Power Series, Taylor Expansion and the Remainder

A convergence test answers whether a given series has a sum. This unit asks the question from the other side: given a function, which series represents it, and how far can a truncation of that series be trusted?

The derivative gave the best linear approximation to $f$ near a point. The Taylor series continues the pattern with every higher derivative, and for $e^{0.5}$ the errors fall by roughly a decimal place per term: $1.5\times {10}^{-1}$, $2.4\times {10}^{-2}$, $2.9\times {10}^{-3}$, $2.8\times {10}^{-4}$.

Two questions have to be kept apart, and the unit turns on the difference.

Where does the series converge? A power series converges on an interval centred at its expansion point, whose half-width is found by the ratio test from the previous unit. The endpoints escape that test and are examined one at a time.

Does it converge to the function? This is a separate question with a separate answer, and it is not automatic. The remainder $R_n=f-T_n$ is what decides it, and there are functions whose Taylor series converges everywhere while equalling the function at exactly one point.

The Lagrange remainder answers both practical needs at once: it bounds the error of a truncation in advance, so a polynomial can be trusted to a stated accuracy, and it is the quantity whose vanishing turns a Taylor polynomial into a representation.

The canonical statement gives the power series, the Taylor coefficients and the remainder. What follows is why each is stated as it is.

Why a power series has a radius rather than an arbitrary domain. Applying the ratio test to $\sum c_n(x-a{)}^n$ gives a condition of the form , symmetric about $a$ by construction. The endpoints escape the test entirely, since there $L=1$, and they must be examined one at a time, often with different verdicts at each end.

Why convergence of a Taylor series is not representation. The coefficients are built from $f$'s derivatives at a single point, so the series is determined by local information. That the resulting series converges says nothing about whether it converges to $f$. The gap is exactly the remainder: $f=T_n+R_n$ always, and $f$ equals the series only where $R_n\to 0$. Establishing that is a separate argument, and for some functions it fails.

Where the Taylor coefficients come from. Suppose $f$ can be written as a power series about $a$:

Setting $x=a$ kills every term but the first, giving $c_0=f(a)$.

Differentiating term by term,

and setting $x=a$ leaves $c_1=f^{\prime }(a)$. Differentiating again,

The pattern is clear: differentiating $n$ times brings down a factor $n!$ from the $n$th term and kills all lower ones, so

The factorial is not a normalising convention; it is what the repeated differentiation produces.

What the argument assumes. It assumes $f$ has such a representation and that term-by-term differentiation is valid. Both hold inside the radius of convergence, but the argument establishes only what the coefficients must be if a representation exists, not that it does. That second question is the remainder's business.

Checking the result. For $f(x)=e^x$ about $a=0$, every derivative is $e^x$ with value 1 at 0, so $c_n=1/n!$ and

At $x=0.5$ the partial sums give $1.5$, $1.625$, $1.6458333$, $1.6484375$ against $e^{0.5}=1.6487213$, errors of $1.49\times {10}^{-1}$, $2.37\times {10}^{-2}$, $2.89\times {10}^{-3}$, $2.84\times {10}^{-4}$, each roughly a tenth of the last.

Statement. Let $f$ have $n+1$ continuous derivatives on an interval containing $a$ and $x$. Then

and the remainder has the Lagrange form

for some $c$ strictly between $a$ and $x$.

The equality is exact. $f=T_n+R_n$ holds for every $n$, with no approximation. The remainder is defined as the difference. The theorem's content is the form of the remainder, which is what makes it estimable.

The usable bound. The point $c$ is not known, so bound the derivative instead. If $|f^{(n+1)}(t)|\le M$ for all $t$ between $a$ and $x$, then

Checking it bites. For $f(x)=e^x$ at $a=0$, $x=0.5$, $n=3$: every derivative is $e^t$, which on $[0,0.5]$ is at most $M=e^{0.5}\approx 1.6487$. The bound is

and the actual error is $2.888\times {10}^{-3}$. The bound exceeds the error, as it must, and is the right order of magnitude.

Why $(n+1)!$ makes the series converge for $e^x$. Fix $x$ and let $n$ grow. The bound is $\frac{M|x{|}^{n+1}}{(n+1)!}$, and the factorial eventually outgrows any fixed power, so $R_n\to 0$ for every real $x$. That is the argument establishing

rather than merely that the series converges. The same argument works for $\sin$ and $\cos$, whose derivatives are bounded by 1 everywhere, giving $M=1$ and convergence on the whole line.

Where the argument fails. Consider

Every derivative at 0 is zero, so every Taylor coefficient vanishes and the Maclaurin series is identically 0. It converges everywhere, to the zero function. But for every $x\ne 0$, so the series equals $f$ only at the centre.

Nothing went wrong with the coefficients; the remainder simply does not tend to zero. This is the case that shows why "the series converges" and "the series represents $f$" are different statements, and why the theorem is stated with a remainder rather than as an identity.

How the bound is used in practice. Given a required accuracy $\epsilon$, choose $n$ so that , then truncate. This is how library implementations of $e^x$, $\sin$ and $\cos$ decide how many terms to compute. The error is bounded in advance rather than discovered afterwards.

A Taylor expansion with an error bound: approximate $e^{0.5}$ to within ${10}^{-2}$.

