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Taylor series

Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,
I am working with Taylor series pretty new for me, I am working with a problem from my book
\(\displaystyle f(x)=\sin(x^3)\), find \(\displaystyle f^{(15)}(0).\)
I know that \(\displaystyle \sin(x) = 1-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}....Rest\)
How does this work now =S?

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

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,887
Hello MHB,
I am working with Taylor series pretty new for me, I am working with a problem from my book
\(\displaystyle f(x)=\sin(x^3)\), find \(\displaystyle f^{(15)}(0).\)
I know that \(\displaystyle \sin(x) = 1-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}....Rest\)
How does this work now =S?

Regards,
\(\displaystyle |\pi\rangle\)
Hey Petrus!

What do you get if you substitute $x^3$ in your expansion?
 

chisigma

Well-known member
Feb 13, 2012
1,704
Hello MHB,
I am working with Taylor series pretty new for me, I am working with a problem from my book
\(\displaystyle f(x)=\sin(x^3)\), find \(\displaystyle f^{(15)}(0).\)
I know that \(\displaystyle \sin(x) = 1-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}....Rest\)
How does this work now =S?
Regards,
\(\displaystyle |\pi\rangle\)
The general expression of the Taylor [or McLaurin] formula is...

$\displaystyle f(x)= \sum_{n=0}^{\infty} \frac{f^{(n)} (0)}{n!}\ x^{n}$ (1)

If in the expansion of $\displaystyle \sin x$ You set $x^{3}$ instead of $x$ You have...

$\displaystyle \sin (x^{3}) = 1 - \frac{x^{9}}{3!} + \frac{x^{15}}{5!} - ...$ (2)

Comparing (1) and (2) what do You conclude?...

Kind regards

$\chi$ $\sigma$
 

Petrus

Well-known member
Feb 21, 2013
739
The general expression of the Taylor [or McLaurin] formula is...

$\displaystyle f(x)= \sum_{n=0}^{\infty} \frac{f^{(n)} (0)}{n!}\ x^{n}$ (1)

If in the expansion of $\displaystyle \sin x$ You set $x^{3}$ instead of $x$ You have...

$\displaystyle \sin (x^{3}) = 1 - \frac{x^{9}}{3!} + \frac{x^{15}}{5!} - ...$ (2)

Comparing (1) and (2) what do You conclude?...

Kind regards

$\chi$ $\sigma$
Hello,
I did type it wrong.. It should be \(\displaystyle \sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}...\)
so we got
\(\displaystyle \sin(x^3)=x^3-\frac{x^9}{3!}+\frac{x^{15}}{5!}\)
hmm.. I think I leak the basic with Taylor series.. I will try read the basic more and will be back.

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

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
Since you have found \(\displaystyle \displaystyle \begin{align*} f(x) = x^3 - \frac{x^9}{3!} + \frac{x^{15}}{5!} - \dots + \dots \end{align*}\), surely you can see that this is the same as

\(\displaystyle \displaystyle \begin{align*} 0 + 0x + 0x^2 + 1x^3 + 0x^4 + 0x^5 + 0x^6 + 0x^7 + 0x^8 \\ -\frac{1}{3!}x^9 + 0x^{10} + 0x^{11} + 0x^{12} + 0x^{13} + 0x^{14} + \frac{1}{5!}x^{15} + \dots \end{align*}\)

How does this relate to the MacLaurin Series formula \(\displaystyle \displaystyle \begin{align*} f(x) = \frac{f(0)}{0!} + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots \end{align*}\)?
 

Petrus

Well-known member
Feb 21, 2013
739
Since you have found \(\displaystyle \displaystyle \begin{align*} f(x) = x^3 - \frac{x^9}{3!} + \frac{x^{15}}{5!} - \dots + \dots \end{align*}\), surely you can see that this is the same as

\(\displaystyle \displaystyle \begin{align*} 0 + 0x + 0x^2 + 1x^3 + 0x^4 + 0x^5 + 0x^6 + 0x^7 + 0x^8 \\ -\frac{1}{3!}x^9 + 0x^{10} + 0x^{11} + 0x^{12} + 0x^{13} + 0x^{14} + \frac{1}{5!}x^{15} + \dots \end{align*}\)

How does this relate to the MacLaurin Series formula \(\displaystyle \displaystyle \begin{align*} f(x) = \frac{f(0)}{0!} + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots \end{align*}\)?
Hello,

Hmm this taylor series is kinda confusing (Worried)
If I understand correct \(\displaystyle C=\frac{f^{(n)}(a)}{n!} = \frac{f^{(15)}(0)}{15!}\) and we se that \(\displaystyle f^{15)}=1-\frac{1}{3!}+\frac{1}{5!}\) is that correct?

