Hard Made Easy
[This post is targeted at a Level 1 student.]
In Lagrange Interpolation, we saw how we transformed a hard problem (question 1) into a simple case (question 2), and then used the simple case to build up our understanding of the hard problem. Some students asked if there are any other similar scenarios. Yes, there are several such instances, which we will cover over time. The intention of this (series of) post, is to illustrate how we can use simple cases to build up our understanding of math.
The following is a well-known example, which is extremely basic. As such, there are no Worked Examples to go through, and students are encouraged to do all the Test Yourself problems. To give everyone a chance to work on them, I will edit out any solutions that are posted in the comments.
Consider the question: How many integers are there between 1234 and 9876?
This seems hard to approach, since we do not have any concrete values to work with. The possible integers are . But how many are there? Don’t we end up having to work out the answer to this question in order to proceed?
Well, this is easy. The integers are , so there are of them. We’re done.
Now, how do we go from Question 1 to Question 2? We use a change of variables, by setting . Then, we have that , or equivalently that . By question 2, we know that there are possible values to , hence there must be possible values to .
1. Back to the original question: How many integers are there between 1234 and 9876?
2. Does it matter in the proof if are negative numbers?
3. If are integers such that , how many integers are there which satisfy ? What happens to the formula if ? Why?
4. Consider the Arithmetic Progression given by , where each term is 7 more than the previous term. How many terms are there in this sequence?