Mixed Taylor Construction for Logarithmic and Radical Products

Why Multiply the Series?

A function can be built from familiar pieces even when its Taylor series is not immediately visible. For \(f(x)=\sqrt{1+x}\ln(1+x)\), the square-root factor has a binomial series and the logarithm has a Maclaurin series. Multiplying those expansions produces a series for the product, but the coefficients must be combined carefully.

The central calculation is called coefficient convolution: the coefficient of \(x^n\) in a product receives contributions from every pair of terms whose powers add to \(n\). After finding the coefficients, we must separately determine where the series converges, including at the boundary points.

The Two Series and Their Product

For \(|x|<1\), the binomial expansion with exponent \(1/2\) gives

The generalized binomial coefficient is defined by \(\binom{\alpha}{0}=1\) and, for \(k\geq 1\), by \(\binom{\alpha}{k}=\alpha(\alpha-1)\cdots(\alpha-k+1)/k!\). The logarithm expansion, also valid for \(|x|<1\), is

Write \(b_k=\binom{1/2}{k}\) and \(l_m=(-1)^{m+1}/m\) for \(m\geq1\), with \(l_0=0\). If the product is written as \(\sum_{n=0}^{\infty}c_nx^n\), then

Because \(l_0=0\), \(c_0=0\). For \(n\geq1\), the same formula can be written explicitly as

Worked Example: Computing the First Coefficients

Why Convolution Gives the Coefficients

For finite polynomials, the coefficient of \(x^n\) in a product comes from multiplying a term \(b_kx^k\) by a term \(l_mx^m\) with \(k+m=n\). Therefore that coefficient is the sum of all \(b_kl_m\) with \(m=n-k\). For power series, the same coefficient rule holds when the Cauchy product is valid. Here both series converge absolutely for \(|x|<1\), which is sufficient for the product series to converge to the product there.

It is useful to have a second expression for \(c_n\). For a parameter \(a\), the coefficient of \(x^n\) in \((1+x)^a\) is \(B_n(a)=a(a-1)\cdots(a-n+1)/n!\), with \(B_0(a)=1\). Differentiating with respect to \(a\) gives the coefficient of \(x^n\) in \((1+x)^a\ln(1+x)\). At \(a=1/2\), this is the product under consideration.

The differentiation can be performed for \(|x|<1\), where the binomial series and its parameter derivative are valid locally uniformly for \(a\) near \(1/2\). The resulting formula is also a useful way to estimate the coefficients at the boundary.

Worked Example: Checking the Parameter Formula

The Radius and the Endpoints

The two original series both have radius of convergence \(1\), so their product series certainly represents the function for \(|x|<1\). To settle the full interval of convergence, we need to examine the coefficients and then check the endpoints rather than infer their behavior from the open interval.

For \(n\geq1\), the exact formula above gives a useful bound. The reciprocal sum satisfies

Indeed, after the first term its terms have magnitudes \(1/(j-1/2)\), whose sum grows at most proportionally to the harmonic sum. Also, \(\left|\binom{1/2}{n}\right|=O(n^{-3/2})\): this follows, for example, by expressing its magnitude using a central binomial coefficient and applying Stirling’s estimate. Consequently,

The comparison series \(\sum_{n=1}^{\infty}\log(n+1)/n^{3/2}\) converges, so \(\sum |c_n|\) converges. Thus the product series converges absolutely at both \(x=1\) and \(x=-1\). The same coefficient estimate, together with the reciprocal sum’s logarithmic growth and the binomial coefficient’s \(n^{-3/2}\) scale, shows that the coefficients have \(n\)th-root limit \(1\); hence the radius is exactly \(1\), not larger.

At \(x=1\), the series equals the limit of the product as \(x\) approaches \(1\) from within the interval, namely \(\sqrt{2}\ln 2\). At \(x=-1\), the product has limiting value \(0\). To see the latter, put \(u=1+x\); then \(\sqrt{1+x}\ln(1+x)=\sqrt{u}\ln u\), which tends to \(0\) as \(u\) decreases to \(0\). Absolute convergence at each endpoint allows the sum there to agree with this radial limit. Therefore the interval of convergence is \([-1,1]\), and the series represents the continuous endpoint values as well.

Worked Example: Evaluating the Series at an Endpoint

Common Pitfall: Stopping at the Open Interval

A frequent error is to conclude that the product series is valid only for \(-1<x<1\) because that is where the two input series converge absolutely. That establishes the product inside the interval, but it does not determine convergence at either endpoint. Conversely, substituting an endpoint into each input series and multiplying their sums can be invalid if one of those series does not converge there in the required way.

For this product, the coefficient estimate proves absolute convergence at both endpoints. The endpoint values are \(0\) at \(x=-1\) and \(\sqrt{2}\ln 2\) at \(x=1\). The complete interval is therefore \([-1,1]\), not merely the open interval.

Summary

The construction combines the binomial series for \(\sqrt{1+x}\) and the logarithm series for \(\ln(1+x)\). Their coefficients combine by finite convolution, and the first terms are

The exact coefficient formula and its estimate show that the radius of convergence is \(1\), while the endpoint coefficients are absolutely summable. The series therefore converges throughout \([-1,1]\) and represents the product there, using the limiting value \(0\) at \(x=-1\).