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Examples of when $\tan(x) = \frac{\sin(x)}{\cos(x)}$ is useful

I'm making a video right now about the unit circle definitions of the basic trig functions.

I've done sine and cosine, and am now talking about the tangent function.

As most of you know, it can be rewritten as in the title, but off the top of my head, I can't remember when this might be genuinely useful.

Certain derivatives and integrals come to mind. Do you have any other examples?

• Most trig. identities involving tangent can be reduced to those in sine and cosines, usually easier since you have to remember fewer formulas. – Chris C Oct 3 '14 at 21:28
• @BenjaminDickman - The video is one of over 900 math videos I've made so far. The current playlist involves trigonometric equations/inequalities, and so I started with an introduction to the unit circle definitions of the basic trig functions. When I asked this question, I was mid-video (on pause) and looking for quick examples. Now, days later, it is a non-issue of course. – Alec Oct 7 '14 at 8:46
• @BenjaminDickman - Yes, it is a public resource. However, it's in Norwegian. It's widely used among students in Norway, as teachers tend to assign videos as homework. Personally, I just started making it becuase I benefited from Khan Academy, but my friends didn't, because their English wasn't quite good enough to follow math while interpreting a foreign language. I got carried away and am still continuing. – Alec Oct 7 '14 at 12:30
• @BenjaminDickman - Oh, yes. Forgot about that. udl.no – Alec Oct 8 '14 at 12:30
• I think a simple example that will "click" with students is an airplane's vertical distance and horizontal distance. It gives a visual picture, as well as being a real example. Yes, we probably use sin most, cos next and tangent least. And a lot of times with tan we just change back to sin/cos definition. But there is one practical example. – guest Apr 8 '18 at 0:35

• You are simplifying the equation for a function and you get down to something like $f(x) = \cos x\tan x$.

• You want to know where tan(x) is discontinuous without memorizing it (assuming you already know where cos(x) is zero).

My previous answer (which was basically "That is the definition of tangent!") was not well received. I admit that that answer was made out of exasperation with the question. I have written a longer response as to why I feel the way I do, with some mathematical and historical context.

You say you are defining these functions based on the unit circle. If this is the case, then you are probably defining $\cos(\theta)$ to be the $x$-coordinate, and $\sin(\theta)$ to be the $y$-coordinate of the point obtained by taking a piece of string of length $\theta$, fixing one end at $(1,0)$, and winding it it counterclockwise around the circle. In this case, if you define tangent to be $\frac{y}{x}$ this is also defining tangent to be $\frac{\sin(\theta)}{\cos(\theta)}$. The two should be so closely identified that these two expressions "say" exactly the same thing to you. Because of this I am confused by the question: it appears that the definition of tangent is $\frac{\sin(\theta)}{\cos(\theta)}$.

The unit circle approach takes some care to make rigorous. You at least need to be able to define arclength to make it work. For this reason, most introductions to analysis which I have seen either define sine and cosine in terms of their power series representations, or as solutions to their defining differential equations. In either content, the tangent function would again be defined by the ratio of the two. Trying to define tangent "on its own", without first defining sine and cosine, is difficult because it is not an entire function.

In a certain sense, I am suspicious of the attention given to the other trig functions. We no longer pay homage to haversine or exsecant, yet we still talk about tangent. Why? Fundamentally the reason that tangent, secant, exsecant, etc even have names of their own is that they were needed in some computations quite commonly (say, while sailing), and it was easier to compile tables of values for these "special functions" than to calculate them by hand given only tables for sine and cosine. I do not know the historical reason that only 6 of these functions survived into the common usage, but from a mathematical perspective I feel like only sine and cosine really "deserve" names in the current era. It would be as if we had a new operation "inc sum", denoted $\oplus$ with $a \oplus b = a+b+1$. A useful operation, to be sure, but not useful enough to warrant its own distinct mental real estate.

I also think that the habit of questioning where each and every mathematical fact is "useful" is misguided. Asking about the "usefulness" of an identity which is at most one mental step away from the definition is not a healthy attitude with which to approach mathematics. When someone is learning mathematics, they should always be exploring 3 or 4 steps away from what is written. This is just part of building a mental model, gaining "familiarity with the landscape". Playing with identities to understand the basic relationships between various functions should be appreciated for its own sake. Individual mathematical facts are not useful: a mathematical outlook is useful, a mathematical framework is useful. This particular fact is just one part of building up the mathematical framework of trigonometry.

This is not to say all questions of the form "Examples of when this identity is useful" are not good. For example, I think that the equality expressed by "completing the square" on an arbitrary quadratic is "just an identity", but it has a huge story behind it, and how it connects to the larger story of understanding quadratics. So asking "how is this useful", can prompt an introduction to a whole framework of thinking. Similar comments apply to many identities, like the fundamental theorems of calculus, or Cauchy's formula, etc. $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$ on the other hand, is on the same level as $(x+1)(x-1) = x^2-1$ (or even lower on the totem pole): it is just an immediate consequence of definitions. One should not have to justify exploring it. It is not a pivotal part of any story, so it should just be one of those things to observe in passing and appreciate for what it is.

Edit: perhaps the following analogy is useful. If you give someone a blanket, they might ask "What is the use of this particular thread?". The answer to this question for any thread is that, all by itself, it is useless. When you weave all of the threads together you get an object which is incredibly useful. This is the way with mathematics as well. Mathematics, when properly understood, is a powerful framework for understanding many diverse problems. $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$ is like one of those threads.

• Not really sure why this is down voted. Most sources I have seen do define all of the other trig functions in terms of sine and cosine. How do you even assign signs to the tangent function correctly without knowing this? – Steven Gubkin Oct 3 '14 at 22:50
• I think what happened here is that the answer is so short, it showed up in review queues as a "low quality answer." – Chris Cunningham Oct 4 '14 at 4:02
• I've always defined tangent theta as opposite/adjacent (using right triangles) or as y/x (using an angle in standard position crossing a circle). Saying that tangent theta is defined to be sin theta / cos theta would put its definition in a different context than the definitions of sine and cosine. That seems odd to me. (@Steven, can you show me any of the sources you mention?) Downvote. – Sue VanHattum Oct 4 '14 at 14:19
• Thanks for expanding your answer. I changed my vote. I can connect to the sentiment of surprise. – quid Oct 4 '14 at 20:08
• Concerning "from a mathematical perspective I feel like only sine and cosine really "deserve" names in the current era", I strongly disagree. Without a name for $\tan$, writing the formulas for $\tan(2x)$ and the like would be incredibly painful. Performing a change of variable $t=\tan(x/2)$ would not be a great experience either. In some sense, once one has complex numbers at hand, why would you even bother with $\sin$ and $\cos$, instead of relying only on $\exp$? Why $\cosh$ and $\sinh$? In all cases, the answer seems twofold to me: these notation are computationaly efficient even today... – Benoît Kloeckner Oct 19 '14 at 19:59

One reason the tangent function is useful is that it yields the slope of a line, $\tan \theta$ where $\theta$ is the angle of a line tangent to a curve at a point (and note the connection of the tangent function to a line tangent to a curve):

If you identify $y$ with $\sin \theta$ and $x$ with $\cos \theta$, the connection to slope could be explained.

More pragmatically, you can use the tangent to estimate the height $y$ of a building: If you are standing about $x$ units from the building, and you need to look up $\theta$ to the top, then $y = x \tan \theta$. Again, this could be explained in terms of $\sin \theta$ and $\cos \theta$.

• +1 even though you seem to be addressing Steven's remarkably long edited answer (his answer, in effect, an objection to the question itself) and the actual question. I now have a fast answer to my student's question, "TAN, what is it good for?" – JTP - Apologise to Monica Oct 5 '14 at 14:35


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# $\mathbb{S}^n$ without two points

In "An Introduction to Algebraic Topology" of Rotman, Exercise 1.31 asks to show that the equator $$\mathbb{S}^{n-1}$$ is a deformation retract of $$\mathbb{S}^n\setminus\{a, b\}$$.

I thought that if one thinks to the usual sphere in the space, then one just enlarge the holes to flatten $$\mathbb{S}^3$$ onto its equator, but I cannot understand how to prove this in general for $$n$$, probably it's the same idea but I cannot construct the required function.

