Niech \(\displaystyle{ W(x)= ax^4 + bx^3 +cx^2+dx}\) taki, że
\(\displaystyle{ \begin{cases} a, b, c, d >0 \\ W(x) \in Z \ dla \ x \in \{ -2, -1, 0, 1, 2 \} \\ W(1)= 1 \\ W(5)= 70 \end{cases}}\).
Udowodnić, że:
\(\displaystyle{ \begin{cases} a=\frac{1}{24} \\ b=d = \frac{1}{4} \\ c=\frac{11}{24} \end{cases}}\)
Jakie współczynniki ?
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mol_ksiazkowy
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loitzl9006
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Jakie współczynniki ?
\(\displaystyle{ W \left( 1 \right) =1\to a+b+c+d=1\\ W \left( -1 \right) \in Z\ \to \ a-b+c-d\in Z \\ W \left( 1 \right) -W \left( -1 \right) \in Z\\ W \left( 1 \right) -W \left( -1 \right) =2b+2d=2 \left( b+d \right) \\ 2 \left( b+d \right) \in Z\\ \begin{cases} a+b+c+d=1 \\ a,b,c,d>0 \end{cases}\ \to \ 0<b+d<1\ \to \ b+d\notin Z\\ b+d=\frac k2; k\in Z\\ \begin{cases} 0<b+d<1 \\ b+d=\frac k2 \\ k\in Z \end{cases} \ \to \ k=1 \ \to \ b+d=\frac12\ \to \ a+c=\frac12\\ d=\frac12-b \ \wedge \ c=\frac12-a\\ W \left( -2 \right) \in Z\ \to \ 16a-8b+4c-2d\in Z\\ W \left( 2 \right) \in Z\ \to \ 16a+8b+4c+2d\in Z\\ W \left( -2 \right) +W \left( 2 \right) \in Z\\ W \left( -2 \right) +W \left( 2 \right) =32a+8c=32a+8 \left( \frac12-a \right) =32a+4-8a=24a+4\ \to \ 24a\in Z\\ W \left( 2 \right) -W \left( -2 \right) \in Z\\ W \left( 2 \right) -W \left( -2 \right) =16b+4d=16b+4 \left( \frac12-b \right) =16b+2-4b=12b+2\ \to \ 12b\in Z\\ W \left( 5 \right) =70\ \to \ 625a+125b+25c+5d=70\ \ \ \ \ |\cdot\frac15\\}\)
\(\displaystyle{ 125a+25b+5c+d=14\\ 125a+25b+5 \left( \frac12-a \right) +\frac12-b=14\\ 125a+25b+\frac52-5a+\frac12-b=14\\ 120a+24b=11\ \to \ b=\frac{11}{24}-5a\\ d=\frac12-b\ \to \ d=\frac12-\frac{11}{24}+5a\ \to \ d=\frac1{24}+5a\\ \begin{cases} b+d=\frac12 \\ b,d>0 \end{cases} \ \to \ d<\frac12\\ d<\frac12\ \to \ \frac1{24}+5a<\frac12\\ 5a<\frac{11}{24}\ \ \ \ \ |\cdot\frac15\\ a<\frac{11}{120}\\ \begin{cases} 24a\in Z \\ a<\frac{11}{120} \end{cases} \ \to \ a=\frac{m}{24}=\frac{5m}{120} \ \wedge \ m\in N\\ a<\frac{11}{120}\ \to \ \frac{5m}{120}<\frac{11}{120}\ \to \ m<\frac{11}5 \ \to \ m=1\ \vee \ m=2\\ a=\frac1{24}\ \vee \ a=\frac1{12}\\ 12b\in Z\ \to \ 12 \left( \frac{11}{24}-5a \right) \in Z\\ \frac{11}2-60a\in Z\\ 1^o: a=\frac1{12} \\\frac{11}2-60a=\frac{11}2-60\cdot\frac1{12}=\frac{11}2-5\notin Z\\ 2^o: a=\frac1{24}\\ \frac{11}2-60a=\frac{11}2-60\cdot\frac1{24}=\frac{11}2-\frac52=3\in Z\\ a= \blue \frac1{24} \black \\ b=\frac{11}{24}-5a=\frac{11}{24}-\frac5{24}=\frac6{24}=\blue\frac14\black \\ c=\frac12-a=\frac12-\frac1{24}=\blue \frac{11}{24}\black \\ d=\frac12-b=\frac12-\frac14=\blue\frac14}\)
\(\displaystyle{ 125a+25b+5c+d=14\\ 125a+25b+5 \left( \frac12-a \right) +\frac12-b=14\\ 125a+25b+\frac52-5a+\frac12-b=14\\ 120a+24b=11\ \to \ b=\frac{11}{24}-5a\\ d=\frac12-b\ \to \ d=\frac12-\frac{11}{24}+5a\ \to \ d=\frac1{24}+5a\\ \begin{cases} b+d=\frac12 \\ b,d>0 \end{cases} \ \to \ d<\frac12\\ d<\frac12\ \to \ \frac1{24}+5a<\frac12\\ 5a<\frac{11}{24}\ \ \ \ \ |\cdot\frac15\\ a<\frac{11}{120}\\ \begin{cases} 24a\in Z \\ a<\frac{11}{120} \end{cases} \ \to \ a=\frac{m}{24}=\frac{5m}{120} \ \wedge \ m\in N\\ a<\frac{11}{120}\ \to \ \frac{5m}{120}<\frac{11}{120}\ \to \ m<\frac{11}5 \ \to \ m=1\ \vee \ m=2\\ a=\frac1{24}\ \vee \ a=\frac1{12}\\ 12b\in Z\ \to \ 12 \left( \frac{11}{24}-5a \right) \in Z\\ \frac{11}2-60a\in Z\\ 1^o: a=\frac1{12} \\\frac{11}2-60a=\frac{11}2-60\cdot\frac1{12}=\frac{11}2-5\notin Z\\ 2^o: a=\frac1{24}\\ \frac{11}2-60a=\frac{11}2-60\cdot\frac1{24}=\frac{11}2-\frac52=3\in Z\\ a= \blue \frac1{24} \black \\ b=\frac{11}{24}-5a=\frac{11}{24}-\frac5{24}=\frac6{24}=\blue\frac14\black \\ c=\frac12-a=\frac12-\frac1{24}=\blue \frac{11}{24}\black \\ d=\frac12-b=\frac12-\frac14=\blue\frac14}\)