Practice: Power Series, Taylor Expansion and the Remainder

(a) $T_3$ for $\sin x$, with an error bound at $x=0.5$. The derivatives at 0 cycle: $\sin 0=0$, $\cos 0=1$, $-\sin 0=0$, $-\cos 0=-1$. So the coefficients are $0,1,0,-\frac{1}{3!}$ and

The bound. Every derivative of $\sin$ is $\pm \sin$ or $\pm \cos$, so $|f^{(4)}|\le 1$ everywhere and $M=1$:

The actual error. $T_3(0.5)=0.5-\frac{0.125}{6}=0.479167$, and $\sin (0.5)=0.479426$, so the error is $0.000259$. Comparison. The bound $0.002604$ exceeds the actual error $0.000259$ by about a factor of ten. It must exceed it, a bound that did not would be wrong, and the slack is expected, since $M=1$ is the worst case over the interval while the relevant fourth derivative is $\sin c$ for some small $c$, which is much smaller than 1. The value of the bound is that it was available before computing $T_3$, so the degree could be chosen to meet a required accuracy in advance. (b) $f(x)=e^{-1/x^2}$, $f(0)=0$. The series. Every derivative at the origin is 0, so every coefficient $\frac{f^{(n)}(0)}{n!}$ is 0, and the Maclaurin series is

Where it converges. Everywhere. A series of zeros converges for every real $x$, to the zero function. The radius of convergence is infinite. What it shows. The series converges for all $x$ and equals $f$ only at $x=0$. For every $x\ne 0$, while the series sums to 0. At $x=1$, for instance, $f(1)=e^{-1}\approx 0.368$ against a series value of 0. So converging and representing the function are different claims, and the first does not imply the second. Where the gap lives. Taylor's theorem gives $f=T_n+R_n$ exactly, for every $n$. Here $T_n=0$ for all $n$, so $R_n=f$. The remainder never shrinks. The series represents $f$ precisely where $R_n\to 0$, and that is a separate condition requiring its own argument. Why this is possible at all. The coefficients are built from derivatives at a single point, so they encode only local information at the origin. This function is flat to infinite order there while being positive everywhere else, so the local data at 0 simply does not determine it. The contrast. For $e^x$, $\sin$ and $\cos$ the Lagrange bound $\frac{M|x{|}^{n+1}}{(n+1)!}$ does tend to 0 for every fixed $x$, because the factorial eventually beats any fixed power, which is the argument establishing that those series equal their functions, rather than merely converging.

(c) The radius of convergence. For $\sum _{n=1}^{\mathrm{\infty }}{x}^n/n$ the ratio test gives $|x|\frac{n}{n+1}\to |x|$, so the series converges absolutely for and $R=1$.

The endpoints escape that test, since there $L=1$, and they must be examined separately. At $x=1$ the series is the harmonic series and diverges. At $x=-1$ it is the alternating harmonic series and converges. One endpoint in, one out, which is why an interval of convergence is never read off the radius alone.