ISS 2016 Statistics Paper-1 Solution: Question 65 (Difference Operator on (a+bx)2^x)

Continuing our question-by-question walk through the ISS Statistics Paper-1 archive, we now land on a finite-differences problem from the 2016 paper. If you've been following this series, you'll notice the operator algebra keeps coming back in slightly new disguises — this one is a neat example of why it pays to know the rules cold rather than memorize individual results.

Quick Summary

• Topic:Finite Differences — operator algebra with the shift operator E and the difference operator Δ

• Reference:ISS 2016, Statistics Paper-1, Question 65

• Correct Answer:(d) 0

The Question

If y = (a + bx)2x with difference interval h = 1, then what is (Δ² − 2Δ + 1) y equal to?

• (a) 4

• (b) 2

• (c) 1

• (d) 0

Step 1: Recognise the Operator Expression

Before touching any algebra, look closely at the expression we're asked to evaluate: Δ² − 2Δ + 1. If you replace the symbol Δ with a plain variable, say t, you get t² − 2t + 1, which is nothing but (t − 1)². So the expression we're being asked to apply is actually a perfect square in disguise:

Δ² − 2Δ + 1 = (Δ − 1)²

This single observation turns a two-step differencing exercise into a one-step operator substitution, which is the whole point of learning operator notation in the first place.

Step 2: Connect Δ to the Shift Operator E

In finite difference calculus, the shift operator E is defined by E y(x) = y(x + h). Since h = 1 here, E y(x) = y(x + 1). The forward difference operator Δ is related to E by the basic identity:

Δ = E − 1

Therefore:

Δ − 1 = (E − 1) − 1 = E − 2

which means our target expression becomes:

(Δ − 1)² = (E − 2)²

Expanding this operator binomial exactly as you would expand any (A − B)² term:

(E − 2)² = E² − 4E + 4

And remember, E² y(x) means apply the shift twice, i.e. E² y(x) = y(x + 2). So the operator we actually need to apply to y(x) is:

(Δ² − 2Δ + 1) y(x) = y(x + 2) − 4y(x + 1) + 4y(x)

Pro Tip: The moment you see an expression like Δ² − 2Δ + 1 (or any polynomial in Δ), check first whether it factors nicely. Treating Δ as an ordinary algebraic symbol — then converting the factored form to E, and only then substituting the function — is far less error-prone than expanding differences term by term from scratch. This is exactly the habit that's drilled into students until it becomes automatic: see the operator as algebra first, arithmetic second.

Step 3: Write Out y(x), y(x + 1) and y(x + 2)

We're given y(x) = (a + bx)2x. Since h = 1, shifting x by 1 or 2 just means substituting (x + 1) or (x + 2) in place of x, everywhere x appears — including inside the exponent.

y(x):

y(x) = (a + bx)2x

y(x + 1):replace x with (x + 1):

y(x + 1) = (a + b(x + 1))2x + 1 = (a + bx + b)2x + 1

Since 2x + 1 = 2·2x, we can pull the 2x out:

y(x + 1) = 2(a + bx + b)2x

y(x + 2):replace x with (x + 2):

y(x + 2) = (a + b(x + 2))2x + 2 = (a + bx + 2b)2x + 2

Since 2x + 2 = 4·2x:

y(x + 2) = 4(a + bx + 2b)2x

Step 4: Substitute and Simplify

Now plug these three pieces into y(x + 2) − 4y(x + 1) + 4y(x):

= 4(a + bx + 2b)2x − 4·[2(a + bx + b)2x] + 4(a + bx)2x

= 4(a + bx + 2b)2x − 8(a + bx + b)2x + 4(a + bx)2x

Every term carries a common factor of 2x, so pull it out and focus only on the bracket:

= 2x [ 4(a + bx + 2b) − 8(a + bx + b) + 4(a + bx) ]

Expand each piece inside the bracket, term by term:

4(a + bx + 2b) = 4a + 4bx + 8b

−8(a + bx + b) = −8a − 8bx − 8b

4(a + bx) = 4a + 4bx

Now add these three lines, collecting the a terms, the bx terms, and the plain b terms separately:

a terms: 4a − 8a + 4a = 0

bx terms: 4bx − 8bx + 4bx = 0

b terms: 8b − 8b = 0

So the bracket collapses entirely to 0, and the whole expression is:

(Δ² − 2Δ + 1) y = 2x × 0 =0

Step 5: Cross-Check by Computing Δy and Δ²y Directly

It's always worth confirming an operator-shortcut answer with a slower, direct calculation, especially in an exam setting where a sign slip is easy to make. Let's compute Δy and Δ²y the long way.

