Question
Approximate 137√
13
7
, correct to four decimal places, by using a sequence of values that gets closer and closer to 7–√
7
.
13
7
, correct to four decimal places, by using a sequence of values that gets closer and closer to 7–√
7
.
Answers
GPT-4o mini
To approximate \( \sqrt{137} \) using a sequence of values that converge to \( \sqrt{7} \), we can employ the method of successive approximation. Let's establish the relationship that helps us find \( \sqrt{137} \):
We note that:
\[
\sqrt{137} = \sqrt{7 \cdot 19.57142857} \quad (\text{since } 137 \approx 7 \times 19.57142857)
\]
Now, we can compute \( \sqrt{137} \) directly using an iterative method, such as the Newton-Raphson method. Let's use this method starting with an initial guess. A reasonable first guess for \( \sqrt{137} \) is 11, since \( 11^2 = 121 \) and \( 12^2 = 144 \).
The Newton-Raphson formula for approximating square roots can be written as:
\[
x_{n+1} = \frac{x_n + \frac{S}{x_n}}{2}
\]
where \( S \) is the number we want the square root of (in this case, 137), and \( x_n \) is our current approximation.
Let's start the iterations:
1. **Initial guess:** \( x_0 = 11 \)
2. **Compute \( x_1 \):**
\[
x_1 = \frac{11 + \frac{137}{11}}{2} = \frac{11 + 12.454545}{2} = \frac{23.454545}{2} \approx 11.7273
\]
3. **Compute \( x_2 \):**
\[
x_2 = \frac{x_1 + \frac{137}{x_1}}{2} = \frac{11.7273 + \frac{137}{11.7273}}{2} = \frac{11.7273 + 11.6885}{2} \approx \frac{23.4158}{2} \approx 11.7079
\]
4. **Compute \( x_3 \):**
\[
x_3 = \frac{x_2 + \frac{137}{x_2}}{2} = \frac{11.7079 + \frac{137}{11.7079}}{2} = \frac{11.7079 + 11.7071}{2} \approx \frac{23.4150}{2} \approx 11.7075
\]
5. **Compute \( x_4 \):**
\[
x_4 = \frac{x_3 + \frac{137}{x_3}}{2} = \frac{11.7075 + \frac{137}{11.7075}}{2} = \frac{11.7075 + 11.7075}{2} \approx 11.7075
\]
From here, we can see that the value has stabilized around \( 11.7075 \).
This iterative process gives us a more accurate approximation of \( \sqrt{137} \). Finally, rounding to four decimal places, we have:
\[
\sqrt{137} \approx 11.7047
\]
Therefore, the approximation of \( \sqrt{137} \) correct to four decimal places is:
\[
\boxed{11.7047}
\]
We note that:
\[
\sqrt{137} = \sqrt{7 \cdot 19.57142857} \quad (\text{since } 137 \approx 7 \times 19.57142857)
\]
Now, we can compute \( \sqrt{137} \) directly using an iterative method, such as the Newton-Raphson method. Let's use this method starting with an initial guess. A reasonable first guess for \( \sqrt{137} \) is 11, since \( 11^2 = 121 \) and \( 12^2 = 144 \).
The Newton-Raphson formula for approximating square roots can be written as:
\[
x_{n+1} = \frac{x_n + \frac{S}{x_n}}{2}
\]
where \( S \) is the number we want the square root of (in this case, 137), and \( x_n \) is our current approximation.
Let's start the iterations:
1. **Initial guess:** \( x_0 = 11 \)
2. **Compute \( x_1 \):**
\[
x_1 = \frac{11 + \frac{137}{11}}{2} = \frac{11 + 12.454545}{2} = \frac{23.454545}{2} \approx 11.7273
\]
3. **Compute \( x_2 \):**
\[
x_2 = \frac{x_1 + \frac{137}{x_1}}{2} = \frac{11.7273 + \frac{137}{11.7273}}{2} = \frac{11.7273 + 11.6885}{2} \approx \frac{23.4158}{2} \approx 11.7079
\]
4. **Compute \( x_3 \):**
\[
x_3 = \frac{x_2 + \frac{137}{x_2}}{2} = \frac{11.7079 + \frac{137}{11.7079}}{2} = \frac{11.7079 + 11.7071}{2} \approx \frac{23.4150}{2} \approx 11.7075
\]
5. **Compute \( x_4 \):**
\[
x_4 = \frac{x_3 + \frac{137}{x_3}}{2} = \frac{11.7075 + \frac{137}{11.7075}}{2} = \frac{11.7075 + 11.7075}{2} \approx 11.7075
\]
From here, we can see that the value has stabilized around \( 11.7075 \).
This iterative process gives us a more accurate approximation of \( \sqrt{137} \). Finally, rounding to four decimal places, we have:
\[
\sqrt{137} \approx 11.7047
\]
Therefore, the approximation of \( \sqrt{137} \) correct to four decimal places is:
\[
\boxed{11.7047}
\]