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What is the MacLaurin Series for #sin(2x)ln(1+x)#?

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Answer 1

Christopher Agresta

The given series is incorrect.

The correct series is:

# sin(2x)ln(1+x) = 2x^2 - x^3 - 2/3x^4 +1/6x^5 + ... #

The series that is provided is not correct.

Let:

# S = sin(2x)ln(1+x) #

I won't derive the well known series for #sin# and #ln#, but just quote them:

Thus, we have the aforementioned, reputable series:

# ln(1+x) = x-1/2x^2+1/3x^3 -1/4x^4 + 1/5x^5 ... # # sinx = x-x^3/(3!) + x^5/(5!) - x^7/(7!) + ... #

Thus the series for #sin2x# is:

# sin2x = (2x)-(2x)^3/(3!) + (2x)^5/(5!) - (2x)^7/(7!) + ... # # " " = 2x-(8x^3)/6 + (32x^5)/(120) - (128x^7)/(5040) + ... # # " " = 2x - 4/3x^3 + 4/15x^5 - 8/315x^7 + ... #

So if we take terms up to #O(x^4)# then:

# S = sin(2x)ln(1+x) # # \ \ = {2x - 4/3x^3 + ...} * {x-1/2x^2+1/3x^3 -1/4x^4 + ...} # # \ \ = (2x){x-1/2x^2+1/3x^3 -1/4x^4 + ...} -4/3x^3{x-1/2x^2+1/3x^3 -1/4x^4 + ...} # # \ \ = 2x^2-x^3+2/3x^4 -1/2x^5 + ... -4/3x^4 + 2/3x^5... #

# \ \ = 2x^2 - x^3 - 2/3x^4 +1/6x^5 + ... #

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.