ch1.4.16-approximation

Ch 1.4.16 — Of the Resolution of Equations by Approximation

Summary: When exact algebraic methods are unavailable — especially for degrees above four — Euler presents two iterative numerical techniques: a Newton-like linearization method and a recurrence-series method, both converging to a root from an initial guess.

Sources: chapter-1.4.16

Last updated: 2026-05-03

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Why approximation (§784)

When roots are irrational and cannot be expressed by radicals, or when no radical formula exists (degree ≥ 5), one must approximate. The goal is a sequence of values converging to a root with arbitrarily small error.

Method 1: Newton-like linearization (§785–791)

Given that a root lies between $n$ and $n+1$, write $x=n+p$ where $p$ is a small fraction. Substitute into the equation, drop all powers of $p$ above the first, and solve the resulting linear equation for $p$. This gives a better approximation $n+p$, which can be fed back in as a new $n$ for another iteration.

For $x^2=a$ (§787)

Setting $x=n+p$ and neglecting $p^2$: $n^2+2np=a$, giving $p=\frac{a-n^2}{2n}$ and:

$x\approx \frac{n^2+a}{2n}$

This is the classical Babylonian / Heron’s method for square roots. Starting from $n=1$ for $\sqrt{2}$:

$1\to \frac{3}{2}\to \frac{17}{12}\to \frac{577}{408}\to \cdots$

Each step roughly doubles the number of correct digits.

For $x^3=a$ (§788)

Setting $x=n+p$ and neglecting $p^2,p^3$: $n^3+3n^{2}p=a$, giving:

$x\approx \frac{2n^3+a}{3n^2}$

Example for $\sqrt[3]{2}$: $1\to \frac{4}{3}\to \frac{91}{72}\to \cdots$

For a general cubic $x^3+a{x}^2+bx+c=0$ (§789)

Setting $x=n-p$ (note the minus) and neglecting $p^2$:

$p=\frac{n^3+a{n}^2+bn+c}{3n^2+2an+b}\Rightarrow x=\frac{2n^3+a{n}^2-c}{3n^2+2an+b}$

(source: §789). This formula coincides with one step of Newton’s method applied to the polynomial.

Example (§790): $x^3+2x^2+3x-50=0$. Starting from $n=3$: $x\approx \frac{62}{21}$; successive iterations converge to the real root.

For degree 5 (§791)

$x^5=6x+10:x\approx \frac{4n^5+10}{5n^4-6}$

Starting from $n=2$ gives $x\approx \frac{69}{37}$, which is close to the true root.

Method 2: Recurrence series (§792–800)

Form a sequence $a_0,a_1,a_2,\dots$ from the equation’s recurrence, then $x\approx a_{n+1}/a_n$ for large $n$. The ratio converges to the largest real root of the equation.

Deriving the recurrence

If $q/p=x$ and $r/p=x^2$, etc., substitute into the equation to find a linear recurrence among consecutive terms. Then:

• Any initial values for the first terms generate the whole sequence.
• The series converges to the largest root; successive ratios $a_{n+1}/a_n$ improve.

Fibonacci example (§793)

$x^2=x+1$: recurrence $r=q+p$. Starting from $(0,1)$:

$0,1,1,2,3,5,8,13,21,\dots$

Ratios: $\frac{1}{1},\frac{2}{1},\frac{3}{2},\frac{5}{3},\frac{8}{5},\dots \to \varphi =\frac{1+\sqrt{5}}{2}$ (golden ratio).

Example: $\sqrt{2}$ via $x^2=2x+1$ (§794)

Recurrence $r=2q+p$. Starting from $(0,1)$: $0,1,2,5,12,29,70,169,408,\dots$. The ratios converge to $1+\sqrt{2}$, and subtracting 1 gives successive approximations to $\sqrt{2}$: $\frac{1}{1},\frac{3}{2},\frac{7}{5},\frac{17}{12},\frac{41}{29},\dots$

General cubic example (§795)

$x^3=x^2+2x+1$: recurrence $s=r+2q+p$. Starting from $(0,0,1)$: $0,0,1,1,3,6,13,28,60,129,\dots$. The ratio $\frac{60}{28}=\frac{15}{7}$ already satisfies the equation with error $\frac{13}{343}$.

Limitation: missing second term (§796–797)

When the second term is absent (e.g., $x^2=2$), the recurrence degenerates and ratios oscillate. Fix: substitute $x=y-1$ to introduce a linear term, then apply the method to $y$.

Convergence to largest root (§798–799)

If the series begins with values that “seed” the largest root, the ratios converge only to that root. To find a smaller root, one must choose initial terms that match the smaller root, or the series will converge to the larger one.

Infinite equation (§800)

$x^{\infty }=x^{\infty -1}+x^{\infty -2}+\cdots$ requires each term of the series to equal the sum of all preceding terms. This forces the series $1,1,2,4,8,16,\dots$ and the root is exactly $x=2$ (verified by summing the geometric series $1/x+1/x^2+\cdots =1/(x-1)=1$).

Comparison of methods (§801)

The linearization method (Method 1) is universally applicable. The recurrence method (Method 2) is sometimes more elegant but requires careful setup and fails outright when the second term is missing.

Related pages

• approximation-methods
• quartic-equations
• cubic-equations
• quadratic-equations
• equations
• infinite-series
• geometrical-progressions