Taylor series Formula with Solved Examples | What is Taylor Series

taylor series formula

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:

  1. Taylor series formula
  2. 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) h2}/2! + f”'{(x) h3}/3! + ⋯

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taylor series formula
Taylor Series Formula

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

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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 2nd order polynomial be a0+a1x + a2x2

Using taylor series, f( x+h )=f( x ) + f'(x)h + f”{(x) h2}/2! + f”'{(x) h3}/3! + ⋯

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

Therefore,

y’(2) = 3x2 – x2y

3(2)2 – (2)2y

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)(3x2 – x2y)

6x – (d/dx)(x2y)

Use product Rule of differentiation, xy

product rule differentiation

6x – ( 2xy – x2 d/dx(y))

6(2) – 2(2)(5) – 22(-8) = 24

Therefore, y(2+h)= 5 + (-8)h + 24h2

Y(2+h) = 5 – 8h + 12h2

Y(x) = 5-8(x-2)+12(x-2)2

Y(x)=12x2-56x+69   (Answer.)

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