Is it valid to say that $cos^3(x^{4/3})=cos(x^4)$?
$begingroup$
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 '18 at 3:06
John adamsJohn adams
226
$endgroup$
add a comment |
$begingroup$
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 '18 at 3:06
John adamsJohn adams
226
$endgroup$
1
$begingroup$
No. They are close near $x=0$ though.
$endgroup$
– user587192
Nov 24 '18 at 3:12
$begingroup$
Had it been $cos(x^{frac{4}{3}})^{3}$ then you could have
$endgroup$
– Akash Roy
Nov 24 '18 at 3:16
$begingroup$
Notice that $cos^3(y) neq cos(y^3)$, so there's no reason to expect the equation in the question to be true.
$endgroup$
– littleO
Nov 24 '18 at 3:27
$begingroup$
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.
$endgroup$
– Clement C.
Nov 24 '18 at 3:43
add a comment |
$begingroup$
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 '18 at 3:06
John adamsJohn adams
226
$endgroup$
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 '18 at 3:06
John adamsJohn adams
226
asked Nov 24 '18 at 3:06
John adamsJohn adams
226
edited Nov 24 '18 at 3:17
John adams
asked Nov 24 '18 at 3:06
John adamsJohn adams
226
asked Nov 24 '18 at 3:06
John adamsJohn adams
226
asked Nov 24 '18 at 3:06
John adamsJohn adams
226
226
1
$begingroup$
No. They are close near $x=0$ though.
$endgroup$
– user587192
Nov 24 '18 at 3:12
$begingroup$
Had it been $cos(x^{frac{4}{3}})^{3}$ then you could have
$endgroup$
– Akash Roy
Nov 24 '18 at 3:16
$begingroup$
Notice that $cos^3(y) neq cos(y^3)$, so there's no reason to expect the equation in the question to be true.
$endgroup$
– littleO
Nov 24 '18 at 3:27
$begingroup$
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.
$endgroup$
– Clement C.
Nov 24 '18 at 3:43
add a comment |
1
$begingroup$
No. They are close near $x=0$ though.
$endgroup$
– user587192
Nov 24 '18 at 3:12
$begingroup$
Had it been $cos(x^{frac{4}{3}})^{3}$ then you could have
$endgroup$
– Akash Roy
Nov 24 '18 at 3:16
$begingroup$
Notice that $cos^3(y) neq cos(y^3)$, so there's no reason to expect the equation in the question to be true.
$endgroup$
– littleO
Nov 24 '18 at 3:27
$begingroup$
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.
$endgroup$
– Clement C.
Nov 24 '18 at 3:43
1
1
$begingroup$
No. They are close near $x=0$ though.
$endgroup$
– user587192
Nov 24 '18 at 3:12
$begingroup$
No. They are close near $x=0$ though.
$endgroup$
– user587192
Nov 24 '18 at 3:12
$begingroup$
Had it been $cos(x^{frac{4}{3}})^{3}$ then you could have
$endgroup$
– Akash Roy
Nov 24 '18 at 3:16
$begingroup$
Had it been $cos(x^{frac{4}{3}})^{3}$ then you could have
$endgroup$
– Akash Roy
Nov 24 '18 at 3:16
$begingroup$
Notice that $cos^3(y) neq cos(y^3)$, so there's no reason to expect the equation in the question to be true.
$endgroup$
– littleO
Nov 24 '18 at 3:27
$begingroup$
Notice that $cos^3(y) neq cos(y^3)$, so there's no reason to expect the equation in the question to be true.
$endgroup$
– littleO
Nov 24 '18 at 3:27
$begingroup$
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.
$endgroup$
– Clement C.
Nov 24 '18 at 3:43
$begingroup$
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.
$endgroup$
– Clement C.
Nov 24 '18 at 3:43
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 '18 at 3:15
Seewoo LeeSeewoo Lee
6,267826
$endgroup$
$begingroup$
Although it's pretty much true, I don't think that that's the most helpful approach.
