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Calculus Level 3

b ( c ! ! d ! ! c ! d ! ) + ∑ n = 1 ∞ ( − 1 ) n ( 4 n + 3 ) ! = 1 a e a ( e sin ⁡ ( 1 a − π b ) + sin ⁡ ( 1 a + π b ) ) \begin{aligned} b\left(\large{\frac{\hspace{3mm} \frac{{\color{#3D99F6}c!!}}{{\color{#D61F06}d!!}}\hspace{3mm} }{\frac{{\color{#3D99F6}c!}}{{\color{#D61F06}d!}}}}\right)+\sum_{n=1}^{\infty}\dfrac{(-1)^n}{(4n+3)!} = \dfrac{1}{a\sqrt[\sqrt a]{e}}\left(\sqrt e \sin\left(\dfrac{1}{\sqrt a} -\dfrac{\pi}{b}\right)+\sin \left(\dfrac{1}{\sqrt a} +\dfrac{\pi}{b}\right)\right)\end{aligned}

The equation above holds true for positive integers a a , b b , c c and d d . Find the value of a + b + c + d a+b+c+d .


The answer is 16.

Naren Bhandari by Naren Bhandari , 21, India

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1 solution

Chew-Seong Cheong
Nov 24, 2018

Consider the Maclaurin series of s i n x sin x as follows.

sin ⁡ x = x 1 ! − x 3 3 ! + x 5 5 ! − x 7 7 ! + x 9 9 ! − ⋯ x sin ⁡ x = x 2 1 ! − x 4 3 ! + x 6 5 ! − x 8 7 ! + x 10 9 ! − ⋯ Putting x = i , where i = − 1 i sin ⁡ i = i 1 ! + 1 3 ! − i 5 ! − 1 7 ! + i 9 ! − ⋯ Taking the real part on both sides ℜ ( i sin ⁡ i ) = ∑ n = 1 ∞ ( − 1 ) n ( 4 n + 3 ) ! \begin{aligned} \sin x & = \frac x{1!} - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!} + \frac {x^9}{9!} - \cdots \\ x\sin x & = \frac {x^2}{1!} - \frac {x^4}{3!} + \frac {x^6}{5!} - \frac {x^8}{7!} + \frac {x^{10}}{9!} - \cdots & \small \color{#3D99F6} \text{Putting }x = \sqrt{i} \text{, where }i=\sqrt {-1} \\ \sqrt i \sin \sqrt i & = \frac i{1!} + \frac 1{3!} - \frac i{5!} - \frac 1{7!} + \frac i{9!} - \cdots & \small \color{#3D99F6} \text{Taking the real part on both sides} \\ \Re \left(\sqrt i \sin \sqrt i \right) & = \sum_{n=1}^\infty \frac {(-1)^n}{(4n+3)!} \end{aligned}

