Chapter 23 — Power Series and Taylor Series

23.1 The Idea: Functions That Are Infinite Polynomials

Polynomials are the friendliest functions in mathematics. You can add them, multiply them, differentiate them, and integrate them using nothing but the power rule. You can evaluate them with a finite list of multiplications and additions — the only operations a computer truly knows. Transcendental functions like $e^x$, $\sin x$, and $\ln x$ have none of this convenience. A processor cannot "compute a sine." There is no finite formula in $+, -, \times, \div$ that produces $\sin(0.7)$.

This chapter resolves that tension with a single, remarkable idea: most functions you care about are secretly infinite polynomials. Write

$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$$

and you have turned the inscrutable exponential into something a fourth-grader could evaluate (given enough patience). Keep finitely many terms and you get a polynomial that approximates $e^x$ as closely as you like. This is the machinery inside every calculator and every scientific library, and it is the last great payoff of approximation — the sixth recurring theme of this book. We began with the tangent line in Chapter 11 (a first-order approximation); Taylor series are the same idea pushed to every order at once.

A power series centered at $a$ has the form

$$\sum_{n=0}^\infty c_n (x - a)^n = c_0 + c_1(x - a) + c_2(x - a)^2 + c_3(x - a)^3 + \cdots$$

Power series generalize polynomials to infinite degree. They converge for some range of $x$ — the interval of convergence, governed by the radius of convergence — and on that range they define a genuine function. The most important power series are Taylor series: power-series representations of standard functions, built from their derivatives.

The Key Insight. A Taylor series turns local information — the value of $f$ and all its derivatives at a single point $a$ — into a global formula that reconstructs $f$ across an entire interval. Differentiation, which Chapter 6 taught you was a local operation, secretly carries enough information to rebuild the whole function.

This chapter develops convergence of power series; the Taylor series of $e^x$, $\sin x$, $\cos x$, $\ln(1+x)$, $\arctan x$, $1/(1-x)$, and the binomial $(1+x)^k$; Taylor's theorem with explicit error bounds; the term-by-term operations that let you build new series from old; and two applications — computing transcendental values, and integrating $e^{-x^2}$, the function with no elementary antiderivative that nonetheless governs all of statistics. That last example is the climax of the "area under the normal curve" anchor we have followed since Chapter 13.

Chapter 24 will use the same three series — $e^x$, $\sin x$, $\cos x$ — to derive Euler's identity $e^{i\pi} + 1 = 0$, often called the most beautiful equation in mathematics. Consider this chapter the toolbox; Chapter 24 is the masterpiece.

23.2 Power Series and the Radius of Convergence

A power series

$$\sum_{n=0}^\infty c_n (x - a)^n$$

is a "function" of $x$ in the following sense: fix a value of $x$, and you get an ordinary numerical series of the kind you studied in Chapters 21 and 22, which either converges or diverges. The set of $x$ for which it converges is the interval of convergence.

The structure of that set is governed by a clean theorem.

Theorem (the trichotomy of convergence). For a power series $\sum c_n(x-a)^n$, exactly one of the following holds: 1. The series converges only at $x = a$ (radius $R = 0$). 2. There is a number $R > 0$ — the radius of convergence — such that the series converges absolutely for $|x - a| < R$ and diverges for $|x - a| > R$. 3. The series converges for all $x$ (radius $R = \infty$).

The convergence region is always a symmetric interval about the center $a$: $(a - R, a + R)$, possibly with one or both endpoints attached. The endpoints are special — at $|x - a| = R$ the ratio test is silent, and convergence must be checked by hand using the tools of Chapter 22.

Geometric Intuition. Picture the real line with a marker at the center $a$. The radius of convergence is the half-width of a "zone of trust" around $a$. Inside the zone the infinite polynomial faithfully reproduces the function; outside it, the terms grow faster than they shrink and the sum flies off to infinity. The endpoints sit exactly on the boundary fence — sometimes inside the zone, sometimes outside, and you must inspect each one individually. We will see in §23.9 that the fence's location is set not by anything on the real line but by the function's complex-plane singularities.

Finding the radius with the ratio test

The radius comes straight from the ratio test of Chapter 22, applied to the terms $|c_n (x - a)^n|$:

$$\rho = \lim_{n \to \infty} \left|\frac{c_{n+1} (x-a)^{n+1}}{c_n (x-a)^n}\right| = |x - a| \cdot \lim_{n\to\infty} \left|\frac{c_{n+1}}{c_n}\right|.$$

The ratio test promises absolute convergence when $\rho < 1$, that is, when

$$|x - a| < \frac{1}{\lim_{n\to\infty} |c_{n+1}/c_n|}.$$

Reading off the right-hand side:

$$\boxed{\,R = \lim_{n \to \infty} \left|\frac{c_n}{c_{n+1}}\right|\,}\qquad\text{when this limit exists.}$$

(Equivalently, the Cauchy–Hadamard formula $R = 1/\limsup_n |c_n|^{1/n}$ always works, even when the ratio limit does not — see the Math Major Sidebar.)

