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acb81887
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Problem Set 3: 2A12d: 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 uglierlooking) approximation of y=ln(cosx) when x is near 0. Looking for insight as to if it is incorrect and if so, why?
 2 years ago
 2 years ago
acb81887 Group Title
Problem Set 3: 2A12d: 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 uglierlooking) approximation of y=ln(cosx) when x is near 0. Looking for insight as to if it is incorrect and if so, why?
 2 years ago
 2 years ago

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bhaweshwebmaster Group TitleBest ResponseYou've already chosen the best response.2
actually the formula for quadratic approximation is : f(x) = f(0) + f ' (0) * x + f ''(0) (x^2)/2 For, y = ln (cos x) f(0) = 0 f ' (0)= 0 and, f '' (0)= 1 now, substitute to get , y =  1/2 x^2
 2 years ago

acb81887 Group TitleBest ResponseYou've already chosen the best response.0
I should have mentioned that the question suggests that you solve the problem algebraically using quadratic approximation formulas for basic functions, for instance I used Q(e^x)=1+x+(1/2)x^2 and Q(cosx)=1(1/2)x^2. edit: I know Q(y)=(1/2)x^2(1/2)(ln(cosx))^2 is the wrong answer, which is evident by using the actual quadratic approximation formula, but I don't see why I cannot also solve the problem algebraically using Q(e^x) and Q(cosx).
 2 years ago

bhaweshwebmaster Group TitleBest ResponseYou've already chosen the best response.2
Okay. So remember that ln (x) = ln [1+(x1)] Using quadratic approximation, ln [1+(x1)]= (x1)  (x1)^2 /2 = 2x  x^2 /2 3/2(1) so, ln (cos x) = ln [ 1+ ( cos x 1)] hence, ln (cos x) = 2cos x 1/2 cos^2 x 3/2 [using equation 1] = 2 (1  x^2/2) 1/2 ( 1 x^2 /2) ^2  3/2 [using approximation for cos x] = 2  x^2  1/2 ( 1 x^2) 3/2 [ we don't need to write terms involving higher powers of x than x^2] = 2 x^2  1/2 x^2 /23/2 = 1/2 x^2 In this, only the quadratic approximation formula for ln (1+x) and cos x are used.
 2 years ago

kantochampion Group TitleBest ResponseYou've already chosen the best response.0
The problem with your "Q(y)" approximation is that the term "(1/2)(ln(cosx))^2" is either higher than degree 2 or equal to 0 (depend on whether you choose to approximate ln(cosx) linearly or quadraticly). So, it will be negligible as you already learned from the lecture. Also, the fact that you involve "(1/2)(ln(cosx))^2" as part of your answer for the approximation of ln(cosx) is illogical because if Q(y) is correct, then it's like saying in order to approximate ln(cosx) you had to already know the value of (ln(cosx))^2.
 one year ago
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