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Welcome to NRICH.

Series expansion for cos^-1x

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By Brad Rodgers (P1930) on Wednesday, April 18, 2001 - 07:24 pm:

If

f=f(x)=-x^-1/2

Then

dⁿf/dxⁿ=(-1)ⁿ⁺¹[P_a=1²ⁿ⁺¹(a)] × x^-(2n+1)/2/2ⁿ

From power series,



Letting m=-x², and b=1, then integrating with respect to x, we get



By substituting intitial conditions cos^-1(0), we find that C=p/2. But this expansion doesn't seem to work; try cos^-1(1/2). What have I done wrong?

Thanks,

Brad

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By Brad Rodgers (P1930) on Friday, April 20, 2001 - 03:06 am:

Oops, I should write it as phi from a=1 to 2n-1; And, thats not entirely correct either, what I'm trying to write down is the multiplication of odd numbers. Any suggestions on how to do that?

Brad

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By Michael Doré (Michael) on Friday, April 20, 2001 - 06:10 pm:

Ah, I see, you can write:

P₁ⁿ(2a-1)

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By Brad Rodgers (P1930) on Friday, April 20, 2001 - 09:11 pm:

Does that mean that the following expansion is correct?



Thanks,

Brad

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By Michael Doré (Michael) on Saturday, April 21, 2001 - 02:29 am:

One thing you can do to simpify the P term is write it out in full:

(2n-1)(2n-3)...3 1

Multiply numerator and denominator by (2n)(2n-2)...4 2

(2n)!/[(2n)(2n-2)...4 2] = (2n)!/[2ⁿ×n!]

Therefore your P term is (2n)!/[2ⁿ×n!].

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By Michael Doré (Michael) on Wednesday, April 25, 2001 - 03:30 pm:

I've checked it through and I'm pretty sure it's correct. So we can write it out in slightly more economical form:

cos-1x =  p 2 +  x3 3 - ¥ å n=1   (2n)!x2n+1 22nn!2(2n+1)

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