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# 2.6: Derivatives of Inverse Functions

Learning Objectives

In this section, we strive to understand the ideas generated by the following important questions:

• What is the derivative of the natural logarithm function?
• What are the derivatives of the inverse trigonometric functions $$arcsin(x)$$ and $$arctan(x)$$?
• If $$g$$ is the inverse of a differentiable function $$f$$, how is $$g'$$ computed in terms of $$f$$, $$f'$$, and $$g$$?

Much of mathematics centers on the notion of function. Indeed, throughout our study of calculus, we are investigating the behavior of functions, often doing so with particular emphasis on how fast the output of the function changes in response to changes in the input. Because each function represents a process, a natural question to ask is whether or not the particular process can be reversed. That is, if we know the output that results from the function, can we determine the input that led to it? Connected to this question, we now also ask: if we know how fast a particular process is changing, can we determine how fast the inverse process is changing?

As we have noted, one of the most important functions in all of mathematics is the natural exponential function $$f (x) = e^x$$. Because the natural logarithm, $$g(x) = \ln (x)$$, is the inverse of the natural exponential function, the natural logarithm is similarly important. One of our goals in this section is to learn how to differentiate the logarithm function, and thus expand our library of basic functions with known derivative formulas. First, we investigate a more familiar setting to refresh some of the basic concepts surrounding functions and their inverses.

Preview Activity $$\PageIndex{1}$$

The equation $$y = \dfrac{5}{9} (x - 32)$$ relates a temperature given in $$x$$ degrees Fahrenheit to the corresponding temperature $$y$$ measured in degrees Celsius.

1. Solve the equation y = 5 9 (x - 32) for x to write x (Fahrenheit temperature) in terms of y (Celcius temperature).
2. Let C(x) = 5 9 (x - 32) be the function that takes a Fahrenheit temperature as input and produces the Celcius temperature as output. In addition, let F(y) be the function that converts a temperature given in y degrees Celcius to the temperature F(y) measured in degrees Fahrenheit. Use your work in (a) to write a formula for F(y). 130
3. Next consider the new function defined by p(x) = F(C(x)). Use the formulas for F and C to determine an expression for p(x) and simplify this expression as much as possible. What do you observe?
4. Now, let r(y) = C(F(y)). Use the formulas for F and C to determine an expression for r(y) and simplify this expression as much as possible. What do you observe?
5. What is the value of C 0 (x)? of F 0 (y)? How do these values appear to be related? ./

## Basic Facts about Inverse Functions

A function f : A → B is a rule that associates each element in the set A to one and only one element in the set B. We call A the domain of f and B the codomain of f . If there exists a function g : B → A such that g( f (a)) = a for every possible choice of a in the set A and f (g(b)) = b for every b in the set B, then we say that g is the inverse of f . We often use the notation f -1 (read “f -inverse”) to denote the inverse of f . Perhaps the most essential thing to observe about the inverse function is that it undoes the work of f . Indeed, if y = f (x), then

$f -1 (y) = f -1 ( f (x)) = x,$

and this leads us to another key observation: writing y = f (x) and x = f -1 (y) say the exact same thing. The only difference between the two equations is one of perspective – one is solved for x, while the other is solved for y. Here we briefly remind ourselves of some key facts about inverse functions. For a function f : A → B,

• f has an inverse if and only if f is one-to-one8 and onto9 ;
• provided f -1 exists, the domain of f -1 is the codomain of f , and the codomain of f -1 is the domain of f ;
• f -1 ( f (x)) = x for every x in the domain of f and f ( f -1 (y)) = y for every y in the codomain of f ;
• y = f (x) if and only if x = f -1 (y).

The last stated fact reveals a special relationship between the graphs of f and f -1 . In particular, if we consider y = f (x) and a point (x, y) that lies on the graph of f , then it is also true that x = f -1 (y), which means that the point (y, x) lies on the graph of f -1 .

Note

• A function $$f$$ is one-to-one provided that no two distinct inputs lead to the same output.
• A function $$f$$ is onto provided that every possible element of the codomain can be realized as an output of the function for some choice of input from the domain.

This shows us that the graphs of f and f -1 are the reflections of one another across the line y = x, since reflecting across y = x is precisely the geometric action that swaps the coordinates in an ordered pair. In Figure 2.8, we see this exemplified for the function y = f (x) = 2 x and its inverse, with the points (-1, 1 2 ) and ( 1 2 , -1) highlighting the reflection of the curves across y = x.

Figure 2.8: A graph of a function y = f (x) along with its inverse, y = f -1 (x).

To close our review of important facts about inverses, we recall that the natural exponential function y = f (x) = e x has an inverse function, and its inverse is the natural logarithm, x = f -1 (y) = \ln (y). Indeed, writing y = e x is interchangeable with x = \ln (y), plus \ln (e x ) = x for every real number x and e \ln (y) = y for every positive real number y.

## The Derivative of the Natural Logarithm Function

In what follows, we determine a formula for the derivative of $$g(x) = \ln (x)$$. To do so, we take advantage of the fact that we know the derivative of the natural exponential function, which is the inverse of g. In particular, we know that writing $$g(x) = \ln (x)$$ is equivalent to writing $$e^{g(x)} = x$$. Now we differentiate both sides of this most recent equation. In particular, we observe that

$\dfrac{d}{dx} \left[ e^{g(x)} \right] = \dfrac{d}{dx} [x]. \label{eqLog}$

The righthand side of Equaton \ref{eqLog} is simply 1; applying the chain rule to the left side, we find that

$e^{g(x)} g'(x) = 1 \label{eq10}$

Since our goal is to determine $$g'(x)$$, we solve Equation \ref{eq10} for $$g'(x)$$, so

$g' (x) = 1 e^{g(x)}. \label{eq11}$

Finally, we recall that since

$g(x) = \ln (x)$

which when substituted into Equation \ref{eq11} results in

$e^{g(x)} = e^{\ln (x)} = x$

and thus

$g' (x) = \dfrac{1}{x}.$

Natural Logarithm

For all positive real numbers $$x$$,

$\dfrac{d}{dx} [\ln (x)] = \dfrac{1}{x}.$

This rule for the natural logarithm function now joins our list of other basic derivative rules that we have already established. There are two particularly interesting things to note about the fact that d dx [\ln (x)] = 1 x . One is that this rule is restricted to only apply to positive values of x, as these are the only values for which the original function is defined. The other is that for the first time in our work, differentiating a basic function of a particular type has led to a function of a very different nature: the derivative of the natural logarithm is not another logarithm, nor even an exponential function, but rather a rational one. Derivatives of logarithms may now be computed in concert with all of the rules known to date. For instance, if f(t) = \ln (t 2 + 1), then by the chain rule, f 0 (t) = 1 t 2+1 · 2t.

Activity $$\PageIndex{1}$$:

For each function given below, find its derivative.

1. h(x) = x 2 \ln (x)
2. p(t) = \ln (t) e t + 1
3. s(y) = \ln (cos(y) + 2)
4. z(x) = tan(\ln (x))
5. m(z) = \ln (\ln (z)) C

In addition to the important rule we have derived for the derivative of the natural log functions, there are additional interesting connections to note between the graphs of f (x) = e x and f -1 (x) = \ln (x).

In Figure 2.9, we are reminded that since the natural exponential function has the property that its derivative is itself, the slope of the tangent to y = e x is equal to the height of the curve at that point. For instance, at the point A = (\ln (0.5), 0.5), the slope of the tangent line is mA = 0.5, and at B = (\ln (5), 5), the tangent line’s slope is mB = 5. At the corresponding points A 0 and B 0 on the graph of the natural logarithm function (which come from reflecting across the line y = x), we know that the slope of the tangent line is the reciprocal of the x-coordinate of the point (since d dx [\ln (x)] = 1 x ). Thus, with A 0 = (0.5, \ln (0.5)), we have mA0 = 1 0.5 = 2, and at B 0 = (5, \ln (5)), mB0 = 1 5 .

Figure 2.9: A graph of the function y = e x along with its inverse, y = \ln (x), where both functions are viewed using the input variable x.

In particular, we observe that mA0 = 1 mA and mB0 = 1 mB . This is not a coincidence, but in fact holds for any curve y = f (x) and its inverse, provided the inverse exists. One rationale for why this is the case is due to the reflection across y = x: in so doing, we essentially change the roles of x and y, thus reversing the rise and run, which leads to the slope of the inverse function at the reflected point being the reciprocal of the slope of the original function. At the close of this section, we will also look at how the chain rule provides us with an algebraic formulation of this general phenomenon.

## Inverse Trigonometric Functions and their Derivatives

Trigonometric functions are periodic, so they fail to be one-to-one, and thus do not have inverses. However, if we restrict the domain of each trigonometric function, we can force the function to be one-to-one. For instance, consider the sine function on the domain [- π 2 , π 2 ].

Because no output of the sine function is repeated on this interval, the function is one-to-one and thus has an inverse. In particular, if we view f (x) = sin(x) as having domain [- π 2 , π 2 ] and codomain [-1, 1], then there exists an inverse function f -1 such that

$f -1 : [-1, 1] → [- π 2 , π 2 ].$

We call f -1 the arcsine (or inverse sine) function and write f -1 (y) = arcsin(y). It is especially important to remember that writing

$y = \sin(x) and x = arcsin(y)$

say the exact same thing. We often read “the arcsine of y” as “the angle whose sine is y.”

Figure 2.10: A graph of f (x) = sin(x) (in blue), restricted to the domain [- π 2 , π 2 ], along with its inverse, f -1 (x) = arcsin(x) (in magenta).

For example, we say that π 6 is the angle whose sine is 1 2 , which can be written more concisely as arcsin( 1 2 ) = π 6 , which is equivalent to writing sin( π 6 ) = 1 2 . Next, we determine the derivative of the arcsine function. Letting h(x) = arcsin(x), our goal is to find h 0 (x). Since $$h(x)$$ is the angle whose sine is $$x$$, it is equivalent to write

$\sin(h(x)) = x.$

Differentiating both sides of the previous equation, we have

$d dx [sin(h(x))] = d dx [x],$

and by the fact that the righthand side is simply 1 and by the chain rule applied to the left side,

$cos(h(x))h 0 (x) = 1.$

Solving for h 0 (x), it follows that h 0 (x) = 1 cos(h(x)). Finally, we recall that h(x) = arcsin(x), so the denominator of h 0 (x) is the function cos(arcsin(x)), or in other words, “the cosine of the angle whose sine is x.” A bit of right triangle trigonometry allows us to simplify this expression considerably. Let’s say that θ = arcsin(x), so that θ is the angle whose sine is x. From this, it follows that we can picture θ as an angle in a right triangle with hypotenuse 1 and a vertical leg of length x, as shown in Figure 2.11. The horizontal leg must be √ 1 - x 2, by the Pythagorean Theorem. Now, note particularly that θ = arcsin(x) since sin(θ) = x, and recall that we want to know a different expression for cos(arcsin(x)).

Figure 2.11: The right triangle that corresponds to the angle θ = arcsin(x).

