Finding the 6th Derivative of a function of $sin(x)$
I am currently studying Maclaurin Series, and have come across a question that has me stumped. The function is:
$$f(x) = frac{1}{1+sin(x^2)}$$
The question asks me to find the 6th derivative of this function evaluated at $0$. Now I am trying to solve this with a Maclaurin Expansion of $sin(x)$, but I get:
$$frac{1}{1+sin(x^2)} = frac{1}{1+x^2-frac{x^6}{6!}+frac{x^{10}}{5!}}$$
I am struggling to find the 6th derivative, as the coefficient of $x^6$ is not obvious.
calculus sequences-and-series derivatives taylor-expansion
add a comment |
I am currently studying Maclaurin Series, and have come across a question that has me stumped. The function is:
$$f(x) = frac{1}{1+sin(x^2)}$$
The question asks me to find the 6th derivative of this function evaluated at $0$. Now I am trying to solve this with a Maclaurin Expansion of $sin(x)$, but I get:
$$frac{1}{1+sin(x^2)} = frac{1}{1+x^2-frac{x^6}{6!}+frac{x^{10}}{5!}}$$
I am struggling to find the 6th derivative, as the coefficient of $x^6$ is not obvious.
calculus sequences-and-series derivatives taylor-expansion
1
THIS might be of interest, especially with high-order derivatives.
– Mark Viola
Dec 9 '16 at 4:24
add a comment |
I am currently studying Maclaurin Series, and have come across a question that has me stumped. The function is:
$$f(x) = frac{1}{1+sin(x^2)}$$
The question asks me to find the 6th derivative of this function evaluated at $0$. Now I am trying to solve this with a Maclaurin Expansion of $sin(x)$, but I get:
$$frac{1}{1+sin(x^2)} = frac{1}{1+x^2-frac{x^6}{6!}+frac{x^{10}}{5!}}$$
I am struggling to find the 6th derivative, as the coefficient of $x^6$ is not obvious.
calculus sequences-and-series derivatives taylor-expansion
I am currently studying Maclaurin Series, and have come across a question that has me stumped. The function is:
$$f(x) = frac{1}{1+sin(x^2)}$$
The question asks me to find the 6th derivative of this function evaluated at $0$. Now I am trying to solve this with a Maclaurin Expansion of $sin(x)$, but I get:
$$frac{1}{1+sin(x^2)} = frac{1}{1+x^2-frac{x^6}{6!}+frac{x^{10}}{5!}}$$
I am struggling to find the 6th derivative, as the coefficient of $x^6$ is not obvious.
calculus sequences-and-series derivatives taylor-expansion
calculus sequences-and-series derivatives taylor-expansion
edited Dec 9 '16 at 5:22
user223391
asked Dec 9 '16 at 4:21
Felicio Grande
499619
499619
1
THIS might be of interest, especially with high-order derivatives.
– Mark Viola
Dec 9 '16 at 4:24
add a comment |
1
THIS might be of interest, especially with high-order derivatives.
– Mark Viola
Dec 9 '16 at 4:24
1
1
THIS might be of interest, especially with high-order derivatives.
– Mark Viola
Dec 9 '16 at 4:24
THIS might be of interest, especially with high-order derivatives.
– Mark Viola
Dec 9 '16 at 4:24
add a comment |
3 Answers
3
active
oldest
votes
You're on the right track.
Continuing, we see that
$$begin{align}
frac{1}{1+sin(x^2)}&=frac{1}{1+left(x^2-frac16x^6+Oleft(x^{10}right)right)}\\
&=1-left(x^2-frac16x^6+Oleft(x^{10}right)right)+left(x^2+Oleft(x^6right)right)^2-left(x^2+Oleft(x^6right)right)^3+O(x^8)\\
&=1-x^2+x^4-frac56x^6+O(x^8)
end{align}$$
Therefore, the sixth derivative is $-(5/6)6!=-600$.
Would the cowardly down voter care to comment??
– Mark Viola
Dec 9 '16 at 5:24
Two cowardly down voters. Seriously? The answer is not useful? That is a dubious vote cast.
– Mark Viola
Dec 9 '16 at 5:32
Hey, Mark ! Good to see you. Please look at my comment to FelicioGrande; I never had the feeling that we compete (at the contrary !)
– Claude Leibovici
Dec 9 '16 at 5:57
1
@ClaudeLeibovici Hello my friend. I hope that you're enjoying this holiday season! And I believe that we support each other in bringing the best answers we can.
