[music] Just for fun I'd like to know a
root of the polynomial x to the 5th, plus
x squared, minus 1.
So here's the function where I want to
find a root, right?
I want to find some input that makes this
function equal to 0.
What do I know about this function?
It's continuous and I know that f of 0, if
I plug in 0.
It's 0 plus 0 minus 1.
So f of zero is minus 1.
And f of 1 is 1 to the 5th plus 1 squared
minus 1.
That's 1 plus 1 minus 1, that's 1.
So what I know here is I've got a
continuous function.
And its value at 0 is minus 1 and its
value at 1 is 1.
So there has to be some input between 0
and 1 where this function's output is
equal to 0.
So I know there's a root there between 0
and 1.
Let's find that root.
Now, if we were just trying to find where
some linear term was equal to 0.
You can just solve for x, in that case.
If I were just trying to figure out where
some quadratic was equal to 0.
There's the quadratic formula.
If I were trying to figure out where ax
cubed plus bx squared plus cx plus d was
equal to 0.
I could use a more complicated formula.
It's the cubic formula.
It's way more complicated than the
quadratic formula.
But there is such a formula.
If I were trying to solve an equation like
this that started with a X to the 4th
term, I could use the quartic formula to
do that.
The bad news is that here I'm trying to
solve a quintic and there is no quintic
formula in terms of the usual arithmetic
operations plus, minus, times, and
divides, and taking roots.
Of course some quintics can be solved I
mean, it's true.
Right?
If I had instead been trying to find a
solution to this equation, x to the 5th
minus 2 equals 0, I'm looking for some
value of x, take its 5th power and
subtract 2 and I'd get 0.
Yes, I could solve this quintec.
Right?
I mean, its solution is the 5th root of 2.
2.
But in contrast this particular example X
to the 5th plus X squared minus 1 can't be
solved in the sense that I can't write
down using fractions, plus, minus, times,
divides, and the taking of roots.
A number which, if I take it's 5th power,
add it to its square, and subtract 1, I
get back 0.
Right?
I mean, that's the sense in which I can't
solve it.
I can't just write that thing down in
terms of the operations that I have at
hand.
Knowing that requires something called the
Galois theory.
So in light of Galois theory, what does it
even mean to find a, A root.
Well, I'm not going to be able to write
down the root.
But I can at least approximate the root by
using Newton's method.
So I can use Newton's method to at least
approximate a root.
So here's the function whose root I want
to approximate.
Right?
I want to find some input so that f is
close to 0.
And I'll first differentiate f.
So the derivative of f, well the
derivative of x to the 5th is 5 x to the
4th, derivative of x squared is 2 x, and
the derivative of 1 is 0.
So I've got my original function.
I've got my derivative, and then Newton's
method tells me that I should make a first
guess.
My first guess will be 1.
So I've made my initial guess.
What's my next guess?
So my next guess according to Newton's
method.
We'll call that x of 1, would be my old
guess minus the functions value of my old
guess, divided by the function's
derivative at my old guess.
And now my old guess is 1.
The function's value at 1, that's 1 to the
5th plus 1 squared minus 1.
That's 1 plus 1 minus 1, that's 1.
And the derivative at 1, now that's 5
times 1 to the 4th plus 2 times 1.
That's 5 plus 2, that's 7.
So, my new guess is 1 minus 1 over 7.
That's 6 7ths, so that's my new guess.
And 6 7ths really isn't that bad of a
guess.
So my original guess, right, x0, that was
1.
And f of 1 was just 1.
My next guess, x1, that was 6 7ths, right?
And what's f of 6 7ths?
Well that's 6 7ths to the 5th plus 6 7ths
squared minus 1.
That turns out to be about .197.
Right?
So certainly this 6 7ths is a better guess
than 1.
Now armed with my better guess I can apply
Newton's method again and hopefully arrive
at an even better approximation to the
rut.
So I get a new guess, x sub 2, by taking
my old guess, x sub 1, and subtracting the
functions value there, divided by the
functions derivative there.
And when I do that I get something which
is a about .812.
X of 2 is an even better guess.
So I evaluate my function F at X 2, I get
well it turns out to be about .014.
So this new guess for the root of my
function is even better then my old guess,
right?
My old guess was giving me .197 as the
output This new guess is getting me closer
to a 0 of the function looks like.
Well don't stop.
I'll do it again.
I can do Newton's method again, right.
And I'll get a newer guess even.
X of 3 will be x of 2 minus f of x of 2
divided by the derivative.
Of, of f at x sub 2 n.
This turns out to be about .809.
This last guess is even better.
And this is really working quite well.
F at this new guess, x sub 3, turns out to
be about 0.000085.
And I'm getting much closer now to the
actual 0 of this function.
These sorts of examples of polynomials
where I can't explicitly write down the
root.
They really raise a fascinating question
as to what it even means to find the root.
All right, the intermediate value theorem
promises me that there is a root.
In this case, between 0 and 1.
And Newton's method, or this bi-section
algorithm permits me to get better and
better approximations to that root.
But it's only in that sense that I can
ever find the root.
It's in the sense that I can get as close
as you want to that root.