WEBVTT
00:00:00.000 --> 00:00:07.740
JOEL LEWIS: Hi.
00:00:07.740 --> 00:00:09.360
Welcome back to recitation.
00:00:09.360 --> 00:00:10.930
In lecture you've
started learning
00:00:10.930 --> 00:00:12.560
about quadratic approximation.
00:00:12.560 --> 00:00:16.540
So today we're just going
to do a quick example of it.
00:00:16.540 --> 00:00:18.870
So I have a question
written here on the board:
00:00:18.870 --> 00:00:21.400
What is the quadratic
approximation of the function
00:00:21.400 --> 00:00:25.470
f of x equals e to the x plus x
squared-- so here x and plus x
00:00:25.470 --> 00:00:28.820
squared is the exponent,
so it's e to the quantity
00:00:28.820 --> 00:00:31.340
x plus x squared--
near x equals 0.
00:00:31.340 --> 00:00:34.780
So why don't you take a minute
or two, pause the video,
00:00:34.780 --> 00:00:36.710
work this out on your
own and then come back
00:00:36.710 --> 00:00:37.793
and we can do it together.
00:00:43.620 --> 00:00:44.120
All right.
00:00:44.120 --> 00:00:44.860
Welcome back.
00:00:44.860 --> 00:00:47.509
So there are two different
ways we can do this problem.
00:00:47.509 --> 00:00:49.800
Let's first just do it the
totally straightforward way,
00:00:49.800 --> 00:00:51.410
which is that you
have this formula
00:00:51.410 --> 00:00:53.240
for quadratic
approximations in terms
00:00:53.240 --> 00:00:56.680
of the derivatives
of your function.
00:00:56.680 --> 00:01:00.350
And so, so we can just
apply that formula.
00:01:00.350 --> 00:01:04.800
So here, so the formula is that
the quadratic approximation
00:01:04.800 --> 00:01:08.220
of the function f-- so
here, near the point 0--
00:01:08.220 --> 00:01:18.570
is equal to f of 0 plus f prime
of 0 x plus f double prime of 0
00:01:18.570 --> 00:01:22.030
over 2 times x squared.
00:01:22.030 --> 00:01:22.640
All right.
00:01:22.640 --> 00:01:26.780
So in order to use
this formula we just
00:01:26.780 --> 00:01:30.060
need to know what the
derivatives of our function
00:01:30.060 --> 00:01:33.540
are and their values at 0.
00:01:33.540 --> 00:01:38.470
So first we can do the
first derivative of f.
00:01:38.470 --> 00:01:41.140
So for that it's just a
straightforward application
00:01:41.140 --> 00:01:42.190
of the chain rule.
00:01:42.190 --> 00:01:45.360
Our outer function is e to
the x and our inner function
00:01:45.360 --> 00:01:46.820
is x plus x squared.
00:01:46.820 --> 00:01:51.835
So the derivative then, applying
the chain rule, is e to the x
00:01:51.835 --> 00:02:01.260
plus x squared times 1 plus 2x,
which I can also write as e--
00:02:01.260 --> 00:02:05.270
well, yeah, let me
just reorder it-- 2x
00:02:05.270 --> 00:02:11.481
plus 1 times e to
the x plus x squared.
00:02:11.481 --> 00:02:11.980
OK.
00:02:11.980 --> 00:02:13.290
So that's the first derivative.
00:02:13.290 --> 00:02:16.750
And for the second derivative I
can apply just the product rule
00:02:16.750 --> 00:02:17.250
here.
00:02:17.250 --> 00:02:17.750
Right?
00:02:17.750 --> 00:02:20.580
So I've got this-- you
know, the second derivative
00:02:20.580 --> 00:02:22.330
is the derivative of
the first derivative,
00:02:22.330 --> 00:02:24.170
so here I have a product.
00:02:24.170 --> 00:02:29.180
So f double prime of
x is equal to-- well,
00:02:29.180 --> 00:02:33.430
so we take the derivative of
the first one, which is just 2,
00:02:33.430 --> 00:02:39.720
times the second plus the
derivative of the second one.
00:02:39.720 --> 00:02:42.920
Well, the second one is e
to the x plus x squared.
00:02:42.920 --> 00:02:45.740
It's actually f of x, so we
already computed it once.
00:02:45.740 --> 00:02:51.730
So the derivative of the
second is 2x plus 1 e to the x
00:02:51.730 --> 00:02:54.390
plus x squared times the first.
00:02:54.390 --> 00:02:59.370
So times another 2x plus 1.
00:02:59.370 --> 00:03:02.950
OK, and if I multiply these
two together and combine
00:03:02.950 --> 00:03:10.560
all my terms, this is 4x
squared plus 4x plus 3 times
00:03:10.560 --> 00:03:13.340
e to the x plus x squared.
00:03:13.340 --> 00:03:15.340
So these are the first
and second derivatives,
00:03:15.340 --> 00:03:17.720
and what I need to plug
them into my formula
00:03:17.720 --> 00:03:19.800
is I need their values at 0.
00:03:19.800 --> 00:03:22.380
So I need the
function value at 0.
00:03:22.380 --> 00:03:26.990
So f of 0, well that's e
to the 0 plus 0 squared.
00:03:26.990 --> 00:03:31.570
So it's e to the 0,
so that's just 1.
00:03:31.570 --> 00:03:36.470
f prime at 0, let's see, I go
over to my formula for f prime
00:03:36.470 --> 00:03:40.480
and I plug in x equals 0,
so I have 2 times 0 plus 1,
00:03:40.480 --> 00:03:43.430
so that's 1, times e to the 0.
