Another derivation using mathematical induction
derivatives of the
D
=
1
/
f
{\displaystyle D=1/f}
edit
Let's define a new function
D
=
1
/
f
(
x
)
{\displaystyle D=1/f(x)}
1
=
f
(
x
)
D
(
x
)
{\displaystyle 1=f(x)D(x)}
By defferentiating
0
=
f
′
(
x
)
D
(
x
)
+
f
(
x
)
D
′
(
x
)
{\displaystyle 0=f^{\prime }(x)D(x)+f(x)D^{\prime }(x)}
−
f
D
′
=
D
f
′
{\displaystyle -fD^{\prime }=Df^{\prime }}
Differentiating multiple times, we get
−
f
D
′
′
=
2
D
′
f
′
+
D
f
′
′
{\displaystyle -fD^{\prime \prime }=2D^{\prime }f^{\prime }+Df^{\prime \prime }}
−
f
D
′
′
′
=
3
D
′
′
f
′
+
3
D
′
f
′
′
+
D
f
′
′
′
{\displaystyle -fD^{\prime \prime \prime }=3D^{\prime \prime }f^{\prime }+3D^{\prime }f^{\prime \prime }+Df^{\prime \prime \prime }}
−
f
D
′
′
′
′
=
4
D
′
′
′
f
′
+
6
D
′
′
f
′
′
+
4
D
′
f
′
′
′
+
D
f
′
′
′
′
{\displaystyle -fD^{\prime \prime \prime \prime }=4D^{\prime \prime \prime }f^{\prime }+6D^{\prime \prime }f^{\prime \prime }+4D^{\prime }f^{\prime \prime \prime }+Df^{\prime \prime \prime \prime }}
The coefficients of the above equations are those of the Pascal's triangle . Taking
C
(
n
,
k
)
{\displaystyle C(n,k)}
binomial coefficient , we would get
−
f
D
(
n
)
=
C
(
n
,
1
)
D
(
n
−
1
)
f
′
+
C
(
n
,
2
)
D
(
n
−
2
)
f
′
′
+
C
(
n
,
3
)
D
(
n
−
3
)
f
′
′
′
+
.
.
.
+
C
(
n
,
n
)
D
f
(
n
)
{\displaystyle -fD^{(n)}=C(n,1)D^{(n-1)}f^{\prime }+C(n,2)D^{(n-2)}f^{\prime \prime }+C(n,3)D^{(n-3)}f^{\prime \prime \prime }+...+C(n,n)Df^{(n)}}
Though some of followings would not be used at the derivation,
let's expand some of above equations to see what forms they have,
−
f
D
=
−
1
{\displaystyle -fD=-1}
−
f
D
′
=
f
′
/
f
{\displaystyle -fD^{\prime }=f^{\prime }/f}
−
f
D
′
′
=
2
D
′
f
′
+
f
′
′
D
=
2
(
−
f
′
/
f
2
)
f
′
+
f
′
′
/
f
=
(
1
/
f
2
)
(
−
2
f
′
2
+
f
f
′
′
)
{\displaystyle {\begin{array}{rl}-fD^{\prime \prime }=&2D^{\prime }f^{\prime }+f^{\prime \prime }D\\=&2\left(-f^{\prime }/{f^{2}}\right)f^{\prime }+f^{\prime \prime }/f\\=&\left(1/f^{2}\right)\left(-2{f^{\prime }}^{2}+ff^{\prime \prime }\right)\end{array}}}
−
f
D
′
′
′
=
(
1
/
f
3
)
(
6
f
′
3
−
6
f
f
′
f
′
′
+
f
2
f
′
′
′
)
{\displaystyle -fD^{\prime \prime \prime }=(1/f^{3})(6{f^{\prime }}^{3}-6ff^{\prime }f^{\prime \prime }+f^{2}f^{\prime \prime \prime })}
−
f
D
′
′
′
′
=
(
1
/
f
4
)
(
−
24
f
′
4
+
36
f
f
′
2
f
′
′
−
8
f
2
f
′
f
′
′
′
−
6
f
2
f
′
′
2
+
f
3
f
′
′
′
′
)
{\displaystyle -fD^{\prime \prime \prime \prime }=(1/f^{4})(-24{f^{\prime }}^{4}+36f{f^{\prime }}^{2}f^{\prime \prime }-8f^{2}f^{\prime }f^{\prime \prime \prime }-6f^{2}{f^{\prime \prime }}^{2}+f^{3}f^{\prime \prime \prime \prime })}
...
