Taylor Series For Ln X 1

Taylor Series for ln x around x = 1 (Equivalently the Series for ln (1 + x))

The Taylor series provides a powerful way to approximate transcendental functions using polynomials. One of the most frequently encountered series in calculus is the expansion of the natural logarithm about the point x = 1, which can also be written as the series for ln (1 + x). Understanding this expansion not only strengthens analytical skills but also opens doors to numerical methods, signal processing, and solving differential equations. Below is a detailed, step‑by‑step treatment of the Taylor series for ln x centered at x = 1, including its derivation, interval of convergence, practical examples, and common pitfalls.

1. Introduction

The natural logarithm, ln x, is undefined at x ≤ 0 and grows slowly for large x. On top of that, direct evaluation of ln x for arbitrary x can be computationally expensive, especially in early computing environments or embedded systems. By expressing ln x as an infinite sum of simple polynomial terms—its Taylor series—we can approximate the function to any desired accuracy using only addition, subtraction, multiplication, and division.

When we expand ln x about x = 1, we set u = x − 1, so that x = 1 + u and the function becomes ln (1 + u). The resulting series is:

[ \ln(1+u)=u-\frac{u^{2}}{2}+\frac{u^{3}}{3}-\frac{u^{4}}{4}+\cdots =\sum_{n=1}^{\infty}(-1)^{n+1}\frac{u^{n}}{n}, \qquad |u|<1. ]

Replacing u by x − 1 gives the Taylor series for ln x around x = 1:

[ \boxed{\displaystyle \ln x = (x-1)-\frac{(x-1)^{2}}{2}+\frac{(x-1)^{3}}{3} -\frac{(x-1)^{4}}{4}+\cdots =\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(x-1)^{n}}{n}} ]

The series converges for 0 < x ≤ 2 (i.e.Day to day, , |x − 1| < 1) and also at the endpoint x = 2 (alternating harmonic series). At x = 0 the series diverges, reflecting the logarithm’s vertical asymptote.

2. Derivation of the Series

2.1 Starting from the Geometric Series

The derivation begins with the well‑known geometric series valid for |t| < 1:

[ \frac{1}{1+t}=1-t+t^{2}-t^{3}+t^{4}-\cdots =\sum_{n=0}^{\infty}(-1)^{n}t^{n}. ]

2.2 Integrating Term‑by‑Term

Recall that the derivative of ln (1 + t) is 1/(1 + t). Integrating both sides of the geometric series from 0 to u (with |u| < 1) yields:

[ \int_{0}^{u}\frac{1}{1+t},dt =\int_{0}^{u}\Bigl(1-t+t^{2}-t^{3}+\cdots\Bigr)dt. ]

The left‑hand side evaluates to ln (1 + u) − ln (1) = ln (1 + u). Performing the integration term‑by‑term on the right gives:

[ \ln(1+u)=\Bigl[t-\frac{t^{2}}{2}+\frac{t^{3}}{3} -\frac{t^{4}}{4}+\cdots\Bigr]_{0}^{u} =u-\frac{u^{2}}{2}+\frac{u^{3}}{3} -\frac{u^{4}}{4}+\cdots . ]

Since the constant term vanishes (ln 1 = 0), we obtain the series shown earlier.

2.3 Shifting to ln x

Setting u = x − 1 produces the expansion about x = 1:

[ \ln x = (x-1)-\frac{(x-1)^{2}}{2} +\frac{(x-1)^{3}}{3} -\frac{(x-1)^{4}}{4}+\cdots . ]

Each term involves a power of (x − 1) divided by its exponent, with alternating signs.

3. Radius and Interval of Convergence ### 3.1 Applying the Ratio Test

For the general term aₙ = (−1)^{n+1}(x − 1)^{n}/n, the ratio test gives:

[ \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right| =\lim_{n\to\infty}\frac{|x-1|^{n+1}}{n+1}\cdot\frac{n}{|x-1|^{n}} =|x-1|;\lim_{n\to\infty}\frac{n}{n+1}=|x-1|. ]

The series converges when this limit is < 1, i.Plus, e. , |x − 1| < 1. Hence the radius of convergence is R = 1.

3.2 Endpoint Behavior

• At x = 0 (u = −1): the series becomes −1 − ½ − ⅓ − … , the negative harmonic series, which diverges to −∞. This matches the fact that ln 0 is undefined (→ −∞).
• At x = 2 (u = 1): the series is the alternating harmonic series 1 − ½ + ⅓ − ¼ + …, which converges conditionally to ln 2.

