NOTE III.
ON THE NUMERICAL SOLUTION OF EQUATIONS
_______________
Continued from N03A numerical.doc
/403
are two quantities of contrary signs, and that the root $a$ is found bound between the two
limits
,   2 .
$\xi ,$ $\xi + 2\alpha .$
-- new text starts here -As for the second part of the proposition stated above, it is an immediate
consequence of scholium II, because the quantity $G$ evidently lies below, making an
abstraction of the sign, polynomial (62), that is to say of the expansion of F1   z  $F_1
\left( {\xi + z} \right)$, as long as the numerical value of $z$ does not exceed that of
$2\alpha$, and consequently is less than the quantity F1 a , which we deduce from
F1   z  $F_1 \left( {\xi + z} \right)$ by putting
z  a
$z = a - \xi $.
Thus it follows from this second part that the real roots greater than $a$ are all greater
than the limit
(63)
a
G
2k m1
$a + \frac{G}{{\left( {2k} \right)^{m - 1} }}$
and the roots smaller than $a$ are less than the limit
(64)
a
G
2k m1
$a - \frac{G}{{\left( {2k} \right)^{m - 1} }}$.
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Cours d'Analyse – Note III – Numerical solutions
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Problem III. – To find values approaching as closely as we might want the real
roots of equation (27).
Solution. – We will begin by determining, with the aid of the preceding problem,
two limits, one greater than and one less than, each real positive root. Suppose in
particular that the root $a$ is of this kind, and denote by $x_0$, $X$ the two limits,
below and above this root. If we form two different sums, the first with the positive
terms of the polynomial $F(x)$, the second with the negative terms taken with the
contrary sign, the first of which will be at its smallest for $x=x_0$ and will become its
largest for $x=X$. Represent the first by  x  $\varphi \left( x \right)$ and the second
by  x  $\chi \left( x \right)$. The two integer functions  x   x  $\varphi \left( x
\right)$, $\chi \left( x \right)$ enjoy the properties stated in theorems II and III, and
consequently, if the function  x  $\varphi \left( x \right)$ is such that we can easily
solve equations of the form
 x   const.
$\varphi \left( x \right) = {\rm{const}}.$,
then formulas (7) and (16) will immediately furnish values /404 above and approaching
the root $a$. This is what will happen, for example, whenever the function  x 
presents itself in the form
n
B x  C   D
$B\left( {x + C} \right)^n + D$
$B$, $C$, $D$ being any three whole numbers, and $n$ a whole number less than or
equal to $m$, because then we will obtain the successive terms of series (6) and (15) by
the extraction of roots of degree $n$. If the function  x $\varphi \left( x \right)$ is not
of the form that we have just indicated, we can easily put it into that form [literally, easily
bring it back to that] by adding to both sides of the equation
 x    x 
$\varphi \left( x \right) = \chi \left( x \right)$
an integer polynomial  x  $\psi \left( x \right)$ where the terms are positive. Indeed, it
is clear that the values of  x $\varphi \left( x \right)$ and of  x  $\chi \left( x
\right)$, modified by the addition of such a polynomial, will conserve all of the same
properties. We can, [au reste = with the remainder?] assign to the polynomial  x  $\psi
\left( x \right)$ an infinity of different values. Suppose, for example, that
 x   x 3  3x 2  8
$\varphi \left( x \right) = x^3 + 3x^2 + 8$.
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The value of  x  $\varphi \left( x \right)$ modified by the addition of the polynomial
 x  $\psi \left( x \right)$ will become
x  13  7,
$\left( {x + 1} \right)^3 + 7,$
if we suppose
 x   3x,
$\psi \left( x \right) = 3x,$
or as well,
x  23
$\left( {x + 2} \right)^3 $
if we suppose
 x   3x 2  12x,
$\psi \left( x \right) = 3x^2 + 12x,$
etc. It is good to remark on this subject, first, that we can always choose the integer
function  x  $\psi \left( x \right)$ so that the number $B$ will be unity, and second,
that in many cases, one of the numbers $C$ or $D$ can be reduced to zero.
After determining by the preceding method the real positive roots of equation
(27), it will evidently be sufficient to obtain the negative roots as well by seeking by the
same method the positive roots of the equation
(65)
$F\left( { - x} \right) = 0.$
F x   0
Scholium. – There exist several methods of approximation other than the one we
have just described, among which we must mention that of Newton. It supposes that we
already know a value approaching  $\xi $, the /405 root that we are seeking, and it
consists of taking as a correction to this value the quantity $\alpha$ determined by the
equation
(55)1
1

F  
.
F1  
This formula is not out of sequence. See the bottom of page 400.
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$\alpha = - \frac{{F\left( \xi \right)}}{{F_1 \left( \xi \right)}}.$
However, because this last method is not always applicable, it is important to examine in
what cases it can be used. We are going to establish this in the following propositions:
Theorem IV. – Let $a$ denote any one of the real roots, positive or negative, of
equation (27) and that  $\xi $ is a value approaching this root. [note recasting of
sentence.] Suppose that we determine  $\alpha $ by means of equation (55). If
$\alpha$ is small enough, making abstraction of the sign, that in polynomial (56) tne
numerical value of the first term exceeds the sum of the numerical values of all the
others. Then the two quantities
  
$\xi $, $\xi + \alpha $,
the second will be closer to $a$ than the first.
Proof. – We have already seen (problem II, scholium IV) that, under the given
hypotheses, the root $a$ is the only root between the limits
  