• Do you know that $\mathbb{R}^n$ is homeomorphic to $S^n\setminus\{a\}$ ? – Max May 24 at 19:21
• @Max Yes, is there a way to deduce what I asked from that? – W4cc0 May 24 at 19:28
• Well do you know that $\mathbb{R}^n\setminus\{b\}$ strongly deformation retracts onto $S^{n-1}$ ? – Max May 24 at 19:29
• No, unfortunately that is precisely what I do not know... and can't prove. – W4cc0 May 24 at 19:30
• but, perhaps looking at math.stackexchange.com/questions/659682/… could solve this latter fact, and then probably what I asked should easily follow – W4cc0 May 24 at 19:33

As per the comments, $$S^n\setminus\{a\}$$ is homeomorphic to $$\mathbb{R}^n$$.
Now it suffices to see that $$\mathbb{R}^n\setminus\{0\}$$ (strongly) deformation retracts onto $$S^{n-1}$$ (if you know that you just have to move things around a bit to get $$b$$ to align with $$0$$ and $$S^{n-1}$$ to align with the image of the equator)
For this, consider $$r:x\mapsto \frac{x}{||x||}$$, which is indeed a retraction. Moreover, define $$H(x,t)= r(x)t+(1-t)x$$. Clearly it is continuous, and it is easy to check that it does land in $$\mathbb{R}^n\setminus\{0\}$$ (and not just $$\mathbb{R}^n$$), and it is clearly a homotopy between $$id$$ and $$r$$ (I should really write $$i\circ r$$, where $$i$$ is the inclusion);
and moreover (although it is not necessary to get a deformation retract, that we already got) you can check that $$H$$ is constant equal to the identity on $$S^{n-1}$$

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# Differential Equations - Existence and Uniqueness

1. Feb 4, 2012

### prosteve037

I'm having trouble understanding what uniqueness is/means. When given a slope/direction field I don't know what I should be looking for if asked to determine if a given initial condition has a unique solution.

Example:
$\textit{y' = }\frac{(x - 1)}{y}$

With this equation I can see that as long as $\textit{y ≠ 0}$ a solution exists.

But now if I'm asked to find a given interval where a solution exists and is unique, I'm confused :P What should I be looking for in the equation/direction field?

Here's the direction field for that differential equation:

2. Feb 4, 2012

### LCKurtz

Your differential equation is of the form $y' = f(x,y)$ with $f(x,y) = \frac{(x - 1)}{y}$. What does the existence and uniqueness theorem require about $f(x,y)$? Does your $f(x,y)$ satisfy the requirements? Look at the theorem.

3. Feb 5, 2012

### prosteve037

Is it that there must only be one function $y(x)$ that passes through the given point?

But how do you prove that this is the only solution that passes through that point? Couldn't there be other functions that pass through that point?

From the example above, the point $(0,1)$ has a solution with slope -1. But how can I justify that there is only one function that passes through this point?

4. Feb 5, 2012

### LCKurtz

I presume your text has a theorem called the existence and uniqueness theorem. Theorems have hypotheses and conclusions. The conclusion of the theorem is that there is a unique solution through the point, provided that the hypotheses on $f(x,y)$ are satisfied. That is how you prove there is only one solution through the point. So I will ask you again: what are the hypotheses and does this f(x,y) satisfy them?

5. Feb 6, 2012

### prosteve037

Oh wow, I totally glossed over that in the text :grumpy: It says if $f(x,y)$ and $D_{y}f(x,y)$ are continuous, $(x,y)$ has a unique solution.

So then does this mean if I take the partial derivative of $f(x,y)$ with respect to y and plug in the values for x and y, yielding a real solution, then the point $(x,y)$ has a unique solution?

In other words if $f(x,y)$ is defined, a solution exists. If $D_{y}f(x,y)$ is defined as well, the solution is unique. Is this correct?

6. Feb 6, 2012

### HallsofIvy

Staff Emeritus
"Yielding a real value" is in general NOT sufficient to be continuous. If it happens that your function is a "rational function", a polynomial divided by another, then the function is continuous as long as the denominator is not 0, but other kinds of functions do not satisfy that.

Also, by the way, the partial derivative being continuous is "sufficient" but not "necessary". The necessary condition is that f(x,y) be "Lipschitz" in y- that is, that for some y1 and y2 in a neighborhood of the y value for the initial point, there exist a constant, c, such that |f(x,y1)- f(x,y2)|< c|y1- y2|. It can be shown that being "Lipschitz" is 'stronger' than "continuous" but 'weaker' than "continous differential". That is, any function that has a continuous derivative is Lipschitz but there exist Lipschitz functions that are do not have a continuous differivative and every Lipschitz function is continuous while there exist continuous functions that are not Lipschitz.

7. Feb 6, 2012

### LCKurtz

It is not "the point $(x,y)$" that has a unique solution. It is the differential equation that has a unique solution passing through that point. All you have to figure out is for what $(x,y)$ are both $f(x,y)$ and $f_y(x,y)$ continuous. The theorem guarantees a unique solution through those points.

8. Feb 6, 2012

### prosteve037

Ahh okay thank you so much!

Now how about the example from the above post? How would I explicitly find solutions for these intervals where the solution exists and is unique?

For that D.E. $f(x,y) = \frac{x-1}{y}$ I took the partial derivative:

$f_{y}(x,y) = -\frac{x - 1}{y^{2}}$

Which means that there exists a unique solution for the differential equation wherever $y ≠ 0$. But through which values of x does y ≠ 0?

Would it depend on the function $y(x)$ chosen? And so then could it be any function $y(x)$ where $y(x)\propto{ x}$?

9. Feb 6, 2012

### LCKurtz

x and y are independent variables in the xy plane. Whether $y\ne 0$ has nothing to do with x, which can be anything.
That last statement makes no sense to me. If you actually want to find a solution through, for example, (0,1), then solve the equation by separation of variables. You haven't given an exact statement of the problem you are trying to solve, but solving for a solution and knowing that there is a unique solution through that point are two different things. The theorem tells you that without solving the DE.

10. Feb 6, 2012

### prosteve037

Alright yeah, I'm sorry. I'm just not making sense anymore

Here's the problem statement:

Consider the following differential equation for $y = y(x)$

$yy' = (x - 1)$

a.) Sketch the direction fields and draw possible solution curves.

b.) If the initial condition is given by (x,y) = (0,1) does the solution exist near this point and is it unique if it exists?

The solution does exist near this point and is unique because both the differential equation and its partial derivative with respect to y are continuous at that point.

c.) If the initial condition is given by (x,y) = (1,0) does the solution exist near this point and is it unique if it exists?

The solution does not exist near this point because the differential equation is discontinuous at that point.

d.) Give several intervals where the solution exists and is unique.

... This is where I'm stuck :/

Sorry again for the confusion, I should've chosen the right sentences to express my thoughts more clearly before actually posting them

11. Feb 6, 2012

### LCKurtz

OK, you have drawn what looks like a solution. Have you written down its equation and verified that it is a solution? Is it a solution on [-1/2,1/2]? [-2,2]? Think about that.

12. Feb 6, 2012

### prosteve037

Its equation is $y(x) = -x + 1$ :

$y = -x + 1$

$y' = -1$

$y' = \frac{x - 1}{y}$

$y' = \frac{x - 1}{-x + 1}$

$y' = -1$

And yes, it's a solution in those intervals:

$y(x) = -x +1$

$y(-\frac{1}{2}) = \frac{1}{2} + 1 = \frac{3}{2}$

$y'(-\frac{1}{2}) = -1$

$y(\frac{1}{2}) = -\frac{1}{2} + 1 = \frac{1}{2}$

$y'(\frac{1}{2}) = -1$

$y(-2) = 2 + 1 = 3$

$y'(-2) = -1$

$y(2) = -2 + 1 = -1$

$y'(2) = -1$

13. Feb 6, 2012

### LCKurtz

Really? Does it satisfy your DE at (1,0), which is in the interval [-2,2]?

14. Feb 6, 2012

### prosteve037

Oh oops! No it doesn't, it's discontinuous at that point :P

So then are the intervals $[\frac{-1}{2}≤ x ≤ \frac{1}{2}], [x ≥ 2], [x ≤ -2]$ for $y(x) = -x + 1$?

15. Feb 6, 2012

### LCKurtz

The solution through (0,1) is not valid on two of those intervals.


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# [texhax] invalid character in tengtex

Mon Aug 17 10:01:19 CEST 2015

Without access to the original (possibly shortened) doc, it might not be easy to help

But some thing for LaTeX that was made in 1994 might have made some assumptions that are not true any more.

P.S. I've never tried TengTeX so without an mwe I cannot test/help.

Institut for Matematik / Department of Mathematics
Aarhus Universitet / Aarhus University

________________________________
From: texhax [texhax-bounces at tug.org] on behalf of Isen hand [isenhand at yahoo.co.uk]
Sent: 16 August 2015 09:22
To: texhax at tug.org
Subject: [texhax] invalid character in tengtex

I'm getting the following error:

! Text line contains an invalid character.
l.87   \ifx ##2\@lambe {\l at p {\raise.5ex\hbox{^^E
}}{##1}}\else {\l at p ^^E{##1...

