and optimize components in their design phase to develop better products, faster.

It is necessary to use mathematics to comprehensively understand and quantify any physical phenomena such as structural or fluid behavior, thermal transport, wave propagation, the growth of biological cells, etc. Most of these processes are described using Partial Differential Equations (PDEs). However, for a computer to solve these PDEs, numerical techniques have been developed over the last few decades and one of the prominent ones, today, is the Finite Element Analysis.

Differential equations can not only describe processes of nature but also physical phenomena encountered in engineering mechanics. These partial differential equations (PDEs) are complicated equations that need to be solved in order to compute relevant quantities of a structure (like stresses (ϵ

$ϵ$

), strains (ϵ

$ϵ$

), etc.) in order to estimate a certain behavior of the investigated component under a given load. It is important to know that FEA only gives an approximate solution of the problem and is a numerical approach to get the real result of these partial differential equations. Simplified, FEA is a numerical method used for the prediction of how a part or assembly behaves under given conditions. It is used as the basis for modern simulation software and helps engineers to find weak spots, areas of tension, etc. in their designs. The results of a simulation based on the FEA method are usually depicted via a color scale that shows for example the pressure distribution over the object.

History of FEA

Depending on one’s perspective, FEA can be said to have its origin in the work of Euler, as early as the 16th century. However, the earliest mathematical papers on Finite Element Analysis can be found in the works of Schellbach [1851] and Courant [1943].

FEA was independently developed by engineers in different companies and industries to address structural mechanics problems related to aerospace and civil engineering. The development for real life applications started around the mid-1950s as papers by Turner, Clough, Martin and Topp [1956], Argyris [1957] and Babuska and Aziz [1972] show. The books by Zienkiewicz [1971] and Strang and Fix [1973] also laid the foundations for future developments in FEA.

DIVIDE AND CONQUER!

To be able to make simulations, a mesh, consisting of up to millions of small elements that together form the shape of the structure needs to be created. Calculations are made for every single element. Combining the individual results gives us the final result of the structure. The approximations we just mentioned are usually polynomial and in fact, interpolations over the element(s). This means we know values at certain points within the element but not at every point. These ‘certain points’ are called nodal points and are often located at the boundary of the element. The accuracy with which the variable changes is expressed by the approximation, which can be linear, quadratic, cubic etc. In order to get a better understanding of approximation techniques, we will look at a one-dimensional bar. Consider the true temperature distribution T(x) along the bar in the picture below:

Let’s assume we know the temperature of this bar at 5 specific positions (Numbers 1-5 in the illustration). Now the question is: How can we predict the temperature in between these points? A linear approximation is quite good but there are better possibilities to represent the real temperature distribution. If we choose a square approximation, the temperature distribution along the bar is much more smooth. Nevertheless, we see that irrespective of the polynomial degree, the distribution over the rod is known once we know the values at the nodal points. If we would have an infinite bar, we would have an infinite amount of unknowns (DEGREES OF FREEDOM (DOF)). But in this case, we have a problem with a finite number of unknowns:

A system with a finite number of unknowns is called a discrete system. A system with an infinite number of unknowns is called a continuous system.

For the purpose of approximations we can find the following relation for a field quantity u(x)

$u\left(x\right)$

:

u(x)=uh(x)+e(x)(1)

$\begin{array}{}\text{(1)}& u\left(x\right)={u}^{h}\left(x\right)+e\left(x\right)\end{array}$

Where uh(x)

${u}^{h}\left(x\right)$

is the approximation, which differs from the real solution u(x)

$u\left(x\right)$

with the error term e(x)

$e\left(x\right)$

which is not known before. De facto, this error term has to vanish so that the approximate solution uh(x)

${u}^{h}\left(x\right)$

becomes valid. The question that arises is how uh(x)

${u}^{h}\left(x\right)$

can be parametrized in a discrete function space. One possible solution is to express uh(x)

${u}^{h}\left(x\right)$

as a sum of shape function ϕi(x)

${\varphi }_{i}\left(x\right)$

with coefficients αi(x)

${\alpha }_{i}\left(x\right)$

.

uh(x)=i=1nαiϕi(x)(2)

$\begin{array}{}\text{(2)}& {u}^{h}\left(x\right)=\sum _{i=1}^{n}{\alpha }_{i}{\varphi }_{i}\left(x\right)\end{array}$

The line illustrated at the top shows this principle for a 1D problem. u

$u$

can represent the temperature along the length of a rod that is heated in an nonuniform manner. In our case, there are four elements along the x-axis, where the function defines the linear approximation of the temperature illustrated by dots along the line.

One of the biggest advantages we have when using the Finite Element Analysis is that we can either vary the discretization per element or discretize the corresponding basis functions. De facto, we could use smaller elements in regions where high gradients of u

$u$

are expected. For the purpose of modelling the steepness of the function we need to make approximations.

Partial Differential Equations

Before proceeding with the Finite Element Analysis itself, it is important to understand the different types of PDE’s and their suitability for FEA. Understanding this is important to everyone, irrespective of one’s motivation to using finite element analysis. One should constantly remind oneself that FEA is a tool and any tool is only as good as its user.

PDE’s can be categorized as elliptic (are quite smooth), hyperbolic (support solutions with discontinuities), and parabolic (describe time dependent diffusion problems). When solving these differential equations boundary and/or initial conditions need to be provided. Based on the type of PDE, the necessary inputs can be evaluated. Examples for PDE’s in each category include Poisson equation (Elliptic), Wave equation (Hyperbolic) and Fourier law (Parabolic).

There are two main approaches to solving elliptic PDE’s – Finite Difference Analysis (FDA) and Variational (or Energy) Methods. FEA falls into the second category of variational methods. Variational approaches are primarily based on the philosophy of energy minimization.

Hyperbolic PDE’s are commonly associated with jumps in solutions. The wave equation for instance is a hyperbolic PDE. Owing to the existence of discontinuities (or jumps) in solutions, the original FEA technology (or Bubnov-Galerkin Method) was believed to be unsuitable for solving hyperbolic PDE’s. However, over the years, modifications have been developed to extend the applicability of FEA technology.

It is important to consider the consequence of using a numerical framework that is unsuitable for the type of PDE that is chosen. Such usage leads to solutions that are known as “improperly posed”. This could mean that small changes in the domain parameters lead to large oscillations in the solutions or the solutions exist only on a certain part of the domain or time. These are not reliable. Well-posed solutions are defined with a unique one, that exists continuously for the defined data. Hence, considering reliability, it is extremely important to obtain them.

The Weak and Strong Formulation

The mathematical models of heat conduction and elastostatics covered in this series consist of (partial) differential equations with initial conditions as well as boundary conditions. This is also referred to as the so-called Strong Form of the problem. A few examples of “strong forms” are given in the illustration below.

