Well, this isn't any function
A real-valued quintic monic polynomial f ( x ) satisfies the following conditions.
f ( 1 ) = 1 2 , f ( 2 ) = 4 0 , f ( 3 ) = 9 0 , f ( 4 ) = 1 6 8 , f ( 5 ) = 2 8 0
Find the value of f ( 6 ) .
The answer is 552.
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2 solutions
Well, a stupid approach - we can assume the monic quintic as f ( x ) = x 5 + b x 4 + c x 3 + d x 2 + e x + f .
So we'll obtain 5 equations in 5 variables with the conditions given. Then, we can obtain f ( x ) and thus f ( 6 ) can be found thereafter. :D
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Haha. I used to solve similar problems like mine, till Pranjal Jain taught me a simpler method
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Method of differences is much simpler method!
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@Swapnil Das – He taught the method to tackle problems like this, like the problem "Photoelectric Primes" by Shivamani Patil.
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@Vishwak Srinivasan – Yeah, the same method is used there. It would be great if you a number of problems like this on Brilliant. They will surely get popular.
Truly inspiring problem and an awesome solution...I did the first part but could not get the end part as it is....can you please post a note on the topic generating functions..well i just wanted to know the basic idea of the topic generating function...thnkx
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Yeah sure.
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thnkx ...i will be waiting
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@Manish Bhargao – Umm....I am a little caught up with personal work now...could you give your email ID...so that I can send you the pic of the explanation instead?
How did you know that the roots of g(x) are 1,2,3,4,5 simply based on the fact that you know g(x) = f(x)-12 - 28(x-1) - ... - 6(x-1)(x-2)(x-3)/3! ?
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Because g(1),g(2),g(3),g(4),g(5) is 0
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No, that's not the main point I'm asking.
What motivates you to substitute x = 1,2,3,4,5 to see that g(x) = 0? Or did you just reverse engineer your question?
It's like asking whether some specific number is a composite number. Then you show a factor out of nowhere and prove that it's composite. It's as if you know what values to put in beforehand.
Do you know the logic behind this series expansion?
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I read it from a book called "Hall and Knight - Higher Algebra".
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You are talking about book by arihant publishers? Do they have written the logic or written it directly?
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@Abhay Tiwari – My cousin had that book a long time back. I don't know if it is published by Arihant.
@Abhay Tiwari – Ya! It's of Arihant.
12 = 3×4 = 1×3×4
40 = 4×10 = 2×4×5
90 = 5×18 = 3×5×6
168 = 6×28 = 4×6×7
280 = 7×40 = 5×7×8
f(x) = (x-1)(x-2)(x-3)(x-4)(x-5) + x(x+2)(x+3)
f(6) = 5×4×3×2×1 + 6×8×9 = 120 + 432 = 5 5 2
Let us find a generating function for the series 1 2 , 4 0 , 9 0 , 1 6 8 , 2 8 0 .
Using Method of Differences, the successive orders of difference are:
2 8 2 2 5 0 6 2 8 7 8 6 3 4 1 1 2
Hence the n t h term of this series is:
1 2 + 2 8 ( n − 1 ) + 2 2 2 ! ( n − 1 ) ( n − 2 ) + 6 3 ! ( n − 1 ) ( n − 2 ) ( n − 3 )
Let us define a function g ( x ) where,
g ( x ) = f ( x ) − 1 2 − 2 8 ( x − 1 ) − 2 2 2 ! ( x − 1 ) ( x − 2 ) − 6 3 ! ( x − 1 ) ( x − 2 ) ( x − 3 )
The roots of g ( x ) are 1 , 2 , 3 , 4 , 5
⇒ g ( x ) = ( x − 1 ) ( x − 2 ) ( x − 3 ) ( x − 4 ) ( x − 5 )
⇒ f ( x ) = ( x − 1 ) ( x − 2 ) ( x − 3 ) ( x − 4 ) ( x − 5 ) + 1 2 + 2 8 ( x − 1 ) + 2 2 2 ! ( x − 1 ) ( x − 2 ) + 6 3 ! ( x − 1 ) ( x − 2 ) ( x − 3 )
⇒ f ( 6 ) = ( 5 ) ( 4 ) ( 3 ) ( 2 ) ( 1 ) + 1 2 + ( 2 8 ) ( 5 ) + 1 1 ( 5 ) ( 4 ) + ( 5 ) ( 4 ) ( 3 ) = 1 2 0 + 1 2 + 1 4 0 + 2 2 0 + 6 0 = 5 5 2