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# If y + |y| = 0 which of the following must be true?

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If y + |y| = 0 which of the following must be true?  [#permalink]

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Updated on: 03 Jul 2013, 07:24
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If y + |y| = 0 which of the following must be true?

A. y > 0
B. y >= 0
C. y < 0
D. y <= 0
E. y = 0

Originally posted by bibha on 07 May 2010, 22:05.
Last edited by Bunuel on 03 Jul 2013, 07:24, edited 1 time in total.
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08 May 2010, 14:55
bibha wrote:
If y + |y| = 0 which of the following must be true?
a. y >0
b. y >= 0
c. y< 0
d. y<= 0
e. y = 0

Thanks

$$y+|y|=0$$ means either $$y=0$$ or $$y<0$$ (in this case $$|y|=-y$$ and $$y + |y| = 0$$ becomes $$y-y= 0$$). So must be true that $$y\leq{0}$$.

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22 Jun 2010, 16:09
How did u know that y was less than 0?

I am confused on the modules.

Isnt the process of doing these problems is x<0 and X>0?

In this case y + y =0 , 2y=0 , y=0 for x>0

and

y-y=0 , 0=0 , 0 for x<0

I am missing something here. My foundation for understanding these problems is wrong...
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23 Jun 2010, 05:50
ksharma12 wrote:
How did u know that y was less than 0?

I am confused on the modules.

Isnt the process of doing these problems is x<0 and X>0?

In this case y + y =0 , 2y=0 , y=0 for x>0

and

y-y=0 , 0=0 , 0 for x<0

I am missing something here. My foundation for understanding these problems is wrong...

You're missing the concept of absolute values. Absolute value is basically, "distance from zero". So |-3|=3, |-2|=2, |2|=2, etc.

Understand this and the problem will seem very simple to you.
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Re: If y + |y| = 0 which of the following must be true?  [#permalink]

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26 Apr 2015, 12:40
1
Hi All,

This question is perfect for TESTing VALUES.

We're told that Y + |Y| = 0. We're asked which of the following MUST be true (which means "which of the following is ALWAYS TRUE no matter how many different examples we can come up with?")

The obvious value for Y is 0 (since 0 + |0| = 0)
Eliminate A and C.

Since there are still 3 values remaining, there MUST be other options to consider....

IF....
Y = -1
-1 + |-1| = 0
Eliminate B and E.

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Re: If y + |y| = 0 which of the following must be true?  [#permalink]

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Re: If y + |y| = 0 which of the following must be true? &nbs [#permalink] 12 Aug 2017, 10:00
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Computing rational cohomology of smooth (not necessarily compact) toric varieties

The title pretty much says it all. I am looking for references (lecture notes, books, readable articles, suggestions), preferably example laden, that explain how to compute the rational cohomology of a smooth toric variety. I am particularly interested in methods for doing this when the variety is not compact. Implicit here is the assumption that this can be done in a practical way (please correct me if I am wrong, I do not know much about the subject).

Some background: I have a bunch of varieties whose cohomology I would really like to compute. I have reasonably explicit descriptions of these varieties as subsets of $\mathbb{C}^n$. These descriptions are along the lines of (but more involved) the variety in this question:

A cohomology computation request.

I realized earlier today that my varieties are toric, and I am hoping this observation will be the answer to my prayers.

If it helps, I can compute the equivariant cohomology of of my varieties (mainly because the Hodge structure on it is pure), but I am presuming this doesn't really completely determine the ordinary cohomology (apart from Hodge-Euler polynomials etc.).

-

I have the following reference in mind:

Franz, M. "The integral cohomology of toric manifolds." Tr. Mat. Inst. Steklova 252 (2006), Geom. Topol., Diskret. Geom. i Teor. Mnozh., 61--70.

the paper is available on arxiv. In this paper the author explains how to compute the cohomology thanks to the Stanley-Reisner ring (theorem 1.2). This is nice but for the cup product structure theorem 1.2 gives the right algebra structure only in some special cases, the general case is a conjecture.

-

Here is a (co-)homology computation for singular and noncompact toric varieties. It gives in particular a generalization of a Jurkiewicz-Danilov theorem (smooth compact case): Jordan,Arno : Homology and cohomology of toric varieties. (Konstanz:Hartong-Gorre (1998),108 p. ; see Zentralblatt 0919.14033).

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transmission coefficient greater than 1
in

All was well untill I plugged the analytic solution for the transmission coefficient into Python to render a graph and got this. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. The reflection coefficient c may be positive or negative so the transmission coefficient t may be greater than unity. In a waveguide or a fiber are TE-TM modes same as horizontal and vertical polarization? You can choose values of parameters which will give transmission coefficients greater than 1, and that would appear to violate conservation of energy. --University of Arizona. Help us to install and run Boltztrap 1.2.5 software. where $\mathrm{Ai}(z)$ and $\mathrm{Bi}(z)$ are the standard Airy functions of the first and second kind. How to install and run Boltztrap 1.2.5 freely downloadable from internet? 0 \\ Use MathJax to format equations. If R 0 equals 1, each existing infection causes one new infection. -\frac{ika}{q_0}e^{-ika} & \frac{ika}{q_0}e^{ika} & \mathrm{Ai}'(-\xi_a) & \mathrm{Bi}'(-\xi_a) & 0 & 0 \\ It's nothing to be alarmed about. If your transmission coefficient is the ratio of power in to power out then Marek's answer is correct, unless you have a medium or boundary with gain (e.g. If some light is absorbed by the substance, then the transmitted light will be a combination of the wavelengths of the light that was transmitted and not absorbed. So maybe you should give the details of your analytical solution, otherwise it is difficult to say anything for sure. Not a surprise. Why is Soulknife's second attack not Two-Weapon Fighting? This is one more example that the raw result of measurement, without estimated uncertainty ("error bounds") is next to worthless. Should I discard the $T>1$ solutions and leave energy gaps in the graph? It only takes a minute to sign up. I'm trying to find the reflection and transmission coefficients for a stream of electrons coming from $x=-\infty$ (so $E>0$, no bound states involved). If white light is shone through the filter, the light transmitted also appears blue because of the absorption … We are not able to install and run the softare. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. Reflection Coefficients … Does a metal have One, Zero or Infinite Relative Permittivity? $$\frac{d^2\psi_{III}}{dx^2}+(\frac{x}{a}+\epsilon^2-1)k_0^2\psi_{III}=0$$ \end{matrix} \right)=\left( \begin{matrix} $$\frac{d^2\psi_{III}}{d\xi^2}=-\xi\psi_{III}$$, $$\psi_{II}(x)=C\mathrm{Ai}(-\xi)+D\mathrm{Bi}(-\xi)$$ e^{-ika} & e^{ika} & -\mathrm{Ai}(-\xi_a) & -\mathrm{Bi}(-\xi_a) & 0 & 0 \\ The reflection coefficient of any device (S11), can it be greater than one? What if the P-Value is less than 0.05, but the test statistic is also less than the critical value? I was recently reading this study. At the seashore we see waves approaching the shore and they get larger as they arrive. \end{split} If your transmission coefficient is the ratio of power in to power out then Marek's answer is correct, unless you have a medium or boundary with gain (e.g. Eine der vier fundamentalen Wechselwirkungskräfte ist die elektromagnetische Wechselwirkungskraft. Asking for help, clarification, or responding to other answers. Using of the rocket propellant for engine cooling. Sometimes, as in this very case, our estimates of uncertainty can and should be improved, here T should be always contained in interval [0,1]. What would it even mean for the coefficient to be higher than 1? This does not violate any physical laws. $$\psi_{IV}(x)=Ge^{ikx},$$. Now, I've gone over my calculations a dozen times, and I still can't spot a mistake, so I want to know wheather it's possible to have a transmission coefficient greater than one for certain values of energy, like in this scenario. It also needs to be pumped (energy put in) to amplify the beam. $\begingroup$ A laser amplifier has a transmission coefficient greater than 1 (aka gain), for some range of wavelengths. reply from a potential PhD advisor? See this, National Institute of Science Education and Research. Long story short, they list a Pearson coefficient of 1.802. Leider geht das nicht: Während sich die Wärmelehre bruchlos in die Mechanik eingliedern lässt, ist This does not violate any physical laws. Should we leave technical astronomy questions to Astronomy SE? What confuses me more is the sings of the elements in your matrix. Meanwhile I've also solved the problem numerically, and got pretty expected results, like the one below. C \\ What is the right way to calculating the acoustic reflection/transmission coefficients? By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. \begin{equation*}\label{parametara} MathJax reference. Now, I've gone over my calculations a dozen times, and I still can't spot a mistake, so I want to know wheather it's possible to have a transmission coefficient greater than one for certain values of energy, like in this scenario. D \\ Is there anyone can help me with this, thanks. \begin{split} I would suspect that your analytical solution does not satisfy the boundary conditions. Has it something to do with the change of variables discussed in akhmeteli's answer and comments? F How do I figure out the probability of finding a particle between two barriers? Have you checked that your solutions are applicable independent of the value of $E$? When they are nearly equal, then the apparent result may easily exceed 1. Also did you mention that for one of $\psi_{II}$ and $\psi_{III}$ : $\frac{d}{dx} = \frac{d}{d\xi}$ and for the other $\frac{d}{dx} = -\frac{d}{d\xi}$, Meaning of transmission coefficient greater than one in a potential well problem, mathworld.wolfram.com/AiryDifferentialEquation.html, “Question closed” notifications experiment results and graduation, MAINTENANCE WARNING: Possible downtime early morning Dec 2/4/9 UTC (8:30PM…. That seems pretty arbitrary arbitrary to me, but it may be the case here. In such cases, how do you define transmission coefficient? -G\frac{ika}{q_0}e^{ika} We have downloaded Boltztrap 1.2.5 from internet. For compressible flows, the pressure coefficient can be greater than 1. $\endgroup$ – Jon Custer Jul 25 '16 at 12:55 Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. that should work well. Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. If your transmission coefficient is the ratio of amplitudes then, Institute of Physics of the Polish Academy of Sciences. Or someone can show me how to write the input parameter. Reflectivity and transmissivity greater than 1. Here are three different methods for factorising harder quadratics, you only need to know one of them.


