Quarto esercizio

Esercizio 7.4

Calcolare $\phi _{n}(x)$$\phi _{n}(x)$ per $n=1\dots ,20$$n=1\dots ,20$.

 

Per calcoli precedenti sappiamo che

$$\phi _{1}(x)=x-1$$

$$\phi _{1}(x)=x-1$$

$$\phi _{2}(x)=x+1$$

$$\phi _{2}(x)=x+1$$

$$\phi _{4}(x)=x^{2}+1$$

$$\phi _{4}(x)=x^{2}+1$$

$$\phi _{6}(x)=x^{2}-x+1$$

$$\phi _{6}(x)=x^{2}-x+1$$

$$\phi _{8}(x)=x^{4}+1$$

$$\phi _{8}(x)=x^{4}+1$$

Inoltre, ricordiamo che in generale, per $p$$p$ primo

$$\phi _{p}(x)={\frac {x^{p}-1}{x-1}}=x^{p-1}+x^{p-2}+\dots +x+1$$

$$\phi _{p}(x)={\frac {x^{p}-1}{x-1}}=x^{p-1}+x^{p-2}+\dots +x+1$$

quindi conosciamo anche $\phi _{3}(x),\,\phi _{5}(x),\,\phi _{7}(x),\,\phi _{11}(x),\,\phi _{13}(x),\,\phi _{17}(x),\,\phi _{19}(x)$$\phi _{3}(x),\,\phi _{5}(x),\,\phi _{7}(x),\,\phi _{11}(x),\,\phi _{13}(x),\,\phi _{17}(x),\,\phi _{19}(x)$.

Calcoliamo i polinomi rimanenti:

1. $\phi _{9}={\frac {x^{9}-1}{\phi _{1}(x)*\phi _{3}(x)}}$$\phi _{9}={\frac {x^{9}-1}{\phi _{1}(x)*\phi _{3}(x)}}$e $\phi _{1}(x)*\phi _{3}(x)=x^{3}-1$$\phi _{1}(x)*\phi _{3}(x)=x^{3}-1$, quindi
   $$\phi _{9}(x)={\frac {x^{9}-1}{x^{3}-1}}$$

   $$\phi _{9}(x)={\frac {x^{9}-1}{x^{3}-1}}$$
   Aggiungendo e togliendo $x^{6}$$x^{6}$ al numeratore:
   $$\phi _{9}(x)={\frac {x^{9}-x^{6}+x^{6}-1}{x^{3}-1}}$$

   $$\phi _{9}(x)={\frac {x^{9}-x^{6}+x^{6}-1}{x^{3}-1}}$$
   $$\phi _{9}(x)={\frac {x^{6}(x^{3}-1)+(x^{3}+1)(x^{3}-1)}{x^{3}-1}}$$

   $$\phi _{9}(x)={\frac {x^{6}(x^{3}-1)+(x^{3}+1)(x^{3}-1)}{x^{3}-1}}$$
   $$\phi _{9}(x)={\frac {(x^{6}+x^{3}+1)(x^{3}-1)}{x^{3}-1}}$$

   $$\phi _{9}(x)={\frac {(x^{6}+x^{3}+1)(x^{3}-1)}{x^{3}-1}}$$
   $$\phi _{9}(x)=x^{6}+x^{3}+1$$

   $$\phi _{9}(x)=x^{6}+x^{3}+1$$
2. $\phi _{10}(x)={\frac {x^{10}-1}{\phi _{1}(x)*\phi _{2}(x)*\phi _{5}(x)}}$$\phi _{10}(x)={\frac {x^{10}-1}{\phi _{1}(x)*\phi _{2}(x)*\phi _{5}(x)}}$
   $$={\frac {x^{10}-1}{(x^{5}-1)\phi _{2}(x)}}$$

   $$={\frac {x^{10}-1}{(x^{5}-1)\phi _{2}(x)}}$$
   $$={\frac {(x^{5}-1)(x^{5}+1)}{(x^{5}-1)\phi _{2}(x)}}$$

   $$={\frac {(x^{5}-1)(x^{5}+1)}{(x^{5}-1)\phi _{2}(x)}}$$
   $$={\frac {x^{5}+1}{x+1}}$$