Therefore
D
=
(
−
1
/
f
)
(
−
1
)
{\displaystyle D=(-1/f)(-1)}
D
′
=
(
−
1
/
f
2
)
f
′
{\displaystyle D^{\prime }=(-1/f^{2})f^{\prime }}
D
′
′
=
(
−
1
/
f
3
)
(
−
2
f
′
2
+
f
f
′
′
)
{\displaystyle D^{\prime \prime }=(-1/f^{3})(-2{f^{\prime }}^{2}+ff^{\prime \prime })}
D
′
′
′
=
(
−
1
/
f
4
)
(
6
f
′
3
−
6
f
f
′
f
′
′
+
f
2
f
′
′
′
)
{\displaystyle D^{\prime \prime \prime }=(-1/f^{4})(6{f^{\prime }}^{3}-6ff^{\prime }f^{\prime \prime }+f^{2}f^{\prime \prime \prime })}
D
′
′
′
′
=
(
−
1
/
f
5
)
(
−
24
f
′
4
+
36
f
f
′
2
f
′
′
−
8
f
2
f
′
f
′
′
′
−
6
f
2
f
′
′
2
+
f
3
f
′
′
′
′
)
{\displaystyle D^{\prime \prime \prime \prime }=(-1/f^{5})(-24{f^{\prime }}^{4}+36f{f^{\prime }}^{2}f^{\prime \prime }-8f^{2}f^{\prime }f^{\prime \prime \prime }-6f^{2}{f^{\prime \prime }}^{2}+f^{3}f^{\prime \prime \prime \prime })}
Relation with
f
{\displaystyle f}
edit
By the Taylor expansion , we get (assuming
Δ
=
x
−
x
0
{\displaystyle \Delta =x-x_{0}}
f
(
x
)
=
f
(
x
0
)
+
f
′
(
x
0
)
Δ
+
(
1
/
2
!
)
f
′
′
(
x
0
)
Δ
2
+
(
1
/
3
!
)
f
′
′
′
(
x
0
)
Δ
3
+
.
.
.
{\displaystyle f(x)=f(x_{0})+f^{\prime }(x_{0})\Delta +(1/2!)f^{\prime \prime }(x_{0}){\Delta }^{2}+(1/3!)f^{\prime \prime \prime }(x_{0}){\Delta }^{3}+...}
where,
x
0
{\displaystyle x_{0}}
f
(
x
)
=
0
{\displaystyle f(x)=0}
Let approximate f(x) by dropping higher orders of the right hand side;
f
(
x
)
=
f
(
x
0
)
+
f
′
(
x
0
)
(
x
−
x
0
)
{\displaystyle f(x)=f(x_{0})+f^{\prime }(x_{0})(x-x_{0})}
x
1
{\displaystyle x_{1}}
f
(
x
)
=
0
{\displaystyle f(x)=0}
x
{\displaystyle x}
f
(
x
1
)
=
0
=
f
(
x
0
)
+
f
′
(
x
0
)
(
x
1
−
x
0
)
{\displaystyle f(x_{1})=0=f(x_{0})+f^{\prime }(x_{0})(x_{1}-x_{0})}
x
1
=
x
0
−
f
(
x
0
)
/
f
′
(
x
0
)
{\displaystyle x_{1}=x_{0}-f(x_{0})/f^{\prime }(x_{0})}
This is the Newton's method .
