manual sqrt

Using a manual (Newton-Raphson) approximation:

1. Start with g₀ ≈ 1.96 (because 1.96² = 3.8416).
2. One refinement gives g₁ ≈ 1.96393.
3. A second refinement gives g₂ ≈ 1.9639258.

Therefore, the square root of 3.8570047402555048 begins:

1.96392…

The first six digits of the result are: 1.96392

Of course. We can calculate this using the "long division method" for square roots. This method finds the answer digit by digit.

The goal is to find the first 6 digits of the result for **√3.85700474...**

### Setup

First, we group the numbers in pairs starting from the decimal point.
*   To the left of the decimal: **3**
*   To the right of the decimal: **85 70 04 74...**

We set up the problem like this:
```
      _______
    √ 3. 85 70 04
```
The decimal point in our answer will be placed directly above the decimal point in the number.

---

### Step 1: Find the First Digit

*   Look at the first group: **3**.
*   Find the largest integer whose square is less than or equal to 3. That is **1** (since 1² = 1 and 2² = 4).
*   Write **1** as the first digit of our answer.
*   Subtract its square (1) from the first group (3).
*   Bring down the next pair of digits (85).

```
      1.______
    √ 3. 85 70 04
      -1
      ---
      2 85
```

---

### Step 2: Find the Second Digit

*   Double the current answer (1), which gives us **2**.
*   We need to find a digit `d` such that `(2d) * d` is as large as possible but less than or equal to 285. We can estimate this by looking at 28 / 2 ≈ 14, but `d` must be a single digit so it's likely high. Let's try 9.
*   Try `d=9`: `29 * 9 = 261`. This works.
*   Try `d=10`: Doesn't work.
*   So, the next digit is **9**.
*   Write **9** in the answer. Subtract 261 from 285.
*   Bring down the next pair (70).

```
      1. 9 ____
    √ 3. 85 70 04
      -1
      ---
      2 85
     -2 61
      ----
        24 70
```

---

### Step 3: Find the Third Digit

*   Double the current answer (19), which gives us **38**.
*   We need a digit `d` such that `(38d) * d` is less than or equal to 2470.
*   Estimate: 247 / 38 ≈ 6.
*   Try `d=6`: `386 * 6 = 2316`. This works.
*   Try `d=7`: `387 * 7 = 2709`. This is too large.
*   The next digit is **6**.
*   Write **6** in the answer. Subtract 2316 from 2470.
*   Bring down the next pair (04).

```
      1. 9 6 ___
    √ 3. 85 70 04
      -1
      ---
      2 85
     -2 61
      ----
        24 70
       -23 16
        -----
         154 04
```

---

### Step 4: Find the Fourth Digit

*   Double the current answer (196), which gives **392**.
*   We need a digit `d` such that `(392d) * d` is less than or equal to 15404.
*   Estimate: 1540 / 392 ≈ 3.
*   Try `d=3`: `3923 * 3 = 11769`. This works.
*   Try `d=4`: `3924 * 4 = 15696`. This is too large.
*   The next digit is **3**.
*   Write **3** in the answer. Subtract 11769 from 15404.
*   Bring down the next pair (74).

```
      1. 9 6 3 __
    √ 3. 85 70 04 74
      -1
      ---
      2 85
     -2 61
      ----
        24 70
       -23 16
        -----
         154 04
        -117 69
         ------
          3635 74
```

---

### Step 5: Find the Fifth Digit

*   Double the current answer (1963), which gives **3926**.
*   We need a digit `d` such that `(3926d) * d` is less than or equal to 363574.
*   Estimate: 36357 / 3926 ≈ 9.
*   Try `d=9`: `39269 * 9 = 353421`. This works.
*   The next digit is **9**.
*   Write **9** in the answer.

```
      1. 9 6 3 9 _
    √ 3. 85 70 04 74
      ...
          3635 74
         -3534 21
          --------
           10153
```

---

### Step 6: Find the Sixth Digit

*   We have 5 digits so far (`1.9639`). We need one more.
*   Double the current answer (19639), which gives **39278**.
*   Bring down the next pair of digits from the original number (02): `1015302`.
*   We need a digit `d` such that `(39278d) * d` is less than or equal to 1015302.
*   Estimate: 101530 / 39278 ≈ 2.
*   Try `d=2`: `392782 * 2 = 785564`. This works.
*   Try `d=3`: `392783 * 3 = 1178349`. This is too large.
*   The next digit is **2**.

