10.12 Lagrange Error Bound

The Lagrange error bound (Taylor’s theorem with remainder) gives a max possible error when you approximate f(x) by its n-th degree Taylor polynomial P_n about a (Maclaurin if a=0). The remainder R_{n}(x) in Lagrange form satisfies |R_{n}(x)| = |f(x) − P_n(x)| ≤ M · |x−a|^{n+1} / (n+1)!, where M is a number so that |f^{(n+1)}(t)| ≤ M for every t between a and x. How to use it: 1. Compute (or write) f^{(n+1)}(t). 2. Find a simple bound M on |f^{(n+1)}(t)| over the interval between a and x. 3. Plug M and |x−a| into M·|x−a|^{n+1}/(n+1)! to get the error bound. Tip: if the Taylor series terms alternate and decrease, the alternating series error bound (next term magnitude) may give a sharper / easier bound. This is a BC-only concept (CED LIM-8.C). For a worked walkthrough, see the Topic 10.12 study guide (https://library.fiveable.me/ap-calculus/unit-10/lagrange-error-bound/study-guide/aDmFnzX0SHl90c5gK9Ol). For more unit review and practice problems, check the unit page (https://library.fiveable.me/ap-calculus/unit-10) and the 1000+ practice questions (https://library.fiveable.me/practice/ap-calculus).

Formula: If P_n(x) is the n-th Taylor polynomial for f about a, Taylor’s theorem in Lagrange form says the remainder R_n(x) equals R_n(x) = f^{(n+1)}(z) / (n+1)! · (x − a)^{\,n+1} for some z between a and x. For an error bound you replace f^{(n+1)}(z) by an upper bound M on |f^{(n+1)}(t)| on the interval between a and x to get |R_n(x)| ≤ M · |x − a|^{\,n+1} / (n+1)!. When to use it: use the Lagrange error bound whenever you approximate a function by a Taylor (or Maclaurin) polynomial and need a guaranteed maximum error (e.g., “approximate e^2 to within 0.001” or justify truncation on the AP BC exam). If the series is alternating with decreasing terms, the alternating series error bound can sometimes give a simpler (and usually smaller) bound—also allowed by the CED. For a focused review see the Topic 10.12 study guide (https://library.fiveable.me/ap-calculus/unit-10/lagrange-error-bound/study-guide/aDmFnzX0SHl90c5gK9Ol) and practice problems (https://library.fiveable.me/practice/ap-calculus).

Short answer: they’re different tools for different situations. Lagrange error bound (Taylor’s theorem with remainder) gives a guaranteed maximum size of the remainder Rn(x) for a Taylor (or Maclaurin) polynomial. If your n-th Taylor polynomial is centered at a and you can bound the (n+1)-th derivative by M on the interval, then |Rn(x)| ≤ M |x−a|^(n+1)/(n+1)!. Use it when you can bound that (n+1)-th derivative—it’s the general AP method (CED LIM-8.C). Alternating series error bound applies only when the Taylor series (or series you’re using) is alternating with terms that decrease in absolute value to 0. Then the error after n terms is at most the absolute value of the next term. It’s often simpler and sharper for alternating Maclaurin series. When to pick which: if the series is alternating with monotone decreasing terms → use the alternating series error bound. If not, or you need a rigorous max error over an interval and you can bound derivatives → use Lagrange remainder. Both appear on the BC exam (Unit 10); for a quick refresher see the Lagrange study guide (https://library.fiveable.me/ap-calculus/unit-10/lagrange-error-bound/study-guide/aDmFnzX0SHl90c5gK9Ol) and practice problems (https://library.fiveable.me/practice/ap-calculus).

