Bài toán : Cho $P(x) = a{x}^{3} + b{x}^{2} + cx + d$ thỏa mãn $\left|P(x) \right| \leq 1$ với mọi $\left|x \right| \leq 1$. Chứng minh rằng $\left|a \right| + \left|b \right| + \left|c \right| + \left|d\right| \leq 7$ Lời giải : Áp dụng công thức nội suy $Lagrange$ cho $P(x)$ với bốn số $-1,1,\frac{-1}{2},\dfrac{1}{2}$ : \[P(x)=P(-1).\dfrac{(x-1)(x-\frac{1}{2})(x+\frac{1}{2})}{(-1-1)(-1-\frac{1}{2})(-1+\frac{1}{2})}+P(1).\dfrac{(x+1)(x-\frac{1}{2})(x+\frac{1}{2})}{(1+1)(1-\frac{1}{2})(1+\frac{1}{2})}+P\left ( \dfrac{1}{2} \right ).\dfrac{(x-1)(x+1)(x+\frac{1}{2})}{(\frac{1}{2}-1)(\frac{1}{2}+1)(\frac{1}{2}+\frac{1}{2})}+P\left ( \dfrac{-1}{2} \right ).\dfrac{(x-\frac{1}{2})(x-1)(x+1)}{(-\frac{1}{2}-\frac{1}{2})(-\frac{1}{2}-1)(-\frac{1}{2}+1)}=P(-1).\dfrac{(x-1)(4x^{2}-1)}{-6}+P(1).\dfrac{(x+1)(4x^{2}-1)}{6}+P\left ( \dfrac{1}{2} \right ).\dfrac{(4x+2)(x^{2}-1)}{-3}+P\left ( -\dfrac{1}{2} \right ).\dfrac{(4x-2)(x^{2}-1)}{3}=\left [ \dfrac{-2}{3}P(-1)+\dfrac{2}{3}P(1)-\dfrac{4}{3}P\left ( \dfrac{1}{2} \right )+\dfrac{4}{3}P\left ( -\dfrac{1}{2} \right ) \right ]x^{3}+\left [ \dfrac{2}{3}P(-1)+\dfrac{2}{3}P(1)-\dfrac{2}{3}P\left ( \dfrac{1}{2} \right )-\dfrac{2}{3}P\left ( -\dfrac{1}{2} \right ) \right ]x^{2}+\left [ \dfrac{1}{6}P(-1)-\dfrac{1}{6}P(1)+\dfrac{4}{3}P\left ( \dfrac{1}{2} \right )-\dfrac{4}{3}P\left ( -\dfrac{1}{2} \right ) \right ]x+\left [ -\dfrac{1}{6}P(-1)-\dfrac{1}{6}P(1)+\dfrac{2}{3}P\left ( \dfrac{1}{2} \right )+\dfrac{2}{3}P\left ( -\dfrac{1}{2} \right ) \right ]\] Đồng nhất hệ số : \[\bigstar\] \[a= \dfrac{-2}{3}P(-1)+\dfrac{2}{3}P(1)-\dfrac{4}{3}P\left ( \dfrac{1}{2} \right )+\dfrac{4}{3}P\left ( -\dfrac{1}{2} \right )\] \[\Rightarrow \left | a \right |\leq \left | \dfrac{2}{3}P(1)-\dfrac{2}{3}P(-1) \right |+\left | \dfrac{4}{3}P\left ( -\dfrac{1}{2} \right )-\dfrac{4}{3}P\left ( \dfrac{1}{2} \right ) \right |\] \[\bigstar\] \[b=\dfrac{2}{3}P(-1)+\dfrac{2}{3}P(1)-\dfrac{2}{3}P\left ( \dfrac{1}{2} \right )-\dfrac{2}{3}P\left ( -\dfrac{1}{2} \right )\] \[\Rightarrow \left | b \right |\leq \left | \dfrac{2}{3}P(1)+\dfrac{2}{3}P(-1) \right |+\left | \dfrac{2}{3}P\left ( \dfrac{1}{2} \right )+\dfrac{2}{3}P\left ( -\dfrac{1}{2} \right ) \right |\] \[\bigstar\] \[c= \dfrac{1}{6}P(-1)-\dfrac{1}{6}P(1)+\dfrac{4}{3}P\left ( \dfrac{1}{2} \right )-\dfrac{4}{3}P\left ( -\dfrac{1}{2} \right )\] \[\Rightarrow \left | c \right |= \dfrac{1}{6}P(-1)-\dfrac{1}{6}P(1)+\dfrac{4}{3}P\left ( \dfrac{1}{2} \right )-\dfrac{4}{3}P\left ( -\dfrac{1}{2} \right )\leq \left | \dfrac{1}{6}P(-1) -\dfrac{1}{6}P(1)\right |+2.\left | \dfrac{2}{3}P\left ( \dfrac{1}{2} \right ) -\dfrac{2}{3}P\left ( -\dfrac{1}{2} \right )\right |\] \[\bigstar\] \[d=-\dfrac{1}{6}P(-1)-\dfrac{1}{6}P(1)+\dfrac{2}{3}P\left ( \dfrac{1}{2} \right )+\dfrac{2}{3}P\left ( -\dfrac{1}{2} \right )\] \[\Rightarrow \left | d \right |=-\dfrac{1}{6}P(-1)-\dfrac{1}{6}P(1)+\dfrac{2}{3}P\left ( \dfrac{1}{2} \right )+\dfrac{2}{3}P\left ( -\dfrac{1}{2} \right )\leq \left | \dfrac{1}{6}P(-1)+\dfrac{1}{6}P(1) \right |+\left | \dfrac{2}{3}P\left ( \dfrac{1}{2} \right )+\dfrac{2}{3}P\left ( -\dfrac{1}{2} \right ) \right |\] Từ đó chú đến bất đẳng thức $\left | a+b \right |+\left | a-b \right |\leq max\left \{ 2\left | a \right |,2\left | b \right | \right \}$ và $\left | P(1) \right |\leq 1,\left | P(-1) \right |\leq 1,\left | P\left ( \dfrac{1}{2} \right ) \right |\leq 1,\left | P\left ( -\dfrac{1}{2} \right ) \right |\leq 1$, ta có : \[\left | a \right |+\left | b \right |+\left | c \right |+\left | d \right |\leq 3.max\left \{ 2\left | \dfrac{2}{3}P(-1) \right |,2\left | \dfrac{2}{3}P(1) \right | \right \}+max\left \{ 2\left | \dfrac{1}{6}P(-1) \right |,2\left | \dfrac{1}{6}P(1) \right | \right \}+ \left | \dfrac{4}{3}P\left ( -\dfrac{1}{2} \right ) -\dfrac{4}{3}P\left ( \dfrac{1}{2} \right )\right | \leq 3.2.\dfrac{2}{3}.1+2.\dfrac{1}{6}.2.\dfrac{4}{3}=7\] Đây là điều phải chứng minh.