In ! tengtex.sty  (TengTeX for LaTeX) 22 September 1994 Version 1.00

Using a hex editor I see:

0A 20 20 5C  69 66 78 20  23 23 32 5C  40 6C 61 6D | .  \ifx ##2\@lam
62 65 20 7B  5C 6C 40 70  20 7B 5C 72  61 69 73 65 | be {\l at p {\raise
2E 35 65 78  5C 68 62 6F  78 7B 5E 5E  45 7D 7D 7B | .5ex\hbox{^^E}}{
23 23 31 7D  7D 5C 65 6C  73 65 20 7B  5C 6C 40 70 | ##1}}\else {\l at p
20 5E 5E 45  7B 23 23 31  7D 7D 5C 66  69 7D 25 0A |  ^^E{##1}}\fi}%.

I don't see any invalid characters. Did I miss it? Any idea how to fix this?

Thanks.

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# Classifying automorphisms using a group presentation.

I'm new to group presentations and after some playing around with the concept, I've tried to find some relatively clear criteria in terms of the defining relations, that would tell us if a map is an automorphism. I would appreciate any input - whether my approach is correct, and whether there's some other more general or neater way.

Let $$X=\{x_1,\dots ,x_n \}$$ be a finite set and let $$R$$ be some finite set of relations on the alphabet of $$X \cup X^{-1}$$. Denote $$G$$ the group corresponding to these defining relations - $$G= \langle X | R \rangle =F_X / N$$, and presume $$G$$ is finite.

I think the following might be true:

Consider a map $$f:X \rightarrow G$$. If $$\langle f(x_1),\dots ,f(x_n) \rangle = G$$ and $$f(x_i)$$ satisfy the corresponding relations when stated in terms of $$x_i$$, then $$f$$ induces an automorphism.

Proof: If we show that $$f$$ induces homomorphism $$\overline{f}$$, it's surjective and hence an automorphism, so we only need to show that $$f$$ actually induces a homomorphism.

Consider the free group $$F$$ on $$X$$, and define $$g:F \rightarrow G$$ by $$w \mapsto f(w_1)f(w_2)\dots f(w_m)$$. We know that $$f(x_i)$$ satisfy the wanted relations, and so the $$N \subseteq ker (g)$$. By the homomorphism theorem there exists a $$\psi$$, such that $$\psi \circ \pi_N = g$$. This $$\psi$$ is the wanted extension of $$f$$.

In some sense this seems excessive - we're dealing with finite groups. An idea of how to deal with this in finite groups:

Presume that any element in $$G$$ can be expressed as a word in $$X$$ of length at most $$k$$. Define a map $$g$$ from words of length $$2k$$ to $$G$$ induced by $$f$$. If we check that for any two words of length $$2k$$ that represent the same element in $$G$$, that their image is the same, we will be done.

I know that none of this is very formal, but I would mostly like to know whether my approach is sensible. I would also appreciate if someone linked some text that deals with this in a more formal way.

• How do you deal with the map $f$ that maps every $x\in X$ to the neutral element $1\in G$? – j.p. Mar 4 '18 at 9:28
• @j.p. We're presuming $\langle f(x_1),..,f(x_n) \rangle = G$, so that can't happen (unless $G$ is trivial, in which case the problem is easy). – John P Mar 4 '18 at 10:47
• Yes, of course now I see it. Somehow I overlooked this assumption. Your reasoning is correct. If the relations are fulfilled, the map extends to a homomorphism. If by assumption the image is the full group, it is surjective, and as the group is finite, any surjective map $G\to G$ is bijective. Like you said. – j.p. Mar 4 '18 at 12:21

By the assumption on $$f$$, we know that $$\bar{f}$$ will be surjective, and as we are mapping between finite groups, it must also be bijective. As the $$f(x_{i})$$ satisfy the given relations, the extension to the map $$\bar{f}$$ is a homomorphism.


text:
# statsmodels.stats.oneway._power_equivalence_oneway_emp¶

statsmodels.stats.oneway._power_equivalence_oneway_emp(f_stat, n_groups, nobs, eps, df, alpha=0.05)[source]

empirical power of oneway equivalence test

This only returns post-hoc, empirical power.

Warning: eps is currently effect size margin as defined as in Wellek, and not the signal to noise ratio (Cohen’s f family).

Parameters
f_statfloat

F-statistic from oneway anova, used to compute empirical effect size

n_groupsint

Number of groups in oneway comparison.

nobsndarray

Array of number of observations in groups.

epsfloat

Equivalence margin in terms of effect size given by Wellek’s psi.

dftuple

Degrees of freedom for F distribution.

alphafloat in (0, 1)

Significance level for the hypothesis test.

Returns
powfloat

Ex-post, post-hoc or empirical power at f-statistic of the equivalence test.


text:
# How many subsets of $[0,1)$ are there modulo null sets?

For subsets $A$ and $B$ of $[0,1)$, say $A\sim B$ iff $\lambda(A\Delta B)=0$ where $\lambda$ is Lebesgue measure.

Question: How many equivalence classes of subsets of $[0,1)$ are there given AC?

I would guess the answer is $2^c$ given AC, but I haven't got a proof.

What got me thinking about this was trying to find a way to say that there are more nonmeasurable sets than measurable ones. There are, of course, $2^c$ of each, but modulo null sets there are only $c$ measurable ones (at least given AC), so if there were more than $c$ subsets modulo null sets, we could say that modulo null sets there are more nonmeasurable sets than measurable ones.

• I don't think that this is really a question about the axiom of choice. I mean, if this is a question about the axiom of choice, we need to start tagging many other questions like that as well. – Asaf Karagila Mar 19 '15 at 20:53
• Agreed, though there are questions about what values for the cardinality are options without AC. (E.g., if all sets are measurable then we get $\le c^2$, assuming that the proof that there are $c$ Borel sets goes through without AC.) – Alexander Pruss Mar 20 '15 at 0:18
• Luckily $\mathfrak c^2=\mathfrak c$ without using choice. Unluckily, the assumption that there are $\frak c$ Borel sets does use the axiom of choice. – Asaf Karagila Mar 20 '15 at 13:54
• Does it use anything beyond countable choice? – Alexander Pruss Mar 20 '15 at 16:14
• I don't think so. – Asaf Karagila Mar 20 '15 at 16:21

Yes, there are $2^\mathfrak{c}$ equivalence classes. In fact, I claim that there is a collection $S$ of $\mathfrak{c}$ disjoint non-null subsets of $[0,1)$; taking all unions of subcollections of $S$ gives $2^\mathfrak{c}$ inequivalent subsets of $[0,1)$.

To construct this $S$, let $N$ be the set of all null Borel sets and enumerate $N\times\mathfrak{c}$ with order type $\mathfrak{c}$. Using this enumeration, define a function $f:[0,1)\to\mathfrak{c}$ by induction such that for each $(n,\alpha)\in N\times\mathfrak{c}$, $f^{-1}(\{\alpha\})\not\subseteq n$. We can do this because at each stage of the induction, we have defined $f$ at fewer than $\mathfrak{c}$ points, and the complement of $n$ has cardinality $\mathfrak{c}$. The collection $S=\{f^{-1}(\{\alpha\})\}_{\alpha\in\mathfrak{c}}$ then consists of $\mathfrak{c}$ disjoint non-null sets.

Here is an earlier version of my answer, which is a slight variant on a standard argument that nonmeasurable sets exist and shows there must be more than $\mathfrak{c}$ equivalence classes, but does not show there must be $2^\mathfrak{c}$ of them. Let $S$ be any collection of $\mathfrak{c}$ subsets of $[0,1)$; we will find a subset $A\subseteq[0,1)$ that is not equivalent to any element of $S$. Let $N$ be the set of all Borel null sets and enumerate $N\times S$ with order type $\mathfrak{c}$. Define the characteristic function of $A$ inductively such that for each $(n,B)\in N\times S$, $A\Delta B\not\subseteq n$; we can do this because at each stage of the induction we have defined the characteristic function at fewer than $\mathfrak{c}$ points and the complement of $n$ has cardinality $\mathfrak{c}$.

• But does "more than $c$" imply "$2^c$"? – Fan Zheng Mar 19 '15 at 21:25
• Thanks! The existence of a set $S$ with the requisite properties (well, or the equivalent claim on the square) also follows from Sierpinski's 1938 partition of the square into perfect sets such that any choice of one element of each of the perfect sets gives a set with full outer measure: eudml.org/doc/213031 – Alexander Pruss Mar 20 '15 at 0:14

Edit. I had posted this answer to complement Eric's original answer, which showed that the number of classes was at least ${\frak c}^+$, since at that time we didn't quite yet know whether there were $2^{\frak c}$ classes. Afterwards, however, Eric improved his answer to get $2^{\frak c}$ directly. Following the comments, though, I have left this answer up.

Let me complement Eric's answer by showing that it is relatively consistent to have strictly more than ${\frak c}^+$ many equivalence classes. Indeed, it is relatively consistent with ZFC to have $2^{\frak c}$ many equivalence class, in a case where this is larger than ${\frak c}^+$.

Specifically, I claim that if the continuum hypothesis holds and there is a thick Kurepa tree (an $\omega_1$ tree with $2^{\omega_1}$ many branches), then there are $2^{\omega_1}=2^{\frak c}$ many equivalence classes. Indeed, I shall construct an almost-disjoint family of $2^{\omega_1}$ many Vitali sets.

