Is it valid to say that $cos^3(x^{4/3})=cos(x^4)$?
As the title says, is it valid to insert the power of the cosine to its angle?
Edit : Is it valid when x is very small ?
trigonometry
asked Nov 24 at 3:06
John adams
206
add a comment |
As the title says, is it valid to insert the power of the cosine to its angle?
Edit : Is it valid when x is very small ?
trigonometry
asked Nov 24 at 3:06
John adams
206
1
No. They are close near $x=0$ though.
– user587192
Nov 24 at 3:12
Had it been $cos(x^{frac{4}{3}})^{3}$ then you could have
– Akash Roy
Nov 24 at 3:16
Notice that $cos^3(y) neq cos(y^3)$, so there's no reason to expect the equation in the question to be true.
– littleO
Nov 24 at 3:27
When $xto 0$, $$cos^3(x^{4/3})=1-frac{3}{2}x^{8/3} + o(x^4)$$ while $$cos(x^4) = 1-frac{1}{2}x^8 + o(x^8)$$ so the first non-constant terms do not even closely match up.
– Clement C.
Nov 24 at 3:43
add a comment |
As the title says, is it valid to insert the power of the cosine to its angle?
Edit : Is it valid when x is very small ?
trigonometry
asked Nov 24 at 3:06
John adams
206
As the title says, is it valid to insert the power of the cosine to its angle?
Edit : Is it valid when x is very small ?
trigonometry
trigonometry
asked Nov 24 at 3:06
John adams
206
asked Nov 24 at 3:06
John adams
206
edited Nov 24 at 3:17
asked Nov 24 at 3:06
John adams
206
asked Nov 24 at 3:06
John adams
206
asked Nov 24 at 3:06
John adams
206
206
1
No. They are close near $x=0$ though.
– user587192
Nov 24 at 3:12
Had it been $cos(x^{frac{4}{3}})^{3}$ then you could have
– Akash Roy
Nov 24 at 3:16
Notice that $cos^3(y) neq cos(y^3)$, so there's no reason to expect the equation in the question to be true.
– littleO
Nov 24 at 3:27
When $xto 0$, $$cos^3(x^{4/3})=1-frac{3}{2}x^{8/3} + o(x^4)$$ while $$cos(x^4) = 1-frac{1}{2}x^8 + o(x^8)$$ so the first non-constant terms do not even closely match up.
– Clement C.
Nov 24 at 3:43
add a comment |
1
No. They are close near $x=0$ though.
– user587192
Nov 24 at 3:12
Had it been $cos(x^{frac{4}{3}})^{3}$ then you could have
– Akash Roy
Nov 24 at 3:16
Notice that $cos^3(y) neq cos(y^3)$, so there's no reason to expect the equation in the question to be true.
– littleO
Nov 24 at 3:27
When $xto 0$, $$cos^3(x^{4/3})=1-frac{3}{2}x^{8/3} + o(x^4)$$ while $$cos(x^4) = 1-frac{1}{2}x^8 + o(x^8)$$ so the first non-constant terms do not even closely match up.
– Clement C.
Nov 24 at 3:43
1
1
No. They are close near $x=0$ though.
– user587192
Nov 24 at 3:12
No. They are close near $x=0$ though.
– user587192
Nov 24 at 3:12
Had it been $cos(x^{frac{4}{3}})^{3}$ then you could have
– Akash Roy
Nov 24 at 3:16
Had it been $cos(x^{frac{4}{3}})^{3}$ then you could have
– Akash Roy
Nov 24 at 3:16
Notice that $cos^3(y) neq cos(y^3)$, so there's no reason to expect the equation in the question to be true.
– littleO
Nov 24 at 3:27
Notice that $cos^3(y) neq cos(y^3)$, so there's no reason to expect the equation in the question to be true.
– littleO
Nov 24 at 3:27
When $xto 0$, $$cos^3(x^{4/3})=1-frac{3}{2}x^{8/3} + o(x^4)$$ while $$cos(x^4) = 1-frac{1}{2}x^8 + o(x^8)$$ so the first non-constant terms do not even closely match up.
– Clement C.
Nov 24 at 3:43
When $xto 0$, $$cos^3(x^{4/3})=1-frac{3}{2}x^{8/3} + o(x^4)$$ while $$cos(x^4) = 1-frac{1}{2}x^8 + o(x^8)$$ so the first non-constant terms do not even closely match up.
– Clement C.
Nov 24 at 3:43
add a comment |
2 Answers
2
active
oldest
votes
For $x = pi^{3/4}$, $cos^{3}(x^{4/3}) = cos^{3}(pi) = -1$. However, $cos(pi^{4})neq -1$, since $pi^{4}$ can't be a rational multiple of $pi$ (since $pi$ is a transcendental number!).
answered Nov 24 at 3:15
Seewoo Lee
6,110826
Although it's pretty much true, I don't think that that's the most helpful approach.
