Error Estimates For The Exponential Near A Point

Why the center and input both matter

A Taylor polynomial for \(e^x\) is built from the function’s derivatives at a chosen center \(a\). Its accuracy at an input \(x\) depends not only on the distance \(|x-a|\), but also on how large the relevant derivative can become between \(a\) and \(x\). For the exponential, every derivative is again the exponential, so the largest derivative value occurs at the larger endpoint.

That endpoint detail matters when the input lies to the right of the center: the derivative bound must then use \(e^x\), not \(e^a\). When the input lies to the left, it uses \(e^a\). We will turn that observation into a bound, check it in several examples, and use it to select a degree for a required accuracy.

The Taylor polynomial and its error bound

For a function with \(n+1\) continuous derivatives between \(a\) and \(x\), Taylor’s theorem with the Lagrange remainder writes the function value as its degree-\(n\) Taylor polynomial plus a remainder. For \(f(t)=e^t\), all derivatives equal \(e^t\). In particular, \(f^{(k)}(a)=e^a\) for every nonnegative integer \(k\).

Definition: The degree-\(n\) Taylor polynomial for \(e^x\) centered at \(a\) is \[ P_n(x)=e^a\sum_{k=0}^{n}\frac{(x-a)^k}{k!}. \] At a fixed input \(x\), the approximation error is \(R_n(x)=e^x-P_n(x)\).

If \(x\ne a\), Taylor’s theorem gives a point \(\xi\) strictly between \(a\) and \(x\) such that

R_n(x)=\frac{e^\xi(x-a)^{n+1}}{(n+1)!}.

The exponential is increasing, since its derivative \(e^t\) is positive for every real \(t\). Thus its maximum on the interval with endpoints \(a\) and \(x\) is \(e^{\max(a,x)}\). Taking absolute values in the remainder formula gives the bound below.

Theorem: For \(x\ne a\), the error in approximating \(e^x\) by \(P_n(x)\) satisfies \[ |R_n(x)|\leq B_n =\frac{e^{\max(a,x)}|x-a|^{n+1}}{(n+1)!}. \] If \(x=a\), then \(P_n(a)=e^a\) and \(R_n(a)=0\) exactly, so the error bound is \(0\).

The condition \(x\ne a\) in the formula for \(B_n\) is useful when comparing bounds for successive degrees. If \(a=x\), every bound \(B_n\) is zero, and a ratio of successive bounds would be \(0/0\), which is undefined. The exact case is simpler: the polynomial already gives the function value at its center.

Why the endpoint maximum gives the bound

The only function-specific step in the Lagrange estimate is bounding \(e^\xi\). Since \(\xi\) lies between \(a\) and \(x\), it cannot be greater than the larger endpoint. Monotonicity therefore gives \(e^\xi\leq e^{\max(a,x)}\). The other factor in the absolute remainder formula is exactly \(|x-a|^{n+1}/(n+1)!\). Combining these facts proves the stated bound.

More explicitly, when \(x>a\), the largest derivative value on the interval is \(e^x\), and the bound is \(e^x(x-a)^{n+1}/(n+1)!\). When \(x<a\), the largest derivative value is \(e^a\), and the bound is \(e^a(a-x)^{n+1}/(n+1)!\). This dependence on direction is why replacing the endpoint maximum by \(e^a\) in every case can give a false bound.

Worked examples

Worked Example: Approximate \(e^{0.5}\) from the center zero

Use the degree-two Taylor polynomial centered at \(a=0\). Here \(x=0.5\), \(x-a=0.5\), and the polynomial is

P_2(0.5) =1+\frac{0.5}{1!}+\frac{(0.5)^2}{2!} =1+0.5+0.125 =1.625.

Since \(x>a\), the maximum exponential value between the endpoints is \(e^{0.5}\). The remainder bound is

|R_2(0.5)| \leq \frac{e^{0.5}(0.5)^3}{3!} =\frac{e^{0.5}(0.125)}{6} =\frac{e^{0.5}}{48} \approx 0.034348.

As a numerical check, \(e^{0.5}\approx1.648721\), so the actual absolute error is approximately \[ |1.648721-1.625|=0.023721. \] It is less than the bound \(0.034348\), as required. The polynomial value and the error bound are therefore consistent with the theorem.

Worked Example: Use a nonzero center to approximate \(e^{1.2}\)

Take \(a=1\), \(x=1.2\), and \(n=3\). The displacement is \(x-a=0.2=1/5\), so

P_3(1.2) =e\left(1+\frac15+\frac{(1/5)^2}{2!}+\frac{(1/5)^3}{3!}\right) =e\left(1+\frac15+\frac1{50}+\frac1{750}\right) =\frac{458e}{375}.

