Solving equations by Hermite type inverse interpolation

by
C. Iancu and I. Păvăloiu

1.

In this work we study some numerical methods for solving equations of the form:

(1)

f\left(x\right)=0,

Or $f:I\rightarrow\mathbb{R}$ is a real function of a real variable, and $I$ is an interval of the real axis.

We designate by $AND$ a set of $n+1$ distinct real numbers of the set $I,$ that's to say, $E=\{x_{1},x_{2},\ldots,x_{n+1}\}$ Or $x_{i}\neq x_{j}$ For $i\neq j,\ i,j=1,2,\ldots,n+1.$

Let us designate by

(2)

\alpha_{1},\alpha_{2},\ldots,\alpha_{n+1}

$n+1$ natural numbers that satisfy the condition:

(3)

\alpha_{1}+\alpha_{2}+,\ldots+\alpha_{n+1}=m+1

Or $m\in\mathbb{N}$ , and by

(4)

y_{i}^{j};\;\;j=0,1,\ldots,\alpha_{i}-1;\ i=1,2,\ldots,n+1,

$m+1$ given real numbers.

The following theorem is well known.

Theorem 1 .

The $and$ has a single polynomial $P$ of at least degree $m,$ which verifies the following conditions:

(5)

P^{\left(j\right)}\left(x_{i}\right)=y_{i}^{j},\;\;\ j=0,1,\ldots,\alpha_{i}-1% ;\ i=1,2,\ldots,n+1;

The polynomial $P$ has the following form:

(6)

P\left(x\right)=\sum_{i=1}^{n+1}\sum_{j=0}^{\alpha_{i}-1}\sum_{k=0}^{\alpha_{i% }-j-1}y_{i}^{j}\tfrac{1}{k!j!}\left[\tfrac{\left(x-x_{i}\right)^{\alpha_{i}}}{% \omega\left(x\right)}\right]_{x=x_{i}}^{\left(k\right)}\cdot\tfrac{\omega\left% (x\right)}{\left(x-x_{i}\right)^{\alpha_{i}-j-k}},

(7)

\omega\left(x\right)=\prod\limits_{i=1}^{n+1}\left(x-x_{i}\right)^{\alpha_{i}}

If we assume that the function $f$ admits derivatives up to order $m+1$ inclusive in the interval $I$ and if $y_{i}^{j}=f^{\left(j\right)}\left(x_{i}\right),\ j=0,1,\ldots,\alpha_{i}-1,\ i% =1,2,\ldots,n+1,$ then the remainder of the Hermite interpolation formula satisfies the following equality:

(8)

f\left(x\right)-P\left(x\right)=\tfrac{f^{\left(m+1\right)}\left(\xi\right)}{% \left(m+1\right)!}\cdot\omega\left(x\right),

Or $\xi\in I.$

Let's suppose that $f^{\prime}\left(x\right)\neq 0,$ for each $x\in I$ and designate by $F=f\left(I\right),$ the image of the interval $I$ by function $f .$ From the above hypothesis it follows that the function $f:I\rightarrow F$ is bijective, that the function $f^{-1}:F\rightarrow I$ admits derivatives up to order $m+1$ inclusive in the interval $F .$ The order derivative $k$ $\left(k\in\mathbb{N},\ k\leq m+1\right)$ at one point $y_{0}=f\left(x_{0}\right),\ y_{0}\in F$ has the following form [ 4 ] , [ 8 ] :

(9)

\displaystyle\left[f^{-1}\left(y_{0}\right)\right]^{\left(k\right)}

\displaystyle=\sum\tfrac{\left(2k-2-i_{1}\right)!\left(-1\right)^{k-1+i_{1}}}{% i_{2}!\ldots i_{k}!\left(f^{\prime}\left(x_{0}\right)\right)^{2k-1}}\left(% \tfrac{f^{\prime}\left(x_{0}\right)}{1!}\right)^{i_{1}}\left(\tfrac{f^{\prime% \prime}\left(x_{0}\right)}{2!}\right)^{i}\ldots\left(\tfrac{f^{\left(k\right)}% \left(x_{0}\right)}{k!}\right)^{i_{k}}

where the above sum extends to all integer, non-negative solutions of the system of equations:

(10)

\left\{\begin{array}[c]{c}i_{2}+2i_{3}+\ldots+\left(k-1\right)i_{k}=k-1\\ i_{1}+i_{2}+\ldots+i_{k}=k-1\end{array}\right.

