High Order Derivative

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Find the ninth derivative of the following function at $x=0$:
$$f(x) =fraccosleft(4x^4right)-1x^7$$




So I did all the manipulations and I got the following Maclaurin Series:



$$sum _n=0^infty ,(-1)^nfrac16x^8n-7(2n)!-1$$



So to have $8n-7=9$, I got that $n=2$. Therefore, I thought that:



$$(-1)^2frac16x^924-1=frac16x^924-1$$



And, therefore, the ninth derivative would be:



$$frac16cdot 9!24-1=241919$$



But this was incorrect. Any help?






calculus derivatives taylor-expansion







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  • 2




    I think your Maclaurin series is not right
    – Zubin Mukerjee
    20 mins ago














up vote
2
down vote

favorite













Find the ninth derivative of the following function at $x=0$:
$$f(x) =fraccosleft(4x^4right)-1x^7$$




So I did all the manipulations and I got the following Maclaurin Series:



$$sum _n=0^infty ,(-1)^nfrac16x^8n-7(2n)!-1$$



So to have $8n-7=9$, I got that $n=2$. Therefore, I thought that:



$$(-1)^2frac16x^924-1=frac16x^924-1$$



And, therefore, the ninth derivative would be:



$$frac16cdot 9!24-1=241919$$



But this was incorrect. Any help?






calculus derivatives taylor-expansion







share|cite|improve this question



















  • 2




    I think your Maclaurin series is not right
    – Zubin Mukerjee
    20 mins ago












up vote
2
down vote

favorite









up vote
2
down vote

favorite












Find the ninth derivative of the following function at $x=0$:
$$f(x) =fraccosleft(4x^4right)-1x^7$$




So I did all the manipulations and I got the following Maclaurin Series:



$$sum _n=0^infty ,(-1)^nfrac16x^8n-7(2n)!-1$$



So to have $8n-7=9$, I got that $n=2$. Therefore, I thought that:



$$(-1)^2frac16x^924-1=frac16x^924-1$$



And, therefore, the ninth derivative would be:



$$frac16cdot 9!24-1=241919$$



But this was incorrect. Any help?






calculus derivatives taylor-expansion







share|cite|improve this question
















Find the ninth derivative of the following function at $x=0$:
$$f(x) =fraccosleft(4x^4right)-1x^7$$




So I did all the manipulations and I got the following Maclaurin Series:



$$sum _n=0^infty ,(-1)^nfrac16x^8n-7(2n)!-1$$



So to have $8n-7=9$, I got that $n=2$. Therefore, I thought that:



$$(-1)^2frac16x^924-1=frac16x^924-1$$



And, therefore, the ninth derivative would be:



$$frac16cdot 9!24-1=241919$$



But this was incorrect. Any help?






calculus derivatives taylor-expansion




calculus derivatives taylor-expansion






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 23 mins ago









Gibbs

3,8322624




3,8322624










asked 32 mins ago









sktsasus

867314




867314







  • 2




    I think your Maclaurin series is not right
    – Zubin Mukerjee
    20 mins ago












  • 2




    I think your Maclaurin series is not right
    – Zubin Mukerjee
    20 mins ago







2




2




I think your Maclaurin series is not right
– Zubin Mukerjee
20 mins ago




I think your Maclaurin series is not right
– Zubin Mukerjee
20 mins ago










2 Answers
2






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4
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$$ cosleft(4x^4right)-1=sum_n=1^infty(-1)^nfrac (4x^4)^2n(2n)!$$



So



$$f(x) = sum_n=1^infty(-1)^nfrac 16^n x^8n-7(2n)! $$






share|cite|improve this answer
















  • 2




    Yes. Three things were wrong with OP's Maclaurin series. First, the lower bound started at $n=0$ instead of $n=1$. Second, $16$ needed to be replaced with $16^n$. Third, there was a random $-1$ outside the sum that was incorrect.
    – Zubin Mukerjee
    14 mins ago

















up vote
1
down vote













When you substitute $(4x^4)$ for $x$ you get $(4x^4)^2n$ =$16^nx^8n$
You forgot the power for $16$






share|cite|improve this answer


















  • 1




    OP also forgot two other things, see @WW1's answer
    – Zubin Mukerjee
    14 mins ago










