Tag 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

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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The square root of minus one asked me “Do I exist?”

Complex number.
“complex” as opposed to “simple” ?
“number” for what ?
Not for counting !
Not for measuring ! We’ll see about that !
“Square root of -1”, maybe, if that means anything at all !

Who needs the “i” ? It’s not essential.
Here goes…..

They say that (a+ib)(p+iq) = ap – bq + (bp + aq)i
But only if i is the square root of -1.

Getting rid of the i
Let us put the a and the b in a+ib together in brackets, as (a,b), and call this “thing” a “pair”.
This gets rid of the (magic) i straightaway.

Let us define an operation * to combine pairs:
(a,b)*(p,q) = (ap-bq, bp+aq)
This is the “pair” version of the “multiplication of complex numbers”.

It’s more interesting to read this as “(a,b) is applied to (p,q)”, and even better if we think of (p,q) as a “variable” and “apply (a,b)” as a function.
Ok, so we will write (x,y) instead of (p,q), and then
(a,b)*(x,y) = (ax-by, bx+ay)
Let us call the output of the “apply (a,b)” function the pair (X,Y)
Then
X = ax-by
Y = bx+ay
Now we can see this as a transformation of points in the plane:
The function “apply (a,b)” sends the point (x,y) to the point (X,Y)

Looking at some simple points we see that
(1,0)*(x,y) = (x,y)….no change at all
(-1,0)*(x,y) = (-x,-y)…the “opposite” of (x,y),
so doing (-1,0)* again gets us back to no change at all.
(0,1)*(x,y) = (-y,x)….which you may recognize as a rotation through 90 deg.
and doing (0,1)* again we get
(0,1)*(0,1)*(x,y) = (0,1)*(-y,x) = (-x,-y)….a rotation through 180 deg.

So with a bit of faith we can see that (0,1)*(0,1) is the same as (-1,0), and also that (-1,0)*(-1,0) = (1,0)…check it!
Consequently we have a system in which there are three interesting operations:
(1,0)* has no effect, it is like multiplying by 1
(-1,0)* makes every thing negative, it is like multiplying by -1, and
(0,1)*(0,1)* has the same effect as (-1,0)*

So we have found something which behaves like the square root of -1, and it is expressed as a pair of ordinary numbers.
It is then quite reasonable to give the name “i” to this “thing”, and use “i squared = -1”.

And generally, a complex number can be seen as a pair of normal (real) numbers, and bye-bye the magic !

When you think about it a fraction also needs two numbers to describe it.

Next post : matrix representation of “apply (a,b) to (x,y)”.

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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

Real problems with conic sections (ellipse, parabola) part two

So suppose we have a parabolic curve and we want to find out stuff about it.

Its equation … Oh, we have no axes.

Its focus … That would be nice, but it is a bit out of reach.

Its axis, in fact its axis of symmetry … Fold it in half? But how?

Try the method of part one, with the ellipse. (previous post)

parabola find the focus1

This looks promising. I even get another axis, for my coordinate system, if I really want the equation.

Now, analysis of the standard equation for a parabola (see later) says that a line at 45 deg to the axis, as shown, cuts the parabola at a point four focal lengths from the axis. In the picture, marked on the “vertical”axis, this is the length DH

parabola find the focus2

So I need a point one quarter of the way from D to H. Easy !

parabola find the focus3

and then the circle center D, with radius DH/4 cuts the axis of the parabola at the focal point (the focus).

Even better, we get the directrix as well …

parabola find the focus4

and now for the mathy bit (well, you do the algebra, I did the picture)

parabola find the focus math

Yes, I know that this one points up and the previous one pointed to the right !

All diagrams were created with my geometrical construction program, GEOSTRUCT

You will find it here:

www.mathcomesalive.com/geostruct/geostructforbrowser1.html

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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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Angle between two lines in the plane……Vector product in 3D…….connections???????

