40th VTRMC, 2018, Solutions

  1. Let I = ∫12x-1arctan(1 + x) dx. First we integrate by parts to obtain

    I = [ln(x)arctan(1 + x)]12 - ∫12(ln x)/(1 + (1 + x)2) dx    
      = ln(2)arctan(3) - ∫12(ln x)/(2 + 2x + x2) dx.    

    Now let J = ∫12(ln x)/(2 + 2x + x2) dx and make the substitution x = 2/y. We obtain

    J = ∫21(-2/y2)(ln 2 - ln y)/(2 + 4/y + 4/y2) dy = ∫12(ln 2)/(1 + (1 + y)2) dy -J.

    Therefore 2J = ∫12(ln 2)/(1 + (1 + y)2) dy = [ln(2)arctan(1 + y)]12 = ln(2)(arctan(3) - arctan(2)) and we deduce that I = ln(2)(arctan(3) + arctan(2))/2. Now tan(arctan(3) + arctan(2)) = (3 + 2)(1 - 6) = -1, which shows that arctan(3) - arctan(2) = 3π/4. Therefore I = 3πln(2)/8, and the answer is q = 3/8.

  2. First we'll show that if X, Y∈M6(ℤ), X≡I≡Y mod 3, and XYX = Y, then X = I. Suppose X≠I and write X = I + pC where p is a positive power of 3 and C≢ 0 mod 3. Note that XYrX = Yr for all odd integers r. Write Y = I + 3D where D∈M6(ℤ). Then Yp≡I mod 3p, so X2≡I mod 3p. Therefore I + 2pC + p2C≡I mod 3p which is not the case. Thus X = I and we conclude that A3 = I. Now write A = I + qD where q is a positive power of 3 and D≢ 0 mod 3. Then (I + qD)3≡I mod 9q, which shows that 3qD≡ 0 mod 9q which is not the case.

  3. Let 𝕄 = {2, 3,...} = ℕ∖{1}. Then f2(ℕ) = 𝕄 and therefore f (ℕ) = ℕ or 𝕄. The former yields f2(ℕ) = ℕ, which is not the case, so we must have the latter which yields f (𝕄) = 𝕄. It follows that f2(𝕄) = 𝕄 and we have a contradiction, so there is no such f, as required.

  4. Let d = gcd(m, n). Then d = an + bm for some integers a and b. Now
     (
     n
     m
    		)
    =
     n/m (
     n - 1
     m - 1
    		),
    therefore

     (d /n) (
     n
     m
    		)
    =
     (a + bm/n) (
     n
     m
    		)
    =
     a (
     n
     m
    		)
    +
     b (
     n - 1
     m - 1
    		).

    Since
     (
     n
     m
    		)
    and
     (
     n - 1
     m - 1
    		)
    are integers, the result follows.

  5. We'll show that (an) is unbounded. We have an-1 = ∫01/√(n-1)(| 1 -enit|)/(| 1 -eit|) dt. Note that | 1 -eit|≤t for t≥ 0. To see this, by squaring both sides, this is equivalent to 2 - 2 cos t≤t2, i.e. t2 +2 cos t - 2≥ 0, which is true because we have equality when t = 0, and the derivative of the left hand side is non-negative for t≥ 0 by using the inequality sin t≤t for t≥ 0. Therefore it will be sufficient to show that bn ≔∫01/√(n-1)| 1 -enit|/t dt is unbounded (because π/4 < 1). However for n∈ℤ,

    ∫πr/nπ(r+1)/n| 1 -enit| dt = ∫πr/nπ(r+1)/n√2 - 2cos nt = 4/n.

    Let k = [√n-1/π], so k is the greatest positive integer such that kπ < √n-1. Note that k→∞ as n→∞. Then bn≥4(1 + 1/2 + ... +1/k)/π, which is unbounded because the harmonic series is divergent.

  6. First we show that an -bn≥ 0 for all n≥1. This is equivalent to proving

    (n + 1)(1/2 + 1/4 + ... +1/(2n))/n≤1 + 1/3 + ... + 1/(2n - 1),

    that is

    1 + 1/2 + 1/3 + ... +1/n≤n((2 - 1) + (2/3 - 2/4) + ... + (2/(2n - 1) - 2/(2n))).

    Since 1 + 1/2 + ... +1/n≤n, the assertion follows. Since a1 -b1 = 0, we see that the minimum of an -bn is zero.

    Next we show that an -bn is decreasing for n sufficiently large. We have

    (an -bn) - (an+1 -bn+1) = an -an+1 - (bn -bn+1) = (1 + 1/3 + ... +1/(2n - 1)) - 1/((n + 2)(2n + 1)) - (1/2 + 1/4 + ... + 1/(2n))/((n(n + 1)) + 1/((n + 1)(2n + 2)).

    Now 1/((n + 1)(2n + 2)) - 1/((n + 2)(2n + 1)) > 0 for all n≥1, so we need to prove

    (1 + 1/3 + ... +1/(2n - 1))/((n(n + 1)) > (1/2 + 1/4 + ... + 1/(2n))/((n(n + 1))

    for n sufficiently large. Multiplying by n(n + 1)(n + 2) and then subtracting n(1/2 + 1/4 + ... + 1/(2n)) from both sides, means we want to prove

    n(1/2 + 1/12 + ... +1/((2n - 1)2n)) > 1 + 1/2 + ... + 1/n

    for sufficiently large n. However this is clear for n≥4. Therefore an -bn takes its maximum value for some n≤4. By inspection, the maximum value occurs when n = 3, which is 7/90.

  7. Note that if g and h are continuous piecewise-monotone functions on [a, b], then ℓ(gh)≤ℓ(g)ℓ(h). Thus ℓ(fn)≤(ℓ(f ))n for all n∈ℕ. Now fix a positive integer k. Given n∈ℕ, there are integers q and r such that n = qk + r with 0≤r < k. Then ℓ(fn)≤(ℓ(fk))q(ℓ(f ))r, consequently

    n√(ℓ(fn))≤(ℓ(fk))q/n(ℓ(f ))r/n.

    Since k is fixed, r/n→ 0 and q/n→1/k as n→∞. Therefore limsup  n√(ℓ(fn))≤  k√(ℓ(fk)) and we deduce that

    limsup  n√(ℓ(fn))≤inf  k√(ℓ(fk))≤liminf k√(ℓ(fk))

    and the result follows.