Question

expanded form of log 15 is​

Answer 1

Expansion :

• Expansion by Taylor's series

$\longrightarrow \: \log(1 + x) = x - \frac{ {x}^{2} }{2} + \frac{ {x}^{3} }{3} - \frac{ {x}^{4} }{4} + ... \: \: \: for \: \: (1 + x) > 0$

$\longrightarrow \: \log(1 - x) = - x - \frac{ {x}^{2} }{2} - \frac{ {x}^{3} }{3} - ... \: \: \: for \: \: (1 - x) > 0$

let's expand this now

$\longrightarrow \: \log(1 + 14) = 14 - \frac{ {14}^{2} }{2} + \frac{ {14}^{3} }{3} - \frac{ {14}^{4} }{4} + ...$

$\bull \: \: \: \: \: \underline{ \underline{ \mathfrak{What \: \: Taylor \: \: Series \: \: says}} \: }_{ \bigstar \star}$

$\longrightarrow f(x) = \: \: { \sum ^{ \infty }}_{n = 0} {f}^{(n)}(a) \frac{ {(x - a)}^{n} }{n!}$