Ngô Quốc Anh

November 25, 2007

arcsinh = ln?

Filed under: Các Bài Tập Nhỏ — Ngô Quốc Anh @ 22:19

Có 2 công thức sau đây mà chúng ta nên biết.

\text{arcsinh}\,x=\ln\,(x+\sqrt{1+x^2})

\operatorname{arcsinh}x =  - \ln \left( {\sqrt {x^2  + 1}  + x} \right).

Chứng minh. By definition, \sinh x = \frac{e^x - e^{-x}}{2}. Now let y = \sinh x, and let’s solve the equation for x:

y = \frac{e^x - e^{-x}}{2} => e^x - e^{-x} = 2y => e^{2x} - 1 = 2ye^x => e^{2x} - 2ye^x - 1 = 0

and now solving for e^x we get

e^x = \frac{2y \pm \sqrt{4y^2+4}}{2} => e^x = y + \sqrt{1+y^2} 

(we take the + sign because the exponential is always positive). Therefore we may conclude).

Since

\sqrt {x^2  + 1}  - x = \frac{{\left( {\sqrt {x^2  + 1}  - x} \right)\left( {\sqrt {x^2  + 1}  + x} \right)}}  <br> {{\sqrt {x^2  + 1}  + x}} = \frac{1}  <br> {{\sqrt {x^2  + 1}  + x}}

then

\ln \left( {\sqrt {x^2  + 1}  - x} \right) = \ln \left( {\frac{1}  <br> {{\sqrt {x^2  + 1}  + x}}} \right) =  - \ln \left( {\sqrt {x^2  + 1}  + x} \right).

 Thus

\operatorname{arcsinh}x =  - \ln \left( {\sqrt {x^2  + 1}  + x} \right).

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