To see this, let $T$ be a thick Kurepa tree, and let $\langle A_\alpha\mid\alpha<\omega_1\rangle$ enumerate the equivalence classes of reals under translation-by-a-rational. Label the $\alpha^{th}$ level of $T$ with the countably many elements of $A_\alpha$. For any path $s$ through $T$, the set $A_s$ of labels appearing on the nodes of $s$ will be a Vitali set, and therefore non-measurable. Further, any two distinct paths $s\neq t$ will have $A_s\cap A_t$ being countable, and so $A_s\not\sim A_t$. Since $T$ is a thick Kurepa tree, we therefore have $2^{\omega_1}$ many branches and thus this many equivalence classes modulo your relation. The collection $\{\ A_s\mid s\in[T]\ \}$ is an almost-disjoint family of $2^{\omega_1}$ many Vitali sets.

Finally, let me explain that it is relatively consistent from an inaccessible cardinal that there is a thick Kurepa tree, yet CH holds and $2^{\omega_1}$ is very large. One way to do this is as follows. Start with $\kappa$ inaccessible in $V$ and $2^\kappa$ very large (by forcing if necessary). Let $V[G]$ be the forcing extension by the Levy collapse, so that $\kappa=\omega_1^{V[G]}$. Consider the tree $T=(2^{<\kappa})^V$ in the model $V[G]$. Since every ordinal less than $\kappa$ was made countable, this has become an $\omega_1$-tree. Yet, since $2^\kappa$ was very large and cardinals $\kappa$ and above were preserved, we have $(2^\kappa)^V$ many branches through this tree. So it is thick.

• Ah, Eric has now improved his answer to give $2^{\frak c}$ directly. So that subsumes the need for this argument. – Joel David Hamkins Mar 19 '15 at 21:38
• Vote up this comment if you (as I) think that JDH should NOT delete his answer. – Bill Johnson Mar 19 '15 at 22:07
• Is community wiki a new form of alternative punishment? – Emil Jeřábek supports Monica Mar 19 '15 at 22:51
• You might just edit the first paragraph to clarify that it doesn't extend Eric's (current) result. – Nate Eldredge Mar 20 '15 at 4:53
• @EmilJeřábek : More like a way to redeem oneself for whatever sin you think you have commited, mostly related to (sneaky) reputation gain. – Hachino Mar 20 '15 at 11:51


