Binomial Series

Generalized Binomial Coefficients

For \(x>-1\) and a real exponent \(\alpha\), the power \((1+x)^\alpha\) is defined as \(e^{\alpha\ln(1+x)}\). We want a power series centered at \(x=0\) for this function. When \(\alpha\) is a nonnegative integer, the binomial theorem gives a finite expansion. For other exponents the expansion has infinitely many terms, and its coefficients are the generalized binomial coefficients.

For example, \(\binom{\alpha}{1}=\alpha\) and \(\binom{\alpha}{2}=\alpha(\alpha-1)/2\). The coefficients satisfy a useful recurrence: for every \(n\geq0\), \(\binom{\alpha}{n+1}=(\alpha-n)\binom{\alpha}{n}/(n+1)\). This follows directly from the product definition when none of the factors vanishes; the recurrence also holds when a factor is zero, since the sequence of coefficients then remains zero.

Common Binomial Series Expansions

Each row below follows from the recurrence \(c_{n+1}=c_n(\alpha-n)/(n+1)\), and each is valid for \(|x|<1\).

Replacing \(x\) by \(-x\) gives series for \((1-x)^\alpha\); for example \(1/\sqrt{1-x}=1+\tfrac x2+\tfrac{3x^2}{8}+\tfrac{5x^3}{16}+\cdots\). Replacing \(x\) by \(x^2\) or \(-x^2\) gives series such as \(1/\sqrt{1-x^2}\), which leads to the series for \(\arcsin x\).

Termination and Radius of Convergence

If \(\alpha=m\) is a nonnegative integer, then the recurrence gives \(\binom{m}{m+1}=0\), followed by zero for every higher coefficient. The series terminates and is a polynomial. If \(\alpha\) is not a nonnegative integer, none of the coefficients vanishes: a zero factor would require \(\alpha\) to be one of \(0,1,\ldots,n-1\). In that case, the ratio of consecutive coefficient magnitudes tends to \(1\).

Proof. The termination claim follows from the recurrence just described. Now suppose that \(\alpha\) is not a nonnegative integer, so every coefficient is nonzero. The recurrence gives

For any fixed \(x\neq0\), the ratio of the magnitudes of consecutive terms in the series is therefore \[ \frac{\left|\binom{\alpha}{n+1}x^{n+1}\right|} {\left|\binom{\alpha}{n}x^n\right|} =|x|\frac{|\alpha-n|}{n+1} \longrightarrow |x|. \] The Ratio Test gives absolute convergence when \(|x|<1\) and divergence when \(|x|>1\). Thus the radius is \(1\). The Ratio Test does not decide convergence at \(|x|=1\); those boundary points require separate analysis for a given exponent. \(\square\)

This result identifies where the series can represent a function, but it does not yet identify its sum. We next prove that inside the radius the sum is exactly \((1+x)^\alpha\). The recurrence is especially effective here because it translates directly into a differential equation.

Why the Series Sums to \((1+x)^\alpha\)

Write \(c_n=\binom{\alpha}{n}\), and let \(F(x)=\sum_{n=0}^{\infty}c_nx^n\) for \(|x|<1\), including the polynomial case. The power series has positive radius of convergence, so the term-by-term differentiation theorem established earlier in this course applies inside its radius. The coefficient recurrence implies that \(F\) satisfies \((1+x)F'(x)=\alpha F(x)\). The function \((1+x)^\alpha\) satisfies the same equation for \(-1<x<1\). Their common value at \(x=0\) then determines their equality.

Proof. First suppose that \(\alpha\) is not a nonnegative integer. By the radius result, \(F(x)=\sum_{n=0}^{\infty}c_nx^n\) converges for \(|x|<1\), and it can be differentiated term by term there. Multiplying the derivative by \(1+x\) gives

For every \(n\geq0\), the recurrence says \((n+1)c_{n+1}=(\alpha-n)c_n\). Thus the coefficient of \(x^0\) in the expression above is \(\alpha c_0\), and for \(n\geq1\) the coefficient of \(x^n\) is \[ (n+1)c_{n+1}+nc_n=(\alpha-n)c_n+nc_n=\alpha c_n. \] Consequently, \((1+x)F'(x)=\alpha F(x)\).

Now let \(G(x)=(1+x)^\alpha=E(\alpha\ln(1+x))\) on \((-1,1)\). The chain rule and \((\ln y)'=1/y\) give \[ G'(x)=\frac{\alpha}{1+x}G(x), \] so \((1+x)G'(x)=\alpha G(x)\). Also \(F(0)=c_0=1=G(0)\). Since \(G(x)>0\), the quotient \(H(x)=F(x)/G(x)\) is differentiable on \((-1,1)\). Using the two differential equations,

For any two points in \((-1,1)\), the Mean Value Theorem therefore shows that \(H\) takes the same value at both points. Since \(H(0)=1\), we have \(H(x)=1\) throughout the interval, and hence \(F(x)=G(x)\). If \(\alpha\) is a nonnegative integer, the recurrence terminates the series; the same differential-equation argument applies to its polynomial sum, or equivalently the finite binomial theorem gives the equality for every real \(x\). \(\square\)

Working with the Coefficients

The recurrence is often more convenient than the product definition when calculating a few terms. It also makes the change in signs visible: once \(n\) exceeds a noninteger \(\alpha\), the factor \(\alpha-n\) is negative. The examples below illustrate a terminating expansion, an infinite expansion with a simple closed form, and a noninteger exponent.

Convergence at the Endpoints \(x=\pm1\)

The radius result only guarantees convergence for \(|x|<1\). At \(x=\pm1\) the behavior depends on \(\alpha\). The following facts are standard; the first two require an additional argument (Abel’s theorem for the value of the sum) that is not proved here.

For example, \(\alpha=-1\) gives \(1-1+1-1+\cdots\) at \(x=1\) and \(1+1+1+\cdots\) at \(x=-1\), and both diverge. For \(\alpha=\tfrac12\) the series at \(x=1\) converges to \(\sqrt2\), though slowly. A nonnegative integer exponent gives a polynomial, so there is no restriction at all.

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