GILBERT STRANG: OK, I
thought I would talk today about power series. These are powers of x. I'm going to keep going. All powers, all those x to the
fourth, x to the fifth, they'll all come in too. And my idea is combine them,
add them up to get a function of x. So we're doing calculus, but a
new part of it, with these infinite series. So what do I mean
by combine them? I mean I'll multiply those
powers by some numbers. Let me call those numbers a0. So this first guy will be
an a0, and then I'll add on an a1 x. I'm out of x. I'll add on some a2 x squared,
some a3 x cubed and onwards. So now I have a function. And that function, let
me call it f of x. So here's my starting
plan here. Well, we've seen this
for e to the x. Let me remember how e to the x
could come, the series for that particular function. So here's the plan. I'm going to choose those
a's so as to match-- let me put these words down. I'll match at x equals 0. The function, its derivative,
its next derivative, its third derivative, and onwards. Each a, like a3, will be chosen
so that that right-hand side has the correct derivative,
third derivative, at x equals 0. So this Taylor series-- Taylor's name is associated
with series like this-- everything's happening
at x equals 0. So in the case of e
to the x, all its derivatives were the same. Still e to the x. And they all equal
1 at x equals 0. So I want that function
to give me 1 for every derivative. That doesn't mean that the
a's should all be 1. Why not? Because when I take the
derivative, for example, of this guy, that x cubed,
the first derivative, will be 3x squared. The next derivative 6x. The next derivative 6. That's the one I want. That third derivative, but it'll
be 6 so a3 will have to be 1/6 to give me the
correct answer 1. Let me write those
things down. So what we just did is the
derivative of x to the n-th. The n-th derivative. What's the n-th derivative
of x to the n? We get to use our nice formula
for derivatives. So the first derivative is
nx to the n minus 1. The derivative of that will be
n times n minus 1 x to one lower power. Keep going, do it n times,
and what have you got? You finally got down to the 0-th
power of x, a constant. But what is that constant? It's n times n minus 1, so that
n-th derivative will be n from the first, n minus
1 from the second. Keep multiply until you
finally get down to 1. And of course, that's called
because it comes up often enough to have its
own special name. That name is n factorial. And it's written n with
an exclamation mark. So that's n factorial and that's
the n-th derivative of x to the n. So for the particular function
e to the x, if I worked out its series, all the derivatives
I'm trying to get are all 1's. But what the powers of x gives
me these n factorials. The a's had better divide
by the n factorial. So let me recall the series for
e to the x, and then go onto new functions. That's the point
of my lecture. So we're getting e to the x in
a slightly different way from the original way, but
this is a good way. e to the x at x equals 0 is 1. At the first derivative of e to
the x is 1, so I divide by 1 factorial. That's 1. But here I have to divide by 2
because the second derivative is 2 and I want those
to cancel. And here I divide by-- what do I divide by? 6 because the third
derivative is 6. And a typical term is I have
to divide by n factorial because when I take n
derivatives I get n factorial. The n-th derivative of
that thing we just worked out is n factorial. So I divide by n factorial and
I've got the derivative to come out 1. And that's correct
for e to the x. So that's the plan, matching
derivatives at x equals 0 by each power of x. And now I'm ready for
a new function. And a nice choice is sine x. So now on this board, if I can
come here, I'm going take a different function. No longer e to the x. My function is going
to be sine x. Well, I better figure out
all its derivatives. And they're nice, of course. Sine x, its derivative. Can I just list them all? These are the things that
I have to match. I'll plug in x equals 0. But let me first find
the derivatives. The derivative of
sine is cosine. The derivative of cosine
is minus the sine. The derivative of minus the
sine is minus the cosine. And then I'm back to sine again,
and repeating forever. That's a list of the derivatives
of sine x. This is my f of x here. This guy, first one. OK, now I plug in x equals
0 because I want all the derivatives at 0. The whole series is being
