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So I was exploring some math the other day... and I came across the following neat identity:

Given $y$ is a function of $x$ ($y(x)$) and $$ y = 1 + \frac{\mathrm{d}}{\mathrm{d}x} \left(1 + \frac{\mathrm{d}}{\mathrm{d}x} \left(1 + \frac{\mathrm{d}}{\mathrm{d}x} \left(1 + \frac{\mathrm{d}}{\mathrm{d}x} \left(1 + \frac{\mathrm{d}}{\mathrm{d}x} \left( \cdots \right) \right) \right) \right) \right) \text{ (repeated differential)} $$

then we can solve this equation as follows: $$ y - 1 = \frac{\mathrm{d}}{\mathrm{d}x} \left(1 + \frac{\mathrm{d}}{\mathrm{d}x} \left( \cdots \right) \right) \iff \int y - 1 \, \mathrm{d} x = 1 + \frac{\mathrm{d}}{\mathrm{d}x} \left( 1 + \frac{\mathrm{d}}{\mathrm{d}x} \left( \cdots \right) \right) $$ $$ \implies \int y - 1 \, \mathrm{d} x = y \iff y - 1 = \frac{\mathrm{d} y }{ \mathrm{d} x} $$

So

$$ \ln \left( y - 1 \right) = x + C \iff y = Ce^x + 1 $$

This problem reminded me a lot of nested radical expressions such as: $$ x = 1 + \sqrt{1 + \sqrt{ 1 + \sqrt{ \cdots }}} \iff x - 1 = \sqrt{1 + \sqrt{ 1 + \sqrt{ \cdots }}} $$ $$ \implies (x - 1)^2 = x \iff x^2 - 3x + 1 = 0 $$

and so

$$ x = \frac{3}{2} + \frac{\sqrt{5}}{2} $$

This reminded of the Ramanujan nested radical which is:

$$ x = 0 + \sqrt{ 1 + 2 \sqrt{ 1 + 3 \sqrt{1 + 4 \sqrt{ \cdots }}}} $$

whose solution cannot be done by simple series manipulations but requires knowledge of general formula found by algebraically manipulating the binomial theorem...

This made me curious...

say $y$ is a function of $x$ ($y(x)$) and

$$ y = 0 + \frac{\mathrm{d}}{\mathrm{d}x} \left(1 + 2\frac{\mathrm{d}}{\mathrm{d}x} \left(1 + 3\frac{\mathrm{d}}{\mathrm{d}x} \left(1 + 4\frac{\mathrm{d}}{\mathrm{d}x} \left(1 + 5\frac{\mathrm{d}}{\mathrm{d}x} \left( \cdots \right) \right) \right) \right) \right) $$

What would the solution come out to be?

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    $\begingroup$ fyi the ramanujan radical evaluates to 3... and its solution can be found here: en.wikipedia.org/wiki/Srinivasa_Ramanujan (go to attention from mathematicians) $\endgroup$ – frogeyedpeas May 2 '13 at 21:40
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    $\begingroup$ It really helps readability to format questions using MathJax (see FAQ). Regards $\endgroup$ – Amzoti May 2 '13 at 21:44
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    $\begingroup$ @Amzoti I tried to reformat it $\endgroup$ – DanZimm May 2 '13 at 22:09
  • $\begingroup$ @frogeyedpeas np, I really like the question $\endgroup$ – DanZimm May 2 '13 at 22:13
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    $\begingroup$ How do you exactly define that repeated differential equation? If we cut-off at $n$, don't we get $y_n = n! \frac{d^n y_n}{dx^n}$? $\endgroup$ – Aryabhata May 3 '13 at 0:53
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If the operator is meant to derive what follows, then we have $$y(x) = \lim_{n \to \infty} n! \cdot y^{(n)}(x) \qquad,\qquad \forall\ x \in X$$

