If-y-x-find-approximate-value-of-101-

Question Number 59733 by otchereabdullai@gmail.com last updated on 14/May/19

If y = \sqrt{x}, find approximate value of

\sqrt{101}

Commented by prakash jain last updated on 14/May/19

{(101)}^{1 / 2} = 100^{1 / 2} {(1 + \frac{1}{100})}^{1 / 2}

= 10 (1 + \frac{1}{200} + \frac{\frac{1}{2} (\frac{1}{2} - 1)}{100^{2}} + \dots)

ignore \frac{1}{100^{2}} and smaller terms

\approx 10.05

Commented by otchereabdullai@gmail.com last updated on 14/May/19

thanks sir

Answered by tanmay last updated on 14/May/19

y = \sqrt{x}

\frac{d y}{d x} = \frac{1}{2 \sqrt{x}}

\frac{d y}{d x} \approx \frac{△ y}{△ x}

△ y = \frac{d y}{d x} \times △ x

n o w x = 100 y = 10

x + △ x = 101 y + △ y = ?

△ x = 1

△ y = \frac{d y}{d x} \times △ x

△ y = \frac{1}{2 \sqrt{x}} \times △ x

△ y = \frac{1}{2 \times \sqrt{100}} = 0.05

s o y + △ y = 10 + 0.05 = 10.05

s o \sqrt{101}

= 10.05

Commented by otchereabdullai@gmail.com last updated on 14/May/19

fantastic solution prof

Answered by MJS last updated on 14/May/19

{Heron}^{'} s Algorithm for \sqrt[n]{a} :

x_{i + 1} = \frac{(n - 1) x_{i}^{n} + a}{{n x}_{i}^{n - 1}}

{we}^{'} re free to choose any x_{1} > 0

for n = 2 :

x_{i + 1} = \frac{x_{i}^{2} + a}{2 x_{i}}

a = 101; x_{1} = 10

x_{2} = \frac{10^{2} + 101}{2 \times 10} = \frac{201}{20} = 10.050 000

x_{3} = \frac{{(\frac{201}{20})}^{2} + 101}{2 \times \frac{201}{20}} = \frac{80 801}{8 040} \approx 10.049 876

Commented by otchereabdullai@gmail.com last updated on 14/May/19

fantastic solution prof