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# The objects below collide inelastically on a frictionless tabletop. Determine the = speed Vz (in m/s) of the 5.0-kg

###### Question:

The objects below collide inelastically on a frictionless tabletop. Determine the = speed Vz (in m/s) of the 5.0-kg object after collision: BEFORE AFTER 2skg S.0kg 4.5 Qs 2skg 302 30" at fest s.ke None of these

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ns: You will have 3 hours to complete the quiz once you begin. Do iate calculations. Round probabilities to the fourth decimal. Question 8 1 pt While conducting experiments, a marine biologist selects water depths from a uniformly distributed collection that varies between 2.00 meters and 7.00 meter...


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# Graphing Polar Equations on a Polar Plane

In the first lesson of this unit, you were lightly introduced to graphing polar coordinates. This lesson revisits what you learned earlier, and extends those ideas to graphing polar equations.

In the first of three videos below, you will learn how to graph polar coordinates with negative values, namely when:

• the angle is negative
• the radius is negative
• both properties are negative

You’ll discover that when the angle is negative angle, the terminal side moves clockwise from the polar axis, and vice versa for when its positive. In addition, a negative radius gives a direction of the terminal side opposite to the angle.

Perhaps the greatest challenge to this section comes when graphing polar equations. Generally, graphing polar equations isn’t an easy task, and depending on your level of study, it can quite complicated when taking into account symmetry. Luckily, our analyse won’t exceed what we already know about negative/positive angles and radii. Starting with the first video, you’ll graph two trigonometric equations – one that’s in degrees and the other that’s in radians. To find the polar coordinates of an equation, you’ll create a table of values using angles from θ = 0° to 360°. Recall that a point with a radius of (−r) is plotted in the opposite direction to (+r).

Sometimes a polar equation can be written as a pair of parametric rectangular equations. If you encounter a question like that, one major thing to keep in mind is to follow the previous point when connecting to form your curve.


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# How many ions are there in 4.5 g of (NH_4)_3PO_4?

Approx. $0.121 \cdot m o l$ of phosphate and ammonium ions.
The formula mass of ammonium phosphate is $149.09 \cdot g \cdot m o {l}^{-} 1$. A mass of $4.5 \cdot g$ corresponds to a molar quantity of $\frac{4.5 \cdot g}{149.09 \cdot g \cdot m o {l}^{-} 1}$ $=$ $3.02 \times {10}^{- 2} \cdot m o l$ with respect to ${\left(N {H}_{4}\right)}_{3} P {O}_{4}$, and thus $3.02 \times {10}^{- 2} \times 4 = 0.121 \cdot m o l$ of ions, given that there are 4 individual ion per formula unit of ammonium phosphate.


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### Section 4-3 : Minimum and Maximum Values

Many of our applications in this chapter will revolve around minimum and maximum values of a function. While we can all visualize the minimum and maximum values of a function we want to be a little more specific in our work here. In particular, we want to differentiate between two types of minimum or maximum values. The following definition gives the types of minimums and/or maximums values that we’ll be looking at.

#### Definition

1. We say that $$f\left( x \right)$$ has an absolute (or global) maximum at $$x = c$$ if$$f\left( x \right) \le f\left( c \right)$$ for every $$x$$ in the domain we are working on.

2. We say that $$f\left( x \right)$$ has a relative (or local) maximum at $$x = c$$ if $$f\left( x \right) \le f\left( c \right)$$ for every $$x$$ in some open interval around $$x = c$$.

3. We say that $$f\left( x \right)$$ has an absolute (or global) minimum at $$x = c$$ if $$f\left( x \right) \ge f\left( c \right)$$ for every $$x$$ in the domain we are working on.

4. We say that $$f\left( x \right)$$ has a relative (or local) minimum at $$x = c$$ if$$f\left( x \right) \ge f\left( c \right)$$ for every $$x$$ in some open interval around $$x = c$$.

Note that when we say an “open interval around$$x = c$$” we mean that we can find some interval $$\left( {a,b} \right)$$, not including the endpoints, such that $$a < c < b$$. Or, in other words, $$c$$ will be contained somewhere inside the interval and will not be either of the endpoints.

Also, we will collectively call the minimum and maximum points of a function the extrema of the function. So, relative extrema will refer to the relative minimums and maximums while absolute extrema refer to the absolute minimums and maximums.

Now, let’s talk a little bit about the subtle difference between the absolute and relative in the definition above.

We will have an absolute maximum (or minimum) at $$x = c$$ provided $$f\left( c \right)$$ is the largest (or smallest) value that the function will ever take on the domain that we are working on. Also, when we say the “domain we are working on” this simply means the range of $$x$$’s that we have chosen to work with for a given problem. There may be other values of $$x$$ that we can actually plug into the function but have excluded them for some reason.

A relative maximum or minimum is slightly different. All that’s required for a point to be a relative maximum or minimum is for that point to be a maximum or minimum in some interval of $$x$$’s around $$x = c$$. There may be larger or smaller values of the function at some other place, but relative to $$x = c$$, or local to $$x = c$$, $$f\left( c \right)$$ is larger or smaller than all the other function values that are near it.

Note as well that in order for a point to be a relative extrema we must be able to look at function values on both sides of $$x = c$$ to see if it really is a maximum or minimum at that point. This means that relative extrema do not occur at the end points of a domain. They can only occur interior to the domain.

There is actually some debate on the preceding point. Some folks do feel that relative extrema can occur on the end points of a domain. However, in this class we will be using the definition that says that they can’t occur at the end points of a domain. This will be discussed in a little more detail at the end of the section once we have a relevant fact taken care of.

It’s usually easier to get a feel for the definitions by taking a quick look at a graph.