Δ
=
x
1
−
x
0
{\displaystyle \Delta =x_{1}-x_{0}}
Let's call above
Δ
{\displaystyle \Delta }
Δ
N
e
w
t
o
n
{\displaystyle {\Delta }_{Newton}}
Δ
(
1
)
{\displaystyle \Delta _{(1)}}
Δ
N
e
w
t
o
n
=
Δ
(
1
)
=
−
f
(
x
0
)
/
f
′
(
x
0
)
=
−
f
/
f
′
=
(
−
1
)
/
(
f
′
/
f
)
=
=
(
−
f
D
)
/
(
−
f
D
′
)
=
D
/
D
′
{\displaystyle {\begin{array}{rl}{\Delta }_{Newton}=\Delta _{(1)}=&-f(x_{0})/f^{\prime }(x_{0})=-f/f^{\prime }=(-1)/(f^{\prime }/f)=\\[0.7em]=&(-fD)/(-fD^{\prime })=D/D^{\prime }\end{array}}}
Therefore the Newton's method is the first kind of Householder's method.
Now by taking three therms of the original Taylor series ,
f
(
x
1
)
=
0
=
f
(
x
0
)
+
f
′
(
x
0
)
Δ
+
(
1
/
2
!
)
f
′
′
(
x
0
)
Δ
2
{\displaystyle f(x_{1})=0=f(x_{0})+f^{\prime }(x_{0})\Delta +(1/2!)f^{\prime \prime }(x_{0}){\Delta }^{2}}
Δ
=
−
f
(
x
0
)
/
(
f
′
(
x
0
)
+
(
1
/
2
!
)
f
′
′
(
x
0
)
Δ
{\displaystyle \Delta =-f(x_{0})/(f^{\prime }(x_{0})+(1/2!)f^{\prime \prime }(x_{0})\Delta }
Δ
{\displaystyle \Delta }
Δ
N
e
w
t
o
n
{\displaystyle {\Delta }_{Newton}}
Δ
=
−
f
(
x
0
)
/
[
f
′
(
x
0
)
+
(
1
/
2
!
)
f
′
′
(
x
0
)
(
−
f
(
x
0
)
/
f
′
)
]
=
−
f
f
′
/
(
f
′
2
−
(
1
/
2
)
f
′
′
f
)
=
2
f
f
′
/
(
f
′
′
f
−
2
f
′
2
)
{\displaystyle {\begin{array}{rl}\Delta =&-f(x_{0})/\left[f^{\prime }(x_{0})+(1/2!)f^{\prime \prime }(x_{0})(-f(x_{0})/f^{\prime })\right]\\[0.7em]=&-ff^{\prime }/({f^{\prime }}^{2}-(1/2)f^{\prime \prime }f)\\[0.7em]=&2ff^{\prime }/(f^{\prime \prime }f-2{f^{\prime }}^{2})\end{array}}}
This is the Halley's method . And Let's call
Δ
{\displaystyle \Delta }
Δ
H
a
l
l
e
y
{\displaystyle {\Delta }_{Halley}}
Δ
(
2
)
{\displaystyle \Delta _{(2)}}
Δ
H
a
l
l
e
y
=
Δ
(
2
)
=
2
f
f
′
/
(
f
′
′
f
−
2
f
′
2
)
=
2
(
f
′
/
f
)
/
[
(
1
/
f
2
)
(
f
′
′
f
−
2
f
′
2
)
]
=
2
(
−
f
D
′
)
/
(
−
f
D
′
′
)
=
2
D
′
/
D
′
′
{\displaystyle {\begin{array}{rl}{\Delta }_{Halley}=\Delta _{(2)}=&2ff^{\prime }/(f^{\prime \prime }f-2{f^{\prime }}^{2})\\[0.7em]=&2(f^{\prime }/f)/\left[(1/f^{2})(f^{\prime \prime }f-2{f^{\prime }}^{2})\right]\\[0.7em]=&2(-fD^{\prime })/(-fD^{\prime \prime })=2D^{\prime }/D^{\prime \prime }\\[0.7em]\end{array}}}
As we progress, we get
Δ
(
3
)
=
(
−
f
)
/
[
f
′
+
(
1
/
2
!
