- #36
Physicsissuef
- 908
- 0
But didn't we said that we need to find
the geometric sum cos2x+cos4x+...+cosnx ?
the geometric sum cos2x+cos4x+...+cosnx ?
Physicsissuef said:But didn't we said that we need to find
the geometric sum cos2x+cos4x+...+cosnx ?
Physicsissuef said:I don't understand what you mean. Can you explain how we should find the solution, starting from the beginning, please?
Physicsissuef said:[tex]\frac{{e}^{2ix}-{e}^{i2(n+1)x}}{1-{e}^{2ix}} \circ \frac{1+{e}^{2ix}}{1+{e}^{2ix}}[/tex]
Like this?
Physicsissuef said:[tex]
\frac{{e}^{2ix}-{e}^{i2(n+1)x}}{1-{e}^{2ix}} \circ \frac{1-{e}^{-2ix}}{1-{e}^{-2ix}}
[/tex]
How will I multiply all of this ? I have never learn to compute or multiply with Euler's formula.
DavidWhitbeck said:Dick is about to earn the 2008 award for homework helper just for this thread! What patience! :-)
Physicsissuef said:In my textbook results it is:
[tex]\frac{cos(n+1)xsin(nx)}{sinx}+i\frac{sin(n+1)xsin(nx)}{sinx}[/tex]
I am wondering how did they found this results...
DavidWhitbeck said:Dick is about to earn the 2008 award for homework helper just for this thread! What patience! :-)
Physicsissuef said:Am I right?
D H said:Does helping the exact same person on the exact same question with the exact same frustration level but in another forum count?