Differential Equations Physical problem Quantities
ddx(AkdTdx)+Q=0

$\frac{d}{dx}\left(Ak\frac{dT}{dx}\right)+Q=0$

One-dimensional heat flow T

$T$

=temperature, A

$A$

=Area, k

$k$

=thermal conductivity, q

$q$

=heat supply

ddx(AEdudx)+b=0

$\frac{d}{dx}\left(AE\frac{du}{dx}\right)+b=0$

$u$

=displacement, A

$A$

=Area, E

$E$

=Young’s modulus, b

$b$

Sd2wdx2+p=0

$S\frac{d{}^{2}w}{dx{}^{2}}+p=0$

$w$

=deflection, S

$S$

=String force, p

$p$

$\frac{d}{dx}\left(AD\frac{dc}{dx}\right)+Q=0$

One-dimensional diffusion c

$c$

=iron contraction, A

$A$

=area, D

$D$

=diffusion coefficient, Q

$Q$

=ion supply

ddx(AγdVdx)+Q=0

$\frac{d}{dx}\left(A\gamma \frac{dV}{dx}\right)+Q=0$

One-dimensional electric current V

$V$

=voltage, A

$A$

=area, γ

$\gamma$

=electric conductivity, Q

$Q$

=electric charge supply

$\frac{d}{dx}\left(A\frac{D{}^{2}}{32\mu }\frac{dp}{dx}\right)+Q=0$

Laminar flow in pipe (Poiseuille flow) p

$p$

=pressure, A

$A$

=area, D

$D$

=diameter, μ

$\mu$

=viscosity, Q

$Q$

=fluid supply

Table 1: Second order differential equations

Second order partial differential equations demand a high degree of smoothness for the solution u(x)

$u\left(x\right)$

. That means that the second derivative of the displacement has to exist and has to be continuous! This also implies requirements for parameters that can not be influenced like the geometry (sharp edges) and material parameters (different Young’s modulus in a material).

To develop the finite element formulation, the partial differential equations must be restated in an integral form called the weak form. The weak form and the strong form are equivalent! In stress analysis, the weak form is called the principle of virtual work.

l0dwdxAEdudxdx=(wAt¯)x=0+l0wbdx   w with w(l)=0(3)

The given equation is the so-called weak form (in this case the weak formulation for elastostatics). The name states that solutions to the weak form do not need to be as smooth as solutions of the strong form, which implies weaker continuity requirements.

You have to keep in mind that the solution satisfying the weak form is also the solution of the strong counterpart of the equation. Also remember that the trial solutions u(x)

$u\left(x\right)$

must satisfy the displacement boundary conditions. This is an essential property of the trial solutions and this is why we call those boundary conditions essential boundary conditions.

If you want to read about the equivalence between the weak and strong formulation, please read more in the forum topic about the equivalence between the weak and strong formulation of PDEs for FEA.

Minimum Potential Energy

The Finite Element Analysis can also be executed with a variation principle. In the case of one-dimensional elastostatics, the minimum of potential energy is resilient for conservative systems. The equilibrium position is stable if the potential energy of the system Π

$\mathrm{\Pi }$

is minimum. Every infinitesimal disturbance of the stable position leads to an energetic unfavorable state and implies a restoring reaction. An easy example is a normal glass bottle that is standing on the ground, where it has minimum potential energy. If it falls over, nothing is going to happen, except for a loud noise. If it is standing on the corner of a table and falls over to the ground, it is rather likely that the bottle breaks since it carries more energy towards the ground. For the variation principle we make use of this fact. The lower the energy level, the less likely it is to get a wrong solution. The total potential energy Π

$\mathrm{\Pi }$

of a system consists of the work of the inner forces (strain energy)

Ai=l012E(x)A(x)(dudx)212σϵA(x)dx(4)

$\begin{array}{}\text{(4)}& {A}_{i}={\int }_{0}^{l}\underset{\frac{1}{2}\sigma ϵA\left(x\right)}{\underset{⏟}{\frac{1}{2}E\left(x\right)A\left(x\right){\left(\frac{du}{dx}\right)}^{2}}}dx\end{array}$

and the work of the external forces

Aa=A(x)t¯(x)u(x)|Γt(5)

$\begin{array}{}\text{(5)}& {A}_{a}=A\left(x\right)\overline{t}\left(x\right)u\left(x\right){|}_{{\mathrm{\Gamma }}_{t}}\end{array}$

The total energy is:

Π=AiAa(6)

$\begin{array}{}\text{(6)}& \mathrm{\Pi }={A}_{i}-{A}_{a}\end{array}$

Find more about the minimum potential energy in our related forum topic.

Mesh Convergence

One of the most overlooked issues in computational mechanics that affect accuracy, is mesh convergence. This is related to how small the elements need to be to ensure that the results of an analysis are not affected by changing the size of the mesh.

The figure above shows the convergence of quantity with increase in degrees of freedom. As depicted in the figure, it is important to first identify the quantity of interest. At least three points need to be considered and as the mesh density increases, the quantity of interest starts to converge to a particular value. If two subsequent mesh refinements do not change the result substantially, then one can assume the result to have converged.

Going into the question of mesh refinement, it is not always necessary that the mesh in the entire model is refined. St. Venant’s Principle enforces that the local stresses in one region do not affect the stresses elsewhere. Hence, from a physical point of view, the model can be refined only in particular regions of interest and further have a transition zone from coarse to fine mesh. There are two types of refinements (h- and p-refinement) as shown in the figure above. h-refinement relates to the reduction in the element sizes, while p-refinement relates to increasing the order of the element.

Here it is important to distinguish between geometric effect and mesh convergence, especially when meshing a curved surface using straight (or linear) elements will require more elements (or mesh refinement) to capture the boundary exactly. Mesh refinement leads to a significant reduction in errors:

Refinement like this can allow an increase in the convergence of solutions without increasing the size of the overall problem being solved.

How to measure convergence?

So now that the importance of convergence has been discussed, how can convergence be measured? What is a quantitative measure for convergence? The first way would be to compare with analytical solutions or experimental results.

Error of the displacements:

eu=uuh(7)

$\begin{array}{}\text{(7)}& {e}_{u}=u-{u}^{h}\end{array}$

where u

$u$

is the analytical solution for the field.

Error of the strains:

eϵ=ϵϵh(8)

$\begin{array}{}\text{(8)}& {e}_{ϵ}=ϵ-{ϵ}^{h}\end{array}$

where ϵ

$ϵ$

is the analytical solution for the strain field.

Error of the stresses:

eσ=σσh(9)

$\begin{array}{}\text{(9)}& {e}_{\sigma }=\sigma -{\sigma }^{h}\end{array}$

where σ

$\sigma$

is the analytical solution for the stress field.

As shown in the equations above, several errors can be defined for displacement, strains and stresses. These errors could be used for comparison and they would need to reduce with mesh refinement. However, in a FEA mesh, the quantities are calculated at various points (nodal and Gauss). So in this case, what needs to be determined is where and at how many points should the error be calculated.

||eu||?=chp(10)

$\begin{array}{}\text{(10)}& ||{e}_{u}|{|}_{?}=c{h}^{p}\end{array}$

Applications of FEA

The Finite Element Analysis started with significant promise in modeling several mechanical applications related to aerospace and civil engineering. The applications of Finite Element Method are just starting to reach their potential. One of the most exciting prospects is its application to coupled problems like fluid-structure interaction; thermo-mechanical, thermo-chemical, thermo-chemo-mechanical problems piezoelectric, ferroelectric, electromagnetics and other relevant areas:

Static

With static analysis, you can analyze linear static and nonlinear quasi-static structures. In a linear case with an applied static load, only a single step is needed to determine the structural response. Geometric, contact and material nonlinearity can be taken into account. An example is a bearing pad of a bridge.

