Let's do some calculus! (41)

Calculus Level 5

n = 1 ( 1 ) n 1 n 8 n ! = k e \large \displaystyle \sum_{n=1}^{\infty} {(-1)}^{n-1} \dfrac{n^8}{n!} = \dfrac ke

What is k | k | ?

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The answer is 50.

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1 solution

Tapas Mazumdar
Jan 23, 2017

Credits for the approach applied to my solution go to Anubhav Tyagi .


Using the series expansion of e x e^x ,

e x = 1 + x 1 ! + x 2 2 ! + x 3 3 ! + x 4 4 ! + . . . e^x = 1 + \dfrac{x}{1!} + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!} + \ ... \

We differentiate both sides w.r.t. x x once to obtain,

e x = 1 1 ! + 2 x 2 ! + 3 x 2 3 ! + 4 x 3 4 ! + . . . e^x = \dfrac{1}{1!} + \dfrac{2x}{2!} + \dfrac{3x^2}{3!} + \dfrac{4x^3}{4!} + \ ... \

Multiplying by x x on both sides of the equation yields,

x e x = x 1 ! + 2 x 2 2 ! + 3 x 3 3 ! + 4 x 4 4 ! + . . . xe^x = \dfrac{x}{1!} + \dfrac{2x^2}{2!} + \dfrac{3x^3}{3!} + \dfrac{4x^4}{4!} + \ ... \

Differentiate again to get,

e x ( x + 1 ) = 1 1 ! + 2 2 x 2 ! + 3 2 x 2 3 ! + 4 2 x 3 4 ! + . . . e^x (x+1) = \dfrac{1}{1!} + \dfrac{2^2 x}{2!} + \dfrac{3^2 x^2}{3!} + \dfrac{4^2 x^3}{4!} + \ ... \

We observe that the following can be written as,

e x ( x + 1 ) = n = 1 x n 1 n 2 n ! e^x (x+1) = \displaystyle \sum_{n=1}^{\infty} x^{n-1} \dfrac{n^2}{n!}

From here, we've obtained a sum where the exponent of n n in the numerator is 2 2 . So, by similar approach, if we continue to differentiate w.r.t. x x and multiply by x x alternatively, we'll obtain a sum where the exponent of n n in the numerator is 8 8 .

After some tedious work , we obtain,

e x ( x 7 + 28 x 6 + 266 x 5 + 1050 x 4 + 1701 x 3 + 966 x 2 + 127 x + 1 ) = 1 1 ! + 2 8 x 2 ! + 3 8 x 2 3 ! + 4 8 x 3 4 ! + . . . = n = 1 x n 1 n 8 n ! e^x (x^7 + 28x^6 + 266x^5 + 1050x^4 + 1701x^3 + 966x^2 + 127x + 1) = \dfrac{1}{1!} + \dfrac{2^8 x}{2!} + \dfrac{3^8 x^2}{3!} + \dfrac{4^8 x^3}{4!} + \ ... \ = \displaystyle \sum_{n=1}^{\infty} x^{n-1} \dfrac{n^8}{n!}

Putting x = 1 x = -1 , we obtain,

50 e = 1 1 ! 2 8 2 ! + 3 8 3 ! 4 8 4 ! + . . . = n = 1 ( 1 ) n 1 n 8 n ! -\dfrac{50}{e} = \dfrac{1}{1!} - \dfrac{2^8}{2!} + \dfrac{3^8}{3!} - \dfrac{4^8}{4!} + \ ... \ = \displaystyle \sum_{n=1}^{\infty} (-1)^{n-1} \dfrac{n^8}{n!}

Thus,

k = 50 k = 50 k = -50 \implies |k| = \boxed{50}


The part (tedious work) skipped form the main solution.

e x ( x + 1 ) = 1 1 ! + 2 2 x 2 ! + 3 2 x 2 3 ! + 4 2 x 3 4 ! + . . . e^x (x+1) = \dfrac{1}{1!} + \dfrac{2^2 x}{2!} + \dfrac{3^2 x^2}{3!} + \dfrac{4^2 x^3}{4!} + \ ... \

Multiplying both sides by x x ,

e x x ( x + 1 ) = x 1 ! + 2 2 x 2 2 ! + 3 2 x 3 3 ! + 4 2 x 4 4 ! + . . . e^x x(x+1) = \dfrac{x}{1!} + \dfrac{2^2 x^2}{2!} + \dfrac{3^2 x^3}{3!} + \dfrac{4^2 x^4}{4!} + \ ... \

Differentiating both sides w.r.t. x x ,

e x ( x 2 + 3 x + 1 ) = 1 1 ! + 2 3 x 2 ! + 3 3 x 2 3 ! + 4 3 x 3 4 ! + . . . e^x (x^2+3x+1) = \dfrac{1}{1!} + \dfrac{2^3 x}{2!} + \dfrac{3^3 x^2}{3!} + \dfrac{4^3 x^3}{4!} + \ ... \

Multiplying both sides by x x ,

e x x ( x 2 + 3 x + 1 ) = x 1 ! + 2 3 x 2 2 ! + 3 3 x 3 3 ! + 4 3 x 4 4 ! + . . . e^x x(x^2+3x+1) = \dfrac{x}{1!} + \dfrac{2^3 x^2}{2!} + \dfrac{3^3 x^3}{3!} + \dfrac{4^3 x^4}{4!} + \ ... \

