JOEL LEWIS: Hi.
Welcome back to recitation.
In lecture you've started learning about quadratic
approximation.
So today we're just going to do a quick example of it.
So I have a question written here on the board.
What is the quadratic approximation of the function
f of x equals e to the x plus x squared-- so here x and plus
x squared is the exponent, so it's e to the quantity x plus
x squared--
near x equals 0.
So why don't you take a minute or two, pause the video, work
this out on your own and then come back and
we can do it together.
All right.
Welcome back.
So there are two different ways we can do this problem.
Let's first just do it the totally straightforward way,
which is that you have this formula for quadratic
approximations in terms of the derivatives of your function.
And so, so we can just apply that formula.
So here, so the formula is that the quadratic
approximation of the function f--
so here, near the point 0--
is equal to f of 0 plus f prime of 0 x plus f double
prime of 0 over 2 times x squared.
All right.
So in order to use this formula we just need to know
what the derivatives of our function are and
their values at 0.
So first we can do the first derivative of f.
So for that it's just a straightforward application of
the chain rule.
Our outer function is e to the x and our inner function is x
plus x squared.
So the derivative, then applying the chain rule is e
to the x plus x squared times 1 plus 2x, which I can also
write as e--
well, yeah, let me just reorder it--
2x plus 1 times e to the x plus x squared.
OK.
So that's the first derivative.
And for the second derivative I can apply just the
product rule here.
Right?
So I've got this, you know--
the second derivative is the derivative of the first
derivative--
so here I have a product.
So f double prime of x is equal to-- well, so we take
the derivative of the first one, which is just--
2 times the second plus the derivative of the second one.
Well, the second one is e to the x plus x squared.
It's actually f of x, so we already computed it once.
So the derivative of the second is 2x plus 1 e to the x
plus x squared times the first. So times
another 2x plus 1.
OK, and if I multiply these two together and combine all
my terms, this is 4x squared plus 4x plus 3 times e to the
x plus x squared.
So these are the first and second derivatives, and what I
need to plug them into my formula is I need
their values at 0.
So I need the function value at 0.
So f of 0, well that's e to the 0 plus 0 squared.
So it's e to the 0, so that's just 1.
f prime at 0, let's see, I go over to my formula for f prime
and I plug in x equals 0, so I have 2 times 0 plus 1, so
that's 1 times e to the 0.
So that's 1 times 1, that's also 1.
And for f double time at 0, I go to my formula for f double
prime and I put in 0.
And so this is 0 and that's 0.
So I have a 3 times 1 so that's 3.
And now I just take these three values and I plug them
right into my formula.
So the quadratic approximation is Q of f equals 1 plus x plus
3x squared over 2.
All right.
Great.
So that was one way to do this problem.
Another way to do this problem is the following.
And in this case, I'm not sure which way is simpler, but in
some cases one way is clearly easier than the other.
So if we illustrate both then you'll have twice as many
tools to work with.
So the other way is to notice--
so the exponential function is nice when you have, you know,
one of your exponential rules--
is that the exponential of a sum is the product of the
exponentials.
So we can rewrite f of x equals e to the x times e to
the x squared.
Now this may seem a little bit silly to you, but if you
watched Christine's recitation video you saw that to find the
quadratic approximation to a product, it's enough to find
the quadratic approximations to each piece separately,
multiply together, and then take that quadratic
approximation.
So in this case the quadratic approximations are things you
might already know.
So in particular, we saw in-- sorry-- in lecture that the
quadratic approximation of e to the x is 1 plus x plus x
squared over 2.
And I'm going to tell you--
if you haven't seen it in recitation, there are a bunch
of different ways you could work it out for yourself--
but the quadratic approximation for e to the x
squared is equal to 1 plus x squared.
All right, so if you don't believe me, by all means work
that out for yourself to check it.
So in this case, so that means that the quadratic
approximation of f is equal to--
by the rule Christine showed you--
it's the, so it's equal to the quadratic approximation of the
product of the quadratic approximations.
So that's 1 plus x plus x squared over 2
times 1 plus x squared.
Now if you multiply these two out, that's not hard.
There are six terms. So this is the quadratic approximation
of 1 plus x plus x squared over 2.
So multiply 1 through here and now x squared.
So that's plus x squared plus x cubed plus--
OK, some fourth degree term-- x to the fourth over 2.
And now the quadratic approximation to this we get
just by dropping the cubic and quartic terms. So, and OK, and
we can add these two square terms together So this is
equal to 1 plus x plus 3x squared over 2.
All right.
And luckily, doing the same problem two different ways we
get the same answer, which is what should happen.
So that's that.