Problem Set 3: 2A-12d: Give the quadratic approximation for y=ln(cosx).
I understand how the solution is -(1/2)x^2 (because L(ln(1+u))=u), but when I first attempted the problem, I did the following:
y=ln(cosx)
e^y=e^(ln(cosx))
e^y=cosx
approximating,
1+y+(1/2)y^2=1-(1/2)x^2
y=-(1/2)x^2-(1/2)y^2
plugging in ln(cosx) for y,
Q(y)=-(1/2)x^2-(1/2)(ln(cosx))^2
This solution seems to be a more accurate (although uglier-looking) approximation of y=ln(cosx) when x is near 0. Looking for insight as to if it is incorrect and if so, why?

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