- #1
MidgetDwarf
- 1,480
- 616
In order to construct a geometric series we do the following:
chose a number a such that a does not equal 0 and a second number r that is between (-1,1). We call r the ration because it is the ratio, the progression of each term to its predecessor.
We have An=a+ar+ar^2+ar^3...ar^n
We multiply An by r. rAn=ar+ar^2+ar^3+ar^4...ar^n+1.
We subtract rAn from An and we end with the terms: An-rAn=a-ar^n+1
An-rAn=a(1-r^n+1).
Here is were i cannot follow the argument. In the book we have An-rAn=a(1-r^n+1)/(1-r).
Where did the 1-r in the denominator come from? Is it the same idea we invoke to compare two physical or more physical quantities, i.e., ratios of sides of triangles?
ignoring my lack of understanding of this step. I can see why, s equals the limit as n->infinity of An-rAn
is a/(1-r). we can pick a value for r between -1 and 1, say 0.5 and raise it to the 1000000 power which will make it go to zero.
chose a number a such that a does not equal 0 and a second number r that is between (-1,1). We call r the ration because it is the ratio, the progression of each term to its predecessor.
We have An=a+ar+ar^2+ar^3...ar^n
We multiply An by r. rAn=ar+ar^2+ar^3+ar^4...ar^n+1.
We subtract rAn from An and we end with the terms: An-rAn=a-ar^n+1
An-rAn=a(1-r^n+1).
Here is were i cannot follow the argument. In the book we have An-rAn=a(1-r^n+1)/(1-r).
Where did the 1-r in the denominator come from? Is it the same idea we invoke to compare two physical or more physical quantities, i.e., ratios of sides of triangles?
ignoring my lack of understanding of this step. I can see why, s equals the limit as n->infinity of An-rAn
is a/(1-r). we can pick a value for r between -1 and 1, say 0.5 and raise it to the 1000000 power which will make it go to zero.