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?

Hey! We 've verified this expert answer for you, click below to unlock the details :)

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

Looking for something else?

Not the answer you are looking for? Search for more explanations.