Raíces de $f$ y $f'$ $1+\sum_{k=0}^{100}\frac{(-1)^{k+1}}{(k+1)!}x(x-1)(x-2)\cdots (x-k)$

Question

Tengo esta pregunta de un examen de admisión. Dado que $$f(x)=1+\sum_{k=0}^{100}\frac{(-1)^{k+1}}{(k+1)!}x(x-1)(x-2)\cdots (x-k)$$ find S(f(x))-S(f'(x))$where$S$denotes the sum of the real roots for$f(x)$respectively$f'(x)

Aquí es cómo trató, desde $x=1,2,\ldots,100, 101$ la parte de la suma se desvanece poco a poco y
$$f(1)=1-1=0$ $$$ f(2)=1-2+\frac{1}{2!}2(2-1)=0 $$f(3)=1-3+\frac{1}{2!}3(3-1)-\frac{1}{3!}3(3-2)(3-1)=\binom{3}{0}-\binom{3}{1}+\binom{3}{2}-\binom{3}{3}$$ and we can see a pattern in the above equation so that we can rewrite $f 3$ as $(1-1) ^3=0\,$ and indeed that $f (101) =(1-1) ^ {101} = 0$

Puesto que el polinomio es también de orden $101$, las raíces son $x=1,2, \ldots, 101$ giving: $$S(f(x))=\sum_{j=1}^{101}j =5151$$ Is this correct? And how can $S(f'(x))$ be evaluated ? This doesnt seem so obvious.. Also now we can rewrite $%$ $f(x)=-\frac{1}{101!}(x-1)(x-2)\cdots(x-101)$

Answer 1

Tenemos

$$f(x) = a{101}x^{101} + a{100}x^{100} + \cdots$$

y sabemos que $a{101} = -\frac1{101!}$. Fórmulas de Vieta dan$$5151 = \text{ sum of roots of } f = -\frac{a{100}}{a_{101}}

así $a_{100} = \frac{5151}{101!}$.

El derivado es

$$f'(x) = 101a{101}x^{100} + 100a{100}x^{99} + \cdots$$

así las fórmulas de Vieta que %#% $ #%

Answer 2

De acuerdo a Vieta dado el polinomio

$$p_n(x) = a_n x^n+a_{n-1}x^{n-1}+\cdots + a_0\\ p'_n(x) = n a_n x^{n-1} + (n-1)a_{n-1}x^{n-2}+\cdots+a_1$$

ahora

$$S_{100}(x) = -\frac{a_{100}}{a_{101}}\\ S'_{100}(x) = -\frac{(101-1)a_{100}}{101 a_{101}}$$

Aquí

$$a_{101} = \frac{(-1)^{101}}{101!}\\ a_{100} = -a_{101}\frac{100(100+1)}{2}+\frac{(-1)^{100}}{100!}$$

entonces

$$S_n(x) = 5151\\ S'_n(x) = 5100$$

y, finalmente,

$$S_{100}(x)-S'_{100}(x) = 5151-5100 = 51$$