Math Problem Statement

Consider the task of approximating sin(x) over the interval [0, 1]. If one uses a polynomial interpolant based upon n + 1 equispaced data points on [0, 1], xi = ih, h = 1/n, i = 0, 1, . . . , n. How many points are required so that the error bound for the interpolant is less than 1 × 10−6?

Solution

To approximate sin(x)\sin(x) over the interval [0,1][0, 1] using polynomial interpolation with n+1n + 1 equispaced points, we can analyze the error associated with polynomial interpolation using the formula for the error bound:

En(x)=M(n+1)!xx0xx1xxnE_n(x) = \frac{M}{(n+1)!} |x - x_0||x - x_1| \cdots |x - x_n|

where MM is the maximum value of the (n+1)(n+1)-th derivative of sin(x)\sin(x) over the interval [0,1][0, 1], and x0,x1,,xnx_0, x_1, \ldots, x_n are the interpolation nodes.

Step 1: Determine MM

The nn-th derivative of sin(x)\sin(x) oscillates between -1 and 1, so the maximum MM is:

M=1M = 1

Step 2: Error Bound Calculation

The maximum value of xxi|x - x_i| for x[0,1]x \in [0, 1] is achieved at the endpoints. For n+1n + 1 points, the largest distance between any two nodes is at most h=1nh = \frac{1}{n}. Therefore, we can express the error as:

En(x)1(n+1)!1(n+1)n+1(1)=1(n+1)!(1n)n|E_n(x)| \leq \frac{1}{(n+1)!} \cdot \frac{1}{(n+1)^{n+1}} \cdot (1) = \frac{1}{(n+1)!} \cdot \left(\frac{1}{n}\right)^{n}

Step 3: Set Up the Inequality

We want this error bound to be less than 1×1061 \times 10^{-6}:

1(n+1)!(1n)n<1×106\frac{1}{(n+1)!} \cdot \left(\frac{1}{n}\right)^{n} < 1 \times 10^{-6}

Step 4: Simplifying and Estimating nn

To find the smallest nn satisfying this inequality, we can calculate (n+1)!(n+1)! for various nn:

  1. For small values of nn, compute 1(n+1)!\frac{1}{(n+1)!} and (1n)n\left(\frac{1}{n}\right)^{n} until the product is less than 1×1061 \times 10^{-6}.

Let’s compute this for increasing values of nn:

  • n=5n = 5: 16!=1720,(15)5=13125\frac{1}{6!} = \frac{1}{720}, \quad \left(\frac{1}{5}\right)^{5} = \frac{1}{3125} 1720131254.55×107 (valid)\frac{1}{720} \cdot \frac{1}{3125} \approx 4.55 \times 10^{-7} \text{ (valid)}

  • n=4n = 4: 15!=1120,(14)4=1256\frac{1}{5!} = \frac{1}{120}, \quad \left(\frac{1}{4}\right)^{4} = \frac{1}{256} 112012563.91×106 (not valid)\frac{1}{120} \cdot \frac{1}{256} \approx 3.91 \times 10^{-6} \text{ (not valid)}

Conclusion

Thus, to achieve an error less than 1×1061 \times 10^{-6}, at least n=5n = 5 points (which corresponds to n+1=6n + 1 = 6 equispaced points) are required.

Follow-Up Questions

  1. How does the choice of equispaced points affect the accuracy of polynomial interpolation?
  2. What are the advantages and disadvantages of using polynomial interpolation compared to other methods like spline interpolation?
  3. Can you explain the concept of the Runge phenomenon in polynomial interpolation?
  4. How would the error bound change if we used Chebyshev nodes instead of equispaced nodes?
  5. What are some applications of polynomial interpolation in numerical analysis?

Tip: Always consider the characteristics of the function you're approximating when choosing the interpolation method, as this can significantly impact the error.

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Math Problem Analysis

Mathematical Concepts

Polynomial Interpolation
Error Analysis
Numerical Methods

Formulas

E_n(x) = \frac{M}{(n+1)!} |x - x_0||x - x_1| ... |x - x_n|
M = 1

Theorems

Error Bound for Polynomial Interpolation

Suitable Grade Level

Grades 11-12