Category Archives: algebra

Area models for completing the square, dynamic approach.

An area model, or a dot array model (same thing really) is one way of illustrating the algebraic completion of a square.

I have used dots as they are easier to create.

The quadratic is viewed initially as the “standard form”, and then rebuilt dynamically line by line into the “square plus a bit over” form, as shown in the following sequence:

area model 1

The odd valued coefficient of x in the original expression can appear as a row and a column of half-dots, or half squares in the area model form.

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Matrix multiplication without tears!

My dad figured this out years ago.


The string method work for all matrices, and it is at least ten times quicker to “do” than to “write about”.

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Negative numbers, the minus sign, abstract algebra.

I was pondering the reality of negative numbers and after figuring out that a sequence of dots on a line can be extended in each of the two directions, and then arbitrarily selecting one dot as “the zero”. The line can be further labelled as 1, 2, 3, … to one side and -1, -2, -3, … on the other side.
(better to label the 1, 2, 3, … as +1, +2, +3, … and consider the lot as “signed numbers”)

Soon proceeding towards arithmetic I concluded that 7-3 is 4, and also 8-4 is 4, and therefore 13-9 is 4, and then 3-7 is -4, and -2-2 is -4. It was then observed that if a-b=c then a-y-(b-y) is also equal to c, regardless of the signs of the specific numbers involved.
This of course is stunningly obvious when looking at the signed difference of the first and the second number as an extended number line diagram.

The outcome of all this was an arithmetic for 0, 1, 2 modulo 3, and  the signed difference x-y is a binary operation diff(x,y) with table:

…x  … 0     1     2
0         0     1     2
1         2      0    1
2         1      2    0

Example: 1-2 is -1, which is 2 modulo 3

So a non abelian, non associative algebra with a not quite identity satisfies the conditions, where A=1, B=2 and C=0
There are three objects and an operation called “doesn’t have a name”.
Two are similar, and the third is a bit different
They are paired to yield a single object as follows:

AA = BB = CC = C
AB = BC = CA = B
AC = CB = BA = A

Notice that BC and CB are different, so non-abelian.
Worse is that (AC)A = C and A(CA) = B are different, so non-associative.
And consequently A and B and C are different.

Interestingly, and maybe separately, the minus sign behaves very differently from the plus sign:

a-(-b) is a+b, but there is no way of writing a-b using only addition.

This means that all expressions can be written with “minus” alone.


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Minimal Abstract Algebra

This is a challenge !!!!!

There are three objects, A, B, and C, and an operation called “doesn’t have a name”.
Two are similar, and the third is a bit different
They are paired to yield a single object as follows:

AA = BB = CC = C
AB = BC = CA = B
AC = CB = BA = A

Notice that BC and CB are different, so non-abelian.
Worse is that (AC)A = C and A(CA) = B are different, so non-associative.
And consequently A and B and C are different.

Your job is to identify (model, in current jargon) the objects and the operation.”

The next post will be “the solution”.

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Adding fractions – phew!

Who needs LCM ?

First, three views of LCM with no comments :

1: Change them to equivalent fractions that will have equal
denominators. As the common denominator, choose the LCM of
the original denominators. Then the larger the numerator, the
larger the fraction.

2: Jun 26, 2011 – If b and d were same it was easy to find LCM
since if denominators are same, we just need to find LCM of
numerators, hence LCM of (a/b) and (c/b) would be LCM(a,c)/b.
So we have to first make denominators of both the fractions same.
Multiply numerator and denominator of first fraction by LCM

3: The GCF and LCM are the underlying concepts for finding
equivalent fractions and adding and subtracting fractions, which
students will do later.


Now we can do fraction addition without LCM. It just needs the use of the distributive law, and the result shows the way in which the divisors combine.


And now using 3/4


But the best one is via multiplication ……


Now for multiplication and division.





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Algebra, algebra, who’s for algebra

x2  – 3x – 4 = 0

x2 – 3x + 2 = 6

(x – 3/2)2 = 6

(x – 3/2) = √6   or   – (x – 3/2) = √6

x = 3/2 + √6   or   x = 3/2 – √6

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A minus times a minus is a plus -Are you sure you know why?

