Category Archives: algebra

Linear transformations, geometrically

 

Following a recent blog post relating a transformation of points on a line to points on another line to the graph of the equation relating the input and output I thought it would be interesting to explore the linear and affine mappings of a plane to itself from a geometrical construction perspective.

It was ! (To me anyway)

These linear mappings  (rigid and not so rigid motions) are usually  approached in descriptive and manipulative  ways, but always very specifically. I wanted to go directly from the transformation as equations directly to the transformation as geometry.

Taking an example, (x,y) maps to (X,Y) with the linear equations

X = x + y + 1 and Y = -0.5x +y

it was necessary to construct a point on the x axis with the value of X, and likewise a point on the y axis with the value of Y. The transformed (x,y) is then the point (X,Y) on the plane.

The construction below shows the points and lines needed to establish the point(X,0), which is G in the picture, starting with the point D as the (x,y)

 

transform of x

The corresponding construction was done for Y, and the resulting point (X,Y) is point J. Point D was then forced to lie on a line, the sloping blue line, and as it is moved along the line the transformed point J moves on another line

gif for lin affine trans1

Now the (x,y) point (B in the picture below, don’t ask why!) is forced to move on the blue circle. What does the transformed point do? It moves on an ellipse, whose size and orientation are determined by the actual transformation. At this point matrix methods become very handy.(though the 2D matrix methods cannot deal with translations)

gif for lin affine trans2

All this was constructed with my geometrical construction program (APP if you like) called GEOSTRUCT and available as a free web based application from

http://www.mathcomesalive.com/geostruct/geostructforbrowser1.html

The program produces a listing of all the actions requested, and these are listed below for this application:

Line bb moved to pass through Point A
New line cc created, through points B and C
New Point D
New line dd created, through Point D, at right angles to Line aa
New line ee created, through Point D, at right angles to Line bb
New line ff created, through Point D, parallel to Line cc
New point E created as the intersection of Line ff and Line aa
New line gg created, through Point E, at right angles to Line aa
New line hh created, through Point B, at right angles to Line bb
New point F created as the intersection of Line hh and Line gg
New line ii created, through Point F, parallel to Line cc
New point G created as the intersection of Line ii and Line aa

G is the X coordinate, from X = x + y + 1 (added by me)

New line jj created, through Point G, at right angles to Line aa
New line kk created, through Point D, at right angles to Line cc
New point H created as the intersection of Line kk and Line bb
New point I created, as midpoint of points H and B
New line ll created, through Point I, at right angles to Line bb
New point J created as the intersection of Line ll and Line jj

J is the Y coordinate, from Y = -x/2 + y  (added by me)
and K is the transformed point (X,Y) Point J chosen as the tracking point (added by me)

New Line mm
Point D moved and placed on Line mm

 

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Filed under algebra, conics, construction, geometrical, geometry app, geostruct, math, ordered pairs, rigid motion, teaching, transformations, Uncategorized

Mathematics in my garden

So there it was, built from one clothesline to another, glistening in the sunlight one morning last week:

spiderweb

A near perfect spiderweb, demanding a better photo:
spiderweb2

Here is  the horizontal strip, showing the variation in spacing of the spiral:
spiderweb2partonly

and again, with a superimposed scale. The spider is at the centre.spiderweb cropped with dots

And now rotated, to give more detail.spiderweb cropped with dots rotated

Now “Get modelling!”.

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Filed under algebra, geometrical, Puerto Rico, series, teaching, tropical garden, Uncategorized

How to murder algebra

Idly poking around the net I found this example:

Mathway example
x+4x=9
Solve for x
Answer
Move all terms that don’t contain x to the right-hand side and solve.
x=9/5

I cried !

Read more from mathway.com

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Filed under algebra, errors, horrors, mistakes

Political correctness in K-12 math

This is definitely worth a read.

Here is a quote:

“High schools focus on elementary applications of advanced mathematics whereas most people really make more use of sophisticated applications of elementary mathematics. This accounts for much of the disconnect noted above, as well as the common complaint from employers that graduates don’t know any math. Many who master high school mathematics cannot think clearly about percentages or ratios.”

