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July 28, 2014

July 28, 2014

Posted by **Sean** on Thursday, June 25, 2009 at 7:55pm.

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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 -
**MathMate**, Thursday, June 25, 2009 at 8:38pm1/(1-x) = Σ

_{0}^{∞}x^{k}

write 1/(2-x) as (1/2)(1/(1-x/2))

then

1/(2-x)

= (1/2)(1/(1-x/2))

= (1/2)Σ_{0}^{∞}x/2^{k}

=Σ_{0}^{∞}x^{k+1}

as you have corrected determined.

- Calculus -
**MathMate**, Thursday, June 25, 2009 at 8:41pmSorry, the last two lines should read:

=(1/2)Σ_{0}^{∞}(x/2)^{k}

=Σ_{0}^{∞}x^{k}/2^{k+1}

- Calculus -
**Sean**, Friday, June 26, 2009 at 1:41amThanks 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**, Friday, June 26, 2009 at 9:21amSorry, 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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