AP Calculus BC · Unit 10: Infinite Sequences & Series
10.12 Lagrange Error Bound
What the Lagrange Error Bound measures
When you approximate a function with a Taylor polynomial of degree n centered at a, the Lagrange Error Bound gives a maximum possible value for the remainder (error). It answers: 'How far off could my approximation be?' This lets you bound error without knowing the exact function value.
The formula
The bound is: |R_n(x)| ≤ [M·|x−a|^(n+1)] / (n+1)!, where M is the maximum value of |f^(n+1)(t)| on the interval between a and x. You need the (n+1)th derivative and its max magnitude on that interval — finding M often requires checking endpoints or critical points.
Using the bound in practice
To guarantee an approximation is accurate within a given tolerance, set the error bound expression ≤ the desired tolerance and solve for n or x. Common on FRQs: 'Show the error is less than 0.001' — you plug in the largest possible M and simplify, rather than computing the actual remainder.
Key terms
- Lagrange Error Bound formula
- |R_n(x)| ≤ M|x−a|^(n+1)/(n+1)!, where M bounds |f^(n+1)(t)| on the interval between a and x.
- What is M in the formula?
- The maximum value of the absolute value of the (n+1)th derivative on the interval between a and x.
- Why use the Lagrange Error Bound?
- To find an upper limit on how far a Taylor polynomial approximation could be from the true function value, without knowing the exact remainder.
- Which derivative order matters for a degree-n polynomial?
- The (n+1)th derivative — one more than the polynomial's degree.
- Typical FRQ use of the bound
- Set the error bound expression ≤ a given tolerance (like 0.001) and show the inequality is satisfied to prove accuracy.
Try an AP-style question
Which is the Lagrange error bound for a degree-n Taylor polynomial about a?
- M(n + 1)! / |x − a|ⁿ⁺¹
- M|x − a|ⁿ⁺¹ / (n + 1)!
- |f(x) − Pₙ(a)|
- M|x − a|ⁿ / n!
Show the answer
B. M bounds |f⁽ⁿ⁺¹⁾| between a and x.
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