Shankar: Ok, so today, it’s again a brand
new topic, so if you like relativity you would be in grief
and mourning. If it got to be really nasty,
you’ll be relieved that that’s behind us now.
And let me remind you one more time that the problem set for
relativity was quite challenging because I had to make up most of
the problems. And they’re not from the book,
so they’re a little more difficult.
The exam will be much more lenient compared to the
homework. That’s generally true,
but particularly true in this case.
Okay. So, today I’m going to
introduce you to some mathematical tricks.
As you’ve probably noticed by now, a lot of physics has to do
with mathematics, and if you’re not good in one,
you’re not going to be good in the other.
And I just thought I would spend some time introducing you
to some tricks of the trade, and then we’ll start using
them. First important trick is to
know what’s called a Taylor series [laughter].Okay,
I see this is greeted by boos and hissing sounds.
I’m not sure what your past experience with Taylor series
was, and why it leaves you so scarred and unhappy.
My experience with Taylor series has been positively
positive, and I don’t think I can carry on any of my things
without knowledge of Taylor series.
But I’ll tell it to you, Taylor series,
as we guys in the Physics Department tend to use it.
The philosophy of the Taylor series is the following:
That if some function f(x)–and I’m going to
draw that function, something like this–but I’m
going to imagine that you don’t have access to the whole
function. You cannot see the whole thing.
You can only zero-in on a tiny region around here.
And the question is, as you try to build a good
approximation to the function, how will you set it up so that
you can write an approximation for this whole function,
valid near the point x=0, where you know something.
So, suppose I block out the whole function–you’ve got to
imagine mentally. I don’t show you anything
except what’s happening here. In other words,
I show you only f(0). That’s all you’re told.
The value of the function is 92. What should we do away from
x=0? Well, you’ve got no information
about this function, you don’t know if it’s going
up, if it’s going down. All I tell you is one number,
what it’s doing now. It’s clear the best
approximation you can make is a flat line.
There’s no reason to tilt it one way or the other,
given the information you have. So, the first approximation of
the function you will say is, f(x) equal to that
[f(0)]. It’s not equal to;
you can put some symbol that means approximately equal to.
It’s like saying, “Temperature today is 92,
now what’s it going to be tomorrow?”
Well, if you don’t know anything else,
you cannot tell what it’s going to be tomorrow.
But if you know that this is fall, that the temperature’s
always going down, and somebody tells you,
“I know the rate of change of temperature.
I know that as of today, the rate of change of
temperature is something.” Then you can use that
information to make a prediction on what it will be tomorrow.
That is to say, for values of x not
equal to 0. And let’s denote by this
following symbol: f prime of 0 is the
derivative, which is df/dx at x=0.
And you’ll multiply it by x, and that’s your best
bet for what the function is, away from x=0.
What that does is to approximate the function by a
straight line with the right intercept and the right slope.
By “right” I mean, matches what you know about the
exact function. And if it turns out the
function really was a straight line, you are done.
It’s not even an approximation, it’ll track the function all
the way to eternity. But it can happen,
of course, that the function decides to curve upwards,
as I’ve shown in this example, and this will not work if you
go too far. For a while,
you’ll be tangent to the function, but then it’ll bend
away from you. So, it’s really good for a very
small x and you can say, “Well, I want to do a little
better when I go further out.” Well, what this doesn’t tell
you, this approximation, is that the rate of change
itself has a rate of change. This assumes the rate of change
is a fixed rate of change, which is the rate of change at
the origin. And the rate of change of the
rate of change is the second derivative.
So, if somebody told you, “Look, here’s one more piece of
information.” I know the second derivative of
the function, which I’m going to write as
f double prime at 0. Well, with that information,
you can build a better approximation to the function,
and you will put that in x^(2).
But the key thing is you’ve got to divide by 2,
but if you like 2 factorial. And the way you divide by the
2, because your goal is take this approximation and make sure
that that’s whatever you know about the function built into
Let’s check this approximation and see if it satisfies those
properties. First of all,
at x=0, let’s compare the two sides.
At x=0, the left hand side is
f(0). On the right hand side,
when you put x=0, you kill this and you kill this
and it matches. So, you’ve certainly got the
right value of the function. Then you say,
“What if I take the derivative of the function at x=
0?” Let’s take the derivative of
this trial function, or the approximate function on
both sides. When I take the derivative,
that being a constant, does not contribute.
This derivative of x is 1.
In the next one, the derivative of x^(2)
is 2x, 2x cancels the 2,
so I get f double prime of 0 times x.
But now, evaluate the derivative at x=0.
That gets rid of this guy, and the derivative of my test
function matches the derivative of the actual function,
because this is the actual derivative of the actual
function. Now, it follows that if you
want to say, “How about the second derivative of this
function?” let’s take the test function I
have, or the approximation. Take a second derivative,
and make sure that comes out right.
I don’t feel like writing this out, but try to do this in your
head. Take two derivatives on the
left-hand side and see what happens on the right hand side.
Well, if you take one derivative, this guy’s gone.
If you take two derivatives, this guy’s gone.
And then, if you take the first derivative, you get 2x,
you take the second derivative you get 2, the 2 cancels the 2.
Now, if you put x=0. If you do, you find that
f(0) here matches the second derivative of the
function on the left-hand side. So, this 2 factorial is there
to make sure that the function you have cooked up has the right
value of the function at the origin,
has the right slope at the origin, has the right rate of
change of the slope at the origin.
Well, it’s very clear what we should do.
If you had a whole bunch of derivatives, then the
approximation we will write will look like, f n-th
derivative. I cannot draw n primes
there, so n means n such primes,
x to the n over n factorial.
And you go as far as you can. If you know 13 derivatives,
put 13 derivatives here. That’s an approximation.
That’s not called a Taylor series, that’s the approximation
to the function. Now, it may happen that
sometimes you hit the jackpot, and you know all the
derivatives. Someone tells you all the
derivatives of the function, then why stop?
Add them all up. Then, you will get the sum
going from the 0 derivative all the way up to infinity.
And if you do the summation of every value of x,
you will get a value, and if that summation is
meaningful and gives you a finite number,
that, in fact, is exactly the function you
were given. That is the Taylor series.
The Taylor series is a series of infinite number of terms
which, if you sum up — and sum up to something sensible — will
actually be as good as the left-hand side.
So, let me give you one example. Here’s a famous example.
1 over 1 minus x is the real function.
You and I know this function, we know how to put it in a
calculator; we know how to plot it.