Differentiating both sides w.r.t. x x ,

e x ( x 3 + 6 x 2 + 7 x + 1 ) = 1 1 ! + 2 4 x 2 ! + 3 4 x 2 3 ! + 4 4 x 3 4 ! + . . . e^x (x^3+6x^2+7x+1) = \dfrac{1}{1!} + \dfrac{2^4 x}{2!} + \dfrac{3^4 x^2}{3!} + \dfrac{4^4 x^3}{4!} + \ ... \

Multiplying both sides by x x ,

e x x ( x 3 + 6 x 2 + 7 x + 1 ) = x 1 ! + 2 4 x 2 2 ! + 3 4 x 3 3 ! + 4 4 x 4 4 ! + . . . e^x x(x^3+6x^2+7x+1) = \dfrac{x}{1!} + \dfrac{2^4 x^2}{2!} + \dfrac{3^4 x^3}{3!} + \dfrac{4^4 x^4}{4!} + \ ... \

Differentiating both sides w.r.t. x x ,

e x ( x 4 + 10 x 3 + 25 x 2 + 15 x + 1 ) = 1 1 ! + 2 5 x 2 ! + 3 5 x 2 3 ! + 4 5 x 3 4 ! + . . . e^x (x^4+10x^3+25x^2+15x+1) = \dfrac{1}{1!} + \dfrac{2^5 x}{2!} + \dfrac{3^5 x^2}{3!} + \dfrac{4^5 x^3}{4!} + \ ... \

Multiplying both sides by x x ,

e x x ( x 4 + 10 x 3 + 25 x 2 + 15 x + 1 ) = x 1 ! + 2 5 x 2 2 ! + 3 5 x 3 3 ! + 4 5 x 4 4 ! + . . . e^x x(x^4+10x^3+25x^2+15x+1) = \dfrac{x}{1!} + \dfrac{2^5 x^2}{2!} + \dfrac{3^5 x^3}{3!} + \dfrac{4^5 x^4}{4!} + \ ... \

Differentiating both sides w.r.t. x x ,

e x ( x 5 + 15 x 4 + 65 x 3 + 90 x 2 + 31 x + 1 ) = 1 1 ! + 2 6 x 2 ! + 3 6 x 2 3 ! + 4 6 x 3 4 ! + . . . e^x (x^5+15x^4+65x^3+90x^2+31x+1) = \dfrac{1}{1!} + \dfrac{2^6 x}{2!} + \dfrac{3^6 x^2}{3!} + \dfrac{4^6 x^3}{4!} + \ ... \

Multiplying both sides by x x ,

e x x ( x 5 + 15 x 4 + 65 x 3 + 90 x 2 + 31 x + 1 ) = x 1 ! + 2 6 x 2 2 ! + 3 6 x 3 3 ! + 4 6 x 4 4 ! + . . . e^x x(x^5+15x^4+65x^3+90x^2+31x+1) = \dfrac{x}{1!} + \dfrac{2^6 x^2}{2!} + \dfrac{3^6 x^3}{3!} + \dfrac{4^6 x^4}{4!} + \ ... \

Differentiating both sides w.r.t. x x ,

e x ( x 6 + 21 x 5 + 140 x 4 + 350 x 3 + 301 x 2 + 63 x + 1 ) = 1 1 ! + 2 7 x 2 ! + 3 7 x 2 3 ! + 4 7 x 3 4 ! + . . . e^x (x^6+21x^5+140x^4+350x^3+301x^2+63x+1) = \dfrac{1}{1!} + \dfrac{2^7 x}{2!} + \dfrac{3^7 x^2}{3!} + \dfrac{4^7 x^3}{4!} + \ ... \

Multiplying both sides by x x ,

e x x ( x 6 + 21 x 5 + 140 x 4 + 350 x 3 + 301 x 2 + 63 x + 1 ) = x 1 ! + 2 7 x 2 2 ! + 3 7 x 3 3 ! + 4 7 x 4 4 ! + . . . e^x x(x^6+21x^5+140x^4+350x^3+301x^2+63x+1) = \dfrac{x}{1!} + \dfrac{2^7 x^2}{2!} + \dfrac{3^7 x^3}{3!} + \dfrac{4^7 x^4}{4!} + \ ... \

Differentiating both sides w.r.t. x x ,

e x ( x 7 + 28 x 6 + 266 x 5 + 1050 x 4 + 1701 x 3 + 966 x 2 + 127 x + 1 ) = 1 1 ! + 2 8 x 2 ! + 3 8 x 2 3 ! + 4 8 x 3 4 ! + . . . e^x (x^7 + 28x^6 + 266x^5 + 1050x^4 + 1701x^3 + 966x^2 + 127x + 1) = \dfrac{1}{1!} + \dfrac{2^8 x}{2!} + \dfrac{3^8 x^2}{3!} + \dfrac{4^8 x^3}{4!} + \ ... \

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.

Syed Shahabudeen - 4 years, 4 months ago

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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.

Tapas Mazumdar - 4 years, 4 months ago

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The formula for the general case is k = 0 ( 1 ) k k n k ! = B n ~ e \sum_{k=0}^{\infty} \frac{(-1)^kk^n}{k!} = \frac{\tilde{B_n}}{e} where B n ~ \tilde{B_n} are the complementary Bell numbers if you are interested.

Romain Bouchard - 3 years, 2 months ago

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