I can't calculate the exact first tree terms $F_0$, $F_1$ and $F_2$ of this continued fraction :
$$F_n=\cfrac{1}{-\text{i$\omega $}\,+A\,\cfrac{(n+1)^2}{{4 (n+1)^2-1}}F_{n+1}}$$ $A$ and $\omega$ are reals.
Please, how to obtain these terms ?
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$$F(0) \text{ is in function } F(1)$$ and $$F(1) \text{ is in function } F(2)$$ then, if you replace, you heve $$F(0) \text{ is in function } F(2)$$ – Wmmoreno Oct 26 '13 at 00:04
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@Wmmoreno, yes i know, but how to obtain the general form of each terms without depending on other? – Betatron Oct 26 '13 at 00:08
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You can not do that. By the form in which the sequence is defined. – Wmmoreno Oct 26 '13 at 00:16
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can you give me more details ? – Betatron Oct 26 '13 at 00:18
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If $n\in \mathbb{N}$, you always will have $F(n)$ depends of $F(n+1)$ – Wmmoreno Oct 26 '13 at 00:21
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so how I can do it? – Betatron Oct 26 '13 at 00:24
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Suppose we know the value of $F(0)$ and calculated on that value. Such that, if $F(0)\in \mathbb{C}$ fix ponit .. – Wmmoreno Oct 26 '13 at 00:29
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Where did you get this, why do you want to know, and why do you not have an initial value? – Will Jagy Oct 26 '13 at 00:58
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Meanwhile, note that it is not difficult to solve for $F_{n+1}$ as a linear fractional transformation in $F_n,$ something of the form $$ \frac{B F_n + C}{D F_n + E}, $$ where your numbers $B,C,D,E$ will be complex numbers and depend on $n,A,\omega.$ – Will Jagy Oct 26 '13 at 01:08
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hint: calculate$F_{-1}$ first,then rest should be easy.
chenbai
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if $n\ge0$,you can only show $F(n）=f(F(n+1))$ or $ F(n+1)=g(F(n)) $, nothing else you can do. ie, you can do nothing. – chenbai Oct 26 '13 at 05:30