Understanding the Geometric Series Formula: Solving for a Complex Variable

[N468989]

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:


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

thankyou very much for any help.

 

[quadraphonics]

Demonstrate the following expression:


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

thankyou very much for any help.

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

 

[stewartcs]

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:


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

thankyou very much for any help.

Precalculus or calculus text.

CS

 

[rbj]

Demonstrate the following expression:


$$\sum_{n=0}^{N-1} \alpha^n = \frac{1-\alpha^N}{1-\alpha}$$

thank you very much for any help.

Well, the case where $\alpha=1$ should be easy enough. To demonstrate the second case, simply multiply the summation by the fraction $\frac{1-\alpha}{1-\alpha}$. 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:

$$(1-\alpha)\sum_{n=0}^{N-1} \alpha^n = \sum_{n=0}^{N-1} \alpha^n - \alpha\sum_{n=0}^{N-1} \alpha^n = 1 - \alpha^N$$

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

 

[N468989]

thanks for the help i found it in a calculus book. it´s mathematical induction.

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

$$(1-\alpha)\sum_{n=0}^{N-1} \alpha^n = \sum_{n=0}^{N-1} \alpha^n - \alpha\sum_{n=0}^{N-1} \alpha^n = 1 - \alpha^N$$

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.