 Welcome to the GVSU Calculus screencasts. In this video, we'll consider the problem of finding Taylor polynomials for functions centered at points other than zero. Now, the nth order Taylor polynomial for a function f centered at x equals a is given by this expression. Notice the difference here between the Taylor polynomial centered at a and the Taylor polynomial centered at zero. We center our Taylor polynomial at a, we're looking at a polynomial in terms of powers of x minus a. When a is zero, we just get powers of x, but here we get powers of x minus a, and that's an important distinction to make. Now, in a previous screencast, we looked at the first few Taylor polynomials centered at zero for the function f defined by f of x equals square root of x plus one. Now, of course, as we see with this formula for the piece of n of x centered at a, Taylor polynomials can be centered at any point, not just zero. So in this screencast, we're going to look at the Taylor polynomials of the same function f of x equals square root of x plus one, but this time centered at x equals three, so our a value will be three. And we should see as we go through this screencast that the problem is essentially the same as it was for finding Taylor polynomials centered at zero. Only we have to make sure we use powers of x minus three instead of powers of x in our polynomials. Now, we know that the first order or degree one Taylor polynomial for a function is just the tangent line, and we've calculated tangent lines many times. Well, let's pause the video for a moment and find the first order Taylor polynomial piece of one of x to this function f at x equals three, and resume the video when you're ready. We know the tangent line, or the first order Taylor polynomial to f at x equals three, is piece of one of x equals f of three plus f prime of three times x minus three. So we just need to find f of three and f prime of three to find this Taylor polynomial. f of three is just the square root of three plus one, which is two. The power and chain rules give us the derivative of f of x. And so f prime of three in this case is just a quarter. That makes piece of one of x equal to two plus one quarter times x minus three. Notice the power here of x minus three. The Taylor polynomials for f centered at x minus three will be good approximations to f as long as we stay close to the base point x equals three. And you can see by the graph of piece of one here and f of x that as long as we stay close to x minus three, piece of one of x is pretty close to f of x. Now we find the second order Taylor polynomial to f at x equals three. Here we have piece of two of x is f of three plus f prime of three times x minus three plus the second derivative of f at three divided by two factorial times x minus three squared. And notice that those first two terms just combine to give us piece of one of x. So to find piece of two of x, we just take piece of one of x, add the second derivative of f evaluated at three divided by two factorial times x minus three squared. We've already calculated piece of one of x. So pause the video for a moment. Use that formula for piece of one of x plus this formula for piece of two of x to figure out exactly what piece of two of x is. This second order Taylor polynomial for f centered at x equals three. And resume the video when you're ready. Now the second order Taylor polynomial piece of two of x to f at x equals three is piece of one of x plus the second derivative of f at three divided by two factorial times this power, the square of x minus three. The power and chain rules show that f double prime of x is negative one quarter x plus one to the negative three halves. So f double prime of three is negative one over 32. This makes piece of two of x equal to piece of one of x minus one over 32 divided by two factorial times x minus three squared or two plus one quarter x minus three minus one over 64 x minus three squared. And again, notice the powers of x minus three that are in this Taylor polynomial. Here's the graph of piece of two of x against f of x. And again, as long as we stay close to x equals three, the second order Taylor polynomial more closely fits the graph of f of x than did the first order Taylor polynomial for f centered at x equals three. Let's find the third order Taylor polynomial to f at x equals three. As before, the third order Taylor polynomial to f at x equals three is the second order Taylor polynomial to f centered at x equals three plus the third derivative of f at three divided by three factorial times x minus three cubed. And again, you just have to remember these powers of x minus three. We already found piece of two of x. All we have to do is find the third derivative of f at three in order to find piece of three of x. So go ahead and pause the video for a moment and find a formula for piece of three of x centered at x equals three. And resume the video when you're ready. Again, to find the third order Taylor polynomial for f centered at x equals three, we already know the second order Taylor polynomial for f centered at x equals three, so we just need the third derivative of f evaluated at three. The third derivative of f at x is three-eighths times x plus one to the negative five-halves. So the third derivative of f evaluated at three