Cyclometric equation $xarcsin x=1$












0












$begingroup$


What method should I use to solve following equation for $x$ $$
xarcsin x=1?
$$
I would be grateful for any comment.






calculus trigonometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    Just a heads up, as everyone is doing it, please refrain from using the tag "real-analysis" for whatever considers calculus. Real analysis is not calculus and not algebra or pre-algebra or whatever. Secondly, do you have any thoughts on the given problem ?
    $endgroup$
    – Rebellos
    Nov 24 '18 at 10:49












  • $begingroup$
    Thank you for your remark. My idea was try to write it as $$sin(frac{1}{x})=x,$$ because $xneq 0$ and $sin$ is inverse to $arcsin$ on $[frac{-pi}{2},frac{pi}{2}]$.
    $endgroup$
    – akap
    Nov 24 '18 at 10:55












  • $begingroup$
    Yes, it's correct. The answer given below is smooth.
    $endgroup$
    – Rebellos
    Nov 24 '18 at 10:58
















0












$begingroup$


What method should I use to solve following equation for $x$ $$
xarcsin x=1?
$$
I would be grateful for any comment.






calculus trigonometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    Just a heads up, as everyone is doing it, please refrain from using the tag "real-analysis" for whatever considers calculus. Real analysis is not calculus and not algebra or pre-algebra or whatever. Secondly, do you have any thoughts on the given problem ?
    $endgroup$
    – Rebellos
    Nov 24 '18 at 10:49












  • $begingroup$
    Thank you for your remark. My idea was try to write it as $$sin(frac{1}{x})=x,$$ because $xneq 0$ and $sin$ is inverse to $arcsin$ on $[frac{-pi}{2},frac{pi}{2}]$.
    $endgroup$
    – akap
    Nov 24 '18 at 10:55












  • $begingroup$
    Yes, it's correct. The answer given below is smooth.
    $endgroup$
    – Rebellos
    Nov 24 '18 at 10:58














0












0








0





$begingroup$


What method should I use to solve following equation for $x$ $$
xarcsin x=1?
$$
I would be grateful for any comment.






calculus trigonometry







share|cite|improve this question











$endgroup$




What method should I use to solve following equation for $x$ $$
xarcsin x=1?
$$
I would be grateful for any comment.






calculus trigonometry




calculus 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 10:49









Rebellos

14.5k31246




14.5k31246










asked Nov 24 '18 at 10:44









akapakap

715




715












  • $begingroup$
    Just a heads up, as everyone is doing it, please refrain from using the tag "real-analysis" for whatever considers calculus. Real analysis is not calculus and not algebra or pre-algebra or whatever. Secondly, do you have any thoughts on the given problem ?
    $endgroup$
    – Rebellos
    Nov 24 '18 at 10:49












  • $begingroup$
    Thank you for your remark. My idea was try to write it as $$sin(frac{1}{x})=x,$$ because $xneq 0$ and $sin$ is inverse to $arcsin$ on $[frac{-pi}{2},frac{pi}{2}]$.
    $endgroup$
    – akap
    Nov 24 '18 at 10:55












  • $begingroup$
    Yes, it's correct. The answer given below is smooth.
    $endgroup$
    – Rebellos
    Nov 24 '18 at 10:58


















  • $begingroup$
    Just a heads up, as everyone is doing it, please refrain from using the tag "real-analysis" for whatever considers calculus. Real analysis is not calculus and not algebra or pre-algebra or whatever. Secondly, do you have any thoughts on the given problem ?
    $endgroup$
    – Rebellos
    Nov 24 '18 at 10:49












  • $begingroup$
    Thank you for your remark. My idea was try to write it as $$sin(frac{1}{x})=x,$$ because $xneq 0$ and $sin$ is inverse to $arcsin$ on $[frac{-pi}{2},frac{pi}{2}]$.
    $endgroup$
    – akap
    Nov 24 '18 at 10:55












  • $begingroup$
    Yes, it's correct. The answer given below is smooth.
    $endgroup$
    – Rebellos
    Nov 24 '18 at 10:58
















$begingroup$
Just a heads up, as everyone is doing it, please refrain from using the tag "real-analysis" for whatever considers calculus. Real analysis is not calculus and not algebra or pre-algebra or whatever. Secondly, do you have any thoughts on the given problem ?
$endgroup$
– Rebellos
Nov 24 '18 at 10:49






$begingroup$
Just a heads up, as everyone is doing it, please refrain from using the tag "real-analysis" for whatever considers calculus. Real analysis is not calculus and not algebra or pre-algebra or whatever. Secondly, do you have any thoughts on the given problem ?
$endgroup$
– Rebellos
Nov 24 '18 at 10:49














$begingroup$
Thank you for your remark. My idea was try to write it as $$sin(frac{1}{x})=x,$$ because $xneq 0$ and $sin$ is inverse to $arcsin$ on $[frac{-pi}{2},frac{pi}{2}]$.
$endgroup$
– akap
Nov 24 '18 at 10:55