Worked examples, graduated

Example 23.2.1 (infinite radius). $\displaystyle\sum_{n=0}^\infty \frac{x^n}{n!}$, with $c_n = 1/n!$. $$\left|\frac{c_{n+1}}{c_n}\right| = \frac{n!}{(n+1)!} = \frac{1}{n+1} \to 0 \quad\Longrightarrow\quad R = \frac{1}{0} = \infty.$$ The factorial in the denominator crushes every power of $x$, so this series converges for all $x$. (It is, of course, $e^x$.)

Example 23.2.2 (the geometric series, $R = 1$). $\displaystyle\sum_{n=0}^\infty x^n$, with $c_n = 1$. $$\left|\frac{c_{n+1}}{c_n}\right| = 1 \quad\Longrightarrow\quad R = 1.$$ Converges for $|x| < 1$, diverges for $|x| > 1$. At $x = 1$ it is $\sum 1$, which diverges; at $x = -1$ it is $\sum (-1)^n$, whose terms do not tend to zero, so it diverges too. Interval of convergence: $(-1, 1)$, both endpoints excluded.

Example 23.2.3 (same radius, different endpoints). $\displaystyle\sum_{n=1}^\infty \frac{x^n}{n}$. $$\left|\frac{c_{n+1}}{c_n}\right| = \frac{n}{n+1} \to 1 \quad\Longrightarrow\quad R = 1.$$ At $x = 1$: $\sum 1/n$, the harmonic series — diverges. At $x = -1$: $\sum (-1)^n/n$, which converges by the alternating series test (Chapter 22). Interval of convergence: $[-1, 1)$. Notice that two series with the same radius can behave completely differently at the endpoints; the radius is decided by the ratio test, but the endpoints demand separate, manual checks.

Example 23.2.4 (zero radius). $\displaystyle\sum_{n=0}^\infty n!\, x^n$, with $c_n = n!$. $$\left|\frac{c_{n+1}}{c_n}\right| = (n+1) \to \infty \quad\Longrightarrow\quad R = 0.$$ The factorials grow so fast that the series converges only at $x = 0$. This series defines no useful function — a cautionary reminder that not every power series is good for anything.

Common Pitfall. Many students compute the radius and then declare victory with an open interval, forgetting the endpoints entirely. But $\sum x^n/n$ converges at one endpoint and not the other, and $\sum x^n/n^2$ converges at both. The ratio test is structurally blind at $|x-a| = R$ (it gives $\rho = 1$, the inconclusive case), so the endpoints are never automatic — always substitute $x = a \pm R$ and test each resulting numerical series with the methods of Chapter 22.

Check Your Understanding. Find the radius of convergence of $\displaystyle\sum_{n=0}^\infty \frac{x^n}{2^n}$.

Answer

Here $c_n = 1/2^n$, so $|c_n/c_{n+1}| = 2^{n+1}/2^n = 2$. Thus $R = 2$: the series converges for $|x| < 2$. (You can also see this directly — it is the geometric series $\sum (x/2)^n$, which converges iff $|x/2| < 1$, i.e. $|x| < 2$. At $x = \pm 2$ the terms are $(\pm1)^n$, which do not tend to zero, so both endpoints diverge and the interval is $(-2, 2)$.)

Math Major Sidebar — why Cauchy–Hadamard is the "real" formula. The ratio formula $R = \lim|c_n/c_{n+1}|$ fails whenever that limit does not exist — for instance when half the coefficients vanish, as in $\sum x^{2n}$. The root-test version $R = 1/\limsup_n |c_n|^{1/n}$ uses the limit superior, which always exists in $[0, \infty]$, and it agrees with the ratio formula whenever the latter applies. This is the Cauchy–Hadamard theorem, and in a real-analysis course it is the definition of the radius of convergence. For the everyday series in this chapter, the ratio test is faster; keep Cauchy–Hadamard in reserve for series with gaps.

23.3 Functions Defined by Power Series

A power series with positive radius $R$ doesn't merely converge — it defines a function on the open interval $(a - R, a + R)$:

$$f(x) = \sum_{n=0}^\infty c_n (x - a)^n.$$

What kind of function? An extraordinarily well-behaved one. Inside the radius of convergence, $f$ is:

• Continuous at every interior point;
• Differentiable, with the derivative obtained by differentiating term-by-term;
• Integrable, with the antiderivative obtained by integrating term-by-term.

In symbols, for $|x - a| < R$:

$$f'(x) = \sum_{n=1}^\infty n\, c_n (x-a)^{n-1}, \qquad \int f(x)\,dx = C + \sum_{n=0}^\infty \frac{c_n}{n+1} (x-a)^{n+1}.$$

Both new series have the same radius of convergence $R$ as the original. (The endpoints can change — differentiation can destroy endpoint convergence, integration can create it — but the radius is invariant.)

