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Divided Differences

DEFINITION 12.2.1 (First Divided Difference)   The ratio

$\displaystyle \frac{f(x_i) - f(x_j)}{x_i - x_j}$

for any two points $ x_i$ and $ x_j$ is called the FIRST DIVIDED DIFFERENCE of $ f(x)$ relative to $ x_i$ and $ x_j.$ It is denoted by $ \delta[x_i, x_j].$

Let us assume that the function $ y=f(x)$ is linear. Then $ \delta[x_i,x_j]$ is constant for any two tabular points $ x_i$ and $ x_j,$ i.e., it is independent of $ x_i$ and $ x_j.$ Hence,

$\displaystyle \delta[x_i, x_j] = \frac{f(x_i) - f(x_j)}{x_i - x_j}= \delta[x_j,x_i].$

Thus, for a linear function $ f(x),$ if we take the points $ x, x_0$ and $ x_1$ then, $ \delta[x_0,x] = \delta[x_0,x_1],$ i.e.,

$\displaystyle \frac{f(x) - f(x_0)}{x - x_0}= \delta[x_0,x_1].$

Thus, $ f(x) = f(x_0) + (x - x_0) \delta[x_0, x_1].$

So, if $ f(x)$ is approximated with a linear polynomial, then the value of the function at any point $ x$ can be calculated by using $ f(x) \approx P_1(x) = f(x_0) + (x - x_0) \delta[x_0, x_1],$ where $ \delta[x_0,x_1]$ is the first divided difference of $ f$ relative to $ x_0$ and $ x_1.$

DEFINITION 12.2.2 (Second Divided Difference)   The ratio

$\displaystyle \delta[x_i, x_j,x_k] = \frac{\delta[x_j,x_k] - \delta[x_i,x_j]}{x_k - x_i}$

is defined as SECOND DIVIDED DIFFERENCE of $ f(x)$ relative to $ x_i, x_j$ and $ x_k.$

If $ f(x)$ is a second degree polynomial then $ \delta[x_0, x]$ is a linear function of $ x.$ Hence,

$\displaystyle \delta[x_i, x_j,x_k] = \frac{\delta[x_j,x_k] - \delta[x_i,x_j]}{x_k - x_i} \;\;{\mbox{ is constant}}.$

In view of the above, for a polynomial function of degree 2, we have $ \delta[x,x_0,x_1] = \delta[x_0,x_1,x_2].$ Thus,

$\displaystyle \frac{\delta[x,x_0] - \delta[x_0,x_1]}{x - x_1} = \delta[x_0,x_1,x_2].$

This gives,

$\displaystyle \delta[x,x_0] = \delta[x_0,x_1] + (x - x_1) \delta[x_0,x_1,x_2].$

From this we obtain,

$\displaystyle f(x) = f(x_0) + (x - x_0) \delta[x_0,x_1] + (x - x_0)(x -x_1)\delta[x_0,x_1,x_2].$

So, whenever $ f(x)$ is approximated with a second degree polynomial, the value of $ f(x)$ at any point $ x$ can be computed using the above polynomial, which uses the values at three points $ x_0, x_1$ and $ x_2.$

EXAMPLE 12.2.3   Using the following tabular values for a function $ y = f(x),$ obtain its second degree polynomial approximation.
$ i$ 0 1 2
$ x_i$ 0.1 0.16 0.2
$ f(x_i)$ 1.12 1.24 1.40
Also, find the approximate value of the function at $ x = 0.13.$
Solution: We shall first calculate the desired divided differences.
$\displaystyle \delta[x_0,x_1]$ $\displaystyle =$ $\displaystyle (1.24 - 1.12)/(0.16 - 0.1) = 2,$  
$\displaystyle \delta[x_1,x_2]$ $\displaystyle =$ $\displaystyle (1.40-1.24)/(0.2-0.16) = 4, \;\; {\mbox{ and }}$  
$\displaystyle \delta[x_0,x_1,x_2]$ $\displaystyle =$ $\displaystyle \frac{\delta[x_1,x_2] - \delta[x_0,x_1]}{x_2 - x_0} = (4-2)/(0.2-0.1) = 20.$  

Thus,

$\displaystyle f(x) \approx P_2(x) = 1.12 + 2(x - 0.1) + 20 (x-0.1)(x-0.16).$

Therefore

$\displaystyle f(0.13) \approx 1.12 + 2(0.13 - 0.1) + 20 (0.13-0.1)(0.13-0.16)=1.162.$

EXERCISE 12.2.4  
  1. Using the following table, which gives values of $ \log(x)$ corresponding to certain values of $ x,$ find approximate value of $ \;\;\log(323.5)$ with the help of a second degree polynomial.
    $ x$ 322.8 324.2 325
    $ \log(x)$ 2.50893 2.51081 2.5118
  2. Show that

    $\displaystyle \delta[x_0,x_1,x_2] = \frac{f(x_0)}{(x_0-x_1)(x_0-x_2)} + \frac{f(x_1)}{(x_1-x_0)(x_1-x_2)}+\frac{f(x_2)}{(x_2-x_0)(x_2-x_1)}.$

