Integral over recurrence relationship
$begingroup$
I'm interested in evaluating the following definite integral
begin{equation}
I_n = int_0^{gamma} F_n(x):dx
end{equation}
Where $gamma gt 0$ and $F_n(x)$ is based on the recurrence relationship:
begin{equation}
F_{n + 1}(x) = frac{1}{1 + F_n(x)}
end{equation}
Here $F_0(x) = f(x)$ where $f$ is a continuous function on $left[0,gammaright]$. My first task was to find a general solution for $F_n(x)$ and this is where I've become unstuck.
I started by letting $F_n(x) = frac{alpha_n(x)}{beta_n(x)}$. Applying it to the recurrence relationship we have:
begin{equation}
F_{n + 1}(x) = frac{alpha_{n+1}(x)}{beta_{n+1}(x)} = frac{beta_n(x)}{alpha_n(x) + beta_n(x)}
end{equation}
And so we have a recurrence relationship over both $alpha_n(x)$ and $beta_n(x)$ with $F_0(x) = f(x) = frac{alpha_0(x)}{beta_0(x)}$.
To begin with I'm focused on $f(x) = sec(x)$ with $alpha_0(x) = 1$ and $beta_0(x) = cos(x)$ and $gamma = frac{pi}{2}$.
With a few iterations via Wolframalpha the pattern that emerges is that $F_n(x)$ takes the form:
begin{equation}
F_n(x) = frac{alpha_n(x)}{beta_n(x)} = frac{a_ncos(x) + b_n}{c_n cos(x) + d_n}
end{equation}
Where $a_n, b_n, c_n, d_n in mathbb{N}$ Here I would like to be able to solve for each but I'm not sure how to start.
Does anyone have any good starting points/references that I can use to begin?
Edit: Changed the definition of $I_n$ to have generalised upper limit of $gamma$.
Update Thanks to hypernova's comment's below, it can be seen that $beta_n(x)$ follows a Fibonacci Sequence:
begin{equation}
beta_{n + 1}(x) = beta_n(x) + alpha_n(x) = beta_n(x) + beta_{n - 1}(x)
end{equation}
And so we can now represent $F_n(x)$ as:
begin{equation}
F_n(x) = frac{alpha_n(x)}{beta_n(x)} = frac{beta_{n-1}(x)}{beta_n(x)}
end{equation}
for $n geq 2$.
For the specific example above we have:
begin{equation}
F_n(x) = frac{alpha_n(x)}{beta_n(x)} = frac{a_{n-1}cos(x) + b_{n-1}}{a_n cos(x) + b_n}
end{equation}
Where $a_n$ and $b_n$ are Fibonacci Sequences with $b_0 = 0$, $b_1 = 1$ and $a_0 = 1$, $a_1 = 1$. We see that $a_n gt b_n$ (this will be important later)
So, we may now evaluate the integral
begin{equation}
I_n = int_0^{tfrac{pi}{2}}frac{a_{n-1}cos(x) + b_{n-1}}{a_n cos(x) + b_n}:dx
end{equation}
I will here employ the Weierstrass substitution $t = tanleft(frac{x}{2} right)$:
begin{align}
I_n &= int_0^{tfrac{pi}{2}}frac{a_{n-1}cos(x) + b_{n-1}}{a_n cos(x) + b_n}:dx = int_0^1 frac{a_{n-1}frac{1 - t^2}{1 + t^2} + b_{n-1}}{a_n frac{1 - t^2}{1 + t^2} + b_n}frac{2:dt}{1 + t^2} \
&= 2int_0^1 frac{a_{n - 1}left(1 - t^2right) + b_{n - 1}left(1 + t^2right)}{left(1 + t^2right)left(a_nleft(1 - t^2right) + b_nleft(1 + t^2right)right)}:dt \
&= 2int_0^1 frac{left(b_{n - 1} - a_{n-1}right)t^2 + left(b_{n-1} + a_{n-1}right)}{left(1 + t^2right)left(left(b_n - a_nright)t^2 + left(b_n + a_nright)right)}:dt \
&= 2left(frac{b_{n - 1} - a_{n-1}}{b_n - a_n}right) int_0^1 frac{t^2 - theta_{n - 1}}{left(1 + t^2right)left(t^2 + theta_nright)}:dt
end{align}
Where
begin{equation}
theta_n = frac{b_n + a_n}{b_n - a_n}
end{equation}
As $a_n gt b_n geq 0$ we see that $theta_n lt 0$. Hence:
begin{align}
I_n &= 2left(frac{b_{n - 1} - a_{n-1}}{b_n - a_n}right) int_0^1 frac{t^2 - left|theta_{n - 1}right|}{left(1 + t^2right)left(t^2 - left| theta_nright)right|}:dt \
&= 2left(frac{b_{n - 1} - a_{n-1}}{b_n - a_n}right)left[ frac{1}{left|theta_nright| + 1} left[left(left|theta_{n-1}right| + 1 right) arctan(x) + frac{left|theta_{n-1}right| - left|theta_{n}right|}{sqrt{left|theta_nright|}}operatorname{arctanh}left(frac{x}{sqrt{left|theta_nright|}} right)right]right]_0^1 \
&=2left(frac{b_{n - 1} - a_{n-1}}{b_n - a_n}right)left(frac{1}{left|theta_nright| + 1} right)left[left(left|theta_{n-1}right| + 1 right) frac{pi}{4} + frac{left|theta_{n-1}right| - left|theta_{n}right|}{sqrt{left|theta_nright|}}operatorname{arctanh}left(frac{1}{sqrt{left|theta_nright|}} right)right]
end{align}
Note $b_n = a_{n - 1}$ and thus:
begin{align}
I_n &=2left(frac{b_{n - 1} - a_{n-1}}{b_n - a_n}right)left(frac{1}{left|theta_nright| + 1} right)left[left(left|theta_{n-1}right| + 1 right) frac{pi}{4} + frac{left|theta_{n-1}right| - left|theta_{n}right|}{sqrt{left|theta_nright|}}operatorname{arctanh}left(frac{1}{sqrt{left|theta_nright|}} right)right] \
&= frac{a_{n - 1}}{a_n}frac{pi}{2} + left[1 - frac{a_{n + 1}left(a_{n - 1} - a_{n - 2} right)}{a_nleft(a_n - a_{n - 1} right)} right]sqrt{frac{a_n - a_{n - 1}}{a_{n + 1}}}operatorname{arccoth}left(sqrt{frac{a_{n + 1}}{a_n - a_{n - 1}}} right)
end{align}
real-analysis integration definite-integrals recurrence-relations continued-fractions
asked Jan 7 at 0:57
DavidGDavidG
1
$endgroup$
add a comment |
$begingroup$
I'm interested in evaluating the following definite integral
begin{equation}
I_n = int_0^{gamma} F_n(x):dx
end{equation}
Where $gamma gt 0$ and $F_n(x)$ is based on the recurrence relationship:
begin{equation}
F_{n + 1}(x) = frac{1}{1 + F_n(x)}
end{equation}
Here $F_0(x) = f(x)$ where $f$ is a continuous function on $left[0,gammaright]$. My first task was to find a general solution for $F_n(x)$ and this is where I've become unstuck.
