I was trying to solve

$$\lim_{n\to\infty} \cos{x\over2}\cos{x\over4}\cos{x\over8}\cdots\cos{x\over2^n}$$

First we can use sine half angle formula from behind and get telescoping series and I got limit $$\lim_{n\to\infty} \cos{x\over2}\cos{x\over4}\cos{x\over8}\cdots\cos{x\over2^n}=\lim_{n\to\infty} \frac{\sin x}{2^n \sin{x\over2^n}}={\sin x\over x}$$

Then I thought using complex number given limit is $$\Re\left[\exp\left(ix\left({1\over2}+{1\over4}+\cdots\right)\right)\right] = \Re[e^{ix}]=\cos x$$

Which answer and solution is correct? What did I do wrong?

  • 1
    $\begingroup$ The limit is a sinc function !! I wonder if there is a deeper interpretation of this problem. It would interest communication engineers for example. $\endgroup$ Oct 16 at 9:11

Your result with the series assumes that

$$\Re(z_1 z_2) = \Re(z_1) \Re(z_2)$$

for $z_1,z_2 \in \Bbb C$. This is not true.

Let $z_1 = a + bi, z_2 = c + di$. Then $z_1z_2 = (ac - bd) + (ad + bc) i$. The lack of multiplicativity is obvious.

  • $\begingroup$ Thanks and sorry I don't know how I made such mistake . So we can't solve this question using complex . $\endgroup$
    – RKK
    Oct 16 at 8:54
  • $\begingroup$ It's a common one, don't worry about it. And you might be able to do it with complex numbers, but it needs to be a little more complicated than this, sadly $\endgroup$ Oct 16 at 8:55

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