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












-1














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




















  • 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














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




















  • 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








-1







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















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 at 3:17

























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
















  • 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












2 Answers
2






active

oldest

votes


















2














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





















  • 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



















0














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























  • 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











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














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





















  • 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
















2














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





















  • 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














2












2








2






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












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 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


















  • 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











0














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























  • 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
















0














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























  • 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














0












0








0






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














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 at 4:24

























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


















  • 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


















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