Dynamic

Dynamic analysis helps you analyze the dynamic response of a structure that experienced dynamic loads over a specific time frame. To model the structural problems in a realistic way, you can also analyze impacts of loads as well as displacements. An example is the impact of a human skull, with or without a helmet.


text:
# Are there any exceptions to Newton's laws?

So I am studying the laws of Newton and I'm wondering, are there any deficiencies related to the laws? I mean, somewhere where I can't use them or anything?

-

For examples of where the laws of Newton are no longer sufficient to describe reality, see the wikipage here. But we don't have to go to extremes like near-$c$ velocities or strong gravitational fields to see that there are some situations where the laws of Newton seem to fail.

Consider an airplane taking off. You can feel yourself being pushed back into your seat, so you are experiencing a force (that is not gravity). However, you remain at rest w.r.t. the coordinate system fixed to the airplane. So the first law of Newton seems to fail: we are at rest, so there is no acceleration and yet a force is exerted on us. When the airplane is traveling at a constant speed on a constant height, however, we feel no force (except for gravity) and the first law seems OK again.

A simpler example still is that of a carousel. When we stand on the carousel we are at rest w.r.t. a coordinate system fixed to the carousel, but we do experience a force trying to push us outward. This again seems to contradict the first law of Newton.

Note that both coordinate systems where the laws seemed to fail in these examples were systems that had a non-zero acceleration themselves. The conclusion is that Newton's laws only seem to work when considered in a non-accelerating reference frames, which we call inertial frames of reference. The forces we feel when we consider a non-inertial frame of reference are called fictitious forces (because they do not arise due to a physical interaction but due to the fact that the reference frame has a finite acceleration) and the wikipage on those has the same example of the carousel that I mentioned, only worked out in more detail (here).

Inertial frames of reference have to be carefully defined. Take, for example, the frame of reference fixed to a lab on earth. When considering most everyday motions, we can use this frame as a good approximation of an inertial frame, despite the fact that the earth is rotating and therefore giving the lab a finite acceleration. However, if we want to describe motions that are a direct consequence of this finite acceleration (such as the deviation to the east of falling objects), we need to find a better frame of reference.

A good question to ask is then: do inertial frames even exist? And the answer is: yes, at least in Newtonian dynamics. I believe this still holds in special relativity but things change when we go to general relativity, though I haven't had any courses on GR yet so I can't be 100% sure.

-
Just about your last paragraph: the "general" of GR refers to being able to do calculations in any frame whatsoever, inertial or not. In some sense free-falling frames are "inertial," but the concept of inertial frame doesn't mean much anymore in GR. –  Chris White Feb 17 '13 at 20:25
@ChrisWhite Just to debunk the popular misconception: you can do accelerating frames perfectly well in SR. (You're no doubt aware of this, but others might not be.) You can use any accelerating frame and any coordinate system you like in SR, no obstruction. The difference between SR and GR is that in SR the curvature vanishes: $R_{\mu\nu\rho\sigma}=0$. In GR the inertial (free-fall) observers are those whose curvature components are as small as possible (second order), so they feel no accelerations, only small tidal forces. In a way they are as close to inertial in the SR sense as possible. –  Michael Brown Feb 18 '13 at 3:34

Newton's second laws are no longer a good approximation to reality for very small systems (like atoms) where quantum mechanical effects become significant, and for certain very large systems where the effects of general relativity become significant. They are no longer a good approximation for at least two reasons:

1. The mathematical model of the physical world in which every system is considered a system of particles that move along well-defined trajectories in three-dimensional space itself breaks down (like in quantum mechanics where the state of a system is described by a vector in a certain space called a Hilbert space).

2. The predictions of Newton's second law no longer hold to sufficient precision in some systems. A famous example is that Newton's law of gravitation coupled with Newton's laws incorrectly predicts the amount which the perihelion of Mercury precesses.

-
+1, but two minor points: galactic dynamics are actually done entirely with Newtonian physics, since gravity isn't overwhelming there and nothing is going too fast; also, you can predict a deflection of Newtonian photons with certain assumptions, and this was done, but it turns out to be off by a factor of exactly 2 from the actual GR result. –  Chris White Feb 17 '13 at 20:30
@ChrisWhite Thanks. When I was writing galaxies, the thought that galactic dynamics might actually be well-described by Newtonian physics occurred to me; I'm certainly no astrophysics/cosmology expert. I took out the parenthetical comparison. I also took out the remark about photon motion since I'm completely unfamiliar with the Newtonian calculation you refer to. –  joshphysics Feb 17 '13 at 20:57
@joshphysics Chris White is probably talking about this ajp.aapt.org/resource/1/ajpias/v56/i5/p413_s1 –  Michael Brown Feb 18 '13 at 3:56
@MichaelBrown Thanks for the reference. –  joshphysics Feb 18 '13 at 6:43


text:
## WT CD28 adopts a bivalent binding conformation when starting from the CTLA-4 dimer orientation.

2014-02-24T03:32:16Z (GMT) by

MD simulations were run with WT CD28 starting from a CTLA-4 dimer orientation. (A) The RMSD of three independent trajectories with respect to the initial conformation are shown over time. There is considerable rearrangement of the subunits at the beginning of the simulations, reflected by an increase in the RMSD. The conformational fluctuations stabilize after the first 100 ns of simulation, although the trajectories adopt different conformations. This instability was not inherent to this dimer orientation, as the RMSD of MD simulations of CTLA-4 remained below 4Å (data not shown). (B) CD80 molecules were docked onto the simulated CD28 dimers to obtain the corresponding CD28-CD80 complexes. The surface area buried between the docked ligands was calculated at various times along each of the three independent trajectories to estimate the fraction of bivalent-competent conformations; the average over all three trajectories was 79%. (C) A representative conformation of CD28 (green) showing docked CD80 ligands (cyan) illustrates the potential for bivalent ligand binding.


text:
# Which of the following process organic peroxides undergo in free radical addition polymerisation?

$\begin{array}{1 1}(a)\;\text{homolytic fusion}&(b)\; \text{homolytic fission}\\(c)\;\text{heterolytic fission}&(d)\;\text{heterolytic fusion}\end{array}$

Organic peroxides undergo homolytic fission to form free radicals which act as initiators in an chain initiation reaction in free radical addition polymerisation.
Hence (b) is the correct answer.


text:
# Excel conditional formula

I need to create a formula dependent on the value of cell U2

If it is between 20 and 24
set cell V2 = "20-24 EE"

or

If it is between 25 and 29
set cell V2 = "25-29 EE"

or

If it is between 30 and 35
set cell V2 = "30-35 EE"

• Are we talking integers only? By "between 20 and 24" do mean "larger or equal to 20 and smaller or equal to 24"? – cybernetic.nomad Mar 21 at 19:08
• I mean larger or equal to 20 and smaller or equal to 24. – Tie Mar 21 at 19:12
• Questions about Excel, such as this one, would be so much cleaner if SuperUser had TeX/MathJax. I'm hoping that StackExchange might consider adopting MathJax more widely in the future. – Nat Mar 21 at 22:12

Try this formula in V2:

=IF(AND($U$2>=20,$U$2<=24),"20-24 EE",IF(AND($U$2>=25,$U$2<=29),"25-29 EE",IF(AND($U$2>=30,$U$2<=35),"30-35 EE","Error")))

This will return "20-24 EE", "25-29 EE", or "30-35 EE" for integer values in the respective range. It will return "Error" for non-integer values or integers less than 20 or greater than 35.

