Infinite sum

  • #1
[tex]\sum{a^ncos(nx)}[/tex]

from zero to infinity

a is a real number -1 < a < 1

I rewrote this as a geometric series involving a complex exponential

Real part of

[tex]\sum{(ae^{ix})^n}[/tex]

Which is a geometric series with common ratio r < 1, so it converges to the sum

(first term)/(1-r)

which seems to be

[tex]\frac{1}{1-ae^{ix}}[/tex]

taking the real part and multiplying top and bottom by (1-acosx), I get


[tex]\frac{1-acos(x)}{1-2acos(x) + a^2cos^2(x))}[/tex]

which is different from the desired result of

[tex]\frac{1-acos(x)}{1-2acos(x) + a^2)}[/tex]

Any help would be appreciated
 

Answers and Replies

  • #2
761
13
[tex]\sum{a^ncos(nx)}[/tex]

from zero to infinity

a is a real number -1 < a < 1

I rewrote this as a geometric series involving a complex exponential

Real part of

[tex]\sum{(ae^{ix})^n}[/tex]

Which is a geometric series with common ratio r < 1, so it converges to the sum

(first term)/(1-r)

which seems to be

[tex]\frac{1}{1-ae^{ix}}[/tex]


Any help would be appreciated
From this point on this go wrong...
 
  • #3
I know i've gone wrong somewhere. I'd like to know what I did wrong.
 
  • #4
761
13
First work out the fraction then take the imaginary part.
 
  • #5
Got it. Thanks.
 

Related Threads on Infinite sum

  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
1
Views
749
  • Last Post
Replies
4
Views
520
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
5
Views
7K
  • Last Post
Replies
4
Views
936
  • Last Post
Replies
6
Views
801
Top