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Let f ( x ) denote a 7 th degree polynomial satisfying f n = 3 n for n = 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 . Find f ( 8 ) .
Inspirations : First link , Second link , Third link .
The answer is 6305.
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4 solutions
Nice! So can we deduce a general form for this? My method was by method of differences.
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I hesitate to call the method "general", but it is an option when fitting a polynomial to exponential data points.
Same, even I used the method of differences.
But i see that the equation you have is not a polynomial one expansion of (1+2)^x so how does this satisfy the question
This was a long time ago, but what exactly is this motivation for coming up with that polynomial?
I used method of differences It is rather longer method though...
Even I used the same :-P
PS-Why don't you display it in L A T E X ?
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I want to show of my nice handwriting hahahaha...
I dont know how to make a diff.table in latex bro...
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OK ! See my solution on this problem.
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@Akshat Sharda – Ah , I can do toggle latex now thanks !!!
'#'SuchMehnat
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Haha , It took me nearly 20 min to draw this table ,,,
MOD is really amazing theorem.
Cheap solution here.
By the Super Automatic Regressionator v1.1.2 , setting 8 points to (0, 1), (1, 3) etc. gives L(x) and L(8) happens to be 6305.
I'm glad I made that stupid calculator >.>
s̶t̶u̶p̶i̶d̶ efficient calculator
I found a veeery weird solution using table of differences... I don't know why it works, though...
The sum is also equal to i = 0 ∑ 7 3 i 2 7 − i = 6 3 0 5
You have got Eagle's eye.....:D
Yes, good observation! You can see that our solutions are the same since 3 8 − 2 8 = ( 3 − 2 ) ∑ i = 0 7 3 i 2 7 − i
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Here is my solution , neatly made Table of differences , in my best handwriting...
Let f ( x ) = ∑ k = 0 7 2 k ( k x ) and find f ( 8 ) = 3 8 − 2 8 = 6 3 0 5