Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.


Math Forum
»
Discussions
»
Courses
»
apcalculus
Notice: We are no longer accepting new posts, but the forums will continue to be readable.
Topic:
[apcalculus] Lagrange Error Bound ln(1+x)
Replies:
1
Last Post:
Apr 12, 2011 4:43 PM




RE: [apcalculus] Lagrange Error Bound ln(1+x)
Posted:
Apr 12, 2011 4:43 PM


Hi Justin,
I think the confusion comes from combining a couple separate parts of the Lagrange error bound into one idea. Let's start with the question of what to put in for your b. We know that Taylor polynomials tend to decrease in accuracy as we move away from the center of the polynomial. This suggests that we should use an endpoint for the value of b. We can actually be more precise about this.
You have R_2 (x) <= M / 3! * b  0^3 which simplifies to R_2 (x) <= M / 3! * b^3.
Notice a couple things about this. First, it really is true that as b increases (in magnitude), the error estimate will grow. The error bound has the form (constant)*b^3. This expression is 0 if b = 0 and increases as b moves away from 0. Second, note that it really doesn't matter which endpoint of the interval you plug in for b. You're going to take the absolute value anyway. While it may not seem this way, it will always not matter in exactly this way... as long as your interval of interest is symmetric about the center of the polynomial. b  c is the distance between b and c, and this distance is the same at either endpoint of the interval (as long as the interval is symmetric). So I would just put in 0.1 for b and call it a day.
Selecting a value for M is a separate issue. M is a bound on the value of the (n+1)st derivative on the interval in question, in this case (0.1, 0.1). In application, M is usually *not* the actual maximum of the relevant derivative on the interval. Usually the max of the relevant derivative is either hard to find, or (more likely) not useful for computation. So we don't actually have to be too careful about matching the maximum value of the 3rd derivative. We just have to make sure we overestimate it. At what xvalue the (n+1)st derivative has its maximum is not really relevant to computing the error bound. You are right that in this particular problem we can actually find the maximum value for the 3rd derivative 2/(x+1)^3 on the given interval since the 3rd derivative is monotonic. (Actually, that's not quite true. The interval you posed is an open interval and so the 3rd derivative doesn't obtain a max on it at all. But let's stay focused on finding M.) I think the fact that we can find and express exactly the least upper bound on f '''(x) in this problem is actually a distraction  if we couldn't find it so exactly, but could only approximate it, I bet you wouldn't be as worried about some of the concerns that have come up. In any event, a choice for M would be 2 / (0.1+1)^3 = 2.74ish. I would just let M = 3 for simplicity. But notice that I haven't paid any attention to the bvalue in this consideration of what M should be, just as I didn't think about M at all when I was considering what to use for b.
Cheers, Benjamin
Benjamin Goldstein The Cambridge School of Weston Weston, MA
Original Message From: Justin OBrien [mailto:OBrienJ@bronxville.k12.ny.us] Sent: Monday, April 11, 2011 3:38 PM To: AP Calculus Subject: [apcalculus] Lagrange Error Bound ln(1+x)
I want to use Lagrange Error Bound for ln(1+x) and I am confused about how to do it. I suppose Alt. Series error bound would be easier to apply but, I want to know how to apply Lagrange also to answer the question below because it raises some questions about the way the formula is stated.
"Find a bound on the maximum error when ln(1+x) is approximated by f(x) = x  (1/2)x^2 on the interval x < 0.1 "
So R(x) <= [ M / (3!) ] x  b  c ^3
c = 0 in this expansion
M = MAX VAL of the 3rd derivative of ln(1+x) on (b,0) or (0,b)
b = ?
Part of my confusion is what to use for b. If I want to maximize the error, I assume I would use b = .1 or b = .1 . But the 3rd derivative of ln(1+x) is 2/(1 + x)^3 . It will attain a maximum value on the interval at the left endpoint , x= .1 . Does that force me to use the interval (.1, 0) ? In other words what determines the value of b in an example such as this where it is not specified explicitly? Does the maximization of the (n+1)th derivative take precedent? Or for instance in this example, can I choose the right endpoint arbitrarily and maximize the (n+1)th derivative on (0, .1)?
Maybe I have answered my own question  would it be correct to say that if an upper bound is sought then one needs to consider both endpoints as candidates for b, then choose the one that gives a higher value for the error?
Thanks Justin
==== Course related websites: http://apcentral.collegeboard.com/calculusab http://apcentral.collegeboard.com/calculusbc  To search the list archives for previous posts go to http://lyris.collegeboard.com/read/?forum=apcalculus To unsubscribe click here: http://lyris.collegeboard.com/read/my_forums/ To change your subscription address or other settings click here: http://lyris.collegeboard.com/read/my_account/edit



