**Literal resolution of aﬀected equations** **by Isaac Newton**

### By

### Naoki Osada ^{∗}

^{∗}

**Abstract**

### In 1669 and 1671, Isaac Newton resolved an algebraic equation *f* (x, y) = 0 by expressing *y* as an infinite series of *x. In this paper, we formulate Newton’s resolution as a contemporary* algorithm along the line of his original text and prove that the series converges asymptotically to the implicit function or one of the branches under certain conditions.

**§** **1.** **Introduction**

**§**

### Isaac Newton gave the literal resolution of an aﬀected equation in *De Analysi* (1669) and *De Methodis* (1671). An aﬀected equation is an algebraic equation that is not binomial, such as *y* ^{3} *−* 2y *−* 5 = 0 or *y* ^{3} + *a* ^{2} *y* *−* 2a ^{3} + *axy* *−* *x* ^{3} = 0. Newton referred to the former equation as numerical and the latter as literal. The literal resolution of an aﬀected equation *f* (x, y) = 0 is to express *y* as an infinite series

### (1.1) *y* =

### ∑

*∞*

*i=k*

*c* _{i} *x*

_{i}

^{±i/r}*,* *c* _{i} *∈* R *, k* *∈* Z *,* and *r* *∈* N *,*

_{i}

### where the double sign is set to + when *x* is close to 0, and it is set to *−* when *x* is suﬃciently large. The series (1.1) is called a Puiseux series if the double sign is +. In *De Analysi, Newton elucidated the two cases, namely, one when* *x* is close to 0 and there exists *c* such that *f* (0, c) = 0 and _{∂y} ^{∂} *f* (0, c) *̸* = 0, and another when *x* is suﬃciently large.

_{∂y}

^{∂}

### In the first case, *k* = 0, c 0 = *c* and *r* = 1 in (1.1). In *De Methodis, Newton improved the* above algorithm so that it can be applied easily even if (0, c) is a singular point, i.e.,

*f(0, c) = 0,* *∂*

*∂x* *f* (0, c) = 0, *∂*

*∂y* *f* (0, c) = 0,

### Received October 20, 2018. Revised January 19, 2019.

### 2010 Mathematics Subject Classification(s): 01A45 41A60 58C15

*Key Words:* *Isaac Newton, aﬀected equation, asymptotic expansion, algebraic equation, Newton* *diagram, implicit function, Puiseux series*

*∗*

### Tokyo Woman’s Christian University 167-8585, Japan.

### e-mail: osada@lab.twcu.ac.jp

*⃝* c 2019 Research Institute for Mathematical Sciences, Kyoto University. All rights reserved.

### using the Newton diagram. For the Newton diagram method, see [3, pp.191-196] or [4, pp.158-164]. Neither *De Analysi* nor *De Methodis* was published at that time, but Newton’s algorithm using the Newton diagram became publicly known because Newton wrote it in the letter *epistola posterior* [9] sent to Henry Oldenburg for Gottfried Wilhelm Leibniz in 1676, and John Wallis reproduced Newton’s algorithm in his *A Treatise of* *Algebra* [11] in 1685.

### In 1850, Victor Puiseux [7, p.401] proved that for a complex algebraic equation *f* (u, z) = 0, *u* can be represented as convergent infinite series

*u* *j* = *b* + *γ* *j* (z *−* *a)*

^{p}^{1}

### + *a* *j* (z *−* *a)*

^{2}

^{p}### + *b* *j* (z *−* *a)*

^{3}

^{p}### + *· · ·* *,* (j = 1, . . . , p).

### See [8, p.194] for more details. Today the following theorem is known:

**Theorem 1.1.** **(Puiseux theorem)**

*Let* C{ *z* *}* *be the ring of all convergent power series. Let* *f(z, y)* *∈* C{ *z* *}* [y] *be a monic* *irreducible polynomial of the form*

*f(z, y) =* *y* ^{n} + *a* *n*

^{n}

*−*

### 1 (z )y ^{n}

^{n}

^{−}^{1} + *· · ·* + *a* 1 (z)y + *a* 0 (z), *a* *j* (z) *∈* C{ *z* *}* *.* *Then, there is* *g(z)* *∈* C{ *z* *}* *such that*

*f* (z, y) =

*n* ∏

*−*

### 1 *j=0*

### (y *−* *g(ζ* ^{j} *z* ^{1/n} )), *where* *ζ* *is a primitive* *n-th root of unity.*

^{j}

*Proof.* See [2, pp.15-26] or [5, pp.235-237].

### In regard to the Puiseux theorem Abyhyankar wrote

### Newton’s theorem was revived by Puiseux in 1850. [...] Puiseux’s proof, be- ing based upon Cauchy’s integral theorems, applies only to convergent power series with complex coeﬃcients. On the other hand, Newton’s proof, being al- gorithmic, applies equally well to power series, whether they converge or not.

### Moreover, and that is the main point, Newton’s algorithmic proof leads to nu- merous other existence theorems while Puiseux’s existential proof does not do

### so. [1, p.417]

### In this paper, we formulate Newton’s algorithm to express an infinite series as a

### contemporary algorithm along the line of his original text. Additionally we prove under

### certain conditions that the infinite series (1.1) is the asymptotic expansion of the implicit

### function or one of the branches.

### Newton’s papers are cited from *Mathematical Papers of Isaac Newton* [12] edited by Whiteside and abbreviated as MP.

**§** **2.** **The asymptotic expansion of the implicit function as** *x* *→* 0 Newton elucidated the literal resolution of aﬀected equations in *De Analysi* as follows.

**§**

### Suppose now that the algebraic equation *y* ^{3} + *a* ^{2} *y* *−* 2a ^{3} + *axy* *−* *x* ^{3} = 0 has to be resolved. First I seek out the value of *y* when *x* is zero, that is, I elicit the root of this equation *y* ^{3} + *a* ^{2} *y* *−* 2a ^{3} = 0, and find it to be +a. And so I write +a in the quotient. Again, supposing *y* = *a* + *p, for* *y* I substitute that value and the terms *p* ^{3} + 3ap ^{2} + 4a ^{2} *p...* which thence result I set in the margin. Out of these I take 4a ^{2} *p* + *a* ^{2} *x, in which* *p* and *x* separately are of least dimension and suppose them nearly equal to zero, that is, *p* = *−* ^{x} _{4} nearly or *p* = *−* ^{x} _{4} + *q.*

^{x}

^{x}

### [...]

*a* *−* ^{1} _{4} *x* + _{64a} ^{x}

^{x}

^{2}

### + ^{131x} _{512a}

^{3}2

### + _{16384a} ^{509x}

^{4}3

*. . .* +a + *p* = *y.)* +y ^{3} +a ^{3} + 3a ^{2} *p* + 3ap ^{2} + *p* ^{3}

### +a ^{2} *y* +a ^{3} + *a* ^{2} *p* +axy +a ^{2} *x* + *axp*

*−* 2a ^{3} *−* 2a ^{3}

*−* *x* ^{3} *−* *x* ^{3}

*−* ^{1} _{4} *x* + *q* = *p.)* +p ^{3} *−* _{64} ^{1} *x* ^{3} + _{16} ^{3} *x* ^{2} *q* *−* ^{3} _{4} *xq* ^{2} + *q* ^{3} +3ap ^{2} + _{16} ^{3} *ax* ^{2} *−* ^{3} _{2} *axq* + 3aq ^{2} +4a ^{2} *p* *−* *a* ^{2} *x* + 4a ^{2} *q*

### +axp *−* ^{1} _{4} *ax* ^{2} + *axq* +a ^{2} *x* +a ^{2} *x*

*−* *x* ^{3} *−* *x* ^{3} + _{64a} ^{x}

^{x}

^{2}

### + *r* = *q.*

### )

### +3aq ^{2} + _{4096a} ^{3x}

^{4}

### + _{32} ^{3} *x* ^{2} *r* + 3ar ^{2} +4a ^{2} *q* + _{16} ^{1} *ax* ^{2} + 4a ^{2} *r*

*−* ^{1} _{2} *axq* *−* _{128} ^{1} *x* ^{3} *−* ^{1} _{2} *axr* + _{16} ^{3} *x* ^{2} *q* + _{1024a} ^{3x}

^{4}

### + _{16} ^{3} *x* ^{2} *r*

*−* _{16} ^{1} *ax* ^{2} *−* _{16} ^{1} *ax* ^{2}

*−* ^{65} _{64} *x* ^{3} *−* ^{65} _{64} *x* ^{3} +4a ^{2} *−* ^{1} _{2} *ax* + _{32} ^{9} *x* ^{2} )

### + ^{131} _{128} *x* ^{3} *−* _{4096a} ^{15x}

^{4}

### (

### +131x

^{3}

### 512a

^{2}

### + _{16384a} ^{509x}

^{4}3

### [. . . ]

### MP II, pp.222-225

### We now formulate the literal resolution of an aﬀected equation *f(x, y) = 0 under* the condition that there exists a root *c* of *f* (0, y) = 0 with _{∂y} ^{∂} *f(0, c)* *̸* = 0.

