Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

                         $f_1(x)=\frac{M}{C}$, where M and C are constants $$f_2(x)=\frac{\int_0^xf_1(y)dy}{C} + \frac{\int_0^x\int_0^yf_1(z)dzdy}{C^2} + \frac{\int_0^x\int_0^y\int_0^zf_1(t)dtdzdy}{C^3} + \cdots$$ $$f_3(x)=\frac{\int_0^xf_2(y)dy}{C} + \frac{\int_0^x\int_0^yf_2(z)dzdy}{C^2} + \frac{\int_0^x\int_0^y\int_0^zf_2(t)dtdzdy}{C^3} + \cdots$$

$$f_{i+1}(x)=\frac{\int_0^xf_i(y)dy}{C} + \frac{\int_0^x\int_0^yf_i(z)dzdy}{C^2} + \frac{\int_0^x\int_0^y\int_0^zf_i(t)dtdzdy}{C^3} + \cdots$$

$$f_{\infty}(x) = ? $$

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

$f_2(x)$ becomes $\frac{M}{C} \left[ e^{\frac{x}{c}}-1 \right]$ when the infinite series is arranged by using the Maclaurin Series.
$f_3(x)$ also becomes $\frac{M}{C} \left[ e^{\frac{x}{c}} \left( e - \frac{x}{c} \right) - e \right]$.
So far I calculated $f_4(x)$, which has too many terms to be written down here.
By writing program codes, I calculated and found that $f_i(x)$ function is getting closer to the function $\frac{2M}{C^2}x$ with increasing i.
I want to mathematically prove this convergence. $$f_{\infty}(x)=\frac{2M}{C^2}x$$ Any tip will be appreciated.

share|cite|improve this question
Your integrals are not well defined ; instead of putting $x$'s everywhere you should put different variables for the upper bounds of the integrals and the $dx$'s. Please modify your text to make it more understandable. – Patrick Da Silva Mar 13 '12 at 8:04
@DidierPiau you are right... thank you for the comment.. I've modified it. – Nate Mar 13 '12 at 8:32
@PatrickDaSilva Thank you for the comment... I've modified it... – Nate Mar 13 '12 at 8:36
Yes, that is better. =) – Patrick Da Silva Mar 13 '12 at 16:32
up vote 4 down vote accepted

For every $i\geqslant0$ and $x\geqslant0$, $$ f_i(x)=\frac{M}C\mathrm e^{x/C}\sum_{k=i}^{+\infty}\frac{(-1)^{k-i}}{C^k}\frac{x^k}{k!}. $$ Hence $f_i(x)\to0$ for every $x\geqslant0$.

To prove the convergence, fix $x_i\gt0$, $i\geqslant3x_i/C$ and some $K_i$ large enough such that $f_i(x)\leqslant K_ix^i$ for every $x\leqslant x_i$. Such a number $K_i$ exists for every $x_i$ because $f_i(x)\sim Mx^i/(C^{i+1}i!)$ when $x\to0$ and $f_i$ is continuous.

Then, for every $j\geqslant i$, $Cf_{j+1}'=f_j+f_{j+1}$ and $f_{j+1}(0)=0$ yield $f_{j+1}(x)\leqslant K_{j+1}x^{j+1}$ for every $x\leqslant x_i$, with $$ K_{j+1}=\frac{K_j}{C\cdot(j+1)-x_i}\leqslant\frac{K_j}{2x_i}. $$ Hence, for every $j\geqslant i$, $K_j\leqslant K_i/(2x_i)^{j-i}$. This implies that $f_j(x)\leqslant K_jx^j\leqslant K_ix^i/2^{j-i}$ for every $x\leqslant x_i$, hence $f_j\to0$ on $(0,x_i)$. Since $x_i$ is as large as desired, the proof is complete.

Note that the limit equation, whatever that means, is $Cf_{\infty}'=2f_\infty$, whose solutions are $f_\infty(x)=K_\infty\mathrm e^{2x/C}$ with $K_\infty=f_\infty(0)$. The convergence above is simply the translation of the fact that $K_\infty=0$.

share|cite|improve this answer
Thank you very much for the answer. I don't know how to thank you enough... Actually I've made a mistake in simplifying the equations before asking this question. Later I found that and I tried to fix it and to edit this page but at that moment you've already answered this question. A few minute ago, I posted the fixed question which was what I originally intended to ask. However, I think that this answer will be also helpful for me to figure out how to solve the fixed problem. Thank you again. – Nate Mar 13 '12 at 12:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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