Logarithmic Equivalent of the Inverse Hybolic Cosine FunctionDate: 10/28/97 at 06:58:22 From: Stephen Gurnell Subject: Logarithmic equivalent of the inverse hybolic cosine function Dr. Math, First, thank you very much for your recent help and swift replies. I have recently attempted a question which I think I have done wrong. I was wondering if there was a set formula for this problem: key --- __ / = square root Q. Use the logarithmic equivalent of the inverse hyperbolic cosine function, to deduce that: __ -1 (70) (35+/69) cosh (--) = ln(------) (68) ( 34 ) A?. -1 cosh 1.029411765 = ln(1.273724231) 0.241945074 = ln(1.273724231) 0.241945074 = 0.241945074 0.242 = 0.242 As you can see from the above I feel that this method of solving the problem must be wrong. as it appears too simple. Many thanks. Stephen Gurnell. Date: 10/28/97 at 10:07:58 From: Doctor Anthony Subject: Re: Logarithmic equivalent of the inverse hybolic cosine function If cosh^(-1)(x) = y, then x = cosh(y) e^y + e^(-y) x = -------------- 2 2x = e^y + e^-y and e^y - 2x + e^(-y) = 0 Multiply by e^y e^(2y) - 2x.e^y + 1 = 0 This is a quadratic in e^y, and applying the quadratic formula 2x +- sqrt(4x^2 - 4) e^y = ---------------------- 2 e^y = x +- sqrt(x^2-1) y = ln(x +-sqrt(x^2-1)) taking + and - sign this gives the two values y = +- ln(x^2+sqrt(x^2-1)) because x-sqrt(x^2-1) = 1/(x+sqrt(x^2-1)) and so cosh^(-1)(x) = +-ln(x+sqrt(x^2-1)) cosh(-1)(35/34) = +-ln(35/34 + sqrt((35/34)^2 - 1)) (35 + sqrt(69) = ln[--------------] 34 -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 10/28/97 at 10:09:26 From: Doctor Mitteldorf Subject: Re: Logarithmic equivalent of the inverse hybolic cosine function Dear Stephen, This is a fine method of showing that the identity is true, but it's not a mathematician's way. The mathematician might say, "How do you know it isn't just a coincidence that the two sides come out approximately equal? Maybe if you evaluated just one more decimal place than your hand calculator is capable of, they would come out different." Not very likely, but there's also a pedagogical reason: if you go back to your definitions, apply some algebra, and find that the two are exactly equivalent, you've practiced using some skills that may help you next time when the numbers are different - or maybe when there are variables in the problem, and not just numbers. So here's a start on the other way: Take cosh of both sides. Cosh of cosh^-1 is just the argument, 70/68. On the right, use the definition of cosh: cosh(x) = (e^x + e^(-x))/2 e^x here is (35+/69) (------) ( 34 ) since e^x is just the inverse of the ln function. What is e^(-x)? It's just the reciprocal of e^x. You take it from here... -Doctor Mitteldorf, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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