MHB Summation #2 Prove: $\sum_{k=1}^n (2^k\sin^2\frac{x}{2^k})^2$

[Saitama]

Prove the following:
$$\sum_{k=1}^n \left(2^k\sin^2\frac{x}{2^k}\right)^2=\left(2^n\sin\frac{x}{2^n}\right)^2-\sin^2x$$

 

[lfdahl]

Spoiler

Proof by means of the induction principle:

\[\sum_{k=1}^{n}\left ( 2^ksin^2\left ( \frac{x}{2^k} \right ) \right )^2=\left ( 2^n sin\left ( \frac{x}{2^n} \right )\right )^2-sin^2(x)\]

The sum holds for $n=1$:
\[\left ( 2sin\left ( \frac{x}{2} \right ) \right )^2-sin^2(x)= 4sin^2\left ( \frac{x}{2} \right )- 4sin^2\left ( \frac{x}{2} \right )cos^2\left ( \frac{x}{2} \right )\\\\ = 4sin^2\left ( \frac{x}{2} \right )- 4sin^2\left ( \frac{x}{2} \right )\left ( 1-sin^2\left ( \frac{x}{2} \right ) \right )=\left ( 2sin^2\left ( \frac{x}{2} \right ) \right )^2\]

Assume the equation holds for some $n>1$. Then it also holds for $n+1$, because:

\[\sum_{k=1}^{n+1}\left ( 2^ksin^2\left ( \frac{x}{2^k} \right ) \right )^2 =\left ( 2^nsin\left ( \frac{x}{2^n} \right ) \right )^2-sin^2(x)+\left ( 2^{n+1}sin^2\left ( \frac{x}{2^{n+1}} \right ) \right )^2\;\;\;\; (1).\]

Rewriting the first term on the RHS:

\[\left ( 2^nsin\left ( \frac{x}{2^n} \right ) \right )^2 =2^{2n}sin^2\left ( 2\cdot \frac{x}{2^{n+1}} \right )= 2^{2n+2}sin^2\left ( \frac{x}{2^{n+1}} \right )cos^2\left ( \frac{x}{2^{n+1}} \right )\\\\ =\left ( 2^{n+1} \right )^2sin^2\left ( \frac{x}{2^{n+1}} \right )\left ( 1-sin^2\left ( \frac{x}{2^{n+1}} \right ) \right ) \;\;\; (2).\]

Inserting $(2)$ into $(1)$:

\[ \sum_{k=1}^{n+1}\left ( 2^ksin^2\left ( \frac{x}{2^k} \right ) \right )^2 = \left ( 2^{n+1} \right )^2sin^2\left ( \frac{x}{2^{n+1}} \right )\left ( 1-sin^2\left ( \frac{x}{2^{n+1}} \right ) \right ) -sin^2(x)+\left ( 2^{n+1}sin^2\left ( \frac{x}{2^{n+1}} \right ) \right )^2\\\\=\left ( 2^{n+1}sin\left ( \frac{x}{2^{n+1}} \right ) \right )^2 - sin^2(x)+\left ( 2^{n+1}sin^2\left ( \frac{x}{2^{n+1}} \right ) \right )^2-\left ( 2^{n+1}sin^2\left ( \frac{x}{2^{n+1}} \right ) \right )^2 \\\\ =\left ( 2^{n+1}sin\left ( \frac{x}{2^{n+1}} \right ) \right )^2 - sin^2(x)\]
I´m sure, there is a more elegant way to prove the identity …

 

[anemone]

Prove the following:
$$\sum_{k=1}^n \left(2^k\sin^2\frac{x}{2^k}\right)^2=\left(2^n\sin\frac{x}{2^n}\right)^2-\sin^2x$$

My solution:

