# ISI2015-MMA-36

1 vote
111 views

For non-negative integers $m$, $n$ define a function as follows

$$f(m,n) = \begin{cases} n+1 & \text{ if } m=0 \\ f(m-1, 1) & \text{ if } m \neq 0, n=0 \\ f(m-1, f(m,n-1)) & \text{ if } m \neq 0, n \neq 0 \end{cases}$$ Then the value of $f(1,1)$ is

1. $4$
2. $3$
3. $2$
4. $1$
in Calculus
recategorized

f(1,1)=f(0,f(1,0))   since m$\neq$0,n$\neq$0

=f(0,f(0,1))           since m$\neq$0,$n=0$

=f(0,2)                 since m$=$0

=3                       since m$=$0

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