solution of equations by divided difference interpolaton

Divided diference interpolation can be used to obtain approximatesolutions to equations and to invert functions numerically. Theidea is that, given an equation $f(y)=x$ which we want to solvefor $y$, we first take several numbers $y_1,\mathrm{\dots },y_n$ andcompute $x_1\mathrm{\dots }{x}_n$ as $x_i=f(y_i)$. Then we compute thedivided 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 transcendentalequation $x+e^{-x}=2$. We note that $2+e^{-2}=2.13533$and $1.5+e^{-1.5}=1.72313$, so there will be a solutionbetween $1.5$ 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:

Next, we form a divided difference table of the $y_i$’s as afunction of the $x_i$’s:

From this table, we form the series

$1.50000$$+1.26952(x-1.72313)-0.19799(x-1.72313)(x-1.80190)$

$+0.12082(x-1.72313)(x-1.80190)(x-1.88268)$

$-0.039609(x-1.72313)(x-1.80190)(x-1.88268)(x-1.96530)$

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

$-0.13457(x-1.72313)(x-1.80190)(x-1.88268)(x-1.96530)(x-2.04957)(x-2.13533)$

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

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

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

Existence follows from the intermediate value theorem. Asnoted above,

Since $x+e^{-x}$ depends continuously on $x$, it followsthat 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 , or$1-e^{-x}>0$. Hence, $x+e^{-x}$ is a strictly increaingfunction of $x$, so there can be at most one $x$ such that$x+e^{-x}=2$.