So I was in the middle of converting my geometry application Geostruct (we used to call them programs) into javascript

 get it here with the introduction .doc file here

when I decided that the “angle between two lines” routine needed a rewrite. Some surprises ensued !

angle between two lines pic1

Two lines,  ax + by + c  = 0  and  px + qy + r = 0

Their slopes (gradients) are  -a/b = tan(θ)  and  -p/q = tan(φ)

The angle between the lines is  φ – θ,

so it would be nice to know something about  tan(φ – θ)

Back to basics, where  tan(φ – θ) = sin(φ– θ)/cos(φ– θ),

and we have the two expansions

sin(φ– θ) = sin(φ)cos(θ) – cos(φ)sin(θ)   and

cos(φ– θ) = cos(φ)cos(θ) + sin(φ)sin(θ)

So we have  tan(φ – θ) = (sin(φ)cos(θ) – cos(φ)sin(θ))/( cos(φ)cos(θ) + sin(φ)sin(θ))

Dividing top and bottom by  cos(φ)sin(θ)  and skipping some tedious algebra we get

tan(φ – θ)   =  (tan(φ) – tan(θ))/(1 +  tan(φ)tan(θ))

This is where the books stop, which turns out to be a real shame !

Going back to the two lines and their equations, the two lines

ax + by = 0  and  px + qy = 0

have the same angle between them (some things are toooo obvious)

Things are simpler if we look at these two lines through the origin when they both have positive slope.

Take b and q as positive and write the equations as   ax – by = 0  and  px – qy = 0

Then the point whose coordinates are (b,a) lies on the first and (q,p) lies on the second.

angle between two lines pic2

Also, the slopes of the two lines are now  a/b , tan(θ)   and  p/q , tan(φ)

Let us put these into the  tan(φ – θ)   equation above, and once more after tedious algebra

tan(φ – θ)  = (bp – aq)/(ap + bq)

which is a very nice formula for the tan of the angle between two lines.

This is ok if we are interested just in “the angle between the lines”,  but if we are considering rotations, and one of the lines is the “first” one, then the tangent is inadequate. We need both the sine and the cosine of the angle to establish size AND direction (clockwise or anticlockwise).

The formula above can be seen as showing  cos(φ– θ)  as  (ap + bq) divided by something

and  sin(φ– θ)  as  (bp – aq) divided by the same something.

Calling the something  M  it is fairly clear that    (ap + bq)2 + (bp – aq) 2 = M2

and more tedious algebra and some “observation and making use of structure” gives

M= (a2 + b2)(p2 + q2)

and we now have

sin(φ– θ)  = (bp – aq)/M  and  cos(φ– θ) = (bq + ap)/M

and M is the product of the lengths of the two line segments, from the origin to (b,a) and from the origin to (q,p)

It was at this point that I saw M times the sine of the “angle between” as twice the well known formula for the area of a triangle. “half a b sin(C)”, or, if you prefer, the area of the parallelogram defined by the two line segments.

Suddenly I saw all this in 2D vector terms, with bq + ap being the dot product of (b,a) and (q,p) , and bp – aq as being part of the definition of the 3D vector or cross product, in fact the only non zero component (and in the z direction), since in 3D terms our two vectors lie in the xy plane.

Why is the “vector product” not considered in the 2D case ??? It is simpler, and looking at the formula for sine , above, we have a 2D interpretation of the “vector”or cross product as twice the area of the triangle formed by (b,a) and (q,p). (just as in the standard 3D definition, but treated as a scalar).

So “bang goes” the common terms, scalar product for c . d  and vector product for  c X d

Dot product and cross product are much better anyway, and a bit of ingenuity will lead you to the reason for the word “cross”.

This is one of the things implemented using this approach:

gif rolling circle

Anyway, the end result of all this, for rotating points on a circle, was a calculation process which did not require the actual calculation of any angle. No arctan( ) !

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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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Complex Numbers via Rigid Motions

https://howardat58.files.wordpress.com/2015/02/complex-numbers-by-rotations.doc

Complex numbers via rigid motions
Just a bit mathematical !

I wrote this in response to a post by Michael Pershan:
http://rationalexpressions.blogspot.com/2015/02/could-this-introduce-kids-to-complex.html?

