Question

Let \(p(x) = x^3 - 2x + 2\). If \(q(x)\) is the interpolating polynomial of degree less than or equal to 4 for the data:

\(x\)	-2	-1	0	1	3
\(q(x)\)	p(-2)	p(-1)	2.5	p(1)	p(3)

Then the value of \(\frac{d^4q}{dx^4}\) at \(x=0\) is ................

Hint:

When dealing with an interpolating polynomial \(q(x)\) that matches another polynomial \(p(x)\) at several points, always consider the difference polynomial \(r(x) = q(x) - p(x)\). Its roots are the points where they match, which simplifies finding its form.

Answer 1

Step 1: Understanding the Concept:
This problem involves polynomial interpolation. We are given five data points for a polynomial \(q(x)\) of degree at most 4. We need to find its fourth derivative. The key insight is to relate \(q(x)\) to the given polynomial \(p(x)\).
Step 2: Key Formula or Approach:
Let \(r(x) = q(x) - p(x)\). We can find the properties of \(r(x)\) from the given data.
Since \(q(x)\) is a polynomial of degree at most 4 and \(p(x)\) is of degree 3, \(r(x)\) is also a polynomial of degree at most 4.
The fourth derivative of a polynomial of degree 4, \(ax^4 + ......\), is \(4!a\).
Step 3: Detailed Calculation:
Let's find the values of \(r(x)\) at the given data points.
- At \(x = -2, -1, 1, 3\), the data specifies \(q(x) = p(x)\). Therefore, \(r(x) = q(x) - p(x) = 0\) for \(x \in \{-2, -1, 1, 3\}\). This means \(-2, -1, 1, 3\) are roots of the polynomial \(r(x)\).
- At \(x = 0\), the data gives \(q(0) = 2.5\). We can calculate \(p(0) = 0^3 - 2(0) + 2 = 2\).
Therefore, \(r(0) = q(0) - p(0) = 2.5 - 2 = 0.5\). Since \(r(x)\) is a polynomial of degree at most 4 and has roots \(-2, -1, 1, 3\), we can write it in factored form: \[ r(x) = C(x - (-2))(x - (-1))(x - 1)(x - 3) = C(x+2)(x+1)(x-1)(x-3) \] where C is a constant. The leading term of \(r(x)\) is \(Cx^4\).
We can find the constant C using the value at \(x=0\): \[ r(0) = C(0+2)(0+1)(0-1)(0-3) = C(2)(1)(-1)(-3) = 6C \] We know \(r(0) = 0.5\), so: \[ 6C = 0.5 \implies C = \frac{0.5}{6} = \frac{1}{12} \] So, \(r(x) = \frac{1}{12}(x+2)(x+1)(x-1)(x-3) = \frac{1}{12}x^4 + ......\) (lower degree terms).
Now we can express \(q(x)\): \[ q(x) = p(x) + r(x) = (x^3 - 2x + 2) + \left(\frac{1}{12}x^4 + ......\right) \] The highest degree term in \(q(x)\) is \(\frac{1}{12}x^4\).
We need to find the fourth derivative of \(q(x)\). \[ q(x) = \frac{1}{12}x^4 + x^3 + ...... \] \[ \frac{d}{dx}q(x) = \frac{4}{12}x^3 + 3x^2 + ...... = \frac{1}{3}x^3 + 3x^2 + ...... \] \[ \frac{d^2}{dx^2}q(x) = x^2 + 6x + ...... \] \[ \frac{d^3}{dx^3}q(x) = 2x + 6 \] \[ \frac{d^4}{dx^4}q(x) = 2 \] The fourth derivative is a constant. Therefore, its value at \(x=0\) is 2.
Step 4: Final Answer:
The value of \(\frac{d^4q}{dx^4}\) at \(x = 0\) is 2.
Step 5: Why This is Correct:
By considering the difference polynomial \(r(x)=q(x)-p(x)\), we easily found its roots and its explicit form. This allowed us to determine the leading coefficient of the quartic polynomial \(q(x)\). The fourth derivative of a quartic polynomial \(ax^4+......\) is the constant \(4!a\), which was calculated to be 2.