    So, $ \delta[x_0,x_1,x_2] = \delta[x_0,x_2,x_1] = \delta[x_1,x_0,x_2] = \delta[x_1,x_2,x_0] = \delta[x_2,x_0,x_1] = \delta[x_2,x_1,x_0].$ That is, the second divided difference remains unchanged regardless of how its arguments are interchanged.
  3. Show that for equidistant points $ x_{0},$ $ x_1$ and $ x_2,$ $ \;\delta[x_0,x_1,x_2] = \displaystyle\frac{\Delta^2 y_0}{2 h^2}=\displaystyle\frac{\nabla^2 y_2}{2 h^2},$ where $ y_k = f(x_k),$ and $ h = x_1 - x_0 = x_2 - x_1.$
  4. Show that for a linear function, the second divided difference with respect to any three points, $ x_{i},x_{j}$ and $ x_{k},$ is always zero.

Now, we define the $ k^{\mbox{th}}$ divided difference.

DEFINITION 12.2.5 ( $ k^{\mbox{th}}$ Divided Difference)   The #MATH5307# MATHEND000# DIVIDED DIFFERENCE of $ f(x)$ relative to the tabular points $ x_0, x_1, \ldots, x_k,$ is defined recursively as

$\displaystyle \delta[x_0,x_1,\ldots,x_k] = \frac{\delta[x_1,x_2,\ldots,x_k] - \delta[x_0,x_1, \ldots,x_{k-1}]}{x_k - x_0}.$

It can be shown by mathematical induction that for equidistant points,

$\displaystyle \delta[x_0,x_1, \ldots, x_k] = \frac{\Delta^k y_0}{k! h^k}=\frac{\nabla^k y_k}{k! h^k}$ (12.2.1)

where, $ y_0 = f(x_0),$ and $ h = x_1 - x_0 = x_2 - x_1 = \cdots = x_k - x_{k-1}.$

In general,

$\displaystyle \delta[x_i,x_{i+1}, \ldots, x_{i+n}] = \frac{\Delta^n y_i}{n! h^n},$

where $ y_i = f(x_i)$ and $ h$ is the length of the interval for $ i=0, 1, 2, \ldots.$

Remark 12.2.6   In view of the remark (11.2.18) and (12.2.1), it is easily seen that for a polynomial function of degree $ n,$ the $ n^{\mbox{th}}$ divided difference is constant and the $ (n+1)^{\mbox{th}}$ divided difference is zero.

EXAMPLE 12.2.7   Show that $ f(x)$ can be written as

$\displaystyle f(x)=f(x_{0})+\delta[x_0,x_1](x-x_{0})+\delta[x,x_0,x_1](x-x_0)(x-x_1).$

Solution:By definition, we have

$\displaystyle \delta[x,x_0,x_1]=\frac{\delta[x,x_0]-\delta[x_0,x_1]}{(x-x_1)},$

so, $ \delta[x,x_0]=\delta[x_0,x_1]+(x-x_0)\delta[x,x_0,x_1].$ Now since,

$\displaystyle \delta[x,x_0]=\frac{f(x)-f(x_0)}{(x-x_0)},$

we get the desired result.

EXERCISE 12.2.8   Show that $ f(x)$ can be written in the following form:

$\displaystyle f(x) = P_{2}(x)+ R_{3}(x), $

where, $ P_{2}(x)=f(x_{0})+\delta[x_0,x_1](x-x_{0})+\delta[x_0,x_1,x_2](x-x_0)(x-x_1)$
and $ R_{3}(x)=\delta[x,x_0,x_1,x_2](x-x_0)(x-x_1)(x-x_2).$

Further show that $ P_{2}(x_{i})=f(x_i)$ for $ i=0 , 1.$

Remark 12.2.9   In general it can be shown that $ f(x)= P_{n}(x)+R_{n+1}(x),$ where,
$\displaystyle P_{n}(x)$ $\displaystyle =$ $\displaystyle f(x_{0})+\delta[x_0,x_1](x-x_{0})+\delta[x_0,x_1,x_2] (x-x_0)(x-x_1) +\cdots$  
    $\displaystyle + \delta[x_0,x_1,x_2,\ldots,x_{n}](x-x_0)(x-x_1)(x-x_2)\cdots(x-x_{n-1}),$  

and $ R_{n+1}(x)=(x-x_0)(x-x_1)(x-x_2)\cdots(x-x_n)\delta[x,x_0,x_1,x_2,\ldots,x_{n}].$

Here, $ R_{n+1}(x)$ is called the remainder term.

It may be observed here that the expression $ P_{n}(x)$ is a polynomial of degree $ 'n'$ and $ P_{n}(x_{i})=f(x_i)$ for $ i=0 , 1, \cdots, (n-1).$

Further, if $ f(x)$ is a polynomial of degree $ n,$ then in view of the Remark 12.2.6, the remainder term, $ R_{n+1}(x)=0, $ as it is a multiple of the $ (n+1)^{\mbox{th}}$ divided difference, which is 0 .

next up previous contents
Next: Lagrange's Interpolation formula Up: Lagrange's Interpolation Formula Previous: Introduction   Contents
A K Lal 2007-09-12