Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Experience the ease of finding quick and accurate answers to your questions from professionals on our platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
To calculate the binomial expansion of [tex]\( \sqrt{1 + x} \)[/tex], we can use the Binomial Theorem for non-integer exponents. The theorem states that:
[tex]\[ (1 + x)^n = 1 + nx + \frac{n(n - 1)x^2}{2!} + \frac{n(n - 1)(n - 2)x^3}{3!} + \cdots \][/tex]
For the function [tex]\( \sqrt{1 + x} \)[/tex], the exponent [tex]\( n \)[/tex] is [tex]\( \frac{1}{2} \)[/tex]. We'll calculate the first four terms of this expansion.
### Step-by-Step Solution:
Step 1: Calculate the constant term (first term):
[tex]\[ \text{Term 1} = 1 \][/tex]
Step 2: Calculate the linear term (second term):
[tex]\[ \text{Term 2} = n \cdot x = \frac{1}{2} \cdot x \][/tex]
Step 3: Calculate the quadratic term (third term):
[tex]\[ \text{Term 3} = \frac{n(n - 1) \cdot x^2}{2!} = \frac{\frac{1}{2} \left(\frac{1}{2} - 1\right) \cdot x^2}{2} = \frac{\frac{1}{2} \cdot -\frac{1}{2} \cdot x^2}{2} = \frac{-\frac{1}{4} \cdot x^2}{2} = -\frac{1}{8} \cdot x^2 \][/tex]
Step 4: Calculate the cubic term (fourth term):
[tex]\[ \text{Term 4} = \frac{n(n - 1)(n - 2) \cdot x^3}{3!} = \frac{\frac{1}{2} \left(\frac{1}{2} - 1\right) \left(\frac{1}{2} - 2\right) \cdot x^3}{6} = \frac{\frac{1}{2} \cdot -\frac{1}{2} \cdot -\frac{3}{2} \cdot x^3}{6} = \frac{\frac{3}{8} \cdot x^3}{6} = \frac{3}{48} \cdot x^3 = \frac{1}{16} \cdot x^3 \][/tex]
So, the first four terms of the binomial expansion of [tex]\( \sqrt{1 + x} \)[/tex] are:
[tex]\[ 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 \][/tex]
ii) Use the expansion to evaluate [tex]\( \sqrt{9} \)[/tex], correct to three decimal places:
To approximate [tex]\( \sqrt{9} \)[/tex], we can use the relationship [tex]\( \sqrt{9} = \sqrt{(1 + x)} \times 3 \)[/tex]. We modify [tex]\( 1 + x \)[/tex] such that the expression inside the square root is near 9. Consider [tex]\( x = -0.1 / 9 \)[/tex]. Thus:
#### Step 1:
Evaluate the expansion for [tex]\( x = -\frac{0.1}{9} \)[/tex]:
[tex]\[ x_{\text{approx}} = -\frac{0.1}{9} \][/tex]
#### Step 2:
Substitute [tex]\( x_{\text{approx}} \)[/tex] into each term calculated in the expansion:
Term 1:
[tex]\[ 1 \][/tex]
Term 2:
[tex]\[ \frac{1}{2} \cdot x_{\text{approx}} = \frac{1}{2} \cdot \left(-\frac{0.1}{9}\right) = -0.005557 \][/tex]
Term 3:
[tex]\[ -\frac{1}{8} (x_{\text{approx}})^2 = -\frac{1}{8} \left(-\frac{0.1}{9}\right)^2 = -1.125 \times 10^{-6} \][/tex]
Term 4:
[tex]\[ \frac{1}{16} (x_{\text{approx}})^3 = \frac{1}{16} \left(-\frac{0.1}{9}\right)^3 = 1.687 \times 10^{-9} \][/tex]
Summing these terms gives the binomial expansion's approximation:
[tex]\[ 1 + (-0.005557) + (-1.125 \times 10^{-6}) + 1.687 \times 10^{-9} \][/tex]
[tex]\[ = 0.994428 - 0.000001 + 0.0000001 \approx 0.994429 \][/tex]
Therefore, the approximation of [tex]\( \sqrt{9} \)[/tex] is:
[tex]\[ 0.994 \text{ (correct to three decimal places)} \][/tex]
This process enables us to use the binomial expansion to approximate the value of the square root to three decimal places.
