Logarithm Series Under Scaling and Shifting

Why Scaling and Shifting Matter

A known power series becomes useful in many settings once its input is changed. Replacing the variable by a multiple of itself gives a series for a scaled logarithm; factoring a constant out of a logarithm gives a series for a shifted logarithm. Logarithm identities then turn ratios into differences of series.

These transformations are simple algebraically, but their conditions matter. A logarithm must have a positive real argument, and a substituted series is valid only where its input lies in the original convergence interval. In particular, the endpoints of that interval do not behave symmetrically.

The Logarithm Series to Be Transformed

From the geometric-series derivation of the logarithm series, for real $u$ we have

\ln(1+u)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}u^n}{n}, \qquad -1<u\leq 1.

The lower endpoint $u=-1$ is excluded: the logarithm is undefined there, and the series becomes the divergent harmonic series with negative terms. The upper endpoint $u=1$ is included: the alternating series converges to $\ln 2$. For $-1<u<1$, the series converges absolutely. At $u=1$, it converges conditionally.

Key rule: After substituting an expression for $u$, retain the full condition $-1<u\leq 1$. Do not replace it automatically with a symmetric condition such as $|u|\leq 1$, because the two endpoints have different behavior.

Scaling the Input: $\ln(1+ax)$

Let $a$ be a real constant. If $a\ne 0$, substitute $u=ax$ into the logarithm series. This gives

\ln(1+ax) =\sum_{n=1}^{\infty}\frac{(-1)^{n+1}a^n x^n}{n}, \qquad -1<ax\leq 1.

The condition also ensures that the logarithm is real, since it implies $1+ax>0$. If $a>0$, it is equivalent to $-1/a<x\leq 1/a$. If $a<0$, dividing reverses the inequalities, so the condition is $1/a\leq x<-1/a$. Keeping $-1<ax\leq 1$ avoids mistakes when $a$ is negative. If $a=0$, the function is identically $\ln 1=0$, and its series is simply zero for every real $x$.

Worked Example: Expanding $\ln(1+3x)$

Here $a=3$, so the required condition is $-1<3x\leq 1$. Dividing by $3$ gives $-1/3<x\leq 1/3$. Substitution into the series gives

\begin{aligned} \ln(1+3x) &=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}(3x)^n}{n}\\ &=3x-\frac{9}{2}x^2+9x^3-\frac{81}{4}x^4+\cdots, \qquad -\frac13<x\leq\frac13. \end{aligned}

For example, the first four coefficients come from $3^1/1=3$, $-3^2/2=-9/2$, $3^3/3=9$, and $-3^4/4=-81/4$. For an interior check, differentiate the series term by term, which is valid for $|3x|<1$:

\begin{aligned} \frac{d}{dx}\left(3x-\frac92x^2+9x^3-\frac{81}{4}x^4+\cdots\right) &=3-9x+27x^2-81x^3+\cdots\\ &=3\left(1-3x+9x^2-27x^3+\cdots\right)\\ &=\frac{3}{1+3x}. \end{aligned}

This is the derivative of $\ln(1+3x)$, and both the series and the logarithm equal zero at $x=0$. At $x=1/3$, the series converges to $\ln 2$. At $x=-1/3$, its terms are $-1/n$, so it diverges; the logarithm is also undefined there.

Shifting the Input: $\ln(c+x)$

To obtain a power series centered at $x=0$ for $\ln(c+x)$, the constant $c$ must be positive. Factor it from the argument:

\ln(c+x)=\ln c+\ln\left(1+\frac{x}{c}\right),\qquad c>0.

Applying the logarithm series with $u=x/c$ gives

\ln(c+x)=\ln c+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\left(\frac{x}{c}\right)^n, \qquad -c<x\leq c,\quad c>0.

The original logarithm is defined for $x>-c$. The series condition additionally imposes $x\leq c$, with the right endpoint included. The resulting interval is therefore exactly $-c<x\leq c$. If $c\leq 0$, then $\ln(c+x)$ is not defined at $x=0$, so this real power-series expansion about zero does not apply.