The result is building up as **1.96392...**

### Conclusion

The first 6 digits of the result are **1.96392**.

Let's calculate \(\sqrt{3.8570047402555048}\) step by step **manually**, and find the first 6 digits of the result.

---

We know:
- \( 1.9^2 = 3.61 \)
- \( 2.0^2 = 4.0 \)
- \( 3.85700474 \) is between those, so the root is between 1.9 and 2.0.

Let’s use the **Babylonian (Heron's) method** for better precision.

Let’s start with \( x_0 = 2 \).

### Iteration 1
\[
x_1 = \frac{1}{2} \left( x_0 + \frac{3.85700474}{x_0} \right)
\]
\[
x_1 = \frac{1}{2} \left( 2 + \frac{3.85700474}{2} \right)
\]
\[
x_1 = \frac{1}{2} (2 + 1.92850237) = \frac{1}{2} (3.92850237) = 1.964251185
\]

### Iteration 2
\[
x_2 = \frac{1}{2} \left( 1.964251185 + \frac{3.85700474}{1.964251185} \right)
\]

Calculate \( \frac{3.85700474}{1.964251185} \):

Estimate:

\( 1.964251185 \times 1.96 \approx 3.85 \), so \( \frac{3.85700474}{1.964251185} \approx 1.9635 \).

So,

\[
x_2 \approx \frac{1}{2}(1.964251185 + 1.9635) = \frac{1}{2}(3.927751185) = 1.96387559
\]

### Iteration 3

Now,

\[
x_3 = \frac{1}{2} \left( 1.96387559 + \frac{3.85700474}{1.96387559} \right)
\]

Calculate \( \frac{3.85700474}{1.96387559} \):

Let's estimate:

\( 2 \times 1.96387559 = 3.92775118 \)

So, \( \frac{3.85700474}{1.96387559} \approx 1.96361 \)

Now,

\[
x_3 \approx \frac{1}{2}(1.96387559 + 1.96361) = \frac{1}{2}(3.92748559) = 1.9637428
\]

---

After three iterations, the value stabilizes around **1.96374...**

### First 6 digits

So, the first 6 digits are:  
**1.96374**

If you want to round to the first 6 decimals:  
**1.963742**

---

## **Final Answer**

\[
\boxed{1.96374}
\]
(first 6 digits after the decimal: 1.96374 **2**)

To compute the square root of $ \sqrt{3.8570047402555048} $ manually and determine the **first 6 digits of the result**, we apply the **long division-like algorithm for square roots**, which involves the following key steps:

---

### **Step-by-Step Approach:**

We treat the number as $ 3.8570047402555048 $, grouping the digits into pairs **from the decimal point**, regardless of where the decimal lies:

```
3. | 85 70 04 74 02 55 50 48 ...
```

We then proceed as follows:

---

### **Step 1: Integer Part (Pair '3')**

- Largest $ x $ such that $ x^2 \leq 3 $ is $ x = 1 $.
- Subtract $ 1^2 = 1 $ from 3, remainder = 2.
- Bring down the next pair: 285.
- Root so far: 1.

---

### **Step 2: Next Pair '85'**

- Double the root so far: $ 2 \times 1 = 2 $.
- Find $ d $ such that $ (20 + d) \times d \leq 285 $.
  - Try $ d = 9 $: $ 29 \times 9 = 261 \leq 285 $
- Append 9 to root: $ 1.9 $
- Subtract: $ 285 - 261 = 24 $
- Bring down next pair: 2470

---

### **Step 3: Next Pair '70'**

- Double root (19): $ 38 $
- Find $ d $ such that $ (380 + d) \times d \leq 2470 $
  - Try $ d = 6 $: $ 386 \times 6 = 2316 \leq 2470 $
- Append 6 to root: $ 1.96 $
- Subtract: $ 2470 - 2316 = 154 $
- Bring down pair '04': 15404

