Maclaurin series for arcsin
Tom Scavo
x^3 3x^5 5x^7
arcsin x = x + ----- + ------ + ------ + ... -1 < x < 1
6 40 112
is giving me some problems. I know how to derive series for arctan
and log (in virtually any calculus book), and suspect this one's
handled similarly (set up as a geometric series and integrate
term-by-term), but I don't quite see how to do it.
1) How is this series derived?
2) What is its general term?
3) Does it converge at the endpoints?
No, this is not homework. This came up recently while considering a
sequence of iteration functions with high-order convergence to pi .
Any help will be appreciated.
--
Tom Scavo
sc...@cie.uoregon.edu
Christopher J. Henrich
>The Maclaurin series for arcsin
>
> x^3 3x^5 5x^7
> arcsin x = x + ----- + ------ + ------ + ... -1 < x < 1
> 6 40 112
>
>is giving me some problems.
>1) How is this series derived?
>
this in even posers of x by the binomial theorem. Then you
can integrate term by term.
>2) What is its general term?
>
It has a closed form, if you don't mind a few factorials.
>3) Does it converge at the endpoints?
>
I think so.
Regards
Chris Henrich
William Mc Ginty
-1 2 -1/2
sin x = Integral from 0 to x of { (1 - t ) wrt t }
Expand the integrand and integrate term by term,
inf 2n 2 2n+1
= Sum { ( (2n)!/[ 2 (n!) (2n+1)] ) x }
n=0
This is uniformly convergent for -1 <= x <= 1. Convergence at the
end points is by Gauss's test.
===============================================================
William McGinty Internet: wmcg...@portal.win.net
U.K. CompuServe User ID: 100113,3206
Tel: (44)0293-560667
William Mc Ginty
>The Maclaurin series for arcsin:
[bit omitted]
>1) How is this series derived?
>
>2) What is its general term?
>
>3) Does it converge at the endpoints?
>
-1 2 -1/2
sin x = integral from 0 to x of { (1-t ) wrt t }
Expand the integrand and integrate term by term,
inf 2n 2 2n+1
= Sum { ( (2n)!/[ 2 (n!) (2n+1) ] ) x }
n=0
This is uniformly convergent for -1 <= x <= 1. The end points converge
William Mc Ginty
Suppose a > 0 for n > an integer. If,
n
a
n+1 b -2
----- = 1 - --- + O( n ) for large n,
a n
n
Then Sum a is convergent if b > 1 and divergent otherwise.
n n
The series we are considering satisfies the conditions for x=1 and the
negative of the series similarly satsiefies the conditions when x = -1.
Larry Riddle
>
>>The Maclaurin series for arcsin:
>[bit omitted]
>>1) How is this series derived?
>>
>>2) What is its general term?
>>
>>3) Does it converge at the endpoints?
>>
>
> -1 2 -1/2
>sin x = integral from 0 to x of { (1-t ) wrt t }
>
>Expand the integrand and integrate term by term,
>
> inf 2n 2 2n+1
> = Sum { ( (2n)!/[ 2 (n!) (2n+1) ] ) x }
> n=0
>
>This is uniformly convergent for -1 <= x <= 1. The end points converge
>by Gauss's test.
>
The limit form of Raabe's test provides a very simple way to test the
series at x=1 for convergence. For a reference, see "An Introduction to
Mathematical Analysis" by Myrtle Lewin and Jonathon Lewin.
Here is the "Limit Form of Raabe's Test": If a[n] positive sequence and p
is some real number, and if
n(1-a[n+1]/a[n]) -> p as n -> infinity
and if p > 1, then the series converges, and if p < 1 the series diverges.
In our case (using Mathematica to help format the output for you to
read):
In[1]:= (2n+2)(2n+1)(2n+1)/(4(n+1)(n+1)(2n+3))
2
(1 + 2 n) (2 + 2 n)
Out[1]= -------------------- (this is a[n+1]/a[n] after simplifying)
2
4 (1 + n) (3 + 2 n)
In[2]:= n (1 - %)
2
(1 + 2 n) (2 + 2 n)
Out[2]= n (1 - --------------------)
2
4 (1 + n) (3 + 2 n)
In[3]:= Limit[%, n -> Infinity]
3
Out[3]= -
2
so p = 3/2 > 1, so this would prove the series converges at x = 1 (and x = -1).
--
Larry Riddle | rid...@mathcs.emory.edu PREFERRED
Agnes Scott College | {rutgers,gatech}!emory!riddle UUCP
Dept of Math | rid...@emory.bitnet NON-DOMAIN BITNET
Decatur, GA 30030 | Phone: Voice 404-371-6222, FAX 404-371-6177
Mark
>
> x^3 3x^5 5x^7
> arcsin x = x + ----- + ------ + ------ + ... -1 < x < 1
> 6 40 112
arcsin x = x + (1/2) x^3/3 + (1/2)(3/4) x^5/5 + (1/2)(3/4)(5/6) x^7/7 + ...
It comes from the derivative series:
(1 - x^2)^(-1/2) = 1 + (1/2) x^2 + (1/2)(3/4) x^4 + (1/2)(3/4)(5/6) x^6 + ...
The arcsin series converges at x = 1, since all the partial sums are always
less than pi.