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Welcome to Calculus, I'm professor Greist.

Â We're about to begin lecture 12 on linearization.

Â Derivatives are useful in so many contexts, moving far beyond computing

Â slopes. In this lesson, we'll cover one of the

Â primal application to linearization. This stems from our understanding of

Â Taylor series. Linearization is the first step in a

Â Taylor series, using the first derivative to approximate.

Â Thinking of derivatives in terms of linear variation can be frustrating, if

Â you're trying to visualize it. It's maybe not so easily visualized as is

Â slope, however in some samples you cant early see the variation.

Â Consider the area A, the square of side length x.

Â What happens if we increase x by small amount?

Â The area becomes x squared plus 2xh plus h squared.

Â We can see those terms by increasing the side length a little bit and decomposing

Â the extra area into a pair or rectangles the area x times h and a small square,

Â whose area is h squared. So, ignoring the terms in big O of h

Â squared, we see that the linear variation is 2x times h.

Â For a triangle, we would get something similar, let's say a right triangle with

Â length x and height x, then the area, one half x squared, has the following

Â variation. A of x plus h is one half x squared plus

Â x times h plus one half H squared. Visualizing the additional area, we can

Â break it up into a parallelogram. Whose width is h, and whose height is x,

Â giving a linear variation term of x times h.

Â The leftover material is in big O of h squared.

Â Lastly, for a circular disc of radius x. We have that the area is pi times x

Â squared. What does the variation look like in this

Â case? I ahve x plus h is pi x squared plus 2 pi

Â x h. Plus pi h squared.

Â that means if we increase The area a little bit then the, the first order term

Â has area two pi X times H. But then you've got this leftover higher

Â order term. How does that work?

Â That's a little harder to see. It's maybe a bit easier if you

Â approximate the circular disk by a regular polygon with many sides.

Â And extend those sides out to a height of h.

Â Then you see that there are a number of small triangles.

Â Left over, whose areas will add up, to that of a disc of radius h in the limit

Â as the number of sides is going to infinity.

Â These first order of variations are the basis for doing linear approximation.

Â For example, let's say you want to estimate the length.

Â Of a bannister for a staircase. If you know something about the

Â dimensions, let's say it's 13 units across and 9 units high, then that length

Â would be square root of 9 squared plus 13 squared or the square root of 250.

Â Let's say you want to come up with an estimate for that.

Â I don't know what that exact number is, and let's say I don't have a calculator.

Â Well, I could consider the function f of x equals square root of x.

Â And I do know what the square root of 256 is.

Â So, that's, 16 squared. To get from there to 250, I can use a

Â value of H, a perturbation of negative six.

Â So that from a Taylor Expansion, f of 250 is going to be f of 256 plus, well, the

Â derivative evaluated at 256 times the difference times h.

Â Negative six. Although the terms in the Taylor

Â expansion are of higher order, well, let's see.

Â The square root of 256, that's 16. Minus six over two times 16.

Â And that gives us 15 and 13 16ths. As as answer, and that's not so bad of an

Â approximation. That's about 15.8.

Â The true answer is the same up to the first three digits.

Â So, in this case, linear approximation works pretty well.

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Consider a different example. Let's try to estimate pi raised to the

Â 20th power without using a calculator. Here's a hint.

Â If you look at pi squared, that's pretty close to ten.

Â it's roughly 9.86. I'll let you do the math without a

Â calculator to verify that one fact. Having done that let's consider the

Â function f of x that raises x to the 10th Power and then, if we tailor expand that

Â function, about x equals 10, then we're not too far off of pie squared to the

Â 10th or pie to the 20th. In this case, h your error, is pi squared

Â minus ten. Which, according to oru computation is

Â about, negative 0.14. All right, so in that case pi to the 20th

Â which is pi squared to the 10th. Is approximately 10 to the 10th plus the

Â derivative of x to the 10th. That's 10 times x to the 9th evaluated at

Â x equals 10 times h negative 0.14. Now if we simply that, a little bit, we

Â see that we get 0.86 times ten to the tenth, or 0.86 times ten to the ninth.

