6.3 Taylor and Maclaurin Series

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Professor: For each of these following functions, find formulas for the Maclaurin polynomials P0, P1, and P3. And we'll go ahead and add P3 in there-- or P2. Find a formula for the n-th Maclaurin polynomial, and write it using sigma notation. use a graphing utility to compare the graphs of the polynomials with f of x. 

Let's begin with just f of x, while we have it given here. So f at 0-- because these are Maclaurin, we are centering at 0-- that will be 1. f prime of x is e to the x, so f prime of 0 is 1. And you might notice, very quickly, this is going to be the same. So f double prime of 0 is 1. f triple prime of 0 equals 1. 

So our polynomials-- we'll go ahead and write these-- our 0-th polynomial is 1. Our first polynomial is 1 plus 1 over 1 factorial times x-- and we'll just have x because we're centering at 0-- so x. So this is 1 plus x. Our second polynomial, 1 plus x plus 1 over 2 factorial x squared, which is just 1/2. P3, 1 plus x plus 1/2x squared plus 1 over 3 factorial times x cubed. 

So in general, the n-th Maclaurin polynomial, our Pn-- and since we're using the n-th polynomial, we better start with a k there. 0 to n of x to the k divided by k factorial-- that should be the closed form of our n-th polynomial there. 

Now, if we carry these out and we graph them, compare them, at 0-- again, all of these agree at 0-- but around 0, close to 0, they are all very good approximations. That's kind of the idea here. We're finding a function that represents our function-- or a power, excuse me, that represents our function at a close range of something. Now, as you take this out to infinity, there's a good chance that our infinite polynomial is going to match up, which is, again, our big goal here. 

All right sin of x-- well, f of x is sin x, so f of 0-- because for Maclaurin polynomials, is 0-- f prime of x is cosine x. f prime of 0 is 1. f double prime of x is a negative sin x. So our f double prime of 0 is 0. Ooh, I'll try that again. f triple prime of x is then negative cosine x, and so evaluating that at 0, we have negative 1. So our 0-th Taylor polynomial will just be 0. 

Our first Taylor polynomial is 1 over-- we're just going to leave off the 0-- 1 over 1 factorial x, which is just x. Our second Taylor polynomial is going to be x plus 1 over-- or 0 over 2 factorial times x squared, which is still going to be x. Our third Taylor polynomial is x plus-- and we will have negative 1, so we'll have a minus there-- and we know we'll have an x cubed. We'll have 1 over 3 factorial. Let's just go ahead and carry this out a couple more, P4 and P5. 

P4-- x minus x cubed over 3 factorial plus-- and that would be 0, for the next one, so actually, that is our polynomial there. Now, the fifth derivative is going to be 1 again, so x minus x cubed over 3 factorial plus 1, or x to the fifth, and this time it'll be 5 factorial. So every other polynomial is actually the same. 

So let's go ahead and write a closed form for Pn. We'll go k equals 0 to n. And this is alternating, so negative 1 to the k. And these are all odd exponents, if you notice-- x, x cubed, x to the fifth, et cetera. So x to the 2k plus 1 divided by 2k plus 1 factorial, because if you notice, our denominators are doing something very similar. They always match. There is our n-th term power series. Now, if we let n go to infinity, then we would be at sin x. 

Now, let's compare these polynomials. And again, close to 0-- close to 0 in this range here, things look very good. 

All right, cosine x, we're going to have a very similar idea here. f of 0 is 1. f prime is negative sin x, so f prime of 0 is 0. Second derivative, negative cosine x, so the second derivative at 0 is 1-- negative 1. Our third derivative will be sin x. So the third derivative, evaluated at 0, is 0. If you notice, we have alternating but in a slightly different form. 

So we'll go ahead and write these out. The 0-th polynomial is 1. It's just f of 0. The first polynomial is 1 plus 0 over 1 factorial x. This one's going to be 1 plus-- we'll do to the second there-- so this will actually be a minus 1 over 2 factorial x squared. The third polynomial-- 1 minus 1/2 x squared plus 0 over 3 factorial x cubed. The fourth polynomial-- I'll go ahead and go out a few more again-- this is going to be 1 minus 1/2 x squared. That would be a positive 1, so plus 1/4 x to the 4. Actually, it would be four factorial. I'll just go ahead and leave the factorial there. 

So in general, it appears that our n-th Taylor polynomial is going to be negative 1-- we are still alternating-- to the k. It's x to the 2k because they are always even. And then we have 2k factorial, again, because they're always even. Now, let's compare these polynomials. Again, in the neighborhood of 0, things are looking very good. 

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