Thus the interval of convergence is (0, 2] — open at 0, closed at 2.

4. Practical Approximation Examples ### 4.1 Approximating ln 1.2

Take x = 1.That's why 2 → u = 0. 2.

[ \ln(1.2)\approx 0.2-\frac{0.2^{2}}{2} +\frac{0.2^{3}}{3} -\frac{0.2^{4}}{4} =0.2-0.0

4. Practical Approximation Examples

4.1 Approximating ln(1.2)

Take x = 1.2 → u = 0.2.

[ \ln(1.2)\approx 0.2-\frac{0.2^{2}}{2} +\frac{0.2^{3}}{3} -\frac{0.2^{4}}{4} =0.2-0.02-0.004-0.0004 =0.1756 ]

This approximation is reasonably accurate, as ln(1.In practice, 2) ≈ 0. Still, 1823. We can see that the series converges quickly, and the terms become increasingly small.

4.2 Approximating ln(1.9)

Let's approximate ln(1.9) with x = 1.9 → u = 0.1.

[ \ln(1.1-0.9)\approx 0.1^{4}}{4} +\frac{0.Because of that, 000004 -0. 1^{3}}{3} -\frac{0.1^{5}}{5} =0.0001\overline{6} +0.1-\frac{0.Consider this: 005-0. 1^{2}}{2} +\frac{0.00000008 =0.

This approximation is closer to the true value of ln(1.6931. Plus, 9) ≈ 0. This leads to while not as accurate as the previous example, it demonstrates the power of the series for approximating logarithms. Notice that the error decreases with each term.

4.3 Using the Series for ln(x) for a Wider Range of x

The series provides a handy tool for approximating the natural logarithm of any x within the interval (0, 2]. So by carefully choosing the number of terms, we can achieve a desired level of accuracy. To give you an idea, for x = 1.That said, 5, we would need to calculate a larger number of terms to achieve a more precise approximation compared to x = 1. Because of that, 2. The accuracy improves as more terms are included, demonstrating the convergence and the decreasing error of the series. This application highlights the utility of the logarithmic series in various fields, including numerical analysis and scientific computing.

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5. Conclusion

The logarithmic series, derived from the geometric series and integrated term-by-term, provides a powerful and efficient method for approximating the natural logarithm of values within the interval (0, 2]. Now, the radius of convergence is 1, and the interval of convergence is (0, 2]. Also, the series exhibits remarkable convergence properties, allowing for accurate approximations with a relatively small number of terms. Practical examples demonstrate its effectiveness in approximating logarithms, offering a valuable tool for numerical calculations and scientific applications. This underscores the importance of understanding and utilizing mathematical series in solving real-world problems.

5.1 Error Bounds and Convergence Speed

When the series

[ \ln(1+u)=\sum_{n=1}^{\infty}(-1)^{,n+1}\frac{u^{n}}{n},\qquad |u|\le 1,;u\neq-1, ]

is truncated after (N) terms, the remainder (the truncation error) can be bounded by the first omitted term, because the series is alternating with monotonically decreasing absolute terms. Hence

[ \bigl|R_{N}(u)\bigr| =\Bigl|\ln(1+u)-\sum_{n=1}^{N}(-1)^{,n+1}\frac{u^{n}}{n}\Bigr| \le \frac{|u|^{,N+1}}{N+1}. ]

For (u=0.Consider this: 2) the bound after four terms is (\frac{0. So 2^{5}}{5}=6. On the flip side, 4\times10^{-5}), which explains why the approximation (0. 1756) is within (7\times10^{-3}) of the true value (\ln(1.Because of that, 2)=0. 182321). Here's the thing — similarly, for (u=0. That said, 1) the bound after five terms is (\frac{0. In practice, 1^{6}}{6}=1. Also, 7\times10^{-7}), justifying the quoted error of order (10^{-4}) for (\ln(1. 9)).

The error therefore shrinks geometrically with the exponent of (|u|). Worth adding: when (|u|) is close to 1 (e. g.Think about it: , (u=0. Because of that, 9) corresponding to (x=1. Think about it: 9)), many more terms are required to achieve the same relative accuracy, which is reflected in the slower convergence observed in the example of (\ln(1. 9)).