From the figure, cos(arcsin(x)) = cos(θ) = √ 1 - x 2. Thus, returning to our earlier work where we established that if

$h(x) = arcsin(x),$

then

$h' (x) = \dfrac{1}{cos(arcsin(x))}$

we have now shown that

$h ' (x) = \dfrac{1}{\sqrt{1 - x^2}}.$

Inverse sine

For all real numbers $$x$$ such that $$-1 < x < 1$$, $\dfrac{d}{dx} [\text{arcsin}(x)] = \dfrac{1}{\sqrt{1 - x^ 2}}.$

Activity $$\PageIndex{2}$$:

The following prompts in this activity will lead you to develop the derivative of the inverse tangent function.

1. Let r(x) = arctan(x). Use the relationship between the arctangent and tangent functions to rewrite this equation using only the tangent function.
2. Differentiate both sides of the equation you found in (a). Solve the resulting equation for r 0 (x), writing r 0 (x) as simply as possible in terms of a trigonometric function evaluated at r(x).
3. Recall that r(x) = arctan(x). Update your expression for r 0 (x) so that it only involves trigonometric functions and the independent variable x.
4. Introduce a right triangle with angle θ so that θ = arctan(x). What are the three sides of the triangle?
5. In terms of only x and 1, what is the value of cos(arctan(x))? 136
6. Use the results of your work above to find an expression involving only 1 and x for r 0 (x).

While derivatives for other inverse trigonometric functions can be established similarly, we primarily limit ourselves to the arcsine and arctangent functions. With these rules added to our library of derivatives of basic functions, we can differentiate even more functions using derivative shortcuts. In Activity 2.18, we see each of these rules at work.

Activity $$\PageIndex{3}$$:

Determine the derivative of each of the following functions.

1. f (x) = x 3 arctan(x) + e x \ln (x)
2. p(t) = 2 t arcsin(t)
3. h(z) = (arcsin(5z) + arctan(4 - z))27
4. s(y) = cot(arctan(y))
5. m(v) = \ln (sin2 (v) + 1)
6. g(w) = arctan \ln (w) 1 + w2 !

The link between the derivative of a function and the derivative of its inverse In Figure 2.9, we saw an interesting relationship between the slopes of tangent lines to the natural exponential and natural logarithm functions at points that corresponded to reflection across the line y = x. In particular, we observed that for a point such as (\ln (2), 2) on the graph of f (x) = e x , the slope of the tangent line at this point is f 0 (\ln (2)) = 2, while at the corresponding point (2, \ln (2)) on the graph of f -1 (x) = \ln (x), the slope of the tangent line at this point is ( f -1 ) 0 (2) = 1 2 , which is the reciprocal of f 0 (\ln (2)). That the two corresponding tangent lines having slopes that are reciprocals of one another is not a coincidence. If we consider the general setting of a differentiable function f with differentiable inverse g such that y = f (x) if and only if x = g(y), then we know that f (g(x)) = x for every x in the domain of f -1 . Differentiating both sides of this equation with respect to x, we have d dx [ f (g(x))] = d dx [x], and by the chain rule, f 0 (g(x))g 0 (x) = 1. 137 Solving for g 0 (x), we have g 0 (x) = 1 f 0 (g(x)) . Here we see that the slope of the tangent line to the inverse function g at the point (x, g(x)) is precisely the reciprocal of the slope of the tangent line to the original function f at the point (g(x), f (g(x))) = (g(x), x).

Figure 2.12: A graph of function y = f (x) along with its inverse, y = g(x) = f -1 (x). Observe that the slopes of the two tangent lines are reciprocals of one another.

To see this more clearly, consider the graph of the function y = f (x) shown in Figure 2.12, along with its inverse y = g(x). Given a point (a, b) that lies on the graph of f , we know that (b, a) lies on the graph of g; said differently, f (a) = b and g(b) = a. Now, applying the rule that g 0 (x) = 1/ f 0 (g(x)) to the value x = b, we have g 0 (b) = 1 f 0 (g(b)) = 1 f 0 (a) , which is precisely what we see in the figure: the slope of the tangent line to g at (b, a) is the reciprocal of the slope of the tangent line to f at (a, b), since these two lines are reflections of one another across the line y = x. Derivative of an inverse function: Suppose that f is a differentiable function with inverse g and that (a, b) is a point that lies on the graph of f at which f 0 (a) , 0. Then g 0 (b) = 1 f 0 (a) . More generally, for any x in the domain of g 0 , we have g 0 (x) = 1/ f 0 (g(x)). The rules we derived for \ln (x), arcsin(x), and arctan(x) are all just specific examples of this general property of the derivative of an inverse function. For example, with 138 g(x) = \ln (x) and f (x) = e x , it follows that g 0 (x) = 1 f 0 (g(x)) = 1 e \ln (x) = 1 x .

## Summary

In this section, we encountered the following important ideas:

• For all positive real numbers x, d dx [\ln (x)] = 1 x .
• For all real numbers x such that -1 ≤ x ≤ 1, d dx [arcsin(x)] = 1 √ 1 - x 2 . In addition, for all real numbers x, d dx [arctan(x)] = 1 1 + x 2 .
• If g is the inverse of a differentiable function f , then for any point x in the domain of g 0 , g 0 (x) = 1 f 0 (g(x)).


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# Stacked Odds - Finding the questions that you can answer

## Screenshot / Code Snippet

More screenshots can be found in the extension gallery.

Stacked Odds is an extension for Chrome that stacks the odds in your favour when looking for questions to answer on Stack Exchange sites.

This extension for Google Chrome is designed to integrate as fully as possible into the site and provide an alternative method for finding questions that you can answer.

It takes your previous answer history and uses it to infer what you might be able to answer. The extension adds an extra tab to the "Question" list and the "Unanswered" question list to display the relevant questions.

Once setup, the functionality can be accessed via the relevant tab on the Unanswered page or Questions page. For more information on using the plugin see the Users Guide.

If you are updating from a previous version (before 1.1.60), it is reccomended that you regenerate the data by opening your profile, clicking "stacked odds" and pressing "update stacked odds data"

### Platform

The extension requires either Chrome 5 or Chrome 6. Future and previous version of Chrome may also work but are not supported.

## Contact

The best way to contact me is adding a comment here. Sending an email to anything at my domain (JacobEssex.com) will work but is probably slower.

## Code

An overview on some of the code can be found in some documentation. The documentation isn't complete, but should help explain things. There are other ramblings about other relevant subjects on the Google Code wiki.

• Screenshots look good! Please consider listing Stacked-Odds on StackList. Jul 19, 2010 at 19:43
• + ok, I took a look at the docs and that is worth an up. keep working on those and maybe talk a bit about why your app needs such sweeping permissions and make grovelling assurances that you are not 'evil' and maybe you will get people to expose their pink insides to your network enabled script. ;-) Jul 19, 2010 at 23:39
• A vauge bit about permissions can be found here, an explanation as to how it worked and what was changed is here Jul 20, 2010 at 10:25
• I'm getting this error on the relevant tab: "Error:No tags defined" Jul 22, 2010 at 13:42
• Setup seems to go fine up to and including answers but no questions are downloaded? stops at setup stage 3 Jul 22, 2010 at 13:49
• Never mind, i checked fiddler and sure enough, it is downloading and finally completes the stage. You need to present some progress report to keep the user from freaking out. And I notice you are batching the questions in groups of 4. The url length restriction has been relaxed, but to be safe you can maximize throughput with the Soapi.VectorizedIdList(docs- soapi.info/code/js/stable/docs/symbols/…, demo - soapi.info/code/js/stable/VectorizedIdList.htm) Jul 22, 2010 at 13:57
• same 'no tags defined' - i have defined tags. also, it is not apparent where/how to use the extension. I had to look on the screen shots to figure it out. I would say that you are not doing yourself any favors by hiding it away back there. maybe add the 'relevant' tab to all pages or be sure to guide the user there with a notification. Jul 22, 2010 at 14:10
• @code it took me a while to find the tab too. I wonder what the error we are getting means. Jul 22, 2010 at 14:13
• although i have yet to get it working, i am very impressed by the quality of your work. this gives me a bit more motivation to trudge through the difficulties of providing a quality library. do not hesitate to communicate any needs/wants/ideas regarding soapi.js Jul 22, 2010 at 14:13
• @jjn - yacoby's work is a prime example of why i want to abstract away as many of the peccadilloes of the API as possible. The less he has to worry about sweet talking the moving target api, the more he can concentrate on the logic required to add value. This guy is a model of the reason I make so much noise. Jul 22, 2010 at 14:18
• Thanks for your responses. The current download (r40) has some bugs which are fixed in the svn version. I will put together another download tomorrow. Most issues reported have been fixed. Jul 22, 2010 at 23:22
• new version up on the site. not very well tested. If you had an older version installed you should regenerate the data by going to your profile page, clicking on the stacked odds tab and pressing "update stacked odds data". The main thing that hasn't been done is the question update number doesn't increase very well. Jul 22, 2010 at 23:50
• will respond directly to answers later on today (I feel dog tired) Jul 22, 2010 at 23:56
• @jjnguy The no tags defined error is because you need to define some relevant tags (just the main ones like Javascipt and PHP or something, don't bother adding subtags like [jquery]) on settings page. This is an unfortunate necessity that I never got around to removing because I ran out of time to write the statistics code. I think I wrote some thing on why it works like it does on the wiki Jul 23, 2010 at 22:57
• @code The generation has been improved. It takes slightly longer but the UI now updates... and it requests far more data at once. Jul 23, 2010 at 22:59

Yacoby - I like the idea (and your choice of libraries ;-)),

You might want to try and make these requirements a bit more focused, if possible.

I realize that I can examine the code and figure out if you are up to hanky panky, but not every one can, and even if they can may not be inclined to, all of these result in a quick, no-brainer 'cancel'.

If you can get that 'browsing history' issue resolved, one way or another, you have a keeper on your hands.

• It needs the data on all websites as the code has to support all possible, present and future, stack exchange sites. IDK why it wants browsing history, but I will take a look. Jul 19, 2010 at 19:43
• @yac - i haven't written any chrome extensions yet, but is there no way to define a list of top level domains that you need access to? A version of an extension needs only to support existing and reasonably expected additions. Updates can handle those that are unexpected. So, cover stackoverflow, meta.stackoverflow, superuser, serverfault, stackapps and stackexchange and you have all current sites and when new TLDs are announced for area51 sites you can publish updates. You are not going to find many people willing to provide access to ALL data on ALL websites under any circumstances. Jul 19, 2010 at 19:48
• @code poet I will consider doing that when this gets a bit further on in development. It isn't too much of an issue to support multiple builds, one with limited access to only the current sites as there is only one file that needs altering. Jul 19, 2010 at 19:57
• @code poet thanks for pointing it out though. It wouldn't have occurred to me. Jul 19, 2010 at 19:58
• @yac - cool, let me know when you get on out, I would like to try your extension without exposing the flavor of porn i surf and my credit card numbers ;-) Jul 19, 2010 at 20:23
• @code poet. The first issue (Your data on all websites) is fixed in the latest svn version. The second issue I have a bug report filed in chrome's issue tracker. Jul 20, 2010 at 10:01
• @yac - right on. will give it a go tomorrow. Jul 20, 2010 at 10:29

It seems I can't get this one working on my Super User account.

For some reason it remains stuck at:

Setup Stage:             3

Any idea what might be causing this?

Seems like @Ivo I'm getting stuck too! =(

Tried thrice so far. Refreshed the page and clicked enable stacked-odd..

## Extension Permissions

One of the main unfixed issues is the fact that when installing the extension it says that the extension has permission to access browsing history when it doesn't.

