Taylor series 2

Petrus

Well-known member
Calculate $$\displaystyle f^{(18)}(0)$$ if $$\displaystyle f(x)=x^2 \ln(1+9x)$$
if we start with $$\displaystyle ln(1+9x)$$ and ignore $$\displaystyle x^2$$ we can calculate that
$$\displaystyle f'(x)= \frac{9}{1+9x} <=> f'(0)=9$$
$$\displaystyle f''(x)= \frac{9^2}{(1+9x)^2} <=> f'(0)=9^2$$
.
.
.
$$\displaystyle f^{n}(x)= \frac{9^n}{(1+9x)^n} <=> f'(0)=9^n$$
how does it work after? Don't we have to use product rule?

Regards,
$$\displaystyle |\pi\rangle$$

Klaas van Aarsen

MHB Seeker
Staff member
Calculate $$\displaystyle f^{(18)}(0)$$ if $$\displaystyle f(x)=x^2 \ln(1+9x)$$
if we start with $$\displaystyle ln(1+9x)$$ and ignore $$\displaystyle x^2$$ we can calculate that
$$\displaystyle f'(x)= \frac{9}{1+9x} <=> f'(0)=9$$
$$\displaystyle f''(x)= \frac{9^2}{(1+9x)^2} <=> f'(0)=9^2$$
.
.
.
$$\displaystyle f^{n}(x)= \frac{9^n}{(1+9x)^n} <=> f'(0)=9^n$$
how does it work after? Don't we have to use product rule?

Regards,
$$\displaystyle |\pi\rangle$$
Can you write down the Taylor expansion of $\ln(1+9x)$?

Petrus

Well-known member
Can you write down the Taylor expansion of $\ln(1+9x)$?
If I got this correct it should be
the same as $$\displaystyle ln(1+x)$$ which my book got it defined.
$$\displaystyle x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}...$$
edit: no it's not.. I am confusing mc lauens polynom and taylor.. My bad

Regards,
$$\displaystyle |\pi\rangle$$

Last edited:

Klaas van Aarsen

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

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

Petrus

Well-known member
Nah, it's not the same.
Either you can apply the Taylor series formula, using a series of $$\displaystyle \frac {x^n} {n!} f^{(n)}(0)$$.
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
$$\displaystyle \ln(1+9x)=9x-\frac{9^2x^2}{2}+\frac{9^3x^3}{3}-\frac{9^4x^4}{4}...$$

Klaas van Aarsen

MHB Seeker
Staff member
I am confusing Maclaurin and Taylor.. I notice it is not the same..
How so?

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

Prove It

Well-known member
MHB Math Helper
Calculate $$\displaystyle f^{(18)}(0)$$ if $$\displaystyle f(x)=x^2 \ln(1+9x)$$
if we start with $$\displaystyle ln(1+9x)$$ and ignore $$\displaystyle x^2$$ we can calculate that
$$\displaystyle f'(x)= \frac{9}{1+9x} <=> f'(0)=9$$
$$\displaystyle f''(x)= \frac{9^2}{(1+9x)^2} <=> f'(0)=9^2$$
.
.
.
$$\displaystyle f^{n}(x)= \frac{9^n}{(1+9x)^n} <=> f'(0)=9^n$$
how does it work after? Don't we have to use product rule?

Regards,
$$\displaystyle |\pi\rangle$$
Notice that if \displaystyle \displaystyle \begin{align*} g(x) = \ln{ \left( 1 + 9x \right) } \end{align*} then \displaystyle \displaystyle \begin{align*} g'(x) = \frac{9}{1 + 9x} \end{align*}.

Now we notice that \displaystyle \displaystyle \begin{align*} \frac{9}{1 + 9x} = 9 \left[ \frac{1}{1 - (-9x)} \right] \end{align*}, which is the closed form of the geometric series \displaystyle \displaystyle \begin{align*} 9 \sum_{k = 0}^{\infty} \left( -9x \right) ^k \end{align*} which is convergent when \displaystyle \displaystyle \begin{align*} \left| -9x \right| < 1 \implies |x| < \frac{1}{9} \end{align*}. So we have

\displaystyle \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*}

So that means \displaystyle \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*} provided \displaystyle \displaystyle \begin{align*} |x| < \frac{1}{9} \end{align*}, and therefore

\displaystyle \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*}

When you compare it to the MacLaurin series \displaystyle \displaystyle \begin{align*} \sum_{k = 0}^{\infty} \frac{f^{(k)}(0)}{k!}\,x^k \end{align*}, we can see that

\displaystyle \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*}

Petrus

Well-known member
How so?

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

Regards,
$$\displaystyle |\pi\rangle$$

Prove It

Well-known member
MHB Math Helper
I love the use of "multiplicate" instead of "multiply"

Klaas van Aarsen

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

Regards,
$$\displaystyle |\pi\rangle$$
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

Well-known member
I love the use of "multiplicate" instead of "multiply"
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 alot better! Many thanks!

Regards,
$$\displaystyle |\pi\rangle$$

ZaidAlyafey

Well-known member
MHB Math Helper
I love the use of "multiplicate" instead of "multiply"
It sounds like "complicate " .