<p>On number 2 C and D, why does the Taylor Series only work exactly for x=2, and I just don't understand the lagrange error thing at all D: </p>
<p>Help! Thanks in advance!</p>
<p>This series is just the first three terms in the Taylor series of this function…the actual Taylor series for the function has infinite terms with powers of (x-2) that go on to infinity. To find the value of the function at x, you need to know every single term in the series so when you plug in x you get an infinite series and add all the terms together to get the value of the function. Unfortunately, since only the first 3 terms of the series are provided, you can’t actually evaluate the function for any x-value. But, if you plug in the value x=2, every term is 0 except for the first one. That means you CAN find the value of the function at x=2, because it’s just the first term.</p>
<p>Ah that makes total sense. Thank you! Now about the lagrange error thing D:</p>
<p>Not sure how familiar you are with the Alternating Series Error bound, but it basically says that the error is less than the absolute value of the next omitted term in the series.</p>
<p>So, for instance, suppose I have the alternating series (sigma, n = 1 to infinity, [(-1)^(n+1)]/n) = 1 - 1/2 + 1/3 - 1/4 - 1/5 + 1/6 - 1/7 + 1/8 - 1/9 + 1/10 - 1/11 + …</p>
<p>The sum of the first four terms is 1 - 1/2 + 1/3 - 1/4 = 7/12. How well does this approximate the entire infinite series? Well, the error is less than the absolute value of the next omitted term, which is 1/5. We can see that the sum of the first five terms would subtract 1/5 from what remains, and by pairing up the remaining terms (1/6 - 1/7), (1/8 - 1/9), (1/10 - 1/11), etc., we can see that we’re actually gaining small amounts thereafter.</p>
<p>Well, the Lagrange Error Bound says something somewhat similar for Taylor Series.</p>
<p>To be more precise, the Lagrange Error Bound says that some value of z exists between the center c and the approximating value x where the next omitted term provides the exact value of the error. Only problem is, we almost never know what exactly z is. So what we tend to do instead is to to find an upper bound for this error. We do that by finding the largest possible value of the (n+1)'st derivative on the interval between x and c, and then just say that the actual error is less than this number. That’s where the information that |f^(4) (x)| <= 6 comes into play; they’re helping you to cap the maximum size of the 4th derivative on the interval [0, 2]. The rest is just finding the coefficient for the 4th degree term of the Taylor polynomial where x = 0 and c = 2.</p>
<p>Ah finally I understand this hahaha thank you! Now what formula would be the one I would need to know for logistic growth on the BC exam?</p>
<p>nevermind I got it :)</p>