If we assume that equation ( 1 ) has a root $\bar{x}\in I,$ then of $f\left(\bar{x}\right)=0$ it follows that $f^{-1}\left(0\right)=\bar{x},$ and of $f^{\prime}\left(x\right)\neq 0$ for each $x\in I$ it follows that $\bar{x}$ is unique.

If in the expression of the Hermite interpolation polynomial ( 6 ) we choose for the interpolation nodes the values $y_{i}=f\left(x_{i}\right),\ i=1,2,\ldots,n+1$ and for the values of the function and its derivatives the numbers $f^{-1}\left(y_{i}\right)=x_{i},\ i=1,2,\ldots,n+1,$ respectively $\left[f^{-1}\left(y_{i}\right)\right]^{\left(j\right)},\ j=1,2,\ldots,\alpha_{% i}-1;\ i=1,2,\ldots,n+1,$ then we obtain the Hermite interpolation polynomial of the function $f^{-1}.$ This polynomial will have the following form:

(11)

P\left(y\right)=\sum_{i=1}^{n+1}\sum_{j=0}^{\alpha_{i}-1}\sum_{k=0}^{\alpha_{i% }-j-1}\left[\left(f^{-1}\left(y_{i}\right)\right)\right]^{\left(j\right)}% \tfrac{1}{k!j!}\cdot\left[\tfrac{\left(y-y_{i}\right)^{\alpha_{i}}}{\omega% \left(y\right)}\right]_{y=y_{i}}^{\left(k\right)}\cdot\tfrac{\omega\left(y% \right)}{\left(y-y_{i}\right)^{\alpha_{i}-1-k}}

(12)

\omega\left(y\right)=\prod\limits_{i=1}^{n+1}\left(y-y_{i}\right)^{\alpha_{i}}.

The polynomial defined by the relation ( 11 ) is called the inverse Hermite interpolation polynomial.

From equality ( 8 ) it follows:

(13)

f^{-1}\left(y\right)-P\left(y\right)=\tfrac{\left[f^{-1}\left(\eta\right)% \right]^{\left(m+1\right)}}{\left(m+1\right)!}\cdot\omega\left(y\right),\ \;\;% \eta\in F.

Taking into account that $f^{-1}\left(0\right)=\bar{x},$ ( 13 ) gives us

(14)

\bar{x}-P\left(0\right)=\tfrac{\left[f^{-1}\left(\eta_{1}\right)\right]^{\left% (m+1\right)}}{\left(m+1\right)!}\cdot\omega\left(0\right)

Or $\eta_{1}$ is contained in the smallest interval that contains the points: $0,f\left(x_{1}\right),\ldots,f\left(x_{n+1}\right).$ If we designate by $F_{1}$ the interval above and by $M_{m+1}$ the number $M_{m+1}=\sup\limits_{y\in F_{1}}\big{|}[f^{-1}\left(y\right)]^{\left(m+1\right% )}\big{|}$ then we deduce from ( 14 )

(15)

\left|\bar{x}-P\left(0\right)\right|\leq\tfrac{{}_{M_{m+1}}}{(m+1)!}\left|f% \left(x_{1}\right)\right|^{\alpha_{1}}\cdot\left|f\left(x_{2}\right)\right|% \ldots\left|f\left(x_{n+1}\right)\right|^{\alpha_{n+1}}.

If the points of the set $AND$ are close enough to $\bar{x}$ , then the numbers $f\left(x_{1}\right),\ f\left(x_{2}\right),\ldots,f\left(x_{n+1}\right)$ are close to zero. Even more so the number $\prod\limits_{i=1}^{n+1}\left|f\left(x_{i}\right)\right|^{\alpha_{i}}$ is close to zero and the choice of $P\left(0\right)$ as an approximation for the root $\bar{x}$ of equation ( 1 ) is justified. Therefore

(16)

\bar{x}\simeq P\left(0\right).

2. Iterative methods obtained using the inverse Hermite interpolation polynomial

Si l’approximation $P\left(0\right)$ of $\bar{x}$ given by ( 16 ) is not suitable, then by successive iterations we can obtain in certain cases approximations closer and closer to the root.

