Given that a and b are distinct positive numbers, find a polynomial P(x) such that the derivative of f(x) = P(x)e^(−x²) is zero for x = 0, x = ±a and x = ±b, but for no other values of x.

This is a question from a STEP II paper.

STEP questions always give you just enough information to solve the problem; we’re looking to use everything that we’re given. The first thing we should do is differentiate f(x) by the product rule. This is a step up from A-level since we’re considering a general function P(x), but the product rule still works just as usual:

f’(x) = P’(x)e^(-x²) - 2xP(x)e^(−x²) = e^(−x²)[P’(x) – 2xP(x)]

Now we can set the derivative we’ve found equal to zero for x = 0, ±a and ±b.

       •   f’(0) = P’(0) = 0

       •   f’(a) = e^(-a²)[P’(a) – 2aP(a)] = 0  ⇒  P’(a) - 2aP(a) = 0

       •   f’(-a) = e^(-a²) [P’(-a) + 2aP(-a)] = 0  ⇒  P’(-a) + 2aP(-a) = 0

       •   f’(b) = e^(-b²)[P’(b)– 2bP(b)] = 0  ⇒  P’(b) - 2bP(b) = 0

       •   f’(-b) = e^(-b²)[P’(-b) + 2bP(-b)] = 0  ⇒  P’(-b) + 2bP(-b) = 0

The only information we’re given that’s left is that the derivative of f(x) isn’t zero for any other values of x. So e^(−x²)[P’(x) – 2xP(x)] ≠ 0 for any other values of x. Since e^(−x²) is always non-zero we can deduce that P’(x) – 2xP(x) ≠ 0 for any other values of x.

We can now combine everything we know together: P’(x) - 2xP(x) = 0 ONLY for x = 0, ±a, ±b.

How do we proceed now? We’ve used all the information given in the question. Let’s look back at what we’re asked to do: we’re asked to find a polynomial P(x) which satisfies the above conditions. The trick here is to equate the polynomial P’(x) – 2xP(x) to a polynomial that we already know equals zero ONLY for x = 0, ±a, ±b. Then by comparing coefficients we can find coefficients for P(x). Let’s use x(x - a)(x + a)(x - b)(x + b) = x(x² - a²)(x² - b²) = x⁵ - (a² + b²)x³ + a²b²x.

What order does P(x) need to be? The order of x⁵ - (a² + b²)x³ + a²b²x is 5, so in order for P’(x) – 2xP(x) to have order 5 as well P(x) needs to have order 4.

Note that x⁵ - (a² + b²)x³ + a²b²x has no x⁴ or x² terms so our P(x) should have no x³ or x terms to avoid x⁴ or x² terms cropping up in P’(x) – 2xP(x).

Thus P(x) = αx⁴ + βx² + γ for some α, β, γ to be determined.

P’(x) – 2xP(x) = (4αx³ + 2βx) – (2αx⁵ + 2βx³ + 2γx) = –2αx⁵ + (4α - 2β)x³ + (2β - 2γ)x

And now we equate coefficients:

–2αx⁵ + (4α - 2β)x³ + (2β - 2γ)x ≡ x⁵ - (a² + b²)x³ + a²b²x

       •   –2α = 1 ⇒ α = -0.5

       •   4α - 2β = - a² - b²  ⇒ -2 - 2β = - a² - b² ⇒ β = (a² + b² - 2)/2

       •   2β - 2γ = a²b² ⇒ a² + b² - 2 - 2γ = a²b² ⇒ γ = (a² + b²-  a²b² - 2)/2

Hence P(x) = -0.5x⁴ + (a² + b² - 2)x²/2 + (a² + b²-  a²b² - 2)/2 is a solution. 

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