Serie de puteri cu coeficienti naturali
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Alin Galatan
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Serie de puteri cu coeficienti naturali
Fie \( a\in (0,1) \) si \( a=0.a_1a_2... \) o scrierea a sa in baza 10. Atunci \( f(x)=\sum_{n=1}^{\infty}a_nx^n \) este o functie rationala (raport de functii polinomiale cu coeficienti intregi) daca si numai daca \( a \) e rational.
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Iulian Cimpean
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Fie\( f(x)=\frac{P(x)}{Q(x)} \). Atunci \( a=f(1/10)=\frac{P(1/10)}{Q(1/10)}\in \mathbb{Q} \).
"<=" Daca a e rational, \( a=0,a_1a_2...a_k(a_{k+1}a_{k+2}...a_{k+t}) \) si seria \( \sum_{n=1}^{\infty}a_nx^n \)are raza de convergenta 1.
Deci\( f(x)=\sum_{n=1}^{\infty}a_nx^n=\sum_{n=1}^{k}a_nx^n + a_{k+1}x^{k+1}\sum_{n=0}^{\infty}x^{nt}+ \)
\( a_{k+2}x^{k+2}\sum_{n=0}^{\infty}x^{nt}+...+a_{k+t}x^{k+t}\sum_{n=0}^{\infty}x^{nt} = \)
\( \sum_{n=1}^{k}a_nx^n +\sum_{n=0}^{\infty}x^{nt}\sum_{i=1}^{t}a_{k+i}x^{k+i} \).
Calculand suma aia infinita, trecand la limita sirul sumelor partiale al carui termen general e o progresie geometrica cu ratia mai mica de 1, apoi aducand la acelasi numitor, rezulta cam ce trebuia.
Sunt in dubii fata de rezolvarea asta, deci astept pareri.
Fie\( f(x)=\frac{P(x)}{Q(x)} \). Atunci \( a=f(1/10)=\frac{P(1/10)}{Q(1/10)}\in \mathbb{Q} \).
"<=" Daca a e rational, \( a=0,a_1a_2...a_k(a_{k+1}a_{k+2}...a_{k+t}) \) si seria \( \sum_{n=1}^{\infty}a_nx^n \)are raza de convergenta 1.
Deci\( f(x)=\sum_{n=1}^{\infty}a_nx^n=\sum_{n=1}^{k}a_nx^n + a_{k+1}x^{k+1}\sum_{n=0}^{\infty}x^{nt}+ \)
\( a_{k+2}x^{k+2}\sum_{n=0}^{\infty}x^{nt}+...+a_{k+t}x^{k+t}\sum_{n=0}^{\infty}x^{nt} = \)
\( \sum_{n=1}^{k}a_nx^n +\sum_{n=0}^{\infty}x^{nt}\sum_{i=1}^{t}a_{k+i}x^{k+i} \).
Calculand suma aia infinita, trecand la limita sirul sumelor partiale al carui termen general e o progresie geometrica cu ratia mai mica de 1, apoi aducand la acelasi numitor, rezulta cam ce trebuia.
Sunt in dubii fata de rezolvarea asta, deci astept pareri.