Substituting Into a Binomial Series
A familiar power series can produce expansions for functions that are not themselves written in the series variable. For example, the generalized binomial series for \((1+u)^\alpha\) can be used with \(u=x^2\), \(u=-x^3\), or another expression in \(x\). The main task is to make the substitution correctly and then determine which values of \(x\) satisfy the convergence conditions.
The substitution changes the argument of the series, not its coefficients. It also changes the way the convergence interval translates to \(x\). In particular, the radius of convergence settles the interior; any boundary values still require their own tests.
The formula applies on the open interval \(-1<u<1\). It does not, by itself, determine whether the series converges at \(u=-1\) or \(u=1\). At those endpoints, substitute the endpoint into the terms and test the resulting series.
Why Substitution Gives the Expansion
Suppose \(g(x)\) is a real-valued expression and \(|g(x)|<1\). The binomial identity can be evaluated at \(u=g(x)\), giving \(\sum_{n=0}^{\infty}\binom{\alpha}{n}g(x)^n=(1+g(x))^\alpha\). This is a direct substitution into a convergent series identity. The convergence requirement is on the whole expression \(g(x)\), not on \(x\) alone.
For a polynomial \(g(x)=cx^k\), the condition \(|g(x)|<1\) becomes \(|c||x|^k<1\). This determines the interior convergence region. The actual series has powers \(x^{kn}\), but those powers do not change the binomial coefficients: the \(n\)-th term is \(\binom{\alpha}{n}c^n x^{kn}\).
Worked Example: Expanding a Square Root With \(x^2\)
Worked Example: The Maclaurin Series for \(\sqrt{1+x^2}\)
Take \(\alpha=\frac12\) and substitute \(u=x^2\) into the binomial series. The condition \(|x^2|<1\) is equivalent to \(|x|<1\).
The displayed coefficients follow from the definition: \(\binom{1/2}{1}=\frac12\), \(\binom{1/2}{2}=\frac{(1/2)(-1/2)}{2!}=-\frac18\), \(\binom{1/2}{3}=\frac{(1/2)(-1/2)(-3/2)}{3!}=\frac1{16}\), and \(\binom{1/2}{4}=-\frac5{128}\).
For a numerical check inside the interval, set \(x=\frac12\). Then \(x^2=\frac14\), and the function value is \(\sqrt{1+\frac14}=\frac{\sqrt5}{2}\). The series becomes \(1+\frac18-\frac1{128}+\frac1{1024}-\frac5{32768}+\cdots\), the binomial series at \(u=\frac14\), whose sum is \(\sqrt{1+\frac14}=\frac{\sqrt5}{2}\). Thus the substitution reproduces the function value.
At \(x=1\) or \(x=-1\), the series is \(\sum_{n=0}^{\infty}\binom{1/2}{n}\), since \(x^{2n}=1\) at both endpoints. For \(n\geq1\), \(\left|\binom{1/2}{n}\right|=\frac{1}{2n-1}\frac{\binom{2n}{n}}{4^n}\). The central binomial coefficient estimate \(\frac{\binom{2n}{n}}{4^n}\sim\frac{1}{\sqrt{\pi n}}\) therefore gives \(\left|\binom{1/2}{n}\right|\sim\frac{1}{2\sqrt{\pi}n^{3/2}}\). The absolute values have a convergent sum, so both endpoints converge absolutely. The sum at each endpoint is \(\sqrt2\), as also follows by continuity of the binomial function and the endpoint convergence theorem for power series.
The exact convergence set is therefore \([-1,1]\). The endpoint check matters: the strict inequality \(|x|<1\) obtained from the interior condition alone would not settle \(x=\pm1\).
Worked Example: A Reciprocal Square Root With \(x^3\)
Worked Example: The Series for \((1-x^3)^{-1/2}\)
Set \(\alpha=-\frac12\) and \(u=-x^3\). Since \(\binom{-1/2}{n}=(-1)^n b_n\), where \(b_n=\frac{\binom{2n}{n}}{4^n}>0\), the two factors of \((-1)^n\) cancel:
For example, \(b_1=\frac12\), \(b_2=\frac38\), and \(b_3=\frac5{16}\). The interior condition \(|-x^3|<1\) gives \(|x|<1\).
At \(x=1\), the terms are \(b_n\), all positive. Since \(b_n\sim1/\sqrt{\pi n}\), the series diverges by comparison with the divergent \(p\)-series \(\sum n^{-1/2}\). At \(x=-1\), the terms are \((-1)^n b_n\). The ratio \(b_{n+1}/b_n=(n+\frac12)/(n+1)\) is less than \(1\), and \(b_n\to0\), so the alternating-series test gives convergence. It is conditional because \(\sum b_n\) diverges.
The sum at \(x=-1\) is \(1/\sqrt2\). Indeed, the binomial series at \(u=1\) converges by the alternating-series test, and its sum agrees with the limit of \((1+u)^{-1/2}\) as \(u\) approaches \(1\) from below (Abel’s theorem). Here \(u=-x^3=1\), and the function itself is \((1-(-1)^3)^{-1/2}=1/\sqrt2\).