$\xi $, $\xi + \alpha $.
This said, if we take
az
(66)
$a = \xi + z$,
then $z$ will be a quantity contained between the limits $0$, $2\alpha$, and will satisfy
the equation
F   z   0
$F\left( {\xi + z} \right) = 0$
or, what amounts to the same thing,
F    zF1    z 2 F2 x   K  0.
(67)
$F\left( \xi \right) + zF_1 \left( \xi \right) + z^2 F_2 \left( x \right) + \ldots = 0.$
For convenience, if we let
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Cours d'Analyse – Note III – Numerical solutions
4
F2    zF3  
F1  
$q = - \frac{{F_2 \left( \xi \right) + zF_3 \left( \xi \right)}}{{F_1 \left( \xi \right)}}$
q
(68)
and consider formula (55), then equation (67) becomes
(69)
$z = \alpha + qz^2 $.
z    qz 2
/406
Consequently, we get
(70)
    z      qz 2 ,
$\alpha = \xi + z = \xi + \alpha + qz^2 ,$
from which it follows that, by taking    $\xi + \alpha $ in place of $\xi$ for the value
approaching $a$, we will commit an error equal to, but not greater than the numerical
value of $z$, but rather than that of qz 2 . [Cauchy seems awkward here. How much
should we untangle it? This may be a popular passage, since everyone will want to look
at Cauchy's analysis of Newton's method.] Moreover, because polynomial (56) has the
same sign as its first term F1   $F_1 \left( \xi \right)$, the two polynomials

 F1    2 2 F2    2 2 zF3    K  1  4 q F1  


 F1    2 2 F2    2 2 zF3    K  1  4 q F1  
$\left\{ \begin{array}{l}
F_1 \left( \xi \right) + 2\left( {2\alpha } \right)F_2 \left( \xi \right) + 2\left( {2\alpha }
\right)zF_3 \left( \xi \right) + \ldots = \left( {1 - 4\alpha q} \right)F_1 \left( \xi \right) \\
F_1 \left( \xi \right) - 2\left( {2\alpha } \right)F_2 \left( \xi \right) - 2\left( {2\alpha }
\right)zF_3 \left( \xi \right) - \ldots = \left( {1 + 4\alpha q} \right)F_1 \left( \xi \right) \\
\end{array} \right.$
(71)
evidently enjoy the same property. [What is required is that ?] the numerical value of
2 q $2\alpha q$ and a fortiori that of $qz$ remains less than 12 ${\textstyle{1 \over
2}}$. We conclude immediately that the numerical value of qz 2 $qz^2 $ is less than that
of 12 z ${\textstyle{1 \over 2}}z$. Thus, the two errors which we make by taking
  
$\xi $, $\xi + \alpha $,
as values approaching $a$, the second [error] is smaller than half of the first.
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Cours d'Analyse – Note III – Numerical solutions
5
Scholium I. – Because we draw from equation (69)
z

1  qz
$z = \frac{\alpha }{{1 - qz}}$
and because the numerical value of $qz$ is less than 12 ${\textstyle{1 \over 2}}$, we are
certain that the value of $z$ will always remain between the limits
2
3
 2
${\textstyle{2 \over 3}}\alpha $, $2\alpha $.
Scholium II. – By solving equation (69) as if the value of $q$ were known, we
find
1  1  4 q
2

.
2q
1 m 1  4 q
$z = \frac{{1 \pm \sqrt {1 - 4\alpha q} }}{{2q}} = \frac{{2\alpha }}{{1 \mp \sqrt {1 4\alpha q} }}.$
z
The radical 1 4 q $\sqrt {1 - 4\alpha q} $ is here given a double sign. But because
the value of $z$ ought to be smaller than that of $2\alpha$, it is clear that we ought to
prefer the inferior sign.2 Thus we will have
2
1  1  4 q
$z = \frac{{2\alpha }}{{1 + \sqrt {1 - 4\alpha q} }}$.
(72)
z
/407
This said, if we call $q_0$, $Q$ two limits, the first less than and the other greater than
the quantity $q$ determined by formula (68), we will conclude from equation (72) that
theexact value of $z$ is contained between the two expressions
2
2
,
1  1  4 q0
1  1  4Q
$\frac{{2\alpha }}{{1 + \sqrt {1 - 4\alpha q_0 } }},$
$\frac{{2\alpha }}{{1 + \sqrt {1 - 4\alpha Q} }}$.
(73)
2
This is an odd choice of words. From what he writes, the "inferior sign" is apparently
the sign on the bottom of the symbol $ \mp $.
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Cours d'Analyse – Note III – Numerical solutions
6
[Consequently, this value will contain all the decimal digits common to the two
expressions reduced to numbers.]
Scholium III. – Suppose that of the two quantities $q_0$, $Q$, the second has the
larger numerical value, and that this numerical value is less than unity. Then, if the
difference a    z $a = \xi - z$ is, making abstraction of the sign, smaller than the unit
decimal of order $n$, that is to say if we have
n
 1
(74)
val. num.z    ,
 10 
${\rm{val}}{\rm{.}}\;{\rm{num}}{\rm{.}}\,z < \left( {\frac{1}{{10}}} \right)^n ,$
then the difference
a       qz 2
$a - \left( {\xi + \alpha } \right) = qz^2 $
will be smaller, making abstraction of the sign, than a unit decimal of order $2n$.
Indeed, we will find that
2n
 1
(75)
val. num.qz   
 10 
${\rm{val}}{\rm{.}}\;{\rm{num}}{\rm{.}}\,qz^2 < \left( {\frac{1}{{10}}}
\right)^{2n} $
2
And so, by taking    $\xi + \alpha $ in place of $\xi$ for the value approaching the
root $a$, we will double the number of exact decimals.
If we suppose that the numerical value of $Q$ is less, not only than $1$ but also
than $0.1$, we will conclue from formula (74) that
2n1
 1
val. num.qz   
 10 
${\rm{val}}{\rm{.}}\;{\rm{num}}{\rm{.}}\,qz^2 < \left( {\frac{1}{{10}}} \right)^{2n
+ 1} $.
2
r
 1
More generally, if we suppose that this value is less than   $\left( {\frac{1}{{10}}}
 10 
\right)^r $, $r$ denoting any whole number, then formula (74) implies the following
(76)
7/29/17
 1
val. num.qz   
 10 
2n r
2
Cours d'Analyse – Note III – Numerical solutions
7
${\rm{val}}{\rm{.}}\;{\rm{num}}{\rm{.}}\,qz^2 < \left( {\frac{1}{{10}}} \right)^{2n
+ r} $.
Finally, if the value if $Q$ is greater than uity, but less than 10  $\left( {10} \right)^r $,
we /408 will find
r
2n r
 1
val. num.qz 2   
.
 10 
$val.\;num.\,qz^2 < \left( {\frac{1}{{10}}} \right)^{2n - r} .$
(77)
Scholion IV. – The error that we make by taking    $\xi + \alpha $ as a value
approaching $a$, or the numerical value of the product qz 2 $qz^2 $, can itself be
calculated by approximation. Indeed, if we consider equation (69), we will find


