Hmm, this looks familiar.

Calculus Level 5

Define a recurrence relation with starting value $x_0$ such that $-1 < x_0 < 1$ and

$x_{n+1}=\sqrt{\frac{1+x_n}{2}}.$

What is the value of

$\cos\left(\frac{\sqrt{1-x_0^2}}{\prod\limits_{n=1}^{\infty}x_n}\right)=\,\,?$

-1

-\frac{1}{x_0}

1 Cannot be determined

x_0

None of the other options

\frac{\sqrt{2}}{2}

1 solution

Isaac Buckley
Jul 13, 2015

We have $x_n=2x_{n+1}^2-1$

Let $x_0=\cos(y)$ for $-1<x_0<1$ .

We can then deduce that $x_n=\cos\left(\frac{y}{2^n}\right)$

We plug it into the expression to get

$\large \cos\left(\frac{\sqrt{1-\cos(y)^2}}{\prod\limits_{n=1}^{\infty}\cos\left(\frac{y}{2^n}\right)}\right)=\cos\left(\frac{\sin(y)}{\frac{\sin(y)}{y}}\right)=\cos(y)=x_0$

Xo is in the interval (-1,1). Hence it could be any value from -1 to 1. Hence the answer can be 1/√2 also

Vishwak Srinivasan - 5 years, 10 months ago

Can you please elaborate on that? I'm not quite sure what you mean.

Isaac Buckley - 5 years, 10 months ago

What I meant was, since the answer is $x_0$ , it simply could be any value between -1 and 1, (-1 and 1 excluded). I just wanted the option $\frac{\sqrt{2}}{2}$ to be removed.

Vishwak Srinivasan - 5 years, 10 months ago

@Vishwak Srinivasan – Oh, I see what you're saying now. It could also be $0$ too. I don't think it makes it that ambiguous by having them as an option but I do see what you mean. I'll see if I can reword it.

Isaac Buckley - 5 years, 10 months ago

Short and sweet. Kudos!

Vishnu Bhagyanath - 5 years, 10 months ago

Did the same

Aakash Khandelwal - 5 years, 11 months ago

Hmm, this looks familiar.

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