1/(1-x) = Σ0∞xk
write 1/(2-x) as (1/2)(1/(1-x/2))
then
1/(2-x)
= (1/2)(1/(1-x/2))
= (1/2)Σ0∞x/2k
=Σ0∞xk+1
as you have corrected determined.
Obtain the MacLaurin series for 1/(2-x) by making an appropriate substitution into the MacLaurin series for 1/(1-x).
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The MacLaurin series for 1/(1-x) = Σ x^k
I substitue (x-1) in for x, because 1/(2-x) = 1/(1-(x-1))
Making the same substitution in the MacLaurin series gives Σ (x-1)^k
If I manually calculate the MacLaurin series for 1/(2-x), 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?
4 answers
Sorry, the last two lines should read:
=(1/2)Σ0∞(x/2)k
=Σ0∞xk/2k+1
=(1/2)Σ0∞(x/2)k
=Σ0∞xk/2k+1
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.
I can't read your symbols, but I can make out what you are doing and it results in the right answer.
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/utf-8
However, utf-8 may not be available automatically on all computers.
I will take more care with the encoding next time. Thanks.
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/utf-8
However, utf-8 may not be available automatically on all computers.
I will take more care with the encoding next time. Thanks.