A problem with sinh where I keep getting 3 different answers

  • Thread starter Thread starter schlynn
  • Start date Start date
AI Thread Summary
The discussion revolves around solving the equation sinh(x) = π, where the user initially calculated a solution using ln(2π)/2 but received conflicting results from calculators and WolframAlpha. The confusion stems from a misunderstanding of logarithmic properties, specifically that ln(a + b) does not equal ln(a) + ln(b). A suggestion is made to substitute u = e^x, transforming the equation into a quadratic form for easier solving. The user acknowledges the mistake regarding logarithmic properties and seeks clarification on the correct approach. The conversation highlights the importance of proper mathematical principles in solving transcendental equations.
schlynn
Messages
88
Reaction score
0

Homework Statement


This isn't a homework problem, just a thought problem. I want to find a solution to the equation sinh(x)=pi.


Homework Equations


sinh(x)=(e^x-e^-x)/2


The Attempt at a Solution


Here is a picture of the work that I did, pretty much, I set the equation for sinh equal to pi, and I found (ln(2pi)/2). Thats all one part, not the natural log of 2pi/2, that's just stupid. Any way, calculators gave my a number along the lines of 1.05 or something like that, and when I had wolframalpha solve sinh(x)=pi, they gave me a number with i and some square roots that I don't know where they came from. I just want an explanation of what's going on. When I plugged (ln(2pi)/2) into the equation for sinh, I got pi, so I showed that this give pi, but wolframalpha gives another answer, and calculators say I'm wrong. What did I do wrong? Sorry about the pictures, but I don't know LaTex.

photo-2.jpg

photo2.jpg
 
Physics news on Phys.org
The first thing you did wrong was saying ln(2pi+e^(-x))=ln(2pi)+ln(e^(-x)). That's completely wrong. ln(a+b) is not ln(a)+ln(b). Take your original equation and substitute u=e^x, so 1/u=e^(-x). Now it's quadratic in u. Solve for u first. Then take the log.
 
Oh, I made the mistake, the property is ln(ab)=ln(a)+ln(b), fail, alright, thanks.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top