Question bank — Maclaurin series of eˣ, sin x, cos x, ln(1+x), (1+x)ⁿ — derive all

False. It only equals the function inside its interval of convergence; outside, the polynomial diverges even where the function is perfectly finite (see $\ln 3$ from $\ln (1+x)$ at $x=2$). Look at figure s01 — the partial sums hug the curve near $0$ and then peel away.

False. $f(x)=e^{-1/x^2}$ (with $f(0)=0$) has all derivatives zero at $0$, so its Maclaurin series is $0$, yet the function is nonzero — the series is right only for analytic functions. Figure s03 shows why: the graph flattens against the axis faster than any polynomial can detect.

True. The factorial denominator $n!$ eventually outgrows $x^n$ for any fixed $x$, so terms shrink to $0$ fast enough — the radius of convergence is infinite.

False. There is nothing to cancel — the odd derivatives of $\cos$ are literally $0$ at $x=0$ ($-\sin 0=0$, etc.), so those coefficients are born as zero.

False. They are factorials $3!,5!,7!=6,120,5040$; the smaller-looking $3$ is just $3!=6$ written out, an easy misread.

False. With $k$ as the term counter, the numerator $n(n-1)\cdots (n-k+1)$ contains the factor $(n-n)=0$ once $k>n$ for a non-negative integer $n$, so every later $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ is zero and the series terminates — the ordinary binomial theorem.

False. Maclaurin is exactly the Taylor series with centre $a=0$; same machinery, special base point.

False. The sign of the error depends on the next term's sign. For $\ln (1+x)$ with $x>0$ the terms alternate, so a partial sum can lie above or below depending on where you stop.

Wrong. The constant term is $a_0=\ln (1+0)=\ln 1=0$; there is no leading $1$. The series starts at the $x$ term: $x-\frac{x^2}{2}+\cdots$.

Missing the $n!$. Here $a_n$ is the weight of $x^n$ from the definition above. Since $\frac{d^n}{d{x}^n}{x}^n=n!$, the surviving term is $n!a_n=f^{(n)}(0)$, giving $a_n=\frac{f^{(n)}(0)}{n!}$.

Sign error. Using the Geometric series with ratio $r=-x$ gives alternating signs: $1-x+x^2-x^3+\cdots$. The all-plus version is $\frac{1}{1-x}$.

Sign of the quadratic term is wrong. The coefficient is $\left(\genfrac{}{}{0ex}{}{1/2}{2}\right)=\frac{\frac{1}{2}(-\frac{1}{2})}{2!}=-\frac{1}{8}$, so it is $1+\frac{1}{2}x-\frac{1}{8}{x}^2+\cdots$ — negative because $n=\frac{1}{2}$ makes $(n-1)$ negative.

The function value exists, but the series only converges for $-1<x\le 1$. At $x=2$ the polynomial diverges; you would need a different centre or identity. Figure s02 puts this interval on a number line.

Wrong sign placement. Substitute $x\to -x$: $e^{-x}=\sum \frac{(-x{)}^n}{n!}=\sum \frac{(-1{)}^n{x}^n}{n!}$, which alternates. Only the odd terms change sign, not every term uniformly.

No contradiction: $2!=2$, so $\frac{x^2}{2}$ is $\frac{x^2}{2!}$. Differentiating the sine series does reproduce the cosine series correctly.

Because $\frac{d^n}{d{x}^n}{x}^n=n!$; without dividing, the $x^n$ piece would contribute $n!a_n$ instead of $a_n$, so we compensate to make coefficients match the derivatives cleanly.

Every term still containing an $x$ vanishes ($0^k=0$), leaving only the constant term of that particular derivative — a clean one-equation-one-unknown extraction.

Because $\frac{d}{dx}{e}^x=e^x$, so every derivative at $0$ equals $e^0=1$; all numerators are the same, leaving only $\frac{1}{n!}$.

$\sin$ is an odd function ($\sin (-x)=-\sin x$), and only odd-power monomials share that symmetry; the even-power coefficients must be zero to preserve it.

Because $\frac{d}{dx}\ln (1+x)=\frac{1}{1+x}$ is already a known geometric series; integrating it term-by-term is far cleaner than computing higher derivatives of a logarithm.

At $x=1$ it becomes the alternating harmonic series (converges to $\ln 2$); at $x=-1$ it becomes the negative harmonic series $-1-\frac{1}{2}-\frac{1}{3}-\cdots$, which diverges — mirroring $\ln 0=-\infty$. See the closed/open endpoints in figure s02.

Substituting $x\to ix$ into $e^x$ splits the terms by parity: real even-power terms rebuild $\cos x$ and imaginary odd-power terms rebuild $i\sin x$, giving $e^{ix}=\cos x+i\sin x$.

Replace $\sin x$ by $x-\frac{x^3}{6}+\cdots$; then $\frac{\sin x}{x}=1-\frac{x^2}{6}+\cdots \to 1$ instantly, no repeated differentiation needed.

It is the polynomial itself — a finite series. Its higher derivatives vanish, so all coefficients $a_n$ beyond degree $2$ are zero; it is already "its own" series.

$\sqrt{2}$ is finite and the series does converge there for $n=\frac{1}{2}$, but convergence is delicate at endpoints and depends on the sign of $n$; the safe interior guarantee is $|x|<1$ (the shaded band in figure s02).

The function is the constant $1$; the series collapses to its first term $\left(\genfrac{}{}{0ex}{}{0}{0}\right)=1$ (recall $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ is the weight of $x^k$, and here only $k=0$ survives) and all further coefficients are zero — consistent with the terminating integer case.

Yes. Setting $x=0$ kills every power term, leaving the constant $1$ — the built-in sanity check that the constant coefficient $a_0$ equals $f(0)$.

It has none centred at $0$: $\ln x$ is undefined at $x=0$, so no derivatives exist there. That is precisely why we expand $\ln (1+x)$ instead, shifting the trouble to $x=-1$.

The series still converges (infinite radius), but you need many terms because $\frac{x^n}{n!}$ only starts shrinking once $n>x\approx 10$; convergence is guaranteed, not necessarily fast.