I started by letting $F_n(x) = frac{alpha_n(x)}{beta_n(x)}$. Applying it to the recurrence relationship we have:
begin{equation}
F_{n + 1}(x) = frac{alpha_{n+1}(x)}{beta_{n+1}(x)} = frac{beta_n(x)}{alpha_n(x) + beta_n(x)}
end{equation}
And so we have a recurrence relationship over both $alpha_n(x)$ and $beta_n(x)$ with $F_0(x) = f(x) = frac{alpha_0(x)}{beta_0(x)}$.
To begin with I'm focused on $f(x) = sec(x)$ with $alpha_0(x) = 1$ and $beta_0(x) = cos(x)$ and $gamma = frac{pi}{2}$.
With a few iterations via Wolframalpha the pattern that emerges is that $F_n(x)$ takes the form:
begin{equation}
F_n(x) = frac{alpha_n(x)}{beta_n(x)} = frac{a_ncos(x) + b_n}{c_n cos(x) + d_n}
end{equation}
Where $a_n, b_n, c_n, d_n in mathbb{N}$ Here I would like to be able to solve for each but I'm not sure how to start.
Does anyone have any good starting points/references that I can use to begin?
Edit: Changed the definition of $I_n$ to have generalised upper limit of $gamma$.
Update Thanks to hypernova's comment's below, it can be seen that $beta_n(x)$ follows a Fibonacci Sequence:
begin{equation}
beta_{n + 1}(x) = beta_n(x) + alpha_n(x) = beta_n(x) + beta_{n - 1}(x)
end{equation}
And so we can now represent $F_n(x)$ as:
begin{equation}
F_n(x) = frac{alpha_n(x)}{beta_n(x)} = frac{beta_{n-1}(x)}{beta_n(x)}
end{equation}
for $n geq 2$.
For the specific example above we have:
begin{equation}
F_n(x) = frac{alpha_n(x)}{beta_n(x)} = frac{a_{n-1}cos(x) + b_{n-1}}{a_n cos(x) + b_n}
end{equation}
Where $a_n$ and $b_n$ are Fibonacci Sequences with $b_0 = 0$, $b_1 = 1$ and $a_0 = 1$, $a_1 = 1$. We see that $a_n gt b_n$ (this will be important later)
So, we may now evaluate the integral
begin{equation}
I_n = int_0^{tfrac{pi}{2}}frac{a_{n-1}cos(x) + b_{n-1}}{a_n cos(x) + b_n}:dx
end{equation}
I will here employ the Weierstrass substitution $t = tanleft(frac{x}{2} right)$:
begin{align}
I_n &= int_0^{tfrac{pi}{2}}frac{a_{n-1}cos(x) + b_{n-1}}{a_n cos(x) + b_n}:dx = int_0^1 frac{a_{n-1}frac{1 - t^2}{1 + t^2} + b_{n-1}}{a_n frac{1 - t^2}{1 + t^2} + b_n}frac{2:dt}{1 + t^2} \
&= 2int_0^1 frac{a_{n - 1}left(1 - t^2right) + b_{n - 1}left(1 + t^2right)}{left(1 + t^2right)left(a_nleft(1 - t^2right) + b_nleft(1 + t^2right)right)}:dt \
&= 2int_0^1 frac{left(b_{n - 1} - a_{n-1}right)t^2 + left(b_{n-1} + a_{n-1}right)}{left(1 + t^2right)left(left(b_n - a_nright)t^2 + left(b_n + a_nright)right)}:dt \
&= 2left(frac{b_{n - 1} - a_{n-1}}{b_n - a_n}right) int_0^1 frac{t^2 - theta_{n - 1}}{left(1 + t^2right)left(t^2 + theta_nright)}:dt
end{align}
Where
begin{equation}
theta_n = frac{b_n + a_n}{b_n - a_n}
end{equation}
As $a_n gt b_n geq 0$ we see that $theta_n lt 0$. Hence:
begin{align}
I_n &= 2left(frac{b_{n - 1} - a_{n-1}}{b_n - a_n}right) int_0^1 frac{t^2 - left|theta_{n - 1}right|}{left(1 + t^2right)left(t^2 - left| theta_nright)right|}:dt \
&= 2left(frac{b_{n - 1} - a_{n-1}}{b_n - a_n}right)left[ frac{1}{left|theta_nright| + 1} left[left(left|theta_{n-1}right| + 1 right) arctan(x) + frac{left|theta_{n-1}right| - left|theta_{n}right|}{sqrt{left|theta_nright|}}operatorname{arctanh}left(frac{x}{sqrt{left|theta_nright|}} right)right]right]_0^1 \
&=2left(frac{b_{n - 1} - a_{n-1}}{b_n - a_n}right)left(frac{1}{left|theta_nright| + 1} right)left[left(left|theta_{n-1}right| + 1 right) frac{pi}{4} + frac{left|theta_{n-1}right| - left|theta_{n}right|}{sqrt{left|theta_nright|}}operatorname{arctanh}left(frac{1}{sqrt{left|theta_nright|}} right)right]
end{align}
Note $b_n = a_{n - 1}$ and thus:
begin{align}
I_n &=2left(frac{b_{n - 1} - a_{n-1}}{b_n - a_n}right)left(frac{1}{left|theta_nright| + 1} right)left[left(left|theta_{n-1}right| + 1 right) frac{pi}{4} + frac{left|theta_{n-1}right| - left|theta_{n}right|}{sqrt{left|theta_nright|}}operatorname{arctanh}left(frac{1}{sqrt{left|theta_nright|}} right)right] \
&= frac{a_{n - 1}}{a_n}frac{pi}{2} + left[1 - frac{a_{n + 1}left(a_{n - 1} - a_{n - 2} right)}{a_nleft(a_n - a_{n - 1} right)} right]sqrt{frac{a_n - a_{n - 1}}{a_{n + 1}}}operatorname{arccoth}left(sqrt{frac{a_{n + 1}}{a_n - a_{n - 1}}} right)
end{align}
real-analysis integration definite-integrals recurrence-relations continued-fractions
asked Jan 7 at 0:57
DavidGDavidG
1
$endgroup$
add a comment |
$begingroup$
I'm interested in evaluating the following definite integral
begin{equation}
I_n = int_0^{gamma} F_n(x):dx
end{equation}
Where $gamma gt 0$ and $F_n(x)$ is based on the recurrence relationship:
begin{equation}
F_{n + 1}(x) = frac{1}{1 + F_n(x)}
end{equation}
Here $F_0(x) = f(x)$ where $f$ is a continuous function on $left[0,gammaright]$. My first task was to find a general solution for $F_n(x)$ and this is where I've become unstuck.
I started by letting $F_n(x) = frac{alpha_n(x)}{beta_n(x)}$. Applying it to the recurrence relationship we have:
begin{equation}
F_{n + 1}(x) = frac{alpha_{n+1}(x)}{beta_{n+1}(x)} = frac{beta_n(x)}{alpha_n(x) + beta_n(x)}
end{equation}
And so we have a recurrence relationship over both $alpha_n(x)$ and $beta_n(x)$ with $F_0(x) = f(x) = frac{alpha_0(x)}{beta_0(x)}$.