Let us designate by

(17)

x_{1},x_{2},\ldots,x_{n+1}

$n+1$ initial approximations of the root $\bar{x}$ of equation ( 1 ). Using these approximations we construct the sequence $\left(x_{p}\right)_{p=1}^{\infty}$ given by the following iterative process:

(18)

x_{n+s+1}=P_{n+s}\left(0\right),

Or $P_{n+1}\left(0\right)=P\left(0\right),\ P$ has the expression given by ( 11 ) and $P_{n+s}$ has the following expression:

(19)		$\displaystyle P_{n+s}\left(y\right)=$
	$\displaystyle=\sum_{i=1}^{n+1}\sum_{j=0}^{\alpha_{i}-1}\sum_{k=0}^{\alpha_{i}-% j-1}\!\!\!\!\left(f^{-1}\left(y_{s+i-1}\right)\right)^{\left(j\right)}\!\!% \tfrac{1}{k!j!}\!\left[\tfrac{\left(y-y_{s+i-1}\right)}{\omega_{s}\left(y% \right)}\right]_{y=y_{s+i+1}}^{\left(k\right)}\!\tfrac{\omega_{s}\left(y\right% )}{\left(y-y_{s+i-1}\right)^{\alpha_{i}-j-k}}$

(20)

\omega_{s}\left(y\right)=\prod\limits_{i=0}^{n}\left(y-y_{s+i}\right)^{\alpha_% {i}+1}.

In the following we study the convergence of the sequence $\left(x_{p}\right)_{p=1}^{\infty}$ generated by the iterative process ( 18 ). We denote by

M=\sup_{y\in F}\left|\left[\left(f^{-1}\left(y\right)\right)\right]^{\left(m+1% \right)}\right|\ \;\;\text{et\ }\beta=\sup_{x\in I}\left|f^{\prime}\left(x% \right)\right|,

then from ( 14 ) and ( 18 ) it results:

(21)

\left|\bar{x}-x_{n+2}\right|\leq\tfrac{M\beta^{m+1}}{\left(m+1\right)!}\cdot% \left|\bar{x}-x_{1}\right|^{\alpha_{1}}\ldots\left|\bar{x}-x_{n+1}\right|^{% \alpha_{n+1}}.

More generally if the elements of the sequence $\left(x_{p}\right)_{p=1}^{\infty}$ are contained in the interval $I,$ we have the following inequalities:

(22)		$\displaystyle\left\|\bar{x}-x_{n+s+1}\right\|$	$\displaystyle\leq\tfrac{M\beta^{m+1}}{\left(m+1\right)!}\left\|\bar{x}-x_{s}% \right\|^{\alpha_{1}}\cdot\left\|\bar{x}-x_{s+1}\right\|^{\alpha_{2}}\ldots\left\|% x-x_{s+n}\right\|^{\alpha_{n+1}}$
	$\displaystyle s$	$\displaystyle=1,2,3,\ldots\ .$

By multiplying the terms of ( 22 ) by $\beta\sqrt[m]{\frac{\beta M}{\left(m+1\right)!}}$ and we designate by $\rho_{i}$ the number

\rho_{i}=\beta\sqrt[m]{\tfrac{\beta M}{\left(m+1\right)!}}\cdot\left|\bar{x}-x% _{i}\right|,\ \;\;i=1,2,\ldots,

the inequalities ( 22 ) become:

(23)

\rho_{n+s+1}\leq\rho_{n+s}^{\alpha_{n+1}}\rho_{n+s-1}^{\alpha_{n}}\cdots\rho_{% s+1}^{\alpha_{2}}\rho_{s}^{\alpha_{1}},\ \;\;s=1,2,\ldots\ .

We assume in the following:

(24)

\rho_{i}\leq d^{\omega^{i}},\ \;\;i=1,2,\ldots,n+1

Or $\omega$ is the positive root of the equation

(25)

t^{n+1}=\alpha_{n+1}t^{n}+\alpha_{n}t^{n-1}+\cdots+\alpha_{2}t+\alpha_{1},

And $0<d<1.$

If we assume

(26)

\rho_{n+s+1}\leq d^{u_{n+s+1}},\ \;\;s=1,2,\ldots,

then taking into account ( 23 ) and ( 25 ) we have

(27)

u_{n+s+1}=\alpha_{n+1}u_{n+s}+\alpha_{n}u_{n+s-1}+\cdots+\alpha_{2}u_{s+1}+% \alpha_{1}u_{s}

It follows from ( 24 ) that

(28)

u_{i}=\omega^{i},\ \;\;i=1,2,\ldots,n+1.

