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ln 2

If we use the integral represenation of the natural logarithm then

ln 2 = Underoverscript[∫, 1, arg3] 1/x d x = 0.693 147 180 559 945 ... (1)

We also have

ln 2 = Underoverscript[∫, 1/2, arg3] 1/x d x = Underoverscript[∫, 2, arg3] 1/x d x = Underoverscript[∫, 4, arg3] 1/x d x = ...

We can also interpret these integrals as the area under the hyperbola  typeset structure with typeset structure going from 1 to 2, from typeset structure to typeset structure, from 2 to 4, from 4 to 8, etc.   

[Graphics:HTMLFiles/Log2_7.gif]

Without mentioning the logarithms the quadrature of the space between the hyperbola and its asymptotes was influenced by Jesuit Grégory de Saint Vincent by his work in Book VII of his Opus geometricum (Antwerp 1647).  He established that if parallels to one asymptote are drawn between the hyperbola and the other asymptote in such a way that the successive areas of the mixtilinear quadrilaterals thus formed are equal, then the length of  these parallels for a geometric progression  [1] .

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Mercator Taylor series expansion for the logarithmic function    with typeset structure yields  

ln 2 = 1 - 1/2 + 1/3 - 1/4 + 1/5 - ...(2)

This a very slowly convergent series. If we take typeset structure instead we get a series with a geometric series convergent rate

ln 2 = Underoverscript[∑, k = 1, arg3] 1/(k 2^k) .(3)

Glaisher found the following expression

ln 2 = 1 - 1/2 Underoverscript[∑, j = 2, arg3] (-1)^(j - 1)/j S _ j ,

where

S _ j = 1/3^j + 1/2 (1/5^j + 1/7^j) + 1/4 (1/9^j + 1/11^j + 1/13^j + 1/15^j) + 1/8 (1/17^j + 1/19^j + ... + 1/31^j) + ...

He also found [2]  that

ln 2 = 1327/1920 + 45/4 Underoverscript[∑, n = 4, arg3] (-1)^n/(n(n^2 - 1) (n^2 - 4) (n^2 - 9)) ,

ln 2 = 5/8 + 1/2 Underoverscript[∑, n = 2, arg3] 1/n(n^2 - 1) = 131/192 + 3/2 Underoverscript[∑, O = O, arg3] (-1)^(n - 1)/(n(n^2 - 1) (n^2 - 4))

ln 2 = 9/13 + 450 Underoverscript[∑, m = 3, arg3] 1/((4 m^2 - 1) (64 m^4 - 16 m^2 + 255))

ln 2 = 2/3 + 24 Underoverscript[∑, m = 3, arg3] (-1)^((m + 1)/2)/((m^2 - 1) (m^4 + 2 m^2 + 9)) - π/(4 k) tan π/(2 k) + 1/(4 k) (ψ((k + 1)/(4 k)) - ψ((k - 1)/(4 k))),

where typeset structure.

Starting from the identity  typeset structure N.Nielsen [3]  proved that

ln 2 = 1/2 · 1/2 + (1 · 3)/(2 · 4) · 1/4 + (1 · 3 · 5)/(2 · 4 · 6) · 1/6 + ...

References

[1]  Cajori, F. (1913, No. 1, Jan.). History of the Exponential and Logarithmic Concepts. Amer. Math. Monthly, 20, 5-14.

[2]  Glaisher, J. W. (1902). Methods of increasing the convergence of certain series of reciprocals. Quart. J., 34, 252-347.

[3]  Nielsen, N. (1894). Om typeset structure og typeset structure. Nyt Tidss. for Math. VB.,  22-25.

Cite this web-page as:

Štefan Porubský: ln 2.

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