text:
# Hyperbolicity¶

Definition :

The hyperbolicity $$\delta$$ of a graph $$G$$ has been defined by Gromov [Gromov87] as follows (we give here the so-called 4-points condition):

Let $$a, b, c, d$$ be vertices of the graph, let $$S_1$$, $$S_2$$ and $$S_3$$ be defined by

$\begin{split}S_1 = dist(a, b) + dist(d, c)\\ S_2 = dist(a, c) + dist(b, d)\\ S_3 = dist(a, d) + dist(b, c)\\\end{split}$

and let $$M_1$$ and $$M_2$$ be the two largest values among $$S_1$$, $$S_2$$, and $$S_3$$. We define $$hyp(a, b, c, d) = M_1 - M_2$$, and the hyperbolicity $$\delta(G)$$ of the graph is the maximum of $$hyp$$ over all possible 4-tuples $$(a, b, c, d)$$ divided by 2. That is, the graph is said $$\delta$$-hyperbolic when

$\delta(G) = \frac{1}{2}\max_{a,b,c,d\in V(G)}hyp(a, b, c, d)$

(note that $$hyp(a, b, c, d)=0$$ whenever two elements among $$a,b,c,d$$ are equal)

Some known results :

• Trees and cliques are $$0$$-hyperbolic
• $$n\times n$$ grids are $$n-1$$-hyperbolic
• Cycles are approximately $$n/4$$-hyperbolic
• Chordal graphs are $$\leq 1$$-hyperbolic

Besides, the hyperbolicity of a graph is the maximum over all its biconnected components.

Algorithms and complexity :

The time complexity of the naive implementation (i.e. testing all 4-tuples) is $$O( n^4 )$$, and an algorithm with time complexity $$O(n^{3.69})$$ has been proposed in [FIV12]. This remains very long for large-scale graphs, and much harder to implement.

Several improvements over the naive algorithm have been proposed and are implemented in the current module.

• Another upper bound on $$hyp(a, b, c, d)$$ has been proved in [CCL15]. It is used to design an algorithm with worse case time complexity in $$O(n^4)$$ but that behaves much better in practice.

Assume that $$S_1 = dist(a, b) + dist(c, d)$$ is the largest sum among $$S_1,S_2,S_3$$. We have

$\begin{split}S_2 + S_3 =& dist(a, c) + dist(b, d) + dist(a, d) + dist(b, c)\\ =& [ dist(a, c) + dist(b, c) ] + [ dist(a, d) + dist(b, d)]\\ \geq &dist(a,b) + dist(a,b)\\ \geq &2dist(a,b)\\\end{split}$

Now, since $$S_1$$ is the largest sum, we have

$\begin{split}hyp(a, b, c, d) =& S_1 - \max\{S_2, S_3\}\\ \leq& S_1 - \frac{S_2+ S_3}{2}\\ \leq& S_1 - dist(a, b)\\ =& dist(c, d)\\\end{split}$

We obtain similarly that $$hyp(a, b, c, d) \leq dist(a, b)$$. Consequently, in the implementation of the ‘CCL’ algorithm, we ensure that $$S_1$$ is larger than $$S_2$$ and $$S_3$$ using an ordering of the pairs by decreasing lengths. Then, we use the best value $$h$$ found so far to stop exploration as soon as $$dist(a, b) \leq h$$.

The worst case time complexity of this algorithm is $$O(n^4)$$, but it performs very well in practice since it cuts the search space. This algorithm can be turned into an approximation algorithm since at any step of its execution we maintain an upper and a lower bound. We can thus stop execution as soon as a multiplicative approximation factor or an additive one is proven.

• The notion of ‘’far-apart pairs’’ has been introduced in [Soto11] to further reduce the number of 4-tuples to consider. We say that the pair $$(a,b)$$ is far-apart if for every $$w$$ in $$V\setminus\{a,b\}$$ we have

$dist(w,a)+dist(a,b) > dist(w,b) \text{ and }dist(w,b)+dist(a,b) > dist(w,a)$

Determining the set of far-apart pairs can be done in time $$O(nm)$$ using BFS. Now, it is proved in [Soto11] that there exists two far-apart pairs $$(a,b)$$ and $$(c,d)$$ satisfying $$\delta(G) = hyp(a, b, c, d)/2$$. For instance, the $$n\times m$$-grid has only two far-apart pairs, and so computing its hyperbolicity is immediate once the far-apart pairs are found. The ‘CCL+FA’ or ‘CCL+’ algorithm improves the ‘CCL’ algorithm since it uses far-apart pairs.

• This algorithm was further improved in [BCCM15]: instead of iterating twice over all pairs of vertices, in the “inner” loop, we cut several pairs by exploiting properties of the underlying graph.

Todo

• Add exact methods for the hyperbolicity of chordal graphs
• Add method for partitioning the graph with clique separators

This module contains the following functions

At Python level :

 hyperbolicity() Return the hyperbolicity of the graph or an approximation of this value. hyperbolicity_distribution() Return the hyperbolicity distribution of the graph or a sampling of it.

REFERENCES:

 [BCCM15] M. Borassi, D. Coudert, P. Crescenzi, and A. Marino. On Computing the Hyperbolicity of Real-World Graphs. Proceedings of the 23rd European Symposium on Algorithms (ESA 2015)
 [CCL15] N. Cohen, D. Coudert, and A. Lancin. On computing the Gromov hyperbolicity. ACM Journal of Experimental Algorithmics, 20(1.6):1-18, 2015. doi:10.1145/2780652 or [https://hal.inria.fr/hal-01182890].
 [FIV12] H. Fournier, A. Ismail, and A. Vigneron. Computing the Gromov hyperbolicity of a discrete metric space. Arxiv 1210.3323.
 [Gromov87] (1, 2, 3) M. Gromov. Hyperbolic groups. Essays in Group Theory, 8:75–263, 1987.
 [Soto11] (1, 2, 3) M. A. Soto Gomez. 2011. Quelques proprietes topologiques des graphes et applications a internet et aux reseaux. Ph.D. Dissertation. Univ. Paris Diderot (Paris 7).

AUTHORS:

• David Coudert (2012): initial version, exact and approximate algorithm, distribution, sampling
• David Coudert (2014): improved exact algorithm using far-apart pairs
• Michele Borassi (2015): cleaned the code and implemented the new algorithm
• Karan Desai (2016): fixed minor typo in documentation

## Methods¶

sage.graphs.hyperbolicity.hyperbolicity(G, algorithm='BCCM', approximation_factor=None, additive_gap=None, verbose=False)

Returns the hyperbolicity of the graph or an approximation of this value.

The hyperbolicity of a graph has been defined by Gromov [Gromov87] as follows: Let $$a, b, c, d$$ be vertices of the graph, let $$S_1 = dist(a, b) + dist(b, c)$$, $$S_2 = dist(a, c) + dist(b, d)$$, and $$S_3 = dist(a, d) + dist(b, c)$$, and let $$M_1$$ and $$M_2$$ be the two largest values among $$S_1$$, $$S_2$$, and $$S_3$$. We have $$hyp(a, b, c, d) = |M_1 - M_2|$$, and the hyperbolicity of the graph is the maximum over all possible 4-tuples $$(a,b, c,d)$$ divided by 2. The worst case time complexity is in $$O( n^4 )$$.

See the documentation of sage.graphs.hyperbolicity for more information.

INPUT:

• G – a connected Graph

• algorithm – (default: 'BCCM'); specifies the algorithm to use among:

• 'basic' is an exhaustive algorithm considering all possible 4-tuples and so have time complexity in $$O(n^4)$$.
• 'CCL' is an exact algorithm proposed in [CCL15]. It considers the 4-tuples in an ordering allowing to cut the search space as soon as a new lower bound is found (see the module’s documentation). This algorithm can be turned into a approximation algorithm.
• 'CCL+FA' or 'CCL+' uses the notion of far-apart pairs as proposed in [Soto11] to significantly reduce the overall computation time of the 'CCL' algorithm.
• 'BCCM' is an exact algorithm proposed in [BCCM15]. It improves 'CCL+FA' by cutting several 4-tuples (for more information, see the module’s documentation).
• 'dom' is an approximation with additive constant four. It computes the hyperbolicity of the vertices of a dominating set of the graph. This is sometimes slower than 'CCL' and sometimes faster. Try it to know if it is interesting for you. The additive_gap and approximation_factor parameters cannot be used in combination with this method and so are ignored.
• approximation_factor – (default: None) When the approximation factor is set to some value (larger than 1.0), the function stop computations as soon as the ratio between the upper bound and the best found solution is less than the approximation factor. When the approximation factor is 1.0, the problem is solved optimally. This parameter is used only when the chosen algorithm is 'CCL', 'CCL+FA', or 'BCCM'.

• additive_gap – (default: None) When sets to a positive number, the function stop computations as soon as the difference between the upper bound and the best found solution is less than additive gap. When the gap is 0.0, the problem is solved optimally. This parameter is used only when the chosen algorithm is 'CCL' or 'CCL+FA', or 'BCCM'.

• verbose – (default: False) is a boolean set to True to display some information during execution: new upper and lower bounds, etc.

OUTPUT:

This function returns the tuple ( delta, certificate, delta_UB ), where:

• delta – the hyperbolicity of the graph (half-integer value).
• certificate – is the list of the 4 vertices for which the maximum value has been computed, and so the hyperbolicity of the graph.
• delta_UB – is an upper bound for delta. When delta == delta_UB, the returned solution is optimal. Otherwise, the approximation factor if delta_UB/delta.

EXAMPLES:

Hyperbolicity of a $$3\times 3$$ grid:

sage: from sage.graphs.hyperbolicity import hyperbolicity
sage: G = graphs.Grid2dGraph(3, 3)
sage: L,C,U = hyperbolicity(G, algorithm='BCCM'); L,sorted(C),U
(2, [(0, 0), (0, 2), (2, 0), (2, 2)], 2)
sage: L,C,U = hyperbolicity(G, algorithm='CCL'); L,sorted(C),U
(2, [(0, 0), (0, 2), (2, 0), (2, 2)], 2)
sage: L,C,U = hyperbolicity(G, algorithm='basic'); L,sorted(C),U
(2, [(0, 0), (0, 2), (2, 0), (2, 2)], 2)

Hyperbolicity of a PetersenGraph:

sage: from sage.graphs.hyperbolicity import hyperbolicity
sage: G = graphs.PetersenGraph()
sage: L,C,U = hyperbolicity(G, algorithm='BCCM'); L,sorted(C),U
(1/2, [6, 7, 8, 9], 1/2)
sage: L,C,U = hyperbolicity(G, algorithm='CCL'); L,sorted(C),U
(1/2, [0, 1, 2, 3], 1/2)
sage: L,C,U = hyperbolicity(G, algorithm='CCL+'); L,sorted(C),U
(1/2, [0, 1, 2, 3], 1/2)
sage: L,C,U = hyperbolicity(G, algorithm='CCL+FA'); L,sorted(C),U
(1/2, [0, 1, 2, 3], 1/2)
sage: L,C,U = hyperbolicity(G, algorithm='basic'); L,sorted(C),U
(1/2, [0, 1, 2, 3], 1/2)
sage: L,C,U = hyperbolicity(G, algorithm='dom'); L,sorted(C),U
(0, [0, 1, 2, 6], 1)

Asking for an approximation in a grid graph:

sage: from sage.graphs.hyperbolicity import hyperbolicity
sage: G = graphs.Grid2dGraph(2, 10)
sage: L,C,U = hyperbolicity(G, algorithm='CCL', approximation_factor=1.5); L,U
(1, 3/2)
sage: L,C,U = hyperbolicity(G, algorithm='CCL+', approximation_factor=1.5); L,U
(1, 1)
sage: L,C,U = hyperbolicity(G, algorithm='CCL', approximation_factor=4); L,U
(1, 4)
sage: L,C,U = hyperbolicity(G, algorithm='CCL', additive_gap=2); L,U
(1, 3)
sage: L,C,U = hyperbolicity(G, algorithm='dom'); L,U
(1, 5)

