ch1.2.12-irrational-powers-infinite-series

Ch 1.2.12 — Of the Expression of Irrational Powers by Infinite Series

Summary: Extends the binomial theorem to fractional exponents, yielding infinite series for square roots and cube roots of compound expressions; uses iterative recentering to approximate irrational numbers by rationals to any desired accuracy.

Sources: chapter-1.2.12

Last updated: 2026-04-29

---

The General Binomial Series (articles 361–362)

The coefficient formula from ch1.2.10-higher-powers-compound is written with an undetermined exponent $n$:

$$(a+b{)}^n=a^n+\frac{n}{1}{a}^{n-1}b+\frac{n(n-1)}{1\cdot 2}{a}^{n-2}{b}^2+\frac{n(n-1)(n-2)}{1\cdot 2\cdot 3}{a}^{n-3}{b}^3+\cdots$$

When $n$ is a positive integer the series terminates after $n+1$ terms. When $n$ is fractional or negative it continues infinitely. For $a-b$ simply negate the even-position terms. (source: chapter-1.2.12)

This is the binomial-theorem in its general form.

Square Root as a Series (articles 363–364)

Since $\sqrt{a+b}=(a+b{)}^{1/2}$, substitute $n=\frac{1}{2}$. The successive coefficient fractions become $\frac{1}{2},-\frac{1}{4},-\frac{3}{6},-\frac{5}{8},\dots$ and the powers of $a$ become $\sqrt{a},\frac{\sqrt{a}}{a},\frac{\sqrt{a}}{a^2},\dots$, giving:

$$\sqrt{a+b}=\sqrt{a}+\frac{b}{2\sqrt{a}}-\frac{b^2}{8a\sqrt{a}}+\frac{b^3}{16a^2\sqrt{a}}-\frac{5b^4}{128a^3\sqrt{a}}+\cdots$$

If $a=c^2$ is a perfect square, $\sqrt{a}=c$ and the series is purely rational:

$$\sqrt{c^2+b}=c+\frac{b}{2c}-\frac{b^2}{8c^3}+\frac{b^3}{16c^5}-\frac{5b^4}{128c^7}+\cdots$$

(source: chapter-1.2.12)

Every number may be written as $c^2+b$ for some perfect square $c^2$, so this formula can approximate any square root. (source: chapter-1.2.12)

Numerical Example: $\sqrt{6}$ (articles 365–366)

Write $6=4+2$, so $c=2$, $b=2$:

$$\sqrt{6}=2+\frac{1}{2}-\frac{1}{16}+\frac{1}{64}-\frac{5}{1024}+\cdots$$

Two terms give $\frac{5}{2}$; three terms give $\frac{39}{16}$. Each partial sum is a rational approximation.

To accelerate convergence Euler recenters: since $\frac{5}{2}$ is close to $\sqrt{6}$, write $6=\frac{25}{4}-\frac{1}{4}$ and apply the formula again with $c=\frac{5}{2}$, $b=-\frac{1}{4}$. Two terms give $\frac{49}{20}$, which differs from $\sqrt{6}$ by less than $\frac{1}{400}$.

A further recentering gives $\frac{4801}{1960}$, with error $\frac{1}{3841600}$. The technique is a forerunner of Newton’s method for square roots. (source: chapter-1.2.12)

Cube Root as a Series (articles 367–368)

With $n=\frac{1}{3}$:

$$\sqrt[3]{a+b}=\sqrt[3]{a}+\frac{b}{3}\cdot \frac{\sqrt[3]{a}}{a}-\frac{b^2}{9}\cdot \frac{\sqrt[3]{a}}{a^2}+\frac{5b^3}{81}\cdot \frac{\sqrt[3]{a}}{a^3}-\cdots$$

If $a=c^3$ then $\sqrt[3]{a}=c$ and:

$$\sqrt[3]{c^3+b}=c+\frac{b}{3c^2}-\frac{b^2}{9c^5}+\frac{5b^3}{81c^8}-\frac{10b^4}{243c^{11}}+\cdots$$

(source: chapter-1.2.12)

Numerical Example: $\sqrt[3]{2}$ (article 369)

Write $2=1+1$, $c=1$, $b=1$:

$$\sqrt[3]{2}\approx 1+\frac{1}{3}-\frac{1}{9}+\frac{5}{81}-\cdots$$

Two terms give $\frac{4}{3}$, whose cube $\frac{64}{27}$ exceeds 2 by $\frac{10}{27}$. Recentering with $c=\frac{4}{3}$, $b=-\frac{10}{27}$ gives two-term approximation $\frac{91}{72}$, with cube $\frac{753571}{373248}$ versus $2=\frac{746496}{373248}$, error $\frac{7075}{373248}$. (source: chapter-1.2.12)

Significance

The key insight is that the same binomial coefficient formula, valid for integer exponents, extends to all rational (and, by continuity, all real) exponents simply by allowing the series to continue infinitely. This observation is the seed of Newton’s generalised binomial series and ultimately of Taylor-series expansions. See binomial-theorem and infinite-series. (source: chapter-1.2.12)

Related pages

• binomial-theorem
• infinite-series
• ch1.2.10-higher-powers-compound
• ch1.2.11-transpositions-and-binomial-proof
• ch1.2.13-negative-powers-infinite-series
• irrational-numbers
• ch1.1.19-fractional-exponents
• powers-and-exponents