Technology Musings

October 24, 2017

Snippets / Calculating an Even Baseline


So, I've been working in Xamarin for cross-platform phone development.  I was using their AbsoluteLayout class to try to get a bar graph working using relative sizes (AbsoluteLayout with relative sizes gives everything an absolute position from 0, top or left of the screen, to 1, the bottom or right of the screen).  The goal was to have the bottom of the bar graph halfway down the screen (to allow room for negative values to point down).  The goal was to look like this:


The problem, though is getting the boxes to align at the center line.  Now, part of the problem may be my inexperience with the platform.  There may be a really easy way to do this.  However, here is what I was running into:

I thought that Xamarin worked by aligning the your specified position to the middle of the box.  Therefore, I started by calculated the box height, and then calculating the origin of the box by setting it above the 0.5 (i.e., 50%) line by half of the box height.  Therefore, when it extended, it should extend down to the 0.5 line.

The problem, though, is that Xamarin doesn't  align to the middle of the box.  It varies the alignment by where in the positioning it is positioned.  So, if you position a box at 0.2, not only is it positioned at 0.2, but the anchoring point is also 0.2 down the box!

Therefore, I needed a way to calculate where I needed to actually position the box to get it to rest on the 0.5 line.

The way I calculated it is this - let's call the y-position that you want the box to rest on as P.  Let's call the y-position which you are going to set your box to as A (for the anchoring position).  Let's call the height of the box itself H.

What we want to happen is for the bottom of the box to hit P.

Now, we can split up the measurements from the top of the box to P into two lengths - the length from the top to the anchoring point (which is actually just A) and the length from the anchoring point to the bottom of the box.

Since the anchoring point sits A down on the box, the length of the top half of the box is given by A*H.  However, this is not the distance we want.  We want the distance down the rest of the box.  That will be (1 - A)*H.

So, our two distances are: A and (1 - A)*H.  They should add together to be P.  Therefore, our equation is:

A + (1 - A)*H = P

However, we want to solve this for A - we already know our box height (H), and we know what position we want this to extend to (P).  Therefore, we have to solve for A:

A + (1 - A)*H = P
A + H - A*H = P
A - A*H = P - H
A*(1 - H) = P - H
A = (P - H) / (1 - H)

And this gives a perfect calculation for A.  In my case, since P is 0.5, I just do A = (0.5 - H) / (1 - H) and it comes out perfectly.

It should be fairly simple to come up with similar calculations for top alignment and center alignment.  Additionally, these work with right-to-left alignments as well.

February 11, 2016

General / Finding Derivatives without Limits or Differentials


I always enjoy finding new ways of doing things.  I was reading Henle/Kleinberg's Infinitesimal Calculus, which is a great book.  It has, throughout the book, a main text and a side-text.  The side-text is infinitely more interesting than the main text.

Anyway, on pages 65-66, it gives a method for finding the slope of y = x^2 at a specific point, let's say its at (2,4).  The way you do it is kind of strange, but it seems to work.  You start by defining a line to intercept the equation:

y = mx + b

Now, we don't know what m and b are, but we *do* know what x and y are.  Therefore,

4 = 2m + b

We don't really care about b in the long run, so we want to find an equation for b to remove it from the mix:

b = 4 - 2m

Now our equation for the line becomes:

y = mx + 4 - 2m

Interestingly, we actually have a value for y - y = x^2, so this becomes:

x^2 = mx + 4 -2m

We can rearrange this into quadratic form with:

x^2 - mx + (2m - 4) = 0

Using the quadratic formula, we can do:

x = (m +- sqrt(m^2 - 4(2m - 4))) / 2

x = (m +- sqrt(m^2 - 8m + 16)) / 2

x = (m +- sqrt((m - 4)^2)) / 2

x = (m +- (m - 4)) / 2

Now, we want to take the "+" side of the "+-" since we are solving for "m".  Otherwise, we will lose "m".  Therefore, this becomes:

x = (2m - 4) / 2

x = m - 2

So, at the point we are looking at, what is the value of x?  At (2, 4), x is 2.  Therefore

2 = m - 2

m = 4

Therefore, the slope at (2, 4) is 4.  