qz 2  q   qz 2  q 2  2 q2 z 2  q 3z 4 .
$qz^2 = q\left( {\alpha + qz^2 } \right)^2 = q\alpha ^2 + \left( {2\alpha } \right)q^2
z^2 + q^3 z^4 .$
2
But suppose that the numerical value of $2\alpha$, and consequently that of $z$, is less
n
 1
than   $\left( {\frac{1}{{10}}} \right)^n $, and that the numerical value of $Q$, and
 10 
consequently that of $q$ is less than 10  $\left( {10} \right)^{ \mp r} $, $n$ and $r$
denoting two whole numbers. We will evidently have
mr
3n  2r
 1
val. num.2 q 2 z 2   
 10 
${\rm{val}}{\rm{.}}\;{\rm{num}}{\rm{.}}\,\left( {2\alpha } \right)q^2 z^2 < \left(
{\frac{1}{{10}}} \right)^{3n \pm 2r} $
and
4 n  3r
 1
val. num.q z   
.
 10 
${\rm{val}}{\rm{.}}\;{\rm{num}}{\rm{.}}\,q^3 z^4 < \left( {\frac{1}{{10}}}
\right)^{4n \pm 3r} .$
3 4
Moreover, if the numerical value of the fraction
(78)
7/29/17
F3    zF4    K
F1  
Cours d'Analyse – Note III – Numerical solutions
8
$\frac{{F_3 \left( \xi \right) + zF_4 \left( \xi \right) + \ldots }}{{F_1 \left( \xi
\right)}}$
is known to be less than 10  $\left( {10} \right)^{ \mp s} $, $s$ denoting again a
whole number, we can take
ms
F2  
F1  
$ - \alpha ^2 \frac{{F_2 \left( \xi \right)}}{{F_1 \left( \xi \right)}}$
 2
for a value approaching the term q 2 $q\alpha ^2 $, without fearing an error more
considerable than
3n  s
 1
 
10
$\left( {\frac{1}{{10}}} \right)^{3n \pm s} $.
F2  
$\xi + \alpha - \alpha ^2 \frac{{F_2 \left(
F1  
\xi \right)}}{{F_1 \left( \xi \right)}}$ in place of    $\xi + \alpha $ for a value
approaching the root $a$, that is to say if we put
Consequently, if we choose      2
F2  
,
F1  
$a = \xi + \alpha - \alpha ^2 \frac{{F_2 \left( \xi \right)}}{{F_1 \left( \xi \right)}},$
a      2
(79)
the error committed upon the root will not suffer more than the decimal units of /409 the
order marked by the largest of the three numbers [awkward, but eloquent]
3n  s
3n  2r
4n  3r
$3n \pm s$,
$3n \pm 2r$,
$4n \pm 3r$.
r
 1
In the particular case where the numerical value of $Q$ is less than   $\left(
 10 
s
 1
{\frac{1}{{10}}} \right)^r $, and that of fraction (78) is less than   $\left(
 10 
{\frac{1}{{10}}} \right)^s $, the new error becomes less than
7/29/17
Cours d'Analyse – Note III – Numerical solutions
9
 1
 
10
3n
$\left( {\frac{1}{{10}}} \right)^{3n} $.
It suffices to substitute the second part of equation (79) for the quantity  $\xi $ to triple
the number of exact decimal digits in the value approaching $a$. This is where we now
arrive, at a very small price, when the number $n$ becomes very considerable. These
results agree with those which M. Nicholson has obtained in a work recently published
in London, which has for its title Essay on involution and evolution, etc.3
Theorem V. – The same things being given as in the preceding theorem, imagine
that the first term of polynomial (56), that is to say the polynomial that represents the
expansion of F1   2  $F_1 \left( {\xi + 2\alpha } \right)$, has a numerical value
greater, not only than the sum of the numerical values of all the other terms, but also
more than the double of this sum. Then if we denote by 1 $\xi _1 $ a quantity contained
between the limits
   2
$\xi $, $\xi + 2\alpha $,
the second of the two quantities
1 
1
F 1 
F1 1 
$\xi _1 $,
$\xi _1 - \frac{{F\left( {\xi _1 } \right)}}{{F_1 \left( {\xi _1 } \right)}}$
will always be closer to $a$ than the first.
Proof. – To establish the proposition that we have just stated, it suffices to
observe that the numerical value of the difference
3
Peter Nicholson (1765-1844) published about April 1820. I've downloaded an article
about Nicholson from JSTOR, but have not been able to find an exact reference to the
original source.
Mr Peter Nicholson, the Practical Builder and Mathematician
E. Keith Lloyd
The Mathematical Gazette, Vol. 66, No. 437 (Oct., 1982), pp. 203-207
doi:10.2307/3616546
This article consists of 5 page(s).
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Cours d'Analyse – Note III – Numerical solutions
10
a  1
$a - \xi _1 $
is greater than that of

