11922 Find all positive n such that both n and n² are palindromes when written in the binary system (without leading zeros).

Solution:

If n = Σ 2^k is a palindrome, the average exponent k is half the highest exponent. But if we square such a palindrome, the average will become more than half the highest exponent, except for small values of n.
This is due to three facts:
1) the double product of 2ⁱ and 2^j is 2^i+j+1,
2) the sum of two terms 2^m is 2^m+1.
3) the product of two terms 2^m is 2^2m.

We find that only n = 1 and n = 3 satisfy the requirements.

11923 Let f_p be the function on (0,π/2) given by f_p(x) = (1 + sin(x))^p - (1 - sin(x))^p - 2 sin(px).
Prove f_p > 0 for 0 < p < 1/2 and f_p < 0 for 1/2 < p < 1.

Solution:

For p = 1/2 we calculate f_p(x) using goniometrical formulas and find f_1/2(x) = 0 for all x.

In general, we calculate the power series and find it's an alternating series with decreasing terms and dominant first term p (1 - 3p + 2p²) x³/3, which is positive for 0 < p < 1/2 and negative for 1/2 < p < 1.

11924 Calculate ∫₀^π/2 {tan(x)}/tan(x) dx, where {u} denotes u - [u].

Solution:

We find π/2 - Σ_k=1^∞ ∫_arctan(k)^arctan(k+1) k/tan(x) dx =
π/2 - Σ_k=1^∞ k(ln((k+1)/√((k+1)²+1)) - ln(k/√(k²+1))) =
π/2 - Σ_k=1^∞ ((k+1)(ln((k+1)/√((k+1)²+1)) - ln(k/√(k²+1)) - ln((k+1)/√((k+1)²+1))) = {telescope series} =
π/2 - ln(2)/2 - (Σ_k=1^∞ ln(1 + 1/(k+1)²))/2 = {using power series, changing summation order, alternating harmonic series has value ln(2)} =
π/2 - ( (Σ_k=1^∞ 1/k²) - (Σ_k=1^∞ 1/k⁴)/2 + (Σ_k=1^∞ 1/k⁶)/3 - (Σ_k=1^∞ 1/k⁸)/4 + ... )/2.

The sums ζ(p) = Σ_k=1^∞ 1/k^p are well known and many of their values listed on the internet. So we find π/2 - S/2, where S is an alternating sum with decreasing terms whose value is between the sum of the first n terms and the first n+1 terms for each n.

11925 Let n be an integer, at least 4. Find the largest k such that for any list (a) of n real numbers that sum to 0, (Σ_j=1ⁿ a_j²)³ ≥ k (Σ_j=1ⁿ a_j³)².

Solution:

Due to symmetry, we find the minimum of (Σ_j=1ⁿ a_j²)³/(Σ_j=1ⁿ a_j³)² under the condition that Σ_j=1ⁿ a_j = 0 for a₁ = a₂ = ... = a_n-1 = a, a_n = -(n-1)a.
Substitution yields n(n-1)/(n-2)², which is the minimum for n=3, too.
(We may first eliminate a_n = -a₁ - a₂ ... - a_n-1, and then equal the partial derivatives to 0.)

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