Taylor Series Remainder — Definition, Formula & Examples

Q: What is the difference between the Taylor series remainder and the Lagrange remainder?

They refer to the same concept. The Lagrange remainder is a specific formula (also called the Lagrange form of the remainder) that gives R_n(x) = f^{(n+1)}(c)(x−a)^{n+1}/(n+1)! for some c between a and x. It is the most commonly used way to express the Taylor series remainder.

Q: How do you find the value of c in the Taylor remainder formula?

You typically do not find the exact value of c. The theorem only guarantees that such a c exists somewhere between a and x. In practice, you find an upper bound for |f^{(n+1)}(c)| over the interval from a to x, which gives you a worst-case error bound without knowing c precisely.

Q: When does the Taylor series remainder go to zero?

If the remainder R_n(x) → 0 as n → ∞ for a particular value of x, the Taylor series converges to f(x) at that point. This is how you prove a Taylor series actually represents the original function, not just that it converges to something. Functions like e^x, sin(x), and cos(x) have remainders that go to zero for all x.

Taylor Series Remainder

A quantity that measures how accurately a Taylor polynomial estimates the sum of a Taylor series.

Formulas for Taylor series, Taylor polynomial, remainder R_n(x), derivative form, and integral form of remainder using f^(n+1)...

Key Formula

R_n(x) = f(x) - P_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x - a)^{n+1}

Where:

$R_n(x)$ = The remainder (error) after using the degree-n Taylor polynomial
$f(x)$ = The actual value of the function at x
$P_n(x)$ = The Taylor polynomial of degree n centered at a
$f^{(n+1)}(c)$ = The (n+1)th derivative of f evaluated at some point c between a and x
$(n+1)!$ = The factorial of (n+1)
$a$ = The center of the Taylor expansion
$c$ = Some value between a and x (guaranteed by the Mean Value Theorem)

Worked Example

Problem: Use the Taylor remainder theorem (Lagrange form) to find an upper bound on the error when approximating e^(0.5) using the degree-3 Taylor polynomial centered at a = 0.

Step 1: Write the Lagrange remainder formula for n = 3.

R_3(x) = \frac{f^{(4)}(c)}{4!}(x - 0)^4

Step 2: Identify the (n+1)th derivative. Every derivative of e^x is e^x, so f^{(4)}(c) = e^c.

f^{(4)}(c) = e^c

Step 3: Since c is between 0 and 0.5, and e^x is increasing, the maximum of e^c on [0, 0.5] is e^{0.5}. Use the safe upper bound e^{0.5} < 2.

|R_3(0.5)| \le \frac{2}{4!}(0.5)^4

Step 4: Compute the numerical bound. Note that 4! = 24 and (0.5)^4 = 0.0625.

|R_3(0.5)| \le \frac{2 \times 0.0625}{24} = \frac{0.125}{24} \approx 0.00521

Answer: The error in approximating e^{0.5} with the degree-3 Taylor polynomial is at most approximately 0.00521.

Another Example

This example differs by using a trigonometric function whose higher derivatives are bounded by 1, making the bound easier to compute without needing to estimate e^c or any growing quantity.

Problem: Find an upper bound on the error when approximating sin(0.1) using the degree-3 Maclaurin polynomial (Taylor polynomial centered at 0).

Step 1: The degree-3 Maclaurin polynomial for sin(x) is P_3(x) = x - x^3/6. The remainder uses n = 3, so we need the 4th derivative.

f^{(4)}(x) = \sin(x)

Step 2: Apply the Lagrange remainder with n = 3 and x = 0.1.

|R_3(0.1)| = \left|\frac{\sin(c)}{4!}(0.1)^4\right|

Step 3: Since |sin(c)| ≤ 1 for all c, we can bound the remainder simply.

|R_3(0.1)| \le \frac{1}{24}(0.0001) = \frac{0.0001}{24} \approx 4.17 \times 10^{-6}

Answer: The error is at most about 0.00000417, so the degree-3 polynomial gives an extremely accurate approximation of sin(0.1).

Frequently Asked Questions

What is the difference between the Taylor series remainder and the Lagrange remainder?

They refer to the same concept. The Lagrange remainder is a specific formula (also called the Lagrange form of the remainder) that gives R_n(x) = f^{(n+1)}(c)(x−a)^{n+1}/(n+1)! for some c between a and x. It is the most commonly used way to express the Taylor series remainder.

How do you find the value of c in the Taylor remainder formula?

You typically do not find the exact value of c. The theorem only guarantees that such a c exists somewhere between a and x. In practice, you find an upper bound for |f^{(n+1)}(c)| over the interval from a to x, which gives you a worst-case error bound without knowing c precisely.

When does the Taylor series remainder go to zero?

If the remainder R_n(x) → 0 as n → ∞ for a particular value of x, the Taylor series converges to f(x) at that point. This is how you prove a Taylor series actually represents the original function, not just that it converges to something. Functions like e^x, sin(x), and cos(x) have remainders that go to zero for all x.

Taylor Series Remainder (Lagrange Form) vs. Remainder of a Series (General)

	Taylor Series Remainder (Lagrange Form)	Remainder of a Series (General)
Definition	Error from truncating a Taylor series at degree n	Sum of all terms after the nth partial sum of any series
Formula	R_n(x) = f^{(n+1)}(c)·(x−a)^{n+1}/(n+1)!	R_n = S − S_n = Σ (from k=n+1 to ∞) a_k
What you need	Derivatives of the function and a bound on them	Knowledge of the series terms or convergence behavior
When to use	Bounding error in polynomial approximations of functions	Estimating the tail of any convergent series

Why It Matters

The Taylor series remainder appears throughout AP Calculus BC and college calculus courses whenever you need to justify how many terms of a Taylor polynomial are sufficient for a desired accuracy. Engineers and scientists use remainder bounds to decide, for example, how many terms of a series expansion to keep in a computation so the error stays below a specified tolerance. Understanding the remainder is also the key to proving that a Taylor series converges to its original function, not merely to some other function.

Common Mistakes

Mistake: Using n instead of n+1 in the remainder formula — writing f^{(n)}(c)(x−a)^n / n! instead of f^{(n+1)}(c)(x−a)^{(n+1)} / (n+1)!.

Correction: The remainder after a degree-n polynomial always involves the (n+1)th derivative, the (n+1)th power of (x−a), and (n+1)! in the denominator. Think of it as the 'next term' beyond what the polynomial includes.

Mistake: Trying to find the exact value of c rather than bounding the derivative over the interval.

Correction: The point c is unknown. Instead, find the maximum of |f^{(n+1)}(t)| for t between a and x. This gives an upper bound on the error, which is what the problem almost always asks for.

Related Terms

Taylor Polynomial — The approximating polynomial whose error the remainder measures
Taylor Series — The infinite series that the polynomial truncates
Remainder of a Series — General concept of error from partial sums
nth Derivative — Required to evaluate the Lagrange remainder formula
Factorial — Appears in the denominator of the remainder formula
Convergence Tests — Used to verify the series converges before bounding error
Accuracy — The remainder directly quantifies approximation accuracy
Convergent Series — Remainder → 0 proves the Taylor series is convergent to f(x)