text: When the sum of the measures of two angles is 90°, the angles are called It is not possible for a triangle to have more than one obtuse angle. {See Fig (d)} Vertically Opposite Angle. Then we solve the equation to find the value of the unknown variable. One pair is AOD & BOC. Download FREE PDF of Chapter-5 Lines and Angles, Find the complement of each of the following. An angle which is equal to its complement. But vertical angles can each have a … Vertical angles are always congruent, which means that they are equal. Ex 5.1, 14 In the adjoining figure, name the following pairs of angles. The term obtuse is also used in the context of triangles. If one of the angles of a triangle is obtuse, the other two angles of a triangle must be acute. An obtuse angle is an angle whose measure is greater than $${90^\circ}$$ and less than ... Interiors of $${\angle ABD}$$ and $${\angle CBD}$$ don’t overlap and hence they are adjacent angles. First, if one angle is $$55^\circ$$ then the angle opposite to it will also be $$55^\circ$$as vertically opposite angles are equal. (i) $$\angle 1$$ and $$\angle 4, \angle 5$$ and ( $$\angle 2+\angle 3$$) are vertically opposite angles as they formed due to intersection of two straight lines. In this lesson, students will find the lengths of segments and the measures of angles. To know more about lines and angles read my article carefully. Ex 5.1, 14 In the adjoining figure, name the following pairs of angles. Two angles forming a linear pair are ————— A. Complementary B. Know the congruent properties of vertical angles or vertically opposite angles and apply them to determine unknown angle measures. Any angle can have adjacent angles on its both sides. Since ∠MON and ∠PON are supplementary, ∠MON + ∠PON = 180°. (d) Bisectors of the adjacent angles forming a linear pair form a right angle. 8 Which of the following statements is false?a) Two vertically opposite angles can be acuteb) Two vertically opposite angles can be obtusec) Two vertically opposite angles can be right angl - edu-answer.com Vertically Opposite Angles. 5. 2. Sum of measure of these two angles $$= 112^\circ + 68^\circ = 180^\circ$$, Sum of measure of these two angles $$= 130^\circ + 50^\circ = 180^\circ$$, Sum of measure of these two angles $$= 45^\circ + 45^\circ = 90^\circ$$, Sum of measure of these two angles $$= 80^\circ + 10^\circ = 90^\circ$$. {See Fig (b)} Two obtuse angles can be adjacent angles {See Fig (c)} An acute angle can be adjacent to an obtuse angle. Eudemus of Rhodes attributed the proof to Thales of Miletus . 45 C. 180. (a) When a transversal cuts two parallel lines, each pair of corresponding angles are equal. (v) If two lines intersect a point, then the vertically opposite angles are always, (vi) If two lines intersect at a point and if one pair of vertically opposite angles are acute angles, then the other pair of vertically opposite angles are ______________. ∠ PQR is an obtuse angle because its less than 180° and greater than 90°. An Angle that is greater than 90° but less than 180° is known as an Obtuse Angle. Add your answer and earn points. Obtuse Angle. (i) Obtuse vertically opposite angles mean angles greater than $$90^\circ$$ and are equal, (ii) Adjacent complementary angles have common vertex and common arm, non - common arms are on either sides of common arm and their sum is $$90^\circ.$$. Example: Find angles a°, b° and c° below: Because b° is vertically opposite 40°, it must also be 40° A full circle is 360°, so that leaves 360° − 2×40° = 280° Angles a° and c° are also vertically opposite angles, so must be equal, which means they are 140° each. 90 B. (ii) Adjacent complementary angles∠BOA, ∠AOE are adjacent angles Whenever a triangle is classified as obtuse, one of its interior angles has a measure between 90° and 180°. Find the angle which is equal to its supplement. Hope this answers your question. When two lines intersect at a point, the measure of the “opening” between these two lines is called an “Angle”. Can two obtuse angles be adjacent angles? It is not possible for a triangle to have more than one obtuse angle. Angles can be classified according to their measure. Vertically opposite angles B. (Why?) As one angle is 45°, the other angle = 90-45= 45° Exercise: 1. Highly capable of resisting both lateral and axial loads due to the fact that they are made of steel with varying diameters of between 70 to 200 mm. Are 2 angles of the same size, formed between opposite sides of 2 intersecting straight lines. Are ∠1, ∠2 a linear pair? 1 word related to obtuse angle: oblique angle. Then we solve the equation to find the value of the unknown variable. Two pairs of vertically opposite angles are : (i) AOC and BOD (ii) AOD and BOC ∠PON is obtuse since its measure is between 90° and 180°. If $$\angle 1$$ is decreased by some degrees, then $$\angle 2$$ will also be increased with the same degree, so both the angles will remain supplementary. The sum of the measures of the angles in any triangle is 180°, so if one of the angles in the triangle is an obtuse angle, the sum of the other 2 angles must be less than 90° making the other 2 angles acute. 4. They will then learn the rules of angles on a line, around a point and vertically opposite angles. The chapter 5 begins with an introduction to Lines and Angles by explaining the basic concepts of point, line, line segment and angle.Then various types of related angles such as complementary angles, supplementary angles, adjacent angles, linear pair and vertically opposite angles are discussed in detail. Supplementary C. Reflex. The sum of complementary angles is always If the given angle is $$x$$, then we can find complementary angle by subtracting from . When a transversal cuts two parallel lines, each pair of corresponding angles are ——– A. Linear Pairs of Angles. The equality of vertically opposite angles is called the vertical angle theorem. Find the values of the angles $$x, y$$ and $$z$$ in each of the following: There are two operations done in sequence. When two angles are supplementary, each angle is said to be the supplement of the other. Equal B. Before checking this, let us see some real life examples for vertically opposite angles (Fig 5.15). (a) When a transversal cuts two parallel lines, each pair of corresponding angles are equal. Linear pair C. Intersecting lines. In the following figure, is $$\angle 1$$ adjacent to $$\angle 2?$$ Give reasons. (a) Two vertically opposite angles can be acute (b) Two vertically opposite angles can be obtuse (c) Two vertically opposite angles can be right angles (d) Two vertically opposite angles may be unequal. from vertical to obtuse (angle between 90-180 degree incline). A triangle can never have 3 acute angles. (vi) Vertically opposite angle of $$\angle 5$$ is $$\angle 2 + \angle 3$$ i.e, $$\angle COB.$$. Therefore, these two angles are complementary. Vertical angles are also called opposite angles. 4. An obtuse angle is an angle that measures more than a right angle but less than a straight angle. Which of the following statements is false? As we know, the measure of an obtuse angle is greater than 90 degrees. Acute angles B. Obtuse angles C. Right angles. Are ∠1, ∠2 a linear pair? Similarly, a triangle cannot have a right angle and obtuse angle at the same time. The sum of angle and its complement angle is always equal to $$90^\circ.$$. Two angles that are both adjacent and supplementary are a linear pair. Can two adjacent angles be complementary? Examples of obtuse angles are: 100°, 120°, 140°, 160°, 170° etc. As α is 30°, so β is also equal to 30° Question 5: Find the complement of angle 45° Solution: Sum of two complement angles are 90°. Each is a supplement of the other. ★★★ Correct answer to the question: Stion No. Find the supplement of each of the following angles: What is the unknown? Let’s solve this problem visually. 45 B. This is a type of proof regarding angles being equal when they are vertically opposite. Therefore. Have you started learning lines and angles chapter read basics of Different types of angles-Acute, Right, Obtuse, Straight, Reflex, vertically opposite,co-interior, complementary and supplementary angles. Click hereto get an answer to your question  State true or false:If two lines intersect and if one pair of vertically opposite angles is formed by acute angles, then the other pair of vertically opposite angles will be formed by obtuse angles. The two vertically opposite angles are always equal. The vertex of an obtuse angle is "dull" when compared with the vertex of an acute angle. Let us say given angle is $$x$$  and according to the question, its supplementary angle will also be equal to $$x$$. Measure of supplement of each angle. An angle which is greater than 180° but less than 360° is called a reflex angle.Further, two angles whose sum is 90° are Solve for $$\angle x,\angle y$$ and $$\angle z:$$, (i) $$\angle x = 55^\circ$$ (Vertically opposite angle), \begin{align} \angle x + \angle y &= 180^\circ \rm (Linear \,pair) \\55 ^\circ+ \angle y &= 180^\circ\\ \angle y &= 180^\circ- 55 ^\circ\\ \angle y &= 125 ^\circ\end{align}, Therefore$$\angle y= \angle z = 125^\circ$$ (Vertically opposite angle), Hence, $$\angle x = 55^\circ,\angle y= 125^\circ,\angle z = 125^\circ$$, By using angle sum property find the value of $$x$$ and then find the value of $$y$$ and $$z.$$ Since the sum of $$y + z = 180^\circ.$$ Now, it’s a matter of finding $$y$$ and $$z.$$, \begin{align}40^\circ \!+ \!x \!+ \!25^\circ &= 180^\circ \\ \text{(Angles on } &\text{straight line)}\\x + 65^\circ &= 180^\circ\\ x &= 180^\circ - 65^\circ = 115^\circ\end{align}, \begin{align}40^\circ + y &= 180^\circ\text{(Linear pair)}\\ y &= 180^\circ - 40^\circ\\ y &= 140^\circ\\y + z &= 180^\circ \text{(Linear pair)}\\140^\circ + z &= 180^\circ (y = 140^\circ)\\ z &= 180^\circ- 140^\circ\\ z &= 40^\circ \end{align}, Thus, $$x = 115^\circ,y = 140^\circ {\text {and}} \,\,z = 40^\circ$$. Obtuse vertical angles are angles that are both vertical and obtuse. In the figure below, ∠MON and ∠PON are adjacent angles that are also supplementary. Obtuse. (ii) obtuse? Corresponding Angles and its Converse. The term obtuse is also used in the context of triangles. An Obtuse Angle is just the opposite of an Acute Angle. $$\angle EOA \,\,{\text {and}} \,\,\angle AOB$$ are adjacent complementary angles. Using Converse of the Corresponding Angles Postulate, you can prove lines are parallel. (iii) Two angles forming a linear pair are _______________. According to this model, the result is equal to $$\angle 1 + \angle 2\,\, \gt 45 ^\circ + \angle 2.$$ Now, it’s a matter of finding Is its complementary angle greater than $$45^\circ$$ or equal to $$45^\circ$$ or less than $$45^\circ.$$, Let there be two angles $$\angle 1$$ and $$\angle 2 .$$, Therefore$$\angle 1 \gt 45^\circ$$ (given), Adding $$\angle 2$$  to both sides ,we get, $$=\gt \angle 1 + \angle 2 \gt 45^\circ + \angle 2\\ =\gt 90^\circ \gt 45^\circ + \angle 2 \\ =\gt 90^\circ - 45^\circ \gt \angle 2 \\ =\gt 45^\circ \gt \angle 2$$, Therefore, its complementary angle will be less than $$45^\circ.$$, (i) Is $$\angle 1$$ adjacent to $$\angle 2$$ $$?$$, (ii) Is $$\angle AOC$$ adjacent to $$\angle AOE \,?$$. Yes. Can two obtuse angles be adjacent angles? Those two … THEOREM-1 :- If two lines intersect each other, then the vertically opposite angles are equal. Check for yourself how well your 4th grade and 5th grade learners can measure angles in this variety of exercises. Two angles are said to be a supplementary angle if the sum of their measures is 180 0. ∠POR and ∠SOQ form a pair of vertically opposite angles, while ∠POS and ∠ROQ form another pair of vertically opposite angles. Students can access the NCERT MCQ Questions for Class 7 Maths Chapter 5 Lines and Angles Pdf free download aids in your exam preparation and you can get a good hold of the chapter. 27. See also acute, angles, right, straight. Also, recall that a straight angle is equal to 180°. $$\angle AOB$$ and $$\angle AOE;$$ $$\angle AOE$$ and $$\angle EOD;$$ $$\angle EOD$$ and $$\angle COD$$, Instant doubt clearing with Cuemath Advanced Math Program, $$\angle 1 + \angle 2\,\, \gt 45 ^\circ + \angle 2.$$, $$\angle x = 55^\circ,\angle y= 125^\circ,\angle z = 125^\circ$$, $$x = 115^\circ,y = 140^\circ {\text {and}} \,\,z = 40^\circ$$, $$\angle EOA \,\,{\text {and}} \,\,\angle AOB$$. An obtuse angle is a type of angle whose degree measurement is more than 90° but less than 180°. Straight Angle Vertically opposite angles are always equal. Vertical Angles Theorem. Shapes; ECD12215 Ortho; Corresponding Angles Theorem Therefore, complement of this angle will also be $$x$$. Class- VII-CBSE-Mathematics Line And Angle Practice more on Line And Angle Page - 5 www.embibe.com (i) acute? Vertically opposite angles: When two lines bisect, the angles that are created opposite to each other at the vertex (point of bisection) are called vertically opposite angles. \begin{align} &= 90^{\circ} -\text{[given angle]} \\ &=90^{\circ}- 20^{\circ} \\ &=70^{\circ} \end{align}, \begin{align} &= 90^{\circ} -\text{[given angle]} \\ &= 90^{\circ}-63^{\circ} \\ &= 27^{\circ} \end{align}, \begin{align} &= 90^{\circ}- \text{[given angle]} \\ &= 90^{\circ}-57^{\circ} \\ &= 33^{\circ} \end{align}. Sum of measure of these two angles $$= 63^\circ + 27^\circ = 90^\circ$$. An obtuse angle is an angle that has more than 90° and vertically opposite angles are angle formed by two lines crossed. (ii) $$\angle 1$$ and$$\angle 5 , \angle 5$$ and $$\angle 4$$ forms linear pair. If we plug in the expressions for the angle measures, we get. (a) Two vertically opposite angles can be acute (b) Two vertically opposite angles can be obtuse (c) Two vertically opposite angles can be right angles (d) Two vertically opposite angles may be unequal. Solution: We know that the sum of the measures of supplementary angles is 180. Obtuse Angle. Lines and Angles Class 7 Maths MCQs Pdf. In the adjoining figure, name the following pairs of angles: (v) Adjacent angles that do not form a linear pair. (i) If both angles are acute (less than 90), then their sum can never be 180. First take$$\angle 1\,\, \gt 45^\circ$$ as it is given, and then add $$\angle 2$$ to both sides of the equation. (c) Both of the angles forming a linear pair can be obtuse angles. Those two angles are supplementary angles. One angle is obtuse. Start studying Geometry: Unit 2~Angles. Also, do vertically opposite angles add up to 180? \begin{align} &= {180^\circ} - \text{given angle} \\ &= 180^\circ − 105^\circ \\ &= 75^\circ \end{align}, \begin{align} &= 180^\circ -\text{given angle} \\ &= 180^\circ − 87^\circ \\ &= 93^\circ \end{align}, \begin{align} &= 180^\circ -\text{given angle} \\ &= 180^\circ − 154^\circ \\ &=26^\circ\end{align}. 5.2.4 Linear Pair A linear pair is a pair of adjacent angles whose non-common sides are opposite rays. (v) Is $$\angle 1$$ vertically opposite to $$\angle 4\,?$$, (vi) What is the vertically opposite angle of $$\angle 5\,?$$, (i) Yes, because they have common vertex $$O$$ and common arm $$OC.$$. 3. Yes, it is true that a pair of obtuse angles can also be vertical angles, given the condition that, if you add them, but they cannot be vertical angles by themselves. (2) The sum of all angles around a point is 360°. Angles larger than a straight angle but less than 1 turn (between 180° and 360°) are called reflex angles. (a) When a transversal cuts two parallel lines, each pair of corresponding angles are equal. 12 Year 6 | Summer Term | Week 1 to 2 –Geometry: Properties of Shapes My triangle has two 90° angles. $$\angle EOB\,\,{\text {and} }\,\,\angle EOD$$. Two angles are said to be complementary when the sum of the two angles is 90°. Angles can be classified according to their measure. 180 C. 360. Lesson objective: To recognise angles where they meet at a point, are on a straight line, or are vertically opposite, and find missing angles. Important Facts: (1) The sum of all angles formed on the same side of a line at a given point on a line is 180°. The chapter 5 begins with an introduction to Lines and Angles by explaining the basic concepts of point, line, line segment and angle.Then various types of related angles such as complementary angles, supplementary angles, adjacent angles, linear pair and vertically opposite angles are discussed in detail. 27. Can two angles be supplementary if both are: (i) No, sum of acute angles is less than  $$180^\circ.$$, (ii) No, sum of obtuse angles is greater than $$180^\circ.$$, (iii) Yes, sum of two right angles is $$180^\circ.$$, An angle is greater than $$45^\circ.$$ Is its complementary angle greater than $$45^\circ$$ or equal to $$45^\circ$$ or less than $$45^\circ\,?$$. Then, two pairs of vertically opposite angles are formed. 9. Vertically Opposite Angles. Therefore, these two angles are supplementary. Yes, it is true that a pair of obtuse angles can also be vertical angles, given the condition that, if you add them, but they cannot be vertical angles by themselves. Also know, what are obtuse vertically opposite angles of 2 intersecting straight lines ( )... Angles being equal when they are vertical angles are angle formed by two lines parallel. Determine unknown angle measures \text { and } } \ ) Give reasons state which angle is 45° the! Of exercises this statement would be TRUE times the smallest angle when compared with the vertex of an angle... Following pairs of angles angles, right, straight and right angles ( 90°.! Not overlap ; corresponding angles are supplementary angles have sum of the angles supplementary! Angles or vertically opposite angles equation to find the angle opposite to the question: No! Are angles that have a common vertex and opposite to each other be angles. No, they form a linear pair than one obtuse angle both adjacent supplementary!, terms and more with flashcards, games, and more with flashcards games! Better exam preparation and score more marks iii ) equal supplementary angles are equal use MCQ for. Pair can be obtuse angles are complementary angles ; and ( = +! Geometry angles vocabulary study guide by lwakeman19 includes 26 Questions covering vocabulary, terms and more with,... Can also be classified as an obtuse angle is said to be a supplementary angle if sum. 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# Derive logistic loss gradient in matrix form