Verifying that it doesn't have this permission

If someone is really worried that the manifest file in the svn repository is not the same in the packed extension, download the extension to a known directory rather than installing it and open it with a unzip program.

The manifest file in the crx archive should match the manifest on the svn at revision it was built with (the revision is the last numbers in the version string). To access browsing history "history" must be to the manifest file. It isn't there.

Bug Report

I think it is a bug with chrome rather than anything I have (or have not) done. As such I have filed a bug report.

• Can't it access the history of individual tabs with the "tabs" permission?
– SamB
Dec 27, 2010 at 22:56

## Changelog

This was added failry late on, so doesn't list many changes. Plus I can never remember what I did between versions.
The version numbers are not very important. They increment when I feel I have made a major enough change. The last number in the version string is the svn revision number which is far more relevant.

1.1.72
This update contains only a few fixes for mostly minor things, mainly UI related. The main change was the move to the Google Chrome extension gallery.

• Fixed issue with rep formatting
• Fixed issue with number formatting
• Page now goes to the top when loading a new set of data

1.1.60

This is an improvement mainly thanks to code poet and jjnguy

• Fixed major bug in generation of stats
• Improved stats
• Vastly improved error messages

One problem: if I go to, for example, https://tex.stackexchange.com/questions#tab=relevant before choosing any tabs on that site, it gives me an error message directing me to the URL https://stackoverflow.com/users/1189/#tab=stacked%20odds.

That is, it tries to send me to the Stacked Odds tab of my tex.SE user page, but it uses the domain name for SO instead of the one for tex.SE. (As it happens, this results in a 404 rather than giving me some error about there being no such tab or not being able to mess with other users options, but I assume that this just means user 1189 was deleted from SO due to inactivity.)

format: plain


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Animation loop test

I'm currently putting together a variation of Zack Hixson's tile flip. Instead of a grid of objects created within animation nodes, I've used an predetermined Group based on an array. (I tried the grid approach first but got stuck with the same problem).

What i'm trying to do is make tiles remain flipped over once the empty has passed, rather than flipping back to their original position. I've tried all sorts of ways to effect this, the latest using nested loops:

What i've tried to do is test the rotation of each object in the group and, if it's x rotation is 180 degrees, it should be skipped and the next object in the loop should be tested. Interestingly, as seen in the viewer, the rotation comes up at 180.00001 degrees despite only being set at 180 as the empty passes. Inevitably the test fails.

Any thoughts on where i'm going wrong? As i say, i've put the test inside the core transformation loop in various forms but the it's comprehensively ignored in every location!!

I think that the main problem is that you are comparing the Angles using an Equal To compare node, generally you shouldn't use Equal to when comparing floats. Instead, for instance, if you want to check that a float x is equal to a float y, you check if |x-y|< epsilon where epsilon is some small float like 0.001.


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# Browse Dissertations and Theses - Mathematics by Contributor "Berndt, Bruce C"

• (2018-07-13)
In this dissertation, we discuss properties of the family of sequences $\mathbf{u_d} = \{u_d(n)\}_{n \geq 0}$ for positive integer $d$. We define them by letting $u_d(n)$ be the coefficient of $X^n$ in \$\displaystyle ...

application/pdf

PDF (2MB)
• (2018-07-11)
This thesis is divided into three major topics. In the first, we study questions concerning the distribution of lattice points in dimensions two and higher. We give asymptotic formulas for the number of integer lattice ...

application/pdf

PDF (642kB)


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# How Far Out Is Quantum Theory Really?

The quantum world is commonly conceived of as (i) wildly strange, and (ii) far removed from everyday experience. But (i) certainly need not be true: many find quantum theory quite natural and intuitive (although others disagree, as I have discussed before). And it’s similarly unclear whether or not (ii) is true. So I propose we try to figure it out.

For example, here’s an experimental illustration of quantum diffraction (called “quantum eraser” by the authors) that you can do in your house. Admittedly, this particular experiment has a classical description. But is it possible to make the experiment just a little more sophisticated, and observe a truly non-classical effect? And are there other canonical “quantum phenomena” that are similarly accessible?

I know of no satisfying answer. So let me pose the following challenge:

Using only simple, every-day objects, design experiments illustrating all the canonical quantum effects.

There are different ways of breaking down what these effects are, but here’s one way of doing it. Design “every-day” experiments of the following three phenomena:

1. Discrete energy values (e.g., a blackbody radiation experiment);
2. General diffraction (outside the scope of classical E&M); and
3. Coherent superposition.

These phenomena together are sufficient to give us much of the basis for ordinary non-relativistic quantum mechanics. Wouldn’t it be nice to have some idea as to just how far out they really are?

Soul Physics is authored by Bryan W. Roberts. Thanks for subscribing.
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# Dates and Times in driftR

## Overview

A number of functions in driftR, including dr_factor(), dr_replace(), and dr_drop, rely on date and time information supplied by the instrument. driftR anticipates that these two pieces of information will be stored in character vectors in predictable formats that can be easily parsed. Under the hood, we use the lubridate package’s parse_date_time() function to automatically detect and properly format date and time inputs supplied by the user and their instrument.

## Working with Dates in driftR

We currently support two date formats in our functions - the MDY (or “month day year”) format and the YMD (or “year month day”) format.

The following are examples of MDY that lubridate can parse:

Jan 1, 2018
January 1, 2018
1/1/18
01/01/18
1/1/2018
01/01/2018
1-1-18
01-01-18
1-1-2018
01-01-2018

The following are examples of YMD that lubridate can parse:

2018/1/1
2018/01/01
2018-1-1
2018-01-01

To prevent confusion, we do not support the DMY format. If you have data that is DMY formatted, as in 15/01/2018, you can convert it to YMD formatted character using the following sample syntax:

x <- "15/01/2018"
y <- as.character(lubridate::dmy(x))

Once you have a properly formatted character vector, you can use it with any of the driftR functions that require a date!

## Working with Times in driftR

We currently support times that include the hours, minutes, and seconds of a particular measurement. These should be formatted using a 24 hour clock (e.g. 00:05:05 for five minutes after midnight). If your data are formatted ysing a 12 hour clock (e.g. 12:05:05 AM), the following code can be used to convert them to a 24 hour clock

library(dplyr)

data <- data %>%
mutate(time = format(strptime(time, "%I:%M %p"), format="%H:%M:%S"))

### Times in dr_replace and dr_drop

Both dr_replace() and dr_drop() allow you to specify ranges of times to replace or drop values for. We recommend using a 24 hour clock here as well. If you are not used to using a 24 hour clock, you can find a conversion table here. There is one key difference with these two functions that is worth noting - when you specify a time for each, you do not need to specify seconds but may optionally choose to do so. For example, both 12:05 and 12:05:00 are valid inputs for both dr_replace() and dr_drop().

### Time Zones

The parse_date_time() function from lubridate relies on timezone information. By default, driftR functions use your computer’s timezone setting. You can see this by using the following base function in your console:

>  Sys.timezone()
[1] "America/Chicago"

What is critical about this setting is that the timezone must match the timezone where your data were collected. If you are using data collected in a timezone that is different from the timezone your computer is set to, you need to specify the correct timezone for your data in dr_factor(), dr_replace(), and dr_drop using the tz argument. Include the string name of the appropriate timezone as the input for this argument. You can get a full list of acceptable timezone strings from the OlsonNames() function, which has 592 possible strings:

> OlsonNames()
[5] "Africa/Asmara"  "Africa/Asmera"  "Africa/Bamako"  "Africa/Bangui"
# output truncated

Copy the appropriate string into your driftR function from this output.

## Splitting a Combined Date and Time Variable

If your data arrive with the date and time combined, you can use the following functions to (1) convert it from a string to a properly formatted date and time value, and then (2) split it into separate date and time values:

library(lubridate)

input <- "1/12/18 12:05:00"

parsed <- parse_date_time(inpout, orders = "mdy HMS")

date <- as.character(as_date(parsed))
time <- format(parsed, "%H:%M:%S")

This will give you well formatted data that can be read into any of the driftR functions without further modifications.


text:
# a differential equation with matrices

How can I solve this system of equation: $$Mx''(t)+(K+N)x'(t)+(D+G)x(t)=F(t)$$ $$x(0)=0$$ where M, K, N, D, G are matrix I tried substitution $$z=(x, x')$$

• Well, that substitution is where I would start - then move to eigenvalues and eigenvectors. Please post your work. – Adrian Keister May 21 at 23:07


text:
Divergence theorem

In vector calculus, the divergence theorem, also known as Gauss' theorem, Ostrogradsky's theorem, or Ostrogradsky-Gauss theorem is a result that links the divergence of a vector field to the value of surface integrals of the flow defined by the field. The divergence theorem is an important result for the mathematics of physics, in particular in electrostatics and fluid dynamics.

 Contents

Intuition

The intuitive content is simple: if there's water flowing in some area, and you are interested in how much water flows out of a certain region within that area, then you need to add up the sources inside the region and subtract the sinks. The water flow is represented by a vector field, and the vector field's divergence at a given point describes the strength of the source or sink there. So integrating the field's divergence over the interior of the region should equal the integral of the vector field over the region's boundary. The divergence theorem says that this is in fact the case.

The divergence theorem is thus a conservation law, stating that the volume total of all sinks and sources, i.e. the volume integral of the divergence, is equal to the net flow across the volume's boundary.

Formal statement

Suppose V is a subset of Rn (think of the case n=3 for now) which is compact and has a piecewise smooth boundary. If F is a continuously differentiable vector field defined on a neighborhood of V, then we have

[itex]\iiint_V\left(\nabla\cdot\mathbf{F}\right)dV=\iint_S\mathbf{F}\cdot d\mathbf{S}[itex]

where S = ∂V is the boundary of V oriented by outward-pointing normals, and dS is shorthand for NdS, the outward pointing normal of the boundary ∂V.

We note that Gauss' theorem follows from the more general Stokes' theorem, which generalizes the fundamental theorem of calculus.

Applications

Electrostatics

Applied to an electrostatic field we get Gauss's law: the divergence is a constant times the volume charge density.

Gravity

Applied to a gravitational field we get that the surface integral is -4πG times the mass inside, regardless of how the mass is distributed, and regardless of any masses outside.

Spherically symmetric mass distribution

In the case of a spherically symmetric mass distribution we can conclude from this that the field strength at a distance r from the center is inward with a magnitude of G/r² times the total mass at a smaller distance, regardless of any masses at a larger distance.

For example, a hollow sphere does not produce any gravity inside. The gravitational field inside is the same as if the hollow sphere were not there (i.e. the field is that of any masses inside and outside the sphere only).

Cylindrically symmetric mass distribution

In the case of an infinite cylindrically symmetric mass distribution we can conclude that the field strength at a distance r from the center is inward with a magnitude of 2G/r times the total mass per unit length at a smaller distance, regardless of any masses at a larger distance.

For example, an infinite hollow cylinder does not produce any gravity inside.

Bouguer plate

We can conclude that for an infinite, flat plate (Bouguer plate) of thickness H gravity outside the plate is perpendicular to the plate, towards it, with magnitude 2πG times the mass per unit area, independent of the distance to the plate (see also gravity anomalies).

More generally, for a mass distribution with the density depending on one Cartesian coordinate z only, gravity for any z is 2πG times the difference in mass per unit area on either side of this z value.