Asking for an approximation in a cycle graph:

sage: from sage.graphs.hyperbolicity import hyperbolicity
sage: G = graphs.CycleGraph(10)
sage: L,C,U = hyperbolicity(G, algorithm='CCL', approximation_factor=1.5); L,U
(2, 5/2)
sage: L,C,U = hyperbolicity(G, algorithm='CCL+FA', approximation_factor=1.5); L,U
(2, 5/2)
sage: L,C,U = hyperbolicity(G, algorithm='CCL+FA', additive_gap=1); L,U
(2, 5/2)

Comparison of results:

sage: from sage.graphs.hyperbolicity import hyperbolicity
sage: for i in range(10): # long time
....:     G = graphs.RandomBarabasiAlbert(100,2)
....:     d1,_,_ = hyperbolicity(G, algorithm='basic')
....:     d2,_,_ = hyperbolicity(G, algorithm='CCL')
....:     d3,_,_ = hyperbolicity(G, algorithm='CCL+')
....:     d4,_,_ = hyperbolicity(G, algorithm='CCL+FA')
....:     d5,_,_ = hyperbolicity(G, algorithm='BCCM')
....:     l3,_,u3 = hyperbolicity(G, approximation_factor=2)
....:     if (not d1==d2==d3==d4==d5) or l3>d1 or u3<d1:
....:        print("That's not good!")

sage: from sage.graphs.hyperbolicity import hyperbolicity
sage: import random
sage: random.seed()
sage: for i in range(10): # long time
....:     n = random.randint(2, 20)
....:     m = random.randint(0, n*(n-1) / 2)
....:     G = graphs.RandomGNM(n, m)
....:     for cc in G.connected_components_subgraphs():
....:         d1,_,_ = hyperbolicity(cc, algorithm='basic')
....:         d2,_,_ = hyperbolicity(cc, algorithm='CCL')
....:         d3,_,_ = hyperbolicity(cc, algorithm='CCL+')
....:         d4,_,_ = hyperbolicity(cc, algorithm='CCL+FA')
....:         d5,_,_ = hyperbolicity(cc, algorithm='BCCM')
....:         l3,_,u3 = hyperbolicity(cc, approximation_factor=2)
....:         if (not d1==d2==d3==d4==d5) or l3>d1 or u3<d1:
....:             print("Error in graph ", cc.edges())

The hyperbolicity of a graph is the maximum value over all its biconnected components:

sage: from sage.graphs.hyperbolicity import hyperbolicity
sage: G = graphs.PetersenGraph() * 2
sage: L,C,U = hyperbolicity(G); L,sorted(C),U
(1/2, [6, 7, 8, 9], 1/2)

sage.graphs.hyperbolicity.hyperbolicity_distribution(G, algorithm='sampling', sampling_size=1000000)

Return the hyperbolicity distribution of the graph or a sampling of it.

The hyperbolicity of a graph has been defined by Gromov [Gromov87] as follows: Let $$a, b, c, d$$ be vertices of the graph, let $$S_1 = dist(a, b) + dist(b, c)$$, $$S_2 = dist(a, c) + dist(b, d)$$, and $$S_3 = dist(a, d) + dist(b, c)$$, and let $$M_1$$ and $$M_2$$ be the two largest values among $$S_1$$, $$S_2$$, and $$S_3$$. We have $$hyp(a, b, c, d) = |M_1 - M_2|$$, and the hyperbolicity of the graph is the maximum over all possible 4-tuples $$(a, b, c, d)$$ divided by 2.

The computation of the hyperbolicity of each 4-tuple, and so the hyperbolicity distribution, takes time in $$O( n^4 )$$.

INPUT:

• G – a Graph.
• algorithm – (default: ‘sampling’) When algorithm is ‘sampling’, it returns the distribution of the hyperbolicity over a sample of sampling_size 4-tuples. When algorithm is ‘exact’, it computes the distribution of the hyperbolicity over all 4-tuples. Be aware that the computation time can be HUGE.
• sampling_size – (default: $$10^6$$) number of 4-tuples considered in the sampling. Used only when algorithm == 'sampling'.

OUTPUT:

• hdict – A dictionary such that hdict[i] is the number of 4-tuples of hyperbolicity i.

EXAMPLES:

Exact hyperbolicity distribution of the Petersen Graph:

sage: from sage.graphs.hyperbolicity import hyperbolicity_distribution
sage: G = graphs.PetersenGraph()
sage: hyperbolicity_distribution(G,algorithm='exact')
{0: 3/7, 1/2: 4/7}

Exact hyperbolicity distribution of a $$3\times 3$$ grid:

sage: from sage.graphs.hyperbolicity import hyperbolicity_distribution
sage: G = graphs.GridGraph([3,3])
sage: hyperbolicity_distribution(G,algorithm='exact')
{0: 11/18, 1: 8/21, 2: 1/126}


text:
# TikZ: redefine/update node label

I need to draw many nodes with labels that I can calculate from their indices. However, a small amount of them do not follow the rule, thus I need to assign their labels manually. Can I do that somehow?

MWE:

\documentclass[tikz,border=3.14mm]{standalone}
\usepackage{amssymb}
\begin{document}
\begin{tikzpicture}
\foreach \i in {1,...,30}
{
\node at (\i,0) [rectangle,label=above:\i] (v\i) {$i$};
}
% here I want to make the label of node v7 to become $\varnothing$
\end{tikzpicture}
\end{document}

Note: this has nothing to do with animation, I just want to create the nodes in a loop and then change labels for some of them.

UPD: I corrected MWE, as the original one was not exactly what I implied.

## 2 Answers

You can use \node also, which is described on p. 250 of the pgfmanual, for that.

\documentclass[tikz,border=3.14mm]{standalone}
\usepackage{amssymb}
\begin{document}
\begin{tikzpicture}
\foreach \i in {1,...,10}
{
\node at (\i,0) [rectangle,label={above:$\i$}] (v\i) {$i$};
}
\foreach \i in {3,6,8}{
\node also [label={[fill=white]above:$\varnothing$}] (v\i);
}
\end{tikzpicture}
\end{document}

BIG THANKS TO MAX for the edit.

Just in case you ever have wider labels: give the labels names and use their width for Max' fill=white trick.

\documentclass[tikz,border=3.14mm]{standalone}
\usetikzlibrary{calc}
\usepackage{amssymb}
\begin{document}
\begin{tikzpicture}
\foreach \i in {1,2,...,10}
{
\node at (\i,0) [rectangle,label={[name=lab\i]above:$\i\i$}] (v\i) {$i$};
}
\foreach \i in {3,6,10}{
\path let \p1=($(lab\i.north east)-(lab\i.south west)$) in node also
[label={[fill=white,minimum width=\x1,minimum height=\y1]above:$\varnothing$}] (v\i);
}
\end{tikzpicture}
\end{document}

ADDENDUM: Just for curiosity I was wondering if there is a simple way to make Max' nice answer work with lists. I am sure there is and leave it to others to use some xparse or other magic. Here I just want to report an irony of fate. If one goes for the built-in LaTeX check whether or not something is an element of a list, then my naive attempt fails for two digits, precisely where the simplest version of the above also starts to go wrong. Rather funny and ironic, I'd say. ;-)

\documentclass[tikz]{standalone}
\makeatletter
% https://tex.stackexchange.com/a/260921/121799 and https://tex.stackexchange.com/a/287094/121799
\newcommand\ifmember[2]{%
\in@{#1}{#2}%
\ifin@
\expandafter\@firstoftwo
\else
\expandafter\@secondoftwo
\fi
}
\makeatother
\usepackage{amssymb}

\begin{document}

\begin{tikzpicture}
\foreach \i in {1,...,10}
{
\edef\temp{\noexpand\ifmember{\i}{3,6,9}{\varnothing}{\i}}
\node at (\i,0) [rectangle,label=above:$\temp$] (v\i) {$i$};
}
\end{tikzpicture}
\end{document}

• Perhaps, my MWE was not exactly what I meant. I assumed that every node got a label in a loop, and then I want to change it. I.e. it is \node at (\i,0) [rectangle,label=above:\i] (v\i) {$i$}; If I try to do that with \node also I end up with two labels, one over another. – Yauhen Yakimenka Aug 13 '18 at 12:31
• I updated MWE to reflect what I mean. – Yauhen Yakimenka Aug 13 '18 at 12:36
• @YauhenYakimenka You could of course do label={[fill=white]above:$\varnothing$}, then you wouldn't see the underlying label. That would be simpler than my answer if you have some more different labels. – Max Aug 13 '18 at 12:50
• @marmot I took the liberty of editing your answer, if you don't agree, please roll it back or edit it further. – Max Aug 13 '18 at 13:36
• @Max I actually think the conditionals in the loop are cleaner. If one were to replace the label 10, there would be some leftover. Why don't you undelete your nice answer? (Having said this, I also realize that one could make the \node also work then by giving the labels also names and then using their width.) – user121799 Aug 13 '18 at 16:22

You can use conditionals inside the \foreach loop:

\documentclass[tikz]{standalone}

\usepackage{amssymb}

\begin{document}
\begin{tikzpicture}
\foreach \i in {1,...,10}
{
\ifnum\i=7
\node at (\i,0) [rectangle,label=above:$\varnothing$] (v\i) {$i$};
\else
\node at (\i,0) [rectangle,label=above:$\i$] (v\i) {$i$};
\fi
}
% here I want to make the label of node v7 to become $\varnothing$
\end{tikzpicture}
\end{document}

If you have more labels that should be different, the code stays more readable if you use \pgfmathparse to do check the conditional:

\documentclass[tikz,border=3.14mm]{standalone}
\usepackage{amssymb}
\begin{document}
\begin{tikzpicture}
\foreach \i in {1,2,...,10}
{
\pgfmathparse{
ifthenelse(\i==3,
"varnothing",
ifthenelse(\i==7,
"varnothing",
ifthenelse(\i==10,
"varnothing",
"i"
)
)
)
}
\def\myLabel{\csname\pgfmathresult\endcsname}
\node at (\i,0) [rectangle,label={above:$\myLabel$}] (v\i) {$i$};
}
\end{tikzpicture}
\end{document}

format: plain


text:
## Fitting stellar spectra with some help from Gaussian processes

ExoStat

Wednesday June 18th, 2014

Ian Czekala

### Introduction

• the fundamental parameters of stars
• some problems with fitting spectra
• how Gaussian processes can help
• some interesting applications

### Fundamental parameters of stars

• $M_\star$
• $T_{\rm eff}$
• $\log g$
• $[{\rm Fe}/{\rm H}]$

##### Stellar evolution

Models make predictions of stellar properties as a function of mass and age. This relationship is rather uncertain for young stars.

##### Exoplanet properties

Determined relative to host star properties, and biases or uncertainties strongly affect habitable zone and $\eta_\bigoplus$ (Kane11)

#### Part I: Mass

Sub-mm interferometric observations of stars with circumstellar disks. Use Keplerian rotation to derive a very precise and accurate stellar mass ($< 5$%) (Rosenfeld+12)

#### Part II: The other properties

There are many ways to determine
$\{ T_{\rm eff}, \log g, [{\rm Fe}/{\rm H}] \}$

Depending on your desired precision and the nature of your target, you can use photometric colors, astroseismology, optical/IR interferometry, or spectroscopy.

By far, spectroscopic inference is the most prevalent and general purpose (e.g., spectral typing).