This is a genuinely surprising license. With ordinary infinite sums you may not interchange limits and derivatives freely; doing so can give nonsense. But inside the radius of convergence, a power series converges so strongly (uniformly on every closed sub-interval) that term-by-term calculus is fully legal. We treat the infinite polynomial exactly like a finite one. This single permission is what makes the rest of the chapter possible: it lets us build a whole library of series from a tiny number of known ones, never touching a derivative formula again.

Real-World Application — power-series solutions of differential equations (physics & engineering). Many of the differential equations of mathematical physics — Bessel's equation (vibrating drumheads and circular waveguides), Legendre's equation (the hydrogen atom, satellite gravity models), Airy's equation (optics near a caustic, quantum tunneling) — have no solution expressible in elementary functions. The standard escape is the power-series method: assume $y = \sum a_n x^n$, substitute into the equation, and the term-by-term differentiation rule above turns the differential equation into a recursion for the coefficients $a_n$. The solution is then computed to any precision by summing enough terms. Whole families of "special functions" in physics are defined this way.

23.4 Taylor Series: From a Function to Its Series

So far we started with coefficients and asked what function they sum to. Now reverse it: start with a function $f$ and ask which power series represents it. The answer is forced by the term-by-term differentiation rule.

Suppose $f(x) = \sum_{n=0}^\infty c_n (x-a)^n$ on some interval around $a$. Differentiate repeatedly and evaluate at $x = a$, where every term with a positive power of $(x-a)$ vanishes:

$$f(a) = c_0,\quad f'(a) = c_1,\quad f''(a) = 2!\,c_2,\quad f'''(a) = 3!\,c_3,\quad \ldots,\quad f^{(n)}(a) = n!\,c_n.$$

Solving for the coefficients gives $c_n = f^{(n)}(a)/n!$ — there is no choice in the matter. This forced formula is the Taylor series of $f$ centered at $a$:

$$\boxed{\,f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}\,(x - a)^n\,}$$

When the center is $a = 0$, it carries the special name Maclaurin series. The degree-$N$ truncation is the Taylor polynomial

$$T_N(x) = \sum_{n=0}^N \frac{f^{(n)}(a)}{n!}(x - a)^n,$$

the unique polynomial of degree $\le N$ that matches $f$ and its first $N$ derivatives at $a$. (Compare Chapter 11: $T_1(x) = f(a) + f'(a)(x-a)$ is exactly the tangent-line linearization. Taylor polynomials are linearization continued to higher order — $T_2$ matches the curvature, $T_3$ the jerk, and so on.)

Warning. Having a Taylor series and being equal to it are not the same thing. The Taylor series is built purely from derivatives at one point; it might converge to a different function, or diverge. We address this subtlety in §23.10. For now, take on faith that for the standard functions below, the series really does converge to the function on the stated interval — Taylor's theorem in §23.5 is what proves it.

Deriving the key series

Exponential. For $f(x) = e^x$, every derivative is $e^x$ again, so $f^{(n)}(0) = 1$ for all $n$. Therefore $$e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots, \qquad (-\infty < x < \infty).$$

Sine. For $f(x) = \sin x$, the derivatives cycle $\sin, \cos, -\sin, -\cos$ with period 4, giving $f^{(n)}(0) = 0, 1, 0, -1, 0, 1, \ldots$ Only the odd powers survive: $$\sin x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots, \qquad (-\infty < x < \infty).$$ This is an odd function, and only odd powers appear — geometry and algebra agreeing, as they must.

Cosine. Likewise $f(x) = \cos x$ gives $f^{(n)}(0) = 1, 0, -1, 0, 1, \ldots$ — only the even powers survive: $$\cos x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots, \qquad (-\infty < x < \infty).$$ An even function gets only even powers. Differentiate the cosine series term-by-term and you recover $-\sin x$ exactly — a free consistency check.

The remaining standard series — $1/(1-x)$, $\ln(1+x)$, $\arctan x$, and the binomial $(1+x)^k$ — are most easily obtained not from the derivative formula but by manipulating the three above. That is the subject of §23.6. Here is the full reference table; commit the first six to memory.

The binomial series uses the generalized binomial coefficient

$$\binom{k}{n} = \frac{k(k-1)(k-2)\cdots(k-n+1)}{n!},$$

valid for any real exponent $k$. When $k$ is a positive integer the series terminates and reproduces the ordinary binomial theorem; for fractional $k$ it is genuinely infinite. For example, with $k = \tfrac12$, $$\binom{1/2}{2} = \frac{\tfrac12\cdot(-\tfrac12)}{2!} = -\frac{1}{8}, \qquad\text{so}\qquad \sqrt{1+x} = 1 + \tfrac12 x - \tfrac18 x^2 + \tfrac{1}{16}x^3 - \cdots$$ This is the workhorse behind every "for small $x$" approximation in physics — we will see it produce special relativity's kinetic-energy correction in §23.8.

Check Your Understanding. Use the table to write the first three nonzero terms of the Maclaurin series for $\cos(2x)$.

