Question-7127

Question Number 7127 by Tawakalitu. last updated on 11/Aug/16

Commented by Yozzii last updated on 11/Aug/16

L e t u = x \sqrt{1 + (x + 1) + \sqrt{1 + (x + 2) + \sqrt{1 + (x + 3) + \sqrt{1 + \dots .}}}}

\Rightarrow \frac{u}{x} = \sqrt{1 + (x + 1) + \sqrt{1 + (x + 2) + \sqrt{1 + (x + 3) + \sqrt{1 + \dots .}}}}

l e t n = x + 1 \Rightarrow x = n - 1 .

∴ \frac{u}{n - 1} = \sqrt{1 + n + \sqrt{1 + (n + 1) + \sqrt{1 + (n + 2) + \sqrt{1 + (n + 3) + \sqrt{1 + \dots . .}}}}}

\frac{u^{2}}{{(n - 1)}^{2}} - n - 1 = \sqrt{1 + (n + 1) + \sqrt{1 + (n + 2) + \sqrt{1 + (n + 3) + \sqrt{1 + \dots .}}}}

B u t f o r a n y x \neq 0, \frac{u}{x} h a s t h e f o r m \sqrt{1 + (x + 1) + \sqrt{1 + (x + 2) + \sqrt{1 + (x + 3) + \sqrt{1 + \dots .}}}}

T h e r e f o r e, i f \sqrt{1 + (x + 1) + \sqrt{1 + (x + 2) + \sqrt{1 + (x + 3) + \sqrt{1 + \dots .}}}} c o n v e r g e s t o \frac{u}{x}

\Rightarrow \sqrt{1 + (n + 1) + \sqrt{1 + (n + 2) + \sqrt{1 + (n + 3) + \sqrt{1 + \dots . .}}}} c o n v e r g e s t o \frac{u}{n} .

\frac{u^{2}}{{(n - 1)}^{2}} - n - 1 = \frac{u}{n}

{n u}^{2} - n (n + 1) {(n - 1)}^{2} = u {(n - 1)}^{2}

{n u}^{2} - {(n - 1)}^{2} u - n (n + 1) {(n - 1)}^{2} = 0

∴ u = \frac{{(n - 1)}^{2} \pm \sqrt{{(n - 1)}^{4} + 4 \times n \times n (n + 1) {(n - 1)}^{2}}}{2 n}

u = \frac{{(n - 1)}^{2} \pm {(n - 1)}^{2} \sqrt{{(n - 1)}^{2} + 4 n^{2} (n + 1)}}{2 n}

n = x + 1

∴ u = x [\frac{1 \pm \sqrt{x^{2} + 4 {(x + 1)}^{2} (x + 2)}}{2 (x + 1)}]

u = x [\frac{1 \pm \sqrt{x^{2} + 4 (x^{2} + 2 x + 1) (x + 2)}}{2 (x + 1)}]

u = x [\frac{1 \pm \sqrt{x^{2} + 4 (x^{3} + 2 x^{2} + 2 x^{2} + 4 x + x + 2)}}{2 (x + 1)}]

u = x [\frac{1 \pm \sqrt{4 x^{3} + 17 x^{2} + 20 x + 8}}{2 (x + 1)}]

L e t r (x) = 4 x^{3} + 17 x^{2} + 20 x + 8

∴ i f r^{'} (x) = 0 \Rightarrow 12 x^{2} + 34 x + 20 = 0

6 x^{2} + 17 x + 10 = 0

∴ x = \frac{- 17 \pm \sqrt{289 - 4 \times 6 \times 10}}{12} = \frac{- 17 \pm 7}{12}

x = \frac{- 24}{12}, \frac{- 10}{12} o r x = - 2, \frac{- 5}{6}

r (- 2) = 4 \times (- 8) + 17 \times 4 - 20 \times 2 + 8

r (- 2) = - 32 + 68 - 40 + 8 = 4 > 1 > 0

I f r (x) ⩾ 1 \Rightarrow 4 x^{3} + 17 x^{2} + 20 x + 7 ⩾ 0

4 x^{2} (x + 1) + 20 x (x + 1) - 7 (x^{2} - 1) ⩾ 0

(x + 1) (4 x^{2} + 20 x - 7 (x - 1)) ⩾ 0

(x + 1) (4 x^{2} + 13 x + 7) ⩾ 0

x = \frac{- 13 \pm \sqrt{169 - 4 \times 4 \times 7}}{8}

x = \frac{- 13 \pm \sqrt{57}}{8}

(x + 1) (x + \frac{13 + \sqrt{57}}{8}) (x + \frac{13 - \sqrt{57}}{8}) ⩾ 0

- - - - - - - - (- 1) + + + + + + + + + +

- - - - r_{1} + + + + + + + + + + + + + + +

- - - - - - - - - - - - - - r_{2} + + + + +

r_{1} ⩽ x ⩽ - 1, x ⩾ r_{2}

I f x = r_{1} o r r_{2} (n o t - 1), u = x [\frac{1 \pm 1}{2 (x + 1)}] = 0 o r \frac{x}{x + 1}

B u t, u = 0 i f f x = 0 a n d x \neq 0 ∴ u = \frac{x}{x + 1}

i f x = r_{1} o r r_{2} .

I f r_{1} < x < - 1 o r x > r_{2}, 1 - \sqrt{4 x^{3} + 17 x^{2} + 20 x + 8} < 1 - 1 = 0

B u t, \sqrt{1 + (x + 1) + \sqrt{1 + (x + 2) + \sqrt{1 + (x + 3) + \sqrt{1 + \dots}}}} > 0

∴ u = \frac{x (1 - \sqrt{4 x^{3} + 17 x^{2} + 20 x + 8})}{2 (x + 1)} f o r \frac{- 13 - \sqrt{57}}{8} = r_{1} < x < - 1

a n d u = \frac{x (1 + \sqrt{4 x^{3} + 17 x^{2} + 20 x + 8})}{2 (x + 1)} f o r x > r_{2} = \frac{- 13 + \sqrt{57}}{8}

Commented by Tawakalitu. last updated on 11/Aug/16

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