$begingroup$
Thank you for your remark. My idea was try to write it as $$sin(frac{1}{x})=x,$$ because $xneq 0$ and $sin$ is inverse to $arcsin$ on $[frac{-pi}{2},frac{pi}{2}]$.
$endgroup$
– akap
Nov 24 '18 at 10:55














$begingroup$
Yes, it's correct. The answer given below is smooth.
$endgroup$
– Rebellos
Nov 24 '18 at 10:58




$begingroup$
Yes, it's correct. The answer given below is smooth.
$endgroup$
– Rebellos
Nov 24 '18 at 10:58










1 Answer
1






active

oldest

votes


















3












$begingroup$

$$xarcsin x = 1 implies arcsin x = frac{1}{x} implies sin frac{1}{x} = x$$



There isn't a good way to solve this, so I guess you can estimate using the MacLaurin expansion of $sin x$.



$$sin x = x-frac{x^3}{3!}+frac{x^5}{5!}-... implies sin frac{1}{x} = frac{1}{x}-frac{1}{3!x^3}+frac{1}{5!x^5}-...$$



You can truncate the series at a point to estimate for $x$. For example, you can solve



$$frac{1}{x}-frac{1}{3!x^3} = x$$



which gives $x approx pm0.888$ (true) and $x approx pm0.46$ (extraneous). You can use more terms for a more accurate estimate.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The value $pm 0.46$ doesn't hold.
    $endgroup$
    – Rebellos
    Nov 24 '18 at 10:59










  • $begingroup$
    Yeah, I labelled it as extraneous.
    $endgroup$
    – KM101
    Nov 24 '18 at 10:59






  • 2




    $begingroup$
    It needs some justification on why it shall not be accepted (my opinion). Generally it's a bad exercise from the start, your solution elaboration is smooth (+1).
    $endgroup$
    – Rebellos
    Nov 24 '18 at 11:00






  • 2




    $begingroup$
    No problem! (And yes, it's not a very good exercise.) I also accidentally forgot to edit the expansion for $sin frac{1}{x}$, which I've now done. (In case it caused any confusion or anything.)
    $endgroup$
    – KM101
    Nov 24 '18 at 11:01













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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

$$xarcsin x = 1 implies arcsin x = frac{1}{x} implies sin frac{1}{x} = x$$



There isn't a good way to solve this, so I guess you can estimate using the MacLaurin expansion of $sin x$.



$$sin x = x-frac{x^3}{3!}+frac{x^5}{5!}-... implies sin frac{1}{x} = frac{1}{x}-frac{1}{3!x^3}+frac{1}{5!x^5}-...$$



You can truncate the series at a point to estimate for $x$. For example, you can solve



$$frac{1}{x}-frac{1}{3!x^3} = x$$



which gives $x approx pm0.888$ (true) and $x approx pm0.46$ (extraneous). You can use more terms for a more accurate estimate.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The value $pm 0.46$ doesn't hold.
    $endgroup$
    – Rebellos
    Nov 24 '18 at 10:59










  • $begingroup$
    Yeah, I labelled it as extraneous.
    $endgroup$
    – KM101
    Nov 24 '18 at 10:59






  • 2




    $begingroup$
    It needs some justification on why it shall not be accepted (my opinion). Generally it's a bad exercise from the start, your solution elaboration is smooth (+1).
    $endgroup$
    – Rebellos
    Nov 24 '18 at 11:00






  • 2




    $begingroup$
    No problem! (And yes, it's not a very good exercise.) I also accidentally forgot to edit the expansion for $sin frac{1}{x}$, which I've now done. (In case it caused any confusion or anything.)
    $endgroup$
    – KM101
    Nov 24 '18 at 11:01


















3












$begingroup$

$$xarcsin x = 1 implies arcsin x = frac{1}{x} implies sin frac{1}{x} = x$$



There isn't a good way to solve this, so I guess you can estimate using the MacLaurin expansion of $sin x$.



$$sin x = x-frac{x^3}{3!}+frac{x^5}{5!}-... implies sin frac{1}{x} = frac{1}{x}-frac{1}{3!x^3}+frac{1}{5!x^5}-...$$



You can truncate the series at a point to estimate for $x$. For example, you can solve



$$frac{1}{x}-frac{1}{3!x^3} = x$$



which gives $x approx pm0.888$ (true) and $x approx pm0.46$ (extraneous). You can use more terms for a more accurate estimate.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The value $pm 0.46$ doesn't hold.
    $endgroup$
    – Rebellos
    Nov 24 '18 at 10:59










  • $begingroup$
    Yeah, I labelled it as extraneous.
    $endgroup$
    – KM101
    Nov 24 '18 at 10:59






  • 2




    $begingroup$
    It needs some justification on why it shall not be accepted (my opinion). Generally it's a bad exercise from the start, your solution elaboration is smooth (+1).
    $endgroup$
    – Rebellos
    Nov 24 '18 at 11:00






