Finding a power series for arcsin(x)

[chouse12]

Homework Statement

1. Find a power series for (1-x)^(-1/2)
2. Find a power series for (1-x^2)^(-1/2)
3. Find a power series for arcsin(x)

Homework Equations

Binomial series, (1+x)^k= 1 + kx + k(k-1)x^2/2!+ k(k-1)(k-2)x^3/3!+...

The Attempt at a Solution

for 1., I have 1+ (-1/2)(-x) + (-1/2)(-1/2-1)(-x)^2/2! + (-1/2)(-1/2-1)(-1/2-2)(-x)^3/3!+...
Simplified, 1+ x/2 +3x^2/8+ 15x^3/48 + 35x^4/128+...
From here, I am having trouble finding the general term.
Once I can find that, the rest is easy...just use x^2 instead of x for (2) and integrate the series to get arcsin(x) for (3).
I know I could look up a power series for arcsin(x) in a table or website but the point is to get it using the binomial series.

 

[tiny-tim]

Welcome to PF!

Hi chouse12! Welcome to PF!

1. Find a power series for (1-x)^(-1/2)
2. Find a power series for (1-x^2)^(-1/2)
3. Find a power series for arcsin(x)
…
I know I could look up a power series for arcsin(x) in a table or website but the point is to get it using the binomial series.

Quite right!

hmm … a series of three questions …

Hint: integrate ​

 

[lanedance]

Homework Statement

1. Find a power series for (1-x)^(-1/2)
2. Find a power series for (1-x^2)^(-1/2)
3. Find a power series for arcsin(x)

Homework Equations

Binomial series, (1+x)^k= 1 + kx + k(k-1)x^2/2!+ k(k-1)(k-2)x^3/3!+...

The Attempt at a Solution

for 1., I have 1+ (-1/2)(-x) + (-1/2)(-1/2-1)(-x)^2/2! + (-1/2)(-1/2-1)(-1/2-2)(-x)^3/3!+...

how about only simplifying the subtractions first to look for a pattern, then multiply out the negative signs to either + or -
$$= 1+ (-\frac{1}{2})(-x) + (-\frac{1}{2})(-\frac{3}{2})(-x)^2\frac{1}{2!} + (-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})(-x)^3\frac{1}{3!}!+...$$

Simplified, 1+ x/2 +3x^2/8+ 15x^3/48 + 35x^4/128+...
From here, I am having trouble finding the general term.
Once I can find that, the rest is easy...just use x^2 instead of x for (2) and integrate the series to get arcsin(x) for (3).
I know I could look up a power series for arcsin(x) in a table or website but the point is to get it using the binomial series.

 

[chouse12]

tiny-tim, Thanks for the welcome and the hint. The main problem for me seems to be getting the first series though..

lanedance, so far I have the series expressed as (-1)^0/2^0/0!(-x)^0 + (((-1)/2^1)/1!)(-x)^1+(((-1)(-3)/2^2)/2!)(-x)^2+(((-1)(-3)(-5)/2^3)/1!)(-x)^3+...+ (((-1)(-3)(-5)...(2n-1)/2^n)/n!) (-x)^n+...
The trouble is I don't know how to express the (-1)(-3)(-5)...(2n-1) as a single term containing "n's".

Sorry for the newbformatting, I need to learn how to use latex

 

[chouse12]

Ok, I messed around with it a little more and ended up with this:
1+ (1/2)(x) + ((1)(3)/(2)(4))x^2 + ((1)(3)(5)/(2)(4)(6))x^3+...
So the coefficients on each term is with consecutive odd integers in the numerator, always multiplying by the next integer in the next term, and consecutive even integers in the denominator, always multiplying by the next integer in the next term (I multiplied all the negative signs to give all positive terms). Now, from this, how would I find the general term of the series?

 

[Dick]

Look up the definition of the 'double factorial', e.g. (2n-1)!.

 

[tiny-tim]

Hi chouse12!

(try using the X² tag just above the Reply box )

1+ (1/2)(x) + ((1)(3)/(2)(4))x^2 + ((1)(3)(5)/(2)(4)(6))x^3+...
…
Now, from this, how would I find the general term of the series?

How about ∑(2n)!/(n!)²2²ⁿ, = ²ⁿC_n/2²ⁿ ?

 

[chouse12]

Dick, I did not know about double factorials or other multiple factorials before so thanks for that. However, how do I make the top odd and the bottom even? On one website I looked at, it just has n! as a piecewise term, making the terms all even for n>0 and even and all odd for n>0 and odd. Or would it just be, as you stated, (2n-1)!?

Tiny-Tim, I believe that is the correct answer but could you explain how you arrived there?

Thanks to everyone for all the help.

 

[chouse12]

Dick, from Tiny-Tim's answer, I notice that it is just (2n-1)! simplified into terms with single factorials (i.e. (2n-1)!=(2n)!/2ⁿn!) but instead of in the denominator 2ⁿn!, it is that quantity squared (2²ⁿ(n!)²). Does this make the terms in the denominator the even integers?

 

[lanedance]

how about this
$$1.3.5.7...(2n-1)$$
$$=\frac{1.2.3.4.5.6.7...(2n-1)}{2.4.6...(2n-2)}$$
$$=\frac{(2n-1)!}{2(1).2(2).2(3)...2(n-1)}$$
$$=\frac{(2n-1)!}{2^{n-1}(n-1)!}$$

 

[chouse12]

lanedance, that is almost correct. That gives coefficients of 1, 2/2, 24/8, 720/48...
which simplify, to 1, 1, 3, 15...
The denominator is correct but the numerator is not. The coefficients should be 1, 1/2, 3/8, 15/48... so if the bottom term was squared then it would be correct. So any ideas on how to find the formula with the denominator squared (while showing work, not just squaring the bottom since it isn't right)?