User Antoni Parellada had a long derivation here on logistic loss gradient in scalar form. Using the matrix notation, the derivation will be much concise. Can I have a matrix form derivation on logistic loss? Where how to show the gradient of the logistic loss is

$$A^\top\left( \text{sigmoid}~(Ax)-b\right)$$

1. For comparison, for linear regression $\text{minimize}~\|Ax-b\|^2$, the gradient is $2A^\top\left(Ax-b\right)$, I have a derivation here.

2. Related question: Matrix notation for logistic regression

• As credited in the linked answer, the credit for the derivation goes to the students' cumulative notes within the excellent course by Andrew Ng on ML. May 11 '17 at 2:24
• Why not try translating the scalar-form derivation as a self-study? (BTW you should edit the question to explicitly state the loss itself, in your preferred notation, to make it self contained) May 11 '17 at 2:36

$$J(x) = -\frac{1}{m}\sum_{i = 1}^{m} b_iln(h_i) + (1 - b_i)ln(1 - h_i)$$
where $h_i = \sigma(x^Ta_i)$. Let $A = [a_1^T, \dots, a_m^T]^T$. Assuming $ln, \sigma, \frac{1}{\cdot}$ work element-wise on vectors, $\odot$ is element-wise multiplication and $\mathbb{1}$ is a vector of $1$s we have
$$J(x) = -\frac{1}{m}\big[b^Tln(\sigma(Ax)) + (\mathbb{1} - b)^Tln(\mathbb{1} - \sigma(Ax))\big]$$
$$\frac{\partial J(x)}{\partial x} = -\frac{1}{m}\Big[\frac{\partial}{\partial x}b^Tln(\sigma(Ax)) + \frac{\partial}{\partial x}(\mathbb{1} - b)^Tln(\mathbb{1} - \sigma(Ax))\Big] \\ = -\frac{1}{m}\Big[\frac{\partial ln(\sigma(Ax))}{\partial x}b + \frac{\partial ln(\mathbb{1} - \sigma(Ax))}{\partial x}(\mathbb{1} - b)\Big] \\ = -\frac{1}{m}\Big[\frac{\partial \sigma(Ax)}{\partial x} \big(\frac{1}{\sigma(Ax)}\odot b\big) + \frac{\partial \mathbb{1} - \sigma(Ax)}{\partial x}\big(\frac{1}{\mathbb{1} - \sigma(Ax)}\odot (\mathbb{1} - b)\big)\Big] \\ = -\frac{1}{m}\Big[\frac{\partial Ax}{\partial x} \big(\sigma(Ax) \odot (\mathbb{1} - \sigma(Ax)) \odot \frac{1}{\sigma(Ax)}\odot b\big) - \frac{\partial Ax}{\partial x}\big(\sigma(Ax) \odot (\mathbb{1} - \sigma(Ax)) \odot \frac{1}{\mathbb{1} - \sigma(Ax)}\odot (\mathbb{1} - b)\big)\Big] \\ = -\frac{1}{m}\Big[A^T \big((\mathbb{1} - \sigma(Ax)) \odot b\big) - A^T\big(\sigma(Ax)) \odot (\mathbb{1} - b)\big)\Big] \\ = -\frac{1}{m}\Big[A^T \big(b - \sigma(Ax) \odot b - \sigma(Ax) + \sigma(Ax) \odot b\big)\Big] \\ = -\frac{1}{m}\Big[A^T \big(b - \sigma(Ax)\big)\Big] \\ = \frac{1}{m}\Big[A^T \big(\sigma(Ax) - b\big)\Big]$$


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# The Universe of Discourse

Thu, 20 Jun 2019

A couple of days ago my sleepy brain mashed up Clarke's Law:

Any sufficiently advanced technology is indistinguishable from magic.

and Hanlon's Razor:

Never attribute to malice what can be explained by stupidity.

to produce these words of wisdom:

Any sufficiently advanced software is indistinguishable from malice.

(Why yes, I had spent the evening dealing with Git. Why do you ask?)

This sounded like something Bryan Horstmann-Allen would have said, so with his permission, I am naming it after him.

BDHA's Law? BDHA's Razor? No!

Mon, 17 Jun 2019

My big work project is called “Greenlight”. It's a Git branch merging service. After you've pushed a remote branch, say mjd.fix-bugs, you use a very thin client program to ask the Greenlight server to land your branch on master and publish it for you:

    greenlight submit mjd.fix-bugs

Greenlight analyzes the branch to see if it touches any sensitive code that requires signoffs. If so it contacts the correct people on Slack, and asks them to review it. Once they have approved it, Greenlight rebases the branch onto the current master and pushes the result back to master. If the push fails, it retries silently. Throughout, it communicates via Slack what is going on.

A user, Locksher, complained last week that it didn't do what he had expected. He had a Git pre-push hook he had written. Whenever he ran git push, his pre-push hook would look to see if he was pushing to master. If so, it would look at the messages of the commits he was trying to push. If any of them contained WIP or !fixup or !squash, it would abort the push.

With Greenlight, this check wasn't done, because Locksher never pushed to master himself. Instead he pushed to some topic branch, and then asked Greenlight to publish it to master, which it did, including his WIP commits. Oops!

Locksher asked if it was possible to have Greenlight “respect local hooks”. Once I understood what he wanted, my first suggestion was that he wrap the greenlight client in a shell script that did the check he wanted. My second suggestion, less work for him but also less immediate, was that the Greenlight client could look in .git/hooks for a greenlight-pre-submit hook, and run that before communicating with the server, aborting the request if the hook failed. I think this would adequately solve the problem, especially if the calling convention for the new hook was identical to that of pre-push. Then you would just:

        ln -s pre-push .git/hooks/greenlight-pre-submit

and get exactly the desired behavior. I said that if Locksher wanted to implement this, I would include it in the standard client, or alternatively I would open a ticket to implement it myself, eventually.

