Integrating infinitely nested radical

Calculus Level 4

0 3 x + x + x + d x = 1 a ( b + c d ) \large \int_0^3 \sqrt{x+\sqrt{x+\sqrt{x+\dots}}}\ dx = \frac1a\left(b+c\sqrt{d}\right)

If the equation above holds true for some positive integers a a , b b , c c , and d d such that gcd ( a , b , c ) = 1 \gcd{(a,b,c)}=1 and d d is square-free, then find the value of a + b + c + d a+b+c+d .


The answer is 55.

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1 solution

Let y = x + x + x + y = \sqrt{x+ \sqrt{x+ \sqrt{x+\cdots}}} , then we have:

y = x + y y 2 = x + y y 2 y x = 0 y = 1 + 1 + 4 x 2 Note that y > 0 \begin{aligned} y & = \sqrt{x+y} \\ y^2 & = x + y \\ y^2 - y - x & = 0 \\ \implies y & = \frac {1+\sqrt{1+4x}}2 & \small \color{#3D99F6} \text{Note that }y > 0 \end{aligned}

Then, we have:

I = 0 3 x + x + x + d x = 0 3 y d x = 0 3 1 + 1 + 4 x 2 d x = 1 2 [ x + 1 6 ( 1 + 4 x ) 3 2 ] 0 3 = 1 2 [ 3 + 1 6 ( 1 + 12 ) 3 2 1 6 ( 1 ) 3 2 ] = 1 12 ( 17 + 13 13 ) \begin{aligned} I & = \int_0^3 \sqrt{x+ \sqrt{x+ \sqrt{x+\cdots}}} \ dx \\ & = \int_0^3 y \ dx \\ & = \int_0^3 \frac {1+\sqrt{1+4x}}2 \ dx \\ & = \frac 12 \left[x + \frac 16 \left(1+ 4x\right)^\frac 32 \right]_0^3 \\ & = \frac 12 \left[3 + \frac 16 \left(1+ 12\right)^\frac 32 - \frac 16(1)^\frac 32 \right] \\ & = \frac 1{12}\left(17 + 13\sqrt{13}\right) \end{aligned}

a + b + c + d = 12 + 17 + 13 + 13 = 55 \implies a+b+c+d = 12+17+13+13 = \boxed{55}

There's a small typo in your solution. You mistyped 'y' in the 8th equation line.

Atomsky Jahid - 2 years, 10 months ago

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Thanks. I have deleted it.

Chew-Seong Cheong - 2 years, 10 months ago

1 12 ( 17 + 13 13 ) = 1 24 ( 34 + 13 52 ) \frac{1}{12}(17+13\sqrt{13}) = \frac{1}{24}(34+13\sqrt{52})

We can found that the g c d ( 24 , 34 , 13 , 52 ) = 1 gcd(24,34,13,52)=1 too,

The question should change it to "Find the smallest value of a + b + c + d a+b+c+d ."

Jian Hau Chooi - 2 years, 10 months ago

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Please report this problem.

Atomsky Jahid - 2 years, 10 months ago

I have changed the wording.

Chew-Seong Cheong - 2 years, 10 months ago

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