To leverage the high quality synthetic spectra that now exist, we develop a forward model of the observed stellar spectrum.

## fitting spectra

 Desire use as much of the spectrum as possible explore lots of parameters Problem spectral synthesis is computationally difficult Solution use pre-synthesized stellar grids $\{ T_{\rm eff}, \log g, [{\rm Fe}/{\rm H}] \}$ interpolate spectrum from the grid and post-process to make it look like the data

### The problem in a nutshell

Spectroscopic fits are different from "normal" $\chi^2$ fits

### The problem in a nutshell

Existing stellar synthesis is amazing, but not perfect.

1. low scale, mild residual structure
2. "outlier" spectral lines with the wrong strength

How can you fit ~1,000 or more lines
with only a few parameters?

a "good" fit
autocorrelation

Czekala+14

### Fitting spectra $\ne$ vanilla $\chi^2$

• the fundamental "data point" of a spectrum is a spectral line
• when the strength of a line is incorrect, it causes several correlated pixel residuals
• you are saved only if your model is perfect
• with any large swath of spectrum,
failure perfection is not an option
Apollo 13
or at least very difficult to attain in practice

## Gaussian processes

### How to cope with imperfection

the residuals are a spectrum $\textbf{R}$

the pixel "variances" are now a covariance matrix $C$

$$p(D | \theta) = \frac{1}{\sqrt{(2 \pi)^N \det(C)}} \exp\left ( -\frac{1}{2} \textbf{R}^T C^{-1} \textbf{R} \right )$$

$$\ln[p(D | \theta)] = -\frac{1}{2} \textbf{R}^T C^{-1} \textbf{R} - \frac{1}{2} \ln \det C - \frac{N}{2} \ln 2 \pi$$

the structure of $C$ is set by a kernel function analogous, say, to the two-point correlation function

Matérn $\nu = 3/2$

$k(\lambda_i, \lambda_j | a, l) = a \left (1 + \frac{\sqrt{3} r_{ij}}{l} \right ) \exp \left ( - \frac{\sqrt{3} r_{ij}}{l} \right )$

This covariance matrix accounts for the "global" correlated structure of the residuals

$\sim 5$ pixels correlated

### Spectral line outliers

A strong outlier generally looks like a Gaussian $\textbf{R}_\lambda(a, \mu, \sigma) = \frac{a}{\sqrt{2 \pi} \sigma} \exp \left ( - \frac{r^2(\lambda,\, \mu)}{2 \sigma^2} \right )$

covariance kernel is

$k(\lambda_i, \lambda_j) = \langle \textbf{R}_i \textbf{R}_j \rangle$

$k(\lambda_i, \lambda_j | a, \mu, \sigma) = \frac{1}{2 \pi} \left ( \frac{a}{\sigma} \right)^2 \exp \left ( - \frac{r^2(\lambda_i,\, \mu) + r^2(\lambda_j,\, \mu)}{2 \sigma^2} \right )$

### Stellar posteriors

(courtesy dfm/triangle.py and dfm/emcee)

• sample the stellar and covariance parameters with MH/Gibbs sampler
• posteriors naturally inflate to capture uncertainty in model
• preserves parameter degeneracies
• tested against main sequence stars (Cushing+05, Rayner+09, Torres+12)

### "Global" Hyperparameters

• covariance hyperparameters are self-consistently determined
• higher quality libraries of synthetic spectra yield lower "noise"
• each outlier line also has its own set of hyperparameters (hierarchical)

### Applications

M dwarf molecular bands are hard to model

...but, some bands tell you a lot about stellar properties

We want to include these areas but weight them appropriately

Workshop topic: custom design of covariance kernels?

(Mann+13)

## The Holy Grail

### and the quest for it

1. fit many stellar spectra ($N > 50$)
2. learn the covariance structure; the locations of the outlier lines
3. use the information in the covariance matrix and the ensemble of real stars to infer (learn) what the models should be

This is not unlike how $UBVRI$ photometric passbands
are regressed out (Bessell & Simon 2012)

Pass this along to the stellar modellers.

If you're impatient/a DIY-er, you can create your own self-calibrated semi-empirical libraries of pseudo-synthetic spectra.

### Conclusions

• new framework for unbiased inference of stellar parameters from spectra
• covariance kernels can help account for systematic model-data mismatch
• applications from FKG stars to M dwarfs
• given many stars, this framework presents a way to improve the stellar models while you fit
• code will be released with the upcoming paper

### Outlier line hyperparameters

• covariance hyperparameters are sampled self-consistently
• better models have fewer outliers
• $h$ is a dummy parameter at this point


text:
# What are the important points to graph y=3cos x?

$0$, $\frac{\pi}{2}$, $\pi$, $3 \frac{\pi}{2}$, $2 \pi$ Radians


text:
OpenStudy (anonymous):

Anyone here to help me:S ten dipoles, each consisting of two charges +- 19 uC, are located within a spherical surface, the radius of which is 2 mm. What is the flux density through the sphere i got 4.1*10^6 N-m^2/C ???????????

4 years ago
OpenStudy (anonymous):

flux is zero man!!!

4 years ago
OpenStudy (anonymous):

since there are dipoles inside......net charge inside is zero! apply gauss law......flux is zero!

4 years ago
OpenStudy (anonymous):

getting?

4 years ago
OpenStudy (anonymous):

no:S

4 years ago
OpenStudy (anonymous):

ok....do u know gauss law for electrostatics?

4 years ago
OpenStudy (anonymous):

yes

4 years ago
OpenStudy (anonymous):

gauss law states that net flux through a closed surface is NET CHARGE ENCLOSED by that closed surface...divided by epsilon nought......isnt it right?

4 years ago
OpenStudy (anonymous):

so...u have dipoles here.....dipole means 2 charges , 1+ve,1-ve....so net charge of a dipole is zero......so even if u have 100 dipoles..net charge wud be zerro...so since there are only dipoles inside ur closed surface....net charge is zero..

4 years ago
OpenStudy (anonymous):

since net charge enclosed by the closed surface is zero.....flux is zero by the gauss law....get it??????

4 years ago
OpenStudy (anonymous):

yes thanks

4 years ago


text:
# Hydrophilic vs water soluble?

Is the term 'hydrophilic' a synonym of 'water soluble' or can you have a molecule that is water insoluble but hydrophilic (or water soluble and not hydrophilic)?

## migrated from biology.stackexchange.comFeb 3 '16 at 15:49

This question came from our site for biology researchers, academics, and students.

• Interesting question, but I think this is more on-topic in chemistry. There are substances as cellulose which are hydrophilic but not water-soluble. – Chris Feb 3 '16 at 9:16
• Possible duplicate of biology.stackexchange.com/q/23763/3340 – WYSIWYG Feb 3 '16 at 9:55
• Consider this example: glass is hydrophilic but not soluble. – Zhe May 22 '17 at 22:22

Hydrophilic substances are not necessarily water soluble, and the two terms are not synonymous.

On a molecular scale, "hydrophilic" is defined by the IUPAC Gold Book$^1$ as:

'Water loving'. The capacity of a molecular entity or of a substituent to interact with polar solvents, in particular with water, or with other polar groups.

Note that the term hydrophilic does not necessarily refer to an entire molecule, and that interacting strongly with a molecule or a part of a molecule does not imply that it must be soluble in water.

Also, entire macroscopic surfaces can be said to be hydrophilic, meaning that they interact strongly with water but certainly are not soluble in water. In this context, the degree of hydrophilicity is measured by the contact angle, which can be simply stated as the angle formed when a droplet of water is placed on a surface, where a small angle (as seen from inside the droplet) means that the droplet has spread out over the surface, and that the surface is therefore hydrophilic.

So, although small, hydrophilic molecules are frequently water soluble, solubility plays no part in the definition of "hydrophilic" on any scale or context.

1) IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). XML on-line corrected version: http://goldbook.iupac.org (2006-) created by M. Nic, J. Jirat, B. Kosata; updates compiled by A. Jenkins. ISBN 0-9678550-9-8

Yes, they mean the same thing actually. Going through any one of the several online sources available, you can clearly understand this: A Hydrophilic molecule is one in which interaction with water is thermodynamically more favorable than interaction with oil or other hydrophobic solvents. Such substances are generally capable of forming hydrogen bonding with water, which makes them soluble in water. One can in fact say that these types of molecules are soluble in all polar solvents, like alcohol and ammonia.

• Solubility depends on polarity but that doesn't mean that all polar substances will dissolve in water or any other polar solvent. – WYSIWYG Feb 3 '16 at 18:54
• @WYSIWYG That is the general rule: like dissolves like. Sure, there might be some exceptions. – ShankRam Feb 4 '16 at 1:55

Seems to me that a drop of oil hitting water beads up, showing no tendency to increase its area of contact with the water. On the other hand, a drop of gasoline, e.g. Heptane or Octane, will immediately spread over the water surface to such a degree that the actual thickness of the film will be one molecule. Such a tendency to maximize the area of contact sounds to me like hydrophilic while not soluble.


text:
• March 30th 2006, 08:04 AM
mazdatuner
A question I have to answer is from the given equasion
2x^2 - 3x - 14
----------------- =F(x)
x^2 - 2x -8

I need to find the Vertical asymtope
Horozontal asymtope, slant asymtope , any holes and the x and y intercepts.

So far I have divided and come up with x^2 - x - 24
I am not sure If I am correct in doing that or even close any help would be greatly apreciated
• March 30th 2006, 09:34 AM
topsquark
Quote:

Originally Posted by mazdatuner
A question I have to answer is from the given equasion
2x^2 - 3x - 14
----------------- =F(x)
x^2 - 2x -8

I need to find the Vertical asymtope
Horozontal asymtope, slant asymtope , any holes and the x and y intercepts.

So far I have divided and come up with x^2 - x - 24
I am not sure If I am correct in doing that or even close any help would be greatly apreciated

Umm...no, that's not what you get when you divide them.

Horizontal asymptotes:
You want to take $\lim_{x \to \infty}\frac{2x^2-3x-14}{x^2-2x-8}$
Note that as x blows up, the most important terms are the x^2. So:
$\lim_{x \to \infty}\frac{2x^2-3x-14}{x^2-2x-8} \approx \frac{2x^2}{x^2} = 2$
(Or you can just plug a large number, say x=10000, and see what you get.)
So we have a horizontal asymptote at y=2 on the +x side.

Similarly:
$\lim_{x \to -\infty}\frac{2x^2-3x-14}{x^2-2x-8} \approx \frac{2x^2}{x^2} = 2$
(Squaring takes away the "-" sign, so the expressions are the same.)
So we have the same horizontal asymptote on both sides of the y-axis: y=2.

Vertical asymptotes:
It's a good idea to factor your expression here before we figure these. I'll show you why in a minute.
$F(x)=\frac{2x^2-3x-14}{x^2-2x-8}=\frac{(2x-7)(x+2)}{(x-4)(x+2)}$

Generally you find vertical asymptotes where the denominator of the expression goes to zero. This means that x = 4 is a vertical asymptote. Normally x = -2 would be another, but the x+2 in the denominator is cancelled by the x+2 in the numerator. That's why you ALWAYS want to factor these things.

Slant asymptotes:
You get a slant asymptote only when the degree of the numerator is one more than the degree of the denominator. In this case they are both quadratics, so we don't have any slant asymptotes.

Holes:
Not sure what you mean by this. Perhaps we are speaking of x = -2? This is a point in the domain that is not allowed.

x intercepts:
This is where F(x) is zero. This will happen when the numerator goes to zero (assuming that none of these x values are where we have a zero in the denominator!) The only non-cancelling term in the numerator is the 2x-7, so if we set that to zero we find an x intercept at the point (7/2,0).

y intercepts:
These are easy. Just set x = 0 in your function:
$F(0)=\frac{2*0^2-3*0-14}{0^2-2*0-8}=\frac{-14}{-8}=\frac{7}{4}$
So we have a y intercept at (0, 7/4).