_{∂y}

^{∂}

**Algorithm 2.1.** (The asymptotic expansion as **x** **→** **0.) Let** *f(x, y) be a* polynomial of the form

**x**

**→**

*f* (x, y) =

### ∑ *l* *i=0*

*a* _{i,0} *x* ^{i} +

_{i,0}

^{i}

### ∑ *n* *j=1*

### ( _{m}

_{m}

### ∑

*i=0*

*a* _{i,j} *x* ^{i} )

_{i,j}

^{i}

*y* ^{j} *.*

^{j}

### Suppose *f* (0, y) = 0 has a root *c* with _{∂y} ^{∂} *f(0, c)* *̸* = 0.

_{∂y}

^{∂}

### (i) Put *f* _{0} (x, y) = *f(x, y) and* *d* _{0} (x) = *c.*

### (ii) Repeat below for *ν* = 1, 2, . . . , N : calculate

*f* _{ν} (x, y) = *f* _{ν}

_{ν}

_{ν}

_{−}_{1} (x, d _{ν}

_{ν}

_{−}_{1} (x) + *y)*

### =

*l*

_{ν}### ∑

*i=i*

*ν*

*a* ^{(ν)} _{i,0} *x* ^{i} +

_{i,0}

^{i}

### ∑ *n* *j=1*

### ( _{m}

_{m}

### ∑

*ν*

*i=0*

*a* ^{(ν)} _{i,j} *x* ^{i} )

_{i,j}

^{i}

*y* ^{j} *,* *a* ^{(ν)} _{i}

^{j}

_{i}

*ν*

*,0* *̸* = 0 and

*d* _{ν} (x) = *−* *a* ^{(ν)} _{i}

_{ν}

_{i}

*ν*

*,0*

*a* ^{(ν)} _{0,1} *x* ^{i}

^{i}

^{ν}*.*

### Then, *y* _{N} (x) = *d* _{0} (x) + *· · ·* + *d* _{N}

_{N}

_{N}

_{−}_{1} (x) + *d* _{N} (x) satisfies *f* (x, y *N* (x)) = *o(x* ^{i}

_{N}

^{i}

^{N}### ) as *x* *→* 0.

### It is further assumed that there exists a positive integer *µ* *∈* N such that

### (2.1) *f* (x, y) =

### ∑ *l/µ* *i=0*

*a* *iµ,0* *x* ^{iµ} +

^{iµ}

### ∑ *n* *j=1*

###

###

*m/µ* ∑

*i=0*

*a* *iµ,j* *x* ^{iµ}

^{iµ}

###

### *y* ^{j} *.*

^{j}

### (iii) Let *N* (x) be the first *N* terms of

*l* ∑

*N*

*/µ* *i=i*

*N*

*/µ*

*a* ^{(N} _{iµ,0} ^{)} *x* ^{iµ} , and let *D* (x) be the first *N* terms of

_{iµ,0}

^{iµ}

*m* ∑

*N*

*/µ* *i=0*

*a* ^{(N)} _{iµ,1} *x* ^{iµ} . Expand *−N* (x)/ *D* (x) to *N* terms by division, and put this expansion as ˜ *d* _{N} (x).

_{iµ,1}

^{iµ}

_{N}

### Then, ˜ *y* *N* (x) = *d* 0 (x) + *· · ·* + *d* *N*

*−*

### 1 (x) + ˜ *d* *N* (x) satisfies

*f* (x, *y* ˜ *N* (x)) = *O(x* ^{i}

^{i}

^{N}^{+N µ} ) as *x* *→* 0.

**Example 2.2.** We now apply Algorithm 2.1 to Newton’s example *f(x, y) =* *y* ^{3} + *a* ^{2} *y* *−* 2a ^{3} + *axy* *−* *x* ^{3} = 0 with *N* = 3. We take *c* = *a* as the root of the equation *f* (0, y) = *y* ^{3} + *a* ^{2} *y* *−* 2a ^{3} = 0. Then, _{∂y} ^{∂} *f* (0, a) = 4a ^{2} *̸* = 0.

_{∂y}

^{∂}

**ν** **= 0** Put *f* 0 (x, y) = *y* ^{3} + *a* ^{2} *y* *−* 2a ^{3} + *axy* *−* *x* ^{3} and *d* 0 (x) = *a.*

**ν**

**ν** **= 1** Since *f* 1 (x, y) = *f* 0 (x, a + *y) =* *a* ^{2} *x* *−* *x* ^{3} + (4a ^{2} + *ax)y* + 3ay ^{2} + *y* ^{3} , we have *i* 1 = 1, a ^{(1)} _{1,0} = *a* ^{2} *, a* ^{(1)} _{0,1} = 4a ^{2} and thus, *d* 1 (x) = *−* ^{1} _{4} *x.*

**ν**

**ν** **= 2** Since

**ν**

*f* 2 (x, y) = *−* 1

### 16 *ax* ^{2} *−* 65 64 *x* ^{3} +

### (

### 4a ^{2} *−* 1

### 2 *ax* + 3 16 *x* ^{2}

### ) *y* +

### (

### 3a *−* 3 4 *x*

### )

*y* ^{2} + *y* ^{3} ,

### we have *i* _{2} = 2, a ^{(2)} _{2,0} = *−* _{16} ^{1} *a, a* ^{(2)} _{0,1} = 4a ^{2} and thus, *d* _{2} (x) = _{64a} ^{1} *x* ^{2} *.* **ν** **= 3** Since

**ν**

*f* _{3} (x, y) = *−* 131

### 128 *x* ^{3} + 15

### 4096a *x* ^{4} *−* 3

### 16384a ^{2} *x* ^{5} + 1

### 262144a ^{3} *x* ^{6} +

### (

### 4a ^{2} *−* 1

### 2 *ax* + 9

### 32 *x* ^{2} *−* 3

### 128a *x* ^{3} + 3 4096a ^{2} *x* ^{4}

### ) *y*

### + (

### 3a *−* 3

### 4 *x* + 3 64a *x* ^{2}

### )

*y* ^{2} + *y* ^{3} , we have *N* (x) = *−* 131

### 128 *x* ^{3} + 15

### 4096a *x* ^{4} *−* 3

### 16384a ^{2} *x* ^{5} and *D* (x) = 4a ^{2} *−* 1

### 2 *ax* + 9

### 32 *x* ^{2} . Expanding *N* (x)/ *D* (x), we obtain *N* (x)

*D* (x) = *−* 131

### 512a ^{2} *x* ^{3} *−* 509

### 16384a ^{3} *x* ^{4} + 1843 131072a ^{4} *x* ^{5} and

*d* ˜ _{3} (x) = 131

### 512a ^{2} *x* ^{3} + 509

### 16384a ^{3} *x* ^{4} *−* 1843 131072a ^{4} *x* ^{5} *.*

### Thus, *f* _{3} (x, *d* ˜ _{3} (x)) = *N* (x) + *D* (x) ˜ *d* _{3} (x) = *O(x* ^{6} ) as *x* *→* 0.

### Therefore,

### ˜

*y* _{3} (x) =d _{0} (x) + *d* _{1} (x) + *d* _{2} (x) + ˜ *d* _{3} (x)

### =a *−* 1

### 4 *x* + *x* ^{2}

### 64a + 131x ^{3}

### 512a ^{2} + 509x ^{4}

### 16384a ^{3} *−* 1843x ^{5} 131072a ^{4} *,* *f* (x, *y* ˜ 3 (x)) =f 3 (x, *d* ˜ 3 (x)) = *O(x* ^{6} ) as *x* *→* 0.

### The next theorem shows that the sequence of functions *{* *f* (x, y *ν* (x)) *}* generated by

### Algorithm 2.1 (i)(ii) asymptotically converges to 0, where *y* *ν* (x) = *d* 0 (x) + *· · ·* + *d* *ν* (x).

**Theorem 2.3.** *Let* *f(x, y)* *be a polynomial of the form*

*f* (x, y) =

### ∑ *l* *i=0*

*a* _{i,0} *x* ^{i} +

_{i,0}

^{i}

### ∑ *n* *j=1*

### ( _{m}

_{m}

### ∑

*i=0*

*a* _{i,j} *x* ^{i} )