What exactly are negative numbers?
A reference , from Wikipedia:
In A.D. 1759, Francis Maseres, an English mathematician, wrote that negative numbers “darken the very whole doctrines of the equations and make dark of the things which are in their nature excessively obvious and simple”.
He came to the conclusion that negative numbers were nonsensical.[25]

A minus times a minus is a plus
Two minuses make a plus
Dividing by a negative, especially a negative fraction !!!!
(10 – 2) x (7 – 3) = 10 x 7 – 2 x 7 + 10 x -3 + 2 x 3, really? How do we know?
Or we use “the area model”, or some hand waving with the number line.

It’s time for some clear thinking about this stuff.

Mathematically speaking, the only place that requires troublesome calculations with negative numbers is in algebra, either in evaluation or in rearrangement, but what about the real world ?
Where in the real world does one encounter negative x negative ?
I found two situations, in electricity and in mechanics:

1: “volts x amps = watts”, as it it popularly remembered really means “voltage drop x current flowing = power”
It is sensible to choose a measurement system (scale) for each of these so that a current flowing from a higher to a lower potential point is treated as positive, as is the voltage drop.

Part of simple circuit A———–[resistors etc in here]————–B
Choosing point A, at potential a, as the reference, and point B, at potential b, as the “other” point, then the potential drop from A to B is a – b
If b<a then a current flows from A to B, and its value is positive, just as a – b is positive
If b>a then a current flows from B to A, and its value is negative, just as a – b is negative

In each case the formula for power, voltage drop x current flowing = power, must yield an unsigned number, as negative power is a nonsense. Power is an “amount”.
So when dealing with reality minus times minus is plus (in this case nosign at all).

The mechanics example is about the formula “force times distance = work done”
You can fill in the details.

Now let’s do multiplication on the number line, or to be more precise, two number lines:
Draw two number lines, different directions, starting together at the zero. The scales do not have to be the same.
To multiply 2 by three (3 times 2):
1: Draw a line from the 1 on line A to the 2 on line B
2: Draw a line from the 3 on line A parallel to the first line.
3: It meets line B at the point 6
4: Done: 3 times 2 is 6
numberlines mult pospos
Number line A holds the multipliers, number line B holds the numbers being multiplied.

To multiply a negative number by a positive number we need a pair of signed number lines, crossing at their zero points.

So to multiply -2 by 3 (3 times -2) we do the same as above, but the number being multiplied is now -2, so 1 on line A is joined to -2 on line B

numberlines mult posneg
The diagram below is for -2 times 3. Wow, it ends in the same place.
numberlines mult posneg

Finally, and you can see where this is going, we do -2 times -3.

Join the 1 on line A to the -3 on line B, and then the parallel to this line passing through the -2 on line A:

numberlines mult negneg

and as hoped for, this line passes through the point 6 on the number line B.

Does this “prove” the general case? Only in the proverbial sense. The reason is that we do not have a proper definition of signed numbers. (There is one).

Incidentally, the numbering on the scales above is very poor. The positive numbers are NOT NOT NOT the same things as the unsigned numbers 1, 1.986, 234.5 etc

Each of them should have a + in front, but mathematicians are Lazy. More on this another day.

Problem for you: Show that (a-b)(c-d) = ac – bc – ad + bd without using anything to do with “negative numbers”


Reference direction for current
Since the current in a wire or component can flow in either direction, when a variable I is defined to represent
that current, the direction representing positive current must be specified, usually by an arrow on the circuit
schematic diagram. This is called the reference direction of current I. If the current flows in the opposite
direction, the variable I has a negative value.

Yahoo Answers: Reference direction for potential difference
Best Answer: Potential difference can be negative. It depends on which direction you measure the voltage – e.g.
which way round you connect a voltmeter. (if this is the best answer, I hate to think of what the worst answer is)


Filed under algebra, arithmetic, definitions, education, geometrical, math, meaning, negative numbers, Number systems, operations, subtraction, teaching, Uncategorized

Part 4: Tuning the feedback controller

Our first order process is described by the equation yn+1 = ayn + kxn , where yn is the process output now (at time n),  xn is the process input between time n and time n+1, and yn+1 is the process output at time n+1.
a is the coefficient determining how quickly the process settles after a change in the input, and k is related to the process steady state gain (ratio of settled output to constant input).

If we set the input to be a constant x and the output settled value to be a constant y, then

y = ay + kx, and solving for y/x we get y/x = k/(1-a), the actual steady state gain.

