HOW TO FIND APPROXIMATE VALUE USING BINOMIAL EXPANSION

Example 1 :

Find 3√1001 approximately (two decimal places).

Solution :

3√1001  =  (1001)1/3  =  (1001)1/3 

  =  (1000 + 1)1/3

  =  (1000)1/3 (1 + (1/1000))1/3

  =  10 (1 + (1/1000))1/3

  =  10 [1 + (1/3) (1/1000) + ...................]

  =  10 [1 + (0.333)(0.001) ...................] 

  =  10 [1 + 0.000333 ...................] 

  =  10 [1.000333 ...................] 

  =  10.0033..........

Hence the approximate value of 3√1001 is 10.0033..........

Example 2 :

Prove that 3√(x3 + 6) − 3√(x3 + 3) is approximately equal to 1/x2 when x is sufficiently large.

Solution :

3√(x3 + 6)  =  (x3 + 6)1/3  =  (x3)1/3 [1 + (6/x3)]1/3

  =  x [1 + (1/3)(6/x3) + ((1/3)(-2/3))/2 (6/x3)2 + ..............]

  =  x [1 + (2/x3) + (4/x6) + ..............]

  =  x + (2/x2) - (4/x5) + ..............   ------(1)

3√(x3 + 3)  =  (x3 + 3)1/3  =  (x3)1/3 [1 + (3/x3]1/3

  =  x [1 + (1/3)(3/x3) + ((1/3)(-2/3)/2)(3/x3)2 + ..............]

  =  x [1 + (1/x3) + (-1/x6) + ..............]

  =  x + (1/x2) - (1/x5) + ..............   ------(2)

3√(x3 + 6) − 3√(x3 + 3)

  =   (x+(2/x2)-(4/x5)+ ..............)-(x+(1/x2)-(1/x5) + ..............)

  =  x - x + (2/x2) - (1/x2)

  =  1/x2

Example 3 :

Prove that √(1−x)/(1+x) is approximately equal to 1 − x + x2 when x is very small.

Solution :

√(1−x)/(1+x)  =  [(1 - x)/(1 + x)]1/2

  =  [(1 - x)1/2/(1 + x)]1/2

  =  (1 - x)1/2(1 + x)-1/2

(1 - x)n

(1 - x)1/2  =  1 - (1/2)x + ((1/2)(-1/2)/2!)x2+....................

(1 + x)-1/2  =  1 - (x/2)x - (x2/8)+....................

(1 + x)-n

(1 + x)-1/2  =  1 - (1/2)x + ((1/2)(3/2)/2!)x2+....................

(1 + x)-1/2  =  1 - (x/2) + (3x2/8) +....................

(1 + x)-1/2 (1 + x)-1/2 

  =  (1-(x/2)-(x2/8)+..........)(1 - (x/2)+(3x2/8) +..............)

  = 1 - x/2 + 3x2/8 - x/2 + x2/4 - 3x3/16 - x2/8 + x3/16 +  ..............

  =  1 - [(x/2) + (x/2)] + [(3x2/8) + (x2/4)- (x2/8)

] + .........

  =  1 - x + [(3x2+ 2x2-x2)/8]

  =  1 - x + (4x2/8)

  =  1 - x + x2/2

Hence proved.

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