Use the alternating-series error bound when the Taylor (or Maclaurin) series you’re using is an alternating series whose terms decrease in absolute value to 0. Then the remainder after n terms satisfies |R_n| ≤ |next term| (the (n+1)th term). That’s quick and tight when the series meets the Alternating Series Test conditions (signs alternate, |a_k| ↓ 0). Use the Lagrange error bound (Taylor’s theorem with remainder) when the series/approximation isn’t alternating or when you need a guaranteed bound based on derivatives. Lagrange’s form says |R_n(x)| ≤ M · |x−a|^{n+1}/(n+1)!, where M bounds the (n+1)th derivative on the interval. It’s more general (CED LIM-8.C.1) but can be messier if M is hard to estimate. For AP BC problems: check first whether the series is alternating and decreasing—if yes, use the alternating-series bound (often gives a simpler bound); otherwise apply the Lagrange remainder using a derivative bound (CED LIM-8.C.2). See the Topic 10.12 study guide for examples (https://library.fiveable.me/ap-calculus/unit-10/lagrange-error-bound/study-guide/aDmFnzX0SHl90c5gK9Ol). For extra practice, try problems at the unit page (https://library.fiveable.me/ap-calculus/unit-10) or the AP practice bank (https://library.fiveable.me/practice/ap-calculus).

You pick M as any upper bound for |f^(n+1)(z)| on the closed interval between the center a and the point x (Taylor’s theorem/Lagrange form requires a max on that interval). Practically: - State the interval: I = [min(a,x), max(a,x)]. - Compute or estimate f^(n+1)(t) and find a simple bound on |f^(n+1)(t)| for all t in I. That bound is your M. - Use M in the Lagrange remainder: |R_n(x)| ≤ M · |x−a|^(n+1)/(n+1)!. Common shortcuts: for e^t, every derivative = e^t, so M = e^{max(I)}; for sin or cos, |f^(k)| ≤ 1 so M = 1; for rational powers or (1−x)^{-m} use monotonicity or known growth to bound derivatives. If the series is alternating with decreasing terms, you can sometimes use the alternating series error bound (smaller/simple). For AP alignment see LIM-8.C and the Topic 10.12 study guide (https://library.fiveable.me/ap-calculus/unit-10/lagrange-error-bound/study-guide/aDmFnzX0SHl90c5gK9Ol) and grab extra practice at (https://library.fiveable.me/practice/ap-calculus).

Use Taylor's theorem with the Lagrange remainder: for the Maclaurin polynomial P_n(x) of f, the remainder R_n(x) satisfies |R_n(x)| ≤ M · |x|^(n+1) / (n+1)!, where M is a bound on |f^(n+1)(t)| for t between 0 and x. For f(x)=sin x: - Every derivative is ±sin x or ±cos x, so |f^(k)(t)| ≤ 1 for all t. Thus M = 1. - So the Lagrange error bound about 0 is |sin x − P_n(x)| ≤ |x|^(n+1)/(n+1)!. Example: use the cubic Maclaurin (n = 3), P_3(x) = x − x^3/3!. The error bound is |sin x − P_3(x)| ≤ |x|^4/4!. So at x = 0.5, error ≤ (0.5)^4/24 = 0.0625/24 ≈ 0.0026. Note: for alternating Taylor series (like sin x with odd terms) you can sometimes use the alternating-series error bound (next term in magnitude) which is often sharper. This is exactly the Lagrange remainder idea in the AP CED (LIM-8.C). For more examples and practice, see the Topic 10.12 study guide (https://library.fiveable.me/ap-calculus/unit-10/lagrange-error-bound/study-guide/aDmFnzX0SHl90c5gK9Ol) and AP Calc practice problems (https://library.fiveable.me/practice/ap-calculus).

The Lagrange error bound gives a guaranteed maximum size for the remainder when you approximate a function by its n-th degree Taylor (or Maclaurin) polynomial. If Pn(x) is the Taylor polynomial about a and f has (n+1) continuous derivatives, the remainder Rn(x) = f(x) − Pn(x) satisfies |Rn(x)| ≤ M |x − a|^(n+1) / (n+1)! where M is an upper bound on |f^(n+1)(t)| for t between a and x. To use it: (1) pick n and a, (2) find or estimate M on the interval of interest, (3) plug into the inequality to get a guaranteed error ≤ that value. This is what the CED labels the Lagrange remainder / remainder estimation theorem (LIM-8.C). For alternating Taylor series you can sometimes use the alternating series error bound instead (it says the error ≤ the first omitted term). For a short AP review, see the Topic 10.12 study guide (https://library.fiveable.me/ap-calculus/unit-10/lagrange-error-bound/study-guide/aDmFnzX0SHl90c5gK9Ol) and try practice problems at (https://library.fiveable.me/practice/ap-calculus).