To begin with I'm focused on $f(x) = sec(x)$ with $alpha_0(x) = 1$ and $beta_0(x) = cos(x)$ and $gamma = frac{pi}{2}$.
With a few iterations via Wolframalpha the pattern that emerges is that $F_n(x)$ takes the form:
begin{equation}
F_n(x) = frac{alpha_n(x)}{beta_n(x)} = frac{a_ncos(x) + b_n}{c_n cos(x) + d_n}
end{equation}
Where $a_n, b_n, c_n, d_n in mathbb{N}$ Here I would like to be able to solve for each but I'm not sure how to start.
Does anyone have any good starting points/references that I can use to begin?
Edit: Changed the definition of $I_n$ to have generalised upper limit of $gamma$.
Update Thanks to hypernova's comment's below, it can be seen that $beta_n(x)$ follows a Fibonacci Sequence:
begin{equation}
beta_{n + 1}(x) = beta_n(x) + alpha_n(x) = beta_n(x) + beta_{n - 1}(x)
end{equation}
And so we can now represent $F_n(x)$ as:
begin{equation}
F_n(x) = frac{alpha_n(x)}{beta_n(x)} = frac{beta_{n-1}(x)}{beta_n(x)}
end{equation}
for $n geq 2$.
For the specific example above we have:
begin{equation}
F_n(x) = frac{alpha_n(x)}{beta_n(x)} = frac{a_{n-1}cos(x) + b_{n-1}}{a_n cos(x) + b_n}
end{equation}
Where $a_n$ and $b_n$ are Fibonacci Sequences with $b_0 = 0$, $b_1 = 1$ and $a_0 = 1$, $a_1 = 1$. We see that $a_n gt b_n$ (this will be important later)
So, we may now evaluate the integral
begin{equation}
I_n = int_0^{tfrac{pi}{2}}frac{a_{n-1}cos(x) + b_{n-1}}{a_n cos(x) + b_n}:dx
end{equation}
I will here employ the Weierstrass substitution $t = tanleft(frac{x}{2} right)$:
begin{align}
I_n &= int_0^{tfrac{pi}{2}}frac{a_{n-1}cos(x) + b_{n-1}}{a_n cos(x) + b_n}:dx = int_0^1 frac{a_{n-1}frac{1 - t^2}{1 + t^2} + b_{n-1}}{a_n frac{1 - t^2}{1 + t^2} + b_n}frac{2:dt}{1 + t^2} \
&= 2int_0^1 frac{a_{n - 1}left(1 - t^2right) + b_{n - 1}left(1 + t^2right)}{left(1 + t^2right)left(a_nleft(1 - t^2right) + b_nleft(1 + t^2right)right)}:dt \
&= 2int_0^1 frac{left(b_{n - 1} - a_{n-1}right)t^2 + left(b_{n-1} + a_{n-1}right)}{left(1 + t^2right)left(left(b_n - a_nright)t^2 + left(b_n + a_nright)right)}:dt \
&= 2left(frac{b_{n - 1} - a_{n-1}}{b_n - a_n}right) int_0^1 frac{t^2 - theta_{n - 1}}{left(1 + t^2right)left(t^2 + theta_nright)}:dt
end{align}
Where
begin{equation}
theta_n = frac{b_n + a_n}{b_n - a_n}
end{equation}
As $a_n gt b_n geq 0$ we see that $theta_n lt 0$. Hence:
begin{align}
I_n &= 2left(frac{b_{n - 1} - a_{n-1}}{b_n - a_n}right) int_0^1 frac{t^2 - left|theta_{n - 1}right|}{left(1 + t^2right)left(t^2 - left| theta_nright)right|}:dt \
&= 2left(frac{b_{n - 1} - a_{n-1}}{b_n - a_n}right)left[ frac{1}{left|theta_nright| + 1} left[left(left|theta_{n-1}right| + 1 right) arctan(x) + frac{left|theta_{n-1}right| - left|theta_{n}right|}{sqrt{left|theta_nright|}}operatorname{arctanh}left(frac{x}{sqrt{left|theta_nright|}} right)right]right]_0^1 \
&=2left(frac{b_{n - 1} - a_{n-1}}{b_n - a_n}right)left(frac{1}{left|theta_nright| + 1} right)left[left(left|theta_{n-1}right| + 1 right) frac{pi}{4} + frac{left|theta_{n-1}right| - left|theta_{n}right|}{sqrt{left|theta_nright|}}operatorname{arctanh}left(frac{1}{sqrt{left|theta_nright|}} right)right]
end{align}
Note $b_n = a_{n - 1}$ and thus:
begin{align}
I_n &=2left(frac{b_{n - 1} - a_{n-1}}{b_n - a_n}right)left(frac{1}{left|theta_nright| + 1} right)left[left(left|theta_{n-1}right| + 1 right) frac{pi}{4} + frac{left|theta_{n-1}right| - left|theta_{n}right|}{sqrt{left|theta_nright|}}operatorname{arctanh}left(frac{1}{sqrt{left|theta_nright|}} right)right] \
&= frac{a_{n - 1}}{a_n}frac{pi}{2} + left[1 - frac{a_{n + 1}left(a_{n - 1} - a_{n - 2} right)}{a_nleft(a_n - a_{n - 1} right)} right]sqrt{frac{a_n - a_{n - 1}}{a_{n + 1}}}operatorname{arccoth}left(sqrt{frac{a_{n + 1}}{a_n - a_{n - 1}}} right)
end{align}
real-analysis integration definite-integrals recurrence-relations continued-fractions
asked Jan 7 at 0:57
DavidGDavidG
1
$endgroup$
I'm interested in evaluating the following definite integral
begin{equation}
I_n = int_0^{gamma} F_n(x):dx
end{equation}
Where $gamma gt 0$ and $F_n(x)$ is based on the recurrence relationship:
begin{equation}
F_{n + 1}(x) = frac{1}{1 + F_n(x)}
end{equation}
Here $F_0(x) = f(x)$ where $f$ is a continuous function on $left[0,gammaright]$. My first task was to find a general solution for $F_n(x)$ and this is where I've become unstuck.
I started by letting $F_n(x) = frac{alpha_n(x)}{beta_n(x)}$. Applying it to the recurrence relationship we have:
begin{equation}
F_{n + 1}(x) = frac{alpha_{n+1}(x)}{beta_{n+1}(x)} = frac{beta_n(x)}{alpha_n(x) + beta_n(x)}
end{equation}
And so we have a recurrence relationship over both $alpha_n(x)$ and $beta_n(x)$ with $F_0(x) = f(x) = frac{alpha_0(x)}{beta_0(x)}$.
To begin with I'm focused on $f(x) = sec(x)$ with $alpha_0(x) = 1$ and $beta_0(x) = cos(x)$ and $gamma = frac{pi}{2}$.
With a few iterations via Wolframalpha the pattern that emerges is that $F_n(x)$ takes the form:
begin{equation}
F_n(x) = frac{alpha_n(x)}{beta_n(x)} = frac{a_ncos(x) + b_n}{c_n cos(x) + d_n}
end{equation}
Where $a_n, b_n, c_n, d_n in mathbb{N}$ Here I would like to be able to solve for each but I'm not sure how to start.