F 1  
F a   F 1 
a  1 
  a  1  
F1 1 
F1 1 

$a - \left[ {\xi _1 - \frac{{F\left( {\xi _1 } \right)}}{{F_1 \left( {\xi _1 } \right)}}}
\right] = \left( {a - \xi _1 } \right) - \frac{{F\left( a \right) - F\left( {\xi _1 }
\right)}}{{F_1 \left( {\xi _1 } \right)}}$,
or, what amounts to the same thing, that the fraction
F a   F 1 
a  1
F1 1 
$\frac{{F_1 \left( {\xi _1 } \right) - \frac{{F\left( a \right) - F\left( {\xi _1 } \right)}}{{a \xi _1 }}}}{{F_1 \left( {\xi _1 } \right)}}$
F1 1  
/410
has a numerical value less than unity. Let us represent this fraction by
u
$\frac{u}{v}$.
v
It will be enough to prove that
$v-u$ and $v+u$,
That is to say in other words,
(80)
F a   F 1 
2F1 1  
F a   F 1 
a  1
a  1
$\frac{{F\left( a \right) - F\left( {\xi _1 } \right)}}{{a - \xi _1 }}$
and
$2F_1 \left( {\xi _1 } \right) - \frac{{F\left( a \right) - F\left( {\xi _1 } \right)}}{{a - \xi
_1 }}$
are two expressions with the same sign. But, if we make
(81)
$a = \xi + z$
and
7/29/17
az
1    
Cours d'Analyse – Note III – Numerical solutions
11
$\xi _1 = \xi + \beta $,
$z$ and $\beta$ being two quantities of the same sign contained between the limits $0$,
$2\alpha$, and expressions (80), after the expansions of the functions
F 1  z 
$F\left( {\xi _1 + z} \right)$,
$F\left( {\xi _1 + \beta } \right)$,
$F_1 \left( {\xi + \beta } \right)$,
F 1   
F1    
become, respectively,


F1      z F2     2   z  z 2 F3   K ,
$F_1 \left( \xi \right) + \left( {\beta + z} \right)F_2 \left( \xi \right) + \left( {\beta ^2 +
\beta z + z^2 } \right)F_3 \left( \xi \right) + \ldots ,$
F1      z  4  F2     2   z  z 2  6 2 F3   K .
$F_1 \left( \xi \right) - \left( {\beta + z - 4\beta } \right)F_2 \left( \xi \right) - \left(
{\beta ^2 + \beta z + z^2 - 6\beta ^2 } \right)F_3 \left( \xi \right) + \ldots .$


Because in each of these last polynomials the coefficient of Fn   $F_n \left( \xi \right)$
has a numerical value evidently less than one of the quantities
nz n 1
2n n1
$nz^{n - 1} $,
$2n\beta ^{n - 1} $,
and consequently less than the double of the numerical value of the product
n 2 
n1
$n\left( {2\alpha } \right)^{n - 1} $,
it is clear that they will both be of the same sign as F1   $F_1 \left( \xi \right)$ if the
condition stated in theorem V finds itself satisfied. Thus, etc.
Scholium I. – The errors committed when we take successively
1
1 
F 1 
F1 1 
$\xi _1 $ and $\xi _1 - \frac{{F\left( {\xi _1 } \right)}}{{F_1 \left( {\xi _1 } \right)}}$
for values approaching the root $a$ are respectfully equal to the numerical values of the
two quantities
7/29/17
Cours d'Analyse – Note III – Numerical solutions
12
a  1 a  1 
F 1 
F1 1 
$a - \xi _1 $ and $a - \xi _1 + \frac{{F\left( {\xi _1 } \right)}}{{F_1 \left( {\xi _1 }
\right)}}$.
/411
We will find also, considering formulas (81),
a  1  z  
(82)
$a - \xi _1 = z - \beta $
and
F 1 
F a   F 1 
F   z   F    
F1 1 
F1 1 
F1    
$a - \xi _1 + \frac{{F\left( {\xi _1 } \right)}}{{F_1 \left( {\xi _1 } \right)}} = a - \xi _1 \frac{{F\left( a \right) - F\left( {\xi _1 } \right)}}{{F_1 \left( {\xi _1 } \right)}} = z \beta - \frac{{F\left( {\xi + z} \right) - F\left( {\xi + \beta } \right)}}{{F_1 \left( {\xi +
\beta } \right)}}$
a  1 
 a  1 
 z
then, by expanding the functions F   z  F     F1    
$F\left( {\xi + z} \right)$, $F\left( {\xi + \beta } \right)$, $F_1 \left( {\xi + \beta }
\right)$,

F 1 
a  1 
F1 1 

(83)

2
2
F4    K
2 F2    z  2  F3    z  2  z  3



z





F1    2  F2    3 2 F3    K

$\left\{ \begin{array}{l}
a - \xi _1 + \frac{{F\left( {\xi _1 } \right)}}{{F_1 \left( {\xi _1 } \right)}} \\
\quad = - \left( {z - \beta } \right)^2 \frac{{F_2 \left( \xi \right) + \left( {z + 2\beta }
\right)F_3 \left( \xi \right) + \left( {z^2 + 2\beta z + 3\beta ^2 } \right)F_4 \left( \xi
\right) + \ldots }}{{F_1 \left( \xi \right) + 2\beta F_2 \left( \xi \right) + 3\beta ^2 F_3
\left( \xi \right) + \ldots }} \\
\end{array} \right.$.