-Dan


text:
# Is the infimum of a continuous function reached?

I am stuck on this problem for a while now, any help would be appreciated. I am working on the proof of a Network Calculus theorem, and I would like to show that the infimum is reached in the horizontal deviation of two functions. I simplified the problem, and left only what is essential I think :

Let $f$ : $\mathbb{R}^+ \rightarrow \mathbb{R}^+ \:;\: a\in\mathbb{R}^+ \:;\: t\in\mathbb{R}^+ \:;\: d\in\mathbb{R}^+$. Here are a few assumptions that I have :

• $f$ is non-decreasing: $x\leq y \rightarrow f(x) \leq f(y)$.
• $f(0)=0$.
• $\inf \{\tau: \tau\ge0 \textrm{ and } a \leq f(t+\tau)\} \leq d$.
• $\exists \theta\geq 0,\ a \leq f(t+\theta)$.
• Also, I am pretty sure that we need : $f$ is continuous.

Let $A = \{\tau: \tau\ge0 \textrm{ and } a \leq f(t+\tau)\}$

I want to prove that: $\inf A \in A$.

Thank you

I did some work and came up with the following. But there are still some gaps left.... $$0 \le \inf A \textrm{ and } a = f(t+ \inf A) \longrightarrow \inf A \in A$$

Proving $0 \leq \inf A$ is trivial, since all elements of A are positive.

Next, let us show that $a = f(t+ \inf A)$ by contradiction:

Let us assume that $$a \neq f(t+ \inf A) \, ,$$

hence: $$0 < |f(t + \inf A) - a| \, ,$$

with the density of reals we can obtain an $ε$ where $$0 < ε \text{ and } ε < |f(t + \inf A) - a| \, . \qquad (1)$$

Moreover, $$a \in \mathbb{R}^+ \text{ and } f(0) = 0 \longrightarrow f(0) \leq a$$

With this and the assumption about the existence of $\theta$, we can use the theorem of intermediate values to obtain $c$ where $$0 \leq c \text{ and }c \leq t + \theta \text{ and } f(c) = a \, .$$

With (1), we have: $$ε < |f(t + \inf A) - f(c)| \, . \qquad (2)$$

And from the continuity of f, we can obtain an $η$ where : $$∀x. |(t + \inf A) - x| < η ⟶ |f (t + \inf A) - f(x)| < ε$$

Now, I am not sure if euclidean division can be used with real, so I used this construction which is an equivalent of the reminder of an euclidean division: $$x_η = c - \lfloor {c - (t+\inf A) \over η} \rfloor \times η \, ,$$

We then obtain: $$t + \inf A \leq x_η\, \qquad (3.1)$$ $$x_η \leq c \, \qquad (3.2)$$ $$|(t + \inf A) - x_η| < η \, \qquad (3.3).$$

With the definition of $η$, we have: $$|f (t + \inf A) - f(x_η)| < ε$$

I need to prove $f(x_η) = f(c)$, which would allow us to conclude, with (2).

From 3.2 and f incresing, we easily obtain $f(x_η) \leq f(c)$.

But from 3.1 and f increasing, all I can get, is that $a \leq f(t + x_η)$. I need to prove that $$t + \inf A = \inf \{\tau: t \leq \tau \text{ and } a \leq f(\tau)\}$$ But I have no clue on how to prove this...

-
Is $R'$ supposed to be the derivative of $R$, or is it supposed to be just another function? – Thomas Andrews Dec 5 '12 at 14:31
What exactly does $\inf_{0\le \tau} \{\tau.R(t) \le R'(t + \tau)\}$ mean? The infimum of the set of all $\tau$ with $\tau \ge 0$ and $R(t) \le R'(t+ \tau)$, I suppose? – martini Dec 5 '12 at 14:35
To Thomas : they are two different functions, To martini: yes, i edited it, is the notation better now ? – E. Mabille Dec 5 '12 at 14:48
I think "true propositions" is incorrect mathematical language here because a proposition is something you test to see if it is true. I think you mean "assumptions" (or something similar). – Adam Rubinson Dec 5 '12 at 15:05
@E.Mabille: I have changed the wedge symbol ($\wedge$) to the word "and", as that symbol is also overloaded with the meaning of minimum (i.e. $a\wedge b=\min(a,b)$). I have also replaced the double infimum ($\inf\inf$) in the statement you want to show to single infimum ($\inf$). I hope I didn't misunderstand your question. – user1551 Dec 5 '12 at 17:23

All you need to understand is that, $f$ being order-preserving and continuous (and actually you can replace continuity by upper semicontinuity), one has $\inf_{x \in X} f(x) = f(\inf_{x \in X})$ for every $X \subset \mathbb{R}_+$. Then $\inf_{\tau \in A} f(t + \tau) = f(\inf_{\tau \in A} (t + \tau)) = f(t + \inf_{\tau \in A} \tau) = f(t + \inf A)$, which implies $\inf A \in A$.


text:
SMS scnews item created by Alex Casella at Tue 18 Apr 2017 1741
Type: Seminar
Distribution: World
Expiry: 18 Jul 2017
Calendar1: 24 Apr 2017 1700-1800
CalLoc1: Carslaw 535A
CalTitle1: MaPSS: Mathematical Postgraduate Seminar Series
Auth: casella@10.17.25.244 (acas5565) in SMS-WASM

# MaPSS: Mathematical Postgraduate Seminar Series: Alexander Kerschl (Sydney University) -- Solving polynomial equations with radicals or why there are no general solutions for polynomial equations of degree 5 and higher

Dear All,

We are delighted to present the MaPSS Seminar topic of Monday 24/04; please see the
abstract below.

**This Semester the Seminar will always run on Monday, at 5:00pm in 535A**

Following the talk, there will be pizza on offer.

Speaker: Alexander Kerschl (Sydney University)

Title: Solving polynomial equations with radicals or why there are no general solutions
for polynomial equations of degree 5 and higher

Abstract: The history of solving polynomial equations dates back to about 2000 BC for
equations.  Throughout the centuries people tried to formulize and solve these equations
in general.  Finally, in Italy during the 16th century scholars discovered the general
solutions for cubic and quartic equations but the general quintic could not be solved.
Nowadays we know that there is no solution using radicals for the general quintic and
higher degree polynomial equations but historically it took until the early 19th century
to give a proof for this fact.  In 1799 Ruffini and GauÃŸ were the first to formulate
that there is no general solution for degree 5 and higher.  Following them Cauchy,
Wantzel, and especially Abel worked to help to finish Ruffini’s first draft of a proof
and led to the famous Abel-Ruffini Theorem in 1824.  Independently and without knowing
about Abel’s proof a young Frenchman named Ã‰variste Galois laid the groundwork of what
is known today as Galois theory.  Galois gave us a beautiful general approach to deal
with solvability of polynomial equations of any kind and, moreover, his work led to
solve two of the three classical problems of ancient mathematics.  Unfortunately, he
died way to young at the age of 20 after being severly injured in a duel.  My talk will
aim to lead throughout the centuries of the quest to solve polynomial equations and
explain why there can’t be a solution for the general quintic.


