# Taylor series

#### Petrus

##### Well-known member
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
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
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
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
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
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 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
Suppose you differentiate $$\displaystyle \frac {x^{15}}{5!}$$ once.
What do you get?
And a second time?
... and 15 times?

#### Petrus

##### Well-known member
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
$$\displaystyle \frac{n!x^{15-n}}{5!}$$

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
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 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
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
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
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
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
$$\displaystyle \frac{15!}{5!}$$

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
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
$$\displaystyle \frac{15!}{5!}$$

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
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
Hello Chisigma,
What do you exactly mean with that?

Regards,
$$\displaystyle |\pi\rangle$$
Just curiosity... no more! ...

Kind regards

$\chi$ $\sigma$