– rafa11111
Nov 24 at 3:25
@rafa11111 I just want to show the simple counter example that just arose in my head :)
– Seewoo Lee
Nov 24 at 3:26
Yes, of course! I just wanted to point out that, IMHO, that's not exactly an answer...
– rafa11111
Nov 24 at 3:29
add a comment |
Although
$$
(x^{4/3})^3 = x^4,
$$
one does not have
$$
[f(x^{4/3})]^3=f(x^4)
$$
in general.
For your added question "Is it valid when $x$ is very small?":
I assume that you mean when $|x|$ is very small.
No. If these two functions are identical near $x=0$, then they must have the same Taylor expansion. But it is not difficult to see by comparing a few terms that they don't have the same Taylor expansion near $x=0$.
answered Nov 24 at 3:19
user587192
1,486112
So this means that this is valid when x is very small ?
– John adams
Nov 24 at 3:21
1
@Johnadams: "close" does not necessarily mean they are "identical". But one can tell that they are identical at $x=0$, of course.
– user587192
Nov 24 at 3:41
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
For $x = pi^{3/4}$, $cos^{3}(x^{4/3}) = cos^{3}(pi) = -1$. However, $cos(pi^{4})neq -1$, since $pi^{4}$ can't be a rational multiple of $pi$ (since $pi$ is a transcendental number!).
answered Nov 24 at 3:15
Seewoo Lee
6,110826
Although it's pretty much true, I don't think that that's the most helpful approach.
– rafa11111
Nov 24 at 3:25
@rafa11111 I just want to show the simple counter example that just arose in my head :)
– Seewoo Lee
Nov 24 at 3:26
Yes, of course! I just wanted to point out that, IMHO, that's not exactly an answer...
– rafa11111
Nov 24 at 3:29
add a comment |
For $x = pi^{3/4}$, $cos^{3}(x^{4/3}) = cos^{3}(pi) = -1$. However, $cos(pi^{4})neq -1$, since $pi^{4}$ can't be a rational multiple of $pi$ (since $pi$ is a transcendental number!).
answered Nov 24 at 3:15
Seewoo Lee
6,110826
Although it's pretty much true, I don't think that that's the most helpful approach.
– rafa11111
Nov 24 at 3:25
@rafa11111 I just want to show the simple counter example that just arose in my head :)
– Seewoo Lee
Nov 24 at 3:26
Yes, of course! I just wanted to point out that, IMHO, that's not exactly an answer...
– rafa11111
Nov 24 at 3:29
add a comment |
For $x = pi^{3/4}$, $cos^{3}(x^{4/3}) = cos^{3}(pi) = -1$. However, $cos(pi^{4})neq -1$, since $pi^{4}$ can't be a rational multiple of $pi$ (since $pi$ is a transcendental number!).
answered Nov 24 at 3:15
Seewoo Lee
6,110826
For $x = pi^{3/4}$, $cos^{3}(x^{4/3}) = cos^{3}(pi) = -1$. However, $cos(pi^{4})neq -1$, since $pi^{4}$ can't be a rational multiple of $pi$ (since $pi$ is a transcendental number!).
answered Nov 24 at 3:15
Seewoo Lee
6,110826
answered Nov 24 at 3:15
Seewoo Lee
6,110826
answered Nov 24 at 3:15
Seewoo Lee
6,110826
answered Nov 24 at 3:15
Seewoo Lee
6,110826
6,110826
Although it's pretty much true, I don't think that that's the most helpful approach.
– rafa11111
Nov 24 at 3:25
@rafa11111 I just want to show the simple counter example that just arose in my head :)
– Seewoo Lee
Nov 24 at 3:26
Yes, of course! I just wanted to point out that, IMHO, that's not exactly an answer...
– rafa11111
Nov 24 at 3:29
add a comment |
Although it's pretty much true, I don't think that that's the most helpful approach.
– rafa11111
Nov 24 at 3:25
@rafa11111 I just want to show the simple counter example that just arose in my head :)
– Seewoo Lee
Nov 24 at 3:26
Yes, of course! I just wanted to point out that, IMHO, that's not exactly an answer...
– rafa11111
Nov 24 at 3:29
Although it's pretty much true, I don't think that that's the most helpful approach.
– rafa11111
Nov 24 at 3:25
Although it's pretty much true, I don't think that that's the most helpful approach.
– rafa11111
Nov 24 at 3:25
@rafa11111 I just want to show the simple counter example that just arose in my head :)
– Seewoo Lee
Nov 24 at 3:26
@rafa11111 I just want to show the simple counter example that just arose in my head :)
– Seewoo Lee
Nov 24 at 3:26
Yes, of course! I just wanted to point out that, IMHO, that's not exactly an answer...