Spoiler

Notice that

$\begin{align*}\left(2^k\sin^2\frac{x}{2^k}\right)^2&=2^{2k}\sin^2\dfrac{x}{2^k}\left(\sin^2\dfrac{x}{2^k}\right)\\&=2^{2k}\sin^2\dfrac{x}{2^k}\left(1-\cos^2\dfrac{x}{2^k}\right)\\&=2^{2k}\sin^2\dfrac{x}{2^k}-2^{2k}\sin^2\dfrac{x}{2^k}\cos^2\dfrac{x}{2^k}\\&=2^{2k}\sin^2\dfrac{x}{2^k}-2^{2k-2}\left(\sin^2\dfrac{x}{2^{k-1}}\right)\end{align*}$

Hence,

$$\sum_{k=1}^n \left(2^k\sin^2\frac{x}{2^k}\right)^2$$$$=\sum_{k=1}^n \left(2^{2k}\sin^2\dfrac{x}{2^k}-2^{2k-2}\left(\sin^2\dfrac{x}{2^{k-1}}\right) \right)$$$$=\left(4\sin^2\dfrac{x}{2}-\sin^2x\right)+\left(16\sin^2\dfrac{x}{4}-4\sin^2\dfrac{x}{2}\right)+\left(64\sin^2\dfrac{x}{8}-16\sin^2\dfrac{x}{4}\right)+\cdots$$$$+\left(2^{2(n-1)}\sin^2\dfrac{x}{2^{n-1}}-2^{2n-4}\sin^2\dfrac{x}{2^{n-3}}\right)+\left(2^{2n}\sin^2\dfrac{x}{2^n}-2^{2(n-1)}\sin^2\dfrac{x}{2^{n-1}}\right)$$$$=\left(\cancel{4\sin^2\dfrac{x}{2}}-\sin^2x\right)+\left(\cancel{16\sin^2\dfrac{x}{4}}-\cancel{4\sin^2\dfrac{x}{2}}\right)+\left(\cancel{64\sin^2\dfrac{x}{8}}-\cancel{16\sin^2\dfrac{x}{4}}\right)+\cdots$$$$+\left(\cancel{2^{2(n-1)}\sin^2\dfrac{x}{2^{n-1}}}-\cancel{2^{2n-4}\sin^2\dfrac{x}{2^{n-3}}}\right)+\left(2^{2n}\sin^2\dfrac{x}{2^n}-\cancel{2^{2(n-1)}\sin^2\dfrac{x}{2^{n-1}}}\right)$$$$=\left(2^n\sin\dfrac{x}{2^n}\right)^2-\sin^2x\text{(QED)}$$

 

[Saitama]

My solution:

Spoiler

Notice that

$\begin{align*}\left(2^k\sin^2\frac{x}{2^k}\right)^2&=2^{2k}\sin^2\dfrac{x}{2^k}\left(\sin^2\dfrac{x}{2^k}\right)\\&=2^{2k}\sin^2\dfrac{x}{2^k}\left(1-\cos^2\dfrac{x}{2^k}\right)\\&=2^{2k}\sin^2\dfrac{x}{2^k}-2^{2k}\sin^2\dfrac{x}{2^k}\cos^2\dfrac{x}{2^k}\\&=2^{2k}\sin^2\dfrac{x}{2^k}-2^{2k-2}\left(\sin^2\dfrac{x}{2^{k-1}}\right)\end{align*}$

Hence,

$$\sum_{k=1}^n \left(2^k\sin^2\frac{x}{2^k}\right)^2$$$$=\sum_{k=1}^n \left(2^{2k}\sin^2\dfrac{x}{2^k}-2^{2k-2}\left(\sin^2\dfrac{x}{2^{k-1}}\right) \right)$$$$=\left(4\sin^2\dfrac{x}{2}-\sin^2x\right)+\left(16\sin^2\dfrac{x}{4}-4\sin^2\dfrac{x}{2}\right)+\left(64\sin^2\dfrac{x}{8}-16\sin^2\dfrac{x}{4}\right)+\cdots$$$$+\left(2^{2(n-1)}\sin^2\dfrac{x}{2^{n-1}}-2^{2n-4}\sin^2\dfrac{x}{2^{n-3}}\right)+\left(2^{2n}\sin^2\dfrac{x}{2^n}-2^{2(n-1)}\sin^2\dfrac{x}{2^{n-1}}\right)$$$$=\left(\cancel{4\sin^2\dfrac{x}{2}}-\sin^2x\right)+\left(\cancel{16\sin^2\dfrac{x}{4}}-\cancel{4\sin^2\dfrac{x}{2}}\right)+\left(\cancel{64\sin^2\dfrac{x}{8}}-\cancel{16\sin^2\dfrac{x}{4}}\right)+\cdots$$$$+\left(\cancel{2^{2(n-1)}\sin^2\dfrac{x}{2^{n-1}}}-\cancel{2^{2n-4}\sin^2\dfrac{x}{2^{n-3}}}\right)+\left(2^{2n}\sin^2\dfrac{x}{2^n}-\cancel{2^{2(n-1)}\sin^2\dfrac{x}{2^{n-1}}}\right)$$$$=\left(2^n\sin\dfrac{x}{2^n}\right)^2-\sin^2x\text{(QED)}$$

