Areas and Series

Calculus Level 4

Let ∣ x ∣ < 1 |x| < 1 and e e be Euler's number .

Let m ( x ) = lim ⁡ n → ∞ [ ( ∑ j = 1 n x j ∑ j = 1 n ( j n ) n x n − j ) ∗ ( ∑ j = 1 n ( − 1 ) n − j ( j n ) n x n − j ) ∑ j = 1 n ( − 1 ) j + 1 x j ) ] m(x) = \lim_{n \rightarrow \infty} \left[\left(\dfrac{\sum_{j = 1}^{n} x^{j}}{\sum_{j = 1}^{n} (\dfrac{j}{n})^n x^{n - j}}\right) * \left(\dfrac{\sum_{j = 1}^{n} (-1)^{n - j} (\dfrac{j}{n})^n x^{n - j})}{\sum_{j = 1}^{n} (-1)^{j + 1} x^{j}}\right)\right] and p ( x ) = − m ( − x ) p(x) = -m(-x) .

If the area A A of the region bounded by m ( x ) m(x) and p ( x ) p(x) on [ 2 − e , e − 2 ] [2 - e,e - 2] can be expressed as A = ( α α ) ( e − α + ( e − β β + e ) ln ⁡ ( β ( λ − e ) ( e − β ) e − β ) ) A = (\alpha^{\alpha})(e - \alpha + (\dfrac{e - \beta}{\beta + e})\ln(\dfrac{\beta}{(\lambda - e)(e - \beta)^{e - \beta}})) , where α , β \alpha, \beta and λ \lambda are coprime positive integers, find α + β + λ \alpha + \beta + \lambda .


The answer is 6.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Jun 4, 2018

lim ⁡ n → ∞ ∑ j = 1 n ( j n ) n x n − j = ∑ j = 0 n − 1 ( 1 − j n ) n x j = ∑ n = 0 ∞ ( x e ) n = e e − x \lim_{n \rightarrow \infty} \sum_{j = 1}^{n} (\dfrac{j}{n})^n x^{n - j} = \sum_{j = 0}^{n - 1} (1 - \dfrac{j}{n})^{n} x^{j} = \sum_{n = 0}^{\infty} (\dfrac{x}{e})^n = \dfrac{e}{e - x} on ∣ x ∣ < e |x| < e

⟹ lim ⁡ n → ∞ ∑ j = 1 n x j ∑ j = 1 n ( j n ) n x n − j = ( x 1 − x ) ( e − x e ) = x ( e − x ) e ( 1 − x ) \implies \lim_{n \rightarrow \infty} \dfrac{\sum_{j = 1}^{n} x^{j}}{\sum_{j = 1}^{n} (\dfrac{j}{n})^n x^{n - j}} = (\dfrac{x}{1 - x})(\dfrac{e - x}{e}) = \boxed{\dfrac{x(e - x)}{e(1 - x)}} on ∣ x ∣ < 1 |x| < 1

⟹ ∑ n = 1 ∞ ( − 1 ) n + 1 x n ∑ n = 0 ∞ ( − 1 ) n ( x e ) n = \implies \dfrac{\sum_{n = 1}^{\infty} (-1)^{n + 1} x^n}{\sum_{n = 0}^{\infty} (-1)^n (\dfrac{x}{e})^n} = − ∑ n = 1 ∞ ( − x ) n ∑ n = 0 ∞ ( − x e ) n = \dfrac{-\sum_{n = 1}^{\infty} (-x)^n}{\sum_{n = 0}^{\infty} (\dfrac{-x}{e})^n} = x ( e + x ) e ( 1 + x ) \dfrac{x(e + x)}{e(1 + x)} on ∣ x ∣ < 1 |x| < 1

⟹ ( ∑ j = 1 n ( − 1 ) n − j ( j n ) n x n − j ) ∑ j = 1 n ( − 1 ) j + 1 x j ) = e ( 1 + x ) x ( e + x ) \implies (\dfrac{\sum_{j = 1}^{n} (-1)^{n - j} (\dfrac{j}{n})^n x^{n - j})}{\sum_{j = 1}^{n} (-1)^{j + 1} x^{j}}) = \boxed{\dfrac{e(1 + x)}{x(e + x)}} on ∣ x ∣ < 1 |x| < 1

