Problem 71 - Calculus for Biology and Medicine

Question

\(\lim _{n \rightarrow \infty} a_{n}=a\). Find the limit \(a\), and determine \(N\) so that \(\left|a_{n}-a\right|<\epsilon\) for all \(n>N\) for the given value of \(\epsilon\) $$ a_{n}=2^{-n}, \epsilon=0.01 $$

Step-by-Step Solution

Answer

The limit is 0, and \(N = 7\) where \(|a_n - 0| < 0.01\).

Step 1: Determine the Limit

The term given is \(a_n = 2^{-n}\). As \(n\) approaches infinity, \(2^{-n}\), which is the same as \(\frac{1}{2^n}\), approaches zero because the denominator increases without bound. Therefore, \(\lim_{n \to \infty} a_n = 0\).

Step 2: Set Up the Inequality

We need to find an \(N\) such that for all \(n > N\), \(|a_n - 0| < \epsilon\). This simplifies to \(|2^{-n}| < 0.01\).

Step 3: Convert Inequality to Solve for n

\(2^{-n} < 0.01\) can be rewritten as \(\frac{1}{2^n} < 0.01\). By taking the reciprocal, this becomes \(2^n > 100\) (since \(0.01\) is the reciprocal of 100).

Step 4: Calculate N

To find \(N\), solve \(2^n > 100\). We take the base-2 logarithm of both sides: \(n > \log_2(100)\). Approximating the base-2 logarithm, \(\log_2(100)\approx 6.644\). Therefore, \(n > 6.644\) means the smallest integer \(N\) where the inequality holds is \(N = 7\).

Step 5: Conclusion

For all \(n > 7\), the inequality \(|2^{-n} - 0| < 0.01\) is satisfied, meaning \(\left|a_{n}-a\right|<\epsilon\) for this \(N\). Therefore, \(N = 7\) works for the given \(\epsilon\).

Key Concepts

Epsilon-N Definition

The epsilon-N definition plays a crucial role in understanding limits in calculus and mathematical sequences. This definition gives us a rigorous approach to define the limit of a sequence, essentially quantifying how closely terms of the sequence approach a specific number. Given a sequence \(a_n\) converging to a limit \(L\), for every positive number \(\epsilon\), no matter how small, there exists a corresponding natural number \(N\). For all sequence terms \(n>N\), the terms \(a_n\) will lie within \(\epsilon\) distance from \(L\). In simpler terms:

• The sequence \(a_n\) must get arbitrarily close to the limit \(L\) as \(n\) becomes very large.
• \(\epsilon\) represents just how close we want \(a_n\) to be to \(L\).
• The number \(N\) tells us from which point onwards all terms are within that desired closeness, \(\epsilon\).

This framework is essential, as it establishes a formal way to speak of the behavior of sequences as they extend towards infinity.

Exponential Decay

Exponential decay describes a process where a quantity decreases rapidly at first, and then levels off as it approaches zero. In our exercise, the sequence \(a_n = 2^{-n}\) demonstrates exponential decay. Here's why:

• The exponent \(-n\) indicates that as \(n\) increases, the base 2 is raised to increasingly negative powers.
• This results in numbers such as \(\frac{1}{2}, \frac{1}{4}, \frac{1}{8}\), which are fractions that get smaller with each step.
• Consequently, \(2^{-n}\) rapidly diminishes towards zero.

Exponential decay is critical in many natural phenomena and scientific calculations, including radioactive decay, cooling of substances, and depreciation in finance. The key characteristic is its speed: changes happen swiftly at first and slow down over time, making it predictable as \(n\) increases.

Logarithms

Logarithms are the mathematical inverses of exponentials, and understanding their role is essential when solving inequalities involving exponential functions. They simplify expressions and allow us to solve for variables trapped in an exponential form. In our problem, we encountered the inequality \(2^n > 100\). Here is how logarithms help:

• By applying logarithms, we can "bring down" the exponent, making the inequality \(n > \log_2(100)\).
• Logarithms convert multiplicative processes into additive ones, simplifying calculations.
• Using the property that \(\log_b(a) = x\) means \(b^x = a\), where \(b\) is the base of the logarithm, provides a direct path to solve for \(n\).

In the context of the exercise, calculating \(\log_2(100) \approx 6.644\) gave a precise threshold above which the sequence met the required condition. Logarithms thus serve as powerful tools not only in mathematics but also across sciences, engineering, and finance, where understanding rates of change and scaling is crucial.