_{i,j}

^{i}

*y* ^{j} *.*

^{j}

*Suppose* *f* (0, y) = 0 *has a root* *c* *with* _{∂y} ^{∂} *f* (0, c) *̸* = 0. Let the polynomials *f* _{ν} (x, y) *and* *the monomials* *d* *ν* (x) *be the same as Algorithm 2.1 (i)(ii). Then, the following (1), (2)* *and (3) hold for* *ν* = 1, 2, . . . , N *:*

_{∂y}

^{∂}

_{ν}

*(1)* *a* ^{(ν)} _{0,1} = *a* ^{(1)} _{0,1} = *∂*

*∂y* *f* (0, c) *̸* = 0, *(2)* 1 *≤* *i* _{1} *<* *· · ·* *< i* _{ν−1} *< i* _{ν} *,*

_{ν−1}

_{ν}

*(3)* *y* *ν* (x) = *d* 0 (x) + *d* 1 (x) + *· · ·* + *d* *ν* (x) *satisfies*

### (2.2) *f* (x, y *ν* (x)) = *o(x* ^{i}

^{i}

^{ν}### ) *as* *x* *→* 0.

*Proof.* We prove by mathematical induction on *ν. By Taylor’s theorem,* *f* _{1} (x, y) = *f* (x, c + *y) =* *f* (x, c) + *∂*

*∂y* *f* (x, c)y +

### ∑ *n* *j=2*

### 1 *j* !

*∂* ^{j}

^{j}

*∂y* ^{j} *f(x, c)y* ^{j} *.* When *x* = 0, we have

^{j}

^{j}

*f* 1 (0, y) = *f(0, c) +* *∂*

*∂y* *f* (0, c)y +

### ∑ *n* *j=2*

### 1 *j!*

*∂* ^{j}

^{j}

*∂y* ^{j} *f* (0, c)y ^{j} *.* Therefore, *a* ^{(1)} _{0,0} = *f* (0, c) = 0 and *a* ^{(1)} _{0,1} = _{∂y} ^{∂} *f* (0, c) *̸* = 0, and thus, *i* _{1} *≥* 1 and

^{j}

^{j}

_{∂y}

^{∂}

*d* 1 (x) = *−* *a* ^{(1)} _{i}

_{i}

1

*,0*

*a* ^{(1)} _{0,1} *x* ^{i}

^{i}

^{1}

*.*

### Then,

*f* (x, y _{1} (x)) =f(x, d _{0} (x) + *d* _{1} (x)) = *f* _{1} (x, d _{1} (x))

### =

*l*

1
### ∑

*i=i*

1
*a* ^{(1)} _{i,0} *x* ^{i} +

_{i,0}

^{i}

### ∑ *n* *j=1*

### ( _{m}

_{m}

### ∑

1*i=0*

*a* ^{(1)} _{i,j} *x* ^{i} )

_{i,j}

^{i}

*d* 1 (x) ^{j}

^{j}

### = (

*a* ^{(1)} _{i}

_{i}

1

*,0* *x* ^{i}

^{i}

^{1}

### + *a* ^{(1)} _{0,1} *d* 1 (x) )

### +

*l*

1
### ∑

*i=i*

1### +1

*a* ^{(1)} _{i,0} *x* ^{i}

_{i,0}

^{i}

### + ( _{m}

_{m}

### ∑

1*i=1*

*a* ^{(1)} _{i,1} *x* ^{i} )

_{i,1}

^{i}

*d* 1 (x) +

### ∑ *n* *j=2*

### ( _{m}

_{m}

### ∑

1*i=0*

*a* ^{(1)} _{i,j} *x* ^{i} )

_{i,j}

^{i}

*d* 1 (x) ^{j}

^{j}

### =o(x ^{i}

^{i}

^{1}

### ) as *x* *→* 0.

### It holds for *ν* = 1.

### Next we assume that it holds for *ν(ν >* 1). Let *f* _{ν} (x, y) =

_{ν}

*l*

*ν*

### ∑

*i=i*

*ν*

*a* ^{(ν)} _{i,0} *x* ^{i} +

_{i,0}

^{i}

### ∑ *n* *j=1*

### ( _{m}

_{m}

### ∑

*ν*

*i=0*

*a* ^{(ν)} _{i,j} *x* ^{i} )

_{i,j}

^{i}

*y* ^{j} *,* *a* ^{(ν)} _{i}

^{j}

_{i}

*ν*

*,0* *̸* = 0.

### By the induction hypothesis, *a* ^{(ν)} _{0,1} = *a* ^{(1)} _{0,1} *̸* = 0 and *d* *ν* (x) = *−* *a* ^{(ν)} _{i}

_{i}

*ν*

*,0*

*a* ^{(1)} _{0,1} *x* ^{i}

^{i}

^{ν}*.* Since

*f* _{ν+1} (x, y) =f _{ν} (x, d _{ν} (x) + *y)*

_{ν+1}

_{ν}

_{ν}

### =

*l*

*ν*

### ∑

*i=i*

*ν*

*a* ^{(ν)} _{i,0} *x* ^{i} +

_{i,0}

^{i}

### ∑ *n* *j=1*

### ( _{m}

_{m}

### ∑

*ν*

*i=0*

*a* ^{(ν)} _{i,j} *x* ^{i} )

_{i,j}

^{i}

### (d *ν* (x) + *y)* ^{j}

^{j}

### = (

*a* ^{(ν)} _{i}

_{i}

*ν*

*,0* *x* ^{i}

^{i}

^{ν}### + *a* ^{(1)} _{0,1} *d* *ν* (x) )

### +

*l*

*ν*

### ∑

*i=i*

*ν*

### +1

*a* ^{(ν)} _{i,0} *x* ^{i} + ( _{m}

_{i,0}

^{i}

_{m}

### ∑

*ν*

*i=1*

*a* ^{(ν} _{i,1} ^{)} *x* ^{i} )

_{i,1}

^{i}

*d* *ν* (x)

### + ( _{m}

_{m}

### ∑

*ν*

*i=0*

*a* ^{(ν)} _{i,1} *x* ^{i} )

_{i,1}

^{i}

*y* +

### ∑ *n* *j=2*

### ( _{m}

_{m}

### ∑

*ν*

*i=0*

*a* ^{(ν)} _{i,j} *x* ^{i} )

_{i,j}

^{i}

### (d *ν* (x) + *y)* ^{j}

^{j}

### and *a* ^{(ν)} _{i}

_{i}

*ν*

*,0* *x* ^{i}

^{i}

^{ν}### + *a* ^{(1)} _{0,1} *d* _{ν} (x) = 0, it holds that *i* _{ν+1} *> i* _{ν} and *a* ^{(ν+1)} _{0,1} = *a* ^{(ν)} _{0,1} = *a* ^{(1)} _{0,1} *̸* = 0. By the definition of *i* _{ν+1} ,

_{ν}

_{ν+1}

_{ν}

_{ν+1}

*i*

_{ν+1}### ∑

*−*

*i*

_{ν}*−*

### 1 *i=1*

### ( *a* ^{(ν)} _{i+i}

_{i+i}

*ν*

*,0* *x* ^{i+i}

^{i+i}

^{ν}### + *a* ^{(ν)} _{i,1} *x* ^{i} *d* *ν* (x) )

_{i,1}

^{i}

### = 0.

### Thus,

*f* _{ν+1} (x, y) =

_{ν+1}

*l*

_{ν}### ∑

*i=i*

_{ν+1}*a* ^{(ν)} _{i,0} *x* ^{i} +

_{i,0}

^{i}

###

### ∑ ^{m}

^{m}

^{ν}*i=i*

_{ν+1}*−*

*i*

_{ν}*a* ^{(ν)} _{i,1} *x* ^{i}

_{i,1}

^{i}

###

### *d* _{ν} (x)

_{ν}

### + ( _{m}

_{m}

### ∑

*ν*

*i=0*

*a* ^{(ν)} _{i,1} *x* ^{i} )