Answer

Substitute $2x$ for $x$ in the cosine series: $\cos(2x) = 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \cdots = 1 - 2x^2 + \frac{2}{3}x^4 - \cdots$ (Since $\frac{16}{24} = \frac{2}{3}$.) Substitution is faster than computing four derivatives of $\cos(2x)$ by hand — a preview of §23.6.

23.5 Taylor's Theorem: Bounding the Error

A Taylor polynomial only approximates $f$. The decisive question for any computation is: how wrong is it? Taylor's theorem answers exactly, by writing the error as an explicit term.

Taylor's Theorem (Lagrange form of the remainder). If $f$ has $N+1$ continuous derivatives on an interval containing $a$ and $x$, then $$f(x) = T_N(x) + R_N(x), \qquad R_N(x) = \frac{f^{(N+1)}(c)}{(N+1)!}(x - a)^{N+1}$$ for some (unknown) point $c$ strictly between $a$ and $x$.

The remainder $R_N(x)$ looks exactly like "the next term of the series," except that the derivative is evaluated at a mysterious intermediate point $c$ rather than at the center $a$. We rarely know $c$, and we don't need to — we only need to bound it. If $M$ is any upper bound for $|f^{(N+1)}|$ on the interval between $a$ and $x$, then

$$\boxed{\,|R_N(x)| \le \frac{M}{(N+1)!}\,|x - a|^{N+1}\,}$$

This is the single most useful inequality in numerical analysis. It tells you, before you compute anything, exactly how many terms guarantee a target accuracy.

Geometric Intuition. Taylor's theorem is the Mean Value Theorem (Chapter 9) grown up. The MVT says $f(x) = f(a) + f'(c)(x-a)$ — a degree-0 polynomial plus one error term with a hidden point $c$. Taylor's theorem is the same statement at degree $N$: match $N$ derivatives at $a$, and the leftover is a single $(N{+}1)$-th-order term evaluated at a hidden $c$. As $N$ grows and $(N+1)!$ explodes in the denominator, that leftover is squeezed toward zero — provided the derivatives $f^{(N+1)}$ don't grow even faster. For $e^x$, $\sin x$, and $\cos x$ the derivatives stay bounded, so the squeeze always wins and the series converges to the function everywhere.

Worked Example 23.5.1 — estimating $e^{0.5}$ and certifying the error

Approximate $e^{0.5}$ with the degree-4 Maclaurin polynomial: $$T_4(0.5) = 1 + 0.5 + \frac{0.5^2}{2} + \frac{0.5^3}{6} + \frac{0.5^4}{24} = 1 + 0.5 + 0.125 + 0.0208\overline{3} + 0.0026041\overline{6} = 1.6484375.$$

Now certify it. The fifth derivative of $e^x$ is again $e^x$, increasing, so on $[0, 0.5]$ it is bounded by $M = e^{0.5} < 2$. Taylor's theorem gives $$|R_4(0.5)| \le \frac{2}{5!}\,(0.5)^5 = \frac{2}{120}\cdot 0.03125 \approx 0.000521.$$ The true value is $e^{0.5} = 1.6487212\ldots$, so the actual error is $0.000284$ — comfortably inside our guaranteed bound of $0.000521$. We knew the approximation was good to three decimals before we ever looked up the true value. That is the power of the Lagrange remainder: it is a certificate, not a guess.

Worked Example 23.5.2 — how many terms for $\sin(1)$ to six decimals?

For $\sin x$, every derivative is $\pm\sin$ or $\pm\cos$, so $|f^{(N+1)}| \le 1 = M$ everywhere. The error of the degree-$N$ polynomial at $x = 1$ is therefore bounded by $1/(N+1)!$. We want this below $5\times10^{-7}$: $$\frac{1}{(N+1)!} < 5\times 10^{-7} \;\Longrightarrow\; (N+1)! > 2{,}000{,}000.$$ Since $9! = 362{,}880$ and $10! = 3{,}628{,}800$, we need $(N+1)! \ge 10!$, i.e. $N+1 = 10$, so $N = 9$. The degree-9 Taylor polynomial of $\sin$ — which, because only odd powers appear, is just the terms through $x^9/9!$ — nails $\sin(1)$ to six decimals. No experimentation, no trial and error: the remainder bound told us in advance.

23.6 Building New Series from Old

Section 23.3 licensed term-by-term differentiation, integration, substitution, and arithmetic inside the radius of convergence. These four moves let you generate dozens of series from the handful you memorized — almost never by computing derivatives.

Substitution

Replace $x$ by any expression in a known series. $$e^{x^2} = \sum_{n=0}^\infty \frac{(x^2)^n}{n!} = \sum_{n=0}^\infty \frac{x^{2n}}{n!} = 1 + x^2 + \frac{x^4}{2!} + \cdots$$ Computing this from the derivative formula would require differentiating $e^{x^2}$ nine times; substitution does it in one line.

