Understanding the Geometric Series Formula: Solving for a Complex Variable

In summary: in summary, this is a problem about calculating derivatives. you need to be familiar with the homogeneous property to solve it.
  • #1
N468989
92
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hi everyone,

i had a class where my teacher was talking about this but i don't know where this comes from, if anyone knows could you please give hint on the type of material this is. Like what type of mathbook i should be looking for.


the problem was:

Demonstrate the following expression:


[tex]\[\displaystyle\sum_{n=0}^{N-1} \alpha^n = \left\{
\begin{array}{l l}
N & \quad \mbox{, $\alpha$ = 1}\\
\frac{1-\alpha^N}{1-\alpha} & \quad \mbox{, for any complex, $\alpha$ $\neq$ 1}\\ \end{array} \right. \][/tex]

thankyou very much for any help.
 
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  • #2
tko_gx said:
Demonstrate the following expression:


[tex]\[\displaystyle\sum_{n=0}^{N-1} \alpha^n = \left\{
\begin{array}{l l}
N & \quad \mbox{, $\alpha$ = 1}\\
\frac{1-\alpha^N}{1-\alpha} & \quad \mbox{, for any complex, $\alpha$ $\neq$ 1}\\ \end{array} \right. \][/tex]

thankyou very much for any help.

Well, the case where [itex]\alpha=1[/itex] should be easy enough. To demonstrate the second case, simply multiply the summation by the fraction [itex]\frac{1-\alpha}{1-\alpha}[/itex]. This will give you the denominator you need, and some subsequent simplifications to the numerator give you the final answer.
 
  • #3
tko_gx said:
hi everyone,

i had a class where my teacher was talking about this but i don't know where this comes from, if anyone knows could you please give hint on the type of material this is. Like what type of mathbook i should be looking for.


the problem was:

Demonstrate the following expression:


[tex]\[\displaystyle\sum_{n=0}^{N-1} \alpha^n = \left\{
\begin{array}{l l}
N & \quad \mbox{, $\alpha$ = 1}\\
\frac{1-\alpha^N}{1-\alpha} & \quad \mbox{, for any complex, $\alpha$ $\neq$ 1}\\ \end{array} \right. \][/tex]

thankyou very much for any help.

Precalculus or calculus text.

CS
 
  • #4
tko_gx said:
Demonstrate the following expression:


[tex]\sum_{n=0}^{N-1} \alpha^n = \frac{1-\alpha^N}{1-\alpha} [/tex]

thank you very much for any help.

quadraphonics said:
Well, the case where [itex]\alpha=1[/itex] should be easy enough. To demonstrate the second case, simply multiply the summation by the fraction [itex]\frac{1-\alpha}{1-\alpha}[/itex]. This will give you the denominator you need, and some subsequent simplifications to the numerator give you the final answer.

in other words, ask yourself why the following is true:

[tex](1-\alpha)\sum_{n=0}^{N-1} \alpha^n = \sum_{n=0}^{N-1} \alpha^n - \alpha\sum_{n=0}^{N-1} \alpha^n = 1 - \alpha^N [/tex]

if you can figger out why that is true, you've solved your problem.
 
  • #5
thanks for the help i found it in a calculus book. it´s mathematical induction.

rbj said:
in other words, ask yourself why the following is true:

[tex](1-\alpha)\sum_{n=0}^{N-1} \alpha^n = \sum_{n=0}^{N-1} \alpha^n - \alpha\sum_{n=0}^{N-1} \alpha^n = 1 - \alpha^N [/tex]

if you can figger out why that is true, you've solved your problem.

i can use the homogeneous property...and thus get the anwser

anyway thankyou all for helping out.
 
Last edited:

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