Question

How do you use a linear approximation or differentials to estimate `1/1002`?

Answer 1

`1/1002 = 1/(1000 + 2)`

`=1/1000 cdot 1/(1 + 2/1000)`

Now we can work with that last term if we write it as:

`=1/1000 cdot 1/(1 + x), absx " << " 1`

The easiest thing is to go with a Binomial Expansion which is the Taylor Expansion: `(1 + x)^alpha = 1 + alpha x + O(x^2)` and here `alpha = -1` and `x = 2/1000`

`implies 1/1000 (1 - x + O(x^2))`

`= 1/1000 (1 - 2/1000 ) = 1/1000 cdot 998/1000`

`approx 9.98 xx 10^(-4)`

You'll get just about that if you type the original into your calculator.

Answer 2

`1/1002~~0.000998`

Explanation:

Let `f(x)=1/x, x ne 0, " so that, "f'(x)=-1/x^2.`

We know that, `f(x+deltax)~~f(x)+f'(x)*deltax.`

Taking, `x=1000, and, deltax=2,` we have,

`f(1000+2)~~f(1000)+2f'(1000), i.e.,`

`1/1002~~1/1000+2(-1/1000^2)=0.001-0.000002`

`:. 1/1002~~0.000998`