Using the power series of $sin x^3$, the value of $f^{(15)}(0)$ is equal to $kcdot11!$. Find the value of...












-1












$begingroup$


I have the following question:




Using the power series of $sin x^3$, the value of $f^{(15)}(0)$ is equal to $kcdot11!$. Find the value of $k$.




I tried to write the power series using the one from $sin(x)$:
$$sin(x^{3})=sum_{n=1}^{+infty}(-1)^{n}frac{x^{6n+3}}{(2n+1)!}$$



So, since $f^{(15)}(0)$ is related to $x^{15}$, so I got $n=2$. But I really don't know how can I use it.






power-series







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  • $begingroup$
    What's $f{}{}$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 9 '18 at 17:56










  • $begingroup$
    Well $f(x)=sin(x^{3})$.
    $endgroup$
    – mvfs314
    Dec 9 '18 at 17:56


















-1












$begingroup$


I have the following question:




Using the power series of $sin x^3$, the value of $f^{(15)}(0)$ is equal to $kcdot11!$. Find the value of $k$.




I tried to write the power series using the one from $sin(x)$:
$$sin(x^{3})=sum_{n=1}^{+infty}(-1)^{n}frac{x^{6n+3}}{(2n+1)!}$$



So, since $f^{(15)}(0)$ is related to $x^{15}$, so I got $n=2$. But I really don't know how can I use it.






power-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    What's $f{}{}$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 9 '18 at 17:56










  • $begingroup$
    Well $f(x)=sin(x^{3})$.
    $endgroup$
    – mvfs314
    Dec 9 '18 at 17:56
















-1












-1








-1





$begingroup$


I have the following question:




Using the power series of $sin x^3$, the value of $f^{(15)}(0)$ is equal to $kcdot11!$. Find the value of $k$.




I tried to write the power series using the one from $sin(x)$:
$$sin(x^{3})=sum_{n=1}^{+infty}(-1)^{n}frac{x^{6n+3}}{(2n+1)!}$$



So, since $f^{(15)}(0)$ is related to $x^{15}$, so I got $n=2$. But I really don't know how can I use it.






power-series







share|cite|improve this question











$endgroup$




I have the following question:




Using the power series of $sin x^3$, the value of $f^{(15)}(0)$ is equal to $kcdot11!$. Find the value of $k$.




I tried to write the power series using the one from $sin(x)$:
$$sin(x^{3})=sum_{n=1}^{+infty}(-1)^{n}frac{x^{6n+3}}{(2n+1)!}$$



So, since $f^{(15)}(0)$ is related to $x^{15}$, so I got $n=2$. But I really don't know how can I use it.






power-series




power-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 4:34









Nosrati

26.5k62354




26.5k62354










asked Dec 9 '18 at 17:54









mvfs314mvfs314

446211




446211












  • $begingroup$
    What's $f{}{}$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 9 '18 at 17:56










  • $begingroup$
    Well $f(x)=sin(x^{3})$.
    $endgroup$
    – mvfs314
    Dec 9 '18 at 17:56




















  • $begingroup$
    What's $f{}{}$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 9 '18 at 17:56










  • $begingroup$
    Well $f(x)=sin(x^{3})$.
    $endgroup$
    – mvfs314
    Dec 9 '18 at 17:56


















$begingroup$
What's $f{}{}$?
$endgroup$
– Lord Shark the Unknown
Dec 9 '18 at 17:56




$begingroup$
What's $f{}{}$?
$endgroup$
– Lord Shark the Unknown
Dec 9 '18 at 17:56












$begingroup$
Well $f(x)=sin(x^{3})$.
$endgroup$
– mvfs314
Dec 9 '18 at 17:56






$begingroup$
Well $f(x)=sin(x^{3})$.
$endgroup$
– mvfs314
Dec 9 '18 at 17:56












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$begingroup$

Since $dfrac{f^{(15)}(0)}{15!}=dfrac{(-1)^2}{5!}=dfrac1{5!}$, you have $f^{(15)}(0)=dfrac{15!}{5!}$. You can get the value of $k$ from this.






share|cite|improve this answer









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    $begingroup$

    Since $dfrac{f^{(15)}(0)}{15!}=dfrac{(-1)^2}{5!}=dfrac1{5!}$, you have $f^{(15)}(0)=dfrac{15!}{5!}$. You can get the value of $k$ from this.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Since $dfrac{f^{(15)}(0)}{15!}=dfrac{(-1)^2}{5!}=dfrac1{5!}$, you have $f^{(15)}(0)=dfrac{15!}{5!}$. You can get the value of $k$ from this.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Since $dfrac{f^{(15)}(0)}{15!}=dfrac{(-1)^2}{5!}=dfrac1{5!}$, you have $f^{(15)}(0)=dfrac{15!}{5!}$. You can get the value of $k$ from this.






        share|cite|improve this answer









        $endgroup$



        Since $dfrac{f^{(15)}(0)}{15!}=dfrac{(-1)^2}{5!}=dfrac1{5!}$, you have $f^{(15)}(0)=dfrac{15!}{5!}$. You can get the value of $k$ from this.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 9 '18 at 18:01









        José Carlos SantosJosé Carlos Santos

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        156k22125227






























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