Course Syllabus AP English-Literature and Composition

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Course Syllabus AP English - Literature and Composition Advanced Placement English Literature and Composition engages students in the practice of reading a variety of texts with the purpose of performing critical, literary analysis. This full year course is composed of three distinct sections, each preparing the student to pass the AP test in the Spring. Be assured, this is not just a test prep course; we will also examine many seminal works of World Literature.

significant works of literature
formal writing
conventions of epic poetry
special assignment
epic
summer reading
poetry
literary analysis
essay
class

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Publié par	mevang
Nombre de lectures	34
Langue	English

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18.06 Linear Algebra, Spring 2010
Transcript – Lecture 1
Hi. This is the first lecture in MIT's course 18.06, linear algebra, and I'm Gilbert
Strang. The text for the course is this book, Introduction to Linear Algebra.

And the course web page, which has got a lot of exercises from the past, MatLab

codes, the syllabus for the course, is web.mit.edu/18.06.

And this is the first lecture, lecture one.

So, and later we'll give the web address for viewing these, videotapes. Okay, so

what's in the first lecture? This is my plan.

The fundamental problem of linear algebra, which is to solve a system of linear

equations.

So let's start with a case when we have some number of equations, say n equations

and n unknowns.

So an equal number of equations and unknowns.

That's the normal, nice case.

And what I want to do is -- with examples, of course -- to describe, first, what I call

the Row picture. That's the picture of one equation at a time. It's the picture you've

seen before in two by two equations where lines meet.

So in a minute, you'll see lines meeting.

The second picture, I'll put a star beside that, because that's such an important one.

And maybe new to you is the picture -- a column at a time.

And those are the rows and columns of a matrix.

So the third -- the algebra way to look at the problem is the matrix form and using a

matrix that I'll call A.

Okay, so can I do an example? The whole semester will be examples and then see

what's going on with the example.

So, take an example. Two equations, two unknowns. So let me take 2x -y =0, let's

say. And -x +2y=3.

