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
add a comment |Â
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
2
I think your Maclaurin series is not right
â Zubin Mukerjee
20 mins ago
add a comment |Â
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
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
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
â Zubin Mukerjee
20 mins ago
add a comment |Â
2
I think your Maclaurin series is not right
â Zubin Mukerjee
20 mins ago
2
2
I think your Maclaurin series is not right
â Zubin Mukerjee
20 mins ago
I think your Maclaurin series is not right
â Zubin Mukerjee
20 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
$$ cosleft(4x^4right)-1=sum_n=1^infty(-1)^nfrac (4x^4)^2n(2n)!$$
So
$$f(x) = sum_n=1^infty(-1)^nfrac 16^n x^8n-7(2n)! $$
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.
â Zubin Mukerjee
14 mins ago
add a comment |Â
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 $16$
1
OP also forgot two other things, see @WW1's answer
â Zubin Mukerjee
14 mins ago
add a comment |Â
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
$$f(x) = sum_n=1^infty(-1)^nfrac 16^n x^8n-7(2n)! $$
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.
â Zubin Mukerjee
14 mins ago
add a comment |Â
up vote
4
down vote
accepted
$$ cosleft(4x^4right)-1=sum_n=1^infty(-1)^nfrac (4x^4)^2n(2n)!$$
So
$$f(x) = sum_n=1^infty(-1)^nfrac 16^n x^8n-7(2n)! $$
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.
â Zubin Mukerjee
14 mins ago
add a comment |Â
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
$$f(x) = sum_n=1^infty(-1)^nfrac 16^n x^8n-7(2n)! $$
$$ cosleft(4x^4right)-1=sum_n=1^infty(-1)^nfrac (4x^4)^2n(2n)!$$
So
$$f(x) = sum_n=1^infty(-1)^nfrac 16^n x^8n-7(2n)! $$
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.
â Zubin Mukerjee
14 mins ago
add a comment |Â
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.
â 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.
â 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.
â Zubin Mukerjee
14 mins ago
add a comment |Â
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 $16$
1
OP also forgot two other things, see @WW1's answer
â Zubin Mukerjee
14 mins ago
add a comment |Â
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 $16$
1
OP also forgot two other things, see @WW1's answer
â Zubin Mukerjee
14 mins ago
add a comment |Â
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 $16$
When you substitute $(4x^4)$ for $x$ you get $(4x^4)^2n$ =$16^nx^8n$
You forgot the power for $16$
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
â Zubin Mukerjee
14 mins ago
add a comment |Â
1
OP also forgot two other things, see @WW1's answer
â Zubin Mukerjee
14 mins ago
1
1
OP also forgot two other things, see @WW1's answer
â Zubin Mukerjee
14 mins ago
OP also forgot two other things, see @WW1's answer
â Zubin Mukerjee
14 mins ago
add a comment |Â
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2
I think your Maclaurin series is not right
â Zubin Mukerjee
20 mins ago