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Taylor series 2

Petrus

Well-known member
Feb 21, 2013
739
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
Mar 5, 2012
8,774
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
Feb 21, 2013
739
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
Mar 5, 2012
8,774
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
Feb 21, 2013
739
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
Mar 5, 2012
8,774

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
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
Feb 21, 2013
739
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
Jan 26, 2012
1,403
I love the use of "multiplicate" instead of "multiply" :p
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
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
Feb 21, 2013
739
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 alot better:)! Many thanks!

Regards,
\(\displaystyle |\pi\rangle\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775