From Approximation Error to a Derivative Bound
A Taylor polynomial replaces a function by a polynomial built from its derivatives at a chosen center. The difference between the function and that polynomial is the remainder, or approximation error. To know how accurate the approximation is, it helps to express that error using a derivative evaluated somewhere between the center and the point of interest.
The Lagrange form of Taylor’s theorem does exactly this. Its proof is a careful use of Rolle’s theorem: we construct a function with several vanishing derivatives at the center and a further zero at the evaluation point. Repeated applications of Rolle’s theorem force one higher derivative to vanish at an intermediate point.
The polynomial matches the function’s value and derivatives through order \(n\) at the center. The theorem below explains how the next derivative controls the remainder at another point.
The Lagrange Remainder Theorem
When \(x=a\), the remainder is zero directly from the definition, so the bound holds as well. The conditions in the theorem ensure that Rolle’s theorem can be applied repeatedly on the interval between the two points. If \(x<a\), we still work on the interval with endpoints \(x\) and \(a\); the point \(c\) lies between them, and the absolute value in the bound handles the reversed direction.
Why Repeated Rolle Applications Give the Formula
We prove the exact remainder formula first. The bound then follows by taking absolute values. Fix \(x\ne a\), and write \(R_n(x)=f(x)-T_n(x)\). Define the constant \(K=R_n(x)/(x-a)^{n+1}\), and introduce the auxiliary function \(G(t)=f(t)-T_n(t)-K(t-a)^{n+1}\).
At \(t=x\), the choice of \(K\) gives \(G(x)=0\). At \(t=a\), both \(f(t)-T_n(t)\) and its derivatives through order \(n\) vanish. This follows by differentiating the Taylor polynomial: for each \(k\) from \(0\) through \(n\), \(T_n^{(k)}(a)=f^{(k)}(a)\). The term \((t-a)^{n+1}\) and its derivatives through order \(n\) also vanish at \(a\). Consequently, \(G(a)=G'(a)=\cdots=G^{(n)}(a)=0\).
\(G(a)=G(x)=0\), so Rolle’s theorem gives an interior point \(u_1\) with \(G'(u_1)=0\).
Since \(G'(a)=0\) as well, Rolle’s theorem applied to \(G'\) between \(a\) and \(u_1\) gives a point \(u_2\) between them with \(G''(u_2)=0\).
At each stage, \(G^{(j)}(a)=0\) and the point found at the preceding stage is a zero of \(G^{(j)}\). Rolle’s theorem therefore gives a new interior zero of \(G^{(j+1)}\).
After \(n+1\) applications, there is a point \(c\) strictly between \(a\) and \(x\) such that \(G^{(n+1)}(c)=0\).
This iteration is valid because each derivative used is continuous on the relevant closed subinterval and differentiable in its interior. Now differentiate the definition of \(G\) \(n+1\) times. The Taylor polynomial has degree \(n\), so its \((n+1)\)st derivative is zero. The \((n+1)\)st derivative of \(K(t-a)^{n+1}\) is \(K(n+1)!\). Hence \(G^{(n+1)}(c)=f^{(n+1)}(c)-K(n+1)!=0\), which implies \(K=f^{(n+1)}(c)/(n+1)!\). Substituting the definition of \(K\) proves the exact formula.
To obtain the error bound, take absolute values in that formula. Since \(c\) lies between \(a\) and \(x\), the assumed derivative bound applies at \(c\): \(\lvert f^{(n+1)}(c)\rvert\le M\). Therefore \(\lvert R_n(x)\rvert\le M\lvert x-a\rvert^{n+1}/(n+1)!\), as claimed.
Worked Examples
Worked Example: A Cubic Expanded to Degree One
Let \(f(t)=t^3\), center the Taylor polynomial at \(a=2\), and evaluate it at \(x=1\), using \(n=1\). First, \(f(2)=8\) and \(f'(t)=3t^2\), so \(f'(2)=12\). Thus \(T_1(t)=8+12(t-2)\), and \(T_1(1)=8+12(-1)=-4\). Since \(f(1)=1\), the remainder is \(R_1(1)=1-(-4)=5\).
Here \(f''(t)=6t\). On \([1,2]\), \(\lvert f''(t)\rvert\le12\), so \(M=12\) gives \(\lvert R_1(1)\rvert\le12\lvert1-2\rvert^2/2!=6\). The computed error satisfies \(5\le6\).
