Unit 4

Power Series & Taylor Expansions

A power series turns a function into an infinite polynomial. Polynomials are the only functions we can really compute with — add, multiply, differentiate, integrate — so this unit is how every other function becomes tractable. For an electrical engineer it is also the entire justification for small-signal analysis and for phasors.

1 · Power series and radius of convergence

\[ \sum_{n=0}^{\infty} c_n (x-a)^n \] \(a\) is the centre. The series always converges at \(x=a\) — every term but the first vanishes — and the question is how far out from \(a\) it keeps working.

There are exactly three possibilities, and no others:

Converges only at \(x=a\) (\(R = 0\))
Converges for all real \(x\) (\(R = \infty\))
Converges on \(|x-a| < R\), diverges outside it, and behaves unpredictably at the two endpoints

Finding \(R\)

Apply the ratio test to the series with \(x\) carried along, then solve \(L < 1\) for \(x\).

Endpoints must be checked separately

The ratio test gives \(L = 1\) at both endpoints and says nothing there. Substitute each endpoint into the original series and test it by hand — usually the alternating series test on one side and a p-series comparison on the other. The interval can be open, closed, or half-open, and only this check tells you which. Losing marks here is routine and entirely avoidable.

A geometric way to see \(R\)

The radius is the distance from the centre to the nearest point where the function misbehaves. \(1/x\) expanded about \(a=1\) has \(R=1\), because there is a singularity at \(x=0\), exactly distance 1 away. No series centred at 1 could possibly reach past it.

2 · Building series from known ones

Deriving a Taylor series from scratch — differentiating \(n\) times and spotting the pattern — is slow and error-prone. Almost always, manipulate a series you already know. All of these operations are legal inside the radius of convergence:

Operation	Example	New radius
Substitute for \(x\)	\(e^{-x^2} = \sum \dfrac{(-x^2)^n}{n!}\)	Adjust: \(u = -x^2\) must stay in range
Multiply by a power	\(x\sin x = \sum \dfrac{(-1)^n x^{2n+2}}{(2n+1)!}\)	Unchanged
Differentiate term by term	\(\dfrac{1}{(1-x)^2} = \sum n x^{n-1}\)	Same \(R\) (endpoints may change)
Integrate term by term	\(\ln(1-x) = -\sum \dfrac{x^{n+1}}{n+1}\)	Same \(R\) (endpoints may change)

The workhorse is the geometric series \(\dfrac{1}{1-x} = \sum x^n\). Substituting \(-x\) gives \(1/(1+x)\); substituting \(-x^2\) and integrating gives \(\arctan x\). A great many series come from that one identity plus a substitution.

3 · Taylor and Maclaurin series

\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}\,(x-a)^n \] A Maclaurin series is just the case \(a = 0\). Nothing else distinguishes them.

The idea: match the function's value, slope, curvature, and every higher derivative at a single point \(a\). The \(n!\) is exactly what makes the \(n\)th derivative of the series come out right — differentiating \(x^n\) that many times produces \(n!\), which the denominator cancels.

Taylor polynomials of \(\sin x\) about \(0\). Each hugs the curve near the centre and eventually peels away; adding terms widens the good region but never makes a polynomial behave like a sine forever. Near the centre even \(T_1 = x\) is excellent — that is the small-angle approximation, and it is why linearisation works.

4 · The seven series worth memorising

These cover essentially every problem you will be handed.

Function	Series	Valid for
\(\dfrac{1}{1-x}\)	\(1 + x + x^2 + x^3 + \cdots = \displaystyle\sum_{n=0}^{\infty} x^n\)	\(\|x\| < 1\)
\(e^{x}\)	\(1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots = \displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n!}\)	all \(x\)
\(\sin x\)	\(x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots = \displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}\)	all \(x\) (odd powers only)
\(\cos x\)	\(1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots = \displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!}\)	all \(x\) (even powers only)
\(\ln(1+x)\)	\(x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots\)	\(-1 < x \le 1\)
\(\arctan x\)	\(x - \dfrac{x^3}{3} + \dfrac{x^5}{5} - \cdots\)	\(\|x\| \le 1\)
\((1+x)^{k}\)	\(1 + kx + \dfrac{k(k-1)}{2!}x^2 + \cdots\)	\(\|x\| < 1\) (binomial series)

Structural checks

\(\sin\) is odd, so its series has only odd powers; \(\cos\) is even, so only even powers. If your sine series produces an \(x^2\), you have made an error — no computation needed to know that.

