Tackling Taylor Series with f(x)=sin(x^3)

[Petrus]

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

 

[I like Serena]

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]

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]

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]

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]

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

 

[I like Serena]

Suppose you differentiate \(\displaystyle \frac {x^{15}}{5!}\) once.
What do you get?
And a second time?
... and 15 times?

 

[Petrus]

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

 

[I like Serena]

\(\displaystyle \frac{n!x^{15-n}}{5!}\)

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]

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

 

[I like Serena]

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]

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 \(\displaystyle \displaystyle f^{(15)}(0)\).

 

[I like Serena]

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]

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

 

[I like Serena]

\(\displaystyle \frac{15!}{5!}\)

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]

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]

\(\displaystyle \frac{15!}{5!}\)

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

A 'very big number' of the order of $10^{10}$... that did discourage You?...

Kind regards

$\chi$ $\sigma$

 

[Petrus]

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]

Hello Chisigma,
What do you exactly mean with that?

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

Just curiosity... no more!:D...

Kind regards

$\chi$ $\sigma$