Asymptotic approximation/expansion for arccosine function?
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Trying to find a 3-term asymptotic expansion for $z=cos^{-1}(x)$, as $xrightarrow1^-$. Found a lot of examples online for inverse tangent, cosine, etc. but have yet to find any guidance on the inverse cosine function. Any help would be greatly appreciated.
asymptotics trigonometric-series
asked Nov 15 at 20:17
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up vote
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down vote
favorite
Trying to find a 3-term asymptotic expansion for $z=cos^{-1}(x)$, as $xrightarrow1^-$. Found a lot of examples online for inverse tangent, cosine, etc. but have yet to find any guidance on the inverse cosine function. Any help would be greatly appreciated.
asymptotics trigonometric-series
asked Nov 15 at 20:17
1460
11
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Trying to find a 3-term asymptotic expansion for $z=cos^{-1}(x)$, as $xrightarrow1^-$. Found a lot of examples online for inverse tangent, cosine, etc. but have yet to find any guidance on the inverse cosine function. Any help would be greatly appreciated.
asymptotics trigonometric-series
asked Nov 15 at 20:17
1460
11
Trying to find a 3-term asymptotic expansion for $z=cos^{-1}(x)$, as $xrightarrow1^-$. Found a lot of examples online for inverse tangent, cosine, etc. but have yet to find any guidance on the inverse cosine function. Any help would be greatly appreciated.
asymptotics trigonometric-series
asymptotics trigonometric-series
asked Nov 15 at 20:17
1460
11
asked Nov 15 at 20:17
1460
11
asked Nov 15 at 20:17
1460
11
asked Nov 15 at 20:17
1460
11
asked Nov 15 at 20:17
1460
11
11
add a comment |
add a comment |
1 Answer
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This is equivalent to the expansion of $arccos(1-t)$ at $t=0^{+}$.
Ask Wolfram Alpha Series[Arccos[1-t)],{t,0,3}]and re-substitute $t=1-x$ in the result
$$arccos(1-t) =sqrt{2} sqrt{t} + frac{t^{3/2}}{6 sqrt{2}} + frac{3t^{5/2}}{80 sqrt{2}}
+ frac{5 t^{7/2}}{448 sqrt{2}} + O(t^{9/2})$$
or see 1.
Yet another approach would be do use the trigonometric identity from
2 and get
$$arccos(1-t) = 2arcsinleft(sqrt{t/2}right),$$
Then use the expansion of $arcsin.$
answered Nov 15 at 21:01
gammatester
16.3k21630
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
This is equivalent to the expansion of $arccos(1-t)$ at $t=0^{+}$.
Ask Wolfram Alpha Series[Arccos[1-t)],{t,0,3}]and re-substitute $t=1-x$ in the result
$$arccos(1-t) =sqrt{2} sqrt{t} + frac{t^{3/2}}{6 sqrt{2}} + frac{3t^{5/2}}{80 sqrt{2}}
+ frac{5 t^{7/2}}{448 sqrt{2}} + O(t^{9/2})$$
or see 1.
Yet another approach would be do use the trigonometric identity from
2 and get
$$arccos(1-t) = 2arcsinleft(sqrt{t/2}right),$$
Then use the expansion of $arcsin.$
answered Nov 15 at 21:01
gammatester
16.3k21630
add a comment |
up vote
1
down vote
This is equivalent to the expansion of $arccos(1-t)$ at $t=0^{+}$.
Ask Wolfram Alpha Series[Arccos[1-t)],{t,0,3}]and re-substitute $t=1-x$ in the result
$$arccos(1-t) =sqrt{2} sqrt{t} + frac{t^{3/2}}{6 sqrt{2}} + frac{3t^{5/2}}{80 sqrt{2}}
+ frac{5 t^{7/2}}{448 sqrt{2}} + O(t^{9/2})$$
or see 1.
Yet another approach would be do use the trigonometric identity from
2 and get
$$arccos(1-t) = 2arcsinleft(sqrt{t/2}right),$$
Then use the expansion of $arcsin.$
answered Nov 15 at 21:01
gammatester
16.3k21630
add a comment |
up vote
1
down vote
up vote
1
down vote
This is equivalent to the expansion of $arccos(1-t)$ at $t=0^{+}$.
Ask Wolfram Alpha Series[Arccos[1-t)],{t,0,3}]and re-substitute $t=1-x$ in the result
$$arccos(1-t) =sqrt{2} sqrt{t} + frac{t^{3/2}}{6 sqrt{2}} + frac{3t^{5/2}}{80 sqrt{2}}
+ frac{5 t^{7/2}}{448 sqrt{2}} + O(t^{9/2})$$
or see 1.
Yet another approach would be do use the trigonometric identity from
2 and get
$$arccos(1-t) = 2arcsinleft(sqrt{t/2}right),$$
Then use the expansion of $arcsin.$
answered Nov 15 at 21:01
gammatester
16.3k21630
This is equivalent to the expansion of $arccos(1-t)$ at $t=0^{+}$.
Ask Wolfram Alpha Series[Arccos[1-t)],{t,0,3}]and re-substitute $t=1-x$ in the result
$$arccos(1-t) =sqrt{2} sqrt{t} + frac{t^{3/2}}{6 sqrt{2}} + frac{3t^{5/2}}{80 sqrt{2}}
+ frac{5 t^{7/2}}{448 sqrt{2}} + O(t^{9/2})$$
or see 1.
Yet another approach would be do use the trigonometric identity from
2 and get
$$arccos(1-t) = 2arcsinleft(sqrt{t/2}right),$$
Then use the expansion of $arcsin.$
answered Nov 15 at 21:01
gammatester
16.3k21630
edited Nov 15 at 22:06
answered Nov 15 at 21:01
gammatester
16.3k21630
answered Nov 15 at 21:01
gammatester
16.3k21630
answered Nov 15 at 21:01
gammatester
16.3k21630
16.3k21630
add a comment |
add a comment |
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