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Homesolution of equations by divided difference interpolaton

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# solution of equations by divided difference interpolaton

Divided diference interpolation can be used to obtain approximate solutions to equations and to invert functions numerically. The idea is that, given an equation $f(y)=x$ which we want to solve for $y$, we first take several numbers $y_{1},\ldots,y_{n}$ and compute $x_{1}\ldots x_{n}$ as $x_{i}=f(y_{i})$. Then we compute the divided differences of the $y_{i}$’s regarded as functions of the $x_{i}$’s and form the divided difference series. Substituting $x$ in this series provides an approximation to $y$.

To illustrate how this works, we will examine the transcendental equation $x+e^{{-x}}=2$. We note that $2+e^{{-2}}=2.13533$ and $1.5+e^{{-1.5}}=1.72313$, so there will be a solution between $1.5$ and $2$, likely closer to $2$ than $1.5$. Therefore, as our values of the $y_{i}$’s, we shall take $1.5$, $1.6$, $1.7$, $1.8$, $1.9$, $2.0$, $2.1$. We now tabulate $x_{i}=y_{i}+e^{{-y_{i}}}$ for those values:

$y_{i}$ | $x_{i}$ |

$1.5$ | $1.72313$ |

$1.6$ | $1.80190$ |

$1.7$ | $1.88268$ |

$1.8$ | $1.96530$ |

$1.9$ | $2.04957$ |

$2.0$ | $2.13533$ |

$2.1$ | $2.22246$ |

Next, we form a divided difference table of the $y_{i}$’s as a function of the $x_{i}$’s:

$\begin{matrix}1.72313&1.50000&&&&&&\\ &&1.26952&&&&&\\ 1.80190&1.60000&&-0.19799&&&&\\ &&1.23793&&0.12082&&&\\ 1.88268&1.70000&&-0.16873&&-0.039609&&\\ &&1.21036&&0.10789&&0.091553&\\ 1.96530&1.80000&&-0.14201&&-0.077347&&-0.13457\\ &&1.18666&&0.08210&&0.024360&\\ 2.04957&1.90000&&-0.12127&&-0.067102&&\\ &&1.16604&&0.05930&&&\\ 2.13533&2.00000&&-0.10602&&&&\\ &&1.14771&&&&&\\ 2.22246&2.10000&&&&&&\end{matrix}$ |

From this table, we form the series

$\displaystyle 1.50000$ | $\displaystyle+1.26952(x-1.72313)-0.19799(x-1.72313)(x-1.80190)$ | ||

$\displaystyle+0.12082(x-1.72313)(x-1.80190)(x-1.88268)$ | |||

$\displaystyle-0.039609(x-1.72313)(x-1.80190)(x-1.88268)(x-1.96530)$ | |||

$\displaystyle+0.091553(x-1.72313)(x-1.80190)(x-1.88268)(x-1.96530)(x-2.04957)$ | |||

$\displaystyle-0.13457(x-1.72313)(x-1.80190)(x-1.88268)(x-1.96530)(x-2.04957)(x% -2.13533)$ |

Substituting $2.00000$ for $x$, we obtain $1.84140$. Given that

$1.84140+e^{{-1.84140}}<2<1.84141+e^{{-1.84141}},$ |

this answer is correct to all $5$ decimal places.

In the presentation above, we tacitly assumed that there was a solution to our equation and focussed our attention on finding that answer numerically. To complete the treatment we will now show that there indeed exists a unique solution to the equation $x+e^{{-x}}=2$ in the interval $(0,\infty)$.

Existence follows from the intermediate value theorem. As noted above,

$1.5+e^{{-1.5}}<2<3+e^{{-2}}.$ |

Since $x+e^{{-x}}$ depends continuously on $x$, it follows that there exists $x\in(1.5,2)$ such that $x+e^{{-x}}=2$.

As for uniqueness, note that the derivative of $x+e^{{-x}}$ is $1-e^{{-x}}$. When $x>0$, we have $e^{{-x}}<1$, or $1-e^{{-x}}>0$. Hence, $x+e^{{-x}}$ is a strictly increaing function of $x$, so there can be at most one $x$ such that $x+e^{{-x}}=2$.

## Mathematics Subject Classification

39A70*no label found*

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