Cyclometric equation $xarcsin x=1$
up vote
0
down vote
favorite
What method should I use to solve following equation for $x$ $$
xarcsin x=1?
$$
I would be grateful for any comment.
calculus trigonometry
edited Nov 24 at 10:49
Rebellos
14.1k31244
asked Nov 24 at 10:44
akap
715
add a comment |
up vote
0
down vote
favorite
What method should I use to solve following equation for $x$ $$
xarcsin x=1?
$$
I would be grateful for any comment.
calculus trigonometry
edited Nov 24 at 10:49
Rebellos
14.1k31244
asked Nov 24 at 10:44
akap
715
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 ?
– Rebellos
Nov 24 at 10:49
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 $[frac{-pi}{2},frac{pi}{2}]$.
– akap
Nov 24 at 10:55
Yes, it's correct. The answer given below is smooth.
– Rebellos
Nov 24 at 10:58
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
What method should I use to solve following equation for $x$ $$
xarcsin x=1?
$$
I would be grateful for any comment.
calculus trigonometry
edited Nov 24 at 10:49
Rebellos
14.1k31244
asked Nov 24 at 10:44
akap
715
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
edited Nov 24 at 10:49
Rebellos
14.1k31244
asked Nov 24 at 10:44
akap
715
edited Nov 24 at 10:49
Rebellos
14.1k31244
asked Nov 24 at 10:44
akap
715
edited Nov 24 at 10:49
Rebellos
14.1k31244
edited Nov 24 at 10:49
Rebellos
14.1k31244
edited Nov 24 at 10:49
Rebellos
14.1k31244
14.1k31244
asked Nov 24 at 10:44
akap
715
asked Nov 24 at 10:44
akap
715
asked Nov 24 at 10:44
akap
715
715
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 ?
– Rebellos
Nov 24 at 10:49
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 $[frac{-pi}{2},frac{pi}{2}]$.
– akap
Nov 24 at 10:55
Yes, it's correct. The answer given below is smooth.
– Rebellos
Nov 24 at 10:58
add a comment |
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 ?
– Rebellos
Nov 24 at 10:49
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 $[frac{-pi}{2},frac{pi}{2}]$.
– akap
Nov 24 at 10:55
Yes, it's correct. The answer given below is smooth.
– Rebellos
Nov 24 at 10:58
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 ?
– Rebellos
Nov 24 at 10:49
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 ?
– Rebellos
Nov 24 at 10:49
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 $[frac{-pi}{2},frac{pi}{2}]$.
– akap
Nov 24 at 10:55
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 $[frac{-pi}{2},frac{pi}{2}]$.
– akap
Nov 24 at 10:55
Yes, it's correct. The answer given below is smooth.
– Rebellos
Nov 24 at 10:58
Yes, it's correct. The answer given below is smooth.
– Rebellos
Nov 24 at 10:58
add a comment |
1 Answer
1
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up vote
3
down vote
$$xarcsin x = 1 implies arcsin x = frac{1}{x} implies sin frac{1}{x} = x$$
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.
answered Nov 24 at 10:56
KM101
3,945417
The value $pm 0.46$ doesn't hold.
– Rebellos
Nov 24 at 10:59
Yeah, I labelled it as extraneous.
– KM101
Nov 24 at 10:59
2
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).
– Rebellos
Nov 24 at 11:00
2
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.)
– KM101
Nov 24 at 11:01
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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active
oldest
votes
up vote
3
down vote
$$xarcsin x = 1 implies arcsin x = frac{1}{x} implies sin frac{1}{x} = x$$
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.
answered Nov 24 at 10:56
KM101
3,945417
The value $pm 0.46$ doesn't hold.
– Rebellos
Nov 24 at 10:59
Yeah, I labelled it as extraneous.
– KM101
Nov 24 at 10:59
2
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).
– Rebellos
Nov 24 at 11:00
2
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.)
– KM101
Nov 24 at 11:01
add a comment |
up vote
3
down vote
$$xarcsin x = 1 implies arcsin x = frac{1}{x} implies sin frac{1}{x} = x$$
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.
answered Nov 24 at 10:56
KM101
3,945417
The value $pm 0.46$ doesn't hold.
– Rebellos
Nov 24 at 10:59
Yeah, I labelled it as extraneous.
– KM101
Nov 24 at 10:59
2
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).
– Rebellos
Nov 24 at 11:00
2
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.)
– KM101
Nov 24 at 11:01
add a comment |
up vote
3
down vote
up vote
3
down vote
$$xarcsin x = 1 implies arcsin x = frac{1}{x} implies sin frac{1}{x} = x$$
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.
answered Nov 24 at 10:56
KM101
3,945417
$$xarcsin x = 1 implies arcsin x = frac{1}{x} implies sin frac{1}{x} = x$$
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.
answered Nov 24 at 10:56
KM101
3,945417
edited Nov 24 at 11:02
answered Nov 24 at 10:56
KM101
3,945417
answered Nov 24 at 10:56
KM101
3,945417
answered Nov 24 at 10:56
KM101
3,945417
3,945417
The value $pm 0.46$ doesn't hold.
– Rebellos
Nov 24 at 10:59
Yeah, I labelled it as extraneous.
– KM101
Nov 24 at 10:59
2
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).
– Rebellos
Nov 24 at 11:00
2
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.)
– KM101
Nov 24 at 11:01
add a comment |
The value $pm 0.46$ doesn't hold.
– Rebellos
Nov 24 at 10:59
Yeah, I labelled it as extraneous.
– KM101
Nov 24 at 10:59
2
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).
– Rebellos
Nov 24 at 11:00
2
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.)
– KM101
Nov 24 at 11:01
The value $pm 0.46$ doesn't hold.
– Rebellos
Nov 24 at 10:59
The value $pm 0.46$ doesn't hold.
– Rebellos
Nov 24 at 10:59
Yeah, I labelled it as extraneous.
– KM101
Nov 24 at 10:59
Yeah, I labelled it as extraneous.
– KM101
Nov 24 at 10:59
2
2
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).
– Rebellos
Nov 24 at 11:00
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).
– Rebellos
Nov 24 at 11:00
2
2
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.)
– KM101
Nov 24 at 11:01
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.)
– KM101
Nov 24 at 11:01
add a comment |
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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 ?
– Rebellos
Nov 24 at 10:49
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 $[frac{-pi}{2},frac{pi}{2}]$.
– akap
Nov 24 at 10:55
Yes, it's correct. The answer given below is smooth.
– Rebellos
Nov 24 at 10:58