Dictionary meaning of interpolation is the estimation of an unknown quantity between two known quantities. If the trends, seasonality and longer term cycles are known then interpolation is easy. The concept of interpolation can be shown in series analysis and regression analysis in statistics.

It is a special case of curve fitting. The solution can be approximated by a low degree when the observation has fairly smooth behaviour between the tabular data and an assumption can be made.

The formula of quadratic interpolation is given as:

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

Even though now that electronic and technology is so advanced that the computers and calculators have a built in function where the algorithm for interpolation is available readily, but, this formula is still important if available only in tabular form and it serves as an introduction for wider application of finite differences.

### 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 |

**Solution:**

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 |

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

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

$cos 80^{\circ}35’$ = 0.1636