Finite Complex Exponential Geometric Series with Negative Exponents

Question

So I want to know if I am doing my manipulations correctly. I have the following expression:

$$ X(\omega) = A\frac{1 - e^{-j \frac{1}{2} \omega N}}{1 - e^{j \frac{1}{2} \omega}} $$

Using the geometric series formula:

$$ \displaystyle\sum\limits_{n=0}^{N-1} A r^{n} = \frac{1 - r^{N}}{1 - r}$$

Can I rewrite the first expression as:

$$ X(\omega) = A\frac{1 - \big(e^{j \frac{1}{2} \omega }\big)^{-N}}{1 - e^{j \frac{1}{2} \omega}} $$ $$ X(\omega) = A \displaystyle\sum\limits_{n=0}^{-(N-1)} e^{j \frac{1}{2} \omega n} $$ $$ X(\omega) = A \displaystyle\sum\limits_{n=0}^{N-1} e^{-j \frac{1}{2} \omega n} $$

Somehow I don't think this is right, because if we reuse the geometric series expansion, the denominator will be different. But, I can't see to convince myself of it.

Or this is manipulation correct?

Answer 1

The simpler way first $$ \eqalign{ & {{X(\omega )} \over A} = {{1 - e^{\, - jw/2N} } \over {1 - e^{\,jw/2} }} = {{1 - e^{\, - jw/2N} } \over {e^{\,jw/2} \left( {e^{\, - jw/2} - 1} \right)}} = \cr & = - e^{\, - jw/2} {{1 - e^{ - jw/2N} } \over {\left( {1 - e^{\, - jw/2} } \right)}} = - e^{\, - jw/2} \sum\limits_{0\, \le \,k\, \le \,N - 1} {e^{\, - jw/2\,k} } = \cr & = - \sum\limits_{1\, \le \,k\, \le \,N} {e^{\, - jw/2\,k} } \cr} $$

Following instead your way, to take the geometric series from $0$ to $-N$, you shall write $$ \eqalign{ & {{X(\omega )} \over A} = {{1 - e^{\, - jw/2N} } \over {1 - e^{\,jw/2} }} = {{1 - \left( {e^{\,jw/2} } \right)^{\, - N} } \over {1 - e^{\,jw/2} }} = \cr & = \sum\nolimits_{k = 0}^{ - N} {e^{\,jw/2\;k} } = - \sum\nolimits_{k = - N}^0 {e^{\,jw/2\;k} } = - \sum\limits_{ - N\, \le \,k\, \le \, - 1} {e^{\,jw/2\,k} } = \cr & = - \sum\limits_{ - N\, \le \,k\, \le \, - 1} {e^{\,jw/2\,k} } = - \sum\limits_{1\, \le \, - k\, \le \,N} {e^{\,jw/2\,k} } = - \sum\limits_{1\, \le \,k\, \le \,N} {e^{\, - \,jw/2\,k} } \cr} $$

where $$ \sum\nolimits_{k = 0}^X {f(k)} $$ indicates the "Indefinite" Sum computed within the indicated bounds. In summary we have $$ \eqalign{ & f(k) = \Delta _{\,k} \,F(k) = F(k + 1) - F(k)\quad \Leftrightarrow \cr & \Leftrightarrow \quad F(k) = \Delta _{\,k} ^{\left( { - 1} \right)} \,f(k) = \sum\nolimits_k {f(k)} \quad \Leftrightarrow \cr & \Leftrightarrow \quad \sum\nolimits_{k = 0}^X {f(k)} = F(X) - F(0)\quad \Leftrightarrow \cr & \Leftrightarrow \quad \sum\nolimits_{k = 0}^X {f(k)} \quad \left| {\,X \in Z} \right. = \left\{ {\matrix{ { - \sum\limits_{X\, \le \,k\, \le \, - 1} {f(k)} } & {X < 0} \cr {\sum\limits_{0\, \le \,k\, \le \,X - 1} {f(k)} } & {0 \le X} \cr } } \right. \cr} $$ --- in reply to your comment ----