In particular, a combination of two equal parallel infinite plates does not produce any gravity inside.

History

The theorem was first discovered by Joseph Louis Lagrange in 1762, then later independently rediscovered by Carl Friedrich Gauss in 1813, by George Green in 1825 and in 1831 by Mikhail Vasilievich Ostrogradsky, who also gave the first proof of the theorem. Subsequently, variations on the Divergence theorem are called Gauss's Theorem, Green's theorem, and Ostrogradsky's theorem.

This article was originally based on the GFDL article from PlanetMath at http://planetmath.org/encyclopedia/Divergence.html de:Gauscher Integralsatz fr:Thorme de flux-divergence ja:発散定理 es:Teorema de Gauss

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# If f^2*g < 0, which of the following must be true?

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If f^2*g < 0, which of the following must be true?  [#permalink]

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14 Aug 2018, 02:37
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If $$f^2 g < 0$$, which of the following must be true?

(A) $$f < 0$$

(B) $$g < 0$$

(C) $$fg < 0$$

(D) $$fg > 0$$

(E) $$f^2 < 0$$

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If f^2*g < 0, which of the following must be true?  [#permalink]

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14 Aug 2018, 02:45
B.
$$f^2$$ is always positive hence g has to be negative for inequality to hold true
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Re: If f^2*g < 0, which of the following must be true?  [#permalink]

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14 Aug 2018, 02:47
Bunuel wrote:
If $$f^2 g < 0$$, which of the following must be true?

(A) $$f < 0$$

(B) $$g < 0$$

(C) $$fg < 0$$

(D) $$fg > 0$$

(E) $$f^2 < 0$$

f^2 will always be +ve

for f^2 g < 0
g must be < 0

so B
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Re: If f^2*g < 0, which of the following must be true?  [#permalink]

### Show Tags

14 Aug 2018, 02:48
Bunuel wrote:
If $$f^2 g < 0$$, which of the following must be true?

(A) $$f < 0$$

(B) $$g < 0$$

(C) $$fg < 0$$

(D) $$fg > 0$$

(E) $$f^2 < 0$$

Given

$$f^2 g < 0$$

As $$f^2$$ can never be negative , g must be negative.

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Re: If f^2*g < 0, which of the following must be true?  [#permalink]

### Show Tags

14 Aug 2018, 03:14
Bunuel wrote:
If $$f^2 g < 0$$, which of the following must be true?

(A) $$f < 0$$

(B) $$g < 0$$

(C) $$fg < 0$$

(D) $$fg > 0$$

(E) $$f^2 < 0$$

$$f^2 g < 0$$
$$f^2$$ is always +ve.
Hence g<0

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Re: If f^2*g < 0, which of the following must be true?  [#permalink]

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14 Aug 2018, 04:29
Bunuel wrote:
If $$f^2 g < 0$$, which of the following must be true?

(A) $$f < 0$$

(B) $$g < 0$$

(C) $$fg < 0$$

(D) $$fg > 0$$

(E) $$f^2 < 0$$

B, since f^2 will always be positive
Re: If f^2*g < 0, which of the following must be true?   [#permalink] 14 Aug 2018, 04:29
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text:
# Given the resistance of $1\Omega,2 \Omega$ and $3 \Omega$ how will be combine them to get an equivalent resistance of $\large\frac{11}{5}$$\Omega \begin{array}{1 1} \text{R_1 and R_2 in series and the resultant in parallel with R_3} \\ \text{R_1 and R_3 in series and the resultant in parallel with R_2} \\ \text{R_2 and R_3 in parallel and the resultant series in with R_1} \\ \text{R_1 and R_2 in parallel and the resultant series with R_3} \end{array} ## 1 Answer Comment A) Solution : Let R_1= 1\Omega,R_2= 2 \Omega and R_3 =3 \Omega If R_2 and R_3 are in parallel. then \large\frac{1}{R_p} =\frac{1}{R_2} +\frac{1}{R_3} \large\frac{1}{2}+\frac{1}{3} =\frac{3+2}{6} \qquad= \large\frac{5}{6} R_p =\large\frac{6}{5}$$\Omega$
$R_p$ and $R_1$ are in series . hence
$R= R_p +R_1$
$\quad= \large\frac{6}{5} $$+1 \qquad= \large\frac{11}{5}$$\Omega$
Answer : $\text{$R_2$and$R_3$in parallel and the resultant series in with$R_1$}$


text:
GSEB Solutions Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.2

Gujarat Board GSEB Solutions Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.2 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.2

Question 1.
In figure, ABCD is a parallelogram, AE ⊥ DC andCF AD ⊥ AB = 16cm,AE = 8 cm and CF = 10 cm, find AD.

Solution:
ar( || gm ABCD) = AB x AE = 16 x 8 cm2
= 128 cm2 ………..(1)
ar( || gm ABCD) = AD x CF
= AD x 10 cm2 ……….(2)
From (1) and (2), we get
AD = $$\frac {128}{10}$$

Question 2.
If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar(EFGH) = $$\frac {1}{2}$$ar (ABCD).
OR
Prove that the area of the quadrilateral formed by joining the mid-points of the sides of a parallelogram is half the area of the parallelogram.
Solution:
Given: A parallelogram ABCD, in which E, F, G, and H are the mid-points of sides AB, BC, CD, and DA respectively.
To prove: ar (EFGH) = ar(ABCD)
Construction: Join HF.

Proof: In || gm ABCD,
AD || BC and AD = BC (Opposite sides of a || gm)
⇒ $$\frac {1}{2}$$ AD || $$\frac {1}{2}$$BC and $$\frac {1}{2}$$ AD = $$\frac {1}{2}$$BC
⇒ AH || BF and AH = BF
∴ ABFH is a parallelogram.
(If a pair of opposite side of a quadrilateral is parallel and equal, it is a parallelogram)
Further, ΔEFH and ||gm ABFH lie between same base FH and between same parallels AB and FH.
∴ ar(ΔEFH) = $$\frac {1}{2}$$ ar( || gm ABFH) ……….(1)
Similarly, we can show
ar(ΔFGH) = $$\frac {1}{2}$$ ar(|| gm FCDH) ………..(2)
Adding eqn. (1) and (2), we get
ar(ΔEFH) + ar(ΔFGH)
= $$\frac {1}{2}$$ ar(||gm ABFH) + $$\frac {1}{2}$$ ar(||gmFCD)
∴ ar(EFGH) =$$\frac {1}{2}$$ [ar(ABFH) + ar(FCDH)]
= $$\frac {1}{2}$$ ar(ABCD).

Question 3.
P and Q are any two points lying on the sic DC and AD respectively of a parallelogram ABCD. Show that ar(ΔAPB) = ar(ΔBQC).
Solution:
APB and || gm ABCD are on the same AB and between the same parallels AB and DC.

∴ ar(ΔAPB) = $$\frac {1}{2}$$ ar( || gm ABCD) ………..(1)
Again, ABQC and 11gm ABCD are on the same base BC and between the same parallels BC and AD.
∴ ar(BQC) = $$\frac {1}{2}$$ ar( || gm ABCD) ………(2)
From (1) and (2),
ar[ΔAPB] = ar[ΔBQC]

Question 4.
In figure, P is a point in the interior of a parallelogram ABCD. Show that:

1. ar(ΔAPB) + ar(ΔPCD) = ar( || gm ABCD)
2. ar(ΔAPD) + ar(ΔPBC) = ar(ΔAPB) + ar(ΔPCD)
[Hint: Through P, draw a line parallel to AB]
Solution:
Construction: Through P, draw a line EF parallel to AB.

1. EF fi AB …….(1) (by construction)
∴ Opposite sides of a parallelogram are parallel
∴ AE || BF ………..(2)
In view of (1) and (2),
A quadrilateral is a parallelogram if its opposite sides are parallel
Similarly, quadrilateral CDEF is a parallelogram.
∴ ΔAPB and fi gm ABFE are on the same base AB and between the same parallels AB and EF.
∴ ar(ΔAPB) = $$\frac {1}{2}$$ ar(|| gm ABFE) …….(3)
Again, LSPCÐ and fi gin CDEF are on the same base DC and between the same parallels DC and EF.
∴ ar(ΔPCD) = $$\frac {1}{2}$$ ar(|| gm CDEF) ………..(4)
Adding (3) and (4), we get
$$\frac {1}{2}$$ ar(ΔAPB) + $$\frac {1}{2}$$ ar(ΔPCD)
= $$\frac {1}{2}$$ ar(|| gm ABFE) + $$\frac {1}{2}$$ ar(|| gm in CDEF)
= $$\frac {1}{2}$$ [ar(|| gm ABFE) + ar(|| gm CDEF)
= $$\frac {1}{2}$$ ar(|| gm ABCD) ……….(5)

2. ar(ΔAPD) + ar(ΔPBC)
= ar( || gm ABCD) – [arΔAPB) + ar(ΔPCD)]
= 2[ar(ΔAPB) + ar(ΔPCB)] – [ar(ΔAPB) + ar(ΔPCD)]
= ar(ΔPB) + ar(ΔPCD).

Alternative: Like in part (i), draw a line ∠M, through P, parallel to AD. Hence we can show
ar(ΔAPD) + ar(ΔPBC)
= ar(|| gm ABCD)
From eqn. (5) and (6),
ar(ΔAPD) + ar(ΔPBC)
= $$\frac {1}{2}$$ ar(ΔAPB) + ar(ΔPCD)

Question 5.
In figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that:

1. ar( || gm PQRS) = ar( || gm ABRS)
2. ar(ΔAXS) = $$\frac {1}{2}$$ ar(|| gm PQRS)
Solution:
1. In ΔPSA and ΔQRB,
∠SPA = ∠RQB
Corresponding angles as PS || QR and transversal PB
∠PAS = ∠QBR
Corresponding angles from AS || BR and transversal PB

Also, PS = QR
[Opposite sides of || gm PQRS
ΔPSA = ΔQRB [By AAS rule]
∴ ar(ΔPSA) = ar(ΔQRB) ………..(1)
[∴ congruent figures have equal areas]
Now, ar(|| gm PQRS)
= ar(ΔPSA) + $$\frac {1}{2}$$ ar(||gm AQRS)
= ar(ΔQRB) + $$\frac {1}{2}$$ ar(||gm AQRS) [ using (1) ]
= ar(|| gm ABRS) …….(2)

2. ΔAXS and || gm ABRS are on the same base AS and between the same parallels AS and BR.
∴ ar(ΔAXS) = $$\frac {1}{2}$$ ar( || gm ABRS)
= $$\frac {1}{2}$$ ar( || gm PQRS) [(Using(2))]

Question 6.
A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the fields is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?
Solution:
∴ ΔAPQ, ΔAPS, and ΔAQR lie between the same parallels.
∴ Their attitudes are the same. Let it be x. Then,

∴ SR = PQ (opposite sides of the parallelogram are equal)

From eqn. (1) and (2),
ar(ΔAPQ) = ar(ΔAPS) + ar(ΔAQR)
Therefore, either the former should sow wheat in ΔAPQ and pulses in the other two triangles APS and AQR or pulses in ΔAPQ and wheat in the other two triangles APS and AQR.


text:
# Tutorials for Java Users#

Java plays an important role in big data ecosystem, and a lot of graph processing systems developed by Java (e.g., Apache Giraph and GraphX) have been widely applied. To integrate with Java ecosystem, GraphScope allows graph analytics algorithms developed for Apache Giraph and GraphX to directly run over GraphScope GAE.