It is easy to show that equation ( 25 ) has a single real and positive root $\omega>1.$ Then from ( 27 ), ( 26 ), ( 28 ) and taking into account that $\omega$ is the root of the equation ( 25 ) we deduce the following equalities:

(29)

u_{n+s+1}=\omega^{n+s+1},\ \;\;s=1,2,3,\ldots,

that's to say

(30)

\rho_{n+s+1}\leq d^{\omega^{n+s+1}},\ \;\;s=1,2,3,\ldots

From equality ( 3 ) it follows that $\omega>1$ And $\lim\limits_{s\rightarrow\infty}\rho_{n+s+1}=0.$ Therefore

\lim_{s\rightarrow\infty}\left|\bar{x}-x_{n+s+1}\right|=0,

what needed to be demonstrated.

The expression of $\rho_{n+s+1}$ provides the following assessment:

(31)

\left|\bar{x}-x_{n+s+1}\right|\leq\tfrac{1}{\beta\sqrt[m]{\frac{\beta M}{\left% (m+1\right)!}}}\cdot d^{\omega^{n+s+1}}

3. The special case $n=1.$

Let us designate by $a_{1},a_{2}$ two numbers of $N$ such as $a_{1}+a_{2}=m+1$ , and by $x_{1},x_{2}$ two points of the interval $I .$ Suppose that in each point $x_{i},\ i=1,2$ we know the values of the function $f$ and its successive derivatives up to the order $a_{1}-1,\ a_{2}-1$ respectively.

Using formula ( 9 ) we can calculate the values of the function $f^{-1}$ in the points $y_{1}=f\left(x_{1}\right),\ y_{2}=f\left(x_{2}\right)$ and the values of these successive derivatives up to the order $a_{1}-1,\ a_{2}-1$ respectively.

The inverse Hermite interpolation polynomial over knots $y_{1},\ y_{2}$ will have the following form:

(32)

\displaystyle P\left(y\right)

\displaystyle=\sum_{i=1}^{2}\sum_{j=0}^{a_{i}-1}\sum_{k=0}^{a_{i}-j-1}\left(f^% {-1}\left(y_{i}\right)\right)^{\left(j\right)}\cdot\tfrac{1}{k!j!}\cdot\left[% \tfrac{\left(y-y_{i}\right)^{\alpha_{i}}}{\omega\left(y\right)}\right]_{y=y_{i% }}^{\left(k\right)}\cdot\tfrac{\omega\left(y\right)}{\left(y-y_{i}\right)^{a_{% i}-j-k}},

(33)

\omega\left(y\right)=\left(y-y_{1}\right)^{a_{1}}\cdot\left(y-y_{2}\right)^{a_% {2}}.

Taking in ( 32 ) $y=0$ we obtain a first approximation of the root $\bar{x}$ of equation ( 1 ). If we denote by $x_{3}$ this approximation then we have:

(34)		$\displaystyle\bar{x}$	$\displaystyle\approx x_{3}=\sum_{i=1}^{2}\sum_{j=0}^{a_{i}-1}\sum_{k=0}^{a_{i}% -j-1}\left(f^{-1}\left(y_{i}\right)\right)^{\left(j\right)}\cdot\tfrac{1}{k!j!% }\cdot\left[\tfrac{\left(y-y_{i}\right)^{a_{i}}}{\omega\left(y\right)}\right]_% {y-y_{i}}^{\left(k\right)}\cdot\tfrac{\left(-1\right)^{a_{i}-j-k}\cdot\omega% \left(0\right)}{y_{i}^{a_{i}-j-k}}$
		$\displaystyle\cdot$

From ( 34 ) and ( 14 ) we deduce

(35)

\left|\bar{x}-x_{3}\right|\leq M\left|f\left(x_{1}\right)\right|^{a_{1}}\cdot% \left|f\left(x_{2}\right)\right|^{a_{2}},\;\;M=\sup_{\eta\in f\left(I\right)}% \tfrac{\left|\left(f^{-1}\left(\eta\right)\right)^{\left(m+1\right)}\right|}{% \left(m+1\right)!}