Does anyone have any good starting points/references that I can use to begin?
Edit: Changed the definition of $I_n$ to have generalised upper limit of $gamma$.
Update Thanks to hypernova's comment's below, it can be seen that $beta_n(x)$ follows a Fibonacci Sequence:
begin{equation}
beta_{n + 1}(x) = beta_n(x) + alpha_n(x) = beta_n(x) + beta_{n - 1}(x)
end{equation}
And so we can now represent $F_n(x)$ as:
begin{equation}
F_n(x) = frac{alpha_n(x)}{beta_n(x)} = frac{beta_{n-1}(x)}{beta_n(x)}
end{equation}
for $n geq 2$.
For the specific example above we have:
begin{equation}
F_n(x) = frac{alpha_n(x)}{beta_n(x)} = frac{a_{n-1}cos(x) + b_{n-1}}{a_n cos(x) + b_n}
end{equation}
Where $a_n$ and $b_n$ are Fibonacci Sequences with $b_0 = 0$, $b_1 = 1$ and $a_0 = 1$, $a_1 = 1$. We see that $a_n gt b_n$ (this will be important later)
So, we may now evaluate the integral
begin{equation}
I_n = int_0^{tfrac{pi}{2}}frac{a_{n-1}cos(x) + b_{n-1}}{a_n cos(x) + b_n}:dx
end{equation}
I will here employ the Weierstrass substitution $t = tanleft(frac{x}{2} right)$:
begin{align}
I_n &= int_0^{tfrac{pi}{2}}frac{a_{n-1}cos(x) + b_{n-1}}{a_n cos(x) + b_n}:dx = int_0^1 frac{a_{n-1}frac{1 - t^2}{1 + t^2} + b_{n-1}}{a_n frac{1 - t^2}{1 + t^2} + b_n}frac{2:dt}{1 + t^2} \
&= 2int_0^1 frac{a_{n - 1}left(1 - t^2right) + b_{n - 1}left(1 + t^2right)}{left(1 + t^2right)left(a_nleft(1 - t^2right) + b_nleft(1 + t^2right)right)}:dt \
&= 2int_0^1 frac{left(b_{n - 1} - a_{n-1}right)t^2 + left(b_{n-1} + a_{n-1}right)}{left(1 + t^2right)left(left(b_n - a_nright)t^2 + left(b_n + a_nright)right)}:dt \
&= 2left(frac{b_{n - 1} - a_{n-1}}{b_n - a_n}right) int_0^1 frac{t^2 - theta_{n - 1}}{left(1 + t^2right)left(t^2 + theta_nright)}:dt
end{align}
Where
begin{equation}
theta_n = frac{b_n + a_n}{b_n - a_n}
end{equation}
As $a_n gt b_n geq 0$ we see that $theta_n lt 0$. Hence:
begin{align}
I_n &= 2left(frac{b_{n - 1} - a_{n-1}}{b_n - a_n}right) int_0^1 frac{t^2 - left|theta_{n - 1}right|}{left(1 + t^2right)left(t^2 - left| theta_nright)right|}:dt \
&= 2left(frac{b_{n - 1} - a_{n-1}}{b_n - a_n}right)left[ frac{1}{left|theta_nright| + 1} left[left(left|theta_{n-1}right| + 1 right) arctan(x) + frac{left|theta_{n-1}right| - left|theta_{n}right|}{sqrt{left|theta_nright|}}operatorname{arctanh}left(frac{x}{sqrt{left|theta_nright|}} right)right]right]_0^1 \
&=2left(frac{b_{n - 1} - a_{n-1}}{b_n - a_n}right)left(frac{1}{left|theta_nright| + 1} right)left[left(left|theta_{n-1}right| + 1 right) frac{pi}{4} + frac{left|theta_{n-1}right| - left|theta_{n}right|}{sqrt{left|theta_nright|}}operatorname{arctanh}left(frac{1}{sqrt{left|theta_nright|}} right)right]
end{align}
Note $b_n = a_{n - 1}$ and thus:
begin{align}
I_n &=2left(frac{b_{n - 1} - a_{n-1}}{b_n - a_n}right)left(frac{1}{left|theta_nright| + 1} right)left[left(left|theta_{n-1}right| + 1 right) frac{pi}{4} + frac{left|theta_{n-1}right| - left|theta_{n}right|}{sqrt{left|theta_nright|}}operatorname{arctanh}left(frac{1}{sqrt{left|theta_nright|}} right)right] \
&= frac{a_{n - 1}}{a_n}frac{pi}{2} + left[1 - frac{a_{n + 1}left(a_{n - 1} - a_{n - 2} right)}{a_nleft(a_n - a_{n - 1} right)} right]sqrt{frac{a_n - a_{n - 1}}{a_{n + 1}}}operatorname{arccoth}left(sqrt{frac{a_{n + 1}}{a_n - a_{n - 1}}} right)
end{align}
real-analysis integration definite-integrals recurrence-relations continued-fractions
real-analysis integration definite-integrals recurrence-relations continued-fractions
asked Jan 7 at 0:57
DavidGDavidG
1
asked Jan 7 at 0:57
DavidGDavidG
1
edited Jan 8 at 0:44
DavidG
asked Jan 7 at 0:57
DavidGDavidG
1
asked Jan 7 at 0:57
DavidGDavidG
1
asked Jan 7 at 0:57
DavidGDavidG
1
1
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since the recurrence relation is fully nonlinear, it might be better to try the first few terms with the initial condition $F_0=f$. By Mathematica, it is easy to find the general expression should be of the form
$$
F_n=frac{a_n+b_nf}{a_n+b_n+a_nf}
$$
for $nge 1$, where $a_n$ and $b_n$ are constants. By the recurrence relation, these constants must follow
$$
left(
begin{array}{c}
a_{n+1}\
b_{n+1}
end{array}
right)=left(
begin{array}{cc}
1&1\
1&0
end{array}
right)left(
begin{array}{c}
a_n\
b_n
end{array}
right),
$$
with
$$
left(
begin{array}{c}
a_1\
b_1
end{array}
right)=left(
begin{array}{c}
1\
0
end{array}
right).
$$
Thanks to linear algebra, you may solve this Fibonacci-like $a_n$ and $b_n$ immediately.
Hope this could be somewhat helpful for you.
answered Jan 7 at 1:41
hypernovahypernova
5,009414
$endgroup$
$begingroup$
This is great! thank you so much!
$endgroup$
– DavidG
Jan 7 at 1:49
$begingroup$
@DavidG: You are welcome!
$endgroup$
– hypernova
Jan 7 at 1:50
$begingroup$
Would this be a pseudo styled Markov Chain?
$endgroup$
– DavidG
Jan 7 at 1:53
1
$begingroup$
@DavidG: For me, it is hard to try using results from Markov chains to your original question. While there might be useful results to use, it might be hard to avoid calculating $A^n$. You may try this built-in function "JordanDecomposition" in Mathematica to figure out $P$ and $J$ such that $A=PJP^{-1}$, where $J$ is the Jordan Normal Form of $A$. Then $A^n=PJ^nP^{-1}$. $J^n$ is very easy to calculate for your question, because $J$ is 2-by-2 and upper-triangular.