Because \(1.2>1\), the largest derivative value is \(e^{1.2}\). Thus

|R_3(1.2)| \leq\frac{e^{1.2}(0.2)^4}{4!} =\frac{e^{1.2}(1/625)}{24} =\frac{e^{1.2}}{15000} \approx 0.000221341.

For a check, \(458e/375\approx3.319928206\), while \(e^{1.2}\approx3.320116923\). Their absolute difference is about \(0.000188717\), which is below \(0.000221341\). The endpoint maximum \(e^{1.2}\), rather than \(e\), is essential to this certified bound.

Worked Example: Approximate to within one millionth

Find the smallest degree certified by this bound to approximate \(e^{1.1}\) using the center \(a=1\), with error less than \(10^{-6}\). Write \(d=|1.1-1|=0.1\). For every degree \(n\),

B_n=\frac{e^{1.1}(0.1)^{n+1}}{(n+1)!}.

First test degrees three and four. Since \(e^{1.1}>1\),

B_3 =\frac{e^{1.1}(0.1)^4}{4!} >\frac{0.0001}{24} \approx 0.000004167 >0.000001.

So degree three does not meet the requested tolerance. To check degree four, use \(1.1<2\) and \(e<3\), which imply \(e^{1.1}<e^2<9\). Therefore

B_4 =\frac{e^{1.1}(0.1)^5}{5!} <\frac{9(0.00001)}{120} =0.00000075 <0.000001.

To see that no lower degree succeeds, for \(x\ne a\) the successive bounds have ratio

\frac{B_{n+1}}{B_n} =\frac{|x-a|}{n+2}.

Here this ratio is \(0.1/(n+2)<1\), so the bounds decrease as the degree increases. Every degree below three has a bound at least as large as \(B_3\), which already exceeds the tolerance. Degree four is consequently the smallest degree certified by this bound.

The corresponding polynomial is

P_4(1.1) =e\left(1+0.1+\frac{0.01}{2}+\frac{0.001}{6}+\frac{0.0001}{24}\right) =e\left(\frac{265241}{240000}\right).

Its error is less than \(0.00000075\), hence less than \(10^{-6}\). This verifies the selected degree against the prescribed tolerance.

Worked Example: The input lies to the left of the center

Approximate \(e^{1.5}\) with the degree-one polynomial centered at \(a=2\). Here \(x-a=-0.5\), so

P_1(1.5)=e^2\left(1-\frac12\right)=\frac{e^2}{2}.

Since \(1.5<2\), the maximum derivative value is \(e^2\), not \(e^{1.5}\). The bound is

|R_1(1.5)| \leq \frac{e^2(0.5)^2}{2!} =\frac{e^2}{8} \approx 0.923632.

Numerically, \(P_1(1.5)\approx3.694528\) and \(e^{1.5}\approx4.481689\), giving actual error about \(0.787161\). This is below \(0.923632\), checking the bound for a negative displacement.

Using the bound carefully

For a fixed input different from the center, increasing the degree decreases the bound once \(|x-a|/(n+2)<1\). This condition is often satisfied quickly when the input is near the center. The exact ratio is useful for checking how successive bounds change, but it must be used only when \(a\ne x\); at the center, each bound is simply zero.

When selecting a degree for a tolerance \(\varepsilon>0\), the aim is to find an integer \(n\geq0\) for which \(B_n<\varepsilon\). A bound below the tolerance certifies the approximation. If a particular bound is above the tolerance, that fact alone does not prove that the actual error exceeds the tolerance; it says only that this bound does not certify it. In the degree-selection example, minimality was established because the bounds decrease with degree and the bound at degree three itself exceeded the tolerance.

Common pitfall: For \(x>a\), using \(e^a\) as the maximum derivative value is incorrect. For example, when approximating \(e^{1.2}\) about \(a=1\) with degree three, the incorrect bound would be \(e(0.2)^4/24\). The derivative can reach \(e^{1.2}>e\) on the interval, so this is not a valid upper bound. The correct bound is \(e^{1.2}(0.2)^4/24\).

The method is valuable because it turns a local polynomial approximation into a quantitative guarantee. The center determines the coefficients, the distance from the center contributes a power of \(|x-a|\), and the largest derivative value controls the scale of the remainder. Keeping all three features straight makes the estimate useful for both error certification and degree selection.

Check Your Understanding

Use the endpoint maximum carefully, and distinguish a certified error bound from the actual error.

Write the degree-two Taylor polynomial for \(e^x\) centered at \(a=2\).
For \(x=2.1\), which endpoint supplies the maximum exponential value in the Lagrange error bound?
State the degree-\(n\) error bound for \(e^x\) when \(x<a\).
Why is the ratio \(B_{n+1}/B_n\) not defined when \(x=a\), and what is the error in that case?
If a calculated bound exceeds a requested tolerance, does that prove the actual error exceeds the tolerance? Explain.