Quadratic Interpolation Formula with Problem Solution & Solved Example

Table of Contents

Interpolation is a technique for calculating values between the lines within a table. This is one of the simplest process that is based on Quadratic approximation polynomial. Interpolation is a popular for tabular form function. It is applicable on polynomials even with approximately low degrees. This is an integral part of numerical analysis where values are obtained with the help of an algorithm in mathematics. Also, you can use calculators or built-in function if necessary.

Interpolation is still important for functions that are stored in the tabular format and they are introduced to understand the wider application of finite differences. If you will check the dictionary meaning of word interpolation then this is the estimation of unknown quantities between any two known quantities.

If you know the long-term cycles, seasonality or trends then calculating interpolation values is easy. It is frequently used for regression analysis or series analysis in mathematics. This is a special case of curve fitting and solutions are easily to identify with fairly low degrees.

The formula of quadratic interpolation in mathematics is given as below:

\[\large f(x_{j}+\theta h)\approx f_{j}+ \theta \triangle f_{j}+ \frac{1}{2}\theta (\theta -1)\triangle^{2}f_{j}\]

When you are living in the computer age then working on interpolation technique is easy with pre-built functions and algorithms. But formula is still important when data is arranged in tabular format and applicable for finite differences. During the construction of interpolation polynomials, there is a trade-off between better fit and smooth well-behaved functions. More data points used for the interpolation, higher the degree of resultant polynomials and accuracy of data points will be perfect.

Solved Example

Question: Find the estimate of $cos(80^{\circ}: {35}’)$ by quadratic interpolation ?

Table of $cos x$:

x $80^{\circ}$
0′ 0.1736
10′ 0.1708
20′ 0.1679
30′ 0.1650
40′ 0.1622


Given Table,

x f(x) = $cos x$ $\delta$ $\delta^{2}$
$80^{\circ}$ 0′ 0.1736
$80^{\circ}$ 10′ 0.1708 -28 -1
$80^{\circ}$ 20′ 0.1679 -29 0
$80^{\circ}$ 30′ 0.1650 -29 1
$80^{\circ}$ 40′ 0.1622 -28 -1
Using the formula

$cos 80^{\circ}35’$ = $f(80^{\circ}30′) + 0.5\delta f(80^{\circ}30′) + \frac{1}{2} 0.5(0.5 – 1)\delta^{2}f(80^{\circ}30′)$

$cos 80^{\circ}35’$ = 0.1650 + 0.5(-0.0028) + (0.5)(0.5)(-0.5)(-0.0001)

$cos 80^{\circ}35’$ = 0.1636