View Full Version : summation of infinite series

grav

2009-May-29, 06:57 PM

Does anyone know what trig function the summation of the infinite series 1 + x^2 / 2 +3 x^2 / 8 + 5 x^6 / 16 + 35 x^8 / 128 + ... is equal to? Or better yet, 1 + (cos x)^2 / 2 + 3 (cos x)^4 / 8 + 5 (cos x)^6 / 16 + 35 (cos x)^8 / 128 + ... ? It is basically just the same thing as taking the trig function for asin x = x (1 + x^2 / 6 + 3 x^4 / 40 + 5 x^6 / 112 + 35 x^8 / 1152 + ...) = x (a + b + c + d + e + ...) and multiplying each term with consecutive odd numbers, so that we get x (a + 3b + 5c + 7d + 9e + ...).

grav

2009-May-29, 07:14 PM

I have also noticed it is similar to the infinite series for a square root, sqrt(1 + x) = 1 + x^2 / 2 - x^4 / 8 + x^6 / 16 - 5 x^8 / 128 + ... If we multiply each term with consecutive odd integers with alternating signs, then we get 1 + (1) x^2 / 2 - (-3) x^4 / 8 + (5) x^6 / 16 - (-7) x^8 / 128 + ... = 1 + x^2 / 2 + 3 x^4 / 8 + 5 x^6 / 16 + 35 x^8 / 128 + ... this way also.

Celestial Mechanic

2009-May-29, 07:58 PM

The first function, 1 + (1/2)x2 + (3/8)x4 + (5/16)x6 + (35/128)x8 + ... is (1-x2)-1/2. What you have observed and rediscovered above is the fact that (d/dx)arcsin(x) = (1-x2)-1/2.

Of course the above formulas are only true for -1 < x < 1.

grav

2009-May-29, 08:59 PM

The first function, 1 + (1/2)x2 + (3/8)x4 + (5/16)x6 + (35/128)x8 + ... is (1-x2)-1/2. What you have observed and rediscovered above is the fact that (d/dx)arcsin(x) = (1-x2)-1/2.

Of course the above formulas are only true for -1 < x < 1.Thank you very much, CM. I wish I had seen this earlier. I finally just ended up writing a program using the summation formula for asin, starting with the summation for the square root and transforming it according to the posts above into that for asin (mainly because the ones I found directly for asin dealt with ellipticals unnecessarily for some reason or functions I didn't recognize) with asin(x) = x [1 + summation (2n)! x^(2n) / (2n + 1) / (n!)^2 / 4^n], which worked out in the program, so then I changed it accordingly for what I needed with 1 + summation (2n)! x^(2n) / (n!)^2 / 4^n and got 1.1547 for x = .5, 1.25 for x = .6, and 1.6666 for x = .8, all of which work out with 1 / sqrt(1 - x^2), as you have. Thanks again, CM. :)

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