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$\int_0^x \cfrac{1}{1+\int_0^t \cfrac{1}{2+\int_0^{t_1} \cfrac{1}{3+\int_0^{t_2} \cfrac{1}{\cdots} dt_3} dt_2} dt_1} dt =f(x)$

$\int_{0}^{x} \frac{1}{n+h_{n+1}(t)}{d} t=h_n(x)$

$h'_n(x)(n+h_{n+1}(x))=1$ $h'_{n+1}(x)(n+1+h_{n+2}(x))=1$

I need to find $ h_1(x)=f(x)$

Please help me how to express $f(x)$ as known functions or power series?

Thanks a lot for answers

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    @MichaelHardy : Really it is so. Thanks a lot for format change.2012-09-11

1 Answers 1

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It seems that additionally you require $h_n(0)=0$ for all $n$ (because $\int_0^0=0$). This allows us to compute the power series of $h_1$ to arbitrary precision. For example working down from $n=4$: $h_4(x)=O(x)$ $h_3'(x)=\frac1{3+O(x)}=\frac13+O(x)$ $h_3(x)=\frac13x+O(x^2)$ $h_2'(x) = \frac1{2+\frac13x+O(x^2)}=\frac12-\frac1{12}x+O(x^2)$ $h_2(x)=\frac12x-\frac1{24}x^2+O(x^3)$ $h_1'(x) = \frac1{1+\frac12x-\frac1{24}x^2+O(x^3)}=1-\frac12x+\frac7{24}x^2+O(x^3)$ $h_1(x) = x-\frac14x^2+\frac7{72}x^3+O(x^4)$ Starting with $n=10$, I get $h_1(x) = x - \frac{1}{4} x^2 + \frac{7}{72} x^3 - \frac{149}{3456} x^4 + \frac{21193}{1036800} x^5 - \frac{235619}{23328000} x^6 + \frac{1408454377}{274337280000} x^7 - \frac{1227854784917}{460886630400000} x^8 + \frac{6524483827239647}{4645737234432000000} x^9+ O(x^{10}). $ I don't know if the sequence of coefficients should be well-known.

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    I calculated the coefficients up to $x^{1000}$. Apparently they are alternating and their absolute value grows slowly (reaching about 0.59 there, if I remember yesterdays output correctly). It might still tend to $\phi$, $\frac23$, $1$ or $\infty$ - who knows?2012-09-12