Taylor and Maclaurin expansions

Taylor's theorem. If $f$ has $n+1$ continuous derivatives on an interval containing $a$ and $x$, then $f(x)$ equals the degree-$n$ Taylor polynomial at $a$ plus a remainder:

where: $f^{(k)}(a)$ is the $k$-th derivative evaluated at $a$, $(x-a)$ is the displacement from the expansion point, and $R_n$ is the remainder after $n$ terms — the exact gap between the polynomial and the true value of $f(x)$.

Two forms of the remainder. Both are proved using a Cauchy-mean-value-theorem-style auxiliary function, and both say the same qualitative thing — the remainder involves the $(n+1)$-th derivative evaluated at some unknown point $c$ strictly between $a$ and $x$:

where: $c$ is some point strictly between $a$ and $x$, not one you can pin down exactly — its role is only to bound the size of $R_n$. Setting $n=0$ collapses either form to $f(x) = f(a) + f'(c)(x-a)$, which is exactly the Lagrange mean value theorem (LMVT): Taylor's theorem is nothing but a higher-order generalization of the MVT you already proved.

Maclaurin's theorem is the special case $a=0$:

It is not a separate theorem with different hypotheses — it is simply Taylor's theorem read at the origin, which is why $e^x$, $\sin x$, $\cos x$ and $\ln(1+x)$ are always expanded "about $0$."

The four expansions worth memorizing (each valid for all real $x$ unless noted):

Expand $f(x) = \ln x$ about $a=1$ up to the term in $(x-1)^3$, with the Lagrange remainder, then use it to estimate $\ln(1.1)$.

Step 1. Differentiate. $f(x) = \ln x$ gives $f'(x) = 1/x$, $f''(x) = -1/x^2$, $f'''(x) = 2/x^3$, and $f^{(4)}(x) = -6/x^4$.

Step 2. Evaluate at $a=1$. $f(1) = 0$, $f'(1) = 1$, $f''(1) = -1$, $f'''(1) = 2$.

Step 3. Substitute into Taylor's theorem to third order:

Step 4. Plug in $x = 1.1$. The polynomial part gives $0.1 - 0.005 + 0.0003\overline{3} = 0.0953\overline{3}$, extremely close to the true value $\ln(1.1) \approx 0.095310$, with the tiny gap accounted for entirely by $R_3$. The third-order Taylor estimate is $\ln(1.1) \approx 0.09533$.

Competitive stretch

The same series-substitution trick evaluates limits that would otherwise need repeated L'Hôpital differentiation. For $\displaystyle\lim_{x\to 0} \frac{e^x - 1 - x}{x^2}$, substitute $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$, so $e^x - 1 - x = \frac{x^2}{2} + \frac{x^3}{6} + \cdots$, and dividing by $x^2$ gives $\frac{1}{2} + \frac{x}{6} + \cdots$. As $x \to 0$ every term beyond $\frac12$ vanishes, so the limit is $\boxed{\dfrac{1}{2}}$ — read straight off the series with no differentiation at all.

• Treating the truncated polynomial as exactly equal to $f(x)$: Taylor's theorem is an equality only when $R_n$ is kept in; the polynomial alone is an approximation, and dropping $R_n$ silently is a scoring error.
• Confusing Maclaurin's theorem with a separate result: it is Taylor's theorem at $a=0$, nothing more, so its hypotheses are identical.
• Sign errors while differentiating $\ln x$, $1/x$, or trig functions repeatedly — one flipped sign cascades through every later coefficient.
• Using an infinite series outside its interval of validity, e.g. plugging $x=2$ into the $\ln(1+x)$ series (valid only for $-1 < x \le 1$); even the finite polynomial with remainder can become inaccurate once $x$ strays far from $a$.

Ask the tutor: if you want the full Cauchy-MVT-style proof of the Lagrange remainder, or practice choosing how many terms to keep for a target accuracy, ask the AI tutor to walk through it step by step.