– Mark Viola
Dec 9 '16 at 6:18
1
@copper.hat Thank you! Much appreciative.
– Mark Viola
Dec 9 '16 at 6:51
add a comment |
You properly wrote $$frac{1}{1+sin(x^2)} = frac{1}{1+x^2-frac{x^6}{6!}+frac{x^{10}}{5!}+Oleft(x^{12}right)}$$ Now perform the long division (by increasing powers of $x$) to get $$frac{1}{1+sin(x^2)} = 1-x^2+x^4-frac{5 x^6}{6}+frac{2 x^8}{3}-frac{61
x^{10}}{120}+Oleft(x^{12}right)$$ Therefore
- the sixth derivative is $-frac 5 6times 6!=-600$
- the eighth derivative is $frac 2 3times 8!=26880$
- the tenth derivative is $-frac {61}{120}times 10!=-1844640$
Tenacious! (+1)
– Mark Viola
Dec 9 '16 at 5:25
This comment is great (+1), but Dr.MV had answered is first, thus my decision to give the checkmark to him. But thank you for this great answer!
– Felicio Grande
Dec 9 '16 at 5:26
@FelicioGrande. You are perfectly correct ! In any manner, I am not competing with my good friend Dr. MV. I was just proposing a way starting from what you wrote. Cheers.
– Claude Leibovici
Dec 9 '16 at 5:55
add a comment |
Use the geometric series i.e. $$frac{1}{1 + sin(x^2)} = 1 + Big( -x^2 + frac{x^6}{3!} - frac{x^{10}}{5!} + ... Big) + Big( - x^2 + frac{x^6}{3!} - frac{x^{10}}{5!} + ... Big)^2 + ...$$
@Dr.MV I think this is enough of a hint for now
– user392576
Dec 9 '16 at 4:28
@Dr.MV Of course it's correct. I already took that into account.
– user392576
Dec 9 '16 at 4:33
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
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votes
You're on the right track.
Continuing, we see that
$$begin{align}
frac{1}{1+sin(x^2)}&=frac{1}{1+left(x^2-frac16x^6+Oleft(x^{10}right)right)}\\
&=1-left(x^2-frac16x^6+Oleft(x^{10}right)right)+left(x^2+Oleft(x^6right)right)^2-left(x^2+Oleft(x^6right)right)^3+O(x^8)\\
&=1-x^2+x^4-frac56x^6+O(x^8)
end{align}$$
Therefore, the sixth derivative is $-(5/6)6!=-600$.
Would the cowardly down voter care to comment??
– Mark Viola
Dec 9 '16 at 5:24
Two cowardly down voters. Seriously? The answer is not useful? That is a dubious vote cast.
– Mark Viola
Dec 9 '16 at 5:32
Hey, Mark ! Good to see you. Please look at my comment to FelicioGrande; I never had the feeling that we compete (at the contrary !)
– Claude Leibovici
Dec 9 '16 at 5:57
1
@ClaudeLeibovici Hello my friend. I hope that you're enjoying this holiday season! And I believe that we support each other in bringing the best answers we can.
– Mark Viola
Dec 9 '16 at 6:18
1
@copper.hat Thank you! Much appreciative.
– Mark Viola
Dec 9 '16 at 6:51
add a comment |
You're on the right track.
Continuing, we see that
$$begin{align}
frac{1}{1+sin(x^2)}&=frac{1}{1+left(x^2-frac16x^6+Oleft(x^{10}right)right)}\\
&=1-left(x^2-frac16x^6+Oleft(x^{10}right)right)+left(x^2+Oleft(x^6right)right)^2-left(x^2+Oleft(x^6right)right)^3+O(x^8)\\
&=1-x^2+x^4-frac56x^6+O(x^8)
end{align}$$
Therefore, the sixth derivative is $-(5/6)6!=-600$.
Would the cowardly down voter care to comment??
– Mark Viola
Dec 9 '16 at 5:24
Two cowardly down voters. Seriously? The answer is not useful? That is a dubious vote cast.
– Mark Viola
Dec 9 '16 at 5:32
Hey, Mark ! Good to see you. Please look at my comment to FelicioGrande; I never had the feeling that we compete (at the contrary !)
– Claude Leibovici
Dec 9 '16 at 5:57
1
@ClaudeLeibovici Hello my friend. I hope that you're enjoying this holiday season! And I believe that we support each other in bringing the best answers we can.
