Why Exponential Error Bounds Are Convenient
When a Taylor polynomial approximates \(e^x\), the approximation is useful only if we also understand how far it could be from the true value. Taylor’s theorem provides a bound on that error. For the exponential function, the calculation is especially direct: every derivative of \(e^x\) is \(e^x\).
The important choice is not just the degree of the polynomial. We must also bound the relevant derivative over the whole interval where the theorem says its value could occur. Choosing a derivative value only at the input, without checking that it controls the entire interval, can produce a bound that is too small.
The Taylor Polynomial and Its Remainder
For a function \(f\) with sufficiently many derivatives, the degree-\(n\) Taylor polynomial centered at \(a\) is
Taylor’s theorem with the Lagrange remainder says that, if \(f\) has \(n+1\) continuous derivatives on the interval between \(a\) and \(x\), then for some number \(c\) between \(a\) and \(x\),
For \(f(x)=e^x\), every derivative is \(e^x\), so the remainder takes the form
where \(c\) lies between \(a\) and \(x\). Since \(e^c\) is positive, taking absolute values gives
Why the Derivative Maximum Must Cover the Interval
The remainder formula does not say that the unknown \(c\) equals \(x\), or that it equals \(a\). It says only that \(c\) lies between them. Therefore, a valid choice of \(M\) must satisfy \(e^t\leq M\) throughout that interval. Because \(e^t\) is increasing, its maximum on an interval occurs at the right endpoint.
For a particular input \(x\), the interval runs from \(a\) to \(x\). Thus the pointwise choice \(M=e^{\max(a,x)}\) is valid. If we need one bound that works for every \(x\) in a specified interval \([L,U]\), and the center \(a\) is also in that interval, then every relevant \(c\) is in \([L,U]\). The single choice \(M=e^U\) works for all those inputs.
Deriving the Error Bound
The error estimate follows directly from the remainder formula and the maximum just described. This short derivation makes clear both the source of each factor and the condition needed for the estimate.
For some \(c\) between \(a\) and \(x\), Taylor’s theorem gives \(R_n(x)=e^c(x-a)^{n+1}/(n+1)!\).
Since \(e^c>0\), this gives \(|R_n(x)|=e^c|x-a|^{n+1}/(n+1)!\).
If \(e^t\leq M\) for every \(t\) between \(a\) and \(x\), then \(e^c\leq M\).
Substituting \(e^c\leq M\) yields \(|R_n(x)|\leq M|x-a|^{n+1}/(n+1)!\).
This argument also explains the two parts of the estimate: \(M\) controls the derivative, while \(|x-a|^{n+1}/(n+1)!\) records the distance from the center and the degree of the polynomial.
Worked Examples
Worked Example: A Uniform Bound on the Interval from 0 to 1
Use the degree-4 Taylor polynomial for \(e^x\) centered at \(0\) to bound the approximation error for every \(x\in[0,1]\).
At \(a=0\), the degree-4 polynomial is
The fifth derivative is \(e^x\). For every \(x\in[0,1]\), the intermediate value \(c\) is also in \([0,1]\), where \(e^c\leq e\). Also, \(|x|\leq1\). Therefore,
For example, at \(x=1\), the polynomial gives
The actual error there is \(e-\frac{65}{24}\approx0.00995\), which is less than \(e/120\approx0.02265\). The bound is valid throughout \([0,1]\), not only at \(x=1\).
Worked Example: A Pointwise Bound at \(x=0.5\)
Approximate \(e^{0.5}\) with the degree-3 Taylor polynomial centered at \(0\), and find a valid error bound for this input.
The polynomial is
Substituting \(x=0.5=\frac12\) gives
The intermediate value \(c\) lies between \(0\) and \(0.5\). Since \(e^t\) is increasing, \(e^c\leq e^{0.5}\). The degree is \(3\), so the remainder involves the fourth power of the distance:
Using \(e^{0.5}\approx1.64872\), this bound is \(1.64872/384\approx0.00429\). The actual error is approximately \(e^{0.5}-79/48\approx0.00289\), which is below the bound. This checks both the polynomial value and the error estimate.
Worked Example: A Negative Input and the Correct Derivative Maximum
Use the degree-2 Taylor polynomial centered at \(0\) to approximate \(e^{-1}\). Find an error bound, taking account of the fact that the intermediate value lies on the negative side of the center.
The polynomial and its value at \(-1\) are
Here \(c\) lies between \(-1\) and \(0\). On that interval, \(e^c\leq e^0=1\); the maximum derivative value is \(1\), not \(e\). Since \(|-1-0|=1\),
Numerically, \(e^{-1}\approx0.36788\), so the actual error is \(\left|0.36788-0.5\right|\approx0.13212\), less than \(1/6\). The bound uses the maximum of \(e^c\) over the full interval \([-1,0]\), as required.
A Common Mistake: Using the Derivative at the Input
In the negative-input example, it would be incorrect to set \(M=e^{-1}\) merely because the input is \(-1\). That choice produces the unsupported estimate
But the actual error is about \(0.13212\), so this claimed bound is false. The mistake is treating the derivative value at the input as if it were an upper bound for \(e^c\). The theorem does not guarantee \(c=-1\); it allows any \(c\) between \(-1\) and \(0\), and \(e^c\) can be as large as \(1\) there.
Pointwise and Uniform Bounds
A pointwise bound is calculated for one specified input \(x\). Its interval for \(c\) runs only between the center \(a\) and that input, so it can use a relatively small maximum. A uniform bound must hold for every input in a stated interval. It must use a derivative maximum valid across all the corresponding intervals for \(c\).
For instance, when the center is \(0\) and the inputs lie in \([-1,1]\), the intermediate values can lie anywhere in \([-1,1]\). Thus a uniform derivative maximum is \(e\), giving
For a single negative input, the pointwise maximum may instead be \(1\), as in the example at \(x=-1\). The uniform estimate is broader because it must cover positive inputs too. Both estimates are valid when their derivative maxima cover the required intervals; they simply answer different questions.
Check Your Understanding
For each question, identify the interval where the intermediate value can lie before choosing a derivative maximum.
- For the degree-2 Taylor polynomial of \(e^x\) centered at \(0\), what uniform error bound applies on \([0,1]\)?
- For a pointwise estimate at \(x=-0.4\) with center \(0\), what is a valid maximum for \(e^c\)?
- Why does a uniform estimate on \([-1,1]\) require a larger derivative maximum than the pointwise estimate at \(x=-1\)?
- What is the power of \(|x-a|\) in the Lagrange remainder for a degree-5 Taylor polynomial?
- Explain why using \(e^x\) as the derivative maximum for a negative input \(x\) may fail.
