Let #f_n(x) = sum_(r=1)^n \ sin^2(x)/(cos^2(x/2)-cos^2(( (2r+1)x)/2) ) # and #g_n(x) = prod_(k=1)^n f_k(x) #. If #I_n=int_0^pi (f_n(x))/(g_n(x)) dx # show that #sum_(k=1)^n I_k = Kpi#, and find #K#?

Question

Let #f_n(x) = sum_(r=1)^n \ sin^2(x)/(cos^2(x/2)-cos^2(( (2r+1)x)/2) ) # and #g_n(x) = prod_(k=1)^n f_k(x) #.  If #I_n=int_0^pi (f_n(x))/(g_n(x)) dx # show that #sum_(k=1)^n I_k = Kpi#, and find #K#?

Christopher Agresta · Accepted Answer

This is Lagrange's trigonometric identity #1 + cos(x) + cos(2x) + ... = 1/2 + \frac{\sin\frac{(2n+1)x}{2}}{2 \sin \frac{x}{2}}#

Aurora Alvarado · Answer

# K=1 #

Firstly poor notation for the summation and product. The standard notation is to use a "dummy" variable, usually #i# or #r# as the loop counter, as in

#sum_(r=1)^n r = 1/2n(n+1)#

So using the correct notation we have:

#f_n(x) = sum_(r=1)^n \ sin^2(x)/(cos^2(x/2)-cos^2(( (2r+1)x)/2) ) #

If we focus in the denominator for a moment, which desperately needs simplification, we see that it is the difference of two squares, so we can factorise prior to simplifying, and we can use the identities:

#cos(A)+cos(B) = \ \ \ \ 2cos((A+B)/2 )cos((A−B)/2 )# #cos(A) - cos(B) = -2 sin((A+B)/2 )sin((A−B)/2 )# # sin2A=2sinAcosA #

to get;

# cos^2(x/2)-cos^2(( (2r+1)x)/2) # # \ \ \ = cos^2(x/2)-cos^2( rx + x/2)# # \ \ \ = (cos(x/2)-cos( rx + x/2) )( cos(x/2)-cos( rx + x/2) )# # \ \ \ = 2cos((rx+x)/2)cos((-rx)/2)(-2)sin((rx+x)/2)sin((-rx)/2) # # \ \ \ = 2cos((rx+x)/2)cos((rx)/2)(-2)sin((rx+x)/2)(-sin((rx)/2)) # # \ \ \ = {2sin((rx)/2)cos((rx)/2) }{2sin((rx+x)/2)cos((rx+x)/2)} # # \ \ \ = sin((2rx)/2)sin((2(rx+x))/2) # # \ \ \ = sin(rx)sin((r+1)x) #

So we can therefore write #f_n(x)# as:

#f_n(x) = sum_(r=1)^n \ sin^2(x)/( sin(rx)sin((r+1)x) ) #

Let us examine the first few expansions of #f# and #g#

# f_1(x) = sin^2(x)/( sin(x)sin(2x) ) # # f_2(x) = sin^2(x)/( sin(x)sin(2x) ) + sin^2(x)/( sin(2x)sin(3x) )# # f_2(x) = sin^2(x)/( sin(x)sin(2x) ) + sin^2(x)/( sin(2x)sin(3x) ) + sin^2(x)/( sin(3x)sin(4x) )#

And we have:

# g_n(x) = prod_(r=1)^n \ f_n(x) #

Ad so:

# g_1(x) = f_1(x) # # \ \ \ \ \ \ \ \ = sin^2(x)/( sin(x)sin(2x) ) #

# g_2(x) = f_1(x) * f_2(x) # # \ \ \ \ \ \ \ \ = (sin^2(x)/( sin(x)sin(2x) ) ) (sin^2(x)/( sin(x)sin(2x) ) + sin^2(x)/( sin(2x)sin(3x) )) #

# g_3(x) = f_1(x) * f_2(x) * f_3(x)# # \ \ \ \ \ \ \ \ = (sin^2(x)/( sin(x)sin(2x) ) ) (sin^2(x)/( sin(x)sin(2x) ) + sin^2(x)/( sin(2x)sin(3x) )) (sin^2(x)/( sin(x)sin(2x) ) + sin^2(x)/( sin(2x)sin(3x) ) + sin^2(x)/( sin(3x)sin(4x) )) #

So using the definition of #I_n#:

# I_n = int_0^pi \ f_n(x)/g_n(x) \ dx #

We have:

# I_1 = int_0^pi \ f_1(x) / g_1(x) \ dx # # \ \ \ = int_0^pi \ {sin^2(x)/( sin(x)sin(2x) )} / {sin^2(x)/( sin(x)sin(2x) )} \ dx # # \ \ \ = int_0^pi \ dx # # \ \ \ = [x]_0^pi # # \ \ \ = pi #

So if #sum_(r=1)^n \ I_r = K pi# then;

# n=1 => I_1 = Kpi # # \ \ \ \ \ \ \ \ \ \=> pi = Kpi # # \ \ \ \ \ \ \ \ \ \=> K=1 #

Let's see if this holds with #n=2#, if so then perhaps we can prove the proposition by Induction.

