Not sure, and positive quantities bite me.

Calculus Level 5

Let a a and b b are positive quantities such that 0 < a < b 0 < a < b . Define the sequences ( a i ) , ( b i ) (a_i), (b_i) as follows:

  • a 1 = a + b 2 a_1 = \frac{a+b}{2} and b 1 = a 1 ⋅ b b_1 = \sqrt{a_1 \cdot b}
  • a n + 1 = a n + b n 2 a_{n+1} = \frac{a_n + b_n}{2} and b n + 1 = a n + 1 ⋅ b n b_{n+1} = \sqrt{a_{n+1} \cdot b_n} for all natural number n n

Which of the following is correct?

Please share!!

For other problems by me, click here

lim ⁡ n → ∞ b n = b 2 − a 2 cos ⁡ − 1 a b \lim_{n \to \infty} b_n = \frac{\sqrt{b^{2} - a^{2}}}{\cos^{-1}\frac{a}{b}} lim ⁡ n → ∞ b n = a 2 − b 2 cos ⁡ − 1 b a \lim_{n \to \infty} b_n = \frac{\sqrt{a^{2} - b^{2}}}{\cos^{-1}\frac{b}{a}} None of these lim ⁡ n → ∞ b n = b 2 − a 2 cos ⁡ − 1 a b \lim_{n \to \infty} b_n = \frac{ b^{2} - a^{2} }{\cos^{-1}\frac{a}{b}}

Vishwak Srinivasan by Vishwak Srinivasan , 24, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Let a = b cos ⁡ θ a = b\cos\theta

Then, a 1 = a + b 2 = b × 1 + cos ⁡ θ 2 = b × cos ⁡ 2 θ 2 a_{1} = \frac{a+b}{2} = \frac{b \times 1+\cos\theta}{2} = b \times \cos^{2}\frac{\theta}{2}

Hence b 1 = a 1 × b = b × cos ⁡ θ 2 b_{1} = \sqrt{a_{1} \times b} = b \times \cos\frac{\theta}{2}

Similarly:

a 2 = b × cos ⁡ θ 2 × cos ⁡ 2 θ 4 a_{2} = b \times \cos\frac{\theta}{2} \times \cos^{2}\frac{\theta}{4}

and b 2 = b × cos ⁡ θ 2 × cos ⁡ θ 2 2 b_{2} = b \times \cos\frac{\theta}{2} \times \cos\frac{\theta}{2^{2}}

Hence by careful observation,

b n = ∏ i = 1 n b × cos ⁡ θ 2 i \displaystyle b_{n} = \prod_{i=1}^n b \times \cos\frac{\theta}{2^{i}}

Hence,

b ∞ = lim ⁡ n → ∞ b n = lim ⁡ n → ∞ b × cos ⁡ θ 2 × cos ⁡ θ 2 2 . . . . . . . . × cos ⁡ θ 2 n \displaystyle b_{\infty} = \lim_{n\to\infty} b_{n} = \lim_{n\to\infty} b \times \cos\frac{\theta}{2} \times \cos\frac{\theta}{2^{2}} ........ \times \cos\frac{\theta}{2^{n}}

Multiplying and dividing by sin ⁡ θ 2 \sin\frac{\theta}{2} , sin ⁡ θ 2 2 \sin\frac{\theta}{2^{2}} , ........ till sin ⁡ θ 2 n \sin\frac{\theta}{2^{n}}

You get,

lim ⁡ n → ∞ b n = lim ⁡ n → ∞ b × sin ⁡ θ 2 n × sin ⁡ θ 2 n \displaystyle \lim_{n\to\infty} b_{n} = \lim_{n\to\infty} b \times \frac{\sin\theta }{2^{n} \times \sin\frac{\theta}{2^{n}}}

Multiplying and dividing θ \theta in the denominator, lim ⁡ n → ∞ b n = lim ⁡ n → ∞ b × sin ⁡ θ θ × sin ⁡ θ / 2 n θ / 2 n = b × sin ⁡ θ θ \displaystyle \lim_{n\to\infty} b_{n} = \lim_{n\to\infty} \frac{b \times \sin\theta}{\theta \times \frac {\sin\theta / 2^{n}}{\theta /2^{n}}} = b \times \frac{\sin\theta}{\theta}

Substituting,

sin ⁡ θ = 1 − cos ⁡ 2 θ \sin\theta =\sqrt{1 - \cos^{2}\theta}

and

θ = cos ⁡ − 1 a b \theta = \cos^{-1}\frac{a}{b}

We get the answer!

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice solution, but I see one problem with it. You'll need to put constraints on a a and b b . Otherwise, you won't always find θ \theta such that a = b cos ⁡ ( θ ) a = b \cos (\theta) , For example ( a , b ) = ( 4 , 2 ) (a,b) = (4,2) would imply that c o s ( θ ) = 2 \ cos (\theta) = 2 , which is not possible.

Siddhartha Srivastava - 6 years, 2 months ago

Log in to reply

Nice catch there, Siddhartha.

Vishwak Srinivasan - 6 years, 2 months ago

I have also made the necessary changes.

Vishwak Srinivasan - 6 years, 2 months ago

Thanks! I see that you addressed the concern that I had initially. I wasn't sure if you wanted people to work in complex numbers (both for b 2 − a 2 \sqrt{b^2-a^2} and cos ⁡ − 1 a b \cos ^{-1} \frac{a}{b} .

Note that all that you need is 0 < a < b 0 < a < b , and there is no need to restrict it by 1 above.

Calvin Lin Staff - 6 years, 2 months ago

Log in to reply

What is a ∞ a_\infty ? Option (a) should also be a correct choice.

Abhishek Sinha - 6 years, 2 months ago

Log in to reply

@Abhishek Sinha – @Vishwak Srinivasan I agree that a ∞ = b ∞ a_ \infty = b _ \infty . If so there would be multiples answers right? Let me know if I should remove the a n a_n option.

Calvin Lin Staff - 6 years, 2 months ago

Log in to reply

@Calvin Lin – @ @Calvin Lin : Sir, please do the needful.

Vishwak Srinivasan - 6 years, 2 months ago

Log in to reply

@Vishwak Srinivasan – Thanks. Those who chose the a n a_n option have been marked correct.

I've replaced that option with one about b n b _ n .

Calvin Lin Staff - 6 years, 2 months ago

Log in to reply

@Calvin Lin – I have been causing a little trouble lately. Sorry about that.

Vishwak Srinivasan - 6 years, 2 months ago

Log in to reply

@Vishwak Srinivasan – No worries. I'm glad that you're helping us fix the problem, so that others can enjoy it :)

Calvin Lin Staff - 6 years, 2 months ago

Is this problem original? If yes then ,great work,and yeah i also did the same thing :)

Siddharth Bhatnagar - 6 years, 2 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...