6.3 Taylor and Maclaurin Series

[STARTS AT 20:23]

Professor: f of x equals x to the 1/3, so f of 8 is 2. f prime of x is 1/3x to the negative 2/3, so f prime of 8 is 1/12. f double prime of x is going to be negative 2/9x to the negative 5/3, meaning the second derivative evaluated at 8 is negative 1/144. 

So let's attempt to write our polynomials. P1 is f of 8 plus f prime of 8 times x minus 8, which means we have 2 plus 1/12 x minus 8. P2 of x is f of 8 plus f prime of 8 times x minus 8 plus f double prime evaluated at 8 times x minus 8 squared, which means we have 2 plus 1/12 x minus 8 minus-- once we divide by-- oops-- our 2 factorial there-- so it will be minus 1 over-- 1 over 288 times x minus 8 squared. Actually, I'm not going to go ahead-- I'm not going to compare the graphs here. We did that plenty before, so ignore that. 

All right, now let's use those two polynomials to estimate the cube root of 11. So what we'll do is we'll evaluate each of those-- our P1, we'll evaluate that at 11. That is 2 plus 1/12 x minus-- well, our x is 11-- so 11 minus 8, which is 2.25. There's our approximation. So P2 of 11 is 2 plus 1/12 11 minus 8 plus-- actually, in this case, it is minus-- 1 over 288 times 11 minus 8 squared. And that is equal to 2.21875. 

Now, the idea, though, that we want here is, can we bound the error? Can we guarantee that we're within a certain range? So with Taylor's theorem, the error, the remainder of our first term-- in this case, we'll go ahead and make this at 11-- by Taylor's theorem, this is equal to the second derivative evaluated at c over 2 factorial-- we're going from the first to the second derivative there-- times 11 minus 8 squared for some c in the interval between x, 8 and 11. 

And now, we don't know what that c value is, so what we want to do is find the maximum value. So if you look at the second derivative, the second derivative, we said that was negative 2/9x to the negative 5/3-- now, that is a decreasing function, so the maximum-- mm-- so the maximum occurs at x equals 8. There we go-- so it's decreasing. So our maximum value is x equals 8. So at that point f double prime of 8 is negative 1 over 144. So that is the highest, or the greatest value, of that function at that point. 

So what we can then do, is we're going to take this to be our M. f double prime of c is less than or equal to 1 over 144. So this means our error, R1 at 11, is less than or equal to-- let's see, 1 over 144 divided by 2 factorial times 11 minus 8 squared-- so our error is less than 0.03125. That is our error bound. So at most, we are 0.03125 off from the actual answer. 

Now let's find our error bound for our second remainder there, evaluated at 11. All right, that would be f triple prime of some value c divided by 3 factorial 11 minus 8 cubed. well, f triple prime of x is equal to 10 over 27x to the 8/3. It also is a decreasing function. So our maximum value-- a maximum value of 0.0014468 at x equals 8, so that is going to be our M value. The absolute value of f, the third derivative, evaluated at 8, or any point in there, is less than that M, which means that our remainder error is less than or equal to 0.00144683 factorial-- and really, we're taking the absolute value here-- x, so 11 minus 8 cubed. And that error bound is then going to be 0.0065104. That is our error bound. And that is what we're after. 

[ENDS AT 28:07]