Worked Example: Expanding $\ln(5+x)$

Factor out $5$, which is positive:

\begin{aligned} \ln(5+x) &=\ln 5+\ln\left(1+\frac{x}{5}\right)\\ &=\ln 5+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n5^n}\\ &=\ln 5+\frac{x}{5}-\frac{x^2}{50}+\frac{x^3}{375}-\frac{x^4}{2500}+\cdots, \qquad -5<x\leq 5. \end{aligned}

The displayed coefficients follow from $1/5$, $-1/(2\cdot25)=-1/50$, $1/(3\cdot125)=1/375$, and $-1/(4\cdot625)=-1/2500$. For $-5<x<5$, differentiating term by term gives

\begin{aligned} \frac{d}{dx}\left(\ln 5+\frac{x}{5}-\frac{x^2}{50}+\frac{x^3}{375}-\cdots\right) &=\frac15-\frac{x}{25}+\frac{x^2}{125}-\frac{x^3}{625}+\cdots\\ &=\frac15\left(1-\frac{x}{5}+\frac{x^2}{25}-\frac{x^3}{125}+\cdots\right)\\ &=\frac15\cdot\frac{1}{1+x/5} =\frac{1}{5+x}. \end{aligned}

This agrees with the derivative of $\ln(5+x)$, and setting $x=0$ gives $\ln 5$ on both sides. The endpoint $x=5$ is included by the alternating logarithm series. The endpoint $x=-5$ is excluded, where the logarithm is undefined and the substituted series has terms proportional to $-1/n$.

Logarithmic Ratios

For positive real numbers $A$ and $B$, the identity $\ln(A/B)=\ln A-\ln B$ allows a ratio to be expanded by subtracting two logarithm series. Each series must satisfy its own convergence condition. Consequently, the formula obtained by this method is guaranteed on the intersection of the two valid intervals.

Theorem: If $a,b$ are real constants and $-1<ax\leq1$ and $-1<bx\leq1$, then $$ \ln\left(\frac{1+ax}{1+bx}\right) =\sum_{n=1}^{\infty}\frac{(-1)^{n+1}(a^n-b^n)}{n}x^n. $$ The conditions ensure both $1+ax$ and $1+bx$ are positive, and hence the logarithm identity applies.

To justify the coefficient formula, write the numerator and denominator logarithms separately, then subtract corresponding terms. For each $n$, their difference is

\frac{(-1)^{n+1}a^n x^n}{n} -\frac{(-1)^{n+1}b^n x^n}{n} =\frac{(-1)^{n+1}(a^n-b^n)}{n}x^n.

Inside the common open interval of convergence, both series converge absolutely, so subtracting them term by term is valid. At a permitted endpoint, the original logarithm series converge as well; subtracting their partial sums and taking limits yields the same identity.

Worked Example: Expanding $\ln\left(\frac{1+2x}{1-x}\right)$

Use the identity $\ln(A/B)=\ln A-\ln B$. The first logarithm has $u=2x$, requiring $-1<2x\leq1$, or $-1/2<x\leq1/2$. For the second, write $\ln(1-x)=\ln(1+u)$ with $u=-x$; its condition is $-1<-x\leq1$, or $-1\leq x<1$. The intersection is $-1/2<x\leq1/2$. Thus

\begin{aligned} \ln\left(\frac{1+2x}{1-x}\right) &=\ln(1+2x)-\ln(1-x)\\ &=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}(2^n-(-1)^n)}{n}x^n\\ &=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}2^n+1}{n}x^n\\ &=3x-\frac32x^2+3x^3-\frac{15}{4}x^4+\cdots. \end{aligned}

For the first four coefficients, the general numerator and denominator give $(2+1)/1=3$, $(-4+1)/2=-3/2$, $(8+1)/3=3$, and $(-16+1)/4=-15/4$. To check the series in the interior, differentiate its series form, where $|2x|<1$:

\begin{aligned} \frac{d}{dx}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}2^n+1}{n}x^n &=\sum_{n=1}^{\infty}\bigl((-1)^{n+1}2^n+1\bigr)x^{n-1}\\ &=\frac{2}{1+2x}+\frac{1}{1-x}\\ &=\frac{3}{(1+2x)(1-x)}. \end{aligned}

Direct differentiation of the logarithm gives $2/(1+2x)+1/(1-x)$, the same expression. Both sides are zero at $x=0$. At $x=1/2$, the first logarithm series reaches its included endpoint and the second remains inside its interval, so the representation holds. At $x=-1/2$, the first logarithm has zero argument and its series diverges, so that endpoint is excluded.