built focused on that point x equals 0. So at x equals 0, that's
easy to plug in. The sine is 0, the cosine is
1, the minus the sine is 0. Minus the cosine is minus 1. The sine is 0. And repeat. 0, 1, 0, minus 1 forever. OK, so I know the derivatives
that I have to match. Now can I construct the power
series that matches that? OK, so that power series will
give me sine x, and what will it have? It starts with 0. The constant term is 0 because
the sine of 0 when x is 0-- of course, we want to get
the answer is 0. Then, the next term, the x,
its coefficient is 1. 1x. No x squared's in sine x. No x squared's. But now minus. Do I have minus 1x cubed? Not quite. Minus x cubed, but I have to
divide by 6 because when I take that three derivatives,
it will produce 6. So I have to divide by 6,
which is 3 factorial. That's really the number
that's there. 3 times 2 times 1. 6. Now the fourth degree
term, the x to the fourth is not there. x to the fifth is going to
come in with a plus. So there's a plus
from this guy. This is x to the 0,
1, 2, 3, 4, 5. x to the fifth. And now what do I
divide by now? 5 factorial. 120. And then minus and so on. Minus an x to the seventh
over 7 factorial. We have created the power series
around 0, focused on 0. And let me remove that 1 because
just waste of space. x minus x cubed. All odd powers and
that's because sine x is an odd function. If I change from x to minus x,
everything will change sign. What would happen if I plugged
in x equal pi? Suppose I took x equal pi in
this formula for sine x. This infinite formula,
keeps going forever. Well, I would get pi minus pi
cubed over 6 plus pi to the fifth over 120. It would look ridiculous. But you and I know that the
answer would have to come out. The correct sine of pi? 0. I don't plan to do it,
but it has to work. OK, so that's the sine. That's the sine. And it's an odd series. Now OK, good example. Its twin has got to
show up here. The cosine. What's the series
for the cosine? These are the two series
that are worth knowing. You notice here that slope of
1, the big deal about the slope of sine x at x equals
0, the slope is 1. And that does have
a slope of 1. OK, what about the cosine? Well, now I have to plug in. All right, the cosine is
going to start here. Cosine minus sine
minus cosine. Now my f of x is going
to be cosine x. And I need its derivatives. I'm going to have three lines
again that are going to look just like these three lines. But they'll be for the cosine. So they start with a cosine. Its derivative is
minus the sine. Its derivative is minus
the cosine. Its derivative is what? Plus sine and then cosine, and
forever, minus the sine. And let me plug in now
at x equals 0. This is our system. Find the derivatives,
plug in 0. So find the derivative at 0. Well, the function itself,
the 0-th derivative is 1. The first derivative is 0. The second derivative
is minus 1. The third is 0. The fourth derivative is
plus 1, 0, and so on. It's the same line as we have,
but just starting over by 1. Starting with the cosine. I know what derivatives I
want, now I just have to create my series for cosine x,
which matches these numbers. One more time. Just match those numbers with
the coefficients that I originally called a0, a1, a2,
a3, but now we have numbers. OK, at x equals 0. So how does this series start? At x equals 0, the
cosine of 0 is 1. It starts with a 1. That's the constant term
sitting there. The coefficient of x, the
linear term is 0. Because the cosine has
0 slope at the start. Then we come to something
that shows up. Minus. This will be-- now what
are we in to? This is the constant, the
first power is gone. The second power minus
x squared. But you know if I'm looking to
match the second derivative to make it b minus 1. Right now it's minus 2. Differentiating would give
me a 2x and a 2. So I have to divide by that
2 or 2 factorial. Now it's good. Now it matches the correct
second derivative minus 1. Then there's no third
derivative. The fourth derivative
is plus 1 x to the fourth over 4 factorial. And then minus and so on, x to
the sixth over 6 factorial. All even powers, so this
is an even powers. The 0-th, second, fourth,
sixth power. So it's an even function. That means that the cosine of
minus x is exactly the same as the cosine of x. We get a nice little insight
on these two special groups for which the sine is the
perfect example of an odd function and the cosine
is the perfect example of an even function. Well, there's so much here. What happens if I cut