You give me an x and I stick it in the denominator,
subtract it from 1, invert it;
that’s the function. But instead,
suppose this function was revealed to us in stages.
We were told f(0)–what is f(0) here?
Put the x equal to 0, f(0) is 1.
Now, let’s take the derivative of this function,
df/dx. df/dx has a minus 1
because it’s 1 minus x to the minus 1, times a square;
then the derivative of what’s inside this, that gives you a
minus 1. This is simple calculus,
how to take the derivative of this fellow here.
Having taken the derivative, evaluate it at x=0,
this vanishes, that becomes 1,
and you get 1. If you now take the second
derivative of this function, which I don’t feel like doing,
and I evaluate that at x=0, then I’ll find it is equal
to 2 factorial. In fact, I’ll find the
nth derivative at the origin to be n factorial.
Well, that’s very nice. Because then,
my approximation to the function begins as 1 plus
x times the first derivative,
which happens to be 1 plus x^(2) over 2 factorial
times the second derivative, which happens to be 2 factorial
plus x^(3) over 3 factorial times 3 factorial,
so you can see where this guy is going.
It looks like 1 + x + x^(2), et cetera.
The Taylor series is its infinite sum.
In practice, you may be happy to just keep a
couple of terms. So, let’s get a feeling for
what those couple of terms can do for us.
So, let me take x=0.1. x equal to point 1;
the real answer is 1 over 1 minus point 1,
which is 1 over point 9, which is 1.1111,
etcetera. That’s the target.
What do you do with the series? The series starts at 1,
times one-tenth, plus 1 over 100,
plus 1 over 1000, and so on.
And you can see, as I keep more and more terms,
I keep just filling up these ones.
If you stop at 1 over 1000, you stop right there.
So, this is the exact answer, this is the approximation.
But it’s clear to you, perhaps in a simple example,
that if you kept all the terms of the series,
you really will get this infinite number of recurring
1’s. So, that is the function.
That is the Taylor approximation.
We’ll just chop it off whenever you want, and it’s a good
approximation, and that’s the Taylor series.
The series means sum everything. Now, summing an infinite number
of numbers is a delicate issue. I don’t want to go there at
all, I discuss that in this math book.
But sometimes a sum makes no sense, then you’ve got to quit.
For example, put x=2. The correct function is 1 – 2,
which is -1. Our approximation for x
=2 looks like 1 + 2 + 4 + 8. Fist of all,
this sum is going to grow to infinity, because the numbers
are getting bigger and bigger. This sum seems to be all
positive; that is the correct answer is
negative. Obviously, the series doesn’t
work. So, the next lesson of our
Taylor series is, you can write down the series,
but it may not sum up to anything sensible beyond a
certain range. So, if you’re doing a Taylor
series at x=0, and you go to x=2,
it just doesn’t work. So, you can ask,
“How far can I go from the origin?”
Well, in this simple example, we know that x=1,
the function just going to infinity, that’s why you
couldn’t go there. And you cannot go to the
right-hand side of that point. The function is well defined on
the other side, but this series,
this knowledge of the function here, is not enough to get you
on the other side. So, this is a case where there
are obvious problems at x=1.
But if I wrote a function like 1 over 1 plus x^(2),
that’s a nice function, got no troubles anywhere.
And yet, if you took the Taylor series for it,
you will find if you go beyond absolute value at x=1,
the series makes no sense. So, I don’t want to do that
mathematical theory of series. I just want to tell you that
functions can be approximated by series.
And if you’re lucky, you can do the whole sum if you
know all the derivatives, and the whole sum may converge
to give a finite answer. In which case,
it’s as good as the function. One guy can use 1over 1 minus
x; the other one can use the
infinite series, and they’re morally and
mathematically in every sense equal,
as long as they don’t stray outside the region of validity
of the infinite series. Ok, so the most popular example
that I’ve been using in the class, remember,
is 1 + x^(n). That’s a function we can do
with Taylor series. f(0) is 1.
What’s the derivative of the function?
It’s n times 1 plus x^(n) minus 1.
The x is n times 1 plus x^(n) minus 1;
if I evaluate it at x=0, it is just n.
That’s why we get this famous result we’ve been using all the
time. If x is small enough you
stop there, because the next time it’s going to involve an
x^(2) and an x^(3),
then if x is tiny, we have no respect for
x^(2) and x^(3), we just cut it off.
But if you want the next term–one of you guys even asked
me, “What happens next?” we’ll take the second
derivative. You can see it’s n times n
minus 1, times 1 plus x^(n) minus 2.
If you put x=0, you’ll get n times
n minus 1, x^(2),
and don’t forget the 2 factorial.
So, what I’ve been doing is I’ve been saying 1 plus
x^(n), approximately equal to this.
But even if you keep that, it’s still approximate.
But it’s a good approximation; how many terms you want to keep
depends on how tiny x is. Ok, so it’s good to know
there’s an infinite series, but it’s also good to know you
can chop it off and do business with a few terms.
In fact, all of relativity we reduce as follows.
The energy of a particle is this, which we can write as
mc^(2) times 1 minus v^(2) over c^(2)
to the minus ½. If you expand this in a power
series, you’re going to get mc^(2) + ½
mv^(2), plus stuff involving v
to the 4th over c to the 4th times c^(2).
We dropped all this, and for 300 years we did
mechanics keeping just this term.
We just kept that first non-trivial term,
and all of our collisions and so on that we did,
used only that up to that point.
So, the approximations have really been useful,
and they describe nature approximately.
If you say, “Well, I want to be exact,” you can go
back and use this. Unfortunately,
generally somebody tells you, “That’s not exactly.
There in quantum mechanics, tells you the whole thing is
wrong.” At every stage,
you will have to give up something.
So, I have a lot of respect for approximations.
If you could not describe the world approximately,
you couldn’t have come where you’ve come.
Because no one knows the exact answer to a single question you
can pose. Ask me a question,
and the answer is, “I don’t know.”
Second part of the answer is, “Nobody knows.”
Because if your question says give an answer to arbitrary
precision, any question asked, we just don’t know.
Newtonian mechanics works for small velocities.
All of relativistic mechanics works for any velocity,
but not for really tiny objects.
Then you’ve got to use quantum mechanics.
So, always theories give way to new theories,
and so approximations are very important.
So, you have to learn this. Okay.
