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`?

Answer 1

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

Answer 2

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}}`

Answer 3

`K=1`

Explanation:

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.