find cubic polynomial using lagrange's interpolation x= 0 1 2 3 f(x)= 1 0 1 10

To find a cubic polynomial using Lagrange's interpolation, we need to construct a polynomial that passes through all the given points. The given points are:

\[
(x_0, f(x_0)) = (0, 1), \quad (x_1, f(x_1)) = (1, 0), \quad (x_2, f(x_2)) = (2, 1), \quad (x_3, f(x_3)) = (3, 10)
\]

The Lagrange interpolation polynomial \( P(x) \) of degree \( n \) that passes through \( n + 1 \) points is given by:

\[
P(x) = \sum_{i=0}^{n} f(x_i) \cdot L_i(x)
\]

where each \( L_i(x) \) is the Lagrange basis polynomial defined as:

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

### Step-by-Step Calculation

Given that we have 4 points, our polynomial \( P(x) \) will be a cubic polynomial (degree 3).

Let's compute each \( L_i(x) \) for \( i = 0, 1, 2, 3 \):

1. **Compute \( L_0(x) \):**

\[
L_0(x) = \frac{(x - x_1)(x - x_2)(x - x_3)}{(x_0 - x_1)(x_0 - x_2)(x_0 - x_3)} = \frac{(x - 1)(x - 2)(x - 3)}{(0 - 1)(0 - 2)(0 - 3)}
\]

\[
L_0(x) = \frac{(x - 1)(x - 2)(x - 3)}{-1 \cdot -2 \cdot -3} = \frac{(x - 1)(x - 2)(x - 3)}{-6}
\]

2. **Compute \( L_1(x) \):**

\[
L_1(x) = \frac{(x - x_0)(x - x_2)(x - x_3)}{(x_1 - x_0)(x_1 - x_2)(x_1 - x_3)} = \frac{(x - 0)(x - 2)(x - 3)}{(1 - 0)(1 - 2)(1 - 3)}
\]

\[
L_1(x) = \frac{(x)(x - 2)(x - 3)}{1 \cdot -1 \cdot -2} = \frac{(x)(x - 2)(x - 3)}{2}
\]

3. **Compute \( L_2(x) \):**

\[
L_2(x) = \frac{(x - x_0)(x - x_1)(x - x_3)}{(x_2 - x_0)(x_2 - x_1)(x_2 - x_3)} = \frac{(x - 0)(x - 1)(x - 3)}{(2 - 0)(2 - 1)(2 - 3)}
\]

\[
L_2(x) = \frac{(x)(x - 1)(x - 3)}{2 \cdot 1 \cdot -1} = \frac{(x)(x - 1)(x - 3)}{-2}
\]

4. **Compute \( L_3(x) \):**

\[
L_3(x) = \frac{(x - x_0)(x - x_1)(x - x_2)}{(x_3 - x_0)(x_3 - x_1)(x_3 - x_2)} = \frac{(x - 0)(x - 1)(x - 2)}{(3 - 0)(3 - 1)(3 - 2)}
\]

\[
L_3(x) = \frac{(x)(x - 1)(x - 2)}{3 \cdot 2 \cdot 1} = \frac{(x)(x - 1)(x - 2)}{6}
\]

Now, construct the polynomial \( P(x) \):

\[
P(x) = f(x_0)L_0(x) + f(x_1)L_1(x) + f(x_2)L_2(x) + f(x_3)L_3(x)
\]

Substitute \( f(x_0) = 1 \), \( f(x_1) = 0 \), \( f(x_2) = 1 \), and \( f(x_3) = 10 \):

\[
P(x) = 1 \cdot L_0(x) + 0 \cdot L_1(x) + 1 \cdot L_2(x) + 10 \cdot L_3(x)
\]

\[
P(x) = L_0(x) + L_2(x) + 10L_3(x)
\]

Now, we substitute the expressions for \(L_0(x)\), \(L_2(x)\), and \(L_3(x)\):

\[
P(x) = \frac{(x - 1)(x - 2)(x - 3)}{-6} + \frac{(x)(x - 1)(x - 3)}{-2} + 10 \cdot \frac{(x)(x - 1)(x - 2)}{6}
\]

Let's simplify this polynomial.

### Simplification of the Polynomial

Let's simplify \( P(x) \):

\[
P(x) = -\frac{1}{6}(x - 1)(x - 2)(x - 3) - \frac{1}{2}(x)(x - 1)(x - 3) + \frac{10}{6}(x)(x - 1)(x - 2)
\]

Calculating the expanded terms, we can find the coefficients of \(P(x)\).

\[
P(x) = x^3 - 3x^2 + 2x + 1
\]

### Final Cubic Polynomial

The cubic polynomial that interpolates the given points is:

\[
\boxed{P(x) = x^3 - 3x^2 + 2x + 1}
\]

Would you like more details or have any questions?

### Related Questions

1. How does Lagrange interpolation differ from Newton's interpolation method?
2. Can Lagrange interpolation be used for extrapolation, and what are its limitations?
3. What is the significance of the Lagrange basis polynomials in the interpolation process?
4. How would the interpolation change if we had an additional data point?
5. What are the computational complexities of Lagrange interpolation for large datasets?

**Tip:** When performing polynomial interpolation, always consider the number of points and ensure that the polynomial degree matches the number of data points for accurate results.

Learn how to find a cubic polynomial using Lagrange's interpolation method with a step-by-step solution. This problem covers key mathematical concepts including interpolation and Lagrange's formula. Suitable for undergraduate level students.

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