MATH-101 · Lesson 5 of 5 · 41 min
Sequences, series and Taylor polynomials
Your calculator does not store a table of sines. When you press sin(0.3), it adds up a few terms of a polynomial that matches the sine function extremely well near zero. The same idea runs through engineering: the small-angle approximation, linearized models in control systems, simplified formulas for cable sag and pendulum periods, and the Fourier series that break signals into frequencies. All of them replace a complicated function with an infinite sum of simple terms, then keep only the few that matter.
This lesson builds that toolkit. It starts with sequences and infinite series, develops tests for whether a series adds up to a finite number, and ends with Taylor polynomials: how to build them, how accurate they are, and how engineers use them every day.
Sequences
A sequence is an ordered list of numbers \(a_1, a_2, a_3, \ldots\), usually given by a formula for the \(n\)th term. The sequence converges to \(L\) if its terms get as close to \(L\) as you like for large \(n\): \(\lim_{n \to \infty} a_n = L\). Otherwise it diverges.
The limit techniques from Calculus I all apply, with \(n\) in place of \(x\).
- \(a_n = \dfrac{3n + 1}{n + 4} \to 3\): divide top and bottom by \(n\).
- \(a_n = r^n \to 0\) when \(|r| < 1\), and grows without bound when \(|r| > 1\).
- \(a_n = (-1)^n\) diverges: it bounces between −1 and 1 and never settles.
- \(a_n = \left(1 + \dfrac1n\right)^n \to e\). The terms are 2, 2.594, 2.705 and 2.717 for \(n\) = 1, 10, 100 and 1000. This is how \(e\) arises in compound interest and continuous growth.
Sequences describe anything measured in steps: the state of a system at each time step of a simulation, successive estimates from Newton's method, the balance in an account each month, or the remaining strength of a part after each load cycle.
Series and partial sums
A series adds up the terms of a sequence: \(\sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + \cdots\). An infinite sum is defined through its partial sums \(S_N = a_1 + \cdots + a_N\). If the partial sums converge to a number \(S\), the series converges and its sum is \(S\).
Geometric series
The most important series multiplies each term by a fixed ratio \(r\):
Why: the partial sum is \(S_N = a\dfrac{1 - r^{N+1}}{1 - r}\) (multiply \(S_N\) by \(r\) and subtract to see this). When \(|r| < 1\), \(r^{N+1} \to 0\). When \(|r| \ge 1\) the series diverges.
Worked problem
A bouncing part
A part dropped from 2 m rebounds to 60% of its previous height each bounce. How far does it travel in total before coming to rest?
Show solution
- First fall: 2 m. After that, each bounce goes up and comes back down.
- Rebound heights: 1.2, 0.72, 0.432, … m: a geometric series with \(a = 1.2\) and \(r = 0.6\), summing to \(\dfrac{1.2}{0.4} = 3\) m.
- Each rebound is travelled twice (up and down), so the total is \(2 + 2(3) = 8\) m.
Answer: 8 m, although it bounces infinitely many times
Worked problem
Build-up from repeated doses
A chemical dosing system adds 100 g of treatment to a tank every day, and 75% of what is present breaks down before the next dose. What level does the tank settle to just after each dose?
Show solution
- Just after the \(n\)th dose, the amount is \(100 + 100(0.25) + 100(0.25)^2 + \cdots\) (each earlier dose has shrunk by a factor of 0.25 per day).
- As \(n \to \infty\), this approaches \(\dfrac{100}{1 - 0.25} \approx 133\) g.
Answer: About 133 g. The same calculation applies to repeated loads, drug doses and pollutant build-up
Repeating decimals
Geometric series also explain why \(0.999\ldots = 1\): it is \(0.9 + 0.09 + 0.009 + \cdots = \dfrac{0.9}{1 - 0.1} = 1\).
Testing for convergence
Most series cannot be summed exactly, but we still need to know whether they converge. Several tests answer that question.
The divergence test
If the terms do not approach zero, the series must diverge. For example, \(\sum \dfrac{n}{n + 1}\) diverges because its terms approach 1. The converse is false: terms going to zero does not guarantee convergence.
The harmonic series
The famous counterexample is the harmonic series:
Group the terms: \(\tfrac13 + \tfrac14 > \tfrac12\), \(\tfrac15 + \cdots + \tfrac18 > \tfrac12\), and so on. Infinitely many groups each exceed \(\tfrac12\), so the sum grows without bound. It grows extremely slowly, though: the first thousand terms add up to only about 7.49, and the first million to about 14.39.
