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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NYT Confronts Conformity?

j giambrone
facebook, and the rest, they almost led me astray

J. Giambrone

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How Facebook Warps Our Worlds

So it goes with the fiction we read, the movies we watch, the music we listen to and, scarily, the ideas we subscribe to. They’re not challenged. They’re validated and reinforced. By bookmarking given blogs and personalizing social-media feeds, we customize the news we consume and the political beliefs we’re exposed to as never before. And this colors our days, or rather bleeds them of color, reducing them to a single hue.

…“Facebook allows people to react to each other so quickly that they are really afraid to step out of line,” he said.

I could add plenty, but won’t. Am I afraid of stepping out of line here?


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Andy Hargreaves: Why England’s Schools are in Trouble

The same old mess, again.

Diane Ravitch's blog

Andy Hargreaves, Professor at Boston College and recipient of many honors, including the Grawemeyer Award, writes here about the problems of English schools, which he attributes to its reckless pursuit of free-market policies, akin to those now dominant in the U.S. In this article, which appeared in the Times Education Supplement (U.K.), Hargreaves blames the free-market  strategy of “reform,” which demoralizes teachers and damages the profession.

He writes:

Britain has a teacher recruitment crisis. But it is not truly British. The complaint is much more spectacular in England. In Scotland, teaching is an attractive profession and while recruitment levels are disappointing, the issue is not as profound. The Scottish system is creaking; the English system has fallen over. What explains the difference?

The answer is simple. Scotland values a strong state educational system run by 32 local authorities that is staffed by well-trained and highly valued professionals who stay and…

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‘Is everything OK? You haven’t photographed your food.’

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Can @PARCCPlace BAN You From Seeing?

Any use would be fair use at the end of the testing period.

Cloaking Inequity

I think the students in our leadership and education policy classes at California State University Sacramento (scholarly and academic purposes) and the readers of Cloaking Inequity (news reporting) will be very interested in this new, ongoing case study where a PARCC, a testing company, is trying to limit the fair use of copyrighted material. Here is a case study that was contributed to by several anonymous and on-the-record authors that I believe is fair use for research, scholarship and news reporting:

scantronCelia Oyler, professor at Teachers College, Columbia University, posted a biting commentary by an anonymous teacher about the flaws of PARCC. She received a letter from PARCC threatening legal action unless she removed the post because it contained copyrighted material —and divulged the name of the author. Oyler left the post on her blog but removed anything that might be copyrighted. She has not given up the name of the author. Many…

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More from my tropical garden

I don’t know the name of this one. The flower is about 2 inches high. The plant is a bush.

DSC00989

Eleconia. The plants grow to about 10 ft.

eleconia

The fancy bromelia. The pink parts are leaves.

fancy bromelia 3

Flowers from my neighbor’s garden.

neighbours flowers

Mini sweet pimiento plant. It just appeared and is now about 3 ft tall.

mini pimiento

Oregano in a pot.

oregano

A tree bromelia which fell out of the tree. Now it is soon going to flower. About 4 ft high.

DSC01009

A Puertorrican Name (nee-amey) with a new shoot. Too late to cook this one.DSC01005

And a visiting beetle. Yesterday we had a visit from a 4-5 ft boa constrictor.

black beetle

That’s all. folks.

 

 

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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”

*******************************************

References.
Wikipedia:
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)
********************************************

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Part 5: James Watt, the steam engine, and the first automatic control system

I was going to do my own version of the James Watt flyball governor, but I found a very readable, illustrated account of the life and inventions of James Watt. Here it is:

http://www.mainlesson.com/display.php?author=bachman&book=inventors&story=watt

In the picture below you can see the flyball governor, driven by the output shaft of the engine. The link from the governor to the steam valve however is not clear, but it is there.

watt steam engine

Here is a detailed view of the governor.regodis

See also the previous four posts.

 

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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

contpic1

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

contpic4

contpic5

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:

contpic6

contpic7
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:  http://mathcomesalive.com/mathsite/firstordersiml.html

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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