• Hello! Can you give me the formula for the below! I tried to use the above format but I kept getting an error. – Tie Mar 22 at 5:32
• If cell f2 says "Core State Inc. - USA Semi-Monthly 1" then I want 000 to be entered into the cell If cell f2 says "Core States Construction Services - USA Semi-Monthly 2" then I want 001 to be entered into the cell If cell f2 says "Integrated Images" then I want 002 to be entered into the cell Thank you, – Tie Mar 22 at 5:33
• For that scenario, use the following: '=IF($F$2="Core State Inc. - USA Semi-Monthly 1","000",IF($F$2="Core States Construction Services - USA Semi-Monthly 2","001",IF($F$2="Integrated Images","002","Error")))' I assumed an absolute reference for F2, but you might have to adjust that depending on how you're using your spreadsheet. Note: Setting up the formula in this manner (using an if/else cascade) works well enough for small amounts of data and a small number of cases (yours had 3). If you plan on adding more than 3 ranges to check, you should definitely use cybernetic.nomad's method. – user2800 Mar 22 at 11:55
• Thank you! You’ve been a great help! – Tie Mar 22 at 12:22

Assuming you are dealing with integers only, you can use INDEX and MATCH:

=INDEX({"","20-24 ""","25-29 EE","30-35 EE",""},MATCH(U2,{0,20,25,29,35,10000000},1))

NB: The formula will return "" if the number is less than 20 or more than 35

• Hello the numbers in cell u2 are ages. I have to turn them into ranges so for people who are aged 20-24, I need it to enter in 20-24 EE and if they are aged between 25-29, I need it to enter in 25-29 EE and so on. – Tie Mar 21 at 19:30

I'll take a different approach, because this IS the use case for which VLOOKUP is designed. INDEX MATCH is indubitably more powerful, but it's needlessly complex here.

## 1. Create and name your lookup table

I put the lookup table in X:Y. It can be a different sheet or workbook altogether, if you like. Put your delimiters in the first column and the group names you want in the second column. Highlight the table (no headers) and select 'Define Name':

## 2. Formula

=VLOOKUP(U2,groups,2,TRUE)

Copy that formula into V2. That formula copies down.

This says look up U2 in the table called groups (you could also skip step 1 and give this function a range in the form of cell references, X2:Y6 in this case), and bring back the value from the 2nd column. The TRUE argument is what makes it work as a lookup table.

## NB.

This is dead easy, as, again, it's exactly why VLOOKUP exists. The advantage of using a lookup table over a series of nested IFs or an INDEX MATCH array formula (both totally viable solutions for this request) is 1) its ease of use due to the simplicity of the formula itself, but even more importantly 2) its flexibility moving forward. If the groupings change, or the group names change, or you need MORE groups or fewer, all you have to do is change the values in the lookup table; no formula entry is required at all.

• There's something unsatisfying about needing to build an extra table on the spreadsheet, as recommended by this answer. As a consequence, I suspect that a lot of users wouldn't like this approach. Still, this approach seems a lot more maintainable, legible, and just generally well-constructed. As such, I probably wouldn't bother doing it this way for a quick calculation that's just going to be tossed anyway, but for creating a professional spreadsheet that'll endure and be shared, this seems like the way to go. – Nat Mar 21 at 21:58
• @Nat I totally agree. If I needed a one-off and was doing it myself for myself, there's zero chance I'd set it up this way. Probably three IFs and definitely fewer, I'd just write nested IFs; more than that and I'd use the INDEX MATCH with arrays. Almost exactly how the other two answers here have it. I thought it'd be nice to present the entire range of approaches - and, as you say, if the use case calls for repeated use, or other people as users, this is definitely the way to go. – Alex M Mar 21 at 22:02
• @Tie very happy to help. Out of curiosity, what solution did you go with, and how has it worked out for you? – Alex M Mar 21 at 22:41
• I used =IF(AND($U$2>=20,$U$2<=24),"20-24 EE",IF(AND($U$2>=25,$U$2<=29),"25-29 EE",IF(AND($U$2>=30,$U$2<=35),"30-35 EE","Error"))) – Tie Mar 21 at 23:12
• Thank you User280! – Tie Mar 21 at 23:13


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# Solution - Solutions of Quadratic Equations by Completing the Square

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ConceptSolutions of Quadratic Equations by Completing the Square

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Solve the following quadratic equation for x: 4x^2 + 4bx – (a^2 – b^2) = 0

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• # question_answer The value of ${{(1+i)}^{8}}+{{(1-i)}^{8}}$ is  [RPET 2001; KCET  2001] A) 16 B) - 16 C) 32 D) - 32

$\because \,{{(1+i)}^{2}}=\,1+{{i}^{2}}+2i=2i$ and ${{(1-i)}^{2}}=1+{{i}^{2}}-2i=-2i$ $\Rightarrow {{(1+i)}^{8}}+{{(1-i)}^{8}}={{(2i)}^{4}}+{{(-2i)}^{4}}$$={{2}^{4}}({{i}^{4}}+{{i}^{4}})$                             $={{2}^{5}}=32.$


text:
# How to format currency units combined with other units?

If I have a value 25.5 Which of the following is correct?

1. 25,5 $/h 2. 25,5$ /h

As you can see, I'm stressed out about the space after the currency symbol. What is the usual format for these kinds of units in French-language documents?

25,5 $/h is correct, it means "25,5 dollar per hour". • To add to that, in french, when it comes to punctuation, there's no space for unit separation. As in m/s,$/h, €/h.. There's a space before the unit, like you used in your example. Spaces for punctuation follow the rule of 1 after when it has one part (like end-of-sentence dots, commas, dashes, parenthesis, etc.), and 1 space before and after when it has two parts (as in ':', ';', etc.). That space before punctuation is insecable. If there's multiple punctuation marks following one another, there's only a space after the last one. – LuckyPenny Apr 25 '16 at 4:11


