Question

A fifth-degree polynomial in \(x\) is defined for \(x > 0\). All coefficients of the polynomial are positive. The first derivative of the polynomial is obtained numerically at a point by using the first-order forward as well as the first-order backward difference methods. Identical step lengths are used for both the methods.
Following statements are made.
(I) Forward difference method underestimates the true derivative.
(II) Backward difference method overestimates the true derivative.
Which one of the following options is CORRECT?

• A. Both statements (I) and (II) are FALSE.
• B. Both statements (I) and (II) are TRUE.
• C. Statement (I) is TRUE and statement (II) is FALSE.
• D. Statement (I) is FALSE and statement (II) is TRUE.

Hint:

Since all coefficients are positive and x is positive, the second derivative is positive, so the function is convex. Use this to judge the sign of the forward and backward difference errors.

Answer 1

The Correct Option is A

Step 1: Find the sign of the second derivative for x greater than 0.
Let the polynomial be \(f(x)=a_5x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+a_0\) with every coefficient \(a_0,\dots,a_5>0\). Differentiating twice:
\[ f''(x)=20a_5x^3+12a_4x^2+6a_3x+2a_2 \]
For \(x>0\), every term here is a positive coefficient times a positive power of \(x\), so every term is positive, meaning \(f''(x)>0\) for all \(x>0\). The function is convex (curves upward) on this domain.

Step 2: Write the Taylor expansion for the forward difference.
The forward difference at a point \(x\) with step \(h\) is \(\dfrac{f(x+h)-f(x)}{h}\). Expand \(f(x+h)\) using Taylor's series:
\[ f(x+h)=f(x)+hf'(x)+\frac{h^2}{2}f''(x)+O(h^3) \]
So:
\[ \frac{f(x+h)-f(x)}{h}=f'(x)+\frac{h}{2}f''(x)+O(h^2) \]
Since \(f''(x)>0\), the extra term \(\frac{h}{2}f''(x)\) is positive, so the forward difference is larger than the true derivative, it OVERestimates it, not underestimates. Statement (I) is FALSE.

Step 3: Write the Taylor expansion for the backward difference.
The backward difference is \(\dfrac{f(x)-f(x-h)}{h}\). Expand \(f(x-h)\):
\[ f(x-h)=f(x)-hf'(x)+\frac{h^2}{2}f''(x)-O(h^3) \]
So:
\[ \frac{f(x)-f(x-h)}{h}=f'(x)-\frac{h}{2}f''(x)+O(h^2) \]
Since \(f''(x)>0\), the term \(-\frac{h}{2}f''(x)\) is negative, so the backward difference is smaller than the true derivative, it UNDERestimates it, not overestimates. Statement (II) is also FALSE.

Step 4: Combine the two results.
Both statements claim the opposite of what actually happens for a convex function: forward difference overestimates, and backward difference underestimates. So both statements are FALSE.

Final Answer:
Both statement (I) and statement (II) are FALSE, option (A). \[ \boxed{\text{Both (I) and (II) are FALSE}} \]