5 · Taylor's theorem and error bounds

Truncating after the \(n\)th-degree term leaves a remainder, and Taylor's theorem says exactly what it looks like:

\[ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} \] for some unknown \(c\) between \(a\) and \(x\). You never find \(c\) — you bound the derivative over the interval instead.

\[ |R_n(x)| \le \frac{M}{(n+1)!}\,|x-a|^{n+1} \] \(M\) is any bound on \(|f^{(n+1)}|\) across the interval. For \(\sin\) and \(\cos\), \(M = 1\) always — which makes their error estimates unusually easy.

Read the structure: error shrinks fast as you move closer to the centre (the power of \(|x-a|\)) and fast as you add terms (the factorial). Both levers are available, and near the centre the first is dramatically more effective.

6 · What series are actually for

Integrals with no closed form

Expand the integrand and integrate term by term. This is how \(\int e^{-x^2}dx\), \(\int \frac{\sin x}{x}dx\), and every other "impossible" integral gets evaluated in practice.

Limits without L'Hôpital

Replace each function by its first few terms and cancel. Often faster and much less error-prone than three rounds of differentiation:

\[ \lim_{x\to 0}\frac{\sin x - x}{x^3} = \lim_{x\to 0}\frac{\big(x - \tfrac{x^3}{6} + \cdots\big) - x}{x^3} = \lim_{x\to 0}\frac{-x^3/6 + \cdots}{x^3} = -\frac16 \]

Approximation — the engineering use

Keep one or two terms and get a formula you can reason about. These five appear constantly in circuit analysis:

All of these are just the first one or two terms of the corresponding Maclaurin series. Each is accurate to a few percent for \(|x|\) below about 0.1.
Exact	Small-\(x\) approximation	Where it shows up
\(\sin x\)	\(x\)	Small-angle phase, pendulum, PLL analysis
\(\cos x\)	\(1 - \tfrac{x^2}{2}\)	Second-order phase error, power factor near unity
\(\dfrac{1}{1+x}\)	\(1 - x\)	Loop gain, loading corrections, tolerance stack-up
\(\sqrt{1+x}\)	\(1 + \tfrac{x}{2}\)	Impedance magnitude near resonance
\(e^{x}\)	\(1 + x\)	Diode equation, thermal drift

7 · Small-signal linearisation

This section is the reason Taylor series appear in an EE curriculum at all. Nearly every real device is nonlinear. Every analysis technique you know — superposition, phasors, transfer functions — requires linearity. Taylor series is the bridge between those two facts.

Expand about a bias point (operating point, Q-point) \(V_Q\), and keep only the linear term:

\[ i(V_Q + v) \;\approx\; \underbrace{i(V_Q)}_{\text{DC bias}} \;+\; \underbrace{\left.\frac{di}{dv}\right|_{V_Q}\!\cdot v}_{\text{small-signal}} \] That split — solve DC and AC separately — is the entire structure of amplifier analysis.

Worked case: the diode

The Shockley equation is thoroughly nonlinear:

\[ i = I_S\left(e^{v/V_T} - 1\right), \qquad V_T \approx 26\ \text{mV at room temperature} \]

Differentiate and evaluate at the bias point:

\[ g_d = \left.\frac{di}{dv}\right|_{V_Q} = \frac{I_S}{V_T}e^{V_Q/V_T} \approx \frac{I_D}{V_T} \] since \(I_D = I_S e^{V_Q/V_T}\) when the diode is well forward-biased and the \(-1\) is negligible.

So the small-signal resistance is \(r_d = V_T/I_D \approx 26\,\text{mV}/I_D\). At 1 mA bias that is 26 Ω.

You have seen this formula before

\(r_e = 26\,\text{mV}/I_E\) for a BJT is the same first-order Taylor coefficient, and \(g_m = I_C/V_T\) is its reciprocal. These are not empirical rules of thumb — they are the linear term of a Maclaurin expansion, and the "small signal" condition (\(v \ll 26\) mV) is precisely the condition that makes the quadratic remainder negligible.