Thus the series converges for \(-1\leq x<1\): it converges conditionally at \(x=-1\), but diverges at \(x=1\). At \(x=1\), the function is undefined over the reals, and its limit from below is \(+\infty\).
As an interior check, take \(x=\frac12\). Then \(u=-\frac18\), and \((1-\frac18)^{-1/2}=\sqrt{\frac87}=\frac{2\sqrt{14}}{7}\). The series gives \(1+\frac12(\frac18)+\frac38(\frac1{64})+\frac5{16}(\frac1{512})+\cdots =1+\frac1{16}+\frac3{512}+\frac5{8192}+\cdots\). This is the binomial series at \(u=-\frac18\), so its sum is exactly \(\frac{2\sqrt{14}}{7}\).
Worked Example: Scaling a Reciprocal Square Root
Worked Example: The Series for \((1+4x^2)^{-1/2}\)
Use \(\alpha=-\frac12\) and \(u=4x^2\). The binomial series gives \(\binom{-1/2}{n}=(-1)^n b_n\), so
For the interior, \(|4x^2|<1\), so \(|x|<\frac12\). At either endpoint \(x=\pm\frac12\), \(4x^2=1\), and the terms are \((-1)^n b_n\). As \(b_n\) decreases to zero, the alternating-series test gives convergence; it is not absolute because \(\sum b_n\) diverges. At each endpoint the sum is \(2^{-1/2}\), agreeing with \((1+4(\frac12)^2)^{-1/2}=1/\sqrt2\). The convergence set is \([-\frac12,\frac12]\).
For an interior check, at \(x=\frac14\) we have \(4x^2=\frac14\). The function value is \((1+\frac14)^{-1/2}=\frac2{\sqrt5}\), and the series becomes \(1-\frac18+\frac3{128}-\frac5{1024}+\frac{35}{32768}-\cdots\). This is the binomial series evaluated at \(u=\frac14\), so its sum is \(\frac2{\sqrt5}\), as required.
Endpoint Convergence Needs Its Own Argument
A radius of convergence identifies where a power series converges absolutely in its interior and diverges outside its radius. It does not automatically classify the boundary. After substitution, different boundary values of \(x\) can even send the binomial argument to opposite endpoints. The signs of the resulting terms must be calculated from the full product \(\binom{\alpha}{n}g(x)^n\).
The reciprocal-square-root example shows why this detail is consequential. With \(g(x)=-x^3\), the terms are \(b_nx^{3n}\): at \(x=1\), they are positive \(b_n\) and the series diverges; at \(x=-1\), they alternate and converge conditionally. Interchanging those endpoint signs reverses the correct convergence set.
The distinction between the function’s domain and the series’ convergence set also matters. At \(x=1\), \((1-x^3)^{-1/2}\) has no finite real value, and the associated series diverges. In contrast, at \(x=-1\) both the function and the conditionally convergent series have value \(1/\sqrt2\). A power series must be tested at the endpoint; it is not enough to infer its behavior from a nearby function value.
A Reliable Substitution Procedure
Write the target as \((1+g(x))^\alpha\), and record both the exponent \(\alpha\) and the substituted expression \(g(x)\).
Use \(\sum_{n=0}^{\infty}\binom{\alpha}{n}g(x)^n\), keeping the signs and powers in \(g(x)^n\) intact.
Solve \(|g(x)|<1\) to obtain the open interval on which the binomial identity applies directly.
Substitute each endpoint into the terms, then use an appropriate convergence test. Do not assume both endpoints behave alike.
Inside the interval, the binomial identity gives equality. At a convergent endpoint, use the endpoint sum or an endpoint convergence theorem before claiming equality.
Why These Expansions Are Useful
Radical composites often resist direct manipulation as infinite series, but a substitution can place them within a known binomial expansion. Once the series is in powers of \(g(x)\), its coefficients are available from the generalized binomial formula, and its convergence can be analyzed through the size and endpoint behavior of \(g(x)\).
The method also makes clear why two similar-looking expressions can have different convergence sets. The exponent determines the binomial coefficients, while the substituted expression determines both the powers of \(x\) and which boundary values are reached. Checking these features separately is a reliable way to avoid sign errors and unjustified endpoint conclusions.
Check Your Understanding
For each question, keep the substituted binomial argument visible so that its sign is clear.
- Write the first four nonzero terms of the binomial expansion of \((1+x^4)^{1/2}\).
- For \((1+x^2)^{1/2}\), what series results at \(x=-1\), and does it converge absolutely?
- For \((1-x^3)^{-1/2}\), write the \(n\)-th term at \(x=-1\). Is the resulting series absolute or conditional?
- Find the interior convergence interval for \((1+9x^2)^{-1/2}\), then state which endpoint test is needed.
- Why does the condition \(|g(x)|<1\) not, by itself, settle convergence at values where \(|g(x)|=1\)?