$endgroup$
– rafa11111
Nov 24 '18 at 3:25
$begingroup$
@rafa11111 I just want to show the simple counter example that just arose in my head :)
$endgroup$
– Seewoo Lee
Nov 24 '18 at 3:26
$begingroup$
Yes, of course! I just wanted to point out that, IMHO, that's not exactly an answer...
$endgroup$
– rafa11111
Nov 24 '18 at 3:29
add a comment |
$begingroup$
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 '18 at 3:19
user587192user587192
1,757215
$endgroup$
$begingroup$
So this means that this is valid when x is very small ?
$endgroup$
– John adams
Nov 24 '18 at 3:21
1
$begingroup$
@Johnadams: "close" does not necessarily mean they are "identical". But one can tell that they are identical at $x=0$, of course.
$endgroup$
– user587192
Nov 24 '18 at 3:41
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 '18 at 3:15
Seewoo LeeSeewoo Lee
6,267826
$endgroup$
$begingroup$
Although it's pretty much true, I don't think that that's the most helpful approach.
$endgroup$
– rafa11111
Nov 24 '18 at 3:25
$begingroup$
@rafa11111 I just want to show the simple counter example that just arose in my head :)
$endgroup$
– Seewoo Lee
Nov 24 '18 at 3:26
$begingroup$
Yes, of course! I just wanted to point out that, IMHO, that's not exactly an answer...
$endgroup$
– rafa11111
Nov 24 '18 at 3:29
add a comment |
$begingroup$
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 '18 at 3:15
Seewoo LeeSeewoo Lee
6,267826
$endgroup$
$begingroup$
Although it's pretty much true, I don't think that that's the most helpful approach.
$endgroup$
– rafa11111
Nov 24 '18 at 3:25
$begingroup$
@rafa11111 I just want to show the simple counter example that just arose in my head :)
$endgroup$
– Seewoo Lee
Nov 24 '18 at 3:26
$begingroup$
Yes, of course! I just wanted to point out that, IMHO, that's not exactly an answer...
$endgroup$
– rafa11111
Nov 24 '18 at 3:29
add a comment |
$begingroup$
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 '18 at 3:15
Seewoo LeeSeewoo Lee
6,267826
$endgroup$
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 '18 at 3:15
Seewoo LeeSeewoo Lee
6,267826
answered Nov 24 '18 at 3:15
Seewoo LeeSeewoo Lee
6,267826
answered Nov 24 '18 at 3:15
Seewoo LeeSeewoo Lee
6,267826
answered Nov 24 '18 at 3:15
Seewoo LeeSeewoo Lee
6,267826
6,267826
$begingroup$
Although it's pretty much true, I don't think that that's the most helpful approach.
$endgroup$
– rafa11111
Nov 24 '18 at 3:25
$begingroup$
@rafa11111 I just want to show the simple counter example that just arose in my head :)
$endgroup$
– Seewoo Lee
Nov 24 '18 at 3:26
$begingroup$
Yes, of course! I just wanted to point out that, IMHO, that's not exactly an answer...
$endgroup$
– rafa11111
Nov 24 '18 at 3:29
add a comment |
$begingroup$
Although it's pretty much true, I don't think that that's the most helpful approach.
$endgroup$
– rafa11111
Nov 24 '18 at 3:25
$begingroup$
@rafa11111 I just want to show the simple counter example that just arose in my head :)
$endgroup$
– Seewoo Lee
Nov 24 '18 at 3:26
$begingroup$
Yes, of course! I just wanted to point out that, IMHO, that's not exactly an answer...
$endgroup$
– rafa11111
Nov 24 '18 at 3:29
$begingroup$
Although it's pretty much true, I don't think that that's the most helpful approach.
$endgroup$
– rafa11111
Nov 24 '18 at 3:25
$begingroup$
Although it's pretty much true, I don't think that that's the most helpful approach.