text: Practice – Non Right-Angled Triangle Trigonometry 117 June 12, 2020 1. 2. At the corner, a park is being built in the shape of a triangle. How did we get an acute angle, and how do we find the measurement of$\,\beta ?\,$Let’s investigate further. Assuming that the street is level, estimate the height of the building to the nearest foot. Find the distance of the plane from point$\,A\,$to the nearest tenth of a kilometer. In order to estimate the height of a building, two students stand at a certain distance from the building at street level. A triangle with two given sides and a non-included angle. To determine how far a boat is from shore, two radar stations 500 feet apart find the angles out to the boat, as shown in (Figure). As the GCSE mathematics curriculum increasingly challenges students to solve multiple step problems it is important for students to understand how to prove, apply and link together the various formulae associated to non-righ… There are three possible cases: ASA, AAS, SSA. The cosine rule can be used to find a missing side when all sides and an angle are involved in the question. Find$\,AD\,$in (Figure). A yield sign measures 30 inches on all three sides. Round each answer to the nearest tenth. There are three possible cases: ASA, AAS, SSA. To do so, we need to start with at least three of these values, including at least one of the sides. A pilot is flying over a straight highway. Calculate the area of the triangle ABC. A pole leans away from the sun at an angle of$\,7°\,$to the vertical, as shown in (Figure). We will investigate three possible oblique triangle problem situations: Knowing how to approach each of these situations enables us to solve oblique triangles without having to drop a perpendicular to form two right triangles. The Law of Sines can be used to solve oblique triangles, which are non-right triangles. Trigonometry in Non-Right Angled Triangles Area of a Triangle You need to know 2 sides and the angle in between. The Law of Sines can be used to solve oblique triangles, which are non-right triangles. Roll over or tap the triangle to see what that means … Students learn how to derive the Sine, Cosine and Area formulae for non-right-angled triangles. It may also be used to find a missing angle if all the sides of a non-right angled triangle are known. These formulae represent the cosine rule. Read more. Since a must be positive, the value of c in the original question is 4.54 cm. How long is the pole? See, The general area formula for triangles translates to oblique triangles by first finding the appropriate height value. Solve applied problems using the Law of Sines. This formula represents the sine rule. (Hint: Draw a perpendicular from$\,N\,$to$\,LM).\,$Round each answer to the nearest tenth. (Figure) shows a satellite orbiting Earth. This is a good indicator to use the sine rule in a question rather than the cosine rule. In this case, we know the angle$\,\gamma =85°,\,$and its corresponding side$\,c=12,\,$and we know side$\,b=9.\,$We will use this proportion to solve for$\,\beta .$. Note that the angle of elevation is the angle up from the ground; for example, if you look up at something, this angle is the angle between the ground and your line of site.. Round to the nearest tenth. See, The Law of Sines can be used to solve triangles with given criteria. 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The inverse sine will produce a single result, but keep in mind that there may be two values for$\,\beta .\,$It is important to verify the result, as there may be two viable solutions, only one solution (the usual case), or no solutions. Find the diameter of the circle in (Figure). Solve both triangles in (Figure). A street light is mounted on a pole. This angle is opposite the side of length 20, allowing us to set up a Law of Sines relationship. Observing the two triangles in (Figure), one acute and one obtuse, we can drop a perpendicular to represent the height and then apply the trigonometric property$\,\mathrm{sin}\,\alpha =\frac{\text{opposite}}{\text{hypotenuse}}\,$to write an equation for area in oblique triangles. The angle of elevation from the tip of the man’s shadow to the top of his head of 28°. Use the Law of Sines to find angle$\,\beta \,$and angle$\,\gamma ,\,$and then side$\,c.\,$Solving for$\,\beta ,\,$we have the proportion. We see in (Figure) that the triangle formed by the aircraft and the two stations is not a right triangle, so we cannot use what we know about right triangles. Points$\,A\,$and$\,B\,$are on opposite sides of a lake. The angle of inclination of the hill is$\,67°.\,$A guy wire is to be attached to the top of the tower and to the ground, 165 meters downhill from the base of the tower. If the man and woman are 20 feet apart, how far is the street light from the tip of the shadow of each person? Students tend to memorise the bottom one as it is the one that looks most like Pythagoras. Given$\,\alpha =80°,a=100,\,\,b=10,\,$find the missing side and angles. Triangles classified as SSA, those in which we know the lengths of two sides and the measurement of the angle opposite one of the given sides, may result in one or two solutions, or even no solution. When the known values are the side opposite the missing angle and another side and its opposite angle. To find the elevation of the aircraft, we first find the distance from one station to the aircraft, such as the side$\,a,$ and then use right triangle relationships to find the height of the aircraft,$\,h.$. Again, it is not necessary to memorise them all – one will suffice (see Example 2 for relabelling). Trigonometry The three trigonometric ratios; sine, cosine and tangent are used to calculate angles and lengths in right-angled triangles. The interior angles of a triangle always add up to 180° while the exterior angles of a triangle are equal to the sum of the two interior angles that are not adjacent to it. In this case, if we subtract$\,\beta \,$from 180°, we find that there may be a second possible solution. Solve the triangle shown in (Figure) to the nearest tenth. Using the sine and cosine rules in non right angled triangles to find the missing sides and angles, and a brief look at the ambiguity in the Sine rule. To find the area of this triangle, we require one of the angles. Find$\,AB\,$in the parallelogram shown in (Figure). Read about Non-right Triangle Trigonometry (Trigonometry Reference) in our free Electronics Textbook According to the Law of Sines, the ratio of the measurement of one of the angles to the length of its opposite side equals the other two ratios of angle measure to opposite side. Non-right angled triangles - cosine and sine rule - StudyWell For right-angled triangles, we have Pythagoras’ Theorem and SOHCAHTOA. The rule also stands if you write the entire thing the other way up. To find the remaining missing values, we calculate$\,\alpha =180°-85°-48.3°\approx 46.7°.\,$Now, only side$\,a\,$is needed. [/latex], The formula for the area of an oblique triangle is given by. Using trigonometry: tan=35=tan−135=30.96° Labelling Sides of Non-Right Angle Triangles. Solving for a side in right … Entering sides of values 1.00, 2.00, and 2.00 will yield much more acurate results of 75.5, 75.5, and 29.0. Trigonometry – Non-Right-Angled Triangles Lessons Powerpoint comes with two assessments, a homework and revision questions. Round the answer to the nearest tenth. They then move 300 feet closer to the building and find the angle of elevation to be 50°. For the following exercises, find the area of the triangle with the given measurements. Two search teams spot a stranded climber on a mountain. If there is more than one possible solution, show both. Are you ready to test your Pure Maths knowledge? Non - Right angled Trigonometry. Here are some types of word problems (applications) that you might see when studying right angle trigonometry.. In order to estimate the height of a building, two students stand at a certain distance from the building at street level. We can stop here without finding the value of$\,\alpha .\,$Because the range of the sine function is$\,\left[-1,1\right],\,$it is impossible for the sine value to be 1.915. 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The Greeks focused on the calculation of chords, while mathematicians in India … A man and a woman standing$\,3\frac{1}{2}\,$miles apart spot a hot air balloon at the same time. Round each answer to the nearest tenth. In fact, inputting$\,{\mathrm{sin}}^{-1}\left(1.915\right)\,$in a graphing calculator generates an ERROR DOMAIN. There are three possible cases that arise from SSA arrangement—a single solution, two possible solutions, and no solution. The Bermuda triangle is a region of the Atlantic Ocean that connects Bermuda, Florida, and Puerto Rico. 1. PRO Features : 1) View calculation steps 2) View formulas 3) No ads • Giving solution based on your input. Access these online resources for additional instruction and practice with trigonometric applications. Solving for a side in a right triangle using the trigonometric ratios. Brian’s house is on a corner lot. This is equivalent to one-half of the product of two sides and the sine of their included angle. 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Similar to an angle of elevation, an angle of depression is the acute angle formed by a horizontal line and an observer’s line of sight to an object below the horizontal. Measurement involves the application of knowledge, skills and understanding of numbers and geometry to quantify and solve problems in practical situations. Use the Law of Sines to solve for$\,a\,$by one of the proportions. When the elevation of the sun is$\,55°,\,$the pole casts a shadow 42 feet long on the level ground. Note that it is not necessary to memorise all of them – one will suffice, since a relabelling of the angles and sides will give you the others. Example:- Calculate the area of this triangle. about$\,8.2\,\,\text{square}\,\text{feet}$. This is different to the cosine rule since two angles are involved. Round each answer to the nearest tenth. Trigonometry: Non Right-Angled Triangles MichaelExamSolutionsKid 2020-03-11T23:34:40+00:00 Trigonometry Non Right Angled Triangles When finding the area of a segment you will often need to find the area of a triangle given two sides and an included angle when the angle is given in degrees or radians. In choosing the pair of ratios from the Law of Sines to use, look at the information given. In any triangle, we can draw an altitude, a perpendicular line from one vertex to the opposite side, forming two right triangles. What type of triangle results in an ambiguous case? The angle of elevation from the second search team to the climber is 22°. Sketch the two possibilities for this triangle and find the two possible values of the angle at Y to 2 decimal places. It follows that the area is given by. The distance from the satellite to station$\,A\,$is approximately 1716 miles. ), it is very obvious that most triangles that could be constructed for navigational or surveying reasons would not contain a right angle. Naomi bought a modern dining table whose top is in the shape of a triangle. If we rounded earlier and used 4.699 in the calculations, the final result would have been x=26.545 to 3 decimal places and this is incorrect. The angle of elevation measured by the first station is 35 degrees, whereas the angle of elevation measured by the second station is 15 degrees. Dropping a perpendicular from$\,\gamma \,$and viewing the triangle from a right angle perspective, we have (Figure). Find the radius of the circle in (Figure). The most important thing is that the base and height are at right angles. This formula works for a right triangle as well, since the since of 90 is one. However, these methods do not work for non-right angled triangles. Designed to solve triangle trigonometry problem with well explanation. A communications tower is located at the top of a steep hill, as shown in (Figure). See. Find the area of the Bermuda triangle if the distance from Florida to Bermuda is 1030 miles, the distance from Puerto Rico to Bermuda is 980 miles, and the angle created by the two distances is 62°. The sides of a triangle are in arithmetic sequence and the greatest angle is double the smallest angle. The roof of a house is at a$\,20°\,$angle. Assuming that the street is level, estimate the height of the building to the nearest foot. Find$\,m\angle ADC\,$in (Figure). A 6-foot-tall man is standing on the street a short distance from the pole, casting a shadow. The diagram shown in (Figure) represents the height of a blimp flying over a football stadium. The angle used in calculation is$\,{\alpha }^{\prime },\,$or$\,180-\alpha . The angle of elevation from the tip of her shadow to the top of her head is 28°. We will work on three key rules. They can often be solved by first drawing a diagram of the given information and then using the appropriate equation. To solve an oblique triangle, use any pair of applicable ratios. MS-M6 Non-right-angled trigonometry. Trigonometry and Non-Right-Angled Triangles. Note that when using the sine rule, it is sometimes possible to get two answers for a given angle\side length, both of which are valid.