 Welcome to the GVSU Calculus Screencasts. In this video, we're going to consider the problem of finding Taylor polynomials centered at zero. The nth order Taylor polynomial for a function centered at zero is found by taking the function and its derivatives evaluated at zero, dividing by the appropriate factorial, and multiplying by the corresponding power of x, as seen here. In this screencast, we'll apply this formula to find the first few Taylor polynomials for the function f defined by f of x equals the square root of x plus one, centered at zero. Now the first order, or degree one Taylor polynomial for f, centered at zero, is just the tangent line for f, centered at zero. And recall that we calculated tangent lines many times in calculus one. So use your knowledge of tangent lines, pause this video for a moment, and find the first order Taylor polynomial for this function f, centered at zero. Then resume the video when you're ready. Well the first order Taylor polynomial to f at x equals zero is f of zero plus f prime of zero times x. Now f of zero is the square root of zero plus one, which is just one. And the power and chain rules show that f prime of x is one half the quantity x plus one to the minus one half. So f prime of zero is a half. And this makes p sub one of x equal to one plus one half x. Here's a graph of our function f in red along with our Taylor polynomial in blue. This is just the tangent line to f at zero. And you can see that as long as x is fairly close to zero, the tangent line is a pretty good approximation to the graph of f. Now we find the second order Taylor polynomial to f at x equals zero. The second order Taylor polynomial, p sub two of x, is f of zero plus f prime of zero times x plus f double prime of zero divided by two factorial times x squared. And notice that the terms f of zero plus f prime of zero times x give us just p sub one of x. So p sub two of x is p sub one of x plus f double prime of zero or two factorial times x squared. Now we've already found p sub one of x, so use that along with this formula for p sub two of x to find p sub two of x. Pause the video for a moment, find p sub two of x at x equals zero, and resume when you're ready. Now since p sub two of x is p sub one of x plus f double prime of zero over two factorial times x squared, and we already know p sub one of x, to find p sub two of x we just need to find the second derivative of f at zero. Since we've already found p sub one of x, all we have to do is find the second derivative of f at zero and then use this formula. The power and chain rules show that f double prime of x is negative one-quarter quantity x plus one to the negative three-halves power. Now using the power rule and the chain rules again, we get f double prime of x is negative one-quarter times x plus one to the negative three-halves power. And this makes f double prime of zero equal negative one-quarter, and this makes f double prime of zero equal to negative a quarter. And so, p sub 2 of x is just p sub 1 of x, 1 plus 1 half x, minus 1 quarter times 1 over 2 factorial x squared. So, p sub 2 of x is 1 plus 1 half x minus 1 eighth x square. And here we show the graph of p sub 2 of x, which is now quadratic in blue, against the graph of f. And you can see that the graph of p sub 2 of x kind of molds a little bit better around the graph of f of x at zero, than did p sub 1 of x. Let's go ahead now and find the third-order Taylor polynomial to f at x equals zero. As in the previous case with p sub 2, p sub 3 of x is just p sub 2 of x, plus the third derivative of f at zero divided by 3 factorial times x cubed. So again, pause the video for a moment, use the formula we found for p sub 2 of x, plus the third derivative of f at zero to find the third-order Taylor polynomial for f at zero, and resume when you're ready. Now since p sub 3 of x is p sub 2 of x plus the third derivative of f at zero divided by 3 factorial x cubed, you already found p sub 2 of x. To find p sub 3 of x, we just need to find the third derivative of f at zero. Now to find the third-order Taylor polynomial to f at zero, we already know the second-order Taylor polynomial, p sub 2, so we just have to find the third derivative of f at zero. Now the power rule and the chain rules show that the third derivative of f of x is 3 eighths times the quantity x plus 1 to the negative 5 halves. Again, using the power and the chain rules, we get the third derivative of f at x is 3 eighths times the quantity x plus 1 to the negative 5 halves. So the third derivative of f at zero is 3 eighths. So the third derivative of f at zero is 3 eighths. And this makes p3 of x equal to p sub 2 of x plus the third derivative of f of 0, which is 3 eighths, divided by 3 factorial