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PROFESSOR: What did
we have last time?

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We had a constant copper
that reflected the energy

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so a unit free
version of the energy,

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was greater than 0, because we
were looking at bound states,

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so the energies were negative.

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And we found that this
[INAUDIBLE] this quantized 1

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over 2 copper was
N plus l plus 1--

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something that we call n.

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And n was the principal
quantum number.

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So this is the principle
quantum number n

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was the degree of a
polynomial in the solution.

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And l was an important
quantum number,

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because it gave you the
amount of angular momentum

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the system had.

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And given that we think of it as
principal quantum number coming

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first, once you have n and you
fix it because the energy just

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depends on n--

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Once you fix n,
you'll have that l

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can go from 0 up to n minus
1, and those corresponds

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to the various
values of capital N.

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But we don't have to focus
on it up to n minus 1.

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And at the same time in
terms of quantum numbers,

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m goes from minus l up to l.

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So the order of thinking
is fixed co-- at little m,

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and the principal
quantum number.

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Then fix an l can go
from up 0 to n minus 1.

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Once you fix a little
l, you fix the m.

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And m can go from my l to l.

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And those are the states
of the hydrogen atom.

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The energies in
terms of n are mine

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z squared, e squared over
2 knot, 1 over n little n

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squared.

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And the solutions-- psi nlm
that depend on r theta and phi

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were of the form of a
normalization constant, A r

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to the l that you can isolate.

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That's a behavior for small
r, then a polynomial n r

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over a0 of degree,
capital N, which

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is n minus L plus 1 times an
exponential decay with radius,

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which goes like z over r.

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It has to be
dimensionless, the argument

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inside the exponential.

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And it turns out to
depend on a0, in fact na0

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and finally, the
spherical harmonic.

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So that was the total solution.

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We didn't investigate
the polynomial in detail,

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because it takes time.

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It's not necessary
for many things.

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Only very detailed calculations
require this polynomial.

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And it's a Laguerre polynomial.

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And if you needed--

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if you needed to construct
the quadratic polynomial,

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it would be a fight
between looking up

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some table of
Laguerre polynomials

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and spending time checking
that the conventions they use

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are the same than
the ones you choose

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versus taking the
recursion relation

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and building it up yourself
of the third coefficient.

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So that's what it is.

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Now the hydrogen atom, there
is a classical description,

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a diagram, for the
hydrogen atom and in fact,

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for any central potential.

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So if you're looking
at bound states,

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the way we do bound
states and represent them

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for central potentials
is by a diagram

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in which you put the energy
on the vertical line.

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It's a negative energy.

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So you can put 0 in here.

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The way the hydrogen atom
works is better, of course,

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as usual to use something
dimensionless here.

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And the thing we put is
minus 1 over n squared,

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the integer the principle
quantum number n minus 1

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over n squared,
because this ratio is,

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in fact, equal to
the energy divided

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by this dimensionless quantity
z squared e squared over 2a0.

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So there is n squared,
and the levels go like 1

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over n squared, indeed.

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So there is minus 1 here,
and then goes to minus 1/4

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for n equals 2 minus
1/9 minus 1/16.

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And they crowd here.

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That's why I don't do
everything in scale.

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I cut the scale here, otherwise
the diagram is very long.

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And you can put
say the minus 1/4

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here, though, 1/9 would be here.

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minus 1/9 minus 1/16--

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somewhere here.

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Those are places where
you have energy levels,

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and here it comes.

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Let's look at what we can get.

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Well, for n equals
1, you're in here.

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So what are we going
to plot in this axis.

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The idea is to plot
a quantum number.

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So actually to say here
is l, but we don't do it

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in marking the values of l.

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We'll put here l
equals 0 and list

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all the states that we get.

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So it's like a histogram
or something like that, l

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equal 1, l equals 2, equals 3.

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And then we put dashes here,
and each dash is a state.

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If you look at it and you
say, what's the value of l?

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So this corresponds to
the idea that you already

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know that if you're solving
a central potential problem,

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you have to solve
a radial equation

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for different values of l.

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Each time another l, l equals
0, l equals 1, l equals 2.

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You go on with them.

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So m, so when n is equal
to 1, l can be only 0.

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So you have 1 states here.

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l is equal to 0.

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And is equal to 1.

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And capital N is
it's equal to 0.

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So the only thing that I
cannot read immediately,

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I know that n is equal
to 1, because I'm here.

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I know that l is equal
to 0, because I'm here.

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But I put the extra information,
the capital N equal to 0 here.

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And that's it for this level.

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So this is that little
n equal 1 level.

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Then we go to that
little n equals 2 level.

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And when little n is equal to
2, l can be 0, or it can be 1.

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When l is equal to
0, capital N would

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have to be 1 so that capital N,
which is 1 plus little l, which

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is 0 plus 1 is equal to 2.

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And this is a level n,
little n equals to 2,

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principal quantum
number equal to 2.

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So here we have n equals to 2.

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Since l is 0, capital N is 1.

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Here you have capital
N is equal to 0.

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And then we go to the next
level little n equals to 3.

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Well, little l can now go
from l equals 0, l equals 1

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and l calls 2.

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Since little l plus
1 plus capital N

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is equal to the
principal quantum number,

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here you'll get N equal to 2,
N equals 1, and it goes down,

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N equals 0.

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These are the states.

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Let's do one more.

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Little n equals to 4.

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And, yes, there's a state
for l equal 0, 1 for equal 1,

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1 for equal 2, and this
time we get to l equal 3.

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Each time you get
one more l, because l

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can go up to n minus 1.

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And what is capital N?

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Here is 3, 2, 1, and 0.

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So this is your diagram.

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This is a very nice diagram.

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And it has actually is some
sort of mystery in this diagram.

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Let me emphasize first one
point that is not mysterious.

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It's this growth from an equal
0 to an equal 1 to n equal 2

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to n equal 3.

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There seems to be
a pattern here.

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Here n equals 0, 1.

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2.

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And it will go up.

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So why is that necessary?

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Why did that happen?

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Remember, that we were solving
a radial equation, which

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was like a
one-dimensional potential.

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And if you are solving a
one-dimensional potential,

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the node theorem works.

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So there should be no
nodes for the ground state.

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And the fact that the wave
function vanishes at r equals 0

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is not a node,
because that's the end

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of the world at r equals 0.

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But then you're solving
this radial equation.

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And let's look at
this polynomial.

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Well, you're not going to
get a 0 of the wave function,

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because of this factor.

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It's an overall factor.

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And the exponential never
vanishes just at infinity,

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but never doesn't
vanish at the point.

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So all the 0's of
the wave function

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have to arise from
this polynomial.

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And there shouldn't be any
0 for the ground state.

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So good.

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N equal 0 means no nodes
here, no nodes for the--

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this is a different problem.

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This is an l equals 1.

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It's again a new potential.

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So you solve the
radial equation again.

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Yes, the ground state, when l is
equal to 1 must have no nodes.

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So you should, remember, you're
solving the radial question.

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One's here for one potential,
here for another potential,

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because the effective
potential depends on l, here

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for another potential.

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There for another potential.

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So each time, it's a new
one-dimensional problem,

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which must have a
ground state and a state

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with 1 node, which is
possible, because a degree 1

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polynomial has one 0.

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A degree two polynomial
can have two 0's.

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Why it has another
degree n polynomial

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00:13:29,000 --> 00:13:32,930
may have n 0's But
it may have less

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00:13:32,930 --> 00:13:34,970
0's if the 0's are complex.

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But if better be that this
polynomials don't do that,

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because they would
violate the known theorem.