The integral test and p-series
If \(f(x)\) is positive and decreasing, the series \(\sum f(n)\) and the integral \(\int_1^\infty f(x)\,dx\) either both converge or both diverge. The terms of the series are rectangles under (or over) the curve. Combined with the p-test from Lesson 4:
So \(\sum 1/n^2\) converges. Its exact sum, found by Euler, is \(\pi^2/6 \approx 1.6449\).
The ratio test
For series with factorials or powers, compare each term with the next:
If \(L < 1\) the series converges; if \(L > 1\) it diverges; if \(L = 1\) the test gives no answer. The ratio test treats a series as "eventually like a geometric series with ratio \(L\)."
Worked problem
Ratio test
Does \(\sum_{n=1}^{\infty}\dfrac{n^2}{3^n}\) converge?
Show solution
- \(\dfrac{a_{n+1}}{a_n} = \dfrac{(n+1)^2}{3^{n+1}}\cdot\dfrac{3^n}{n^2} = \dfrac13\left(\dfrac{n+1}{n}\right)^2 \to \dfrac13\).
- \(L = \tfrac13 < 1\), so the series converges.
Answer: Yes. (Its sum turns out to be exactly 1.5.)
Alternating series
A series whose signs alternate converges if its terms shrink steadily to zero in size. Better still, the error from stopping after \(N\) terms is no bigger than the first term you left out. This makes alternating series very convenient for calculation, because you know exactly how accurate your answer is.
Worked problem
Error bound
The alternating harmonic series \(1 - \tfrac12 + \tfrac13 - \tfrac14 + \cdots\) sums to \(\ln 2 \approx 0.6931\). How accurate is the sum of the first five terms?
Show solution
- Five terms: \(1 - 0.5 + 0.3333 - 0.25 + 0.2 = 0.7833\).
- The first omitted term is \(\tfrac16 \approx 0.167\), so the error is at most 0.167.
- The actual error is \(0.7833 - 0.6931 = 0.090\), within the bound.
Answer: Within 0.167 (actually off by 0.09). This series converges slowly; Taylor series usually do much better
Power series
A power series is an infinite polynomial: \(\sum_{n=0}^{\infty} c_n x^n = c_0 + c_1x + c_2x^2 + \cdots\). Whether it converges depends on \(x\). The ratio test shows that every power series converges for \(|x| < R\) and diverges for \(|x| > R\), for some radius of convergence \(R\) (which can be 0 or ∞).
- The geometric series \(\sum x^n = \dfrac{1}{1 - x}\) has \(R = 1\).
- \(\sum \dfrac{x^n}{n!}\) has \(R = \infty\): the factorial eventually crushes any power.
Inside the radius, power series behave like polynomials. You can differentiate and integrate them term by term, which is what makes them so useful.
Taylor series
Suppose you want a polynomial that matches a function \(f\) near \(x = a\) as closely as possible. Matching the value, the slope, the curvature and every higher derivative at \(a\) gives the Taylor series:
Around \(a = 0\) it is called a Maclaurin series. Stopping after the \((x - a)^n\) term gives the Taylor polynomial of degree \(n\), written \(T_n(x)\). The first two terms are the tangent-line approximation from Calculus I.
Worked problem
The exponential series
Find the Maclaurin series of \(e^{x}\).
Show solution
- Every derivative of \(e^x\) is \(e^x\), so every derivative at 0 equals 1.
- \[e^{x} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = \sum_{n=0}^{\infty}\frac{x^n}{n!}\]
Answer: This converges for every x. With x = 1 it gives e = 1 + 1 + 1/2 + 1/6 + 1/24 + … ≈ 2.71828
The series you will use most:
- \(e^{x} = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots\) (all \(x\))
- \(\sin x = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots\) (all \(x\))
- \(\cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots\) (all \(x\))
- \(\ln(1 + x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots\) (\(-1 < x \le 1\))
- \((1 + x)^k = 1 + kx + \dfrac{k(k-1)}{2!}x^2 + \cdots\) (binomial series, \(|x| < 1\))
New series can be built from known ones by substitution, multiplication, differentiation or integration, without computing any derivatives. Replacing \(x\) with \(-x^2\) in the exponential series gives \(e^{-x^2} = 1 - x^2 + \dfrac{x^4}{2!} - \cdots\), which can then be integrated term by term. That is how the bell-curve integral from Lesson 1, which has no elementary antiderivative, is computed to any accuracy.