text: Survivalstan is a library of Survival Models written in Stan. The point estimate for mu is 9.98 and the true value is contained within the 95% credible interval [9.92, 10.05]. Since our model is fairly simple and all checks are in order, I won’t describe them here. In addition, we welcome contributions of: There are several examples included in the example-notebooks, roughly one corresponding to each model. The project is licensed under the Apache 2.0 license. A Bayesian competitor to the Lasso makes use of the “Horseshoe prior” (which I’ll call “the Horseshoe” for symmetry). In csetraynor/predsurv: Build and compare diverse prediction survival models. Prepare data & fit a survival model using Stan This function wraps a number of steps into one function: Prepare input data dictionary for Stan - calls SurvivalStanData with user-provided formulas & df - (can be overridden using the input_data parameter) Compiles & optionally caches compiled stan code If nothing happens, download Xcode and try again. Likelihood Ratio Test. Survival model 1 = Alive 2 = Dead y y Competing risks model 1 = Alive 2 = Dead (heart disease) 3 = Dead (other cause) y y Illness-death model with death from any cause 1. The variable N_Cis the number of covariates, in our case equal to $$1$$, since we only use metastized. Or model survival as a function of covariates using Cox, Weibull, lognormal, and other regression models. Below is the Stan model for Weibull distributed survival times. We use optional third-party analytics cookies to understand how you use GitHub.com so we can build better products. Learn more. survival data18, survHE19proposes survival models for health economics studies, rstanarm constructs regression models using Stan including a joint survival model, and SemiCompRisks estimates hierarchical multistate models for the analysis of independent or clustered semicompeting risks data20. For versions 2.18 and later, this is titled Stan User’s Guide. Bayesian Survival Analysis 1: Weibull Model with Stan; by Kazuki Yoshida; Last updated about 2 years ago Hide Comments (–) Share Hide Toolbars Documentation http://jburos.github.io/survivalstan_, http://github.com/hammerlab/survivalstan/issues, http://github.com/survivalstan/survivalstan, Weibull, Exponential, and Gamma parameterization, A variety of semi-parametric and non-parametric baseline hazards. GitHub is home to over 50 million developers working together to host and review code, manage projects, and build software together. Illness 3. The scale is inverted as Stan uses inverse-scale parameterization. For help, please reach out to us on gitter. The variance components model has less variation at level 2 than the random intercept model. The PWE survival model described earlier divided the time scale into a sequence of intervals, under the assumption that the hazard function was constant within each of these intervals. The Stan user’s guide provides example models and programming techniques for coding statistical models in Stan. Fit stan survival model Fit stan survival model Usage Stan code for survival models; Worked examples, as jupyter notebooks or markdown documents; Usage examples. This leads to Weibull, generalized gamma, log-normal or log-logistic models for T. 8 A popular model making this bet is the Lasso or, less handily, L1-regularized regression. Stan has an amazing array of diagnostics to check the quality of the fitted model. Flexibility is instead supported by including more Stan files (roughly one per baseline hazard type) and by supporting direct editing of any of these Stan files. Level 2 variance can increase House prices in 4 di erent areas of a city Size Price u1 u2 u3 u4. Death x x x Chris Jackson Multi-state Models: Methods and Software 5/ 72. For versions 2.17 and earlier, this is part of the Stan Reference Manual. – The survival function gives the probability that a subject will survive past time t. – As t ranges from 0 to ∞, the survival function has the following properties ∗ It is non-increasing ∗ At time t = 0, S(t) = 1. This topic is called reliability theory or reliability analysis in engineering, duration analysis or duration modelling in economics, and event history analysis in sociology. If nothing happens, download the GitHub extension for Visual Studio and try again. In brms: Bayesian Regression Models using 'Stan' Description Usage Format Source Examples. Documentation http://jburos.github.io/survivalstan_ is available online. This dataset, originally discussed in McGilchrist and Aisbett (1991), describes the first and second (possibly right censored) recurrence time of infection in kidney patients using portable dialysis equipment. The following shows how to use the Horseshoe in Stan. You can always update your selection by clicking Cookie Preferences at the bottom of the page. Survival modeling is a core component of any clinical data analysis toolset. Library of Stan Models for Survival Analysis. The model was designed for mice, so the prior on the missing data assumes survival of 500 units (days for me) with SD 150. The C index of discrimination varied from 0.8 (SE=0.093) at year 1, to 0.75 (SE=0.034) at year 7 for the continuous time model; from 0.81 (SE=0.07) at year 1, to 0.75 (SE=0.033) at year 7 for the discrete time model. Cox Proportional Hazards Model is a semi-parametric model in the sense that the baseline hazard function does not have to be specified i.e it can vary, allowing a different parameter to be used for each unique survival time. JAGS_survival_model.txt contains code for a JAGS model of survival for trees at least 1 cm DBH. Many of the other notebooks are sparse on explanation, but do illustrate variations on the different models. Please contribute to survivalstan development by letting us know if you encounter any bugs or have specific feature requests. If you are not sure where to start, Test pem_survival_model with simulated data.ipynb contains the most explanatory text. Use Git or checkout with SVN using the web URL. Description. Millions of developers and companies build, ship, and maintain their software on GitHub — the largest and most advanced development platform in the world. Predict hazard ratios, mean survival time, and survival probabilities. A stanmodel object can then be used to draw samples from the model. Consider an ordinary regression model for log survival time, of the form Y = logT= x0+ ˙W; where the error term Whas a suitable distribution, e.g. Learn more, We use analytics cookies to understand how you use our websites so we can make them better, e.g. Extensible framework - bring your own Stan code, or edit the models provided, Plot posterior estimates of key parameters using seaborn, Annotated posterior draws of parameter estimates, as pandas dataframes, Supports caching via stancache or pystan-cache. For more information, see our Privacy Statement. If nothing happens, download GitHub Desktop and try again. We will start with model code adapted from wei_bg.stan within the github repo accompanying Peltola et al, 2014 ’s nice paper describing a bayesian approach to biomarker evaluation. If you are having issues or questions, please let us know. In fitting the PWE survival model, each subject's duration of exposure (or at‐risk time) during the interval is taken into account (as an offset variable). extreme value, gen- eralized extreme value, normal or logistic. Here we will work through an example of fitting a survival model in Stan, using as an example data from TCGA on patients with Bladder Urothelial Carcinoma. Work fast with our official CLI. I was just surprised to see that using Survivalstan to simulate data (simple exponential model) and then infer parameters of an exponential model results in a posterior distribution with such bad neighbourhoods. The federal agencies that make critical decisions on how this money is spent and on how the whole Columbia river is operated regularly use a model (a variant of the Cormack-Jolly-Seber model) which will generate survival probabilities greater than 1. 6.1. Monotone splines. If you are not sure where to start, Test pem_survival_model with simulated data.ipynb contains the most explanatory text. The censored observations are modeled as missing data with a lower bound at the censoring threshold and unknown (modeled) additional contribution. View source: R/predsurv.R. they're used to gather information about the pages you visit and how many clicks you need to accomplish a task. This results in increased flexibility of the model. “Monotone regression splines in action.” Statistical science 3.4 (1988): 425-441. We introduce here some common tests that we can use to examine the goodness-of-fit of a discrete-time survival model, including the likelihood ratio test, Akaike information criterion (AIC) and deviance residuals. In survival model terminology, the former are uncensored instances, for which death (the event or endpoint of interest) was observed, and the latter are censored instances, for which no event was observed in the observation time time. Survival analysis is a branch of statistics for analyzing the expected duration of time until one or more events happen, such as death in biological organisms and failure in mechanical systems. There are several examples included in the example-notebooks, roughly one corresponding to each model. It also serves as an example-driven introduction to Bayesian modeling and inference. Note in the transformed parameters block we specify the canonical accelerated failure time (AFT) parameterization – modeling the scale as a function of the shape parameter, $$\alpha$$, and covariates. Well 2. It also contains a number of utility functions helpful when doing survival analysis. When is a variable a level? Construct an instance of S4 class stanmodel from a model specified in Stan's modeling language. test the models on a real survival analysis problem, and we show that both models exhibit good discrimination and calibration capabilities. We can also plot all the samples from our posterior. We use essential cookies to perform essential website functions, e.g. download the GitHub extension for Visual Studio, update long-surv data structure to alphabetize columns, Update local copy with latest code from hammerlab (, Test pem_survival_model with simulated data.ipynb, Weibull, Exponential, and Gamma parameterizations, PEM models with variety of baseline hazards, PEM model with varying-coefficients (by group), Extensible framework - bring your own Stan code, or edit the models above, Graphical posterior predictive checking (currently PEM models only), Plot posterior estimates of key parameters using, Annotate posterior draws of parameter estimates, format as, Worked examples, as jupyter notebooks or markdown documents. Survivalstan is a library of Survival Models written in Stan _ It also contains a number of utility functions helpful when doing survival analysis. Do you have groups of individuals in your study? Description Usage Arguments. Stan_survival_model.txt contains code for a Stan model of seedling survival. In other words, the probability of surviving past time 0 is 1. The ﬁrst is to model the instantaneous rate of the event (known as the hazard) as a function of time. they're used to log you in. Adjust for within-group correlation using a random-effects or shared-frailty model. This model assumes that the time to event x follows a Weibull distribution. Learn more. I'm aware that Stan models sometimes need reparametrization and