x^{2 }– 3x – 4 = 0

x^{2} – 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

x^{2 }– 3x – 4 = 0

x^{2} – 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

Filed under algebra, Uncategorized

j giambrone

facebook, and the rest, they almost led me astray

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

afraidto step out of line,” he said.

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

*Gilding the Allegory (Free)*

Free -e-Book

Filed under Uncategorized

The same old mess, again.

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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Filed under Uncategorized

, and it’s not just california.

Filed under Uncategorized

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

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:

Celia 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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Filed under Uncategorized

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

Eleconia. The plants grow to about 10 ft.

The fancy bromelia. The pink parts are leaves.

Flowers from my neighbor’s garden.

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

Oregano in a pot.

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

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

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

That’s all. folks.

Filed under caribbean, Puerto Rico, tropical garden, Uncategorized

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

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

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

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:

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)

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

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

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.

Here is a detailed view of the governor.

See also the previous four posts.

Filed under dynamic systems, engineering, flyball governor, James Watt, Uncategorized

Our first order process is described by the equation **y _{n+1} = ay_{n} + kx_{n }**, where

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

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

**y _{n+1} = ay_{n} + kx_{n }** and

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 **A **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 x_{n} from the two equations to get **y _{n+1} = ay_{n} + k(Ad_{n} + h(y_{n} – d_{n}))**

which rearranged is **y _{n+1} = (a + kh)y_{n} + k(A – h)d_{n} **

Substituting **(1-a)/k for A, as found above, gives y_{n+1} = (a + kh)y_{n} + (1 – a – kh)d_{n} **

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: 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…..

Filed under algebra, computer, control systems, discrete model, engineering, math, tuning, Uncategorized

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”.

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

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

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.

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 **y _{n}**, the desired speed

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), **x _{n} = Ad_{n} + h(y_{n} – d_{n})** , it is “obvious” that if the coefficient

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 x_{n} from the two equations to get **y _{n+1} = ay_{n} + k(Ad_{n} + h(y_{n} – d_{n}))**

which rearranged is **y _{n+1} = (a + h)y_{n} + k(A – h)d_{n} **

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.

Filed under algebra, computer, discrete model, dynamic systems, engineering, forecasting, math, Uncategorized