R_n(x) is the Lagrange remainder (error) after approximating f(x) by its n-th degree Taylor polynomial about a. Taylor’s theorem (Lagrange form) says there’s some z between x and a with R_n(x) = f^{(n+1)}(z)/(n+1)! · (x−a)^{\,n+1}. You usually can’t find z, so use the remainder estimation theorem: if |f^{(n+1)}(t)| ≤ M on the interval between a and x, then |R_n(x)| ≤ M · |x−a|^{\,n+1} / (n+1)!. That gives a guaranteed error bound (CED LIM-8.C). If the Taylor series at a is alternating with decreasing terms, you can sometimes use the alternating series error bound instead (also in the CED). For worked examples and AP-style practice on using these bounds, see the Topic 10.12 study guide (https://library.fiveable.me/ap-calculus/unit-10/lagrange-error-bound/study-guide/aDmFnzX0SHl90c5gK9Ol) and lots of practice problems (https://library.fiveable.me/practice/ap-calculus).

In the Lagrange error bound Rn(x) ≤ M · |x − a|^(n+1) / (n+1)!, the symbol M means a number that bounds the absolute value of the (n+1)th derivative on the interval between the center a and the point x. Concretely: - M = a number so that |f^(n+1)(t)| ≤ M for every t between a and x. - You don’t need the exact maximum—pick any valid upper bound (often from a simple inequality or monotonicity of f^(n+1)). - Then the bound gives a guaranteed maximum error for the n-th degree Taylor (or Maclaurin) approximation. Example: For e^x centered at 0, every derivative is e^t and on [0,1] we have e^t ≤ e, so M = e and Rn(1) ≤ e · 1^(n+1)/(n+1)!. On the AP BC exam you’ll use this (CED LIM-8.C) to show a guaranteed error; sometimes the alternating-series error bound is easier if the series alternates. For more review, see the Topic 10.12 study guide (https://library.fiveable.me/ap-calculus/unit-10/lagrange-error-bound/study-guide/aDmFnzX0SHl90c5gK9Ol) and practice problems (https://library.fiveable.me/practice/ap-calculus).

You need an interval where you can bound the (n+1)th derivative by some constant M. Taylor’s theorem (Lagrange form of the remainder) says |R_n(x)| ≤ M · |x − a|^(n+1) / (n+1)! provided |f^(n+1)(t)| ≤ M for every t between a and x. So to find the interval: 1. Pick the center a and the degree n. 2. Compute f^(n+1)(x) symbolically. 3. Find an interval around a (say |x−a| ≤ r) on which you can prove |f^(n+1)(t)| ≤ M (choose the smallest convenient M). 4. Then the Lagrange bound is valid for all x with |x−a| ≤ r (because every t between a and x stays in that interval). Example: for e^x about a=0, f^(k)(x)=e^x. On |x| ≤ 1 you can take M = e^1, so |R_n(x)| ≤ e · |x|^(n+1)/(n+1)! for |x| ≤ 1. If you can’t easily bound the derivative but the series is alternating with decreasing terms, use the alternating series error bound instead. For more worked examples and AP-aligned explanations see the Topic 10.12 study guide (https://library.fiveable.me/ap-calculus/unit-10/lagrange-error-bound/study-guide/aDmFnzX0SHl90c5gK9Ol) and unit resources (https://library.fiveable.me/ap-calculus/unit-10). For extra practice try the AP practice problems (https://library.fiveable.me/practice/ap-calculus).