Interestingly, we can also generalize this to get the full derivative for y = x^2.  To do this, we will introduce the variables p and q to be the specific x and y values at a specific point.  Therefore, the slope at (p, q) for y = x^2 is:

y = mx + b, therefore q = pm + b, therefore b = q - pm

q = p^2, therefore b = p^2 - pm

y = mx + p^2 - pm

y = x^2, therefore:

x^2 = mx + p^2 - pm

x^2 - mx + (pm - p^2) = 0

x = (m +- sqrt(m^2 - 4(pm - p^2)))/2

x = (m +- sqrt(m^2 - 4pm + 4p^2))/2

x = (m +- sqrt((m - 2p)^2))/2

x = (m +- (m - 2p))/2

x = (2m - 2p) / 2

x = m - p

m = x + p

Since p = x, this gives

m = x + x = 2x, which is the derivative of x^2

I don't know how many classes of functions you can do this successfully for, but I imagine it should cover quadratics at least.


January 13, 2016

General / Ratios of Infinity


A lot of people do a lot of thinking about whether or not infinities exist, and wonder how real numbers can add up to infinity.

Sometimes I wonder if that sort of thought might be the inverse of reality - perhaps the infinities are the foundational realities, and all finite numbers are merely infinities in ratio.

January 01, 2016

General / A Meaning for d^2y/d^2x?


I have recently been randomly curious about random calculus-y things.  Anyway, the notion for the second derivative of a function is d^2y/dx^2.  This is the second differential of y divided by the first differential of x squared.

However, there is, technically, also a d^2x, though it doesn't get much attention.  And, d^2y and d^2x can be put into ratio with each other, but I don't really know what it means.  But it is an interesting operation nonetheless.

So, the derivative of an equation is dy/dx and the derivative of its inverse is dx/dy.

The second derivative of an equation is d^2y/dx^2 and the second derivative of its inverse is d^2x/dy^2.

Therefore, to get the d^2y/d^2x you just do:

(2nd Derivative of y wrt x / 2nd derivative of x wrt y) * (First derivative of the inverse)^2


(d^2y/dx^2) / (d^2x/dy^2) * (dx/dy)^2

I will post an example later when I have a good one.

December 27, 2015

General / The derivative of u^v


I don't know why this isn't listed as a standard rule.  The differential of the exponent function, u^v (u raised to the v power), is pretty basic, and you can use it to formulate the other exponent rules for differentiation. However, for some reason it seems to be left off of most rules for differentiation.

The basic rule is this:

d(u^v) = v*u^(v-1)*du + u^v*ln(u)*dv

This can be clearly seen to be a more general form of u^n, because if n is a constant, this becomes:

d(u^n) = n*u^(n - 1)*du + u^v*ln(u)*0

And that zero drops it to the rule we all know and love:

d(u^n) = n*u^(n - 1)*du

This can be derived as follows:

z = u^v
ln(z) = ln(u^v)

Using log rules, we get:

ln(z) = v*ln(u)

Now take the differential of both sides:

d(ln(z) = d(v*ln(u))

Differentiating both sides, we get:

dz/z = v*du/u + ln(u) * dv

Then we multiply by z:

dz = z*v*du/u + z * ln(u) * dv

Substituting in z = u^v:

dz = u^v * v * du / u + u^v * ln(u) * dv

Now, u^v/u simplifies to u^(v-1), giving us:

dz = v*u^(v-1)*du + u^v*ln(u)*dv

Since z = u^v, dz = d(u^v), so

d(u^v) = v*u^(v-1)*du + u^v*ln(u)*dv

December 22, 2015

Platforms / Rails and Updated Postgres not Performing Migrations


Apparently, there was a change to something in postgres 9.5, or maybe just my install of it.  If you upgrade your Postgres database version and suddenly your rails app stops doing migrations, add the following line to config/database.yml database config to get it working again:


  schema_search_path: public

The error message i was getting was:


ActiveRecord::StatementInvalid: PG::DuplicateTable: ERROR:  relation "schema_migrations" already exists

Basically, sometimes Rails was finding the schema_migrations table, and sometimes it wasn't, but setting the schema_search_path fixed it.

