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#

Related questions
See all questions in Using Newton's Method to Approximate Solutions to Equations
Impact of this question
8523 views around the world
You can reuse this answer
Creative Commons License