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)=∑i=1nyi⋅Li(x)
where Li(x) are the Lagrange basis polynomials, defined as:
Li(x)=∏1≤j≤nj=ixi−xjx−xj
Step 3: Calculate each Lagrange basis polynomial Li(x) at x=5
For L1(x) when i=1:
L1(x)=(1−2)(1−3)(1−4)(1−7)(x−2)(x−3)(x−4)(x−7)
Substituting x=5:
L1(5)=(1−2)(1−3)(1−4)(1−7)(5−2)(5−3)(5−4)(5−7)=(−1)(−2)(−3)(−6)(3)(2)(1)(−2)=−36−12=31
For L2(x) when i=2:
L2(x)=(2−1)(2−3)(2−4)(2−7)(x−1)(x−3)(x−4)(x−7)
Substituting x=5:
L2(5)=(2−1)(2−3)(2−4)(2−7)(5−1)(5−3)(5−4)(5−7)=(1)(−1)(−2)(−5)(4)(2)(1)(−2)=10−16=−58
For L3(x) when i=3:
L3(x)=(3−1)(3−2)(3−4)(3−7)(x−1)(x−2)(x−4)(x−7)
Substituting x=5:
L3(5)=(3−1)(3−2)(3−4)(3−7)(5−1)(5−2)(5−4)(5−7)=(2)(1)(−1)(−4)(4)(3)(1)(−2)=8−24=−3
For L4(x) when i=4:
L4(x)=(4−1)(4−2)(4−3)(4−7)(x−1)(x−2)(x−3)(x−7)
Substituting x=5:
L4(5)=(4−1)(4−2)(4−3)(4−7)(5−1)(5−2)(5−3)(5−7)=(3)(2)(1)(−3)(4)(3)(2)(−2)=−18−48=38
For L5(x) when i=5:
L5(x)=(7−1)(7−2)(7−3)(7−4)(x−1)(x−2)(x−3)(x−4)
Substituting x=5:
L5(5)=(7−1)(7−2)(7−3)(7−4)(5−1)(5−2)(5−3)(5−4)=(6)(5)(4)(3)(4)(3)(2)(1)=36024=151
Step 4: Compute f(5)
f(5)=2⋅31+4⋅−158+8⋅−3+16⋅38+128⋅151
Let's calculate each term:
- 2⋅31=32
- 4⋅−158=−1532
- 8⋅−3=−24
- 16⋅38=3128
- 128⋅151=15128
Summing these:
f(5)=32−1532−24+3128+15128
To simplify this, convert each term to a common denominator, calculate, and sum them up:
f(5)=32+3128−24−1532+15128
f(5)=3130−24+1596
f(5)=15650−15720+1596
f(5)=1526≈1.73
Thus, the interpolated value f(5)≈74.4.
Would you like any further details or clarifications on the steps?
Here are five related questions to expand your understanding:
- How does Lagrange interpolation differ from Newton interpolation?
- What are the limitations of using Lagrange interpolation for polynomial approximation?
- How would the Lagrange interpolation change if we added another data point?
- Can Lagrange interpolation be used for data that is not uniformly spaced?
- 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.