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# Maclaurin expansion watch

1. I am stuck on part d). I know how to manually find the first 4 non-zero terms (as the question asks) but how would I expand ln(1-x) using the series expansion of ln(1 + x). ? I mean the series expansion given in the formula booklet only gives the first 3 terms..

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2. (Original post by GPODT)

I am stuck on part d). I know how to manually find the first 4 non-zero terms (as the question asks) but how would I expand ln(1-x) using the series expansion of ln(1 + x). ? I mean the series expansion given in the formula booklet only gives the first 3 terms..

Posted from TSR Mobile
Note that

and replace x with -x.
3. ln(1 - x) = ln(1 + (-x) )
4. (Original post by Indeterminate)
Note that

and replace x with -x.
But doesn't that suggest the first term in the expansion of ln(1+x) = (-1)2 x2 = x2 (If you sub in n=1), which is incorrect?
5. (Original post by GPODT)
But doesn't that suggest the first term in the expansion of ln(1+x) = (-1)2 x2 = x2 (If you sub in n=1), which is incorrect?
Typo Edited now, sorry
6. (Original post by Indeterminate)
Typo Edited now, sorry
Ah thank you! I understand now!

One quick last question if you don't mind..

Using the formula you gave, I found the fourth term in the expansion of ln(1-x) to be:

-x4/4

However, how do I know what the sign in front of the -x is? I mean if its a + sign, the -x will remain but if there is a negative sign it will turn into --x = +x ?

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