And here is the link:

http://www.stolaf.edu/people/steen/Papers/07carnegie.pdf

Link found on f(t)’s blog

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Filed under abstract, algebra, CCSS, Common Core, depth, education, K-12, math, standards, teaching, tests

Halving a triangle, follow-up number three, bisecting an angle

While looking at the bisection of area formula (see previous posts)
a’b’ = (1/2)ab
where a’ and b’ are the distances from the vertex of points on the sides of the triangle, and a and b are the lengths of the sides I remembered another formula about triangles, the bisection of the angle formula, with a” and b” being the lengths of the two parts into which the opposite side is divided, namely
a”/b” = a/b

These are like Cuba and Puerto Rico, “Two wings of the same bird”, Jose Marti (in Spanish)
Neither involves the angle itself, and so is very general. I decided that there must be a connection, and after a futile look for some duality in the situation I suddenly saw the connection, in simple algebraic terms:
a’b’ = a’/(1/b’)
and so a triangle with sides a’ and 1/b’ will give a”/b” = a’b’
and then a”/b” = (1/2)ab as well.

This is the construction for the halving a triangle.
half triangle and bisector0

This is the extended construction for the bisector. The opposite side is in brown.
half triangle and bisector1
and this is a gif showing how as the point a’ (E) is moved the brown line (OE, opposite side) moves parallel to itself, thus preserving the value of the ratio a”/b”

gif halving bisecting

And in case you got this far, some light relief. Bullfrog eats dog food, this morning. The dish is 10 inches across.
bullfrog dogdish

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Filed under algebra, bisecting, geometry, halving

Halving a triangle, follow-up number two, pursuing the hyperbola

Halving the triangle, any triangle, led to the equation XY = 2 as the condition on the points on two sides of the triangle, distant X and Y from the vertex.
The envelope of this set of lines turned out to be a hyperbola.
But XY = 2 defines a hyperbola – what is the connection ?

I took xy = 1 for the condition, on a standard xy grid, and wrote it as representing a function x —-> y, namely y = 1/x
The two points of interest are then (x,0) on the x-axis and (0,1/x) on the y-axis.
We need the equation of the line joining these two points, so first of all we have to see that our x, above, is telling us which line we are talking about, and so it is a parameter for the line.
We had better give it a different name, say p.
Now we can find the equation of the line in x,y form, using (p,0) and (0,1/p) for the two points:
(y – 0)/(1/p – 0) = (x – p)/(0 – p)
which is easier to read as yp = -x/p + 1, and easier to process as yp2 = -x + p

Now comes the fun bit !
To find the envelope of a set of straight lines we have to find the points of intersection of adjacent lines (really? adjacent?). To do this we have to find the partial derivative (derivative treating almost everything as constant) of the line equation with respect to the parameter p. A later post will reveal all about this mystifying procedure).
So do it and get  2yp = 1

And then eliminate p from the two equations, the line one and the derived one:
From the derived equation we get p = 1/(2y), so substituting in the line equation gives 1 = 2xy
This is the equation of the envelope, and written in functional form it is
y = 1/(2x), or (1/2)(1/x)
Yes ! Another rectangular hyperbola, with the same asymptotes.
(write it as xy = 1/2 if you like)

Now I thought “What will this process do with y = x2 ?”
So off I go, and to cut a long story short I found the following:
For y = x2 the envelope was y = (-1/4)x2, also a multiple of the original, with factor -1/4
parabola by axa track point
parabola by axa track point and line

Some surprise at this point, so I did it for 1/x2 and for x3
Similar results: Same function, with different factors.
Try it yourself ! ! ! ! ! ! !

This was too much ! No stopping ! Must find the general case ! (y = xk)
Skipping the now familiar details (left to the reader, in time honoured fashion) I found the following:

Original equation: y = xk

Equation of envelope: y = xk multiplied by -(k-1)k-1/kk

which I did think was quite neat.
—————————————————————
The next post will be the last follow-up to the triangle halving.

