+0

# One more :)

+1
951
8
+676

Note: Continuous Fraction.

Good Luck Boys :)

May 3, 2017

#1
+569
+1

$$\frac{1+\sqrt{5}}{2}$$

It's the Fibonacci sequence! The geometric difference between the numbers is asymptotic to the above: 1/1, 2/1, 3/2, 5/3, 8/5, etc. To get the next number in the series, simply take the sum of the previous 2.

May 3, 2017
#2
+109813
0

Interesting :)

Let

$$y=x+\frac{1}{x+\frac{1}{x+\frac{1}{x+....}}}\\ so\\ y=x+\frac{1}{y}\\ x=y-y^{-1}\\ \frac{dx}{dy}=1+y^{-2}\\ dx=(1+y^{-2})dy\\when \quad x=0\quad \\ y=\frac{1}{y}\\ y^2=1\\ y=\pm1 \qquad \text{this is a stumbling block, I only want one value} \\when\quad x=1\\ 1=y-y^{-1}\\ y=y^2-1\\y^2-y-1=0 \\y=\frac{1\pm\sqrt{5}}{2} \quad \text{two answers again :(}$$

$$\displaystyle\int_0^1 x+\frac{1}{x+\frac{1}{x+\frac{1}{x+....}}}dx\\ =\displaystyle\int_?^? y(1+y^{-2})dy\\ =\displaystyle\int_?^? y+y^{-1}dy\\ =\left[ \frac{y^2}{2}+lny \right ]_?^?$$

Now I have to think about those questions marks

To be continued   :)

I can see Heureka's answer underneath ..... I'll have to think about it more but I will leave this here for now.

Maybe Heuarka, you might like to commnent on what I have done ??

May 3, 2017
#3
+109813
+1

You cannot find the log of a negative number so I am goinf for the positive values.( this is not great mathematical reasoning here:

I'd believe Heureka if I were you   LOL

$$\displaystyle\int_0^1 x+\frac{1}{x+\frac{1}{x+\frac{1}{x+....}}}dx\\ =\displaystyle\int_1^{\frac{1+\sqrt5}{2}} y(1+y^{-2})dy\\ =\displaystyle\int_1^{\frac{1+\sqrt5}{2}} y+y^{-1}dy\\ =\left[ \frac{y^2}{2}+lny \right ]_1^{\frac{1+\sqrt5}{2}}\\ =\left[ ({\frac{1+\sqrt5}{2})^2}\div2+ln(\frac{1+\sqrt5}{2}) -\frac{1}{2}\right ]\\ =\left[ {\frac{(1+\sqrt5)^2}{8}}+ln(\frac{1+\sqrt5}{2}) -\frac{4}{8}\right ]\\ =\left[ \frac{6+2\sqrt5-4}{8}+ln(\frac{1+\sqrt5}{2}) \right ]\\ =\left[ \frac{2+2\sqrt5}{8}+ln(\frac{1+\sqrt5}{2}) \right ]\\ = \frac{1+\sqrt5}{4}+ln(\frac{1+\sqrt5}{2}) \\$$

Melody  May 3, 2017
#4
+109813
0

Oh dear, these two regions are not the same area.  What a pity.

I guess I better keep thinking!

May 3, 2017
#5
+310
+2

Ok. I think i got this-

Lets call the expression f(x). we know that x+1/f(x)=f(x). We can multiply by f(x) and get-

x*f(x)+1=f(x)    |  subtract x*f(x)

(f(x)-x/2)2=1+x2/4

f(x)=x/2+(1+x2/4)1/2.

Im terrible at finding integrals, so i used an integral calculator:

its integral is:

1/4 (x (sqrt(x^2 + 4) + x) + 4 sinh^(-1)(x/2)) + constant

May 3, 2017
edited by Ehrlich  May 3, 2017
#6
+25353
+1

One more :)

Continuous Fraction.

$$\begin{array}{|rcll|} \hline \begin{equation*} sum=x+\cfrac{1}{x+\cfrac{1}{x+\cfrac{1}{x+\cdots}}} \end{equation*}\\\\ sum &=& x + \frac{1}{sum} \\ sum - \frac{1}{sum} &=& x \\ \frac{sum^2-1}{sum} &=& x \\ sum^2-1 &=& x\cdot sum \\ sum^2-x\cdot sum -1 &=& 0 \\ sum &=& \frac{x\pm\sqrt{x^2-4\cdot(-1)} }{2} \\ \mathbf{sum} & \mathbf{=} & \mathbf{ \frac{x\pm\sqrt{x^2+4} }{2} } \\ \hline \end{array}$$