$, $A\approx 47.8°\,$or$\,{A}^{\prime }\approx 132.2°$, Find angle$\,B\,$when$\,A=12°,a=2,b=9.$. Knowing Base and Height. Trigonometry: Right and Non-Right Triangles Area of a Triangle Using Sine We can use sine to determine the area of non-right triangles. Our mission is to provide a free, world-class education to anyone, anywhere. 3. Answering the question given amounts to finding side a in this new triangle. Round to the nearest tenth. The sine and cosine rules calculate lengths and angles … Need to know one pair (angle and side) plus Round each answer to the nearest tenth. With this, we can utilize the Law of Cosines to find the missing side of the obtuse triangle—the distance of the boat to the port. In the acute triangle, we have$\,\mathrm{sin}\,\alpha =\frac{h}{c}\,$or$c\mathrm{sin}\,\alpha =h.\,$However, in the obtuse triangle, we drop the perpendicular outside the triangle and extend the base$\,b\,$to form a right triangle. (Hint: Draw a perpendicular from$\,H\,$to$\,JK).\,$Round each answer to the nearest tenth. Here we take trigonometry to the next level by working with triangles that do not have a right angle. The angle of depression is the angle that comes down from a … When the satellite is on one side of the two stations, the angles of elevation at$\,A\,$and$\,B\,$are measured to be$\,86.2°\,$and$\,83.9°,\,$respectively. See (Figure). In the triangle shown in (Figure), solve for the unknown side and angles. Round to the nearest tenth of a mile. $\,\angle m\,$is obtuse. See. Use the Law of Sines to solve oblique triangles. For non-right angled triangles, we have the cosine rule, the sine rule and a new expression for finding area. Compare right triangles and oblique triangles. The Law of Sines is based on proportions and is presented symbolically two ways. The angle supplementary to$\,\beta \,$is approximately equal to 49.9°, which means that$\,\beta =180°-49.9°=130.1°.\,$(Remember that the sine function is positive in both the first and second quadrants.) © Copyright of StudyWell Publications Ltd. 2020. Similarly, to solve for$\,b,\,$we set up another proportion. Visit our Practice Papers page and take StudyWell’s own Pure Maths tests. The sine rule can be used to find a missing angle or a missing side when two corresponding pairs of angles and sides are involved in the question. The ambiguous case arises when an oblique triangle can have different outcomes. Therefore, no triangles can be drawn with the provided dimensions. The more we study trigonometric applications, the more we discover that the applications are countless. However, in the diagram, angle$\,\beta \,$appears to be an obtuse angle and may be greater than 90°. You can round when jotting down working but you should retain accuracy throughout calculations. 4.3 4 customer reviews. If there is more than one possible solution, show both. They then move 250 feet closer to the building and find the angle of elevation to be 53°. Round each answer to the nearest tenth. Right-Angled Triangles: h Non-Right-Angled Triangles: Note that to maintain accuracy, store values on your calculator and leave rounding until the end of the question. In the Law of Sines, what is the relationship between the angle in the numerator and the side in the denominator? Note the standard way of labeling triangles: angle$\,\alpha \,$(alpha) is opposite side$\,a;\,$angle$\,\beta \,$(beta) is opposite side$\,b;\,$and angle$\,\gamma \,$(gamma) is opposite side$\,c.\,$See (Figure). As is the case with the sine rule and the cosine rule, the sides and angles are not fixed. Round the distance to the nearest tenth of a foot. They’re really not significantly different, though the derivation of the formula for a non-right triangle is a little different. It is not possible for a triangle to have more than one vertex with internal angle greater than or equal to 90°, or it would no longer be a triangle. Preview and details Files included (6) pdf, 136 KB. There are three possible cases: ASA, AAS, SSA. Round the altitude to the nearest tenth of a mile. The three trigonometric ratios; sine, cosine and tangent are used to calculate angles and lengths in right-angled triangles. The sine rule will give us the two possibilities for the angle at Z, this time using the second equation for the sine rule above: Solving gives or . The Sine rule, the Cosine rule and the formula for th area of a triangle. For oblique triangles, we must find$\,h\,$before we can use the area formula. Round answers to the nearest whole mile. The distance from one station to the aircraft is about 14.98 miles. It's a PRO app and easy to use with eye-catching User Interface. However, these methods do not work for non-right angled triangles. Find the length of the side marked x in the following triangle: The triangle PQR has sides PQ=6.5cm, QR=9.7cm and PR = c cm. Round your answers to the nearest tenth. Because the angles in the triangle add up to 180 degrees, the unknown angle must be 180°−15°−35°=130°. Area of Triangles. We can use the Law of Sines to solve any oblique triangle, but some solutions may not be straightforward. For the following exercises, assume$\,\alpha \,$is opposite side$\,a,\beta \,$is opposite side$\,b,\,$and$\,\gamma \,$is opposite side$\,c.\,$Solve each triangle, if possible. \hfill \\ \text{ }\,\frac{\mathrm{sin}\,\alpha }{a}=\frac{\mathrm{sin}\,\beta }{b}\hfill & \hfill \end{array}[/latex], $\frac{\mathrm{sin}\,\alpha }{a}=\frac{\mathrm{sin}\,\gamma }{c}\text{ and }\frac{\mathrm{sin}\,\beta }{b}=\frac{\mathrm{sin}\,\gamma }{c}$, $\frac{\mathrm{sin}\,\alpha }{a}=\frac{\mathrm{sin}\,\beta }{b}=\frac{\mathrm{sin}\,\lambda }{c}$, $\frac{\mathrm{sin}\,\alpha }{a}=\frac{\mathrm{sin}\,\beta }{b}=\frac{\mathrm{sin}\,\gamma }{c}$, $\frac{a}{\mathrm{sin}\,\alpha }=\frac{b}{\mathrm{sin}\,\beta }=\frac{c}{\mathrm{sin}\,\gamma }$, $\begin{array}{l}\begin{array}{l}\hfill \\ \beta =180°-50°-30°\hfill \end{array}\hfill \\ \,\,\,\,=100°\hfill \end{array}$, $\begin{array}{llllll}\,\,\frac{\mathrm{sin}\left(50°\right)}{10}=\frac{\mathrm{sin}\left(30°\right)}{c}\hfill & \hfill & \hfill & \hfill & \hfill & \hfill \\ c\frac{\mathrm{sin}\left(50°\right)}{10}=\mathrm{sin}\left(30°\right)\hfill & \hfill & \hfill & \hfill & \hfill & \text{Multiply both sides by }c.\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c=\mathrm{sin}\left(30°\right)\frac{10}{\mathrm{sin}\left(50°\right)}\hfill & \hfill & \hfill & \hfill & \hfill & \text{Multiply by the reciprocal to isolate }c.\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c\approx 6.5\hfill & \hfill & \hfill & \hfill & \hfill & \hfill \end{array}$, $\begin{array}{ll}\begin{array}{l}\hfill \\ \,\text{ }\frac{\mathrm{sin}\left(50°\right)}{10}=\frac{\mathrm{sin}\left(100°\right)}{b}\hfill \end{array}\hfill & \hfill \\ \text{ }b\mathrm{sin}\left(50°\right)=10\mathrm{sin}\left(100°\right)\hfill & \text{Multiply both sides by }b.\hfill \\ \text{ }b=\frac{10\mathrm{sin}\left(100°\right)}{\mathrm{sin}\left(50°\right)}\begin{array}{cccc}& & & \end{array}\hfill & \text{Multiply by the reciprocal to isolate }b.\hfill \\ \text{ }b\approx 12.9\hfill & \hfill \end{array}$, $\begin{array}{l}\begin{array}{l}\hfill \\ \alpha =50°\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a=10\hfill \end{array}\hfill \\ \beta =100°\,\,\,\,\,\,\,\,\,\,\,\,b\approx 12.9\hfill \\ \gamma =30°\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c\approx 6.5\hfill \end{array}$, $\begin{array}{r}\hfill \frac{\mathrm{sin}\,\alpha }{a}=\frac{\mathrm{sin}\,\beta }{b}\\ \hfill \frac{\mathrm{sin}\left(35°\right)}{6}=\frac{\mathrm{sin}\,\beta }{8}\\ \hfill \frac{8\mathrm{sin}\left(35°\right)}{6}=\mathrm{sin}\,\beta \,\\ \hfill 0.7648\approx \mathrm{sin}\,\beta \,\\ \hfill {\mathrm{sin}}^{-1}\left(0.7648\right)\approx 49.9°\\ \hfill \beta \approx 49.9°\end{array}$, $\gamma =180°-35°-130.1°\approx 14.9°$, ${\gamma }^{\prime }=180°-35°-49.9°\approx 95.1°$, $\begin{array}{l}\frac{c}{\mathrm{sin}\left(14.9°\right)}=\frac{6}{\mathrm{sin}\left(35°\right)}\hfill \\ \text{ }c=\frac{6\mathrm{sin}\left(14.9°\right)}{\mathrm{sin}\left(35°\right)}\approx 2.7\hfill \end{array}$, $\begin{array}{l}\frac{{c}^{\prime }}{\mathrm{sin}\left(95.1°\right)}=\frac{6}{\mathrm{sin}\left(35°\right)}\hfill \\ \text{ }{c}^{\prime }=\frac{6\mathrm{sin}\left(95.1°\right)}{\mathrm{sin}\left(35°\right)}\approx 10.4\hfill \end{array}$, $\begin{array}{ll}\alpha =80°\hfill & a=120\hfill \\ \beta \approx 83.2°\hfill & b=121\hfill \\ \gamma \approx 16.8°\hfill & c\approx 35.2\hfill \end{array}$, $\begin{array}{l}{\alpha }^{\prime }=80°\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{a}^{\prime }=120\hfill \\ {\beta }^{\prime }\approx 96.8°\,\,\,\,\,\,\,\,\,\,\,\,\,{b}^{\prime }=121\hfill \\ {\gamma }^{\prime }\approx 3.2°\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{c}^{\prime }\approx 6.8\hfill \end{array}$, $\begin{array}{ll}\,\,\,\frac{\mathrm{sin}\left(85°\right)}{12}=\frac{\mathrm{sin}\,\beta }{9}\begin{array}{cccc}& & & \end{array}\hfill & \text{Isolate the unknown}.\hfill \\ \,\frac{9\mathrm{sin}\left(85°\right)}{12}=\mathrm{sin}\,\beta \hfill & \hfill \end{array}$, $\begin{array}{l}\beta ={\mathrm{sin}}^{-1}\left(\frac{9\mathrm{sin}\left(85°\right)}{12}\right)\hfill \\ \beta \approx {\mathrm{sin}}^{-1}\left(0.7471\right)\hfill \\ \beta \approx 48.3°\hfill \end{array}$, $\alpha =180°-85°-131.7°\approx -36.7°,$, $\begin{array}{l}\begin{array}{l}\hfill \\ \begin{array}{l}\hfill \\ \frac{\mathrm{sin}\left(85°\right)}{12}=\frac{\mathrm{sin}\left(46.7°\right)}{a}\hfill \end{array}\hfill \end{array}\hfill \\ \,a\frac{\mathrm{sin}\left(85°\right)}{12}=\mathrm{sin}\left(46.7°\right)\hfill \\ \text{ }\,\,\,\,\,\,a=\frac{12\mathrm{sin}\left(46.7°\right)}{\mathrm{sin}\left(85°\right)}\approx 8.8\hfill \end{array}$, $\begin{array}{l}\begin{array}{l}\hfill \\ \alpha \approx 46.7°\text{ }a\approx 8.8\hfill \end{array}\hfill \\ \beta \approx 48.3°\text{ }b=9\hfill \\ \gamma =85°\text{ }c=12\hfill \end{array}$, $\begin{array}{l}\,\frac{\mathrm{sin}\,\alpha }{10}=\frac{\mathrm{sin}\left(50°\right)}{4}\hfill \\ \,\,\mathrm{sin}\,\alpha =\frac{10\mathrm{sin}\left(50°\right)}{4}\hfill \\ \,\,\mathrm{sin}\,\alpha \approx 1.915\hfill \end{array}$, $\text{Area}=\frac{1}{2}\left(\text{base}\right)\left(\text{height}\right)=\frac{1}{2}b\left(c\mathrm{sin}\,\alpha \right)$, $\text{Area}=\frac{1}{2}a\left(b\mathrm{sin}\,\gamma \right)=\frac{1}{2}a\left(c\mathrm{sin}\,\beta \right)$, $\begin{array}{l}\text{Area}=\frac{1}{2}bc\mathrm{sin}\,\alpha \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{2}ac\mathrm{sin}\,\beta \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{2}ab\mathrm{sin}\,\gamma \hfill \end{array}$, $\begin{array}{l}\text{Area}=\frac{1}{2}ab\mathrm{sin}\,\gamma \hfill \\ \text{Area}=\frac{1}{2}\left(90\right)\left(52\right)\mathrm{sin}\left(102°\right)\hfill \\ \text{Area}\approx 2289\,\,\text{square}\,\,\text{units}\hfill \end{array}$, $\begin{array}{l}\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \end{array}\hfill \\ \text{ }\frac{\mathrm{sin}\left(130°\right)}{20}=\frac{\mathrm{sin}\left(35°\right)}{a}\hfill \end{array}\hfill \\ a\mathrm{sin}\left(130°\right)=20\mathrm{sin}\left(35°\right)\hfill \\ \text{ }a=\frac{20\mathrm{sin}\left(35°\right)}{\mathrm{sin}\left(130°\right)}\hfill \\ \text{ }a\approx 14.98\hfill \end{array}$, $\begin{array}{l}\mathrm{sin}\left(15°\right)=\frac{\text{opposite}}{\text{hypotenuse}}\hfill \\ \mathrm{sin}\left(15°\right)=\frac{h}{a}\hfill \\ \mathrm{sin}\left(15°\right)=\frac{h}{14.98}\hfill \\ \text{ }\,\text{ }h=14.98\mathrm{sin}\left(15°\right)\hfill \\ \text{ }\,h\approx 3.88\hfill \end{array}$, http://cnx.org/contents/[email protected], $\begin{array}{l}\frac{\mathrm{sin}\,\alpha }{a}=\frac{\mathrm{sin}\,\beta }{b}=\frac{\mathrm{sin}\,\gamma }{c}\,\hfill \\ \frac{a}{\mathrm{sin}\,\alpha }=\frac{b}{\mathrm{sin}\,\beta }=\frac{c}{\mathrm{sin}\,\gamma }\hfill \end{array}$, $\begin{array}{r}\hfill \text{Area}=\frac{1}{2}bc\mathrm{sin}\,\alpha \\ \hfill \text{ }=\frac{1}{2}ac\mathrm{sin}\,\beta \\ \hfill \text{ }=\frac{1}{2}ab\mathrm{sin}\,\gamma \end{array}$. The satellite is approximately 1706 miles above the ground. [/latex], Find side$\,a$ when$\,A=132°,C=23°,b=10. The satellite passes directly over two tracking stations[latex]\,A\,$and$\,B,\,$which are 69 miles apart. In this section, we will find out how to solve problems involving non-right triangles. In triangle XYZ, length XY=6.14m, length YZ=3.8m and the angle at X is 27 degrees. The altitude extends from any vertex to the opposite side or to the line containing the opposite side at a 90° angle. (See (Figure)). Therefore, the complete set of angles and sides is, $\begin{array}{l}\alpha ={98}^{\circ }\,\,\,\,\,\,\,\,\,\,\,\,a=34.6\\ \beta ={39}^{\circ }\,\,\,\,\,\,\,\,\,\,\,\,b=22\\ \gamma ={43}^{\circ }\,\,\,\,\,\,\,\,\,\,\,\,\,\,c=23.8\end{array}$.


