Why Taylor Approximation Errors Matter
A Taylor polynomial can make a difficult function value manageable, but the result is trustworthy only if the polynomial is constructed correctly and the input lies in a valid range. A small-looking algebra mistake can change the approximation; an unjustified error bound can make an unreliable answer appear precise.
The main checks are systematic: verify the coefficients, include every term through the stated degree, check the interval of convergence when using a power series, and confirm that the hypotheses of the chosen remainder estimate are satisfied. Finally, keep the approximation separate from the exact value it estimates.
What a Taylor Polynomial Includes
For a function \(f\) with the needed derivatives at \(a\), its degree-\(n\) Taylor polynomial centered at \(a\) is
The index runs from \(k=0\) through \(k=n\). Thus, degree \(n\) means that every power through \((x-a)^n\) is accounted for; it does not mean that the polynomial must have \(n\) nonzero terms. Some coefficients can be zero.
When \(a=0\), the Taylor polynomial is called the Maclaurin polynomial. The coefficient of \((x-a)^k\) is \(f^{(k)}(a)/k!\). The factorial is essential: the coefficient is generally not just the \(k\)th derivative.
Why the Coefficient Formula Has a Factorial
Suppose a polynomial centered at \(a\) is written as \(P_n(x)=\sum_{k=0}^{n}c_k(x-a)^k\). Differentiating \(k\) times and evaluating at \(a\) isolates the \(k\)th coefficient: terms with powers above \(k\) still contain a factor of \(x-a\), while terms below \(k\) have differentiated to zero. The surviving term is \(k!c_k\). Therefore, matching the \(k\)th derivative of \(f\) at \(a\) requires \(k!c_k=f^{(k)}(a)\), or \(c_k=f^{(k)}(a)/k!\).
This also explains a common source of coefficient errors: even when the derivatives are correct, forgetting to divide by \(k!\) produces the wrong polynomial.
Worked Example: Correcting a Missing Factorial
Find the degree-3 Maclaurin polynomial for \(e^x\), evaluate it at \(x=1/5\), and compare it with a calculation that omits a factorial.
Every derivative of \(e^x\) is \(e^x\), so \(f^{(k)}(0)=1\) for \(k=0,1,2,3\). The required coefficients are \(1/0!=1\), \(1/1!=1\), \(1/2!=1/2\), and \(1/3!=1/6\). Thus,
At \(x=1/5\), the terms are \(1\), \(1/5\), \(1/50\), and \(1/750\). Converting to denominator \(750\) gives
If the \(x^2/2\) term is incorrectly written as \(x^2\), the result is \(1+1/5+1/25+1/750=931/750\approx1.2413333\). The difference is \(15/750=1/50=0.02\), exactly the extra \(x^2/2\) that was added.
The Lagrange remainder theorem gives \(|R_3(1/5)|\le M(1/5)^4/4!\), where \(M\) bounds \(|e^x|\) between \(0\) and \(1/5\). Since \(e^x\le e^{1/5}<1.23\) there, the error is less than \(1.23/15000=0.000082\). This supports the correct polynomial’s accuracy; it does not validate the polynomial with the wrong coefficient.
Do Not Skip a Nonzero Term
Degree and term count are different ideas. A degree-4 Taylor polynomial includes powers through the fourth power, even if some of their coefficients vanish. Conversely, omitting a nonzero term of lower degree cannot be justified by using an error bound that assumes that term was included.
Worked Example: A Missing Term in a Sine Approximation
Construct the degree-4 Maclaurin polynomial for \(\sin x\), evaluate it at \(x=1/2\), and examine the mistake of dropping a nonzero term.
The Maclaurin series begins \(\sin x=x-x^3/3!+x^5/5!-\cdots\). Through degree 4, the \(x^2\) and \(x^4\) coefficients are zero, but the \(x^3\) term is not. Therefore,
Writing \(P_4(x)=x\) because “the later terms are small” is not the degree-4 Taylor polynomial: it omits the nonzero cubic term. The omitted amount at \(x=1/2\) is \(1/48\approx0.0208333\).
For the correct polynomial, the alternating-series remainder is at most the first omitted nonzero term, \((1/2)^5/5!=1/3840\approx0.0002604\). This bound applies because the sine-series terms alternate and decrease in magnitude at \(x=1/2\). It does not bound the error of the shortened approximation \(x\), whose error also includes the omitted cubic term.
As a check, \(\sin(1/2)\approx0.4794255\), so its difference from \(23/48\) is about \(0.0002589\), less than \(1/3840\).
Check the Input Against the Convergence Interval
A Taylor polynomial can be formed from derivatives at a center even when the corresponding infinite power series does not converge at a particular input. When using an infinite series as a representation of a function, first check that the input is inside its interval of convergence; endpoints need separate tests. A familiar algebraic pattern is not enough to justify substituting any value.