The exact formula requires \(5=f''(c)(1-2)^2/2!=6c/2=3c\), so \(c=5/3\). This lies strictly between \(1\) and \(2\). Substitution checks the formula: \(f''(5/3)(-1)^2/2!=10/2=5\), matching the remainder.
Worked Example: A Reciprocal Function at One-Half
Let \(f(t)=1/(1+t)\), \(a=0\), \(x=1/2\), and \(n=2\). The derivatives at the center are \(f(0)=1\), \(f'(0)=-1\), and \(f''(0)=2\). Therefore \(T_2(t)=1-t+t^2\), and \(T_2(1/2)=1-1/2+1/4=3/4\). The exact function value is \(f(1/2)=2/3\), so \(R_2(1/2)=2/3-3/4=8/12-9/12=-1/12\).
The third derivative is \(f'''(t)=-6/(1+t)^4\), so \(\lvert f'''(t)\rvert\le6\) on \([0,1/2]\). Taking \(M=6\), the bound is \(6(1/2)^3/3!=6(1/8)/6=1/8\). Indeed, \(\lvert-1/12\rvert=1/12\le1/8\).
The exact formula also identifies a point giving this error: \(-1/12=[-6/(1+c)^4](1/2)^3/6=-1/[8(1+c)^4]\). Thus \((1+c)^4=3/2\), or \(c=(3/2)^{1/4}-1\). Since \(1<3/2<(3/2)^4\), it follows that \(0<c<1/2\). Substituting this value makes the exact remainder formula equal to \(-1/12\), as required.
Worked Example: A Fourth-Degree Polynomial Expanded to Degree Two
Let \(f(t)=t^4\), \(a=0\), \(x=1\), and \(n=2\). The values \(f(0)\), \(f'(0)\), and \(f''(0)\) are all zero, so \(T_2(t)=0\). Hence \(R_2(1)=f(1)-T_2(1)=1\).
Here \(f'''(t)=24t\), whose absolute value is at most \(24\) on \([0,1]\). The bound is \(24\lvert1-0\rvert^3/3!=24/6=4\), and the error satisfies \(1\le4\).
The exact formula gives \(1=f'''(c)/6=24c/6=4c\), so \(c=1/4\), which lies between \(0\) and \(1\). Checking, \(f'''(1/4)(1)^3/3!=6/6=1\), equal to the computed remainder.
Reading the Bound Correctly
The estimate has two separate ingredients: a bound \(M\) for the absolute value of the \((n+1)\)st derivative over the interval, and the factor \(\lvert x-a\rvert^{n+1}/(n+1)!\). The derivative bound controls how rapidly the function can depart from its degree-\(n\) approximation; the distance factor shows how the error depends on how far the evaluation point is from the center.
The exact formula and the inequality have different roles. The exact formula says that the error equals the next derivative at some intermediate point times the distance factor. The inequality replaces that unknown derivative value with a uniform bound \(M\). It need not be an equality: the actual derivative at the intermediate point may be much smaller than \(M\).
Why the Result Is Useful
The theorem turns a local polynomial approximation into a quantitative statement about its error. Once a suitable \(M\) is known on the interval, the remainder bound tells us how far the function’s value can be from the Taylor polynomial’s value. The factorial in the denominator can make the bound small as the degree increases, while the power of the distance makes the choice of center important.
The theorem does not say that every Taylor series converges to its function, nor does it require an infinite series: it concerns a finite Taylor polynomial and its remainder under the stated differentiability conditions. Its proof rests on a finite chain of Rolle applications, and its estimate follows directly from the exact remainder formula.
Check Your Understanding
Answer these questions using the theorem and its proof.
- For the exact Lagrange remainder formula, where must the point \(c\) lie?
- Why does the auxiliary function \(G\) have its first \(n+1\) derivative values, from order zero through order \(n\), equal to zero at \(a\)?
- If \(\lvert f^{(n+1)}(t)\rvert\le M\) between \(a\) and \(x\), what expression bounds \(\lvert R_n(x)\rvert\)?
- Why is the absolute value \(\lvert x-a\rvert^{n+1}\) needed when stating the bound?
- In the reciprocal-function example, what are the computed remainder and the Lagrange bound?