This said, imagine that, for all values of $\beta$ and $z$ contained between $0$ and
$2\alpha$, the numerical value of the polynomial
(84)
7/29/17


F2    z  2 F3    z 2  2 z  3 2 F4   K
Cours d'Analyse – Note III – Numerical solutions
13
$F_2 \left( \xi \right) + \left( {z + 2\beta } \right)F_3 \left( \xi \right) + \left( {z^2 +
2\beta z + 3\beta ^2 } \right)F_4 \left( \xi \right) + \ldots $
is always less than the limit $M$, and that of the polynomial
F1    2  F2    3 2 F3    K
(85)
$F_1 \left( \xi \right) + 2\beta F_2 \left( \xi \right) + 3\beta ^2 F_3 \left( \xi \right) +
\ldots $
is always greater than the limit $M$. If we have
n
 1
val. num. z      
 10 
${\rm{val}}{\rm{.}}\;{\rm{num}}{\rm{.}}\;\left( {z - \beta } \right) < \left(
{\frac{1}{{10}}} \right)^n $
(86)
and
M
mr
 10  ,
N
$\frac{M}{N} < \left( {10} \right)^{ \mp r} ,$
(87)
$n$ and $r$ denoting any two whole numbers, we will conclude from equation (83) that
2n mr

F 1    1 
val. num.  a  1 

.
  
F

10


1
1


${\rm{val}}{\rm{.}}\;{\rm{num}}{\rm{.}}\;\left[ {a - \xi _1 + \frac{{F\left( {\xi _1 }
\right)}}{{F_1 \left( {\xi _1 } \right)}}} \right] < \left( {\frac{1}{{10}}} \right)^{2n \mp
r} .$
(88)
It is essential to remark that to obtain convenient [convenables = appropriate?] values of
$M$ and $N$, it suffices, first, to replace $z$ and $\beta$ for $2\alpha$ in polynomial
(84), then to calculate the sum of the numerical values of all the terms, and second, to
replac $\beta$ by $2\alpha$ in polynomial (85) and then to look for the difference
between the numerical value of the first term and the sum of the numerical values of all
the others.
\412
Scholium II. – The same things being assumed as in theorem V, if we successively
make
(89)
7/29/17
1   
F  
,
F1  
2  1 
F 1 
,
F1 1 
3  2 
F 2 
,
F1 2 
Cours d'Analyse – Note III – Numerical solutions
14
$\xi _1 = \xi - \frac{{F\left( \xi \right)}}{{F_1 \left( \xi \right)}},$
$\xi _2 = \xi _1 - \frac{{F\left( {\xi _1 } \right)}}{{F_1 \left( {\xi _1 } \right)}},$
$\xi _3 = \xi _2 - \frac{{F\left( {\xi _2 } \right)}}{{F_1 \left( {\xi _2 } \right)}},$
$\ldots$,
the quantities 1, 2 , 3 , $\xi _1 ,$ $\xi _2 ,$ $\xi _3 ,$, $\ldots$ will be values that are
closer and closer to the root $a$. Moreover, if we give $M$ and $N$ the same values as
in scholium I, then by supposing
n
 1
val. num. a       ,
 10 
${\rm{val}}{\rm{.}}\;{\rm{num}}{\rm{.}}\;\left( {a - \xi } \right) < \left(
{\frac{1}{{10}}} \right)^n ,$
we will then conclude that
2n  r
 1
val. num. a  1    
,
 10 
${\rm{val}}{\rm{.}}\;{\rm{num}}{\rm{.}}\;\left( {a - \xi _1 } \right) < \left(
{\frac{1}{{10}}} \right)^{2n \pm r} ,$
4 n  3r
 1
val. num. a  2    
,
 10 
${\rm{val}}{\rm{.}}\;{\rm{num}}{\rm{.}}\;\left( {a - \xi _2 } \right) < \left(
{\frac{1}{{10}}} \right)^{4n \pm 3r} ,$
8n  7r
 1
val. num. a  3    
,
 10 
${\rm{val}}{\rm{.}}\;{\rm{num}}{\rm{.}}\;\left( {a - \xi _3 } \right) < \left(
{\frac{1}{{10}}} \right)^{8n \pm 7r} ,$
$\ldots$.
These last formulas contain the proposition stated by M. Fourier in the Bulletin de
la Société philomathique (read in May 1818), relative to the number of exact decimal
places furnished by each new operation of Newton's method.
M
$\frac{M}{N}$ is less than unity, we can take $r=0$
N
and consequently the successive differences between the root $a$ and its nearby values
Every time the fraction
, 1, 2 , 3 ,
$\xi ,$ $\xi _1 ,$ $\xi _2 ,$ $\xi _3 ,$, $\ldots$
7/29/17
Cours d'Analyse – Note III – Numerical solutions
15
are respectively less than the numbers
n
2n
4n
 1
 1
 1
  ,
  ,
  ,
10
10
10
$\left( {\frac{1}{{10}}} \right)^n ,$
$\left( {\frac{1}{{10}}} \right)^{2n} ,$
$\left( {\frac{1}{{10}}} \right)^{4n} ,$
$\left( {\frac{1}{{10}}} \right)^{8n} ,$
$\ldots$.
8n
 1
  ,
10
Thus we find that the number of exact decimal places at least doubles for each new
operation.
The preceding researches furnish several methods for solving numerical
equations. In order to give a better sense of the advantages that these methods present, I
will apply them to the two equations
(90)
$x^2 - 2x - 5 = 0$
x 2  2x  5  0
and
(91)
$x^3 - 7x + 7 = 0$
x 3  7x  7  0
/413
which Lagrange chose as his examples (Résolution des equations numeriques, Chap. IV)
and the first of which was earlier treated by Newton.
If we initially consider equation (90), we will find (theorem III, scholium II) that
it has a single positive root contained between the two limits
3
22  2
2  5  2.15K .
$\sqrt {2 \cdot 2} = 2$ and $\sqrt[3]{{2 \cdot 5}} = 2.15 \ldots .$4
Moreover, the positive value of $x$ that satisfies the equation
4
Cauchy writes his multiplication symbols at ground level, thus what he writes $2.5$, we
would write 2  5 $2 \cdot 5$ or 2  5 $2 \times 5$. In general, we try to preserve
Cauchy's notation, but in this case we defer to modern conventions in order to avoid
confusion.
7/29/17
Cours d'Analyse – Note III – Numerical solutions
16
2x  5  x 2
$2x + 5 = x^2 $
will satisfy (problem I, scholium IV) the conditon
2 5  2x  x2
$2\sqrt {5 \cdot 2x} < x^2 $,
or, what amounts to the same thing, the following:
1
x  40 5  2.09K .
$x > \left( {40} \right)^{\frac{1}{5}} = 2.09 \ldots .$
The root under consideration is thus contained between the numbers $2.09\ldots$ and
$2.15\ldots$; indeed, its value, to less than a tenth part, will be $2.1$. To obtain a more
exact value, we will observe that in the present case,
F x   x 3  2x  5, F1 x   3x 2  2,
$F\left( x \right) = x^3 - 2x - 5,$
$F_1 \left( x \right) = 3x^2 - 2,$
$F_2 \left( x \right) = 3x,$
$F_3 \left( x \right) = 1,$
F2 x   3x,
F3 x   1,
and that, if we take
  2.1,
$\xi = 2.1,$
the condition stated in theorem IV will be satisfied. This said, because we get from
equation (55)
5  43  
  0.005431878K ,
3 2  2 3
$\alpha = \frac{{5 + {\textstyle{4 \over 3}}\xi }}{{3\xi ^2 - 2}} = \frac{\xi }{3} = 0.005431878 \ldots ,$

