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# How do you use linear approximation or differentials to estimate $\dfrac{1}{{1002}}$?

Last updated date: 03rd Mar 2024
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Hint: The given number is in the form of fraction. The solution is obtained by dividing the value of the numerator by the value of the denominator. But here in this question we have to solve or simplify the given fraction by using the linear approximation or by the differentials. Hence we obtain the required solution.

Complete step by step explanation:
The number is in the form of fraction. The number which contains above the line is known as numerator and the number below the line is known as
denominator.
Now consider the given question $\dfrac{1}{{1002}}$.
In the denominator, the number 1002 can be written as 1000 + 2. So the question is rewritten as
$\Rightarrow \dfrac{1}{{1000 + 2}}$
Take 1000 as a common in the denominator, the equation is written as
$\Rightarrow \dfrac{1}{{1000\left( {1 + \dfrac{2}{{1000}}} \right)}}$
Let we consider $\dfrac{2}{{1000}}$ as x and the equation is written as
$\Rightarrow \dfrac{1}{{1000\left( {1 + x} \right)}}$
In the denominator the product of two terms are written as
$\Rightarrow \dfrac{1}{{1000}}.\dfrac{1}{{\left( {1 + x} \right)}}$
Take (1+x) to the numerator we have
$\Rightarrow \dfrac{1}{{1000}}.{\left( {1 + x} \right)^{ - 1}}$
By the formula of binomial expansion ${(1 + x)^n} = 1 + nx + \dfrac{{n(n - 1)}}{{2!}}{x^2} \mp ...$.
By using this formula the equation is written as
$\Rightarrow \dfrac{1}{{1000}}.\left( {1 + ( - 1)x + \dfrac{{( - 1)(( - 1) - 1)}}{{2!}}{x^2} + ...} \right)$
Substituting the value of x to the above equation we have
$\Rightarrow \dfrac{1}{{1000}}.\left( {1 + ( - 1)\dfrac{2}{{1000}} + \dfrac{{( - 1)(( - 1) - 1)}}{{2!}}{{\left( {\dfrac{2}{{1000}}} \right)}^2} + ...} \right)$
We consider only two terms and we neglect the others terms. Since we have high value in the denominator, if we divide the small number by the big number we get a very very small number and if we multiply it we get still a smaller number so we are neglecting the terms. So the equation is written as
$\Rightarrow \dfrac{1}{{1000}}.\left( {1 - \dfrac{2}{{1000}}} \right)$
Simplify the term which is in the brace.
$\Rightarrow \dfrac{1}{{1000}}.\left( {\dfrac{1}{1} - \dfrac{2}{{1000}}} \right)$
Take LCM for the number 1000 and 1, the LCM of 1000 and 1 is 1000.
So we have
$\Rightarrow \dfrac{1}{{1000}}.\left( {\dfrac{{1000 - 2}}{{1000}}} \right)$
On simplifying we get
$\Rightarrow \dfrac{1}{{1000}}.\left( {\dfrac{{998}}{{1000}}} \right)$
By dividing the number 998 by 1000 we get
$\Rightarrow \dfrac{1}{{1000}}.\left( {0.998} \right)$
The $\dfrac{1}{{1000}}$ in the exponential form is ${10^{ - 3}}$. So we have
$\Rightarrow \left( {0.998} \right){10^{ - 3}}$
This can be written as
$\Rightarrow \left( {9.98} \right){10^{ - 4}}$

Note: the number which is in the form of fraction we can divide directly we obtain the same answer. But dividing the small number by a huge number is very difficult. The binomial expansion for the linear term is defined as ${(1 + x)^n} = 1 + nx + \dfrac{{n(n - 1)}}{{2!}}{x^2} \mp ...$ where n is power term. By using this we can solve the problem.