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Topic: [ap-calculus] Lagrange Error Bound ln(1+x)
Replies: 1   Last Post: Apr 12, 2011 4:43 PM

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Benjamin Goldstein

Posts: 434
Registered: 9/6/09
RE: [ap-calculus] Lagrange Error Bound ln(1+x)
Posted: Apr 12, 2011 4:43 PM
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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
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 x-value 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 b-value 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.


Benjamin Goldstein
The Cambridge School of Weston
Weston, MA

-----Original Message-----
From: Justin OBrien []
Sent: Monday, April 11, 2011 3:38 PM
To: AP Calculus
Subject: [ap-calculus] 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


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