_{i,1}

^{i}

*y* +

### ∑ *n* *j=2*

### ( _{m}

_{m}

### ∑

*ν*

*i=0*

*a* ^{(ν)} _{i,j} *x* ^{i} )

_{i,j}

^{i}

### (d *ν* (x) + *y)* ^{j}

^{j}

### =a ^{(ν+1)} _{i}

_{i}

*ν+1*

*,0* *x* ^{i}

^{i}

^{ν+1}### +

*l*

*ν*

### ∑

*i=i*

*ν+1*

### +1

*a* ^{(ν)} _{i,0} *x* ^{i} +

_{i,0}

^{i}

###

### ∑ ^{m}

^{m}

^{ν}*i=i*

*ν+1*

*−*

*i*

*ν*

### +1

*a* ^{(ν)} _{i,1} *x* ^{i}

_{i,1}

^{i}

###

### *d* *ν* (x)

### + ( _{m}

_{m}

### ∑

*ν*

*i=0*

*a* ^{(ν)} _{i,1} *x* ^{i} )

_{i,1}

^{i}

*y* +

### ∑ *n* *j=2*

### ( _{m}

_{m}

### ∑

*ν*

*i=0*

*a* ^{(ν)} _{i,j} *x* ^{i} )

_{i,j}

^{i}

### (d _{ν} (x) + *y)* ^{j} *,*

_{ν}

^{j}

### where

*a* ^{(ν+1)} _{i}

_{i}

*ν+1*

*,0* = *a* ^{(ν)} _{i}

_{i}

*ν+1*

*,0* + *a* ^{(ν)} _{i}

_{i}

*ν+1**−*

*i*

*ν*

*,1*

### (

*−* *a* ^{(ν} _{i} ^{)}

_{i}

*ν*

*,0*

*a* ^{(1)} _{0,1} )

*.*

### Thus,

*d* _{ν+1} (x) = *−* *a* ^{(ν+1)} _{i}

_{ν+1}

_{i}

*ν+1*

*,0*

*a* ^{(1)} _{0,1}

*x* ^{i}

^{i}

^{ν+1}*.*

### The coeﬃcient of *x* ^{i}

^{i}

^{ν+1}### in *f* _{ν+1} (x, d _{ν+1} (x)) is *a* ^{(ν+1)} _{i}

_{ν+1}

_{ν+1}

_{i}

*ν+1*

*,0* + *a* ^{(ν)} _{0,1} (

*−* *a* ^{(ν+1)} _{i}

_{i}

*ν+1*

*,0*

*a* ^{(1)} _{0,1} )

### = 0.

### Therefore,

*f* (x, y *ν+1* (x)) = *f* 1 (x, d 1 (x) + *· · ·* + *d* *ν+1* (x)) = *f* 2 (x, d 2 (x) + *· · ·* + *d* *ν+1* (x)) = *· · ·*

### = *f* _{ν+1} (x, d _{ν+1} (x)) = *o(x* ^{i}

_{ν+1}

_{ν+1}

^{i}

^{ν+1}### ) as *x* *→* 0.

### This theorem has been proved by mathematical induction.

### The core of Algorithm 2.1 is to take *d* *ν* (x) so that *a* ^{(ν)} _{i}

_{i}

*ν*

*,0* *x* ^{i}

^{i}

^{ν}### + *a* ^{(1)} _{0,1} *d* _{ν} (x) = 0, thereby increasing the order *i* _{ν+1} of *f* _{ν+1} (x, 0). The equation

_{ν}

_{ν+1}

_{ν+1}

*a* ^{(ν)} _{i}

_{i}

*ν*

*,0* *x* ^{i}

^{i}

^{ν}### + *a* ^{(1)} _{0,1} *y* = 0

### corresponds with “in which *p* and *x* separately are of least dimension and suppose them nearly equal to zero.”

### By the implicit function theorem, there exist open intervals *I* = ( *−* *δ, δ),* (δ > 0), *J* = (c *−* *η, c* + *η),* (η > 0) and the unique function *ϕ* : *I* *→* *J* such that

### (i) *ϕ(0) =* *c,*

### (ii) *f* (x, ϕ(x)) = 0 for *∀* *x* *∈* *I* , (iii) *ϕ(x) is of class* *C*

^{∞}### (I).

### Since *y* _{ν} (0) = *c, the asymptotic formula (2.2) means that* *y* _{ν} (x) asymptotically converges to *ϕ(x) as* *ν* *→ ∞* . We can say that Newton gave an algorithm to construct the implicit function *ϕ(x) of an algebraic equation* *f* (x, y) = 0 when *f* (0, c) = 0 with _{∂y} ^{∂} *f* (0, c) *̸* = 0.

_{ν}

_{ν}

_{∂y}

^{∂}

### The next theorem shows that Algorithm 2.1 (iii) is valid.

**Theorem 2.4.** *Let* *f(x, y)* *be a polynomial of the form*

*f* (x, y) =

### ∑ *l/µ* *i=0*

*a* _{iµ,0} *x* ^{iµ} +

_{iµ,0}

^{iµ}

### ∑ *n* *j=1*

###

###

*m/µ* ∑

*i=0*

*a* _{iµ,j} *x* ^{iµ}

_{iµ,j}

^{iµ}

###

### *y* ^{j} *,* *µ* *∈* N *.*

^{j}

*Suppose* *f* (0, y) = 0 *has a root* *c* *with* _{∂y} ^{∂} *f* (0, c) *̸* = 0. Let integers *ν, N, i* _{ν} *, l* _{ν} *,* *and* *m* _{ν} *be* *the same as Algorithm 2.1. Let polynomials or monomials* *f* _{ν} (x, y), f _{N} (x, y), d _{ν} (x), *d* ˜ _{N} (x), *y* _{ν} (x), *y* ˜ _{N} (x), *N* (x), *and* *D* (x) *be the same as Algorithm 2.1. Then, the following asymp-* *totic formulas hold:*

_{∂y}

^{∂}

_{ν}

_{ν}

_{ν}

_{ν}

_{N}

_{ν}

_{N}

_{ν}

_{N}

*f* *N* (x, *d* ˜ *N* (x)) =O(x ^{i}

^{i}

^{N}^{+N µ} ) *as* *x* *→* 0, *f* (x, *y* ˜ _{N} (x)) =O(x ^{i}

_{N}

^{i}

^{N}^{+N µ} ) *as* *x* *→* 0.

*Proof.* As in the proof of Theorem 2.3, it holds that *f* _{ν} (x, y) =

_{ν}

*l* ∑

*ν*

*/µ* *i=i*

*ν*

*/µ*

*a* ^{(ν)} _{iµ,0} *x* ^{iµ} +

_{iµ,0}

^{iµ}

### ∑ *n* *j=1*

###

###

*m* ∑

*ν*

*/µ* *i=0*

*a* ^{(ν)} _{iµ,j} *x* ^{iµ}

_{iµ,j}

^{iµ}

###

### *y* ^{j} *,* *a* ^{(ν)} _{i}

^{j}

_{i}

*ν*

*,0* *̸* = 0.

### Thus, *N* (x) and *D* (x) can be written as *N* (x) = *a* ^{(N} _{i} ^{)}

_{i}

*N*

*,0* *x* ^{i}

^{i}

^{N}### + *a* ^{(N} _{i} ^{)}

_{i}

*N*

### +µ,0 *x* ^{i}

^{i}

^{N}^{+µ} + *· · ·* + *a* ^{(N)} _{i}

_{i}

*N*

### +(N

*−1)µ,0*

*x* ^{i}

^{i}

^{N}^{+(N}

^{−}^{1)µ} *,* *D* (x) = *a* ^{(N} _{0,1} ^{)} + *a* ^{(N} _{µ,1} ^{)} *x* ^{µ} + *· · ·* + *a* ^{(N} _{(N} ^{)}

_{µ,1}

^{µ}

_{−}_{1)µ,1} *x* ^{(N}

^{−}^{1)µ} *,*

### respectively. Then, *N* (x)/ *D* (x) can be written as *N* (x)

*D* (x) = *e* _{i}

_{i}

_{N}*x* ^{i}

^{i}

^{N}### + *e* _{i}

_{i}

_{N}_{+µ} *x* ^{i}

^{i}

^{N}^{+µ} + *· · ·* + *e* _{i}

_{i}

_{N}_{+(N}

_{−}_{1)µ} *x* ^{i}

^{i}

^{N}^{+(N}

^{−}^{1)µ} (2.3)

### + *O(x* ^{i}

^{i}

^{N}^{+N µ} ) as *x* *→* 0.