– Mark Viola
Dec 9 '16 at 6:18
1
@copper.hat Thank you! Much appreciative.
– Mark Viola
Dec 9 '16 at 6:51
add a comment |
You're on the right track.
Continuing, we see that
$$begin{align}
frac{1}{1+sin(x^2)}&=frac{1}{1+left(x^2-frac16x^6+Oleft(x^{10}right)right)}\\
&=1-left(x^2-frac16x^6+Oleft(x^{10}right)right)+left(x^2+Oleft(x^6right)right)^2-left(x^2+Oleft(x^6right)right)^3+O(x^8)\\
&=1-x^2+x^4-frac56x^6+O(x^8)
end{align}$$
Therefore, the sixth derivative is $-(5/6)6!=-600$.
You're on the right track.
Continuing, we see that
$$begin{align}
frac{1}{1+sin(x^2)}&=frac{1}{1+left(x^2-frac16x^6+Oleft(x^{10}right)right)}\\
&=1-left(x^2-frac16x^6+Oleft(x^{10}right)right)+left(x^2+Oleft(x^6right)right)^2-left(x^2+Oleft(x^6right)right)^3+O(x^8)\\
&=1-x^2+x^4-frac56x^6+O(x^8)
end{align}$$
Therefore, the sixth derivative is $-(5/6)6!=-600$.
edited Dec 9 '16 at 4:43
answered Dec 9 '16 at 4:37
Mark Viola
130k1273170
130k1273170
Would the cowardly down voter care to comment??
– Mark Viola
Dec 9 '16 at 5:24
Two cowardly down voters. Seriously? The answer is not useful? That is a dubious vote cast.
– Mark Viola
Dec 9 '16 at 5:32
Hey, Mark ! Good to see you. Please look at my comment to FelicioGrande; I never had the feeling that we compete (at the contrary !)
– Claude Leibovici
Dec 9 '16 at 5:57
1
@ClaudeLeibovici Hello my friend. I hope that you're enjoying this holiday season! And I believe that we support each other in bringing the best answers we can.
– Mark Viola
Dec 9 '16 at 6:18
1
@copper.hat Thank you! Much appreciative.
– Mark Viola
Dec 9 '16 at 6:51
add a comment |
Would the cowardly down voter care to comment??
– Mark Viola
Dec 9 '16 at 5:24
Two cowardly down voters. Seriously? The answer is not useful? That is a dubious vote cast.
– Mark Viola
Dec 9 '16 at 5:32
Hey, Mark ! Good to see you. Please look at my comment to FelicioGrande; I never had the feeling that we compete (at the contrary !)
– Claude Leibovici
Dec 9 '16 at 5:57
1
@ClaudeLeibovici Hello my friend. I hope that you're enjoying this holiday season! And I believe that we support each other in bringing the best answers we can.
– Mark Viola
Dec 9 '16 at 6:18
1
@copper.hat Thank you! Much appreciative.
– Mark Viola
Dec 9 '16 at 6:51
Would the cowardly down voter care to comment??
– Mark Viola
Dec 9 '16 at 5:24
Would the cowardly down voter care to comment??
– Mark Viola
Dec 9 '16 at 5:24
Two cowardly down voters. Seriously? The answer is not useful? That is a dubious vote cast.
– Mark Viola
Dec 9 '16 at 5:32
Two cowardly down voters. Seriously? The answer is not useful? That is a dubious vote cast.
– Mark Viola
Dec 9 '16 at 5:32
Hey, Mark ! Good to see you. Please look at my comment to FelicioGrande; I never had the feeling that we compete (at the contrary !)
– Claude Leibovici
Dec 9 '16 at 5:57
Hey, Mark ! Good to see you. Please look at my comment to FelicioGrande; I never had the feeling that we compete (at the contrary !)
– Claude Leibovici
Dec 9 '16 at 5:57
1
1
@ClaudeLeibovici Hello my friend. I hope that you're enjoying this holiday season! And I believe that we support each other in bringing the best answers we can.
– Mark Viola
Dec 9 '16 at 6:18
@ClaudeLeibovici Hello my friend. I hope that you're enjoying this holiday season! And I believe that we support each other in bringing the best answers we can.
– Mark Viola
Dec 9 '16 at 6:18
1
1
@copper.hat Thank you! Much appreciative.