Worked Example: Using a Geometric Series Outside Its Interval
The geometric expansion
is valid for \(|x|<1\). Its ratio is \(-x\), so the geometric series converges only when \(|-x|<1\). At \(x=2\), the function value is \(1/3\), but the series terms are \(1,-2,4,-8,\ldots\), which do not approach zero.
The partial sum through \(x^3\) is \(1-2+4-8=-5\), not a reliable approximation to \(1/3\). In general, the partial sum through \(x^n\) is
As \(n\) increases, \((-2)^{n+1}\) does not approach zero, so these partial sums do not converge to \(1/3\). The correction is to respect the condition \(|x|<1\); the series cannot be used at \(x=2\).
Use Remainder Bounds Only Under Their Hypotheses
For a Taylor polynomial of degree \(n\), Taylor’s theorem with the Lagrange remainder states that, when \(f\) has \(n+1\) continuous derivatives on the interval between \(a\) and \(x\),
for some \(c\) between \(a\) and \(x\). If \(|f^{(n+1)}(t)|\le M\) throughout that interval, it follows that
The value \(M\) must bound the derivative throughout the interval. Evaluating the derivative only at the center is not enough. The Lagrange formula also identifies an unknown intermediate point \(c\); the remainder is not generally obtained by substituting \(a\) for \(c\). An alternating-series error bound has different hypotheses: the series must alternate, its terms must decrease in magnitude, and the terms must approach zero.
Worked Example: A Zero Derivative at the Center Does Not Give Zero Error
Use the degree-2 Maclaurin polynomial for \(\cos x\) at \(x=1/2\), and find a valid Lagrange error bound.
Since \(\cos x=1-x^2/2+\cdots\), the polynomial is \(P_2(x)=1-x^2/2\), so
Here \(f'''(x)=\sin x\). Taylor’s theorem gives \(R_2(1/2)=f'''(c)(1/2)^3/3!\) for some \(c\) between \(0\) and \(1/2\). On this interval, \(|\sin c|\le1\), so a valid bound is
Using the sharper fact that \(\sin c\le\sin(1/2)<1/2\) gives the stronger bound \(1/96\approx0.0104167\). But setting the remainder to zero because \(f'''(0)=\sin 0=0\) is invalid: the theorem uses \(f'''(c)\) at an unspecified intermediate point, not necessarily at \(0\). Indeed, \(\cos(1/2)\approx0.8775826\), so the actual error is about \(0.8775826-0.875=0.0025826\), which is nonzero and below either valid bound.
An Approximation Is Not an Exact Value
The statement \(f(x)\approx P_n(x)\) means that \(P_n(x)\) is being used as an estimate. It does not assert equality. The difference \(R_n(x)=f(x)-P_n(x)\) is the remainder, and an error bound gives a limit on its magnitude. Unless the remainder is proved to be zero, the polynomial value should not be presented as the exact function value.
For example, in the cosine calculation, \(7/8\) is the Taylor approximation and \(|R_2(1/2)|\le1/48\) is a bound on its error. Neither statement says \(\cos(1/2)=7/8\), nor does the bound claim that the error equals \(1/48\). A bound is a ceiling, not a measurement of the actual error.
A Reliable Error-Checking Routine
Write down \(a\), \(n\), and the requested input \(x\). A degree-\(n\) polynomial includes all powers from \(0\) through \(n\).
Use \(f^{(k)}(a)/k!\) for the coefficient of \((x-a)^k\), including the factorial.
Verify the input lies in the interval of convergence, and test an endpoint separately if needed.
For a Lagrange bound, bound the next derivative on the whole interval; for an alternating-series bound, verify alternation and decreasing term magnitudes.
Report the polynomial value as an approximation and the remainder estimate as a bound, not as an exact error unless it is known exactly.
Common Mistakes to Avoid
- Forgetting the factorial: use \(f^{(k)}(a)/k!\), not just \(f^{(k)}(a)\).
- Confusing degree with number of terms: a degree-4 polynomial includes zero coefficients through degree 4 and every nonzero coefficient in between.
- Skipping a small-looking term: a bound for a correctly formed Taylor polynomial does not automatically bound the error after an additional term has been omitted.
- Ignoring the convergence interval: a series formula need not represent the function at every input.
- Evaluating a derivative only at the center: a Lagrange bound requires a valid bound across the interval.
- Calling a bound the actual error: \(|R_n|\le B\) does not mean \(|R_n|=B\), and does not mean \(R_n=0\).
Check Your Understanding
For each question, identify the check that would prevent the stated error.
- What coefficient belongs to \((x-a)^4\) in a Taylor polynomial for \(f\)?
- Why can a degree-5 Taylor polynomial have fewer than six nonzero terms?
- What must be checked before substituting a value into an infinite power series?
- For a Lagrange remainder bound, where must the derivative be bounded?
- If a remainder has magnitude at most \(0.01\), what does that say—and not say—about the actual error?
