Problem 5
Question
Do the following. (a) Compute the fourth degree Taylor polynomial for \(f(x)\) at \(x=0 .\) (b) On the same set of axes, graph \(f(x), P_{1}(x), P_{2}(x), P_{3}(x)\), and \(P_{4}(x)\). (c) Use \(P_{1}(x), P_{2}(x), P_{3}(x)\), and \(P_{4}(x)\) to approximate \(f(0.1)\) and \(f(0.3) .\) Compare these approximations to those given by a calculator. $$ f(x)=\sqrt{1+x} $$
Step-by-Step Solution
Verified Answer
The fourth degree Taylor polynomial is \(P_{4}(x) = 1 + 0.5x - 0.125x^2 + 0.03125x^3 - 0.015625x^4\). For \(f(0.1)\), the approximations using the Taylor polynomials are 1.05, 1.04875, 1.0484375 and 1.0484375 and the actual function value is about 1.048808848. For \(f(0.3)\), the approximations are 1.15, 1.1425, 1.140625 and 1.14078125 and the function value is approximately 1.140175425.
1Step 1: Compute the derivatives of \(f(x)\) at \(x=0\)
First, you have to calculate the \(n\)th derivative of the function \(f(x) = \sqrt{1+x}\) at \(x = 0\) in order to use the Taylor series formula. The derivatives of \(f(x)\) at \(x=0\) for \(n=0, 1, 2, 3, 4\) are: \(f(0) = 1, f'(0) = 0.5, f''(0) = -0.25, f'''(0) = 0.375, f''''(0) = -1.875\).
2Step 2: Compute the Taylor Polynomials
Next, compute the Taylor series up to the fourth degree using the formula for the Taylor series: \( P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(0) \cdot x^k}{k!} \) which gives: \( P_{1}(x) = 1 + 0.5x, P_{2}(x) = 1+ 0.5x - 0.125x^2, P_{3}(x) = 1 + 0.5x -0.125x^2 + 0.03125x^3, P_{4}(x) = 1 + 0.5x - 0.125x^2 + 0.03125x^3 - 0.015625x^4 \).
3Step 3: Graph the Function and the Polynomials}
Using any graphing tool at your disposal, graph the function \(f(x) = \sqrt{1+x}\) and the Taylor polynomials \(P_{1}(x), P_{2}(x), P_{3}(x), P_{4}(x)\). This can help you visualize the function and its approximations by the Taylor polynomials.
4Step 4: Approximate the Function Values}
You can approximate the function values \(f(0.1)\) and \(f(0.3)\) using the Taylor polynomials. Again, to do this, substitute \(x\) with 0.1 and 0.3 in the polynomials. Compare these approximations with the actual function value. Notice that as the degree of the polynomial increases, the approximation becomes more accurate.
Key Concepts
Taylor polynomialfunction approximationgraphing functionsderivatives
Taylor polynomial
A Taylor polynomial is a powerful mathematical tool used to approximate complex functions. It's based on the idea of expanding a function into an infinite series of polynomials. For our case with the function \( f(x) = \sqrt{1+x} \), we calculated its Taylor polynomial up to the fourth degree at \( x=0 \). Here's how it works:
- Start with a function you want to approximate.
- Find the function's derivatives at the point of approximation, which is \( x=0 \) here.
- Use the derivatives to build the Taylor series: \( P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(0) \cdot x^k}{k!} \).
function approximation
Function approximation using Taylor polynomials is straightforward but essential for analyzing complex behaviors. Taylor polynomials provide a way to estimate the output of a function without calculating its exact value.
For example, to approximate \( f(0.1) \) and \( f(0.3) \) for \( f(x) = \sqrt{1+x} \), we substitute these values into each Taylor polynomial from \( P_1(x) \) to \( P_4(x) \). As we go from \( P_1(x) \) being a simple linear approximation to \( P_4(x) \) as more complex polynomial, our estimates become more precise:
For example, to approximate \( f(0.1) \) and \( f(0.3) \) for \( f(x) = \sqrt{1+x} \), we substitute these values into each Taylor polynomial from \( P_1(x) \) to \( P_4(x) \). As we go from \( P_1(x) \) being a simple linear approximation to \( P_4(x) \) as more complex polynomial, our estimates become more precise:
- \( P_1(x) \) might give a rough estimate.
- \( P_2(x) \) adds a quadratic term for more accuracy.
- \( P_3(x) \) includes a cubic term, refining the approximation now further.
- \( P_4(x) \) provides the best approximation with a quartic term.
graphing functions
Visualizing functions and their approximations is an insightful exercise. When graphing \( f(x) = \sqrt{1+x} \) alongside its Taylor polynomials, we observe how well each polynomial configuration matches the actual function.
Graphing steps include:
Graphing steps include:
- Plot the original function \( f(x) = \sqrt{1+x} \).
- Overlay the Taylor polynomials \( P_1(x), P_2(x), P_3(x), \) and \( P_4(x) \) on the same axes.
- Compare their proximity to \( f(x) \) within the region of interest, like near \( x=0 \).
derivatives
Derivatives play a pivotal role in forming Taylor polynomials. In essence, they measure how a function changes, which is crucial for constructing accurate approximations.
For the function \( f(x) = \sqrt{1+x} \), calculating its first several derivatives at \( x=0 \) is essential for determining each term in its Taylor series. Here's a quick breakdown:
For the function \( f(x) = \sqrt{1+x} \), calculating its first several derivatives at \( x=0 \) is essential for determining each term in its Taylor series. Here's a quick breakdown:
- The first derivative \( f'(x) \) tells us the slope at a point, which helps confine the linear approximation.
- The second derivative \( f''(x) \) provides insight into the function's curvature, aiding in building the quadratic term.
- Further derivatives \( f'''(x) \) and \( f''''(x) \) refine our understanding of how the function changes, leading to higher degree terms.
Other exercises in this chapter
Problem 5
The second degree Taylor polynomial for \(f(x)=(1+x)^{p}\) is \(1+p x+\frac{p(p-1)}{2 !} x^{2} .\) If the second degree Taylor polynomial is used to approximate
View solution Problem 5
Find the Taylor series for \(f(x)\) centered at the indicated value of \(b\). $$ f(x)=2 \cos x, \quad b=\frac{\pi}{2} $$
View solution Problem 6
Determine whether the series converges or diverges. It is possible to solve Problems 4 through 19 without the Limit Comparison, Ratio, and Root Tests. \(\sum_{k
View solution Problem 6
For each series, determine whether the series converges absolutely, converges conditionally, or diverges. $$ \sum_{k=0}^{\infty}\left(-\frac{11}{12}\right)^{k}
View solution