ISS 2016 Statistics Paper-1 Solution: Question 62 (Error Order in Fourth-Order Runge-Kutta Method)

Continuing our question-by-question walk through the ISS Statistics Paper-1 previous year papers, today we pick up right where we left off — Question 62 of the 2016 paper, which sits in the Numerical Analysis portion of the syllabus. This one trips up a lot of aspirants because it hides a classic terminology trap inside a seemingly simple fact.

Quick Summary

• Topic:Numerical solution of ordinary differential equations — order of truncation error in the fourth-order Runge-Kutta (RK4) method

• Question Reference:ISS 2016, Statistics Paper-1, Question 62

• Correct Answer:(d) (equal interval)5

The Question, As Asked

The error in fourth order Runge-Kutta formula for solving the differential equation dy/dx = f(x, y) is of the order

• (a) (equal interval)2

• (b) (equal interval)3

• (c) (equal interval)4

• (d) (equal interval)5

Here "equal interval" is just the paper's wording for the step size, usually written as h, since in RK methods the independent variable is advanced in equal steps of length h from xn to xn + h.

Step 1: Set Up What We're Actually Solving

We are given the first-order differential equation

dy/dx = f(x, y), with y(x0) = y0,

and we want to march the solution forward from xn to xn+1 = xn + h using the classical fourth-order Runge-Kutta formula:

k1 = h f(xn, yn)

k2 = h f(xn + h/2, yn + k1/2)

k3 = h f(xn + h/2, yn + k2/2)

k4 = h f(xn + h, yn + k3)

yn+1 = yn + (1/6)(k1 + 2k2 + 2k3 + k4)

The question is asking: when we take a single step of size h with this formula, how far off is yn+1 from the true value y(xn + h), in terms of powers of h? This quantity is called the local truncation error (LTE) of the method.

Step 2: Write the Exact Solution as a Taylor Series

This is the manual, no-shortcuts part — don't skip it. The exact solution y(x), expanded around xn using Taylor's theorem, gives the true value at xn + h as:

y(xn + h) = y(xn) + h y'(xn) + (h2/2!) y''(xn) + (h3/3!) y'''(xn) + (h4/4!) y''''(xn) + (h5/5!) y(5)(xn) + ...

This is just the ordinary Taylor expansion of a function around a point — every term is a derivative of y at xn, multiplied by h raised to a matching power and divided by the matching factorial. Nothing Runge-Kutta specific has entered yet; this is simply "what the true answer looks like."

Step 3: Expand the RK4 Formula in Powers of h

Now take the RK4 formula from Step 1 and expand each of k1, k2, k3, k4 as a Taylor series in h as well (substituting f and its partial derivatives at (xn, yn), and using the chain rule repeatedly since y' = f, y'' = fx + f·fy, and so on). When you carry this expansion out fully — which is a long but entirely mechanical algebra exercise — and substitute back into yn+1 = yn + (1/6)(k1 + 2k2 + 2k3 + k4), something remarkable happens: every term up to and including the h4 term matches the Taylor expansion of the true solution from Step 2, coefficient for coefficient.

That matching is not an accident — it is exactly how the coefficients 1/6, 2/6, 2/6, 1/6 and the midpoint evaluations at h/2 were chosen by the method's designers in the first place. They solved a system of equations forcing agreement with the Taylor series through the h4 term.

Step 4: Compare the Two Expansions to Get the Error

Since yn+1 (from RK4) and y(xn + h) (the true value) agree in every term through h4, their difference — the local truncation error — can only come from the first term where they disagree. That is the h5 term:

Local truncation error = y(xn + h) − yn+1 = C·h5 + higher order terms, for some constant C depending on derivatives of f.

In Big-O notation, the error introduced in a single step of the fourth-order Runge-Kutta method is therefore O(h5)— order 5 in the step size, not order 4.

Pro Tip: This is the single most common trap in this topic, and it's worth drilling until it's automatic rather than something you have to re-derive under exam pressure. "Fourth order Runge-Kutta" refers to the

Step 5: Match to the Options and Confirm Against the Key

Our derivation gives local truncation error of order h5, which is exactly option (d): (equal interval)5. Checking against the official ISS 2016 Statistics Paper-1 answer key (Series A booklet, as used in this question paper), the key also lists option (d)as correct for Question 62. Our independent derivation and the official key agree — there is no discrepancy to flag here.

Why This Question Matters

Numerical solution of ordinary differential equations is a recurring, high-weightage topic in ISS Statistics Paper-1, and Runge-Kutta methods specifically show up across multiple years in slightly different forms — sometimes asking for the error order directly (as here), sometimes asking you to compute k1 through k4 numerically, and sometimes testing the related concept of truncation error in Euler's method. Understanding the Taylor-series derivation behind the "order p method ⇒ local error O(hp+1), global error O(hp)" rule means you can answer all of these variants from one piece of understanding instead of memorizing each as a separate fact.

Frequently Asked Questions

What is the difference between local truncation error and global truncation error?

Local truncation error is the error introduced in a single step, assuming the starting value for that step was exact. Global truncation error is the total accumulated error after many steps, which includes both the local errors from each step and how those errors propagate and compound through subsequent steps.

Why is RK4 called "fourth order" if its local error is O(h⁵)?

The naming convention for numerical ODE methods is based on the global error, not the local error. Since roughly 1/h steps are needed to cover a fixed interval, and each step contributes an O(h⁵) local error, the total accumulated error scales as O(h⁴) — hence "fourth order."

Does halving the step size h always reduce the error in RK4?

In principle yes — halving h should reduce the global error by roughly a factor of 2⁴ = 16, since the global error is O(h⁴). In practice, making h extremely small can introduce round-off error from floating-point arithmetic, so there is a practical lower limit below which further reducing h stops helping and can even hurt accuracy.

How is the order of a Runge-Kutta method actually determined?

It is determined by expanding the method's formula as a Taylor series in h and finding the highest power of h through which it matches the Taylor expansion of the true solution. The first power at which the two expansions disagree gives the order of the local truncation error.

Is RK4 always more accurate than lower-order methods like RK2 or Euler's method?

For a given, sufficiently small step size h, yes — RK4's error shrinks much faster as h decreases because it is O(h⁴) globally compared to O(h²) for RK2 and O(h) for Euler's method. However, RK4 requires four function evaluations per step compared to one for Euler's method, so there is a computational cost trade-off to consider.

Does this Taylor-series approach work for finding the error order of any numerical ODE method?

Yes. This Taylor-series matching technique is general and applies to any one-step method for solving ordinary differential equations, including Euler's method, the midpoint method, and other Runge-Kutta variants — only the amount of algebra involved changes with the complexity of the formula.

If anything in this derivation felt unclear, or you'd like to see the full k1 through k4 Taylor expansion worked out term by term, drop a comment below. And if you know a fellow ISS aspirant who's stuck on numerical analysis, feel free to pass this along — questions like this one come up often enough that it's worth getting fully comfortable with.