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===Answer 1=== | ===Answer 1=== | ||
| − | + | First we know the summation of an infinity geometric series: | |
| + | <math> \sum_{n=0}^{\infty} \alpha^n = \frac{1}{(1 - \alpha)} , \left| \alpha \right| < 1</math>; (eq1) | ||
| + | |||
| + | so we can compute | ||
| + | |||
| + | <math> \sum_{n=M}^{\infty} \alpha^n = \left( \alpha \right)^M \frac{1}{(1 - \alpha)} , \left| \alpha \right| < 1</math>; (eq2) | ||
| + | |||
| + | similarly, | ||
| + | |||
| + | <math> \sum_{n=N+1}^{\infty} \alpha^n = \left( \alpha \right)^{N+1} \frac{1}{(1 - \alpha)} , \left| \alpha \right| < 1</math>; (eq3) | ||
| + | |||
| + | then we can substract eq3 from eq2, if N+1> M | ||
| + | |||
| + | <math> \sum_{n=M}^{\infty} \alpha^n - \sum_{n=N+1}^{\infty} \alpha^n = \frac{{\left( \alpha \right)^M } - {\left( \alpha \right)^{N+1}}}{(1 - \alpha)} , \left| \alpha \right| < 1</math>; | ||
| + | |||
| + | for N larger or equal to M, <math>\left| \alpha\right| < 1</math>, the equation above holds. | ||
| + | |||
| + | //Did I make any mistake in the N+1 part or it's just a typo? | ||
| + | |||
| + | |||
===Answer 2=== | ===Answer 2=== | ||
Write it here. | Write it here. | ||
Revision as of 08:38, 10 September 2011
Contents
When is this super duper geometric series formula valid?
A student in ECE301 once wrote the following formula on his exam:
$ \sum_{n = M}^N \alpha^n = \frac{\alpha^M - \alpha^{N-1}}{(1 - \alpha)} $
Is this formula correct? For what values of the parameters is the formula valid? Please comment.
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
Answer 1
First we know the summation of an infinity geometric series: $ \sum_{n=0}^{\infty} \alpha^n = \frac{1}{(1 - \alpha)} , \left| \alpha \right| < 1 $; (eq1)
so we can compute
$ \sum_{n=M}^{\infty} \alpha^n = \left( \alpha \right)^M \frac{1}{(1 - \alpha)} , \left| \alpha \right| < 1 $; (eq2)
similarly,
$ \sum_{n=N+1}^{\infty} \alpha^n = \left( \alpha \right)^{N+1} \frac{1}{(1 - \alpha)} , \left| \alpha \right| < 1 $; (eq3)
then we can substract eq3 from eq2, if N+1> M
$ \sum_{n=M}^{\infty} \alpha^n - \sum_{n=N+1}^{\infty} \alpha^n = \frac{{\left( \alpha \right)^M } - {\left( \alpha \right)^{N+1}}}{(1 - \alpha)} , \left| \alpha \right| < 1 $;
for N larger or equal to M, $ \left| \alpha\right| < 1 $, the equation above holds.
//Did I make any mistake in the N+1 part or it's just a typo?
Answer 2
Write it here.
Answer 3
Write it here
