Why Composition Requires More Than a Valid Inner Series
A Taylor expansion of a composite function is built in layers. An inner function may have a valid power series, but that does not automatically mean an outer series can be applied to it. The inner series must converge, and its value must lie within the convergence range of the outer series.
For instance, the geometric expansion of a rational function can converge at a point where the logarithm series applied to that function does not. With radicals, logarithms, and rational expressions nested together, each layer contributes a condition. A reliable interval is obtained by listing and combining all those conditions.
The interval found this way is often a guaranteed interval for the chosen composition method. It need not be the largest interval on which the final Taylor series converges: algebraic simplification or cancellation can sometimes extend convergence. Endpoint behavior also requires separate checking.
Series Conditions to Track
Write \(t=x-c\) when expanding about the center \(c\). The standard geometric, logarithmic, and binomial expansions used here have the following conditions.
For the interval guarantee in a composition, the strict inequalities shown are convenient: they ensure absolute convergence and make rearranging the resulting sums legitimate. Some of these series also converge at boundary points, but a boundary value must be examined separately.
Why Absolute Convergence Makes Substitution Safe
Suppose an inner series is \(g(t)=\sum_{n=0}^{\infty}b_nt^n\), and an outer series is \(F(y)=\sum_{k=0}^{\infty}a_ky^k\). The outer series converges absolutely when its input has magnitude less than its radius \(S\). To substitute the inner series and regroup terms into powers of \(t\), it is enough to ensure that the absolute values of the inner terms sum to less than \(S\).
The key quantity is the absolute majorant \(B(r)=\sum |b_n|r^n\), not merely the value \(g(r)\). The theorem follows from a direct bound.
For \(|t|\le r\), the sum of the absolute values of its terms is at most \(B(r)\).
Since \(B(r)<S\), the sum \(\sum_k |a_k|B(r)^k\) converges.
The absolute convergence of the full substituted sum permits its terms to be collected by powers of \(t\), without changing its value.
This gives a practical rule: find where each inner series converges, calculate an absolute majorant for its input to the next layer, and require that majorant to fit inside the next series’ convergence range.
Worked Example: A Rational Expression Inside a Logarithm
Worked Example: Finding the Guaranteed Interval for a Logarithmic Composition
Expand \(f(x)=\ln\left(1-\frac{x}{2-x}\right)\) about \(x=0\), and determine the interval obtained from the nested series conditions.
First expand the rational expression geometrically:
The outer expansion is \(\ln(1-u)=-\sum_{k=1}^{\infty}u^k/k\), which requires \(|u|<1\). For substitution into the power series, the absolute majorant of the rational series at \(|x|=r<2\) is
Require this majorant to be less than \(1\):
Thus the nested expansion is justified for \(|x|<1\). To identify the resulting series, simplify the function near the center:
Both logarithms now have valid Taylor expansions for \(|x|<1\). Subtracting them gives
The result checks at the center: both the original function and the series have value \(0\) at \(x=0\). Differentiating the series for \(|x|<1\) gives
So the series represents the stated function on its guaranteed open interval \((-1,1)\). In this example, the series also converges at \(x=-1\): its terms alternate and their magnitudes decrease to zero. At \(x=1\), the function has a logarithmic singularity and the series diverges. The full real interval of convergence is therefore \([-1,1)\).
Worked Example: A Radical Inside a Logarithm
Worked Example: Expanding \(\ln(2+\sqrt{1+x})\)
Expand \(f(x)=\ln(2+\sqrt{1+x})\) about \(x=0\), using the radical as the inner series.
The binomial expansion of the radical is
Set \(v=(\sqrt{1+x}-1)/3\). Then \(2+\sqrt{1+x}=3(1+v)\), so \(f(x)=\ln 3+\ln(1+v)\). For \(r=|x|<1\), the absolute majorant of \(\sqrt{1+x}-1\) is
Consequently, the absolute majorant for \(v\) is \((1-\sqrt{1-r})/3<1\), satisfying the logarithm condition. The composed expansion is therefore justified for \(|x|<1\).
For the first terms,
Substitute into \(\ln(1+v)=v-v^2/2+v^3/3-\cdots\). Through degree three, \(v^2=x^2/36-x^3/72+\cdots\) and \(v^3=x^3/216+\cdots\). Hence
At \(x=0\), this gives \(f(0)=\ln 3\), as required. Also, differentiating the original function gives \(f'(0)=1/(2\cdot1\cdot3)=1/6\), matching the coefficient of \(x\). The conditions above guarantee convergence on \((-1,1)\); they do not, by themselves, decide either endpoint.
Worked Example: Conditions at Three Nested Layers
Worked Example: A Rational Input to a Radical and a Logarithm
Consider
about \(x=0\). First, \(w=x/(4-x)=\sum_{n=1}^{\infty}x^n/4^n\), which converges for \(|x|<4\). Its absolute majorant at \(r=|x|<4\) is \(r/(4-r)\). For the radical expansion in \(w\), require that majorant to be less than \(1\):
For \(0\le t<1\), the absolute majorant of \(\sqrt{1+w}-1\) at \(|w|=t\) is \(1-\sqrt{1-t}<1\). After division by \(2\), the input to the logarithm has majorant less than \(1/2\), so the logarithm condition is satisfied. All three layers are therefore valid for \(|x|<2\).
The first terms illustrate the nested substitution. Since \(w=x/4+x^2/16+x^3/64+\cdots\),
Applying \(\ln(1+q)=q-q^2/2+q^3/3-\cdots\) to this input \(q\), with \(q^2=x^2/256+3x^3/2048+\cdots\) and \(q^3=x^3/4096+\cdots\), gives
The constant term is \(F(0)=\ln(1+0)=0\). Differentiating the original expression at zero gives \(F'(0)=1/16\), agreeing with its linear coefficient. The interval \((-2,2)\) is guaranteed by the nested conditions; it should not be mistaken for a proof that the final series cannot converge farther.
A Common Mistake: Checking Only the Rational Series
For \(g(x)=\ln\left(1+\frac{x}{2-x}\right)\), the rational series \(x/(2-x)=\sum_{n=1}^{\infty}x^n/2^n\) converges for \(|x|<2\). It is tempting to conclude that the logarithmic substitution is valid throughout that interval. At \(x=3/2\), however, the rational series sums to
The outer logarithm series \(\ln(1+z)=\sum_{k=1}^{\infty}(-1)^{k+1}z^k/k\) cannot be used at \(z=3\): its terms do not tend to zero. In fact, the function itself is defined there, since \(1+3=4\), but this particular logarithmic expansion is not valid there.
The correction is to check the outer condition as well as the inner one. For a justified power-series substitution, the rational series has majorant \(r/(2-r)\), and requiring \(r/(2-r)<1\) yields \(r<1\). The function may have a Taylor series with a larger interval by a different argument, but the nested expansion alone guarantees only the interval its conditions establish.
Check Your Understanding
For each question, identify the relevant layer conditions before deciding what interval is guaranteed.
- For \(\ln(1+x/(3-x))\) about zero, what is the convergence radius of the geometric series for the rational input?
- For the same composition, what absolute-majorant condition must hold for the logarithm series to be safely substituted?
- Why is \(|g(t)|<S\) not always enough to justify regrouping the terms after substituting a series for \(g\)?
- In a radical-logarithm composition, which convergence condition should be checked first, and which should be checked after finding the radical’s range or majorant?
- Why can a guaranteed interval obtained from nested conditions be smaller than the full interval of convergence of the resulting Taylor series?
