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

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The discussion revolves around finding the 15th derivative of the function f(x) = sin(x^3) at x = 0 using Taylor series. Participants clarify that substituting x^3 into the Taylor series expansion of sin(x) leads to the series for sin(x^3), which includes terms like x^3, x^9, and x^15. The key point is that the 15th term in this expansion corresponds to f^(15)(0)/15!, equating it to 1/5! allows for solving f^(15)(0) as 15!/5!. The conversation emphasizes understanding the Maclaurin series and the differentiation process to grasp the concept better. Overall, the participants are encouraged to continue exploring Taylor series problems for deeper comprehension.
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 alot!:) 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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