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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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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) –  frogeyedpeas May 2 '13 at 21:40
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It really helps readability to format questions using MathJax (see FAQ). Regards –  Amzoti May 2 '13 at 21:44
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@Amzoti I tried to reformat it –  DanZimm May 2 '13 at 22:09
    
@DanZimm thank you! –  frogeyedpeas May 2 '13 at 22:11
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How do you exactly define that repeated differential equation? If we cut-off at $n$, don't we get $y_n = n! \frac{d^n y_n}{dx^n}$? –  Aryabhata May 3 '13 at 0:53

1 Answer 1

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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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