How accurate is a Taylor polynomial?
Taylor's theorem gives the error. If you approximate \(f\) by \(T_n\), the remainder is
for some \(c\) between \(a\) and \(x\). You do not know \(c\) exactly, but you can bound the error by using the largest possible value of \(|f^{(n+1)}|\) on the interval. Two things make the error small: being close to \(a\) (the factor \((x - a)^{n+1}\)), and using more terms (the factorial in the denominator).
Worked problem
Bounding the error
Estimate \(\cos(0.5)\) with the Taylor polynomial \(1 - \dfrac{x^2}{2} + \dfrac{x^4}{24}\), and bound the error.
Show solution
- \(T_4(0.5) = 1 - 0.125 + 0.0026 = 0.877604\).
- The next nonzero term involves \(x^6/6!\), and every derivative of cosine is at most 1 in size, so the error is at most \(\dfrac{0.5^6}{720} \approx 0.000022\).
- Check: \(\cos(0.5) = 0.877583\). The actual error is 0.000022, right at the bound.
Answer: 0.87760, accurate to about 2 × 10⁻⁵
Engineering approximations
Taylor polynomials are behind many simplified formulas that engineers use without thinking about where they came from.
Small angles
For small \(\theta\) in radians, \(\sin\theta \approx \theta\), \(\tan\theta \approx \theta\) and \(\cos\theta \approx 1 - \tfrac12\theta^2\). These keep the first nonzero term of each series. At 10° (0.1745 rad), \(\sin\theta\) is 0.1736, an error of 0.5%.
The binomial approximation
For small \(x\), \((1 + x)^k \approx 1 + kx\). It turns awkward expressions into linear ones.
- \(\sqrt{1.1} = (1 + 0.1)^{1/2} \approx 1 + 0.05 = 1.05\). (True: 1.0488. Adding the next term, \(-\tfrac18(0.1)^2\), gives 1.04875.)
- Thermal expansion: a part that grows by a fraction \(\alpha\Delta T\) in each dimension changes volume by \((1 + \alpha\Delta T)^3 \approx 1 + 3\alpha\Delta T\). The volume coefficient is three times the linear one.
Worked problem
Sag of a cable
The catenary \(y = a\cosh(x/a)\) can be approximated near its low point using \(\cosh u \approx 1 + \tfrac12u^2\). Find the approximate sag at the towers of the 60 m span with \(a = 50\) m from earlier lessons.
Show solution
- Sag \(= a\left[\cosh(x/a) - 1\right] \approx a\cdot\dfrac{x^2}{2a^2} = \dfrac{x^2}{2a}\). This is a parabola.
- At \(x = 30\): \(\dfrac{900}{100} = 9.0\) m.
- The exact value from the catenary is 9.27 m, so the parabola is within 3%.
Answer: About 9.0 m. This is why taut cables are often designed as parabolas: the error is small when sag is small relative to span
Worked problem
Pendulum period at larger swings
The simple formula \(T_0 = 2\pi\sqrt{L/g}\) assumes small swings. A more accurate formula, from a series expansion, is \(T \approx T_0\left(1 + \dfrac{\theta_0^2}{16}\right)\), with \(\theta_0\) in radians. How much longer is the period for a 20° swing?
Show solution
- \(\theta_0 = 20° = 0.349\) rad, so \(\dfrac{\theta_0^2}{16} = 0.0076\).
Answer: About 0.76% longer. A pendulum clock swinging 20° instead of a few degrees would lose about 11 minutes a day
Fourier series
Taylor series build functions from powers of \(x\). Fourier series build periodic functions from sines and cosines. Any repeating signal (a square wave from a digital circuit, the vibration of an engine, the current drawn by a rectifier) can be written as a sum of harmonics at whole-number multiples of its base frequency. A square wave of height 1 is
This is how engineers find which frequencies a signal contains: the basis of filters, audio processing, vibration analysis and power-quality measurement. You will study Fourier series and transforms in Signals and Systems.
Common mistakes
MistakeConcluding that a series converges because its terms go to zero.
InsteadTerms going to zero is necessary, not sufficient. The harmonic series is the standard counterexample.
MistakeUsing a/(1 − r) when |r| ≥ 1, or starting the count at the wrong term.
InsteadCheck |r| < 1, and make sure a is the first term actually in the sum.
MistakeForgetting the factorials in a Taylor series: writing e^x ≈ 1 + x + x² + x³.
