Calculus II · Unit 4B · lesson

Logarithm and Arctangent Series

Concept

Learning objectives

derive and use the standard series for $\ln(1+x)$ and $\arctan x$ .

Logarithm and Arctangent Series

Explanation

Integration turns rational geometric forms into inverse functions

The derivatives of $\ln(1+x)$ and $\arctan x$ are rational functions that can be expanded geometrically. Integrating those expansions produces denominators $n$ or $2n+1$ . This is often easier and more revealing than differentiating the original functions repeatedly.

Endpoint behavior becomes especially important. The logarithm series converges conditionally at $x=1$ and diverges at $x=-1$ . The arctangent series converges at both $x=\pm1$ , producing formulas for $\pi/4$ . These endpoint values connect power series with famous numerical constants.

Bridge

Integration creates series that derivative cycles do not reveal easily

The series for logarithm and arctangent are most naturally derived by integrating geometric-series variants. This shows why power-series algebra is more than a catalog: a known representation can be transformed into a new function whose derivatives would otherwise be cumbersome to organize.

Endpoint behavior becomes mathematically interesting. The interior representation follows from termwise integration, while values such as $x=1$ may converge only conditionally. That boundary evaluation connects power series back to alternating-series theory.

Proof idea

A kernel identity becomes a new function identity

Inside $|x|<1$ , integrate

\frac1{1+t}=1-t+t^2-t^3+\cdots

from $0$ to $x$ . Uniform convergence on smaller closed intervals justifies the termwise integral and yields the logarithm series with its constant fixed automatically.

Concept

Two derived series

For $-1<x\le1$ ,

\ln(1+x)=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n}.

For $|x|\le1$ ,

\arctan x=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{2n+1}.

Guided walkthrough

Approximate $\pi$

At $x=1$ ,

\frac\pi4=1-\frac13+\frac15-\frac17+\cdots.

The formula is elegant but converges slowly, so it is historically important and computationally inefficient without acceleration.

Worked example

Approximate ln(1.2) with a certified alternating series

Using

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

set $x=0.2$ . Four terms give

0.2-0.02+\frac{0.008}{3}-\frac{0.0016}{4} =0.182266\ldots.

The next term has magnitude $0.2^5/5=0.000064$ , so the true value differs from this approximation by at most $0.000064$ .

Common mistake

Track the constant of integration with an initial value

Termwise integration produces an arbitrary constant. Use a known value such as $\ln(1)=0$ or $\arctan(0)=0$ to determine it.

Interactive checku4b-logarithm_and_arctangent_series-01

Evaluate the arctangent series at $x=1$ .

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Show hint

Use $\arctan(1)$ .

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Exercise

Derive the logarithm series from $1/(1+x)$ .

Exercise

Use four terms to approximate $\ln(1.5)$ .

Exercise

State endpoint behavior of $\ln(1+x)$ .

Exercise

Explain why the arctangent approximation to $\pi$ is slow.

After the explanation

Use the section idea

Reading lens

Match value and derivatives at one center, then separate the polynomial approximation from the infinite-series convergence claim.

Mental model

Taylor coefficients encode local derivative data as a polynomial of increasing degree.

Decision

Choose the center, compute the derivative pattern, divide by factorials, and state whether you need a polynomial or an infinite series.

Common trap

Assuming every smooth-looking function equals its Taylor series everywhere.

Check yourself

Can you verify the first coefficients directly from derivatives at the center?

Source & rights

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