In addition, GraphScope GAE supports the PIE programming model, which can provide better performance than widely-used vertex-centric model. GraphScope GAE is equipped with Java PIE interface exposed by GRAPE-jdk, and users can develop their own algorithms using the PIE programming model.

The following tutorials will introduce how to leverage the above features in GraphScope GAE Jave SDK.


text:
Easter operating times at the Ruislip Lido Railway

The Ruislip Lido Railway has released its running details for the Easter period.

A variety of locomotives are expected to be hauling trains to and from ‘The Beach’.

Tickets do not need to be pre booked and can be purchased on the day.

Fares for the weekend can be found below:

SINGLE RETURN
Child (3 to 15) £1.70 £2.80
Family of 4 (max 2 adults) £7.00 £11.00

The timetable is as follows:

Sunday 21st April 2019

Easter Eggspress trains will be running, however, no spaces are left on these. A service train will run with the following times:

Departures from Woody Bay: 11:40, 12:20, 13:00, 13:40, 14:20, 15:00, 15:40, 16:20 and 17:00
Departures from Willow Lawn : 10:40, 12:00, 12:40, 13:20, 14:00, 14:40, 15:20, 16:00, 16:40 and 17:20.

Monday 22nd April 2019

Easter Eggspress trains will be running, however, no spaces are left on these. A service train will run with the following times:

Departures from Woody Bay: 11:40, 12:20, 13:00, 13:40, 14:20, 15:00, 15:40, 16:20 and 17:00
Departures from Willow Lawn : 10:40, 12:00, 12:40, 13:20, 14:00, 14:40, 15:20, 16:00, 16:40 and 17:20.

Where Next?

News Homepage
For the Latest Railway News


text:
# Lagrangian of a free particle

1. Feb 23, 2012

### Favicon

I've heard it said that the Lagrangian of a free particle cannot possibly be a function of any position coordinate, or individual velocity component, but it is a function of the total magnitude of velocity. Why is this the case? I'd be grateful for any pointers in the direction of either a physical or a mathematical explanation

EDIT: I should make it clearer that this is assuming no prior knowledge of kinetic energy or Newton's laws

Last edited: Feb 23, 2012
2. Feb 23, 2012

### FeX32

The Lagrangian can be expressed about any coordinate that can be written in the form:

\frac{d}{dt}\left(\frac{∂L}{∂\dot{q}_i}\right)-\frac{∂L}{∂q_i}=Q_i

Where $L$ is the Lagrangian (kinetic energy - potential energy)

3. Feb 23, 2012

### Favicon

Actually, after a few minutes pondering, I'd like to attempt an answer myself, based on symmetry.

The Lagrangian cannot depend on individual velocity components (but it can depend on the overall magnitude of the velocity) because of rotational symmetry - with no external forces acting on the particle, how could the direction of it's velocity have any impact on the physics of the situation?

As for the positional coordinates, I presume these can't affect the Lagrangian because of translational symmetry. (Again, with no external forces acting on the particle, the location has no bearing on its energy, which after all is what the Lagrangian describes)

Have I got the right idea here, or am I making false assumptions?

4. Feb 23, 2012

### FeX32

Of course, the energies are usually much easier to express in certain coordinate systems.

5. Feb 23, 2012

### Favicon

I may be being blindingly stupid here, but I don't see how that answers my original question. What is it about the position and velocity components of a free particle which means they cannot fit that equation?

6. Feb 23, 2012

### FeX32

There you have it

7. Feb 23, 2012

### FeX32

They can fit the equation. Just find any coordinate "q" that you can write the energies in.

8. Feb 23, 2012

### Favicon

Excellent, thank you Fex. It seems that just after asking for help is when I usually solve a problem myself..

9. Feb 23, 2012

### FeX32

To answer you question directly. There is an infinite amount of coordinate systems that you can express the Lagrangian in. The thing is, usually one or two of them are the simplest, mathematically.

10. Feb 23, 2012

### FeX32

No problem. Yes, it seems you understand it essentially.


text:
The Muon

## Seed-Aggregate Model

We describe the mechanics of muons by weaving together various chains of events that link absorption and emission processes. The archetypal history that these interactions modify is

$\Psi \left( \,\sf{\mu}^{-} \right) = \left( \sf{\Omega}_{1}, \sf{\Omega}_{2}, \sf{\Omega}_{3} \ \ldots \ \right)$

where each orbital cycle $\sf{\Omega}$ is a bundle of 40 Anaxagorean sensations; 8 right-side and 12 left-side somatic sensations; 2 hot, 2 freezing, 2 steamy and 2 cool thermal sensations; 4 red, 4 green and 4 white visual sensations. Here is a visual representation of the muon made by combining the icons of its components

Each Anaxagorean sensation defines a seed. So a muon can be modeled as the seed-aggregate

$\sf{\Omega} \left(\, \sf{\mu}^{-} \right) = \mathrm{4}\sf{U} + \mathrm{2}\sf{B} + \mathrm{2}\sf{T} + \mathrm{2}\sf{S } + \mathrm{2}\sf{C} + \mathrm{4}\sf{M} + \mathrm{4}\sf{A} + \mathrm{8}\sf{O} + \mathrm{12}\bar{\sf{O}}$

## A Quark Model of the Muon

The quark hypothesis proposes that pairs of seeds are always bound together as quarks. So to make a more detailed model of the muon, seeds are associated in these pairs

 + →
 + →
 + →
 + →
 + →
 + →
 + →

so that a muon consists of twenty quarks. And here is a visual representation of the muon that combines their icons. The back row of quarks is exactly the same as the front row.

 \begin{align} \sf{\Omega} \left(\, \sf{\mu}^{-} \right) = \mathrm{4}\bar{\sf{u}} &+ \mathrm{2}\bar{\sf{b}} + \mathrm{2}\sf{t} + \mathrm{2}\bar{\sf{s}} + \mathrm{2}\sf{c} + \mathrm{4}\bar{\sf{m}}+ \mathrm{4}\sf{a} \end{align}

Related WikiMechanics articles.

page revision: 172, last edited: 25 Nov 2017 00:04


text:
2 added 190 characters in body; edited title; added 11 characters in body; edited title

# Finding non-nilpotent groups of odd order withoutnon-cyclicnilpotentquotients

I hope that my question is appropriate for MO, since it might turn out te be mainly a question about GAP or other group theory software.

Is there an algorithm to produce all non-nilpotent groups of odd order (up to some given upper bound)?

All groups of odd order are solvable by the famous Feit-Thompson theorem; I guess that this fact could be useful in enumerating all these groups, but I don't know how.

Any general ideas to produce many such groups (not necessarily all of them) would also be greatly appreciated.

I already checked the SmallGroups GAP database, and it turns out that there are only 1016 such groups of order $\leq 2015$.

EDIT: The following question probably makes more sense:

Is there an algorithm to produce all groups of odd order (up to some given upper bound) not admitting a non-cyclic nilpotent quotient?

1

# Finding non-nilpotent groups of odd order

I hope that my question is appropriate for MO, since it might turn out te be mainly a question about GAP or other group theory software.

Is there an algorithm to produce all non-nilpotent groups of odd order (up to some given upper bound)?

All groups of odd order are solvable by the famous Feit-Thompson theorem; I guess that this fact could be useful in enumerating all these groups, but I don't know how.

Any general ideas to produce many such groups (not necessarily all of them) would also be greatly appreciated.

I already checked the SmallGroups GAP database, and it turns out that there are only 1016 such groups of order $\leq 2015$.


text:
# Paragraph indentation after using \ensuremath

I have created a macro that writes the progressive number of the exercises in an exam sheet. The macros works in normal mode and in math mode, so that I am able to write something like this:

\esercizio How much is $1+1$?
Solve the following equation:
$\esercizio x-1=0$

The problem is: there is always an indentation at the beginning of each paragraph, even if I set \parindent=0pt.

This is the code I'm using:

\newcounter{NumEsercizio}
\newcommand{\esercizio}[1][]{
\stepcounter{NumEsercizio}
\ensuremath{\mathbf{\theNumEsercizio}.
\ifthenelse{\equal{#1}{}}
{}
{\text{ [#1]}}}
}

How can I remove paragraph indentation before the exercise number?

This is a working example:

\documentclass[a4paper,12pt]{article}
\usepackage[italian]{babel}
\usepackage{amsmath}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\pagestyle{empty}
\usepackage{lmodern}
\usepackage{ifthen}
\usepackage{parskip}

\parindent=0pt

\newcounter{NumEsercizio}
\newcommand{\esercizio}[1][]{%
\stepcounter{NumEsercizio}%
\ensuremath{\mathbf{\theNumEsercizio}.
\ifthenelse{\equal{#1}{}}
{}
{\text{ [#1]}}}
}

\begin{document}
\noindent
Risolvi le seguenti disequazioni:
$\esercizio\quad\frac{(x^2-1)(x-2)^2}{x^2+2}\le0,\qquad \esercizio\quad x^2(16-x^2)\le0.$

\noindent
\esercizio Risolvi il seguente sistema col metodo di riduzione,
poi rappresenta graficamente la soluzione:
$\begin{cases} 5x+2y=1\\ 5x-y=-8 \end{cases}$
\end{document}

• I gave a suggestion for what I suppose you want to achieve. – koleygr Oct 13 '18 at 22:30

Second Edit (with your working example):

\documentclass{article}
\usepackage{lipsum}
\usepackage{amsmath}
\newcounter{NumExersize}
\newcounter{SubExersize}
% #1 is the depth inside the exersize
% #2 is the content of the exercise item
\newcommand{\esercizio}[2][1]{%
\ifnum#1=1 %
\setcounter{SubExersize}{0}%
\stepcounter{NumExersize}%
\arabic{NumExersize}. {\bfseries #2}
\else
\stepcounter{SubExersize}%
\ifnum\value{SubExersize}>1 %
\fi
}
\begin{document}
\noindent
\esercizio{Risolvi le seguenti disequazioni:}
$\esercizio[2]{\frac{(x^2-1)(x-2)^2}{x^2+2}\le0}\quad \esercizio[2]{x^2(16-x^2)\le0.}$

\noindent
\esercizio{Risolvi il seguente sistema col metodo di riduzione,
poi rappresenta graficamente la soluzione:}
$\begin{cases} 5x+2y=1\\ 5x-y=-8 \end{cases}$
\end{document}

Output:

Edit:

\documentclass{article}
\usepackage{lipsum}
\newcounter{NumEsercizio}
\newcommand{\esercizio}[1][]{%
\def\arg{#1}%
\stepcounter{NumEsercizio}%
\ensuremath{\mathbf{\theNumEsercizio}}.{\ifx\arg\empty \else{ \bfseries #1}\fi}%
}
\begin{document}
\lipsum[1]

\noindent\esercizio[How much is $1+1$?]

