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I found this question (linked here) which asks to find what this infinite series converges to

$$ \sum_{n=1}^{\infty} \int_0^{\pi} f_n(x) dx $$

where $f_{n+1}(x) = \sin(f_n(x)) $ and $f_1 = \sin(x)$. I actually wrote a few scripts which I figure could be optimized but I'm looking for a better approach or if the type of problem is known. I've linked the script here and posted it

import scipy.integrate as integrate
import math

def repeated(f,  n):
  if n < 1:
    raise ValueError()
  elif n == 1:
    return f
  else:
    return lambda x: repeated(f, n-1)(f(x))

def get_sum(n):
    ### return nth sum

    sum_ = 0
    lower_bound = 0 
    upper_bound = math.pi
    
    for n in range(1,n+1):
        sum_ += integrate.quad(repeated(lambda x: math.sin(x), n), lower_bound, upper_bound)[0]
    return sum_

tolerance = 10**-4
initial_val = get_sum(1)
k = 2
while True:
    new_val = get_sum(k)
    absolute_error = abs(new_val-initial_val)
    print("K : {k}\nKth sum : {s}\nAbsolute Error : {e}".format(k=k,s=new_val, e=absolute_error))
    if absolute_error < tolerance:
        break
    initial_val = new_val
    k+=1

I can run this until around k=500 and it starts to get slow. I modified it and increased the step size and I hit the max recursion depth for compose at around k=2500. I then attempted to interpolate the integral values with an exponential function instead. This seemed to be a way around the recursion issue but it doesn't seem accurate. Any suggestions would be great.

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  • $\begingroup$ Are you sure it converges? I appears that sequence formed by the function evaluated at $\pi/2$ (the maximum of the function) grows monotonically. $\endgroup$
    – nicoguaro
    Jun 21 at 23:25
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    $\begingroup$ Sorry, I meant "the sequence formed by the partial sums". I tried with 1000 000 terms. I think that sequence gives an estimate for a lower bound since you can come the area of the triangle that lies below your convex function as $f_{\max} \pi/2$. $\endgroup$
    – nicoguaro
    Jun 22 at 0:03
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    $\begingroup$ I didn't used a recursive function. I stored the last two values and accumulated them. $\endgroup$
    – nicoguaro
    Jun 22 at 0:20
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    $\begingroup$ I don't believe there is a finite limit here because for each term the integral $\int_0^{\pi} f_n(x) dx > \int_0^{\epsilon} f_n(x) dx$, for any positive $\epsilon < \pi$. So if we take $\epsilon \ll 1$ then $f_n(x) \to x$ and the sum becomes $\sum_1^n \epsilon^2/2 = n \epsilon^2/2$ which diverges for $n \to \infty$. $\endgroup$ Jun 22 at 2:46
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    $\begingroup$ On the function sequence involved, the iterated sine, or any other of the form $y_{k+1}=y_k-ay_k^3+...$, see math.stackexchange.com/questions/1449281/…, math.stackexchange.com/questions/1609995/…, math.stackexchange.com/questions/105452/… $\endgroup$ Jun 25 at 7:18
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There is a lot to unpack here, and probably this is a better question for math.SE. TL; DR version is: in exact arithmetic, this does not converge; see Maxim Umansky's answer. In FP arithmetic, it will converge. I don't know to what, see the long version for an attempt.

Let's assume that it is possible to compute $\int_0^{\pi} f_n(x)\text{d}x$ exactly. I don't want to deal with the extra arithmetic coming from there, and it doesn't change the final answer much anyways.

After a while $\int_0^{\pi} f_n(x)\text{d}x$ will become small enough that $$fl(\sum_{n=1}^N \int_0^{\pi} f_n(x) \text{d}x+ \int_0^{\pi} f_{N+1}(x)\text{d}x)= fl(\sum_{n=1}^N \int_0^{\pi} f_n(x)\text{d}x)$$ where $fl$ is the rounding function. This is a common phenomenon in computational mathematics that you should be careful about.

(A well-known example is $\sum_{n=1}^{\infty} 1/n$ which diverges in exact arithmetic, but converges in FP arithmetic. If you reverse the ordering of the sum $\sum_{n=M}^{1} 1/n$ for $M\gg 1$, then you get the expected divergence result.)

Also, $f_n(x)$ itself can be thought as a fixed point iteration, and the integral $\int_0^{\pi} f_n(x)$ can be interpreted as testing the convergence of the fixed point iterations for all the initial guesses in the interval $[0,\pi]$ at the $n$-th iteration. The fixed point iteration $x=\sin(x)$ has the solution $x=0$ for any initial guess $x_0\in [0,\pi]$ (not too hard to prove, so I will skip). Hence, the integral $\int_0^{\pi} f_n(x)\text{d}x \to 0$ as $n\to\infty$. The upper bound for this integral is $\pi f_n(\pi/2)$.

Now, we can naively bound $\sin^n(\pi/2)$ ($\sin^n$ means $\sin$ composited with itself $n$ times) from below by $1/n$. Here is a proof by induction:

$n=1$ case: $\sin^1(\pi/2) \geq 1/1$. Hence, we are good here.