I'm still not sure where exactly the problem came from, but setting the setting fixed it.

December 13, 2015

General / Explanation of Liebniz Notation of the Second Derivative


The notation for the second (and higher) derivatives in Liebniz notation has always troubled me.  The second derivative is usually notated as d^y/dx^2.  And I always wondered, why is the the 2 in relation to the d on the top, but in relation to the whole term on the bottom?

This puzzled me for a while, and I looked through at least 10 Calculus textbooks to find the answer, all to no avail.  Finally, I put pen to paper and figured it out.  The answer was straightforward, and I am sure that whoever invented the 2nd derivative Liebniz notation knew why they did it that way, it's just that every Calculus book since then seems to have forgotton.

Anyway, the usual notation for the derivative operation id d/dx.  I eventually came to realize that this is not a single operation, but TWO operations in one.  The first is to take the differential (not derivative) of the equation.  For that, I mean, let's say you have the equation y = 2x.  The differential of that equation is dy = 2xdx.  Doing a differential instead of a derivative is powerful, because it allows a more natural approach to both implicit differentiation and multivariable differentiation.

Now, the derivative is simply the differential divided by dx.  So, if our differential is dy = 2xdx, then when we divide by dx we get our normal notation dy/dx = 2x.

Now, let's leave off dividing by dx, and just look at the differential.  What happens if we take a second differential?

In this case, we have to treat dy as a separate variable.  So, if the differential of y is dy, what is the differential of dy?  This is where the d^2y comes in.  The 2 is by the "d" not because it is "d squared", but it is "d applied twice".  So, if we take our initial differential dy = 2xdx and take another differential, what do we have?

By the product rule, we get:

d^2y = 2((x)(d^2x) + (dx)(dx))

Now, since we want the second derivative, we divide both sides by dx twice (once for the first derivative, and once for the second), which is the same as dividing by dx^2.

That gives us:

d^2y / dx^2 = 2((x)(d^2x) + (dx)(dx)) / dx^2

The (dx)(dx) simplifies to dx^2, and we can separate this out at the plus sign to give us:

d^2y / dx^2 = 2(x)(d^2x)/dx^2 + 2(dx^2)/dx^2

This further simplifies to:

d^2y / dx^2 = 2(x)(d^2x)/dx^2 + 2

Now, we know that the second derivative is 2, so that means that 2(x)(d^2x)/dx^2 must equal zero.  But why?

Well, let's rewrite this term a little:

2x * (d^2x)/(dx^2)

d^2x/dx^2 is simply the second derivative of x with respect to itself!

To understand why this must be zero, first imagine the first derivative of x with respect to itself: dx/dx

This must always be 1!  Another way of stating this is that "for every change in x, the corresponding change in x is equivalent", which seems obvious.

Now, 1 is a constant, so the second derivative must be zero!  This zeroes out the whole term, leaving:

d^2y/dx^2 = 2

Thus, we have the notation of the derivative from first principles.

Another way to look at this is to use the quotient rule.

If we start with dy/dx, then take the differential, what do we get?

((dx)(d^2y) - (dy)(d^2x)) / dx^2

If we separate, we get:

(dx)(d^2y) / dx^2 - (dy)(d^2x)/dx^2

The first term simplifies to d^2y / dx.  The second term is our second derivative of x with respect to itself again, so it becomes (dy)*0, giving us:

d^2y/dx - (dy)*0

Which gives us just


Now, the second step of the derivative is to divide by dx.  Doing so gives us:


Which is our second derivative!

November 14, 2015

General / A General Solution for Line Integrals


Update: I found that this is a standard formula, but it is referred to as an "arc length" formula, not a line integral.  I knew it had to be somewhere, I was just searching the wrong terms.

I was playing around last night, and I came up with, what appears to be, a general solution for line integrals.  I have looked around and I have not found this equation anywhere, although I am certain in must be well-known somewhere.  In any case, I will give it here, because it is definitely not well-known on the Internet.

This is an equation for determining the line integral of y = f(x), where f(x) is a true function of x (i.e. it passes the vertical line test), and is differentiable on the portion of the line where you want to find the length of the line segment.  Most line integrals work by adding in an auxiliary parameter, t, to the equation.  This operates directly on the function itself.