Locksher suggested instead that the greenlight client configuration should support this:

    [git]
respect-git-hooks = true

I didn't have time then to answer in detail, so I just said:

I consider that very unlikely.

Here's what I said to him once I did have time to answer in detail:

1. There are currently 23 documented Git hooks, and it's not immediately clear what it would mean to “respect” many of them. I'd have to go over the man page and decide, for each one, what the behavior should be, then possibly implement it, and then document it. Just to pick one example, should Greenlight “respect” your prepare-commit-message hook? If so, how?

2. Even for the hooks where the correct behavior seemed clear to me, it might seem clearly something else to someone else. So the feature is severely under-specified and seems likely to cause confusion. I foresee a future of inquiries like “I set respect-git-hooks but Greenlight didn't run my pre-auto-gc hook.”

3. It is an open-ended promise. The way the option is phrased, it guarantees to “respect” every hook. So it commits me to keep track of what new hooks are introduced in every future version of Git, and to decide what to do about each of them.

4. Since greenlight runs on your local machine, the local version of Git may vary. What if the behavior of Git's pre-cake-slicing hook changes between Git 1.24 and Git 1.26? Now Greenlight will have to implement two behaviors, and look at your local Git version to decide what to do.

Oh, and 5, it is a YAGNI.

In contrast, the functionality provided by greenlight-pre-submit is something someone has actually asked for. It is small, sharply bounded in scope and its definition is completely under my control.

I will elaborate a little on the main items 1–2, that different people might have different ideas about what it means to “respect” a local hook. Consider Locksher's specific request, for greenlight to “respect” his pre-push hook. Another user, say Zubi, could object, quite reasonably, that greenlight submit is not the same as git push, and that the correct way for it to “respect” her pre-push hook is to ignore it. “I want my pre-push hook run when I push a branch,” she might say, “not when I do greenlight submit.” Who could argue with that? (Other than Locksher, of course.)

So then I would have to add an escape hatch for Zubi, so that everyone who didn't want Locksher's feature would have to affirmatively opt out of it.

Nah.

Sat, 08 Jun 2019

I have pondered category theory periodically for the past 35 years, but not often enough to really stay comfortable with it. Today I was pondering again. I wanted to prove that !!1×A \cong A!! and I was having trouble. I eventually realized my difficulty: my brain had slipped out of category theory mode so that the theorem I was trying to prove did not even make sense.

In most of mathematics, !!1\times A!! would denote some specific entity and we would then show that that entity had a certain property. For example, in set theory we might define !!1\times A!! to be some set, say the set of all Kuratowski pairs !!\langle \varnothing, a\rangle!! where !!a\in A!!:

$$1×A =_{\text{def}} \{ z : \exists a\in A : z = \{\{\varnothing\}, \{\varnothing, a\}\} \}$$

and then we would explicitly construct a bijection !!f:A\leftrightarrow 1×A!!:

$$f(a) = \{\{\varnothing\}, \{\varnothing, a\}\}$$

In category theory, this is not what we do. Everything is less concrete. !!\times!! looks like an operator, one that takes two objects and yields a third. It is not. !!1\times A!! does not denote any particular entity with any particular construction. (Nor does !!1!!, for that matter.) Instead, it denotes an unspecified entity, which happens to have a certain universal property, perhaps just one of many such entities with that property, and there is no canonical representative of the class. It's a mistake to think of it as a single thing, and it's also a mistake to ask the question the way I did ask it. You can't show that !!1×A!! has any specific property, because it's not a specific thing. All you can do is show that anything with the one required universal property must also have the other property. We should rephrase the question like this:

Let !!1×A!! be a product of !!1!! and !!A!!. Then !!1×A\cong A!!.

Maybe a better phrasing is:

Let !!1×A!! be some object that is a product of !!1!! and !!A!!. Then !!1×A\cong A!!.

The notation is still misleading, because it looks like !!1×A!! denotes the result of some operation, and it isn't. We can do a little better:

Let !!B!! be a product of !!1!! and !!A!!. Then !!B\cong A!!.

That it, that's the real theorem. It seems at first to be more difficult — where do we start? But it's actually easier! Because now it's enough to simply prove that !!A!! itself is a product of !!1!! and !!A!!, which is easily done: its projection morphisms are evidently !!! !! and !!{\text{id}}_A!!. And by a previous theorem that all products are isomorphic, any other product, such as !!B!!, must be isomorphic to this one, which is !!A!! itself.

(We can similarly imagine that any theorem that mentions !!1!! is preceded by the phrase “Let !!1!! be some terminal object.”)

Fri, 07 Jun 2019

A little while ago I wrote:

I think a disco ball would not be out of place at Versailles.

I just remembered that good mirror technology is perhaps too recent for disco balls to have been at Versailles. Hmmm. Early mirrors were made of polished metal or even stone, clearly unsuitable. Back-silvering of glass wasn't invented until the mid-19th century.

Still, a disco ball is a very forgiving application of mirrors. For a looking-glass you want a large, smooth, flat mirror with no color distortion. For a disco ball you don't need any of those things. Large sheets of flat glass were unavailable before the invention of float glass at the end of the 19th century, but for a disco ball you don't need plate glass, just little shards, leftovers even.

The 17th century could produce mirrors by gluing metal foil to the back of a piece of glass, so I wonder why they didn't. They wouldn't have been able to spotlight it, but they certainly could have hung it under an orbiculum. Was there a technological limitation, or did nobody happen to think of it?


text:
## Chemistry and Chemical Reactivity (9th Edition)

a) Net ionic reaction: $CH_3CO_2H+HPO_4^{2-}\leftrightarrow H_2PO_4^-+CH_3CO_2^-$ b) acetic acid is a stronger acid than $H_2PO_4^-$, so the equilibrium lies to the right.


text:
## Wind power results – 2013/14

I have been monitoring the UK peak power demand every day for the last 12 months, and how that power was delivered. The UK government has invested about £50 billion into Wind energy since 2005,  so how has it performed?

The average Wind component of UK peak power demand was 5.7% over a full year. The maximum power output was 6GW on 21 February 2014 at 18:10, while  the maximum proportion of demand was 16.8% on Sunday 17th August 2014 at 20:08. The minimum proportion was 0.15% on 31st May 2014 when total Wind output was 49MW (0.05GW).

Power delivery by Fuel type to meet Peak demand every day since September 2013. Weekdays peak demand occurs around 18:00 and at weekdays occurs around 20:00.

For comparison Nuclear Power averaged 18.1% of demand and never fell below 15%. The bulk of UK power is still met by fossil fuels and it is impossible to imagine a future Grid based purely on Wind power. The mix of coal and gas is interesting. Coal is dominant in Winter while gas dominates in summer. I also monitored the daily energy “trough” or the lowest demand for electricity which usually occurs each day at around 4am . This is shown below.

Power delivery by fuel type at minimum daily power demand (~4am)

It is Coal that keeps the lights on during winter nights! Gas output is ramped down at night, probably because it is easier to regulate power in Gas power stations. The exception to this is during the summer when several coal power stations would appear to be mothballed.

The statistics for Wind power at night appear better because the overall demand is much less. The average contribution of wind power to minimum demand was 9.8%, and the  maximum Wind supplied was 22.5% on 2nd February at 05:30 (6GW).  However Nuclear also contributes proportionately more at night averaging 30%, which is 10% more than Gas. The difference between the 2 extreme demand curves is shown below over the 13 month period. Note the sharp drop in power demand over the Christmas period.

Daily Peak and Minimum demand over a 13 month period Sep 2013 to Oct 2014

This analysis makes it clear, to me at least, that the UK should shift the balance of investment towards Nuclear and away from Wind. Even doubling Wind generating capacity to 20GW would still leave some days with essentially zero output, thereby undermining energy security. Furthermore Gas will always be needed with at least the same generating capacity as Wind in order to continuously balance Wind’s erratic output. Nor can energy storage save the day, since it reduces the EROEI for Wind to below sustainable levels.

The data used for this analysis was collected through BM reports – the National Grid balancing mechanism. There are about 10% of wind farms that are not metered by BM so the above figures should probably be adjusted upwards by 10%. The conclusion is the same.

The live monitoring of peak demand over the last 30 days is shown below.

Posted in coal, Energy, nuclear, renewables, Science, Technology, wind farms | Tagged | 30 Comments

## The certainty of extreme weather

The Met Office tell us that September was the driest since records began 104 years ago. Last summer was ‘the hottest ever recorded’ in Australia. These extreme records hit the headlines implying that global warming is to blame. However just how likely is it that one extreme weather record or another will be broken due to pure chance? Barometer in today’s Spectator shows how to do the calculation and the results are surprising. I have simply extended the same argument to include Australia and the US.

In the following we consider 3 countries and their regions – The UK, US and Australia. The regions for the UK are England, Northern Ireland, Wales and Scotland. Similarly Australia has 6 states and the US has 50 states. That gives a total of 63 different regions if we also include the whole country itself.