That also tells you when the model fails: push the amplitude up and the \(x^2/2\) term stops being negligible. Its effect is second-harmonic distortion. Taylor series does not just give you the model — it predicts the distortion when you exceed it.

8 · Euler's formula

Take the series for \(e^{x}\), substitute \(x = i\theta\), and sort the terms by whether they carry an \(i\). Using \(i^2 = -1\), \(i^3 = -i\), \(i^4 = 1\):

\[ e^{i\theta} = 1 + i\theta - \frac{\theta^2}{2!} - i\frac{\theta^3}{3!} + \frac{\theta^4}{4!} + i\frac{\theta^5}{5!} - \cdots \]

Group the real terms and the imaginary terms:

\[ = \underbrace{\left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots\right)}_{\cos\theta} + i\underbrace{\left(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots\right)}_{\sin\theta} \]

Those are exactly the cosine and sine series from the table above:

\[ \boxed{\,e^{i\theta} = \cos\theta + i\sin\theta\,} \] Setting \(\theta = \pi\) gives \(e^{i\pi} + 1 = 0\).

Why this is the most useful identity in your degree

It converts trigonometry into algebra. Multiplying complex exponentials adds their exponents, so phase shifts become addition rather than angle-sum identities. That is the whole reason phasors work: a sinusoid \(V\cos(\omega t + \phi)\) becomes a static complex number \(Ve^{j\phi}\), differentiation becomes multiplication by \(j\omega\), and differential equations collapse into ordinary algebra.

It also gives you the inverse forms, which turn up whenever you integrate a product of sinusoids:

\[ \cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2} \qquad\qquad \sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i} \]

Notation warning: electrical engineers write \(j\), not \(i\), because \(i\) is current. Mathematicians write \(i\). Same object; expect to switch between conventions constantly.

9 · Worked examples

1. Radius and interval of convergence

Find the interval of convergence of \(\displaystyle\sum_{n=1}^{\infty}\frac{(x-2)^n}{n\,3^n}\).

Ratio test, carrying \(x\):

\[ \left|\frac{a_{n+1}}{a_n}\right| = \frac{|x-2|^{n+1}}{(n+1)3^{n+1}}\cdot\frac{n\,3^n}{|x-2|^n} = \frac{|x-2|}{3}\cdot\frac{n}{n+1} \] \[ L = \frac{|x-2|}{3} < 1 \implies |x-2| < 3 \implies R = 3,\quad -1 < x < 5 \]

Now the endpoints, one at a time.

\(x = 5\): the series becomes \(\sum \frac{3^n}{n\,3^n} = \sum \frac1n\) — the harmonic series, divergent.

\(x = -1\): it becomes \(\sum \frac{(-3)^n}{n\,3^n} = \sum \frac{(-1)^n}{n}\) — alternating harmonic, convergent (conditionally).

\[ \text{Interval of convergence: } [-1,\,5) \]

Half-open, the most common outcome and the one you can only get by checking. Reporting \((-1,5)\) or \([-1,5]\) both lose marks.

2. Build a series by substitution and integration

Find the Maclaurin series for \(\arctan x\), starting from the geometric series.

Step 1. Start from \(\frac{1}{1-u} = \sum u^n\) and substitute \(u = -x^2\):

\[ \frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 + \cdots \]

Step 2. That is exactly the derivative of \(\arctan\), so integrate term by term:

\[ \arctan x = \int_0^x \frac{dt}{1+t^2} = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots \]

The constant of integration is 0 because \(\arctan 0 = 0\). Radius: \(u = -x^2\) needs \(|x^2| < 1\), so \(R = 1\).

A famous consequence: at \(x = 1\) this gives \(\frac{\pi}{4} = 1 - \tfrac13 + \tfrac15 - \cdots\) — the Leibniz formula. Beautiful, and useless for computing \(\pi\): by the alternating-series bound you would need about \(10^6\) terms for six digits.

3. Approximate an integral to a stated accuracy

Estimate \(\displaystyle\int_0^1 e^{-x^2}dx\) with error below \(0.001\).