So now, I’m going to consider the following function:
e^(x). Now ,this guy is something you
all know and love, because it’s one nice thing
about the function, is every derivative is
Every child knows e^(x) has got e^(x) as its
derivative. Why do we like that here?
That means all the derivatives are sitting in front of you.
They’re e^(x), and at x=0,
e to the 0 is 1, so every derivative is 1.
And the function is very easy to write down.
It is really 1 + x + x^(2) over 2 factorial,
plus x to the 3 over 3 factorial, x to the 4
over 4 factorial. You can go on like this,
and if you like to use a compact notation,
it’s x^(n) over n factorial,
and going from 0 to infinity, that 0 factorial is defined to
be 1. That’s e^(x).
If you ever forget the value of e, and I need to know the
value of e, because when I lock my suitcase
and check it into the airport, I use either e or
π, because they’re the only two
numbers I can remember–So, if I forget the value of
e, I just say e is the same
as e to the 1, so I’ve got 1 + 1 + 1 + 2 + 1
over 6, and pretty soon I’m telling the
guy scanning my luggage, “That’s my code.”
And it’s much better than any other code because I cannot
reconstruct, say, my grandfather’s name.
I may not get it right. I may not construct my house
address, or my phone address. This doesn’t change.
I’ve been moving around all over the world,
this is a very reliable number. Now, π is a good
number, but rules for computing π are somewhat more
difficult. π also can be computed
as an infinite series, but this guy is very easy.
There’s 2 point in 7-7 something, that is e.
Okay. Now, here is the very nice
property of this series. It is good for any x.
You remember the series for 1 over 1 minus x crashed
and burned at x=1. This series — always good.
You put x=37 million, you’ve got 37 million,
37 million cubed, 37 million at the 4th power;
don’t worry. These factorials downstairs
will tame it down and make it converge, and will give you
e to the whatever number I give.
That’s something I’m not proving, but the series for
e^(x), no limits on validity.
All right. Then, we take the function,
cos x. Now, cos x,
we all know and love as this guy.
That’s got a period of 2π.
But we can write a series for it because it’s a function.
And what do I need to know to write the series for it?
I need to know the value of the function at the origin.
So, we know cosine of 0 is 1. If you take the derivative you
get minus sine, and its value at 0 is 0.
You take one more derivative, you get minus cosine 0,
which is minus 1, and I think by now you get the
point. Every other derivative will
vanish, and the remaining derivatives will alternate in
sine from 1 to minus 1. That means, if you crank it
out, you will get 1, you won’t have anything linear
in x because the coefficient of that is
sin 0 which vanishes; then you’ll get x^(2)
over 2 factorial plus x to the 4 over 4 factorial,
and so on. This is cos x.
If you cut it off after some number of terms,
it’s not going to work. In the beginning,
1 minus x^(2) looks very good.
This is 1 minus x^(2) over 2, looks like this.
But eventually, this approximation will go bad
on you, because x to the 4th is just x to the 4th.
It’ll grow without limit. No one’s telling you this
approximation has a property that cosine is less than 1.
It doesn’t satisfy that, because you’re not supposed to
use it that far out. But if you keep all the terms,
then in fact, even though x to the 4th
is taking off, x to the minus 6 is
saying hey, come down, x to the 8th is saying,
go up. You add them all together,
remarkably, you will reproduce this function.
Very hard to imagine that the cosine that we know as
oscillatory is really this function.
And I hope all of you know enough to know how to find the
series. You should know the derivative
of cosine is minus sine, the derivative of sine is
cosine. And evaluate them at the origin;
I say you can build up the series.
Then, if you do similar tricks for sin x,
sin of 0 is 0, derivative of sin is cosine,
and of course, n of 0 is 1.
So, the first derivative is 1; that gives you this term.
Next one won’t give you anything, the one after that
will give you this, and this guy will go on like
this one. Okay.
So, these are series for e^(x),
cos x, and sin x,
good for all x. So, there’s no restriction on
this. One of the really interesting
things to do when you are bored, or stranded somewhere in an
airport, is to take cosine squared plus sine squared.
Squared the whole thing, add the squared of the whole
thing and add them up. You’ll find,
by miracle, you will get the one from squaring the one and
every other power of x will keep on vanishing.
Of course, you’ll have to collect powers of x by
expanding the bracket. But it will all cancel.
That’s what I meant by saying this series is as good as this
function. It will obey all the identities
Now, for the punch line. I think everyone–Some of you
may know what the punch line is, the punch line is the
following. Let us introduce,
without any preamble right now, the number i,
which is the squared root of minus 1.
All I expect of i is that when I squared it,
I get minus 1. If I cube it,
i^(3) is i^(2) times i,
so that’s minus i, and i to the 4th is back
to plus 1 because i to the 4th is i^(2) times
i^(2), and i^(2) is minus 1.
So, if you take powers of i, it’ll keep jumping
between these values. It’ll be i,
it’ll become minus 1, it’ll become minus i,
then it’ll become plus 1. That’s the set of possible
values for powers of i. It’ll be one of these four
Then we are told, if you now consider the
following rather strange object, e^(ix).
Now, e^(ix), you have to agree,
is really bizarre. e is some number,
you want to raise it to 2, that’s fine.
Number times number. i is a strange number,
right? It’s the squared root of minus
1, and you wanted me to raise e to a complex power.
What does that even mean? Multiply e by itself
ix times? Well, that definition of powers
is no good. But the series for e^(x)
defines it for all x, then we boldly define
e^(x) for even complex values of x as simply
this thing too, with ix plugged in place
of x. And by that fashion,
we define exponential. So, the exponential function is
simply defined as, suppose you write e
raised to dog. e raised to dog
is 1 plus dog plus dog^(2) over 2.
e raised to anything is a code for this series.
And now we raise e to various things.
Real numbers, complex numbers,
matrices, whatever you want. Your pet, you can put that up.
Of course, you’ve got to be a little careful.
You cannot raise e raised to dog,
because the units don’t match. But there’s a dog here
and dog^(2), so you should divide it by some
standard dog, like President’s dog
[laughter]. Take some standard.
Napoleon’s dog, divide it, then you’ve got
something dimensionless [laughter].
Then it will work. So, as long as it’s some
dimensionless object, this is what we mean by
e raised to that. That’s a fantastic leap of
imagination. How to raise e to
whatever power you like; take the series.
