$f_1(x)=\frac{M}{C}$, where M and C are constants $$h_1(x)=\frac{\int_0^xf_1(y)dy}C + \frac{\int_0^x\int_0^yf_1(z)dzdy}{C^2} + \frac{\int_0^x\int_0^y\int_0^zf_1(t)dtdzdy}{C^3} + \cdots$$ $$f_2(x)=h_1(C)-h_1(x)$$ $$h_2(x)=\frac{\int_0^xf_2(y)dy}C + \frac{\int_0^x\int_0^yf_2(z)dzdy}{C^2} + \frac{int_0^x\int_0^y\int_0^zf_2(t)dtdzdy}{C^3} + \cdots$$ $$f_3(x)=h_2(C)-h_2(x)$$ $$\cdots$$
$$h_{i}(x)=\frac{\int_0^xf_i(y)dy}C + \frac{\int_0^x\int_0^yf_i(z)dzdy}{C^2} + \frac{\int_0^x\int_0^y\int_0^zf_i(t)dtdzdy}{C^3} + \cdots$$ $$f_{i+1}(x)=h_i(C)-h_i(x)$$
$$h_{\infty}(x) = ? $$

I want to examine the convergence of the function of $h_{\infty}(x)$. Each function $h_i(x)$ can be represented with $e^{\frac{x}C}$ function as a shorter version by using the maclaurin Series.

$h_1(x)$ becomes $ \frac{M}{C} \left( e^{\frac{x}C}-1 \right) $ when the infinite series is arranged by using the Maclaurin Series.
$h_2(x)$ becomes $\frac{M}{C} \left( e^{\frac{x}C} \left(-\frac{x}C+e \right) - e \right)$.
$h_3(x)$ becomes $\frac{M}{C} \left( e^x\left(\frac{1}{2}{\frac{x}{C}}^{2}-e{\frac{x}{C}}+e^2-e\right)-\left(e^2-e\right) \right)$

By writing program codes, I calculated and found that $h_i(x)$ function is getting closer to the function $2\frac{M}{C}\left(\frac{x}{C}\right)$ with increasing i, I want to mathematically prove this convergence. $$h_{\infty}(x)=2\frac{M}{C}\left(\frac{x}{C}\right)$$

Any tip will be appreciated. Thank you for reading this.

For the recently posted question, the $f(i)$ mentioned in the question is the same as the following variant of $h_i(x)$ when x = 1. $$ f(i) = h_i(1) $$ The variant of $h_i(x)$ is defined as follows.

$$g_1(x)=1$$ $$h_1(x)=\int_0^xg_1(y)dy + \int_0^x\int_0^yg_1(z)dzdy + \int_0^x\int_0^y\int_0^zg_1(t)dtdzdy + \cdots$$ $$g_2(x)=h_1(1)-h_1(x)$$ $$h_2(x)=\int_0^xg_2(y)dy + \int_0^x\int_0^yg_2(z)dzdy + \int_0^x\int_0^y\int_0^zg_2(t)dtdzdy + \cdots$$ $$g_3(x)=h_2(1)-h_2(x)$$ $$\cdots$$ $$h_{i}(x)=\int_0^xg_i(y)dy + \int_0^x\int_0^yg_i(z)dzdy + \int_0^x\int_0^y\int_0^zg_i(t)dtdzdy + \cdots$$ $$g_{i+1}(x)=h_i(1)-h_i(x)$$

In the variant of $h_i(x)$, the constant M is removed and C is substituted with 1 from the original $h_i(x)$.

  • $\begingroup$ Have you seen the formula for repeated integrals? $\endgroup$ – J. M. is a poor mathematician Jul 14 '12 at 16:12
  • $\begingroup$ @J.M. I have just seen it. I have calculated them by myself. I have found that h1(x) is a Maclaurin Series so it becomes e^x - 1. $\endgroup$ – Nate Jul 15 '12 at 1:54
  • 1
    $\begingroup$ Nate: You just modified the notations and the setting of your question, thus my answer now refers to a question which is no more visible. This is bad practice. You should (1.) revert to the original version of the question, (2.) possibly accept an answer, and (3.) ask the new question on another page. $\endgroup$ – Did Jul 15 '12 at 5:22
  • $\begingroup$ @did I am so sorry for the mistake... Since I thought that asking many questions at once might not be appropriate, I have changed this question. I will revert it and make a new question instead. $\endgroup$ – Nate Jul 15 '12 at 9:28
  • 1
    $\begingroup$ Let us stick to the present question: my answer proves that there is no possible limit $(f,h)$ of $(f_i,h_i)$ other than $f\equiv0$ and $h\equiv h(C)$. Thus, I suggest you try to reconcile the assertions in your other question with this established fact. $\endgroup$ – Did Jul 15 '12 at 10:13

The constants $M$ and $C$ can be absorbed by scaling the functions by $M$ and the independent variables by $C$, so I'll use $M=C=1$.

Then for any $a\in\mathbb R$, the functions $h(x)=ax$ and $f(x)=a(1-x)$ form a fixed point of the iteration. To show that these are the only fixed points, we merely need to correct a small mistake in Didier's answer. The correct equation is

$$h_i(x)=\int_0^x(f_i(y)+h_i(y))\,\mathrm dy\;,$$

and with $f_{i+1}(x)=h_i(1)-h_i(x)$ this yields,

$$h_{i+1}(x)=\int_0^x(h_i(1)-h_i(y)+h_{i+1}(y))\,\mathrm dy\;,$$

from which

$$h(x)=\int_0^xh(1)\,\mathrm dy=h(1)x=:ax$$



Thus, if the sequence converges, it converges to one of these fixed points, and it only remains to determine $a$. To do so, note that the iteration preserves the normalization of $f$:

$$ \begin{align} \int_0^1f_{i+1}(x)\,\mathrm dx &= \int_0^1(h_i(1)-h_i(x))\,\mathrm dx \\ &= h_i(1)-\int_0^1h_i(x)\,\mathrm dx \\ &= \int_0^1(f_i(x)+h_i(x))\,\mathrm dx-\int_0^1h_i(x)\,\mathrm dx \\ &= \int_0^1f_i(x)\,\mathrm dx\;. \end{align} $$

Thus, since $\int_0^1f_1(x)\,\mathrm dx=1$ and $\int_0^1f(x)\,\mathrm dx=a/2$, we must have $a=2$, and thus, if the sequence converges, it converges to $h(x)=2x$ and $f(x)=2(1-x)$.

Here's a plot of the first three iterates and the fixed point, showing quite rapid convergence.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.