)
f
′
′
Δ
H
a
l
l
e
y
+
(
1
/
3
!
)
f
′
′
′
Δ
H
a
l
l
e
y
Δ
N
e
w
t
o
n
]
=
(
−
f
)
/
[
f
′
+
(
1
/
2
!
)
f
′
′
(
2
D
′
/
D
′
′
)
+
(
1
/
3
!
)
f
′
′
′
(
2
D
′
/
D
′
′
)
(
D
/
D
′
)
]
=
(
−
3
f
)
/
[
3
f
′
+
3
f
′
′
(
D
′
/
D
′
′
)
+
f
′
′
′
(
D
/
D
′
′
)
]
=
(
−
3
f
D
′
′
)
/
[
3
f
′
D
′
′
+
3
f
′
′
D
′
+
f
′
′
′
D
]
=
(
−
3
f
D
′
′
)
/
(
−
f
D
′
′
′
)
=
3
D
′
′
/
D
′
′
′
{\displaystyle {\begin{array}{rl}\Delta _{(3)}=&(-f)/\left[f^{\prime }+(1/2!)f^{\prime \prime }{\Delta }_{Halley}+(1/3!)f^{\prime \prime \prime }{\Delta }_{Halley}{\Delta }_{Newton}\right]\\[0.7em]=&(-f)/\left[f^{\prime }+(1/2!)f^{\prime \prime }(2D^{\prime }/D^{\prime \prime })+(1/3!)f^{\prime \prime \prime }(2D^{\prime }/D^{\prime \prime })(D/D^{\prime })\right]\\[0.7em]=&(-3f)/\left[3f^{\prime }+3f^{\prime \prime }(D^{\prime }/D^{\prime \prime })+f^{\prime \prime \prime }(D/D^{\prime \prime })\right]\\[0.7em]=&(-3fD^{\prime \prime })/\left[3f^{\prime }D^{\prime \prime }+3f^{\prime \prime }D^{\prime }+f^{\prime \prime \prime }D\right]\\[0.7em]=&(-3fD^{\prime \prime })/(-fD^{\prime \prime \prime })\\[0.7em]=&3D^{\prime \prime }/D^{\prime \prime \prime }\\[0.7em]\end{array}}}
This is the third kind of Householder's method.
Δ
(
3
)
=
−
6
f
f
′
2
−
3
f
2
f
′
′
6
f
′
3
−
6
f
f
′
f
′
′
+
f
2
f
′
′
′
{\displaystyle \Delta _{(3)}=-{\frac {6f{f^{\prime }}^{2}-3f^{2}f^{\prime \prime }}{6{f^{\prime }}^{3}-6ff^{\prime }f^{\prime \prime }+f^{2}f^{\prime \prime \prime }}}}
Δ
(
n
)
=
(
−
f
)
/
[
f
′
+
(
1
/
2
!
)
f
′
′
Δ
(
n
−
1
)
+
(
1
/
3
!
)
f
′
′
′
Δ
(
n
−
1
)
Δ
(
n
−
2
)
+
.
.
.
]
=
(
−
f
)
/
[
f
′
+
(
1
/
2
!
)
f
′
′
(
(
n
−
1
)
D
(
n
−
2
)
/
D
(
n
−
1
)
)
+
(
1
/
3
!
)
f
′
′
′
(
(
n
−
1
)
D
(
n
−
2
)
/
D
(
n
−
1
)
)
(
(
n
−
2
)
D
(
n
−
3
)
/
D
(
n
−
2
)
)
+
(
1
/
4
!
)
f
′
′
′
′
(
(
n
−
1
)
D
(
n
−
2
)
/
D
(
n
−
1
)
)
(
(
n
−
2
)
D
(
n
−
3
)
/
D
(
n
−
2
)
)
(
(
n
−
3
)
D
(
n
−
4
)
/
D
(
n
−
3
)
)
+
.