Differentiation — Worked Example 23.6.1

From the geometric series $\sum_{n=0}^\infty x^n = \dfrac{1}{1-x}$, differentiate both sides: $$\frac{d}{dx}\sum_{n=0}^\infty x^n = \sum_{n=1}^\infty n x^{n-1} = \frac{1}{(1-x)^2}, \qquad |x| < 1.$$ Re-indexing, $\displaystyle\sum_{n=0}^\infty (n+1)x^n = \frac{1}{(1-x)^2}$. This exact series computes the expected value of a geometric probability distribution — a recurring appearance in data science and queueing theory.

Integration — Worked Example 23.6.2 (deriving $\ln(1+x)$)

Substitute $-x$ into the geometric series: $\dfrac{1}{1+x} = \sum_{n=0}^\infty (-1)^n x^n$. Integrate from $0$ to $x$: $$\ln(1+x) = \int_0^x \frac{dt}{1+t} = \sum_{n=0}^\infty \frac{(-1)^n x^{n+1}}{n+1} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$$ The constant of integration is fixed by $\ln(1+0) = 0$. Integration shifted the radius's endpoint behavior: the geometric series diverges at $x = 1$, but its integral converges there (to $\ln 2 = 1 - \tfrac12 + \tfrac13 - \cdots$, the alternating harmonic series of Chapter 22) — integration smooths the boundary.

Integration — Worked Example 23.6.3 (deriving $\arctan x$)

Since $\dfrac{d}{dx}\arctan x = \dfrac{1}{1+x^2}$, substitute $-x^2$ into the geometric series and integrate: $$\frac{1}{1+x^2} = \sum_{n=0}^\infty (-1)^n x^{2n} \;\Longrightarrow\; \arctan x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots$$ Setting $x = 1$ gives the famous Leibniz formula $\dfrac{\pi}{4} = 1 - \dfrac13 + \dfrac15 - \dfrac17 + \cdots$. It is beautiful but useless for computation — it converges agonizingly slowly, needing hundreds of terms for two decimals. Machin's identity $\frac{\pi}{4} = 4\arctan\frac15 - \arctan\frac{1}{239}$ feeds small arguments into the same series and converges in a handful of terms; it computed $\pi$ to hundreds of digits by hand in the 18th century.

Arithmetic — Worked Example 23.6.4 (a limit without L'Hôpital)

Series make hard limits transparent. Consider $\displaystyle\lim_{x\to 0}\frac{\sin x - x}{x^3}$. Substitute the sine series: $$\sin x - x = \left(x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots\right) - x = -\frac{x^3}{6} + \frac{x^5}{120} - \cdots$$ Divide by $x^3$: $$\frac{\sin x - x}{x^3} = -\frac{1}{6} + \frac{x^2}{120} - \cdots \;\xrightarrow[x\to 0]{}\; -\frac{1}{6}.$$ L'Hôpital's rule would require differentiating top and bottom three times. The series sees the answer at a glance — the limit is just the leading coefficient.

Common Pitfall. When multiplying or dividing series, students often forget that the operation mixes terms of different degrees. To find the $x^4$ coefficient of $e^x \sin x$, you cannot multiply the $x^4$ terms alone — you must collect every product whose powers sum to 4: $(1)(\,\cdot x^4 \text{term of }\sin) + (x)(x^3\text{ of }\sin) + \cdots$. This bookkeeping is the Cauchy product, $\left(\sum a_n x^n\right)\!\left(\sum b_n x^n\right) = \sum_n\!\left(\sum_{k=0}^n a_k b_{n-k}\right)\!x^n$. Skipping the cross terms is the most common series-arithmetic error.

23.7 The Anchor Payoff: Integrating the Normal Curve

We now reach the moment this book has been building toward since Chapter 13. The area under the normal curve — the central object of all statistics — finally yields to calculus, and the key that opens it is Taylor series.

Recall the chain of the anchor. In Chapter 13 we met the bell curve $\phi(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}$ and learned that probabilities are areas under it. In Chapter 14 the Fundamental Theorem told us those areas are net changes in the cumulative function $\Phi$, with $\Phi' = \phi$ — but that $\Phi$ has no elementary antiderivative. Liouville's theorem (mentioned in Chapter 14, §14.12) proves no finite formula in elementary functions equals $\int e^{-x^2}\,dx$. The Fundamental Theorem applies perfectly; we simply could not write the antiderivative. For nine chapters that gap has stood open. Taylor series close it.

The trick is to never seek an elementary antiderivative at all. Expand the integrand as a series, then integrate term-by-term — each term a harmless power of $x$.

Start from $e^x = \sum x^n/n!$ and substitute $-x^2$: $$e^{-x^2} = \sum_{n=0}^\infty \frac{(-x^2)^n}{n!} = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{n!} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots$$ This series converges for all $x$ (radius $\infty$). Now integrate term-by-term from $0$ to $a$: $$\int_0^a e^{-x^2}\,dx = \sum_{n=0}^\infty \frac{(-1)^n a^{2n+1}}{n!\,(2n+1)} = a - \frac{a^3}{3} + \frac{a^5}{10} - \frac{a^7}{42} + \cdots$$ There it is — a closed expression for the "impossible" integral, valid for every $a$, computable to any precision. This series is the error function (up to the conventional scaling $\operatorname{erf}(a) = \frac{2}{\sqrt\pi}\int_0^a e^{-t^2}\,dt$). The function that defied antidifferentiation is, all along, just an infinite polynomial.