The way I have presented it is showing how mathematicians think. Get an idea, try it out, if it appears to work then attempt to produce a logical and mathematically sound derivation.
(This last part I have not included)
The idea is that wherever you have operations on things, and one operation can be followed by another of the same type, then you can consider the combinations of the operations separately from the things being operated on. The result is a new type of algebra, in this case the algebra of rotations.
Read on . . .

Rotations around the origin.
angle 180 deg or pi
Y = -y, and X = -x —> coordinate transformation
so (1,0) goes to (-1,0) and (-1,0) goes to (1,0)
Let us call this transformation H (for a half turn)

angle 90 deg or pi/2
Y = x, and X = -y
so (1,0) goes to (0,1) and (-1,0) goes to (0,-1)
and (0,1) goes to (-1,0) and (0,-1) goes to (1,0)
Let us call this transformation Q (for a quarter turn)

Then H(x,y) = (-x,-y)
and Q(x,y) = (-y,x)

Applying H twice we have H(H(x,y)) = (x,y) and if we are bold we can write HH(x,y) = (x,y)
and then HH = I, where I is the identity or do nothing transformation.
In the same way we find QQ = H

Now I is like multiplying the coodinates by 1
and H is like multiplying the coordinates by -1
This is not too outrageous, as a dilation can be seen as a multiplication of the coordinates by a number <> 1

So, continuing into uncharted territory,
we have H squared = 1 (fits with (-1)*(-1) = 1
and Q squared = -1 (fits with QQ = H, at least)

So what is Q ?
Let us suppose that it is some sort of a number, definitely not a normal one,
and let its value be called k.
What we can be fairly sure of is that k does not multiply each of the coordinates.
This appears to be meaningful only for the normal numbers.

Now the “number” k describes a rotation of 90, so we would expect that the square root of k to describe a rotation of 45

At this point it helps if the reader is familiar with extending the rational numbers by the introduction of the square root of 2 (a surd, although this jargon seems to have disappeared).

Let us assume that sqrt(k) is a simple combination of a normal number and a multiple of k:
sqrt(k) = a + bk
Then k = sqr(a) + sqr(b)*sqr(k) + 2abk, and sqr(k) = -1
which gives k = sqr(a)-sqr(b) + 2abk and then (2ab-1)k = sqr(a) – sqr(b)

From this, since k is not a normal number, 2ab = 1 and sqr(a) = sqr(b)
which gives a = b and then a = b = 1/root(2)

Now we have a “number” representing a 45 degree rotation. namely
(1/root(2)*(1 + k)

If we plot this and the other rotation numbers as points on a coordinate axis grid with ordinary numbers horizontally and k numbers vertically we see that all the points are on the unit circle, at positions corresponding to the rotation angles they describe.

OMG there must be something in this ! ! !

The continuation is left to the reader (as in some Victorian novels)

ps. root() and sqrt() are square root functions, and sqr() is the squaring function .

pps. Diagrams may be drawn at your leisure !

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Filed under abstract, algebra, education, geometry, operations, teaching

The angle bisector theorem and a locus

The theorem states that if we bisect an angle of a triangle then the two parts of the side cut by the bisector are in the same ratio as the two other sides of the triangle. So I thought, what if we move the point, G in the gif below, so that GB always bisects the angle AGC, then where does it go ? Or, more mathematically, what is its locus. Geometry failed me at this point so I did a coordinate geometry version of the ratios and found that the locus was a circle ( a little surprised !). The algebra is fairly simple.

gif bisector locus

What was a real surprise was that if we take the origin of measurements to be the other point where the circle cross the line ABC, then point C is at the harmonic mean of points A and B. There is YOUR problem from this post.

Getting the construction to show all this was tricky. It is shown below.

locus angle bisector point full

I did the construction on my geometry program, Geostruct (see my Software page).

Line bb passes through A. Point D is on line bb. Point E is on this line and also on the line through C parallel to line cc joining B and D, So AD and DE are in the same ratio as AC and CB. When point D is moved the ratios are unchanged (feature of the program). So the point of interest is the intersection of a circle centre A, radius AD, and a circle centre C and radius DE (= FC)

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