It's All Tangents 2

Calculus Level 4

Let ∣ x ∣ < 1 |x| < 1 and ( 0 < a < 1 ) (0 < a < 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) .

(1): Show d d x ( m ( x ) ) ∣ x = a = d d x ( p ( x ) ) ∣ x = − a \dfrac{d}{dx}(m(x))|_{x = a} = \dfrac{d}{dx}(p(x))|_{x = -a} .

(2): If A C ↔ \overleftrightarrow{AC} is tangent to m ( x ) m(x) at A : ( e − 2 , m ( e − 2 ) ) A: (e - 2,m(e -2)) and B D ↔ \overleftrightarrow{BD} is tangent to p ( x ) p(x) at B : ( 2 − e , p ( 2 − e ) ) B: (2 - e,p(2 - e)) , find the tangent lines to both curves and find the distance B C ‾ \overline{BC} to 6 decimal places.


The answer is 0.834912.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Jun 3, 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 ) 1 + e ( ( 1 − x ) − 1 − e ( e + x ) − 1 ) ⟹ d d x ( ( m ( x ) ) = 2 ( e − 1 ) 1 + e ( 1 ( 1 − x ) 2 + e ( e + x ) 2 ) m(x) = 1 + \dfrac{2(e - 1)}{1 + e}((1 - x)^{-1} - e(e + x)^{-1}) \implies \dfrac{d}{dx}((m(x)) = \dfrac{2(e - 1)}{1 + e}(\dfrac{1}{(1 - x)^2} + \dfrac{e}{(e + x)^2})

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

For ( 0 < a < 1 ) ⟹ d d x ( m ( x ) ) ∣ x = a = 2 ( e − 1 ) 1 + e ( 1 ( 1 − a ) 2 + e ( e + a ) 2 ) = d d x ( p ( x ) ) ∣ x = − a (0 < a < 1) \implies \dfrac{d}{dx}(m(x))|_{x = a} = \dfrac{2(e - 1)}{1 + e}(\dfrac{1}{(1 - a)^2} + \dfrac{e}{(e + a)^2}) = \dfrac{d}{dx}(p(x))|_{x = -a}

Let x = e − 2 ⟹ d d x ( m ( x ) ) ∣ x = e − 2 = d d x ( p ( x ) ) ∣ x = 2 − e = ( e 2 − 3 e + 4 ) 2 ( 3 − e ) 2 ( e − 1 ) x = e - 2 \implies \dfrac{d}{dx}(m(x))|_{x = e - 2} = \dfrac{d}{dx}(p(x))|_{x = 2 - e} = \dfrac{(e^2 - 3e + 4)}{2(3 - e)^2(e - 1)} and A : ( e − 2 , 1 3 − e ) A: (e - 2, \dfrac{1}{3 - e}) and B : ( 2 − e , − 1 3 − e ) B: (2 - e, -\dfrac{1}{3 - e}) .

Using point A : ( e − 2 , 1 3 − e ) ⟹ y − 1 3 − e = e 2 − 3 e + 4 2 ( 3 − e ) 2 ( e − 1 ) ( x − ( e − 2 ) ) A: (e - 2, \dfrac{1}{3 - e}) \implies y - \dfrac{1}{3 - e} = \dfrac{e^2 - 3e + 4}{2(3 - e)^2(e - 1)}(x - (e - 2))

Using point B : ( 2 − e , − 1 3 − e ) ⟹ y + 1 3 − e = e 2 − 3 e + 4 2 ( 3 − e ) 2 ( e − 1 ) ( x − ( 2 − e ) ) B: (2 - e, -\dfrac{1}{3 - e}) \implies y + \dfrac{1}{3 - e} = \dfrac{e^2 - 3e + 4}{2(3 - e)^2(e - 1)}(x - (2 - e))

Rewriting the tangent line to m ( x ) m(x) at A : ( e − 2 , 1 3 − e ) A: (e - 2, \dfrac{1}{3 - e}) as − ( e 2 − 3 e + 4 ) x + 2 ( 3 − e ) 2 ( e − 1 ) y + ( e − 2 ) ( e 2 − 3 e + 4 ) − 2 ( 3 − e ) ( e − 1 ) = 0 -(e^2 - 3e + 4)x + 2(3 - e)^2(e - 1)y + (e - 2)(e^2 - 3e + 4) - 2(3 - e)(e - 1) = 0 and using point B : ( 2 − e , − 1 3 − e ) B: (2 - e, -\dfrac{1}{3 - e}) ⟹ \implies the distance B C BC from point B : ( 2 − e , − 1 3 − e ) B: (2 - e, -\dfrac{1}{3 - e}) to line − ( e 2 − 3 e + 4 ) x + 2 ( 3 − e ) 2 ( e − 1 ) y + ( e − 2 ) ( e 2 − 3 e + 4 ) − 2 ( 3 − e ) ( e − 1 ) = 0 -(e^2 - 3e + 4)x + 2(3 - e)^2(e - 1)y + (e - 2)(e^2 - 3e + 4) - 2(3 - e)(e - 1) = 0 is:

B C = 2 ( e 3 − 3 e 2 + 2 e − 2 ) 4 e 6 − 56 e 5 + 317 e 4 − 918 e 3 + 1421 e 2 − 1104 e + 340 ≈ 0.834912 BC = \dfrac{2(e^3 - 3e^2 + 2e - 2)}{\sqrt{4e^6 - 56e^5 + 317e^4 -918e^3 + 1421e^2 - 1104e + 340}} \approx \boxed{0.834912} .

In General: If y = f ( x ) y = f(x) and u = − f ( − x ) ⟹ d y d x = f ′ ( x ) u = -f(-x) \implies \dfrac{dy}{dx} = f'(x) and d u d x = f ′ ( − x ) ⟹ d y d x ∣ x = a = f ′ ( a ) = d u d x ∣ x = − a \dfrac{du}{dx} = f'(-x) \implies \dfrac{dy}{dx}|_{x = a} = f'(a) = \dfrac{du}{dx}|x = -a .

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