.
.
]
=
(
−
f
)
/
[
f
′
+
(
1
/
2
!
)
f
′
′
(
(
n
−
1
)
D
(
n
−
2
)
/
D
(
n
−
1
)
)
+
(
1
/
3
!
)
f
′
′
′
(
(
n
−
1
)
(
n
−
2
)
D
(
n
−
3
)
/
D
(
n
−
1
)
)
+
(
1
/
4
!
)
f
′
′
′
′
(
(
n
−
1
)
(
n
−
2
)
(
n
−
3
)
D
(
n
−
4
)
/
D
(
n
−
1
)
)
+
.
.
.
+
(
(
n
−
1
)
!
/
n
!
)
f
(
n
)
(
D
/
D
(
n
−
1
)
)
]
=
(
−
n
f
D
(
n
−
1
)
)
/
[
C
(
n
,
1
)
f
′
D
(
n
−
1
)
+
C
(
n
,
2
)
f
′
′
D
(
n
−
2
)
+
C
(
n
,
3
)
f
′
′
′
D
(
n
−
3
)
+
C
(
n
,
4
)
f
′
′
′
′
D
(
n
−
4
)
+
.
.
.
+
C
(
n
,
n
)
f
(
n
)
D
]
=
(
−
n
f
D
(
n
−
1
)
/
(
−
f
D
(
n
)
)
=
n
D
(
n
−
1
)
/
D
(
n
)
{\displaystyle {\begin{array}{rl}\Delta _{(n)}=&(-f)/\left[f^{\prime }+(1/2!)f^{\prime \prime }{\Delta }_{(n-1)}+(1/3!)f^{\prime \prime \prime }{\Delta }_{(n-1)}{\Delta }_{(n-2)}+...\right]\\[0.7em]=&(-f)/\left[f^{\prime }+(1/2!)f^{\prime \prime }((n-1)D^{(n-2)}/D^{(n-1)})+(1/3!)f^{\prime \prime \prime }((n-1)D^{(n-2)}/D^{(n-1)})((n-2)D^{(n-3)}/D^{(n-2)})+(1/4!)f^{\prime \prime \prime \prime }((n-1)D^{(n-2)}/D^{(n-1)})((n-2)D^{(n-3)}/D^{(n-2)})((n-3)D^{(n-4)}/D^{(n-3)})+...\right]\\[0.7em]=&(-f)/\left[f^{\prime }+(1/2!)f^{\prime \prime }((n-1)D^{(n-2)}/D^{(n-1)})+(1/3!)f^{\prime \prime \prime }((n-1)(n-2)D^{(n-3)}/D^{(n-1)})+(1/4!)f^{\prime \prime \prime \prime }((n-1)(n-2)(n-3)D^{(n-4)}/D^{(n-1)})+...+((n-1)!/n!)f^{(n)}(D/D^{(n-1)})\right]\\[0.7em]=&(-nfD^{(n-1)})/\left[C(n,1)f^{\prime }D^{(n-1)}+C(n,2)f^{\prime \prime }D^{(n-2)}+C(n,3)f^{\prime \prime \prime }D^{(n-3)}+C(n,4)f^{\prime \prime \prime \prime }D^{(n-4)}+...+C(n,n)f^{(n)}D\right]\\[0.7em]=&(-nfD^{(n-1)}/(-fD^{(n)})\\[0.7em]=&nD^{(n-1)}/D^{(n)}\\[0.7em]\end{array}}}
Derivation
edit
Following is not Gauge00's derivation, it is from the original derivation Householder's method .
An exact derivation of the Householder's methods starts from the Padé approximation of order (d+1) , where the approximant with linear numerator of the form
x
−
x
1
{\displaystyle x-x_{1}}
The Padé approximation has the form
f
(
x
)
=
x
−
x
1
b
0
+
b
1
(
x
−
x
0
)
+
.
.
.
