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JOEL LEWIS: Hi.

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Welcome back to recitation.

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In lecture you've
started learning

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about quadratic approximation.

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So today we're just going
to do a quick example of it.

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So I have a question
written here on the board:

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What is the quadratic
approximation of the function

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f of x equals e to the x plus x
squared-- so here x and plus x

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squared is the exponent,
so it's e to the quantity

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x plus x squared--
near x equals 0.

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So why don't you take a minute
or two, pause the video,

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work this out on your
own and then come back

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and we can do it together.

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All right.

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Welcome back.

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So there are two different
ways we can do this problem.

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Let's first just do it the
totally straightforward way,

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which is that you
have this formula

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for quadratic
approximations in terms

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of the derivatives
of your function.

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And so, so we can just
apply that formula.

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So here, so the formula is that
the quadratic approximation

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of the function f-- so
here, near the point 0--

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is equal to f of 0 plus f prime
of 0 x plus f double prime of 0

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over 2 times x squared.

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All right.

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So in order to use
this formula we just

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need to know what the
derivatives of our function

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are and their values at 0.

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So first we can do the
first derivative of f.

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So for that it's just a
straightforward application

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of the chain rule.

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Our outer function is e to
the x and our inner function

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is x plus x squared.

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So the derivative then, applying
the chain rule, is e to the x

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plus x squared times 1 plus 2x,
which I can also write as e--

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well, yeah, let me
just reorder it-- 2x

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plus 1 times e to
the x plus x squared.

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OK.

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So that's the first derivative.

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And for the second derivative I
can apply just the product rule

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here.

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Right?

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So I've got this-- you
know, the second derivative

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is the derivative of
the first derivative,

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so here I have a product.

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So f double prime of
x is equal to-- well,

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so we take the derivative of
the first one, which is just 2,

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times the second plus the
derivative of the second one.

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Well, the second one is e
to the x plus x squared.

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It's actually f of x, so we
already computed it once.

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So the derivative of the
second is 2x plus 1 e to the x

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plus x squared times the first.

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So times another 2x plus 1.

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OK, and if I multiply these
two together and combine

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all my terms, this is 4x
squared plus 4x plus 3 times

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e to the x plus x squared.

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So these are the first
and second derivatives,

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and what I need to plug
them into my formula

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is I need their values at 0.

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So I need the
function value at 0.

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So f of 0, well that's e
to the 0 plus 0 squared.

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So it's e to the 0,
so that's just 1.

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f prime at 0, let's see, I go
over to my formula for f prime

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and I plug in x equals 0,
so I have 2 times 0 plus 1,

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so that's 1, times e to the 0.

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So that's 1 times
1, that's also 1.

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And for f double time
at 0, I go to my formula

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for f double prime
and I put in 0.

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And so this is 0 and that's 0.

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So I have a 3 times
1 so that's 3.

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And now I just take
these three values

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and I plug them right
into my formula.

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So the quadratic
approximation is

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Q of f equals 1 plus x
plus 3 x squared over 2.

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All right.

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Great.

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So that was one way
to do this problem.

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Another way to do this
problem is the following.

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And in this case, I'm not
sure which way is simpler,

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but in some cases one way is
clearly easier than the other.

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So if we illustrate
both then you'll

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have twice as many
tools to work with.

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So the other way is to notice--
so the exponential function is

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nice.

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When you have-- you know,
one of your exponential rules

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is that the exponential
of a sum is the product

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of the exponentials.

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So we can rewrite f of x
equals e to the x times

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e to the x squared.

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Now this may seem a
little bit silly to you,

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but if you watched
Christine's recitation

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video you saw that to find
the quadratic approximation

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to a product, it's
enough to find

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the quadratic
approximations to each piece

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separately, multiply
together, and then take

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that quadratic approximation.

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So in this case the
quadratic approximations

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are things you
might already know.

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So in particular, we
saw in recit-- sorry,

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in lecture that the quadratic
approximation of e to the x

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is 1 plus x plus
x squared over 2.

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And I'm going to tell you--
if you haven't seen it

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in recitation, there are
a bunch of different ways

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you could work it
out for yourself--

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but the quadratic approximation
for e to the x squared

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is equal to 1 plus x squared.

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All right, so if you
don't believe me,

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by all means work that out
for yourself to check it.

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So in this case,
so that means that

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the quadratic approximation
of f is equal to-- by the rule

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Christine showed you--
it's the, so it's

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equal to the quadratic
approximation

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of the product of the
quadratic approximations.

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So that's 1 plus
x plus x squared

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over 2 times 1 plus x squared.

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Now if you multiply these
two out, that's not hard.

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There are six terms.

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So this is the quadratic
approximation of 1

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plus x plus x squared over 2.

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So multiply 1 through
here and now x squared.

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So that's plus x
squared plus x cubed

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plus-- OK, some fourth degree
term-- x to the fourth over 2.

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And now the quadratic
approximation to this we

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get just by dropping the
cubic and quartic terms.

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So-- and OK, and we can
add these two square terms

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together.

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So this is equal to 1 plus
x plus 3 x squared over 2.

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All right.

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And luckily, doing the same
problem two different ways we

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get the same answer, which
is what should happen.

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So that's that.