Problem 20
Question
Find the third-degree Taylor polynomial for \(f(x)=\) $$x^{3}+7 x^{2}-5 x+1\( about \)x=0 .$$ What do you notice?
Step-by-Step Solution
Verified Answer
The Taylor polynomial is identical to the original function, as both are third-degree polynomials.
1Step 1: Understand Taylor Polynomial Formula
The third-degree Taylor polynomial for a function \( f(x) \) centered at \( x = a \) is given by \( T_3(x) = f(a) + f'(a)x + \frac{f''(a)x^2}{2!} + \frac{f'''(a)x^3}{3!} \). Since we are interested in \( x = 0 \), we substitute \( a = 0 \) into the formula.
2Step 2: Compute Function and Derivatives at x = 0
First, calculate \( f(0) \), \( f'(0) \), \( f''(0) \), and \( f'''(0) \) for \( f(x) = x^3 + 7x^2 - 5x + 1 \). - First derivative: \( f'(x) = 3x^2 + 14x - 5 \), so \( f'(0) = -5 \).- Second derivative: \( f''(x) = 6x + 14 \), so \( f''(0) = 14 \).- Third derivative: \( f'''(x) = 6 \), so \( f'''(0) = 6 \).- Function at \( x=0 \): \( f(0) = 1 \).
3Step 3: Apply Taylor Polynomial Formula
Now, apply the values into the Taylor polynomial formula:\[T_3(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} = 1 - 5x + \frac{14x^2}{2} + \frac{6x^3}{6}.\]Simplify the expression to:\[T_3(x) = 1 - 5x + 7x^2 + x^3.\]
4Step 4: Analyze the Result
The Taylor polynomial \( T_3(x) = 1 - 5x + 7x^2 + x^3 \) is exactly the same as the original polynomial \( f(x) \). This happens because \( f(x) \) is itself a third-degree polynomial, which means the Taylor polynomial of the same degree will match the original function entirely.
Key Concepts
third-degree polynomialderivativespolynomialscalculus
third-degree polynomial
A third-degree polynomial is a mathematical expression of the highest degree term with exponent three. In simpler terms, it typically has the form:
- \( ax^3 + bx^2 + cx + d \)
derivatives
In calculus, derivatives represent the rate at which a function is changing at any given point. Think of them as the mathematical tool used for determining slopes of functions. For the function \( f(x) = x^3 + 7x^2 - 5x + 1 \), the derivative process involves
- First finding the first derivative: \( f'(x) = 3x^2 + 14x - 5 \).
- The second derivative: \( f''(x) = 6x + 14 \).
- The third derivative: \( f'''(x) = 6 \).
polynomials
Polynomials are expressions in which variables are raised to whole-number exponents. They are often written in a standard form where the terms are ordered from the highest degree to the lowest. To illustrate, the polynomial expressed as \( x^3 + 7x^2 - 5x + 1 \) is already in its standard form.
- Each term in the polynomial (like \( x^3 \) or \( 7x^2 \)) is called a monomial.
- A polynomial can have one or more terms.
calculus
Calculus is a branch of mathematics focused on change and motion. It is split into two main branches: differential calculus and integral calculus. Derivatives are part of differential calculus and are integral in the development of Taylor polynomials. The exercise here centers around expanding a function using its Taylor series around a point, relying on its derivatives and coefficients.
- Differential calculus concerns calculating instantaneous rates of change (like slopes).
- Integral calculus focuses on finding areas under curves and accumulations.
Other exercises in this chapter
Problem 19
Find the second-degree Taylor polynomial for \(f(x)=\) \(4 x^{2}-7 x+2\) about \(x=0 .\) What do you notice?
View solution Problem 20
For \(|x| \leq 0.1,\) graph the error $$E_{0}=\cos x-P_{0}(x)=\cos x-1$$ Explain the shape of the graph, using the Taylor expansion of \(\cos x .\) Find a bound
View solution Problem 20
using known Taylor series, find the first four nonzero terms of the Taylor series about 0 for the function. $$\sqrt{(1+t)} \sin t$$
View solution Problem 20
Find an expression for the general term of the series and give the range of values for the index \((n \text { or } k\) for example). $$\sin x=x-\frac{x^{3}}{3 !
View solution