Hello math sir, Can anyone please help me solve this question?I m banging my head on table but cuuldnt solve it.Detailled answer would be much appreciated please.Sorry for pics as on website you cannot upload big size pic so i had to crop it. Thank you.
1 Answer
# ln(1+cos theta) = ln2 - theta^2/4 - theta^4/96 #
Explanation:
The questions asks that we Expand
# ln(1+cos theta) = ln2 - theta^2/4 - theta^4/96 #
Let us start by forming the Maclaurin Series
The Maclaurin series is given by
# f(x) = f(0) + (f'(0))/(1!)x + (f''(0))/(2!)x^2 + (f'''(0))/(3!)x^3 + ... (f^((n))(0))/(n!)x^n + ...#
We seek two non-zero terms:
We start with the function
# f(x) = ln(cosx) #
# => f(0) =ln (1) = 0 #
Then, we compute the first derivative (using the chain rule):
# f^((1))(x) = 1/(cosx)d/(dx) (cosx)#
# \ \ \ \ \ \ \ \ \ \ \ \ = (-sin x)/(cosx)#
# \ \ \ \ \ \ \ \ \ \ \ \ = -tanx#
# => f^((1))(0) = 0/(0) = 0#
So we must continue and compute the second derivative:
# f^((2))(x) = -sec^2x#
# => f^((2))(0) = -1/(1) = -1#
And the third derivative (chain rule):
# f^((3))(x) = -2secxd/dx(secx)#
# \ \ \ \ \ \ \ \ \ \ \ \ = -2secx secx tanx#
# \ \ \ \ \ \ \ \ \ \ \ \ = -2sec^2x tanx#
# => f^((3))(0) = 1*0 = 0#
And the fourth derivative (product rule):
# f^((4))(x) = -2{(sec^2x)(sec^2x) + (2sec^2x tanx)(tanx)}#
# \ \ \ \ \ \ \ \ \ \ \ \ = -2sec^4x-4sec^2xtan^2x#
# => f^((4))(0) = -2-0 = -2#
So we can write the Maclaurin series as:
# ln(cosx) = 0 + 0 + (-1)/(2)x^2 + 0 -2/(24)x^4 + ...#
And if we truncate at the first two non-zero
# ln(cosx) ~~-x^2/2 -x^4/12#
For the second part we use the identity:
# cos 2A -= cos^2A - sin^2A = 2cos^2A-1 #
From which we get with A=theta/2#:
# cos theta -= 2cos^2(theta/2)-1 #
And so we have:
# 1+cos theta -= 2cos^2(theta/2) #
Taking (Natural) logarithms of both sides:
# ln(1+cos theta) = ln{2cos^2(theta/2)} #
# " " = ln2 + 2lncos(theta/2) #
# " " ~~ ln2 + 2{-(theta/2)^2/2 -(theta/2)^4/12} #
# " " = ln2 + 2{-(theta^2/4)/2 -(theta^4/16)/12} #
# " " = ln2 -(theta^2/4) -(theta^4/16)/6 #
# " " = ln2 -theta^2/4 -theta^4/96 \ \ \ # QED