You don’t need the exact value of c—just a bound for the (n+1)th derivative on the interval between the center a and the point x. Taylor’s theorem says the remainder R_{n}(x) = f^{(n+1)}(c) (x−a)^{n+1}/(n+1)! for some c between a and x. On the AP exam you use the Lagrange error bound by finding M = max{|f^{(n+1)}(t)| : t between a and x}, then |R_n(x)| ≤ M · |x−a|^{n+1}/(n+1)!. So pick M as a convenient upper bound of |f^{(n+1)}(t)| on [min(a,x), max(a,x)]. Example: for e^x all derivatives = e^t, so M = e^{max endpoint}; for sin or cos use M = 1. If the series is alternating with decreasing terms, you can instead use the alternating-series error bound (next term). This is the BC-only Lagrange remainder approach in the CED (LIM-8.C). For more examples see the Topic 10.12 study guide (https://library.fiveable.me/ap-calculus/unit-10/lagrange-error-bound/study-guide/aDmFnzX0SHl90c5gK9Ol) and practice problems (https://library.fiveable.me/practice/ap-calculus).

Most likely you over-estimated the (n+1)th derivative M or forgot the effect of the distance from the center. The Lagrange bound |R_n(x)| ≤ M · |x−a|^(n+1)/(n+1)! is valid but often very crude if you pick a huge M (take the max of |f^(n+1)(t)| on the whole interval). Common mistakes that give a huge (and useless) number: - using a global upper bound instead of the actual max on the interval (e.g., using e^10 when you only need max on [0,2]); - expanding about the wrong center a (Maclaurin vs a shifted center) so |x−a| is large; - small n (factorial not big enough) while derivatives grow rapidly (like e^x or high powers); - arithmetic slip in computing f^(n+1) or plugging the endpoints. Fixes: recompute the correct (n+1)th derivative, find a tighter M on the specific interval, consider expanding about a point closer to x, or use the alternating series error bound when the series meets its conditions (often gives a much smaller bound). For AP guidance see LIM-8.C in the CED and the Topic 10.12 study guide (https://library.fiveable.me/ap-calculus/unit-10/lagrange-error-bound/study-guide/aDmFnzX0SHl90c5gK9Ol). For extra practice try problems at (https://library.fiveable.me/practice/ap-calculus).

Actual error = the true remainder Rn(x) = f(x) − Pn(x): it’s what your Taylor (or Maclaurin) polynomial misses exactly. You usually can’t compute it directly because it needs the unknown value of a higher derivative at some point. The Lagrange error bound is a guaranteed maximum size for that error. Using Taylor’s theorem with remainder, if |f (n+1)(t)| ≤ M on the interval between the center and x, then |Rn(x)| ≤ M |x − a|^(n+1)/(n+1)!. So the bound uses an easily found constant M (a worst-case) and factorial growth to give a safe upper limit on the true error. On the AP BC exam you should be able to set up and use this Lagrange bound (CED LIM-8.C); sometimes an alternating-series error bound gives a tighter/simpler bound when the series alternates. For a quick study guide see Fiveable’s topic page (https://library.fiveable.me/ap-calculus/unit-10/lagrange-error-bound/study-guide/aDmFnzX0SHl90c5gK9Ol) and use practice problems (https://library.fiveable.me/practice/ap-calculus) to get comfortable applying M and the (n+1)! term.

You can only use the alternating-series error bound when the Taylor series (or the tail you’re using) actually is an alternating series with terms that decrease in absolute value to 0. If your Taylor polynomial’s individual terms have both + and − signs but do not form a true alternating sequence, the alternating bound does not apply—use the Lagrange remainder instead. How to check and proceed: 1. Look at the general term a_k(x) of the series. Does sign(a_k) = (−1)^k (or (−1)^{k+1}) for all k beyond some index? If yes, it’s alternating. 2. Verify |a_k(x)| is eventually decreasing and → 0. If both hold, the error after n terms satisfies |R_n(x)| ≤ |next term| = |a_{n+1}(x)|. 3. If either condition fails, estimate |R_n(x)| with the Lagrange bound: |R_n(x)| ≤ M·|x−c|^{n+1}/(n+1)!, where M bounds the (n+1)-th derivative on the interval (CED LIM-8.C.1). Example: sin x at 0 is alternating and decreasing, so error ≤ |x|^{n+1}/(n+1)!; e^x is not alternating, so use Lagrange. For AP-aligned support and practice, see the Topic 10.12 study guide (https://library.fiveable.me/ap-calculus/unit-10/lagrange-error-bound/study-guide/aDmFnzX0SHl90c5gK9Ol) and practice questions (https://library.fiveable.me/practice/ap-calculus).

