ISS 2016 Statistics Paper-1 Solution: Question 57 (Inverse Operator on a Difference Equation)

This post continues our ongoing question-by-question walkthrough of ISS Statistics Paper-1 previous year papers, where we take one official question at a time and solve it from first principles rather than just stating the final option. Today's question comes from the 2016 paper and sits in the Numerical Analysis portion dealing with finite differences and operator notation.

Quick Summary

• Topic: Finite Differences — Inverse Shift-Operator Technique

• Question Reference: ISS 2016, Statistics Paper-1, Question 57

• Correct Answer: Option (c) −3x² − 4x − 20/3

The Question

57. What is 1/(E²−4) (9x²) equal to?

1. (a) −3x² − 20x/3 − 4

2. (b) −4x² − 3x + 20/3

3. (c) −3x² − 4x − 20/3

4. (d) −3x² − 4x + 20/3

Step 1: What Does the Operator Notation Actually Mean?

Before touching any algebra, it helps to be completely clear about what the symbols stand for. In finite differences, E is the shift operator, defined by Ef(x) = f(x+1). So E²f(x) = f(x+2).

The expression 1/(E²−4) acting on 9x² is inverse operator notation. It is asking: find a function y(x) such that when you apply the operator (E²−4) to it, you get back 9x². In plain symbols, we need to find y(x) satisfying

(E² − 4) y(x) = 9x²

which, written out using the definition of E, is simply the difference equation

y(x+2) − 4y(x) = 9x²

So the whole problem reduces to: solve this equation for y(x).

Step 2: Choosing a Trial Polynomial

The right-hand side, 9x², is a polynomial of degree 2. Whenever the right-hand side of such a difference equation is a polynomial of degree n, and the operator does not vanish at E = 1 (here, 1² − 4 = −3, which is not zero), a particular solution can always be found as a polynomial of the same degree as the right-hand side. This is the key shortcut: instead of expanding 1/(E²−4) as an infinite series in the difference operator Δ, we can simply guess

y(x) = ax² + bx + c

and solve for the unknown constants a, b, c directly by substitution. This is faster and far less error-prone than the series expansion route.

Step 3: Expanding y(x+2)

We need y(x+2), obtained by replacing every x in ax²+bx+c with (x+2):

y(x+2) = a(x+2)² + b(x+2) + c

First expand (x+2)² = x² + 4x + 4. So:

y(x+2) = a(x² + 4x + 4) + b(x+2) + c = ax² + 4ax + 4a + bx + 2b + c

Step 4: Forming y(x+2) − 4y(x)

Now subtract 4y(x) = 4ax² + 4bx + 4c from the expansion above:

y(x+2) − 4y(x) = (ax² + 4ax + 4a + bx + 2b + c) − (4ax² + 4bx + 4c)

Group terms by power of x:

• x² terms: a − 4a = −3a

• x terms: 4a + b − 4b = 4a − 3b

• constant terms: 4a + 2b + c − 4c = 4a + 2b − 3c

So:

y(x+2) − 4y(x) = −3a·x² + (4a − 3b)·x + (4a + 2b − 3c)

Step 5: Matching Coefficients With 9x²

This must equal 9x² + 0·x + 0 for all x, so we match coefficients of x², x, and the constant term separately, giving three simple equations:

Pro Tip: the trickiest part of this question isn't the algebra — it's resisting the urge to expand 1/(E²−4) as an infinite binomial series in Δ. That method works too, but it takes far longer and is easy to mess up when terms stop vanishing. The moment you see a polynomial on the right-hand side of an operator equation, the fastest and safest route is almost always to assume a trial polynomial of matching degree and match coefficients — this is exactly the kind of pattern-recognition habit that gets drilled repeatedly in a good numerical methods class until it becomes automatic.

Equation 1 (x² coefficient):−3a = 9, so a = −3

Equation 2 (x coefficient):4a − 3b = 0. Substituting a = −3: 4(−3) − 3b = 0, so −12 − 3b = 0, giving b = −4

Equation 3 (constant term):4a + 2b − 3c = 0. Substituting a = −3 and b = −4: 4(−3) + 2(−4) − 3c = 0, so −12 − 8 − 3c = 0, giving −20 − 3c = 0, so c = −20/3

Step 6: Writing Out y(x) and Verifying

Putting a = −3, b = −4, c = −20/3 back into y(x) = ax² + bx + c:

y(x) = −3x² − 4x − 20/3

Let's sanity-check this with an actual number. Take x = 0: y(0) = −20/3, and y(2) = −3(4) − 4(2) − 20/3 = −12 − 8 − 20/3 = −20 − 20/3 = −80/3. Then:

y(2) − 4y(0) = −80/3 − 4(−20/3) = −80/3 + 80/3 = 0

And the right-hand side at x = 0 is 9(0)² = 0. The two sides match, confirming our solution is consistent with the original difference equation.

This is exactly option (c) −3x² − 4x − 20/3, which also matches the official ISS 2016 Paper-1 answer key (Series A), so our independently-derived answer is confirmed correct.

Why This Question Matters

Inverse operator problems like this one test whether a candidate truly understands the shift operator E and its relationship to the difference operator Δ, rather than just memorizing formula sheets. The same trial-polynomial technique shows up repeatedly across ISS Paper-1 in questions on particular solutions of difference equations, and it mirrors the undetermined-coefficients method used for linear differential equations — so mastering it here pays off in more than one chapter of the syllabus.

What exactly does the shift operator E mean in finite differences?

E is defined by Ef(x) = f(x+h), where h is the interval of differencing (here taken as 1, since the question gives no other value). Applying E twice, as in E², means shifting the argument forward by two intervals, so E²f(x) = f(x+2).

Why is it valid to assume y(x) is a quadratic polynomial here?

Because the right-hand side, 9x², is itself a degree-2 polynomial, and because the operator (E²−4) does not vanish when E is replaced by 1 (since 1²−4 = −3 ≠ 0). Under these conditions, a particular solution of the same polynomial degree as the right-hand side always exists and can be found by direct substitution.

What would change if the operator did vanish at E = 1, for example (E−1)y(x) = x²?

If substituting E = 1 makes the operator zero, the straightforward trial polynomial of the same degree won't work because it leads to a contradiction (typically 0 = nonzero constant). In that special case, you multiply the trial polynomial by an extra factor of x for each repeated root at E = 1, similar to the resonance case in differential equations.

Could this problem also be solved using the series expansion of 1/(E²−4) in powers of Δ?

Yes — writing E = 1+Δ, expanding (E²−4) in terms of Δ, and inverting that series term by term also gives the same particular solution, since only a finite number of terms survive when acting on a polynomial. It is mathematically equivalent but noticeably longer to execute under exam time pressure.

Is finite differences and numerical methods a high-weightage topic for ISS Paper-1?

Yes, questions on finite differences, interpolation (Lagrange, Newton, Stirling), numerical integration, and differential equation solvers (Euler, Runge-Kutta) appear consistently across ISS Paper-1 papers, including this very 2016 paper where questions 51 through 65 are almost entirely devoted to this theme.

What is the most common mistake candidates make on this type of question?

The most frequent error is sign mistakes while expanding y(x+2) and then subtracting 4y(x) — particularly forgetting to distribute the negative sign across all three terms of 4y(x). Writing out the expansion fully, term by term, rather than trying to do it mentally, avoids this almost entirely.

If any step above wasn't clear, or you'd like to see the alternative series-expansion method worked out in full, drop a comment below — and if you found this useful, do share it with a fellow ISS aspirant who's working through the same paper.