Calculate Taylor Series of f^{(18)}(0)

[Petrus]

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
$$f'(x)= \frac{9}{1+9x} <=> f'(0)=9$$
$$f''(x)= \frac{9^2}{(1+9x)^2} <=> f'(0)=9^2$$
.
.
.
$$f^{n}(x)= \frac{9^n}{(1+9x)^n} <=> f'(0)=9^n $$
how does it work after? Don't we have to use product rule?

Regards,
$$|\pi\rangle$$

 

[I like Serena]

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
$$f'(x)= \frac{9}{1+9x} <=> f'(0)=9$$
$$f''(x)= \frac{9^2}{(1+9x)^2} <=> f'(0)=9^2$$
.
.
.
$$f^{n}(x)= \frac{9^n}{(1+9x)^n} <=> f'(0)=9^n $$
how does it work after? Don't we have to use product rule?

Regards,
$$|\pi\rangle$$

Can you write down the Taylor expansion of $\ln(1+9x)$?

 

[Petrus]

Can you write down the Taylor expansion of $\ln(1+9x)$?

If I got this correct it should be
the same as $$ln(1+x)$$ which my book got it defined.
$$x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}...$$
edit: no it's not.. I am confusing mc lauens polynom and taylor.. My bad

Regards,
$$|\pi\rangle$$

 

[I like Serena]

If I got this correct it should be
the same as $$ln(1+x)$$ which my book got it defined.
$$x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}...$$

Regards,
$$|\pi\rangle$$

Nah, it's not the same.
Either you can apply the Taylor series formula, using a series of $$\frac {x^n} {n!} f^{(n)}(0)$$.
Or you can use the expansion you already have and substitute $x \to 9x$.

 

[Petrus]

Nah, it's not the same.
Either you can apply the Taylor series formula, using a series of $$\frac {x^n} {n!} f^{(n)}(0)$$.
Or you can use the expansion you already have and substitute $x \to 9x$.

I am confusing Maclaurin and Taylor.. I notice it is not the same..
We got
$$\ln(1+9x)=9x-\frac{9^2x^2}{2}+\frac{9^3x^3}{3}-\frac{9^4x^4}{4}...$$

 

[I like Serena]

I am confusing Maclaurin and Taylor.. I notice it is not the same..

How so?

We got
$$\ln(1+9x)=9x-\frac{9^2x^2}{2}+\frac{9^3x^3}{3}-\frac{9^4x^4}{4}...$$

Yep!
So...?

 

[Prove It]

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
$$f'(x)= \frac{9}{1+9x} <=> f'(0)=9$$
$$f''(x)= \frac{9^2}{(1+9x)^2} <=> f'(0)=9^2$$
.
.
.
$$f^{n}(x)= \frac{9^n}{(1+9x)^n} <=> f'(0)=9^n $$
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]

 

[Petrus]

How so?
Yep!
So...?

I multiplicate with $$x^2$$? So I got
$$9x^3- \frac{9^2x^4}{2}+\frac{9^3x^6}{3}...-\frac{9^{16}x^{18}}{16}$$
is this correct? I just did think wrong with maclaurin etc..
is that correct?

Regards,
$$|\pi\rangle$$

 

[Prove It]

I love the use of "multiplicate" instead of "multiply" :P

 

[I like Serena]

I multiplicate with $$x^2$$? So I got
$$9x^3- \frac{9^2x^4}{2}+\frac{9^3x^6}{3}...-\frac{9^{16}x^{18}}{16}$$
is this correct? I just did think wrong with maclaurin etc..
is that correct?

Regards,
$$|\pi\rangle$$

Yes, that is correct, although the series does continue after the last term.

So now you can either differentiate 18 times, or you can match the series with a Taylor/MacLaurin series.

 

[Petrus]

I love the use of "multiplicate" instead of "multiply" :P

I am trying to improve the English dictionary;)

Yes, that is correct, although the series does continue after the last term.

So now you can either differentiate 18 times, or you can match the series with a Taylor/MacLaurin series.

Thanks for the help from you both!:) This did take a while understanding Talyor series and now I feel a lot better:)! Many thanks!

Regards,
$$|\pi\rangle$$

 

[alyafey22]

I love the use of "multiplicate" instead of "multiply" :P

It sounds like "complicate " :).

 

[MarkFL]

I am trying to improve the English dictionary;)...

(Rofl) I will add this to my vocabulary. (Rock)