—————————————————————
Constructions made with GEOSTRUCT, an online browser application:

To get geostruct from the net click
http://www.mathcomesalive.com/geostruct/geostructforbrowser1.html

and to download the .doc instructions file
http://www.mathcomesalive.com/geostruct/geostruct basics.doc

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Filed under algebra, calculus, construction, envelope, geometrical

Mathematical discrimination in the brave new world of Common Core

Merriam Webster
discriminating
adjective dis·crim·i·nat·ing

: liking only things that are of good quality

: able to recognize the difference between things that are of good quality and those that are not

Full Definition of DISCRIMINATING

1: making a distinction : distinguishing
2: marked by discrimination:
a : discerning, judicious
b : discriminatory

Take your pick !

In the UK we used to call it prejudice, as in “racial prejudice”.

Today the perfectly good word “discrimination” has been hijacked and the old meaning has all but disappeared.

Now you ask “What’s this got to do with math?”.

We have to look at quadratic equations for an answer.

The standard quadratic equation is ax2 + bx + c = 0
It may or may not have real roots
and the sign of D = b2 – 4ac resolves this uncertainty.

D < 0 … no real roots, D = 0 … equal real roots D > 0 … different real roots

What is this D ? It is the “discriminant” of the equation.

It is used to discriminate between equations with real roots and equations without real roots.

So WHY isn’t it in the Common Core math standards. It’s about as standard a thing as is possible?

Clearly the CCSSM authors are guilty of serious discrimination in discriminating against the word “discriminant”, in order not to be accused of discrimination language. Neither “discrimination” nor “discriminant” appear in the CCSSM doc, and the result is that the poor kids ( in the little darlings sense) have to complete the square from scratch every time they have a quadratic equation ( instead of once only in order to get the discriminant formula).

So sad !

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Filed under algebra, CCSS, Common Core, education, teaching

What is Algebra really for ?

An example tells a good tale.

Translation of a line in an x-y coordinate system:

Take a line  y = 2x -3, and translate it by 4 up and 5 to the right.

Simple approach : The point P = (2, 1) is on the line (so are some others!). Let us translate the point to get Q = (2+5, 1+4), which is Q = (7, 5), and find the line through Q parallel to the original line.  The only thing that changes is the c value, so the new equation is  y = 2x + c, and it must pass through Q.  So we require  5 = 14 + c, giving the value of c as -9.

Not much algebra there, but a horrible question remains – “What happened to all the other points on the line ?”

We try a more algebraic approach – with any old line  ax + by + c = 0, and any old translation, q up and p to the right.

First thing is to find a point on the line – “What ? We don’t know ANYTHING about the line.”

This is where algebra comes to the rescue. Let us suppose (state) that a point P = (d, e) IS on the line.

Then ad + be + c = 0

Now we can move the point P to Q = (d + p, e + q)  (as with the numbers earlier), and make the new line pass through this point:  This requires a new constant c (call it newc) and we then have  a(d + p) + b(e + q) + ‘newc’ = 0

Expand the parentheses (UK brackets, and it’s shorter) to get  ad + ap + be + bq + ‘newc’ = 0

Some inspired rearranging gives  ‘newc’ = -ap – bq – ad – be, which is equal to -(ap + bq) – (ad + be)

“Why did you do that last step ?” – “Because I looked back a few lines and figured that  (ad + be) = -c, which not only simplifies the expression, it also disposes of the unspecified point  P.

End result is:  Translated line equation is  ax + by + ‘newc’ =0,  that is,  ax + by + c – (ap + bq) = 0

and the job is done for ALL lines, even the vertical ones, and ALL translations. Also we can be sure that we know what has happened to ALL the points on the line.

I am not going to check this with the numerical example, you are !

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Filed under algebra, geometry, math, teaching

A fun math/computing problem

I found this on http://www.playwithyourmath.com/ and adapted it a little.

The number 25 can be broken up in many ways, like 1+4+4+7+9

Let’s multiply the parts together,  getting 504 (or something near)

Problem 1: Find the break-up which gives the max product of the parts. 1+1+1+…+1 is not much use.

Problem 2: Find a rule for doing this for any whole number.

Problem 3: Put this rule in the form of a computer algorithm (pseudocode is OK)

Problem 4: Write the rule as a single calculation (formula)

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Filed under algebra, math, operations

The Future

“How’s your Mary doing?”.

“She’s doing well. She’s 8 now. She’s in Grade 3. She really enjoys the Pre-Algebra and the Pre-Textual Analysis.”.

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Filed under algebra, education, language in math, teaching