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Chapter 11: Problem 3

How would you approximate \(e^{-a^{6}}\) using the Taylor series for \(e^{x} ?\)

Short Answer

Expert verified
Answer: The Taylor series expansion of the function \(e^{-a^6}\) around \(x=0\) is given by: $$ e^{-a^6} = \sum_{n=0}^{\infty}\frac{(-1)^n a^{6n}}{n!} = 1 - a^6 + \frac{a^{12}}{2} - \frac{a^{18}}{6} + \cdots $$ The approximation criteria depend on the desired accuracy and the magnitudes of \(a\) and the subsequent terms in the series. A few terms can already provide a reasonably good approximation, but for higher accuracy, more terms should be included in the approximation.

Step by step solution

01

Recall the Taylor series for \(e^x\)

The Taylor series expansion for the exponential function \(e^x\) around \(x=0\) (Maclaurin series) is given by: $$ e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots $$
02

Replace \(x\) with \(-a^6\) in the Taylor series formula

To approximate \(e^{-a^6}\) using the Taylor series for \(e^x\), we need to replace \(x\) with \(-a^6\) in the formula: $$ e^{-a^6} = \sum_{n=0}^{\infty}\frac{(-a^6)^n}{n!} = 1 - a^6 + \frac{(-a^6)^2}{2!} - \frac{(-a^6)^3}{3!} + \cdots $$
03

Simplify the terms in the series

Simplify the terms in the series expression: $$ e^{-a^6} = \sum_{n=0}^{\infty}\frac{(-1)^n a^{6n}}{n!} = 1 - a^6 + \frac{a^{12}}{2} - \frac{a^{18}}{6} + \cdots $$
04

Determine the approximation criteria

The approximation criteria depend on the desired accuracy and depend on the magnitudes of \(a\) and the subsequent terms in the series. The first few terms should already provide a reasonably good approximation for \(e^{-a^6}\). For higher accuracy, more terms should be included in the approximation.

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Key Concepts

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

Maclaurin series
A Maclaurin series is a special case of a Taylor series, which approximates a function through an infinite sum of terms. These terms are derived from the function's derivatives at a single point, namely the origin \(x = 0\). When using a Maclaurin series for approximation, you're essentially building the polynomial representation of the function at the origin.

The expansion of the Maclaurin series looks like this:
  • The 0th term is the function evaluated at zero: \(f(0)\)
  • The 1st term involves the first derivative and is \(x \cdot f'(0)\)
  • Subsequent terms are derived similarly through higher derivatives, divided by factorial numbers: \(\frac{f''(0)}{2!}x^2, \frac{f^3(0)}{3!}x^3, \ldots\)
This method is particularly useful because it can morph complex functions into manageable polynomial approximations, which are much easier to handle, especially for calculating exponential functions around zero.
exponential function
The exponential function \(e^x\) is a mathematical constant, which is key in many areas across math and science. This function is special because of its unique property: its rate of change is proportional to its current value.

It is expressed in the form of the equation:
  • \(e^x\)
  • where \(e\) is Euler's number, approximately equal to 2.71828
The exponential function grows rapidly and is useful in modeling growth processes like populations, investments, and radioactive decay among other phenomena. When expanding \(e^x\) into a series, every 'n' term is given by \(\frac{x^n}{n!}\), forming the basis for many approximations and computations.
series expansion
Any function can be approximated effectively using a series expansion, which breaks the function down into a sum of simpler polynomial terms. Series expansions can turn challenging, nonlinear functions into series of infinitely many simpler terms.

The process involves:
  • Identifying the function to expand (in this case, the exponential function)
  • Using the derivatives of the function to construct polynomial terms in its expansion
  • Adding these terms to create an entire series which converges to the original function
By taking derivatives and evaluating at the target point \(x = 0\), you obtain successive terms which increase in power and factorial divisor. Each additional term enhances the accuracy of the function approximation, converting complex calculations into simpler arithmetic.
approximation accuracy
Approximation accuracy refers to how closely a series expansion represents the true value of the function being approximated. It depends on several factors including the number of terms in the series and the nature of the function.

To determine the accuracy of an approximation:
  • Consider the size of the additional terms added: smaller terms signify higher precision
  • Evaluate the difference between the last two terms: if this difference is negligible, your approximation is near precise
  • Determine the level of precision you need: sometimes fewer terms suffice, but for more challenging functions, more terms might be necessary
In this context, the more terms used from the Maclaurin series, the closer the approximation will align with \(e^{-a^6}\). Understanding approximation accuracy is crucial for deciding how many terms to include given the required precision.

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Most popular questions from this chapter

a.Use any analytical method to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. You do not need to use the definition of the Taylor series coefficients. b.Determine the radius of convergence of the series. $$f(x)=\left\\{\begin{array}{ll}\frac{\sin x}{x} & \text { if } x \neq 0 \\\1 & \text { if } x=0\end{array}\right.$$

Let $$f(x)=\sum_{k=0}^{\infty} c_{k} x^{k} \text { and } g(x)=\sum_{k=0}^{\infty} d_{k} x^{k}$$a. Multiply the power series together as if they were polynomials, collecting all terms that are multiples of \(1, x,\) and \(x^{2} .\) Write the first three terms of the product \(f(x) g(x)\) b. Find a general expression for the coefficient of \(x^{n}\) in the product series, for \(n=0,1,2, \ldots\)

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. Only even powers of \(x\) appear in the Taylor polynomials for \(f(x)=e^{-2 x}\) centered at 0. b. Let \(f(x)=x^{5}-1 .\) The Taylor polynomial for \(f\) of order 10 centered at 0 is \(f\) itself. c. Only even powers of \(x\) appear in the \(n\)th-order Taylor polynomial for \(f(x)=\sqrt{1+x^{2}}\) centered at 0. d. Suppose \(f^{\text {- }}\) is continuous on an interval that contains \(a\), where f has an inflection point at \(a\). Then the second-order Taylor polynomial for \(f\) at \(a\) is linear.

Use the Maclaurin series $$(1+x)^{-2}=1-2 x+3 x^{2}-4 x^{3}+\cdots, \text { for }-1

By comparing the first four terms, show that the Maclaurin series for \(\cos ^{2} x\) can be found (a) by squaring the Maclaurin series for \(\cos x,\) (b) by using the identity \(\cos ^{2} x=\frac{1+\cos 2 x}{2},\) or \((c)\) by computing the coefficients using the definition.
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