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You can put this solution on YOUR website!
It turns out the ‘?’ can be whatever we want it to be!\r
\n" ); document.write( "\n" ); document.write( "Allow me to illustrate. Take for example, f(x)=(1/40)x^4-(13/20)x^3+(291/40)x^2-(553/20)x+42, and evaluate f(8), f(7), f(6), f(5), f(3). It turns out that:\r
\n" ); document.write( "\n" ); document.write( "f(8)=56,
\n" ); document.write( "f(7)=42,
\n" ); document.write( "f(6)=30,
\n" ); document.write( "f(5)=20,
\n" ); document.write( "f(3)=9.\r
\n" ); document.write( "\n" ); document.write( "Wait, what sorcery is this? Turns out that although the popular rule f(x)=x(x-1), which gives f(x)=6, satisfies the known values in the sequence, that f(x)=(1/40)x^4-(13/20)x^3+(291/40)x^2-(553/20)x+42 also satisfies them–except with a different value of f(3)!\r
\n" ); document.write( "\n" ); document.write( "Here’s another one that also works but gives f(3)=12:
\n" ); document.write( "f(x)=(1/20)x^4-(13/10)x^3+(271/20)x^2-(543/10)x+84\r
\n" ); document.write( "\n" ); document.write( "And here is one where f(3)=π
\n" ); document.write( "f(x)=(1/120)(π-6)x^4-(13/60)(π-6)x^3+(1/120)(251π-1386)x^2+(1/60)(3138-533π)x+14(π-6)\r
\n" ); document.write( "\n" ); document.write( "Finally, in general if you want the ‘?’=k, i.e., f(3)=k where k is the value of your choice, then
\n" ); document.write( "(1/120)(k-6)x^4-(13/60)(k-6)x^3+(1/120)(251k-1386)x^2+(1/60)(3138-533k)x+14(k-6)\r
\n" ); document.write( "\n" ); document.write( "More details here: https://www.scribd.com/doc/260182194/Elementary-Sequences
\n" ); document.write( "For a non-polynomial rule see here: http://i.imgur.com/BHkg0Ad.png
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