Using equality

(36)

\left|\bar{x}-x_{3}\right|=\tfrac{\left|f\left(x_{3}\right)\right|}{\left|% \left[\bar{x},x_{3};f\right]\right|},

where by $\left[\bar{x},x_{3};f\right]$ we denote the divided difference of the function $f$ on the knots $\bar{x},\ x_{3};$ we deduce from ( 35 ) the following inequality:

(37)

\left|f\left(x_{3}\right)\right|\leq K\left|f\left(x_{1}\right)\right|^{a_{1}}% \cdot\left|f\left(x_{2}\right)\right|^{a_{2}}

Or $K\geq\big{|}M\cdot\left[x,y;f\right]\big{|},\ x,y\in I.$

If we consider the numbers $x_{1}$ And $x_{2}$ as initial approximations for the root $\bar{x}$ from equation ( 1 ), we obtain from ( 19 ) for $y=0$ a sequel $\left(x_{n}\right)_{n=1}^{\infty}$ , with

(38)		$\displaystyle x_{s+1}=$
	$\displaystyle=\sum_{i=1}^{2}\sum_{j=0}^{a_{i}-1}\sum_{k=0}^{a_{i}-j-1}\left(f^% {-1}\left(y_{s+i-1}\right)\right)^{\left(j\right)}\tfrac{1}{k!j!}\left[\tfrac{% \left(y-y_{s+i-1}\right)^{a_{i}}}{\omega_{s}\left(y\right)}\right]_{y=y_{s+i-1% }}^{\left(k\right)}\!\!\!\tfrac{\left(-1\right)^{a_{i}-j-k}\omega_{s}\left(0% \right)}{y_{s+i-1}^{a_{i}-j-k}}$

(39)

\omega_{s}\left(y\right)=\left(y-y_{s-1}\right)^{a_{1}}\cdot\left(y-y_{s}% \right)^{a_{2}},\ \;\;s=2,3,\ldots

Assuming that all elements of the sequence $\left(x_{n}\right)_{n=1}^{\infty}$ are contained in the interval $I,$ we have the following inequalities:

(40)

\left|f\left(x_{n+2}\right)\right|\leq K\left|f\left(x_{n}\right)\right|^{a_{1% }}\cdot\left|f\left(x_{n+1}\right)\right|^{a_{2}},\ \;\;n=1,2,\ldots

If we multiply the inequalities ( 40 ) by $K^{\frac{1}{m}}$ and if we take into account that $a_{1}+a_{2}=m+1$ we get:

(41)

\displaystyle K^{\frac{1}{m}}\left|f\left(x_{n+2}\right)\right|

\displaystyle\leq\left[K^{\frac{1}{m}}\left|f\left(x_{n}\right)\right|\right]^% {a_{1}}\cdot\left[K^{\frac{1}{m}}\left|f\left(x_{n+1}\right)\right|\right]^{a_% {2}},\quad n=1,2,\ldots

In the following we designate by $\rho_{i}$ the names

(42)

\rho_{i}=K^{\frac{1}{m}}\left|f\left(x_{i}\right)\right|,\ \;\;i=1,2,\ldots

Then the inequalities ( 41 ) become:

(43)

\rho_{n+2}\leq\rho_{n}^{a_{1}}\cdot\rho_{n+1}^{a_{2}},\ \;\;n=1,2,\ldots

We will assume that

(44)

\left\{\begin{array}[c]{c}\rho_{1}\leq d^{\omega_{1}}\\ \rho_{2}\leq d^{\omega_{2}}\end{array}\right.

Or $0<d<1$ And $\omega_{1}$ is the positive root of the equation

(45)

t^{2}-a_{2}t-a_{1}=0.

From ( 43 ) we deduce:

(46)

\rho_{n}\leq d^{u_{n}}

where the sequel $\left(u_{n}\right)_{n=1}^{\infty}$ checks for equalities

(47)		$\displaystyle u_{n+2}$	$\displaystyle=a_{2}u_{n+1}+a_{1}u_{n},\ \;\;n=1,2,\ldots$
	$\displaystyle u_{1}$	$\displaystyle=\omega_{1},u_{2}=\omega_{1}^{2}.$

From ( 45 ) we deduce:

(48)

\omega_{1}=\tfrac{a_{2}+\sqrt{a_{2}^{2}+4a_{1}}}{2}

in which case ( 47 ) gives

(49)

u_{n}=\omega_{1}^{n},\ \;\;n=1,2,\ldots\ .