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
4
down vote



accepted










$$ cosleft(4x^4right)-1=sum_n=1^infty(-1)^nfrac (4x^4)^2n(2n)!$$



So



$$f(x) = sum_n=1^infty(-1)^nfrac 16^n x^8n-7(2n)! $$






share|cite|improve this answer
















  • 2




    Yes. Three things were wrong with OP's Maclaurin series. First, the lower bound started at $n=0$ instead of $n=1$. Second, $16$ needed to be replaced with $16^n$. Third, there was a random $-1$ outside the sum that was incorrect.
    – Zubin Mukerjee
    14 mins ago














up vote
4
down vote



accepted










$$ cosleft(4x^4right)-1=sum_n=1^infty(-1)^nfrac (4x^4)^2n(2n)!$$



So



$$f(x) = sum_n=1^infty(-1)^nfrac 16^n x^8n-7(2n)! $$






share|cite|improve this answer
















  • 2




    Yes. Three things were wrong with OP's Maclaurin series. First, the lower bound started at $n=0$ instead of $n=1$. Second, $16$ needed to be replaced with $16^n$. Third, there was a random $-1$ outside the sum that was incorrect.
    – Zubin Mukerjee
    14 mins ago












up vote
4
down vote



accepted







up vote
4
down vote



accepted






$$ cosleft(4x^4right)-1=sum_n=1^infty(-1)^nfrac (4x^4)^2n(2n)!$$



So



$$f(x) = sum_n=1^infty(-1)^nfrac 16^n x^8n-7(2n)! $$






share|cite|improve this answer












$$ cosleft(4x^4right)-1=sum_n=1^infty(-1)^nfrac (4x^4)^2n(2n)!$$



So



$$f(x) = sum_n=1^infty(-1)^nfrac 16^n x^8n-7(2n)! $$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 17 mins ago









WW1

6,7071712




6,7071712







  • 2




    Yes. Three things were wrong with OP's Maclaurin series. First, the lower bound started at $n=0$ instead of $n=1$. Second, $16$ needed to be replaced with $16^n$. Third, there was a random $-1$ outside the sum that was incorrect.
    – Zubin Mukerjee
    14 mins ago












  • 2




    Yes. Three things were wrong with OP's Maclaurin series. First, the lower bound started at $n=0$ instead of $n=1$. Second, $16$ needed to be replaced with $16^n$. Third, there was a random $-1$ outside the sum that was incorrect.
    – Zubin Mukerjee
    14 mins ago







2




2




Yes. Three things were wrong with OP's Maclaurin series. First, the lower bound started at $n=0$ instead of $n=1$. Second, $16$ needed to be replaced with $16^n$. Third, there was a random $-1$ outside the sum that was incorrect.
– Zubin Mukerjee
14 mins ago




Yes. Three things were wrong with OP's Maclaurin series. First, the lower bound started at $n=0$ instead of $n=1$. Second, $16$ needed to be replaced with $16^n$. Third, there was a random $-1$ outside the sum that was incorrect.
– Zubin Mukerjee
14 mins ago










up vote
1
down vote













When you substitute $(4x^4)$ for $x$ you get $(4x^4)^2n$ =$16^nx^8n$
You forgot the power for $16$






share|cite|improve this answer


















  • 1




    OP also forgot two other things, see @WW1's answer
    – Zubin Mukerjee
    14 mins ago














up vote
1
down vote













When you substitute $(4x^4)$ for $x$ you get $(4x^4)^2n$ =$16^nx^8n$
You forgot the power for $16$






share|cite|improve this answer


















  • 1




    OP also forgot two other things, see @WW1's answer
    – Zubin Mukerjee
    14 mins ago












up vote
1
down vote










up vote
1
down vote









When you substitute $(4x^4)$ for $x$ you get $(4x^4)^2n$ =$16^nx^8n$
You forgot the power for $16$






share|cite|improve this answer














When you substitute $(4x^4)$ for $x$ you get $(4x^4)^2n$ =$16^nx^8n$
You forgot the power for $16$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 14 mins ago

























answered 14 mins ago









Mohammad Riazi-Kermani

33.5k41855




33.5k41855







  • 1




    OP also forgot two other things, see @WW1's answer
    – Zubin Mukerjee
    14 mins ago












  • 1




    OP also forgot two other things, see @WW1's answer
    – Zubin Mukerjee
    14 mins ago







1




1




OP also forgot two other things, see @WW1's answer
– Zubin Mukerjee
14 mins ago




OP also forgot two other things, see @WW1's answer
– Zubin Mukerjee
14 mins ago

















 

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