Among the basic properties of the Indefinite Sum, also called Antidelta, we have $$ \left\{ \matrix{ \sum\nolimits_a^a {f(k)} = 0 \hfill \cr \sum\nolimits_a^b {f(k)} + \sum\nolimits_b^c {f(k)} = \sum\nolimits_a^c {f(k)} \hfill \cr} \right. $$ which implies $$ \eqalign{ & \sum\nolimits_a^b {f(k)} + \sum\nolimits_b^a {f(k)} = \sum\nolimits_a^a {f(k)} = 0 \cr & \sum\nolimits_b^a {f(k)} = - \sum\nolimits_a^b {f(k)} \cr} $$

The Indefinite Sum is the discrete analog of the Integral and mimicks many (not all) of its properties.

Take, in our case $$ F(x) = - {{r^{\,x} } \over {1 - r}} = - r^{\,x} \sum\limits_{0\, \le \,k} {r^{\,k} } = - \sum\limits_{0\, \le \,k} {r^{\,k + x} } $$ then $$ f(x) = \Delta _{\,x} F(x) = F(x + 1) - F(x) = {{r^{\,x} - r^{\,x + 1} } \over {1 - r}} = r^{\,x} $$ and $$ \sum\nolimits_a^b {f(k)} = \sum\nolimits_a^b {r^{\,k} } = - {{r^{\,b} } \over {1 - r}} + {{r^{\,a} } \over {1 - r}} = {{r^{\,a} - r^{\,b} } \over {1 - r}} $$ which allows to define the sum even for real bounds $a$ and $b$.

In particular, for a non-negative integer $N$ $$ \eqalign{ & \sum\nolimits_0^{ - N} {r^{\,k} } = {{1 - r^{\, - N} } \over {1 - r}} = \left( {1 - r^{\, - N} } \right)\sum\limits_{0\, \le \,k} {r^{\,k} } = \sum\limits_{0\, \le \,k} {r^{\,k} } - \sum\limits_{0\, \le \,k} {r^{\,k - N} } = \cr & = \sum\limits_{0\, \le \,k} {r^{\,k} } - \sum\limits_{ - N\, \le \,k - N} {r^{\,k - N} } = - \sum\limits_{ - N\, \le \,j\; \le \, - 1} {r^{\,j} } = - \sum\nolimits_{ - N}^0 {r^{\,k} } = \cr & = - \sum\limits_{1\, \le \,j\; \le \,N} {\left( {r^{\, - 1} } \right)^{\,j} } = - {{1 - \left( {r^{\, - 1} } \right)^{\,N} } \over {1 - 1/r}} + 1 = {{ - r\left( {1 - r^{\, - N} } \right) + r - 1} \over {r - 1}} = \cr & = {{1 - r^{\, - N} } \over {1 - r}} \cr} $$ that is to say $$ {{1 - r^{\, - N} } \over {1 - r}} = - \sum\limits_{1\, \le \,j\; \le \,N} {\left( {r^{\, - 1} } \right)^{\,j} } = - \sum\limits_{1\, \le \,j\; \le \,N} {\left( {1/r} \right)^{\,j} } = 1 - \sum\limits_{0\, \le \,j\; \le \,N} {\left( {1/r} \right)^{\,j} } $$

Answer 2

The manipulation is not correct in general.

Starting with OP's last line and applying the finite geometric series formula we obtain \begin{align*} \color{blue}{A\sum_{n=0}^{N-1}e^{-j\frac{1}{2}\omega n}}&=A\sum_{n=0}^{N-1}e^{\left(-\frac{j\omega}{2}\right)^n}\\ &=A\frac{1-e^{\left(-\frac{j\omega}{2}\right)^N}}{1-e^{-\frac{j\omega}{2}}}\\ &=A\frac{1-e^{-\frac{j\omega N}{2}}}{1-e^{-\frac{j\omega}{2}}}\\ &=-\frac{A}{e^{-\frac{j\omega}{2}}}\cdot\frac{1-e^{-\frac{j\omega N}{2}}}{1-e^\frac{j\omega}{2}}\\ &=\color{blue}{-e^{\frac{j\omega}{2}}A\frac{1-e^{-\frac{j\omega N}{2}}}{1-e^{\frac{j\omega}{2}}}}\tag{1} \end{align*} we observe that (1) coincides with OP's expression $A\frac{1-e^{-\frac{j\omega N}{2}}}{1-e^{\frac{j\omega}{2}}}$ in the first line iff $-e^{\frac{j\omega}{2}}=1$.