5.2 Transformations to Extend the Range

The basic series converges only for (0<u\le 1) (i.Think about it: e. , (0<x\le 2)).

[ \ln(x)=\ln!\bigl(\tfrac{x}{2^{k}}\bigr)+k\ln 2, ]

where an appropriate integer (k) brings the reduced argument into the interval ((0,2]). As an example, to compute (\ln(5)) we may write

[ \ln(5)=\ln!\bigl(\tfrac{5}{4}\bigr)+\ln 4 =\ln(1.25)+2\ln 2, ]

and then apply the series to (u=0.25). This technique—sometimes called argument reduction—preserves the rapid convergence of the series while allowing the logarithm of essentially any positive real number to be obtained.

5.3 Numerical Illustration: (\ln(7)) via Reduction

Let (x=7). Choose (k=2) so that (\frac{7}{2^{2}}=\frac{7}{4}=1.75). Then

[ \ln(7)=2\ln 2+\ln(1.75),\qquad u=0.75. ]

Using the first eight terms of the series for (u=0.75) we obtain

[ \begin{aligned} \ln(1.75)&\approx 0.047416-0.Now, 75^{3}}{3} -\frac{0. Even so, 75^{2}}{2} +\frac{0. In real terms, 75^{6}}{6} +\frac{0. So 0292969+0. 75^{4}}{4} +\frac{0.Which means 75^{7}}{7} -\frac{0. On the flip side, 75-\frac{0. 75^{8}}{8} \ &=0.140625-0.Practically speaking, 0791016+0. 28125+0.75-0.0183105-0.75^{5}}{5} -\frac{0.Plus, 0113906 \ &=0. 575512.

Adding (2\ln 2\approx1.386294) yields

[ \ln(7)\approx 1.961806, ]

which differs from the exact value (\ln 7\approx1.6\times10^{-2}). 945910) by less than (1.Including a few more terms would push the error below (10^{-4}).

Beyond simple argument reduction, the convergence of the alternating series can be further accelerated by exploiting the symmetry of the logarithm. Writing

[ \ln x = 2,\operatorname{artanh}!\left(\frac{x-1}{x+1}\right) = 2\sum_{n=0}^{\infty}\frac{1}{2n+1}\left(\frac{x-1}{x+1}\right)^{2n+1}, ]

the series now involves only odd powers of the reduced variable

[ y=\frac{x-1}{x+1}\in(-1,1), ]

and its absolute value is always smaller than (|u|) for the same (x). So naturally, fewer terms are needed to reach a prescribed tolerance. Take this: to evaluate (\ln(7)) we have

[ y=\frac{7-1}{7+1}=\frac{6}{8}=0.75, ]

and the same eight‑term truncation of the artanh series gives

[ 2\Bigl(y-\frac{y^{3}}{3}+\frac{y^{5}}{5}-\frac{y^{7}}{7} +\frac{y^{9}}{9}-\frac{y^{11}}{11}+\frac{y^{13}}{13} -\frac{y^{15}}{15}\Bigr)=1.945910;, ]

which matches the true value to better than (10^{-6}) without any extra terms.

When very high precision is required, binary splitting or the Brent‑McGregor algorithm can be employed to evaluate the partial sums of either the original or the artanh series in quasi‑linear time. These methods compute the numerator and denominator of the truncated series recursively, avoiding the loss of significance that occurs when adding many floating‑point terms of varying magnitude.

Finally, for applications where the logarithm must be computed millions of times (e.g.Even so, , in probabilistic simulations or graphics shading), a hybrid approach is common: a small lookup table provides (\ln(2^{k})) for integer (k), argument reduction brings the mantissa into a narrow interval such as ([0. 5,1)) or ([1,2)), and a low‑order polynomial or rational approximation (often derived from a Padé approximant of the artanh series) supplies the final correction. This combination yields both the theoretical guarantee of the alternating‑series error bound and the practical speed of hardware‑friendly arithmetic.

To keep it short, the Mercator series (\displaystyle\ln(1+u)=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{u^{n}}{n}) provides a clear, elementary way to understand the convergence behavior of the logarithm near 1. Also, by pairing it with simple argument‑reduction identities, the artanh transformation, and modern summation techniques, one can obtain (\ln x) for any positive (x) with rigorously controlled error and excellent computational efficiency. These tools together form the backbone of many mathematical libraries and continue to be relevant in both theoretical analysis and practical numerical work.