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I have the following expression $\forall n \ge 1 \in \Bbb N, f(n) \ge 2 \in \Bbb N$:

$$E = f(n)-1$$

$f(n)$ is a increasing function. The value $f(1)$ is a special case, its value can be $2$ or $3$. In the case that $f(1)=2$ I just want $E = f(1) = 2$, and in the case that $f(1)=3$ I would like to use the general expression $E = f(n)-1$, so $E=f(1)-1=2$. So basically I would like $f(1)$ to be always $2$ independently of $f(1)$.

So if possible I would like to have only one expression, specifically avoiding definitions-by-case, like this one:

$E = f(n)+$(some nice notation able to unify the general case and the special case)

Valid for any value of $n$ including the special cases of $f(1)$.

I was thinking about something like this:

$$E = f(n) + \frac{\mid 2-f(n) \mid}{2-f(n)}$$

That will work for $n \gt 2$ and also for the case $n=1$ when $f(1)=3$ but I am not sure if it is valid for the case $n=1$ when $f(1)=2$ (not sure if $\frac{\mid 2-f(n) \mid}{2-f(n)}$ can be evaluated as $0$ for $f(1)=2$).

Is there a notation able to unify the expressions into one? $f(1)=2$ is an special case. Initially I do not think it is possible but there might be a notation capable of doing it and I do not see it. Thank you!

iadvd
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  • @dxiv good suggestion, thank you! I have added that I am not looking for definitions-by-case, just one single expression covering all the possible cases. – iadvd Sep 26 '16 at 04:23

2 Answers2

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$$E=f(n)-1\,+\,\delta_{n,1}\, \delta_{f(n),2}$$

This uses the Kronecker delta.

However, I don't necessarily recommend doing this. The point of writing is clarity, and it's probably clearer to write something like this in cases.

Mitchell Spector
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    wow! many thanks! nice trick. I see, so an expression like the one I was writing in the question is not possible... By the way, I was trying to explain in my question that $f(n)$ can only be $2$ when $n=1$, so your nice expression finally will be $E=f(n)-1,+, \delta_{f(n),n+1}$ – iadvd Sep 26 '16 at 04:45
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    If $f(n)$ can only be 2 when $n=1,$ you can just write $E=f(n)-1+\delta_{f(n),2}.$ – Mitchell Spector Sep 26 '16 at 04:54
  • that is totally right! this was the question I was writing, thank you for helping me to wrap-up the special case! http://math.stackexchange.com/q/1941765/189215 – iadvd Sep 26 '16 at 05:04
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    I should have added that your formula with $\frac{\mid 2-f(n) \mid}{2-f(n)}$ doesn't work because when $f(n)=2,$ you get $\frac{0}{0},$ which is undefined. – Mitchell Spector Sep 26 '16 at 05:04
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$$E = \mathbf{1}_{\{1\}} \cdot 2 + (1-\mathbf{1}_{\{1\}}) \cdot (f(n) - 1)$$ will do that (where $\mathbf{1}_X$ is the indicator function of $X$), though honestly I wouldn't recommend it.

dxiv
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    thank you for the proposal, I am writing another bigger question and I needed to wrap-up the special case before asking it. – iadvd Sep 26 '16 at 04:46