we will find for the new values approaching the unknown $x$
    2.094568121K
$\xi + \alpha = 2.094568121 \ldots $
and
7/29/17
Cours d'Analyse – Note III – Numerical solutions
17
F2  
3
     2 2
 2.0945515K .
F1  
3  2
$\xi + \alpha - \alpha ^2 \frac{{F_2 \left( \xi \right)}}{{F_1 \left( \xi \right)}} = \xi +
\alpha - \alpha ^2 \frac{{3\xi }}{{3\xi ^2 - 2}} = 2.0945515 \ldots .$
    2
Finally, because the exact value of $x$ is presented in the form x    z $x = \xi + z$,
/414 $z$ will be a quantity contained between the limits $0$, $2\alpha$ and that
consequently we will evidently have
q 
F2    zF3   6.3  z

 0.6  1,
F1  
11.23

F3  
1

 0.1,
F1   1.23

val. num. z  val. num.   qz 2  val. num.   2  val. num. q  0.01.
$ - q = \frac{{F_2 \left( \xi \right) + zF_3 \left( \xi \right)}}{{F_1 \left( \xi \right)}} =
\frac{{6.3 + z}}{{11.23}} < 0.6 < 1,$
$\frac{{F_3 \left( \xi \right)}}{{F_1 \left( \xi \right)}} = \frac{1}{{1.23}} < 0.1,$
${\rm{val}}{\rm{.}}\;{\rm{num}}{\rm{.}}\;z =
{\rm{val}}{\rm{.}}\;{\rm{num}}{\rm{.}}\;\left( {\alpha + qz^2 } \right) <
{\rm{val}}{\rm{.}}\;{\rm{num}}{\rm{.}}\;\alpha + \left( {2\alpha } \right)^2
{\rm{val}}{\rm{.}}\;{\rm{num}}{\rm{.}}\;q < 0.01.$
2
and we will conclude (theorem IV, scholia III and IV) that in taking
$x=2.0945681,$
we commit an error smaller than $0.0001$, and in taking
$x=2.0945515$
an error smaller than $0.000001$.
Instead of using the general formulas, we could carry out the calculation in the
following matter. After finding the value 2.1 close to $x$, we put into equation (90)
$x=2.1+z$,
and we will find, by dividing all the terms by the coefficient of $z$,
0.005431878K  z  0.560997328K z 2  0.089047195K z 3  0
(92)
$0.005431878 \ldots + z + 0.560997328 \ldots z^2 + 0.089047195 \ldots z^3 = 0$
or, what amounts to the same thing,
7/29/17
Cours d'Analyse – Note III – Numerical solutions
18
(93)
z  0.005431878K  qz 2 ,
$z = - 0.005431878 \ldots + qz^2 ,$
the value of $q$ being determined by the formula
q  0.560997328K  0.089047195K z.
(94)
$q = - 0.560997328 \ldots - 0.089047195 \ldots z.$
The double of the first term of equation (92) is almost 0.01, and because the left hand
side of this equation furnishes two results of contrary signs when we successively make
$z=0$, $z=-0.01$,
we can be sure it has a real root contained between the limits $0$ and $-0.01$. To prove
that this root is unique, it suffices to observe that by virtue of formula (60) the equation
F1 2.1  z   0
$F_1 \left( {2.1 + z} \right) = 0$
/415
reduces to
1 2  0.560997328K z  3  0.089047195K z2  0,
$1 + 2 \times 0.560997328 \ldots z + 3 \times 0.089047195 \ldots z^2 = 0,$
and this last equation will not be satisfied by any value of $z$ between the limits that
concern us. Moreover, it is clear that for a similar value of $z$, the quantity $q$
determined by formula (94) remains contained between $-0.560$ and $-0.561$, and
because we draw from equation (93)

