Get Expert Help with Equation Homework - Step-by-Step Guide

  • Thread starter Thread starter APHELION
  • Start date Start date
AI Thread Summary
The discussion revolves around a homework equation f(x+ba)-q(2b-3f)x and seeks clarification on its structure. Participants point out potential misplacement of parentheses and question whether the second "f" represents a function or a constant. There is also a request for the complete equation to understand what it is equal to. The original poster acknowledges the need to clarify the equation and promises to repost it soon. Clear communication of the equation's format is essential for effective assistance.
APHELION
Messages
25
Reaction score
0

Homework Statement


Homework Equations


The Attempt at a Solution

 
Last edited:
Physics news on Phys.org
APHELION said:

Homework Statement



f(x+ba)-q(2b-3f)x=

Homework Equations





The Attempt at a Solution



Hello everyone, I'm sure this is probably very easy algebra but I'm trying to figure out what type of equation this is and how to work toward a solution. Any help would be much appreciated.
The parentheses are strangely placed. Do you mean f(x+ba)- q(2b-3f(x)) or do you mean f(x+ ba)- q(3b-3f) where the second "f" is a number and not the same as the function f? In either case, you seem to have dropped the right side of the equation. What is it equal to?

Please post the correct equation.
 
...will repost soon.
 
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