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[music] Just for fun I'd like to know a
root of the polynomial x to the 5th, plus

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x squared, minus 1.
So here's the function where I want to

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find a root, right?
I want to find some input that makes this

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function equal to 0.
What do I know about this function?

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It's continuous and I know that f of 0, if
I plug in 0.

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It's 0 plus 0 minus 1.
So f of zero is minus 1.

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And f of 1 is 1 to the 5th plus 1 squared
minus 1.

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That's 1 plus 1 minus 1, that's 1.
So what I know here is I've got a

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continuous function.
And its value at 0 is minus 1 and its

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value at 1 is 1.
So there has to be some input between 0

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and 1 where this function's output is
equal to 0.

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So I know there's a root there between 0
and 1.

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Let's find that root.
Now, if we were just trying to find where

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some linear term was equal to 0.
You can just solve for x, in that case.

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If I were just trying to figure out where
some quadratic was equal to 0.

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There's the quadratic formula.
If I were trying to figure out where ax

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cubed plus bx squared plus cx plus d was
equal to 0.

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I could use a more complicated formula.
It's the cubic formula.

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It's way more complicated than the
quadratic formula.

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But there is such a formula.
If I were trying to solve an equation like

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this that started with a X to the 4th
term, I could use the quartic formula to

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do that.
The bad news is that here I'm trying to

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solve a quintic and there is no quintic
formula in terms of the usual arithmetic

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operations plus, minus, times, and
divides, and taking roots.

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Of course some quintics can be solved I
mean, it's true.

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Right?
If I had instead been trying to find a

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solution to this equation, x to the 5th
minus 2 equals 0, I'm looking for some

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value of x, take its 5th power and
subtract 2 and I'd get 0.

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Yes, I could solve this quintec.
Right?

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I mean, its solution is the 5th root of 2.
2.

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But in contrast this particular example X
to the 5th plus X squared minus 1 can't be

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solved in the sense that I can't write
down using fractions, plus, minus, times,

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divides, and the taking of roots.
A number which, if I take it's 5th power,

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add it to its square, and subtract 1, I
get back 0.

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Right?
I mean, that's the sense in which I can't

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solve it.
I can't just write that thing down in

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terms of the operations that I have at
hand.

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Knowing that requires something called the
Galois theory.

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So in light of Galois theory, what does it
even mean to find a, A root.

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Well, I'm not going to be able to write
down the root.

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But I can at least approximate the root by
using Newton's method.

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So I can use Newton's method to at least
approximate a root.

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So here's the function whose root I want
to approximate.

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Right?
I want to find some input so that f is

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close to 0.
And I'll first differentiate f.

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So the derivative of f, well the
derivative of x to the 5th is 5 x to the

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4th, derivative of x squared is 2 x, and
the derivative of 1 is 0.

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So I've got my original function.
I've got my derivative, and then Newton's

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method tells me that I should make a first
guess.

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My first guess will be 1.
So I've made my initial guess.

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What's my next guess?
So my next guess according to Newton's

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method.
We'll call that x of 1, would be my old

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guess minus the functions value of my old
guess, divided by the function's

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derivative at my old guess.
And now my old guess is 1.

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The function's value at 1, that's 1 to the
5th plus 1 squared minus 1.

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That's 1 plus 1 minus 1, that's 1.
And the derivative at 1, now that's 5

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times 1 to the 4th plus 2 times 1.
That's 5 plus 2, that's 7.

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So, my new guess is 1 minus 1 over 7.
That's 6 7ths, so that's my new guess.

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And 6 7ths really isn't that bad of a
guess.

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So my original guess, right, x0, that was
1.

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And f of 1 was just 1.
My next guess, x1, that was 6 7ths, right?

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And what's f of 6 7ths?
Well that's 6 7ths to the 5th plus 6 7ths

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squared minus 1.
That turns out to be about .197.

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Right?
So certainly this 6 7ths is a better guess

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than 1.
Now armed with my better guess I can apply

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Newton's method again and hopefully arrive
at an even better approximation to the

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rut.
So I get a new guess, x sub 2, by taking

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my old guess, x sub 1, and subtracting the
functions value there, divided by the

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functions derivative there.
And when I do that I get something which

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is a about .812.
X of 2 is an even better guess.

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So I evaluate my function F at X 2, I get
well it turns out to be about .014.

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So this new guess for the root of my
function is even better then my old guess,

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right?
My old guess was giving me .197 as the

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output This new guess is getting me closer
to a 0 of the function looks like.

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Well don't stop.
I'll do it again.

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I can do Newton's method again, right.
And I'll get a newer guess even.

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X of 3 will be x of 2 minus f of x of 2
divided by the derivative.

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Of, of f at x sub 2 n.
This turns out to be about .809.

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This last guess is even better.
And this is really working quite well.

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F at this new guess, x sub 3, turns out to
be about 0.000085.

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And I'm getting much closer now to the
actual 0 of this function.

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These sorts of examples of polynomials
where I can't explicitly write down the

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root.
They really raise a fascinating question

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as to what it even means to find the root.
All right, the intermediate value theorem

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promises me that there is a root.
In this case, between 0 and 1.

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And Newton's method, or this bi-section
algorithm permits me to get better and

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better approximations to that root.
But it's only in that sense that I can

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ever find the root.
It's in the sense that I can get as close

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as you want to that root.