Power Series, Taylor Expansion and the Remainder

${\displaystyle \frac{1}{1-x}}=\sum _{n=0}^{\mathrm{\infty }}{x}^n$ 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.