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In this segment, we're going to calculate the two-point Gauss quadrature rule, we're going to do the
derivation, we're going to continue the derivation from the previous segment.|So let's go ahead and see what
we had in the previous segment.|We had c1 plus c2 equal to 2,
we had c1 x1, plus c2 x2 equal to
0, then we had c1 x1 squared,
plus c2 x2 squared equal to 2/3,
and then we had c1 x1 cubed, plus c2 x2 cubed
equal to 0.|So we had these four equations, four unknowns, so
let me call this equation 1, 2, 3, and 4.
So these are four simultaneous nonlinear equations, where the unknowns are c1, c2,
x1, and x2, and I need to find out if there is an acceptable solution
from these four simultaneous nonlinear equations.|So what I'm going to do it I'm going to take this
second equation, I'll multiply it by x1 squared and then I'm going to subtract it from equation 4,
and I'm going to see that whether that will allow me to find out the solutions.
So if you look at, this is my equation 4, c2 x2 cubed
equal to 0, this is my equation 4 being written again, and then I'm going
to write down this equation 2, the only difference is that I'm multiplying it by x1 squared.
So I'll get c1 x1 cubed, plus c2 x1
squared x2 equal to 0, that's what I'm going to get
from equation 2, which is being modified here
by multiplying by x1 squared.|Now, if I subtract these two equations,
I'll get minus here, minus here, minus here, this will cancel out, I'll get c2
x2 cubed, minus c2 x1 squared
x2 equal to 0, that's what I'm going to get if I subtract now this modified
equation 2, which is obtained by multiplying it by x1 squared, as you see here,
and then if I subtract the two, this is the equation which I am going to get, and if I
rewrite this, I'll get c2 x2
here, and I'll get x2 squared minus x1 squared
equal to 0.|So that's what I am going to get as
the equation which I get from by using equation 4 and the modified form of equation 2.
Now, you can see that if you look at this particular equation here, which is the
solution here, part of the solution, that if I want this to be 0,
there are several possibilities, I can have c2 equal to 0,
or/and x2 equal to 0, or/and I can have
x2 equal to x1, or/and I can have x2 equal to minus x1
So I'm going to write down those four possibilities, and try to see that which possibility
is acceptable, which one is not, and that's how I'm going to be able to figure out what I should be doing.
So if I look at this here, I have c2 x2
times x1 squared . . . x2 squared minus x1 squared equal to 0,
and I am saying that c2 equal to 0 is a possibility, x2 is equal to 0
is a possibility, and keep mind that I am saying and/or, that's very important,
and/or, or x2 is equal to x1, and/or,
because either one of them has to be true, or more than one can be
correct, or can be possible.|So these are the four choices
which I have, by which making this to be equal to 0, and again, keep mind these are and/or
or possibilities which I have there, and what I'm going to show to you is that which ones are
acceptable, which ones are not acceptable, and that's how we'll be able to find the solution to these four simultaneous
nonlinear equations.|So let's look at the first case.
Let me put it on the next board here, so let's look at the first case, where
we're saying that c2 is equal to 0.|Let's see if c2 is equal to 0 is a possibility
for part . . . for the solution.|If c2 is equal to 0, then if you
look at equation 1, equation 1 will turn out, we have c1 plus c2 equal to
2, that will become c1 equal to 0 . . . sorry, c1 equal to 2,
that's what I'm going to get if I choose c2 equal to 0, I'm going to get c2 equal to 2.
Now, if I look at the next equation, I have
c1 x1 . . . c1 x1, plus c2 x2
equal to 0, this is the second equation.|Now, if I choose c2 equal to 0, I'll get c1
times x1 is equal to 0.|So from
here, I'll get . . . so, since c1 is equal to 2, from here, what will I get
if I choose these two?|I'll get x1 equal to 0.|Now,
if I choose x1 equal to 0, the third equation which I will have is that c1
x1 squared, plus c2 x2 squared is equal to 2/3,
but c2 is already equal to 0, this is already equal to 0, I'll get
c1 x1 squared equal to 2/3.|But c1 x1 squared
is equal to 2/3 is not possible.|Why?|Because I just obtained x1 equal to 0.
So it does not allow . . .
you cannot allow x1 equal to 0 when your c1 times x1 squared is equal to
2/3.|So c2 is equal to 0 is unacceptable.
c2 is equal to 0
is an unacceptable choice, so that's not going to work out.|So the first
choice, which we said, hey, in order to make that equal to 0, that c2 is equal to 0 is a possible choice,
we have just shown that c2 equal to 0 is unacceptable.|How have we shown that?|We have basically taken the
first equation, and said that since c2 is equal to 0, I'm getting c1 equal to 2.
Then the second equation, c1 x1, plus c2 x2 is equal to 0, since c2 is equal to 0,
I'll be left with c1 x1 equal to 0, and since c1 is a nonzero quantity, as
established here, x1 has to be 0, and I put
x1 equal to 0 . . . sorry, this should be
x1 is 0, so since . . . sorry, yeah, this is 0, because
c2 is 0, so since c2 is 0, then c1 x1 squared is equal to 2/3,
but that's not possible, because we just showed that, hey, in order for that to be possible, x1
has to be 0, so c2 equal to 0 is an unacceptable choice.
So let's go ahead and see if the other choices are unacceptable or not.|So I'm going to show you one more,
I'll leave the other ones as homework.|So let's look at the second choice.|The second choice is
x2 equal to 0.|Is this a good choice to use?
Now, if you look at x2 equal to 0, we have c1 plus c2
equal to 2, which is the first choice, so we are okay,
and then the second equation will reduce down to c1 x1, plus c2 x2 equal
to 0, and since x2 is equal to 0, this will reduce to
c1 x1 equal to 0, okay?
Now we have another, the next equation is c1 x1 squared,
plus c2 x2 squared is equal to 2/3, and since we are assuming x2
equal to 0, I'm going to get c1 x1 squared equal to
2/3.|Now, again, you are
already seeing the unacceptable nature of x2 equal to 0, because this
one is giving you c1 times x1 squared is equal to 2/3, and this is giving you c1 times x1 is equal to 0,
because if c1 x1 has to be 0, then what has to take
place here?|Either the c1 has to be 0, or x1 has to be 0,
or both of them have to be 0.|So if either one of them is 0, that means that
this is also 0, because you are assuming c1 or x1 to be equal to 0, but that's going to violate
this, that it's equal to 2/3.|So you can again see here that
x2 equal to 0 is unacceptable.
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x2 is equal to 0 is unacceptable.|So you are finding out little by little that
what is acceptable and what is not . . . what is not acceptable.|And the third choice
which we had was x2 equal to x1, and it is unacceptable,
and I am just telling you that this is unacceptable, and I would like you to
do it as homework to see that how . . . why is x2 equal to x1
unacceptable?|And the fourth thing which we have, x2 is equal to
-x1, that is acceptable,
and that's, again, I'm going to assign it as homework.|So please don't think that it's because
we had four choices and the first three choices are unacceptable, that the fourth has to
automatically be acceptable, because when you are solving simultaneous nonlinear equations,
you may have no solution, unique solution, or multiple solutions, so it
has to be shown that this particular choice, x2 equal to -x1, is acceptable.
So once you have proven that x2 equal to -x1 is acceptable,
let's go ahead and see that what does that lead us to?
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