Shifting a Ratio and Checking the Conditions

The same reasoning handles ratios of shifted expressions. If $c>0$ and $d>0$, factor each constant:

\ln\left(\frac{c+x}{d+x}\right) =\ln\left(\frac{c}{d}\right) +\ln\left(1+\frac{x}{c}\right) -\ln\left(1+\frac{x}{d}\right).

Using the logarithm series gives

\ln\left(\frac{c+x}{d+x}\right) =\ln\left(\frac{c}{d}\right) +\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} \left(\frac{1}{c^n}-\frac{1}{d^n}\right)x^n, \qquad -c<x\leq c,\quad -d<x\leq d.

Both inequalities are required. Their intersection ensures that each logarithm series is valid and that $c+x$ and $d+x$ are positive. This method gives a representation on the intersection of the two intervals; it does not claim that no other expansion or larger interval can be obtained by a different argument.

Common Mistakes and Why the Conditions Matter

A frequent error is to treat logarithm-series endpoints as if both were included. For instance, the rule requires $-1<u\leq1$, not $-1\leq u\leq1$. At $u=-1$, $\ln(1+u)$ is not a real number, and the series does not converge. Another error is to ignore the sign of a scaling constant when converting the condition into an interval for $x$. Retaining the condition in terms of $ax$ until the final step is a reliable safeguard.

Coefficient arithmetic also deserves a direct check. In the ratio example, the cubic coefficient is obtained from $((-1)^4 2^3+1)/3$. Since $2^3=8$, this is $(8+1)/3=3$, not $17/3$. The correct beginning of the expansion is therefore

3x-\frac32x^2+3x^3-\frac{15}{4}x^4+\cdots.

A useful final check is to differentiate the series inside its radius of convergence and compare with the derivative of the original logarithmic expression. This catches many sign and coefficient errors, but it does not replace checking the real domain or the endpoints.

Check Your Understanding

For each question, state the relevant condition as well as the series or conclusion.

Find the power series for $\ln(1-4x)$, and give its full interval of validity.
Write the first four nonconstant terms of the series for $\ln(7+x)$, and state which endpoint is included.
For $\ln\left(\frac{1+x}{1+3x}\right)$, identify the common interval on which subtracting the two logarithm series is justified.
Why does the series for $\ln(1+u)$ include $u=1$ but exclude $u=-1$?
Differentiate the first three terms of your answer to Question 1 and compare them with the derivative of $\ln(1-4x)$.

Logarithm Series Under Scaling and Shifting

What You'll Learn

Why Scaling and Shifting Matter

The Logarithm Series to Be Transformed

Scaling the Input: \(\ln(1+ax)\)

Worked Example: Expanding \(\ln(1+3x)\)

Shifting the Input: \(\ln(c+x)\)

Worked Example: Expanding \(\ln(5+x)\)

Logarithmic Ratios

Worked Example: Expanding \(\ln\left(\frac{1+2x}{1-x}\right)\)

Shifting a Ratio and Checking the Conditions

Common Mistakes and Why the Conditions Matter

Check Your Understanding

Logarithm Series Under Scaling and Shifting

What You'll Learn

Why Scaling and Shifting Matter

The Logarithm Series to Be Transformed

Scaling the Input: \(\ln(1+ax)\)

Worked Example: Expanding \(\ln(1+3x)\)

Shifting the Input: \(\ln(c+x)\)

Worked Example: Expanding \(\ln(5+x)\)

Logarithmic Ratios

Worked Example: Expanding \(\ln\left(\frac{1+2x}{1-x}\right)\)

Shifting a Ratio and Checking the Conditions

Common Mistakes and Why the Conditions Matter

Check Your Understanding

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