If f ( x ) is a 5th degree monic polynomial such that :
f ( 1 ) = 1
f ( 2 ) = 2
f ( 3 ) = 5
f ( 4 ) = 1 3
f ( 5 ) = 3 4
then find the value of ( f ( 6 ) + f ( 7 ) + f ( 8 ) )
This problem is original
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
All right! I can see now. I think I agree with this method and it is the best way to solve this question.
good one sir........ i also followed the same but yours approach is an easier one
For Monic Quintic Polynomial, only five points are required to fix the curve:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 |
|
With f (x) = x 5 + p x 4 + q x 3 + r x 2 + s x + t , of 5 unknowns found from Cramer's method.
f (6) + f (7) + f (8) = 201 + 892 + 2850 = 3943.
Answer: 3 9 4 3
it would be immediate using using method of differences
Log in to reply
Would it be 100% reliable or could it need some special coincidence using the method you mentioned?
Log in to reply
Method of difference is a nice way and you can read it on Brilliant wiki
Log in to reply
@Dev Sharma – Hi Dev , I took nth diff. as n! because given poly. is monic..
@Dev Sharma – There is one I know called Gregory-Newton interpolation formula using forward finite differences for finding range of x in between the given five points. Interpolation is an on the line prediction whereby regression is analytical scatters. Actually, an interpolation as shown in my solution force to find for f (6), f (7) and f (8) by presuming that they are also points on a stated Monic Quintic Polynomial. Otherwise, this prediction is not a good prediction sometime for given points with many bending compared to regression.
Can you show the Method of difference as a solution wished by people at here?
Log in to reply
@Lu Chee Ket – I posted a solution using the Method of Differences.
Problem Loading...
Note Loading...
Set Loading...
Method of differences:
1 1 2 2 3 3 5 5 5 8 8 1 2 0 1 3 1 3 1 2 5 2 1 1 3 3 1 2 0 3 4 1 4 6 2 4 5 1 6 7 3 7 8 1 2 0 2 0 1 5 2 4 3 6 5 6 9 1 7 4 3 8 9 2 1 2 6 7 1 9 5 8 2 8 5 0
Because this is a fifth degree polynomial, all 5th differences are equal. Because it is monic, the 5th differences are equal to 5 ! = 1 2 0 .
Solution: 2 0 1 + 8 9 2 + 2 8 5 0 = 3 9 4 3 .