– rafa11111
Nov 24 at 3:29
Yes, of course! I just wanted to point out that, IMHO, that's not exactly an answer...
– rafa11111
Nov 24 at 3:29
add a comment |
Although
$$
(x^{4/3})^3 = x^4,
$$
one does not have
$$
[f(x^{4/3})]^3=f(x^4)
$$
in general.
For your added question "Is it valid when $x$ is very small?":
I assume that you mean when $|x|$ is very small.
No. If these two functions are identical near $x=0$, then they must have the same Taylor expansion. But it is not difficult to see by comparing a few terms that they don't have the same Taylor expansion near $x=0$.
answered Nov 24 at 3:19
user587192
1,486112
So this means that this is valid when x is very small ?
– John adams
Nov 24 at 3:21
1
@Johnadams: "close" does not necessarily mean they are "identical". But one can tell that they are identical at $x=0$, of course.
– user587192
Nov 24 at 3:41
add a comment |
Although
$$
(x^{4/3})^3 = x^4,
$$
one does not have
$$
[f(x^{4/3})]^3=f(x^4)
$$
in general.
For your added question "Is it valid when $x$ is very small?":
I assume that you mean when $|x|$ is very small.
No. If these two functions are identical near $x=0$, then they must have the same Taylor expansion. But it is not difficult to see by comparing a few terms that they don't have the same Taylor expansion near $x=0$.
answered Nov 24 at 3:19
user587192
1,486112
So this means that this is valid when x is very small ?
– John adams
Nov 24 at 3:21
1
@Johnadams: "close" does not necessarily mean they are "identical". But one can tell that they are identical at $x=0$, of course.
– user587192
Nov 24 at 3:41
add a comment |
Although
$$
(x^{4/3})^3 = x^4,
$$
one does not have
$$
[f(x^{4/3})]^3=f(x^4)
$$
in general.
For your added question "Is it valid when $x$ is very small?":
I assume that you mean when $|x|$ is very small.
No. If these two functions are identical near $x=0$, then they must have the same Taylor expansion. But it is not difficult to see by comparing a few terms that they don't have the same Taylor expansion near $x=0$.
answered Nov 24 at 3:19
user587192
1,486112
Although
$$
(x^{4/3})^3 = x^4,
$$
one does not have
$$
[f(x^{4/3})]^3=f(x^4)
$$
in general.
For your added question "Is it valid when $x$ is very small?":
I assume that you mean when $|x|$ is very small.
No. If these two functions are identical near $x=0$, then they must have the same Taylor expansion. But it is not difficult to see by comparing a few terms that they don't have the same Taylor expansion near $x=0$.
answered Nov 24 at 3:19
user587192
1,486112
edited Nov 24 at 4:24
answered Nov 24 at 3:19
user587192
1,486112
answered Nov 24 at 3:19
user587192
1,486112
answered Nov 24 at 3:19
user587192
1,486112
1,486112
So this means that this is valid when x is very small ?
– John adams
Nov 24 at 3:21
1
@Johnadams: "close" does not necessarily mean they are "identical". But one can tell that they are identical at $x=0$, of course.
– user587192
Nov 24 at 3:41
add a comment |
So this means that this is valid when x is very small ?
– John adams
Nov 24 at 3:21
1
@Johnadams: "close" does not necessarily mean they are "identical". But one can tell that they are identical at $x=0$, of course.
– user587192
Nov 24 at 3:41
So this means that this is valid when x is very small ?
– John adams
Nov 24 at 3:21
So this means that this is valid when x is very small ?
– John adams
Nov 24 at 3:21
1
1
@Johnadams: "close" does not necessarily mean they are "identical". But one can tell that they are identical at $x=0$, of course.
– user587192
Nov 24 at 3:41
@Johnadams: "close" does not necessarily mean they are "identical". But one can tell that they are identical at $x=0$, of course.
– user587192
Nov 24 at 3:41
add a comment |
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1
No. They are close near $x=0$ though.
– user587192
Nov 24 at 3:12
Had it been $cos(x^{frac{4}{3}})^{3}$ then you could have
– Akash Roy
Nov 24 at 3:16
Notice that $cos^3(y) neq cos(y^3)$, so there's no reason to expect the equation in the question to be true.
– littleO
Nov 24 at 3:27
When $xto 0$, $$cos^3(x^{4/3})=1-frac{3}{2}x^{8/3} + o(x^4)$$ while $$cos(x^4) = 1-frac{1}{2}x^8 + o(x^8)$$ so the first non-constant terms do not even closely match up.
– Clement C.
Nov 24 at 3:43