Why a Taylor Polynomial Needs an Error Bound
A Taylor polynomial gives a local polynomial approximation to a function. Knowing the polynomial alone, however, does not tell us how accurate it is at a particular input. Taylor’s theorem supplies a precise answer: under suitable smoothness conditions, the error is controlled by the function’s next derivative somewhere between the center and the input.
This result is useful whenever an approximation must come with a guarantee. Rather than relying only on how close a few computed values appear, we can bound the maximum possible error using a derivative bound over the relevant interval.
The Taylor Polynomial and Its Remainder
Let \(f\) have derivatives through order \(n\) at a center \(a\). Its degree-\(n\) Taylor polynomial centered at \(a\) is
The remainder is the difference between the function and this polynomial:
Thus \(f(x)=T_n(x)+R_n(x)\). The polynomial is the approximation; the remainder is its actual signed error. The absolute error is \(|R_n(x)|\).
The point \(c\) is generally not known. The theorem’s practical value is that we do not need to find it: a bound on the absolute value of the next derivative throughout the segment gives a bound on the remainder.
Why the Error Bound Follows
The reasoning from Taylor’s theorem to the bound is short and important. The theorem expresses the remainder as a product involving the next derivative at an intermediate point. Taking absolute values and using the assumed bound on that derivative gives the result.
For some \(c\) between \(a\) and \(x\), \(R_n(x)=f^{(n+1)}(c)(x-a)^{n+1}/(n+1)!\).
\(|R_n(x)|=|f^{(n+1)}(c)||x-a|^{n+1}/(n+1)!\).
Since \(c\) is between \(a\) and \(x\), \(|f^{(n+1)}(c)|\le M\). Therefore \(|R_n(x)|\le M|x-a|^{n+1}/(n+1)!\).
This proves the error bound from the theorem. The factor \(|x-a|^{n+1}\) shows why the distance from the center matters, while the factorial and the derivative maximum also affect the size of the guarantee. This is an upper bound, not usually the exact error.
Applying the Bound
A reliable application follows the same sequence each time: identify the degree and center, differentiate one order beyond the polynomial, bound that derivative on the full segment, and substitute into the formula. The interval is the segment connecting the center to the target, not necessarily a larger interval suggested by the function’s domain.
Worked Example: Approximating \(e^{1/2}\) with a Quadratic
Use the degree-2 Taylor polynomial for \(f(x)=e^x\) centered at \(a=0\) to approximate \(e^{1/2}\), and bound the error.
First, \(f(0)=1\), \(f'(x)=e^x\), and \(f''(x)=e^x\), so \(f'(0)=f''(0)=1\). Therefore
At \(x=1/2\), the approximation is
The next derivative is \(f^{(3)}(t)=e^t\). On the interval \([0,1/2]\), \(e^t\) is increasing, so its maximum is \(M=e^{1/2}\). The theorem gives
As a numerical check, \(e^{1/2}\approx1.64872\), so the actual absolute error is approximately \(1.64872-1.625=0.02372\), which is less than \(0.03435\). The derivative bound guarantees the inequality; the decimal comparison is a check of this particular approximation.
Worked Example: Bounding a Quadratic Approximation to \(\sin 1\)
Find the degree-2 Taylor approximation to \(f(x)=\sin x\) centered at \(0\), evaluate it at \(x=1\), and bound its error.
The needed values are \(f(0)=0\), \(f'(x)=\cos x\), and \(f''(x)=-\sin x\). Thus \(f'(0)=1\) and \(f''(0)=0\), giving
At \(x=1\), \(T_2(1)=1\). The next derivative is \(f^{(3)}(t)=-\cos t\), and \(|\cos t|\le1\) throughout \([0,1]\), so \(M=1\) is valid. Therefore
Numerically, \(\sin 1\approx0.84147\), so the actual absolute error is approximately \(|0.84147-1|=0.15853\), which is below \(0.16667\). The approximation and the bound are both checked.
Worked Example: Approximating \(\ln(1.2)\) at the Center \(1\)
Use the degree-2 Taylor polynomial for \(f(x)=\ln x\) centered at \(a=1\), then bound the error at \(x=6/5\).
The derivatives are \(f(1)=0\), \(f'(x)=1/x\), and \(f''(x)=-1/x^2\), so \(f'(1)=1\) and \(f''(1)=-1\). Hence
Since \(6/5-1=1/5\), the approximation is
The next derivative is \(f^{(3)}(t)=2/t^3\). For \(1\le t\le6/5\), this is positive and decreasing, so its maximum absolute value is \(M=2\), attained at \(t=1\). The remainder bound is
As a numerical check, \(\ln(1.2)\approx0.182322\), so the actual error is approximately \(0.182322-0.18=0.002322\), which is less than \(0.002667\).
A Common Mistake: Using the Wrong Derivative
The error after a degree-\(n\) Taylor polynomial depends on the derivative of order \(n+1\), not the derivative of order \(n\). Using the wrong order can produce a bound that is too small to be true.
For the quadratic approximation \(T_2(x)=x\) to \(\sin x\) at \(x=1\), an incorrect calculation might use \(f''(t)=-\sin t\). On \([0,1]\), \(|f''(t)|\le\sin 1\), so this mistake would give the alleged bound
But the actual error is approximately \(0.15853\), larger than \(0.14025\). The calculation cannot be a valid error bound. The correction is to use the next derivative, \(f^{(3)}(t)=-\cos t\), and the valid bound \(|f^{(3)}(t)|\le1\). This gives the correct guarantee \(1/6\approx0.16667\).
Choosing \(M\) Carefully
The number \(M\) must bound the absolute value of the next derivative at every point between the center and the target. A derivative value at the center alone is not enough. When the derivative is continuous, its absolute value has a maximum on a closed, bounded segment, so a maximum exists; sometimes it is simpler to use any convenient upper bound instead.
Also keep the role of the result precise: Taylor’s theorem guarantees that the error is no greater than the calculated bound. It does not say the error equals that bound, and it does not generally determine the sign of the error. A larger valid \(M\) still gives a true estimate, but it may be less informative.
Check Your Understanding
For each question, identify the derivative order and interval before applying a remainder bound.
- For a degree-3 Taylor approximation centered at \(a\), which derivative determines the Lagrange remainder?
- What interval must be considered when bounding the error at a target \(x\) using a Taylor polynomial centered at \(a\)?
- If \(|f^{(4)}(t)|\le 5\) between \(a=2\) and \(x=3\), what upper bound does Taylor’s theorem give for the error of a degree-3 polynomial?
- Why is the value of the next derivative at the center alone insufficient to establish a remainder bound?
- Does the Lagrange remainder bound generally give the exact error or only an upper bound for its absolute value?
