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

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Discussion Overview

The discussion revolves around finding the 15th derivative at zero of the function $$f(x)=\sin(x^3)$$ using Taylor series. Participants explore the application of Taylor series, specifically the MacLaurin series, and how to derive the necessary coefficients from the series expansion.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant expresses confusion about how to apply the Taylor series to $$f(x)=\sin(x^3)$$ and seeks clarification on the process.
  • Another participant suggests substituting $$x^3$$ into the Taylor series expansion of $$\sin(x)$$ to derive the series for $$\sin(x^3)$$.
  • A participant corrects an earlier statement about the series expansion of $$\sin(x)$$ and provides the correct form, leading to a revised series for $$\sin(x^3)$$.
  • Discussion includes the relationship between the derived series and the MacLaurin series formula, with participants attempting to relate the coefficients to the derivatives at zero.
  • Some participants explore the implications of differentiating terms in the series multiple times and how that affects the coefficients.
  • There is a suggestion to equate the 15th term of the MacLaurin series with the derived series to solve for $$f^{(15)}(0)$$.
  • One participant emphasizes understanding the underlying principles of the MacLaurin series rather than just applying formulas.
  • Participants discuss the numerical value of $$\frac{15!}{5!}$$ and its significance in the context of the problem.

Areas of Agreement / Disagreement

Participants generally agree on the method of using Taylor series to find the derivatives, but there is some confusion and uncertainty about the application and interpretation of the series. Multiple viewpoints on the best approach to understand the series and its terms are present, indicating that the discussion remains somewhat unresolved.

Contextual Notes

Some participants express uncertainty about the basic concepts of Taylor series, indicating potential gaps in foundational knowledge. The discussion also reflects varying levels of familiarity with the topic, which may influence the clarity of the arguments presented.

Who May Find This Useful

This discussion may be useful for students learning about Taylor series, particularly those seeking to understand the application of series expansions in calculus and their derivatives.

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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