【Math】2. Infinite Product、A Constrained Polynomial
🔢 Math
Problem 1 (Easy–Medium): Infinite Product
Evaluate
∏n=2∞n3−1n3+1\prod_{n=2}^{\infty} \frac{n^3 - 1}{n^3 + 1}n=2∏∞n3+1n3−1
Hint: Factor both numerator and denominator using a3±1a^3 \pm 1a3±1, then look for telescoping patterns.
Solution
Let ω=ei2π3\omega=e^{i\frac{2\pi}{3}}ω=ei32π, then n3−1=(n−1)(n−ω)(n+1+ω),n3+1=(n+1)(n+ω)(n−1−ω)n^3-1=(n-1)(n-\omega)(n+1+\omega), n^3+1=(n+1)(n+\omega)(n-1-\omega)n3−1=(n−1)(n−ω)(n+1+ω),n3+1=(n+1)(n+ω)(n−1−ω), so ∏n=2∞n3−1n3+1=∏n=2∞(n−1)(n−ω)(n+1+ω)(n+1)(n+ω)(n−1−ω)=2×12+ω×11−ω=23\prod_{n=2}^{\infty} \frac{n^3 - 1}{n^3 + 1}=\prod_{n=2}^{\infty} \frac{(n-1)(n-\omega)(n+1+\omega)}{(n + 1)(n+\omega)(n-1-\omega)}=2\times \frac{1}{2+\omega}\times \frac{1}{1-\omega}=\frac{2}{3}n=2∏∞n3+1n3−1=n=2∏∞(n+1)(n+ω)(n−1−ω)(n−1)(n−ω)(n+1+ω)=2×2+ω1×1−ω1=32
Problem 2 (Medium–Hard): A Constrained Polynomial
Let p(x)p(x)p(x) be a monic polynomial of degree 444 satisfying p(1)=1p(1) = 1p(1)=1, p(2)=2p(2) = 2p(2)=2, and p(3)=3p(3) = 3p(3)=3. Note that these three constraints do not uniquely determine ppp — there is a one-parameter family of solutions.
Show that the value of p(0)+p(4)p(0) + p(4)p(0)+p(4) is nonetheless the same for every such polynomial, and find it.
Hint: Consider q(x)=p(x)−xq(x) = p(x) - xq(x)=p(x)−x. What do you know about its roots?
Solution
Consider f(x)=p(x)−xf(x)=p(x)-xf(x)=p(x)−x, it has roots 1,2,31,2,31,2,3 thus f(x)=(x−1)(x−2)(x−3)(x−x0)f(x)=(x-1)(x-2)(x-3)(x-x_0)f(x)=(x−1)(x−2)(x−3)(x−x0). f(0)=6x0,f(4)=6(4−x0)f(0)=6x_0, f(4)=6(4-x_0)f(0)=6x0,f(4)=6(4−x0), thus f(0)+f(4)=p(0)+p(4)−4=24f(0)+f(4)=p(0)+p(4)-4=24f(0)+f(4)=p(0)+p(4)−4=24, p(0)+p(4)=28p(0)+p(4)=28p(0)+p(4)=28.
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