$endgroup$
– rafa11111
Nov 24 '18 at 3:25
$begingroup$
@rafa11111 I just want to show the simple counter example that just arose in my head :)
$endgroup$
– Seewoo Lee
Nov 24 '18 at 3:26
$begingroup$
@rafa11111 I just want to show the simple counter example that just arose in my head :)
$endgroup$
– Seewoo Lee
Nov 24 '18 at 3:26
$begingroup$
Yes, of course! I just wanted to point out that, IMHO, that's not exactly an answer...
$endgroup$
– rafa11111
Nov 24 '18 at 3:29
$begingroup$
Yes, of course! I just wanted to point out that, IMHO, that's not exactly an answer...
$endgroup$
– rafa11111
Nov 24 '18 at 3:29
add a comment |
$begingroup$
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 '18 at 3:19
user587192user587192
1,757215
$endgroup$
$begingroup$
So this means that this is valid when x is very small ?
$endgroup$
– John adams
Nov 24 '18 at 3:21
1
$begingroup$
@Johnadams: "close" does not necessarily mean they are "identical". But one can tell that they are identical at $x=0$, of course.
$endgroup$
– user587192
Nov 24 '18 at 3:41
add a comment |
$begingroup$
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 '18 at 3:19
user587192user587192
1,757215
$endgroup$
$begingroup$
So this means that this is valid when x is very small ?
$endgroup$
– John adams
Nov 24 '18 at 3:21
1
$begingroup$
@Johnadams: "close" does not necessarily mean they are "identical". But one can tell that they are identical at $x=0$, of course.
$endgroup$
– user587192
Nov 24 '18 at 3:41
add a comment |
$begingroup$
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 '18 at 3:19
user587192user587192
1,757215
$endgroup$
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 '18 at 3:19
user587192user587192
1,757215
edited Nov 24 '18 at 4:24
answered Nov 24 '18 at 3:19
user587192user587192
1,757215
answered Nov 24 '18 at 3:19
user587192user587192
1,757215
answered Nov 24 '18 at 3:19
user587192user587192
1,757215
1,757215
$begingroup$
So this means that this is valid when x is very small ?
$endgroup$
– John adams
Nov 24 '18 at 3:21
1
$begingroup$
@Johnadams: "close" does not necessarily mean they are "identical". But one can tell that they are identical at $x=0$, of course.
$endgroup$
– user587192
Nov 24 '18 at 3:41
add a comment |
$begingroup$
So this means that this is valid when x is very small ?
$endgroup$
– John adams
Nov 24 '18 at 3:21
1
$begingroup$
@Johnadams: "close" does not necessarily mean they are "identical". But one can tell that they are identical at $x=0$, of course.
$endgroup$
– user587192
Nov 24 '18 at 3:41
$begingroup$
So this means that this is valid when x is very small ?
$endgroup$
– John adams
Nov 24 '18 at 3:21
$begingroup$
So this means that this is valid when x is very small ?
$endgroup$
– John adams
Nov 24 '18 at 3:21
1
1
$begingroup$
@Johnadams: "close" does not necessarily mean they are "identical". But one can tell that they are identical at $x=0$, of course.
$endgroup$
– user587192
Nov 24 '18 at 3:41
$begingroup$
@Johnadams: "close" does not necessarily mean they are "identical". But one can tell that they are identical at $x=0$, of course.
$endgroup$
– user587192
Nov 24 '18 at 3:41
add a comment |
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1
$begingroup$
No. They are close near $x=0$ though.
$endgroup$
– user587192
Nov 24 '18 at 3:12
$begingroup$
Had it been $cos(x^{frac{4}{3}})^{3}$ then you could have
$endgroup$
– Akash Roy
Nov 24 '18 at 3:16
$begingroup$
Notice that $cos^3(y) neq cos(y^3)$, so there's no reason to expect the equation in the question to be true.
$endgroup$
– littleO
Nov 24 '18 at 3:27
$begingroup$
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.
$endgroup$
– Clement C.
Nov 24 '18 at 3:43