[tex]\[ (1 + x)^n = 1 + nx + \frac{n(n - 1)x^2}{2!} + \frac{n(n - 1)(n - 2)x^3}{3!} + \cdots \][/tex]
For the function [tex]\( \sqrt{1 + x} \)[/tex], the exponent [tex]\( n \)[/tex] is [tex]\( \frac{1}{2} \)[/tex]. We'll calculate the first four terms of this expansion.
### Step-by-Step Solution:
Step 1: Calculate the constant term (first term):
[tex]\[ \text{Term 1} = 1 \][/tex]
Step 2: Calculate the linear term (second term):
[tex]\[ \text{Term 2} = n \cdot x = \frac{1}{2} \cdot x \][/tex]
Step 3: Calculate the quadratic term (third term):
[tex]\[ \text{Term 3} = \frac{n(n - 1) \cdot x^2}{2!} = \frac{\frac{1}{2} \left(\frac{1}{2} - 1\right) \cdot x^2}{2} = \frac{\frac{1}{2} \cdot -\frac{1}{2} \cdot x^2}{2} = \frac{-\frac{1}{4} \cdot x^2}{2} = -\frac{1}{8} \cdot x^2 \][/tex]
Step 4: Calculate the cubic term (fourth term):
[tex]\[ \text{Term 4} = \frac{n(n - 1)(n - 2) \cdot x^3}{3!} = \frac{\frac{1}{2} \left(\frac{1}{2} - 1\right) \left(\frac{1}{2} - 2\right) \cdot x^3}{6} = \frac{\frac{1}{2} \cdot -\frac{1}{2} \cdot -\frac{3}{2} \cdot x^3}{6} = \frac{\frac{3}{8} \cdot x^3}{6} = \frac{3}{48} \cdot x^3 = \frac{1}{16} \cdot x^3 \][/tex]
So, the first four terms of the binomial expansion of [tex]\( \sqrt{1 + x} \)[/tex] are:
[tex]\[ 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 \][/tex]
ii) Use the expansion to evaluate [tex]\( \sqrt{9} \)[/tex], correct to three decimal places:
To approximate [tex]\( \sqrt{9} \)[/tex], we can use the relationship [tex]\( \sqrt{9} = \sqrt{(1 + x)} \times 3 \)[/tex]. We modify [tex]\( 1 + x \)[/tex] such that the expression inside the square root is near 9. Consider [tex]\( x = -0.1 / 9 \)[/tex]. Thus:
#### Step 1:
Evaluate the expansion for [tex]\( x = -\frac{0.1}{9} \)[/tex]:
[tex]\[ x_{\text{approx}} = -\frac{0.1}{9} \][/tex]
#### Step 2:
Substitute [tex]\( x_{\text{approx}} \)[/tex] into each term calculated in the expansion:
Term 1:
[tex]\[ 1 \][/tex]
Term 2:
[tex]\[ \frac{1}{2} \cdot x_{\text{approx}} = \frac{1}{2} \cdot \left(-\frac{0.1}{9}\right) = -0.005557 \][/tex]
Term 3:
[tex]\[ -\frac{1}{8} (x_{\text{approx}})^2 = -\frac{1}{8} \left(-\frac{0.1}{9}\right)^2 = -1.125 \times 10^{-6} \][/tex]
Term 4:
[tex]\[ \frac{1}{16} (x_{\text{approx}})^3 = \frac{1}{16} \left(-\frac{0.1}{9}\right)^3 = 1.687 \times 10^{-9} \][/tex]
Summing these terms gives the binomial expansion's approximation:
[tex]\[ 1 + (-0.005557) + (-1.125 \times 10^{-6}) + 1.687 \times 10^{-9} \][/tex]
[tex]\[ = 0.994428 - 0.000001 + 0.0000001 \approx 0.994429 \][/tex]
Therefore, the approximation of [tex]\( \sqrt{9} \)[/tex] is:
[tex]\[ 0.994 \text{ (correct to three decimal places)} \][/tex]
This process enables us to use the binomial expansion to approximate the value of the square root to three decimal places.
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.