$endgroup$
– hypernova
Jan 7 at 3:15
1
$begingroup$
@DavidG: Sure! Thanks for your message. marty cohen's solution is really beautiful. I seems to have misunderstood your expectations. While the coefficients $a_n$ and $b_n$ diverge, their resulted $F_n$ converges. marty cohen's solution definitely clarifies. Thank you!
$endgroup$
– hypernova
Jan 8 at 10:35
|
show 4 more comments
$begingroup$
This is copied from
another answer of mine:
Let
$f_1 = frac{1}{1 + g(x) }
$
where
$g(x) > 0,
$,
and let
$f_n(x)
=frac{1}{1+f_{n-1}(x)}
$.
Then
$f_n(x)
to dfrac{sqrt{5}-1}{2}
$.
Note:
I doubt that any of this
is original,
but this was all done
just now by me.
Proof.
$begin{array}\
f_n(x)
&=frac{1}{1+frac{1}{1+f_{n-2}(x)}}\
&=frac{1+f_{n-2}(x)}{1+f_{n-2}(x)+1}\
&=frac{1+f_{n-2}(x)}{2+f_{n-2}(x)}\
end{array}
$
Therefore,
if $f_{n-2}(x) > 0$
then
$frac12 < f_n(x)
lt 1$.
Similarly,
if $f_{n-1}(x) > 0$
then
$0 < f_n(x)
lt 1$.
$begin{array}\
f_n(x)-f_{n-2}(x)
&=dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)}-f_{n-2}(x)\
&=dfrac{1+f_{n-2}(x)-f_{n-2}(x)(2+f_{n-2}(x))}{2+f_{n-2}(x)}\
&=dfrac{1-f_{n-2}(x)-f_{n-2}^2(x)}{2+f_{n-2}(x)}\
end{array}
$
$begin{array}\
f_n(x)+f_n^2(x)
&=dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)}+(dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)})^2\
&=dfrac{(1+f_{n-2}(x))(2+f_{n-2}(x))}{(2+f_{n-2}(x))^2}+dfrac{1+2f_{n-2}(x)+f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
&=dfrac{2+3f_{n-2}(x)+f_{n-2}^2(x)+1+2f_{n-2}(x)+f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
&=dfrac{3+5f_{n-2}(x)+2f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
text{so}\
1-f_n(x)-f_n^2(x)
&=dfrac{4+4f_{n-2}(x)+f_{n-2}^2(x)-(3+5f_{n-2}(x)+2f_{n-2}^2(x))}{(2+f_{n-2}(x))^2}\
&=dfrac{1-f_{n-2}(x)-f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
end{array}
$
Therefore
$1-f_n(x)-f_n^2(x)$
has the same sign as
$1-f_{n-2}(x)-f_{n-2}^2(x)$.
Also,
$|1-f_n(x)-f_n^2(x)|
lt frac14|1-f_{n-2}(x)-f_{n-2}^2(x)|
$
so
$|1-f_n(x)-f_n^2(x)|
to 0$.
Let
$p(x) = 1-x-x^2$
and
$x_0 = frac{sqrt{5}-1}{2}
$
so
$p(x_0) = 0$,
$p'(x) < 0$ for $x ge 0$.
Since
$f_n(x) > 0$,
$f_n(x)
to x_0$.
answered Jan 8 at 0:54
marty cohenmarty cohen
74.6k549129
$endgroup$
$begingroup$
Thanks for your post - I actually was playing around with a few different cases last night and noticed that it converged as you spoke to. I wasn't sure how to go about proving it, so thanks heaps for your post!
$endgroup$
– DavidG
Jan 8 at 1:01
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since the recurrence relation is fully nonlinear, it might be better to try the first few terms with the initial condition $F_0=f$. By Mathematica, it is easy to find the general expression should be of the form
$$
F_n=frac{a_n+b_nf}{a_n+b_n+a_nf}
$$
for $nge 1$, where $a_n$ and $b_n$ are constants. By the recurrence relation, these constants must follow
$$
left(
begin{array}{c}
a_{n+1}\
b_{n+1}
end{array}
right)=left(
begin{array}{cc}
1&1\
1&0
end{array}
right)left(
begin{array}{c}
a_n\
b_n
end{array}
right),
$$
with
$$
left(
begin{array}{c}
a_1\
b_1
end{array}
right)=left(
begin{array}{c}
1\
0
end{array}
right).
$$
Thanks to linear algebra, you may solve this Fibonacci-like $a_n$ and $b_n$ immediately.
Hope this could be somewhat helpful for you.
answered Jan 7 at 1:41
hypernovahypernova
5,009414
$endgroup$
$begingroup$
This is great! thank you so much!
$endgroup$
– DavidG
Jan 7 at 1:49
$begingroup$
@DavidG: You are welcome!
$endgroup$
– hypernova
Jan 7 at 1:50
$begingroup$
Would this be a pseudo styled Markov Chain?
$endgroup$
– DavidG
Jan 7 at 1:53
1
$begingroup$
@DavidG: For me, it is hard to try using results from Markov chains to your original question. While there might be useful results to use, it might be hard to avoid calculating $A^n$. You may try this built-in function "JordanDecomposition" in Mathematica to figure out $P$ and $J$ such that $A=PJP^{-1}$, where $J$ is the Jordan Normal Form of $A$. Then $A^n=PJ^nP^{-1}$. $J^n$ is very easy to calculate for your question, because $J$ is 2-by-2 and upper-triangular.
$endgroup$
– hypernova
Jan 7 at 3:15
1
$begingroup$
@DavidG: Sure! Thanks for your message. marty cohen's solution is really beautiful. I seems to have misunderstood your expectations. While the coefficients $a_n$ and $b_n$ diverge, their resulted $F_n$ converges. marty cohen's solution definitely clarifies. Thank you!
$endgroup$
– hypernova
Jan 8 at 10:35
|
show 4 more comments
$begingroup$
Since the recurrence relation is fully nonlinear, it might be better to try the first few terms with the initial condition $F_0=f$. By Mathematica, it is easy to find the general expression should be of the form
$$
F_n=frac{a_n+b_nf}{a_n+b_n+a_nf}
$$
for $nge 1$, where $a_n$ and $b_n$ are constants. By the recurrence relation, these constants must follow
$$
left(
begin{array}{c}
a_{n+1}\
b_{n+1}
end{array}
right)=left(
begin{array}{cc}
1&1\
1&0
end{array}
right)left(
begin{array}{c}
a_n\
b_n
end{array}
right),
$$
with
$$
left(
begin{array}{c}
a_1\
b_1
end{array}
right)=left(
begin{array}{c}
1\
0
end{array}
right).
$$
Thanks to linear algebra, you may solve this Fibonacci-like $a_n$ and $b_n$ immediately.
Hope this could be somewhat helpful for you.
answered Jan 7 at 1:41
hypernovahypernova
5,009414
$endgroup$
$begingroup$
This is great! thank you so much!
$endgroup$
– DavidG
Jan 7 at 1:49
$begingroup$
@DavidG: You are welcome!
$endgroup$
– hypernova
Jan 7 at 1:50
$begingroup$
Would this be a pseudo styled Markov Chain?