Let’s do the series. So, I’m going to get 1 plus
ix, plus (ix)^(2) over 2 factorial,
plus (ix)^(3) over 3 factorial, and I’m going to stop
after this last term, (ix)^(4) over 4
factorial. Keep going, then what do you
get? Let’s take the 1 from here,
leave this alone for a minute. You write i times
x, go here, i^(2) is minus 1,
so I get minus x^(2) over 2 factorial,
then I come here. I cubed is minus i,
so I write here, minus x^(3) over 3
factorial. Then, I go back here and write
x to the 4th over 4 factorial.
By now, everybody knows what’s going on.
Everybody knows that this is just cos x + i sin x. Okay, this is a super duper
formula. Life cannot go on without this.
You want to draw a box around this one, draw it.
Because without this formula, we cannot do so many things we
do. It says that the trigonometric
functions and the exponential functions are actually very
intimately connected. And if they’re all defined by
the power series, you’re able to prove it.
So, this is the formula proven by Euler.
And it’s considered a very beautiful formula,
and here is the particularly beautiful special case of this
formula. If you put x=π,
I get the result e^(iπ) equal the cos π,
which is minus 1, plus i sin π,
which is 0. I get the result e^(iπ)
+ 1=0. Now, everybody agrees this has
got to be one of the most beautiful formulas you can
imagine involving numbers or in mathematics.
Why is it such a great formula? Look at this formula here.
π, defined from the Egyptian times,
is a ratio of circle to diameter.
i is square root to minus 1.
1 is the basis for all integers. e is the basis for the
logarithm. Here is a formula in which all
the key numbers appear in one single compact formula.
The fact that these numbers would have a relationship at all
is staggering, and this is the nature of the
relationship. This is voted the formula most
likely to be remembered, this formula.
Now, we are going to use not only the special case,
but we are going to use this all the time.
So, how many people have seen this?
Okay. If you have not seen this,
then you’ve got more work than people who have seen this.
You’ve got to go and you’ve got to do all the intermediate steps
so you get used to this thing. Now, let’s do the following two
other variations, and then I’ll move on.
If I change x to minus x, then I get
e^(-ix). Cosine of minus x is
cosine x, and sin of minus x is
minus sin x. So, I can get that relation. Then, I’m going to combine
those two to do the following: I’m going to say that cos
x=e^(ix) + e^(-ix) over 2,
you can check that. And sin x=
e^(ix) – e^(-ix) over 2i.
You just add and subtract these two formulas and you get that. Okay.
So, what does this mean? It means that trigonometric
functions, you don’t need them. If you’ve got exponential
functions, you can manufacture trigonometric functions out of
them, provided you’re not afraid to
go to exponents with complex number i in them.
And all the identities about sines and cosines will follow
from this. For example,
if you take cosine squared plus sine squared,
you’re supposed to get 1. Well, you can square the right
hand side, you can square the left hand side,
and you will get 1. To get the 1,
you better remember the following: e^(ix) times
e^(-ix) is what? You know how you come by an
exponent? You raise it to one power times
the same thing raised to another power, is e raised to ix minus
x, which is e raised to 0, which is 1.
So, when you raise the number to a power, you multiply it by
the number times a different power, the product is a number
to the sum of the two powers. Powers combine.
2 cubed times 2 to the 4th is 2 to the 6th.
That’s true for 2; it’s true for e;
it’s true for everything. Exponents add when you multiply
them, that’s why e^(ix) and e^(-ix) combine to
give you 1. Once you know that,
you can prove this. Okay.
Now, I’m going to do a little more of complex numbers,
so maybe I’ll leave this alone for now.
I introduced you the number i by saying it’s the
square root of minus 1, and complex numbers entered our
life, even though we didn’t go looking for them.
You can write down equations with real numbers with no
intention of invoking anything fancy,
like this, and you find there is no solution to this equation.
Even though everything there is completely real.
So, you can say, “Well, x^(2) is minus
1,” and you can manufacture a number i,
with a property i^(2) is minus 1;
then, of course, you can have x equal to
plus and minus i as your answer.
So, complex numbers arose by trying to solve quadratic
equations. So, let me write you a slightly
more interesting quadratic equation, x^(2) + x + 1=
0. So, there’s no funny business,
all real numbers. We want to solve this equation,
so we go back to our good old sandbox days when we knew what
the answer for this was–minus 3 over 2.
And we already have a problem because we don’t know what to do
with square root of minus 3. So, we will write it as square
root of minus 1 times square root of 3.
Square root of minus 1 we will call i,
and we will say this equation has two roots.
x plus/minus (there are ) two roots, which are minus 1
plus or minus square root of 3i over 2.
So, one root is 1 minus 1, plus root 3i over 2.
And the other root is minus 1 minus root 3i over 2.
These are solutions formerly to this equation in the following
sense. Take this x that I’m
giving you; put that into that equation,
square the x and add the x to it,
and add the 1, it’ll in fact satisfy the
equation. All you will have to know when
you manipulate it is that i^(2) is minus 1.
Using the one property, you can now solve the quadratic
equation. So, people realize if you
enlarge numbers to include complex numbers,
then you can solve any polynomial equation with that
many number of roots. If it’s quadratic,
it’ll have two roots, if it’s cubic,
it’ll have three roots. They may not all be real;
they may involve the complex number i.
Now, a very important point to notice is that this whole thing,
minus 1 plus root 3i over 2, the whole thing is a
single complex number. Don’t think of it as the sum of
two numbers; it’s a single complex number.
For example, if this had been plus instead
of minus, you’d get something like 1 plus or minus root 3 over
2. Take the positive root,
you don’t think of it as two numbers.
This is a single number. And that continues to be true
even if it’s a complex number. So, you should think of the
whole combination as a single entity which you can add to
others as entities and square, and so on.
Okay. So, that’s what we are going to
do now. We’re going to generalize this
particular case and introduce now a complex number z.
And we’re going to write every complex number z as a
part that has no i in it, and a part that has an i
in it. The example I had here,
x is equal to–in this example, x was minus 1
over 2 and y was plus or minus root 3 over 2.
And we’re going to visualize that complex number as a point
in the xy plane. We just measure x
horizontally and y vertically, put a dot there,
that’s your complex number, z. Then, we say the new complex
number called z star [*] and it’s defined to be x –
iy, is called the complex conjugate; complex conjugate of z.
It’s obtained by changing the sign of i.
