n = 1 ∑ ∞ ( − 1 ) n − 1 n ! n 8 = e k
What is ∣ k ∣ ?
Notations:
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When I was solving this problem I thought that there might be some pattern on this recurring polynomials . Unfortunately I end up by doing differentiation.
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Unfortunately I wasn't able to find a pattern as well. Maybe there's a more better solution to this problem for even higher powers.
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The formula for the general case is ∑ k = 0 ∞ k ! ( − 1 ) k k n = e B n ~ where B n ~ are the complementary Bell numbers if you are interested.
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Credits for the approach applied to my solution go to Anubhav Tyagi .
Using the series expansion of e x ,
e x = 1 + 1 ! x + 2 ! x 2 + 3 ! x 3 + 4 ! x 4 + . . .
We differentiate both sides w.r.t. x once to obtain,
e x = 1 ! 1 + 2 ! 2 x + 3 ! 3 x 2 + 4 ! 4 x 3 + . . .
Multiplying by x on both sides of the equation yields,
x e x = 1 ! x + 2 ! 2 x 2 + 3 ! 3 x 3 + 4 ! 4 x 4 + . . .
Differentiate again to get,
e x ( x + 1 ) = 1 ! 1 + 2 ! 2 2 x + 3 ! 3 2 x 2 + 4 ! 4 2 x 3 + . . .
We observe that the following can be written as,
e x ( x + 1 ) = n = 1 ∑ ∞ x n − 1 n ! n 2
From here, we've obtained a sum where the exponent of n in the numerator is 2 . So, by similar approach, if we continue to differentiate w.r.t. x and multiply by x alternatively, we'll obtain a sum where the exponent of n in the numerator is 8 .
After some tedious work , we obtain,
e x ( x 7 + 2 8 x 6 + 2 6 6 x 5 + 1 0 5 0 x 4 + 1 7 0 1 x 3 + 9 6 6 x 2 + 1 2 7 x + 1 ) = 1 ! 1 + 2 ! 2 8 x + 3 ! 3 8 x 2 + 4 ! 4 8 x 3 + . . . = n = 1 ∑ ∞ x n − 1 n ! n 8
Putting x = − 1 , we obtain,
− e 5 0 = 1 ! 1 − 2 ! 2 8 + 3 ! 3 8 − 4 ! 4 8 + . . . = n = 1 ∑ ∞ ( − 1 ) n − 1 n ! n 8
Thus,
k = − 5 0 ⟹ ∣ k ∣ = 5 0
The part (tedious work) skipped form the main solution.
e x ( x + 1 ) = 1 ! 1 + 2 ! 2 2 x + 3 ! 3 2 x 2 + 4 ! 4 2 x 3 + . . .
Multiplying both sides by x ,
e x x ( x + 1 ) = 1 ! x + 2 ! 2 2 x 2 + 3 ! 3 2 x 3 + 4 ! 4 2 x 4 + . . .
Differentiating both sides w.r.t. x ,
e x ( x 2 + 3 x + 1 ) = 1 ! 1 + 2 ! 2 3 x + 3 ! 3 3 x 2 + 4 ! 4 3 x 3 + . . .
Multiplying both sides by x ,
e x x ( x 2 + 3 x + 1 ) = 1 ! x + 2 ! 2 3 x 2 + 3 ! 3 3 x 3 + 4 ! 4 3 x 4 + . . .
Differentiating both sides w.r.t. x ,
e x ( x 3 + 6 x 2 + 7 x + 1 ) = 1 ! 1 + 2 ! 2 4 x + 3 ! 3 4 x 2 + 4 ! 4 4 x 3 + . . .
Multiplying both sides by x ,
e x x ( x 3 + 6 x 2 + 7 x + 1 ) = 1 ! x + 2 ! 2 4 x 2 + 3 ! 3 4 x 3 + 4 ! 4 4 x 4 + . . .
Differentiating both sides w.r.t. x ,
e x ( x 4 + 1 0 x 3 + 2 5 x 2 + 1 5 x + 1 ) = 1 ! 1 + 2 ! 2 5 x + 3 ! 3 5 x 2 + 4 ! 4 5 x 3 + . . .
Multiplying both sides by x ,
e x x ( x 4 + 1 0 x 3 + 2 5 x 2 + 1 5 x + 1 ) = 1 ! x + 2 ! 2 5 x 2 + 3 ! 3 5 x 3 + 4 ! 4 5 x 4 + . . .
Differentiating both sides w.r.t. x ,
e x ( x 5 + 1 5 x 4 + 6 5 x 3 + 9 0 x 2 + 3 1 x + 1 ) = 1 ! 1 + 2 ! 2 6 x + 3 ! 3 6 x 2 + 4 ! 4 6 x 3 + . . .
Multiplying both sides by x ,
e x x ( x 5 + 1 5 x 4 + 6 5 x 3 + 9 0 x 2 + 3 1 x + 1 ) = 1 ! x + 2 ! 2 6 x 2 + 3 ! 3 6 x 3 + 4 ! 4 6 x 4 + . . .
Differentiating both sides w.r.t. x ,
e x ( x 6 + 2 1 x 5 + 1 4 0 x 4 + 3 5 0 x 3 + 3 0 1 x 2 + 6 3 x + 1 ) = 1 ! 1 + 2 ! 2 7 x + 3 ! 3 7 x 2 + 4 ! 4 7 x 3 + . . .
Multiplying both sides by x ,
e x x ( x 6 + 2 1 x 5 + 1 4 0 x 4 + 3 5 0 x 3 + 3 0 1 x 2 + 6 3 x + 1 ) = 1 ! x + 2 ! 2 7 x 2 + 3 ! 3 7 x 3 + 4 ! 4 7 x 4 + . . .
Differentiating both sides w.r.t. x ,
e x ( x 7 + 2 8 x 6 + 2 6 6 x 5 + 1 0 5 0 x 4 + 1 7 0 1 x 3 + 9 6 6 x 2 + 1 2 7 x + 1 ) = 1 ! 1 + 2 ! 2 8 x + 3 ! 3 8 x 2 + 4 ! 4 8 x 3 + . . .