⟹ m ( x ) = ( e − x ) ( 1 + x ) ( 1 − x ) ( e + x ) \implies m(x) = \dfrac{(e - x)(1 + x)}{(1 - x)(e + x)}

⟹ p ( x ) = − m ( − x ) = − ( e + x ) ( 1 − x ) ( 1 + x ) ( e − x ) \implies p(x) = -m(-x) = -\dfrac{(e + x)(1 - x)}{(1 + x)(e - x)}

m ( x ) = 1 + 2 ( e − 1 ) x ( 1 − x ) ( e + x ) m(x) = 1 + \dfrac{2(e - 1)x}{(1 - x)(e + x)} and p ( x ) = − 1 + 2 ( e − 1 ) x ( 1 + x ) ( e − x ) p(x) = -1 + \dfrac{2(e - 1)x}{(1 + x)(e - x)}

From here using partial fractions we can write m ( x ) m(x) and p ( x ) p(x) as:

m ( x ) = 1 + 2 ( e − 1 ) 1 + e ( 1 1 − x − e e + x m(x) = 1 + \dfrac{2(e - 1)}{1 + e}(\dfrac{1}{1 - x} - \dfrac{e}{e + x} and p ( x ) = − 1 + 2 ( e − 1 ) 1 + e ( e ( e − x ) 2 + 1 ( 1 + x ) 2 ) p(x) = -1 + \dfrac{2(e - 1)}{1 + e}(\dfrac{e}{(e - x)^2} + \dfrac{1}{(1 + x)^2})

⟹ ∫ 2 − e e − 2 m ( x ) − p ( x ) d x = 2 ∫ 2 − e e − 2 ( 1 + e − 1 1 + e ( 1 1 − x − e e + x − e e − x + 1 1 + x ) ) d x = \implies \int_{2 - e}^{e - 2} m(x) - p(x) dx = 2\int_{2 - e}^{e - 2} (1 + \dfrac{e - 1}{1 + e}(\dfrac{1}{1 - x} - \dfrac{e}{e + x} - \dfrac{e}{e - x} + \dfrac{1}{1 + x})) dx =

2 ( x + e − 1 1 + e ( ln ⁡ ( ( 1 + x ) ( e − x ) e ( 1 − x ) ( e + x ) e ) ) ) ∣ 2 − e e − 2 = 2(x +\dfrac{e - 1}{1 + e}(\ln(\dfrac{(1 + x)(e - x)^e}{(1 - x)(e + x)^e})))|_{2 - e}^{e - 2} = 2 ( 2 e − 4 + e − 1 1 + e ( ln ⁡ ( e − 1 ( 3 − e ) ( e − 1 ) e ) − ln ⁡ ( ( 3 − e ) ( e − 1 ) e e − 1 ) ) ) 2(2e - 4 + \dfrac{e - 1}{1 + e}(\ln(\dfrac{e - 1}{(3 - e)(e - 1)^{e}}) - \ln(\dfrac{(3 - e)(e - 1)^{e}}{e - 1})))

= 2 ( 2 e − 4 + 2 ( e − 1 1 + e ) ln ⁡ ( 1 ( 3 − e ) ( e − 1 ) e − 1 ) ) = = 2(2e - 4 + 2(\dfrac{e - 1}{1 + e})\ln(\dfrac{1}{(3 - e)(e - 1)^{e - 1}})) = 4 ( e − 2 + ( e − 1 1 + e ) ln ⁡ ( 1 ( 3 − e ) ( e − 1 ) e − 1 ) ) = ( α α ) ( e − α + ( e − β β + e ) ln ⁡ ( β ( λ − e ) ( e − β ) e − β ) ) ⟹ α + β + λ = 6 4(e - 2 + (\dfrac{e - 1}{1 + e})\ln(\dfrac{1}{(3 - e)(e - 1)^{e - 1}})) = (\alpha^{\alpha})(e - \alpha + (\dfrac{e - \beta}{\beta + e})\ln(\dfrac{\beta}{(\lambda - e)(e - \beta)^{e - \beta}})) \implies \alpha + \beta + \lambda = \boxed{6} .

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