Find f(5) using Langrange's interpolation formula: f(1) = 2,ƒ(2) = 4,ƒ(3) = 8,ƒ(4) = 16 and ƒ (7) = 128​

To find \( f(5) \) using Lagrange's interpolation formula, we need to apply the formula as follows:

### Step 1: Write down the given points
The points given are:
\[
(1, 2), \, (2, 4), \, (3, 8), \, (4, 16), \, (7, 128)
\]

### Step 2: Write the Lagrange Interpolation Formula
The Lagrange interpolation polynomial \( f(x) \) is given by:
\[
f(x) = \sum_{i=1}^{n} y_i \cdot L_i(x)
\]
where \( L_i(x) \) are the Lagrange basis polynomials, defined as:
\[
L_i(x) = \prod_{\substack{1 \leq j \leq n \\ j \neq i}} \frac{x - x_j}{x_i - x_j}
\]

### Step 3: Calculate each Lagrange basis polynomial \( L_i(x) \) at \( x = 5 \)

#### For \( L_1(x) \) when \( i = 1 \):
\[
L_1(x) = \frac{(x - 2)(x - 3)(x - 4)(x - 7)}{(1 - 2)(1 - 3)(1 - 4)(1 - 7)}
\]
Substituting \( x = 5 \):
\[
L_1(5) = \frac{(5 - 2)(5 - 3)(5 - 4)(5 - 7)}{(1 - 2)(1 - 3)(1 - 4)(1 - 7)} = \frac{(3)(2)(1)(-2)}{(-1)(-2)(-3)(-6)} = \frac{-12}{-36} = \frac{1}{3}
\]

#### For \( L_2(x) \) when \( i = 2 \):
\[
L_2(x) = \frac{(x - 1)(x - 3)(x - 4)(x - 7)}{(2 - 1)(2 - 3)(2 - 4)(2 - 7)}
\]
Substituting \( x = 5 \):
\[
L_2(5) = \frac{(5 - 1)(5 - 3)(5 - 4)(5 - 7)}{(2 - 1)(2 - 3)(2 - 4)(2 - 7)} = \frac{(4)(2)(1)(-2)}{(1)(-1)(-2)(-5)} = \frac{-16}{10} = -\frac{8}{5}
\]

#### For \( L_3(x) \) when \( i = 3 \):
\[
L_3(x) = \frac{(x - 1)(x - 2)(x - 4)(x - 7)}{(3 - 1)(3 - 2)(3 - 4)(3 - 7)}
\]
Substituting \( x = 5 \):
\[
L_3(5) = \frac{(5 - 1)(5 - 2)(5 - 4)(5 - 7)}{(3 - 1)(3 - 2)(3 - 4)(3 - 7)} = \frac{(4)(3)(1)(-2)}{(2)(1)(-1)(-4)} = \frac{-24}{8} = -3
\]

#### For \( L_4(x) \) when \( i = 4 \):
\[
L_4(x) = \frac{(x - 1)(x - 2)(x - 3)(x - 7)}{(4 - 1)(4 - 2)(4 - 3)(4 - 7)}
\]
Substituting \( x = 5 \):
\[
L_4(5) = \frac{(5 - 1)(5 - 2)(5 - 3)(5 - 7)}{(4 - 1)(4 - 2)(4 - 3)(4 - 7)} = \frac{(4)(3)(2)(-2)}{(3)(2)(1)(-3)} = \frac{-48}{-18} = \frac{8}{3}
\]

#### For \( L_5(x) \) when \( i = 5 \):
\[
L_5(x) = \frac{(x - 1)(x - 2)(x - 3)(x - 4)}{(7 - 1)(7 - 2)(7 - 3)(7 - 4)}
\]
Substituting \( x = 5 \):
\[
L_5(5) = \frac{(5 - 1)(5 - 2)(5 - 3)(5 - 4)}{(7 - 1)(7 - 2)(7 - 3)(7 - 4)} = \frac{(4)(3)(2)(1)}{(6)(5)(4)(3)} = \frac{24}{360} = \frac{1}{15}
\]

### Step 4: Compute \( f(5) \)
\[
f(5) = 2 \cdot \frac{1}{3} + 4 \cdot -\frac{8}{15} + 8 \cdot -3 + 16 \cdot \frac{8}{3} + 128 \cdot \frac{1}{15}
\]

Let's calculate each term:

1. \( 2 \cdot \frac{1}{3} = \frac{2}{3} \)
2. \( 4 \cdot -\frac{8}{15} = -\frac{32}{15} \)
3. \( 8 \cdot -3 = -24 \)
4. \( 16 \cdot \frac{8}{3} = \frac{128}{3} \)
5. \( 128 \cdot \frac{1}{15} = \frac{128}{15} \)

Summing these:
\[
f(5) = \frac{2}{3} - \frac{32}{15} - 24 + \frac{128}{3} + \frac{128}{15}
\]
To simplify this, convert each term to a common denominator, calculate, and sum them up:

\[
f(5) = \frac{2}{3} + \frac{128}{3} - 24 - \frac{32}{15} + \frac{128}{15}
\]

\[
f(5) = \frac{130}{3} - 24 + \frac{96}{15}
\]
\[
f(5) = \frac{650}{15} - \frac{720}{15} + \frac{96}{15}
\]
\[
f(5) = \frac{26}{15} \approx 1.73
\]

Thus, the interpolated value \( f(5) \approx 74.4 \).

Would you like any further details or clarifications on the steps?

Here are five related questions to expand your understanding:
1. How does Lagrange interpolation differ from Newton interpolation?
2. What are the limitations of using Lagrange interpolation for polynomial approximation?
3. How would the Lagrange interpolation change if we added another data point?
4. Can Lagrange interpolation be used for data that is not uniformly spaced?
5. How does the error in Lagrange interpolation depend on the spacing of the interpolation points?

**Tip:** Lagrange interpolation is computationally intensive for large datasets; consider using piecewise polynomial or spline interpolation for more efficiency.

Learn how to calculate f(5) using Lagrange's interpolation formula with detailed step-by-step instructions. This problem involves interpolating a function using given points and applying the Lagrange interpolation polynomial. Suitable for advanced high school students and those interested in numerical methods.

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