$endgroup$
– DavidG
Jan 7 at 1:53
1
$begingroup$
@DavidG: For me, it is hard to try using results from Markov chains to your original question. While there might be useful results to use, it might be hard to avoid calculating $A^n$. You may try this built-in function "JordanDecomposition" in Mathematica to figure out $P$ and $J$ such that $A=PJP^{-1}$, where $J$ is the Jordan Normal Form of $A$. Then $A^n=PJ^nP^{-1}$. $J^n$ is very easy to calculate for your question, because $J$ is 2-by-2 and upper-triangular.
$endgroup$
– hypernova
Jan 7 at 3:15
1
$begingroup$
@DavidG: Sure! Thanks for your message. marty cohen's solution is really beautiful. I seems to have misunderstood your expectations. While the coefficients $a_n$ and $b_n$ diverge, their resulted $F_n$ converges. marty cohen's solution definitely clarifies. Thank you!
$endgroup$
– hypernova
Jan 8 at 10:35
|
show 4 more comments
$begingroup$
Since the recurrence relation is fully nonlinear, it might be better to try the first few terms with the initial condition $F_0=f$. By Mathematica, it is easy to find the general expression should be of the form
$$
F_n=frac{a_n+b_nf}{a_n+b_n+a_nf}
$$
for $nge 1$, where $a_n$ and $b_n$ are constants. By the recurrence relation, these constants must follow
$$
left(
begin{array}{c}
a_{n+1}\
b_{n+1}
end{array}
right)=left(
begin{array}{cc}
1&1\
1&0
end{array}
right)left(
begin{array}{c}
a_n\
b_n
end{array}
right),
$$
with
$$
left(
begin{array}{c}
a_1\
b_1
end{array}
right)=left(
begin{array}{c}
1\
0
end{array}
right).
$$
Thanks to linear algebra, you may solve this Fibonacci-like $a_n$ and $b_n$ immediately.
Hope this could be somewhat helpful for you.
answered Jan 7 at 1:41
hypernovahypernova
5,009414
$endgroup$
Since the recurrence relation is fully nonlinear, it might be better to try the first few terms with the initial condition $F_0=f$. By Mathematica, it is easy to find the general expression should be of the form
$$
F_n=frac{a_n+b_nf}{a_n+b_n+a_nf}
$$
for $nge 1$, where $a_n$ and $b_n$ are constants. By the recurrence relation, these constants must follow
$$
left(
begin{array}{c}
a_{n+1}\
b_{n+1}
end{array}
right)=left(
begin{array}{cc}
1&1\
1&0
end{array}
right)left(
begin{array}{c}
a_n\
b_n
end{array}
right),
$$
with
$$
left(
begin{array}{c}
a_1\
b_1
end{array}
right)=left(
begin{array}{c}
1\
0
end{array}
right).
$$
Thanks to linear algebra, you may solve this Fibonacci-like $a_n$ and $b_n$ immediately.
Hope this could be somewhat helpful for you.
answered Jan 7 at 1:41
hypernovahypernova
5,009414
answered Jan 7 at 1:41
hypernovahypernova
5,009414
answered Jan 7 at 1:41
hypernovahypernova
5,009414
answered Jan 7 at 1:41
hypernovahypernova
5,009414
5,009414
$begingroup$
This is great! thank you so much!
$endgroup$
– DavidG
Jan 7 at 1:49
$begingroup$
@DavidG: You are welcome!
$endgroup$
– hypernova
Jan 7 at 1:50
$begingroup$
Would this be a pseudo styled Markov Chain?
$endgroup$
– DavidG
Jan 7 at 1:53
1
$begingroup$
@DavidG: For me, it is hard to try using results from Markov chains to your original question. While there might be useful results to use, it might be hard to avoid calculating $A^n$. You may try this built-in function "JordanDecomposition" in Mathematica to figure out $P$ and $J$ such that $A=PJP^{-1}$, where $J$ is the Jordan Normal Form of $A$. Then $A^n=PJ^nP^{-1}$. $J^n$ is very easy to calculate for your question, because $J$ is 2-by-2 and upper-triangular.
$endgroup$
– hypernova
Jan 7 at 3:15
1
$begingroup$
@DavidG: Sure! Thanks for your message. marty cohen's solution is really beautiful. I seems to have misunderstood your expectations. While the coefficients $a_n$ and $b_n$ diverge, their resulted $F_n$ converges. marty cohen's solution definitely clarifies. Thank you!
$endgroup$
– hypernova
Jan 8 at 10:35
|
show 4 more comments
$begingroup$
This is great! thank you so much!
$endgroup$
– DavidG
Jan 7 at 1:49
$begingroup$
@DavidG: You are welcome!
$endgroup$
– hypernova
Jan 7 at 1:50
$begingroup$
Would this be a pseudo styled Markov Chain?
$endgroup$
– DavidG
Jan 7 at 1:53
1
$begingroup$
@DavidG: For me, it is hard to try using results from Markov chains to your original question. While there might be useful results to use, it might be hard to avoid calculating $A^n$. You may try this built-in function "JordanDecomposition" in Mathematica to figure out $P$ and $J$ such that $A=PJP^{-1}$, where $J$ is the Jordan Normal Form of $A$. Then $A^n=PJ^nP^{-1}$. $J^n$ is very easy to calculate for your question, because $J$ is 2-by-2 and upper-triangular.
$endgroup$
– hypernova
Jan 7 at 3:15
1
$begingroup$
@DavidG: Sure! Thanks for your message. marty cohen's solution is really beautiful. I seems to have misunderstood your expectations. While the coefficients $a_n$ and $b_n$ diverge, their resulted $F_n$ converges. marty cohen's solution definitely clarifies. Thank you!
$endgroup$
– hypernova
Jan 8 at 10:35
$begingroup$
This is great! thank you so much!
$endgroup$
– DavidG
Jan 7 at 1:49
$begingroup$
This is great! thank you so much!
$endgroup$
– DavidG
Jan 7 at 1:49
$begingroup$
@DavidG: You are welcome!
$endgroup$
– hypernova
Jan 7 at 1:50
$begingroup$
@DavidG: You are welcome!
$endgroup$
– hypernova
Jan 7 at 1:50
$begingroup$
Would this be a pseudo styled Markov Chain?
$endgroup$
– DavidG
Jan 7 at 1:53
$begingroup$
Would this be a pseudo styled Markov Chain?
$endgroup$
– DavidG
Jan 7 at 1:53
1
1
$begingroup$
@DavidG: For me, it is hard to try using results from Markov chains to your original question. While there might be useful results to use, it might be hard to avoid calculating $A^n$. You may try this built-in function "JordanDecomposition" in Mathematica to figure out $P$ and $J$ such that $A=PJP^{-1}$, where $J$ is the Jordan Normal Form of $A$. Then $A^n=PJ^nP^{-1}$. $J^n$ is very easy to calculate for your question, because $J$ is 2-by-2 and upper-triangular.
$endgroup$
– hypernova
Jan 7 at 3:15
$begingroup$
@DavidG: For me, it is hard to try using results from Markov chains to your original question. While there might be useful results to use, it might be hard to avoid calculating $A^n$. You may try this built-in function "JordanDecomposition" in Mathematica to figure out $P$ and $J$ such that $A=PJP^{-1}$, where $J$ is the Jordan Normal Form of $A$. Then $A^n=PJ^nP^{-1}$. $J^n$ is very easy to calculate for your question, because $J$ is 2-by-2 and upper-triangular.
