Lets push trigonometry to limits

Calculus Level 5

f k ( θ ) = ∑ r = 1 n ( tan ⁡ ( θ 2 r ) 2 r ) k + 1 3 ∑ r = 1 n ( tan ⁡ ( θ 2 r ) 8 r ) \displaystyle{{ f^{ k }\left( \theta \right) =\sum _{ r=1 }^{ n }{ { \left( \cfrac { \tan { \left( \cfrac { \theta }{ { 2 }^{ r } } \right) } }{ { 2 }^{ r } } \right) }^{ k } } }+\cfrac { 1 }{ 3 } \sum _{ r=1 }^{ n }{ \left( \cfrac { \tan { \left( \cfrac { \theta }{ { 2 }^{ r } } \right) } }{ 8^{ r } } \right) } }

Consider a function f f defined as above. And we again define a function L L below.

L ( θ ) = lim ⁡ n → ∞ ( ∑ k = 1 3 f k ( θ ) ) \displaystyle{L(\theta )=\lim _{ n\rightarrow \infty }{ \left( \sum _{ k=1 }^{ 3 }{ f^{ k }\left( \theta \right) } \right) } }

If L ( π 6 ) = a π 3 − b π 2 + c π − 1 d + p ( 1 − q ) − q \displaystyle{{ { L\left( \cfrac { \pi }{ 6 } \right) =\cfrac { a }{ { \pi }^{ 3 } } -\cfrac { b }{ { { \pi } }^{ 2 } } +\cfrac { c }{ { \pi } } -\cfrac { 1 }{ d } +p(1-\sqrt { q } ) }-\sqrt { q } }}

Find a + b + c + d + p + q a+b+c+d+p+q .

Details and assumptions :

  • Here a , b , c , d , p , q a,b,c,d,p,q are positive integers .

  • Hint: Do you know how to convert product into sum? \text{Hint: Do you know how to convert product into sum?}

Original


The answer is 268.

View wiki
Deepanshu Gupta by Deepanshu Gupta , 25, 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

Deepanshu Gupta
May 28, 2015

I will give you hint's and expressions , try to prove and kill at your own .

Hints

H i n t Hint -1)- Try to prove this elementary and standard result:

∏ k = 1 n cos ⁡ ( θ 2 k ) = sin ⁡ θ 2 n sin ⁡ ( θ 2 n ) \displaystyle{\prod _{ k=1 }^{ n }{ \cos { \left( \cfrac { \theta }{ { 2 }^{ k } } \right) } } =\cfrac { \sin { \theta } }{ { 2 }^{ n }\sin { \left( \cfrac { \theta }{ { 2 }^{ n } } \right) } } }

H i n t Hint -2)- Think how to convert product into sum ? (Yes Take logarithm to base e)

H i n t Hint -3)- Yes after converting ito sum , Diffrentiate 3 times (with some smart manupilations) , and note down each derivatives .

Expressions: (Try to Prove it)

Exp-1)- ∑ r = 1 n ( tan ⁡ ( θ 2 r ) 2 r ) = 1 2 n tan ⁡ ( θ 2 n ) − 1 tan ⁡ θ \displaystyle{{ \sum _{ r=1 }^{ n }{ { \left( \cfrac { \tan { \left( \cfrac { \theta }{ { 2 }^{ r } } \right) } }{ { 2 }^{ r } } \right) } } =\cfrac { 1 }{ { 2 }^{ n }\tan { \left( \cfrac { \theta }{ { 2 }^{ n } } \right) } } -\cfrac { 1 }{ \tan { \theta } } }}

Exp-2)- ∑ r = 1 n ( tan ⁡ ( θ 2 r ) 2 r ) 2 = 1 sin ⁡ 2 θ − 1 4 n sin ⁡ 2 ( θ 2 n ) − 4 n − 1 3 × 4 n \displaystyle{{\sum _{ r=1 }^{ n }{ { \left( \cfrac { \tan { \left( \cfrac { \theta }{ { 2 }^{ r } } \right) } }{ { 2 }^{ r } } \right) }^{ 2 } } =\cfrac { 1 }{ \sin ^{ 2 }{ \theta } } -\cfrac { 1 }{ { 4 }^{ n }\sin ^{ 2 }{ \left( \cfrac { \theta }{ { 2 }^{ n } } \right) } } -\cfrac { { 4 }^{ n }-1 }{ 3\times { 4 }^{ n } } } }

