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

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SUMMARY

The forum discussion centers on calculating the 15th derivative of the function $$f(x) = \sin(x^3)$$ at zero, denoted as $$f^{(15)}(0)$$. Participants clarify the Taylor series expansion for $$\sin(x)$$ and substitute $$x^3$$ into the series, resulting in $$\sin(x^3) = x^3 - \frac{x^9}{3!} + \frac{x^{15}}{5!} - \dots$$. The conclusion is that $$f^{(15)}(0) = \frac{15!}{5!}$$, derived from the MacLaurin series, which is a specific case of the Taylor series centered at zero.

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

Regards,
$$|\pi\rangle$$
 
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Petrus said:
Hello MHB,
I am working with Taylor series pretty new for me, I am working with a problem from my book
$$f(x)=\sin(x^3)$$, find $$f^{(15)}(0).$$
I know that $$\sin(x) = 1-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}...Rest$$
How does this work now =S?

Regards,
$$|\pi\rangle$$

Hey Petrus!

What do you get if you substitute $x^3$ in your expansion?
 
Petrus said:
Hello MHB,
I am working with Taylor series pretty new for me, I am working with a problem from my book
$$f(x)=\sin(x^3)$$, find $$f^{(15)}(0).$$
I know that $$\sin(x) = 1-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}...Rest$$
How does this work now =S?
Regards,
$$|\pi\rangle$$

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

$\chi$ $\sigma$
 
chisigma said:
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 $$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}...$$
so we got
$$\sin(x^3)=x^3-\frac{x^9}{3!}+\frac{x^{15}}{5!}$$
hmm.. I think I leak the basic with Taylor series.. I will try read the basic more and will be back.

Regards,
$$|\pi\rangle$$
 
Since you have found [math]\displaystyle \begin{align*} f(x) = x^3 - \frac{x^9}{3!} + \frac{x^{15}}{5!} - \dots + \dots \end{align*}[/math], surely you can see that this is the same as

[math]\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*}[/math]

How does this relate to the MacLaurin Series formula [math]\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*}[/math]?
 
Prove It said:
Since you have found [math]\displaystyle \begin{align*} f(x) = x^3 - \frac{x^9}{3!} + \frac{x^{15}}{5!} - \dots + \dots \end{align*}[/math], surely you can see that this is the same as

[math]\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*}[/math]

How does this relate to the MacLaurin Series formula [math]\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*}[/math]?
Hello,

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

Regards,
$$|\pi\rangle$$
 
Last edited:
Suppose you differentiate $$\frac {x^{15}}{5!}$$ once.
What do you get?
And a second time?
... and 15 times?
 
I like Serena said:
Suppose you differentiate $$\frac {x^{15}}{5!}$$ once.
What do you get?
And a second time?
... and 15 times?
$$\frac{n!x^{15-n}}{5!}$$

Regards,
$$|\pi\rangle$$
 
Petrus said:
$$\frac{n!x^{15-n}}{5!}$$

Regards,
$$|\pi\rangle$$

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 $$\frac {x^9} {3!}$$ also 15 times?
And $$\frac {x^{21}} {7!}$$?
 
  • #10
I like Serena said:
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 $$\frac {x^9} {3!}$$ also 15 times?
And $$\frac {x^{21}} {7!}$$?
Ohh I thought you wanted n times:P sorry
$$\frac{15!}{5!}$$
$$\frac {x^9} {3!}= 0$$
$$\frac {x^{21}} {7!}= \frac{21!x^{6}}{6!7!}$$

Regards,
$$|\pi\rangle$$
 
  • #11
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?
 
  • #12
You're all making life difficult upon yourselves. The "n"th term of a MacLaurin Series is [math]\displaystyle \frac{f^{(n)}(0)}{n!}\,x^n [/math], so the 15th term will be [math]\displaystyle \frac{f^{(15)}(0)}{15!}\,x^{15}[/math]. Equating these gives

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

Now solve for [math]\displaystyle f^{(15)}(0)[/math].
 
  • #13
Prove It said:
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.
 
  • #14
I like Serena said:
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?
$$\frac{15!}{5!}$$

Regards,
$$|\pi\rangle$$
 
  • #15
Petrus said:
$$\frac{15!}{5!}$$

Regards,
$$|\pi\rangle$$

There you go! ;)

So if you differentiate the Taylor expansion of $\sin x^3$ for 15 times, you'll get $$\frac{15!}{5!}$$ 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.
 
  • #16
I like Serena said:
There you go! ;)

So if you differentiate the Taylor expansion of $\sin x^3$ for 15 times, you'll get $$\frac{15!}{5!}$$ 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 a lot!:) I will keep work with more kind of this problem :)

Regards,
$$|\pi\rangle$$
 
  • #17
Petrus said:
$$\frac{15!}{5!}$$

Regards,
$$|\pi\rangle$$

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

Kind regards

$\chi$ $\sigma$
 
  • #18
chisigma said:
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,
$$|\pi\rangle$$
 
  • #19
Petrus said:
Hello Chisigma,
What do you exactly mean with that?

Regards,
$$|\pi\rangle$$

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

Kind regards

$\chi$ $\sigma$
 

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