times x cubed. So p sub 3 of x is 1 plus 1 half x minus 1 eighth x squared plus 1 sixteenth x cubed. Here we show the graph of p sub 3 of x against the graph of f of x. It's getting a little harder to see, but the graph of p sub 3 of x is actually fitting the graph of f of x better around x equals 0 than did p sub 1 of x or p sub 2 of x. Let's repeat the process one more time and find the fourth-order Taylor polynomial to f at x equals 0. Now the pattern should emerge here that p sub 4 of x is just p sub 3 of x plus the fourth derivative of f at 0 divided by 4 factorial times x to the 4. So again, pause the video for a moment and find the fourth-order Taylor polynomial for f at 0 and then resume the video when you're ready. So to find the fourth-order Taylor polynomial for f at 0, we've already found the third-order Taylor polynomial for f at 0. So we just need to add the fourth derivative of f at 0 divided by 4 factorial times x to the 4. Differentiating the third derivative of f at x gives us the fourth derivative of f at x is negative 15 sixteenths times the quantity x plus 1 to the negative 7 halves power. And so the fourth derivative of f at 0 is just negative 15 sixteenths. This makes p sub 4 of x equal to p sub 3 of x minus 15 sixteenths divided by 4 factorial times x to the 4. So p sub 4 of x is 1 plus 1 half x minus 1 eighth x squared plus 1 sixteenth x cubed minus 5 over 128 x to the 4. And here's a graph of the fourth-order Taylor polynomial p sub 4 of x against f And again, it's hard to see based on this picture that but the fourth-order Taylor polynomial for f molds around the graph of f more closely than do the earlier Taylor polynomials, at least if we stay close to x equals 0. And we could keep going and find the fifth-order Taylor polynomial for f at 0 shown here, the sixth-order Taylor polynomial for f shown here. And the seventh-order Taylor polynomial p sub 7 of x to f at x equals 0 shown here. Now, it would be worth your while to go through and verify these formulas for the fifth, sixth, and seventh-order Taylor polynomials. And to illustrate the increasing accuracy of these Taylor polynomials for f at 0, we're going to show the graphs successively. So here's the graph of f of x against p sub 1 of x. Here's the graph of f against p sub 2 of x, the graph of f against p sub 3 of x. Here's p sub 4 of x, p sub 5 of x, p sub 6 of x, and p sub 7 of x. And we can see that as we increase the order of the Taylor polynomial, the polynomials fit the graph of f better and better for more values of x around x equals 0. Now these Taylor polynomials are useful for approximating values of functions that are somewhat complicated. For example, if we wanted to approximate the square root of 2, we could use our Taylor polynomial, p sub 7 of x, to do that. So pause the video for a moment and use the seventh-order Taylor polynomial for f at x equals 0 to approximate the square root of 2. Compare this to a result you get from your calculator and see how good your approximation is. And then illustrate what you've done on this graph. Resume the video when you're ready. The first thing to notice here is that the square root of 2 is the square root of 1 plus 1, and that's f of 1. And as long as we stay close to 0, p sub 7 of x is a good approximation to f of x. So we can approximate f of 1 with p sub 7 of 1. Now going back to our formula for p sub 7, substituting 1 in for x, we get p sub 7 of 1 is 2,911 divided by 2,048, which is about 1.4214. And using my calculator, I get the square root of 2 is about 1.414. And that's reasonably close to the approximation we got for the square root of 2 using p sub 7 of 1. And if we look at the points on the graph of p and the graph of f, corresponding to the input x equals 1, shown right here, we can see that p sub 7 of 1 and f of 1 are really hard to tell apart on this picture. Just confirming the fact that p sub 7 of 1 is a pretty good approximation to f of 1. So to summarize, we have this formula for the nth order Taylor polynomial for a function centered at 0. We can use our Taylor polynomial centered at 0 to approximate functions, as long as we stay close to this base point 0. And the accuracy of our approximations increase as we increase the value then. The polynomial approximations are very useful because to evaluate a polynomial, we just have to add, subtract, multiply, and divide, just using the most basic of operations. And this makes polynomial approximations especially useful for calculators, for example, who can only do those basic operations. So we can use these Taylor polynomials to approximate complicated functions by just using these very simple operations. That concludes the screencast on Taylor polynomial centered at 0. Please come back again soon.