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                          $f_1(x)=\frac{M}{C}$, 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 $h_i(x)$.

  • 0
    Have you seen the formula for [repeated integrals](http://mathworld.wolfram.com/RepeatedIntegral.html)?2012-07-14
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    @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 - 1.2012-07-15
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    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.2012-07-15
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    @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.2012-07-15
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    Mar 13 and Jul 15 = *at once*??2012-07-15
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    @did This question is dealing with the same problem. Since I just substituted a constant $C$ of the original problem with 1 for simpler expression. Before reverting it, I want you to check another question [http://math.stackexchange.com/questions/170749/how-to-prove-convergence-of-a-power-series-of-e-eulers-number]. As I mentioned in this question, $h_i(x)$ converges to $h_\infty(x) = 2x$ and $h_\infty(1)=2$. For the original version $h_i(x)$ converges to $h_\infty(x)=2\frac{x}{C}$ and $h_\infty(C)=2$. I've made an awful mistake. If you see this, please leave me a comment. I will revert this.2012-07-15
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    @did I am a novice here. Please understand my awful mistakes. I didn't know how to accept an answer even though you gave me a right answer for the first question that I asked Mar 13. Now, I've just known how to accept it, so I've chosen your answer for [the first question] http://math.stackexchange.com/questions/119605/series-of-series-of-integrals-where-it-can-converge/119629#1196292012-07-15
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    @did After spending several months for another task (implementing my ideas), I recently began to try to understand your answer for this question, but as I mentioned in [the new question] http://math.stackexchange.com/questions/170749/how-to-prove-convergence-of-polynomials-in-e-eulers-number#comment392798_170749 , I found that $h_i(C)$ (or $h_i(1)$ for now) converges to 2. I have no idea how to prove it. Therefore, I asked the new question yesterday and am waiting for an answer.2012-07-15
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    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**.2012-07-15
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    @did I have reverted this to the original problem, and appended several comments. Sorry for the mistake again.2012-07-15
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    @did: I understand your frustration; it does happen that people just go on arguing against established facts -- but in this case the boldface turned out to be slightly too bold -- there was a small mistake in your answer, and the same approach can actually be used to find the fixed point.2012-07-31
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    @joriki: Oh my. Well done.2012-07-31

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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

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

and

$$f(x)=a-h(x)=a(1-x)\;.$$

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.