Posted by Sean on .
Obtain the MacLaurin series for 1/(2x) by making an appropriate substitution into the MacLaurin series for 1/(1x).

The MacLaurin series for 1/(1x) = Σ x^k
I substitue (x1) in for x, because 1/(2x) = 1/(1(x1))
Making the same substitution in the MacLaurin series gives Σ (x1)^k
If I manually calculate the MacLaurin series for 1/(2x), I get Σ x^k/2^(k+1)
Those two don't match. What did I do wrong? Or does that substitution method not work?

Calculus 
MathMate,
1/(1x) = Î£_{0}^{âˆž}x^{k}
write 1/(2x) as (1/2)(1/(1x/2))
then
1/(2x)
= (1/2)(1/(1x/2))
= (1/2)Î£_{0}^{âˆž}x/2^{k}
=Î£_{0}^{âˆž}x^{k+1}
as you have corrected determined. 
Calculus 
MathMate,
Sorry, the last two lines should read:
=(1/2)Î£_{0}^{âˆž}(x/2)^{k}
=Î£_{0}^{âˆž}x^{k}/2^{k+1} 
Calculus 
Sean,
Thanks so much!
I can't read your symbols, but I can make out what you are doing and it results in the right answer. 
Calculus 
MathMate,
Sorry, I was accidentally on unicode.
You would be able to read the symbols with unicode encoding. If you are on Windows XP, you can use the menu to go
view/character encoding/utf8
However, utf8 may not be available automatically on all computers.
I will take more care with the encoding next time. Thanks.