For the function shown in this graph we have relative maximums at $$x = b$$ and $$x = d$$. Both of these points are relative maximums since they are interior to the domain shown and are the largest point on the graph in some interval around the point. We also have a relative minimum at $$x = c$$ since this point is interior to the domain and is the lowest point on the graph in an interval around it. The far-right end point, $$x = e$$, will not be a relative minimum since it is an end point.

The function will have an absolute maximum at $$x = d$$ and an absolute minimum at $$x = a$$. These two points are the largest and smallest that the function will ever be. We can also notice that the absolute extrema for a function will occur at either the endpoints of the domain or at relative extrema. We will use this idea in later sections so it’s more important than it might seem at the present time.

Let’s take a quick look at some examples to make sure that we have the definitions of absolute extrema and relative extrema straight.

Example 1 Identify the absolute extrema and relative extrema for the following function. $f\left( x \right) = {x^2}\hspace{0.25in}{\mbox{on}}\hspace{0.25in}\left[ { - 1,2} \right]$
Show Solution

Since this function is easy enough to graph let’s do that. However, we only want the graph on the interval $$\left[ { - 1,2} \right]$$. Here is the graph,

Note that we used dots at the end of the graph to remind us that the graph ends at these points.

We can now identify the extrema from the graph. It looks like we’ve got a relative and absolute minimum of zero at $$x = 0$$ and an absolute maximum of four at $$x = 2$$. Note that $$x = - 1$$ is not a relative maximum since it is at the end point of the interval.

This function doesn’t have any relative maximums.

As we saw in the previous example functions do not have to have relative extrema. It is completely possible for a function to not have a relative maximum and/or a relative minimum.

Example 2 Identify the absolute extrema and relative extrema for the following function. $f\left( x \right) = {x^2}\hspace{0.25in}{\mbox{on}}\hspace{0.25in}\left[ { - 2,2} \right]$
Show Solution

Here is the graph for this function.

In this case we still have a relative and absolute minimum of zero at $$x = 0$$. We also still have an absolute maximum of four. However, unlike the first example this will occur at two points, $$x = - 2$$ and $$x = 2$$.

Again, the function doesn’t have any relative maximums.

As this example has shown there can only be a single absolute maximum or absolute minimum value, but they can occur at more than one place in the domain.

Example 3 Identify the absolute extrema and relative extrema for the following function. $f\left( x \right) = {x^2}$
Show Solution

In this case we’ve given no domain and so the assumption is that we will take the largest possible domain. For this function that means all the real numbers. Here is the graph.

In this case the graph doesn’t stop increasing at either end and so there are no maximums of any kind for this function. No matter which point we pick on the graph there will be points both larger and smaller than it on either side so we can’t have any maximums (of any kind, relative or absolute) in a graph.

We still have a relative and absolute minimum value of zero at $$x = 0$$.

So, some graphs can have minimums but not maximums. Likewise, a graph could have maximums but not minimums.

Example 4 Identify the absolute extrema and relative extrema for the following function. $f\left( x \right) = {x^3}\hspace{0.25in}{\mbox{on}}\hspace{0.25in}\left[ { - 2,2} \right]$
Show Solution

Here is the graph for this function.

This function has an absolute maximum of eight at $$x = 2$$ and an absolute minimum of negative eight at $$x = - 2$$. This function has no relative extrema.

So, a function doesn’t have to have relative extrema as this example has shown.

Example 5 Identify the absolute extrema and relative extrema for the following function. $f\left( x \right) = {x^3}$
Show Solution

Again, we aren’t restricting the domain this time so here’s the graph.

In this case the function has no relative extrema and no absolute extrema.

As we’ve seen in the previous example functions don’t have to have any kind of extrema, relative or absolute.

Example 6 Identify the absolute extrema and relative extrema for the following function. $f\left( x \right) = \cos \left( x \right)$
Show Solution

We’ve not restricted the domain for this function. Here is the graph.

Cosine has extrema (relative and absolute) that occur at many points. Cosine has both relative and absolute maximums of 1 at

$x = \ldots - 4\pi ,\, - 2\pi ,\,\,0,\,\,2\pi ,\,\,4\pi , \ldots$

Cosine also has both relative and absolute minimums of -1 at

$x = \ldots - 3\pi ,\, - \pi ,\,\,\pi ,\,\,3\pi , \ldots$

As this example has shown a graph can in fact have extrema occurring at a large number (infinite in this case) of points.

We’ve now worked quite a few examples and we can use these examples to see a nice fact about absolute extrema. First let’s notice that all the functions above were continuous functions. Next notice that every time we restricted the domain to a closed interval (i.e. the interval contains its end points) we got absolute maximums and absolute minimums. Finally, in only one of the three examples in which we did not restrict the domain did we get both an absolute maximum and an absolute minimum.

These observations lead us the following theorem.

#### Extreme Value Theorem

Suppose that $$f\left( x \right)$$ is continuous on the interval $$\left[ {a,b} \right]$$ then there are two numbers $$a \le c,d \le b$$ so that $$f\left( c \right)$$ is an absolute maximum for the function and $$f\left( d \right)$$ is an absolute minimum for the function.

So, if we have a continuous function on an interval $$\left[ {a,b} \right]$$ then we are guaranteed to have both an absolute maximum and an absolute minimum for the function somewhere in the interval. The theorem doesn’t tell us where they will occur or if they will occur more than once, but at least it tells us that they do exist somewhere. Sometimes, all that we need to know is that they do exist.

This theorem doesn’t say anything about absolute extrema if we aren’t working on an interval. We saw examples of functions above that had both absolute extrema, one absolute extrema, and no absolute extrema when we didn’t restrict ourselves down to an interval.

The requirement that a function be continuous is also required in order for us to use the theorem. Consider the case of

$f\left( x \right) = \frac{1}{{{x^2}}}\hspace{0.25in}{\mbox{on}}\hspace{0.25in}[ - 1,1]$

Here’s the graph.

This function is not continuous at $$x = 0$$ as we move in towards zero the function is approaching infinity. So, the function does not have an absolute maximum. Note that it does have an absolute minimum however. In fact the absolute minimum occurs twice at both $$x = - 1$$ and $$x = 1$$.

If we changed the interval a little to say,

$f\left( x \right) = \frac{1}{{{x^2}}}\hspace{0.25in}{\mbox{on}}\hspace{0.25in}\left[ {\frac{1}{2},1} \right]$

the function would now have both absolute extrema. We may only run into problems if the interval contains the point of discontinuity. If it doesn’t then the theorem will hold.

We should also point out that just because a function is not continuous at a point that doesn’t mean that it won’t have both absolute extrema in an interval that contains that point. Below is the graph of a function that is not continuous at a point in the given interval and yet has both absolute extrema.

This graph is not continuous at $$x = c$$, yet it does have both an absolute maximum ($$x = b$$) and an absolute minimum ($$x = c$$). Also note that, in this case one of the absolute extrema occurred at the point of discontinuity, but it doesn’t need to. The absolute minimum could just have easily been at the other end point or at some other point interior to the region. The point here is that this graph is not continuous and yet does have both absolute extrema

The point of all this is that we need to be careful to only use the Extreme Value Theorem when the conditions of the theorem are met and not misinterpret the results if the conditions aren’t met.

In order to use the Extreme Value Theorem we must have an interval that includes its endpoints, often called a closed interval, and the function must be continuous on that interval. If we don’t have a closed interval and/or the function isn’t continuous on the interval then the function may or may not have absolute extrema.

We need to discuss one final topic in this section before moving on to the first major application of the derivative that we’re going to be looking at in this chapter.

#### Fermat’s Theorem

If $$f\left( x \right)$$ has a relative extrema at $$x = c$$ and $$f'\left( c \right)$$ exists then $$x = c$$ is a critical point of $$f\left( x \right)$$. In fact, it will be a critical point such that $$f'\left( c \right) = 0$$.

