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Chapter 9: Problem 13

Find the Maclaurin polynomials of orders \(n=0,1,2,3\), and 4, and then find the \(n\) th Maclaurin polynomials for the function in sigma notation. $$ \cosh x $$

Short Answer

Expert verified
P_n(x) = \sum_{k=0}^{\lfloor n/2 \rfloor} \frac{x^{2k}}{(2k)!} is the nth Maclaurin polynomial for \( \cosh x \).

Step by step solution

01

Understand Maclaurin Series

The Maclaurin series of a function \( f(x) \) is a Taylor series expansion about 0: \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \ldots \). For \( n \)-th order polynomial, use terms up to \( f^{(n)}(0)x^n \).
02

Evaluate Derivatives of \( \cosh x \)

The function is \( \cosh x = \frac{e^x + e^{-x}}{2} \). Calculate derivatives as: - \( f(0) = \cosh(0) = 1 \) - \( f'(x) = \sinh x, \ f'(0) = 0 \) - \( f''(x) = \cosh x, \ f''(0) = 1 \) - \( f'''(x) = \sinh x, \ f'''(0) = 0 \) - \( f^{(4)}(x) = \cosh x, \ f^{(4)}(0) = 1 \)
03

Construct Maclaurin Polynomial for \( n=0 \)

For \( n=0 \), the Maclaurin polynomial is simply the value of the function at zero: \( P_0(x) = 1 \).
04

Construct Maclaurin Polynomial for \( n=1 \)

The first order polynomial is: \( P_1(x) = 1 + 0\cdot x = 1 \), since the first derivative at zero is 0.
05

Construct Maclaurin Polynomial for \( n=2 \)

The second order polynomial is: \( P_2(x) = 1 + 0\cdot x + \frac{1}{2!}x^2 = 1 + \frac{x^2}{2} \).
06

Construct Maclaurin Polynomial for \( n=3 \)

For the third order polynomial: \( P_3(x) = 1 + 0\cdot x + \frac{1}{2}x^2 + 0 \cdot x^3 = 1 + \frac{x^2}{2} \).
07

Construct Maclaurin Polynomial for \( n=4 \)

The fourth order polynomial is: \( P_4(x) = 1 + 0 \cdot x + \frac{1}{2}x^2 + 0 \cdot x^3 + \frac{1}{4!}x^4 = 1 + \frac{x^2}{2} + \frac{x^4}{24} \).
08

Find General Maclaurin Polynomial in Sigma Notation

The nth Maclaurin polynomial in sigma notation is: \[ P_n(x) = \sum_{k=0}^{\lfloor n/2 \rfloor} \frac{x^{2k}}{(2k)!} \], as only even powers contribute due to odd derivatives being zero.

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Key Concepts

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

Taylor Series
A Taylor series is essentially a powerful way of approximating complex functions with an infinite sum of polynomial terms. This technique involves taking the derivatives of a function at a particular point 鈥� often zero, which is specifically known as a Maclaurin series.
The general form for Taylor series of a function \( f(x) \) about a value \( a \) is expressed as:
  • \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots \).
In simpler words, it's like building a recipe for the function starting from the evaluation of the function and its derivatives up to the desired number of terms. If \( a = 0 \), this becomes the Maclaurin series, which simplifies writing and calculations, especially for functions centered at zero.
Hyperbolic Functions
Hyperbolic functions are analogs of trigonometric functions but are based on hyperbolas instead of circles. A key hyperbolic function is the hyperbolic cosine, \( \cosh x \). This function can be expressed as:
  • \( \cosh x = \frac{e^x + e^{-x}}{2} \).
This representation makes \( \cosh x \) symmetric and is noteworthy for its use in solving problems involving hyperbolic geometry, special relativity, and complex numbers.
Just like trigonometric functions, hyperbolic functions also have derivatives:
  • Derivative of \( \cosh x \) is \( \sinh x \).
  • Derivative of \( \sinh x \) is \( \cosh x \).
These relationships simplify many calculations involving these functions.
Polynomial Approximation
Polynomial approximation allows us to represent complex functions as sums of polynomials up to a certain degree. The idea is to use simpler polynomial functions to approximate more complex functions.
When constructing a Maclaurin polynomial, we only need to consider terms up to a certain order, \( n \), in order to approximate the function closely. For instance, when approximating \( \cosh x \), we can express it as:
  • \( P_0(x) = 1 \) 鈥� first term for \( n=0 \)
  • \( P_1(x) = 1 \) 鈥� for \( n=1 \), due to zero first derivative
  • \( P_2(x) = 1 + \frac{x^2}{2} \)
  • \( P_3(x) = 1 + \frac{x^2}{2} \)
  • \( P_4(x) = 1 + \frac{x^2}{2} + \frac{x^4}{24} \)
Notice how terms involving odd powers do not appear due to zero derivatives at those orders, and hence only even-powered terms contribute.
Derivatives
Derivatives measure how a function changes as its input changes. When constructing a Taylor or Maclaurin series, derivatives play a key role in determining each term of the approximation. They provide the coefficients for each polynomial term in the series expansion.
For \( \cosh x \), we compute several derivatives at \( x = 0 \):
  • \( f(0) = 1 \)
  • First derivative: \( f'(x) = \sinh x, \ f'(0) = 0 \)
  • Second derivative: \( f''(x) = \cosh x, \ f''(0) = 1 \)
  • Third derivative: \( f'''(x) = \sinh x, \ f'''(0) = 0 \)
  • Fourth derivative: \( f^{(4)}(x) = \cosh x, \ f^{(4)}(0) = 1 \)
These derivatives help us determine which terms appear in the Maclaurin polynomial and thus affect the overall approximation of the function.

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Most popular questions from this chapter

Use the Maclaurin series for \(\sinh x\) and \(\cosh x\) to obtain the first four nonzero terms in the Maclaurin series for tanh \(x\).

Find the radius of convergence and the interval of convergence. $$ \sum_{k=0}^{\infty} \frac{\pi^{k}(x-1)^{2 k}}{(2 k+1) !} $$

Suppose that the function \(f\) is represented by the power series $$ f(x)=1-\frac{x}{2}+\frac{x^{2}}{4}-\frac{x^{3}}{8}+\cdots+(-1)^{k} \frac{x^{k}}{2^{k}}+\cdots $$ (a) Find the domain of \(f\). (b) Find \(f(0)\) and \(f(1)\).

We showed by Formula (19) of Section \(8.2\) that if there are \(y_{0}\) units of radioactive carbon- 14 present at time \(t=0\), then the number of units present \(t\) years later is $$ y(t)=y_{0} e^{-0.000121 t} $$ (a) Express \(y(t)\) as a Maclaurin series. (b) Use the first two terms in the series to show that the number of units present after 1 year is approximately \((0.999879) y_{0}\) (c) Compare this to the value produced by the formula for \(y(t)\)

Find the first four nonzero terms of the Maclaurin series for the function by multiplying the Maclaurin series of the factors. (a) \(e^{x} \sin x\) (b) \(\sqrt{1+x} \ln (1+x)\)
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