00:03:43.430 --> 00:03:46.920
So that's 1 times
1, that's also 1.
00:03:46.920 --> 00:03:53.080
And for f double time
at 0, I go to my formula
00:03:53.080 --> 00:03:55.850
for f double prime
and I put in 0.
00:03:55.850 --> 00:03:57.190
And so this is 0 and that's 0.
00:03:57.190 --> 00:04:03.120
So I have a 3 times
1 so that's 3.
00:04:03.120 --> 00:04:04.940
And now I just take
these three values
00:04:04.940 --> 00:04:06.790
and I plug them right
into my formula.
00:04:06.790 --> 00:04:10.050
So the quadratic
approximation is
00:04:10.050 --> 00:04:21.820
Q of f equals 1 plus x
plus 3 x squared over 2.
00:04:21.820 --> 00:04:22.320
All right.
00:04:22.320 --> 00:04:23.300
Great.
00:04:23.300 --> 00:04:25.490
So that was one way
to do this problem.
00:04:25.490 --> 00:04:28.410
Another way to do this
problem is the following.
00:04:28.410 --> 00:04:31.650
And in this case, I'm not
sure which way is simpler,
00:04:31.650 --> 00:04:34.820
but in some cases one way is
clearly easier than the other.
00:04:34.820 --> 00:04:36.320
So if we illustrate
both then you'll
00:04:36.320 --> 00:04:40.800
have twice as many
tools to work with.
00:04:40.800 --> 00:04:43.950
So the other way is to notice--
so the exponential function is
00:04:43.950 --> 00:04:44.450
nice.
00:04:44.450 --> 00:04:46.750
When you have-- you know,
one of your exponential rules
00:04:46.750 --> 00:04:50.010
is that the exponential
of a sum is the product
00:04:50.010 --> 00:04:50.960
of the exponentials.
00:04:50.960 --> 00:04:57.700
So we can rewrite f of x
equals e to the x times
00:04:57.700 --> 00:04:59.360
e to the x squared.
00:04:59.360 --> 00:05:01.740
Now this may seem a
little bit silly to you,
00:05:01.740 --> 00:05:05.190
but if you watched
Christine's recitation
00:05:05.190 --> 00:05:08.120
video you saw that to find
the quadratic approximation
00:05:08.120 --> 00:05:10.120
to a product, it's
enough to find
00:05:10.120 --> 00:05:12.580
the quadratic
approximations to each piece
00:05:12.580 --> 00:05:14.790
separately, multiply
together, and then take
00:05:14.790 --> 00:05:16.370
that quadratic approximation.
00:05:16.370 --> 00:05:21.130
So in this case the
quadratic approximations
00:05:21.130 --> 00:05:22.910
are things you
might already know.
00:05:22.910 --> 00:05:26.040
So in particular, we
saw in recit-- sorry,
00:05:26.040 --> 00:05:31.620
in lecture that the quadratic
approximation of e to the x
00:05:31.620 --> 00:05:36.460
is 1 plus x plus
x squared over 2.
00:05:36.460 --> 00:05:39.137
And I'm going to tell you--
if you haven't seen it
00:05:39.137 --> 00:05:41.220
in recitation, there are
a bunch of different ways
00:05:41.220 --> 00:05:42.720
you could work it
out for yourself--
00:05:42.720 --> 00:05:46.716
but the quadratic approximation
for e to the x squared
00:05:46.716 --> 00:05:50.386
is equal to 1 plus x squared.
00:05:50.386 --> 00:05:53.000
All right, so if you
don't believe me,
00:05:53.000 --> 00:05:56.710
by all means work that out
for yourself to check it.
00:05:56.710 --> 00:05:58.180
So in this case,
so that means that
00:05:58.180 --> 00:06:03.590
the quadratic approximation
of f is equal to-- by the rule
00:06:03.590 --> 00:06:06.080
Christine showed you--
it's the, so it's
00:06:06.080 --> 00:06:08.100
equal to the quadratic
approximation
00:06:08.100 --> 00:06:10.860
of the product of the
quadratic approximations.
00:06:10.860 --> 00:06:15.190
So that's 1 plus
x plus x squared
00:06:15.190 --> 00:06:19.420
over 2 times 1 plus x squared.
00:06:19.420 --> 00:06:21.580
Now if you multiply these
two out, that's not hard.
00:06:21.580 --> 00:06:23.100
There are six terms.
00:06:23.100 --> 00:06:27.570
So this is the quadratic
approximation of 1
00:06:27.570 --> 00:06:31.310
plus x plus x squared over 2.
00:06:31.310 --> 00:06:34.340
So multiply 1 through
here and now x squared.
00:06:34.340 --> 00:06:40.200
So that's plus x
squared plus x cubed
00:06:40.200 --> 00:06:46.314
plus-- OK, some fourth degree
term-- x to the fourth over 2.
00:06:46.314 --> 00:06:48.230
And now the quadratic
approximation to this we
00:06:48.230 --> 00:06:51.380
get just by dropping the
cubic and quartic terms.
00:06:51.380 --> 00:06:54.590
So-- and OK, and we can
add these two square terms
00:06:54.590 --> 00:06:55.120
together.
00:06:55.120 --> 00:07:02.580
So this is equal to 1 plus
x plus 3 x squared over 2.
00:07:02.580 --> 00:07:03.080
All right.
00:07:03.080 --> 00:07:05.480
And luckily, doing the same
problem two different ways we
00:07:05.480 --> 00:07:08.190
get the same answer, which
is what should happen.
00:07:08.190 --> 00:07:09.900
So that's that.