For the standard normal CDF itself, the same move on $e^{-t^2/2}$ gives

$$\Phi(z) = \frac12 + \frac{1}{\sqrt{2\pi}}\sum_{n=0}^\infty \frac{(-1)^n z^{2n+1}}{2^n\, n!\,(2n+1)}.$$

Let's verify the whole pipeline numerically — the series should reproduce $\Phi(1) \approx 0.8413$, the probability that a standard normal value falls below one standard deviation.

import math
from statistics import NormalDist

def Phi_series(z: float, N: int) -> float:
    """Standard-normal CDF via term-by-term integration of e^{-t^2/2}."""
    s = sum((-1)**n * z**(2*n + 1) / (2**n * math.factorial(n) * (2*n + 1))
            for n in range(N))
    return 0.5 + s / math.sqrt(2 * math.pi)

print(f"series  Phi(1) = {Phi_series(1.0, 40):.12f}")  # 0.841344746069
print(f"library Phi(1) = {NormalDist().cdf(1.0):.12f}")  # 0.841344746069
# The "impossible" normal-curve area, computed to 12 digits by an infinite polynomial.

The series and the library agree to twelve digits. The area under the normal curve — the quantity that determines every confidence interval, every p-value, every Six Sigma tolerance — is computed by integrating a Taylor series term-by-term. The anchor that opened in Chapter 13 is now closed.

Real-World Application — statistics and quality control. Every "z-table" in the back of a statistics textbook, and every scipy.stats.norm.cdf call, ultimately evaluates this kind of series (in practice via fast rational approximations to $\operatorname{erf}$, but the series is the conceptual engine). When an engineer certifies that a process runs at "Six Sigma" — fewer than 3.4 defects per million — that number is $\int e^{-x^2/2}\,dx$ out in the tail, delivered by Taylor's term-by-term integration. There is no other way to get it: the antiderivative does not exist in elementary form, yet the series computes it to arbitrary precision.

Computational Note. Term-by-term integration of $e^{-x^2}$ is numerically excellent for small $a$ but degrades for large $a$: the partial sums involve large alternating terms that nearly cancel, and floating-point round-off corrupts the difference (catastrophic cancellation). For $a \gtrsim 4$, libraries switch to an asymptotic expansion of the tail, which is divergent if summed forever but gives superb accuracy when truncated early. Two different series, each ruling its own region — a recurring pattern in numerical computing.

23.8 Visualizing Convergence, and a Physics Application

Watching Taylor polynomials close in on the function

The phrase "the partial sums approach the function" becomes vivid when you plot it. Each Taylor polynomial $T_N$ hugs $\sin x$ near the origin and then peels away; as $N$ grows, the point of departure marches outward, the polynomial swallowing more of the curve.

import math
import numpy as np
import matplotlib.pyplot as plt

x = np.linspace(-2*np.pi, 2*np.pi, 600)

def taylor_sin(x: np.ndarray, N: int) -> np.ndarray:
    """Degree-(2N+1) Maclaurin polynomial of sin, summed term-by-term."""
    total = np.zeros_like(x)
    for n in range(N + 1):
        total += (-1)**n * x**(2*n + 1) / math.factorial(2*n + 1)
    return total

plt.plot(x, np.sin(x), 'k', lw=2, label='sin x')
for N, style in [(0, '--'), (1, '-.'), (3, ':'), (6, '-')]:
    plt.plot(x, taylor_sin(x, N), style, label=f'$T_{{{2*N+1}}}$')
plt.ylim(-2, 2); plt.legend(); plt.title('Figure 23.1: Taylor polynomials converging to sin x')
# Each successive polynomial tracks sin x over a wider interval before diverging.
plt.show()

Figure 23.1 shows the story: $T_1 = x$ is the tangent line, good only near $0$; $T_3, T_7, T_{13}$ each grip the sine curve over a wider stretch before flinging off to $\pm\infty$. Because $\sin$ has infinite radius of convergence, any fixed point $x_0$ is eventually captured — choose $N$ large enough and $T_{2N+1}(x_0)$ is as close to $\sin x_0$ as you wish. The animation in your mind's eye is the geometric meaning of "radius $= \infty$."

Check Your Understanding. From Figure 23.1, why does no finite Taylor polynomial of $\sin x$ stay close to the curve forever?

Answer

A polynomial of degree $d$ has $|T(x)| \to \infty$ as $x\to\pm\infty$ (its leading term dominates), whereas $\sin x$ stays trapped in $[-1, 1]$. So every finite polynomial must eventually escape the bounded sine curve. Convergence is pointwise — for each fixed $x$ the error $\to 0$ as $N \to \infty$ — but no single polynomial works on the whole line at once. Increasing $N$ pushes the escape point farther out, never eliminating it.

