Exponential and Logarithmic Series

In higher mathematics, many functions cannot be expressed in finite algebraic terms but can be represented using infinite series expansions. Two of the most important are:

• Exponential Series – Expansion of in powers of x.
• Logarithmic Series – Expansion of ln⁡(1+x) in powers of x.

1.0Exponential Series

Definition

The exponential function is defined as: $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots ,\forall x\in \mathbb{R}$

General Expansion Formula

$e^x={\sum }_{n=0}^{\infty }\frac{x^n}{n!}$

Properties of Exponential Series

1. Convergence: The series converges for all absolute values of x.
2. Differentiability: Differentiating or integrating the series term by term gives back the exponential function. $\frac{d}{dx}(e^x)=e^x,\int e^{x}dx=e^x+C$
3. Symmetry:

• $e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\cdots$
• Hence, even terms remain positive, odd terms alternate.

4. Euler’s Identity: Using complex numbers: $e^{ix}=cosx+isinx$

Standard Results

• $e^0=1$
• ${\lim }_{n\to \infty }{\left(1+\frac{1}{n}\right)}^n=e$
• $e^x\approx 1+x\text{for small }x$

2.0Solved Examples of Exponential Series

Example 1: Approximate $e^{0.1}$ up to 3 decimal places using series.

Sol: $e^{0.1}=1+0.1+\frac{(0.1{)}^2}{2!}+\frac{(0.1{)}^3}{3!}=1+0.1+0.005+0.000167=1.105167\approx 1.105$

Example 2: If $S=1+\frac{1}{1!}+\frac{1}{2!}+\cdots +\infty$, show that S=e.

Sol: By definition, S is the exponential series with x=1. Hence, S=e.

3.0Logarithmic Series

Definition

The natural logarithm can be expanded as: $\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots \text{for }-1<x\le 1$

General Expansion Formula

$\ln (1+x)={\sum }_{n=1}^{\infty }(-1{)}^{n+1}\frac{x^n}{n}$

Properties of Logarithmic Series

1. Domain of Convergence: −1< x ≤1.
2. Alternating Nature: Positive and negative terms alternate.
3. First Term Approximation: For small x, ln⁡(1+x)≈x.
4. Special Cases:

• $\ln (2)=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$
• $\ln \left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots \right)$

Standard Results

• ln⁡(1)=0.
• ln⁡(1+x) expansion is valid only for ∣x∣<1.
• Useful for solving problems involving limits like: ${\lim }_{x\to 0}\frac{\ln (1+x)}{x}=1$

4.0Solved Examples of Logarithmic Series

Example 1: Expand ln⁡(1.1) using logarithmic series.

Sol: $\ln (1.1)=0.1-\frac{(0.1{)}^2}{2}+\frac{(0.1{)}^3}{3}=0.1-0.005+0.000333=0.0953(\text{approx})$

Example 2: Prove that ${\lim }_{n\to \infty }{\left(\frac{n}{n+1}\right)}^n=\frac{1}{e}$

Sol: Taking log: $\ln L={\lim }_{n\to \infty }n\ln \left(1-\frac{1}{n+1}\right)$

Using ln⁡(1−x)≈−x:

$\ln L={\lim }_{n\to \infty }n\left(-\frac{1}{n+1}\right)=-1$

So, $L=e^{-1}=\frac{1}{e}$

5.0Comparison Between Exponential and Logarithmic Series