Chapter 4: Problem 58
(a) Show that \(e^{x} \geqslant 1+x\) for \(x \geqslant 0\) . (b) Deduce that \(e^{x} \geqslant 1+x+\frac{1}{2} x^{2}\) for \(x \geqslant 0\) . (c) Use mathematical induction to prove that for \(x \geqslant 0\) and any positive integer \(n\) $$e^{x} \geqslant 1+x+\frac{x^{2}}{2 !}+\cdots+\frac{x^{n}}{n !}$$
Short Answer
Step by step solution
Understanding the Function
Proof for e^x ">=" 1 + x (Taylor Expansion)
Starting From Basic Inequality
Verify the Base Case for Mathematical Induction
Inductive Step
Conclusion for Mathematical Induction
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Taylor Series
- \[ e^{x} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \]
This series allows us to approximate \( e^x \) with any desired level of accuracy by including more terms.
When \( x \geq 0 \), each of these additional terms \( \frac{x^n}{n!} \) is positive, reinforcing the approximation to be always greater than or equal to the truncated sum. This property is crucial when proving inequalities involving the exponential function as it provides a straightforward way to compare \( e^x \) with polynomial expressions.
Mathematical Induction
Induction is often likened to dominoes falling; once the base case is proven, and you show the inductive step works, all subsequent numbers "fall" because they follow logically from the assumptions.
- Base Case: Establish the truth of the statement for an initial value.
- Inductive Step: Assuming the proposition is true for some \( n = k \), prove it holds for \( n = k + 1 \).
Polynomial Approximation
For the exponential function, a natural polynomial approximation is provided by utilizing the Taylor series. By truncating the series after a finite number of terms, you derive polynomial approximations, such as:
- \( P_1(x) = 1 + x \)
- \( P_2(x) = 1 + x + \frac{x^2}{2!} \)
Because each term added through the Taylor series represents a higher degree of accuracy, these polynomial forms approach the behavior of \( e^x \) more closely as more terms are included.
Polynomial approximations are practical for calculations when the exact value of \( e^x \) is cumbersome or unnecessary at high precision.
Exponential Function
Characterized by its smooth and continuous nature, \( e^x \) is continuously increasing for all real \( x \), and has a derivative equal to itself, meaning that \( \frac{d}{dx} [e^x] = e^x \).
- Grows rapidly as \( x \) increases.
- \( e^0 = 1 \), illustrating its fixed point at zero.
- Increases infinitely without asymptotes in the positive direction.
This behavior is crucial in proving exponential inequalities, where minimizing expressions with lower power terms underestimates the true value of \( e^x \) but never surpasses it, hence comparisons like \( e^x \geq 1+x \) hold true.