  • 2




    $begingroup$
    No problem! (And yes, it's not a very good exercise.) I also accidentally forgot to edit the expansion for $sin frac{1}{x}$, which I've now done. (In case it caused any confusion or anything.)
    $endgroup$
    – KM101
    Nov 24 '18 at 11:01
















3












3








3





$begingroup$

$$xarcsin x = 1 implies arcsin x = frac{1}{x} implies sin frac{1}{x} = x$$



There isn't a good way to solve this, so I guess you can estimate using the MacLaurin expansion of $sin x$.



$$sin x = x-frac{x^3}{3!}+frac{x^5}{5!}-... implies sin frac{1}{x} = frac{1}{x}-frac{1}{3!x^3}+frac{1}{5!x^5}-...$$



You can truncate the series at a point to estimate for $x$. For example, you can solve



$$frac{1}{x}-frac{1}{3!x^3} = x$$



which gives $x approx pm0.888$ (true) and $x approx pm0.46$ (extraneous). You can use more terms for a more accurate estimate.






share|cite|improve this answer











$endgroup$



$$xarcsin x = 1 implies arcsin x = frac{1}{x} implies sin frac{1}{x} = x$$



There isn't a good way to solve this, so I guess you can estimate using the MacLaurin expansion of $sin x$.



$$sin x = x-frac{x^3}{3!}+frac{x^5}{5!}-... implies sin frac{1}{x} = frac{1}{x}-frac{1}{3!x^3}+frac{1}{5!x^5}-...$$



You can truncate the series at a point to estimate for $x$. For example, you can solve



$$frac{1}{x}-frac{1}{3!x^3} = x$$



which gives $x approx pm0.888$ (true) and $x approx pm0.46$ (extraneous). You can use more terms for a more accurate estimate.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 24 '18 at 11:02

























answered Nov 24 '18 at 10:56









KM101KM101

5,9261523




5,9261523












  • $begingroup$
    The value $pm 0.46$ doesn't hold.
    $endgroup$
    – Rebellos
    Nov 24 '18 at 10:59










  • $begingroup$
    Yeah, I labelled it as extraneous.
    $endgroup$
    – KM101
    Nov 24 '18 at 10:59






  • 2




    $begingroup$
    It needs some justification on why it shall not be accepted (my opinion). Generally it's a bad exercise from the start, your solution elaboration is smooth (+1).
    $endgroup$
    – Rebellos
    Nov 24 '18 at 11:00






  • 2




    $begingroup$
    No problem! (And yes, it's not a very good exercise.) I also accidentally forgot to edit the expansion for $sin frac{1}{x}$, which I've now done. (In case it caused any confusion or anything.)
    $endgroup$
    – KM101
    Nov 24 '18 at 11:01




















  • $begingroup$
    The value $pm 0.46$ doesn't hold.
    $endgroup$
    – Rebellos
    Nov 24 '18 at 10:59










  • $begingroup$
    Yeah, I labelled it as extraneous.
    $endgroup$
    – KM101
    Nov 24 '18 at 10:59






  • 2




    $begingroup$
    It needs some justification on why it shall not be accepted (my opinion). Generally it's a bad exercise from the start, your solution elaboration is smooth (+1).
    $endgroup$
    – Rebellos
    Nov 24 '18 at 11:00






  • 2




    $begingroup$
    No problem! (And yes, it's not a very good exercise.) I also accidentally forgot to edit the expansion for $sin frac{1}{x}$, which I've now done. (In case it caused any confusion or anything.)
    $endgroup$
    – KM101
    Nov 24 '18 at 11:01


















$begingroup$
The value $pm 0.46$ doesn't hold.
$endgroup$
– Rebellos
Nov 24 '18 at 10:59




$begingroup$
The value $pm 0.46$ doesn't hold.
$endgroup$
– Rebellos
Nov 24 '18 at 10:59












$begingroup$
Yeah, I labelled it as extraneous.
$endgroup$
– KM101
Nov 24 '18 at 10:59




$begingroup$
Yeah, I labelled it as extraneous.
$endgroup$
– KM101
Nov 24 '18 at 10:59




2




2




$begingroup$
It needs some justification on why it shall not be accepted (my opinion). Generally it's a bad exercise from the start, your solution elaboration is smooth (+1).
$endgroup$
– Rebellos
Nov 24 '18 at 11:00




$begingroup$
It needs some justification on why it shall not be accepted (my opinion). Generally it's a bad exercise from the start, your solution elaboration is smooth (+1).
$endgroup$
– Rebellos
Nov 24 '18 at 11:00




2




2




$begingroup$
No problem! (And yes, it's not a very good exercise.) I also accidentally forgot to edit the expansion for $sin frac{1}{x}$, which I've now done. (In case it caused any confusion or anything.)
$endgroup$
– KM101
Nov 24 '18 at 11:01






$begingroup$
No problem! (And yes, it's not a very good exercise.) I also accidentally forgot to edit the expansion for $sin frac{1}{x}$, which I've now done. (In case it caused any confusion or anything.)
$endgroup$
– KM101
Nov 24 '18 at 11:01




















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