f ( x ) = n = 0 ∑ ∞ F n x n
Define f ( x ) as the series above with F n as the Fibonacci sequence with F 0 = 0 , F 1 = 1 . Calculate f ( 2 1 ) !
If you think the answer diverges, submit your answer as -1000.
Bonus : Can you generalize the formula f ( x ) = n = 0 ∑ ∞ a n x n , which the sequence a n + 2 = a n + 1 + a n , a 0 = p , a 1 = q ? And what is the limit/boundaries of the generalization?
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Uh, you forgot to factorial it.
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Thank you for pointing it out Stewart, but i believe it was only the excitement of the question poster and not really the factorial that was asked for since the function clearly does not always take integer values, I am cynical about how much sense it makes to ask for the factorial value of a real value function.
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Hmm. I suppose what it boils down to is that by being told to take the factorial, you are basically being told that the result happens to be an integer. In a fractional expression, often the denominator isn't a nonzero-valued function, but the value happens to be nonzero and so the expression is considered valid. So I suppose the same would apply here.
In this instance, luckily 2! = 2 and so it doesn't make a difference. But if it did, then the problem setter would have to be careful not to include a punctuation mark where it can validly be interpreted as a mathematical symbol with a totally different meaning.
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@Stewart Gordon – I understand your view about the fractional expression, but I would like to argue that the factorial operator works only in the domain of positive integers and the function at hand here is a real value function which only happens to take a real value that is also an integer. That is why I challenge the notion of finding the factorial. Anyway, we will just have to live with the happy coincidence of 2!=2 unless the poster chooses to comment.
Please look at the last but one step. Will it be ( -1/x) or (1/x) ?Pl clarify.. Thanks for splendid solution.
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Let us define f ( x ) ( p , q ) as the summation with a 0 = p and a 1 = q We can then express f ( x ) ( p , q ) = p + ( p x + q ) ( f ( x ) ( 0 , 1 ) ) We require f ( x ) ( 0 , 1 ) in this question, Let us consider f ( x ) ( 1 , 1 ) , We observe that the value of this sequence is the same as f ( x ) ( 0 , 1 ) since the first term of the sequence, which makes the only difference between the two sequences, is zero. f ( x ) ( 1 , 1 ) = f ( x ) ( 0 , 1 ) Substituting this result in the earlier equation, f ( x ) ( 1 , 1 ) = 1 + ( x + 1 ) ( f ( x ) ( 0 , 1 ) ) f ( x ) ( 0 , 1 ) = 1 + ( x + 1 ) ( f ( x ) ( 0 , 1 ) ) f ( x ) ( 0 , 1 ) = x 1 Therefore, f ( 1 / 2 ) ( 0 , 1 ) = 2