The Maclaurin series is $\sum x^n/n!$. For the bound, every derivative of $e^x$ is $e^x$, at most $M=e^{0.5}\approx 1.6487$ on $[0,0.5]$:

So $n=3$ suffices for ${10}^{-2}$. The polynomial is

and $e^{0.5}=1.6487213$, an error of $2.888\times {10}^{-3}$, inside the predicted bound.

Here $n$ was chosen before computing, from the bound alone.

For a power series. Apply the ratio test to $|c_n(x-a{)}^n|$ and solve for $x$. That gives the radius $R$. Then test $x=a-R$ and $x=a+R$ separately, since the ratio test is inconclusive there and the two endpoints often disagree.

For a Taylor expansion.

1. Compute derivatives at $a$ until the pattern is clear.
2. Form $T_n(x)=\sum _{k\le n}\frac{f^{(k)}(a)}{k!}(x-a{)}^k$.
3. Bound $|f^{(n+1)}|$ by $M$ on the interval between $a$ and $x$.
4. Use $|R_n|\le \frac{M|x-a{|}^{n+1}}{(n+1)!}$ to choose $n$ for a required accuracy, before computing the polynomial.
5. To claim the series represents $f$, show $R_n\to 0$ as $n$ grows. Convergence of the series alone is not enough.

Checks. For a Taylor approximation, compare the actual error against the bound. The bound must be the larger, and if it is not, one of the two is wrong.

${\displaystyle \frac{1}{1-x}}=\sum _{n=0}^{\mathrm{\infty }}{x}^n$ for . The geometric series, and the only one with an elementary closed form. Everything else in this list can be obtained from it by differentiating, integrating or substituting. At $x=0.5$ the sum is 2, at $x=0.9$ it is 10, and at $x=1.5$ the terms grow and the series diverges.

$e^x=\sum _{n=0}^{\mathrm{\infty }}{\displaystyle \frac{x^n}{n!}}$ for all $x$. Every derivative is $e^x$, so every coefficient is $1/n!$. The factorial in the denominator is what gives an infinite radius of convergence. It eventually outgrows $x^n$ for any fixed $x$.

$\sin x=x-{\displaystyle \frac{x^3}{3!}}+{\displaystyle \frac{x^5}{5!}}-\cdots$ for all $x$. Odd powers only, because $\sin$ is odd. At $x=0.5$ the successive truncations give $0.5$, $0.47917$, $0.47943$ against the exact $0.47943$, three terms already agree to five decimals. At $x=1$ they give $1$, $0.83333$, $0.84167$ against $0.84147$, converging more slowly since $x$ is larger.

$\cos x=1-{\displaystyle \frac{x^2}{2!}}+{\displaystyle \frac{x^4}{4!}}-\cdots$ for all $x$. Even powers only, being the derivative of the $\sin$ series term by term.

$\ln (1+x)=x-{\displaystyle \frac{x^2}{2}}+{\displaystyle \frac{x^3}{3}}-\cdots$ for . Obtained by integrating the geometric series for $\frac{1}{1+x}$. The radius is 1, and the endpoints differ: at $x=1$ the series is the alternating harmonic, converging to $\ln 2$; at $x=-1$ it is the negative harmonic series and diverges. One endpoint in, one out, which is why endpoints are always checked separately.

$\sum _{n=1}^{\mathrm{\infty }}{\displaystyle \frac{x^n}{n}}$, radius 1. The ratio test gives $|x|\frac{n}{n+1}\to |x|$, so $R=1$. Inside, the sum is $-\ln (1-x)$: at $x=0.5$ a 2000-term partial sum gives $0.69314718$ against $\ln 2=0.69314718$, and at $x=0.99$ it gives $4.60517019$ against $-\ln (0.01)=4.60517019$.

$\sum {\displaystyle \frac{1}{n^p}}$ — the $p$-series. Converges exactly when , by the integral test. At $p=2$ the sum is ${\pi }^2/6$; at $p=1$ it is the harmonic series and diverges. Useful mainly as the comparison standard, since the threshold is easy to remember and many series can be bounded against it.

Three have infinite radius because of a factorial; three have radius 1 because their coefficients decay only polynomially. The factorial is the difference between a series that converges everywhere and one that converges on an interval, and reading which case applies is usually a one-line ratio test.

A power series says nothing about its endpoints. For $\sum x^n/n$ the ratio test gives $R=1$, and the two endpoints behave differently: at $x=1$ the series is harmonic and diverges, at $x=-1$ it is alternating harmonic and converges. Reporting convergence on $[-1,1]$ or on $(-1,1)$ without checking is wrong in one case each way.

A Taylor series may converge without representing the function. Let

Every derivative at 0 is zero, so every Maclaurin coefficient is zero and the series is identically 0. It converges for every $x$, to the zero function, while for all $x\ne 0$. The series represents $f$ at exactly one point.

Nothing miscomputed: the coefficients are correct and the series converges. What fails is $R_n\to 0$, which is the separate condition that turns a Taylor polynomial into a representation. This is why the theorem carries a remainder rather than an equals sign.

A Taylor polynomial is local. $T_3$ for $e^x$ about 0 has error $2.9\times {10}^{-3}$ at $x=0.5$ and about $7.1$ at $x=3$. The bound $\frac{M|x{|}^4}{4!}$ grows with the fourth power of the distance. An approximation validated near the centre says nothing about behaviour far from it.

Comparison needs the inequality in the right direction. Bounding $a_n$ below by a convergent series establishes nothing, and bounding it above by a divergent one establishes nothing either. Only above-by-convergent and below-by-divergent carry information.