Date: 11/21/2001 at 05:16:59 From: Henning Arnor Ulfarsson Subject: Taylor Expansion Can you give me the proof of this statement? The basis of the calculation is a Taylor series: arcsin(x) = x + 1/2 (x^3/3) + (1/2)(3/4)(x^5/5) + (1/2)(3/4)(5/6)(x^7/7) + ... Best regards, Henning
Date: 11/21/2001 at 12:10:17 From: Doctor Pete Subject: Re: Taylor Expansion Hi, This series can be found by using the generalized Binomial Theorem, which says (1+x)^n = a + ax + ax^2 + ... + a[k]x^k + ... where a[k] = Binomial[n,k] = n!/(k!(n-k)!). However, n is not an integer. Since the derivative of the arcsine is (1-x^2)^(-1/2), we see that n = -1/2, and we substitute -x^2 for x in the formula. Then integrating term by term gives the final result. You may wonder how to define the binomial coefficient when n is not an integer. Since n!/(k!(n-k)!) = n(n-1)(n-2)...(n-k+1)/k!, it is natural to use the right-hand side when n is not an integer. For n = -1/2, we have Binomial[1/2,k] = (-1/2)(-3/2)(-5/2)...(-(2k-1)/2)/k! = (-1)^k (1*3*5*...*(2k-1))/(k!2^k). We can express this as factorials of integers by noting (2k)! = (1*3*5*...*(2k-1))(2*4*6*...*(2k)) = (1*3*5*...*(2k-1))(2^k)(1*2*3*...*k) = (1*3*5*...*(2k-1))(2^k)k!, therefore a[k] = (-1)^k (2k)!/((2^k)k!)/(k!2^k) = (-1)^k (2k)!/(k!2^k)^2. It follows that the k(th) term of the expansion of (1-x^2)^(-1/2) is a[k](-x^2)^k = (2k)!/(k!2^k)^2 x^(2k). Integrating with respect to x, we obtain the k(th) term of the expansion of arcsin(x), (2k)!/(k!2^k)^2 x^(2k+1)/(2k+1), where k = 0 to infinity. It is not difficult to see that this agrees with the orignal statement of the problem. By the uniqueness of the series expansion, we can be assured that this is the series representation of arcsin(x). It is interesting to note that we did not rely on Taylor's formula, which says that for F[x] = arcsin(x), a[k] = F(k)/k!, where F(k)[x] is the k(th) derivative of F, evaluated at x. - Doctor Pete, The Math Forum http://mathforum.org/dr.math/
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