Mínimo de $\sum_{l=0}^{n} \frac{1}{(l!)^2((n-l)!)^2} x^{2l} (1-x)^{2(n-l)} $.

Question

$\forall n \in \mathbf{N}$, demostrar que la función de $f(x)=\sum_{l=0}^{n} \frac{1}{(l!)^2((n-l)!)^2} x^{2l} (1-x)^{2(n-l)} $ alcanza su mínimo en $x=\frac{1}{2}$.

Ahora basta probar que \begin{equation*} \sum_{l=0}^{n} \binom{n}{l}^2 (x+y)^{2l} (x-y)^{2(n-l)} = \sum_{l=0}^{n} \binom{2l}{l} \binom{2(n-l)}{n-l} x^{2l}y^{2(n-l)} \end{ecuación*}

Answer 1

\begin{align*} n &> 0 \text{.} \\ f'(1/2) &= \left. \sum_{l=0}^{n} \frac{1}{(l!)^2((n-l)!)^2} \left( (2l)x^{2l-1}(1-x)^{2(n-l)} - 2(n-l)x^{2l}(1-x)^{2(n-l)-1} \right) \right|_{x = 1/2} \\ &= \sum_{l=0}^{n} \frac{1}{(l!)^2((n-l)!)^2} \left( (2l)(1/2)^{2l-1}(1/2)^{2(n-l)} - 2(n-l)(1/2)^{2l}(1/2)^{2(n-l)-1} \right) \\ &= (1/2)^{2n-1}\sum_{l=0}^{n} \frac{1}{(l!)^2((n-l)!)^2} \left( (2l) - 2(n-l) \right) \\ &= 0 \text{.} \\ f(x) &= \sum_{l=0}^{n} \frac{1}{(l!)^2((n-l)!)^2} x^{2l} (1-x)^{2(n-l)} \\ &= \frac{(1-x)^{2n}}{(n!)^2} + \frac{x^{2n}}{(n!)^2} + \sum_{l=1}^{n-1} \frac{1}{(l!)^2((n-l)!)^2} x^{2l} (1-x)^{2(n-l)} \text{.} \\ f''(x) &= \frac{2n(2n-1)\left(x^{2(n-1)} + (1-x)^{2(n-1)}\right)}{(n!)^2} + \\ {}&+ \sum_{l=1}^{n-1} \frac{1}{(l!)^2((n-l)!)^2} x^{2(l-1)}(1-x)^{2(n-l-1)} \left( 2n(2n-1)x^2 - 4l(2n-1)x + 2l(2l-1)\right) \\ &> 0 + \sum_{l=1}^{n-1} \frac{1}{(l!)^2((n-l)!)^2} x^{2(l-1)}(1-x)^{2(n-l-1)} \frac{2l(l-n)}{n} \\ &\geq 0 \text{.} \blacksquare \end{align*}

Answer 2

Sugerencia:$$f(x)=\sum_{l=0}^{n} \frac{1}{(l!)^2((n-l)!)^2} x^{2l} (1-x)^{2(n-l)} $$

$$f(x) = \frac{1}{(n!)^2}\sum_{l=0}^{n} \left(\frac{n!}{l!(n-l)!}x^l(1-x)^{n-l}\right)^2 = \frac{1}{(n!)^2}\sum_{l=0}^{n} \left(\binom{n}{l}x^l(1-x)^{n-l}\right)^2$$

El factor de $\dfrac{1}{(n!)^2}$ es irrelevante para el mínimo de $f(x)$.

Pero, nosotros sabemos acerca de los coeficientes binomiales que $\displaystyle\binom{n}{l} = \binom{n}{n-l}$.