200 Followers Problem

Algebra Level 5

If $f(x)$ is a 5th degree monic polynomial such that :

$f(1) = 1$

$f(2) = 2$

$f(3) = 5$

$f(4) = 13$

$f(5) = 34$

then find the value of $(f(6) + f(7) + f(8))$

This problem is original

The answer is 3943.

2 solutions

Arjen Vreugdenhil
Nov 27, 2015

Method of differences:

$\begin{array}{ccccccccccccccc} 1 && 2 && 5 && 13 && 34 && 201 && 892 && 2850 \\ & 1 && 3 && 8 && 21 && 167 && 691 && 1958 & \\ && 2 && 5 && 13 && 146 && 524 && 1267 && \\ &&& 3 && 8 && 133 && 378 && 743 &&& \\ &&&& 5 && 125 && 245 && 365 &&&& \\ &&&&& 120 && 120 && 120 &&&&& \end{array}$

Because this is a fifth degree polynomial, all 5th differences are equal. Because it is monic, the 5th differences are equal to $5! = 120$ .

Solution: $201 + 892 + 2850 = \boxed{3943}$ .

All right! I can see now. I think I agree with this method and it is the best way to solve this question.

Lu Chee Ket - 5 years, 6 months ago

good one sir........ i also followed the same but yours approach is an easier one

Abhisek Mohanty - 5 years, 4 months ago

Lu Chee Ket
Nov 25, 2015

For Monic Quintic Polynomial, only five points are required to fix the curve:

1       1   1   1   1   1   1   =   1
32      16  8   4   2   1   p   =   2
243     81  27  9   3   1   q   =   5
1024    256 64  16  4   1   r   =   13
3125    625 125 25  5   1   s   =   34
                            t       

1   1   1   1   1   p   =   0
16  8   4   2   1   q   =   -30
81  27  9   3   1   r   =   -238
256 64  16  4   1   s   =   -1011
625 125 25  5   1   t   =   -3091

1   1   1   1   1   288
16  8   4   2   1   
81  27  9   3   1   
256 64  16  4   1   
625 125 25  5   1   

0       1   1   1   1   -4260
-30     8   4   2   1   
-238    27  9   3   1   
-1011   64  16  4   1   
-3091   125 25  5   1   

1   0       1   1   1   24024
16  -30     4   2   1   
81  -238    9   3   1   
256 -1011   16  4   1   
625 -3091   25  5   1   

1   1   0       1   1   -63276
16  8   -30     2   1   
81  27  -238    3   1   
256 64  -1011   4   1   
625 125 -3091   5   1   

1   1   1   0       1   76920
16  8   4   -30     1   
81  27  9   -238    1   
256 64  16  -1011   1   
625 125 25  -3091   1   

1   1   1   1   0      -33408
16  8   4   2   -30 
81  27  9   3   -238    
256 64  16  4   -1011   
625 125 25  5   -3091   

p   =   -355/24
q   =   1001/12
r   =   -5273/24
s   =   3205/12
t   =   -116

6   201 3943
7   892 
8   2850

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: $\boxed{3943}$

it would be immediate using using method of differences

Dev Sharma - 5 years, 6 months ago

Would it be 100% reliable or could it need some special coincidence using the method you mentioned?

Lu Chee Ket - 5 years, 6 months ago

Method of difference is a nice way and you can read it on Brilliant wiki

Dev Sharma - 5 years, 6 months ago

@Dev Sharma – Hi Dev , I took nth diff. as n! because given poly. is monic..

A Former Brilliant Member - 5 years, 6 months ago

@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?

Lu Chee Ket - 5 years, 6 months ago

@Lu Chee Ket – I posted a solution using the Method of Differences.

Arjen Vreugdenhil - 5 years, 6 months ago

200 Followers Problem

The answer is 3943.

2 solutions

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