$endgroup$
– hypernova
Jan 7 at 3:15
1
1
$begingroup$
@DavidG: Sure! Thanks for your message. marty cohen's solution is really beautiful. I seems to have misunderstood your expectations. While the coefficients $a_n$ and $b_n$ diverge, their resulted $F_n$ converges. marty cohen's solution definitely clarifies. Thank you!
$endgroup$
– hypernova
Jan 8 at 10:35
$begingroup$
@DavidG: Sure! Thanks for your message. marty cohen's solution is really beautiful. I seems to have misunderstood your expectations. While the coefficients $a_n$ and $b_n$ diverge, their resulted $F_n$ converges. marty cohen's solution definitely clarifies. Thank you!
$endgroup$
– hypernova
Jan 8 at 10:35
|
show 4 more comments
$begingroup$
This is copied from
another answer of mine:
Let
$f_1 = frac{1}{1 + g(x) }
$
where
$g(x) > 0,
$,
and let
$f_n(x)
=frac{1}{1+f_{n-1}(x)}
$.
Then
$f_n(x)
to dfrac{sqrt{5}-1}{2}
$.
Note:
I doubt that any of this
is original,
but this was all done
just now by me.
Proof.
$begin{array}\
f_n(x)
&=frac{1}{1+frac{1}{1+f_{n-2}(x)}}\
&=frac{1+f_{n-2}(x)}{1+f_{n-2}(x)+1}\
&=frac{1+f_{n-2}(x)}{2+f_{n-2}(x)}\
end{array}
$
Therefore,
if $f_{n-2}(x) > 0$
then
$frac12 < f_n(x)
lt 1$.
Similarly,
if $f_{n-1}(x) > 0$
then
$0 < f_n(x)
lt 1$.
$begin{array}\
f_n(x)-f_{n-2}(x)
&=dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)}-f_{n-2}(x)\
&=dfrac{1+f_{n-2}(x)-f_{n-2}(x)(2+f_{n-2}(x))}{2+f_{n-2}(x)}\
&=dfrac{1-f_{n-2}(x)-f_{n-2}^2(x)}{2+f_{n-2}(x)}\
end{array}
$
$begin{array}\
f_n(x)+f_n^2(x)
&=dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)}+(dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)})^2\
&=dfrac{(1+f_{n-2}(x))(2+f_{n-2}(x))}{(2+f_{n-2}(x))^2}+dfrac{1+2f_{n-2}(x)+f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
&=dfrac{2+3f_{n-2}(x)+f_{n-2}^2(x)+1+2f_{n-2}(x)+f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
&=dfrac{3+5f_{n-2}(x)+2f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
text{so}\
1-f_n(x)-f_n^2(x)
&=dfrac{4+4f_{n-2}(x)+f_{n-2}^2(x)-(3+5f_{n-2}(x)+2f_{n-2}^2(x))}{(2+f_{n-2}(x))^2}\
&=dfrac{1-f_{n-2}(x)-f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
end{array}
$
Therefore
$1-f_n(x)-f_n^2(x)$
has the same sign as
$1-f_{n-2}(x)-f_{n-2}^2(x)$.
Also,
$|1-f_n(x)-f_n^2(x)|
lt frac14|1-f_{n-2}(x)-f_{n-2}^2(x)|
$
so
$|1-f_n(x)-f_n^2(x)|
to 0$.
Let
$p(x) = 1-x-x^2$
and
$x_0 = frac{sqrt{5}-1}{2}
$
so
$p(x_0) = 0$,
$p'(x) < 0$ for $x ge 0$.
Since
$f_n(x) > 0$,
$f_n(x)
to x_0$.
answered Jan 8 at 0:54
marty cohenmarty cohen
74.6k549129
$endgroup$
$begingroup$
Thanks for your post - I actually was playing around with a few different cases last night and noticed that it converged as you spoke to. I wasn't sure how to go about proving it, so thanks heaps for your post!
$endgroup$
– DavidG
Jan 8 at 1:01
add a comment |
$begingroup$
This is copied from
another answer of mine:
Let
$f_1 = frac{1}{1 + g(x) }
$
where
$g(x) > 0,
$,
and let
$f_n(x)
=frac{1}{1+f_{n-1}(x)}
$.
Then
$f_n(x)
to dfrac{sqrt{5}-1}{2}
$.
Note:
I doubt that any of this
is original,
but this was all done
just now by me.
Proof.
$begin{array}\
f_n(x)
&=frac{1}{1+frac{1}{1+f_{n-2}(x)}}\
&=frac{1+f_{n-2}(x)}{1+f_{n-2}(x)+1}\
&=frac{1+f_{n-2}(x)}{2+f_{n-2}(x)}\
end{array}
$
Therefore,
if $f_{n-2}(x) > 0$
then
$frac12 < f_n(x)
lt 1$.
Similarly,
if $f_{n-1}(x) > 0$
then
$0 < f_n(x)
lt 1$.
$begin{array}\
f_n(x)-f_{n-2}(x)
&=dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)}-f_{n-2}(x)\
&=dfrac{1+f_{n-2}(x)-f_{n-2}(x)(2+f_{n-2}(x))}{2+f_{n-2}(x)}\
&=dfrac{1-f_{n-2}(x)-f_{n-2}^2(x)}{2+f_{n-2}(x)}\
end{array}
$
$begin{array}\
f_n(x)+f_n^2(x)
&=dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)}+(dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)})^2\
&=dfrac{(1+f_{n-2}(x))(2+f_{n-2}(x))}{(2+f_{n-2}(x))^2}+dfrac{1+2f_{n-2}(x)+f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
&=dfrac{2+3f_{n-2}(x)+f_{n-2}^2(x)+1+2f_{n-2}(x)+f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
&=dfrac{3+5f_{n-2}(x)+2f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
text{so}\
1-f_n(x)-f_n^2(x)
&=dfrac{4+4f_{n-2}(x)+f_{n-2}^2(x)-(3+5f_{n-2}(x)+2f_{n-2}^2(x))}{(2+f_{n-2}(x))^2}\
&=dfrac{1-f_{n-2}(x)-f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
end{array}
$
Therefore
$1-f_n(x)-f_n^2(x)$
has the same sign as
$1-f_{n-2}(x)-f_{n-2}^2(x)$.
Also,
$|1-f_n(x)-f_n^2(x)|
lt frac14|1-f_{n-2}(x)-f_{n-2}^2(x)|
$
so
$|1-f_n(x)-f_n^2(x)|
to 0$.
Let
$p(x) = 1-x-x^2$
and
$x_0 = frac{sqrt{5}-1}{2}
$
so
$p(x_0) = 0$,
$p'(x) < 0$ for $x ge 0$.
Since
$f_n(x) > 0$,
$f_n(x)
to x_0$.
answered Jan 8 at 0:54
marty cohenmarty cohen
74.6k549129
$endgroup$
$begingroup$
Thanks for your post - I actually was playing around with a few different cases last night and noticed that it converged as you spoke to. I wasn't sure how to go about proving it, so thanks heaps for your post!