Then

∑ n = 0 ∞ ( − 1 ) n ( 4 n + 3 ) ! = ℜ ( i sin ⁡ i ) = ℜ ( i 2 i ( e i i − e − i i ) ) = ℜ ( e π 4 i 2 i ( e e 3 π 4 i − e − e 3 π 4 i ) ) = ℜ ( e π 4 i 2 i ( e − 1 2 + i 2 − e 1 2 − i 2 ) ) = ℜ ( 1 2 i ( e − 1 2 e ( π 4 + 1 2 ) i − e 1 2 e ( π 4 − 1 2 ) i ) ) = ℜ ( 1 2 i ( e − 1 2 ( cos ⁡ ( π 4 + 1 2 ) + i sin ⁡ ( π 4 + 1 2 ) ) − e 1 2 ( cos ⁡ ( π 4 − 1 2 ) + i sin ⁡ ( π 4 − 1 2 ) ) ) ) = 1 2 e 2 ( 2 sin ⁡ ( 1 2 − π 4 ) + sin ⁡ ( 1 2 + π 4 ) ) \begin{aligned} \sum_{\color{#D61F06}n=0}^\infty \frac {(-1)^n}{(4n+3)!} & = \Re \left(\sqrt i \sin \sqrt i \right) \\ & = \Re \left(\frac {\sqrt i}{2i} \left(e^{i\sqrt i} - e^{-i\sqrt i}\right)\right) \\ & = \Re \left(\frac {e^{\frac \pi 4 i}}{2i} \left(e^{e^{\frac {3\pi}4 i}} - e^{-e^{\frac {3\pi}4 i}}\right)\right) \\ & = \Re \left(\frac {e^{\frac \pi 4 i}}{2i} \left(e^{-\frac 1{\sqrt 2} + \frac i{\sqrt 2}} - e^{\frac 1{\sqrt 2} - \frac i{\sqrt 2}}\right)\right) \\ & = \Re \left(\frac 1{2i} \left(e^{-\frac 1{\sqrt 2}} e^{\left(\frac \pi 4 + \frac 1{\sqrt 2}\right)i} - e^{\frac 1{\sqrt 2}} e^{\left(\frac \pi 4 - \frac 1{\sqrt 2}\right)i} \right)\right) \\ & = \Re \left(\frac 1{2i} \left(e^{-\frac 1{\sqrt 2}} \left(\cos \left(\frac \pi 4 + \frac 1{\sqrt 2} \right) + i \sin \left(\frac \pi 4 + \frac 1{\sqrt 2} \right) \right) - e^{\frac 1{\sqrt 2}} \left(\cos \left(\frac \pi 4 - \frac 1{\sqrt 2} \right) + i \sin \left(\frac \pi 4 - \frac 1{\sqrt 2} \right) \right) \right) \right) \\ & = \frac 1{2\sqrt[\sqrt 2]e} \left(\sqrt 2 \sin \left(\frac 1{\sqrt 2} - \frac \pi 4 \right) + \sin \left(\frac 1{\sqrt 2}+\frac \pi 4 \right)\right) \end{aligned}

Implying that a = 2 a=2 , b = 4 b=4 and:

1 3 ! + ∑ n = 1 ∞ ( − 1 ) n ( 4 n + 3 ) ! = 1 2 e 2 ( 2 sin ⁡ ( 1 2 − π 4 ) + sin ⁡ ( 1 2 + π 4 ) ) ⟹ b ( c ! ! d ! ! c ! d ! ) = 1 3 ! c ! ! d ! ! ⋅ d ! c ! = 1 24 ( c − 1 ) ! ! ( d − 1 ) ! ! = 24 = 6 ⋅ 4 ⋅ 2 2 \begin{aligned} \frac 1{3!} + \sum_{\color{#3D99F6}n=1}^\infty \frac {(-1)^n}{(4n+3)!} & = \frac 1{2\sqrt[\sqrt 2]e} \left(\sqrt 2 \sin \left(\frac 1{\sqrt 2} - \frac \pi 4 \right) + \sin \left(\frac 1{\sqrt 2}+\frac \pi 4 \right)\right) \\ \implies b \left(\frac {\frac {c!!}{d!!}}{\frac {c!}{d!}} \right) & = \frac 1{3!} \\ \frac {c!!}{d!!} \cdot \frac {d!}{c!} & = \frac 1{24} \\ \frac {(c-1)!!}{(d-1)!!} & = 24 = \frac {6 \cdot 4 \cdot 2}2 \end{aligned}

Therefore, c = 7 c = 7 and d = 3 d=3 , and a + b + c + d = 2 + 4 + 7 + 3 = 16 a+b+c+d = 2+4+7+3 = \boxed{16} .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice solution(same ) Sir !! I was really inspired by previous problem ,so I came to write up this problem. Do you think this problem is good ?

Naren Bhandari - 2 years, 6 months ago

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Again, the part b c ! ! d ! ! c ! d ! b\dfrac {\frac {c!!}{d!!}}{\frac {c!}{d!}} is not necessary. I think you have added it so that you can post it as your own. The previous one the original should be ( n − 1 ) ! (n-1)! instead of ( n − 2 ) ! (n-2)! , because computation for ( n − 1 ) ! (n-1)! is much easier and formula can be used. I think you should quote who posted them and where.

Chew-Seong Cheong - 2 years, 6 months ago

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The previous problem doesn't have factorial part which is just ∑ n = 0 ∞ ( − 1 ) n 4 n + 3 \sum_{n=0}^{\infty}\dfrac{(-1)^n}{4n+3} and what i do is just add the factorial part to it.

I got that problem(original problem), ∑ n = 0 ∞ ( − 1 ) n 4 n + 3 \sum_{n=0}^{\infty}\dfrac{(-1)^n}{4n+3} in Facebook group

Naren Bhandari - 2 years, 6 months ago

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