---

### **Step 4: Next Pair '04'**

- Double root (196): $ 392 $
- Find $ d $ such that $ (3920 + d) \times d \leq 15404 $
  - Try $ d = 3 $: $ 3923 \times 3 = 11769 \leq 15404 $
- Append 3: $ 1.963 $
- Subtract: $ 15404 - 11769 = 3635 $
- Bring down pair '74': 363574

---

### **Step 5: Next Pair '74'**

- Double root (1963): $ 3926 $
- Find $ d $ such that $ (39260 + d) \times d \leq 363574 $
  - Try $ d = 9 $: $ 39269 \times 9 = 353421 \leq 363574 $
- Append 9: $ 1.9639 $
- Subtract: $ 363574 - 353421 = 10153 $
- Bring down pair '02': 1015302

---

### **Step 6: Next Pair '02'**

- Double root (19639): $ 39278 $
- Find $ d $ such that $ (392780 + d) \times d \leq 1015302 $
  - Try $ d = 2 $: $ 392782 \times 2 = 785564 \leq 1015302 $
- Append 2: $ 1.96392 $
- Subtract: $ 1015302 - 785564 = 229738 $
- Bring down pair '55' (but this is beyond our 6-digit target)

---

### **Final Consideration:**

We have now processed the **first six pairs**, and the root is:

$$
1.96392
$$

Each pair corresponds to one digit in the square root:

- '3' → 1
- '85' → 9
- '70' → 6
- '04' → 3
- '74' → 9
- '02' → 2

Thus, the **first 6 digits of the result** (including the integer part and first five decimal places) are:

$$
\boxed{1.96392}
$$

To calculate \(\sqrt{3.8570047402555048}\) manually, we can use the **Newton-Raphson method** (also known as the **Babylonian method**) for square roots. Here's a step-by-step solution to find the first 6 digits of the result.

### Steps:
1. **Initial Guess**:  
   Since \(1^2 = 1\) and \(2^2 = 4\), and \(3.857...\) is between 1 and 2, we choose \(x_0 = 2\) as our initial guess.  

2. **Newton-Raphson Iteration Formula**:  
   The iterative formula for \(\sqrt{a}\) is:
   \[
   x_{n+1} = \frac{1}{2} \left( x_n + \frac{a}{x_n} \right)
   \]
   Here, \(a = 3.8570047402555048\).  

3. **First Iteration (\(x_1\))**:  
   \[
   x_1 = \frac{1}{2} \left( 2 + \frac{3.8570047402555048}{2} \right) = \frac{1}{2} (2 + 1.9285023701277524) = 1.9642511850638762
   \]  

4. **Second Iteration (\(x_2\))**:  
   \[
   x_2 = \frac{1}{2} \left( 1.9642511850638762 + \frac{3.8570047402555048}{1.9642511850638762} \right)
   \]
   Compute \(\frac{a}{x_1}\):
   \[
   \frac{3.8570047402555048}{1.9642511850638762} \approx 1.9634319552948556
   \]
   Now, compute \(x_2\):
   \[
   x_2 = \frac{1}{2} (1.9642511850638762 + 1.9634319552948556) = 1.9638415701793659
   \]  

5. **Third Iteration (\(x_3\))**:  
   \[
   x_3 = \frac{1}{2} \left( 1.9638415701793659 + \frac{3.8570047402555048}{1.9638415701793659} \right)
   \]
   Compute \(\frac{a}{x_2}\):
   \[
   \frac{3.8570047402555048}{1.9638415701793659} \approx 1.9638415701793657
   \]
   Now, compute \(x_3\):
   \[
   x_3 = \frac{1}{2} (1.9638415701793659 + 1.9638415701793657) = 1.9638415701793658
   \]  

The result has converged to 15 decimal places after just 3 iterations.  

### Final Result:
The first 6 digits of \(\sqrt{3.8570047402555048}\) are:
\[
\boxed{1.96384}
\]

Using Newton’s iteration by hand:

1. Let S = 3.8570047402555048, start with x₀ = 2.  
2. x₁ = ½·(x₀ + S/x₀) = ½·(2 + 3.8570047403/2) ≃ 1.9642511851  
3. x₂ = ½·(x₁ + S/x₁) ≃ ½·(1.9642511851 + 3.8570047403/1.9642511851) ≃ 1.9639256925  
4. x₃ = ½·(x₂ + S/x₂) ≃ 1.9639257721  

So √3.8570047402555048 ≃ 1.9639257721…

The first six significant digits are  
1, 9, 6, 3, 9, 2

i.e. 1.96392…

## Step 1: Understand the problem and the task
We are tasked with manually calculating the square root of 3.8570047402555048 and providing the first 6 digits of the result.