Okay. let me -- I can even say right away -- what's the matrix, that is, what's the
coefficient matrix? The matrix that involves these numbers -- a matrix is just a rectangular array of numbers. Here it's two rows and two columns, so 2 and -- minus
1 in the first row minus 1 and 2 in the second row, that's the matrix.
And the right-hand -- the unknown -- well, we've got two unknowns. So we've got a
vector, with two components, x and x, and we've got two right-hand sides that go
into a vector 0 3.
I couldn't resist writing the matrix form, right -- even before the pictures. So I
always will think of this as the matrix A, the matrix of coefficients, then there's a
vector of unknowns.
Here we've only got two unknowns.
Later we'll have any number of unknowns.
And that vector of unknowns, well I'll often -- I'll make that x -- extra bold. A and
the right-hand side is also a vector that I'll always call b.
So linear equations are A x equal b and the idea now is to solve this particular
example and then step back to see the bigger picture. Okay, what's the picture for
this example, the Row picture? Okay, so here comes the Row picture.
So that means I take one row at a time and I'm drawing here the xy plane and I'm
going to plot all the points that satisfy that first equation. So I'm looking at all the
points that satisfy 2x-y =0. It's often good to start with which point on the horizontal
line -- on this horizontal line, y is zero.
The x axis has y as zero and that -- in this case, actually, then x is zero. So the
point, the origin -- the point with coordinates (0,0) is on the line. It solves that
equation.
Okay, tell me in -- well, I guess I have to tell you another point that solves this same
equation.
Let me suppose x is one, so I'll take x to be one.
Then y should be two, right? So there's the point one two that also solves this
equation.
And I could put in more points. But, but let me put in all the points at once, because
they all lie on a straight line. This is a linear equation and that word linear got the
letters for line in it.
That's the equation -- this is the line that ...
of solutions to 2x-y=0 my first row, first equation.
So typically, maybe, x equal a half, y equal one will work. And sure enough it does.
Okay, that's the first one. Now the second one is not going to go through the origin.
It's always important. Do we go through the origin or not? In this case, yes, because there's a zero over
there. In this case we don't go through the origin, because if x and y are zero, we
don't get three. So, let me again say suppose y is zero, what x do we actually get? If
y is zero, then I get x is minus three.
So if y is zero, I go along minus three.
So there's one point on this second line.
Now let me say, well, suppose x is minus one -- just to take another x. If x is minus
one, then this is a one and I think y should be a one, because if x is minus one, then
I think y should be a one and we'll get that point. Is that right? If x is minus one,
that's a one.
If y is a one, that's a two and the one and the two make three and that point's on
the equation.
Okay. Now, I should just draw the line, right, connecting those two points at -- that
will give me the whole line. And if I've done this reasonably well, I think it's going to
happen to go through -- well, not happen -- it was arranged to go through that
point. So I think that the second line is this one, and this is the all-important point
that lies on both lines. Shall we just check that that point which is the point x equal
one and y was two, right? That's the point there and that, I believe, solves both
equations.
Let's just check this. If x is one, I have a minus one plus four equals three, okay.
Apologies for drawing this picture that you've seen before. But this -- seeing the row
picture -- first of all, for n equal 2, two equations and two unknowns, it's the right
place to start. Okay.
So we've got the solution. The point that lies on both lines. Now can I come to the
column picture? Pay attention, this is the key point. So the column picture.
I'm now going to look at the columns of the matrix.
I'm going to look at this part and this part.
I'm going to say that the x part is really x times -- you see, I'm putting the two --
I'm kind of getting the two equations at once -- that part and then I have a y and in
the first equation it's multiplying a minus one and in the second equation a two, and
on the right-hand side, zero and three. You see, the columns of the matrix, the
columns of A are here and the right-hand side b is there. And now what is the
equation asking for? It's asking us to find -- somehow to combine that vector and
this one in the right amounts to get that one. It's asking us to find the right linear
combination -- this is called a linear combination.
And it's the most fundamental operation in the whole course.
It's a linear combination of the columns.
That's what we're seeing on the left side.
Again, I don't want to write down a big definition. You can see what it is. There's column one, there's column two. I multiply by some
numbers and I add. That's a combination -- a linear combination and I want to make
those numbers the right numbers to produce zero three. Okay.
Now I want to draw a picture that, represents what this -- this is algebra. What's the
geometry, what's the picture that goes with it? Okay. So again, these vectors have
two components, so I better draw a picture like that. So can I put down these
columns? I'll draw these columns as they are, and then I'll do a combination of them.
So the first column is over two and down one, right? So there's the first column.
The first column. Column one.
It's the vector two minus one. The second column is -- minus one is the first
component and up two.
It's here. There's column two.
So this, again, you see what its components are. Its components are minus one,
two. Good.
That's this guy. Now I have to take a combination. What combination shall I take?
Why not the right combination, what the hell? Okay. So the combination I'm going to
take is the right one to produce zero three and then we'll see it happen in the
picture. So the right combination is to take x as one of those and two of these.
It's because we already know that that's the right x and y, so why not take the
correct combination here and see it happen? Okay, so how do I picture this linear
combination? So I start with this vector that's already here -- so that's one of column
one, that's one times column one, right there.
And now I want to add on -- so I'm going to hook the next vector onto the front of
the arrow will start the next vector and it will go this way. So let's see, can I do it
right? If I added on one of these vectors, it would go left one and up two, so we'd go
left one and up two, so it would probably get us to there.
Maybe I'll do dotted line for that.
Okay? That's one of column two tucked onto the end, but I wanted to tuck on two of
column two. So that -- the second one -- we'll go up left one and up two also.
It'll probably end there. And there's another one.
So what I've put in here is two of column two.
Added on. And where did I end up? What are the coordinates of this result? What do
I get when I t