   $$={\frac {x^{5}+1}{x+1}}$$
   Eseguo la divisione:
   $${\frac {x^{5}+1}{x+1}}=$$

   $${\frac {x^{5}+1}{x+1}}=$$
   $$q_{1}=x^{4},\,r_{1}=x^{5}+1-x^{4}(x+1)=-x^{4}+1$$

   $$q_{1}=x^{4},\,r_{1}=x^{5}+1-x^{4}(x+1)=-x^{4}+1$$
   $$q_{2}=-x^{3},\,r_{2}=-x^{4}+1+x^{3}(x+1)=x^{3}+1$$

   $$q_{2}=-x^{3},\,r_{2}=-x^{4}+1+x^{3}(x+1)=x^{3}+1$$
   $$q_{3}=x^{2},\,r_{3}=x^{3}+1-x^{2}(x+1)=-x^{2}+1$$

   $$q_{3}=x^{2},\,r_{3}=x^{3}+1-x^{2}(x+1)=-x^{2}+1$$
   $$q_{4}=-x,\,r_{4}=-x^{2}+1+x(x+1)=x+1$$

   $$q_{4}=-x,\,r_{4}=-x^{2}+1+x(x+1)=x+1$$
   $$q_{5}=1$$

   $$q_{5}=1$$
   Complessivamente,
   $$x^{5}+1=(x+1)*(x^{4}-x^{3}+x^{2}-x+1)$$

   $$x^{5}+1=(x+1)*(x^{4}-x^{3}+x^{2}-x+1)$$
   quindi
   $$\phi _{10}(x)=x^{4}-x^{3}+x^{2}-x+1.$$

   $$\phi _{10}(x)=x^{4}-x^{3}+x^{2}-x+1.$$
3. $\phi _{12}(x)={\frac {x^{12}-1}{\phi _{1}(x)*\phi _{2}(x)*\phi _{3}(x)*\phi _{4}(x)*\phi _{6}(x)}}$$\phi _{12}(x)={\frac {x^{12}-1}{\phi _{1}(x)*\phi _{2}(x)*\phi _{3}(x)*\phi _{4}(x)*\phi _{6}(x)}}$
   $$={\frac {(x^{6}+1)(x^{6}-1)}{\phi _{4}(x)*(x^{6}-1)}}$$

   $$={\frac {(x^{6}+1)(x^{6}-1)}{\phi _{4}(x)*(x^{6}-1)}}$$
   $$={\frac {x^{6}+1}{x^{2}+1}}$$

   $$={\frac {x^{6}+1}{x^{2}+1}}$$
   Eseguo la divisione
   $${\frac {x^{6}+1}{x^{2}+1}}=$$

   $${\frac {x^{6}+1}{x^{2}+1}}=$$
   $$q_{1}=x^{4},\,r_{1}=x^{6}+1-x^{4}(x^{2}+1)=-x^{4}+1$$

   $$q_{1}=x^{4},\,r_{1}=x^{6}+1-x^{4}(x^{2}+1)=-x^{4}+1$$
   $$q_{2}=-x^{2},\,r_{2}=-x^{4}+1+x^{2}(x^{2}+1)=x^{2}+1$$

   $$q_{2}=-x^{2},\,r_{2}=-x^{4}+1+x^{2}(x^{2}+1)=x^{2}+1$$
   $$q_{3}=1,\,r_{3}=0$$

   $$q_{3}=1,\,r_{3}=0$$
   Quindi
   $$x^{6}+1=(x^{4}-x^{2}+1)*(x^{2}+1)$$

   $$x^{6}+1=(x^{4}-x^{2}+1)*(x^{2}+1)$$
   $$\longrightarrow \,\phi _{12}(x)=x^{4}-x^{2}+1$$

   $$\longrightarrow \,\phi _{12}(x)=x^{4}-x^{2}+1$$
4. $\phi _{14}={\frac {x^{14}-1}{\phi _{1}(x)*\phi _{2}(x)*\phi _{7}(x)}}$$\phi _{14}={\frac {x^{14}-1}{\phi _{1}(x)*\phi _{2}(x)*\phi _{7}(x)}}$
   $$={\frac {x^{7}+1}{\phi _{2}(x)}}$$

