+633 = \(\frac{1+\sqrt{5}}{2}\)
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.
+118667
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 ??
+118667 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
\(\displaystyle\int_0^1 x+\frac{1}{x+\frac{1}{x+\frac{1}{x+....}}}dx\\ =\displaystyle\int_1^{\frac{1+\sqrt5}{2}} y(1+y^{-2})dy\\ =\displaystyle\int_1^{\frac{1+\sqrt5}{2}} y+y^{-1}dy\\ =\left[ \frac{y^2}{2}+lny \right ]_1^{\frac{1+\sqrt5}{2}}\\ =\left[ ({\frac{1+\sqrt5}{2})^2}\div2+ln(\frac{1+\sqrt5}{2}) -\frac{1}{2}\right ]\\ =\left[ {\frac{(1+\sqrt5)^2}{8}}+ln(\frac{1+\sqrt5}{2}) -\frac{4}{8}\right ]\\ =\left[ \frac{6+2\sqrt5-4}{8}+ln(\frac{1+\sqrt5}{2}) \right ]\\ =\left[ \frac{2+2\sqrt5}{8}+ln(\frac{1+\sqrt5}{2}) \right ]\\ = \frac{1+\sqrt5}{4}+ln(\frac{1+\sqrt5}{2}) \\ \)
+118667 Oh dear, these two regions are not the same area. What a pity.
I guess I better keep thinking!
+312 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)2 | subtract x*f(x)
f(x)2-x*f(x)=1 | add x2/4
(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
+26388 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} }\)
\(\begin{array}{|rcll|} \hline = \frac14 + 1.0402288194345508 \qquad &\text{or}&\qquad = \frac14 - 1.0402288194345508 \\ = 1.2902288194345508 \qquad &\text{or}& \qquad =-0.7902288194345508 \\ \hline \end{array}\)
