Maclaurin series for 1/sqrt(1+x^3)

Question

To write the maclaurin series for 1/sqrt(1+x^3), why is it incorrect to multiply top and bottom by sqrt(1+x^3) and write the series as if it were in the form 1/(1-x), thus making the series sqrt(1+x^3) * Summation (-x^3)^n?

 

Answer 1

Good question, and the honest answer is that your manipulation is not wrong — it is just not a Maclaurin series. Those are different complaints, and the distinction is the whole lesson here.

What your expression actually is

Multiplying top and bottom by √(1+x³) gives

1/√(1+x³) = √(1+x³) / (1+x³)

and expanding the 1/(1+x³) as a geometric series with ratio −x³ gives exactly what you wrote:

√(1+x³) · ∑ (−x³)ⁿ

That identity is true for |x| < 1. Plug in x = 0.5 and it checks out. So nothing you did was illegal.

Why it does not answer the question

A Maclaurin series has to be a power series — a sum of the form ∑ c_nxⁿ, where every c_n is a constant. Your expression has √(1+x³) sitting out front, and that factor is not a constant and not a polynomial. It still contains the very square root you were trying to get rid of.

Look at what the algebra accomplished: it moved the radical from the denominator to the numerator. That is a genuinely useful move in other contexts — rationalizing a denominator to evaluate a limit, for instance — but here the obstacle was never where the radical sat. It was that a radical is not a polynomial. Relocating it does not remove it.

You could still finish from your expression: expand √(1+x³) = (1+x³)^1/2 as its own binomial series, then multiply the two series together and collect like powers. That works, but it is strictly more labor than the direct route below — and it requires the binomial series anyway, so you have not avoided anything.

The direct way

Write the function with a negative exponent, which is the form the binomial series wants:

1/√(1+x³) = (1 + x³)^−1/2

The binomial series says (1+u)^k = ∑ C(k,n) uⁿ, where C(k,n) = k(k−1)…(k−n+1)/n! and k may be any real number. Here k = −1/2 and u = x³.

Working out the first few coefficients:

n = 0: 1

n = 1: (−1/2)/1! = −1/2

n = 2: (−1/2)(−3/2)/2! = 3/8

n = 3: (−1/2)(−3/2)(−5/2)/3! = −5/16

n = 4: (−1/2)(−3/2)(−5/2)(−7/2)/4! = 35/128

So

1/√(1+x³) = 1 − (1/2)x³ + (3/8)x⁶ − (5/16)x⁹ + (35/128)x¹² − …

and in closed form, since those coefficients simplify nicely,

= ∑_n=0^∞ (−1)ⁿ · [ C(2n, n) / 4ⁿ ] · x³ⁿ

Convergence needs |u| < 1, that is |x³| < 1, so the radius of convergence is 1.

The takeaway

The geometric series ∑ uⁿ = 1/(1−u) is only one special case of the binomial series — it is what you get when k = −1. Students reach for it reflexively because it is the first one they learn, and then try to bend the problem into 1/(1−u) shape by any algebra available. But the binomial series handles every exponent, including −1/2, so there is no bending required. When you see a root or a fractional power, go straight to (1+u)^k and read off k.