– Mark Viola
Dec 9 '16 at 6:51
@copper.hat Thank you! Much appreciative.
– Mark Viola
Dec 9 '16 at 6:51
add a comment |
You properly wrote $$frac{1}{1+sin(x^2)} = frac{1}{1+x^2-frac{x^6}{6!}+frac{x^{10}}{5!}+Oleft(x^{12}right)}$$ Now perform the long division (by increasing powers of $x$) to get $$frac{1}{1+sin(x^2)} = 1-x^2+x^4-frac{5 x^6}{6}+frac{2 x^8}{3}-frac{61
x^{10}}{120}+Oleft(x^{12}right)$$ Therefore
- the sixth derivative is $-frac 5 6times 6!=-600$
- the eighth derivative is $frac 2 3times 8!=26880$
- the tenth derivative is $-frac {61}{120}times 10!=-1844640$
Tenacious! (+1)
– Mark Viola
Dec 9 '16 at 5:25
This comment is great (+1), but Dr.MV had answered is first, thus my decision to give the checkmark to him. But thank you for this great answer!
– Felicio Grande
Dec 9 '16 at 5:26
@FelicioGrande. You are perfectly correct ! In any manner, I am not competing with my good friend Dr. MV. I was just proposing a way starting from what you wrote. Cheers.
– Claude Leibovici
Dec 9 '16 at 5:55
add a comment |
You properly wrote $$frac{1}{1+sin(x^2)} = frac{1}{1+x^2-frac{x^6}{6!}+frac{x^{10}}{5!}+Oleft(x^{12}right)}$$ Now perform the long division (by increasing powers of $x$) to get $$frac{1}{1+sin(x^2)} = 1-x^2+x^4-frac{5 x^6}{6}+frac{2 x^8}{3}-frac{61
x^{10}}{120}+Oleft(x^{12}right)$$ Therefore
- the sixth derivative is $-frac 5 6times 6!=-600$
- the eighth derivative is $frac 2 3times 8!=26880$
- the tenth derivative is $-frac {61}{120}times 10!=-1844640$
Tenacious! (+1)
– Mark Viola
Dec 9 '16 at 5:25
This comment is great (+1), but Dr.MV had answered is first, thus my decision to give the checkmark to him. But thank you for this great answer!
– Felicio Grande
Dec 9 '16 at 5:26
@FelicioGrande. You are perfectly correct ! In any manner, I am not competing with my good friend Dr. MV. I was just proposing a way starting from what you wrote. Cheers.
– Claude Leibovici
Dec 9 '16 at 5:55
add a comment |
You properly wrote $$frac{1}{1+sin(x^2)} = frac{1}{1+x^2-frac{x^6}{6!}+frac{x^{10}}{5!}+Oleft(x^{12}right)}$$ Now perform the long division (by increasing powers of $x$) to get $$frac{1}{1+sin(x^2)} = 1-x^2+x^4-frac{5 x^6}{6}+frac{2 x^8}{3}-frac{61
x^{10}}{120}+Oleft(x^{12}right)$$ Therefore
- the sixth derivative is $-frac 5 6times 6!=-600$
- the eighth derivative is $frac 2 3times 8!=26880$
- the tenth derivative is $-frac {61}{120}times 10!=-1844640$
You properly wrote $$frac{1}{1+sin(x^2)} = frac{1}{1+x^2-frac{x^6}{6!}+frac{x^{10}}{5!}+Oleft(x^{12}right)}$$ Now perform the long division (by increasing powers of $x$) to get $$frac{1}{1+sin(x^2)} = 1-x^2+x^4-frac{5 x^6}{6}+frac{2 x^8}{3}-frac{61
x^{10}}{120}+Oleft(x^{12}right)$$ Therefore
- the sixth derivative is $-frac 5 6times 6!=-600$
- the eighth derivative is $frac 2 3times 8!=26880$
- the tenth derivative is $-frac {61}{120}times 10!=-1844640$
answered Dec 9 '16 at 5:20
Claude Leibovici
118k1156132
118k1156132
Tenacious! (+1)
– Mark Viola
Dec 9 '16 at 5:25
This comment is great (+1), but Dr.MV had answered is first, thus my decision to give the checkmark to him. But thank you for this great answer!
– Felicio Grande
Dec 9 '16 at 5:26
@FelicioGrande. You are perfectly correct ! In any manner, I am not competing with my good friend Dr. MV. I was just proposing a way starting from what you wrote. Cheers.
