Welcome 2016! Part 19
The function f ( x ) is a monic quartic polynomial satisfying
f ( 1 ) = 1 , f ( 2 ) = 4 , f ( 3 ) = 9 , f ( 4 ) = 1 6 .
Find f ( 5 ) .
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2 solutions
another way is Method of Finite Differences
This question can also be solved using Method of Differences.
f D1 D2 D3 D4
1| 1 3 2 0 24
2|4 5 2 24 24
3|9 7 26
4|16 33
5|49
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I Have a doubt . in D 2 we have got our constant differnece 2. But you have further subtracted and got 0.But then where comes the number 24 ?? Why have you written 24 in D 4 ?? Is this some reule or what ?? Please answer.
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If a polynomial f ( x ) has degree k , then the k t h difference is constant. Furthermore, the k t h difference is equal to k ! times the leading coefficient of f ( x ) .... It is well explained here
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@Rishabh Jain – Thank you very much :) !
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@Chirayu Bhardwaj – You are in class 12th na ?? What's your score in JEE ??
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@Chirayu Bhardwaj – Screwed.... Something abt 210-220... :-(
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@Rishabh Jain – I think that's a very nice score !! I did'nt know much about the paper . Btw by the talks of senior around they your score is great !. ;)
question says that f(x) is quadratic polynomial..... but according to your solution it is 4.... :(
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it says quartic .
f(x) = (x-1)(x-2)(x-3)(x-4) + x^{2}. As we need five parameters to determine a quartic polynomial and also it is given that the polyomial is monic. Hence, f(5) = 4 \times 3 \times 2 \times 1 + 5^{2} = 49
define a new poly: p ( x ) = f ( x ) − x 2 note that the roots of p(x) are 1,2,3,4. as it is monic: p ( x ) = ( x − 1 ) ( x − 2 ) ( x − 3 ) ( x − 4 ) f ( x ) = ( x − 1 ) ( x − 2 ) ( x − 3 ) ( x − 4 ) + x 2 so: f ( 5 ) = ( 5 − 1 ) ( 5 − 2 ) ( 5 − 3 ) ( 5 − 4 ) + 5 2 = 2 4 + 2 5 = 4 9