try this one!

Find : 1212+121212+1212+121212+1212+1212+1212 \sqrt{ \frac{1}{2}}\sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}} \sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}}}\sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2} +\frac{1}{2}\sqrt{\frac{1}{2}}}}} \\ \ldots \infty

Note by Kushagraa Aggarwal
6 years, 1 month ago

No vote yet
13 votes

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Let a1=12a_1 = \sqrt{\dfrac{1}{2}} and let an+1=12+12ana_{n+1} = \sqrt{\dfrac{1}{2} + \dfrac{1}{2}a_n} for integers n1n \ge 1. We want the value of n=1an\displaystyle\prod_{n = 1}^{\infty}a_n.

Note that a1=cos(π4)a_1 = \cos \left(\dfrac{\pi}{4}\right) and if an=cos(π2n+1)a_n = \cos\left(\dfrac{\pi}{2^{n+1}}\right) then an+1=12+12cos(π2n+1)=cos(π2n+2)a_{n+1} = \sqrt{\dfrac{1}{2} + \dfrac{1}{2}\cos\left(\dfrac{\pi}{2^{n+1}}\right)} = \cos\left(\dfrac{\pi}{2^{n+2}}\right).

Therefore, by induction, we have an=cos(π2n+1)a_n = \cos\left(\dfrac{\pi}{2^{n+1}}\right) for all integers n1n \ge 1.

Let PN=n=1Nan=n=1Ncos(π2n+1)P_N = \displaystyle \prod_{n = 1}^{N} a_n = \displaystyle \prod_{n = 1}^{N}\cos\left(\dfrac{\pi}{2^{n+1}}\right). Using the identity sinθcosθ=12sin2θ\sin \theta \cos \theta = \dfrac{1}{2}\sin 2\theta repeatedly, we have:

PNsin(π2N+1)=sin(π2N+1)n=1Ncos(π2n+1)=12Nsin(π2)=12NP_N \sin\left(\dfrac{\pi}{2^{N+1}}\right) = \sin\left(\dfrac{\pi}{2^{N+1}}\right)\displaystyle \prod_{n = 1}^{N}\cos\left(\dfrac{\pi}{2^{n+1}}\right) = \dfrac{1}{2^N}\sin\left(\dfrac{\pi}{2}\right) = \dfrac{1}{2^N}.

Therefore, PN=12Ncsc(π2N+1)P_N = \dfrac{1}{2^N}\csc\left(\dfrac{\pi}{2^{N+1}}\right) for all integers N1N \ge 1.

For x0x \approx 0 we have cscx=1x+O(x)\csc x = \dfrac{1}{x} + O(x). Hence, PN=12N[2N+1π+O(π2N+1)]=2π+O(22N)P_N = \dfrac{1}{2^N}\left[\dfrac{2^{N+1}}{\pi} + O\left(\dfrac{\pi}{2^{N+1}}\right)\right] = \dfrac{2}{\pi} + O(2^{-2N}).

As NN \to \infty we have 22N02^{-2N} \to 0. Therefore, n=1an=limNPN=2π\displaystyle\prod_{n = 1}^{\infty}a_n = \lim_{N \to \infty}P_N = \dfrac{2}{\pi}.

Jimmy Kariznov - 6 years, 1 month ago

Log in to reply

thanks a lot!. it was actually not easy to get to substitute 12 \frac{1}{\sqrt{2}} by cos(π4) cos(\frac{\pi}{4})

kushagraa aggarwal - 6 years, 1 month ago

Log in to reply

If you don't mind, can you please share the source of this problem?

Thanks!

Pranav Arora - 6 years, 1 month ago

Log in to reply

@Pranav Arora it was asked in a mock test.

kushagraa aggarwal - 6 years, 1 month ago

Log in to reply

@Kushagraa Aggarwal Are you talking about an IIT-JEE mock test? If so, FIITJEE?

Pranav Arora - 6 years, 1 month ago

Log in to reply

@Pranav Arora Nope , Vidyamandir Classes.

kushagraa aggarwal - 6 years, 1 month ago

Log in to reply

Oh yeah, 2/pi, Viete's formula.

Jeremy Shuler - 6 years, 1 month ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...