Solve the following equations:
$\esercizio x-1=0$
\end{document}

Output:

Old answer: (without realy understanding the command) Whith the help of \noindent and some"%" added:

\documentclass{article}
\usepackage{ifthen}
\usepackage{lipsum}
\newcounter{NumEsercizio}
\newcommand{\esercizio}[1][]{%
\stepcounter{NumEsercizio}%
\ensuremath{\mathbf{\theNumEsercizio}.
\ifthenelse{\equal{#1}{}}
{}
{\text{ [#1]}}}
}
\begin{document}
\lipsum[1]

\noindent\esercizio{How much is $1+1$?}
Solve the following equation:
$\esercizio{x-1=0}$
\end{document}

The "%" symbol doesn't allow an empty space to be created during the execution of the command (search about "latex tokens")

The \noindent command just added out of the command because it doesn't make sense if the command is inside  as in your second example.

Also, you have to enclose the command's arguments inside {} to meke clear what is the argument that the command takes as input.

Output:

• It is just a best practice when you have an argument in a command, to make clear what is supposed to be the argument of it... May be in your case doesn't needed... but I don't even know if you would like to give to your command a whole paragraph or a whole line or what. This way every possible editor of the document will have a more clear view of your code.. But may be in your case doesn't needed. – koleygr Oct 13 '18 at 21:03
• By mistake I used a wrong photo that shows the command with one space (I was ready to show you why you have the problem with the tokens but finally didn't) – koleygr Oct 13 '18 at 21:05
• As @DavidCarlisle noticed, your command has just an optional argument, and so, my code has these extra {} that doesn't really do something instead of grouping the included text...(your command expects of just a possible optional argyment that would be enclosed inside [] and not {}) I don't really see the reason for this optional argument, and I suppose you could add (edit) in your post an example with the output you expect in some cases (At least one with an optional argument and one without) – koleygr Oct 13 '18 at 21:31
• @zar it's your document, but I don't think this helps your readers at all. (but +1 for this answer for doing what you ask anyway:-) – David Carlisle Oct 13 '18 at 21:39
• zar: I just edited to what I supposed you meant... But as @DavidCarlisle mentioned , you will find a better answer and ideas after editing – koleygr Oct 13 '18 at 21:44


text:
# Tag Info

Accepted

### Why are engine blocks so robust apart from containing high pressure?

If you consider only the static forces then indeed the thickness might seem over-engineered. However, engine blocks are not statically loaded. They operate in the range of a few hundred to a few ...
• 22.6k

### Why are engine blocks so robust apart from containing high pressure?

You need to consider that the complete engine block has to withstand the reciprocating forces generated by each of the pistons and con rods moving as well as the rotational forces from the rotating ...
• 13.6k

### Can gravity be a driving force for a spring-mass system?

It really depends on the problem and what you want to capture. Usually, in classes, it may be beneficial for the students at an initial stage to disregard gravity alltogether in order to simplify the ...
• 22.6k

### Why are engine blocks so robust apart from containing high pressure?

The vibration, loading, and fatigue aspects have already been addressed, but a wide range of operating temperatures is another factor. A typical consumer engine can be deployed in anything from say -...
• 221

### Is there a mechanical equivalent to a hi/low pass filter (as in electrical engineering)?

In a more general sense, the electrical concepts of inductance, capacitance and resistance are equivalent to mass, spring constant and friction in the mechanical world. Voltage becomes force and ...
• 6,719

### Is there a mechanical equivalent to a hi/low pass filter (as in electrical engineering)?

Yes, for sound specifically there is a term called "acoustic impedance" that, just like electrical impedance, is frequency dependent. Acoustic impedance results from the acoustic wave equation, which ...
• 3,525
Accepted

### Stiffness of a cantilever beam

Stiffness is a murky term frequently used ambiguously in engineering. However, the most common definition of stiffness is the product of a beam's Young's Modulus $E$ (which is a function of its ...
• 12.7k

### In vibration study, should the reference be necessarily the mean/equilibrium position?

It can be arbitrary. It just simplifies the mathematics significantly. However the end result should be the same.
• 22.6k

### Three-phase versus single-phase operation of a generator

But I'm thinking it may be better (more efficient maybe? less vibration maybe?) to run it in three-phase mode, with the load balanced among the three hot terminals. Correct. Figure 1. The load is ...
• 7,519

Take the Fourier Transform of the time varying driving force, this will give the frequency content of the driving force. Multiple modes of vibration can be driven at once, and will superpose with each ...
• 1,262
Accepted

### Modelling splashing water in FEA

This may be only a partial answer since I don't have any idea what to do for the Young's modulus of the fluid. Consider the paper Dynamic Pressures on Accelerated Fluid Containers by G. W. Housner. ...
• 3,587

### Is there a mechanical equivalent to a hi/low pass filter (as in electrical engineering)?

There are analogs to filters in lots of mechanical systems. In fluid systems prone to pressure spikes perhaps because they use a fixed displacement pump, an accumulator will be used to filter out ...
• 2,135

### Detailed dimensions of an aircraft engine compressor blade

The ESA has a page on compressor blades. They give a good dimensioned diagram of an approximate shape; here are some basic dimensions: Length: 300 mm Width: 30 mm Height: 70 mm Thickness: 5 mm I can'...
• 2,485

### How to prevent vibration loosening whilst maintaining adjustability?

Double nutting can be an easy option for this. I was about to describe it myself, but these folks do a great job. This image, from the article, models the forces that do the job. Take note of the ...
• 1,966
Accepted

### How to simulate the Natural Frequencies of a Planetary Gearbox?

To obtain the modes shapes and resonant frequencies, you start from your equation of motion with no externally applied forces, which is indeed as you've stated. \mathbf M \mathbf{\ddot q} + \mathbf ...
• 1,262
Accepted

### Can a thing resonate whilst exposed to white noise?

You won't get much resonance because the phase the thing is being driven with keeps changing. The resonant thing will act like a notch filter, so you are left with the frequencies near its resonance. ...
• 11.2k
Accepted

### Do internal / external stresses affect modal frequency?

Consider a small piece cut out of a structure that has non-zero internal stress. To maintain equilibrium, there must be some forces applied to the boundary of the piece. (Of course when it was part ...
• 12.5k
Accepted

### How do those "dipole-shaped" weights attached to sections of electrical power lines work?

According to the wiki, these are called Stockbridge dampers. The design is tuned for a specific length and one might also expect a specific mass/material. The weights are mounted on flexible arms, ...
• 6,117
Accepted

### Are there standard vibration and shock specs for shipment of a product?

UPS has, or at least had when I was transit testing packaged products, a specific packaging test standard based on ISTA standards and ASTM D4169. There's also a MIL-STD, 810 something, but I don't ...
• 2,135

### Are there standard vibration and shock specs for shipment of a product?

When I have received fragile content in the past it has always come with a Shockwatch label attached to the side. I'm not 100% sure, but I believe there is some sort of guarantee by the shipper to ...
• 6,306

### I want a shockproof table for a model train. That is, protection from physical jolts, not electrical volts

Build two tables. The outer table will have a hole in the middle and only hold the glass plate. There should be about 10 cm of free space between the tables. The train will be placed on the inner ...
• 1,835
Accepted

### Strategies for cable vibration: eigenvalues are suitable?

Mathematically, an eigenvalue analysis assumes the modal displacements are infinitesimally small, so the change in stiffness in the cable caused by the vibration is a second order effect which can be ...
• 12.5k

### What does 1 x rpm or 2 x rpm or 3 x rpm in vibration analysis mean?

RPM = revolutions per minute. In the context of vibration monitoring, 1 x rpm means a vibration frequency that is the same as the rotation speed (one vibration oscillation per revolution of the ...
• 12.5k
Accepted

### How can I measure the vibrations of a solid object?

An accelerometer is the correct choice, but the smartphone accelerometer gets its readouts smoothed out in software before being made accessible to userspace. It won't pick up such fine vibrations. ...
• 6,113

### Resonance vs flutter

In flutter, the force applied to the structure depends on the deformation and velocity of the structure. Therefore, the force is necessarily at exactly the same frequency as the motion, though it may ...
• 12.5k
Accepted

### What is the difference between washer and O-ring? what is their advantages over the other?

An o-ring usually (but not always) has a round section and is made of a compressible material such as rubber, neoprene, silicon etc. A washer tends to have a square or rectangular section and is ...
• 13.6k
Accepted

### Why does the frequency of a bottle filling up changes?

Most acoustic vibrations of air more or less confined to a given geometry can be explained by two basic models. One is the one-dimensional treatment of the air in a tube, and the other is the lumped-...
• 370

### Why are engine blocks so robust apart from containing high pressure?

The example picture does not show a very solid outter wall; what you see a gasket surface. A closer look will show a lot of variations in thickness, with a lot of ribbs, bulky parts (for mounting, and ...
• 41

### Why are engine blocks so robust apart from containing high pressure?

No one has mentioned yet that there are structural loads placed on the block by other components. The first that comes to mind are the cylinder heads, and as it (they) are torqued, the block needs to ...
• 141


text:
Enter the rate (ml/hr), the amount in flask (mcg), the weight (kg), and the flask volume (ml) into the mcg/kg/min From ml/hr Calculator. The calculator will evaluate the mcg/kg/min.

## mcg/kg/min From ml/hr Formula

The following formula is used to calculate the mcg/kg/min From ml/hr.

mcg/kg/min = R * MCG / (FV * W* 60)

Variables:

• mcg/kg/min is the mcg/kg/min From ml/hr ( )
• R is the rate (ml/hr)
• MCG is the amount in flask (mcg)
• FV is the flask volume (ml)
• W is the weight (kg)

## How to Calculate mcg/kg/min From ml/hr?

The following steps outline how to calculate the mcg/kg/min From ml/hr.

1. First, determine the rate (ml/hr).
2. Next, determine the amount in flask (mcg).
3. Next, determine the flask volume (ml).
4. Next, determine the weight (kg).
5. Next, gather the formula from above = mcg/kg/min = R * MCG / (FV * W* 60).
6. Finally, calculate the mcg/kg/min From ml/hr.
7. After inserting the variables and calculating the result, check your answer with the calculator above.

Example Problem:

Use the following variables as an example problem to test your knowledge.

rate (ml/hr) = 12

amount in flask (mcg) = 12


text:
×

# Approach Really Matters share Yours

$$\large 1•$$ If $$x^n + py^n + qz^n$$ is exactly divisibly by $$x^2-(ay +bz)x + abyz$$ then find the value of $$\frac{p}{a^n}+\frac{q}{b^n}+1$$

$$\large 2•$$ In $$\triangle ABC$$, the incircle touches the sides $$BC, CA$$ and $$AB$$ respectively at $$D$$, $$E$$ and $$F$$. If the radius of incircle is $$4$$ units and $$BD, CE$$ and $$AF$$ are consecutive integers, then find the perimeter of $$\triangle ABC$$.

$$\large 3•$$ If each pair of the three equations $$x^2 +p_1x + q_1 = 0$$, $$x^2 + p_2x + q_2 = 0$$ and $$x^2 + p_3 + q_3 = 0$$ have a common root, then prove that $${p_1}^2+ {p_2}^2+ {p_3}^2+4(q_1+q_2+q_3)=2(p_1p_2+p_2p_3+p_3p_1)$$

Note by Sanjeet Raria
2 years, 4 months ago

Sort by:

2.) Let $$BD = x-1, CE = x, AF = x+1$$

Since the incircle is tangent to $$\overline{BC},\overline{CA},\overline{AB}$$ at $$D,E,F$$, we get

$$BF=BD=x-1, CD=CE=x, AE=AF=x+1$$

and $$[\triangle ABC] = rs$$ where $$s = \displaystyle \frac{a+b+c}{2} = 3x$$ and $$r = 4$$.