z  0.005431878K  0.000029505K q 

2
 0.000000320K q  K ,


$\left\{ \begin{array}{l}
z = - 0.005431878 \ldots - 0.000029505 \ldots \left( { - q} \right) \\
\quad \quad \quad \quad \quad \quad \quad \quad \;\, - 0.000000320 \ldots \left( { - q}
\right)^2 - \ldots , \\
\end{array} \right.$
(95)
and we conclude, first supposing that $-q=0.560 $,
$z=-0.00544850\ldots,$
and second supposing that $-q=0.561$,
7/29/17
Cours d'Analyse – Note III – Numerical solutions
19
$z=-0.00544853\ldots.$
Consequently, the real positive value that satisfies equation (90) will be contained
between the limits
$2.1-0.00544850\ldots=2.09455150$
and
$2.1-0.00544854\ldots=2.09455146$.
Thus this equation has a unique positive real root very nearly equal to
$2.0945515$.
Moreover, it is easy to assure ourselves that it does not have any reasons. For if it had
one, we would be able to satisfy with a positive value of $x$ the forluma
x 3  2x  5  0
(96)
$x^3 - 2x + 5 = 0$;
and this value of $x$ (see scholium V of problem I) will be at the same time less than the
positive root of the equation
x 3  2x  0,
$x^3 - 2x = 0,$
that is to say, than
2  1.414K ,
$\sqrt 2 = 1.414 \ldots ,$
and greater than the root of the equation
$5-2x=0$,
/416
that is to say than
5
 2.5;
2
$\frac{5}{2} = 2.5;$
which is absurd.
7/29/17
Cours d'Analyse – Note III – Numerical solutions
20
Let us move on to equation (91) and begin by looking for its positive roots. To
have a limit greater than the roots of this kind, it will suffice to observe that the equation
under consideration can be put into the form
x 3  7  7x
$x^3 + 7 = 7x$,
and we find (problem I, scholium IV) by supposing $x$ is positive, that
2 7x 3  7x
$2\sqrt {7x^3 } < 7x$
and consequently
x
7
.
4
$x < \frac{7}{4}.$
7
$\frac{7}{4}$ as a value near the largest positive root. This said, if
4
in equation (91) we make
Thus we can take
x
7
 z,
4
$x = \frac{7}{4} + z,$
we will find
32 3
z 0
70
$0.5 + z + 2.40z^2 + \frac{{32}}{{70}}z^3 = 0$
0.5  z  2.40z 2 
(97)
or, what amounts to the same thing,
(98)
$z = - 0.05 + qz^2 ,$
z  0.05  qz 2 ,
the value of $q$ being determined by the formula
q  2.40 
(99)
32
z.
70
$q = - 2.40 - \frac{{32}}{{70}}z.$
7/29/17
Cours d'Analyse – Note III – Numerical solutions
21
The double of the first term of equation (97) is $0.1$ and, as the left hand side changes
sign as it passes from $z=0$ to $z=-0.1$, while the polynomial
32 2
z
70
$1 + 2 \times 2.40z + 3 \times \frac{{32}}{{70}}z^2 $
1  2  2.40z  3 
remains always positive in this interval, it results that it has a real real root, but only one,
contained between the limits $0$ and $-0.1.$ The /417 value corresponding to $q$ is
evidently contained between the two quantities
$-2.354\ldots$ and $-2.40$.
We find from equation (98) that
0.1

 z   1  1  0.2q
(100) 
  0.05  0.0025 q   0.00025 q 2  0.0003125 q 3  K .