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### Introduction

If an ionic compound is dissolved in water, it dissociates into ions and the resulting solution will conduct electricity. Dissolving solid sodium chloride in water releases ions according to the equation:

${\text{NaCl(s)}} \to {\text{N}}{{\text{a}}^{\text{+ }}}{\text{(aq) + C}}{{\text{l}}^{\text{- }}}{\text{(aq)}}$

In this experiment, you will study the effect of increasing the concentration of an ionic compound on conductivity. Conductivity will be measured as concentration of the solution is gradually increased by the addition of concentrated NaCl drops. The same procedure will be used to investigate the effect of adding solutions with the same concentration (1.0 M), but different numbers of ions in their formulas: aluminum chloride, AlCl3, and calcium chloride, CaCl2. A Conductivity Probe will be used to measure conductivity of the solution. Conductivity is measured in microsiemens per centimeter (μS/cm).

### Objectives

In this experiment, you will

• Use a Conductivity Probe to measure the conductivity of solutions.
• Investigate the relationship between the conductivity and concentration of a solution.
• Investigate the conductivity of solutions resulting from compounds that dissociate to produce different number of ions.


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# Dispersion parameter for Gamma family

I have ran a glm in R, and near the bottom of the summary() output, it states:

(Dispersion parameter for Gamma family taken to be 1.680014)

What does this mean/represent?

• Do you mean what implications it has for your scientific question or do you concerned about what it means fr the gamma distribution? – mdewey Nov 24 '16 at 12:25

Gamma distribution defined by two parameters - shape ($\alpha$) and rate ($\beta$).
There is alternative parameterization through mean ($\mu$) and shape, which is used in GLM.
We take $\mu = \alpha/\beta$ and put it into place of rate (as $\beta = \alpha/\mu$), resulting in function $Gamma(\mu,\alpha)$.
In R GLM assumes shape to be a constant (as linear regression assumes constant variance). To satisfy this assumption dispersion ($\phi$) is introduced: $$\phi = \frac{1}{\alpha}$$
For the simple case glm(x ~ 1, family = Gamma(link = 'identity)), summary.glm gives you $\text{estimate}$, that is equal to $\mu$ (note that default link is 'inverse' and estimate = $1/\mu$) and $\text{dispersion}$ is $\phi$.
• @tatami Well, in my example $\mu$ is the mean of x. But in general effects estimates of explanatory variables would change in accordance with model. – Andrey Kolyadin Oct 23 '17 at 7:00


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## LATEX-L@LISTSERV.UNI-HEIDELBERG.DE

 Options: Use Classic View Use Monospaced Font Show Text Part by Default Show All Mail Headers Topic: [<< First] [< Prev] [Next >] [Last >>]

 Re: l3doc, was Re: xparse Lars Hellström <[log in to unmask]> Wed, 2 Sep 2009 17:18:01 +0200 text/plain (70 lines) Joseph Wright skrev: > I'd imagine that the code needs a careful overhaul at some stage, as > things are very much a mish-mash at the the moment. However, that > depends on what we want, as you say. For example, the gmdoc approach of > not needing \begin{macrocode} ... \end{macrocode} is interesting: I > wonder if it makes it easier for new users to write documented files? I've found that an advantage of having an explicit environment around blocks of codelines is that this environment can be varied according to the type of code. For single-language projects that is perhaps not so important, but for mixed-language projects it can be used to improve the code typography. Also, one can skip the expensive code indexing when typesetting for languages that doesn't have a TeX-style \ character in front of every keyword. It's another matter that doc's macrocode is perhaps not the friendliest of environments, requiring exactly four spaces between % and \end{macrocode} (might it be that this originates in a preferrence of Frank's for an environment indentation step of 3 spaces?) since it uses the same end-detection mechanism as verbatim. A perfectly workable alternative (that I've used in the tcl environment of the tclldoc package) is to mostly restore normal LaTeX catcodes after a % at the beginning of a line, so that the \end can be a normal end of environment (this also allows users to define shorter aliases of the environment name). Having looked a bit at gmdoc today, I must say the games played there with active line-ends throughout the document and special escapes for breaking command arguments over several lines seem most worrying. (Is it even possible in gmdoc to have a linebreak between two arguments of a command?) If the general direction in LaTeX3 is to have less \catcode-tricks and active characters, then it would be a strange move to rely more on them in the LaTeX3 documentation... Manuel Pégourié-Gonnard skrev: > > and I find Paul Isambert's CodeDoc very nice > > which extracts package code while typesetting the documentation. > > > Well, I don't like this very much: I prefer package code to be the primary > content Package code as primary content has the disadvantage of restricting you to one "target file" at a time. When having to make two pieces of software communicate with each other, it can be much better to interleave the code producing output with the code that then has to parse this output. > and documentation to be extracted from comments. I find it much easier > for changing the code and testing it without needing ot extract it first. (Same > problem as dtx.) For what it's worth, I have for about a decade been using an add-on to my text editor that lets me select a piece of code in a .dtx file and then press a key (well, a modifier and a key) to have the comment lines stripped out and the whole thing sent to an interpreter for evaluation; this makes testing small fixes extremely convenient, and it wasn't all that difficult to implement. I'd have to admit I don't use it for developing LaTeX code though; the reason for this being that TeX tends to operate more as a batch utility than as an interactive application, so by the time I've corrected the bug there usually isn't anything I can sent the corrected definition to anymore. On the other hand, a simple Makefile should go a long way towards automating extraction: run your test as "make mytest", and make sure the dependencies of mytest.tex on .sty and of .sty on .dtx are explicit. Lars Hellström


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# How do you write an equation in slope intercept form of the line with slope 1/2 that contains the point (-4, -3)?

Oct 10, 2015

We know that this is a straight line, so it will follow the slope intercept equation:

$y = m x + b$.

To solve this problem, we need to plug in the x and y values for one point to get b, and then we will know the equation for the line, which is valid for all points.

$x = - 4$
$y = - 3$
$m = \frac{1}{2}$

Let's start by solving for b:
$b = y - m x$

Now we can plug in numbers:
$b = - 3 - \left(\frac{1}{2} \times - 4\right)$

$b = - 3 - \left(- 2\right)$
$b = - 3 + 2$
$b = - 1$

Now that we know m and b, we can plug into our original equation:

The equation is $y = \frac{1}{2} x - 1$.


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# How to define a custom function in Scilab

Scilab is very flexible and versatile in terms of custom functions definitions. If the user is not satisfied with the already existing embedded Scilab function, he/she has the option of defining custom Scilab functions. In order to make the function reusable, the best option is to define it as a script in the SciNotes editor.

In order to create a custom function in Scilab, we need to define it following a certain structure:

Image: Scilab custom function structure

where:

function – is a required keyword which must be added to the function definition; it defines the start of the function definition;

y1, y2 – return values of the function; they have to be separated by a comma (,) and enclosed in right brackets [ ]; the function can have none, one or multiple return values;

myFunction – the name of the custom function; it needs to be different than the existing embedded Scilab functions and it should not have the same name as a Scilab script file (*.sce) in the same working directory;

x1, x2 – function arguments, input variables; they need to be separated by a comma (,) and enclosed in round brackets ( ); the function can have none, one or multiple arguments;

function content – is the sum of all the Scilab instructions that represents the definition of the function; it is also named the body of the function;

endfunction – is a required keyword which must be added to the function definition; it defines the end of the function definition;

After the function is defined, the script file is saved with the extension *.sci. This extension informs Scilab that the script file contains a function definition.

The next step is to load the function in the Scilab environment. To do this we need to use the exec() function. If the function is saved in the working directory, the argument of the exec() function is the name of the custom function file as a string:

exec('myFunction.sci')

After the function is loaded, it can be called as any embedded Scilab function.

As an example, let’s define the following mathematical function is Scilab:

$f(x,y)=\left\{\begin{matrix} \frac{x^2-y^2}{x+y}, x+y>0, y>0\\ \frac{x^2+y^2}{x+y}, x+y>0, y \le 0\\ \frac{x-y}{x+y}, x+y<0, y \in \mathbb{R} \end{matrix}\right.$

Step 1. Open a new file in SciNotes and define the function.

We can observe that the function has only one return variable and two arguments. Also, according to the above definition, the function f is not defined for x = y = 0. One way of implementing it is by using if-elseif conditional statements:

function [z] = f(x,y)
if ((x+y>0) & (y>0)) then
z = (x^2-y^2)/(x+y);
elseif ((x+y>0) & (y<=0)) then
z = (x^2+y^2)/(x+y);
elseif (x+y<0) then
z = (x-y)/(x+y);
else
z = %nan;
end
endfunction

Step 2. Save the file as f.sci in the working Scilab directory.

Step 3. Load the function in the Scilab environment:

-->exec('f.sci')

After running the instruction, all the content of the function will be executed in the Scilab console.

Step 4. Use the function in the same way as an embedded Scilab function:

-->k = f(2,3)
k =

- 1.

-->

Notice that the return of the function is associated with the variable k,which has nothing to do with the z, x and y variables from the function definition. Bear in mind that the function arguments (x, y) and the return variable (z) are only visible locally in the function, they can not be accessed from outside the function.

Once the f function is loaded in the Scilab environment, we can use it inside other Scilab scripts. Let us create a script which evaluates the f function for x and y between -5 and 5. To make the exercise more interesting, we are going to use the mprintf() function to display the result in a matrix form:

for i=-2:2
for j=-2:2
mprintf("%1.3f\t",f(i,j));
end
mprintf("\n");
end

Running the above script will return in the Scilab console the following result:

-0.000 0.333  1.000 3.000  Nan
-0.333 -0.000 1.000 Nan    -3.000
-1.000 -1.000 Nan   -1.000 -2.000
-3.000 Nan    1.000 0.000  -1.000
Nan    5.000  2.000 1.000  0.000

If we want to delete the function from the Scilab environment, we can use the clear() function:

-->clear f

We can also define custom Scilab function without any arguments or return variables. For example, we are going to define a function which displays in the Scilab console the current day, date and hour. The content of the function will be saved in a *.sci file:

function dayDateTime()
today = datenum();
[dayNumber,dayString] = weekday(today,'long');
mprintf("\nToday is %s, the %dth day of the week.\n", dayString, dayNumber);
mprintf("The date is: %s\n", date());
t = clock();
mprintf("The time is: %d:%d:%d\n",t(4), t(5), t(6));
endfunction

Save the file as dayDateTime.sci and load it in the Scilab environment:

-->exec('dayDateTime.sci')

Call the function and check the result:

-->dayDateTime()

Today is Friday, the 6th day of the week.
The date is: 13-Jan-2017
The time is: 0:57:29

-->

This tutorial should give the reader a good, solid overview regarding the definition of custom functions in Scilab. Try the above examples and ask for more details, if needed, using the comment form below.

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 Question from Kenneth: Hello: An investor wants to divide $1000.00 into four amounts or accounts -- one account earning 2%, the second account earning 5%, the third earning 9%, and the fourth earning 12%. He wants the overall total return to equal 5%. What are some of the monetary amounts that can be invested into these four accounts to earn him the 5% total return, 5% of$1,000.00? ------------------------------- My question is as follows: Can the below calculation be reversed engineered into a more simple or basic arithmetic type calculation--something less algebraic? I thank you for your reply. ------------------------------- Use the variables w, x, y, z to represent the amount of money in each of the accounts (w is amount in 2%, x in 5%, etc.). (1) w + x + y + z = 1000 (2) 0.02w + .05x + .09y + .12z = .05(1000) There are only two equations and four unknowns, there will be lots (in fact, infinitely many) possible solutions. Lets see why. Using (1) solve for w: w = 1000 - x - y - z Plug this into 2.: .02(1000 - x - y - z) + .05x + .09y + .12z = 50 20 - .02x - .02y - .02z + .05x + .09y + .12 z = 50 Collect all like terms: .03x + .07y + .1z = 30 (3) Any combination of x, y, z that satisfy this equation (3) will give you the desired return of 5%. Remember, take the values that you obtain and use them to reconstruct w = 1000 - x - y - z. An example of how to do this would be to try any old (reasonable) value for x and y, and solve for z: Say x = 100, y = 200: .03(100) + .07(200) + .1z = 30 3 + 14 + .1z = 30 .1z = 17 z = 170 w = 1000 - (100 + 200 + 170) = 530

Hi Kenneth,

My first reaction to this problem was to suggest investing the entire $1000 in the second account that earns 5% as required. What if you want to diversify some and invest in two accounts. One must earn more than 5% and the other less than 5% (why?) and hence you have two choices, 9% and 2% or 12% and 2%. Set up the algebra in each of these two situations as you did above, and solve. In each case you get one solution. Before you solve you might want to make a guess as to how much you invest at each rate and see how close you come to making 5%. What about three accounts. There are 4 possible choices: 2%, 5%, 9%; 2%, 5%, 12%; 5%, 9%, 12%; and 2%, 9%, 12%. You can immediately rule out 5%, 9% and 12% (why?) and hence set up the algebra in each of the three remaining situations, for example 2%, 5% and 9%. Now you can see why there might be many solutions. You can choose some amount to invest at 2%, say$100 and then try to solve for the other two amounts. What if you invest $200 at 5%? What about$900?