Â That's our approximation to pi to the 20th.

Â In scientific notation that would be 8.6E9.

Â The true answer is approximately 8.77E9. We're not terribly close but we at least

Â got the the number of digits correct and the first digit correct as well.

Â We're within 2%. Now I can hear you wondering why we don't

Â simply use a calculator. But what is the calculator?

Â How does the calculator compute a number? Well, a calculator is using linear

Â approximation as well, often through the following algorithm.

Â Newton's method is an algorithm for finding the root of a function f of x.

Â The first step in this method is to guess.

Â Pick some value let's call it x where we can compute the function and its

Â derivative Then, let's say that the root is nearby at some value x plus h, for

Â some h. We're going to linearize f at x, to get f

Â at x plus h equals f of x, plus f prime at x times h, plus Terms and [UNKNOWN]

Â another h squared, ignoring those terms and approximating we can solve for the

Â root. Set that equal to zero and solve for h.

Â That gives us minus f of x over f prime of x...

Â Now, h is not the root, x plus h is the root.

Â And so we get the following formula. And the genius of Newton's method is not

Â to stop there, but to repeat, or iterate. And do it again, getting a better

Â approximation. And then, your last step is to hope that

Â this repetition will converge to the desired root.

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sequence of points x sub n where the next term in the sequence x sub n plus 1 is

Â given by x sub n minus f of x n over f prime at x n.

Â This is Newton's method. Now, hopefully it converges.

Â Let's consider the problem of trying to compute the reciprocol of a number, a.

Â One over a can be realized as the root of f of x equals a minus 1 over x.

Â What would it take to compute that root? Well the derivative of this function is

Â easy doc compute, it's one over x squared, and taking Neuton's update law,

Â substituting in for f and f prime. I'm doing a little bit of

Â simplification... Gives that x n plus 1 equals twice x n

Â minus a times xn squared. What's remarkable, in this formula is

Â that there's no division involved, only multiplication and subtraction.

Â So, let's, fire up the computer. And see if we can't, solve this for a

Â value of a equal to e 2.71828, et cetera. We'll make an initial guess of x not

Â equals 1/2. And then, we'll start computing using

Â Newton's updated law that we have derived.

Â And what we see is that after a few short steps, just four steps, seems to be

Â giving us 6 digits of accuracy. That's pretty good convergence.

Â Likewise, if we want to compute a cube root of A, we can realize that as the

Â root of F of X equals X cubed minus A. In this case also the derivative is

Â trivial to compute... And inserting these into Newton's update

Â law gives us xn plus 1 equals 2 3rds xn plus a over 3 times xn squared.

Â Now again, if we say evaluate this For a = 100.

Â What's the cube root of 100? We need to make a guess.

Â I know the cube root of 125 is five, so let's use that for our initial guess.

Â And then doing the computations Gives a value that seems to be converging

Â extremely quickly. Even after three steps, we've got as many

Â digits of accuracy as represented on this computer.

Â So what we see is that sometimes Newton's method converges very quickly.

Â But other things can happen. You have to be careful.

Â Let's say you pick an initial condition and start applying Newton's method.

Â One of the things that can happen is that you do not converge to the root you

Â wanted to get. You may converge to a different root.

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Worse things can happen still. Depending on your initial condition, you

Â might wind up getting a value XN for which the derivative there vanishes.

Â As you can see either graphically or from the update law, this leads to a situation

Â where XN plus one is undefined. And this is really bad.

Â You might not converge to anything. But despite those dangers, Newton's

Â method and linearization are extremely useful in practice.

Â Linearization pervades mathematics, and there are all manner of things that you

Â can linearize. You can linearize data.

Â You can linearize differential equations and dynamics.

Â You can even linearize operators. We'll get to a little bit of this later

Â on in this course, but some of it is going to have to wait for you to take

Â more math. And so, we see the power of linearization

Â using the first derivative to approximate.

Â But this is really just the first step in a broader tailor expansion.

Â In our next lesson, we'll consider higher derivatives and their uses across a

Â variety of applications.

Â