In what follows the k will represent the actual steady state gain, the old k divided by (1-a)

The fist two plots show the process alone, with input set to 6 at time zero. In the first the steady state gain is set to 1, and in the second it is set to 2


contpic 2

Now we look a t the process under “direct” control, where the input is determined only by the chosen setpoint value. The two equations are

yn+1 = ayn + kxn  and  xn = Adn  (direct control: h is zero)

To obtain a controlled system with overall steady state gain equal to 1 (settled output  equal to desired output) it is easy to see that  has to be equal to (1-a)/k



It is not so obvious how the choice of h affects the performance of the controlled system. To do this we observe that the complete system is described entirely by the process equation and the controller equation together, and we can eliminate the xn from the two equations to get yn+1 = ayn + k(Adn + h(yn – dn))

which rearranged is yn+1 = (a + kh)yn + k(A – h)dn 

Substituting   (1-a)/k for A, as found above, gives  yn+1 = (a + kh)yn + (1 – a – kh)dn 

which has the required steady state gain of 1.

This final equation has the SAME structure as the process equation,
with a + kh in place of a

So now we will see how the value of the “a” coefficient affects the dynamic response of the system.

If h > 0 the controlled system will respond slower than with h = 0, and if h < 0 it will respond faster:


Setpoint changes were made at time 20 and at time 40

Congratulations if you got this far. This introduction to computer controlled processes has been kept as simple as possible, while using just the minimum amount of really basic math. The difficulties are in the interpretation and meaning of the various equations, and this something which is studiously avoided in school math. Such a shame.

Now you can run the program yourself, and play with the coefficients. It is a webpage with javascript:

Aspects and theoretical stuff which follow this (not here !) include the  backward shift operator z and its use in forming the transfer function of the system, behaviour of systems with wave form inputs to assess frequency response, representation of systems in matrix form (state space), non-linear systems and limit cycles, optimal control, adaptive control, and more…..

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Part 3: Computer control of real dynamic systems

Let us return to the industrial steam engine driving a range of equipment, each piece in the charge of an operator. In order to keep the factory working properly it is necessary to keep the speed of the steam engine reasonably constant, at what is called the desired output, or the set point. A human controller can achieve this fairly well, by direct observation of the speed and experience of the effect of changing the steam flow rate, but automatically? Well, James Watt solved the problem with a mechanical device (details later). We will now see how to “do it with a computer”.

steam physical
Diagram of the physical system. There will be a valve on the steam line (not shown).

steam informational
The information flow diagram. The load is not measured, and may vary.

steam controlled
Now we have a controller in the picture, fed with two pieces of information, the current speed and the desired speed. The controller can be human, mechanical or computer based. It has a set of rules to figure out the appropriate steam flow rate.

steam controlled with equations

Now we have computer control. The output of the system is measured (sampled) at regular intervals, the computer calculates the required flow rate , sends the corresponding value to the valve actuating mechanism, which holds this value until the next sampling time, when the process is repeated. Using n=1, 2, 3, ….. for the times at which the output is sampled and the controller does its bit, we have at time n the speed measure yn, the desired speed dn and the difference between them (or error in the speed), and they are used to calculate the system input xn with the formula shown in the controller box.

The equation in the steam engine box is our fairly simple first order linear system model as described in the previous post. It is a good idea to have a model of the process dynamics for many reasons (!), one of which is that we can do experiments on the whole controlled system by simulation rather than on the real thing.

Looking at the controller function (formula), xn = Adn + h(yn – dn) , it is “obvious” that if the coefficient h is zero we have direct control (no feedback), and so the value of the coefficient A must be chosen accordingly (see next post).

It is not so obvious how the choice of h affects the performance of the controlled system. To do this we observe that the complete system is described entirely by the process equation and the controller equation together, and we can eliminate the xn from the two equations to get yn+1 = ayn + k(Adn + h(yn – dn))

which rearranged is yn+1 = (a + h)yn + k(A – h)dn 

and has the SAME structure as the process equation, with a + h in place of a

In the next post, on tuning our controller, we will see how the value of the “a” coefficient affects the dynamic response of the system.

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Time Series and Discrete Processes, or Calculus Takes a Back Seat


In the business and manufacturing world data keeps on coming in. What is it telling me? What is it telling me about the past? Can it tell me something believable about the future. These streams of data form the food for “time series analysis”, procedures for condensing
the data into manageable amounts.

