Prove a^n u(n) + a^{-n} u(-n-1) = a^{-|n|}

Understand the Problem

The question requires proving the given equation involving the term a raised to the power of n, multiplied by the unit step function u(n), and a term involving a raised to the power of -n, multiplied by a shifted and inverted unit step function u(-n-1). The result is expected to be equal to a raised to the power of negative absolute value of n. To solve this, we need to analyze the behavior of the unit step functions and the exponential terms for different values of n (positive, negative, and zero), and show that they simplify to the given expression.

Answer

The equation $a^n u(n) + a^{-n} u(-n-1) = a^{-|n|}$ is not universally true.

Steps to Solve

Define the Unit Step Function

The unit step function $u(n)$ is defined as:

$u(n) = \begin{cases} 1, & n \geq 0 \ 0, & n < 0 \end{cases}$

Analyze the term $a^n u(n)$

If $n \geq 0$, then $u(n) = 1$, so $a^n u(n) = a^n$.

If $n < 0$, then $u(n) = 0$, so $a^n u(n) = 0$.

Analyze the term $a^{-n} u(-n-1)$

First, let's consider $u(-n-1)$. It is 1 when $-n-1 \geq 0$ which simplifies to $-n \geq 1$ or $n \leq -1$. It is 0 when $-n-1 < 0$ which simplifies to $-n < 1$ or $n > -1$.

So, $u(-n-1) = \begin{cases} 1, & n \leq -1 \ 0, & n > -1 \end{cases}$

Now, let's consider $a^{-n}u(-n-1)$:

If $n \leq -1$, then $u(-n-1) = 1$, so $a^{-n} u(-n-1) = a^{-n}$.

If $n > -1$, then $u(-n-1) = 0$, so $a^{-n} u(-n-1) = 0$.

Combine both terms $a^n u(n) + a^{-n} u(-n-1)$

We want to show that $a^n u(n) + a^{-n} u(-n-1) = a^{-|n|}$.

We'll analyze this for different values of $n$:

Case 1: $n > 0$. Then $a^n u(n) = a^n$ and $a^{-n} u(-n-1) = 0$. Therefore, $a^n u(n) + a^{-n} u(-n-1) = a^n$. Since $n > 0$, $a^{-|n|} = a^{-n}$. So we must confirm that $a^n = a^{-|n|}=0$, which only holds at $a=0$. So this identity doesn't hold true, unless we assume $a=0$.

Case 2: $n = 0$. Then $a^n u(n) = a^0 u(0) = 1 \cdot 1 = 1$ and $a^{-n} u(-n-1) = a^0 u(-1) = 1 \cdot 0 = 0$. Therefore, $a^n u(n) + a^{-n} u(-n-1) = 1$. Also, $a^{-|n|} = a^{-|0|} = a^0 = 1$. So the equation holds for $n = 0$.

Case 3: $n < 0$. Let $n = -k$ where $k > 0$. Then $a^n u(n) = a^{-k} u(-k) = a^{-k} \cdot 0 = 0$ and $a^{-n} u(-n-1) = a^{-(-k)} u(-(-k)-1) = a^k u(k-1)$. Since $k > 0$, $k-1 \geq 0$ only when $k \geq 1$, i.e., $n \leq -1$. If $k=1$ then $n=-1$ and it is $a u(0) = a$. Then $a^k u(k-1)= a^{-n} u(-n-1)= a^{-n}$. Then $a^n u(n) + a^{-n} u(-n-1) = 0+a^{-n} = a^{-n}$. $a^{-|n|} = a^{-|-k|} = a^{-k}$. Then $a^{-n} = a^{-|n|}$ only when $-n = -k$, so $n=-k$, which is what we defined.

Revised approach Let's consider when $n \leq -1$. $a^n u(n) + a^{-n} u(-n-1) = a^n(0) + a^{-n}(1) = a^{-n}$. $a^{-|n|} = a^{-(-n)} = a^n$ (Because $n$ is negative). Therefore: $a^{-n} = a^{n}$ if the initial hypothesis holds.

The given equation is thus invalid.

The given equation $a^n u(n) + a^{-n} u(-n-1) = a^{-|n|}$ is not universally true. It holds for $n=0$ in all cases. It holds for the case when $n \leq -1$, AND $n \geq 1$ only if: $a = \frac{1}{a}$ which means that $a^2 = 1$, or $a = \pm 1$.

More Information

The unit step function is a fundamental concept in signal processing and discrete-time systems. The given equation tries to relate exponential functions with the unit step, but the equation is incorrect except for certain conditions such as $n = 0$ or $a = \pm 1$. The absolute value in the exponent makes the expression symmetric around $n = 0$.

Tips

A common mistake is to not properly analyze the regions defined by the unit step functions (i.e., when they are 0 or 1). Also, it is important to keep track of the different cases based on the sign of $n$ (positive, negative, and zero). Another common mistake is assuming the equation is valid for all values of 'a' and 'n', without considering specific constraints this implies.

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