So, if you like, if z is sitting here,
z star is sitting there, reflected around the x
axis. So, how does one work with
complex numbers? You add them,
you subtract them, and you multiply them,
and you divide them. If you know how to do that,
you can do with complex numbers everything you did with real
numbers. So if z_1 is a
complex number that is equal to x_1 +
iy_1, and z_2 is a
complex number that is x_2 +
iy_2, we will define
z_1 + z_2 to be
x_1 + x_2 times
i times y_1 plus
y_2. If you draw pictures,
then if z_1 is that number and
z_2 is that number [pointing to the board],
then z_1 + z_2 it really
is like adding vectors. This number is z_1
+ z_2. It’s exactly like adding
vectors. But you don’t think of this
plus that as two disjoined numbers, but as forming a single
entity. In a complex number z,
x is called a real part of z, and is denoted by
symbol “real z.” If you want to get z,
if you want to get the real part of a complex number
z, you add to its complex conjugate and divide by 2.
Because you are adding x + iy to x – iy,
you get 2x and you divide by 2, you get that.
If you want to get the y, which is called the
imaginary part of z, you take z – z star and divide
by 2i. It’s like saying find the
x component of a vector. It’s whatever multiplies
i. y component of a vector
is whatever multiplies j. Similarly, a complex number can
be broken down into the part which is real and the part that
multiplies the i. To extract the real part,
add to it the conjugate divided by 2, to extract the imaginary
part, subtract the conjugate and divide by 2i.
Again, if you have seen this, you don’t need this.
If you haven’t seen it, it may look a little fast.
But that’s why I posted all the notes on the website,
chapter 5 from this math book, which has got all this stuff.
So, you can go home and you can read it.
Ok, the next question is what is z_1 times
z_2? Well, that’s very easy to
calculate. x_1 +
iy_1 times x_2 +
iy_2. Just open all the brackets and
remember i^(2) is minus 1.
That’s all you’ve got to do. So, that gives me
then, let me multiply the iy_1 times the
iy_2. And that gives me minus
because i^(2) is minus 1 plus i times
x_1y _2 +
This is how you multiply two complex numbers. That’s something very nice that
happens when you multiply a number z by its complex
conjugate. What happens when you multiply
z by its conjugate, you are saying,
take x + iy, multiply it by x minus
iy; a + b times a – b=
a^(2) – b^(2). But remember,
b is iy. So, when you take the square of
that, you should keep track of all the signs and you will find
it’s x^(2) + y^(2). We denote that as this,
and this thing is called the modulus of z [|z|].
It’s the length of the complex number z,
just given by Pythagoras’ theorem. You should know,
whenever you take a complex number, multiply by each
conjugate, by its complex conjugate,
the result will be a real number equal to the square of
the length of the complex number.
Because I’m going to use that now to do the last of the
manipulations, which is, what is
z_1 divided by z_2?
How do you divide these crazy numbers?
So, on the top I write x_1 +
iy_1, the bottom, I write
x_2 + iy_2.
If I only had x_2,
I know how to divide. It’s just a number.
Divide that by x_2 and that by
x_2. But I’ve got the sum of these
two numbers in the bottom, and you can ask,
“How do you do the division?” So, the trick,
always, when you run into this problem, is to multiply the top
and bottom by the complex conjugate of the bottom.
Multiply it by x_2 – iy_2,
top and bottom. Then, something nice happens to
the denominator, because the denominator then
becomes x_2^(2) + y_2^(2),
which is an ordinary real number, nothing complex about
that. The numerator,
you can open out the brackets. I don’t know if I want to do
that; it’s just going to be
x_1x_2 + y_1y_2 +
i times y_1x _2 –
Now, don’t worry about all the details.
All it means is, if you know how to multiply two
complex numbers, you can also divide by a
complex number. That’s the key point.
Why? Because if you’ve got a complex
number denominator, you don’t like it,
multiply top and bottom by the complex conjugate of this guy,
the denominator turns into a purely real number.
So, this whole thing could be 36, for example.
Well, we know how to divide the numerator by 36,
right? x_1x_2 +
y_1y_2 some number, maybe 9;
then you divide by 36. One-fourth.
This could be i times 18. It would be 36,
you get i over 2. So, this number could have been
one fourth plus i over 2. So, dividing by a complex
number is not a problem. Now, why did I do all that
stuff today about e^(ix)? You’re going to see that now.
The rationale for going through that math is the following.
Let’s take this complex number that goes from here to here,
that is x + iy. And let’s introduce,
just like we would for an ordinary vector,
this angle θ and that length r.
Whenever you have a vector, you can talk about the
Cartesian components x and y,
or you can talk about the length of the vector and the
angle it makes. Then, it’s clear that x=r
cos θ, and y=r sin θ, where θ is now
an angle associated with a complex number which tells you
at what angle it’s located, and r is the length of
the complex number. It’s called the polar form of
the complex number. Then, you can see that
z, which is x + iy, is equal to r
times cos θ + i sin θ which you can now write as
r times e^(iθ). That’s why we did all that
work, to tell you that a complex number can be written in this
form. You can either write this x
+ iy, or else re^(iθ).
They both talk about the same number.
One talks about how much real part and how much imaginary part
it has, other way of writing it talks about how long the vector
is, and at what angle it is located.
It contains the same information.
So, that r is equal to– the inverse of this formula is
that r is square root of x squared plus y
squared, and θ is tan
inverse y over x. In other words,
tan θ is y over x.
This is y, and this is x. Now, the advantage of writing a
complex number in polar form is the following.
Suppose I give you two complex numbers, z_1 is
and z_2 is r_2
The product z_1z _2 is very easy to
calculate in this form because r_1r
_2 multiply to give you this,
and the exponentials combine. It’s a lot easier to multiply
them in this form than when I multiplied them somewhere here.
See, if I write it in this Cartesian form,
it’s a big mess. In polar form, it’s very easy.
So, the rule is, if you want to multiply two
numbers, multiply the lengths to get the length of the new number
and add the angles to get the angle of the product.
So, what I’m telling you is that if z_1
looks like this guy with some angle θ 1,
and z_2 is that guy with an angle θ
2, the number z_1z _2 has a length
equal to the product of these two lengths,
and that angle equal the sum of these two angles,
so it will look like that [writing graph on board].
To go from this to that, you multiply the lengths and
you add the angles. So, the polar form is very well
suited for multiplying, and it’s even better suited for
dividing. If I say, “Give me
z_1 over z_2,” well,
you can all do that in your head.
It’s r_1, e^(iθ
)_1, divided by
)_2. The modulus of the new number
is just r_1 over r_2.