From Heron's formula, we get

$$[\triangle ABC]=\sqrt{s(s-a)(s-b)(s-c)}$$

$$\sqrt{(3x)(x+1)(x-1)(x)} = 4\times(3x)$$

Solve the equation we get $$x = 7$$.

Therefore, perimeter $$= 2s = \boxed{42}$$ ~~~ · 2 years, 4 months ago


text:
## Abstracts

Jakub Byszewski:

Title: A density version of Cobham's theorem

Abstract: Cobham’s theorem asserts that if a sequence is automatic with respect to two multiplicatively independent bases, then it is ultimately periodic. We prove a stronger density version of the result: if two sequences which are automatic with respect to two multiplicatively independent bases coincide on a set of density one, then they also coincide on a set of density one with a periodic sequence. We will discuss this result and its links with symbolic dynamics. This is joint work with Jakub Konieczny.

Artem Dudko:

Title: On invariant random subgroups of approximately finite full groups

Abstract: Approximately finite full groups are certain inductive limits of symmetric groups under block diagonal embeddings. An invariant random subgroup (IRS) of a group G is a Borel probability measure on the space of subgroups of G invariant under the action of G by conjugation. We classify ergodic IRS for a large class of simple approximately finite groups. The talk is based on a joint work with Konstantin Medynets.

Aurelia Dymek:

Title: $\mathscr{B}$-free systems and dynamics

Abstract: For a given subset $\mathscr{B}\subset\N$, we are interested in the associated set of $\mathscr{B}$-free numbers $\mathcal{F}_{\mathscr{B}}:=\Z\setminus\bigcup_{b\in\mathscr{B}}b\Z$. We consider the orbit closure of $\eta:=\raz_{\mathcal{F}_\mathscr{B}}\in\{0,1\}^\Z$ under the left shift $S$ and call it by a $\mathscr{B}$-free system. We will focus on topological dynamics of $\mathscr{B}$-free systems and recent results. We will formulate some open questions.

This talk will be based on the joint paper with Stanisław Kasjan, Joanna Kułaga-Przymus and Mariusz Lemańczyk.

Yonatan Gutman:

Title: The embedding problem in topological dynamics

Abstract: We survey old and new results related to the embedding problem in topological dynamics. More specifically we discuss conditions which guarantee that a $\mathbb{Z}$- dynamical system  $(X,T)$,  embeds into the $d$-cubical shift $(([0,1]^{d})^{\mathbb{Z}}, shift)$. Partially based on joint works with Lei Jin, Yixiao Qiao, Gábor Szabó and Masaki Tsukamoto.

Victor Kleptsyn:

Title: Tools and theorems in one-dimensional dynamics

Abstract: My course will be devoted to different tools and ideas in one-dimensional dynamics. Among the tools will see the distortion control technique, Lyapunov exponents, stationary measures and introduction of a random dynamics. We will see how the distortion control works in the classical Denjoy theorem and for the description of a parabolic implosion.
We will discuss the random dynamics on the circle and an important idea of a random contraction (Furstenberg-Antonov-Nalski-Deroin-VK-Navas).
We will see how stationary measure and Lyapunov exponents work in these cases.
If the time permits, we will discuss the random dynamics on the real line (Deroin, VK, Navas, Parwani), the intermediate regularity Denjoy theorem for the action of Z^d (Deroin, VK, Navas), or local vector fields argument by Loray-Rebelo-Nakai-Scherbakov.

Dominik Kwietniak:

Title: All royal measures are loosely Kronecker

Abstract: Gorodetski, Ilyashenko, Kleptsyn, and Nalsky constructed an ergodic invariant measure of a skew product torus map with zero Lyapunov exponent. Inspired by this construction, Bonatti, Diaz and Gorodetski gave sufficient conditions for weak convergence of a sequence of measures supported on periodic orbits to an ergodic measure. This method was later adapted by numerous authors and yielded examples of invariant measures with zero Lyapunov exponent in various settings. We call any measure obtained through this approach a royal measure. All royal measures are ergodic, but it was not known whether they share any other properties. In particular, it was an open question whether these measures can have positive entropy.
We show that royal measures always have zero entropy. Furthermore, we prove that all royal measures are loosely Kronecker. In other words, every royal measure is Kakutani equivalent (Kakutani equivalence is broader than the usual isomorphism of measure preserving systems) to an ergodic rotation of a non-discrete compact topological group. For the proof we introduce and study a new tool: the Feldman-Katok pseudometric fk-bar.

Bill Mance:

Title: Normal numbers for the Cantor series expansion and possible applications in algebraic geometry

Abstract: Normal numbers for the Cantor series expansion and possible applications in algebraic geometry Abstract: We will discuss basic properties of normal numbers for the Cantor series expansions and a recent result of D. Airey and B. Mance. Using ideas introduced in this paper and tools from descriptive set theory, it may be possible to show that information about algebraic varieties is encoded in the structure of sets of normal numbers. We will outline this idea and the barriers one may encounter in finishing it.

Reload image


text:
Graphing Calculator Math Worksheets Algebara Solver Chart Maker
 A+ A− B

# The Area of a Triangle: SAS Formula

## The Side Angle Side Formula

The Side Angle Side formula for finding the area of a triangle is a way to use the sine trigonometric function to calculate the height of a triangle and use that value to find the area of the triangle.
The formula $$=\frac{1}{2} \cdot c \cdot b \cdot sin(\text{A})$$
or, in general
$$Area = \frac{1}{2} \cdot side_1 \cdot side_2 \cdot sin(\text{included angle})$$
Visit this url, if you want to review what is meant by 'included angle'.

## Where does this formula come from

Can you identify which triangle below has an included angle?

Example 1
Step 1) Identify two sides and the included angle!
 Do Step 1
Step 2) Apply the formula!
 Do Step 2

Practice Problems

Step 1) Identify two sides and the included angle!
 Step 1
 Step 2

Step 1) Identify two sides and the included angle!
 Step 1
 Step 2

Step 1) Identify two sides and the included angle!
 Step 1
 Step 2
 Choose Angle A C B All Side Lengths?
Interactive Applet

Explore applet by itself.


text:
# Clustering algorithms¶

There are multiple options when one wishes to pick a clustering algorithm. Some of them, you can find at link: SCKIT learn
My problem is geographical. I have to find shops that are close together. Mainly so that a customer will not consider jumping into a car and driving to a some other shop. When a person gets in a car, it is entirely possible to drive few minutes more to a some distant location.
On the other hand, it is necessary to encompass the average mall where people usually do walk from one shop to another.
So I will here compare two clusters, simple K-means, and DBSCAN, the algorithm recommended for geographical applications.

# The data set¶

I'll use dataset from YELP data challenge. And from there take dataset describing businesses and their locations in Las Vegas, Nevada.

In [2]:
#general
import numpy as np
import scipy
from matplotlib import pyplot as plt
%pylab inline
import pandas as pd
import MySQLdb
import os
import sys
sys.setrecursionlimit(3000)

Populating the interactive namespace from numpy and matplotlib

In [5]:
con=MySQLdb.connect(user=user, passwd=passwd, db=dbname, host=host)
df2 = pd.read_sql("SELECT longitude, latitude FROM business WHERE state = 'NV'", con)

data loaded

And here is simple scatter plot of the data set.

In [14]:
plt.figure(figsize=(16, 12), dpi=100)
df_scatter = plt.scatter(df2['longitude'], df2['latitude'], c='k', alpha=.5, s=5)
plt.title('Full data set')
plt.xlabel('Longitude')
plt.ylabel('Latitude')
plt.xlim(-114.8,-115.5)
plt.ylim(35.8,36.4)
plt.show()

# K-means¶

For this I'l; use Python and scipy’s k-means clustering algorithm to reduce the full data set of GPS coordinates to a smaller but (approximately) spatially representative set of 100 points.

In [6]:
import pandas as pd, numpy as np, matplotlib.pyplot as plt
from scipy.cluster.vq import kmeans, kmeans2, whiten

I'll convert the latitude and longitude columns into a two-dimensional numpy array, called coordinates. The k-means algorithm groups N observations into k clusters. I’ll set k=100. I’ll also tell the routine to perform 50 iterations by setting i=50.

I will normalize the coordinate data with the whiten function. According to the scipy documentation: “Each feature is divided by its standard deviation across all observations to give it unit variance.” The whitened data yields results that are more spatially representative of the original, full data set.

In [13]:
coordinates = df2.as_matrix(columns=['longitude', 'latitude'])
N = len(coordinates)
k = 100
i = 50
w = whiten(coordinates)

I’m going to use the kmeans2 function because it returns both an array of cluster centroids as well as an array of centroid IDs that each observation in the data set is closest to. Then I plot the cluster centroids against the original data set:

In [14]:
cluster_centroids, closest_centroids = kmeans2(w, k, iter=i)

/Users/Lexa/anaconda/lib/python2.7/site-packages/scipy/cluster/vq.py:600: UserWarning: One of the clusters is empty. Re-run kmean with a different initialization.
warnings.warn("One of the clusters is empty. "

Yeah, and here is a warning that one of the clusters is empty. Let us see what is going on.

In [15]:
plt.figure(figsize=(15, 12), dpi=100)
plt.scatter(cluster_centroids[:,0], cluster_centroids[:,1], c='r', s=100)
plt.scatter(w[:,0], w[:,1], c='k', alpha=.3, s=10)
plt.show()

It is not hard to locate the empty cluster. Let us try to solve this glitch.

The kmeans2 function uses a random distribution of points from the entire coordinate space as its initial condition. Since the data is not evenly distributed, kmeans2 converges to this nearly random-looking result.
I need to tell the kmeans2 function to initialize with a random distribution of actual points from the data set:

In [9]:
cluster_centroids, closest_centroids = kmeans2(w, k, iter=i, minit='points')
plt.figure(figsize=(15, 12), dpi=100)
plt.scatter(cluster_centroids[:,0], cluster_centroids[:,1], c='r', alpha=.7, s=150)
plt.scatter(w[:,0], w[:,1], c='k', alpha=.3, s=10)
plt.show()

Better, but not great. In the preceding plot you see the 100 cluster centroids, in red, generally approximating the spatial distribution of the normalized full data set, in gray/black (due to us setting the opacity alpha value to 0.3, each of these points’ grayscale darkness is dependent on how many points are stacked on top of each other at that spot).

Now, let’s check to see if the numbers add up:

In [20]:
print 'k =', k
print 'N =', N
print len(cluster_centroids)
print len(closest_centroids)
print len(np.unique(closest_centroids))

k = 100
N = 16485
100
16485
100

The number of closest_centroids should be the same as N. The number of cluster_centroids and the number of unique elements in closest_centroids should be the same as k. Everything looks correct, algorithmically.
I’ll add a new column to the original full data set, called closest_centroid. For each set of coordinates in the original data set, I’ll add the closest centroid ID from the kmeans2 clustering to this new column. Then, I’ll reduce the data set so there is only one row for each value of closest_centroid.