   $$={\frac {x^{7}+1}{\phi _{2}(x)}}$$
   $$={\frac {x^{7}+1}{x+1}}$$

   $$={\frac {x^{7}+1}{x+1}}$$
   e questa divisione è simile a quella per il calcolo di $\phi _{10}(x)$$\phi _{10}(x)$, quindi
   $$\phi _{14}(x)=x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+1$$

   $$\phi _{14}(x)=x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+1$$
5. $\phi _{15}(x)={\frac {x^{15}-1}{\phi _{1}(x)*\phi _{3}(x)*\phi _{5}(x)}}$$\phi _{15}(x)={\frac {x^{15}-1}{\phi _{1}(x)*\phi _{3}(x)*\phi _{5}(x)}}$e $\phi _{1}(x)*\phi _{5}(x)=x^{5}-1$$\phi _{1}(x)*\phi _{5}(x)=x^{5}-1$, mentre $\phi _{3}(x)=x^{2}+x+1$$\phi _{3}(x)=x^{2}+x+1$, quindi
   $$={\frac {x^{15}-1}{(x^{2}+x+1)(x^{5}-1)}}$$

   $$={\frac {x^{15}-1}{(x^{2}+x+1)(x^{5}-1)}}$$
   Eseguo la divisione:
   $${\frac {x^{15}-1}{x^{5}-1}}=$$

   $${\frac {x^{15}-1}{x^{5}-1}}=$$
   $$q_{1}=x^{10},\,r_{1}=x^{15}-1-x^{10}(x^{5}-1)=x^{10}-1$$

   $$q_{1}=x^{10},\,r_{1}=x^{15}-1-x^{10}(x^{5}-1)=x^{10}-1$$
   $$q_{2}=x^{5},\,r_{2}=x^{10}-1-x^{5}(x^{5}-1)=x^{5}-1$$

   $$q_{2}=x^{5},\,r_{2}=x^{10}-1-x^{5}(x^{5}-1)=x^{5}-1$$
   $$q_{3}=1,\,r_{3}=0$$

   $$q_{3}=1,\,r_{3}=0$$
   quindi
   $$x^{15}-1=(x^{10}+x^{5}+1)(x^{5}-1)$$

   $$x^{15}-1=(x^{10}+x^{5}+1)(x^{5}-1)$$
   $$\longrightarrow \,\phi _{15}(x)={\frac {x^{10}+x^{5}+1}{x^{2}+x+1}}$$

   $$\longrightarrow \,\phi _{15}(x)={\frac {x^{10}+x^{5}+1}{x^{2}+x+1}}$$
   Ora eseguo la divisione:
   $${\frac {x^{10}+x^{5}+1}{x^{2}+x+1}}=$$

   $${\frac {x^{10}+x^{5}+1}{x^{2}+x+1}}=$$
   $$q_{1}=x^{8},\,r_{1}=x^{10}+x^{5}+1-x^{8}(x^{2}+x+1)=-x^{9}-x^{8}+x^{5}+1$$

   $$q_{1}=x^{8},\,r_{1}=x^{10}+x^{5}+1-x^{8}(x^{2}+x+1)=-x^{9}-x^{8}+x^{5}+1$$
   $$q_{2}=-x^{7},\,r_{2}=-x^{9}-x^{8}+x^{5}+1+x^{7}(x^{2}+x+1)=x^{7}+x^{5}+1$$

   $$q_{2}=-x^{7},\,r_{2}=-x^{9}-x^{8}+x^{5}+1+x^{7}(x^{2}+x+1)=x^{7}+x^{5}+1$$
   $$q_{3}=x^{5},\,r_{3}=x^{7}+x^{5}+1-x^{5}(x^{2}+x+1)=-x^{6}+1$$

   $$q_{3}=x^{5},\,r_{3}=x^{7}+x^{5}+1-x^{5}(x^{2}+x+1)=-x^{6}+1$$
   $$q_{4}=-x^{4},\,r_{4}=-x^{6}+1+x^{4}(x^{2}+x+1)=x^{5}+x^{4}+1$$

   $$q_{4}=-x^{4},\,r_{4}=-x^{6}+1+x^{4}(x^{2}+x+1)=x^{5}+x^{4}+1$$
   $$q_{5}=x^{3},\,r_{5}=x^{5}+x^{4}+1-x^{3}(x^{2}+x+1)=-x^{3}+1$$