+
b
d
−
1
(
x
−
x
0
)
d
−
1
+
O
(
(
x
−
x
0
)
d
+
1
)
.
{\displaystyle f(x)={\frac {x-x_{1}}{b_{0}+b_{1}(x-x_{0})+...+b_{d-1}(x-x_{0})^{d-1}}}+O((x-x_{0})^{d+1}).}
where
x
0
{\displaystyle x_{0}}
b
i
{\displaystyle b_{i}}
x
1
{\displaystyle x_{1}}
x
0
{\displaystyle x_{0}}
f
(
x
)
{\displaystyle f(x)}
Since
f
(
x
1
)
=
0
{\displaystyle f(x_{1})=0}
x
1
{\displaystyle x_{1}}
In Pade approximant, the degrees of numerator and denominator polynomials have to add to the order of the approximant. Therefore, in our approximation of
d
{\displaystyle d}
b
d
=
0
{\displaystyle b_{d}=0}
One could determine the Padé approximant starting from the Taylor polynomial of f using Euclid's algorithm .
However, starting from the Taylor polynomial of 1/f is shorter and leads directly to the given formula.
(
1
/
f
)
(
x
)
=
(
1
/
f
)
(
x
0
)
+
(
1
/
f
)
′
(
x
0
)
(
x
−
x
0
)
+
…
{\displaystyle (1/f)(x)=(1/f)(x_{0})+(1/f)'(x_{0})(x-x_{0})+\dots }
+
1
(
d
−
1
)
!
(
1
/
f
)
(
d
−
1
)
(
x
0
)
(
x
−
x
0
)
d
−
1
+
1
d
!
(
1
/
f
)
(
d
)
(
x
0
)
(
x
−
x
0
)
d
+
O
(
(
x
−
x
0
)
d
+
1
)
{\displaystyle +{\frac {1}{(d-1)!}}(1/f)^{(d-1)}(x_{0}){(x-x_{0})}^{d-1}+{\frac {1}{d!}}(1/f)^{(d)}(x_{0}){(x-x_{0})}^{d}+O((x-x_{0})^{d+1})}
And
x
−
x
0
=
(
x
−
x
0
)
−
(
x
1
−
x
0
)
{\displaystyle x-x_{0}={(x-x_{0})-(x_{1}-x_{0})}}
(
1
/
f
)
(
x
)
∗
(
(
x
−
x
0
)
−
(
x
1
−
x
0
)
)
=
−
(
1
/
f
)
(
x
0
)
∗
(
x
1
−
x
0
)
+
(
x
−
x
0
)
∗
[
(
1
/
f
)
(
x
0
)
−
(
1
/
f
)
′
(
x
0
)
(
x
1
−
x
0
)
]
+
(
x
−
x
0
)
2
∗
[
(
1
/
f
)
′
(
x
0
)
−
(
1
/
f
)
″
(
x
0
)
(
x
1
−
x
0
)
/
2
]
+
.
.
.
+
(
x
−
x
0
)
d
∗
[
(
1
/
f
)
(
d
−
1
)
(
x
0
)
/
(
d
−
1
)
!
−
(
1
/
f
)
(
d
)
(
x
0
)
(
x
1
−
x
0
)
/
(
d
!
)
]
+
O
(
(
x
−
x
0
)
d
+
1
)
{\displaystyle {\begin{array}{rl}(1/f)(x)*&((x-x_{0})-(x_{1}-x_{0}))\\[0.5em]=&-(1/f)(x_{0})*(x_{1}-x_{0})\\[0.5em]+&(x-x_{0})*\left[(1/f)(x_{0})-(1/f)'(x_{0})(x_{1}-x_{0})\right]\\[0.5em]+&(x-x_{0})^{2}*[(1/f)'(x_{0})-(1/f)''(x_{0})(x_{1}-x_{0})/2]\\[0.5em]+&...\\[0.5em]+&(x-x_{0})^{d}*[(1/f)^{(d-1)}(x_{0})/(d-1)!-(1/f)^{(d)}(x_{0})(x_{1}-x_{0})/(d!)]\\[0.5em]+&O((x-x_{0})^{d+1})\end{array}}}
This has to be the denominator of the Pade approximant of f(x) of d th order
of
f
(
x
)
{\displaystyle f(x)}
b
d
=
0
{\displaystyle b_{d}=0}
0
=
b
d
=
(
1
/
f
)
(
d
−
1
)
(
x
0
)
1
(
d
−
1
)
!
