Summer of 69 Part-2

Calculus Level 5

Evaluate :

d 69 ( a r c t a n ( x ) ) d x 69 a t x = 5 \frac { { d }^{ 69 }(arctan(x)) }{ d{ x }^{ 69 } } \quad at\quad x=5

The answer is of the form x ∗ 10 47 x*{ 10 }^{ 47 } .

Find x x .

This question requires lots and lots of patience if not proceeded properly.


The answer is 3.289.

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Aditya Kumar by Aditya Kumar , 22, India

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1 solution

Aditya Kumar
May 3, 2015

d n ( a r c t a n ( x ) ) d x n = ( − 1 ) n − 1 ( n − 1 ) ! ( 1 + x 2 ) n 2 s i n ( n . a r c s i n ( 1 1 + x 2 ) ) W e c a n p r o v e t h e a b o v e t h e o r e m b y i n d u c t i o n . O n s u b s t i t u t i n g v a l u e s w e g e t , d 69 ( a r c t a n ( x ) ) d x 69 ( a t x = 5 ) = 3.289 ∗ 10 47 \frac { { d }^{ n }(arctan(x)) }{ d{ x }^{ n } } \quad =\quad \frac { { (-1) }^{ n-1 }(n-1)! }{ { (1+{ x }^{ 2 }) }^{ \frac { n }{ 2 } } } sin\left( n.arcsin\left( \frac { 1 }{ \sqrt { 1+{ x }^{ 2 } } } \right) \right) \\ We\quad can\quad prove\quad the\quad above\quad theorem\quad by\quad induction.\\ On\quad substituting\quad values\quad we\quad get,\\ \frac { { d }^{ 69 }(arctan(x)) }{ d{ x }^{ 69 } } \quad (at\quad x=5)\quad =\quad 3.289*{ 10 }^{ 47 }

3 Helpful 0 Interesting 0 Brilliant 0 Confused

There is another way to solve this but that requires an insane amount of work (computing power). Take the derivative as S S .

Using the Maclaurin series for arctan ⁡ ( x ) \arctan(x) for x ≥ 1 x\geq 1 and then interchanging derivative operator and summation operator, you can get,

2 5 34 × S = ∑ i = 1 ∞ ( ( − 1 ) i + 1 ( ∏ m = 0 67 ( − 2 i − m ) ) 1 2 5 i ) ⟹ S = 1 2 5 34 ∑ i = 1 ∞ ( ( − 1 ) i + 1 ( 2 i + 67 ) ! ( 2 i − 1 ) ! ⋅ 1 2 5 i ) 25^{34}\times S=\sum_{i=1}^\infty\left((-1)^{i+1}\left(\prod_{m=0}^{67}(-2i-m)\right)\frac{1}{25^i}\right)\\ \implies S=\frac{1}{25^{34}}\sum_{i=1}^\infty\left((-1)^{i+1}\frac{(2i+67)!}{(2i-1)!}\cdot\frac{1}{25^i}\right)

By ratio test, the sum on the right side converges. Now, we can use some computing power (software like Maxima, or W|A) or even manually calculate the partial sums and then conclude our answer. This, obviously requires a lot of manual labor and hence the theorem you presented is a better alternative but it just presents no idea as to how someone gets the motivation to make that claim. It just doesn't seem obvious to me.

EDIT: I think I found the source of your theorem. Here is the link of the pdf with a direct proof of it too.

Prasun Biswas - 6 years, 1 month ago

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Yes there are several ways to solve it @Prasun Biswas .

Aditya Kumar - 6 years, 1 month ago

like u said ... i have another method which forms a recurrence relation b\w nth derivatives at x=5 ..... but solving that relation is horrible!!

Abhinav Raichur - 6 years ago

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