   $$q_{5}=x^{3},\,r_{5}=x^{5}+x^{4}+1-x^{3}(x^{2}+x+1)=-x^{3}+1$$
   $$q_{6}=-x,\,r_{6}=-x^{3}+1+x(x^{2}+x+1)=x^{2}+x+1$$

   $$q_{6}=-x,\,r_{6}=-x^{3}+1+x(x^{2}+x+1)=x^{2}+x+1$$
   $$q_{7}=1,\,r_{7}=0$$

   $$q_{7}=1,\,r_{7}=0$$
   Quindi
   $$x^{10}+x^{5}+1=(x^{2}+x+1)(x^{8}-x^{7}+x^{5}-x^{4}+x^{3}-x+1)$$

   $$x^{10}+x^{5}+1=(x^{2}+x+1)(x^{8}-x^{7}+x^{5}-x^{4}+x^{3}-x+1)$$
   $$\longrightarrow \,\phi _{15}(x)=x^{8}-x^{7}+x^{5}-x^{4}+x^{3}-x+1$$

   $$\longrightarrow \,\phi _{15}(x)=x^{8}-x^{7}+x^{5}-x^{4}+x^{3}-x+1$$
6. $\phi _{16}(x)={\frac {x^{16}-1}{[\phi _{1}(x)*\phi _{2}(x)*\phi _{4}(x)]*\phi _{8}(x)}}$$\phi _{16}(x)={\frac {x^{16}-1}{[\phi _{1}(x)*\phi _{2}(x)*\phi _{4}(x)]*\phi _{8}(x)}}$Il termine tra parentesi quadra è $x^{4}-1$$x^{4}-1$, e uso l'espressione di $\phi _{8}(x)$$\phi _{8}(x)$:
   $$={\frac {(x^{8}+1)(x^{4}+1)(x^{4}-1)}{(x^{4}-1)(x^{4}+1)}}$$

   $$={\frac {(x^{8}+1)(x^{4}+1)(x^{4}-1)}{(x^{4}-1)(x^{4}+1)}}$$
   $$=x^{8}+1$$

   $$=x^{8}+1$$
7. $\phi _{18}(x)={\frac {(x^{9}-1)*(x^{9}+1)}{\phi _{1}(x)*\phi _{2}(x)*\phi _{3}(x)*\phi _{6}(x)*\phi _{9}(x)}}$$\phi _{18}(x)={\frac {(x^{9}-1)*(x^{9}+1)}{\phi _{1}(x)*\phi _{2}(x)*\phi _{3}(x)*\phi _{6}(x)*\phi _{9}(x)}}$
   $$={\frac {(x^{9}-1)*(x^{9}+1)}{[\phi _{1}(x)*\phi _{3}(x)*\phi _{9}(x)]*\phi _{2}(x)*\phi _{6}(x)}}$$

   $$={\frac {(x^{9}-1)*(x^{9}+1)}{[\phi _{1}(x)*\phi _{3}(x)*\phi _{9}(x)]*\phi _{2}(x)*\phi _{6}(x)}}$$
   $$={\frac {x^{9}+1}{(x^{2}-x+1)(x+1)}}$$

   $$={\frac {x^{9}+1}{(x^{2}-x+1)(x+1)}}$$
   Eseguo la divisione:
   $${\frac {x^{9}+1}{x^{2}-x+1}}=$$

   $${\frac {x^{9}+1}{x^{2}-x+1}}=$$
   $$q_{1}=x^{7},\,r_{1}=x^{9}+1-x^{7}(x^{2}-x+1)=x^{8}-x^{7}+1$$

   $$q_{1}=x^{7},\,r_{1}=x^{9}+1-x^{7}(x^{2}-x+1)=x^{8}-x^{7}+1$$
   $$q_{2}=x^{6},\,r_{2}=x^{8}-x^{7}+1-x^{6}(x^{2}-x+1)=-x^{6}+1$$

   $$q_{2}=x^{6},\,r_{2}=x^{8}-x^{7}+1-x^{6}(x^{2}-x+1)=-x^{6}+1$$
   $$q_{3}=-x^{4},\,r_{3}=-x^{6}+1+x^{4}(x^{2}-x+1)=-x^{5}+x^{4}+1$$

