Approximating Functions with Taylor Polynomials
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 6.3: “Taylor and Maclaurin Series” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/6-3-taylor-and-maclaurin-series |
| Textbook used in class | Stewart, Calculus, Section 11.11: “Applications of Taylor Polynomials” (Examples 1, 2, 3) |
Opening Scenario
Some definite integrals like $\int_0^1 e^{-x^2}\,dx$ cannot be computed exactly using elementary antiderivatives. Taylor polynomials provide a practical escape: substitute the power series for the integrand, integrate term by term (which is always possible), and evaluate the resulting polynomial integral. The error is bounded by Taylor’s Inequality.
Quick Reference
Strategy for using Taylor polynomials in applications:
- Replace $f(x)$ by its Taylor polynomial $T_n(x)$ (choosing $n$ to meet the required accuracy).
- Compute with $T_n$ instead of $f$ (integrate it, evaluate it, take its limit, etc.).
- Bound the error using Taylor’s Inequality.
Key Concepts
1. Integrating a Series Term by Term
Example 1. Approximate $\displaystyle\int_0^1 e^{-x^2}\,dx$ to within $0.001$. (Stewart 11.11, Example 1.)
Substitute $u = -x^2$ into the $e^u$ series: \[ e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots \]
Integrate term by term from $0$ to $1$: \[ \int_0^1 e^{-x^2}\,dx = \left[x - \frac{x^3}{3} + \frac{x^5}{10} - \frac{x^7}{42} + \frac{x^9}{216} - \cdots\right]_0^1 \] \[ = 1 - \frac{1}{3} + \frac{1}{10} - \frac{1}{42} + \frac{1}{216} - \cdots \]
This is an alternating series with decreasing terms tending to $0$. By the Alternating Series Estimation Theorem, stopping when a term is less than $0.001$ guarantees error less than $0.001$.
The term $1/216 \approx 0.0046 > 0.001$. The next term is $1/(9 \cdot 7!) \approx 1/45360 \approx 0.000022 < 0.001$.
So stop after five terms: \[ \int_0^1 e^{-x^2}\,dx \approx 1 - \frac{1}{3} + \frac{1}{10} - \frac{1}{42} + \frac{1}{216} \approx 0.7475. \]
Boxed answer: $\displaystyle\int_0^1 e^{-x^2}\,dx \approx 0.7475$, with error less than $0.001$.
2. Evaluating a Difficult Limit Using Series
Series expansions can resolve indeterminate forms faster than repeated L’Hopital applications.
Example 2. Find $\displaystyle\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$.
Substitute the Maclaurin series $e^x = 1 + x + x^2/2! + x^3/3! + \cdots$: \[ e^x - 1 - x = \frac{x^2}{2} + \frac{x^3}{6} + \cdots \] \[ \frac{e^x - 1 - x}{x^2} = \frac{x^2/2 + x^3/6 + \cdots}{x^2} = \frac{1}{2} + \frac{x}{6} + \cdots \to \frac{1}{2}. \]
Boxed answer: $\dfrac{1}{2}$.
Recap. L’Hopital’s rule would require differentiating twice and then evaluating. The series method reads off the limit in one step.
3. The Small-Angle Approximation
For small $|\theta|$ (in radians): \[ \sin\theta \approx \theta, \quad \cos\theta \approx 1 - \frac{\theta^2}{2}. \]
These are the first nonzero terms of the Taylor series for $\sin$ and $\cos$ at $a = 0$. They are widely used in physics and engineering for pendulums, optics, and structural analysis when angles are small.
The error in $\sin\theta \approx \theta$ is at most $|\theta|^3/6$. For $|\theta| \leq 0.1$ radians, the error is at most $(0.1)^3/6 \approx 0.00017$, which is less than $0.02\%$.
assuming the series approximation works for all $x$, not just small $|x-a|$. Taylor polynomials are local approximations, most accurate near the center $a$. Using $T_3$ for $\sin x$ at $x = 10$ would give a wildly inaccurate result. Always check the magnitude of $|x - a|$ before trusting the approximation.
Common Errors Summary
| Error | Correction |
|---|---|
| Integrating the full series as if finite | You can integrate term by term, but be careful to bound the tail (use AST or Taylor’s Inequality) |
| Forgetting to apply the estimation bound | State the error bound; do not just say “approximately” |
| Using degree-$n$ terms but too few for required accuracy | Find $n$ so that the next term is smaller than the required error, then stop |
Leveled Practice
Problem 1. Approximate $\displaystyle\int_0^{0.5} \sin(x^2)\,dx$ to within $0.001$.
Show answer
Substitute $u = x^2$ into $\sin u = u - u^3/6 + \cdots$: $\sin(x^2) = x^2 - x^6/6 + x^{10}/120 - \cdots$
Integrate from $0$ to $0.5$: $\int_0^{0.5} \sin(x^2)\,dx = [x^3/3 - x^7/42 + x^{11}/1320 - \cdots]_0^{0.5}$ $= (0.5)^3/3 - (0.5)^7/42 + \cdots = 0.04167 - 0.000187 + \cdots \approx 0.04148$.
The third term is $(0.5)^{11}/1320 < 0.0001 < 0.001$, so two terms are enough. Answer: $\approx 0.04148$.
Problem 2. Use Taylor series to evaluate $\displaystyle\lim_{x\to 0}\dfrac{\cos x - 1 + x^2/2}{x^4}$.
Show answer
$\cos x = 1 - x^2/2 + x^4/24 - \cdots$, so $\cos x - 1 + x^2/2 = x^4/24 - x^6/720 + \cdots$
$\dfrac{\cos x - 1 + x^2/2}{x^4} = \dfrac{1}{24} - \dfrac{x^2}{720} + \cdots \to \dfrac{1}{24}$.
Mastery Checklist
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