$\left\{ \begin{array}{c}
z = - \frac{{0.1}}{{1 + \sqrt {1 + 0.2q} }} \\
= - 0.05 - 0.0025\left( { - q} \right) - 0.00025\left( { - q} \right)^2 - 0.0003125\left( { q} \right)^3 - \ldots . \\
\end{array} \right.$
[I verified this expansion with Maple because I suspected the coefficient -0.00025. It is
correct.]
If in this last equation we successively make
$q=-2.354$, $q=-2.40$
we will find the corresponding values of $z$,
$z=-0.05788\ldots$, $z=-0.5810\ldots$,
and we conclude that the largest positive root of the given equation is contained between
the limits
7
 0.05788K  1.69211K
4
$\frac{7}{4} - 0.05788 \ldots = 1.69211 \ldots $
and
7/29/17
Cours d'Analyse – Note III – Numerical solutions
22
7
 0.05810K  1.69189K .
4
$\frac{7}{4} - 0.05810 \ldots = 1.69189 \ldots .$
Thus, if we call this largest root $a$, its value approximated to eleven one hundred
thousandths will be given by the formula
(101)
$a=1.6920$.
In speaking of this first approximate value, in just one operation we could obtain a second
in which the error would not be more than decimals of the twelfth order.
In addition to the root $a$ which we have just considered, equation (91) evidently
admits a negative root equal, except for sign, to the unique positive solution of the
equation
(102)
$x^3 - 7x - 7 = 0,$
x 3  7x  7  0,
and consequently (theorem III, scholium II) contained between the limits
 14  3.7416K
 3 14  2.41K .
$ - \sqrt {14} = - 3.7416 \ldots $ and $ - \sqrt[3]{{14}} = - 2.41 \ldots .$
/418
Let us call this root $c$. The third real root $b$ of equation (91) will evidently be real
and positive because the product $abc$ of the three roots will equal the negative of the
third term, that is to say, equal to $-7$. Let us now determine this third root. To find it,
we will first look for a number $G$ less than or equal to the numerical value of F1 a 
$F_1 \left( a \right)$. But because in this case
F x   x 2  7x  7,
F1 x   3x 2  7,
$F\left( x \right) = x^2 - 7x + 7,$
$F_1 \left( x \right) = 3x^2 - 7,$
and we conclude
F1 a   3a 2  7.
$F_1 \left( a \right) = 3a^2 - 7.$
Thus we can take
7/29/17
Cours d'Analyse – Note III – Numerical solutions
23
G  31.69189  7  1.5874K .
$G = 3\left( {1.69189} \right)^2 - 7 = 1.5874 \ldots .$
2
Moreover, by virtue of what has come before, we also have
$a<1.6922$, $-c<3.7417$
and, consequently,
$a-c<5.4339.$
This said, we find (problem II scholium II)
ab
G
1.5874

 0.29212K ,
a  c 5.4339
and as a consequence we will have
$b<1.69211\ldots-0.29214\ldots<1.40.$
After having recognized, as we have just done, that the root $b$ is less than the limit
$1.40$, suppose that
$x=1.40+z.$
Under this hypothesis, equation (91) gives
25 3
z 0
28
$0.05 + z - 3.75z^2 - \frac{{25}}{{28}}z^3 = 0$
0.05  z  3.75z 2 
(103)
or, what amounts to the same thing,
(98)
$z = - 0.05 + qz^2 ,$
z  0.05  qz 2 ,
the value of $q$ being determined by the formula
q  3.75 
(104)
25
z.
28
$q = 3.75 + \frac{{25}}{{28}}z.$
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The double of the first term of equation (103) is $0.1$, and as the left hand side of this
equation changes sign when it passes from $z=0$ to $z=-0.1$, while the polynomial
25 2
z
28
$1 - 2 \times 3.75z - 3 \times \frac{{25}}{{28}}z^2 $
1  2  3.75z  3 
always remains positive in the interval. As a result the polynomial has a single real root
contained between the limits $0$ and $-0.1$. The corresponding value of $q$ is
evidently contained between the two quantities
$3.66$ and $3.75$.
By successively substituting these two quantities in place of the letter $q$ in equation
(100), we obtain two new limits for the unknown $z$, namely
0.1
 0.04317K
1  1.732
$ - \frac{{0.1}}{{1 + \sqrt {1.732} }} = - 0.04317 \ldots $

and
0.1
 0.04305K ,
1  1.750
$ - \frac{{0.1}}{{1 + \sqrt {1.750} }} = - 0.04305 \ldots ,$

so we conclude that the positive root is contained between
$1.40-0.04317\ldots=1.35682\ldots$
and
$1.40-0.04305\ldots=1.35694\ldots$.
Thus we obtain for the value approximating this root to a ten thousandth part if we take
(105)
$b=1.3569.$
As for the negative root $c$ of equation (91), we already know that it is contained
between the limits
$-3.7416\ldots$ and $-2.41\ldots$.
Thus we have its value approximated to within unity if we suppose it equal to $-3$. This
said, take in equation (91)
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25
$x=-3_z$.
We will find
0.05  z  0.45z 2  0.05z 3  0
(106)
$0.05 + z - 0.45z^2 + 0.05z^3 = 0$
/420
or, what amounts to the same thing,
z  0.05  qz 2 ,
(98)
$z = - 0.05 + qz^2 ,$
the value of $q$ being determined by the formula
(107)
$q=0.45-0.05z$.
Moreover, we will easily recognize, first, that equation (105) has a real root, but only one
of them, contained between the limits $0$ and $-0.1$, second, that the corresponding
value of $q$ is contained between the two numbers
$0.45$ and $0.455$,
and third, that these two numbers substituted in place of the letter $q$ in equation (100)
furnish two new values approximating $z$, namely
0.1
 0.048922K
1  1.09
$ - \frac{{0.1}}{{1 + \sqrt {1.09} }} = - 0.048922 \ldots $

and
0.1
 0.048911K
1  1.091
$ - \frac{{0.1}}{{1 + \sqrt {1.091} }} = - 0.048911 \ldots .$

Consequently, the value approximating $c$ to a hundred millionth part [to the hundred
millionths place?] will be
(108)
$c=-3.04892.$
Finally, we would have been able to deduce immediately the value approximating $c$
from formulas (101) and (105). Indeed, because in equation (91) the coefficient of x 2
$x^2 $ is reduced to zero, we conclude
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$a+b+c=0$,
$c=-a-b$,
and consequently, to within a very small margin,
$c=-(1.6920-1.3569)=-3.0489.$
To end this note, we will present here two theorems, the second of which contains
the rule stated by Descartes relative to the determination of the number of positive or
negative roots that pertain to a polynomial of any degree. With this plan, we will begin
by examining the number of variations and [permanences of signs which /421 a sequence
of quantities can offer,] when we suppose that the different terms of the sequence are
compared to each other in the order in which they appear.
Let
(109)
Translators' conventions
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