data sometimes need rescaling. The Stan program (the model expressed in the Stan modeling language) is first translated to C++ code and then the C++ code for the model plus other auxiliary code is compiled into a dynamic shared object (DSO) and then loaded. This prior captures the belief that regression coefficients are rather likely to be zero (the bet on sparsity). By comparison, the Stan code included in SurvivalStan is focused on a particular model and so is only as complex as that model demands. Please submit an issue github or via gitter. Learn more. This session will work through several examples of doing survival analysis using Stan, including both parametric and nonparametric approaches to modeling the baseline hazard. A model has a better fit to the data if the model, compared with a model with fewer predictors, demonstrates an improvement in the fit. We use optional third-party analytics cookies to understand how you use GitHub.com so we can build better products. There are several examples included in the example-notebooks, roughly one corresponding to each model. That’s right, dams don’t kill fish, they resurrect them! Description. ∗ At time t = ∞, S(t) = S(∞) = 0. Stan code for survival models; Worked examples, as jupyter notebooks or markdown documents; Usage examples. Survivalstan is a library of Survival Models written in Stan It also contains a number of utility functions helpful when doing survival analysis. But, it assumes that the rate ratio remains proportional throughout the follow-up period[13]. There are two common approaches to modelling survival data. If you are not sure where to start, Test pem_survival_model with simulated data.ipynb contains the most explanatory text. Fitting survival models in Stan is fairly straightforward. Ramsay, James O. For our first analysis we will work with a parametric Weibull survival model. You signed in with another tab or window. True value is contained within the 95 % credible interval [ 9.92, 10.05 ] random-effects or shared-frailty model follow-up... Survival as a function of covariates, in our case equal to \ 1\. Bayesian regression models using 'Stan ' Description Usage Format Source examples under the Apache 2.0 license can., the probability of surviving past time 0 is 1 to each model inverse-scale parameterization how many clicks need! City Size Price u1 u2 u3 u4 on gitter ’ s guide provides models! To accomplish a task with SVN using the web URL to host and review code, manage,... Draw samples from our posterior the example-notebooks, roughly one corresponding to each model handily! Additional contribution happens, download the GitHub extension for Visual Studio and try.... Intercept model 4 di erent areas of a city Size Price u1 u2 u4! Later, this is titled Stan user ’ s guide provides example models and programming techniques coding... ; Usage examples are rather likely to be zero ( the bet on sparsity ) 2.17 and,... Having issues or questions, please let us know if you are not sure where to start, pem_survival_model! A library of survival for trees at least 1 cm DBH models on a survival! Of utility functions helpful when doing survival analysis problem, and other regression models that Stan models sometimes reparametrization! Variation at level 2 variance can increase House prices in 4 di erent areas of a Size... Specified in Stan it survival model in stan contains a number of utility functions helpful when doing survival analysis data.ipynb!, Weibull, lognormal, and we show that both models exhibit good discrimination and calibration.... Github is home to over 50 million developers working together to host and review code, manage,. Bottom of the Stan Reference Manual or model survival as a function of.... 10.05 ] model has less variation at level 2 variance can increase House in! Stanmodel object can then be used to gather information about the pages you visit how. Versions 2.18 and later, this is titled Stan user ’ s right, don! That Stan models sometimes need rescaling used to draw samples from our posterior using web... Test the models on a real survival analysis of utility functions helpful when doing survival analysis analytics.: Methods and Software 5/ 72 brms: Bayesian regression models then be used to gather information the! Groups of individuals in your study by letting us know surviving past time 0 is 1 'Stan ' Description Format. Clinical data analysis toolset in csetraynor/predsurv: build and compare diverse prediction survival models written in Stan inverse-scale. At time t = ∞, s ( ∞ ) = s ( ∞ ) = s ( ). By clicking Cookie Preferences at the censoring threshold and unknown ( modeled ) additional contribution sparsity. From a model specified in Stan 's modeling language to understand how you use so... 4 di erent areas of a city Size Price u1 u2 u3 u4 you are having issues or,! Are having issues or questions, please reach out to us on gitter is... Of the Stan user ’ s guide, Weibull, lognormal, and build Software together this bet is Stan! Used to draw samples from our posterior the number of utility functions helpful when doing analysis... Github is home to over 50 million developers working together to host and review code, manage projects and! To be zero ( the bet on sparsity ) build and compare diverse survival... Visual Studio and try again specific feature requests to \ ( 1\ ), since we only metastized... How you use GitHub.com so we can build better products and Software 5/.! To be zero ( the bet on sparsity ) 's modeling language, less handily, L1-regularized regression with... That both models exhibit good discrimination and calibration capabilities are not sure where to start, Test pem_survival_model simulated... Rate ratio remains proportional throughout the follow-up period [ 13 ] in,... Then be used to draw samples from our posterior variance can increase House prices in 4 erent! Do you have groups of individuals in your study contributions of: there are several included! Data with a parametric Weibull survival model fit Stan survival model fit Stan survival model Stan! Please let us know and Software 5/ 72 covariates, in our case equal \... Can also plot all the samples from the model a lower bound at the bottom of Stan... Are having issues or questions, please let us know if you are not sure where to start Test! Belief that regression coefficients are rather likely to be zero ( the bet on sparsity ) survival model Stan. Http: //jburos.github.io/survivalstan  _ is available online there are several examples included in the example-notebooks, roughly corresponding! Reach out to us on gitter //jburos.github.io/survivalstan  _ is available online discrimination and capabilities. ( t ) = s ( t ) = s ( t ) = s ( ∞ ) 0! An example-driven introduction to Bayesian modeling and inference at least 1 cm DBH encounter. Model fit Stan survival model with SVN using the web URL over 50 million developers working to... The instantaneous rate of the other notebooks are sparse on explanation, but do illustrate variations on the models! Analysis we will work with a lower bound at the censoring threshold and unknown ( modeled additional. Surviving past time 0 is 1 an amazing array of diagnostics to check the of. To understand how you use GitHub.com so we can build better products your study credible interval [ 9.92, ]. Components model has less variation at level 2 than the random intercept model component any. Sometimes need reparametrization and data sometimes need reparametrization and data sometimes need reparametrization and data sometimes need reparametrization and sometimes. The rate ratio remains proportional throughout the follow-up period [ 13 ] code for a JAGS model of models. That ’ s guide provides example models and programming techniques for coding statistical models in Stan are rather to. Third-Party analytics cookies to understand how you use GitHub.com so we can build better products within-group correlation using a or! S guide only use metastized each model has less variation at level 2 than the random intercept model earlier. And earlier, this is titled Stan user ’ s right, dams don t... Multi-State models: Methods and Software 5/ 72 less variation at level 2 the! Stan has an amazing array of diagnostics to check the quality of Stan. On explanation, but do illustrate variations on the different models and we show that both models exhibit discrimination. Handily, L1-regularized regression they resurrect them roughly one corresponding to each model fit Stan survival model models in! Are rather likely to be zero ( the bet on sparsity ) for models. But, it assumes that the time to event x follows a Weibull.. Component of any clinical data analysis toolset time t = ∞, s ( t ) s... Model fit Stan survival model example-driven introduction to Bayesian modeling and inference selection clicking... A core component of any clinical data analysis toolset several examples included in the example-notebooks roughly. This prior captures the belief that regression coefficients are rather likely to be zero ( the bet sparsity. Horseshoe in Stan model assumes that the rate ratio remains proportional throughout the follow-up [... [ 13 ] you use GitHub.com so we can also plot all the samples from posterior. Contains code for survival models in csetraynor/predsurv: build and compare diverse prediction survival models model the instantaneous rate the! N_Cis the number of covariates using Cox, Weibull, lognormal, and other regression models using '! ( t ) = 0 also contains a number of utility functions helpful doing... And earlier, this is part of the event ( known as the hazard ) a. As an example-driven introduction to Bayesian modeling and inference, Test pem_survival_model with simulated data.ipynb contains the explanatory. Our posterior to perform essential website functions, e.g Description Usage Format examples. Good discrimination and calibration capabilities, manage projects, and build Software together Preferences at the censoring threshold unknown! Host and review code, manage projects, and we show that both models exhibit good discrimination and calibration.... Make them better, e.g check the quality of the page to accomplish a task SVN using the web.! Increase House prices in 4 di erent areas of a city Size Price u1 u3! % credible interval [ 9.92, 10.05 ] distributed survival times stanmodel from a model specified in.. T kill fish, they resurrect them about the pages you visit and how clicks., 10.05 ] to model the instantaneous rate of the page written in Stan 's language... Third-Party analytics cookies to understand how you use GitHub.com so we can build better products Stan survival model of. Source examples and survival probabilities or model survival as a function of time to perform essential website,... The models on a real survival analysis of surviving past time 0 is 1 a. Bayesian regression models using 'Stan ' Description Usage Format Source examples GitHub is home over... Cox, Weibull, lognormal, and we show that both models exhibit discrimination... Any bugs or have specific feature requests or have specific feature requests need rescaling sparsity. 4 di erent areas of a city Size Price u1 u2 u3 u4 stanmodel a. Increase House prices in 4 di erent areas of a city Size Price u1 u3... The time to event x follows a Weibull distribution 10.05 ] random-effects shared-frailty! Distributed survival times analysis problem, and survival probabilities intercept model =,. For a JAGS model of survival models ; Worked examples, as jupyter or...


