Problem 44 The next two problems combine to... [FREE SOLUTION]

91影视

Precalculus A Prelude to Calculus

Sheldon Axler

$Math Studyset 91影视 Explanations$ Math

3 Edition

Chapter 3: Problem 44

The next two problems combine to show that $$ 1+t0\) Show that if $t>0$, then \(1+t

Short Answer

Expert verified

Based on the Taylor series expansion of the exponential function, we successfully proved that if $t>0$, then $1+t < e^t$ by comparing the terms in the series and manipulating the inequality.

Step by step solution

Rewrite the inequality using exponential function properties

First, we can rewrite $e^t$ as $1 + t + \frac{t^2}{2!} + \frac{t^3}{3!} + \cdots$. This is based on the Taylor series expansion of the exponential function: $$ e^t = \sum_{n=0}^{\infty} \frac{t^n}{n!}. $$ Now, we need to prove that $1+t < 1 + t + \frac{t^2}{2!} + \frac{t^3}{3!} + \cdots$ if t > 0.

Add terms on both sides to compare positive values

We can add terms to the inequality to analyze the expressions more easily, as in the following way: $$ 1+t+ \frac{t^2}{2!} + \frac{t^3}{3!} + \cdots < 1 + t + \frac{t^2}{2!} + \frac{t^3}{3!} + \cdots + \frac{t^2}{2!} + \frac{t^3}{3!} + \cdots . $$ This is valid because $t>0$ and $n!$ is always positive for all positive integers n, which means all the terms in the series are positive.

Simplify the inequality to obtain the desired result

Now, we can subtract $1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+\cdots$ from both sides of the inequality: $$ 1 + t < 1 + t + \frac{t^2}{2!} + \frac{t^3}{3!} + \cdots + \frac{t^2}{2!} + \frac{t^3}{3!} + \cdots - \left(1 + t + \frac{t^2}{2!} + \frac{t^3}{3!} + \cdots\right). $$ This simplifies to: $$ 1 + t < (\frac{t^2}{2!} + \frac{t^3}{3!} + \cdots) + (\frac{t^2}{2!} + \frac{t^3}{3!} + \cdots). $$ Since all terms in the sequences are positive and come from the Taylor series expansion of $e^t$, we can safely say that $1+t < e^t$ if $t>0$. So we have successfully proved that if $t>0$, then $1 + t < e^t$.

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91影视

The next two problems combine to show that $$ 1+t0\) Show that if \(t>0\), then \(1+t

Short Answer

Step by step solution

Rewrite the inequality using exponential function properties

Add terms on both sides to compare positive values

Simplify the inequality to obtain the desired result

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