Find the third order Taylor polynomial at x=0 of f(x)=(1−x)1/7.

To find the third-order Taylor polynomial for the function \( f(x) = (1 - x)^{1/7} \) centered at \( x = 0 \), we need to compute the derivatives of \( f(x) \) and then apply the Taylor series expansion formula:

\[
P_3(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3
\]

### Step 1: Compute the derivatives of \( f(x) \)

The function \( f(x) = (1 - x)^{1/7} \) is a power function, so we can use the chain rule and power rule to compute the derivatives.

#### First derivative:
\[
f'(x) = \frac{d}{dx} \left( (1 - x)^{1/7} \right) = \frac{1}{7}(1 - x)^{-6/7} \cdot (-1) = -\frac{1}{7}(1 - x)^{-6/7}
\]

At \( x = 0 \):
\[
f'(0) = -\frac{1}{7}(1 - 0)^{-6/7} = -\frac{1}{7}
\]

#### Second derivative:
\[
f''(x) = \frac{d}{dx} \left( -\frac{1}{7}(1 - x)^{-6/7} \right) = -\frac{1}{7} \cdot \left( -\frac{6}{7} \right) (1 - x)^{-13/7}
\]
\[
f''(x) = \frac{6}{49} (1 - x)^{-13/7}
\]

At \( x = 0 \):
\[
f''(0) = \frac{6}{49}(1 - 0)^{-13/7} = \frac{6}{49}
\]

#### Third derivative:
\[
f^{(3)}(x) = \frac{d}{dx} \left( \frac{6}{49}(1 - x)^{-13/7} \right) = \frac{6}{49} \cdot \left( -\frac{13}{7} \right) (1 - x)^{-20/7}
\]
\[
f^{(3)}(x) = -\frac{78}{343} (1 - x)^{-20/7}
\]

At \( x = 0 \):
\[
f^{(3)}(0) = -\frac{78}{343}(1 - 0)^{-20/7} = -\frac{78}{343}
\]

### Step 2: Construct the third-order Taylor polynomial

Now, substitute the values of \( f(0) \), \( f'(0) \), \( f''(0) \), and \( f^{(3)}(0) \) into the Taylor polynomial formula:

\[
P_3(x) = 1 + \left( -\frac{1}{7} \right)x + \frac{6}{49} \cdot \frac{x^2}{2!} + \left( -\frac{78}{343} \right) \cdot \frac{x^3}{3!}
\]

Simplify the terms:

\[
P_3(x) = 1 - \frac{1}{7}x + \frac{3}{49}x^2 - \frac{13}{343}x^3
\]

Thus, the third-order Taylor polynomial for \( f(x) = (1 - x)^{1/7} \) at \( x = 0 \) is:

\[
P_3(x) = 1 - \frac{1}{7}x + \frac{3}{49}x^2 - \frac{13}{343}x^3
\]

Would you like further clarification or details on this? Here are 5 related questions to explore further:

1. What is the general form of the Taylor series for any function?
2. How would you find the radius of convergence for this Taylor series?
3. How does the Taylor polynomial approximate the function near \( x = 0 \)?
4. Can you generalize the formula for any exponent \( n \) in \( (1 - x)^n \)?
5. How can you determine the error in approximating \( f(x) \) using a Taylor polynomial?

**Tip**: The higher the degree of the Taylor polynomial, the more accurate it is near the point of expansion.

Find the third order Taylor polynomial at x=0 of f(x)=(1−x)^{1/7}.

This page explains how to find the third-order Taylor polynomial for the function f(x) = (1−x)^{1/7} centered at x=0. It covers key concepts such as Taylor series expansion, derivatives, and polynomial approximation, with step-by-step calculations. Suitable for undergraduate calculus students.

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