AP Calculus BC

Maclaurin Series vs Binomial Series

Q: Is the binomial series a Maclaurin series?

Yes. It is the Maclaurin series of (1+x)^k, and its coefficients are exactly \frac{f^{(n)}(0)}{n!} for that function. The name marks the one function it applies to, not a different construction.

Q: What happens when k is a nonnegative integer?

The falling product picks up a factor of zero once n passes k, so every later coefficient vanishes. The series stops, becomes the finite binomial theorem, and holds for every x rather than only on \lvert x \rvert < 1.

Q: How do I expand \sqrt{4+x}?

Factor the constant out first: \sqrt{4+x} = 2\sqrt{1 + \frac{x}{4}}. Then use the pattern with k = \frac{1}{2} and \frac{x}{4} in place of x, which puts the interval at \lvert x \rvert < 4.

Every Maclaurin series is built from the derivatives of a function at zero. The binomial series is that construction already carried out for one plus x raised to a power k, so you quote its coefficient pattern instead of differentiating, and it holds for x strictly between negative 1 and 1.

Maclaurin series

Use when: You can differentiate the function at zero, or you can reach it by substituting into one of the standard series you already know.

Binomial series

Use when: The function is one plus x raised to a power that is not a nonnegative integer, such as a square root or a reciprocal of a power.

Side by side

	Maclaurin series	Binomial series
Applies to	Any $f$ with derivatives of every order at $0$	Only $(1+x)^k$
Coefficient of $x^n$	$\frac{f^{(n)}(0)}{n!}$	$\frac{k(k-1)\cdots(k-n+1)}{n!}$
How you build it	Differentiate at $0$ , or substitute into a known series	Quote the pattern with your $k$ dropped in
Where it holds	Depends on the function, from a single point to every real number	$\lvert x \rvert < 1$ , unless $k$ is a nonnegative integer, where the sum stops and holds everywhere
Worked case	$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$	$\sqrt{1+x} = 1 + \frac{x}{2} - \frac{x^2}{8} + \cdots$

Nothing new is happening in the binomial series. Differentiating $f(x) = (1+x)^k$ repeatedly gives $f^{(n)}(x) = k(k-1)\cdots(k-n+1)(1+x)^{k-n}$ , so at $x = 0$ the derivative is the falling product $k(k-1)\cdots(k-n+1)$ . Divide by $n!$ , as the Maclaurin definition demands, and the binomial coefficients fall out.

(1+x)^k = 1 + kx + \frac{k(k-1)}{2!}x^2 + \frac{k(k-1)(k-2)}{3!}x^3 + \cdots, \qquad \lvert x \rvert < 1

Setting $k = -1$ recovers something familiar: $\frac{1}{1+x} = 1 - x + x^2 - x^3 + \cdots$ , which is the geometric series with ratio $-x$ . That is where the radius of $1$ comes from, and the same radius applies for every $k$ that is not a nonnegative integer. Setting $k = \frac{1}{2}$ gives the square root expansion in the table, $1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16} - \cdots$ .

Know the four, derive the rest

The series expected cold are those for $e^x$ , $\sin x$ , $\cos x$ , and $\frac{1}{1-x}$ . The binomial pattern is not on that list, so when a negative or fractional power shows up, either substitute into $\frac{1}{1-x}$ or differentiate at zero and read off the falling product.

Frequently asked questions

Is the binomial series a Maclaurin series?

Yes. It is the Maclaurin series of $(1+x)^k$ , and its coefficients are exactly $\frac{f^{(n)}(0)}{n!}$ for that function. The name marks the one function it applies to, not a different construction.

What happens when $k$ is a nonnegative integer?

The falling product picks up a factor of zero once $n$ passes $k$ , so every later coefficient vanishes. The series stops, becomes the finite binomial theorem, and holds for every $x$ rather than only on $\lvert x \rvert < 1$ .

How do I expand $\sqrt{4+x}$ ?

Factor the constant out first: $\sqrt{4+x} = 2\sqrt{1 + \frac{x}{4}}$ . Then use the pattern with $k = \frac{1}{2}$ and $\frac{x}{4}$ in place of $x$ , which puts the interval at $\lvert x \rvert < 4$ .

In the CED: Unit 10: Infinite Sequences and Series (BC)