The longer way of writing it out #1

Calculus Level 2

f ( x ) = 1 x n = 1 ( 1 ) n x 2 n ( 2 n 1 ) ! f(x) = -\frac 1x \sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{(2n-1)!}

For f ( x ) f(x) as defined above, what is f ( π 6 ) f ( π 6 ) \dfrac{f'(\frac{\pi}{6})}{f(\frac{\pi}{6})} ?


The answer is 1.732050807568877293527446341505.

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3 solutions

The given sum is f ( x ) = x x 3 3 ! + x 5 5 ! x 7 7 ! + . . . = sin x f ( x ) = cos x f(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...=\sin x\implies f'(x) =\cos x

f ( x ) f ( x ) = cot x \implies \frac{f'(x) }{f(x)}=\cot x .

f ( π 6 ) f ( π 6 ) = cot ( π 6 ) = 3 1.732 \implies \dfrac{f'(\frac{π}{6})}{f(\frac{π}{6})}=\cot (\frac{π}{6})=\sqrt 3\approx \boxed {1.732} .

Chew-Seong Cheong
Jun 11, 2020

f ( x ) = 1 x n = 1 ( 1 ) n x 2 n ( 2 n 1 ) ! = 1 x n = 0 ( 1 ) n x 2 n + 2 ( 2 n + 1 ) ! = n = 0 ( 1 ) n x 2 n + 1 ( 2 n + 1 ) ! = sin x f ( x ) = cos x f ( π 6 ) f ( π 6 ) = cot π 6 = 3 1.73 \begin{aligned} f(x) & = - \frac 1x \sum_\blue{n=1}^\infty \frac {(-1)^n x^{2n}}{(2n-1)!} \\ & = \frac 1x \sum_\red{n=0}^\infty \frac {(-1)^n x^{2n+2}}{(2n+1)!} \\ & = \sum_{n=0}^\infty \frac {(-1)^n x^{2n+1}}{(2n+1)!} \\ & = \sin x \\ \implies f'(x) & = \cos x \\ \frac {f'(\frac \pi 6)}{f(\frac \pi 6)} & = \cot \frac \pi 6 = \sqrt 3 \approx \boxed{1.73} \end{aligned}

Aruna Yumlembam
Jun 10, 2020

Can you rotate the picture, @Aruna Yumlembam ?

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