### Thus,

*f* *N* (x, y) = *N* (x) +

*l* ∑

_{N}*/µ* *i=i*

*N*

*/µ+N*

*a* ^{(N)} _{iµ,0} *x* ^{iµ} + *D* (x)y +

_{iµ,0}

^{iµ}

###

###

*m* ∑

_{N}*/µ* *i=N*

*a* ^{(N} _{iµ,1} ^{)} *x* ^{iµ}

_{iµ,1}

^{iµ}

###

### *y*

### +

### ∑ *n* *j=2*

###

###

*m* ∑

_{N}*/µ* *i=0*

*a* ^{(N} _{iµ,j} ^{)} *x* ^{iµ}

_{iµ,j}

^{iµ}

###

### *y* ^{j} *,*

^{j}

*d* ˜ *N* (x) = *−* (e *i*

_{N}*x* ^{i}

^{i}

^{N}### + *e* *i*

_{N}### +µ *x* ^{i}

^{i}

^{N}^{+µ} + *· · ·* + *e* _{i}

_{i}

_{N}_{+(N}

_{−}_{1)µ} *x* ^{i}

^{i}

^{N}^{+(N}

^{−}^{1)µ} ).

### Since *D* (x) = *O(1) and* *N* (x)/ *D* (x) = *−* *d* ˜ *N* (x) + *O(x* ^{i}

^{i}

^{N}^{+N µ} ) as *x* *→* 0, we have *f* *N* (x, *d* ˜ *N* (x)) = *O(x* ^{i}

^{i}

^{N}^{+N µ} ) as *x* *→* 0.

### This completes the proof.

**Example 2.5.** We apply Algorithm 2.1 with *N* = 3 to *f* (x, y) = *x* ^{2} *−* 2ay+y ^{2} = 0 which satisfies (2.1) with *µ* = 2.

### The roots of *f* (0, y) = *−* 2ay + *y* ^{2} = 0 are 2a and 0. We take *c* = 0 as the root of the equation. Then, _{∂y} ^{∂} *f* (0, 0) = *−* 2a *̸* = 0. The implicit function *ϕ(x) of* *f(x, y) = 0* with *ϕ(0) = 0 is* *ϕ(x) =* *a* *−* *√*

_{∂y}

^{∂}

*a* ^{2} *−* *x* ^{2} .

**ν** **= 0** Put *f* 0 (x, y) = *x* ^{2} *−* 2ay + *y* ^{2} and *d* 0 (x) = 0.

**ν**

**ν** **= 1** Since *f* _{1} (x, y) = *f* _{0} (x, 0 + *y) =* *x* ^{2} *−* 2ay + *y* ^{2} , we have *i* _{1} = 2, a ^{(1)} _{2,0} = 1, a ^{(1)} _{0,1} = *−* 2a, and thus *d* _{1} (x) = _{2a} ^{1} *x* ^{2} *.*

**ν**

**ν** **= 2** Since *f* 2 (x, y) = _{4a} ^{1}

2**ν**

*x* ^{4} + (

*−* 2a + _{a} ^{1} *x* ^{2} )

_{a}

*y* + *y* ^{2} *,* we have *i* 2 = 4, a ^{(2)} _{4,0} =

### 1

### 4a

^{2}

*, a* ^{(2)} _{0,1} = *−* 2a, and thus, *d* _{2} (x) = _{8a} ^{1}

3*x* ^{4} *.* **ν** **= 3** Since *f* 3 (x, y) = 1

**ν**

### 8a ^{4} *x* ^{6} + 1 64a ^{6} *x* ^{8} +

### (

*−* 2a + 1

*a* *x* ^{2} + 1 4a ^{3} *x* ^{4}

### )

*y* + *y* ^{2} *,* we have *i* 3 = 6.

*d* ˜ 3 (x) = *−*

### 1

### 8a

^{4}

*x* ^{6} + _{64a} ^{1}

6*x* ^{8}

*−* 2a + ^{1} _{a} *x* ^{2} + _{4a} ^{1}

3_{a}

*x* ^{4} = 1

### 16a ^{5} *x* ^{6} + 5

### 128a ^{7} *x* ^{8} + 7

### 256a ^{9} *x* ^{10} + *O(x* ^{12} ).

### Therefore,

### ˜

*y* 3 (x) = 0 + 1

### 2a *x* ^{2} + 1

### 8a ^{3} *x* ^{4} + 1

### 16a ^{5} *x* ^{6} + 5

### 128a ^{7} *x* ^{8} + 7

### 256a ^{9} *x* ^{10} + *O(x* ^{12} ), *f* (x, *y* ˜ 3 (x)) = *f* 3 (x, *d* ˜ 3 (x)) = *O(x* ^{12} ).

**§** **3.** **The asymptotic expansion of the implicit function as** *x* *→ ∞* Up to this point, Newton elucidated the literal resolution when *x* is close to zero, but from here elucidated when *x* is suﬃciently large.

**§**

### But if you wish that the value of the area should approach nearer the truth greater *x* is, take this an example:y ^{3} + *axy* + *x* ^{2} *y* *−* *a* ^{3} *−* 2x ^{3} = 0. Accordingly, ready to resolve this, I take out the terms *y* ^{3} + *x* ^{2} *y* *−* 2x ^{3} in which *x* and *y* either separately or multiplied together are of the most and equal dimensions everywhere. From these, set as it were equal to zero, I elicit the root, finding it to be *x, and write it in the quotient: or, what comes to the same thing,* on substituting unity for *x* from *y* ^{3} + *y* *−* 2 I extract the root 1, multiply it by *x* and write the product *x* in the quotient. Then I suppose *x* + *p* = *y* and so proceed as in the former example until I have the quotient *x* *−* *a*

### 4 + *a* ^{2} 64x + 131a ^{3}

### 512x ^{2} + 509a ^{4}

### 16384x ^{3} &c,[...]

### MP II, pp.226-227

### We formulate the literal resolution of the aﬀected equation when *x* is suﬃciently large as a modern algorithm faithful to Newton’s elucidation.

### For a function

*f* (x, y) = ∑

*i,j*

*a* *i,j* *x* ^{q}

^{q}

^{i,j}*y* ^{j} *,* *q* *i,j* *∈* Q *,*

^{j}

### the set of all exponents of *x* is defined by

*P* (f, α) = *{* *q* *i,j* + *αj* *|* *a* *i,j* *̸* = 0 *}* *,* when *y* *∼* *cx* ^{α} as *x* *→ ∞* for some *c* *∈* R .

^{α}

**Algorithm 3.1.** (The asymptotic expansion as **x** **→ ∞** .) Let *f(x, y) =*

**x**

**→ ∞**

### ∑ *n* *j=0*

### ( _{m}

_{m}

### ∑

*i=0*

*a* _{i,j} *x* ^{i} )

_{i,j}

^{i}

*y* ^{j} = 0, be an algebraic equation.

