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For ln(2+x) can this be written as 2ln(1+x) or is it ln2 + ln(1+x)?

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Neither! Your second attempt is closer to the target. Start with ln(2+x)=ln[2(1+x/2)].

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Can i write ln(2+x) as ln[1+(1+x)] ?? would this work the same as ln[2(1+x/2)] ??

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You can certainly write ln(2+x) as ln[1+(1+x)], but then, what will you do with that? That would be only useful to get Taylor representation around a=-1, not so for Maclaurin representation.

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Ok but if I have ln(2+x)=ln[2(1+x/2)] does this equal ln2 + ln(1+x/2)?

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It does.

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So then for ln(2+x)*Tan-(x^2) I get [ln2 + (x + x^2/2 +x^3/3 ...)]*(x^2 + x^6/3 + x^10/5...)

DO I multiply ln2 with (x^2 +...) and end with

x + (x^2/2 + ln2x^2) + x^3/3

or am I missing something?

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You multiply series like polynomials: each term with each term. So, you should not end up with you wrote above.

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Candice, I would check your numbers for expanding ln(2+x)*tan-(x^2) because they are off. ln (2+x) is alternating too.

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In my notes I have ln(1+x)=sum of x^n/n

is this not right?

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ln(1+x)= Sigma (-1)^(n+1) x^n/n (sigma is from 1 to infinity)

also tan-1(x) = (-1)^n x^(2n+1)/ (2n+1)

so they both will be alternating in sign.

I had the same numbers as you for tan-(x^2)but with alternating signs

but for ln(2+x) i had (ln 2 + x/2 - x^2/8 + x^3/24 - ...)

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okay thanks a lot, that helps.