text:
# Cosine similarity with arrays contaning NaN

I am trying to calculate a cosine similarity using Python in order to find similar users basing on ratings they have given to movies. As it can be expected there are a lot of NaN values. I am using movie dataset from Kaggle.

When I use np.dot() on two nd.arrays the outcome is:'nan'. I have checked with np.nansum() that there are some other than 'nan' values.

I do not want to change all 'nan' values to '0' as it would mean that users have given 0 rating to the movies which lead to 'false' similarity between users.

Please, give me some advice regarding how to proceed with this problem.

• Check if the arrays have the same size when you put them in the function. – Juan Esteban de la Calle Apr 27 '19 at 14:14
• Can you provide an example matrix? A smaller version of your actual data. – n1k31t4 Apr 27 '19 at 16:01
• If you need so more control, implement the dot multiplication function on your own in Python. – Shubham Panchal Apr 27 '19 at 16:19

Have you couple of types na's handling. Its sample of code:

def na_handling(df, name_of_strategy):

#list of stategies -> mean, mode, 0, spefic_value, next_row, previous_row

if name_of_strategy=="previous_row":
df.fillna(method="backfill", inplace=True)
return df
elif name_of_strategy=="next_row":
df.fillna(method="ffill", inplace=True)
return df
elif name_of_strategy=="0":
df.fillna(0, inplace=True)
return df

elif name_of_strategy=="mean":
df.fillna(df.mean(), inplace=True)
return df
elif name_of_strategy=="mode":
df.fillna(df.mode(), inplace=True)
return df
else:
print("Wrong specified strategy")

vec1 = na_handling(old_vec, "next_row")

def cosine_sim(df1, df2):

df1na = df1.isna()
df1clean = df1[~df1na]
df2clean = df2[~df1na]

df2na = df2clean.isna()
df1clean = df1clean[~df2na]
df2clean = df2clean[~df2na]

# Compute cosine similarity
distance = cosine(df1clean, df2clean)
sim = 1 - distance

return sim

• This doesn't explain what you are trying to do: consider dimensions only where both are non-NaN, which is reasonable. I think you swapped distance and similarity though. – Sean Owen Mar 22 at 14:48

I think it's rarely meaningful to consider cosine similarity on sparse data like this, not just because of sparsity (because it's only defined for dense data), but because it's not obvious the cosine similarity is meaningful. For example a user that rates 10 movies all 5s has perfect similarity with a user that rates those 10 all as 1. Magnitude doesn't matter in cosine similarity, but it matters in your domain.

It's much more likely that it's meaningful on some dense embedding of users and items, such as what you get from ALS.

To answer the question, you either need to impute the missing ratings (don't assume 0, but a mean value or similar), or ignore dimensions that aren't defined in both.


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Augustus Acevedo

Answered

2022-07-15

Can you express this in form $\frac{1}{a}+\frac{1}{b}$
Can you express the fraction $\frac{1949}{1999}$ in the form $\frac{1}{a}+\frac{1}{b}$? Give reasons supporting your answer.
I think the only way to do this is keep trying numbers but then I will never get the answer. I cry every time.

Answer & Explanation

Karla Hull

Expert

2022-07-16Added 20 answers

$\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab},$
so you would need $ab$ to divide $1999$. But…

Logan Wyatt

Expert

2022-07-17Added 5 answers

Given
$\frac{1}{a}+\frac{1}{b}=\frac{1949}{1999}$
Combining the fraction gives
$\frac{a+b}{ab}=\frac{1949}{1999}$
Setting terms gives the system
$\left\{\begin{array}{l}a+b=1949\\ ab=1999\end{array}$
With $a,b$ can be solved by a quadratic. Namely
${b}^{2}-1949b+1999$
${b}^{2}-1949b+1999$
Where $a=1949-b$
Since $1999$ cannot be factored, the roots are really ugly looking numbers. Namely, the two possible values of $b$ are
${b}_{1}=\frac{1949+\sqrt{3790605}}{2}\phantom{\rule{0ex}{0ex}}{b}_{2}=\frac{1949-\sqrt{3790605}}{2}$
and with the $a$ values as
${a}_{1}=\frac{1949-\sqrt{3790605}}{2}\phantom{\rule{0ex}{0ex}}{a}_{2}=\frac{1949+\sqrt{3790605}}{2}$

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## Encyclopedia > CASA

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# CASA

Construcciones Aeronáuticas S.A. (CASA)

Founded in 1923 in Spain.

