Pages 41-42 QC and DI Study Guide

[hallumchris]

Beginning on the bottom of page 41 the problem states that:

v&= 2v-1

Quantity A= (v&)& Quantity B= 4v

I understand the explanation with one exception. When (2v-1)& is distributed it is found to equal 2(2v-1)-1, however when I do it I can only see how to get to 2(2v-1)-&. Any insight would be much appreciated.

[esledge]

I see what you mean--this function definition is a little unconventional in the way it is presented, with the & after the input variable indicating that something is to be done with the input.

In more conventional function form, this would look like f(v) = 2v - 1, which in words means "double the input (whatever it is) and subtract 1 from the result."

So, that step of (2v-1)& would more conventionally be written as f(2v-1), for which we double the input to get 2(2v - 1) and then subtract 1, yielding 2(2v - 1) - 1.

I think the key is to remember that & is not an amount or a variable, any more than "f" is when you see a typical f(x). Instead, & is a set of instructions of what to do to v, or in the later step, to (2v - 1).

I hope this helps!