Answer :
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.