Inductive hypothesis: Assume for $n=k$, $\sin^n(\pi/2)\geq 1/n$.

$n=k+1$ case: Consider $\begin{align}\sin(\sin^k(\pi/2)) - 1/(k+1) &= \sin(\sin^k(\pi/2)) - 1/k + 1/(k^2+k) \\ &\text{by the inductive hypothesis and since sine is an increasing func. in } [0,1]\\ &\geq \sin(1/k) - 1/k + 1/(k^2+k) \\ &\text{by Taylor's remainder theorem}\\ &\geq 1/k -1/(6k^3) -1/k +1/(k^2+k)\\ &\text{for any } k\geq 1\\ &\geq 0. \end{align}$

So, the lower bound for the sum $\pi\sum_{n=1}^{\infty} f_n(\pi/2)$ is $\pi\sum_{n=1}^{\infty} 1/n$, which will definitely not increase after $n=4.5\times 10^{15}\approx 1/\varepsilon_{\text{mach}}$ since $fl(1+\varepsilon_{\text{mach}}) = fl(1)$. I would expect $\pi\sum_{n=1}^{\infty} f_n(\pi/2)$ to "converge" earlier than that (the word "converge" is in quotation marks, because it is not a real convergence).

On the other hand, we can use $1$ as an upper bound to the upper bound $f_n(\pi/2)$ (since the fixed point iterations converge to zero). Then we see that $$4.5\pi\times 10^{15}\approx\pi\sum_{n=1}^{\infty} 1 \geq \pi\sum_{n=1}^{\infty} f_n(\pi/2) \geq \pi\sum_{n=1}^{\infty} 1/n \approx \pi(\ln(4.5\times 10^{15}) + \gamma),$$ where $\gamma$ is the Euler–Mascheroni constant. (Approximation signs in the equation above mean floating point approximation, and it is non-standard notation. I am abusing the notation here)

Hence, the sum $\sum_{n=1}^{\infty} \int_0^{\pi} f_n(x)$ definitely has a finite limit when calculated on a computer using floating point arithmetic and that value is no larger than $4.5\pi\times 10^{15}$. I would expect it to be slightly larger $\pi\ln(4.5\times 10^{15})$, but I cannot prove that rigorously. That is more of an educated guess.

Edit: If Maxim Umansky's answer is not clear to you, you can look for a lower bound for the integral $\int_{0}^{\pi} f_n(x)\text{d}x$ using quadrature rules. A simple to prove lower bound comes from the observation that $f_n$ is concave (see Theorems 1 and 2) for all finite $n$ so $\int_{0}^{\pi} f_n(x)\text{d}x \geq f_n(\pi/2)\pi/2 \geq \pi/(2n)$. The first inequality basically says that the composite trapezoidal rule applied to the integral using the subintervals $[0,\pi/2]$ and $[\pi/2,\pi]$ will be a lower bound to the value of the integral. Then using the crude lower bound to $f_n(\pi/2)$, we can say that $\int_{0}^{\pi} f_n(x)\text{d}x\geq \frac{\pi}{2} \frac{1}{n}$. Hence, $\sum_{n=1}^{\infty}\int_{0}^{\pi} f_n(x)\text{d}x\geq \frac{\pi}{2} \sum_{n=1}^{\infty}\frac{1}{n}$ which diverges.

Theorem 1: If $h(x)$ is concave and non-decreasing, and $g(x)$ is concave, then $f(x) = h(g(x))$ is concave.

Theorem 2: If $h(x)$ is concave and non-increasing, and $g(x)$ is concave, then $f(x) = h(g(x))$ is concave.

Proof (example): https://math.stackexchange.com/questions/2307500/composition-of-functions-and-concavity

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  • $\begingroup$ Thank you. It wasn't that clear at first. This helps. $\endgroup$
    – Ryan Howe
    Jun 22 at 10:43
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    $\begingroup$ No worries. You understand why your Python code, even it was lightning quick, wouldn't give you the correct answer, right? $\endgroup$ Jun 22 at 15:01
  • $\begingroup$ I do now. I probably jumped the gun trying to calculate it first but it's a lot more complex than I thought. $\endgroup$
    – Ryan Howe
    Jun 22 at 15:28
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There is no finite limit for this series sum. Note that for each $n$ the function $f_n$ is positive definite, $f_n(x) > 0$ within the semi-open interval $(0,\pi]$, and we can construct the lower bound for the sum as follows. Consider some parameter $\epsilon \in (0,\pi]$. Then for each $n$ the integral $\int_0^{\pi} f_n(x) dx \ge \int_0^{\epsilon} f_n(x) dx$. But if we take $\epsilon \ll 1$ then for each $n$ the function $f_n(x) \to x$ and the sum $\sum_1^n \int_0^{\epsilon} f_n(x) dx \to \sum_1^n \epsilon^2/2 = n \epsilon^2/2$ which diverges for $n \to \infty$.

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