The discussion revolves around proving the identity involving the summation of squared terms related to sine functions. The focus is on the mathematical formulation and verification of the expression for a given integer \( n \). The discussion is primarily theoretical and mathematical in nature.
Participants seem to agree on the identity that needs to be proven, but the discussion does not provide any resolution or consensus on the proof itself.
No specific assumptions or limitations are discussed, nor are there any unresolved mathematical steps mentioned in the posts.
Readers interested in mathematical proofs, particularly those involving trigonometric identities and summations, may find this discussion relevant.
Pranav said:Prove the following:
$$\sum_{k=1}^n \left(2^k\sin^2\frac{x}{2^k}\right)^2=\left(2^n\sin\frac{x}{2^n}\right)^2-\sin^2x$$
anemone said:My solution:
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,
$$\sum_{k=1}^n \left(2^k\sin^2\frac{x}{2^k}\right)^2$$$$=\sum_{k=1}^n \left(2^{2k}\sin^2\dfrac{x}{2^k}-2^{2k-2}\left(\sin^2\dfrac{x}{2^{k-1}}\right) \right)$$$$=\left(4\sin^2\dfrac{x}{2}-\sin^2x\right)+\left(16\sin^2\dfrac{x}{4}-4\sin^2\dfrac{x}{2}\right)+\left(64\sin^2\dfrac{x}{8}-16\sin^2\dfrac{x}{4}\right)+\cdots$$$$+\left(2^{2(n-1)}\sin^2\dfrac{x}{2^{n-1}}-2^{2n-4}\sin^2\dfrac{x}{2^{n-3}}\right)+\left(2^{2n}\sin^2\dfrac{x}{2^n}-2^{2(n-1)}\sin^2\dfrac{x}{2^{n-1}}\right)$$$$=\left(\cancel{4\sin^2\dfrac{x}{2}}-\sin^2x\right)+\left(\cancel{16\sin^2\dfrac{x}{4}}-\cancel{4\sin^2\dfrac{x}{2}}\right)+\left(\cancel{64\sin^2\dfrac{x}{8}}-\cancel{16\sin^2\dfrac{x}{4}}\right)+\cdots$$$$+\left(\cancel{2^{2(n-1)}\sin^2\dfrac{x}{2^{n-1}}}-\cancel{2^{2n-4}\sin^2\dfrac{x}{2^{n-3}}}\right)+\left(2^{2n}\sin^2\dfrac{x}{2^n}-\cancel{2^{2(n-1)}\sin^2\dfrac{x}{2^{n-1}}}\right)$$$$=\left(2^n\sin\dfrac{x}{2^n}\right)^2-\sin^2x\text{(QED)}$$
lfdahl said: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 …