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Function f : R → R is such that f ( x + y ) = f ( x ) + f ( y ) − x y − 1 for all x , y ∈ R and that f ( 1 ) = 1 .
Find the number of solutions to f ( n ) = n , where n ∈ N .
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3 solutions
as usual great and perfect solution sir ,, thanks for solving .
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Hey bro , How's life?
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hey bro , it is fine but bit struggling ,, what about you .. are you having gmail account ??
I too am struggling, especially in chemistry. Yes i have a gmail account , you should come to slack.
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@Aditya Chauhan – but i don't know how to get on slack ?? i have not used it earlier ,
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@Rudraksh Sisodia – https://brilliant.slack.com/signup
How do you know f(1)=1 ??? I cant understand this part
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Sorry, I edited the problem without including f ( 1 ) = 1 back.
actually @Kushal Bose it is the editted question by the editors , i have given f(1) =1 in my original question .. but now i don't know how to edit it again ,,
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Sorry, I edited the problem and forgetting about the f ( 1 ) = 1 part.
Also, we get an explicit formula for f ( n ) whenever n ∈ N , which is f ( n ) = 1 − 2 n ( n − 1 ) .
@Samrat Mukhopadhyay , please can you explain the way of using explicit formula ?? and even what is it ?
Damn, I neglected the fact that natural numbers did not include negatives and thought the answer was two since "-2" is a solution as well.
Put y=0 and you will get f ( 0 ) = 1 . Now, put f(0) in place of 1 in the question and divide by 'y' and put the limit y → 0 . Using constraint that f ( 0 ) = f ( 1 ) = 1 , you will get f ( x ) = x / 2 − x 2 / 2 + 1
its not hard to see its a continuous function which map to itself , by applying fixed point theorem ,there must be at least one value for f(n)=n
f ( x + y ) ⟹ f ( n + 1 ) ⟹ f ( n + 1 ) = f ( x ) + f ( y ) − x y − 1 = f ( n ) + f ( 1 ) − n − 1 = f ( n ) + 1 − n − 1 = f ( n ) − n Putting x = n , y = 1
For n ∈ N , we note that f ( n + 1 ) < f ( n ) . Since f ( 1 ) = 1 , f ( n ) ≤ 0 for n > 1 . Therefore, there is only 1 n such that f ( n ) = n , that is f ( 1 ) = 1 .