Question

Sa se dezvolte in serie Maclaurin functia​

Answer 1

Răspuns:

$f(x)=ln(\frac{1+x}{2x+3}) = ln(1+x) - ln(2x+3) = ln(1+x) - ln 3 (1+\frac{2x}{3}) = ln(1+x) - ln (1+\frac{2x}{3}) - ln 3$

Pe de alta parte, $ln(1+x)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}x^n$, de unde

$ln(1+\frac{2x}{3}) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} (\frac{2x}{3})^n =\sum_{n=1}^{\infty} \frac{(-1)^{n-1}2^n}{n\cdot 3^n} x^n$

Atunci $f(x)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}x^n - \sum_{n=1}^{\infty} \frac{(-1)^{n-1}2^n}{n\cdot 3^n} x^n - ln 3$

$f(x)=-ln 3 + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}3^n + (-1)^n 2^n}{n\cdot 3^n} x^n$