To see the proof of this theorem see the Proofs From Derivative Applications section of the Extras chapter.

Also note that we can say that $$f'\left( c \right) = 0$$ because we are also assuming that $$f'\left( c \right)$$ exists.

This theorem tells us that there is a nice relationship between relative extrema and critical points. In fact, it will allow us to get a list of all possible relative extrema. Since a relative extrema must be a critical point the list of all critical points will give us a list of all possible relative extrema.

Consider the case of $$f\left( x \right) = {x^2}$$. We saw that this function had a relative minimum at $$x = 0$$ in several earlier examples. So according to Fermat’s theorem $$x = 0$$ should be a critical point. The derivative of the function is,

$f'\left( x \right) = 2x$

Sure enough $$x = 0$$ is a critical point.

Be careful not to misuse this theorem. It doesn’t say that a critical point will be a relative extrema. To see this, consider the following case.

$f\left( x \right) = {x^3}\hspace{0.25in}\hspace{0.25in}f'\left( x \right) = 3{x^2}$

Clearly $$x = 0$$ is a critical point. However, we saw in an earlier example this function has no relative extrema of any kind. So, critical points do not have to be relative extrema.

Also note that this theorem says nothing about absolute extrema. An absolute extrema may or may not be a critical point.

Before we leave this section we need to discuss a couple of issues.

First, Fermat’s Theorem only works for critical points in which $$f'\left( c \right) = 0$$. This does not, however, mean that relative extrema won’t occur at critical points where the derivative does not exist. To see this consider $$f\left( x \right) = \left| x \right|$$. This function clearly has a relative minimum at $$x = 0$$ and yet in a previous section we showed in an example that $$f'\left( 0 \right)$$ does not exist.

What this all means is that if we want to locate relative extrema all we really need to do is look at the critical points as those are the places where relative extrema may exist.

Finally, recall that at that start of the section we stated that relative extrema will not exist at endpoints of the interval we are looking at. The reason for this is that if we allowed relative extrema to occur there it may well (and in fact most of the time) violate Fermat’s Theorem. There is no reason to expect end points of intervals to be critical points of any kind. Therefore, we do not allow relative extrema to exist at the endpoints of intervals.

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Positive definite matrices and the S-divergenceA small fraction of an initial version of this work was presented at the Advances in Neural Information Processing Systems (NIPS) 2012 conference—see [48].

# Positive definite matrices and the S-divergence††thanks: A small fraction of an initial version of this work was presented at the Advances in Neural Information Processing Systems (NIPS) 2012 conference—see [48].

Suvrit Sra Parts of this paper were written during my stay at Carnegie Mellon University; the initial version was prepared while I was at the Max Planck Institute for Intelligent Systems, Tübingen, Germany.
###### Abstract

Positive definite matrices abound in a dazzling variety of applications. This ubiquity can be in part attributed to their rich geometric structure: positive definite matrices form a self-dual convex cone whose strict interior is a Riemannian manifold. The manifold view is endowed with a “natural” distance function while the conic view is not. Nevertheless, drawing motivation from the conic view, we introduce the S-Divergence as a “natural” distance-like function on the open cone of positive definite matrices. We motivate the S-divergence via a sequence of results that connect it to the Riemannian distance. In particular, we show that (a) this divergence is the square of a distance; and (b) that it has several geometric properties similar to those of the Riemannian distance, though without being computationally as demanding. The S-Divergence is even more intriguing: although nonconvex, we can still compute matrix means and medians using it to global optimality. We complement our results with some numerical experiments illustrating our theorems and our optimization algorithm for computing matrix medians.

Key words. Bregman matrix divergence; Log Determinant; Stein Divergence; Jensen-Bregman divergence; matrix geometric mean; matrix median; nonpositive curvature

## 1 Introduction

Hermitian positive definite (HPD) matrices are a noncommutative generalization of positive reals. They abound in a multitude of applications and exhibit attractive geometric properties—e.g., they form a differentiable Riemannian (also Finslerian) manifold [10, 33] that is a well-studied example of a manifold of nonpositive curvature [17, Ch.10]. HPD matrices possess even more structure: (i) they embody a canonical higher-rank symmetric space [51]; and (ii) their closure forms a closed, self-dual convex cone.

The convex conic view enjoys great importance in convex optimization [43, 6, 44] and in nonlinear Perron-Frobenius theory [40]; symmetric spaces are important in algebra, analysis [51, 32, 39], and optimization [43, 52]; while the manifold view (Riemannian or Finslerian) plays diverse roles—see [10, Ch.6] and [46].

The manifold view is equipped with a with a “natural” distance function while the conic view is not. Nevertheless, drawing motivation from the convex conic view, we introduce the S-Divergence as a “natural” distance-like function on the open cone of positive definite matrices. Indeed, we prove a sequence of results connecting the S-Divergence to the Riemannian distance. Most importantly, we show that (a) this divergence is the square of a distance; and (b) that it has several geometric properties in common with the Riemannian distance, without being numerically as demanding. This builds an informal link between the manifold and conic views of HPD matrices.

### 1.1 Background and notation

We begin by fixing notation. The letter denotes some Hilbert space, usually just . The inner product between two vectors and in is ( denotes ‘conjugate transpose’). The set of Hermitian matrices is denoted as . A matrix is called positive definite if

 ⟨x,Ax⟩>0for allx≠0,also % written asA>0. \hb@xt@.01(1.1)

The set of all positive definite (henceforth positive) matrices is denoted by . We say is positive semidefinite if for all ; denoted . The inequality is the usual Löwner order and means . The Frobenius norm of a matrix is defined as ; and denotes the operator 2-norm. Let be an analytic function on ; for a matrix with eigendecomposition , equals with .

The set is a well-studied differentiable Riemannian manifold, with the Riemannian metric given by the differential form . This metric induces the Riemannian distance (see e.g., [10, Ch. 6]):

 δR(X,Y):=∥log(Y−1/2XY−1/2)∥FforX,Y>0, \hb@xt@.01(1.2)

and where denotes the matrix logarithm.

A counterpart to the distance (LABEL:eq.riem) was formally introduced in [48] under the name S-Divergence111It is a divergence because although nonnegative, definite, and symmetric, it is not a metric.; this divergence is defined as

 δ2S(X,Y):=logdet(X+Y2)−12logdet(XY)forX,Y>0. \hb@xt@.01(1.3)

Our definition above already writes in anticipation of Theorem LABEL:thm.metric that shows to be a metric. This paper suggests S-Divergence as an alternative to (LABEL:eq.riem), and studies several of its properties that may also be of independent interest. The simplicity of (LABEL:eq.ss) is one of the key reasons for using it as an alternative to (LABEL:eq.riem): it is cheaper to compute, as is its derivative, and certain basic algorithms involving it run much faster than corresponding ones that use  [48].

This line of thought actually originates in [24, 25], where for an image search task based on “nearest neighbors,” is used to measure nearness instead of , and is shown to yield large speedups without blighting the quality of search results. Although exact details of this image search are outside the scope of this paper, let us highlight below the two speedups that were crucial to [24, 25].

The first speedup is shown in the left panel of Fig. LABEL:fig.one, which compares times taken to compute and . For computing the latter, we used the dist.m function in the Matrix Means Toolbox (MMT)222Downloaded from http://bezout.dm.unipi.it/software/mmtoolbox/. The second, more dramatic speedup in shown in the right panel which shows time taken to compute the matrix means

 GMℓd:=argminX>0∑mi=1δ2S(X,Ai),andGM:=argminX>0∑mi=1δ2R(X,Ai),

where are HPD matrices. For details on see Section LABEL:sec.mtxmeans; the geometric mean is also known as the “Karcher mean”, and was computed using the MMT via the rich.m script which implements a state-of-the-art method [13, 34].

