Taylor and Maclaurin Series: 17 MCQs with answers

An infinite sum of powers of $(x-a)$ built from derivatives at $a$

A Taylor series expresses a smooth function as an infinite sum of powers of $(x-a)$, where the coefficients come from the derivatives of the function evaluated at the centre $a$.

The Taylor series with centre $a=0$

Maclaurin is not a separate idea — it is the special case of the Taylor series where the centre is $a=0$.

The centre or expansion point of the series

The point $a$ is the centre or expansion point; all derivatives are evaluated at $a$, and the series is most accurate near $x=a$.

Differentiating $(x-a)^n$ repeatedly produces $n!$, which must be cancelled

Repeated differentiation of $(x-a)^n$ leaves $n!$ as a constant factor. Dividing by $n!$ cancels it so the $n$-th derivative of the series matches $f^{(n)}(a)$ exactly.

$1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$

Every derivative of $e^x$ is $e^x$, and $e^0=1$, so $f^{(n)}(0)=1$ for all $n$. This gives $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$.

True

A Taylor series centred at $a$ has the form $f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n$, and the Maclaurin series is the special case with $a=0$.

$\sin x \approx x$

The first term of the Maclaurin series of $\sin x$ is $x$, giving the small-angle approximation $\sin\theta \approx \theta$.

It converges for all real $x$

The Maclaurin series of $e^x$ converges for every real $x$, unlike $\frac{1}{1-x}$ which only converges for $|x|

$f^{(n)}(2)$

Every derivative must be evaluated at the centre $a$, not at $x$ and not at $0$. For $a=2$, the coefficient uses $f^{(n)}(2)$.

The derivatives at $0$ alternate as $1,0,-1,0,\dots$, so even-order terms vanish

The derivatives of $\sin x$ cycle as $\cos x, -\sin x, -\cos x, \sin x$, and at $x=0$ they give $1,0,-1,0,\dots$. Even-order derivatives are zero, so only odd powers survive.

Polynomials are easy to add, multiply, differentiate, and integrate

Taylor and Maclaurin series replace complicated functions with polynomials that are easy to compute and behave almost identically near the chosen point, which is why calculators use them.

The series no longer matches the derivatives of the original function

Dropping the factorial is the most frequent slip. Without it, the series no longer matches the derivatives of the original function, and the approximation collapses.

Both A and R are true, and R correctly explains A

Maclaurin is exactly the Taylor series with centre $a=0$, so A is true. Substituting $a=0$ into $\sum \frac{f^{(n)}(a)}{n!}(x-a)^n$ gives $\sum \frac{f^{(n)}(0)}{n!}x^n$, which is precisely the Maclaurin form — so R explains A.

I and II only

Statement I is true — convergence must always be checked. Statement II is true — the Lagrange remainder bounds the truncation error. Statement III is false: for example, $\frac{1}{1-x}$ about $0$ converges only for $|x|

The claim is false; $\frac{1}{1-x}$ about $0$ converges only for $|x|<1$

Convergence is not automatic. While $e^x$, $\sin x$, and $\cos x$ converge for all real $x$, $\frac{1}{1-x}$ about $0$ converges only for $|x|

Near the centre, the first terms dominate and the remainder is small

The remainder term $R_n(x)$ bounds the error of truncation. Near the centre, the first one or two terms dominate and the remainder is small, which is why the small-angle approximation $\sin\theta \approx \theta$ works.

The first four non-zero terms are \( x - rac{x^2}{2} + rac{x^3}{3} - rac{x^4}{4} \) with interval of convergence \( |x| < 1 \).

The Maclaurin series for \( \ln(1+x) \) is derived from its derivatives evaluated at \( x=0 \). The correct series expansion yields \( x - rac{x^2}{2} + rac{x^3}{3} - rac{x^4}{4} \), and it converges for \( |x| < 1 \) due to the nature of the logarithm function.