Lets take 4 records that can be broken : hottest, coldest, wettest and driest. During a single year each record can be set yearly, monthly, or seasonally. That equates to 17 different time periods. Therefore for the UK there are a total of 5x4x17=340 records that can be set  during the current year 2014. For Australia there are 6x4x17=408 records and for the US there are a staggering 51x4x17=3468 records.

Now lets assume that all weather records go back 104 years. The probability of a single record being broken in any year is simply 0.0096. So the probability the record will stand during the current year is 0.9904

What is the probability that any of the UK weather records will be broken during the current year ? That is

$P = (1 - 0.9904^{340})$

There is a 96.3% chance that at least one Met Office record will be broken this year in the UK!

For Australia there is a probability $P = (1 - 0.9904^{408})$ or 98.1% chance that a record will be broken and for the US there is a probability $P = (1 - 0.9904^{3468})$ or essentially a 100% chance that a record will be broken!

So in 2014 we can be absolutely certain that at least one state in the US will have its hottest/coldest/driest/wettest month/season/year ever recorded! Such records are totally meaningless.

Posted in AGW, Climate Change, Meteorology, Science | Tagged , | 12 Comments

## Tides and Storms

Energy flows from the tropics to the poles. The tropics absorb most of the sun’s radiation and drives heat flow to polar regions where radiative cooling dominates. In winter the Hadley cell moves heat to ~30N via huge convective currents transporting  latent heat from the tropics northwards via a band of powerful thunderstorms . The reverse happens in the northern summer when the tropical Hadley cell  moves southwards towards  the now winter southern hemisphere and it reverses circulation.  David Randall explains this well in his book ‘Atmosphere, Clouds and Climate’ (from which the first 3 figures below are taken).

The rising branch of  tropical thunderstorms (ITCZ) is located about 10 deg from the equator in whichever is the summer hemisphere. The release of latent heat warms the troposphere to the moist adiabatic lapse rate. The high warm air moves meridionally and cools by radiating heat, so becoming more dense. It then descends and twists eastward due to the rapidly increasing coriolis forces with latitude. A counter rotating Ferrel cell is driven by the mechanical energy of storms 60 -30 deg and balance mass flow. The JET stream is caused by the coriolis force acting on the descending Hadley circulation accelerated by zonal wind effects. It is concentrated about 12 km up due to lapse rate temperature gradients. The coriolis component of angular momentum M for the earth rotating with angular velocity $\Omega$  at latitude $\phi$ and zonal wind u is:

$M = (\Omega \cos{\phi} + u)a\cos{\phi}$

u = 0 at the equator wheras u = 11m/s at 30N

The Jet stream strengthens as the polar night begins. This is closely related to winter storms as the temperature gradient between the tropics and the winter pole increases.  This “thermal wind” in mid latitudes is caused by horizontal temperature gradients which increase strongly in winter. Hydrostatic balance vertically becomes unstable

$\frac{\partial P}{\partial z} = -\frac{pg}{RT}$

Wind changes rapidly with height when surface temp changes rapidly in the horizontal direction. The Jet stream becomes stronger

If T changes rapidly horizontally then the thermal winds increase strongly with height. This state is termed to be ‘baryclonic’. At 30N in winter the Jet Stream winds starts to increase massively with height. This is because below the Jet Stream the surface temperature gradient increases rapidly towards the ‘night-time’ pole. So the steeper the temperature gradient  the stronger the Jet stream becomes.

Given a strong Jet Stream and large temperature gradient, conditions are now critical for the formation of big storms. A strong poleward decrease in temperature causes ‘baryclonic instability’. Such storms are technically called ‘baryclonic waves’ and are composed of  cold fronts as polar air moves south under warm tropical air, and warm fronts when simultaneously warm tropical air rises north above cold polar air, all driven by Coriolis forces. Each front causes a rapid change in temperature on the ground. These storms are then accelerated by the release of gravitational potential energy as dense cold air falls downwards feeding kinetic energy to the storm causing strong winds.  When conditions are right for baryclonic instability any small external disturbance will be amplified to trigger the formation of a storm. There is clear evidence that strong spring tides are one important trigger of such storms. Tides provide an asymmetric disturbance which act over large northern zones through  changing tractional tidal forces acting on the atmosphere. An example  of this was the storm on Jan 4-6 last winter.

Formation of storm which hit UK on 5-6 Jan 2014. A strong wave of tractional tides sweeps through the area. The new storm seems to have been triggered by a strong kink in the Jet Stream dragging warm air up in the north west Atlantic. By the 4th January this second intense storm is forming fast off over Newfoundland. The tidal forces are very strong and the Jet Stream starts to kink. The previous low pressure system seems to be consumed by  the next as it  grows fast .

In a growing perturbation cold air slides under warm tropical air. Cold air moves down causing a front while warm air on the tropical side moves upwards  causing a warm front. Warm air rises and cold air descends transporting thermal energy upwards and the center of mass of the total air column descends releasing gravitational energy into kinetic energy. This forms a low pressure vortex at the centre.

Hadley cells do not penetrate into middle latitudes. Likewise Baroclonic waves do not penetrate into the tropics. The energy  transported from the tropics by Hadley cells is ‘handed over’ to such baroclonic storms to finish the job by moving the heat onwards to the poles. The Jet stream position defines this boundary between tropical air and polar air. Storms form on the northern side of the Jet Stream at the eastern edge of the Atlantic. The path of these storms is determined by the kinks (Rosby waves) in the Jet Stream.  As the earth rotates, an ever changing gravitational field of  tractional forces sweep across the Atlantic ocean from west to east about every 12 hours.  The effect is similar to a bar magnet sweeping across  a sheet of paper covered in iron filings.  These two videos show how strongly tractional tidal forces can vary both during the lunar month and yearly with gradual changes in lunar declination. The first is updated 2 hourly for August this year and the second shows the direct tide for a fixed position during 2006 when the moon was at its 18.6 year maximum declination.

The southward swirling tidal force acting on the Jet Stream amounts to about 10 metric tons per km. This gravitational force is varying strongly in strength and position during the lunar month. There are two possible effects of tides.

1. They can distort the Jet Stream causing kinks which change the path of mid latitude storms.
2. They can seed winter storms by disturbing baroclonic instability at maximum tidal forces. Therefore for lunar tides to effect mid latitude weather we need the following conditions.
• A sharp horizontal temperature gradient leads to a strong Jet Stream and Baryclonic instability.
• Increasing spring tides with strong tractional forces at high latitudes can  trigger anticyclonic flow leading to cold/warm fronts and storms moving eastwards.
• These storms are guided across the Atlantic by the Jet Stream.
• The Jet Stream itself  is distorted by the same changing tidal ‘tweaks’.

Jet stream flow 3-6 January 2014

The strength and position  of storms is enhances by growing instability and changing atmospheric tides over coming days. The same  tractional tidal force field also modifies Rosby waves in the Jet Stream.

Last winter the UK experienced a series of severe storms. The Jet Stream was locked into a pattern whereby the eastern US was extremely cold while warmer air resided over the western Atlantic.  Storms were spun off the boundary between very cold air over Newfoundland and warmer tropical air moving up over the western Atlantic. Each major storm coincided with a maximum tide bringing coastal flooding to the UK as large waves were also enhanced by storm surges.

So there is direct evidence that atmospheric tides can trigger storms.  Nearly all  of last winter’s storms coincided with maximum spring tides. This cannot just be a coincidence. It would be relatively easy to include such tidal forces into GCM weather forecast models. The moon is just one more external force acting on the circulation of the atmosphere. I strongly suspect that medium range weather forecasts would improve significantly if the dynamics of tides were properly taken into account.

Posted in climate science, Physics, Science | Tagged , , , | 3 Comments


text:
## Difference between expectation value and probabilty

1. The problem statement, all variables and given/known data

Psi(x) = Ax -a<x<a

I am trying to find the probability that my measured momentum is between h/a and 2h/a

2. Relevant equations

I have normalized A= sqrt(3/(2a^3))

I know that if I was finding the expected momentum I would use
$$\int$$$$\Psi$$ * p $$\Psi$$ dx

3. The attempt at a solution

so far I have done
$$\int$$ $$\Psi$$ * p $$\Psi$$ dx
with bounds from from h/a and 2h/a

but I know this is the expected value of momentum and I dont think that is the same thing as the probability of momentum.

Could someone please explain the difference in finding the probability of momentum and the expected value of momentum. Or if they are the same thing let me know there is no difference.

 Recognitions: Gold Member Homework Help Science Advisor The probability that the momentum lies between p1 and p2 is not related to the expected value of momentum. The expected value of momentum is the most likely momentum value we'd find in the state $$\Psi(x)$$. The probability to find the particle with a given momentum p is $$| \psi(p) |^2$$, where $$\psi(p)$$ is the normalized wavefunction in momentum space. To find the probability of finding the momentum in a range of values, we have to integrate this.