I have a function:
$$f(n) = 1 - \left(\frac{1}{n}\right)$$
I want to add another variable $k$ to this equation such that:
if $k=n$ , then $f(n,k)=1$
if $n=1$, then $f(n,k)=0$
How can we do that ?
Tell me more
×
Mathematics Stack Exchange is a question and answer site for
people studying math at any level and professionals in related fields. It's 100% free, no registration required.
|
|
||||
|
|
|
Consider the function $g(n,k) = 1 - \frac{1}{n} + \frac{1}{k}$ Then if $k = n$, you have $g(n,k) = 1$. (Note that $n$ and $k$ can not be equal to $0$.) As Ross Millikan suggested, for your new question, you can consider the function: $$h(n,k) = \begin{cases} g(n,k) & \quad n \neq 1 \\ 0 & \quad n = 1 \end{cases}$$ $h$ is a perfectly good function and satisfy the property that $n \neq 1$ and $k = n$, then $h(n,k) = 1$; and if $n = 1$, then for all $k$, $h(n,k) = 0$. |
|||||||||
|
|
Assuming that the following statement will never be true - n=k=1 - you can also do this without an "if" statement. Try $(1-\frac{1}{n})^{n-k}$. If n=1, then the fraction $\frac{1}{n}$ will be $\frac{1}{1}=1$, and $1-1=0$. $0^{anything} = 0$. The other issue is if $n=k$. In this case, $n-k = 0$ and $anything^0 = 1$. I think this equation is also good because at $f(1,1)$, your constraints have given both of the values $0$ and $1$, so it is undefined at $f(1,1)$. At $f(1,1)$, my equation gives you $0^0$, which is undefined. |
|||
|
For the first part, a key point is that if $ab=0$, then $a=0$ or $b=0$. So this suggests that whatever you add on should be multiplied by what you already have, since the equation as you give it solves $f(1,k)=0$. Multiplying all that by something won't change the result. Similarly, for the second part, what simple relationship gives $1$ when two numbers are equal? There's a basic operation that does exactly what you want for that part. Stick them together and you should have a suitable solution. |
|||
|
|