the series off? I just want to look at those
first terms to see exactly what they represent. Suppose I stop here after
the linear term. What do I have? What is that x just by itself? It's really 0 plus x because
there was a 0 from the constant term. That is the linear
approximation. That gives me the equation
of the tangent line, y equals x, slope 1. More interesting, cut
this one off. Cut this one off here. That's a very important
estimate. It's not the exact cosine
because the exact cosine has got all these later guys. But don't forget and I should
have said this from the very beginning, these n factorials
grow fast. And all the series that we're talking about,
because those n factorials grow so fast and I'm dividing by
them, I can take any x and I get a reasonable number. If I take x equaled pi, that's
this sine series gave me 0. What do I get if I plug
in x equals pi in the cosine series? So the cosine series, if I
plugged in x equals pi and had patience to go pretty far, my
numbers would be getting near the cosine of pi. Which would be minus 1. I don't see minus
1 coming out. Here is 1, minus 1/2
of pi squared. I don't know, that's around-- 1/2 pi squared might be
around 5 or something. But they knock each other off. They get very small and we
get the answer minus 1. OK, so those are two important
series and now I get to tell you about Euler's
great formula. It connects these three series
that you've seen. But to make that connection
I have to bring in the imaginary number i. Is that OK? Just imagine a number i. And everybody knows what you're
supposed to imagine. You're supposed to imagine that
i squared is minus 1. And we all know there
is no real number. The square of a real number is
always going to be greater or equal to 0. So let's just create a symbol
i with a rule, with the understanding that any time we
see i squared, I'm entitled to write minus 1. OK. So now, what is Euler's
great formula? Euler's great formula, his
brilliant insight was make x in this e to the x series,
make x imaginary. Change x to ix. So make it an imaginary
number. So can I just take Euler's, take
Taylor's series, or oh, maybe Euler's out of this
too, because that letter e is his initial. Probably he did. So I guess that's why he found
this lovely connection. So if I take e to the ix and
instead of x in this series I put in ix, just go for it. Let x be imaginary. OK, can I write out the
series 1 plus-- instead of x I have ix. And then I have 1 over 2
factorial ix squared. And then I have 1 over
3 factorial ix cubes. And 1 over 4 factorial
ix to the fourth. That's e to the ix. OK, you say, you just
changed x to ix. That's all I did. Now, here's the point. Now I'm going to look at this
mess and I'm going to separate out the part that is real from
the part that's imaginary. I'm going to separate it into
its real part and its imaginary part. So what is real in this thing? I see one is certainly
a real number. Do you see the other one, the
next one that's real? It comes from this i squared. That i squared I can replace
by minus 1, perfectly real. So it's minus from the i squared
1 over 2 factorial. x squared is still there. The i squared was minus 1. That's all. And then would come something
from the i to the fourth. Because what is i
to the fourth? It's i squared squared
minus 1 squared. We'd be back to plus 1. So plus sign. Good. Now comes the part that has an
i in it and a single i I have to live with. So that i is multiplied by x. Now I have i cubed. How do I deal with i cubed? i cubed is i squared
minus 1 times i. i squared times i is minus i. So I have a minus i. 1 over 3 factorial and the
x cubed and so on. Do you see what we have? Do you see what this real
part of e to the ix is? It's the cosine. Right there, same thing. So I'm getting cosine x for the
real part and then i times this series. And you can see what
that series is. It's the sine series, x minus
1/6 x cubed plus 1/20 of x to the fifth sine x. There is Euler's great formula
that e to the ix-- oh, I better write it
on a fresh board. Maybe I'll just write
it over here. I'm going to copy from this
board my Euler's great formula that e to the ix comes out to
have a real part cos x. Imaginary part gives
me the i sine x. And I'll write that down. Now let me work here. e to the ix is cos
x plus i sine x. And I want to draw a picture. OK, here's a picture. Actually, Euler often wrote his
formula, or we often write his formula because we're taking
