Chapter 9: Problem 24
Show that the given sequence is eventually strictly increasing or eventually strictly decreasing. $$ \left\\{n^{5} e^{-n}\right\\}_{n=1}^{+\infty} $$
Short Answer
Expert verified
The sequence \( n^5 e^{-n} \) is eventually strictly decreasing for \( n > 5 \).
Step by step solution
01
Understand the sequence
The given sequence is \(a_n = n^5 e^{-n}\). Here, \(n^5\) is a polynomial term and \(e^{-n}\) is an exponential term. Our task is to show that this sequence is eventually strictly increasing or strictly decreasing.
02
Analyze long-term behavior
To determine the long-term behavior of the sequence \(a_n = n^5 e^{-n}\), we know that as \(n\) becomes very large, the exponential term \(e^{-n}\) decreases much faster than the polynomial term \(n^5\) increases. This is because \(e^{-n}\) tends towards zero faster than any power of \(n\) can increase.
03
Differentiate to find critical points
Consider the function \(f(x) = x^5 e^{-x}\). To find its critical points, compute the derivative \(f'(x)\) using the product rule:\[f'(x) = 5x^4 e^{-x} - x^5 e^{-x} = x^4 e^{-x}(5 - x)\]Setting \(f'(x) = 0\), we find critical points at \(x = 5\) because \(x^4 e^{-x} eq 0\) for \(x > 0\). This means the function changes behavior at \(x = 5\).
04
Analyze intervals around critical point
Evaluate the sign of \(f'(x) = x^4 e^{-x}(5-x)\) for intervals around the critical point \(x=5\):- For \(x < 5\), \(5-x > 0\), so \(f'(x) > 0\) and the sequence is increasing.- For \(x > 5\), \(5-x < 0\), so \(f'(x) < 0\) and the sequence is decreasing.Thus, beyond a certain point (after the maximum at \(n=5\)), \(a_n\) is strictly decreasing.
05
Conclusion about the sequence
Since the sequence reaches a maximum at \(n=5\) and then strictly decreases, it is eventually strictly decreasing for \(n > 5\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polynomial Term
A polynomial term is a mathematical expression involving powers of a variable, usually represented as \( n^k \), where \( k \) is a non-negative integer. In the context of our sequence \( a_n = n^5 e^{-n} \), the polynomial term is \( n^5 \). This means that the term has a degree of 5 and as \( n \) increases, \( n^5 \) becomes very large.
Polynomial terms grow at different rates depending on their degree:- For example, \( n^2 \) grows slower than \( n^3 \).
- Raising \( n \) to the 5th power means it grows faster than lower-degree polynomial terms.
Understanding how polynomial terms behave is important in calculus, especially when they are part of a sequence or function that includes other types of terms, like exponential terms. It helps us predict the overall behavior as \( n \) changes.
Polynomial terms grow at different rates depending on their degree:- For example, \( n^2 \) grows slower than \( n^3 \).
- Raising \( n \) to the 5th power means it grows faster than lower-degree polynomial terms.
Understanding how polynomial terms behave is important in calculus, especially when they are part of a sequence or function that includes other types of terms, like exponential terms. It helps us predict the overall behavior as \( n \) changes.
Exponential Term
Exponential terms involve a base raised to a variable exponent, typically expressed as \( b^n \), where \( b > 0 \) is a constant. Common bases include Euler's number, \( e \), which is about 2.718. In our sequence, we have an exponential term \( e^{-n} \).
When \( n \) is positive, \( e^{-n} \) is actually \( \frac{1}{e^n} \), which means it quickly shrinks towards zero as \( n \) increases. This decay rate is much faster than any polynomial can grow, making exponential terms dominate in determining the limit behavior of sequences and functions.
When \( n \) is positive, \( e^{-n} \) is actually \( \frac{1}{e^n} \), which means it quickly shrinks towards zero as \( n \) increases. This decay rate is much faster than any polynomial can grow, making exponential terms dominate in determining the limit behavior of sequences and functions.
- Exponential decay describes processes where quantities reduce over time, like radioactive decay or cooling.
- This rapid reduction is why \( e^{-n} \) becomes negligible compared to \( n^5 \) for very large \( n \).
Derivative
In calculus, the derivative is a measure of how a function changes as its input changes. For our sequence, we consider the derivative of the function \( f(x) = x^5 e^{-x} \) to understand the behavior.
To compute its derivative, we use the product rule because \( f(x) \) is a product of two functions, \( x^5 \) and \( e^{-x} \). The product rule states:\[ (uv)' = u'v + uv' \]For \( f(x) \), let:- \( u = x^5 \) with \( u' = 5x^4 \)
- \( v = e^{-x} \) with \( v' = -e^{-x} \).
Using the product rule, the derivative \( f'(x) = 5x^4 e^{-x} - x^5 e^{-x} \). This tells us how \( f(x) \) increases or decreases.
By setting \( f'(x) = 0 \), we find critical points where the function potentially changes direction. Here, we find that at \( x = 5 \), the sequence reaches its maximum.
To compute its derivative, we use the product rule because \( f(x) \) is a product of two functions, \( x^5 \) and \( e^{-x} \). The product rule states:\[ (uv)' = u'v + uv' \]For \( f(x) \), let:- \( u = x^5 \) with \( u' = 5x^4 \)
- \( v = e^{-x} \) with \( v' = -e^{-x} \).
Using the product rule, the derivative \( f'(x) = 5x^4 e^{-x} - x^5 e^{-x} \). This tells us how \( f(x) \) increases or decreases.
By setting \( f'(x) = 0 \), we find critical points where the function potentially changes direction. Here, we find that at \( x = 5 \), the sequence reaches its maximum.
- For \( x < 5 \), \( f'(x) > 0 \) indicates increasing behavior.
- When \( x > 5 \), \( f'(x) < 0 \) reveals decreasing behavior.