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:

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

From this table, we form the series

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 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,

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 $x+e^{{-x}}=2$.