$endgroup$
– DavidG
Jan 8 at 1:01
add a comment |
$begingroup$
This is copied from
another answer of mine:
Let
$f_1 = frac{1}{1 + g(x) }
$
where
$g(x) > 0,
$,
and let
$f_n(x)
=frac{1}{1+f_{n-1}(x)}
$.
Then
$f_n(x)
to dfrac{sqrt{5}-1}{2}
$.
Note:
I doubt that any of this
is original,
but this was all done
just now by me.
Proof.
$begin{array}\
f_n(x)
&=frac{1}{1+frac{1}{1+f_{n-2}(x)}}\
&=frac{1+f_{n-2}(x)}{1+f_{n-2}(x)+1}\
&=frac{1+f_{n-2}(x)}{2+f_{n-2}(x)}\
end{array}
$
Therefore,
if $f_{n-2}(x) > 0$
then
$frac12 < f_n(x)
lt 1$.
Similarly,
if $f_{n-1}(x) > 0$
then
$0 < f_n(x)
lt 1$.
$begin{array}\
f_n(x)-f_{n-2}(x)
&=dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)}-f_{n-2}(x)\
&=dfrac{1+f_{n-2}(x)-f_{n-2}(x)(2+f_{n-2}(x))}{2+f_{n-2}(x)}\
&=dfrac{1-f_{n-2}(x)-f_{n-2}^2(x)}{2+f_{n-2}(x)}\
end{array}
$
$begin{array}\
f_n(x)+f_n^2(x)
&=dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)}+(dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)})^2\
&=dfrac{(1+f_{n-2}(x))(2+f_{n-2}(x))}{(2+f_{n-2}(x))^2}+dfrac{1+2f_{n-2}(x)+f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
&=dfrac{2+3f_{n-2}(x)+f_{n-2}^2(x)+1+2f_{n-2}(x)+f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
&=dfrac{3+5f_{n-2}(x)+2f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
text{so}\
1-f_n(x)-f_n^2(x)
&=dfrac{4+4f_{n-2}(x)+f_{n-2}^2(x)-(3+5f_{n-2}(x)+2f_{n-2}^2(x))}{(2+f_{n-2}(x))^2}\
&=dfrac{1-f_{n-2}(x)-f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
end{array}
$
Therefore
$1-f_n(x)-f_n^2(x)$
has the same sign as
$1-f_{n-2}(x)-f_{n-2}^2(x)$.
Also,
$|1-f_n(x)-f_n^2(x)|
lt frac14|1-f_{n-2}(x)-f_{n-2}^2(x)|
$
so
$|1-f_n(x)-f_n^2(x)|
to 0$.
Let
$p(x) = 1-x-x^2$
and
$x_0 = frac{sqrt{5}-1}{2}
$
so
$p(x_0) = 0$,
$p'(x) < 0$ for $x ge 0$.
Since
$f_n(x) > 0$,
$f_n(x)
to x_0$.
answered Jan 8 at 0:54
marty cohenmarty cohen
74.6k549129
$endgroup$
This is copied from
another answer of mine:
Let
$f_1 = frac{1}{1 + g(x) }
$
where
$g(x) > 0,
$,
and let
$f_n(x)
=frac{1}{1+f_{n-1}(x)}
$.
Then
$f_n(x)
to dfrac{sqrt{5}-1}{2}
$.
Note:
I doubt that any of this
is original,
but this was all done
just now by me.
Proof.
$begin{array}\
f_n(x)
&=frac{1}{1+frac{1}{1+f_{n-2}(x)}}\
&=frac{1+f_{n-2}(x)}{1+f_{n-2}(x)+1}\
&=frac{1+f_{n-2}(x)}{2+f_{n-2}(x)}\
end{array}
$
Therefore,
if $f_{n-2}(x) > 0$
then
$frac12 < f_n(x)
lt 1$.
Similarly,
if $f_{n-1}(x) > 0$
then
$0 < f_n(x)
lt 1$.
$begin{array}\
f_n(x)-f_{n-2}(x)
&=dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)}-f_{n-2}(x)\
&=dfrac{1+f_{n-2}(x)-f_{n-2}(x)(2+f_{n-2}(x))}{2+f_{n-2}(x)}\
&=dfrac{1-f_{n-2}(x)-f_{n-2}^2(x)}{2+f_{n-2}(x)}\
end{array}
$
$begin{array}\
f_n(x)+f_n^2(x)
&=dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)}+(dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)})^2\
&=dfrac{(1+f_{n-2}(x))(2+f_{n-2}(x))}{(2+f_{n-2}(x))^2}+dfrac{1+2f_{n-2}(x)+f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
&=dfrac{2+3f_{n-2}(x)+f_{n-2}^2(x)+1+2f_{n-2}(x)+f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
&=dfrac{3+5f_{n-2}(x)+2f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
text{so}\
1-f_n(x)-f_n^2(x)
&=dfrac{4+4f_{n-2}(x)+f_{n-2}^2(x)-(3+5f_{n-2}(x)+2f_{n-2}^2(x))}{(2+f_{n-2}(x))^2}\
&=dfrac{1-f_{n-2}(x)-f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\
end{array}
$
Therefore
$1-f_n(x)-f_n^2(x)$
has the same sign as
$1-f_{n-2}(x)-f_{n-2}^2(x)$.
Also,
$|1-f_n(x)-f_n^2(x)|
lt frac14|1-f_{n-2}(x)-f_{n-2}^2(x)|
$
so
$|1-f_n(x)-f_n^2(x)|
to 0$.
Let
$p(x) = 1-x-x^2$
and
$x_0 = frac{sqrt{5}-1}{2}
$
so
$p(x_0) = 0$,
$p'(x) < 0$ for $x ge 0$.
Since
$f_n(x) > 0$,
$f_n(x)
to x_0$.
answered Jan 8 at 0:54
marty cohenmarty cohen
74.6k549129
answered Jan 8 at 0:54
marty cohenmarty cohen
74.6k549129
answered Jan 8 at 0:54
marty cohenmarty cohen
74.6k549129
answered Jan 8 at 0:54
marty cohenmarty cohen
74.6k549129
74.6k549129
$begingroup$
Thanks for your post - I actually was playing around with a few different cases last night and noticed that it converged as you spoke to. I wasn't sure how to go about proving it, so thanks heaps for your post!
$endgroup$
– DavidG
Jan 8 at 1:01
add a comment |
$begingroup$
Thanks for your post - I actually was playing around with a few different cases last night and noticed that it converged as you spoke to. I wasn't sure how to go about proving it, so thanks heaps for your post!
$endgroup$
– DavidG
Jan 8 at 1:01
$begingroup$
Thanks for your post - I actually was playing around with a few different cases last night and noticed that it converged as you spoke to. I wasn't sure how to go about proving it, so thanks heaps for your post!
$endgroup$
– DavidG
Jan 8 at 1:01
$begingroup$
Thanks for your post - I actually was playing around with a few different cases last night and noticed that it converged as you spoke to. I wasn't sure how to go about proving it, so thanks heaps for your post!
$endgroup$
– DavidG
Jan 8 at 1:01
add a comment |
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