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| discontinuous function https://forumdematematica.org/viewtopic.php?f=15&t=170 |
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| Autor: | kinu [ 28 jan 2012, 05:35 ] |
| Título da Pergunta: | discontinuous function |
If \(f(x)\) is a real valued function discontinuous at all integral points lying in \(\left[0,n\right]\;\; n\in\mathbb{N}\) and if (f(x))^2 = 1 forall \(0 \leq x \leq n\) then number of functions \(f(x)\) are I have calculate it for n=2 and Getting 6 no. of function which is discontinuous in [0,2] but how can i generalise it Help required Thanks |
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| Autor: | Rui Carpentier [ 09 fev 2012, 20:23 ] |
| Título da Pergunta: | Re: discontinuous function |
Hint: Given a set \(A\subset\mathbb{R}\), consider the set \(S(A)\) contain any real valued function \(f\) discontinuous at all integral points lying in \(A\) such that \((f(x))^2 = 1\) for all \(x\in A\). You want to know how many elements has the set \(S([0,n])\) (i.e. the cardinality of \(S([0,n])\)). Show that: 1-\(\sharp S([0,n])=\sharp S([0,n[)\) (i.e. there is one and only one way to extent a function in \(S([0,n[)\) to a function in \(S([0,n])\)); 2-\(\sharp S([0,n+1])=3\sharp S([0,n[)\) (i.e. there are three and only three independent ways to extent a function in \(S([0,n[)\) to a function in \(S([0,n+1])\)). Then you can conclude that \(\sharp S([0,n])=2\times 3^{n-1}\) Notation: \([0,n[=\{x\in\mathbb{R}: 0\leq x < n\}\) |
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| Autor: | kinu [ 23 fev 2012, 16:28 ] |
| Título da Pergunta: | Re: discontinuous function |
Thanks ruicarpenter got it |
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| Autor: | Olivio [ 05 mai 2012, 17:00 ] |
| Título da Pergunta: | discontinuous function |
Thank you so much for sharing that You can't look at that and not know these people are deeply mentally ill. |
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