−
(
1
/
f
)
(
d
)
(
x
0
)
1
d
!
∗
(
x
1
−
x
0
)
{\displaystyle 0=b_{d}=(1/f)^{(d-1)}(x_{0}){\frac {1}{(d-1)!}}-(1/f)^{(d)}(x_{0}){\frac {1}{d!}}*(x_{1}-x_{0})}
Now, solving the last equation
x
1
−
x
0
{\displaystyle x_{1}-x_{0}}
x
1
−
x
0
=
(
1
/
f
)
(
d
−
1
)
(
x
0
)
/
(
d
−
1
)
!
(
1
/
f
)
(
d
)
(
x
0
)
/
d
!
=
d
∗
(
1
/
f
)
(
d
−
1
)
(
x
0
)
(
1
/
f
)
(
d
)
(
x
0
)
{\displaystyle x_{1}-x_{0}={\frac {(1/f)^{(d-1)}(x_{0})/(d-1)!}{(1/f)^{(d)}(x_{0})/d!}}=d*{\frac {(1/f)^{(d-1)}(x_{0})}{(1/f)^{(d)}(x_{0})}}}
x
n
+
1
=
x
n
+
d
(
1
/
f
)
(
d
−
1
)
(
x
n
)
(
1
/
f
)
(
d
)
(
x
n
)
{\displaystyle x_{n+1}=x_{n}+d\;{\frac {\left(1/f\right)^{(d-1)}(x_{n})}{\left(1/f\right)^{(d)}(x_{n})}}}
![{\displaystyle {\begin{array}{rl}{\Delta }_{Newton}=\Delta _{(1)}=&-f(x_{0})/f^{\prime }(x_{0})=-f/f^{\prime }=(-1)/(f^{\prime }/f)=\\[0.7em]=&(-fD)/(-fD^{\prime })=D/D^{\prime }\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4c899a9107ae0850b30b7967a8b974837d70e2ef)
![{\displaystyle {\begin{array}{rl}\Delta =&-f(x_{0})/\left[f^{\prime }(x_{0})+(1/2!)f^{\prime \prime }(x_{0})(-f(x_{0})/f^{\prime })\right]\\[0.7em]=&-ff^{\prime }/({f^{\prime }}^{2}-(1/2)f^{\prime \prime }f)\\[0.7em]=&2ff^{\prime }/(f^{\prime \prime }f-2{f^{\prime }}^{2})\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e0dcb2cf97e7cf4a7e8270a5dea9d13081f172b9)
![{\displaystyle {\begin{array}{rl}{\Delta }_{Halley}=\Delta _{(2)}=&2ff^{\prime }/(f^{\prime \prime }f-2{f^{\prime }}^{2})\\[0.7em]=&2(f^{\prime }/f)/\left[(1/f^{2})(f^{\prime \prime }f-2{f^{\prime }}^{2})\right]\\[0.7em]=&2(-fD^{\prime })/(-fD^{\prime \prime })=2D^{\prime }/D^{\prime \prime }\\[0.7em]\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/befa9963043d1e74494141661e9ccdf7324de330)
![{\displaystyle {\begin{array}{rl}\Delta _{(3)}=&(-f)/\left[f^{\prime }+(1/2!)f^{\prime \prime }{\Delta }_{Halley}+(1/3!)f^{\prime \prime \prime }{\Delta }_{Halley}{\Delta }_{Newton}\right]\\[0.7em]=&(-f)/\left[f^{\prime }+(1/2!)f^{\prime \prime }(2D^{\prime }/D^{\prime \prime })+(1/3!)f^{\prime \prime \prime }(2D^{\prime }/D^{\prime \prime })(D/D^{\prime })\right]\\[0.7em]=&(-3f)/\left[3f^{\prime }+3f^{\prime \prime }(D^{\prime }/D^{\prime \prime })+f^{\prime \prime \prime }(D/D^{\prime \prime })\right]\\[0.7em]=&(-3fD^{\prime \prime })/\left[3f^{\prime }D^{\prime \prime }+3f^{\prime \prime }D^{\prime }+f^{\prime \prime \prime }D\right]\\[0.7em]=&(-3fD^{\prime \prime })/(-fD^{\prime \prime \prime })\\[0.7em]=&3D^{\prime \prime }/D^{\prime \prime \prime }\\[0.7em]\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4756ed5e6b9db30c7a5ac436aaa5d9e7d1bb17b5)
![{\displaystyle {\begin{array}{rl}\Delta _{(n)}=&(-f)/\left[f^{\prime }+(1/2!)f^{\prime \prime }{\Delta }_{(n-1)}+(1/3!)f^{\prime \prime \prime }{\Delta }_{(n-1)}{\Delta }_{(n-2)}+...\right]\\[0.7em]=&(-f)/\left[f^{\prime }+(1/2!)f^{\prime \prime }((n-1)D^{(n-2)}/D^{(n-1)})+(1/3!)f^{\prime \prime \prime }((n-1)D^{(n-2)}/D^{(n-1)})((n-2)D^{(n-3)}/D^{(n-2)})+(1/4!)f^{\prime \prime \prime \prime }((n-1)D^{(n-2)}/D^{(n-1)})((n-2)D^{(n-3)}/D^{(n-2)})((n-3)D^{(n-4)}/D^{(n-3)})+...\right]\\[0.7em]=&(-f)/\left[f^{\prime }+(1/2!)f^{\prime \prime }((n-1)D^{(n-2)}/D^{(n-1)})+(1/3!)f^{\prime \prime \prime }((n-1)(n-2)D^{(n-3)}/D^{(n-1)})+(1/4!)f^{\prime \prime \prime \prime }((n-1)(n-2)(n-3)D^{(n-4)}/D^{(n-1)})+...+((n-1)!/n!)f^{(n)}(D/D^{(n-1)})\right]\\[0.7em]=&(-nfD^{(n-1)})/\left[C(n,1)f^{\prime }D^{(n-1)}+C(n,2)f^{\prime \prime }D^{(n-2)}+C(n,3)f^{\prime \prime \prime }D^{(n-3)}+C(n,4)f^{\prime \prime \prime \prime }D^{(n-4)}+...+C(n,n)f^{(n)}D\right]\\[0.7em]=&(-nfD^{(n-1)}/(-fD^{(n)})\\[0.7em]=&nD^{(n-1)}/D^{(n)}\\[0.7em]\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e451a3f725d2b812178bc774d84e8e17ea89604e)
![{\displaystyle {\begin{array}{rl}(1/f)(x)*&((x-x_{0})-(x_{1}-x_{0}))\\[0.5em]=&-(1/f)(x_{0})*(x_{1}-x_{0})\\[0.5em]+&(x-x_{0})*\left[(1/f)(x_{0})-(1/f)'(x_{0})(x_{1}-x_{0})\right]\\[0.5em]+&(x-x_{0})^{2}*[(1/f)'(x_{0})-(1/f)''(x_{0})(x_{1}-x_{0})/2]\\[0.5em]+&...\\[0.5em]+&(x-x_{0})^{d}*[(1/f)^{(d-1)}(x_{0})/(d-1)!-(1/f)^{(d)}(x_{0})(x_{1}-x_{0})/(d!)]\\[0.5em]+&O((x-x_{0})^{d+1})\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fac42a8e9034e0cac78b6fabd0d90c81ec511440)
