Calculus

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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?

• Calculus -

1/(1-x) = Î£0âˆžxk

write 1/(2-x) as (1/2)(1/(1-x/2))
then
1/(2-x)
= (1/2)(1/(1-x/2))
= (1/2)Î£0âˆžx/2k
=Î£0âˆžxk+1

as you have corrected determined.

• Calculus -

Sorry, the last two lines should read:

=(1/2)Î£0âˆž(x/2)k
=Î£0âˆžxk/2k+1

• Calculus -

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 -

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.

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