Taylor Series

A power series centred at $x_0$ is a series of the form:

$$\sum_{n=0}^\infty a_n (x-x_0)^n.$$

The set of $x$ for which the series converges is an interval centred at $x_0$, with radius of convergence $R$. The series converges absolutely for $|x-x_0|<R$ and diverges for $|x-x_0|>R$. At the endpoints $|x-x_0|=R$ the behaviour varies and must be checked separately.

Ratio test for the radius: if $$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = L$$ exists, then $R = 1/L$ (with $R=+\infty$ if $L=0$, $R=0$ if $L=+\infty$).

Example: $\sum x^n/n$ has $a_n=1/n$, so $L=\lim n/(n+1)=1$, hence $R=1$. At $x=1$ diverges (harmonic), at $x=-1$ converges conditionally (alternating harmonic).

Power series can be differentiated and integrated term by term within their interval of convergence, preserving the same radius.

If $f$ is infinitely differentiable at $x_0$, the Taylor series centred at $x_0$ is:

$$f(x) = \sum_{k=0}^\infty \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k.$$

The $n$-th order Taylor polynomial $T_n(x)$ truncates the series at $k=n$. The remainder $R_n(x)=f(x)-T_n(x)$ measures the approximation error.

Peano remainder: $R_n(x)=o((x-x_0)^n)$ as $x\to x_0$ (little-o notation). Sufficient when studying limits.

Lagrange remainder: there exists $\xi$ between $x$ and $x_0$ such that $$R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1}.$$ Useful for bounding the error.

MacLaurin series: the special case with $x_0=0$: $f(x)=\sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!} x^k$.

Fundamental MacLaurin expansions (memorise for exams):

• $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots = \displaystyle\sum_{k=0}^\infty \frac{x^k}{k!},\quad R=\infty$
• $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots = \displaystyle\sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!},\quad R=\infty$
• $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots = \displaystyle\sum_{k=0}^\infty (-1)^k \frac{x^{2k}}{(2k)!},\quad R=\infty$
• $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots = \displaystyle\sum_{k=1}^\infty (-1)^{k+1} \frac{x^k}{k},\quad R=1$
• $(1+x)^\alpha = 1 + \alpha x + \frac{\alpha(\alpha-1)}{2!}x^2 + \frac{\alpha(\alpha-1)(\alpha-2)}{3!}x^3 + \cdots$ (binomial series, $R=1$)
• $\frac{1}{1-x} = \displaystyle\sum_{k=0}^\infty x^k$ (geometric series, $R=1$, valid for $|x|<1$)

Euler's formula from series: substituting $ix$ into the exponential series and separating real/imaginary parts yields $e^{ix}=\cos x+i\sin x$.

Taylor expansions are often more powerful than L'Hôpital's rule for evaluating limits, especially when the limit involves compositions or when L'Hôpital would require repeated differentiations.

Method: replace each function with its Taylor expansion up to the necessary order, simplify algebraically, and read off the limit.

Infinitesimal hierarchy (as $x\to0$):

$$x \gg x^2 \gg x^3 \gg \cdots \gg x^n$$

When performing expansions, keep only terms up to the lowest order that resolves the indeterminacy.

Example: $\displaystyle\lim_{x\to0}\frac{\sin x - x}{x^3}$. Expand $\sin x = x - x^3/6 + o(x^3)$. Then $\frac{x - x^3/6 + o(x^3) - x}{x^3} = \frac{-x^3/6 + o(x^3)}{x^3} \to -1/6$.

Warning: the expansion must be carried to an order sufficient so that the numerator's leading term does not cancel completely.