Held by SEPI, a Spanish holding company (99.8%).

Member of EADS since 1999.

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# lie algebra semisimple?

If L is a semisimple lie algebra then L=[L,L]. Is the opposite true?

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## 1 Answer

No. A Lie algebra satisfying that property is called perfect. For an example of a perfect Lie algebra that isn't semisimple, take a semisimple $L$ and an irreducible representation $V$ of $L$, and define a bracket on $L \times V$ by $$[(X,v),(Y,u)] := ([X,Y],Xu-Yv).$$ This turns $L \times V$ into a perfect Lie algebra with $\text{Rad}(L \times V) = V$.

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Maybe it's also helpful to mention the substantial though often isolated literature on perfect Lie algebras and related structure theory? There are some interesting connections with other questions, as in the paper by Benkart and Zelmanov in Invent. Math. 126 (1996), 1-45. –  Jim Humphreys Apr 5 '11 at 13:08


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# Calculus of variation help

1. Nov 10, 2004

### matpo39

hi, we have just got to the point in my physics course where Newtons laws are now longer that easy to work with anymore and we are now begining to reformulate those using variational methods, and im a little confused on one of the problems.
The shortest path between two points on a curved surface, such as the surface of a sphere is called a geodesic. To find a geodestic, one has first to set up an integral that gives the length of a path on the surface in question. Use sperical polar coordinates (r,theat,phi) to show that the length of a path joining two points on a shere of radius R is L=R*integral(from theta_1 to theta_2)sqrt(1+sin^2(theta)*(phi_prime(theta))^2)*d(theta)

if (theta_1,phi_1) and (theta_2,phi_2) specify two points and we assume that the path is expressed as phi=phi(theta).

I know how to do this problem if it were just (x,y) it would be L=integral(from x_1 to x_2) ds, where ds= sqrt(dx^2+dy^2)= sqrt(1+y_prime(x)^2) dx
but im getting confused on how to impliment the spherical polar cords. For example x=r*sin(phi)*cos(theta), but then i dont know what dx would be because im not sure what to differantiat?

thanks for the help.

2. Nov 10, 2004

### ReyChiquito

use chain rule...

$$dx=\frac{\partial x}{\partial r}dr+\frac{\partial x}{\partial \phi}d\phi+\frac{\partial x}{\partial \theta}d\theta$$

same with $dy$

3. Nov 10, 2004

### matpo39

ok i used the chain rule and came up with these values for my dx and dy

dx= cos(theta)sin(phi)*dr+r*cos(theta)cos(phi)*d(phi) - r*sin(theta)sin(phi)*d(theta)

dy= sin(theta)cos(phi)*dr-r*sin(theta)sin(phi)*d(phi)+r*cos(theta)cos(phi)*d(phi)

I then put those dy,dx values into the equation

ds=Sqrt(dx^2+dy^2) but i still cant get it to look like

L= R integral(from theta_1 to theta_2) (sqrt(1+sin^2(theta)*phi_prime(theta)^2)*d(theta)

also how would the dr,d(phi) work themselves out so all that would be left is the d(theta)?

4. Nov 11, 2004

### ReyChiquito

Ok.. first of all, you are in 3 dimentions so the lenght of a curve is given by

$$L=\int_{t_{1}}^{t_{2}}{\sqrt{{dx}^2+{dy}^2+{dz}^2}}$$

now, given the spherical coordinates

$$x=r\cos\phi\sin\theta$$
$$y=r\sin\phi\sin\theta$$
$$z=r\cos\theta$$

you have that

$${dx}^2+{dy}^2+{dz}^2={dr}^2+r^2\sin^2\theta{d\phi}^2+r^2{d\theta}^2$$

but you are over a sphere right? so $dr=0$ and $r=R$

using all this we have

$$L=R\int_{\theta_{1}}^{\theta_{2}}{\sqrt{\sin^2\theta{d\phi}^2+{d\theta}^2}}=R\int_{\theta_{1}}^{\theta_{2}}{\sqrt{\sin^2\theta{\left(\frac{d\phi}{d\theta}\right)}^2{d\theta}^2+{d\theta}^2}}$$

so

$$L=R\int_{\theta_{1}}^{\theta_{2}}{\sqrt{1+\sin^2\theta{\left(\frac{d\phi}{d\theta}\right)}^2}d\theta}$$

Last edited: Nov 11, 2004


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# Linear Algebra: A matrix vs a determinant

I would like to clarify something in regards to finding a determinant of a matrix. For example say I have this problem where I have to evaluate the determinant:

Is the determinant notation simply a reference to the notion that you are supposed to find the determinant?(as in an operation) So in other words, could I write this matrix with bracket notation if I wanted to? (or if there were no instructions, say)

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Yes: $$\det\left(\begin{array}{ccr}1&7&-3\\1&3&1\\4&8&1\end{array}\right)\quad\text{and}\quad\left|\begin{array}{ccr}1&7&-3\\1&3&1\\4&8&1\end{array}\right|$$mean the same thing; so if the instructions are to "evaluate", then you are to evaluate the determinant of the matrix. –  Arturo Magidin Jun 9 '11 at 19:51
So I can use bracket ( "[ ]" ) notation too right? –  Virtuoso Jun 9 '11 at 19:53
I think the convention is that $||$ is used for the determinant of the matrix, and $()$ or $[]$ for the matrix itself. –  Joel Cohen Jun 9 '11 at 19:53
@Virtuoso: Brackets should not be used for determinants, because a lot of books use them for matrices/vectors. Vertical bars are the standard for determinants. –  Arturo Magidin Jun 9 '11 at 19:54
@Arturo: do you mind if I ask what the {ccr} means in your code for the matrix (etc) above? –  amWhy Jun 9 '11 at 20:50

Frequent notations for a matrix are

$\begin{pmatrix}a& b\\ c&d\end{pmatrix}$ and $\begin{bmatrix}a& b\\ c&d\end{bmatrix}$.

Frequent notations for determinants are

$\left|\begin{matrix}a& b\\ c&d\end{matrix}\right|$ and $\left\|\begin{matrix}a& b\\ c&d\end{matrix}\right\|$ and, of course, $\det\begin{pmatrix}a& b\\ c&d\end{pmatrix}$ and $\det\begin{bmatrix}a& b\\ c&d\end{bmatrix}$.

Brackets are sometimes used for determinants if the matrix is given by its row or column vectors, so if $v$ and $w$ are vectors with 2 coordinates then $[v,w]$ might be used for the determinant of the corresponding matrix.

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...and if we were ever considering the absolute value of a determinant... :D –  Ｊ. Ｍ. Nov 26 '11 at 16:50
Then we take "||" in latex instead of "\|". –  Phira Nov 26 '11 at 17:02