Here is a plot of quarterly Apple I-phone sales. (ignore the obvious error, this is common)


I see several things, one is a pronounced and fairly regular variation according to the quarter, another is an apparent tendency for the sales to be increasing over time, and the third is what I will call the jiggling up and down, more formally called the unpredictable or random behaviour.
The quarterly variation can be removed from the data by “seasonal analysis”, a standard method which I am not going to deal wirh (look it up. Here’s a link:
The gradual increase, if real, can be found by fitting a straight line to the data using linear regression, and then subtracting the line value from the observed value for each data point. This gives the “residuals”, which are the unpredictable bits.

Some time series have no pattern, and can be viewed as entirely residuals.
We cannot presume that the residuals are just “random”, so we need a process of smoothing the data to a) reveal any short term drifting, and b) some help with prediction.
The simple way is that of “moving average”, the average of the last k residuals. Let’s take k = 5, why not !
Then symbolically, with the most recent five data items now as x(n), x(n-1),..,x(n-4), the moving average is

ma = (x(n)+x(n-1)+x(n-2)+x(n-3)+x(n-4))/5

If we write it as ma = x(n)/5 + x(n-1)/5 + x(n-2)/5 + x(n-3)/5 + x(n-4)/5 we can see that this is a weighted sum of the last five items, each with a weight of 1/5 (the weights add up to 1).
This is fine for smoothing the data, but not very good for finding out anything about future short term behaviour.
It was figured out around 60 years ago that if the weights got smaller as the data points became further back in time then things might be better.

Consider taking the first weight, that applied to x(n) as 1/5, or 0.2 so the next one back is 0.2 squared and so on. These won’t add up to 1 which is needed for an average, so we fix it.
0.2 + 0.2^2 + 0.2^3 + … is equal to 0.2/(1-0.2) or 0.2/0.8 so the initial weights all need to be divided by the 0.8, and we get the “exponentially weighted moving average” ewma.

ewma(n) =
x(n)*0.8+x(n-1)*0.8*0.2+x(n-2)*0.8*0.04+x(n-3)*0.8*0.008+x(n-)*0.8*0.0016 + …… where n is the time variable, in the form 1, 2, 3, 4, …

A quick look at this shows that ewma(n-1) is buried in the right hand side, and we can see that ewma(n) = 0.2*ewma(n-1) + 0.8*x(n),
which makes the progressive calculations very simple.
In words, the next value of the ewma is a weighted average of the previous value of the ewma and the new data value. (0.2 + 0.8 = 1)

The weighting we have just worked out is not very useful, as it pays too much attention to the new data, and the ewma will therefore be almost as jumpy as the data. Better values are in the range 0.2 to 0.05 as you can see in the following pictures, in which the k value is the weighting of the data value x, and the weighting of the ewma value is then (1-k):

So the general form is ewma(n) = (1-k)*ewma(n-1) + k*x(n)


With k=0.5 we do not get a very good idea of the general level of the sequence, as the ewma value is halfway between the data and the previous ewma value, so we try k=0.1


Much better, in spite of my having increased the random range from 4 to 10.

Going back to random=4, but giving the data an upward drift the ewma picks up the direction of the drift, but fails to provide a good estimate of current data value. This can be fixed by modifying the ewma formula. A similar situation arises when the data values rise at first and then start to fall (second picture)


Common sense says that to reduce the tracking error for a drifting sequence it is enough to increase the value of k. But that does not get rid of the error. We need a measure of error which gets bigger and bigger as long as the error persists. Well, one thing certainly gets bigger is the total error, so let us use it and see what happens.

Writing the error at time n as err(n), and the sum of errors to date as totalerr(n) we have

err(n) = x(n) – ewma(n), and totalerr(n) = totalerr(n-1) + err(n)

Then we can incorporate the accumulated error into the ewma formula by adding a small multiple of totalerr(n) to get

ewma(n) =  ewma(n) = (1-k)*ewma(n-1) + k*x(n) + h*totalerr(n)

In the first example below h = 0.05, and in the second h = 0.03, as things were too lively with h=0.05.


A good article on moving averages is:

In my next post I will show how the exponentially weighted moving average can be described in the language of discrete time feedback systems models, as used in modern control systems, and with luck I will get as far as the z-transfer function idea.


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