^{j}

### (i) Put *f* _{0} (x, y) = *f(x, y).*

### (ii) Find a rational number *α* _{0} such that there are two or more terms in *g* _{0} (x, y; *α* _{0} ) = ∑

*i+α*

_{0}

*j=max* *P* (f,α

_{0}

### )

*a* _{i,j} *x* ^{i} *y* ^{j} *.*

_{i,j}

^{i}

^{j}

### (iii) Take a root *v* = *c* 0 of the equation *g* 0 (1, v; *α* 0 ) = 0.

### (iv) Put *d* 0 (x) = *c* 0 *x* ^{α}

^{α}

^{0}

### .

### (v) Repeat (1), (2), (3) and (4) below for *ν* = 1, 2, . . . , N : (1) calculate *f* *ν* (x, y) = *f* *ν*

*−*

### 1 (x, d *ν*

*−*

### 1 (x) + *y), say*

*f* *ν* (x, y) =

### ∑ *n* *j=0*

### ( ∑

*i*

*a* ^{(ν)} _{i,j} *x* ^{q}

_{i,j}

^{q}

(ν)
*i,j*

### ) *y* ^{j} *,*

^{j}

### (2) find a rational number *α* *ν* with max *P* (f *ν* *, α* *ν* ) *<* max *P* (f *ν*

*−*

### 1 *, α* *ν*

*−*

### 1 ) such that there are two or more terms in

*g* _{ν} (x, y; *α* _{ν} ) = ∑

_{ν}

_{ν}

*q*

^{(ν)}

_{i,j}### +α

*ν*

*j=max* *P* (f

*ν*

*,α*

*ν*

### )

*a* ^{(ν)} _{i,j} *x* ^{q}

_{i,j}

^{q}

^{(ν)}

^{i,j}*y* ^{j} *,*

^{j}

### (3) take a root *v* = *c* *ν* of the equation *g* *ν* (1, v; *α* *ν* ) = 0,

### (4) put *d* _{ν} (x) = *c* _{ν} *x* ^{α}

_{ν}

_{ν}

^{α}

^{ν}### . Then, the function *y* _{N} (x) = ∑ *N*

_{N}

*ν=0* *d* _{ν} (x) satisfies

_{ν}

*f(x, y* _{N} (x)) = *o(x* ^{max} ^{P} ^{(f}

_{N}

^{P}

^{N}^{,α}

^{,α}

^{N}^{)} ) as *x* *→ ∞* *.*

**Theorem 3.2.** *Keep the notation in Algorithm 3.1. Put* max *P* (f

_{−}### 1 *, α*

_{−}### 1 ) = *∞* *.* *Suppose there exist* *α* _{ν} *and* *c* _{ν} *in Algorithm 3.1, for* *ν* = 0, 1, . . . , N *. Then,*

_{ν}

_{ν}

*1. for* *ν* = 0, 1, . . . , N *,*

*g* *ν* (x, d *ν* (x); *α* *ν* ) = 0,

*f* _{ν} (x, d _{ν} (x)) = *o(x* ^{max} ^{P} ^{(f}

_{ν}

_{ν}

^{P}

^{ν}^{,α}

^{,α}

^{ν}^{)} ) *as* *x* *→ ∞* *,* max *P* (f *ν* *, α* *ν* ) *<* max *P* (f *ν*

*−*

### 1 *, α* *ν*

*−*

### 1 );

*2. the function* *y* *N* (x) = ∑ *N*

*ν=0* *d* *ν* (x) *satisfies*

*f* (x, y *N* (x)) = *o(x* ^{max} ^{P} ^{(f}

^{P}

^{N}^{,α}

^{,α}

^{N}^{)} ) *as* *x* *→ ∞* *.* *Proof.*

### 1. Assume there exists a rational number *α* _{ν} such that there are two or more terms in *g* *ν* (x, y; *α* *ν* ) = ∑

_{ν}

*q*

_{i,j}^{(ν)}

### +α

*ν*

*j=max* *P* (f

*ν*

*,α*

*ν*

### )

*a* ^{(ν)} _{i,j} *x* ^{q}

_{i,j}

^{q}

^{i,j}^{(ν)}

*y* ^{j} *.*

^{j}

### Let *c* *ν* be a non-zero root of

*g* _{ν} (1, v; *α* _{ν} ) = ∑

_{ν}

_{ν}

*q*

_{i,j}^{(ν)}

### +α

*ν*

*j=max* *P* (f

*ν*

*,α*

*ν*

### )

*a* ^{(ν)} _{i,j} *v* ^{j} = 0.

_{i,j}

^{j}

### Then,

*g* *ν* (x, c *ν* *x* ^{α}

^{α}

^{ν}### ; *α* *ν* ) = ∑

*q*

_{i,j}^{(ν)}

### +α

_{ν}*j=max* *P* (f

_{ν}*,α*

_{ν}### )

*a* ^{(ν)} _{i,j} *c* ^{j} _{ν} *x* ^{max} ^{P} ^{(f}

_{i,j}

^{j}

_{ν}

^{P}

^{ν}^{,α}

^{,α}

^{ν}^{)}

### =g _{ν} (1, c _{ν} ; *α* _{ν} )x ^{max} ^{P} ^{(f}

_{ν}

_{ν}

_{ν}

^{P}

^{ν}^{,α}

^{,α}

^{ν}^{)} = 0.

### Since

*f* *ν* (x, c *ν* *x* ^{α}

^{α}

^{ν}### ) = *g* *ν* (x, c *ν* *x* ^{α}

^{α}

^{ν}### ; *α* *ν* ) + *o(x* ^{max} ^{P} ^{(f}

^{P}

^{ν}^{,α}

^{,α}

^{ν}^{)} ) as *x* *→ ∞* *,* we have

*f* _{ν} (x, d _{ν} (x)) = *o(x* ^{max} ^{P} ^{(f}

_{ν}

_{ν}

^{P}

^{ν}^{,α}

^{,α}

^{ν}^{)} ) as *x* *→ ∞* *.* From the way of deciding *α* *ν* , it holds that

### max *P* (f *ν* *, α* *ν* ) *<* max *P* (f *ν*

*−*

### 1 *, α* *ν*

*−*

### 1 ).

### 2. By the definition of *f* _{ν} ,

_{ν}

*f(x, d* 0 (x) + *d* 1 (x) + *· · ·* + *d* *N* (x)) = *f* 1 (x, d 1 (x) + *· · ·* + *d* *N* (x)) = *· · ·*

### =f _{N} (x, d _{N} (x)) = *o(x* ^{max} ^{P} ^{(f}

_{N}

_{N}

^{P}

^{N}^{,α}

^{,α}

^{N}^{)} ) as *x* *→ ∞* *.*

### The function *g* _{ν} (x, y; *α* _{ν} ) is the sum of the terms in which “x and *y* either separately or multiplied together are of the most and equal dimensions everywhere.” The equation *g* *ν* (1, v; *α* *ν* ) = 0 corresponds with “on substituting unity for *x.”*

_{ν}

_{ν}

**Example 3.3.** We apply Algorithm 3.1 to Newton’s example *f* (x, y) = *y* ^{3} + *axy* + *x* ^{2} *y* *−* *a* ^{3} *−* 2x ^{3} = 0.

**ν** **= 0** Two or more exponents of *{* 3α, 1 + *α,* 2 + *α,* 0, 3 *}* are equal to each other and maximized when *α* = 1. Thus, *g* 0 (x, y; 1) = *y* ^{3} + *x* ^{2} *y* *−* 2x ^{3} , and the root of *g* 0 (1, v; 1) = *v* ^{3} + *v* *−* 2 = 0 is *c* = 1. Therefore, *d* 0 (x) = *x* and *f(x, x) =* *ax* ^{2} *−* *a* ^{3} = *o(x* ^{3} ) as *x* *→ ∞* *.*

**ν**

**ν** **= 1** Since *f* 1 (x, y) = *f* (x, x + *y) =* *ax* ^{2} *−* *a* ^{3} + (4x ^{2} + *ax)y* + 3xy ^{2} + *y* ^{3} , two or more exponents of *{* 2, 0, 2 + *α,* 1 + *α,* 1 + 2α, 3α *}* are equal to each other and maximized when *α* = 0. Thus, *g* _{1} (x, y; 0) = *ax* ^{2} + 4x ^{2} *y, c* = *−* ^{a} _{4} *,* and *d* 1 (x) = *−* ^{a} _{4} . Therefore, *f* 1 (x, *−* ^{a} _{4} ) = *−* ^{a} _{16}

**ν**

^{a}

^{a}

^{a}

^{a}

^{2}

*x* *−* ^{65} _{64} *a* ^{3} = *o(x* ^{2} ) as *x* *→ ∞* *.* **ν** **= 2** Since *f* 2 (x, y) = *−* 1

**ν**

### 16 *a* ^{2} *x* *−* 65 64 *a* ^{3} +

### (

### 4x ^{2} *−* 1

### 2 *ax* + 3a ^{2} 16

### )

*y* + (3x *−* 3

### 4 *a)y* ^{2} + *y* ^{3} , two or more exponents of *{* 1, 0, 2 + *α,* 1 + *α, α,* 1 + 2α, 2α, 3α *}* are equal to each other and maximized when *α* = *−* 1. Thus, *g* _{2} (x, y; *−* 1) =

