Maclaurin Series for ln(1 + x) and Its Use in Evaluating ln 3 (the “ln 1 x 2” case)
Here's the thing about the Maclaurin series is a special case of the Taylor series centered at zero. It provides a polynomial approximation of a function that becomes exact as more terms are added—provided the function is analytic at the origin and the argument lies within the series’ radius of convergence. When students see the notation “ln 1 x 2”, they are often asking how to use this series to find the value of ln 3 (because ln (1 + 2) = ln 3). One of the most frequently encountered examples in calculus courses is the expansion of the natural logarithm ln(1 + x). In this article we derive the Maclaurin series for ln(1 + x), discuss its convergence, show why plugging x = 2 directly does not work, and then illustrate a reliable method to compute ln 3 to any desired accuracy using related series.
1. What Is a Maclaurin Series?
A Maclaurin series expresses a function f(x) as an infinite sum of powers of x:
[ f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!},x^{n}, ]
where f^{(n)}(0) denotes the n‑th derivative of f evaluated at 0.
If the function is infinitely differentiable at 0 and the remainder term tends to zero as n → ∞, the series converges to f(x) for all x inside a certain interval (‑R, R), called the radius of convergence Simple as that..
The Maclaurin series is especially useful because:
- It turns transcendental functions (like eˣ, sin x, cos x, ln (1 + x)) into polynomials that are easy to differentiate, integrate, or evaluate numerically.
- It provides a foundation for approximation techniques used in engineering, physics, and computer science.
2. Deriving the Maclaurin Series for ln(1 + x)
We start with the function
[ f(x)=\ln(1+x),\qquad f(0)=\ln(1)=0. ]
Successive derivatives are:
[ \begin{aligned} f'(x) &= \frac{1}{1+x},\[4pt] f''(x) &= -\frac{1}{(1+x)^{2}},\[4pt] f^{(3)}(x) &= \frac{2}{(1+x)^{3}},\[4pt] f^{(4)}(x) &= -\frac{6}{(1+x)^{4}},\ &\ \vdots\ f^{(n)}(x) &= (-1)^{,n-1}\frac{(n-1)!}{(1+x)^{n}}\quad (n\ge 1). \end{aligned} ]
Evaluating each derivative at x = 0 gives:
[ f^{(n)}(0)=(-1)^{,n-1}(n-1)!. ]
Plugging these into the Maclaurin formula:
[ \begin{aligned} \ln(1+x) &= \sum_{n=1}^{\infty}\frac{f^{(n)}(0)}{n!}{n!Still, },x^{n} \ &= \sum_{n=1}^{\infty}\frac{(-1)^{,n-1}(n-1)! },x^{n} \ &= \sum_{n=1}^{\infty}(-1)^{,n-1}\frac{x^{n}}{n}.
Thus the Maclaurin series for ln(1 + x) is
[ \boxed{\displaystyle \ln(1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\frac{x^{5}}{5}-\cdots} \qquad\text{(valid for }|x|<1\text{).} ]
The alternating‑sign pattern and the simple denominator n make this series easy to remember and to manipulate But it adds up..
3. Radius of Convergence – Why x = 2 Is Problematic
To find where the series converges, we apply the ratio test to the absolute term |xⁿ/n|:
[ \lim_{n\to\infty}\left|\frac{x^{n+1}/(n+1)}{x^{n}/n}\right| =|x|\lim_{n\to\infty}\frac{n}{n+1}=|x|. ]
The series converges when this limit is < 1, i.Worth adding: e. , when |x| < 1.
- For x = 1, the series becomes the alternating harmonic series 1 − ½ + ⅓ − ¼ + …, which converges conditionally to ln 2.
- For x = −1, the series is −1 − ½ − ⅓ − ¼ − …, which diverges to −∞.
Hence the interval of convergence is [−1, 1).
Because |2| > 1, substituting x = 2 directly into the series x − x²/2 + x³/3 − … does not yield a meaningful sum; the partial sums grow without bound. This is a common pitfall: students sometimes try to compute ln 3 by plugging 2 into the naïve series and wonder why the answer diverges It's one of those things that adds up..
4. Getting ln
5. Computing ln 3 When the Naïve Series Fails
Even though the Maclaurin expansion of (\ln(1+x)) is perfectly legitimate for (|x|<1), it does not let us evaluate (\ln 3) simply by inserting (x=2). The reason is straightforward: the power‑series representation is only guaranteed to converge on the interval ((-1,1)). Attempting to extrapolate beyond that region leads to divergent partial sums, as the example of (x=2) demonstrates.
A common remedy is to move the centre of the expansion away from the problematic endpoint while staying close enough to the desired point. One convenient trick is to rewrite the argument in a form that fits the known interval. Observe that
[ 3=\frac{2\cdot 3}{2}=2\Bigl(1+\frac12\Bigr), \qquad\text{so}\qquad \ln 3 =\ln 2 +\ln!\Bigl(1+\frac12\Bigr). ]
Now the second term involves (\ln(1+u)) with (u=\tfrac12). Because (|u|=0.5<1), the original Maclaurin series applies safely:
[ \ln!\Bigl(1+\frac12\Bigr) =\sum_{n=1}^{\infty}\frac{(-1)^{,n-1}}{n},\Bigl(\frac12\Bigr)^{!n} =\frac12-\frac{1}{2\cdot2^{2}}+\frac{1}{3\cdot2^{3}} -\frac{1}{4\cdot2^{4}}+\cdots . ]
Adding the constant (\ln 2) (which itself has a well‑known Maclaurin series, (\ln(1+x)=\ln 2,\big|_{x=1}=1-x+\frac{x^{2}}{2}-\frac{x^{3}}{3}+ \dots) evaluated at (x=1)) yields an accurate numerical estimate once a sufficient number of terms have been summed.
If one prefers to stay entirely within the original variable (x), another strategy is to use a Taylor expansion about a different base point. To give you an idea, expanding (\ln(1+x)) about (x=a\neq0) gives
[ \ln(1+x)=\ln(1+a)+\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k},(x-a)^{k}, \qquad |x-a|<|1+a|. ]
Choosing (a=1) makes the expansion valid for (-2<x<0), but not for (x>0). By selecting a centre that lies between the two singularities of the logarithmic function—namely the
