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

2 Answers
Eddie
Apr 17, 2017

#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.

Ratnaker Mehta
Apr 17, 2017

# 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#

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