Deduced for $\rho_{n}$ the following assessments:

(50)

\rho_{n}\leq d^{\omega_{1}^{n}},\ \;\;n=1,2,\ldots

The previous inequality and ( 42 ) imply:

(51)

\left|f\left(x_{n}\right)\right|\leq K^{-\frac{1}{m}}\cdot d^{\omega_{1}^{n}}.

By designating by $\alpha$ the number

(52)

\alpha=\inf_{x,y\in\left[x_{1},x_{2}\right]}\big{|}\left[x,y;f\right]\big{|}

and assuming $\alpha>0,$ we deduce from ( 51 )

(53)

\left|\bar{x}-x_{n}\right|\leq\tfrac{1}{\alpha}\left|f\left(x_{n}\right)\right% |\leq\tfrac{1}{\alpha}K^{-\frac{1}{m}}\cdot d^{\omega_{1}^{n}},\qquad n=1,2,\ldots

which represents an evaluation of the error after $n-2$ no iteration.

Passing to limit when $n\rightarrow\infty$ we deduce from ( 53 )

(54)

\bar{x}=\lim_{n\rightarrow\infty}x_{n}

and from ( 51 ) it results, from the fact that $f$ is a continuous function, $f\left(x\right)=0.$

In the following we will present an analysis of the convergence speed of the sequence $\left(x_{n}\right)_{n=1}^{\infty}$ given by ( 38 ). To do this we note that if we consider in the Hermite interpolation polynomial the knot $x_{1}$ having the order of multiplicity $a_{2},$ and the knot $x_{2}$ having the order of multiplicity $a_{1},$ the equation which gives us the speed of convergence of the sequence $\left(x_{n}\right)_{n=1}^{\infty}$ ( 45 ) will take the form:

(55)

t^{2}-a_{1}t-a_{2}=0.

It can be easily demonstrated that, if $a_{1}\leq a_{2}et$ and $\omega_{2}$ is the positive root of equation ( 55 ), then:

(56)

\omega_{2}\leq\omega_{1}.

From the above considerations we conclude that the maximum order of convergence is obtained when in the two-node Hermite type inverse interpolation the multiplicity order of $x_{1}$ is at least equal to the order of multiplicity of $x_{2}.$ It goes without saying that in the above conclusion we assumed that the elements of the sequence $\left(x_{n}\right)_{n=1}^{\infty}$ are obtained using the relation ( 38 ).

In the following we present two special cases of the Hermite inverse interpolation polynomial and the related iterative methods.

1. $a_{1}=a_{2}=1.$

In this case the inverse Hermite interpolation polynomial has the following form:

(57)

P_{1}\left(y\right)=\tfrac{1}{y_{1}-y_{2}}\left[\left(y-y_{2}\right)f^{-1}% \left(y_{1}\right)-\left(y-y_{1}\right)f^{-1}\left(y_{2}\right)\right].

Taking into account that $f^{-1}\left(y_{1}\right)=x_{1}$ And $f^{-1}\left(y_{2}\right)=x_{2}$ and doing in ( 57 ) $y=0$ we obtain the first approximation of $\bar{x}$ given by the relation:

(58)

x_{3}=x_{1}-\tfrac{x_{2}-x_{1}}{f\left(x_{2}\right)-f\left(x_{1}\right)}\cdot f% \left(x_{1}\right).

Proceeding step by step we obtain the following: $\left(x_{n}\right)_{n=1}^{\infty}$ given by the relation:

(59)

x_{n+2}=x_{n}-\tfrac{x_{n+1}-x_{n}}{f\left(x_{n+1}\right)-f\left(x_{n}\right)}% \cdot f\left(x_{n}\right),\ \;\;n=1,2,\ldots,

which represents the rope method. In this case the following $\left(x_{n}\right)_{n=1}^{\infty}$ has the order of convergence $\omega_{1}=(1+\sqrt{5})/2.$

2. $a_{1}=1,\ a_{2}=2.\$ [2].