A note to OP's line \begin{align*} X(\omega) = A \displaystyle\sum\limits_{n=0}^{-(N-1)} e^{j \frac{1}{2} \omega n} \end{align*}

The usual meaning of $\sum_{n=a}^b f(n)=\sum_{a\leq n \leq b}f(n)=0$ if $b<a$. So, you probably want to write $\sum_{n=-(N-1)}^0e^{j \frac{1}{2} \omega n}$. But this does not match the standard form of the finite geometric sum formula.

Answer 3

Addendum

Let me try and clear your doubts about summing definition and handling.

There are fundamentally three ways to express a sum.

a) over a set

e.g. $$ \sum\limits_{k\, \in \,\left\{ {1,2,3} \right\}} k = 1 + 2 + 3 = 2 + 1 + 3 = \ldots $$

b) under a condition on the index

e.g. $$ \sum\limits_{1\, \le \,k\; \le \,3} k = 1 + 2 + 3 = \ldots \quad \Rightarrow \quad \sum\limits_{1\, \le \,k\; \le \, - 3} k = \sum\limits_\emptyset k = 0 $$ or $$ \sum\limits_{\,k\;\backslash \,4} k = 1 + 2 + 4 $$ etc.

c) as antidelta

that is $$ \eqalign{ & f(x) = \Delta _{\,x} F(x) = F(x + 1) - F(x) = \quad \Rightarrow \cr & \Rightarrow \quad F(x) = \Delta _{\,x} ^{\,\left( { - 1} \right)} f(x) = \sum\nolimits_x {f(x)} \quad \Rightarrow \cr & \Rightarrow \quad \sum\nolimits_a^b {f(k)} = F(b) - F(a) \cr} $$

So, concerning the geometric sum

when the exponent is positive, e.g. $3$, then $$ {{1 - r^{\,3} } \over {1 - r}} = \sum\limits_{k\, \in \,\left\{ {0,1,2} \right\}} {r^{\,k} } = \sum\limits_{0\, \le \,k\; \le \,2} {r^{\,k} } = \sum\limits_{0\, \le \,k\; < \,3} {r^{\,k} } = \sum\nolimits_0^3 {r^{\,k} } $$ the three definitions coincide.

But when it is negative $$ \eqalign{ & {{1 - r^{\, - 3} } \over {1 - r}}\quad \ne \quad \sum\limits_{k\, \in \,\left\{ {0, - 1, - 2} \right\}} {r^{\,k} } = {{1 - \left( {1/r} \right)^{\,3} } \over {1 - \left( {1/r} \right)}} \cr & \quad \quad \quad \;\; \ne \quad \sum\limits_{0\, \le \,k\; \le \, - 2} {r^{\,k} } = 0 \cr} $$

while the antidelta gives $$ \eqalign{ & {{1 - r^{\, - 3} } \over {1 - r}} = \quad \sum\nolimits_0^{ - 3} {r^{\,k} } = - {{r^{\, - 3} } \over {1 - r}} + {{r^{\,0} } \over {1 - r}} = \cr & = - \sum\nolimits_{ - 3}^0 {r^{\,k} } = \sum\limits_{ - 3\, \le \,k\; < \,0} {r^{\,k} } = \sum\limits_{ - 3\, \le \,k\; \le \, - 1} {r^{\,k} } = \cr & = \sum\limits_{1\, \le \,k\; \le \,3} {\left( {1/r} \right)^{\,k} } = \sum\limits_{0\, \le \,k\; \le \,3} {\left( {1/r} \right)^{\,k} } - 1 = {{1 - \left( {1/r} \right)^{\,4} } \over {1 - \left( {1/r} \right)}} - 1 = \cr & = {{r\left( {r^{\,4} - 1} \right)} \over {r^{\,4} \left( {r - 1} \right)}} - 1 = {{r^{\,3} - 1} \over {r^{\,3} \left( {r - 1} \right)}} \cr} $$