Is it valid to say that $cos^3(x^{4/3})=cos(x^4)$?












-1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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


















-1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















-1












-1








-1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 '18 at 3:17







John adams

















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
















  • 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












2 Answers
2






active

oldest

votes


















2












$begingroup$

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!).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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



















0












$begingroup$

Although
$$
(x^{4/3})^3 = x^4,
$$
one does not have
$$
[f(x^{4/3})]^3=f(x^4)
$$
in general.



enter image description here




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 $x=0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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!).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















2












$begingroup$

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!).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














2












2








2





$begingroup$

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!).






share|cite|improve this answer









$endgroup$



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!).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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











0












$begingroup$

Although
$$
(x^{4/3})^3 = x^4,
$$
one does not have
$$
[f(x^{4/3})]^3=f(x^4)
$$
in general.



enter image description here




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 $x=0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















0












$begingroup$

Although
$$
(x^{4/3})^3 = x^4,
$$
one does not have
$$
[f(x^{4/3})]^3=f(x^4)
$$
in general.



enter image description here




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 $x=0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














0












0








0





$begingroup$

Although
$$
(x^{4/3})^3 = x^4,
$$
one does not have
$$
[f(x^{4/3})]^3=f(x^4)
$$
in general.



enter image description here




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 $x=0$.






share|cite|improve this answer











$endgroup$



Although
$$
(x^{4/3})^3 = x^4,
$$
one does not have
$$
[f(x^{4/3})]^3=f(x^4)
$$
in general.



enter image description here




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 $x=0$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 24 '18 at 4:24

























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


















  • $begingroup$
    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


















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