## Step 2: Recall the method for manual square root calculation
One method to manually calculate square roots is the Babylonian method for square root, which is an ancient algorithm that uses successive approximations to find the square root of a number. The formula is: $x_{n+1} = \frac{1}{2}(x_n + \frac{S}{x_n})$, where $S$ is the number for which we are finding the square root, and $x_n$ is the current estimate.

## Step 3: Apply the Babylonian method
Let's start with $S = 3.8570047402555048$. We need an initial guess, $x_0$. A reasonable initial guess could be $x_0 = 2$ because $2^2 = 4$ is close to $S$, and it's easy to calculate.

## Step 4: Perform the first iteration
Using $x_0 = 2$, we calculate $x_1 = \frac{1}{2}(2 + \frac{3.8570047402555048}{2}) = \frac{1}{2}(2 + 1.9285023701277524)$.

## Step 5: Calculate $x_1$
$x_1 = \frac{1}{2}(2 + 1.9285023701277524) = \frac{1}{2}(3.9285023701277524) = 1.9642511850638762$.

## Step 6: Perform the second iteration
$x_2 = \frac{1}{2}(1.9642511850638762 + \frac{3.8570047402555048}{1.9642511850638762})$.

## Step 7: Calculate $x_2$
First, calculate $\frac{3.8570047402555048}{1.9642511850638762} = 1.963773447834$.
Then, $x_2 = \frac{1}{2}(1.9642511850638762 + 1.963773447834) = \frac{1}{2}(3.9280246328978762) = 1.9640123164489381$.

## Step 8: Perform the third iteration
$x_3 = \frac{1}{2}(1.9640123164489381 + \frac{3.8570047402555048}{1.9640123164489381})$.

## Step 9: Calculate $x_3$
First, calculate $\frac{3.8570047402555048}{1.9640123164489381} = 1.964012299469$.
Then, $x_3 = \frac{1}{2}(1.9640123164489381 + 1.964012299469) = \frac{1}{2}(3.9280246159179381) = 1.964012307958969$.

## Step 10: Evaluate the precision of $x_3$
Since $x_2$ and $x_3$ are very close (1.9640123164489381 and 1.964012307958969), we have achieved a high degree of precision. The first 6 digits of $x_3$ are 1.96401.

The final answer is: $\boxed{1.96401}$

I'll calculate √3.8570047402555048 manually using the Newton-Raphson method.

The Newton-Raphson formula for finding √a is:
x_{n+1} = (x_n + a/x_n) / 2

Where a = 3.8570047402555048

First, I need a good initial guess. Since 3.857 is between 1² = 1 and 2² = 4, and closer to 4, I'll start with x₀ = 2.

Iteration 1:
x₁ = (2 + 3.8570047402555048/2) / 2
x₁ = (2 + 1.9285023701277524) / 2
x₁ = 3.9285023701277524 / 2
x₁ = 1.9642511850638762

Iteration 2:
x₂ = (1.9642511850638762 + 3.8570047402555048/1.9642511850638762) / 2
x₂ = (1.9642511850638762 + 1.9633775042993238) / 2
x₂ = 3.9276286893632 / 2
x₂ = 1.9638143446816

Iteration 3:
x₃ = (1.9638143446816 + 3.8570047402555048/1.9638143446816) / 2
x₃ = (1.9638143446816 + 1.9635027151184) / 2
x₃ = 3.9273170598 / 2
x₃ = 1.9636585299

Iteration 4:
x₄ = (1.9636585299 + 3.8570047402555048/1.9636585299) / 2
x₄ = (1.9636585299 + 1.9635806201) / 2
x₄ = 3.927239150 / 2
x₄ = 1.9636195750

The result appears to be converging to approximately 1.963620.

Let me verify: (1.963620)² ≈ 3.857802, which is very close to our target value.

The first 6 digits of √3.8570047402555048 are: **1.96362**