The short version:

To get a line integral from x=a to x=b on y=f(x), we can use the following formula:

Evaluate this from x=a to x=b and you will have the length of the line.  Note, the software I was using gives this as a partial derivative on the inside.  I don't know enough to know if it is only limited to partials or not.

So, if you know the derivate, you can figure out (in theory) the line integral.  Whether or not this actually helps you find a solution to the integral is another story.

Now, how does one arrive at that?

Well, first, remember, that an integral is an infinite sum of infinitesimals.  A curve is simply an infinite sum of tiny, infinitesimal lines.  So, all we need to do is find the length of each line segment and add it up.

How big are our line segments?

Well, if y=f(x), then the line segment is going to be the line between the points x, y and x + dx, y + dy.  The distance formula states that the line length is sqrt( (x1 - x0)^2 + (y1 - y0)^2).  The difference between x + dx and x is simply dx, and the difference between y + dy and y is simply dy.  So, each of our infinitesimal line segments (dL) will be:

dL = sqrt(dx^2 + dy^2)

If we square both sides, we get the following:

dL^2 = dx^2 + dy^2

dL^2 - dx^2 = dy^2

Now we can divide both sides by dx^2 and get:

dL^2/dx^2 - 1 = dy^2/dx^2

dL^2/dx^2 = 1 + dy^2/dx^2

dL^2/dx^2 = 1 + (dy/dx)^2

Now we can square root both sides:

dL/dx = sqrt(1 + (dy/dx)^2)

dL = sqrt(1 + (dy/dx)^2) dx

And this is the final form of our equation.  I tested it out with a few simple integrals (a line and the top half of a circle) and it seems to work as expected.  And, as I said, if you have a line that loops back around, you still need to use the parameterized methods.  But this is an easier way to setup line integrals where f(x) is a true function of x.

May 28, 2015

Snippets / ActiveMerchant, Authorize.Net Root Certificates, and Linux SSL


This week, is starting its rollout of new server certificates, signed by a new certificate authority - Entrust.  If you are using ActiveMerchant, this could break your stuff!  It may or may not affect newer versions of ActiveMerchant - I don't know, I only have sites using old versions (like 1.18).  So, even if it breaks, the solution may be different depending on your version of ActiveMerchant.  Still, I'll try to keep the info detailed enough that you can find the solution yourself.

First, the problem.  This affects you if, when ActiveMerchant tries to connect to, you receive the error:

SSL_connect returned=1 errno=0 state=SSLv3 read server certificate B:
  certificate verify failed

Note that you may not get this error when trying to connect to via manual Ruby code, but only via ActiveMerchant.

The reason for this is ActiveMerchant maintains its own list of Root CAs!

Therefore, to fix, you have to, (a) make sure that the new Root CAs for are stored in your server's root CA list, and (b) make sure that ActiveMerchant is using your server's list rather than its own.  For good measure, you can also copy ActiveMerchant's CA list to yours.

So, first, to add the Entrust root CAs to your Linux server (assuming it is a fairly recent CentOS-based distro), do the following as root:

  1. Enable the trust infrastructure with the following command:update-ca-trust enable
  2. Go to /etc/pki/ca-trust/source/anchors
  3. Grab the .cer files from (just right-click on the links, copy the URLs, and download them to the current directory with wget)
  4. Go to wherever the gem files for your project are, and find the active_utils gem.  In the lib/certs directory, there is a file called or something like that.  Copy that to /etc/pki/ca-trust/source/anchors
  5. Now run the following command to update your CA list:

Now you have all of the certs installed.  Now we need to tell ActiveMerchant to use your own certs rather than its internal list.  However, again, this is in the active_utils gem, not the activemerchant gem.  Within the gem, find the file called lib/active_utils/common/connection.rb.  Now look for a function called configure_ssl.  Copy that function to your clipboard.

Now create a file in your own project called config/initializers/active_merchant_pem_fix.rb (the name doesn't matter as long as it is in config/initializers).  In this file you need to have:

module ActiveMerchant
  class Connection
    # Paste the configure_ssl function you copied from lib/active_utils/common/connection.rb here.

