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[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.