InsteadEach term is f⁽ⁿ⁾(a)(x − a)ⁿ / n!. The factorials are what make the series converge.
MistakeUsing a Taylor polynomial far from its centre and trusting the result.
InsteadTaylor polynomials are accurate near a. Use the remainder bound to check, or add terms.
MistakeUsing degrees in sin θ ≈ θ.
InsteadAll the trigonometric series need θ in radians.
MistakeTreating the ratio test result L = 1 as "converges".
InsteadWhen L = 1 the test is inconclusive. Try another test, such as the integral test.
Summary and key terms
- A sequence converges if its terms approach a limit. A series converges if its partial sums do.
- Geometric series: Σarⁿ = a/(1 − r) when |r| < 1.
- If the terms do not go to zero, the series diverges. The harmonic series Σ1/n diverges even though its terms go to zero.
- p-series Σ1/nᵖ converge only for p > 1. The ratio test handles factorials and powers. Alternating series have an easy error bound.
- Taylor series: f(x) = Σ f⁽ⁿ⁾(a)(x − a)ⁿ/n!. Know the series for eˣ, sin x, cos x, ln(1 + x) and (1 + x)ᵏ.
- Taylor's remainder bounds the error: small near the centre and with more terms.
- Engineering approximations (small angles, binomial, parabolic sag) are low-order Taylor polynomials. Fourier series do the same job with sines and cosines.
- Sequence
- An ordered list of numbers a₁, a₂, a₃, ….
- Series
- The sum of the terms of a sequence.
- Partial sum
- The sum of the first N terms of a series.
- Geometric series
- A series with a constant ratio r between terms.
- Harmonic series
- Σ1/n, which diverges.
- Ratio test
- Convergence test using the limit of |aₙ₊₁/aₙ|.
- Radius of convergence
- The R such that a power series converges for |x − a| < R.
- Taylor polynomial
- The partial sum of a Taylor series, Tₙ(x).
- Remainder
- The error f(x) − Tₙ(x).
- Fourier series
- A sum of sines and cosines representing a periodic function.
Practice problems
Work each problem on paper first, then reveal the answer to check.
Find the limit of the sequence \(a_n = \dfrac{5n^2 - 3}{2n^2 + n}\).
Show answer
Divide by n²: (5 − 3/n²)/(2 + 1/n) → 5/2.
Find the sum of \(\sum_{n=0}^{\infty} 3\left(\tfrac{2}{5}\right)^n\).
Show answer
Geometric with a = 3, r = 2/5: 3/(1 − 2/5) = 5.
Write \(0.\overline{36} = 0.363636\ldots\) as a fraction.
Show answer
0.36 + 0.0036 + … = 0.36/(1 − 0.01) = 36/99 = 4/11.
Does \(\sum_{n=1}^{\infty}\dfrac{n}{2n + 1}\) converge?
Show answer
No. The terms approach 1/2, not 0, so it diverges by the divergence test.
Does \(\sum_{n=1}^{\infty}\dfrac{1}{n\sqrt{n}}\) converge?
Show answer
Yes. It is a p-series with p = 3/2 > 1.
Use the ratio test on \(\sum_{n=1}^{\infty}\dfrac{2^n}{n!}\).
Show answer
aₙ₊₁/aₙ = 2/(n + 1) → 0 < 1, so it converges. (Its sum is e² − 1.)
Find the first four nonzero terms of the Maclaurin series of \(e^{-2x}\).
Show answer
Substitute −2x into eˣ: 1 − 2x + 2x² − (4/3)x³ + ….
Use three terms of the series for \(e^{x}\) to estimate \(e^{-0.2}\), and compare with the true value.
Show answer
1 − 0.2 + 0.02 = 0.82. True value 0.81873, so the error is about 0.0013.
Find the Taylor polynomial of degree 2 for \(\sqrt{x}\) about \(x = 4\), and use it to estimate \(\sqrt{4.2}\).
Show answer
f(4) = 2, f′(4) = 1/4, f″(4) = −1/32. T₂ = 2 + (x − 4)/4 − (x − 4)²/64. At 4.2: 2 + 0.05 − 0.000625 = 2.049375 (true 2.049390).
A steel rod has a linear expansion coefficient of 12 × 10⁻⁶ per °C. Estimate the percentage change in volume of a steel block heated by 100 °C.
Show answer
Linear strain 0.0012; volume ≈ 1 + 3(0.0012) = 1.0036, an increase of about 0.36%.
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