# 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?

Feb 17, 2017

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

Feb 17, 2017

This is Lagrange's trigonometric identity $1 + \cos \left(x\right) + \cos \left(2 x\right) + \ldots = \frac{1}{2} + \setminus \frac{\setminus \sin \setminus \frac{\left(2 n + 1\right) x}{2}}{2 \setminus \sin \setminus \frac{x}{2}}$

Feb 17, 2017

$K = 1$

#### 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

${\sum}_{r = 1}^{n} r = \frac{1}{2} n \left(n + 1\right)$

So using the correct notation we have:

${f}_{n} \left(x\right) = {\sum}_{r = 1}^{n} \setminus {\sin}^{2} \frac{x}{{\cos}^{2} \left(\frac{x}{2}\right) - {\cos}^{2} \left(\frac{\left(2 r + 1\right) x}{2}\right)}$

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​​ )
$\sin 2 A = 2 \sin A \cos A$

to get;

${\cos}^{2} \left(\frac{x}{2}\right) - {\cos}^{2} \left(\frac{\left(2 r + 1\right) x}{2}\right)$
$\setminus \setminus \setminus = {\cos}^{2} \left(\frac{x}{2}\right) - {\cos}^{2} \left(r x + \frac{x}{2}\right)$
$\setminus \setminus \setminus = \left(\cos \left(\frac{x}{2}\right) - \cos \left(r x + \frac{x}{2}\right)\right) \left(\cos \left(\frac{x}{2}\right) - \cos \left(r x + \frac{x}{2}\right)\right)$
$\setminus \setminus \setminus = 2 \cos \left(\frac{r x + x}{2}\right) \cos \left(\frac{- r x}{2}\right) \left(- 2\right) \sin \left(\frac{r x + x}{2}\right) \sin \left(\frac{- r x}{2}\right)$
$\setminus \setminus \setminus = 2 \cos \left(\frac{r x + x}{2}\right) \cos \left(\frac{r x}{2}\right) \left(- 2\right) \sin \left(\frac{r x + x}{2}\right) \left(- \sin \left(\frac{r x}{2}\right)\right)$
$\setminus \setminus \setminus = \left\{2 \sin \left(\frac{r x}{2}\right) \cos \left(\frac{r x}{2}\right)\right\} \left\{2 \sin \left(\frac{r x + x}{2}\right) \cos \left(\frac{r x + x}{2}\right)\right\}$
$\setminus \setminus \setminus = \sin \left(\frac{2 r x}{2}\right) \sin \left(\frac{2 \left(r x + x\right)}{2}\right)$
$\setminus \setminus \setminus = \sin \left(r x\right) \sin \left(\left(r + 1\right) x\right)$

So we can therefore write ${f}_{n} \left(x\right)$ as:

${f}_{n} \left(x\right) = {\sum}_{r = 1}^{n} \setminus {\sin}^{2} \frac{x}{\sin \left(r x\right) \sin \left(\left(r + 1\right) x\right)}$

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

${f}_{1} \left(x\right) = {\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)}$
${f}_{2} \left(x\right) = {\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)} + {\sin}^{2} \frac{x}{\sin \left(2 x\right) \sin \left(3 x\right)}$
${f}_{2} \left(x\right) = {\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)} + {\sin}^{2} \frac{x}{\sin \left(2 x\right) \sin \left(3 x\right)} + {\sin}^{2} \frac{x}{\sin \left(3 x\right) \sin \left(4 x\right)}$

And we have:

${g}_{n} \left(x\right) = {\prod}_{r = 1}^{n} \setminus {f}_{n} \left(x\right)$

${g}_{1} \left(x\right) = {f}_{1} \left(x\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)}$

${g}_{2} \left(x\right) = {f}_{1} \left(x\right) \cdot {f}_{2} \left(x\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left({\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)}\right) \left({\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)} + {\sin}^{2} \frac{x}{\sin \left(2 x\right) \sin \left(3 x\right)}\right)$

${g}_{3} \left(x\right) = {f}_{1} \left(x\right) \cdot {f}_{2} \left(x\right) \cdot {f}_{3} \left(x\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left({\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)}\right) \left({\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)} + {\sin}^{2} \frac{x}{\sin \left(2 x\right) \sin \left(3 x\right)}\right) \left({\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)} + {\sin}^{2} \frac{x}{\sin \left(2 x\right) \sin \left(3 x\right)} + {\sin}^{2} \frac{x}{\sin \left(3 x\right) \sin \left(4 x\right)}\right)$

So using the definition of ${I}_{n}$:

${I}_{n} = {\int}_{0}^{\pi} \setminus {f}_{n} \frac{x}{g} _ n \left(x\right) \setminus \mathrm{dx}$

We have:

${I}_{1} = {\int}_{0}^{\pi} \setminus {f}_{1} \frac{x}{g} _ 1 \left(x\right) \setminus \mathrm{dx}$
$\setminus \setminus \setminus = {\int}_{0}^{\pi} \setminus \frac{{\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)}}{{\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)}} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = {\int}_{0}^{\pi} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = {\left[x\right]}_{0}^{\pi}$
$\setminus \setminus \setminus = \pi$

So if ${\sum}_{r = 1}^{n} \setminus {I}_{r} = K \pi$ then;

$n = 1 \implies {I}_{1} = K \pi$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \implies \pi = K \pi$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \implies K = 1$

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

${I}_{2} = {\int}_{0}^{\pi} \setminus {f}_{2} \frac{x}{g} _ 2 \left(x\right) \setminus \mathrm{dx}$
$\setminus \setminus \setminus = {\int}_{0}^{\pi} \setminus \frac{{\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)} + {\sin}^{2} \frac{x}{\sin \left(2 x\right) \sin \left(3 x\right)}}{\left({\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)}\right) \left({\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)} + {\sin}^{2} \frac{x}{\sin \left(2 x\right) \sin \left(3 x\right)}\right)} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = {\int}_{0}^{\pi} \frac{1}{{\sin}^{2} \frac{x}{\sin \left(x\right) \sin \left(2 x\right)}} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = {\int}_{0}^{\pi} \frac{\sin \left(x\right) \sin \left(2 x\right)}{\sin} ^ 2 \left(x\right) \setminus \mathrm{dx}$
$\setminus \setminus \setminus = {\int}_{0}^{\pi} \frac{\sin \left(x\right) 2 \sin x \cos x}{\sin} ^ 2 \left(x\right) \setminus \mathrm{dx}$
$\setminus \setminus \setminus = {\int}_{0}^{\pi} 2 \cos x \setminus \mathrm{dx}$
$\setminus \setminus \setminus = 2 \setminus {\left[\sin x\right]}_{0}^{\pi} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = 2 \setminus \left(\sin \pi - \sin 0\right)$
$\setminus \setminus \setminus = 0$

So if ${\sum}_{r = 1}^{n} \setminus {I}_{r} = K \pi$ then;

$n = 2 \implies {I}_{1} + {I}_{2} = K \pi$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \implies \pi + 0 = K \pi$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \implies K = 1$, consistent with the above case $n = 1$

I think it's fairly easy to see that ${I}_{k} = 0 \forall k \ge 2$, and if I have a bit more time later I will attempt to prove that by Induction.