FIND MACLAURIN EXPANSION

Taylor’s series and Maclaurin's series expansion of a function which are infinitely differentiable

(i)  e^x

Solution :

Let f(x)  =  e^x

(ii)  sin x

Solution :

Let f(x)  =  sin x

f^v(x)  =  cos x

f^v(0)  =  cos (0)

f^v(0)  =  1

(iii)  cos x

Solution :

Let f(x)  =  cos x

(iv)  log (1-x) ; -1 ≤ x < 1

Solution :

f(x)  =  log (1-x)

f(0)  =  log (1-0)  ==>  0

f'(x)  =  -1/(1-x)

f'(0)  =  -1

f''(x)  =  -1/(1-x)²

f''(0)  =  -1

f'''(x)  =  -2/(1-x)³

f'''(0)  =  -2

(v)  tan^-1 x ; -1 ≤ x ≤ 1

Solution :

f(x)  =  tan^-1 x

f(0)  =  tan^-1 (0)  ==>  0

f'(x)  =  1/(1+x²)

f'(x)  =  (1+x²)^-1

f'(x)  =  1 - x² + x⁴ - x⁶ +...............

f'(0)  =  1

f''(x)  =  0 - 2x + 4x³ - 6x⁵ +...............

f''(0)  =  0

f'''(x)  =  -2 + 12x² - 30x⁴ +...............

f'''(0)  =  -2

f'^v(x)  =  24x - 120x³ +...............

f'^v(0)  =  0

f^v(x)  =  24 - 360x² +...............

f^v(0)  =  24

(vi)  cos² x

Solution :

f(x)  =  cos² x

f(0)  =  cos² (0) 

f(0)  =  1

f'(x)  =  2cos x (-sin x)

f'(x)  =  -sin 2x

f'(0)  =  -sin 2(0)  =  0

f''(x)  =  -cos 2x (2)

f''(x)  =  -2cos 2x

f''(0)  =  -2cos 2(0)  ==>  -2

f'''(x)  =  4sin 2x

f'''(0)  =  4sin 2(0)  ==>  0

f'^v(x)  =  8cos 2x

f'^v(0)  =  8cos 2(0)  ==>  8

cos²x  =  1 + (0x)/1! + (-2)x²/2! + 0x³/3! + 8x⁴/4! +.......

cos²x  =  1 - x² + x⁴/3 +.......

Problem 2 :

Write down the Taylor series expansion, of the function log x about x = 1 upto three non-zero terms for x > 0

Solution :

f(x)  =  log x

f(x)  =  f(1) + [(x-1)/1!] f'(1) + [(x-1)²/2!] f''(1)

+ [(x-1)³/3!] f'''(1) +................

f(x)  =  log x

f(1)  =  log 1  ==>  0

f'(x)  =  1/x

f'(1)  =  1/1  ==>  1

f''(x)  =  -1/x²

f''(1)  =  -1/1²  ==>  -1

f'''(x)  =  1/x³

f'''(1)  =  1/1³  ==>  1

log x  =  0 + [(x-1)/1!](1) + [(x-1)²/2!] (-1)

+ [(x-1)³/3!] (1) +................

log x  =  (x-1) - [(x-1)²/2] + [(x-1)³/6] (1) +................

Problem 3 :

Expand sin x in ascending powers x-π/4 upto three non zero terms.

Solution :

f(x)  =  sin x

x  =  x-π/4

f(π/4)  =  sin π/4  =  1/√2

f'(x)  =  cos x : f'(π/4)  =  1/√2

f''(x)  =  -sin x : f''(π/4)  =  -1/√2

f(x)  =  f(π/4) + [(x-π/4)/1!] f'(π/4) + [(x-π/4)²/2!] f''(π/4) + [(x-π/4)³/3!] f'''(π/4) +................

f(x)  =  (1/√2) + (x-π/4) (1/√2) + [(x-π/4)²/2] (-1/√2)

 + [(x-π/4)³/6] (-1/√2) +................

f(x)  =  (1/√2)[1 + (x-π/4) - [(x-π/4)²/2]

 - [(x-π/4)³/6] +................

Problem 4 :

Expand the polynomial f(x)  =  x²-3x+2 in powers of x-1.

Solution :

f(x)  =  x²-3x+2

f(1)  =  1²-3(1)+2  =  0

f'(x)  =  2x-3

f'(1)  =  2(1)-3  =  -1

f''(x)  =  2

f(x)  =  f(1)+[(x-1)/1!]f'(1)+[(x-1)²/2!]f''(1)

 +[(x-1)³/3!]f'''(1)................

f(x)  =  0+[(x-1)/1!](-1)+[(x-1)²/2!](2)

f(x)  =  -(x-1)+(x-1)²

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