And how about this? e^(iθ
)_1 divided by e^(iθ
)_2. Dividing by
e^(iθ) is the same as multiplying by
e^(-iθ). So, this is really
e^(iθ )_1 – θ 2.
So, I’m assuming you guys can figure this part out,
1 over e^(iθ) is the same as
e^(-iθ). Well, if you doubt me,
multiply–cross multiply, and all I’m telling you is that
1=e^(iθ) times e^(-iθ),
which you know is true when you combine the angles.
When you’ve got e^(iθ)
downstairs, you can take it upstairs with the reverse angle.
And that’s a useful trick; I hope you will remember that
useful trick. So, there is one part of this
thing that I want you to carry in your head,
because it’s very, very important.
When you take a complex number, it’s got a length and it’s got
a direction. Then, you multiply by a second
complex number, you’re able to do two things at
the same time. You’re able to rescale it,
and you’re able to rotate it. You rescale by the length of
the second factor, and you rotate by the angle
carried by the second factor. The fact that two operations
are done in one shot is the reason complex numbers play an
incredibly important role in physics,
and certainly in engineering and mathematical physics. I’m going to now change gears,
but I want to stop a little bit and answer any questions any of
you have. So, how many people have seen
all of this before? So, if you saw something,
you saw the whole thing. But look, that’s not such a big
number, so I am cognizant of the fact that not many of you have
seen it. But you will have to go and
learn this today. I’m going to post the problem
set for this sometime in the day so you can start practicing.
Don’t wait, the problem set has nothing to do with next
Wednesday, it has a lot to do with today.
When I assign a problem today, I imagine you’re going
breathlessly to your room, not able to wait,
jumping into the problem set and working it out.
That’s the only way you’re going to find out what these
things mean if you’ve never seen them.
If you’ve seen them before in high school, or taken a math
course and you’ve got lots of practice, then I’m not talking
to you. But I’m going to use this at
some point, so you should understand complex numbers.
And I’m telling you, there are very few branches of
any science where complex numbers will not be used.
Now, you may not believe this now, but that is true.
If an electrical engineer, it’s an absolute must.
If you solve any kind of differential equation which can
occur in biology and chemistry, that’s a must.
So, it’s very, very important. All right.
So now, I’m going back to physics.
Going back from mathematics to physics, and the physics of this
topic for today has to do with something very different from
the past, so forget all the relativity
now. You’re going back to Newtonian
days. Kinetic energy is ½
mv^(2). It’s a little difficult to go
back and forget what you learned.
On the other hand, for some of you,
it may not be so hard if you didn’t learn anything.
Well then, you are that much ahead of the other guys.
But remember now, the truth is the relativistic
theory. We’re going to go back to
Newtonian days, and the reason we do it that
way is to give you something interesting,
hopefully, compared to what you normally do.
So, what we’re going to study now is what’s called small
oscillations, or simple harmonic motion.
It’s a ubiquitous fact, that if you took any mechanical
system which is in a state of equilibrium, and you give it a
little kick, it vibrates. Now, if you’ve got a pillar,
ceiling to floor, you hit it with a hammer,
it vibrates. If you take a gong and hit it,
it also vibrates. If you’ve got a rod hanging
form the ceiling through a pivot — it’s hanging there — if you
pull it, it goes back and forth. The standard example everyone
likes to use is if you’ve got a particle in a bowl,
it’s very happy sitting in the bottom.
If you push it up a little bit, it’ll go back and forth.
And the example that we’re going to consider is the
following. If you have a mass,
m, connected to a spring, and the spring is not
stretched or contracted, it’s very happy to be there.
That’s what I mean by equilibrium.
Equilibrium means the body has all the forces on it adding up
to 0; it has no desire to move.
The question is, if you give it a little kick,
what’ll happen? Well, there are two situations
you can have. You can have a situation where
the particle’s on top of a hill. That’s called unstable
equilibrium because if you give that a kick, it’s going to come
down and never return to you. That’s equilibrium,
but unstable equilibrium. I’m talking about stable
equilibrium. That’s because there are
restoring forces. If you stray away from the
equilibrium, there are forces bringing you back.
And in the case of the mass and spring system,
the force, md 2.. ma, is equal to
-kx. And what this equation tells
you is if you stray to the right, x is positive,
[I will apply a force to the left.]
Remember, this is ma, and this is F.
F is such that it always sends you back to your normal
position. So, we want to understand the
behavior of such a problem. We want to solve this problem.
How do you solve this problem? Well, I did it for you along
back, when I gave you a typical paradigm for how you apply
Newton’s laws; I gave you this example.
So, I’m going to go through it somewhat fast.
Our job is to find the function x that satisfies this
equation. And we would like to write it
as follows: d^(2)x over dt^(2) is equal to minus
Ω^(2) x, where Ω is the
shorthand for squared root of k over m.
By the way, I’m using big X and small x like
crazy, so it’s just small x.
I don’t mean anything significant between this and
this. They’re all the same.
If you want, you can take this–pardon me?
[inaudible question] This is k,
and this guy is k. And the other equations you use
to follow the lecture today is, [x=X]
these are all the same. Okay?
So, don’t say what happened there, when you move from one to
the other. Alright, so what did I say we
should do? You can make it a word problem
and say, “I’m looking for a function which,
when I take two derivatives, looks like minus itself,
except for this number.” And we just saw that today.
Trigonometric functions have the property that you take two
derivatives, they return to minus themselves.
So, you can take a guess that x looks like cos t
but it won’t work, that I showed you before.
So, I’m not going to go through that again.
If you took this guess–but x is not a number,
x is a function of time [x(t)].
If you took this function of time, it, in fact,
will obey this equation, provided the Ω that you
put in is the Ω that’s in the equation.
Why? Because, take two derivatives.
First time you get -Ω, sin Ω.
The second time you get – Ω ^(2) cos Ωt,
which means it’s minus Ω ^(2),
times your x itself, and it is whatever you like. Then I said,
“Let’s plot this guy.” When you plot this guy it looks
like this. And A is the amplitude. What is Ω?
Well, Ω is related to the time of the frequency of
oscillations as follows: If I start with t=0
over the x=A, how long do I have to wait
’till I come back to A? I think everybody should know
that I have to wait at time capital T so that
Ω times T is 2π.
Because that’s when the cosine returns to 1.
