Boolen alegbra simplification problem

  • Thread starter Thread starter Sadeq
  • Start date Start date
AI Thread Summary
The discussion revolves around a Boolean algebra simplification problem where the original solution is found in one file, while the user's attempt is in another. The user expresses confusion regarding discrepancies between their work and the provided solution, specifically questioning the third line of the solution. Participants request clearer steps and the original logical expression to better understand the problem. The conversation highlights the need for detailed explanations and clarity in mathematical writing. Overall, the focus is on resolving the differences in the Boolean algebra solutions.
Sadeq
Messages
103
Reaction score
0

Homework Statement



The problem with right solution is in file 1




The Attempt at a Solution


My attempt is in file 2

iM REALY DONT KNOW WHY THERE IS DIFFERENCE
Please help me to find the mistake
 

Attachments

  • 1.jpg
    1.jpg
    16.3 KB · Views: 471
  • 2.pdf
    2.pdf
    49.2 KB · Views: 260
Physics news on Phys.org
I don't see how you get the third line in your working. Please write out the steps from the second line in more detail.
 
Your writing is very difficult to read, but I'm trying to help. Are you sure that 2.pdf is your attempt and the attached thumbnail is the correct solution? In fact... would you mind stating the original problem logical expression, as it's not entirely clear?
 
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