We mention here that other alternatives to are also possible, for instance the popular “log-Euclidean” distance [3], given by

 δle(X,Y)=∥logX−logY∥F. \hb@xt@.01(1.4)

Notice that to compute we require two eigenvector decompositions; this makes it more expensive than which requires only 3 Cholesky factorizations. Even though the matrix mean under can be computed in closed form, its dependency on matrix logarithms and exponentials can make it slower than . However, much more importantly, for the applications in [24, 25], and other alternatives to proved to be substantially less competitive than (LABEL:eq.ss), so we limit our focus to ; for more extensive empirical comparisons with other distances, we refer the reader to [24, 25].

While our paper was under review (in 2011), we became aware of a concurrent paper of Chebbi and Moakher (CM) [21], who consider a one parameter family of divergences that generalize (LABEL:eq.ss). Our work differs from CM in the following aspects:

• CM prove to be a distance only for commuting matrices. As per Remark LABEL:rmk.commute, the commuting case essentially reduces to the scalar case. The noncommuting case is much harder, and was conjectured to be true in [21]. We propose and solve the general noncommuting case, independent of CM.

• We establish several theorems connecting to the Riemannian distance . These connections have not been made either by CM or elsewhere.

• A question closely related to metricity of is whether the matrix

 [det(Xi+Xj)−β]mi,j=1,X1,…,Xm∈Pn,

is positive semidefinite for every integer and every scalar . CM considered special cases of this question. We provide a complete characterization of necessary and sufficient for the above matrix to be semidefinite.

• CM analyze the “matrix-mean” , whose solution they obtain by solving . CM’s results essentially imply global optimality333We thank a referee for alerting us to this fact, which ultimately follows from CM’s uniqueness theorem and the observation that the cost function goes to for both and (see Sec. LABEL:sec.mtxmeans for more details).; we provide two different proofs of this fact. One of our proofs is based of establishing geodesic convexity of the S-Divergence—a result that is also interesting because in our previous attempts [48, 49] we oversaw this property. In fact, we show (Theorem LABEL:thm.jointgc) that is jointly geodesically convex.

Other contributions. The present paper substantially extends our initial work [48]; we outline below the key differences from [48].

• Due to lack of space proofs of the lemmas supporting Theorem LABEL:thm.metric did not appear in [48]. In particular, proofs of Corollaries LABEL:corr.mink, LABEL:cor.eigs and Theorems LABEL:thm.diag, LABEL:thm.det are absent from [48] (these results are not difficult though).

• None of our results on similarities between the Riemannian distance and are present in [48]. Although some of these results are mentioned in a summary table in [48], the full theorem statements as well as their proofs are absent. The concerned results are: Theorems LABEL:thm.gm, LABEL:thm.contract, LABEL:thm.contract3, LABEL:thm.cancel, LABEL:thm.tucontract, and LABEL:thm.contract2.

• This paper uncovers a new result (previously unknown): is jointly geodesically convex—Prop. LABEL:prop.mixedmean and Theorem  LABEL:thm.jointgc establish this remarkable fact.

• This paper proves several new “conic” contraction results for and . The appurtenant results are: Proposition LABEL:prop.logdet, Corollary LABEL:cor.ratio, Theorem LABEL:thm.compr, Corollary LABEL:cor.compr.tensor, Theorem LABEL:thm.geig, and Corollary LABEL:corr.riem.compress.

• This paper proves bi-Lipschitz-like inequalities for and (Theorem LABEL:thm.bnds).

• Finally, this paper studies the weighted matrix-medians problem:

 minX>0∑mi=1wiδS(X,Ai),Ai∈Pn, for 1≤i≤m.

This problem was also partially studied by [19], who presented an iterative method that was erroneously claimed to be a fixed-point iteration in the Thompson metric. We present a counterexample to illustrate this error, and rectify it by presenting different analysis that ensures convergence (see §LABEL:sec.median).

## 2 The S-Divergence

We proceed now to formally introduce the S-Divergence. We follow the viewpoint of Bregman divergences. Consider, thus, a differentiable strictly convex function ; then, , with equality if and only if . The difference between the two sides of this inequality defines the Bregman Divergence444Bregman divergences over scalars and vectors have been well-studied; see e.g., [18, 4]. They are called divergences because they are not distances (though they often behave like squared distances, in a sense that can be made precise for certain choices of  [22]).

 Df(x,y):=f(x)−f(y)−f′(y)(x−y). \hb@xt@.01(2.1)

The scalar divergence (LABEL:eq.breg.scalar) can be extended to Hermitian matrices.

###### Proposition 2.1

Let be differentiable and strictly convex on ; let be arbitrary. Then, we have the matrix Bregman Divergence:

 \hb@xt@.01(2.2)

By construction is nonnegative, strictly convex in , and zero if and only if . It is typically asymmetric, and may be viewed as a measure of dissimilarity.

###### Example 2.2

Let . Then, for , , with which (LABEL:eq.breg) yields the squared Frobenius norm

 Df(X,Y)=12∥X−Y∥2F.

If on , then , and (LABEL:eq.breg) yields the (unnormalized) von Neumann Divergence of quantum information theory [45]:

 Dvn(X,Y)=tr(XlogX−XlogY−X+Y),X,Y∈Pn.

For on , , and we obtain the divergence

 Dℓd(X,Y)=tr(Y−1(X−Y))−logdet(XY−1),X,Y∈Pn,

which is known as the LogDet Divergence [37], or more classically as Stein’s loss [50].

The divergence is of key importance to our paper, so we mention some additional noteworthy contexts where it occurs: (i) information theory [26], as the relative entropy between two multivariate gaussians with same mean; (ii) optimization, when deriving the famous Broyden-Fletcher-Goldfarb-Shanno (BFGS) updates [47]; (iii) matrix divergence theory [5, 28, 46]; (iv) kernel learning [37].

Despite the broad applicability of Bregman divergences, their asymmetry is sometimes undesirable. This drawback has leads us to consider symmetric divergences, among which the most popular is the “Jensen-Bregman” divergence555This symmetrization has been largely studied only for divergences over scalars or vectors.

 Sf(X,Y):=Df(X,X+Y2)+Df(X+Y2,Y). \hb@xt@.01(2.3)

This divergence has two attractive and perhaps more useful representations:

 Sf(X,Y) =12(trf(X)+trf(Y))−trf(X+Y2), \hb@xt@.01(2.4) Sf(X,Y) =minZDf(X,Z)+Df(Y,Z). \hb@xt@.01(2.5)

Compare (LABEL:eq.breg) with (LABEL:eq.js1): both formulas define divergence as departure from linearity; the former uses derivatives, while the latter is stated using midpoint convexity. Representation (LABEL:eq.js1) has an advantage over (LABEL:eq.breg), (LABEL:eq.js), and (LABEL:eq.js2), in that it does not need to assume differentiability of .

The reader must have realized by now that the S-Divergence (LABEL:eq.ss) is nothing but the symmetrized divergence (LABEL:eq.js) generated by . Alternatively, the S-Divergence may be essentially viewed as the Jensen-Bregman divergence between two multivariate gaussians [26], or as the Bhattacharya distance between them [12].

Let us now list a few basic properties of .

###### Proposition 2.3

Let be the vector of eigenvalues of , and be a diagonal matrix with as its diagonal. Let . Then,

1. ;

2. , where , ;

3. ;

4. ; and

5. .

Proof. (i) follows from the equality .
(ii) follows upon observing that

 det(PAQ+PBQ)[det(PAQ)]1/2[det(PBQ)]1/2=det(P)⋅det(A+B)⋅det(Q)det(P)⋅[det(A)]1/2[det(B)]1/2⋅det(Q).

(iii) follows upon noting

 det(A−1+B−1)[det(A−1)]1/2[det(B)−1]1/2=det(A)⋅det(A−1+B−1)⋅det(B)[det(A)]1/2[det(B)]1/2.

(iv) follows as , and .
(v) is trivial since .  \qedsymbol The most useful corollary to Prop. LABEL:prop.basic is congruence invariance of .

###### Corollary 2.4

Let , and let be any invertible matrix. Then,

 δS(X∗AX,X∗BX)=δS(A,B).

