What is Lagrange interpolation?

In numerical analysis, Lagrange polynomials are used for polynomial interpolation. For a given set of points with no two values equal, the Lagrange polynomial is the polynomial of lowest degree that assumes at each value the corresponding value, so that the functions coincide at each point. Move to advantages of lagrange's interpolation formula. Move to advantages of lagrange's interpolation method. Move to advantages and disadvantages of lagrange's interpolation method.

## What is Interpolation?

Interpolation is a method of finding new data points within the range of a discrete set of known data points. In other words interpolation is the technique to estimate the value of a mathematical function, for any intermediate value of the independent variable.

## Introduction To Lagrange's Interpolation

We have various interpolation methods like Newton's forward and backward interpolation formulae and central difference formulae like Gauss forward and backward formulae and Bessel's interpolation formulae. All these methods are used when there are equal intervals in the values or arguments. So this is the disadvantage of these methods. So it is necessary to have some formula to for unequally spaced arguments/entries. For unequally spaced arguments there are various methods or formulae. Out of which, some are listed below:
1. Lagrange's Interpolation Formula
2. Newtons Divided Difference Formula
3. Hermite's Interpolation Formula

## Lagrange's Interpolation Formula

### Interpolation With Unequal Intervals

#### Statement:

Let f(x0), f(x1), f(x2),......,f(x0) be the values (entries) of the function f(x) corresponding to the arguments x0,x1,x2,...,xn, not necessarily equally spaced then

Here is the Lagrange's Interpolation formula for unequally spaced arguments
```f(x) = {[(x-x1)(x-x2).....(x-xn)] / [(x0-x1)(x0-x2).....(x0-xn)]}f(x0) +

{[(x-x0)(x-x2).....(x-xn)] / [(x1-x0)(x1-x2).....(x1-xn)]}f(x1) + .....

....{[(x-x0)(x-x1).....(x-xn-1)] / [(xn-x0)(xn-x1).....(xn-xn-1)]}f(xn)
```

#### Proof:

Since we are given (n+1) tabular values, we can fit a polynomial of degree n to the data. Let this polynomial be
```f(x) = (x-x1)(x-x2)(x-x3).....(x-xn)A0 +
(x-x0)(x-x2)(x-x3).....(x-xn)A1 +
(x-x0)(x-x1)(x-x3).....(x-xn)A2 + ....
....(x-x0)(x-x1)(x-x2).....(x-xn-1)An     .......(1)

where A0,A1,A2,....,An are constants to be determined.
```

Now let x=x0 in (1),
∴ f(x0) = (x0-x1)(x0-x2).....(x0-xn)A0
∴ A0 = f(x0) / [(x0-x1)(x0-x2).....(x0-xn)]

For finding A1, let x=x1,
∴ A1 = f(x1) / [(x1-x0)(x1-x2).....(x1-xn)]

Similarly if we put x=x2 in (1), we get
∴ A2 = f(x2) / [(x2-x0)(x2-x1)(x2-x3).....(x2-xn)]

Finally put x=xn in (1) to get An as
An = f(xn) / [(xn-x0)(xn-x1).....(xn-xn-1)]

```Putting the values of these constraints in (1), we get
f(x) = {[(x-x1)(x-x2).....(x-xn)] / [(x0-x1)(x0-x2).....(x0-xn)]}f(x0) +

{[(x-x0)(x-x2).....(x-xn)] / [(x1-x0)(x1-x2).....(x1-xn)]}f(x1) + .....

....{[(x-x0)(x-x1).....(x-xn-1)] / [(xn-x0)(xn-x1).....(xn-xn-1)]}f(xn)

which is Lagrange's interpolation formula for unequal intervals.
```

## Advantages of Lagrange's Interpolation Formula

1. Used in simultaneous optimization of norms of derivatives of lagrange polynomials
2. The answers for higher order polynomials will be more accurate.
3. For higher order polynomials the approximate result converges to the exact solution very quickly.
4. For higher order derivatives where n refers to the order of lagrange polynomial, the error decreases by 2n+1 if we decrease the distance between interpolation points by 2.

## Disadvantages of Lagrange's Interpolation Method

1. It becomes a tedious job to do when the polynomial order increases because the number of points increases and we need to evaluate approximate solutions for each point.

You might want to check out newton raphson method and regula falsi method.