Spoiler

Proof by means of the induction principle:

\[\sum_{k=1}^{n}\left ( 2^ksin^2\left ( \frac{x}{2^k} \right ) \right )^2=\left ( 2^n sin\left ( \frac{x}{2^n} \right )\right )^2-sin^2(x)\]

The sum holds for $n=1$:
\[\left ( 2sin\left ( \frac{x}{2} \right ) \right )^2-sin^2(x)= 4sin^2\left ( \frac{x}{2} \right )- 4sin^2\left ( \frac{x}{2} \right )cos^2\left ( \frac{x}{2} \right )\\\\ = 4sin^2\left ( \frac{x}{2} \right )- 4sin^2\left ( \frac{x}{2} \right )\left ( 1-sin^2\left ( \frac{x}{2} \right ) \right )=\left ( 2sin^2\left ( \frac{x}{2} \right ) \right )^2\]

Assume the equation holds for some $n>1$. Then it also holds for $n+1$, because:

\[\sum_{k=1}^{n+1}\left ( 2^ksin^2\left ( \frac{x}{2^k} \right ) \right )^2 =\left ( 2^nsin\left ( \frac{x}{2^n} \right ) \right )^2-sin^2(x)+\left ( 2^{n+1}sin^2\left ( \frac{x}{2^{n+1}} \right ) \right )^2\;\;\;\; (1).\]

Rewriting the first term on the RHS:

\[\left ( 2^nsin\left ( \frac{x}{2^n} \right ) \right )^2 =2^{2n}sin^2\left ( 2\cdot \frac{x}{2^{n+1}} \right )= 2^{2n+2}sin^2\left ( \frac{x}{2^{n+1}} \right )cos^2\left ( \frac{x}{2^{n+1}} \right )\\\\ =\left ( 2^{n+1} \right )^2sin^2\left ( \frac{x}{2^{n+1}} \right )\left ( 1-sin^2\left ( \frac{x}{2^{n+1}} \right ) \right ) \;\;\; (2).\]

Inserting $(2)$ into $(1)$:

\[ \sum_{k=1}^{n+1}\left ( 2^ksin^2\left ( \frac{x}{2^k} \right ) \right )^2 = \left ( 2^{n+1} \right )^2sin^2\left ( \frac{x}{2^{n+1}} \right )\left ( 1-sin^2\left ( \frac{x}{2^{n+1}} \right ) \right ) -sin^2(x)+\left ( 2^{n+1}sin^2\left ( \frac{x}{2^{n+1}} \right ) \right )^2\\\\=\left ( 2^{n+1}sin\left ( \frac{x}{2^{n+1}} \right ) \right )^2 - sin^2(x)+\left ( 2^{n+1}sin^2\left ( \frac{x}{2^{n+1}} \right ) \right )^2-\left ( 2^{n+1}sin^2\left ( \frac{x}{2^{n+1}} \right ) \right )^2 \\\\ =\left ( 2^{n+1}sin\left ( \frac{x}{2^{n+1}} \right ) \right )^2 - sin^2(x)\]
I´m sure, there is a more elegant way to prove the identity …

Thank you both for your participation and nicely done anemone. :)