Problem 81 - Calculus Early Transcendentals

Question

Find the next two terms of the following Taylor series. $$\frac{1}{\sqrt{1+x}} 1-\frac{1}{2} x+\frac{1 \cdot 3}{2 \cdot 4} x^{2}-\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} x^{3}+\cdots$$

Step-by-Step Solution

Answer

Answer: The next two terms in the Taylor series for the function \(f(x) = \frac{1}{\sqrt{1+x}}\) are \(\frac{1 \cdot 3 \cdot 5 \cdot 7}{2 \cdot 4 \cdot 6 \cdot 8}x^4\) and \(-\frac{1 \cdot 3 \cdot 5 \cdot 7 \cdot 9}{2 \cdot 4 \cdot 6 \cdot 8 \cdot 10} x^5\).

Step 1: 1. Understanding the given Taylor series

The given series is: $$\frac{1}{\sqrt{1+x}} = 1-\frac{1}{2} x+\frac{1 \cdot 3}{2 \cdot 4} x^{2}-\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} x^{3}+\cdots$$ We can see that the coefficients of the Taylor series have alternating signs and a pattern in the terms of the numerators and denominators.

Step 2: 2. Finding the general term of the Taylor series

By analyzing the series' coefficients, we can guess the general term of the Taylor series. We note that every term has a numerator with consecutive odd numbers and a denominator with consecutive even numbers. Additionally, we can see that the series has alternating signs. Hence, we can express the nth term of the series as: $$T_n = (-1)^n \cdot \frac{(2n-1)!!}{(2n)!!}x^n$$ Where \(n=0, 1, 2, \cdots\) and \(!!\) denotes the double factorial function (product of integers spaced by two, e.g., \(5!!=5\cdot3\cdot1\)).

Step 3: 3. Calculating the next term (Term 4) of the Taylor series

To find the fourth term of the series, we will evaluate the general term when \(n=4\), using the expression we found in step 2: $$T_4=(-1)^4 \cdot \frac{(2 \cdot 4 - 1)!!}{(2 \cdot 4)!!}x^4 = \frac{(7)!!}{(8)!!}x^4 = \frac{1 \cdot 3 \cdot 5 \cdot 7}{2 \cdot 4 \cdot 6 \cdot 8} x^4$$

Step 4: 4. Calculating the next term (Term 5) of the Taylor series

To find the fifth term of the series, we will evaluate the general term when \(n=5\), using the expression we found in step 2: $$T_5=(-1)^5 \cdot \frac{(2 \cdot 5 - 1)!!}{(2 \cdot 5)!!}x^5 = -\frac{(9)!!}{(10)!!}x^5 = -\frac{1 \cdot 3 \cdot 5 \cdot 7 \cdot 9}{2 \cdot 4 \cdot 6 \cdot 8 \cdot 10} x^5$$

Step 5: 5. Writing the updated Taylor series with the found terms

Now that we have found the next two terms of the Taylor series, we can write the updated series as: $$\frac{1}{\sqrt{1+x}} = 1-\frac{1}{2} x+\frac{1 \cdot 3}{2 \cdot 4} x^{2}-\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} x^{3} + \frac{1 \cdot 3 \cdot 5 \cdot 7}{2 \cdot 4 \cdot 6 \cdot 8}x^4 -\frac{1 \cdot 3 \cdot 5 \cdot 7 \cdot 9}{2 \cdot 4 \cdot 6 \cdot 8 \cdot 10} x^5+\cdots$$