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Magazine Chapter 7: Problem 9
In Problems \(6-10\), compute the Taylor polynomial of degree \(n\) about \(a=0\) for the indicated functions. $$ f(x)=x^{5}, n=6 $$
Short Answer
Expert verified
The Taylor polynomial of degree 6 for \(f(x) = x^5\) about \(a = 0\) is \(P_6(x) = x^5\).
Step by step solution
01
Understanding a Taylor Polynomial
A Taylor polynomial of degree \(n\) for a function \(f(x)\) about \(x = a\) is given by:\[P_n(x) = f(a) + f'(a)x + \frac{f''(a)}{2!}x^2 + \cdots + \frac{f^{(n)}(a)}{n!}x^n\]where \(f^{(k)}(a)\) is the \(k\)-th derivative of \(f\) evaluated at \(x = a\). In this case, \(a = 0\) and \(f(x) = x^5\), and we need a polynomial up to degree \(n = 6\).
02
Compute Zero and Higher Derivatives
First, evaluate \(f(x) = x^5\) at \(x = 0\). Then, calculate the higher derivatives of \(f(x)\):- \(f(x) = x^5\), so \(f(0) = 0\).- \(f'(x) = 5x^4\), so \(f'(0) = 0\).- \(f''(x) = 20x^3\), so \(f''(0) = 0\).- \(f'''(x) = 60x^2\), so \(f'''(0) = 0\).- \(f^{(4)}(x) = 120x\), so \(f^{(4)}(0) = 0\).- \(f^{(5)}(x) = 120\), so \(f^{(5)}(0) = 120\).- \(f^{(6)}(x) = 0\), so \(f^{(6)}(0) = 0\).
03
Construct the Taylor Polynomial
Using the formula for the Taylor polynomial and the computed derivatives, we form:\[P_6(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \frac{f^{(5)}(0)}{5!}x^5 + \frac{f^{(6)}(0)}{6!}x^6\]Substitute the values:\[P_6(x) = 0 + 0 + 0 + 0 + 0 + \frac{120}{5!}x^5 + 0\]Calculate and simplify:\(5! = 120\), thus:\[P_6(x) = x^5\].
04
Confirm Polynomial Degree and Coefficients
Check that all terms beyond \(x^5\) are zero to ensure \(P_6(x)\) doesn't extend beyond degree 5. Calculate higher-order derivatives and evaluate them, confirming all coefficients after \(x^5\) are zero, finalizing the Taylor polynomial for degree \(n = 6\) as:\[P_6(x) = x^5\].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Derivatives
Derivatives are a fundamental concept in calculus. They represent the rate at which a function changes. In simpler terms, a derivative provides the slope of the function at any given point.
For a function such as the one presented, \(f(x) = x^5\), calculating the derivative involves using the power rule. This rule states that if you have a term \(x^n\), its derivative is \(nx^{n-1}\).
For a function such as the one presented, \(f(x) = x^5\), calculating the derivative involves using the power rule. This rule states that if you have a term \(x^n\), its derivative is \(nx^{n-1}\).
- The first derivative \(f'(x) = 5x^4\) tells us the slope of the original function \(f(x)\).
- Higher derivatives follow similarly: the second derivative \(f''(x) = 20x^3\) shows how the slope itself is changing, and this pattern continues through \(f^{(n)}(x)\).
Polynomial Degree Explored
The degree of a polynomial is the highest power of \(x\) present in the expression.
Understanding polynomial degree is crucial when constructing Taylor polynomials because it dictates how many terms are included. For example, a Taylor polynomial of degree \(n=6\) will potentially have terms ranging up to \(x^6\).
However, in the given problem, after calculating higher-order derivatives, we notice:
Understanding polynomial degree is crucial when constructing Taylor polynomials because it dictates how many terms are included. For example, a Taylor polynomial of degree \(n=6\) will potentially have terms ranging up to \(x^6\).
However, in the given problem, after calculating higher-order derivatives, we notice:
- All terms with powers higher than 5 are zero, simplifying our work.
- The degree of the polynomial directly impacts the accuracy around the point of expansion \(a=0\).
Function Evaluation in Taylor Polynomials
Function evaluation involves substituting the point of interest into the function and its derivatives. Evaluating these at \(x = a = 0\) is the basis of forming the Taylor series.
For any polynomial term \(f^{(k)}(a)x^k\), the value of \(f^{(k)}(a)\) is essential:
For any polynomial term \(f^{(k)}(a)x^k\), the value of \(f^{(k)}(a)\) is essential:
- For \(f(x) = x^5\), evaluating yields \(f(0) = 0\).
- For each derivative up to the fifth order, the evaluations at \(x=0\) reveal significant properties, primarily resulting in zeros till the fifth derivative.
Exploring Higher-Order Derivatives
Higher-order derivatives extend beyond just finding the slope; they reveal the function's curvature and changes in curvature. This progression from the first derivative (slope) to successive derivatives depicts the function's detailed behavior.
With \(f(x) = x^5\), calculating derivatives up to the sixth order, we see:
With \(f(x) = x^5\), calculating derivatives up to the sixth order, we see:
- Higher derivatives such as \(f^{(3)}(x) = 60x^2\), and \(f^{(4)}(x) = 120x\) illustrate how the function grows more dynamically.
- The fifth derivative \(f^{(5)}(x) = 120\) results in a constant, representing a critical part of the Taylor expansion at the given degree.
- The sixth derivative \(f^{(6)}(x)\) becoming zero, affirms the stopping point for higher degrees without new contributions.
