<p>Do we hafta know lagrange error bound for the bc test tomorrow? I don't seem to see it in the PR book. If so, does anyone wanna explain it to me? I'm so lost for series/seq questions :(</p>
<p>Okay, I’m going to try to explain this one because I’m almost positive that we DO need to know this for the BC test. It doesn’t come up every year but it’s there often enough to study it.</p>
<p>Basically, the question will usually be worded like, “Use the Lagrange error bound to show that the nth-degree Taylor polynomial for f about x = c
approximates x = _ with error less than _.”</p>
<p>So, the inequality you basically have to set up is |Rn(x)| >/= the (n+1)th derivative of f evaluated at some value C divided by (n+1)! all multiplied by (x - c)^(n+1)</p>
<p>The value C has to be either the center, the number you’re trying to approximate, or some number in between. Graph the function and whichever one gives you the max. value is your C. Then just plug in that x-value as your C. The lower case c is the center given in the problem and the n is the degree of the series that they gave in the problem. |Rn(x)| is the maximum error that they gave you in the problem and x is the number you’re trying to approximate. When you solve all of that (and yes, you get an actual numerical value), it should be lower than the error they gave you in the problem (the Rn(X)).</p>
<p>Confusing as hell, I know.</p>
<p>There’s a much easier way to explain it.
If it says to get the error bound on a 3rd degree polynomial:
Then you take the fourth derivative of it (as you would if you were expanding a taylor series). Then you multiply it by the |x-c| (c is the value for which the equation is centered, x is what you are approximating). Then take (f’‘’'x times (|x-c|^4))/4!</p>
<p>^Yeah, that’s probably a better explanation, but you’re leaving out an important part. You say “then you take the fourth derivative of it” which wouldn’t necessarily give you a constant value. You would have to take the fourth derivative evaluated at whichever value in the interval of approximation (the number which you are approximating to the number you are using for approximation) will give you the maximum.</p>
<p>Thanks! I hope the test isn’t hard tomorrow :)</p>