Petrus said:
Calculate $$f^{(18)}(0)$$ if $$f(x)=x^2 \ln(1+9x)$$
if we start with $$ln(1+9x)$$ and ignore $$x^2$$ we can calculate that
$$f'(x)= \frac{9}{1+9x} <=> f'(0)=9$$
$$f''(x)= \frac{9^2}{(1+9x)^2} <=> f'(0)=9^2$$
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$$f^{n}(x)= \frac{9^n}{(1+9x)^n} <=> f'(0)=9^n $$
how does it work after? Don't we have to use product rule?
Regards,
$$|\pi\rangle$$
Notice that if [math]\displaystyle \begin{align*} g(x) = \ln{ \left( 1 + 9x \right) } \end{align*}[/math] then [math]\displaystyle \begin{align*} g'(x) = \frac{9}{1 + 9x} \end{align*}[/math].
Now we notice that [math]\displaystyle \begin{align*} \frac{9}{1 + 9x} = 9 \left[ \frac{1}{1 - (-9x)} \right] \end{align*}[/math], which is the closed form of the geometric series [math]\displaystyle \begin{align*} 9 \sum_{k = 0}^{\infty} \left( -9x \right) ^k \end{align*}[/math] which is convergent when [math]\displaystyle \begin{align*} \left| -9x \right| < 1 \implies |x| < \frac{1}{9} \end{align*}[/math]. So we have
[math]\displaystyle \begin{align*} g'(x) &= 9\sum_{k = 0}^{\infty} \left( -9x \right)^k \\ &= 9\sum_{k = 0}^{\infty} \left( -9 \right) ^k \, x^k \\ \\ g(x) &= \int{ 9\sum_{k = 0}^{\infty} \left( -9 \right) ^k \, x^k \, dx} \\ &= 9\sum_{k = 0}^{\infty} \int{ \left( -9 \right) ^k \, x^k \,dx } \\ &= 9\sum_{k=0}^{\infty}{ \frac{\left( -9 \right) ^k}{k + 1} \, x^{k + 1} } \end{align*}[/math]
So that means [math]\displaystyle \begin{align*} \ln{ \left( 1 + 9x \right) } = 9 \sum_{k = 0}^{\infty} { \frac{ \left( -9 \right) ^k }{k + 1} \, x^{k + 1} } \end{align*}[/math] provided [math]\displaystyle \begin{align*} |x| < \frac{1}{9} \end{align*}[/math], and therefore
[math]\displaystyle \begin{align*} f(x) &= x^2 \ln{ \left( 1 + 9x \right) } \\ &= x^2 \cdot 9\sum_{k = 0}^{\infty} { \frac{ \left( -9 \right) ^k }{ k + 1 }\, x^{ k + 1} } \textrm{ where } |x| < \frac{1}{9} \\ &= 9\sum_{k = 0}^{\infty} { \frac{ \left( -9 \right) ^k }{ k + 1 } \, x^{k + 3} } \textrm{ where } |x| < \frac{1}{9} \end{align*}[/math]
When you compare it to the MacLaurin series [math]\displaystyle \begin{align*} \sum_{k = 0}^{\infty} \frac{f^{(k)}(0)}{k!}\,x^k \end{align*}[/math], we can see that
[math]\displaystyle \begin{align*} \frac{ f^{(18)}(0) }{ 18! } \, x^{18} &= 9 \left[ \frac{ \left( -9 \right) ^{15} }{ 16 } \, x^{18} \right] \\ \frac{f^{(18)}(0)}{18!} &= \frac{9 \left( -9 \right) ^{15}}{ 16 } \\ f^{(18)}(0) &= \frac{ 9 \left( -9 \right) ^{15} \, 18!}{ 16 } \end{align*}[/math]