Regards,
\(\displaystyle |\pi\rangle\)
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,887
Suppose you differentiate \(\displaystyle \frac {x^{15}}{5!}\) once.
What do you get?
And a second time?
... and 15 times?
 

Petrus

Well-known member
Feb 21, 2013
739
Suppose you differentiate \(\displaystyle \frac {x^{15}}{5!}\) once.
What do you get?
And a second time?
... and 15 times?
\(\displaystyle \frac{n!x^{15-n}}{5!}\)

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

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,887
\(\displaystyle \frac{n!x^{15-n}}{5!}\)

Regards,
\(\displaystyle |\pi\rangle\)
Hmm... that is not once or twice, nor 15 times.
Oh! It is for "n" times!

Can you fill in n=15 then?

Similarly what do you get if you differentiate \(\displaystyle \frac {x^9} {3!}\) also 15 times?
And \(\displaystyle \frac {x^{21}} {7!}\)?
 

Petrus

Well-known member
Feb 21, 2013
739
Hmm... that is not once or twice, nor 15 times.
Oh! It is for "n" times!

Can you fill in n=15 then?

Similarly what do you get if you differentiate \(\displaystyle \frac {x^9} {3!}\) also 15 times?
And \(\displaystyle \frac {x^{21}} {7!}\)?
Ohh I thought you wanted n times:p sorry
\(\displaystyle \frac{15!}{5!}\)
\(\displaystyle \frac {x^9} {3!}= 0\)
\(\displaystyle \frac {x^{21}} {7!}= \frac{21!x^{6}}{6!7!}\)

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

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,887
Okay!
As you can see you can do this will all the terms in your expansion.
What do you get if you substitute x=0?
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
You're all making life difficult upon yourselves. The "n"th term of a MacLaurin Series is \(\displaystyle \displaystyle \frac{f^{(n)}(0)}{n!}\,x^n \), so the 15th term will be \(\displaystyle \displaystyle \frac{f^{(15)}(0)}{15!}\,x^{15}\). Equating these gives

\(\displaystyle \displaystyle \frac{f^{(15)}(0)}{15!}\,x^{15} = \frac{1}{5!}\,x^{15}\).

Now solve for \(\displaystyle \displaystyle f^{(15)}(0)\).
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,887
You're all making life difficult upon yourselves.
I'm trying to show why the MacLaurin formula works at all.
IMO that's more useful than just applying a formula, that you will first have to learn by heart.
 

Petrus

Well-known member
Feb 21, 2013
739
Okay!
As you can see you can do this will all the terms in your expansion.
What do you get if you substitute x=0?
\(\displaystyle \frac{15!}{5!}\)

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

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,887
\(\displaystyle \frac{15!}{5!}\)

Regards,
\(\displaystyle |\pi\rangle\)
There you go! ;)

So if you differentiate the Taylor expansion of $\sin x^3$ for 15 times, you'll get \(\displaystyle \frac{15!}{5!}\) at $x=0$.

Perhaps you can take a look at the MacLaurin series now, as $\chi\ \sigma$ and Prove It suggested, and check out the term that contains $f^{(15)}(0)$.
Btw, the MacLaurin series is the same as the Taylor series, just expanded around 0, instead of around a generic constant.
 

Petrus

Well-known member
Feb 21, 2013
739
There you go! ;)

So if you differentiate the Taylor expansion of $\sin x^3$ for 15 times, you'll get \(\displaystyle \frac{15!}{5!}\) at $x=0$.

Perhaps you can take a look at the MacLaurin series now, as $\chi\ \sigma$ and Prove It suggested, and check out the term that contains $f^{(15)}(0)$.
Btw, the MacLaurin series is the same as the Taylor series, just expanded around 0, instead of around a generic constant.
I start to understand more, thanks alot!:) I will keep work with more kind of this problem :)

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

chisigma

Well-known member
Feb 13, 2012
1,704
\(\displaystyle \frac{15!}{5!}\)

Regards,
\(\displaystyle |\pi\rangle\)
A 'very big number' of the order of $10^{10}$... that did discourage You?...

Kind regards

$\chi$ $\sigma$
 

Petrus

Well-known member
Feb 21, 2013
739
A 'very big number' of the order of $10^{10}$... that did discourage You?...

Kind regards

$\chi$ $\sigma$
Hello Chisigma,
What do you exactly mean with that?

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

chisigma

Well-known member
Feb 13, 2012
1,704
Hello Chisigma,
What do you exactly mean with that?

Regards,
\(\displaystyle |\pi\rangle\)
Just curiosity... no more!:D...

Kind regards

$\chi$ $\sigma$