since the derivative of a constant is always $0$ , and the derivative of a sum is the sum of derivatives. However, if multiplication is meant, with the “last” term of the nested product presumably being none other than y(x), then we have $$y(x) = \sum_{n=1}^\infty n! \cdot y^{(n)}(x) \qquad,\qquad \forall\ x \in X$$

where $y : X \to Y$ ; either way, since $$\lim_{n \to \infty}n! = \infty$$ then, in order for the function to converge $\forall\ x \in X$ , we must have $$\lim_{n \to \infty} y^{(n)}(x) = 0 \quad,\quad \forall\ x \in X \quad=>\quad y(x)\ =\ P_m(x)\ =\ \sum_{k=0}^m a_k \cdot x^k \quad,\quad m \in \mathbb{N}$$

since the Nth nested integral of $0$ is nothing else than a polynomial function of degree N-$1$ .

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  • $\begingroup$ I don't understand the last equation. $\lim_{n\to\infty}y^{(n)}(x)=0$ does not imply that $y$ must be a polynomial (e.g., $y(x)=\exp(kx)$ with $k<1$ also satisfies this limit). I'm probably missing something though; can you please elaborate? Thanks. $\endgroup$ – AccidentalFourierTransform May 27 '18 at 3:08
  • $\begingroup$ @AccidentalFourierTransform: Without denying that $\lim\limits_{n\to\infty}k^n\to0$ for $k\in(0,1),$ we also have $\lim\limits_{x\to\infty}e^x\to\infty.$ More to the point, for every $n>0$ there is an $x_n\ge\ln a-n\ln k,$ so that $\lim\limits_{n\to\infty}k^ne^{x_n}\ge a,$ for some $a>0.$ $\endgroup$ – Lucian May 27 '18 at 12:00
  • $\begingroup$ Sure: the limits $x\to\infty$ and $n\to\infty$ don't commute. But the limit we are interested in is the latter, so that's the one we must take first. I am still unconvinced polynomials are necessarily the only solution, but nevermind. $\endgroup$ – AccidentalFourierTransform May 27 '18 at 13:12
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Having been quite a bit of time I thought about this some more!

If we cut our original radical at a finite time we have (as pointed out by Aryabhata) we have

$$y = n!\frac{d^n}{dx^n} $$

Solutions to this include $$ y= C_1 e^{x\sqrt[n]{n!}_1} +C_2 e^{x\sqrt[n]{n!}_2}... C_n e^{x\sqrt[n]{n!}_n} $$

For all possible nth roots of unity.

Now consider stirlings approximation of n! which states

$$n! \le e n^{n + \frac{1}{2}} e^{-n} $$

(Note that for $n \ge 0$ these functions are both greater than or equal to 1) hence

$$ |\sqrt[n]{n!}| \le |\sqrt[n]{e n^{n + \frac{1}{2}} e^{-n}} | $$

Which yields

$$ |\sqrt[n]{n!}| \le |\sqrt[n]{e n^{n + \frac{1}{2}} e^{-n}} | $$

Which yields:

$$ |\sqrt[n]{n!}| \le |{e^{\frac{1}{n}} n^{1 + \frac{1}{2n}} e^{-1}} | $$

Which yields:

$$ |\sqrt[n]{n!}| = O(n) $$

As n tends to infinity, so does this.

But before I throw out any hope for this, I would like to point out that as n gets larger, we essentially obtain a basis for functions in terms of complex exponentials, I'm curious if there is a way to pick the constants $C_i$ such that for a given n, you can model the function as closely as possible, and determine then in the limit what the $C_i$ need to be (probably very degenerate piecewise constant functions)

If there is such a scheme, it may very well be that ANY function satisfies this differential equation, assuming the set of exponentials gets closer and closer to forming a basis for all functions (which in the limit as n goes to infinity means it does indeed form a basis). The tricky part is defining, "how well" a basis is. IE if it doesn't cover every case, how do we say that it covers more cases, or relatively more cases than its predecessor.

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