WEBVTT

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PROFESSOR: Welcome
back to recitation.

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In this video what
I'd like us to do

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is practice Taylor series.

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So I want us to write
the Taylor series

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for the following
function, f of x

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equals 3 x cubed plus 4 x
squared minus 2x plus 1.

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So why don't you
pause the video,

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take some time to work on
that, and then I'll come back

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and show you what I get.

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All right, welcome back.

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Well, we want to find the Taylor
series for this polynomial

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f of x equals 3 x cubed plus
4 x squared minus 2x plus 1.

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So what I'm going to do is
I'm just going to write down

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Taylor's-- or the expression
we have for the sum,

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for the Taylor series in general
and then I'm going to start

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computing what I need and
I'm going to see what I get.

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So what do I need to remember?

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Well let's remind ourselves
what the formula is.

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We should get f of x is
equal to the sum from n

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equals 0 to infinity of the
nth derivative of f at 0 over n

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factorial times x to the n.

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So that's what we want.

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So what I obviously
need to start doing

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is figuring out
derivatives of f at 0.

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And so what I'm going
to do is I'm just going

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to make myself a little table.

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So let's see, we're going
to say f 0 at 0, f 1 at 0,

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f 2 at 0, f 3 at 0, f 4 at 0.

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And I'm getting
tired of writing,

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so I'm going to stop there.

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OK, so let's take the function--
the zeroth derivative of f

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is just the function itself,
so let's come back here.

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What is the function if I
evaluate it at x equals 0?

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0, 0, 0, 1.

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I get 1.

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All right.

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What is the first derivative?

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So I'm going to write
out the first derivative

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and then I'm going to say I'm
evaluating it at x equals 0.

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So the first derivative looks
like it's 9 x squared plus 8

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x minus 2.

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So I'm gonna write this down.

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9 x squared plus 8 x minus 2.

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Evaluate it at x equals 0.

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0, 0.

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I get negative 2.

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All right, well, let me
take the second derivative.

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OK let's see what I get here.

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I get 18x plus 8.

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And I'm going to evaluate
that at x equals 0.

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This is just a way
to write, I'm going

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to evaluate what's here
at x equal this number,

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so if you haven't
seen that before.

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So I get 8.

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OK and then the third
derivative is 18, oh just 18.

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Evaluate it at x equals 0.

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I get 18.

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And then the fourth derivative.

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What's the derivative
of a constant function?

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It's 0.

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What do you think the fifth
derivative is evaluated at 0?

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Looks like it'll be 0.

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You take the sixth derivative.

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Looks like everything
bigger than 3--

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so the nth derivative at 0 is
equal to 0 for n bigger than 3.

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So it looks like we should
only have 4 terms in this.

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So that maybe seems a little
weird, but let's keep going

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and see what happens.

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Let's start plugging things in.

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So again, let's
remember the formula.

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I'm going to walk
over here to the right

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and I'm going to start
using that formula

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and using these numbers that
I have and writing it out.

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So the first term is going to
be the function evaluated at 0

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divided by 0 factorial times 1.

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0 factorial is 1,
so it's just going

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to be the function
evaluated at 0 times 1.

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The function evaluated
at 0, we said was 1,

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so that's the first term
in the Taylor series.

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OK what's the next term?

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The next term, remember,
is the first derivative

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evaluated at 0 divided
by 1 factorial, which

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is still 1, times x.

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So the first derivative,
if I come back

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over here, evaluated
at 0, I get negative 2.

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So I'm going to get minus 2x.

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The next term, so I had zeroth
derivative, first derivative,

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now I'm at the
second derivative.

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Now it's getting confusing.

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I'm going to start
writing the things above.

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The second derivative
evaluated at 0

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divided by 2 factorial
times x squared.

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That's what I should have here.

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Let's come over here.

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Second derivative
evaluated at 0 was 8.

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So it's going to be 8 over
2, 'cause 2 factorial is 2,

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x squared.

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So it's going to be
plus 4 x squared.

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And then I have to have the
third derivative evaluated

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at 0 divided by 3
factorial times x cubed.

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What's 3 factorial?

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3 factorial is 6.

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What was the third
derivative evaluated at 0?

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It was 18.

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18 divided by 6 is 3.

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So I get plus 3 x cubed.

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And all the other terms were 0,
so I'll just stop writing them.

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OK now if you watched the
video all the way through here,

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at some point maybe you said
"Christine, this is madness."

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Well why is it madness?

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Because what is this?

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Well this is the
function again, right?

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It's exactly what
we started with.

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The order is opposite of
what it was before 'cause now

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the powers go up
instead of down,

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but it's the same polynomial.

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OK we talked about
this briefly, I think,

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when we were doing some
quadratic approximations.

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And I mentioned way back
that quadratic approximations

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of polynomials at x equals 0
are just the polynomials again.

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This is the exact same
kind of thing happening.

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Because what is
the Taylor series?

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It's just better and
better approximations

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as n gets larger and larger.

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So if I wanted to
have a fourth order

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approximation of this
function f of x at x equals 0,

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I would get the
same function back.

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That's really the idea
of what's happening here.

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So maybe you saw the sort
of trick in this question,

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and when you saw this
problem you laughed at me

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and you said, "Well
I'm just going

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to write down the function
again and I'll be done."

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Maybe you didn't
see that right away,

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and if you didn't see
that right away that's OK.

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I bet you're in good company.

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And it's totally fine
because now you've seen this.

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You've seen how it works out.

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And you know, hey, now any
time I see a polynomial

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and I want to do the Taylor
series for this polynomial,

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I just have to write down
the polynomial again.

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So that was the main
goal of this video.

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It took us a long time to get
there, but I think we got it.

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So the answer to the ultimate
answer to the question of

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write the Taylor series
of this function,

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it's just this function again.

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All right, that's
where I'll stop.