# How do I define the following piecewise function?

For 0<=t<=10, if 5t<=n<=5t+2, then f[n]=4n, and if 5t+3<=n<=5t+4, then f[n]=4n+1?

I would like then to use the specific values of f[n], for all n between 0 and 54 (t here is any integer between 0 and 10).

I tried:

f[n_] := Piecewise[{{4 t,     5 t     <= n <= 5 t + 2},
{4 t + 1, 5 t + 3 <= n <= 5 t + 4}}]


I tried f (after Clear[t]). I got: f is 4t if 5t<=5<=5t+2 and 4t+1 if 5t+3<=5<5t+4.

I would have liked to simply get 4.

• Please post the code you tried, the output it gives you, and the output you would like to get instead. – Szabolcs Feb 1 '15 at 18:16
• So you have actually f[n,t] instead of f[n]? – mgamer Feb 1 '15 at 18:22
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• I added the code... Thanks! I am new to this, Your help would be greatly appreciated. No, f is a function of n as I see it. – Lola Feb 1 '15 at 18:26
• You say, "When I type f I get 0", but this is actually not the case unless t was assigned a value. Can you give a complete example, including all relevant definitions? Do you have a definition for t? This site gives good guidelines on how to ask a question: sscce.org – Szabolcs Feb 1 '15 at 18:31

Let me know if this was what was intended:

c1 = {4 n, 5 # <= n <= 5 # + 2} & /@ Range[0, 10];
c2 = {4 n + 1, 5 # + 3 <= n <= 5 # + 4} & /@ Range[0, 10];
f[u_] := Piecewise[Join[c1, c2] /. n -> u]
DiscretePlot[f[x], {x, 0, 54}, PlotRange -> All] • This is great! It works! (typo: it should be 4# instead of 4n - will edit this myself). Thanks!!! – Lola Feb 2 '15 at 15:35

For 0<=t<=10, if 5t<=n<=5t+2, then f[n]=4t, and if 5t+3<=n<=5t+4, then f[n]=4t+1

f[n_,t_]:=(5t<=n<=5t+2)&&4t||(5t+3<=n<=5t+4)&&(4t+1)


I often use such defenition of condition-depending expressions, if I'm not going to differentiate or integrate them

• Thanks! This is useful. The only thing is that I would really need a function f of n alone, without keeping track of the t. – Lola Feb 1 '15 at 21:12
• I guess you can solve for t in terms of n and plug it in the function f[n,t], so I guess this will do it in this particular example. However, I am still curious on how would one define a function of n without solving for t. This will allow for a more complicated condition instead of 5t<=n<=5t+3... – Lola Feb 1 '15 at 21:31
• f depends on two variables. – mgamer Feb 1 '15 at 22:01
• No, f as I wrote it, is a function of n alone. In order to define it, I used the auxiliary variable t. In other words, what I wrote means: f[n]=0 if 0<=n<2, f[n]=1 if 3<=n<=4, f[n]=4 if 5<=n<=7, f[n]=5 if 8<=n<=9, f[n]=8 if 10<=n<=12, etc – Lola Feb 2 '15 at 2:45
• I wanted to have an efficient way of defining this function. Something that would work when the range is much larger or the conditional more complicated. – Lola Feb 2 '15 at 2:51