 

[lanedance]

i only gave the term for the (1.3.5...(2n-1)) part of the sequence

your sequence is
$$= 1+ \frac{1}{2}x + \frac{1}{2}\frac{3}{2}x^2\frac{1}{2!} + \frac{1}{2}\frac{3}{2}\frac{5}{2}x^3\frac{1}{3!}+...$$

so the coefficient for the nth power of x should be
$$a_n = \frac{1.3.5...(2n-1)}{2^n.n!}$$

so you get the extra terms on the denominator... I've done this pretty quickly, should work, though you may have to be careful with how n is defined - ie which term it relates to

 

[chouse12]

Oh that clears everything up...Thank you so much for the help.

 

[chouse12]

Just another quick question...I need to find the interval of convergence for this series. I set up the ratio test and I think I must be doing it wrong because I am getting convergence for all real values of x but in my textbook it says the binomial series only converges from -1 to 1.

Here is my work:
lim n--> infinity of abs. value (((2n+1)!/(2ⁿ⁺¹(n+1)!)²)/((2n)!/(2ⁿ(n)!)²)) (x)

and I simplified that to lim n--> infinity of abs. value(2n+1/4(n+1)²) abs. value of (x)
and that limit goes to 0

 

[Dick]

If you've settled on a_n=(2n-1)!/(2^(n-1)*n!). And I'll accept Lanedance's opinion on that, since I haven't checked it and it seems about right. Then I get that a_(n+1)/a_n->1. That makes the limit a_(n+1)*x^(n+1)/(a_n*x^n) equal to x. Note: (2(n+1)-1)!/(2n-1)!=(2n+1).

 

[lanedance]

Just another quick question...I need to find the interval of convergence for this series. I set up the ratio test and I think I must be doing it wrong because I am getting convergence for all real values of x but in my textbook it says the binomial series only converges from -1 to 1.

Here is my work:
lim n--> infinity of abs. value (((2n+1)!/(2ⁿ⁺¹(n+1)!)²)/((2n)!/(2ⁿ(n)!)²)) (x)

this should be (2n-1)! in the denominator (instead of (2n!)) which will give you an extra remaining factor of n in the numerator and the limit Dick gievs

and I simplified that to lim n--> infinity of abs. value(2n+1/4(n+1)²) abs. value of (x)
and that limit goes to 0

 

[Dick]

this should be (2n-1)! in the denominator (instead of (2n!)) which will give you an extra remaining factor of n in the numerator and the limit Dick gievs

Nooo. (2(n+1)-1)!/(2n-1)!. The double factorials give you a (2n+1) factor in the numerator. The single factorials give you an n+1 in the denominator and the 2^n gives you a 2 in the denominator. lim (2n+1)/((n+1)*2)=1.

 

[lanedance]

Nooo. (2(n+1)-1)!/(2n-1)!. The double factorials give you a (2n+1) factor in the numerator. The single factorials give you an n+1 in the denominator and the 2^n gives you a 2 in the denominator. lim (2n+1)/((n+1)*2)=1.

Hey Dick, I agree with the double factroial thing which makes life a lot easier, by if you're working with singles

$$a_n = \frac{(2n-1)!}{(2^n.n!)^2}$$
(i haven't checked this either but it should be close based on working)

chouse wrote

Here is my work:
lim n--> infinity of abs. value (((2n+1)!/(2n+1(n+1)!)2)/((2n)!/(2n(n)!)2)) (x)

i meant it should be

$$a_n = \frac{\frac{(2n+1)!}{(2^{n+1}.(n+1)!)^2}} {\frac{(2n-1)!}{(2^n.n!)^2}}$$

which reduces to the required 1 in the limit

 

[tiny-tim]

Ok, I messed around with it a little more and ended up with this:
1+ (1/2)(x) + ((1)(3)/(2)(4))x^2 + ((1)(3)(5)/(2)(4)(6))x^3+...

How about ∑(2n)!/(n!)²2²ⁿ, = ²ⁿC_n/2²ⁿ ?

Tiny-Tim, I believe that is the correct answer but could you explain how you arrived there?

Hi chouse12!

the general term is (n odds)/(n evens),

so I multiplied top and bottom to get (n odds)(n evens)/(n evens)²

(n odds)(n evens) is just (2n)!,

and (n evens) is n!2ⁿ (because eg 2.4.6.8 = 2.1.2.2.2.3.2.4 = 2.2.2.2.1.2.3.4)

Incidentally, ²ⁿC_n/2²ⁿ is the chance of getting exactly n heads on tossing a coin 2n times ​

oh … and how are you doing on arcsin(x)?

 

[Dick]

i meant it should be

$$a_n = \frac{\frac{(2n+1)!}{(2^{n+1}.(n+1)!)^2}} {\frac{(2n-1)!}{(2^n.n!)^2}}$$

which reduces to the required 1 in the limit

Oh. Ok, then.

 

[chouse12]

I see now the mistake I was making with the ratio test...thanks for the help.

I can get the other two parts of the problem pretty easily now. Thanks again for all the help, couldn't have done it without you guys.