i,j positive integers
f(i,i+1)=1/3(for all i)
f(i,j)=f(i,k)+f(k,j)-2f(i,k)(k,j)
Find the value of f(1,100)
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1 Answers
1057this is basically a recursion problem..
in the second statement,
put i=1 j=100 k=99
you'll get f(1,100)=(1/3)f(1,99)
now put j=99,k=98(note i will always be one)
you'll get f(1,99)=(1/3)f(1,98)=(1/32)f(1,100)
carry on....
you'll get f(1,2)=(1/398)f(1,100)
and f(1,2) is already (1/3)
so f(1,100)=397
you can easily generalise this....