cosines and sines. Somehow x isn't such-- those are angles. So it's more natural
to write-- Now that we've showing up with
sines and cosines, it's more natural to write a symbol
that we think of as an angle like theta. So you would more often
see it this way. I'm just changing letters from
x to theta as a way of remembering that
it's an angle. And now I'll draw it. So I have to draw that thing. OK, this is the real direction
and that's the imaginary direction. I just go that's
the real part. I go cos theta across here. So let that be cos theta. And then I go upwards in the
imaginary up or down. So across is the real part,
up/down is the imaginary part. Say sine theta I go up. That height is sine theta
and that angle is theta. Fantastic. That's a picture of
Euler's formula. Well, that's the best
way to see it is that beautiful statement. And this is a picture
to remind us. We would say that's the complex
plane because points have two parts, a real part
and an imaginary part. Nothing so complex about that. Now, before I stop, we've done
three important series. Can I mention two more, just two
more out of a long list of possibilities? One is the most important
series of all, where the coefficients are all 1's. So the coefficients
are all 1's. That's called the geometric
series. Let me write its name here. That's a Taylor series. That's a power series. And the function it comes
from happens to be 1 over 1 minus x. That's the function. And you will see why, if you
multiply both sides by 1 minus x, I'll get 1 here. If you watch, everything will
cancel except the 1. So that's it. Now, there's a significant
difference between that series and e to the x. The biggest difference is
we're not dividing by n factorial anymore. And as a result, these terms
don't get necessarily smaller and smaller and smaller. Unless x is below 1. So we're OK for x below 1. And x could be negative. I can even say absolute value of
x below 1, then these terms gets smaller. But at x equals 1 we're dead. At x equals 1 I have 1
plus 1 plus 1 plus 1. All 1's. I'm getting infinity. And on the left side I'm
getting infinity also. At x equals 1 blows up. OK, one more series,
then we're done. One more. It's a neat one because it
brings in the logarithm. How am I going to get it? I'm going to start with this
series, which is the big one, the geometric series. And I'm going to take the
integral of every term. So if I integrate 1 I get x. If I integrate x I get
x squared over 2. If I integrate x squared
I get x cube over 3. x fourth over 4 and so on. Not 3 factorial, just 3. And if I integrate this, well,
let me put the answer down and then we can take its derivative
and say, yep, it does give that. So the answer is minus. This minus sign shows up
here as a minus the logarithm of 1 minus x. Because if I take the derivative
of that the logarithm always puts this
inside function down to the bottom, and then the derivative
of the inside function, the chain rule brings
out a minus 1, and the minus 1's go away,
and beautiful. So just have a look at that
series then for the logarithm. The logarithm of 1 minus x. Again, we're matching
at x equals 0. At x equals 0, this
function is OK. In fact, at x equals 0,
what is that function? Logarithm of 1, which is 0, and
there's no constant term. Good. OK, what comments to make about
this final example? This one was OK for
x smaller than 1. But then it died
at x equals 1. This one, well, it's getting
a little help dividing by 3 and 4 and 5. But that's puny help. That's no way compared to
dividing by 3 factorial, 4 factorial, and so on, which
will really help. So actually, this series
is also only OK out to x equals 1. At x equals 1, it fails again. At x equals 1, what do I have? When x is 1, I have the log
of 0 minus infinity. I've got infinity
at x equals 1. At x equals 1, this is
1 plus 1/2 plus 1/3 plus 1/4 plus 1/5. Getting smaller, but
not very fast and adding up to infinity. So there's a whole discussion. We could spend hours on that
famous series, 1 plus 1/2 plus 1/3 plus a quarter and other
series of numbers. I wanted to do calculus,
derivatives, integrals, so I took functions and series of
powers, not series of numbers to illustrate this. OK, good. Thanks. ANNOUNCER: This has been
a production of MIT OpenCourseWare and
Gilbert Strang. Funding for this video was
provided by the Lord Foundation. To help OCW continue to provide
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