is three over 256. And therefore, piece of three of x equals piece of two of x plus three over 256 divided by three factorial times x minus three cubed. So piece of three of x is two plus one-quarter x minus three minus one over 64 x minus three squared plus one over 512 x minus three cubed. And again, these are all polynomials in powers of x minus three. Here's a graph of p three of x against f of x. Again, we can see that as long as x is close to three, piece of three of x is a pretty good approximation to f of x. Let's repeat this process one more time and find the fourth order Taylor polynomial to f at x equals three. Piece of four of x is going to be piece of three of x plus the fourth derivative of f evaluated at three divided by four factorial times x minus three to the four. Use the formula we found for piece of three of x plus this formula for piece of four of x to find the formula for piece of four of x, the fourth order Taylor polynomial for f centered at x equals three. Resume the video when you're ready. To find the fourth order Taylor polynomial, the piece of four of x, we've already found piece of three of x, so all we need is the fourth derivative of f evaluated at three. We can find the fourth derivative of f evaluated at x, which tells us that the fourth derivative of f evaluated at three is negative 15 over 2048. This gives us piece of four of x is two plus one quarter x minus three minus 164 times x minus three squared plus one over 512 times x minus three cubed minus five over 16,384 times x minus three to the four. Again, it's important to note that these Taylor polynomials are polynomials in powers of x minus three. Here's the graph of piece of four of x against f of x. And again, notice that as long as x is close to three, piece of four of x is a good approximation to f of x. This is why we want to keep piece of four of x and all of these Taylor polynomials in powers of x minus three because we're interested in what happens when x is close to three. We can similarly calculate the fifth-order Taylor polynomials f centered at x equals three, shown here. The sixth-order Taylor polynomial to f centered at x equals three, shown here. And the seventh-order Taylor polynomial to f at x equals three. It would be a good exercise to go and verify these formulas for these fifth, sixth, and seventh-order Taylor polynomials for f at x equals three. We want to illustrate the increasing accuracy of these Taylor polynomial approximations for x close to three. Let's look at the graphs in succession. Here's the graph of piece of one of x versus f of x. Here's piece of two of x versus f of x. Piece of three of x versus f of x. Piece of four of x. Piece of five of x. Piece of six of x. And piece of seven of x. And we can see that as the order of the Taylor polynomial, the value of n, increases, these Taylor polynomials fit the graph of f better and better for more values of x, extending out from the base point x equals three. Taylor polynomials are used to approximate functions for values of x that are close to the base point. So let's go ahead and use piece of seven of x to approximate the square root of five. Pause the video. Make this calculation. Compare the value you obtain from the value you get in your calculator to the square root of five. And then illustrate your approximation on this graph. Resume the video when you're ready. The first thing to notice here is that the square root of five is the square root of four plus one. That's just f of four. And we can approximate f of four with piece of seven, evaluated at four. Using our formula for piece of seven that we found earlier evaluating at x equals four, we get piece of seven of four is this fraction which has an approximate value of 2.236. When I use my calculator to approximate the square root of five, I get about 2.236. And so piece of seven of four is a pretty good approximation to the square root of five. And we can see this on the graph. We have the graph of piece of seven of x against f of x. And if we evaluate when x is four, we get this point right here. And you can see that it's really difficult to tell piece of seven of four and f of four apart because they're so close together. So to summarize, the nth order Taylor polynomial for a function f centered at x equals a is given by this polynomial expansion in powers of x minus a. And this is the important thing to remember here is that when we're centering our Taylor polynomials a, we get a polynomial in powers of x minus a. We use these Taylor polynomials to approximate values of f of x as long as x stays close to a. And as we increase the order, the value of n, our approximations to f of x get better and better, and they get good for more and more values of x in general. And one advantage to using Taylor polynomials centered at points other than the origin is if we want to approximate a function f at some point c, we look for a value of a to center our Taylor polynomials that's near c, and we need that value to be one at which we can evaluate f and all of its derivatives. And that way, we can use typically a smaller value of n in piece of n of c to approximate f of c than if we centered our polynomials, say, at zero. That concludes our screencast on Taylor polynomials centered at points other than the origin. Please come back soon.