Short answer: error bounds tell you how far your Taylor (or Maclaurin) polynomial could be from the true function value—a guaranteed maximum error. That matters because when you approximate (on tests or in applications) you need to know whether your approximation is good enough. Why it’s useful (AP terms): Taylor’s theorem with the Lagrange remainder gives Rn(x) = f^(n+1)(ξ) (x−a)^(n+1)/(n+1)!, so if you can bound |f^(n+1)(t)| ≤ M on the interval, the Lagrange error bound |Rn(x)| ≤ M |x−a|^(n+1)/(n+1)! (CED: LIM-8.C). Use that to - pick n to guarantee a target precision (e.g., error < 10^−4), - justify answers on the free-response (showing you meet required accuracy), - or use the alternating series error bound when applicable (smaller, simpler bound). Want practice? Review the Topic 10.12 study guide (https://library.fiveable.me/ap-calculus/unit-10/lagrange-error-bound/study-guide/aDmFnzX0SHl90c5gK9Ol) and try problems from the Unit 10 page (https://library.fiveable.me/ap-calculus/unit-10) or the practice bank (https://library.fiveable.me/practice/ap-calculus).

Use Taylor’s Theorem with the Lagrange form of the remainder. If Pn(x) is the degree-n Taylor polynomial for f about a, the error Rn(x) = f(x) − Pn(x) satisfies |Rn(x)| ≤ M · |x − a|^(n+1) / (n+1)!, where M is any upper bound for |f^(n+1)(t)| on the interval between a and x. Steps to check accuracy: 1. Pick the center a and the x value (or interval) where you need the approximation. 2. Choose n (degree you’ll use) and compute the (n+1)th derivative of f. 3. Find M = max{|f^(n+1)(t)|} for t between a and x (or use a simple bound if exact max is hard). 4. Plug into M·|x−a|^(n+1)/(n+1)! to get the guaranteed maximum error. 5. If that bound ≤ your tolerance (e.g., 10^−4), the approximation is “accurate enough.” If not, increase n or shrink the interval. If the series is alternating with decreasing terms → 0, you can instead use the alternating series error bound (error ≤ magnitude of next term). For more examples and AP-aligned practice, see the Topic 10.12 study guide (https://library.fiveable.me/ap-calculus/unit-10/lagrange-error-bound/study-guide/aDmFnzX0SHl90c5gK9Ol) and lots of practice problems (https://library.fiveable.me/practice/ap-calculus).

Short answer: yes—but only when the exam lets you use a calculator. The Lagrange error bound needs an M = max{|f^(n+1)(z)|} for z between the center and your x; if you’re on a calculator-permitted part (the calculator blocks shown in the CED), you may use your graphing calculator to evaluate or numerically maximize that derivative on the interval. If you’re on a calculator-not-permitted part, you must give an analytic bound (use monotonicity, known inequalities, or a simple overestimate). Quick tips: - If f^(n+1)(x) is simple (e.g., trig, exp), often you can identify its max analytically (use derivatives or known ranges). That’s safest for non-calculator parts. - If you use a calculator, still justify the interval you checked and report a conservative M (error bound needs an upper bound). - When stuck, pick a simple bound (like |f^(n+1)(x)| ≤ K on the interval)—the AP rubric accepts conservative M’s. For more practice and worked examples on Lagrange bounds, see the Topic 10.12 study guide (https://library.fiveable.me/ap-calculus/unit-10/lagrange-error-bound/study-guide/aDmFnzX0SHl90c5gK9Ol) and unit resources (https://library.fiveable.me/ap-calculus/unit-10). For extra problems, check the practice set (https://library.fiveable.me/practice/ap-calculus).