Exp-3)- ∑ r = 1 n ( tan ⁡ ( θ 2 r ) 2 r ) 3 + ∑ r = 1 n ( tan ⁡ ( θ 2 r ) 8 r ) = 1 8 n sin ⁡ 2 ( θ 2 n ) tan ⁡ ( θ 2 n ) − csc ⁡ 2 θ cot ⁡ θ \displaystyle{{\sum _{ r=1 }^{ n }{ { \left( \cfrac { \tan { \left( \cfrac { \theta }{ { 2 }^{ r } } \right) } }{ { 2 }^{ r } } \right) }^{ 3 } } +\sum _{ r=1 }^{ n }{ \left( \cfrac { \tan { \left( \cfrac { \theta }{ { 2 }^{ r } } \right) } }{ 8^{ r } } \right) } =\cfrac { 1 }{ { 8 }^{ n }\sin ^{ 2 }{ \left( \cfrac { \theta }{ { 2 }^{ n } } \right) \tan { \left( \cfrac { \theta }{ { 2 }^{ n } } \right) } } } -\csc ^{ 2 }{ \theta } \cot { \theta } } }

Final Result

Closed form of L( θ \theta ) is ,:

L ( θ ) = 1 θ − 1 θ 2 + 1 θ 3 − 1 3 + csc ⁡ 2 θ ( 1 − cot ⁡ θ ) − cot ⁡ θ . L(\theta) = \frac{1}{\theta} - \frac{1}{\theta^2} + \frac{1}{\theta^3} - \frac{1}{3} + \csc^2 \theta (1 - \cot \theta) - \cot \theta.

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

The major steps here are to be able to recognize how to deal with an infinite product as a telescoping product.

Letting n → ∞ n \to \infty , we get

∑ r = 1 ∞ ( tan ⁡ ( θ 2 r ) 2 r ) = 1 θ − 1 tan ⁡ θ , \sum _{ r=1 }^{ \infty }{ { \left( \cfrac { \tan { \left( \cfrac { \theta }{ { 2 }^{ r } } \right) } }{ { 2 }^{ r } } \right) } } = \frac{1}{\theta} - \frac{1}{\tan \theta},

∑ r = 1 ∞ ( tan ⁡ ( θ 2 r ) 2 r ) 2 = 1 sin ⁡ 2 θ − 1 θ 2 − 1 3 , \sum _{ r=1 }^{ \infty }{ { \left( \cfrac { \tan { \left( \cfrac { \theta }{ { 2 }^{ r } } \right) } }{ { 2 }^{ r } } \right) }^{ 2 } } = \frac{1}{\sin^2 \theta} - \frac{1}{\theta^2} - \frac{1}{3},

∑ r = 1 ∞ ( tan ⁡ ( θ 2 r ) 2 r ) 3 + ∑ r = 1 ∞ ( tan ⁡ ( θ 2 r ) 8 r ) = 1 θ 3 − csc ⁡ 2 θ cot ⁡ θ . \sum _{ r=1 }^{ \infty }{ { \left( \cfrac { \tan { \left( \cfrac { \theta }{ { 2 }^{ r } } \right) } }{ { 2 }^{ r } } \right) }^{ 3 } } +\sum _{ r=1 }^{ \infty }{ \left( \cfrac { \tan { \left( \cfrac { \theta }{ { 2 }^{ r } } \right) } }{ 8^{ r } } \right) } = \frac{1}{\theta^3} - \csc^2 \theta \cot \theta.

Adding these up, we get L ( θ ) = 1 θ − 1 θ 2 + 1 θ 3 − 1 3 + csc ⁡ 2 θ ( 1 − cot ⁡ θ ) − cot ⁡ θ . L(\theta) = \frac{1}{\theta} - \frac{1}{\theta^2} + \frac{1}{\theta^3} - \frac{1}{3} + \csc^2 \theta (1 - \cot \theta) - \cot \theta.

Jon Haussmann - 6 years ago

Beautiful question , messed up in exp-2,so got exp-3 incorrect as well !

Parth Lohomi - 6 years ago

Nice question deepanshu.

siddharth bhatt - 6 years ago

Excellent!

Thrayaksha Byroju Thrayaksha Byroju - 10 months, 2 weeks ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...