Taylor's Theorem and the Lagrange Error Bound

Taylor's theorem writes $f(x) = T_n(x) + R_n(x)$: the Taylor polynomial you've been building from derivatives, plus an exact leftover $R_n$. The Lagrange form turns that leftover into a concrete number you can bound — a guarantee on how wrong the approximation can possibly be.

By the end you'll be able to estimate a function value with a Taylor polynomial AND attach a guaranteed error bar to it via $|R_n(x)| \le \dfrac{M}{(n+1)!}|x-a|^{n+1}$ — plus use $R_n \to 0$ to justify that a Taylor series really equals its function.

Predict: move x farther from the center — does the error grow faster than linearly? Slide x to check, then raise n and watch the error collapse.

Blue is $f(x) = \sin x$; dashed green is its Taylor polynomial $T_n$ about $a = 0$. The pink region is the actual error $|f - T_n|$, and the dashed orange envelope is $T_n \pm \frac{|x|^{n+1}}{(n+1)!}$ — the Lagrange bound with $M = 1$. The true curve can never escape the envelope. Hover the chart or focus a dot for exact values.

Degree n = 3

x = 0.40

T₃(0.40) = 0.38933 · sin(0.40) = 0.38942 · actual error = 0.00009 ≤ bound 0.00107

sin x vs. T_n(x) — actual error (pink) inside the Lagrange envelope (orange)

f(x) = sin x Tₙ(x) Lagrange envelope actual error

Taylor's theorem splits $f$ into a degree-$n$ polynomial plus a remainder; the Lagrange form bounds that remainder by $\frac{M}{(n+1)!}|x-a|^{n+1}$, quantifying exactly how accurate the polynomial is.

Visual

Look back at the pink region in the demo: near the center it's invisibly thin, then it fans out — and not at a steady rate. That's the $|x-a|^{n+1}$ in the bound at work: doubling your distance from the center multiplies the worst-case error by $2^{n+1}$, so accuracy is superb close to $a$ and decays fast far away. The other lever is $n$: each extra degree divides the bound by another factor of $(n+1)$ via the factorial, which eventually crushes any power of $|x-a|$ — slide $n$ up and watch the orange envelope clamp down onto the blue curve everywhere.

Formal

If $f$ has $n+1$ continuous derivatives near $a$, then $f(x) = T_n(x) + R_n(x)$, and the Lagrange form of the remainder says there exists some $c$ strictly between $a$ and $x$ with $R_n(x) = \dfrac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$. Note the derivative order: cutting off after degree $n$ leaves an error governed by the $(n+1)$-th derivative, not the $n$-th. Since $c$ is unknown, you trade the equality for an inequality: if $|f^{(n+1)}(t)| \le M$ for every $t$ between $a$ and $x$, then $$|R_n(x)| \le \frac{M}{(n+1)!}|x-a|^{n+1}.$$ This also settles convergence: the Taylor series equals $f(x)$ exactly when $R_n(x) \to 0$ as $n \to \infty$. For $\sin x$ and $\cos x$ every derivative is bounded by $M = 1$, and for $e^x$ by $e^{|x|}$, so the bound $\frac{M}{(n+1)!}|x|^{n+1} \to 0$ for every real $x$ — their Maclaurin series really are the functions.

Worked example

Estimate $\sin(0.4)$ with $T_3$ and bound the error. The degree-3 Maclaurin polynomial is $T_3(x) = x - \frac{x^3}{3!}$, so $T_3(0.4) = 0.4 - \frac{0.064}{6} \approx 0.38933$. For the error, $n = 3$ means the remainder involves $f^{(4)}(c) = \sin c$, and $|\sin c| \le 1$, so take $M = 1$: $|R_3(0.4)| \le \frac{1}{4!}(0.4)^4 = \frac{0.0256}{24} \approx 0.00107$. So $\sin(0.4) = 0.38933 \pm 0.00107$ — and indeed the true value $\sin(0.4) \approx 0.38942$ sits inside that window, just $0.00009$ away. The bound is a worst-case guarantee; the actual error is usually smaller.

Your turn

Estimate $\cos(0.5)$ with $T_2(x) = 1 - \frac{x^2}{2}$ and bound the error. First, $T_2(0.5) = 1 - \frac{0.25}{2} = 0.875$. Now the bound: with $n = 2$, the remainder involves the $3$rd derivative of cosine, which is bounded by $M = 1$, so $|R_2(0.5)| \le \dfrac{1}{3!}(0.5)^{\,?} = $ ____

Reveal the answer

The exponent matches the factorial's order: $|R_2(0.5)| \le \frac{(0.5)^3}{3!} = \frac{0.125}{6} \approx $ 0.02083. So $\cos(0.5) = 0.875 \pm 0.021$ — consistent with the true value $\cos(0.5) \approx 0.87758$, which is only $0.0026$ off. (Sharper trick: since the $x^3$ coefficient of cosine's series is zero, $T_2 = T_3$, so you may bound with $R_3$ instead: $(0.5)^4/4! \approx 0.0026$ — much tighter.)

More info — the remainder is what makes "the series equals the function" honest

Back in the Taylor & Maclaurin series lesson, you built Taylor series by matching derivatives at the center — but matching derivatives at one point doesn't by itself promise anything about values far away. The remainder closes that gap: $f(x) - T_n(x) = R_n(x)$ exactly, so asking "does the series converge to $f$?" is precisely asking "does $R_n(x) \to 0$?" For sine, the demo above shows it happening: fix any $x$, even $x = 4$, and step $n$ upward — the envelope width $\frac{|x|^{n+1}}{(n+1)!}$ shrinks to nothing because the factorial in the denominator outruns the power in the numerator. The OpenStax section in Dive deeper runs this argument in full for $\sin x$, $\cos x$, and $e^x$.

Check your understanding

Question 1 of 4

You approximate $\sin x$ at $x = 1$ with the degree-5 Maclaurin polynomial $T_5$. Taking $M = 1$ (every derivative of sine is bounded by 1), what is the Lagrange bound on $|R_5(1)|$?

Question 2 of 4

The Taylor series of $f$ about $a$ converges at some $x$. Does its sum necessarily equal $f(x)$?

Question 3 of 4

To use the Lagrange error bound $|R_n(x)| \le \dfrac{M}{(n+1)!}|x-a|^{n+1}$, the constant $M$ must satisfy which condition?

Question 4 of 4

From your standard-series toolkit you recall $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$ and truncate it to $T_2(x) = 1 - x^2/2$. With $M = 1$, which expression is the Lagrange bound on the error $|\cos(0.6) - T_2(0.6)|$?

Recap

Taylor's theorem: $f(x) = T_n(x) + R_n(x)$ — the polynomial plus an exact remainder.
Lagrange form: $R_n(x) = \dfrac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$ for some $c$ between $a$ and $x$ — note the $(n+1)$-th derivative, not the $n$-th.
Practical bound: if $|f^{(n+1)}(t)| \le M$ on the whole interval between $a$ and $x$, then $|R_n(x)| \le \dfrac{M}{(n+1)!}|x-a|^{n+1}$. It's an upper bound — the actual error is usually smaller.
Error grows like $|x-a|^{n+1}$ as you leave the center and shrinks factorially as $n$ rises.
The Taylor series converges to $f(x)$ exactly when $R_n(x) \to 0$ — this is how you justify $\sin x$, $\cos x$, and $e^x$ equaling their Maclaurin series for all real $x$.

Dive deeper

OpenStax Calculus Volume 2, §6.3 Taylor and Maclaurin Series State Taylor's theorem and the Lagrange remainder formula.

Sources