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Chapter 7: Problem 7

Compute the Taylor polynomial of degree \(n\) about \(x=0\) for each function. $$ f(x)=(1+x)^{3}, n=5 $$

Short Answer

Expert verified
The Taylor polynomial is \( P_5(x) = 1 + 3x + 3x^2 + x^3 \).

Step by step solution

01

Find the function's derivatives

We need to determine the derivatives of \(f(x) = (1+x)^{3}\)\ up to the fifth derivative because we want a Taylor polynomial of degree 5. 1. The function itself:\[ f(x) = (1+x)^{3}\]2. First derivative:\( f'(x) = 3(1+x)^{2} \)3. Second derivative:\( f''(x) = 6(1+x) \)4. Third derivative:\( f'''(x) = 6 \)\5. Fourth and higher derivatives (until 5):\( f^{(4)}(x), f^{(5)}(x) = 0 \) since the constant derivative remains zero.
02

Evaluate the derivatives at \( x = 0 \)

Now plug \( x = 0 \) into each derivative calculated above to find the specific values for the Taylor polynomial.1. \( f(0) = (1+0)^{3} = 1 \)2. \( f'(0) = 3(1+0)^{2} = 3 \)3. \( f''(0) = 6(1+0) = 6 \)4. \( f'''(0) = 6 \)5. \( f^{(4)}(0) = 0 \)6. \( f^{(5)}(0) = 0 \)
03

Write the Taylor polynomial formula

The general Taylor polynomial of degree \( n \) centered at \( a \) is:\[ P_n(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots + \frac{f^{(n)}(a)}{n!}(x-a)^n \]Since we want \( n = 5 \) and \( a = 0 \), our polynomial becomes:\[ P_5(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \frac{f^{(5)}(0)}{5!}x^5 \]
04

Substitute the derivative values into the polynomial

Substitute the values from Step 2 into the Taylor polynomial:\[P_5(x) = 1 + \frac{3}{1!}x + \frac{6}{2!}x^2 + \frac{6}{3!}x^3 + \frac{0}{4!}x^4 + \frac{0}{5!}x^5 = 1 + 3x + 3x^2 + x^3\]
05

Write final Taylor polynomial

Since higher-degree terms are zero, the Taylor polynomial of degree 5 for \( (1+x)^{3} \) centered at \( x=0 \) is:\( P_5(x) = 1 + 3x + 3x^2 + x^3 \)

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Key Concepts

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

Derivatives
To construct a Taylor polynomial, specifically of degree 5, for the function \(f(x) = (1+x)^{3}\), we first need to determine its derivatives up to the fifth order. Derivatives are essentially the rates at which the function's values change as its input \(x\) varies. Each derivative provides a snapshot of this rate of change at different levels of accuracy:
  • The function itself is \(f(x) = (1+x)^{3}\).
  • The first derivative, \(f'(x) = 3(1+x)^{2}\), tells us the slope or the rate of change of \(f(x)\).
  • The second derivative, \(f''(x) = 6(1+x)\), gives us the rate of change of the first derivative. In simpler terms, it describes how the slope itself is changing.
  • The third derivative, \(f'''(x) = 6\), reveals how the second derivative changes, offering deeper insight into the curvature of \(f(x)\).
  • The fourth and fifth derivatives, \(f^{(4)}(x)\) and \(f^{(5)}(x)\), both equal \(0\), meaning from this point, changes in higher derivatives do not affect the function's shape for our purpose.
Understanding these derivatives allows us to piece together a Taylor polynomial efficiently, as they dictate the polynomial's terms.
Centered at x=0
When we talk about a Taylor polynomial centered at \(x=0\), we're focusing our approximation around this particular point, also known as the expansion point. This concept is pivotal because as we evaluate our derivatives at \(x=0\), we determine how the Taylor series behaves in its vicinity. Here's why it matters:
  • The center \(x=0\) is chosen because it might simplify calculations, a common choice known as a Maclaurin series when the center is zero.
  • Evaluating derivatives at this point gives specific coefficients for our polynomial terms. For instance, \(f(0) = 1\), \(f'(0) = 3\), and so forth.
  • These evaluations fix the polynomial's behavior exactly at \(x=0\), making it a starting point for approximating \(f(x)\) in areas close to \(0\).
By centering at \(x=0\), we essentially tailor our approximation to be most accurate where \(x\) is near \(0\). This approach is optimal when we're dealing with functions defined over the whole of \(\mathbb{R}\) or those centered primarily at \(x=0\).
Degree 5
The degree of a Taylor polynomial determines the level of detail in approximating the original function. For a degree of 5, we retain terms up to \(x^5\), although in this specific case of \(f(x) = (1+x)^3\), any terms beyond \(x^3\) are zero.Having a Taylor polynomial of degree 5 means:
  • We look at how the function behaves not just linearly, but also quadratically, cubically, and so forth, capturing nuances of \(f(x)\).
  • The Taylor polynomial becomes \(P_5(x) = 1 + 3x + 3x^2 + x^3\), effectively stopping after \(x^3\) since higher derivatives \(f^{(4)}(0)\) and \(f^{(5)}(0)\) are zero.
  • Having a higher degree increases the precision of our approximation, but in this case, since our function is a cubic polynomial, a degree 5 means simplicity with zero terms for \(x^4\) and \(x^5\).
The degree effectively dictates how far we carry out terms, ensuring our polynomial approximates \(f(x)\) around our chosen center \(x=0\) to the specified order.

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Most popular questions from this chapter

Use the Table of Integrals to compute each integral after manipulating the integrand in a suitable way. $$ \int \frac{1}{\sqrt{16-9 x^{2}}} d x $$

Show that, for positive constants \(r\) and \(K\), $$ r N\left(1-\frac{N}{K}\right) \approx r N $$ for \(N\) close to 0 .

Compute the Taylor polynomial of degree \(n\) about \(x=0\) for each function and compare the value of the function at the indicated point with the value of the corresponding Taylor polynomial. $$ f(x)=\ln (1+x), n=3, x=0.1 $$

We can show that the Taylor polynomial for \(f(x)=\tan ^{-1} x\) about \(x=0\) converges for \(|x| \leq 1\). (a) Show that the following is true: $$ \tan ^{-1} x=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\cdots+R_{n+1}(x) $$ (b) Explain why the following holds: $$ \frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots $$ (This series converges very slowly, as you would see if you used it to approximate \(\pi .\) )

Compute the Taylor polynomial of degree \(n\) about a and compare the value of the approximation with the value of the function at the given point \(x\). $$ f(x)=\sqrt{x}, a=1, n=3 ; x=2 $$
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