Math Problem Statement

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

Solution

To find f(5)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)(1, 2), \, (2, 4), \, (3, 8), \, (4, 16), \, (7, 128)

Step 2: Write the Lagrange Interpolation Formula

The Lagrange interpolation polynomial f(x)f(x) is given by: f(x)=i=1nyiLi(x)f(x) = \sum_{i=1}^{n} y_i \cdot L_i(x) where Li(x)L_i(x) are the Lagrange basis polynomials, defined as: Li(x)=1jnjixxjxixjL_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 Li(x)L_i(x) at x=5x = 5

For L1(x)L_1(x) when i=1i = 1:

L1(x)=(x2)(x3)(x4)(x7)(12)(13)(14)(17)L_1(x) = \frac{(x - 2)(x - 3)(x - 4)(x - 7)}{(1 - 2)(1 - 3)(1 - 4)(1 - 7)} Substituting x=5x = 5: L1(5)=(52)(53)(54)(57)(12)(13)(14)(17)=(3)(2)(1)(2)(1)(2)(3)(6)=1236=13L_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 L2(x)L_2(x) when i=2i = 2:

L2(x)=(x1)(x3)(x4)(x7)(21)(23)(24)(27)L_2(x) = \frac{(x - 1)(x - 3)(x - 4)(x - 7)}{(2 - 1)(2 - 3)(2 - 4)(2 - 7)} Substituting x=5x = 5: L2(5)=(51)(53)(54)(57)(21)(23)(24)(27)=(4)(2)(1)(2)(1)(1)(2)(5)=1610=85L_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 L3(x)L_3(x) when i=3i = 3:

L3(x)=(x1)(x2)(x4)(x7)(31)(32)(34)(37)L_3(x) = \frac{(x - 1)(x - 2)(x - 4)(x - 7)}{(3 - 1)(3 - 2)(3 - 4)(3 - 7)} Substituting x=5x = 5: L3(5)=(51)(52)(54)(57)(31)(32)(34)(37)=(4)(3)(1)(2)(2)(1)(1)(4)=248=3L_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 L4(x)L_4(x) when i=4i = 4:

L4(x)=(x1)(x2)(x3)(x7)(41)(42)(43)(47)L_4(x) = \frac{(x - 1)(x - 2)(x - 3)(x - 7)}{(4 - 1)(4 - 2)(4 - 3)(4 - 7)} Substituting x=5x = 5: L4(5)=(51)(52)(53)(57)(41)(42)(43)(47)=(4)(3)(2)(2)(3)(2)(1)(3)=4818=83L_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 L5(x)L_5(x) when i=5i = 5:

L5(x)=(x1)(x2)(x3)(x4)(71)(72)(73)(74)L_5(x) = \frac{(x - 1)(x - 2)(x - 3)(x - 4)}{(7 - 1)(7 - 2)(7 - 3)(7 - 4)} Substituting x=5x = 5: L5(5)=(51)(52)(53)(54)(71)(72)(73)(74)=(4)(3)(2)(1)(6)(5)(4)(3)=24360=115L_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)

f(5)=213+4815+83+1683+128115f(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. 213=232 \cdot \frac{1}{3} = \frac{2}{3}
  2. 4815=32154 \cdot -\frac{8}{15} = -\frac{32}{15}
  3. 83=248 \cdot -3 = -24
  4. 1683=128316 \cdot \frac{8}{3} = \frac{128}{3}
  5. 128115=12815128 \cdot \frac{1}{15} = \frac{128}{15}

Summing these: f(5)=23321524+1283+12815f(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)=23+1283243215+12815f(5) = \frac{2}{3} + \frac{128}{3} - 24 - \frac{32}{15} + \frac{128}{15}

f(5)=130324+9615f(5) = \frac{130}{3} - 24 + \frac{96}{15} f(5)=6501572015+9615f(5) = \frac{650}{15} - \frac{720}{15} + \frac{96}{15} f(5)=26151.73f(5) = \frac{26}{15} \approx 1.73

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

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Math Problem Analysis

Mathematical Concepts

Interpolation
Lagrange Interpolation

Formulas

Lagrange Interpolation Formula

Theorems

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Suitable Grade Level

Advanced High School