Now, there will be a line that says something like this:

        http.ca_file = File.dirname(__FILE__) + '/../../certs/cacert.pem'

This is the offending line!!!  Comment that sucker out!!!

For my version of ActiveMerchant, this is what my file looks like:

   def configure_ssl(http)
      return unless endpoint.scheme == "https"

      http.use_ssl = true

      if verify_peer
        http.verify_mode = OpenSSL::SSL::VERIFY_PEER
        # http.ca_file = File.dirname(__FILE__) + '/../../certs/cacert.pem'
        http.verify_mode = OpenSSL::SSL::VERIFY_NONE

Now that you have the updated root CA on your server, and ActiveMerchant is using your server's list rather than your own, you are all set!

June 11, 2014

General / Why You Shouldn't Put Validations on Your Model


This post is about model validators in Ruby on Rails, although it probably applies to validators in most similar MVC systems that allow declarative model validation.

One of the biggest disasters, in my opinion, has been the trend for developers to put more and more validation in their models.  Putting validation in the model always *sounds* like a good idea.  "We want our objects to look like X, therefore, we will stick our validations in the place where we define the object."

However, this leads to many, many problems.

There are two problems, and the most general way of stating them is this:

  1. life never works the way you expect it, but validating on the model assumes that it will.  
  2. while it is cleaner to validate on the model for a given configuration, when that validation needs to change, you don't always have enough information to understand the purpose of the validation in the first place (i.e., if a new developer is making the change)

Example of both -

Let's say that we require all users to enter their first name, last name, birthdate, and email.  Okay, so we'll put validators on the model to make sure that this happens.  

Let's say that six months go by, and your company signed an agreement with XYZ corporation to load in new users.  However, XYZ corp didn't require that their users enter in their birthdate.  However, you don't know that, and the bit of data you looked at, it looks like they are all there.  So, you do a load through the database, and everything loads in cleanly, and you think that everything is good.  

In fact, *Everyone* can view their records.

However, the next week, people start getting errors when they try to update their profile.  Why?  Because the people who don't have a birthdate entered are getting validation errors when they try to update something else.

So here is problem #1 - if an existing record doesn't match the validation, nothing can be saved, and this causes headaches because you are getting errors in places where they were unexpected.  This doesn't just happen with dataloads.  It happens every time you add a validator as well.  So, every time you change policy, even if it seems benign, you have the potential of messing up your entire program, even if all of your tests succeed (because your tests will never have had the bad data in them, since it doesn't have the old version of the code!).

So, we go back and recode, and remove the validator from the model.  Great, but let's say this is a new programmer.  Now the programmer has to figure out where to put the validation in.  Presumably, we still want new users to enter in their birthdate, even if we allow non-birthdate users that we load from outside.  However, now, because we aren't validating on the model, we have to move it somewhere else.  But where?  Since the validator just floated out there declaratively, there is no direct link between the validator and *where* it was to be enforced.  Now the new programmer has to go through and recode and retest all of the situations to figure out the proper place(s) for the validator.  This is especially problematic if you have more than one path for entrance on your system.  It's not so bad if it is a system you wrote yourself, but when having to maintain someone else's code, this is a problem.

Therefore, the policy I follow is this - only use model validation when a failure of the validator would lead to a loss of data *integrity*, not just cleanliness.  The phone number field should be validated somewhere else.  You should validate a uniqueness constraint only if uniqueness is actually required for the successful running of the code.  Otherwise, validate in the controller.  Now, you can make this easier by putting the code to validate in the model, but just don't hook it in to the global validation sequence.  So, go ahead and define a method called "is_proper_record?" or something, but don't hook it up to the validation system - call it directly from your controller when you need it.

NOTE - if I don't get around to blogging about it, there are similar problems with State Machine, and a lot of other declarative-enforcement procedures in the model.  Basically, they are elegant ways of coding, but *terrible* when you need to modify them, especially if the person modifying the code wasn't the one who built it.  Untangling the interaction between declarative and manual processes leads to both delays (as the new developer tries to figure out the code) and bugs (when the new developer doesn't figure out the code perfectly the first time).  When your processes are coded *as processes* and not declarations, it is more visible to future programmers, and therefore more maintainable.