A physics application: relativity reduces to Newton

The binomial series of §23.4 powers one of the most satisfying approximations in physics. The relativistic kinetic energy of a mass $m$ moving at speed $v$ is $E_k = (\gamma - 1)mc^2$, where the Lorentz factor is $\gamma = (1 - v^2/c^2)^{-1/2}$. Apply the binomial series $(1+u)^{-1/2} = 1 - \tfrac12 u + \tfrac38 u^2 - \cdots$ with $u = -v^2/c^2$:

$$\gamma = 1 + \frac{1}{2}\frac{v^2}{c^2} + \frac{3}{8}\frac{v^4}{c^4} + \cdots$$

Therefore

$$E_k = (\gamma - 1)mc^2 = \frac{1}{2}mv^2 + \frac{3}{8}\frac{mv^4}{c^2} + \cdots$$

The leading term is exactly the Newtonian kinetic energy $\tfrac12 mv^2$. The next term is the first relativistic correction, negligible when $v \ll c$ (since $v^4/c^2$ is tiny) but significant near light speed. Einstein's theory doesn't contradict Newton's — it contains it as the first term of a Taylor series, with relativity supplying the higher-order corrections. This is the deepest meaning of the approximation theme: new physics often appears as the next term in an old expansion.

Real-World Application — engineering linearization. Engineers live in the first one or two Taylor terms. A pendulum's equation $\ddot\theta = -\tfrac{g}{L}\sin\theta$ is unsolvable in elementary functions, but for small swings $\sin\theta \approx \theta$ (the degree-1 Taylor polynomial), turning it into the simple harmonic oscillator $\ddot\theta = -\tfrac{g}{L}\theta$ — the source of the familiar period $T = 2\pi\sqrt{L/g}$. Control theory, circuit analysis, and structural mechanics all run on linearization: replace a nonlinear law by its tangent-line (or quadratic) Taylor approximation near an operating point, solve the easy linear problem, and trust Taylor's remainder to bound the error.

23.9 Why the Radius Is What It Is

A genuine puzzle: the series $\dfrac{1}{1+x^2} = \sum_{n=0}^\infty (-1)^n x^{2n}$ has radius of convergence $1$, yet the function $1/(1+x^2)$ is perfectly smooth on all of $\mathbb{R}$ — it never blows up, never misbehaves. Why should its series quit at $x = \pm 1$, where nothing visible goes wrong?

The resolution is one of the most beautiful facts in all of analysis:

The radius of convergence of a Taylor series equals the distance from the center to the nearest singularity of the function — in the complex plane.

For $1/(1-x)$ centered at $0$, the only singularity is the pole at $x = 1$, distance $1$ away; hence $R = 1$. For $\ln(1+x)$, the singularity is at $x = -1$; $R = 1$. But for $1/(1+x^2)$, the function $\frac{1}{1+z^2}$ has poles where $z^2 = -1$, namely $z = \pm i$ — points that live off the real line, at distance $1$ from the origin. The real-axis series can sense those complex poles even though it never reaches them, and it stops dead at radius $1$ accordingly. The real line's good behavior is irrelevant; the complex plane sets the boundary.

This is why $e^x$, $\sin x$, $\cos x$ have infinite radius: extended to the complex plane, they are entire — analytic everywhere, with no singularities at any finite distance, so $R = \infty$. The "zone of trust" from §23.2 is really a disk in the complex plane, and its edge is wherever the function first breaks.

Historical Note. The complex-plane explanation of real radii of convergence is due to Augustin-Louis Cauchy (1789–1857), who built the theory of complex functions in the 1820s–40s. He proved that a function differentiable once in the complex sense is automatically infinitely differentiable and equal to its Taylor series — a rigidity with no analogue for real functions. The phenomenon that real convergence is dictated by invisible complex singularities was, for the 19th century, a stunning vindication of taking $\sqrt{-1}$ seriously.

23.10 When Taylor Series Fail

Smoothness is not enough. A function can have every derivative at a point and still not equal its Taylor series there. The canonical counterexample is

$$f(x) = \begin{cases} e^{-1/x^2}, & x \ne 0,\\ 0, & x = 0.\end{cases}$$

This function is infinitely differentiable, and a patient computation shows that every derivative at $0$ is zero: $f^{(n)}(0) = 0$ for all $n$. Its Maclaurin series is therefore $0 + 0\cdot x + 0\cdot x^2 + \cdots = 0$ — the zero function. The series converges everywhere, but it converges to $0$, not to $f$, except at the single point $x = 0$. The function is invisible to its own Taylor series.

A function that does equal its Taylor series on a neighborhood of every point is called analytic. So we must distinguish three rigor levels of "nice":

1. Continuous — no jumps (Chapter 4).
2. Smooth ($C^\infty$) — infinitely differentiable. Stronger than continuous.
3. Analytic — equals its Taylor series locally. Stronger still: every analytic function is smooth, but $e^{-1/x^2}$ shows the converse fails.

