## What is Taylor Series

In mathematics, sometimes it is really difficult to evaluate some functions. In such cases, we use the approximation formulas where the function is expressed as a series. There are two such approximation formulas:

- Taylor series formula
- Maclaurin series formula

Taylor series formula helps us in writing a function as a series (or sum) of terms involving the derivatives of the function.

This formula helps us in finding the approximate value of the function.

## Taylor series formula

### f( x+h )=f( x ) + f'(x)h + f”{(x) h^{2}}/2! + f”'{(x) h^{3}}/3! + ⋯

or

## 1 Example Solved

Given f(4)=125, f’(4)=74, f’’(4)=30, f’’’(4)=6. Find **f(6)=?**

** Sol:** Here h = 2, and all other derivates are zero.

F(6) = 125 + 74(2) + 30(2)^{2} / 2 + 6(2)^{3} / 3 + 0 + 0 + 0

F(6) = 125 + 148 + 60 + 8

F(6) = **341 Ans.**

## 2 Example Solved

### Given The Ordinary differential dy/dx = 3×2 – x2y, y(2)=5. Find 2nd Order Polynomial for y as a function of x & x=2.

Sol.

Let the 2^{nd} order polynomial be **a _{0}+a_{1}x + a_{2}x^{2}**

Using taylor series, *f( x+h )=f( x ) + f'(x)h + f”{(x) h ^{2}}/2! + f”'{(x) h^{3}}/3! + ⋯*

Given, y(2) = 5 and x =2.

Therefore,

y’(2) = 3x^{2} – x^{2}y

3(2)^{2} – (2)^{2}y

12 – 4y

Y = 12/4

Y=3, put this value of y

Y’(2)=3(2)^{2} – (2)-8

Y’(2) = -8

Now, y’’(2) = (d/dx)(3x^{2} – x^{2}y)

6x – (d/dx)(x^{2}y)

Use product Rule of differentiation, xy

6x – ( 2xy – x^{2} d/dx(y))

6(2) – 2(2)(5) – 22(-8) = 24

Therefore, y(2+h)= 5 + (-8)h + 24h^{2}

Y(2+h) = 5 – 8h + 12h^{2}

Y(x) = 5-8(x-2)+12(x-2)^{2}

**Y(x)=12x ^{2}-56x+69 ** (Answer.)

*Suggested Methods:*