The inverse Hermite interpolation polynomial in this case takes the following form:

(60)		$\displaystyle P_{2}\left(y\right)=$	$\displaystyle\left(f^{-1}\left(y_{2}\right)\right)^{\prime}\cdot\tfrac{\left(y% -y_{1}\right)\left(y-y_{2}\right)}{y_{2}-y_{1}}+$
		$\displaystyle+f^{-1}\left(y_{2}\right)\tfrac{y-y_{1}}{y_{2}-y_{1}}\left(1+% \tfrac{y-y_{2}}{y_{1}-y_{2}}\right)+f^{-1}\left(y_{1}\right)\cdot\tfrac{\left(% y-y_{2}\right)^{2}}{\left(y_{1}-y_{2}\right)^{2}}.$

Taking into account equalities $f^{-1}\left(y_{1}\right)=x_{1},\ f^{-1}\left(y_{2}\right)=x_{2}$ And $\left(f^{-1}\left(y_{2}\right)\right)^{\prime}$ $=1/f^{\prime}\left(x_{2}\right),$ and by doing in ( 60 ) $y=0,$ we deduce:

(61)

x_{3}=x_{1}-\tfrac{x_{2}-x_{1}}{f\left(x_{2}\right)-f\left(x_{1}\right)}f\left% (x_{1}\right)+\tfrac{f\left(x_{2}\right)-f\left(x_{1}\right)-\left(x_{2}-x_{1}% \right)f^{\prime}\left(x_{2}\right)}{\left(f\left(x_{2}\right)-f\left(x_{1}% \right)\right)^{2}f^{\prime}\left(x_{2}\right)}f\left(x_{1}\right)f\left(x_{2}% \right).

Proceeding step by step we obtain the following: $\left(x_{n}\right)_{n=1}^{\infty}$

(62)		$\displaystyle x_{n+2}=$	$\displaystyle x_{n}-\tfrac{x_{n+1}-x_{n}}{f\left(x_{n+1}\right)-f\left(x_{n}% \right)}f\left(x_{n}\right)$
		$\displaystyle+\tfrac{f\left(x_{n+1}\right)-f\left(x_{n}\right)-\left(x_{n+1}-x% _{n}\right)f^{\prime}\left(x_{n+1}\right)}{\left(f\left(x_{n+1}\right)-f\left(% x_{n}\right)\right)^{2}\cdot f^{\prime}\left(x_{n+1}\right)}f\left(x_{n}\right% )f^{\prime}\left(x_{n+1}\right),$

Or $n=1,2,\ldots$

The order of convergence of this method is $\omega_{1}=1+\sqrt{2}.$

Bibliography

[1] I.G. Berezin, P.N. Zidkov, Computation Methods , I. Moscow, 1962.
[2] Gh. Coman, ^†^†margin: clickable $\rightarrow$ Some practical approximation methods for nonlinear equations, Anal. Numér. Théor. Approx., 11, 1-2, (1982), 41–48.
[3] M.A. Ostrowski, Solution of Equations and Systems of Equations, Academic Press New York - London, 1960.
[4] I. Pavaloiu, ^†^†margin: clickable $\rightarrow$
clickable $\rightarrow$
clickable $\rightarrow$ Solving equations by interpolation , Ed. Dacia, 1981.
[5] I. Păvăloiu, Introduction to the approximation of solutions of equations , Ed. Dacia, 1976.
[6] DD Stancu, On Hermite's interpolation formula and some of its applications , Studii si Cercet. Mat. (Cluj), 3-4 VIII, (1957), 339-355.
[7] J.F. Traub, Iterative Methods for the Solution of Equations, Prentice Hall Series in Automatic Computation 1964.
[8] BA Turowicz, On the higher order derivatives of an inverse function , Colloq. Math., (1959), 83–87.

Received on 06.06.1982

Institute of Computing

3400 Cluj-Napoca, Romania

Solving equations by Hermite type inverse interpolation

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Solving equations by Hermite type inverse interpolation

1.

Theorem 1 .

2. Iterative methods obtained using the inverse Hermite interpolation polynomial

3. The special case $n=1.$

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Solving equations by Hermite type inverse interpolation

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Solving equations by Hermite type inverse interpolation

1.

Theorem 1 .

2. Iterative methods obtained using the inverse Hermite interpolation polynomial

3. The special casen=1.

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3. The special case $n=1.$