That means the time that you have to wait is 2π over
Ω. Or you can say Ω is
2π over T, or you can also write it a
2π times frequency, where frequency is what you and
I would call frequency, how many oscillations it does
per second. That’s the inverse of the time
period. So, if you pull a mass and you
let it go, it oscillates with a frequency which is connected to
the force constant and the mass. If the spring is very stiff and
k is very large, frequency is very high.
If the mass is very big and the motion is very sluggish,
f is diminished. So, all that stuff comes out of
the equation. One really remarkable part of
the equation is that you can pick any A you like.
Think about what that means. What is the meaning of A?
A is the amount by which you pulled it when you let it
go. You are told whether you pull
the spring by one inch or by ten inches, the time it takes to
finish a full back and forth motion is independent of
A. This frequency here is
independent of A. Yes?
Professor Ramamurti Shankar: This t here?
[inaudible] Professor Ramamurti
Shankar: Ah, very good.
That’s correct. So, let’s be careful.
This is the t axis [pointing at graph on board];
this is a particular time, capital T. So, the remarkable property of
simple harmonic motion is that the amplitude does not determine
the time period. If you pull it by two inches,
compared to one inch, it’s got a long way to go.
But because you pulled it by two inches, the spring is going
to be that much more tense, and it’s going to exert a
bigger force so that it’ll go faster for most of the time,
that’s very clear. But the fact that it goes
faster in exactly the right way to complete the trip at the same
time, is rather a miraculous property
of the fact that this equation has this particular form.
If you tamper with this, if you add to this a little
bit, like 1 over 100 times x^(3),
then all these bets are off. It’s got to be that equation
for that result to be true. Okay.
So, this is simple harmonic motion.
Then, you can do the following variant of this solution,
which I will write down now. Suppose I set my clock to 0,
right there. You set your clock to 0 here,
but I can say, “You know what?
I just got up and I’m looking at the mass, I set my clock to
0.” When it hits 0,
it’s not at the maximum. But it’s the same physics,
it’s the same equation. And that really comes from the
fact that we had one more latitude here of adding a
certain number φ, which is called a “phase,” to
the oscillator. Whatever you pick for
φ, it’ll still work. And whatever you pick for
capital A, it’ll still work.
So, whenever you have an oscillator, namely,
a mass and spring system, and you want to know what
x is going to be at all times,
you need to somehow determine the amplitude and the phase.
Once you know that, you can calculate x at
all future times. So, let me give you an example.
Suppose an oscillator has x=5 and velocity equal
to 0, at t equal to 0. So what does that mean?
I pulled it by 5 and I let it go.
I tell you the spring constant k and I tell you the
mass, and I say, “What’s the future x?”
You’ve got to come back to this equation.
And you’ve got to say, 5=A cosine of 0 +
φ. And how about the velocity?
Velocity is supposed to be 0; that is minus ΩA sin
Ωt + φ. But t is 0.
So, that tells me 0 is minus Ω a sin φ.
Ω is not 0, A is not 0.
Therefore sin φ is 0, that means φ is 0 if
you get rid of that, so the subsequent motion is
x=5 cos Ωt. This is a problem where we did
not need φ. But it could have been that
when you joined the experiment, you were somewhere here;
then, the mass is actually moving.
Then I will say at t=0, velocity was some other
number. Maybe 6.
Then, you’ve got to go back and write two equations.
One for x: take the derivative which looks
like this, and instead of putting 0, you put 6.
You’ve got two unknowns, A and φ.
You’ve got to juggle them around, and you’ve got to solve
for A and φ, given these two numbers.
It’s a simple exercise to solve for them, and when you do,
you have completely fit the problem.
In other words, simply to be told a mass is
connected to a spring of known mass and force constant is not
enough to determine the future of the spring.
You have to be told–For example, if nobody pulled the
mass, it’s just going to sit there forever.
If you pulled it by 10, it’s going to go back and forth
from 10 to minus 10. So, you have to be told further
information to nail down these two numbers.
They’re called free parameters in the solution.
On the other hand, no matter how much you pulled
it and how you released it, this Ω,
and therefore the frequency, are not variable.
[inaudible] Professor Ramamurti
Shankar: If you start it at rest,
indeed, it will be 0, because the velocity at
t=0 is minus ΩA sin φ.
If you want to kill that, the only way is φ=0.
You could put φ=π if you like, but if you put 0,
that’ll do. Okay.
So, this is the elementary parts of this thing.
And I already told you the following things.
If x(t) is equal to A cos Ωt–let’s
put φ=0. In other words,
we’ll agree that since there’s only one oscillator,
it is perverse to set your clock to 0 at any time other
than when the oscillator is at one extreme point.
If there are two oscillators oscillating out of step,
it’s impossible to make φ=0 for both of them.
Because if you can sync your clock when one of them is at a
maximum, that may not be when the other one’s at a maximum.
But with one oscillator, to choose φ other than
0 is perverse. It’s like saying,
“I’m doing a projectile problem.
I’ll pick the place where I launch the projectile to be x=
y=0.” You can pick some other crazy
origin, but it doesn’t help. So, we’ll assume φ is 0
in this problem. If that is x,
what is velocity at time t?
Take the derivative of this, you get minus Ω A
sin Ωt. That means the velocity is also
oscillating sinusiodally but the amplitude for oscillation is
Ω times A. So, if A is the range of
variation for x, Ω times A is the
range of variation for velocity. Velocity will go all the way
from plus Ω A to minus Ω A.
The acceleration, which is one more derivative,
is minus Ω ^(2)A,
cos Ωt. Which is really minus Ω
^(2) times x itself. So, the amplitude for
oscillation for the acceleration is Ω^(2) times A.
So, I think you should understand if Ω is very
large; then, the velocity will have
very big excursions and the acceleration will have even
bigger excursions, because the range of,
this A really means if you plot x,
it’s going to look like this, going from A to minus
A. This says if you plot the
velocity, it’s going to look like this.
The range will go from Ω A to minus Ω A.
If you plot the acceleration, it’ll go from Ω
^(2)A to minus Ω ^(2)A. Last thing you want to verify
is the Law of Conservation of Energy.
I think I mentioned this to you when I studied energy,
but let me repeat one more time.
If you take, in this problem,
½ mv^(2) +½ kx^(2)–I did that for
you–You can also verify this. Okay?
½ kx^(2) is going to involve A^(2) cosine
squared Ωt. ½ mv^(2) is going to
involve something, something, something sin^(2)
Ωt. And by magic,
sin^(2) and cos^(2) will have the
same coefficient, so you can use the identity to
set them equal to 1, and get an answer that does not
depend on time. That’s the beauty.