The next result reaffirms that is a divergence, while showing that it enjoys some limited convexity and concavity.

###### Proposition 2.5

Let . Then, (i) with equality if and only if ; (ii) for fixed , is convex in for , while for , it is concave.

Proof. Since is a sum of Bregman divergences, property (i) follows from definition (LABEL:eq.js). Alternatively, note that , with equality if and only if . Part (ii) follows upon analyzing the Hessian . This Hessian can be identified with the matrix

 H:=12(A−1⊗A−1)−(A+B)−1⊗(A+B)−1, \hb@xt@.01(2.6)

where is the usual the Kronecker product. Matrix is positive definite for and negative definite for , which proves (ii).

Below we show that is richer than a divergence: its square-root is actually a distance on . This is the first main result of our paper. Previous authors [24, 21] conjectured this result but could not establish it, perhaps because both ultimately sought to map to a Hilbert space metric. This approach fails because HPD matrices do not form even a (multiplicative) semigroup, which renders the powerful theory of harmonic analysis on semigroups [7] inapplicable to . This difficulty necessitates a different path to proving metricity of , and this is the subject of the next section.

## 3 The δS metric

In this section we prove the following main theorem.

###### Theorem 3.1

Let be defined by (LABEL:eq.ss). Then, is a metric on .

The proof of Theorem LABEL:thm.metric depends on several results, which we first establish.666We note that a referee wondered whether the “metrization” results of [22, 23] could yield an alternative proof of Thm. LABEL:thm.metric. Unfortunately, those results rely very heavily on the commutativity and total-order available on , both of which are missing in . Indeed, for the commutative case, Lemma LABEL:lem.cpd alone suffices to prove that is a metric.

###### Definition 3.2 ([7, Def. 1.1])

Let be a nonempty set. A function is said to be negative definite if for all , , and the inequality

 ∑ni,j=1cicjψ(xi,xj)≤0,

holds for all integers , and subsets , with .

###### Theorem 3.3 ([7, Prop. 3.2, Ch. 3])

Let be negative definite. Then, there is a Hilbert space and a mapping from such that one has the relation

 ∥φ(x)−φ(y)∥2H=12(ψ(x,x)+ψ(y,y))−ψ(x,y). \hb@xt@.01(3.1)

Moreover, negative definiteness of is necessary for such a mapping to exist.

Theorem LABEL:thm.schoen helps prove the triangle inequality for the scalar case.

###### Lemma 3.4

Define, the scalar version of as

 δs(x,y):=√log[(x+y)/(2√xy)],x,y>0.

Then, satisfies the triangle inequality, i.e.,

 δs(x,y)≤δs(x,z)+δs(y,z)for all  x,y,z>0. \hb@xt@.01(3.2)

Proof. We show that is negative definite. Since , Theorem LABEL:thm.schoen then immediately implies the triangle inequality (LABEL:eq.11). To prove that is negative definite, by [7, Thm. 2.2, Ch. 3] we may equivalently show that is a positive definite function for , and . To that end, it suffices to show that the matrix

 H=[hij]=[(xi+xj)−β],1≤i,j≤n,

is HPD for every integer , and positive numbers . Now, observe that

 hij=1(xi+xj)β=1Γ(β)∫∞0e−t(xi+xj)tβ−1dt, \hb@xt@.01(3.3)

where is the Gamma function. Thus, with , we see that equals the Gram matrix , whereby .  \qedsymbol Using Lemma LABEL:lem.cpd obtain the following “Minkowski” inequality for .

###### Corollary 3.5

Let ; and let . Then,

 (∑iδps(xi,yi))1/p≤(∑iδps(xi,zi))1/p+(∑iδps(yi,zi))1/p. \hb@xt@.01(3.4)

Proof. Lemma LABEL:lem.cpd implies that for positive scalars , , and , we have

 δs(xi,yi)≤δs(xi,zi)+δs(yi,zi),1≤i≤n.

Exponentiate, sum, and invoke Minkowski’s inequality to conclude (LABEL:eq.36).  \qedsymbol

###### Theorem 3.6

Let be diagonal matrices. Then,

 δS(X,Y)≤δS(X,Z)+δS(Y,Z) \hb@xt@.01(3.5)

Proof. For diagonal matrices and , it is easy to verify that Now invoke Corollary LABEL:corr.mink with .  \qedsymbol

Next, we recall an important determinantal inequality for positive matrices.

###### Theorem 3.7 ([9, Exercise VI.7.2])

Let . Let denote the vector of eigenvalues of sorted in decreasing order; define likewise. Then,

 ∏ni=1(λ↓i(A)+λ↓i(B))≤det(A+B)≤∏ni=1(λ↓i(A)+λ↑i(B)). \hb@xt@.01(3.6)
###### Corollary 3.8

Let . Let denote the diagonal matrix with as its diagonal; define likewise. Then,

 δS(Eig↓(A),Eig↓(B)) ≤ δS(A,B) ≤ δS(Eig↓(A),Eig↑(B)).

Proof. Scale and by 2, divide each term in (LABEL:eq.14) by , and note that is invariant to permutations of ; take logarithms to conclude.  \qedsymbol

The final result we need is well-known in linear algebra (we provide a proof).

###### Lemma 3.9

Let , and let be Hermitian. There is a matrix for which

 P∗AP=I,andP∗BP=D,where D is diagonal. \hb@xt@.01(3.7)

Proof. Let , and define . The the matrix is Hermitian; so let diagonalize it to . Set , to obtain

 P∗AP=V∗S∗U∗UΛU∗USV=V∗U∗Λ−1/2ΛΛ−1/2UV=I;

and by construction, it follows that .  \qedsymbol

Accoutered with the above results, we can finally prove Theorem LABEL:thm.metric.

Proof. (Theorem LABEL:thm.metric).   We need to show that is symmetric, nonnegative, definite, and that is satisfies the triangle inequality. Symmetry is obvious. Nonnegativity and definiteness were shown in Prop. LABEL:prop.scvx. The only hard part is to prove the triangle inequality, a result that has eluded previous attempts [24, 21].

Let be arbitrary. From Lemma LABEL:lem.diag2 we know that there is a matrix such that and . Since is arbitrary, and congruence preserves positive definiteness, we may write just instead of . Also, since (see Prop. LABEL:prop.basic), proving the triangle inequality reduces to showing that

 δS(I,D)≤δS(I,Z)+δS(D,Z). \hb@xt@.01(3.8)

Consider now the diagonal matrices and . Theorem LABEL:thm.diag asserts

 δS(I,D↓)≤δS(I,Eig↓(Z))+δS(D↓,Eig↓(Z)). \hb@xt@.01(3.9)

Prop. LABEL:prop.basic(i) implies that and , while Corollary LABEL:cor.eigs shows that . Combining these inequalities, we immediately obtain (LABEL:eq.35).

We now turn our attention to a connection of importance to machine learning and approximation theory: kernel functions related to . Indeed, some of connections (e.g., Theorem LABEL:thm.wallach) have already been recently useful in computer vision [31].

### 3.1 Hilbert space embedding

Since is a metric, and Lemma LABEL:lem.cpd shows that for scalars, embeds isometrically into a Hilbert space, one may ask if also admits such an embedding. But as mentioned previously, it is the lack of such an embedding that necessitated a different route to metricity. Let us look more carefully at what goes wrong, and what kind of Hilbert space embeddability does admit.

Theorem LABEL:thm.schoen implies that a Hilbert space embedding exists if and only if is a negative definite kernel; equivalently, if and only if the map (cf. Lemma LABEL:lem.cpd)

 e−βδ2S(X,Y)=det(X)βdet(Y)βdet((X+Y)/2)β,

is a positive definite kernel for . It suffices to check whether the matrix

 Hβ=[hij]=[det(Xi+Xj)−β],1≤i,j≤m, \hb@xt@.01(3.10)

is positive definite for every and arbitrary HPD matrices .

