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Chapter 10: Problem 8

Find the Taylor polynomials of degree \(n\) approximating the functions for \(x\) near \(0 .\) (Assume \(p\) is a constant.) $$\tan x, \quad n=3,4$$

Short Answer

Expert verified
Both Taylor polynomials are \(x + \frac{x^3}{3}\).

Step by step solution

01

Understand the Function

Taylor polynomials are used to approximate functions using derivatives. Here, we want to approximate the function \(\tan x\) near \(x = 0\) and up to degree \(n = 3\) and \(n = 4\).
02

Find Derivatives of Function

First, calculate the derivatives of \(\tan x\):- \(f(x) = \tan x\)- \(f'(x) = \sec^2 x\)- \(f''(x) = 2\sec^2 x \tan x\)- \(f'''(x) = 2(\sec^4 x + \sec^2 x \tan^2 x)\)- \(f''''(x) = 4\sec^4 x \tan x (\sec^2 x + 1)\)Evaluate each at \(x = 0\). We find:- \(f(0) = \tan(0) = 0\)- \(f'(0) = \sec^2(0) = 1\)- \(f''(0) = 0\)- \(f'''(0) = 2\)- \(f''''(0) = 0\)
03

Construct the Taylor Polynomial for n=3

Using the derivatives evaluated at \(x = 0\), we construct the third-degree Taylor polynomial:\[ P_3(x) = f(0) + f'(0) \frac{x}{1!} + f''(0) \frac{x^2}{2!} + f'''(0) \frac{x^3}{3!} = 0 + 1 \cdot x + 0 \cdot x^2 + \frac{2x^3}{6} \]Simplifying gives:\[ P_3(x) = x + \frac{x^3}{3} \]
04

Construct the Taylor Polynomial for n=4

Add the fourth derivative term to the polynomial:\[ P_4(x) = P_3(x) + f''''(0) \frac{x^4}{4!} = x + \frac{x^3}{3} + 0 \cdot \frac{x^4}{24} \]Thus, simplifying gives:\[ P_4(x) = x + \frac{x^3}{3} \]
05

Conclusion

Both the third and fourth-degree Taylor polynomials for \(\tan x\) at \(x = 0\) are \(P_3(x) = P_4(x) = x + \frac{x^3}{3}\).

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Key Concepts

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

Understanding Derivatives
Derivatives play a pivotal role when it comes to Taylor polynomials. A derivative represents the rate at which a function is changing at any given point. For a function like \(\tan x\), derivatives help us understand how the function behaves near a point, especially its slope and curvature.
  • The first derivative \(f'(x)\) of \(\tan x\) is \(\sec^2 x\), which tells us how steep or flat the tangent line is at a point.
  • The second derivative \(f''(x) = 2\sec^2 x \tan x\) gives us insight into the function's concavity 鈥� how it bends around a point.
  • Higher-order derivatives, like \(f'''(x)\) and \(f''''(x)\), provide further detailing of the function's behavior.
When constructing a Taylor polynomial, you evaluate these derivatives at a specific point, typically \(x = 0\). This point is known as the center of the polynomial. By using derivatives, the polynomial closely mimics the actual function nearby.
Function Approximation with Taylor Polynomials
Taylor polynomials are a mathematical tool used for approximating functions. Imagine trying to replicate a curve with a series of straight lines and curves鈥攖hat's essentially what a Taylor polynomial does!
  • For \(\tan x\), the goal is to find a polynomial function that behaves like \(\tan x\) when \(x\) is close to \(0\).
  • A Taylor polynomial of degree \(n\) incorporates derivatives of the function up to the \(n\)th order.
  • For degree 3, represented as \(P_3(x)\), the polynomial is \(P_3(x) = x + \frac{x^3}{3}\), which is derived using the derivatives of \(\tan x\).
  • The polynomial is a simpler function to work with and can provide good approximations over a small range of \(x\) values.
The Taylor polynomial thus serves as an approximation, simplifying complex functions into a form that鈥檚 easier to compute and analyze.
Trigonometric Functions in Taylor Series
Trigonometric functions frequently appear in Taylor series, showcasing their periodic nature through polynomial expressions.
  • For \(\tan x\), Taylor series help in expressing it as `polynomial` terms based on its derivatives.
  • At \(x = 0\), \(\tan(0) = 0\), meaning its Taylor series has a simple beginning, often starting off with \(x\) itself.
  • The odd derivatives contribute to the polynomial terms, while even derivatives might reduce some terms to zero, simplifying the expression.
This method is particularly useful for approximations around key points like \(x = 0\) due to the simple evaluation of trigonometric identities. Thus, using Taylor polynomials with trigonometric functions enables solving complex equations and enhancing calculation efficiency.

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Most popular questions from this chapter

Use the third degree Taylor polynomial \(P_{3}(x)=4+2(x-1)-2(x-1)^{2}+(x-1)^{3} / 2\) of \(f(x)\) about \(x=1\) to find the given value, or explain why you can't. $$f(1)$$

Decide if the statements in Problems \(65-71\) are true or false. Assume that the Taylor series for a function converges to that function. Give an explanation for your answer. If \(f\) is an even function, then the Taylor series for \(f\) near \(x=0\) has only terms with even exponents.

Explain what is wrong with the statement. If \(f(x)=\ln (2+x),\) then the second-degree Taylor polynomial approximating \(f(x)\) near \(x=0\) has a negative constant term.

Use a Taylor polynomial with the derivatives given to make the best possible estimate of the value. \(g(1.8),\) given that \(g(2)=-3, g^{\prime}(2)=-2, g^{\prime \prime}(2)=0\) \(g^{\prime \prime \prime}(2)=12\)

Find the value of the integer \(a_{i}\) given that the Taylor series of \(f(x)=\cos \sqrt{x}\) is: $$a_{0}+\frac{1}{a_{1}} x+\frac{1}{a_{2}} x^{2}+\frac{1}{a_{3}} x^{3}+\cdots$$ $$a_{2}$$
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