Chapter 8: Q. 37 (page 680)

In Exercises 31–40 find the Maclaurin series for the specified function. Note: These are the same functions as in Exercises 21–30.
$\sqrt{1 - x}$

Short Answer

Expert verified

Ans:The Maclaurin series of the function $f (x) = 1 - \frac{1}{2} x - \sum_{k = 2}^{\infty} \frac{1 \cdot 3 \cdot 5 \dots (2 k - 3)}{2^{k} k!} x^{k}$

Step by step solution

Step 1. Given information:

$f (x) = \sqrt{1 - x}$

Step 2. Finding the general form Maclaurin series:

Since for any function $f$ with derivatives of all orders at the point $x_{0}=0$, then the Maclurin series is

$f (x) = f (0) + f^{'} (0) x + \frac{f^{''} (0)}{2!} x^{2} + \frac{f^{''} (0)}{3!} x^{3} + \frac{f^{'''} (0)}{4!} x^{4} + \dots$

Or, we can write the general form Maclurin series of the function $f$ is

f (x) = \sum_{n = 0}^{\infty} \frac{f^{n} (0)}{n!} x^{n}

Step 3. Constructing the table of the Maclaurin series for the function :

So, let us first construct the table of the Maclaurin series for the function $f (x) = \sqrt{1 - x}$

localid="1649437869098" $\begin{matrix} n & f^{n} (x) & f^{n} (0) & \frac{f^{n} (0)}{n!} \\ 0 & \sqrt{1 - x} & 1 & 1 \\ 1 & - \frac{1}{2 \sqrt{1 - x}} & - \frac{1}{2} & - \frac{1}{2} \\ 2 & - \frac{1}{2^{2}} (1 - x)^{- \frac{3}{2}} & - \frac{1}{2^{2}} & - \frac{1}{2^{2}} \\ 3 & - \frac{1 \cdot 3}{2^{3}} (1 - x)^{- \frac{5}{2}} & - \frac{1 \cdot 3}{2^{3}} & - \frac{1 \cdot 3}{2^{3}} \\ 4 & - \frac{1 \cdot 3 \cdot 5}{2^{4}} (1 - x)^{- \frac{7}{2}} & - \frac{1 \cdot 3 \cdot 5}{2^{4}} & - \frac{1 \cdot 3 \cdot 5}{2^{4}} \\ \cdot & \cdot & \cdot & 4! \\ 5 & - \frac{1 \cdot 3 \cdot 5 \dots (2 k - 3)}{2^{k}} (1 - x)^{- \frac{2 k - 1}{2}} & - \frac{1 \cdot 3 \cdot 5 \dots (2 k - 3)}{2 k} & - \frac{1 \cdot 3 \cdot 5 \dots (2 k - 3)}{2^{k}} \end{matrix}$

Step 4. Finding the Maclaurin series :

Therefore, the Maclaurin series for the function $f (x) = \sqrt{1 - x}$ is

$1 + - \frac{1}{2} \cdot x + \frac{- \frac{1}{2^{2}}}{2!} x^{2} + \frac{- \frac{1 \cdot 3}{2^{3}}}{3!} x^{3} + \frac{- \frac{1 \cdot 3 \cdot 5}{2^{4}}}{4!} x^{4} + \dots + \frac{- \frac{1 \cdot 3 \cdot 5 \dots (2 k - 3)}{2^{k}}}{k!} x^{k}$

Or, we can write as

f (x) = 1 - \frac{1}{2} x - \sum_{k = 2}^{\infty} \frac{1 \cdot 3 \cdot 5 \dots (2 k - 3)}{2^{k} k!} x^{k}

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In Exercises 31–40 find the Maclaurin series for the specified function. Note: These are the same functions as in Exercises 21–30.
$\sqrt{1 - x}$

Short Answer

Step by step solution

Step 1. Given information:

Step 2. Finding the general form Maclaurin series:

Step 3. Constructing the table of the Maclaurin series for the function :

Step 4. Finding the Maclaurin series :

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In Exercises 31–40 find the Maclaurin series for the specified function. Note: These are the same functions as in Exercises 21–30.1-x

Short Answer

Step by step solution

Step 1. Given information:

Step 2. Finding the general form Maclaurin series:

Step 3. Constructing the table of the Maclaurin series for the function :

Step 4. Finding the Maclaurin series :

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

Discrete Mathematics

Decision Maths

Probability and Statistics

Geometry

Logic and Functions

Study anywhere. Anytime. Across all devices.

In Exercises 31–40 find the Maclaurin series for the specified function. Note: These are the same functions as in Exercises 21–30.
$\sqrt{1 - x}$