Unfortunately, a quick numerical experiment reveals that can be indefinite. A counterexample is given by the following positive definite matrices (, )

 X1=[0.14060.03470.03470.1779], X2=[2.01950.00660.00660.2321], X3=[1.09240.06090.06091.2520],X4=[1.03090.86940.86941.2310], and  X5=[0.2870−0.4758−0.47582.3569], \hb@xt@.01(3.11)

and by setting , for which . This counterexample destroys hopes of embedding the metric space isometrically into a Hilbert space.

Although matrix (LABEL:eq.60) is not HPD in general, we might ask: For what choices of is HPD? Theorem LABEL:thm.wallach answers this question for formed from symmetric real positive definite matrices, and characterizes the values of necessary and sufficient for to be positive definite. We note here that the case was essentially treated in [27], in the context of semigroup kernels on measures.

###### Theorem 3.10

Let be real symmetric matrices in . The matrix defined by (LABEL:eq.60) is positive definite, if and only if satisfies

 β∈{j2:j∈N, and 1≤j≤(n−1)}∪{γ:γ∈R,and γ>12(n−1)}. \hb@xt@.01(3.12)

Proof. We first prove the “if” part. Recall therefore, the Gaussian integral

 ∫Rne−xTXxdx=πn/2det(X)−1/2.

Define the map , where the inner-product is given by

 ⟨fi,fj⟩:=1πn/2∫Rne−xT(Xi+Xj)xdx=det(Xi+Xj)−1/2.

Thus, it follows that . The Schur product theorem says that the elementwise product of two positive matrices is again positive. So, in particular is positive whenever is an integer multiple of . To extend the result to all covered by (LABEL:eq.62), we invoke another integral representation: the multivariate Gamma function, defined as [42, §2.1.2]

 Γn(β):=∫Pne−tr(A)det(A)β−(n+1)/2dA,where β>12(n−1). \hb@xt@.01(3.13)

Let , for some constant ; then, compute the inner product

 ⟨fi,fj⟩:=c′∫Pne−tr(A(Xi+Xj))det(A)β−(n+1)/2dA=det(Xi+Xj)−β,

which exists if . Thus, for all defined by (LABEL:eq.62).

The converse is a deeper result grounded in the theory of symmetric cones. Specifically, since is a symmetric cone, and is decreasing on this cone, an appeal to [30, Thm. VII.3.1] yields the only if part of our claim.

###### Remark 3.11

Let be a set of HPD matrices that commute with each other. Then, can be isometrically embedded into a Hilbert space. This claim follows because a commuting set of matrices can be simultaneously diagonalized, and for diagonal matrices, , which is a nonnegative sum of negative definite kernels and is therefore itself negative definite.

Theorem LABEL:thm.wallach shows that is not a kernel for all , while Remark LABEL:rmk.commute mentions an extreme case for which is always a positive definte kernel. This prompts us to pose the following problem.

Open problem 1. Determine necessary and sufficient conditions on a set , so that is a kernel function on for all .

## 4 Connections with δR

This section returns to our original motivation: using as an alternative to the Riemannian metric . In particular, in this section we show a sequence of results that highlight similarities between and . Table LABEL:tab.summ lists these to provide a quick summary. Thereafter, we develop the details.

### 4.1 Geometric mean

We begin by studying an object that connects and most intimately: the matrix geometric mean (GM). For HPD matrices and , the GM is denoted by , and is given by the formula

 A♯B:=A1/2(A−1/2BA−1/2)1/2A1/2. \hb@xt@.01(4.1)

The GM (LABEL:eq.47) has numerous attractive properties—see for instance [1]—among which, the following variational characterization is very important [11]:

 A♯B=argminX>0δ2R(A,X)+δ2R(B,X),andδR(A,A♯B)=δR(B,A♯B). \hb@xt@.01(4.2)

Surprisingly, the GM enjoys a similar characterization even with .

###### Theorem 4.1

Let . Then,

 A♯B=argminX>0[h(X):=δ2S(X,A)+δ2S(X,B)]. \hb@xt@.01(4.3)

Moreover, is equidistant from and , i.e., .

Proof. If , then clearly minimizes . Assume therefore, that . Ignoring the constraint for the moment, we see that any stationary point of must satisfy . This condition translates into

 ∇h(X)= (X+A2)−112+(X+B2)−112−X−1=0, ⟹X−1=(X+A)−1+(X+B)−1 \hb@xt@.01(4.4) ⟹(X+A)X−1(X+B)=2X+A+B ⟹B=XA−1X.

The last equation is a Riccati equation whose unique, positive definite solution is the geometric mean (see [10, Prop 1.2.13]).

Next, we show that this stationary point is a local minimum, not a local maximum or saddle point. To that end, we show that the Hessian is positive definite at the stationary point . The Hessian of is given by

 2∇2h(X)=X−1⊗X−1−[(X+A)−1⊗(X+A)−1+(X+B)−1⊗(X+B)−1].

Writing , and , upon using (LABEL:eq.33) we obtain

 2∇2h(X)=(P+Q)⊗(P+Q)−P⊗P−Q⊗Q=(P+Q)⊗P+(P+Q)⊗Q−P⊗P−Q⊗Q=(Q⊗P)+(P⊗Q)>0.

Thus, is a strict local minimum of (LABEL:eq.11). This local minimum is actually global as has a unique positive solution and goes to at the boundary.

To prove the equidistance, recall that ; then observe that

 S(A,A♯B) =S(A,B♯A)=S(A,B1/2(B−1/2AB−1/2)1/2B1/2) =S(B−1/2AB−1/2,(B−1/2AB−1/2)1/2) =S(B1/2(B−1/2AB−1/2)1/2B1/2,B) =S(B♯A,B)=S(B,A♯B).\qedsymbol

### 4.2 Geodesic convexity

The above derivation concludes optimality from first principles. It was originally driven by the fact that is not geodesically convex [48]. However, we recently realized that the square is actually geodesically convex. This realization leads to more insightful proof of uniqueness and optimality of the S-mean.

In fact even more is true: Theorem LABEL:thm.jointgc shows that is not only geodesically convex, it is jointly geodesically convex. Before proving Theorem LABEL:thm.jointgc, we recall two useful results; the first immediately implies the second777It is a minor curiosity to note that [41, Thm. 2] proved a mixed-mean inequality for the matrix geometric and arithmetic means; Prop. LABEL:prop.mixedmean includes their result as a special case..

###### Theorem 4.2 ([36])

The GM of is given by the variational formula

###### Proposition 4.3 (Joint-concavity (see e.g. [36]))

Let . Then,

 (A♯B)+(C♯D)≤(A+C)♯(B+D). \hb@xt@.01(4.5)

Proof. From one application of Thm. LABEL:thm.andomax we have

 0⪯[AA♯BA♯BB]+[CC♯DC♯DD]=[A+CA♯B+C♯DA♯


text:
1. ## simple questions need answer # 3

3. Let the joint density of (X,Y) be given by f(x,y)=2, 0<X<1, 0<Y<1-X
a) Calculate marginal density f(x),
b) Calculate P(0<Y<3/4|X=0.5).

Thanks~

2. the constraint is $x+y\le 1$

the two rvs are uniformly distributed over that triangle in the first quadrant

so $f_X(x)=\int_0^{1-x}2dy=2(1-x)$ on 0<x<1

$f(y|x)= {2\over 2(1-x)}={1\over 1-x}$ on 0<x+y<1

THIS IS UNIFORMLY DISTRIBUTED on any line for a fixed x

$f(y|x=.5)= 2$ on 0<y<.5

so P(0<Y<3/4|X=0.5) is a silly question, the answer is 1.

$P(0


text:
# Latex & Vanilla with Axelle

No, I did not go out for ice cream with Axelle after our latex shoot. Although that would be a great idea actually! I must keep that in mind. We shot 2 sets where Axelle was wearing a mix of latex and plain clothes. In the latex scene, plain clothes are often referred to as vanilla. That’s why I choose this title for the blog post.

