Mixing
Powers of trigonometric functions
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
Are formulas known for non-integer powers of trigonometric functions, analgolous to the power reduction identities? Like if you had a root of sine, could you express it in as a finite summation of sine and cosine functions with different inputs?
trigonometry
asked Jul 16 at 13:21
Jacob Lewis
313
add a comment |Â
up vote
2
down vote
favorite
Are formulas known for non-integer powers of trigonometric functions, analgolous to the power reduction identities? Like if you had a root of sine, could you express it in as a finite summation of sine and cosine functions with different inputs?
trigonometry
asked Jul 16 at 13:21
Jacob Lewis
313
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Are formulas known for non-integer powers of trigonometric functions, analgolous to the power reduction identities? Like if you had a root of sine, could you express it in as a finite summation of sine and cosine functions with different inputs?
trigonometry
asked Jul 16 at 13:21
Jacob Lewis
313
Are formulas known for non-integer powers of trigonometric functions, analgolous to the power reduction identities? Like if you had a root of sine, could you express it in as a finite summation of sine and cosine functions with different inputs?
trigonometry
asked Jul 16 at 13:21
Jacob Lewis
313
asked Jul 16 at 13:21
Jacob Lewis
313
asked Jul 16 at 13:21
Jacob Lewis
313
asked Jul 16 at 13:21
Jacob Lewis
313
313
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
That depends on what you understand by different inputs. For instance, defining $phi(x)=arcsinbigl(sqrtsin xbigr)$, $0le xlepi$, we have
$$
sqrtsin x=sin(phi(x)),quad0le xlepi.
$$
If you mean scalar multiples of $x$, then the answer is no: $sin(phi(x))$ is not differentiable at $x=0$ and at $x=pi$, but a finite sum of sines and cosines of scalar multiples of $x$ is infinitely differentiable.
However, $sqrtsin x$ can be written as a Fourier series, like:
$$
sqrtsin x=sum_n=1^infty a_nsin(n,x),quad0le xlepi,
$$
for certain coefficients $a_n$.
answered Jul 16 at 16:15
Julián Aguirre
64.7k23894
add a comment |Â
up vote
0
down vote
No.
The derivative of $sqrtsin x$ goes to infinity at $0$, which is not possible for a finite summation of sinusoids.
The situation is similar with
$$(1+x)^n$$ that has a finite expansion, while $$sqrt1+x$$ not.
answered Jul 16 at 16:25
Yves Daoust
111k665204
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
That depends on what you understand by different inputs. For instance, defining $phi(x)=arcsinbigl(sqrtsin xbigr)$, $0le xlepi$, we have
$$
sqrtsin x=sin(phi(x)),quad0le xlepi.
$$
If you mean scalar multiples of $x$, then the answer is no: $sin(phi(x))$ is not differentiable at $x=0$ and at $x=pi$, but a finite sum of sines and cosines of scalar multiples of $x$ is infinitely differentiable.
However, $sqrtsin x$ can be written as a Fourier series, like:
$$
sqrtsin x=sum_n=1^infty a_nsin(n,x),quad0le xlepi,
$$
for certain coefficients $a_n$.
answered Jul 16 at 16:15
Julián Aguirre
64.7k23894
add a comment |Â
up vote
2
down vote
That depends on what you understand by different inputs. For instance, defining $phi(x)=arcsinbigl(sqrtsin xbigr)$, $0le xlepi$, we have
$$
sqrtsin x=sin(phi(x)),quad0le xlepi.
$$
If you mean scalar multiples of $x$, then the answer is no: $sin(phi(x))$ is not differentiable at $x=0$ and at $x=pi$, but a finite sum of sines and cosines of scalar multiples of $x$ is infinitely differentiable.
However, $sqrtsin x$ can be written as a Fourier series, like:
$$
sqrtsin x=sum_n=1^infty a_nsin(n,x),quad0le xlepi,
$$
for certain coefficients $a_n$.
answered Jul 16 at 16:15
Julián Aguirre
64.7k23894
add a comment |Â
up vote
2
down vote
up vote
2
down vote
That depends on what you understand by different inputs. For instance, defining $phi(x)=arcsinbigl(sqrtsin xbigr)$, $0le xlepi$, we have
$$
sqrtsin x=sin(phi(x)),quad0le xlepi.
$$
If you mean scalar multiples of $x$, then the answer is no: $sin(phi(x))$ is not differentiable at $x=0$ and at $x=pi$, but a finite sum of sines and cosines of scalar multiples of $x$ is infinitely differentiable.
However, $sqrtsin x$ can be written as a Fourier series, like:
$$
sqrtsin x=sum_n=1^infty a_nsin(n,x),quad0le xlepi,
$$
for certain coefficients $a_n$.
answered Jul 16 at 16:15
Julián Aguirre
64.7k23894
That depends on what you understand by different inputs. For instance, defining $phi(x)=arcsinbigl(sqrtsin xbigr)$, $0le xlepi$, we have
$$
sqrtsin x=sin(phi(x)),quad0le xlepi.
$$
If you mean scalar multiples of $x$, then the answer is no: $sin(phi(x))$ is not differentiable at $x=0$ and at $x=pi$, but a finite sum of sines and cosines of scalar multiples of $x$ is infinitely differentiable.
However, $sqrtsin x$ can be written as a Fourier series, like:
$$
sqrtsin x=sum_n=1^infty a_nsin(n,x),quad0le xlepi,
$$
for certain coefficients $a_n$.
answered Jul 16 at 16:15
Julián Aguirre
64.7k23894
answered Jul 16 at 16:15
Julián Aguirre
64.7k23894
answered Jul 16 at 16:15
Julián Aguirre
64.7k23894
answered Jul 16 at 16:15
Julián Aguirre
64.7k23894
64.7k23894
add a comment |Â
add a comment |Â
up vote
0
down vote
No.
The derivative of $sqrtsin x$ goes to infinity at $0$, which is not possible for a finite summation of sinusoids.
The situation is similar with
$$(1+x)^n$$ that has a finite expansion, while $$sqrt1+x$$ not.
answered Jul 16 at 16:25
Yves Daoust
111k665204
add a comment |Â
up vote
0
down vote
No.
The derivative of $sqrtsin x$ goes to infinity at $0$, which is not possible for a finite summation of sinusoids.
The situation is similar with
$$(1+x)^n$$ that has a finite expansion, while $$sqrt1+x$$ not.
answered Jul 16 at 16:25
Yves Daoust
111k665204
add a comment |Â
up vote
0
down vote
up vote
0
down vote
No.
The derivative of $sqrtsin x$ goes to infinity at $0$, which is not possible for a finite summation of sinusoids.
The situation is similar with
$$(1+x)^n$$ that has a finite expansion, while $$sqrt1+x$$ not.
answered Jul 16 at 16:25
Yves Daoust
111k665204
No.
The derivative of $sqrtsin x$ goes to infinity at $0$, which is not possible for a finite summation of sinusoids.
The situation is similar with
$$(1+x)^n$$ that has a finite expansion, while $$sqrt1+x$$ not.
answered Jul 16 at 16:25
Yves Daoust
111k665204
edited Jul 16 at 16:57
answered Jul 16 at 16:25
Yves Daoust
111k665204
answered Jul 16 at 16:25
Yves Daoust
111k665204
answered Jul 16 at 16:25
Yves Daoust
111k665204
111k665204
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2853400%2fpowers-of-trigonometric-functions%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password