$$\small{ \begin{array}{llcl} \int \limits_{x=0}^{1} { x+\cfrac{1}{x+\cfrac{1}{x+\cfrac{1}{x+\cdots}}} \ dx} \\\\ = \int \limits_{x=0}^{1} { sum \ dx} \\ = \int \limits_{x=0}^{1} { \mathbf{ \frac{x\pm\sqrt{x^2+4} }{2} } \ dx} \\ = \frac12 \int \limits_{x=0}^{1} {x \ dx } \pm \int \limits_{x=0}^{1} { \frac{ \sqrt{x^2+4} }{2} \ dx } \\ = \frac12 [\frac{x^2}{2}]_{x=0}^{1} \pm \int \limits_{x=0}^{1} { 2\cdot \frac{ \sqrt{(\frac{x}{2})^2+1} }{2} \ dx } \\ = \frac14 \pm \int \limits_{x=0}^{1} { \sqrt{(\frac{x}{2})^2+1} \ dx } \\ & \text{substitute:}\\ & \boxed{~ \frac{x}{2}=\sinh(z) \qquad z = \text{arsinh}\left(\frac{x}{2}\right)\\ dx = 2\cdot \cosh(z)\ dz ~}\\ & \text{new limits:}\\ & \boxed{~ x=0: \qquad z=\text{arsinh}\left(\frac{0}{2}\right) \Rightarrow z = 0 \\ x=1: \qquad z=\text{arsinh}\left(\frac{1}{2}\right) ~}\\ = \frac14 \pm \int \limits_{x=0}^{\text{arsinh}\left(\frac{1}{2}\right) } { \sqrt{\sinh^2(z)+1} \cdot 2\cdot \cosh(z)\ dz } \\ & \boxed{~ \cosh^2(z) = 1+\sinh^2(z) ~}\\ = \frac14 \pm \int \limits_{x=0}^{\text{arsinh}\left(\frac{1}{2}\right) } { \sqrt{\cosh^2(z)}\cdot 2\cdot \cosh(z)\ dz } \\ = \frac14 \pm \int \limits_{x=0}^{\text{arsinh}\left(\frac{1}{2}\right) } { \cosh(z)\cdot 2\cdot \cosh(z) \ dz } \\ = \frac14 \pm 2 \int \limits_{x=0}^{\text{arsinh}\left(\frac{1}{2}\right) } { \cosh^2(z) \ dz } \\ & \boxed{~ \cosh^2(z) = \frac12+\frac12\cosh(2z) ~}\\ = \frac14 \pm 2 \int \limits_{x=0}^{\text{arsinh}\left(\frac{1}{2}\right) } { \frac12+\frac12\cosh(2z) \ dz } \\ = \frac14 \pm \int \limits_{x=0}^{\text{arsinh}\left(\frac{1}{2}\right) } { 1+ \cosh(2z) \ dz } \\ = \frac14 \pm \Big( \int \limits_{x=0}^{\text{arsinh}\left(\frac{1}{2}\right) } { \ dz } +\int \limits_{x=0}^{\text{arsinh}\left(\frac{1}{2}\right) } { \cosh(2z) \ dz } \Big) \\ = \frac14 \pm \Big( \left[z\right]_{x=0}^{\text{arsinh}\left(\frac{1}{2}\right)} + \left[\frac{\sinh(2z)}{2} \right]_{x=0}^{\text{arsinh}\left(\frac{1}{2}\right) } \Big) \\ & \boxed{~ \sinh(2z) = 2\sinh(z) \cosh(z) ~}\\ = \frac14 \pm \Big( \left[z\right]_{x=0}^{\text{arsinh}\left(\frac{1}{2}\right)} + \left[\frac{2\sinh(z) \cosh(z)}{2} \right]_{x=0}^{\text{arsinh}\left(\frac{1}{2}\right) } \Big) \\ = \frac14 \pm \Big( \left[z\right]_{x=0}^{\text{arsinh}\left(\frac{1}{2}\right)} + \left[ \sinh(z) \cosh(z) \right]_{x=0}^{\text{arsinh}\left(\frac{1}{2}\right) } \Big) \\ = \frac14 \pm \Big( \text{arsinh}\left(\frac{1}{2}\right) + \frac12 \cdot \cosh\left(\text{arsinh}\left(\frac{1}{2}\right)\right) \Big) \\ = \frac14 \pm \Big( 0.4812118250596 + \frac12 \cdot \cosh\left( 0.4812118250596\right) \Big) \\ = \frac14 \pm \Big( 1.0402288194345508 \Big) \\ \end{array} }$$

$$\begin{array}{|rcll|} \hline = \frac14 + 1.0402288194345508 \qquad &\text{or}&\qquad = \frac14 - 1.0402288194345508 \\ = 1.2902288194345508 \qquad &\text{or}& \qquad =-0.7902288194345508 \\ \hline \end{array}$$

May 3, 2017
edited by heureka  May 3, 2017
#7
+109813
0

Thanks Heureka :)

Melody  May 3, 2017
#8
+310
+2

You truly are the LaTeX master

Although i believe the expression cant be

(x-(x2+4)1/2)/2 because that means its negative, and im quite sure it cant be negative. I cant prove it properly right now, but i believe there is only one answer to the question.

Ehrlich  May 3, 2017