( 2 1 ) ! 1 + ( 2 3 ) ! 1 + ( 2 5 ) ! 1 + ( 2 7 ) ! 1 + ⋯ = ?
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Having a good command / familiarity with various types of function expansions can make it easier to manipulate such infinite sums.
Sometimes, it just boils down to knowing what to substitute in.
Just one doubt how u took 2 2 n + 1 .it doesn't seems to include even powers
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You are right. There was an error in my reference Erf -- from Wolfram MathWorld .
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SO is the problem incorrect or it should be modified to 2 1 ! 1 + 2 3 ! 2 + 2 5 ! 4 + 2 7 ! 8 + . . .
now ur method seems to be correct
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@Kyle Finch – The problem is correct. The summation equation given my reference was wrong. See image . Note the the expanded series matches = π 1 ( 1 2 + 1 ˙ 3 2 2 + 1 ˙ 3 ˙ 5 2 3 + 1 ˙ 3 ˙ 5 ˙ 7 2 4 + . . . ) .
Oh really? There is such a series defined for erf(x) ? I 'wasted' my energy computing it(of course using Beta function). But still computing it oneself is nice, made this problem harder.
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We know that the gamma function Γ ( 1 + x ) = x Γ x = x ! and Γ ( 2 1 ) = π . Therefore, the sum:
S = ( 2 1 ) ! 1 + ( 2 3 ) ! 1 + ( 2 5 ) ! 1 + ( 2 7 ) ! 1 + . . . = Γ ( 2 3 ) 1 + Γ ( 2 5 ) 1 + Γ ( 2 7 ) 1 + Γ ( 2 9 ) 1 + . . . = 2 1 Γ ( 2 1 ) 1 + 2 3 Γ ( 2 3 ) 1 + 2 5 Γ ( 2 5 ) 1 + 2 7 Γ ( 2 7 ) 1 + . . . = 2 1 Γ ( 2 1 ) 1 + 2 3 ˙ 2 1 Γ ( 2 1 ) 1 + 2 5 ˙ 2 3 ˙ 2 1 Γ ( 2 1 ) 1 + 2 7 ˙ 2 5 ˙ 2 3 ˙ 2 1 Γ ( 2 1 ) 1 + . . . = π 1 ( 1 2 + 1 ˙ 3 2 2 + 1 ˙ 3 ˙ 5 2 3 + 1 ˙ 3 ˙ 5 ˙ 7 2 4 + . . . ) = π 1 n = 0 ∑ ∞ ( 2 n + 1 ) ! ! 2 2 n + 1
We know that:
e r f ( x ) ⇒ S = π e − x 2 n = 0 ∑ ∞ ( 2 n + 1 ) ! ! ( 2 x ) 2 n + 1 = e e r f ( 1 ) = 2 . 7 1 8 2 × 0 . 8 4 2 7 = 2 . 2 9