Problem 39 Show that if $t>0$, then \(... [FREE SOLUTION]

91影视

Precalculus A Prelude to Calculus

Sheldon Axler

$Math Studyset 91影视 Explanations$ Math

1 Edition

Chapter 4: Problem 39

Show that if $t>0$, then $1+t0 .$

Short Answer

Expert verified

To show that if $t > 0$, then $1 + t < e^t$, we can use the Taylor series expansion of the exponential function $e^t$, given by \[e^t = 1 + t + \frac{t^2}{2!} + \frac{t^3}{3!} + \cdots\] Since the remaining terms are all positive when $t > 0$, this leads to the inequality $1 + t < e^t$ for $t > 0$.

Step by step solution

The Taylor series expansion of the exponential function $e^t$ is given by: \[e^t = \sum_{n=0}^{\infty} \frac{t^n}{n!} = 1 + t + \frac{t^2}{2!} + \frac{t^3}{3!} + \cdots\] #Step 2: Compare the first two terms of the Taylor series with the expression 1 + t#

Considering the first two terms of the Taylor series: \[e^t \approx 1 + t\] If we can show that the remaining terms always contribute a positive value for $t > 0$, then we can establish that $e^t > 1 + t$. #Step 3: Show that the remaining terms of the Taylor series are positive when t > 0#

Since $t > 0$, all the remaining terms in the Taylor series of $e^t$ are positive: \[\frac{t^2}{2!} > 0, \frac{t^3}{3!} > 0, \frac{t^4}{4!} > 0, \cdots\] #Step 4: Conclude that 1 + t < e^t for t > 0#

As $e^t$ is the sum of the first two terms (1 + t) and a series of positive terms, the inequality $1 + t < e^t$ holds when $t > 0$.

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91影视

Show that if \(t>0\), then \(1+t0 .\)

Short Answer

Step by step solution

The Taylor series expansion of the exponential function \(e^t\) is given by: \[e^t = \sum_{n=0}^{\infty} \frac{t^n}{n!} = 1 + t + \frac{t^2}{2!} + \frac{t^3}{3!} + \cdots\] #Step 2: Compare the first two terms of the Taylor series with the expression 1 + t#

Since \(t > 0\), all the remaining terms in the Taylor series of \(e^t\) are positive: \[\frac{t^2}{2!} > 0, \frac{t^3}{3!} > 0, \frac{t^4}{4!} > 0, \cdots\] #Step 4: Conclude that 1 + t < e^t for t > 0#

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