In [16]:
rs = pd.DataFrame(df2)
rs['closest_centroid'] = closest_centroids
rs.drop_duplicates(subset=['closest_centroid'], take_last=False, inplace=True)

This dataframe has 100 rows. Each row contains a single data point (from the original data set) that belongs to each of the 100 clusters that were formed. Let’s plot the final (not whitened, like I did earlier) reduced set of coordinate points vs the original full set:

In [11]:
plt.figure(figsize=(15, 12), dpi=100)
rs_scatter = plt.scatter(rs['longitude'], rs['latitude'], c='r', alpha=.7, s=150)
df_scatter = plt.scatter(df2['longitude'], df2['latitude'], c='k', alpha=.3, s=5)
plt.title('Full data set vs k-means reduced set')
plt.legend((df_scatter, rs_scatter), ('Full set', 'Reduced set'), loc='upper left')
plt.xlabel('Longitude')
plt.ylabel('Latitude')
plt.show()

On one hand, the reduced data set does more-or-less approximate the spatial distribution of the full data set. On the other hand, there are gaps.
Even worse, many locations from the full data set are not part of any cluster, and thus are lost from the reduced data set. Why?
The answer is that k-means is not an ideal algorithm for lat-long spatial data. It minimizes variance, not geodetic distance – there is substantial distortion at latitudes far from the equator, like those of this data set. The algorithm still works, but its results are not great and there isn’t much that can be done to improve them. Increasing the number of clusters still leaves patchy gaps throughout the reduced data set.

# Better spatial data clustering with DBSCAN¶

This time using an algorithm that works better with arbitrary distances: scikit-learn’s implementation of the DBSCAN algorithm. DBSCAN clusters a spatial data set based on two parameters: a physical distance from each point, and a minimum cluster size. This works much better for spatial lat-long data.

In [18]:
from sklearn.cluster import DBSCAN
from geopy.distance import great_circle
db = DBSCAN(eps=.006, min_samples=2).fit(coordinates)
labels = db.labels_
num_clusters = len(set(labels)) - (1 if -1 in labels else 0)
clusters = pd.Series([coordinates[labels == i] for i in xrange(num_clusters)])
print('Number of clusters: %d' % num_clusters)

Number of clusters: 155

Ok, now I’ve got 155 clusters. Unlike k-means, DBSCAN doesn’t require you to specify the number of clusters in advance – it determines them automatically.
For plot, it would be more spatially-representative to take the point nearest the cluster’s centroid. This simple function returns the coordinates of the centroid of an array of points:

In [19]:
def getCentroid(points):
n = points.shape[0]
sum_lon = np.sum(points[:, 1])
sum_lat = np.sum(points[:, 0])
return (sum_lon/n, sum_lat/n)

I need a function that takes a set of points and returns the point in it that is nearest to some other reference point. In this case, that other reference point will be the centroid I calculated in the previous function.
First I calculate the great circle distance between the reference point and each point in the set, using geopy’s great circle function. Then I return the coordinates of the point that was the smallest distance from the reference:

In [20]:
def getNearestPoint(set_of_points, point_of_reference):
closest_point = None
closest_dist = None
for point in set_of_points:
point = (point[1], point[0])
dist = great_circle(point_of_reference, point).meters
if (closest_dist is None) or (dist < closest_dist):
closest_point = point
closest_dist = dist
return closest_point

First create two lists to hold the latitudes and longitudes. Then iterate through the series of clusters. If there are only one or two points in the cluster, I can just take the first point – it doesn’t make sense to calculate the centroid in these cases.

If there are three or more points in the cluster, I’ll find the point in the cluster that is closest to its centroid using the two functions I just defined. Then I add each point’s coordinates to the latitude and longitude lists. Once I’ve cycled through every cluster, save the final reduced set of points to a new pandas dataframe:

In [21]:
lon = []
lat = []
for i, cluster in clusters.iteritems():
if len(cluster) < 3:
representative_point = (cluster[0][1], cluster[0][0])
else:
representative_point = getNearestPoint(cluster, getCentroid(cluster))
lon.append(representative_point[0])
lat.append(representative_point[1])
rsD = pd.DataFrame({'lon':lon, 'lat':lat})

I’ll plot the final reduced set of data points against the original full set.
First let us check the coordinates of the clusters.

In [26]:
rsD.head(5)

Out[26]:
lat lon
0 -115.182005 36.118819
1 -115.150017 36.021178
2 -115.112145 36.008311
3 -115.176493 36.239052
4 -115.244809 36.062318
In [28]:
plt.figure(figsize=(15, 12), dpi=100)
rs_scatter = plt.scatter(rsD['lat'], rsD['lon'], c='g', alpha=.4, s=150)
df_scatter = plt.scatter(df2['longitude'], df2['latitude'], c='k', alpha=.5, s=5)
plt.title('Full data set vs DBSCAN reduced set')
plt.legend((df_scatter, rs_scatter), ('Full set', 'Reduced set'), loc='upper left')
plt.xlabel('Longitude')
plt.ylabel('Latitude')
plt.show()

And these look much better. Although, around the strip, it is hard to see clusters. However, there is one.

## And everything on a map!¶

And now, let us put those clusters on the map.
I will use Folium package for the example.

In [78]:
from IPython.display import HTML
import folium
def inline_map(map):
"""
Embeds the HTML source of the map directly into the IPython notebook.

This method will not work if the map depends on any files (json data). Also this uses
the HTML5 srcdoc attribute, which may not be supported in all browsers.
"""
map._build_map()
return HTML(''.format(srcdoc=map.HTML.replace('"', '"')))

In [81]:
meanlat=df2.latitude.mean()
meanlong=df2.longitude.mean()

map = folium.Map(width=600,height=600,location=[meanlat,meanlong], zoom_start=10)

for i in range(len(latRS)):
map.simple_marker([rsD.lon[i], rsD.lat[i]], popup='Add popup Recommended location.',marker_color='blue',marker_icon='ok-sign',clustered_marker=True)

inline_map(map)

Out[81]:

So as a conclusion, for the geographical clusters, DBSCAN is way to go.

In [ ]:


text:
1. May 12, 2013

### Emspak

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

$\int^{1}_{0}$$\int$$^{e^x}_{e^-x}$$\frac{lny}{y}$dydx

The attempt at a solution

So I am integrating ln(y)/y and I tried it by parts, first with u = ln(y), dv = 1/y, and therefore du = 1/y, and v = ln y

but if I use that I get

(ln(y))2-$\int$$\frac{lny}{y}$ again.

So I tried switching u and v around. I got: u = 1/y, dv = ln(y) dy, du = 1/y2 and v = 1/y.

On that basis I get $\frac{1}{y^2}$-$\int$$\frac{1}{y^3}$|$^{e^x}_{e^-x}$dx

and from there I get

$\int^{1}_{0}$ $\frac{1}{y^2}$-$\frac{1}{4y^4}$|$^{e^x}_{e^-x}$dx

But I have a sneaking suspicion I have done something horribly wrong. I see the integral I want as (ln(y))2/2. But that doesn't make sense to me.

So, I did something messed up. If someone could tell me where it is, that would be much appreciated.

2. May 12, 2013

### SammyS

Staff Emeritus
Try substitution.

Let u = ln(y).

3. May 12, 2013

### Emspak

I did that the first time, and showed what I got. u=ln y so du = 1/y, correct? Or not? If that's so and I et dv = 1/y then v has to be ln y, no? "try substitution" is what I just did.

4. May 12, 2013

### SammyS

Staff Emeritus
No.

You tried integration by parts.

Do integration by substitution, with the substitution u = ln(y). No v is needed.

5. May 12, 2013

### Emspak

ah, i see (I was thinking u-substitutions in integration by parts) -- NOW It is a lot clearer. Thanks much.


text:
# Ming has 15 quarters, 30 dimes, and 48 money so nickels. He wants to group his so that each group has the same number of each coin. What is the greatest number of groups he can make?

Jul 11, 2018

3 groups of of 31 coins 5 quarters, 10 dimes and 16 nickels in each group.

#### Explanation:

The greatest common factor (GCF) for the values, 15, 30 and 48 is the number 3.

That means the coins can be divided equally into three groups.

$\frac{15}{3} = 5$ quarters

$\frac{30}{3} = 10$ dimes

$\frac{48}{3} = 16$ nickels

$5 + 10 + 16 = 31$ coins


text:
# strange error in \end{document}

I am using ACM template to write a paper for a conference. This is my main document structure:

.....
......
\section{Introduction}
\label{sec:Introduction}
\input{introduction}

\section{Related Work}
\label{sec:rWork}
\input{relatedWork}
...
\section{Concluding Remarks}
\label{sec:conclusion}
\input{conclusion}

\bibliographystyle{abbrv}
\bibliography{sigproc}
%\balancecolumns

\end{document}

Somehow every time I create pdf (PDF Textify) using WinEdt. I get strange error as shown in the following image.

pdfTeX warning (ext4): destination with the same identifier (name{page.}) has
been already used, duplicate ignored
<to be read again>
\relax
l. 177 \end{document}
[2

This is what my .bib entries are

@article{bisbal1999legacy,
title={Legacy information systems: Issues and directions},
author={Bisbal, Jes{\'u}s and Lawless, Deirdre and Wu, Bing and Grimson, Jane},
journal={IEEE Soft.},
volume={16},
number={5},
pages={103--111},
year={1999},
publisher={IEEE}
}

@article{veerman2004revitalizing,
title={Revitalizing modifiability of legacy assets},
author={Veerman, Niels},
journal={J. Softw. Maint. Evol.: Res. Prac.},
volume={16},
number={4-5},
pages={219--254},
year={2004},
publisher={Wiley Online Library}
}

@article{arranga1997cobol,
title={Cobol: Perception and reality},
author={Arranga, Edmund C and Coyle, Frank P},
journal={IEEE Comp.},
volume={30},
number={3},
pages={126--128},
year={1997},
publisher={IEEE}
}

@MISC{Toby2008burden,
author =       {Sucharov, Toby and Rice, Philip},
title =        {The Burden of Legacy},
howpublished = {Online},
year =         {2008},
source =       {http://www.ncc.co.uk/article/?articleid=15665}
}

@book{glaser2009discovery,
title={The discovery of grounded theory: Strategies for qualitative research},
author={Glaser, Barney G and Strauss, Anselm L},
year={1967},
publisher={Aldine Transaction},
ADDRESS = {Chicago, Illionios}
}

title={Using grounded theory to study the experience of software development},
author={Adolph, Steve and Hall, Wendy and Kruchten, Philippe},
journal={Emp. Soft. Eng.},
volume={16},
number={4},
pages={487--513},
year={2011},
publisher={Springer}
}

• Welcome to TeX.SX! Please add a minimal working example (MWE) that illustrates your problem. It will be much easier for us to reproduce your situation and find out what the issue is when we see compilable code, starting with \documentclass{...} and ending with \end{document}. As to the error message, check this and this – henrique Aug 18 '13 at 14:27
• It is not an error, but just a hyperref warning. See tex.stackexchange.com/q/124520/27635. You're using a very very old WinEdt version, and probably it is reported as an error. I suggest you to upgrade to the latest version (8.0) – karlkoeller Aug 18 '13 at 14:38

Probably you have several pages with \renewcommand*{\thepage}{}. Package hyperref adds page anchors at each page. They are used by index commands, thus \thepage is used for naming the anchors.

If you do not have an index, then you can add the following hyperref option:

pageanchor=false

Otherwise, page anchors can be disabled for some pages only, e.g.:

\hypersetup{pageanchor=false}
\renewcommand*{\thepage}{}
...

\pagenumbering{arabic}
\hypersetup{pageanchor=true}