Compute cosh(4.25) using appropriate interpolation formula to find the value of y at x=

4.25

4.0

4.1

4.2

4.3

4.4

4.5

Cosh x

27.3081

30.1784

33.3507

36.8567

40.7316

45.0141

To find \(\cosh(4.25)\) using interpolation, we'll apply the **Lagrange interpolation** method. This approach is suitable because it can estimate the value of a function based on given discrete data points.

### Given Data Points:

| \( x \)  | \( \cosh(x) \) |
|----------|----------------|
| 4.0      | 27.3081        |
| 4.1      | 30.1784        |
| 4.2      | 33.3507        |
| 4.3      | 36.8567        |
| 4.4      | 40.7316        |
| 4.5      | 45.0141        |

We want to find the value at \( x = 4.25 \).

### Lagrange Interpolation Formula:

For the data points \((x_0, y_0), (x_1, y_1), \ldots, (x_n, y_n)\), the interpolated value at \(x\) is given by:

\[
P(x) = \sum_{i=0}^{n} y_i \cdot \prod_{\substack{0 \leq j \leq n \\ j \neq i}} \frac{x - x_j}{x_i - x_j}
\]

We'll use the data points closest to \(x = 4.25\), which are:

1. \( (4.0, 27.3081) \)
2. \( (4.1, 30.1784) \)
3. \( (4.2, 33.3507) \)
4. \( (4.3, 36.8567) \)
5. \( (4.4, 40.7316) \)

This will be a 4th-degree polynomial (since we use 5 points).

### Calculation:

Using the Lagrange interpolation formula, the polynomial can be expressed as:

\[
P(4.25) = y_0 \cdot L_0(4.25) + y_1 \cdot L_1(4.25) + y_2 \cdot L_2(4.25) + y_3 \cdot L_3(4.25) + y_4 \cdot L_4(4.25)
\]

Where \( L_i(x) \) are the Lagrange basis polynomials defined as:

\[
L_i(x) = \prod_{\substack{0 \leq j \leq 4 \\ j \neq i}} \frac{x - x_j}{x_i - x_j}
\]

Let's calculate \( L_i(4.25) \) for \( i = 0, 1, 2, 3, 4 \):

1. For \( i = 0 \) (x = 4.0, y = 27.3081):
   \[
   L_0(4.25) = \frac{(4.25-4.1)(4.25-4.2)(4.25-4.3)(4.25-4.4)}{(4.0-4.1)(4.0-4.2)(4.0-4.3)(4.0-4.4)}
   \]

2. For \( i = 1 \) (x = 4.1, y = 30.1784):
   \[
   L_1(4.25) = \frac{(4.25-4.0)(4.25-4.2)(4.25-4.3)(4.25-4.4)}{(4.1-4.0)(4.1-4.2)(4.1-4.3)(4.1-4.4)}
   \]

3. For \( i = 2 \) (x = 4.2, y = 33.3507):
   \[
   L_2(4.25) = \frac{(4.25-4.0)(4.25-4.1)(4.25-4.3)(4.25-4.4)}{(4.2-4.0)(4.2-4.1)(4.2-4.3)(4.2-4.4)}
   \]

4. For \( i = 3 \) (x = 4.3, y = 36.8567):
   \[
   L_3(4.25) = \frac{(4.25-4.0)(4.25-4.1)(4.25-4.2)(4.25-4.4)}{(4.3-4.0)(4.3-4.1)(4.3-4.2)(4.3-4.4)}
   \]

5. For \( i = 4 \) (x = 4.4, y = 40.7316):
   \[
   L_4(4.25) = \frac{(4.25-4.0)(4.25-4.1)(4.25-4.2)(4.25-4.3)}{(4.4-4.0)(4.4-4.1)(4.4-4.2)(4.4-4.3)}
   \]

### Intermediate Calculations:

Let's compute these values using the provided data.

#### \( L_0(4.25) \):

\[
L_0(4.25) = \frac{(4.25-4.1)(4.25-4.2)(4.25-4.3)(4.25-4.4)}{(4.0-4.1)(4.0-4.2)(4.0-4.3)(4.0-4.4)}
\]
\[
= \frac{0.15 \cdot 0.05 \cdot -0.05 \cdot -0.15}{-0.1 \cdot -0.2 \cdot -0.3 \cdot -0.4}
\]
\[
= \frac{0.00005625}{0.0024}
\]
\[
= 0.0234375
\]

#### \( L_1(4.25) \):

\[
L_1(4.25) = \frac{(4.25-4.0)(4.25-4.2)(4.25-4.3)(4.25-4.4)}{(4.1-4.0)(4.1-4.2)(4.1-4.3)(4.1-4.4)}
\]
\[
= \frac{0.25 \cdot 0.05 \cdot -0.05 \cdot -0.15}{0.1 \cdot -0.1 \cdot -0.2 \cdot -0.3}
\]
\[
= \frac{0.00046875}{0.0006}
\]
\[
= 0.78125
\]

#### \( L_2(4.25) \):

\[
L_2(4.25) = \frac{(4.25-4.0)(4.25-4.1)(4.25-4.3)(4.25-4.4)}{(4.2-4.0)(4.2-4.1)(4.2-4.3)(4.2-4.4)}
\]
\[
= \frac{0.25 \cdot 0.15 \cdot -0.05 \cdot -0.15}{0.2 \cdot 0.1 \cdot -0.1 \cdot -0.2}
\]
\[
= \frac{0.00028125}{0.0004}
\]
\[
= 0.703125
\]

#### \( L_3(4.25) \):

\[
L_3(4.25) = \frac{(4.25-4.0)(4.25-4.1)(4.25-4.2)(4.25-4.4)}{(4.3-4.0)(4.3-4.1)(4.3-4.2)(4.3-4.4)}
\]
\[
= \frac{0.25 \cdot 0.15 \cdot 0.05 \cdot -0.15}{0.3 \cdot 0.2 \cdot 0.1 \cdot -0.1}
\]
\[
= \frac{-0.00028125}{-0.0006}
\]
\[
= 0.46875
\]

#### \( L_4(4.25) \):

\[
L_4(4.25) = \frac{(4.25-4.0)(4.25-4.1)(4.25-4.2)(4.25-4.3)}{(4.4-4.0)(4.4-4.1)(4.4-4.2)(4.4-4.3)}
\]
\[
= \frac{0.25 \cdot 0.15 \cdot 0.05 \cdot 0.05}{0.4 \cdot 0.3 \cdot 

Learn how to compute cosh(4.25) using Lagrange interpolation with a detailed step-by-step solution. This problem involves applying the Lagrange interpolation method to estimate the value of a function between discrete data points. Suitable for undergraduate level mathematics.

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