[Bài toán] Đa thức, Công thức nội suy Lagrange

Bài toán : Cho $P (x) = a x^{3} + b x^{2} + c x + d$ thỏa mãn $∣ P (x) ∣ \leq 1$ với mọi $∣ x ∣ \leq 1$ . Chứng minh rằng $∣ a ∣ + ∣ b ∣ + ∣ c ∣ + ∣ d ∣ \leq 7$

Lời giải :

Áp dụng công thức nội suy $L a g r an g e$ cho $P (x)$ với bốn số $- 1, 1, \frac{- 1}{2}, \frac{1}{2}$ :

$P (x) = P (- 1) . \frac{( x - 1 ) ( x - \frac{1}{2} ) ( x + \frac{1}{2} )}{( - 1 - 1 ) ( - 1 - \frac{1}{2} ) ( - 1 + \frac{1}{2} )} + P (1) . \frac{( x + 1 ) ( x - \frac{1}{2} ) ( x + \frac{1}{2} )}{( 1 + 1 ) ( 1 - \frac{1}{2} ) ( 1 + \frac{1}{2} )} + P (\frac{1}{2}) . \frac{( x - 1 ) ( x + 1 ) ( x + \frac{1}{2} )}{( \frac{1}{2} - 1 ) ( \frac{1}{2} + 1 ) ( \frac{1}{2} + \frac{1}{2} )} + P (\frac{- 1}{2}) . \frac{( x - \frac{1}{2} ) ( x - 1 ) ( x + 1 )}{( - \frac{1}{2} - \frac{1}{2} ) ( - \frac{1}{2} - 1 ) ( - \frac{1}{2} + 1 )} = P (- 1) . \frac{( x - 1 ) ( 4 x ^{2} - 1 )}{- 6} + P (1) . \frac{( x + 1 ) ( 4 x ^{2} - 1 )}{6} + P (\frac{1}{2}) . \frac{( 4 x + 2 ) ( x ^{2} - 1 )}{- 3} + P (- \frac{1}{2}) . \frac{( 4 x - 2 ) ( x ^{2} - 1 )}{3} = [\frac{- 2}{3} P (- 1) + \frac{2}{3} P (1) - \frac{4}{3} P (\frac{1}{2}) + \frac{4}{3} P (- \frac{1}{2})] x^{3} + [\frac{2}{3} P (- 1) + \frac{2}{3} P (1) - \frac{2}{3} P (\frac{1}{2}) - \frac{2}{3} P (- \frac{1}{2})] x^{2} + [\frac{1}{6} P (- 1) - \frac{1}{6} P (1) + \frac{4}{3} P (\frac{1}{2}) - \frac{4}{3} P (- \frac{1}{2})] x + [- \frac{1}{6} P (- 1) - \frac{1}{6} P (1) + \frac{2}{3} P (\frac{1}{2}) + \frac{2}{3} P (- \frac{1}{2})]$

