WEBVTT

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PROFESSOR: Hi, and welcome.

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Today we're going to do
a problem about least

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squares approximations.

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Here's the question.

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Find the quadratic equation
through the origin that

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is a best fit for these
three points: (1, 1), (2, 5),

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and (-1, -2).

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I'll give you a minute to
work that out on your own.

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You can hit the
pause button now,

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and we'll be back in a minute to
work the problem out together.

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OK and we're back.

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So what's the first step
in a problem like this?

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The first step is figuring out
what our equation looks like,

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the one that we're
going to find.

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So our equation is going to
look like c*t plus d t squared

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equals y.

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So that's what
we're looking for.

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We're looking for a quadratic
equation through the origin.

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Now, if it were just
any quadratic equation,

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then we would have
a constant term,

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but through the
origin just means

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that the constant term is 0.

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Now, what's the next step?

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Next we need to set
up a matrix equation.

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So we need an A, a
matrix A. And which

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matrix is that going to be?

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Well, let me start with the
first coordinate of these three

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points.

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And I'm going to put them in the
first column of this matrix, 1,

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2, -1.

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And then my second
column is going

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to be the squares of these
coordinates, 1, 4, and 1.

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And my x hat--
that's just c and d.

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And my b-- this is going to be
the second coordinates, 1, 5,

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and -2.

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OK, why did I set this
problem up like that?

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Well, multiply A times x hat.

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The first coordinate of
A times x hat is 1 times

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c plus 1 times d.

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It's just the same as
plugging in this first point

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into the left-hand
side of this equation.

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And similarly, if I
took the second point

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and plugged it in here,
I would just get 2 times

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c plus 4 times d,
which is just the same

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as the second coordinate in the
multiplication A times x hat.

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2 times c plus 4 times d.

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OK, and where did
the b come from?

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Well, b came from
plugging in these points

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to the right-hand side.

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So 1, 5, and -2 are
just the y-coordinates

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of these three points.

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So it would be great if we
could solve A x hat equals b.

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But we can't solve
A x hat equals

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b, because there isn't
a quadratic equation

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through the origin that
contains these three points.

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But we need to find the
best approximation to that,

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so that's the same as
solving A x hat equals

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the projection of b onto
the column space of A.

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Because we only have a chance
of solving A x hat equals

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something if it's in
the column space of A.

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And remember from our
projections part of the class

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that this is the same as solving
A transpose A x hat equals

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A transpose b.

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So that's what we're
really going to do.

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We're really going
to solve A transpose

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A x hat equals A transpose b.

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And now all we have to
do is just a computation.

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So let's do it.

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So what is A transpose A?

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I'm going to do this
kind of quickly,

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because you should be good at
multiplying matrices by now.

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Oh this-- I have the negative
in the wrong place, thank you.

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Thanks for that correction.

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I have the negative
backwards there.

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And what do I get
when I multiply these?

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I'll let you check.

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This is 6, 8; 8, and 10.

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And what is A transpose b?

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Well, if you multiply
this out, I'll

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let you check that
you get 13 and 19.

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So this is just
some computation.

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And so what we're solving
here is [6, 8; 8, 10]

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times x hat equals [13; 19].

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And we remember how to do this
just by using elimination.

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We replace the second row
by 3 times the second row

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minus 4 times the first row.

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Again, I'm going to do this
quickly because you know this

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from other parts of the class.

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You similarly
change the b vector,

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and we backsolve
to get d is -5/2

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and c equals--
let's plug that in

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and see-- think you're
going to get 11 over 2.

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So what's our final equation?

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Our final equation is y equals
11/2 t minus 5/2 t squared.

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OK.

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So this is our best
fit quadratic equation

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through the origin.

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Now before I end, let
me do a couple things.

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First, let me go back and
review what the key steps were.

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Whenever you're faced with such
a best fit equation, first,

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you want to see what the
general form of the equation is.

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Next you want to write
it in terms of matrices.

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Write down your matrix
A and your vector b.

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And lastly, you set up
your projection equation.

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And then all you have to
do is just a computation.

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But it might also
be worth noting

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that these three
points certainly aren't

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on this quadratic equation.

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For instance, if I plug in 1
here, I don't get (1, 1), I get

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(1, 3).

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But that's OK.

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This is as close as we can do.

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Thanks.