Infinite quotient rule?

Calculus Level 3

F ( x ) = x x + x x + x x + x . . . \large F(x) = \frac{x}{x + \frac{x}{x + \frac{x}{x + \frac{x }{._{._.}}} } }

If the value of the first derivative of F ( x ) F(x) at x = 1 x = 1 can be expressed as a b − b c \frac{a\sqrt b - b}{c} for integers a , b , a,b, and c c , find the value of a + b + c . a+b+c.


The answer is 18.

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Partho Roychoudhury by Partho Roychoudhury , 23, India

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3 solutions

F ( 1 ) = 3 5 − 5 10 F(1) = \dfrac{3\sqrt5 - 5}{10}

8 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice problem and solution. I took a slightly different approach.

I first differentiated (implicitly) the equation y 2 + x y − x = 0 y^{2} + xy - x = 0 to find that

2 y y ′ + y + x y ′ − 1 = 0 ⟹ y ′ ( x + 2 y ) = 1 − y ⟹ y ′ = 1 − y x + 2 y . 2yy' + y + xy' - 1 = 0 \Longrightarrow y'(x + 2y) = 1 - y \Longrightarrow y' = \dfrac{1 - y}{x + 2y}.

Now with y ( 1 ) = a y(1) = a we have that a = 1 1 + a ⟹ a 2 + a − 1 = 0 ⟹ a = − 1 + 5 2 , a = \dfrac{1}{1 + a} \Longrightarrow a^{2} + a - 1 = 0 \Longrightarrow a = \dfrac{-1 + \sqrt{5}}{2},

as clearly a > 0. a \gt 0. We then find that

y ′ ( 1 ) = 1 − y ( 1 ) x + 2 y ( 1 ) = 1 − − 1 + 5 2 1 + ( − 1 + 5 ) = 3 − 5 2 5 = 3 5 − 5 10 . y'(1) = \dfrac{1 - y(1)}{x + 2y(1)} = \dfrac{1 - \frac{-1 + \sqrt{5}}{2}}{1 + (-1 + \sqrt{5})} = \dfrac{3 - \sqrt{5}}{2\sqrt{5}} = \dfrac{3\sqrt{5} - 5}{10}.

Brian Charlesworth - 5 years, 6 months ago

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Exactly this method!

Noel Lo - 5 years, 6 months ago

Why does it strictly have to be a + before the square root when you use the quadratic formula? Why can't a minus replace it?

Vladimir Smith - 5 years, 6 months ago

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When we plug x = 1 x = 1 into the expression for y y the result will be a value between 0 0 and 1. 1. The only root that fits this description is the one with the + + sign before the square root, as the other root will be < 0. \lt 0.

Brian Charlesworth - 5 years, 6 months ago

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Thanks! That explains it :)

Vladimir Smith - 5 years, 6 months ago
Fahim Saikat
Jul 9, 2017

Let, F ( x ) = x x + x x + x x + x . . . F ( x ) = x x + F ( x ) F ( x ) 2 + x F ( x ) − x = 0 F(x) = \cfrac{x}{x + \cfrac{x}{x + \cfrac{x}{x + \cfrac{x }{._{._.}}} } }\\ F(x)=\frac{x}{x+F(x)}\\ F(x)^2+x F(x)-x=0

by ignoring the negative value,

F ( x ) = x 2 + 4 x − x 2 F ′ ( x ) = x + 2 − x 2 + 4 x 2 x 2 + 4 x F ′ ( 1 ) = 3 − 5 2 5 = 3 5 − 5 5 2 5 5 = 3 5 − 5 10 ≡ a b − b c F(x)=\frac{\sqrt{x^2+4x}-x}{2}\\ F'(x)=\frac{x+2-\sqrt{x^2+4x}}{2\sqrt{x^2+4x}}\\ F'(1)=\frac{3-\sqrt{5}}{2\sqrt{5}}=\frac{3\sqrt{5}-\sqrt{5}\sqrt{5}}{2\sqrt{5}\sqrt{5}}=\frac{3\sqrt{5}-5}{10}\equiv\frac{a\sqrt{b}-b}{c}

therefore, a = 3 , b = 5 , c = 10 a=3,b=5,c=10 a + b + c = 3 + 5 + 10 = 18 \boxed{a+b+c=3+5+10=18}

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Noel Lo
Nov 30, 2015

Haha...like it!!!

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