In this set she wears a black & white vanilla top over a high waisted latex pencil skirt from William Wilde and a pair of amazing high heels from Pleaser Shoes. I will post pics from the 2nd set pretty soon. Stay tuned!


text:
# Almost conjugation-invariant neighborhoods of units in locally compact groups

Let $G$ be a locally compact topological group with unity $e$ and left Haar measure $m$. Let also $g\in G$ be a given element and $U$ a neighborhood (of compact closure) of $e$.

I am interested to get another, smaller neighborhood $V$, contained in $U$, with the conjugation-invariance property $gVg^{-1}=V$ (with respect to the given element $g$).

This is immediate for Abelian $G$. Also, it is easy if the order of $g$ is finite, say $o(g)=n$, because then we can just take $V:=\cap_{j=0}^{n-1} g^jUg^{-j}$.

If $o(g)=\infty$, this may fail, as relatively easy examples can show. However, I would be satisfied by a quazi-invariance property: for any $0<\varepsilon<1$, find some $V\subset U$ with $m(V\cap gVg^{-})> (1-\varepsilon)m(V)$.

When do we have the first, or the approximative second, invariance property? What generally well-known properties, conditions ensure it and what groups (and for what kind of elements, besides of course having $o(g)=\infty$) or perhaps neighborhoods, (if the property depends on that) are such that it fails?

-
There's a function $\Delta$ defined by $\Delta(x)$ equals the amount by which right multiplication by $x$ scales the left Haar measure. If $G$ is compact or abelian then $\Delta$ is identically 1, but for groups like the upper triangular matrices in $GL(2,k)$ for $k$ your favourite locally compact field (e.g. the reals or the $p$-adics) this function is non-trivial. If $\Delta(g)$ is very big then the measure of $gVg^{-1}$ is much smaller than the measure of $V$, and in this case it seems to me that your quasi-invariance property cannot possibly hold. –  Kevin Buzzard Aug 1 '12 at 22:52
Regarding your first question: groups where you can find such $V$ for every given $U$ and $g$ are the so-called SIN groups (where SIN stands for "small invariant enighbourhoods"). There is quite a lot known about which groups can and can't be SIN, and even a structure theorem for them due to Grosser and Moskowitz. Would further details be informative, or is this too strong a condition for what you are after? –  Yemon Choi Aug 2 '12 at 0:29
Thanks, indeed. I would be very much interested in all you can tell about these SIN groups, references, description, related works etc. I am generalizing a result from already know but rather particular cases to the extent I possibly can: already the extension to the case of locally compact Abelian groups is interesting (to me, at least). But I wish to go as far as one possibly can. Therefore, a better description of this condition - which is shown to be sufficient to the generalization I am after - is certainly useful. –  Szilárd RÉVÉSZ Aug 2 '12 at 7:23


text:
Thread: Integral of Bessel function View Single Post

Integral of Bessel function

Use the recurrence relation:

$$J_{n-1}(x) = \frac{2n}{x} J_{n}(x) - J_{n+1}(x)$$

to write the integral as

$$\int x^3 J_0(x)dx = \int x^3 (\frac{2}{x} J_{1}(x) - J_{2}(x)) dx = \int (2 x^2 J_{1}(x) - x^3 J_{2}(x) ) dx$$

then use the relation

$$x^n J_{n-1}(x) = \frac{d}{dx}[x^n J_{n}(x)]$$

on each of the terms and perform the integration...


text:
# Optocoupler to drive LEDs (maybe with transistor)

I wonder if there is some simple solution to drive LEDs powered from single rectified voltage.

Assume AC is 220 V, each LEDs is 3.0 - 3.2 V 150 mA. I reckon series connection made it simpler but not sure if the optocoupler is able to provide the current.

Adding a BJT/MOSFET at the opto output should be OK but I am not sure how to connect it. Protective part can be advised but I wish to have the basic ON/OFF function working first.

EDIT on 2019-01-28 12:34

Actually, my design consists of two 30xLEDs paths and wish to end them with an optocoupler to Neutral.

Thanks to JonRB and Transistor, they discovered several fatal issues, so i retry and hope this would make more sense.

Refer to this active current limiter, i wonder if the optocoupler can attach to it. Now, with 60 x LEDs (3.0 - 3.2 V 150 mA):

100k Resistor to turn IRF740 FET On.
8 Ohm Resistor with BC817 BJT to limit the current to ~150mA.
Optocoupler PC817 would shutdown both transistors.

I think IRF740((400Vds) should be capable of AC peak voltage and current limiter can save the LEDs from destory as well.

When in doubt, check the datasheet:

https://www.futurlec.com/Datasheet/LED/PC817.pdf

The datasheet has two key pieces of information pertinent to your use-case

1. Collector current - 50mA
2. Collector-Emitter voltage - 35V

Your use case is targeting 150mA and thus this part is not suitable. Likewise you are stating this is powered from 220V (thus a peak of ~310V) yet this part has a Vce of 35V

Basically this part is not suitable for direct blocking.

A suitably rated FET would be. However... It isn't that simple. I have included an example circuit WITHOUT values as working with main potential isn't something you should do, unless you know what you are doing

The FET needs to have a blocking potential greater than the maximum DCLink voltage. It must be able to sink your target LED current.

The Zener is acting as a crude voltage regulator to limit the voltage applied to the gate. The resistive chain is dropping the vast majority of the DCLink potential.

simulate this circuit – Schematic created using CircuitLab

• I understand PC817 won't work. I'd first remove the optocoupler label. I also try to avoid another track from the main. maybe I should add this to my question as well. btw, do you think it is impossible? although PS2533 is not popular, do you think it could do the job? – Simon Jan 27 '19 at 10:40
• Please note that your circuit will destroy the LEDs - you have not included a current-limiting component. While I understand that you were focussing on the LED switching, the OP clearly does not understand the implications of your simplification. – WhatRoughBeast Jan 27 '19 at 19:17
• As it stands yes (unless you say you had 200odd in series and the RDS on is sufficient). – JonRB Jan 27 '19 at 19:19

The PC817 datasheet lists the Absolute Maximum Ratings as

Collector-emitter voltage    80 V
Emitter-collector voltage     6 V
Collector current            50 mA
Collector power dissipation 150 mW

With a 220 V AC supply and a diode rectifier you will have $$\ 200 \sqrt 2 = 350 \ \text {V DC}\$$ on the opto-terminals. It will die instantly.

Meanwhile, you have 30 LEDs in series with a minimum forward voltage drop of 3 V giving 90 V across the LEDs. Therefore the worst case voltage drop across the two 120 Ω parallel resistors is 350 - 90 = 260 V. The resistors will limit the current to $$\ I = \frac {V}{R} = \frac {260}{60} = 4.3 \ \text A \$$. The LEDs will be spectacularly brightly for a very short while.

Note that the power dissipated in each resistor is given by $$\ P = I^2R = 2.15 \times 120 = 555 \ \text W \$$.

Back to the drawing board, I'm afraid.

• ok. thanks. seem I have to vanish this idea....maybe try a relay at rectifier input instead. – Simon Jan 27 '19 at 10:50
• @Simon Your idea isn't impossible. You've just not considered the parts to use properly; you need to think about ratings when selecting parts. – Hearth Jan 27 '19 at 12:08
• @Simon - You seem to have concluded that you only need to use a relay rather than an optoisolator. This is not true. Please pay attention to the calculations involving LED current and resistor power. If you use a relay or a MOSFET (perJonRBs answer) you will destroy your LEDS. If, somehow, you don't, you will instead destroy your 180 ohm resistors. – WhatRoughBeast Jan 27 '19 at 19:16
• ok, I should not give up so easily since I started up this question. please check if this current limiter works? – Simon Jan 28 '19 at 8:26
• Have you calculated the power dissipation in your current limiter yet? (I haven't but I suspect that it's going to be very high.) – Transistor Jan 28 '19 at 13:58