Even though position and velocity are constantly
changing, this combination, if you crank it out,
will not depend on time. And what can it possibly be?
If it doesn’t depend on time, I can calculate it whenever I
like. So, let me go to the instant
when the mass has reached one extremity and is about to swing
back. At that instant it has no
velocity; it only has an x,
which is equal to amplitude, or motion.
So, that’s the energy of an oscillator, ½ kA^(2).
When v=0, x will be plus or minus
A, but any other intermediate point you can find
the velocity, if you give me the x.
We did that too. So, this is all I want to say
about the linear oscillator. But let me mention to you
another kind of oscillation, which is very interesting.
Suppose you suspend a rod from the ceiling like this.
It’s hanging there. If you give it a twist,
then it’ll go twisting back and forth, like that.
That’s also a simple harmonic motion, except what’s varying
with time is not the linear coordinate,
but the angle θ by which the rod has been twisted.
What’s the equation in this case?
The equation here is I, d 2θ over
dt^(2). You guys remember now,
that’s the analog of ma? That’s going to be equal to the
torque. Now, what’ll happen in the
problems we consider is that θ is a small number,
and the torque will be such that it brings you back to
θ=0. So, this number will be
approximately equal to minus some number κ times
θ. That’s the analog of minus
kx. Little k was the
restoring force divided by the displacement,
which produced the restoring force.
Little κ is the restoring torque divided by the
angle at which you twisted it to get that torque.
So, this isn’t always true. For small oscillations,
sin τ vanishes at θ=0;
it will look like some number times θ.
If you like, it’s the Taylor series for
τ for small angles. If you knew this κ you
are done, because mathematically,
this equation is the same as md 2 x over
dt^(2) is minus kx.
Because the Ω for this guy will be square root of
κ over I. You can steal the whole answer,
because mathematically, mathematicians don’t care if
you’re talking about θ or if you’re talking about
x. Now, if you compare this
equation to this equation, md 2 x over
dt^(2) equal to minus kx,
if the Ω there was the root of k over m,
the Ω here would be root of κ over I.
So, you may be thinking, “Give me an example where
τ looks like κ times θ.”
I’m going to give you that example. So, that example is going to be
the following simple pendulum. Take a pendulum like this.
It’s got a mass m. It’s got some length l;
it’s hanging from the ceiling. It’s a rigid rod,
not a string, but a rigid,
massless rod. And the mass m is at the
bottom. So, it’s happy to hang like
this, but suppose you give it a kick, so you push it over there
to an angle θ, which is whatever you like.
What is the torque in this particular case?
Let’s find out. The force is mg,
and the angle between the force–and you’re trying to do
rotations around this point–this angle here is the
same θ. So, torque is really minus
mgl sin θ. That’s the formula for torque.
It’s the force times the distance of the lever arm or the
distance over which the force is from the rotation axis times the
sine of the angle between the direction of the force and the
direction of the separation. So, the equation for the
pendulum really is the following: Id^(2) θ over
dt^(2) is equal to minus mgl sin θ.
Now, we cannot do business with this equation.
This equation is a famous equation, but you and I don’t
know how to solve it with what we know.
But we have learned, even earlier today–look at the
formula for sin x. Sin x begins with
x minus x^(3) over 3 factorial, and so on.
So, if θ is small, then you may write
mglθ. So, for small angles,
sin θ is θ, and the restoring
torque, in fact, is linear in the angle of the
coordinate. So, by comparison now,
we know that this fellow is our κ.
So, κ for the pendulum is mgl.
So, κ may not be universal in every problem,
in each problem you’ve got to find κ.
You find that by moving the system off equilibrium,
finding the restoring torque, and if you’re doing it right,
restoring torque will always look like something,
something, something times θ.
Everything multiplying θ is our κ.
So, let’s now calculate Ω for the simple
pendulum. I’ve shown you this is
κ over I, and κ was mgl.
And what’s I? It’s equal to moment of inertia
of this mass around this point. The rod is massless.
For a single point mass, we know the moment of inertia
is just m times l^(2).
So, you find Ω is equal to square root of g over
l, and that is 2π over T.
So, you get back the famous formula you learned,
T is 2π square root of l over g.
This is the origin of the formula you learned in school.
It comes from small oscillations of a pendulum.
So, think about this problem. If you pulled a pendulum by an
angle so large that sin θ is no longer
approximated by θ but you need a θ^(3) over 6
term — it can happen for large enough θ – then,
the equation to solve is no longer this.
What that means is, if you took a real pendulum
with a massless rod and a bob at the end,
the time it takes to finish an oscillation, in fact,
will depend on the amplitude. But only for small
oscillations, when sin θ can
be approximated by θ, you will find whether you do
that the time is the same. But if you go too far,
the other terms in sin θ will kick in,
and the result is no longer true.
The period of a pendulum, in fact, is not just a function
of l and g, but also a function of the
amplitude. Only for small amplitudes,
you get this marvelous result. There’s one problem I want you
to think about. I may do it myself.
I think you should try to read up on it, and I will try to
explain it next time. I’ll give the problem to you
and I will do it for you next time.
I want you to think about it. Here is a hoop,
like a hula hoop. It’s hanging on the wall with a
nail. And it’s very happy to be where
it is. Now, I come and give it a
little push. It’ll start oscillating.
It’ll look like that, and some time later it’ll look
like that, it’s going back and forth.
Your job is to find the Ω for that hoop. So, you move it a little bit,
and you find the restoring torque.
And you’re done. But you’ve got to do two things
right to get the right answer. You’ve got to find the moment
of inertia for circular loop not around the center,
but around the point and the circumference.
So, you guys know the magic words you’ve got to use.
Parallel Axis theorem. And to find the restoring
torque, here is the part that blows everybody’s mind.
When the loop goes to that position, where is gravity
acting? Gravity’s acting at the center
of mass. The center of mass is somewhere
here in the middle of the loop, even though there is no matter
there. So, center of mass of a body
need not lie on the body. The ring is a perfect example.
So, if you remember the center of mass of a loop is in the
center of the loop and put the mg there,
you will get the right torque, and everything will just
follow. By the way, my offer to you is
always open. If there’s something about the
class you want to change, you find it’s hard,
this, that, you can always write to me.
I want your feedback, because I just don’t know which
parts are easy, which parts are hard,
which are slow, which are fast.
But if you haven’t had enough math and you need some help,
send me some e-mail and I will try my best to answer them.