Finding f(n) for All Multiples of 5

[Dustinsfl]

F is all integers that are multiplies of 5

f(n) = <br /> \begin{cases} <br /> 0, &amp; \mbox{if }n\mbox{ 1} \\<br /> \frac{5n}{2}, &amp; \mbox{if }n\mbox{ is even}\\<br /> ? , &amp; \mbox{if }n\mbox{ is odd}<br /> \end{cases}<br />

N exists in the natural number
I don't know how to obtain the function that will represent the odd integers.

 

[Gib Z]

Are you trying to get a function that produces all the integer multiples of 5?

Whats wrong with f(n) = 5n ?

 

[Dustinsfl]

Are you trying to get a function that produces all the integer multiples of 5?

Whats wrong with f(n) = 5n ?

I forgot to mention n exists in the natural numbers.

 

[Gib Z]

Ok then, the n is even part of your function will give all the positive multiples, now try something similar for n odd. I don't know how to give you a decent hint without giving it away, but basically, the odd case is very similar to the even case, except with a negative sign, and a little change to the n term to make sure it's divisible by 2.