*−* _{16} ^{1} *a* ^{2} + 4x ^{2} *y, c* = ^{a} _{64}

^{a}

^{2}

*, d* 2 (x) = _{64x} ^{a}

^{a}

^{2}

### . Therefore *f* 2 (x, _{64x} ^{a}

^{a}

^{2}

### ) = *−* ^{131a} _{128}

^{3}

### +

### 15a

^{4}

### 4096x *−* _{16384x} ^{3a}

^{5}2

### + _{262144x} ^{a}

^{a}

^{6}3

### = *o(x)* as *x* *→ ∞* . Put *y* 2 (x) = *x* *−* ^{a} _{4} + _{64x} ^{a}

^{a}

^{a}

^{2}

### . Then

*f* (x, y 2 (x)) = *f* 1 (x, *−* *a* 4 + *a* ^{2}

### 64x ) = *f* 2 (x, *a* ^{2}

### 64x ) = *o(x)* as *x* *→ ∞* *.* By Theorem 3.2, the infinite series

*x* *−* *a* 4 + *a* ^{2}

### 64x + 131a ^{3}

### 512x ^{2} + 509a ^{4} 16384x ^{3} *. . .*

### asymptotically converges to the implicit function of *y* ^{3} + *axy* + *x* ^{2} *y* *−* *a* ^{3} *−* 2x ^{3} = 0 as

*x* *→ ∞* *.*

**§** **4.** **The Newton diagram in** **De Methodis**

**§**

**De Methodis**

### The explanation of the literal resolution of aﬀected equations in *De Analysi* is insuﬃcient. Even if *f* (0, y) = 0 has a root *c, the initial quotient can not always be* found when _{∂y} ^{∂} *f* (0, c) = 0. Also, when *x* is large, we cannot obtain the terms of the maximum dimension unless we decide *α* of *y* *∼* *cx* ^{α} . In *De Methodis, Newton solved* these points by using the Newton diagram which he called the parallelogram.

_{∂y}

^{∂}

^{α}

### However, to make this rule still more evident, I thought it fitting to expound it in addition with the aid of the following diagram. Describing the right angle BAC, I divide its sides BA, AC into equal segments and from these raise normals distributing the space between the angle into equal squares or rectangles: these I conceive to be denominated by the powers of the variables *x* and *y, as you* see them entered in figure 1. Next, when some equation is proposed, I mark the rectangles corresponding to each of its terms with some sign and apply a ruler to two or maybe several of the rectangles so marked, one of which it to be the lowest in the left-hand column alongside AB, a second to the right touching the ruler, and all the rest not in contact with the ruler should lie above it. I then choose the terms of the equation which are marked out by the rectangles in contact with the ruler and thence seek the quantity to be added to the quotient.

### 0 *x* *x* ^{2} *x* ^{3} *x* ^{4}

*y* *xy* *x* ^{2} *y* *x* ^{3} *y* *x* ^{4} *y*

*y* ^{2} *xy* ^{2} *x* ^{2} *y* ^{2} *x* ^{3} *y* ^{2} *x* ^{4} *y* ^{2}

*y* ^{3} *xy* ^{3} *x* ^{2} *y* ^{3} *x* ^{3} *y* ^{3} *x* ^{4} *y* ^{3}

*y* ^{4} *xy* ^{4} *x* ^{2} *y* ^{4} *x* ^{3} *y* ^{4} *x* ^{4} *y* ^{4}

### A B

### C fig 1

### *

### *

### *

### *

### *

### *

*•*

*•* A *•*

### B

### C D

### fig 2 E So to extract the root *y* from

*y* ^{6} *−* 5xy ^{5} + (x ^{3} */a)y* ^{4} *−* 7a ^{2} *x* ^{2} *y* ^{2} + 6a ^{3} *x* ^{3} + *b* ^{2} *x* ^{4} = 0,

### I mark the rectangles answering to its terms with some sign *∗* , as you see done

### in the second illustration. I then apply the ruler DE to the lower corner of

### the places marked out in the left-hand column and make it swing to the right

### from bottom to top until in like fashion it begins to touch a second or maybe

### several together of the other marked places. Those so touched I see to be *x* ^{3} *, x* ^{2} *y* ^{2} and *y* ^{6} . Hence from the terms *y* ^{6} *−* 7a ^{2} *x* ^{2} *y* ^{2} + 6a ^{3} *x* ^{3} as through set equal to nothing (and in addition, if it places, reduced to *v* ^{6} *−* 7v ^{2} + 6 = 0 by supposing *y* = *v* *×* *√*

*ax) I seek the value of* *y* and find it to be fourfold, + *√*

*ax,* *−* *√*

*ax,* + *√*

### 2ax and *−* *√*

### 2ax. Any of these may be acceptable as an initial term in the quotient depending on whether the decision is made to extract one or other of the roots.

### MP III, pp.48-53 Modern expression of how to extract the root *y* from *f* (x, y) = 0 when *x* is close to 0 is as follows. Let *f* (x, y) = ∑ *n*

*j=0*

### ∑

*i* *a* _{i,j} *x* ^{i} *y* ^{j} = 0 be an algebraic equation. Each monomial *a* *i,j* *x* ^{i} *y* ^{j} is represented by a lattice point (j, i) in the *j* -i plane with *j* on the horizontal axis and *i* on the vertical axis. The set of all lattice points of *f(x, y) is* denoted by *L(f* ) = *{* (j, i) *|* *a* *i,j* *̸* = 0 *}* . Let *S(f) be the set of line segments whose end* points are the points of the set *L(f* ). Take the lowest point (0, m) of *L(f* ) and the line segment *ℓ* *∈* *S(f) with the equation*

_{i,j}

^{i}

^{j}

^{i}

^{j}

### (4.1) *ℓ* : *i* + *αj* = *m*

### such that ¯ *i* + *α* ¯ *j* *≥* *m* *∀* (¯ *j,* ¯ *i)* *∈* *L(f* ). The line segment *ℓ* corresponds with “the ruler”.

### Let (j 1 *, i* 1 ), . . . , (j *r* *, i* *r* ) be all points of *L(f* ) on *ℓ, then* *i* *k* + *αj* *k* = *m,* (k = 1, . . . , r). Put *g(x, y) =*

### ∑ *r* *k=1*

*a* _{i}

_{i}

_{k}_{j}

_{j}

_{k}*x* ^{i}

^{i}

^{k}*y* ^{j}

^{j}

^{k}*.*

### Let *c* be a root of *g(1, v) = 0. Take* *d(x) =* *cx* ^{α} as a quotient.

^{α}

### The *j-i* plane plotted the lattice points in *L(f) and drawn the lowest line segment* is called the Newton diagram of *f* (x, y) = ∑ *n*

*j=0*

### ∑

*i* *a* *i,j* *x* ^{i} *y* ^{j} = 0, and the above method is called the Newton diagram method. At the time of illustration, the lattice points on the line segment are represented by black circles *•* , and the lattice points outside the line segment are represented by white circles *◦* .

^{i}

^{j}

**Example 4.1.** We apply the Newton diagram method to Newton’s example *f* (x, y) = *y* ^{3} + *a* ^{2} *y* *−* 2a ^{3} + *axy* *−* *x* ^{3} = 0, when *x* is close to 0.

**ν** **= 0** The Newton diagram of *f* 0 (x, y) = *y* ^{3} + *a* ^{2} *y* *−* 2a ^{3} + *axy* *−* *x* ^{3} = 0 is as below.

**ν**

*j* *i*

*◦*

*◦*

*• •* *•*

### The lowest line is *ℓ* : *i* = 0; thus, *α* = 0 and *g(x, y) =* *y* ^{3} +a ^{2} *y* *−* 2a ^{3} . Thus, the first term of the quotient is *y* = *a, which is the root of* *y* ^{3} +a ^{2} *y* *−* 2a ^{3} = 0.

**ν** **= 1** The Newton diagram of *f* 1 (x, y) = *f* (x, a + *y) =* *a* ^{2} *x* *−* *x* ^{3} + (4a ^{2} + *ax)y* + 3ay ^{2} + *y* ^{3} is as below.

**ν**

*j* *i*

*◦*

*◦*

*◦ ◦*

*•*

*•*

### The lowest line is *ℓ* : *i* + *j* = 1; thus, *α* = 1 and *g(x, y) =* *a* ^{2} *x* + 4a ^{2} *y.*

### Thus, the second term of the quotient is *y* = *−* ^{x} _{4} which is the root of *a* ^{2} *x* + 4a ^{2} *y* = 0.

^{x}

### We formulate the Newton diagram method in *De Methodis* as an algorithm similar to Algorithm 3.1. Therefore, the order of a function

*f(x) =* ∑

*i*

*a* *i* *x* ^{q}

^{q}

^{i}*,* *q* *i* *∈* Q *,* is defined by

### ord *f* = min *{* *q* _{i} *|* *a* _{i} *̸* = 0 *}* *.*

_{i}

_{i}

**Algorithm 4.2.** (The asymptotic expansion as **x** **→** **0.) For an algebraic** equation

**x**

**→**

*f(x, y) =*

### ∑ *n* *j=0*

### ( _{m}

_{m}

### ∑

*i=0*

*a* _{i,j} *x* ^{i} )

_{i,j}

^{i}

*y* ^{j} = 0, put *f* 0 (x, y) = *f* (x, y).