Expand by substituting \(-x^2\) into the \(e^x\) series:

\[ e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \frac{x^8}{4!} - \cdots \]

Integrate term by term from 0 to 1:

\[ \int_0^1 e^{-x^2}dx = 1 - \frac13 + \frac{1}{10} - \frac{1}{42} + \frac{1}{216} - \frac{1}{1320} + \cdots \] General term: \(\dfrac{(-1)^n}{n!\,(2n+1)}\).

It alternates, so the error is under the first omitted term. Since \(1/1320 \approx 0.00076 < 0.001\), stopping just before it suffices — keep through \(1/216\):

\[ \approx 1 - 0.33333 + 0.10000 - 0.02381 + 0.00463 = 0.74749 \]

True value \(0.74682\), so the actual error is \(0.00067\) — within tolerance and consistent with the bound. Compare Exam 1 question 7, where Simpson's rule gave \(0.74685\) on the same integral: two completely different methods, same answer.

4. Error bound for a sine approximation

How accurate is \(\sin x \approx x - x^3/6\) for \(|x| \le 0.5\) rad? And \(\sin x \approx x\) for \(|x| \le 0.1\)?

First part. The approximation is \(T_3\), but since the \(x^4\) coefficient is zero it is also \(T_4\) — use \(n=4\) for a sharper bound. All derivatives of sine are bounded by \(M = 1\):

\[ |R_4| \le \frac{1}{5!}(0.5)^5 = \frac{0.03125}{120} \approx 2.6\times10^{-4} \]

Better than four decimal places from two terms.

Second part. \(\sin x \approx x\) is \(T_2\) (again the \(x^2\) term vanishes):

\[ |R_2| \le \frac{1}{3!}(0.1)^3 = \frac{0.001}{6} \approx 1.7\times10^{-4} \]

A relative error near 0.17%, which is why the small-angle approximation is used so freely below about 0.1 rad (roughly 6°). Note how sharply the bound depends on the interval: at \(x = 0.5\) the same one-term approximation would have error about \(0.02\), a hundredfold worse.

5. EE — linearising the diode equation

A diode is biased at \(I_D = 2\) mA. Find its small-signal resistance, and estimate the error the linear model makes for a 5 mV signal.

\[ r_d = \frac{V_T}{I_D} = \frac{26\ \text{mV}}{2\ \text{mA}} = 13\ \Omega \]

How good is the linear model? Write \(x = v/V_T\) and expand:

\[ e^{x} = 1 + x + \frac{x^2}{2} + \cdots \]

For a 5 mV signal, \(x = 5/26 = 0.192\). Compare the term you keep against the first one you drop:

\[ \frac{x^2/2}{x} = \frac{x}{2} = \frac{0.192}{2} = 9.6\% \]

So the quadratic term is nearly 10% of the linear term — the "small signal" is not small enough. That 9.6% shows up as second-harmonic distortion.

Design consequence: to hold distortion under 1% you need \(x/2 < 0.01\), i.e. \(v < 0.02(26\ \text{mV}) \approx 0.5\) mV. That is a hard, quantitative limit on input swing, and it came straight out of a Taylor remainder. This is the difference between "I memorised \(r_e = 26/I_E\)" and "I know when it stops being true."

6. A limit by series

Evaluate \(\displaystyle\lim_{x\to 0}\frac{e^{x} - 1 - x}{x^2}\).

This is \(0/0\). L'Hôpital works but needs two rounds. Series is one step:

\[ e^{x} - 1 - x = \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots\right) - 1 - x = \frac{x^2}{2} + \frac{x^3}{6} + \cdots \] \[ \frac{e^{x}-1-x}{x^2} = \frac12 + \frac{x}{6} + \cdots \;\longrightarrow\; \frac12 \]

The general principle: for a \(0/0\) limit at a point, expand everything about that point and cancel the common power. It also tells you more than the limit does — the \(x/6\) term shows how the ratio approaches \(\tfrac12\), which L'Hôpital never reveals.

Checklist

I find \(R\) with the ratio test and then check both endpoints by hand.
I build new series by substituting into known ones rather than differentiating \(n\) times.
I know all seven series in the table, including where each is valid.
I know sine has only odd powers and cosine only even ones, and I use that as a check.
I can bound a truncation error with \(M|x-a|^{n+1}/(n+1)!\).
I can derive Euler's formula from the three series, not just quote it.
I can explain small-signal linearisation as a first-order Taylor expansion, and say when it breaks.