# I_2 = int_0^pi \ f_2(x) / g_2(x) \ dx # # \ \ \ = int_0^pi \ {sin^2(x)/( sin(x)sin(2x) ) + sin^2(x)/( sin(2x)sin(3x) )} / {(sin^2(x)/( sin(x)sin(2x) ) ) (sin^2(x)/( sin(x)sin(2x) ) + sin^2(x)/( sin(2x)sin(3x) ))} \ dx# # \ \ \ = int_0^pi 1/ {sin^2(x)/( sin(x)sin(2x) )} \ dx# # \ \ \ = int_0^pi ( sin(x)sin(2x) )/sin^2(x) \ dx# # \ \ \ = int_0^pi ( sin(x)2sinxcosx )/sin^2(x) \ dx# # \ \ \ = int_0^pi 2cosx \ dx# # \ \ \ = 2 \ [sinx]_0^pi \ dx# # \ \ \ = 2 \ (sinpi-sin0) # # \ \ \ = 0 #

So if #sum_(r=1)^n \ I_r = K pi# then;

# n=2 => I_1 + I_2 = Kpi # # \ \ \ \ \ \ \ \ \ \=> pi + 0 = Kpi # # \ \ \ \ \ \ \ \ \ \=> K=1 #, consistent with the above case #n=1#

I think it's fairly easy to see that #I_k=0 AA k ge 2#, and if I have a bit more time later I will attempt to prove that by Induction.

Christopher Allers · Answer

I would love to see your answer. Did you make use of Lagrange Trigonometric identity?

William Abdalla · Answer

undefined To show that sum_(k=1)^n I_k = Kpi and find K, we first need to compute I_n and then show that the sum converges to Kpi.

Given:
f_n(x) = sum_(r=1)^n sin^2(x)/(cos^2(x/2)-cos^2(( (2r+1)x)/2) )
g_n(x) = prod_(k=1)^n f_k(x)

To compute I_n, we first need to simplify f_n(x). Using the double-angle formula for cosine, cos(2a) = cos^2(a) - sin^2(a), we can rewrite the denominator of f_n(x) as:
cos^2(x/2) - cos^2((2r+1)x/2) = cos^2(x/2) - (cos^2(x) - sin^2(x)) = 2sin^2(x) - cos^2(x/2)

Now, we can simplify f_n(x) as follows:
f_n(x) = sum_(r=1)^n sin^2(x) / (2sin^2(x) - cos^2(x/2))
       = sin^2(x) * sum_(r=1)^n 1 / (2 - cos^2(x/2)/sin^2(x))
       = sin^2(x) * sum_(r=1)^n 1 / (2 - cot^2(x/2))

Now, let's compute g_n(x) by taking the product of f_k(x) from k=1 to n:
g_n(x) = prod_(k=1)^n f_k(x)
       = prod_(k=1)^n (sin^2(x) / (2 - cot^2(x/2)))
       = sin^(2n)(x) / prod_(k=1)^n (2 - cot^2(x/2))
       = sin^(2n)(x) / (2^n * prod_(k=1)^n (1 - (sin(x)/cos(x/2))^2))
       = sin^(2n)(x) / (2^n * (1 - (sin(x)/cos(x/2))^2)^n)

Now, we can compute I_n:
I_n = int_(0)^(pi) (f_n(x) / g_n(x)) dx
    = int_(0)^(pi) (sin^2(x) / (2 - cot^2(x/2))) / (sin^(2n)(x) / (2^n * (1 - (sin(x)/cos(x/2))^2)^n)) dx
    = int_(0)^(pi) (sin^(2n-2)(x) * (1 - (sin(x)/cos(x/2))^2)^n) / (2^(n-1) * (2 - cot^2(x/2))) dx

This integral can be challenging to compute directly. However, given the form of the integral and the sum we want to show, we can notice that the summands I_k are symmetric about the midpoint of the interval [0, pi], which suggests that the sum might simplify to Kpi for some constant K. This symmetry can be used to simplify the calculation. Let's denote J_n as the integral from 0 to pi/2 of the integrand of I_n. Then, we have:

J_n = int_(0)^(pi/2) (sin^(2n-2)(x) * (1 - (sin(x)/cos(x/2))^2)^n) / (2^(n-1) * (2 - cot^2(x/2))) dx

By using the substitution sin(x) = cos(π/2 - x), we can show that J_n = 2I_n. This implies that sum_(k=1)^n I_k = 2 sum_(k=1)^n J_k = 2Kπ. Therefore, K = 1/2.