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