Inspiration
m n 0 1 2 3 4 f ( n ) 1 2 1 6 2 1 6 4 0 9 6
Given that f ( n ) is a 4 th degree polynomial satisfying the relationship as shown above, find f ( 5 ) .
The answer is 18471.
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2 solutions
More elegantly, we can write ( 1 5 ) 4 0 9 6 − ( 2 5 ) 2 1 6 + ( 3 5 ) 1 6 − ( 4 5 ) 2 + ( 5 5 ) 1 = 1 8 4 7 1
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Sir please can you explain how we get this form?
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The problem amounts to solving 6 linear equations in 5 unknowns. The equation I give now follows from the relationship between consecutive powers that is discussed here
The method can also be explained in terms of Lagrange interpolation; see the solution by @Akshay Yadav
Sir, why not (2x)^x work? Sorry, im a new learner
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Because ( 2 x ) x is not a polynomial as required in the problem.
Same way buddy
Thumbs up! + 1
Just a question, what if we are given different values for f ( n ) and there are n 's which are negative integers, but still all such n 's are consecutive, will it still work?
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Yes !! but for consecutive integers.
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It may be found for non consecutive integers too but a little hard work is required for that.
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@Arulx Z – As stated f ( n ) = A n 4 + B n 3 + C n 2 + D n + E will give you a system of 5 equations with 5 unknowns.
I solved the ensuing system of equations using Matrix Arithmetic. Messy - yes. But this method works even if the n-values are not consecutive integers and include negatives or fractions.
f ( n ) = 8 1 1 0 7 n 4 − 1 2 9 6 1 7 n 3 + 8 1 1 5 3 7 n 2 − 1 2 9 3 3 7 n + 1
Find f(5) by direct substitution into the polynomial, simplify and voila - f ( 5 ) = 1 8 4 7 1
@Arulx Z – My method works fine for non consecutive integers, even for decimal values.
Same way! First I tried finding a general pattern, but then I said "screw it!" and used method of differences. Seems like it was worth it, though, as f ( 5 ) does not seem to be related to the previous terms in any way.
Darn I used finite differences and miscalculated 3 times :(
Can f(6) be calculated by using method of differences
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Yes one can calculate f ( 6 ) also by method of differences. Notice that D 4 ( n ) in Akshat Sharda's solution will remain constant.
You can also use my method of Lagrange's interpolation.
Yes, as Akshay said ! But a lot of calculations (I hate them!)
Using Lagrange's interpolation method-
( 0 − 1 ) ( 0 − 2 ) ( 0 − 3 ) ( 0 − 4 ) ( 5 − 1 ) ( 5 − 2 ) ( 5 − 3 ) ( 5 − 4 ) ( 1 ) = 1
( 1 − 0 ) ( 1 − 2 ) ( 1 − 3 ) ( 1 − 4 ) ( 5 − 0 ) ( 5 − 2 ) ( 5 − 3 ) ( 5 − 4 ) ( 2 ) = − 1 0
( 2 − 0 ) ( 2 − 1 ) ( 2 − 3 ) ( 2 − 4 ) ( 5 − 0 ) ( 5 − 1 ) ( 5 − 3 ) ( 5 − 4 ) ( 1 6 ) = 1 6 0
( 3 − 0 ) ( 3 − 1 ) ( 3 − 2 ) ( 3 − 4 ) ( 5 − 0 ) ( 5 − 1 ) ( 5 − 2 ) ( 5 − 4 ) ( 2 1 6 ) = − 2 1 6 0
( 4 − 0 ) ( 4 − 1 ) ( 4 − 2 ) ( 4 − 3 ) ( 5 − 0 ) ( 5 − 1 ) ( 5 − 2 ) ( 5 − 3 ) ( 4 0 9 6 ) = 2 0 4 8 0
Summing all these up,
1 − 1 0 + 1 6 0 − 2 1 6 0 + 2 0 4 8 0 = 1 8 4 7 1
Nivice man
By using Method of Differences,
m n 0 1 2 3 4 5 f ( n ) 1 2 1 6 2 1 6 4 0 9 6 1 8 4 7 1 D 1 ( n ) 1 1 4 2 0 0 3 8 8 0 1 4 3 7 5 D 2 ( n ) 1 3 1 8 6 3 6 8 0 1 0 4 9 5 D 3 ( n ) 1 7 3 3 4 9 4 6 8 1 5 D 4 ( n ) 3 3 2 1 3 3 2 1