Δy(x) = y(x + 1) − y(x):

= 2(a + bx + b)2x − (a + bx)2x

= 2x[2(a + bx + b) − (a + bx)]

= 2x[2a + 2bx + 2b − a − bx]

= 2x(a + bx + 2b)

Δ²y(x) = Δy(x + 1) − Δy(x):

Δy(x + 1) = 2x + 1(a + b(x+1) + 2b) = 2·2x(a + bx + 3b)

Δ²y(x) = 2·2x(a + bx + 3b) − 2x(a + bx + 2b)

= 2x[2(a + bx + 3b) − (a + bx + 2b)]

= 2x[2a + 2bx + 6b − a − bx − 2b]

= 2x(a + bx + 4b)

Now assemble Δ²y − 2Δy + y:

= 2x(a + bx + 4b) − 2·2x(a + bx + 2b) + 2x(a + bx)

= 2x[(a + bx + 4b) − 2(a + bx + 2b) + (a + bx)]

= 2x[a + bx + 4b − 2a − 2bx − 4b + a + bx]

= 2x[0 + 0 + 0] = 0

Both routes — the operator-factoring shortcut and the brute-force difference-by-difference calculation — agree exactly. The answer is 0.

Final Answer

The correct option is (d) 0. This matches the official ISS 2016 Statistics Paper-1 answer key, which also lists option (d) for Question 65.

Why This Question Matters

Numerical analysis questions in the ISS syllabus frequently test whether a candidate can move fluidly between the Δ-operator world and the E-operator world. A function of the form (polynomial in x)×(exponential in x) is a classic test case, because the exponential part factors out cleanly under any shift, while the polynomial part behaves like an ordinary algebraic expression under substitution. Once you've internalised the identity Δ = E − 1, a whole family of "evaluate this operator expression" questions — not just this one — becomes mechanical rather than mysterious.

Frequently Asked Questions

Why does Δ² − 2Δ + 1 factor as (Δ − 1)²?

Because Δ behaves algebraically like a normal variable when you're combining and factoring operator expressions (as long as you don't need to worry about operators not commuting, which isn't an issue here since we only have powers of a single operator Δ). So t² − 2t + 1 = (t − 1)² carries over directly to Δ² − 2Δ + 1 = (Δ − 1)².

What is the relationship between the shift operator E and the difference operator Δ?

By definition, Δy(x) = y(x + h) − y(x) and Ey(x) = y(x + h). Comparing these two shows Δy(x) = Ey(x) − y(x), or simply Δ = E − 1. This identity is the bridge that lets you convert any Δ-based expression into an E-based one and vice versa.

Why did the answer come out to exactly 0 instead of some function of x?

Because (a + bx) is a first-degree (linear) polynomial in x, and the operator (E − 2)² was specifically matched to the base of the exponential term, 2x. Whenever a function has the form p(x)·cx, applying (E − c)k+1, where k is the degree of the polynomial p(x), always annihilates it completely. Here p(x) = a + bx has degree 1, c = 2, and we applied (E − 2)², i.e. k + 1 = 2, so the result is forced to be zero.

Could this question have been solved without converting to the E operator?

Yes — the cross-check in Step 5 does exactly that, computing Δy and Δ²y directly and then combining them. It takes a few more lines of algebra, but it's a useful safety net during an exam if you're not fully confident with operator substitution, or simply want to verify your shortcut answer before marking the final option.

Is h = 1 important to the final answer, or would it change with a different interval?

The specific value h = 1 matters because it fixes exactly how E acts: Ey(x) = y(x + 1). With a different interval, say h = 2, you would have Ey(x) = y(x + 2), and the whole derivation would shift accordingly. The annihilation result (that the operator eventually gives zero) still holds in general, but the intermediate expressions depend on h, so always read the stated interval before setting up E.

What other ISS topics commonly pair with this kind of operator question?

Numerical analysis questions involving Δ, E, Simpson's rule, Lagrange interpolation, and Newton–Raphson type root finding tend to cluster together in the same section of the paper, as you may have already noticed from the surrounding questions 61 through 64 in this very paper. Being comfortable with operator notation in general pays off across several of these question types, not just this one.

If anything in this derivation felt shaky, drop a comment below and we'll walk through it again — and if you know a fellow ISS aspirant who's stuck on finite differences, go ahead and share this post with them.