– Claude Leibovici
Dec 9 '16 at 5:55
add a comment |
Tenacious! (+1)
– Mark Viola
Dec 9 '16 at 5:25
This comment is great (+1), but Dr.MV had answered is first, thus my decision to give the checkmark to him. But thank you for this great answer!
– Felicio Grande
Dec 9 '16 at 5:26
@FelicioGrande. You are perfectly correct ! In any manner, I am not competing with my good friend Dr. MV. I was just proposing a way starting from what you wrote. Cheers.
– Claude Leibovici
Dec 9 '16 at 5:55
Tenacious! (+1)
– Mark Viola
Dec 9 '16 at 5:25
Tenacious! (+1)
– Mark Viola
Dec 9 '16 at 5:25
This comment is great (+1), but Dr.MV had answered is first, thus my decision to give the checkmark to him. But thank you for this great answer!
– Felicio Grande
Dec 9 '16 at 5:26
This comment is great (+1), but Dr.MV had answered is first, thus my decision to give the checkmark to him. But thank you for this great answer!
– Felicio Grande
Dec 9 '16 at 5:26
@FelicioGrande. You are perfectly correct ! In any manner, I am not competing with my good friend Dr. MV. I was just proposing a way starting from what you wrote. Cheers.
– Claude Leibovici
Dec 9 '16 at 5:55
@FelicioGrande. You are perfectly correct ! In any manner, I am not competing with my good friend Dr. MV. I was just proposing a way starting from what you wrote. Cheers.
– Claude Leibovici
Dec 9 '16 at 5:55
add a comment |
Use the geometric series i.e. $$frac{1}{1 + sin(x^2)} = 1 + Big( -x^2 + frac{x^6}{3!} - frac{x^{10}}{5!} + ... Big) + Big( - x^2 + frac{x^6}{3!} - frac{x^{10}}{5!} + ... Big)^2 + ...$$
@Dr.MV I think this is enough of a hint for now
– user392576
Dec 9 '16 at 4:28
@Dr.MV Of course it's correct. I already took that into account.
– user392576
Dec 9 '16 at 4:33
add a comment |
Use the geometric series i.e. $$frac{1}{1 + sin(x^2)} = 1 + Big( -x^2 + frac{x^6}{3!} - frac{x^{10}}{5!} + ... Big) + Big( - x^2 + frac{x^6}{3!} - frac{x^{10}}{5!} + ... Big)^2 + ...$$
@Dr.MV I think this is enough of a hint for now
– user392576
Dec 9 '16 at 4:28
@Dr.MV Of course it's correct. I already took that into account.
– user392576
Dec 9 '16 at 4:33
add a comment |
Use the geometric series i.e. $$frac{1}{1 + sin(x^2)} = 1 + Big( -x^2 + frac{x^6}{3!} - frac{x^{10}}{5!} + ... Big) + Big( - x^2 + frac{x^6}{3!} - frac{x^{10}}{5!} + ... Big)^2 + ...$$
Use the geometric series i.e. $$frac{1}{1 + sin(x^2)} = 1 + Big( -x^2 + frac{x^6}{3!} - frac{x^{10}}{5!} + ... Big) + Big( - x^2 + frac{x^6}{3!} - frac{x^{10}}{5!} + ... Big)^2 + ...$$
answered Dec 9 '16 at 4:25
user392576
1394
1394
@Dr.MV I think this is enough of a hint for now
– user392576
Dec 9 '16 at 4:28
@Dr.MV Of course it's correct. I already took that into account.
– user392576
Dec 9 '16 at 4:33
add a comment |
@Dr.MV I think this is enough of a hint for now
– user392576
Dec 9 '16 at 4:28
@Dr.MV Of course it's correct. I already took that into account.
– user392576
Dec 9 '16 at 4:33
@Dr.MV I think this is enough of a hint for now
– user392576
Dec 9 '16 at 4:28
@Dr.MV I think this is enough of a hint for now
– user392576
Dec 9 '16 at 4:28
@Dr.MV Of course it's correct. I already took that into account.
– user392576
Dec 9 '16 at 4:33
@Dr.MV Of course it's correct. I already took that into account.
– user392576
Dec 9 '16 at 4:33
add a comment |
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1
THIS might be of interest, especially with high-order derivatives.
– Mark Viola
Dec 9 '16 at 4:24