My reading of the problem suggests that this is the type of exploration expected.

Penny

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.


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# American Institute of Mathematical Sciences

September  2015, 14(5): 1781-1801. doi: 10.3934/cpaa.2015.14.1781

## Finite dimensional global attractor for a Bose-Einstein equation in a two dimensional unbounded domain

 1 Unité de recherche: Ondelettes et Fractals, Faculté des Sciences de Monastir, Av. de l'environnement, 5000 Monastir

Received  September 2014 Revised  March 2015 Published  June 2015

We study the long-time behavior of the solutions to a nonlinear damped driven Schrödinger type equation with quadratic potential on a strip. We prove that this behavior is described by a regular compact global attractor with finite fractal dimension.
Citation: Brahim Alouini. Finite dimensional global attractor for a Bose-Einstein equation in a two dimensional unbounded domain. Communications on Pure & Applied Analysis, 2015, 14 (5) : 1781-1801. doi: 10.3934/cpaa.2015.14.1781
##### References:
 [1] B. Alouini, Long-time behavior of a Bose-Einstein equation in a two dimensional thin domain,, \emph{Communications in Pure and Applied Analysis}, 10 (2011), 1629.  doi: 10.3934/cpaa.2011.10.1629.  Google Scholar [2] B. Alouini and O. Goubet, Regularity of the attractor for a Bose-Einstein equation in a two dimensional unbounded domain,, \emph{Discrete Continuous Dynam. Systems - B}, 19 (2014), 651.  doi: 10.3934/dcdsb.2014.19.651.  Google Scholar [3] J. M. Ball, Global attractors for damped semilinear wave equations,, \emph{Discrete Continuous Dynam. Systems - A}, 10 (2004), 31.  doi: 10.3934/dcds.2004.10.31.  Google Scholar [4] B. Bongioanni and J. L. Torrea, Sobolev spaces associated to the harmonic oscillator,, \emph{Proc. Indian. Acad. Sci. (Math. Sci.)}, 116 (2006), 337.  doi: 10.1007/BF02829750.  Google Scholar [5] C. C. Bradlay, C. A. Sackett and R. G. Hulet, Bose-Einstein condensation of lithium: observation of limited condensate number,, \emph{Phys. Rev. Lett.}, 78 (1997), 985.  doi: 10.1103/PhysRevLett.78.985.  Google Scholar [6] H. Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations,, Universitext, (2011).   Google Scholar [7] R. Carles, Remarks on nonlinear Schrödinger equation with harmonic potential,, \emph{Annales Henri Poincare}, 3 (2002), 757.  doi: 10.1007/s00023-002-8635-4.  Google Scholar [8] T. Cazenave, Semilinear Schrödinger Equations,, Courant Lecture Notes in Mathematics, 10 (2003).   Google Scholar [9] I. D. Chueshov, Introduction to The Theory of Infinite-Dimensional Dissipative Systems,, University Lectures in Contemporary Mathematics, (2002).   Google Scholar [10] I. D. Chueshov and I. Lasiecka, Long-Time Behavior of Second Order Evolution Equations With Nonlinear Damping,, Memoirs of the American Mathematical Society, 195 (2008).   Google Scholar [11] G. B. Folland, Fourier Analysis and Its Applications,, The Wadsworth & Brooks/Cole Mathematics Series. Wadsworth & Brooks/Cole Advanced Books & Software, (1992).   Google Scholar [12] O. Goubet, Regularity of the attractor for a weakly damped nonlinear Schrödinger equation in $R^2$,, \emph{Advances in Differential Equations}, 3 (1998), 337.   Google Scholar [13] O. Goubet and L. Legry, Finite dimensional global attractor for a parametric nonlinear Schrödinger system with a trapping potential,, \emph{Nonlinear Analysis}, 72 (2010), 4397.  doi: 10.1016/j.na.2010.02.013.  Google Scholar [14] A. Hasegawa and F. Tappert, Transmission of stationary nonlinear optical pulses in dispersive dielectric fibers. I. Anormalous dispersion,, \emph{Applied Physics Lettres}, 23 (1973), 14.  doi: 10.1063/1.1654836.  Google Scholar [15] P. Lauren\ccot, Long-time behavior for weakly damped driven nonlinear Schrödinger equations in $R^N,N\leq 3$,, \emph{NoDEA}, 2 (1995), 357.  doi: 10.1007/BF01261181.  Google Scholar [16] Q. Liu, Y. Zhou, J. Zhang and W. Zhang, Sharp condition of global existence for nonlinear Schrödinger equation with a harmonic potential,, \emph{Appl. Math. Comput.}, 177 (2006), 482.   Google Scholar [17] K. Nosaki and N. Bekki, Low-Dimentional chaos in a driven damped nonlinear Schrödinger equation,, \emph{Physica D: Nonlinear Phenomena}, 21 (1986), 381.  doi: 10.1016/0167-2789(86)90012-6.  Google Scholar [18] K. Promislow and J. N. Kutz, Bifurcation and asymptotic stability in the large detuning limit of optical parametric oscillator,, \emph{Nonlinearity}, 13 (2000), 675.  doi: 10.1088/0951-7715/13/3/310.  Google Scholar [19] J.C. Robinson, Infinite-Dimensionel Dynamical Systems, An Introduction To Dissipative Parabolic PDEs And The Theorie Of Global Attractors,, Cambridge Texts in Applied Mathematics. Cambridge University Press, (2001).   Google Scholar [20] B. Ruf, A sharp Trudinger-Moser type inequality for unbounded domains in $R^2$,, \emph{J. Funct. Anal.}, 219 (2005), 340.  doi: 10.1016/j.jfa.2004.06.013.  Google Scholar [21] R. Temam, Infinite-Dimensional Dynamical Systems In Mechanics and Physics,, Springer applied mathmatical sciences, 68 (1997).   Google Scholar [22] X. Wang, An energy equation for the weakly damped driven nonlinear Schrödinger equations and its application to their attractors,, \emph{Physica D: Nonlinear Phenomena}, 88 (1995), 167.  doi: 10.1016/0167-2789(95)00196-B.  Google Scholar

show all references

##### References:
 [1] B. Alouini, Long-time behavior of a Bose-Einstein equation in a two dimensional thin domain,, \emph{Communications in Pure and Applied Analysis}, 10 (2011), 1629.  doi: 10.3934/cpaa.2011.10.1629.  Google Scholar [2] B. Alouini and O. Goubet, Regularity of the attractor for a Bose-Einstein equation in a two dimensional unbounded domain,, \emph{Discrete Continuous Dynam. Systems - B}, 19 (2014), 651.  doi: 10.3934/dcdsb.2014.19.651.  Google Scholar [3] J. M. Ball, Global attractors for damped semilinear wave equations,, \emph{Discrete Continuous Dynam. Systems - A}, 10 (2004), 31.  doi: 10.3934/dcds.2004.10.31.  Google Scholar [4] B. Bongioanni and J. L. Torrea, Sobolev spaces associated to the harmonic oscillator,, \emph{Proc. Indian. Acad. Sci. (Math. Sci.)}, 116 (2006), 337.  doi: 10.1007/BF02829750.  Google Scholar [5] C. C. Bradlay, C. A. Sackett and R. G. Hulet, Bose-Einstein condensation of lithium: observation of limited condensate number,, \emph{Phys. Rev. Lett.}, 78 (1997), 985.  doi: 10.1103/PhysRevLett.78.985.  Google Scholar [6] H. Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations,, Universitext, (2011).   Google Scholar [7] R. Carles, Remarks on nonlinear Schrödinger equation with harmonic potential,, \emph{Annales Henri Poincare}, 3 (2002), 757.  doi: 10.1007/s00023-002-8635-4.  Google Scholar [8] T. Cazenave, Semilinear Schrödinger Equations,, Courant Lecture Notes in Mathematics, 10 (2003).   Google Scholar [9] I. D. Chueshov, Introduction to The Theory of Infinite-Dimensional Dissipative Systems,, University Lectures in Contemporary Mathematics, (2002).   Google Scholar [10] I. D. Chueshov and I. Lasiecka, Long-Time Behavior of Second Order Evolution Equations With Nonlinear Damping,, Memoirs of the American Mathematical Society, 195 (2008).   Google Scholar [11] G. B. Folland, Fourier Analysis and Its Applications,, The Wadsworth & Brooks/Cole Mathematics Series. Wadsworth & Brooks/Cole Advanced Books & Software, (1992).   Google Scholar [12] O. Goubet, Regularity of the attractor for a weakly damped nonlinear Schrödinger equation in $R^2$,, \emph{Advances in Differential Equations}, 3 (1998), 337.   Google Scholar [13] O. Goubet and L. Legry, Finite dimensional global attractor for a parametric nonlinear Schrödinger system with a trapping potential,, \emph{Nonlinear Analysis}, 72 (2010), 4397.  doi: 10.1016/j.na.2010.02.013.  Google Scholar [14] A. Hasegawa and F. Tappert, Transmission of stationary nonlinear optical pulses in dispersive dielectric fibers. I. Anormalous dispersion,, \emph{Applied Physics Lettres}, 23 (1973), 14.  doi: 10.1063/1.1654836.  Google Scholar [15] P. Lauren\ccot, Long-time behavior for weakly damped driven nonlinear Schrödinger equations in $R^N,N\leq 3$,, \emph{NoDEA}, 2 (1995), 357.  doi: 10.1007/BF01261181.  Google Scholar [16] Q. Liu, Y. Zhou, J. Zhang and W. Zhang, Sharp condition of global existence for nonlinear Schrödinger equation with a harmonic potential,, \emph{Appl. Math. Comput.}, 177 (2006), 482.   Google Scholar [17] K. Nosaki and N. Bekki, Low-Dimentional chaos in a driven damped nonlinear Schrödinger equation,, \emph{Physica D: Nonlinear Phenomena}, 21 (1986), 381.  doi: 10.1016/0167-2789(86)90012-6.  Google Scholar [18] K. Promislow and J. N. Kutz, Bifurcation and asymptotic stability in the large detuning limit of optical parametric oscillator,, \emph{Nonlinearity}, 13 (2000), 675.  doi: 10.1088/0951-7715/13/3/310.  Google Scholar [19] J.C. Robinson, Infinite-Dimensionel Dynamical Systems, An Introduction To Dissipative Parabolic PDEs And The Theorie Of Global Attractors,, Cambridge Texts in Applied Mathematics. Cambridge University Press, (2001).   Google Scholar [20] B. Ruf, A sharp Trudinger-Moser type inequality for unbounded domains in $R^2$,, \emph{J. Funct. Anal.}, 219 (2005), 340.  doi: 10.1016/j.jfa.2004.06.013.  Google Scholar [21] R. Temam, Infinite-Dimensional Dynamical Systems In Mechanics and Physics,, Springer applied mathmatical sciences, 68 (1997).   Google Scholar [22] X. Wang, An energy equation for the weakly damped driven nonlinear Schrödinger equations and its application to their attractors,, \emph{Physica D: Nonlinear Phenomena}, 88 (1995), 167.  doi: 10.1016/0167-2789(95)00196-B.  Google Scholar
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