   $$q_{3}=-x^{4},\,r_{3}=-x^{6}+1+x^{4}(x^{2}-x+1)=-x^{5}+x^{4}+1$$
   $$q_{4}=-x^{3},\,r_{4}=-x^{5}+x^{4}+1+x^{3}(x^{2}-x+1)=x^{3}+1$$

   $$q_{4}=-x^{3},\,r_{4}=-x^{5}+x^{4}+1+x^{3}(x^{2}-x+1)=x^{3}+1$$
   $$q_{5}=x,\,r_{5}=x^{3}+1-x(x^{2}-x+1)=x^{2}-x+1$$

   $$q_{5}=x,\,r_{5}=x^{3}+1-x(x^{2}-x+1)=x^{2}-x+1$$
   $$q_{6}=1,\,r_{6}=0$$

   $$q_{6}=1,\,r_{6}=0$$
   quindi
   $$x^{9}+1=(x^{2}-x+1)(x^{7}+x^{6}-x^{4}-x^{3}+x+1)$$

   $$x^{9}+1=(x^{2}-x+1)(x^{7}+x^{6}-x^{4}-x^{3}+x+1)$$
   $$\phi _{18}(x)={\frac {x^{7}+x^{6}-x^{4}-x^{3}+x+1}{x+1}}$$

   $$\phi _{18}(x)={\frac {x^{7}+x^{6}-x^{4}-x^{3}+x+1}{x+1}}$$
   $$\phi _{18}(x)={\frac {x^{6}(x+1)-x^{3}(x+1)+x+1}{x+1}}$$

   $$\phi _{18}(x)={\frac {x^{6}(x+1)-x^{3}(x+1)+x+1}{x+1}}$$
   $$\phi _{18}(x)={\frac {(x^{6}-x^{3}+1)(x+1)}{x+1}}$$

   $$\phi _{18}(x)={\frac {(x^{6}-x^{3}+1)(x+1)}{x+1}}$$
   $$\phi _{18}(x)=x^{6}-x^{3}+1$$

   $$\phi _{18}(x)=x^{6}-x^{3}+1$$
8. $\phi _{20}(x)={\frac {(x^{10}-1)(x^{10}+1)}{\phi _{1}(x)*\phi _{2}(x)*\phi _{4}(x)*\phi _{5}(x)*\phi _{10}(x)}}$$\phi _{20}(x)={\frac {(x^{10}-1)(x^{10}+1)}{\phi _{1}(x)*\phi _{2}(x)*\phi _{4}(x)*\phi _{5}(x)*\phi _{10}(x)}}$
   $$\phi _{20}(x)={\frac {(x^{10}-1)(x^{10}+1)}{\phi _{2}(x)*\phi _{4}(x)*\phi _{10}(x)}}$$

   $$\phi _{20}(x)={\frac {(x^{10}-1)(x^{10}+1)}{\phi _{2}(x)*\phi _{4}(x)*\phi _{10}(x)}}$$
   e $\phi _{4}(x)=x^{2}+1$$\phi _{4}(x)=x^{2}+1$, e $\phi _{10}(x)={\frac {x^{5}+1}{x+1}}$$\phi _{10}(x)={\frac {x^{5}+1}{x+1}}$, allora
   $$={\frac {(x^{5}+1)(x^{10}+1)}{(x^{5}+1)/(x+1)*(x+1)*(x^{2}+1)}}$$

   $$={\frac {(x^{5}+1)(x^{10}+1)}{(x^{5}+1)/(x+1)*(x+1)*(x^{2}+1)}}$$
   $$={\frac {x^{10}+1}{x^{2}+1}}$$

   $$={\frac {x^{10}+1}{x^{2}+1}}$$
   $$=x^{8}-x^{6}+x^{4}-x^{2}+1$$

   $$=x^{8}-x^{6}+x^{4}-x^{2}+1$$
   (il risultato si ottiene facendo la sostituzione $y=x^{2}$$y=x^{2}$, infatti abbiamo già eseguito la divisione ${\frac {y^{5}+1}{y+1}}$${\frac {y^{5}+1}{y+1}}$)