Đồng nhất hệ số :

$★$ $a = \frac{- 2}{3} P (- 1) + \frac{2}{3} P (1) - \frac{4}{3} P (\frac{1}{2}) + \frac{4}{3} P (- \frac{1}{2})$ $\Rightarrow ∣ a ∣ \leq \frac{2}{3} P (1) - \frac{2}{3} P (- 1) + \frac{4}{3} P (- \frac{1}{2}) - \frac{4}{3} P (\frac{1}{2})$

$★$ $b = \frac{2}{3} P (- 1) + \frac{2}{3} P (1) - \frac{2}{3} P (\frac{1}{2}) - \frac{2}{3} P (- \frac{1}{2})$ $\Rightarrow ∣ b ∣ \leq \frac{2}{3} P (1) + \frac{2}{3} P (- 1) + \frac{2}{3} P (\frac{1}{2}) + \frac{2}{3} P (- \frac{1}{2})$

$★$ $c = \frac{1}{6} P (- 1) - \frac{1}{6} P (1) + \frac{4}{3} P (\frac{1}{2}) - \frac{4}{3} P (- \frac{1}{2})$ $\Rightarrow ∣ c ∣ = \frac{1}{6} P (- 1) - \frac{1}{6} P (1) + \frac{4}{3} P (\frac{1}{2}) - \frac{4}{3} P (- \frac{1}{2}) \leq \frac{1}{6} P (- 1) - \frac{1}{6} P (1) + 2. \frac{2}{3} P (\frac{1}{2}) - \frac{2}{3} P (- \frac{1}{2})$

$★$ $d = - \frac{1}{6} P (- 1) - \frac{1}{6} P (1) + \frac{2}{3} P (\frac{1}{2}) + \frac{2}{3} P (- \frac{1}{2})$ $\Rightarrow ∣ d ∣ = - \frac{1}{6} P (- 1) - \frac{1}{6} P (1) + \frac{2}{3} P (\frac{1}{2}) + \frac{2}{3} P (- \frac{1}{2}) \leq \frac{1}{6} P (- 1) + \frac{1}{6} P (1) + \frac{2}{3} P (\frac{1}{2}) + \frac{2}{3} P (- \frac{1}{2})$

Từ đó chú đến bất đẳng thức $∣ a + b ∣ + ∣ a - b ∣ \leq ma x {2 ∣ a ∣, 2 ∣ b ∣}$ và $∣ P (1) ∣ \leq 1, ∣ P (- 1) ∣ \leq 1, P (\frac{1}{2}) \leq 1, P (- \frac{1}{2}) \leq 1$ , ta có :

$∣ a ∣ + ∣ b ∣ + ∣ c ∣ + ∣ d ∣ \leq 3. ma x {2 \frac{2}{3} P (- 1), 2 \frac{2}{3} P (1)} + ma x {2 \frac{1}{6} P (- 1), 2 \frac{1}{6} P (1)} + \frac{4}{3} P (- \frac{1}{2}) - \frac{4}{3} P (\frac{1}{2}) \leq 3.2. \frac{2}{3} .1 + 2. \frac{1}{6} .2. \frac{4}{3} = 7$

Đây là điều phải chứng minh.

[Bài toán] Đa thức, Công thức nội suy Lagrange

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