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Chapter 9: Problem 14

Find a formula for an arbitrary Taylor polynomial of \(f\). $$ f(x)=\cos x $$

Short Answer

Expert verified
The Taylor polynomial of \( \cos x \) at \( c=0 \) is \( T_n(x) = \sum_{k=0}^{\lfloor n/2 \rfloor} (-1)^k \frac{x^{2k}}{(2k)!} \).

Step by step solution

01

Introduction to Taylor Series

A Taylor series is a representation of a function as an infinite sum of terms calculated from the values of its derivatives at a single point. The Taylor series of a function \(f(x)\) centered at \(c\) is given by: \[ f(x) = f(c) + f'(c)(x-c) + \frac{f''(c)}{2!}(x-c)^2 + \frac{f'''(c)}{3!}(x-c)^3 + \, \ldots \] For this problem, we aim to find the Taylor polynomial for \(f(x)=\cos x\) centered at \(c=0\). This is known as the Maclaurin series.
02

Compute the Derivatives

Start by computing the derivatives of \(f(x) = \cos x\) at \(c = 0\):- \( f(x) = \cos x \Rightarrow f(0) = \cos 0 = 1 \)- \( f'(x) = -\sin x \Rightarrow f'(0) = -\sin 0 = 0 \)- \( f''(x) = -\cos x \Rightarrow f''(0) = -\cos 0 = -1 \)- \( f'''(x) = \sin x \Rightarrow f'''(0) = \sin 0 = 0 \)- \( f^{(4)}(x) = \cos x \Rightarrow f^{(4)}(0) = \cos 0 = 1 \)These derivatives start to repeat every four terms.
03

Substitute Derivatives into the Taylor Series

Next, substitute these derivative values into the general Taylor series formula:\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \ldots \]Using the values:\[ \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \]Thus, the general formula for the Taylor polynomial is:\[ T_n(x) = \sum_{k=0}^{\lfloor n/2 \rfloor} (-1)^k \frac{x^{2k}}{(2k)!} \]
04

Conclusion

The Taylor polynomial up to degree \(n\) of \(f(x) = \cos x\) is concluded to be:\[ T_n(x) = \sum_{k=0}^{\lfloor n/2 \rfloor} (-1)^k \frac{x^{2k}}{(2k)!} \] This series includes only the even-power terms due to the derivatives of \(cos(x)\) at zero.

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Key Concepts

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

Maclaurin Series
The Maclaurin series is a special kind of Taylor series that is centered at zero. It's a fantastic way to express complex functions using polynomial terms, making them easier to work with for approximations and calculations. When you hear Maclaurin, just remember: like other Taylor series but simpler because it starts from zero!

Here鈥檚 how it works:
  • Start with a function you want to approximate, like the cosine function.
  • Find the derivatives of your function.
  • Substitute these derivatives into the Maclaurin series formula.
The general formula for a Maclaurin series is \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \ldots \).

In essence, each term involves a derivative evaluated at zero, providing a neat polynomial that approximates functions like trigonometry and exponential ones quite effectively!
Derivative
Derivatives are the backbone of calculus, telling us how a function changes when its input changes. For function calculations, especially when finding a Taylor or Maclaurin series, derivatives are crucial since each term in these series depends on the value of the function's derivatives at a particular point.

When calculating derivatives for the cosine function, you will observe a repeating pattern:
  • The first derivative of \( \cos x \) is \( -\sin x \).
  • The second derivative becomes \( -\cos x \).
  • The third derivative results in \( \sin x \).
  • The fourth derivative returns to \( \cos x \).
This repetitive cycle is because of cosine's symmetry, making it straightforward when adding these into series forms. Recognizing these cycles is very helpful, as it shortens the process of calculating many terms.
Cosine Function
The cosine function, \( \cos x \), is one of the fundamental trigonometric functions you'll encounter. It has a beautiful sinusoidal shape, oscillating between -1 and 1.

When expressed as a series like a Maclaurin series, it yields a series comprising only even-powered terms. This is because all the odd-power derivatives evaluated at zero are zero, leaving us with only even degrees. For example:
  • The constant term is 1 because \( \cos(0) = 1 \).
  • The term with \( x^2 \) is \( -\frac{x^2}{2!} \), representing the second derivative term.
  • Continuing this pattern, the next significant term becomes \( \frac{x^4}{4!} \).
Thus, with the found derivatives repeating every four terms, cosine's Maclaurin series naturally becomes a neat sum of alternative even powers.

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Most popular questions from this chapter

Determine which series diverge, which converge conditionally, and which converge absolutely. $$ \sum_{n=1}^{\infty}(-1)^{n} \frac{1 \cdot 3 \cdot 5 \cdots(2 n+1)}{2 \cdot 5 \cdot 8 \cdots(3 n+2)} $$

Approximate the sum of the given series with an error less than \(0.001\). $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{n^{3}} $$

Find the Taylor series of the given function about \(a\). Use the series already obtained in the text or in previous exercises. $$ f(x)=\cos ^{2} x ; a=0 $$

Approximate the value of the integral with an error less than the given error, first using the Integration Theorem to express the integral as an infinite series and then approximating the infinite series by an appropriate partial sum. $$ \int_{0}^{1 / 2} \frac{x^{2}}{1+x} d x ; 10^{-3} $$

The first partial sum \(1+x / 2\) of the binomial series for \((1+x)^{1 / 2}\) is often used as an approximation of \((1+x)^{1 / 2}\) Using the Lagrange form of the remainder \(r_{1}(x)\), show that the error introduced is at most \((0.172) x^{2}\), provided that \(|x|<0.19 .\)
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