The good news: virtually every function you meet in calculus — polynomials, $e^x$, $\sin$, $\cos$, $\ln$, roots, and their compositions and quotients where defined — is analytic, and Taylor's theorem (§23.5) is precisely the tool that proves it, by showing the remainder $R_N \to 0$. The pathological cases are rare but they mark the exact boundary of the theory, and knowing they exist is what separates a computation from a proof.

Math Major Sidebar — proving a series converges to its function. Constructing a Taylor series is automatic; proving it equals $f$ is not. The rigorous statement is: $f(x) = \sum f^{(n)}(a)(x-a)^n/n!$ if and only if the remainder $R_N(x) \to 0$ as $N \to \infty$ at that $x$. For $e^x$, $\sin x$, $\cos x$ the bound $|R_N| \le M|x-a|^{N+1}/(N+1)!$ with $M$ fixed drives the remainder to zero everywhere (the factorial beats any fixed power), which is the honest proof that these series represent their functions. For $e^{-1/x^2}$, the remainder does not vanish — the series and the function part ways even though the Taylor polynomial machinery runs without complaint. Always check the remainder; existence of the series is never the same as representation.

23.11 Chapter Summary

• A power series $\sum c_n(x-a)^n$ converges on an interval centered at $a$, governed by the radius of convergence $R = \lim|c_n/c_{n+1}|$ (ratio test) or $1/\limsup|c_n|^{1/n}$ (Cauchy–Hadamard). The endpoints $|x-a| = R$ must always be tested separately.
• Inside the radius, a power series defines a function you may differentiate and integrate term-by-term, with the radius preserved.
• The Taylor series of $f$ at $a$ is $\sum f^{(n)}(a)(x-a)^n/n!$; centered at $0$ it is the Maclaurin series. Memorize the series for $e^x$, $\sin x$, $\cos x$, $1/(1-x)$, $\ln(1+x)$, $\arctan x$, and the binomial $(1+x)^k$.
• Taylor's theorem writes the error as a Lagrange remainder $R_N = f^{(N+1)}(c)(x-a)^{N+1}/(N+1)!$, bounded by $M|x-a|^{N+1}/(N+1)!$ — the certificate that tells you how many terms guarantee a target accuracy.
• Build new series from old by substitution, differentiation, integration, and arithmetic — almost never from the raw derivative formula.
• The chapter's climax: term-by-term integration computes $\int e^{-x^2}\,dx$ and the normal-curve area $\Phi(z)$ to arbitrary precision, even though no elementary antiderivative exists — closing the "area under the normal curve" anchor opened in Chapter 13.
• The radius of convergence is the distance to the nearest complex-plane singularity; smooth is not the same as analytic ($e^{-1/x^2}$ is the cautionary tale).

Add to Your Modeling Portfolio. Add a Taylor-series approximation to your model — replace a hard function by a polynomial you can compute and bound. Biology: expand the logistic growth rate or a Michaelis–Menten reaction rate to second order near an operating point, and use Taylor's remainder to state the range over which the linearization is trustworthy. Economics: approximate a utility or production function by its degree-2 Taylor polynomial near current output (the basis of marginal analysis and risk-aversion measures), reporting the error bound. Physics: derive the small-angle pendulum or the relativistic-correction term of §23.8, and quantify when the next Taylor term can no longer be neglected. Data Science: implement your own erf/normal-CDF by term-by-term integration of $e^{-x^2}$ (as in §23.7), and certify its accuracy with the Lagrange remainder — a function you built from a series, not a black box.

Looking Ahead

You can now turn nearly any smooth function into an infinite polynomial, truncate it with a guaranteed error bound, and integrate things that have no antiderivative. Chapter 24 spends this capital on the spectacular: it feeds the imaginary unit $i$ into the series for $e^x$ and watches $\sin$ and $\cos$ fall out, producing Euler's formula $e^{ix} = \cos x + i\sin x$ and its celebrated specialization $e^{i\pi} + 1 = 0$ — the anchor first teased back in Chapter 11. The same machinery yields the solution of the Basel problem and a first glimpse of Fourier series. Everything in Chapter 24 rests on the term-by-term manipulation you just learned; the toolbox is built, and the next chapter shows what it can make.

Reflection

For most of this book, transcendental functions were oracles — you trusted that a calculator could produce $\sin(0.7)$ or $e^{0.5}$, without ever seeing how. This chapter pulled back the curtain. Behind every sine, every exponential, every probability under the bell curve, there is an infinite polynomial, summed term by careful term, with an error you can bound before you begin. The deepest lesson is the one that has run through the whole subject: approximation is not a retreat from exactness but a road to it. A polynomial you can compute, plus a remainder you can bound, equals a transcendental value you can trust to as many digits as you please. That is the quiet machinery of every computation you will ever run — and now you know how it works.

Continue to: Exercises · Quiz · Case Study 1 · Case Study 2 · Key Takeaways · Further Reading