Question

Use the approximation (1+x) ~1+nx,[x]<<1,to find
Approximate value for a)√99 b)1/1.01​

Answer 1

Answer:

99)^1/2

=(100–1)^1/2

=(100)^1/2.( 1- 1/100)^1/2

=10.[1 + ( -1/100) ]^1/2 .

=10[ 1+1/2.(-1/100) +(1/2).(-1/2).(-1/100)^2/(1.2)+…………………].

=10[1 - 1/200- (1/8)/10000+…………]

=10[1–0.005 -0.0000125+…….]

= 10 [1 -0.0050125+………….]

= 10[0.9949875+……..]

= 9.949875 , Answer.

Answer 2

Use the approximation,

$(1+x)^n\approx1+nx,|x|<<1$

We have to find the approximate value for

$(a) \sqrt{99} \: \: \: \: \: \: \: \: (b) \frac{1}{1.01}$

$(a)\sqrt{99}$

$=\sqrt{100-1}\\\\=\sqrt{100\left(1-\frac{1}{100}\right)}\\\\=10(1-0.01)^{\frac{1}{2}}$

here, 0.01 << 1

Using approximation,

$(1+x)^n\approx1+nx,|x|<<1$

≈ 10(1 - 0.01/2)

= 10(1 - 0.005)

= 10 × 0.995

= 9.95

Therefore the approximate value for √99 ≈ 9.95.

$(b)\frac{1}{1.01}$

=$(1.01)^{-1}\\\\=(1+0.01)^{-1}$

here, 0.01 << 1

Using approximation,

$(1+x)^n\approx1+nx,|x|<<1$

≈ (1 -0.01)

= (0.99)

Therefore the approximate value for 1/1.01 is 0.99