Learn How to Solve Equations with Exponential Terms | Homework Help

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The discussion revolves around solving the equation Ae^(-at) + Be^(-bt) = (A+B)/2. Participants express the need for a clearer problem statement and more context to provide effective assistance. There is a call for the poster to share their attempts and thought processes regarding the solution. The emphasis is on understanding what is meant by "solve" in this context. Providing additional details will facilitate better guidance in solving the equation.
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Homework Statement



Ae^(-at) + Be^(-bt) = (A+B)/2

Homework Equations





The Attempt at a Solution

 
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ha.dnri said:

Homework Statement



Ae^(-at) + Be^(-bt) = (A+B)/2

Homework Equations





The Attempt at a Solution

What do you mean "solve"? Perhaps if you provided us with a more complete statement of the problem and detailed your attempt/thoughts we would be able to offer more help.
 
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...

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