Problem 21 Prove that if \(\mu \in(0, \inft... [FREE SOLUTION]

91影视

Numerical Analysis: Mathematics of Scientific Computing

David R. Kincaid, E. Ward Cheney

$Math Studyset 91影视 Explanations$ Math

3 Edition

Chapter 1: Problem 21

Prove that if $\mu \in(0, \infty)$ and $|\lambda|<1$, then $\lim _{n \rightarrow \infty} n^{\mu} \lambda^{n}=0$.

Short Answer

Expert verified

The limit is 0 because exponential decay dominates polynomial growth as $ n \to \infty $.

Step by step solution

Understanding the Limit Expression

We want to evaluate the limit $ \lim_{n \rightarrow \infty} n^{\mu} \lambda^{n} = 0 $, where $ \mu \in (0, \infty) $ and $ |\lambda| < 1 $. This involves the behavior of functions involving both polynomial growth ($ n^{\mu} $) and exponential decay ($ \lambda^{n} $).

Analyzing the Exponential Decay

Since $ |\lambda| < 1 $, we know that $ \lambda^{n} $ tends to 0 as $ n $ goes to infinity. Exponentially decaying terms (like $ \lambda^n $) go to zero much faster than polynomial terms grow, indicating that the exponential part dominates in the limit.

Considering the Polynomial Growth

The term $ n^{\mu} $ grows polynomially, where $ \mu > 0 $. However, the speed of polynomial growth is much slower compared to the exponential decay since any power $ n^{\mu} $ grows slower than the decay of an exponentially decreasing sequence like $ \lambda^n $.

Applying the Limit

As $ n \to \infty $, the exponential decay $ \lambda^n \to 0 $ faster than any polynomial $ n^\mu $ grows. Thus, the combination $ n^{\mu} \lambda^{n} $ converges to 0.

Concluding the Proof

By dominating the power function $ n^{\mu} $ by the exponentially decaying function $ \lambda^n $, the expression $ \lim _{n \rightarrow \infty} n^{\mu} \lambda^{n} $ can be seen to approach zero. Therefore, the limit is 0.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Exponential Decay

In calculus, exponential decay refers to a process where a quantity decreases at a rate proportional to its current value. This creates an exponential function, often expressed as $ \lambda^n $, where $ |\lambda| < 1 $. Here's why exponential decay is crucial:

As $ n $ increases, $ \lambda^n $ approaches zero very quickly. This happens because multiplying a number less than 1 by itself repeatedly makes it shrink very fast.
An exponentially decaying function like $ \lambda^n $ will eventually become negligible compared to other terms that grow slower, like polynomial terms.
This concept helps us understand why in the expression $ n^{\mu} \lambda^{n} $, the exponential part ensures that the whole expression heads towards zero.

Exponential decay plays a pivotal role when discussing limits because it often indicates rapid convergence to zero.

Exploring Polynomial Growth

Polynomial growth, expressed as $ n^{\mu} $ where $ \mu > 0 $, involves terms increasing as a power of $ n $. Here's what to know about polynomial growth compared to other types of growth:

Although $ n^{\mu} $ does grow, the growth is relatively slow compared to exponential functions. It grows by adding powers, not multiplying factors like exponential functions.
Despite having a rate influenced by the power $ \mu $, it cannot surpass the decay rate of $ \lambda^n $ when $ \lambda $ is a fraction less than 1.
This slower growth rate helps in calculating limits, as even if $ n^{\mu} $ becomes large, it cannot offset the zeroing effect of $ \lambda^n $.

In limits involving polynomial and exponential functions, this difference in growth rates is key to determining the resulting behavior of a function as it approaches infinity.

Applying Limit Proof Techniques

Limit proof techniques in calculus help determine how functions behave as they approach infinity or other critical points. Here's how these techniques apply to our problem:

Comparative Growth Analysis: By comparing exponential decay and polynomial growth, one appreciates that exponential decay overwhelms polynomial growth at infinity, pushing the entire expression towards zero.
Squeezing Arguments: Sometimes, finding an upper bound for $ n^{\mu} \lambda^{n} $ helps. Since $ \lambda^n $ rapidly approaches zero, even a relatively large polynomial term can't prevent the product from becoming negligible.
Direct Calculation: In some cases, direct manipulation of the expression is sufficient to show that $ n^{\mu} \lambda^{n} \to 0 $.

Mastering these techniques allows you to tackle various limit problems, equipping you to prove or disprove limits effortlessly.

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91影视

Prove that if \(\mu \in(0, \infty)\) and \(|\lambda|<1\), then \(\lim _{n \rightarrow \infty} n^{\mu} \lambda^{n}=0\).

Short Answer

Step by step solution

Understanding the Limit Expression

Analyzing the Exponential Decay

Considering the Polynomial Growth

Applying the Limit

Concluding the Proof

Key Concepts

Understanding Exponential Decay

Exploring Polynomial Growth

Applying Limit Proof Techniques

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