^{j}

### (i) Find a rational number *α* 0 such that there are two or more terms in *g* 0 (x, y; *α* 0 ) = ∑

*i+α*

0*j=ord* *f*

0### (x,0)

*a* ^{(0)} _{i,j} *x* ^{i} *y* ^{j} *.*

_{i,j}

^{i}

^{j}

### (ii) Take a root *v* = *c* 0 of the equation *g* 0 (1, v; *α* 0 ) = 0.

### (iii) Put *d* _{0} (x) = *c* _{0} *x* ^{α}

^{α}

^{0}

### .

### (iv) Repeat (1), (2), (3) and (4) below for *ν* = 1, 2, . . . , N :

### (1) calculate *f* _{ν} (x, y) = *f* _{ν}

_{ν}

_{ν}

_{−}_{1} (x, d _{ν}

_{ν}

_{−}_{1} (x) + *y), say* *f* *ν* (x, y) =

### ∑ *n* *j=0*

### ( ∑

*i*

*a* ^{(ν)} _{i,j} *x* ^{q}

_{i,j}

^{q}

^{(ν)}

^{i,j}### )

*y* ^{j} *,*

^{j}

### (2) find a rational number *α* *ν* such that there are two or more terms in *g* *ν* (x, y; *α* *ν* ) = ∑

*q*

_{i,j}^{(ν)}

### +α

*ν*

*j=ord* *f*

*ν*

### (x,0)

*a* ^{(ν)} _{i,j} *x* ^{q}

_{i,j}

^{q}

^{(ν)}

^{i,j}*y* ^{j} *,*

^{j}

### (3) take a root *v* = *c* *ν* of the equation *g* *ν* (1, v; *α* *ν* ) = 0, (4) put *d* *ν* (x) = *c* *ν* *x* ^{α}

^{α}

^{ν}### .

### Then, the function *y* *N* (x) = ∑ *N*

*ν=0* *d* *ν* (x) satisfies

*f* (x, y *N* (x)) = *o(x* ^{ord} ^{f}

^{f}

^{N}^{(x,0)} ) as *x* *→* 0.

**Example 4.3.** We apply Algorithm 4.2 to Newton’s example *f* (x, y) = *y* ^{3} + *a* ^{2} *y* *−* 2a ^{3} + *axy* *−* *x* ^{3} = 0.

**ν** **= 0** Two or more exponents of *{* 3α, α, 0, 1 + *α,* 3 *}* are equal to ord *f* (x, 0) = 0 and minimized when *α* = 0. Thus, *g* 0 (x, y; 0) = *y* ^{3} + *a* ^{2} *y* *−* 2a ^{3} . The root of *g* _{0} (x, y; 0) = 0 is *y* = *a,* and thus, *d* _{0} (x) = *a.* Therefore, *f* (x, a) = *a* ^{2} *x* *−* *x* ^{3} = *o(1)* as *x* *→* 0.

**ν**

**ν** **= 1** Since *f* _{1} (x, y) = *f* (x, a + *y) =* *a* ^{2} *x* *−* *x* ^{3} + (4a ^{2} + *ax)y* + 3ay ^{2} + *y* ^{3} , two or more exponents of *{* 1, 3, α, 1 + *α,* 2α, 3α *}* are equal to ord *f* 1 (x, 0) = 1 and minimized when *α* = 1. Thus, *g* 1 (x, y; 1) = *a* ^{2} *x* + 4a ^{2} *y, and* *d* 1 (x) = *−* ^{1} _{4} *x.*

**ν**

### Therefore, *f* 1 (x, *−* ^{1} _{4} *x) =* *−* _{16} ^{1} *ax* ^{2} *−* ^{65} _{64} *x* ^{3} = *o(x)* as *x* *→* 0.

**ν** **= 2** Thereafter, it can be executed in the same way.

**ν**

### Krantz and Parks [6, pp.15-20] determine *α* by diﬀerent method from Algorithm 4.2. They assume *y(0) = 0 and* *y(x) =* *x* ^{α} *y(x) with ˜* ˜ *y(x) a continuous function that does* not vanish when *x* = 0. They substitute *y* = *x* ^{α} *y(x) in* ˜ *f(x, y) = 0, and explain “To be* able to determine ˜ *y(0) from”* *f* (x, x ^{α} *y(x)) = 0, “there must be two or more monomials* ˜ in” *f(x, x* ^{α} *y(x)) = 0 “which have the same power of* ˜ *x* and all other monomials must have a large power of *x.”*

^{α}

^{α}

^{α}

^{α}

**§** **5.** **An improved algorithm as** *x* *→* 0

**§**

### By rewriting Algorithm 3.1 dually, even if (0, c) is a singular point, infinite series

### expansion (Puiseux expansion) of one of the branches can be obtained.

**Algorithm 5.1.** (An improved algorithm as *x* *→* 0.) For an algebraic equa- tion

*f(x, y) =*

### ∑ *n* *j=0*

### ( _{m}

_{m}

### ∑

*i=0*

*a* _{i,j} *x* ^{i} )

_{i,j}

^{i}

*y* ^{j} = 0, put *f* _{0} (x, y) = *f* (x, y).

^{j}

### (i) Find a rational number *α* _{0} such that there are two or more terms in *g* 0 (x, y; *α* 0 ) = ∑

*i+α*

0*j=min* *P* (f

0*,α*

0### )

*a* ^{(0)} _{i,j} *x* ^{i} *y* ^{j} *.*

_{i,j}

^{i}

^{j}

### (ii) Take a root *v* = *c* _{0} of the equation *g* _{0} (1, v; *α* _{0} ) = 0.

### (iii) Put *d* 0 (x) = *c* 0 *x* ^{α}

^{α}

^{0}

### .

### (iv) Repeat (1), (2), (3) and (4) below for *ν* = 1, 2, . . . , N : (1) calculate *f* _{ν} (x, y) = *f* _{ν−1} (x, d _{ν−1} (x) + *y), say*

_{ν}

_{ν−1}

_{ν−1}

*f* _{ν} (x, y) =

_{ν}

### ∑ *n* *j=0*

### ( ∑

*i*

*a* ^{(ν)} _{i,j} *x* ^{q}

_{i,j}

^{q}

^{(ν)}

^{i,j}### )

*y* ^{j} *,*

^{j}

### (2) find a rational number *α* *ν* with min *P* (f *ν* *, α* *ν* ) *>* min *P* (f *ν*

*−*

### 1 *, α* *ν*

*−*

### 1 ) such that there are two or more terms in

*g* *ν* (x, y; *α* *ν* ) = ∑

*q*

_{i,j}^{(ν)}

### +α

*ν*

*j=min* *P* (f

*ν*

*,α*

*ν*

### )

*a* ^{(ν)} _{i,j} *x* ^{q}

_{i,j}

^{q}

^{(ν)}

^{i,j}*y* ^{j} *,*

^{j}

### (3) take a root *v* = *c* *ν* of the equation *g* *ν* (1, v; *α* *ν* ) = 0, (4) put *d* *ν* (x) = *c* *ν* *x* ^{α}

^{α}

^{ν}### .

### Then, the function *y* *N* (x) = ∑ *N*

*ν=0* *d* *ν* (x) satisfies

*f(x, y* *N* (x)) = *o(x* ^{min} ^{P} ^{(f}

^{P}

^{N}^{,α}

^{,α}

^{N}^{)} ) as *x* *→* 0.

### The Newton polygon of an algebraic equation *f* (x, y) = ∑

*i,j*

*a* *i,j* *x* ^{i} *y* ^{j} = 0

^{i}

^{j}

### is defined by the convex hull of the set *{* (j + *x, i* + *y)* *|* *a* _{i,j} *̸* = 0; *x, y* *∈* R ^{+} *}}* in the *j-i*

_{i,j}