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Chapter 11: Problem 13

Determine the \(n\) th Taylor polynomial for \(f(x)=e^{x}\) at \(x=0 .\)

Short Answer

Expert verified
\( T_n(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \text{...} + \frac{x^n}{n!} \)

Step by step solution

01

Understand the Taylor Series Formula

The Taylor series for a function, centered at a point, is given by the formula: \[ T_n(x) = \frac{f(a)}{0!} + \frac{f'(a)}{1!}(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \text{...} + \frac{f^{(n)}(a)}{n!}(x - a)^n \] In this case, the function is \( f(x) = e^x \) and we are focusing on \( x = 0 \).
02

Evaluate the Derivatives of the Function

The function \( f(x) = e^x \) has the unique property that all its derivatives are also \( e^x \). Therefore, we have: \[ f'(x) = e^x, \ f''(x) = e^x, \ f'''(x) = e^x, \ \text{and so on.} \] Evaluating these at \( x = 0 \): \[ f(0) = 1, \ f'(0) = 1, \ f''(0) = 1, \ f'''(0) = 1, \ \text{and so on.} \]
03

Construct the Taylor Polynomial

Substitute the derivatives evaluated at \( x = 0 \) into the Taylor series formula: \[ T_n(x) = \frac{1}{0!}x^0 + \frac{1}{1!}x^1 + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \text{...} + \frac{1}{n!}x^n \] Simplifying, we get: \[ T_n(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \text{...} + \frac{x^n}{n!} \]

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Key Concepts

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

Taylor Polynomial
A Taylor polynomial is used to approximate complex functions. It expands the function into a sum of simpler polynomial terms. This expansion helps to analyze and compute functions more easily. The Taylor series, the infinite sum of these polynomial terms, is represented as:
Exponential Function
The exponential function, denoted as p(x) = e^x, is a crucial mathematical function. It has the unique property that its rate of growth is proportional to its current value. This feature renders it exceptionally useful in a variety of fields such as calculus, biology, and finance.
Derivatives
Derivatives represent the rate of change of a function. They are fundamentally important in calculus, providing insight into the behavior of functions. Calculating derivatives involves determining how a function changes as its input changes. In the context of the Taylor Series, derivatives are evaluated at a specific point to develop the polynomial approximation.

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Most popular questions from this chapter

Rely on the fact that $$ \lim _{n \rightarrow \infty} \frac{|x|^{n+1}}{(n+1) !}=0 $$ The proof of this fact is omitted. Let \(R_{n}(x)\) be the \(n\) th remainder of \(f(x)=e^{x}\) at \(x=0\). (See Section 11.1.) Show that, for any fixed value of \(x\), \(\left|R_{n}(x)\right| \leq e^{|x|} \cdot|x|^{n+1} /(n+1) !\), and hence, conclude that \(\left|R_{n}(x)\right| \rightarrow 0\) as \(n \rightarrow \infty\). This shows that the Taylor series for \(e^{x}\) converges to \(e^{x}\) for every value of \(x\).

Use three repetitions of the Newton-Raphson algorithm to approximate the following: The zero of \(\sin x+x^{2}-1\) near \(x_{0}=0\).

It can be shown that \(\lim _{b \rightarrow \infty} b e^{-b}=0 .\) Use this fact and the integral test to show that \(\sum_{k=1}^{\infty} \frac{k}{e^{k}}\) is convergent.

If \(f(x)=2-6(x-1)+\frac{3}{2 !}(x-1)^{2}-\frac{5}{3 !}(x-1)^{3}+\) \(\frac{1}{4 !}(x-1)^{4}\), what are \(f^{\prime \prime}(1)\) and \(f^{\prime \prime \prime}(1) ?\)

Use three repetitions of the Newton-Raphson algorithm to approximate the following: The zero of \(e^{x}+10 x-3\) near \(x_{0}=0\).
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